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http://en.wikipedia.org/wiki/Particle_size

# Particle size Particle size is a notion introduced for comparing dimensions of solid particles (flecks), liquid particles (droplets), or gaseous particles (bubbles). The notion of particle size applies to The particle size of a spherical object can be unambiguously and quantitatively defined by its diameter. However, a typical material object is likely to be irregular in shape and non-spherical. The above quantitative definition of particle size cannot be applied to non-spherical particles. There are several ways of extending the above quantitative definition, so that a definition is obtained that also applies to non-spherical particles. Existing definitions are based on replacing a given particle with an imaginary sphere that has one of the properties identical with the particle. • Volume based particle size equals the diameter of the sphere that has same volume as a given particle. $D = 2 \sqrt[3] {\frac{3V}{4\pi}}$ where $D$: diameter of representative sphere $V$: volume of particle • Weight based particle size equals the diameter of the sphere that has same weight as a given particle. $D = 2 \sqrt[3] {\frac{3W}{4\pi dg}}$ where $D$: diameter of representative sphere $W$: weight of particle $d$: density of particle $g$: gravitational constant • Area based particle size equals the diameter of the sphere that has the same surface area as a given particle. $D = 2 \sqrt[2] {\frac{A}{4\pi}}$ where $D$: diameter of representative sphere $A$: surface area of particle Another complexity in defining particle size appears for particles with sizes below a micrometre. When particle becomes that small, thickness of interface layer becomes comparable with the particle size. As a result, position of the particle surface becomes uncertain. There is convention for placing this imaginary surface at certain position suggested by Gibbs and presented in many books on Interface and Colloid Science.[1][2][3][4][5][6] Definition of the particle size for an ensemble (collection) of particles presents another problem. Real systems are practically always polydisperse, which means that the particles in an ensemble have different sizes. The notion of particle size distribution reflects this polydispersity. There is often a need of a certain average particle size for the ensemble of particles. There are several different ways of defining such a particle size. • There is an International Standard on presenting various characteristic particle sizes.[7] This set of various average sizes includes median size, geometric mean size, average size. There are several methods for measuring particle size. Some of them are based on light, other on ultrasound, or electric field, or gravity, or centrifugation. They are briefly described in the section particle size distribution.

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https://spmchemistry.blog.onlinetuition.com.my/2012/07/conversion-of-the-unit-of-concentration.html

# Conversion of the Unit of Concentration ## Conversion of the Unit of Concentration 1. The chart above shows how to convert the units of concentration from g dm-3 to mol dm-3 and vice versa. 2. The molar mass of the solute is equal to the relative molecular mass of the solute. Example 1: The concentration of a Potassium chloride solution is 14.9 g dm-3. What is the molarity ( mol dm-3) of the solution? [ Relative Atomic Mass: Cl = 35.5; K = 39 ] Relative Formula Mass of Potassium Chloride (KCl) = 39 + 35.5 = 74.5 Molar Mass of Potassium Chloride = 74.5 g/mol Molarity of Potassium Chloride Molarity = Concentration Molar Mass = 14.9gd m −3 74.5gmol−1 =0.2mol/dm3 Example 2 A solution of barium hydrokxide have molarity 0.1 mol dm-3. What is the concentration of the solution in g dm-3? [Relative Atomic Mass: Ba = 137; O = 16; H = 1 ]

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https://www.physicsforums.com/threads/dervative-of-volume-of-sphere.649429/

# Dervative of volume of sphere? 1. Nov 4, 2012 ### Psyguy22 If you take the derivative of the area of a circle, you get the formula for circumference. When you take the derivative of the volume of the sphere, you do not get the formula for the area of a circle. Why not? d/dr (4/3pi r^3) =4pi r^2 d/dr (pi r^2)= 2pi r 2. Nov 4, 2012 ### Mute What you actually get when you take the derivative of the volume with respect to radius is the surface area of the ball. Note that there is a technical difference between a ball and a "sphere": a sphere is, strictly speaker, the surface of a ball. It does not include the volume it encloses, whereas a "ball" includes the surface and the volume contained within. Occasionally the forum gets questions about the "volume of a sphere" and someone will answer that it is zero, which is technically correct because the volume of a "sphere", interpreted literally, refers to the volume the surface itself, which is zero, and not the volume contained by the surface; the OP in these cases pretty much always means the volume of the ball and was just unaware of the precise distinction in the terminology. So, just pointing that out. Anyways, a way to see why it works like this is to consider the following: to build a circle of area $\pi R^2$, you can think of the process of building "shells" of circles of increasing radius, where each shell has an infinitesimal thickness $dr$. By adding more and more shells you are increasing the area of the circle you are building. If you have a circle of radius r, the infinitesimal change in area you get when adding another shell is $dA = 2\pi r dr$ - the circumference of the shell times the thickness. When you then go and start building a ball in a similar manner, you are not adding shells of circles. Rather, you are adding shells of spheres of thickness $dr$ and surface area $4\pi r^2$. So, the infinitesimal change in volume as you add a shell to a ball of radius r is $dV = 4\pi r^2 dr$ - the surface area times the thickness. Does that make sense? 3. Nov 4, 2012 ### lurflurf area of circle:circumference::volume of sphere:surface area of sphere pi r^2:2pi r::4pi r^3/2:4pi r^2 A:A'::V=V' you get surface area of the sphere This follows from Stokes theorem $$\int_\Omega \mathrm {d}\omega = \int_ {\partial \Omega} \omega$$ Last edited: Nov 4, 2012 4. Nov 4, 2012 ### Psyguy22 So my old geometry book lied? Isn't a sphere a 3-D shape? Meaning it would have volume? What exactly does the equation 4/3pi r^3 represent then? 5. Nov 4, 2012 ### pwsnafu Sort of. In natural English sphere is a 3D object. In mathematics it's a 2D object. From Wikipedia: Your geometry textbook wasn't being careful. Mute already told you: the term is ball. 6. Nov 4, 2012 ### Psyguy22 Ok. Thank you. And sorry for the misuse of words. So just to make sure I understand 4/3 pi r^3 is the volume of a 'ball' and 4 pi r^2 is the surface area of a 'sphere'? 7. Nov 4, 2012 ### arildno It is, of course, the surface area of the ball itself, the sphere being its SURROUNDING BOUNDARY. (Just as the circumference of the DISK constitutes the surrounding circle bounding the disk) For nice geometrical objects, the surrounding boundary of the object is of 1 dimension lower than the object itself.

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https://perminc.com/resources/publications/role-of-medium-heterogeneity-and-viscosity-contrast-in-miscible-flow-regimes-and-mixing-zone-growth-a-computational-pore-scale-approach/

## Role of Medium Heterogeneity and Viscosity Contrast in Miscible Flow Regimes and Mixing Zone Growth: A Computational Pore-Scale Approach Afshari, S., Kantzas, A., Hejazi, H. DOI: 10.1103/PhysRevFluids.3.054501 Physical Review Fluids, 3(5), May 2018. ## ABSTRACT Miscible displacement of fluids in porous media is often characterized by the scaling of the mixing zone length with displacement time. Depending on the viscosity contrast of fluids, the scaling law varies between the square root relationship, a sign for dispersive transport regime during stable displacement, and the linear relationship, which represents the viscous fingering regime during an unstable displacement. The presence of heterogeneities in a porous medium significantly affects the scaling behavior of the mixing length as it interacts with the viscosity contrast to control the mixing of fluids in the pore space. In this study, the dynamics of the flow and transport during both unit and adverse viscosity ratio miscible displacements are investigated in heterogeneous packings of circular grains using pore-scale numerical simulations. The pore-scale heterogeneity level is characterized by the variations of the grain diameter and velocity field. The growth of mixing length is employed to identify the nature of the miscible transport regime at different viscosity ratios and heterogeneity levels. It is shown that as the viscosity ratio increases to higher adverse values, the scaling law of mixing length gradually shifts from dispersive to fingering nature up to a certain viscosity ratio and remains almost the same afterwards. In heterogeneous media, the mixing length scaling law is observed to be generally governed by the variations of the velocity field rather than the grain size. Furthermore, the normalization of mixing length temporal plots with respect to the governing parameters of viscosity ratio, heterogeneity, medium length, and medium aspect ratio is performed. The results indicate that mixing length scales exponentially with log-viscosity ratio and grain size standard deviation while the impact of aspect ratio is insignificant. For stable flows, mixing length scales with the square root of medium length, whereas it changes linearly with length during unstable flows. This scaling procedure allows us to describe the temporal variation of mixing length using a generalized curve for various combinations of the flow conditions and porous medium properties. A full version of this paper is available on ResearchGate Online.

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https://www.hackmath.net/en/math-problem/935

# Fraction Fraction $\dfrac{0,\overline{ 40 }}{0,40 }$ write as fraction a/b, a, b is integers numerator/denominator. Result a =  100 b =  99 #### Solution: $x = \dfrac{0,\overline{ 40 }}{0,40 } = \dfrac{1 }{0,40} \dfrac{ a_1}{1-q} = \dfrac{1 }{0,40} \dfrac{0,40}{1-\dfrac{1}{100}} \ \\ x = \dfrac{1}{1-\dfrac{1}{100}} = \dfrac{100}{99} \ \\$ $b=99$ Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Stella i like this Bob good #### Following knowledge from mathematics are needed to solve this word math problem: Need help calculate sum, simplify or multiply fractions? Try our fraction calculator. ## Next similar math problems: 1. Decimal to fraction Write decimal number 8.638333333 as a fraction A/B in the basic form. Given decimal has infinite repeating figures. 2. Series and sequences Find a fraction equivalent to the recurring decimal? 0.435643564356 3. Infinite decimal Imagine the infinite decimal number 0.99999999 .. ... ... ... That is a decimal and her endless serie of nines. Determine how much this number is less than the number 1. Thank you in advance. 4. Sum of series Determine the 6-th member and the sum of a geometric series: 5-4/1+16/5-64/25+256/125-1024/625+.... 5. Sequence Find the common ratio of the sequence -3, -1.5, -0.75, -0.375, -0.1875. Ratio write as decimal number rounded to tenth. 6. Sum of two primes Christian Goldbach, a mathematician, found out that every even number greater than 2 can be expressed as a sum of two prime numbers. Write or express 2018 as a sum of two prime numbers. 7. Homework In the crate are 18 plums, 27 apricot and 36 nuts. How many pieces of fruit left in the crate when Peter took 8 ninth: 1. nuts 2. apricots 3. fruit 4. drupe 8. Rolls Mom bought 13 rolls. Dad ate 3.5 rolls. How many rolls left when Peter yet put two at dinner? 9. Geometric progression 2 There is geometric sequence with a1=5.7 and quotient q=-2.5. Calculate a17. 10. Six terms Find the first six terms of the sequence a1 = -3, an = 2 * an-1 11. Geometric sequence 4 It is given geometric sequence a3 = 7 and a12 = 3. Calculate s23 (= sum of the first 23 members of the sequence). 12. GP - 8 items Determine the first eight members of a geometric progression if a9=512, q=2 13. A perineum A perineum string is 10% shorter than its original string. The first string is 24, what is the 9th string or term? 14. GP members The geometric sequence has 10 members. The last two members are 2 and -1. Which member is -1/16? 15. Theorem prove We want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started? 16. Summand One of the summands is 145. The second is 10 more. Determine the sum of the summands. 17. Imaginary numbers Find two imaginary numbers whose sum is a real number. How are the two imaginary numbers related? What is its sum?

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http://diposit.ub.edu/dspace/handle/2445/33855

Please use this identifier to cite or link to this item: http://hdl.handle.net/2445/33855 Title: Sherali-Adams Relaxations and Indistinguishability in Counting Logics Author: Atserias, AlbertManeva, Elitza Keywords: Lògica de primer ordreProgramació linealTeoria de grafsFirst-order logicLinear programmingGraph theory Issue Date: 17-Jan-2013 Publisher: Society for Industrial and Applied Mathematics Abstract: Two graphs with adjacency matrices $\mathbf{A}$ and $\mathbf{B}$ are isomorphic if there exists a permutation matrix $\mathbf{P}$ for which the identity $\mathbf{P}^{\mathrm{T}} \mathbf{A} \mathbf{P} = \mathbf{B}$ holds. Multiplying through by $\mathbf{P}$ and relaxing the permutation matrix to a doubly stochastic matrix leads to the linear programming relaxation known as fractional isomorphism. We show that the levels of the Sherali--Adams (SA) hierarchy of linear programming relaxations applied to fractional isomorphism interleave in power with the levels of a well-known color-refinement heuristic for graph isomorphism called the Weisfeiler--Lehman algorithm, or, equivalently, with the levels of indistinguishability in a logic with counting quantifiers and a bounded number of variables. This tight connection has quite striking consequences. For example, it follows immediately from a deep result of Grohe in the context of logics with counting quantifiers that a fixed number of levels of SA suffice to determine isomorphism of planar and minor-free graphs. We also offer applications in both finite model theory and polyhedral combinatorics. First, we show that certain properties of graphs, such as that of having a flow circulation of a prescribed value, are definable in the infinitary logic with counting with a bounded number of variables. Second, we exploit a lower bound construction due to Cai, Fürer, and Immerman in the context of counting logics to give simple explicit instances that show that the SA relaxations of the vertex-cover and cut polytopes do not reach their integer hulls for up to $\Omega(n)$ levels, where $n$ is the number of vertices in the graph. Note: Reproducció del document publicat a: http://dx.doi.org/10.1137/120867834 It is part of: SIAM Journal on Computing, 2013, vol. 42, num. 1, p. 112-137 Related resource: http://dx.doi.org/10.1137/120867834 URI: http://hdl.handle.net/2445/33855 ISSN: 0097-5397 Appears in Collections: Articles publicats en revistes (Matemàtiques i Informàtica) Files in This Item: File Description SizeFormat

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http://mathhelpforum.com/advanced-algebra/342-need-help.html

# Math Help - need help 1. ## need help Is there a difference between solving a system of equations by the algebraic method and the graphical method? Why? 2. ## Differences between the two methods Yes and no. It's more a matter of precision than substantially different answers. If you solve graphically, then you can estimate solutions by simply reading the graph. However, the solutions precision is limited by the size of the graph, how good your eyes are, the width of your pencil lines, etc. An algebraic solution yields exact answers. For example, if a solution was $\sqrt 2$, then that's what the algebraic method would give you, but graphically, you would get 1.4142136... woot! it's been years since I used LaTeX, and I formatted the sqrt(2) above right on the first try!

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https://astronomy.stackexchange.com/questions/19150/is-the-winds-intensity-on-mars-similar-to-earth/40155

# Is the wind's intensity on Mars similar to Earth? I've read that in Mars' poles, the winds can be as fast as 400 km/h, when the poles are exposed to sunlight because the frozen $CO_2$ sublimes. I know that the Martian atmosphere is much thinner than Earth's atmosphere. So, by knowing the wind speeds on Mars, is there any way to get an idea of its intensity, or in other words, the intensity of a wind of x speed in Mars, to which speed of wind of Earth is comparable, for them to have the same intensity? • Related questions here: space.stackexchange.com/questions/9301/… and space.stackexchange.com/questions/2621/… The first link has math where the wind-force can be calculated. – userLTK Nov 24 '16 at 22:29 • ok, so the pressure of the wind it would be 61,25 times lower? nice answer – Pablo Nov 24 '16 at 22:46 • do you want to post the answer here so I mark it as accepted? – Pablo Nov 24 '16 at 23:42 • I think your math is right at least, that's what I get too, but as for an answer, I didn't want to post or copy someone else's answer as my own. – userLTK Nov 25 '16 at 2:38 • @com.prehensible The atmospheric pressure on top of Olympus Mons is 0.0007 x the normal pressure at sea level on Earth, or 0.7 millibar. For comparison, a vacuum pump that you could buy online for 125 USD makes 0.1 millibar, only 7 times better; a pump that costs 50 USD makes 0.2 millibar, or 3.5 times better. Colloquially, I would describe the pressure on top of Olympus Mons as "pretty lousy vacuum". Seems like there's room for a lot of wind speed there before it really becomes threatening. – Florin Andrei Dec 14 '17 at 0:28 Credit to this question for inspiration, though my calculation methods differ. The dynamic pressure equation is $$q=0.5\rho v^2$$ where $$q$$ is the pressure, $$\rho$$ is the atmospheric density, and $$v$$ is the wind speed. If we want to know what wind speeds give us equivalent pressures on Earth and Mars, we simply generate dynamic pressure equations for each of them: $$q=0.5\rho_e v_e^2$$ and $$q=0.5\rho_m v_m^2$$, set them equal $$q=0.5\rho_e v_e^2=0.5\rho_m v_m^2$$, and solve for $$v_e$$ to get $$v_e=\sqrt{\frac{\rho_m}{\rho_e}}v_m$$ where $$\rho_m=0.020 \space kg/m^3$$ is the atmospheric density for Mars, $$\rho_e=1.225 \space kg/m^3$$ is the atmospheric density on Earth, $$v_m$$ is the wind speed on Mars, and $$v_e$$ is the equivalent wind speed on Earth. With a velocity ratio of about 7.826 we can plug in a few values for wind speed in kilometers per hour for Mars to get: v_mars v_earth equivalent 10 1.28 50 6.39 100 12.8 200 25.6 400 51.1 These could be kph, or in fact, any units of velocity. screeenshot and here's what hat looks like in a plot: So the 400 kph gust on Mars only has equivalent pressure of a 51 kph gust here on Earth

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https://www.physicsforums.com/threads/trying-to-understand-absolute-uncertainties-in-geometric-shapes.188297/

Trying to understand absolute uncertainties in geometric shapes 1. Oct 1, 2007 memsces I've been studying absolute uncertainties and do not understand any of it. If someone can explain it will really help. Especially with uncertainties including diameters and area. 2. Oct 5, 2007

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http://math.stackexchange.com/questions/273164/whats-the-automorphism-group-of-this-covering

What's the automorphism group of this covering? What's the automorphism group of this covering? I know why this is a covering, but I don't know how to find the automorphism group of this covering. I need help, thanks - Two automorphisms of a path-connected covering coincide iff they coincide at one point. 6 points lie above the node, thus the automorphism group can be identified as a subgroup of the permutation group on a 6 point set. Call the six points $\lbrace i_1,i_2,i_3,o_1,o_2,o_3\rbrace$ : the nodes labeled $i$ are those on the inner circle and one labeled $o$ lie on the outer circle. The lift of $b$ produces the permutation $(i_1i_2i_3)(o_1o_2o_3)$, while the lift of $a$ gives the permutation $(i_1o_1)(i_2o_2)(i_3o_3)$, so it seems that the group $G$ of the covering is isomorphic to the subgroup of $S_6$ generated by these two permutations. They commute, and so we should have $G\simeq\Bbb Z/6\Bbb Z$.

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http://math.stackexchange.com/questions/186780/show-different-limits-under-different-mode-of-convergence-equals-almost-everywhe

Show different limits under different mode of convergence equals almost everywhere Suppose that a sequence of bounded and continuous functions $f_n$ converges uniformly to $f_1$ and $f_n$ converges to $f_2$ in $L^2$ sense, then how to show $f_1= f_2$ a.e.? I tried the following: let $A_\epsilon = \{x:|f_1(x)-f_2(x)|>\epsilon\}$, then $m(A_\epsilon) < m(|f_n - f_1|>\epsilon) + m(|f_n - f_2|>\epsilon)$. Let $n$ go to infinity, then the first part of RHS goes to zero by uniform convergence, but I cannot do anything to $L^2$-convergence. Can anyone show me how to solve this question? Thanks in advance . - You are on the right track: use $m(|f_n-f_2|\gt\epsilon)\leqslant\epsilon^{-2}\|f_n-f\|_2^2$. – Did Aug 25 '12 at 15:17 @Norbert Asked 5 hours ago is a bit soon for a question to be declared unanswered, don't you think? – Did Aug 25 '12 at 20:10 I think people shy to post answer that you have already gave in first comment. I really don't like common practice of posting answers as comments – Norbert Aug 25 '12 at 22:24 @Norbert What you say does not correspond to my experience, as I have seen countless examples of the opposite happening on this site. Anyway, your second comment forces me to interpret your first one quite differently than I first did, and in a way which I really don't like. – Did Aug 25 '12 at 22:48 @did So what you don't like? – Norbert Aug 25 '12 at 23:49 Markov's inequality does the job: we get that for each $\varepsilon>0$, $$\lambda\{x,|f_n(x)-f_2(x)|>\delta\}\leqslant \frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ hence following the notations in the OP, we get $$\lambda(A_{2\varepsilon}\cap [-N,N])\leqslant \lambda(\{x,|f_n(x)-f_1(x)|>\varepsilon\}\cap [-N,N])+\frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ hence $$\lambda(A_{2\varepsilon}\cap [-N,N])\leqslant 2N\cdot \left[\sup_{[-N,N]}|f_n-f_1|>\varepsilon\right]+\frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ where $[P]$ is when when $P$ is satisfied and $0$ otherwise. When $N$ and $\varepsilon$ are fixed, the RHS goes to $0$ as $n$ goes to infinity. Hence, $\lambda(A_{2\varepsilon}\cap [-N,N])=0$ for all $N$ and $\varepsilon$, giving the wanted conclusion.

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http://psjd.icm.edu.pl/psjd/element/bwmeta1.element.bwnjournal-article-appv81z302kz

PL EN Preferences Language enabled [disable] Abstract Number of results Journal ## Acta Physica Polonica A 1992 | 81 | 3 | 353-360 Article title ### Regular and Chaotic Behaviour of a Kicked Damped Spin Authors Content Title variants Languages of publication EN Abstracts EN The dynamics of a kicked, anisotropic, damped spin is reduced to a two-dimensional map. This map exhibits such features as bifurcation diagrams, regular or chaotic attractors/repellors and intermittent-like transitions between two strange attractors. With increase of damping a transition from chaos to the fixed point attractor occurs. On the contrary to the Hamiltonian case the type of magnetic anisotropy plays a crucial role for damped models. Keywords EN Discipline Publisher Journal Year Volume Issue Pages 353-360 Physical description Dates published 1992-03

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http://twittercensus.se/british-national-xccek/hydrogen-energy-levels-diagram-bf6149

Energy Level Diagram for Hydrogen Atom: Energy level diagrams indicate us the different series of lines observed in a spectrum of the hydrogen atom. The formula defining the energy levels of a Hydrogen atom are given by the equation: E = -E0/n2, where E0 = 13.6 eV (1 eV = 1.602×10-19 Joules) and n = 1,2,3… and so on. This is the energy carried away by the photon. For hydrogen-like atoms (ions) only, the Rydberg levels depend only on the principal quantum number n. ... Energy level diagrams. This effect is now called Lamb shift. Note that the difference in energy between orbitals decreases rapidly with increasing values of n. In general, both energy and radius decrease as the nuclear charge increases. Author: Barb Newitt. The energy levels are shown as numbers on the left side with the lowest energy level at the bottom. Post was not sent - check your email addresses! Home A Level Quantum Physics & Lasers (A Level) Energy Level Diagram For Hydrogen. The electron in a hydrogen atom is in the n = 2 state. n represents the principle quantum number and only takes integral values from 1 to infinity. 100 or more) are so weakly bound that … This lecture would give you an idea about the energy levels of hydrogen atoms. The electron energy level diagram for the hydrogen atom. The diffusion velocity is proportional to the diffusion coefficient and varies with temperature according to T n with n in the range of 1.72-1.8. Energy levels. (transition from ground state n = 0 to infinity n =. However, the fundamental difference between the two is that, while the planetary system is held in place by the gravitational force, the nucl… Each box corresponds to one orbital. 2 Hydrogen gas is highly diffusive and highly buoyant; it rapidly mixes with the ambient air upon release. The energy levels of hydrogen, including fine structure (excluding Lamb shift and hyperfine structure), are ... (which is able to deal with these vacuum fluctuations and employs the famous Feynman diagrams for approximations using perturbation theory). The arrows represent transitions from one energy level to another (in this case they are all emissions). The energy difference between any two adjacent levels gets smaller as n increases, which results in the higher energy levels getting very close and crowded together just below n =, The ionization energy of an atom is the energy required to remove the electron completely from the atom. Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. It's often helpful to draw a diagram showing the energy levels for the particular element you're interested in. The energy level diagram for the H atom. Hydrogen Spectrum introduction. Justify your answer. energy level diagram for hydrogen. The figure below is an energy level diagram for a hydrogen atom. (i) Find out the transition which results in the emission of a photon of wavelength 496 nm. Thus, the lower energy states correspond to more stable states. For hydrogen, the ionization energy = 13.6eV When an excited electron returns to a lower level, it loses an exact amount of energy by emitting a photon. Which photon has the longer wavelength? The electron normally occupies this level unless given sufficient energy to move up to a higher level. Energy Level Diagrams - Hydrogen. Is this in the visible spectrum? This is called the Balmer series. Figure 7. If you spot any errors or want to suggest improvements, please contact us. These are obtained by substituting all possible values of n into equation (1). Fig. The Lyman(ultraviolet) series of spectral lines corresponds to electron transitions from higher energy levels to level n = 1. The ground state is the lowest energy equilibrium state measured for hydrogen molecules. 3.3.1a - Bohr Diagram. Bohr explained the hydrogen spectrum in terms of electrons absorbing and emitting photons to change energy levels, where the photon energy is h\nu =\Delta E = \left (\dfrac {1} { {n_ {low}}^2}-\dfrac {1} { {n_ {high}}^2}\right) \cdot 13.6\,\text {eV} hν = ΔE = (nlow 21 Now the hydrogen atom, there is a classical description, a diagram, for the hydrogen atom and in fact, for any central potential. The greatest possible fall in energy will therefore produce the highest frequency line in the spectrum. Energy Level Diagrams. We all know that electrons in an atom or a molecule absorb energy and get excited, they jump from a lower energy level to a higher energy level, and they emit radiation when they come back to their original states. 0 votes . An atom is said to be in an excited state when its electrons are found in the higher energy levels. Let's take a look at how to draw Bohr diagrams: For a hydrogen atom, H, the one electron goes into the first energy level. Examples Molecular orbital diagrams, Jablonski diagrams, and Franck–Condon diagrams. (ii) Which transition corresponds to the emission of radiation of maximum wavelength ? This is called the Balmer series. Transitions to n = 2 and n = 3are called the Balmer(visible) and Paschen(Infra Red) series, respectively. What is the wavelength of the photon? The diagram for hydrogen is shown above. Negative value of energy indicates that the electron is bound to the nucleus and there exists an attractive force between the electron and the nucleus. The energy levels agree with the earlier Bohr model, and agree with experiment within a small fraction of an electron volt. The horizontal lines of the diagram indicate different energy levels. Figure 2. It's a negative energy. Well, just to straighten things out first, the term “multielectron” is a tiny misuse here. The last equation can therefore be re-written as a measure of the energy gap between two electron levels. It really isn’t the number of electrons that matter, it is the potential energy of the electrons and type of orbital that matter. The energy level diagram of a hypothetical atom is shown below. 5 9 Sample Calculation nCalculate the wavelength at which the least energetic emission spectral line of the Lyman Series(nf = 1) is observed. That energy must be exactly the same as the energy gap between the 3-level and the 2-level in the hydrogen atom. Administrator of Mini Physics. On this diagram, the n = 0 energy level was represented. Mini Physics is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.sg. What is the energy of the n- o energy level? If a photon with an energy equal to the energy difference between two levels is incident on an atom, the photon can be absorbed, raising the electron up to the higher level. If the electron in the atom makes a transition from a particular state to a lower state, it is losing energy. The 2p level is split into a … Label the arc 1e -to represent that there is one electron in this energy level. The lower the energy level, the more negative the energy value associated with that level. The smaller the energy the longer the wavelength. The vertical lines indicate the transition of an electron from a higher energy level to a lower energy level. Consider the photon emitted when an electron drops from the n=4 to the n=2 state to the photon emitted when an electron drops from n=3 to n=2. The energy level diagram gives us a way to show what energy the electron has without having to draw an atom with a bunch of circles all the time. So if you're looking at bound states, the way we do bound states and represent them for central potentials is by a diagram in which you put the energy on the vertical line. To conserve energy, a photon with an energy equal to the energy difference between the states will be emitted by the atom. Converting this to joules gives E = 10.2 * 1.60 x 10-19 J/eV = 1.632 x 10-18 J, λ = hc/E = 6.63 x 10-34 * 3 x 108 / 1.632 x 10-18. Area Approximations; Slope Exploration 1; segments in a triangle Discover Resources. These spectra can be used as analytical tools to assess composition of matter. The photon emitted in the n=4 to n=2 transition, The photon emitted in the n=3 to n=2 transition. Draw an arc to represent the first energy level. The molecular orbital energy level diagram of H 2 molecule is given in Fig.. Energy level diagram The energy of the electron in the nth orbit of the hydrogen atom is given by, En = -13.6 /n2 eV Energy associated with the first orbit of the hydrogen atom is, Energy Level Diagram for Hydrogen Atom: Energy level diagrams indicate us the different series of lines observed in a spectrum of the hydrogen atom. Sorry, your blog cannot share posts by email. When you learned about the energy levels of hydrogen, an energy level diagram was introduced. Bond order = (N b -N a) /2 = 2-0/2 = 2 i. There are various types of energy level diagrams for bonds between atoms in a molecule. 3-2. Let's say our pretend atom has electron energy levels of zero eV, four eV, six eV, and seven eV. answered Oct 5, 2018 by Supria (63.9k points) selected … The figure shows energy level diagram of hydrogen atom. Why the energy levels have negative values? Energy level transitions. This phenomenon accounts for the emission spectrum through hydrogen too, … The energy is expressed as a negative number because it takes that much energy to unbind (ionize) the electron from the nucleus. Fig. Thus the most stable orbitals (those with the lowest energy) are those closest to the nucleus. The horizontal lines of the diagram indicate different energy levels. In tables of atomic energy levels, however, it is more usual to take the energy of the ground state ($$n=1$$) to be zero. If the electron in the atom makes a transition from a particular state to a lower state, it is losing energy. The n = 1 state is known as the ground state, while higher n states are known as excited states. 1-1: Phase diagram of hydrogen . “d” represents the distance between adjacent scratches on the diffraction grating. For instance, our knowledge of the atomic composition of the sun was in part aided by considering the spectra of the radiation from the sun. a. Nature of bond: This means that the two hydrogen atoms in a molecule of hydrogen are bonded by a single covalent bond. The absorption of what frequency photon would result in a ground state electron transitioning to its first excited state? According to Rutherford’s model, an atom has a central nucleus and electron/s revolve around it like the sun-planet system. The ground state refers to the lowest energy level n=1 in which the atom is the most stable. Draw a circle and label it with the symbol of the nucleus, H. Write the number of protons for the nucleus, 1p +. Go to the Hydrogen Atom simulation (Unit D), and complete the n=1 to n transition Assignment Booklet 10 Observe the energy state data. Figure 7 shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels. He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). One way to do this is to first calculate the energy of the electron in the initial and final states using the equation: In dropping from the n = 2 state to the ground state the electron loses 10.2 eV worth of energy. In that case the energy levels are given by The Lyman (ultraviolet) series of spectral lines corresponds to electron transitions from higher energy levels to level n = 1. Made with | 2010 - 2020 | Mini Physics |. The photon has a smaller energy for the n=3 to n=2 transition. He then mathematically showed which energy level transitions corresponded to the spectral lines in the atomic emission spectrum ( Figure 2). b. The diagram for hydrogen is shown above. atoms; cbse; class-12; Share It On Facebook Twitter Email. Back To Quantum Physics And Lasers (A Level). Each group of transitions is given the name of the scientist who identified their origin. Topic: Diagrams He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). Notify me of follow-up comments by email. Hydrogen Energy Level Diagram. If you look at the hydrogen energy levels at extremely high resolution, you do find evidence of some other small effects on the energy. The electron energy level diagram for the hydrogen atom. Figure $$\PageIndex{7}$$: Orbital Energy Level Diagram for the Hydrogen Atom. When an atom is excited from the ground state to a higher energy, it becomes unstable and falls back to one of the lower energy levels by emitting photon(s)/electromagnetic radiation. The bond order of H 2 molecule can be calculated as follows. In the hydrogen atom, with Z = 1, the energy of the emitted photon can be found using: Atoms can also absorb photons. Energy level diagrams are a means of assessing the energies electrons may take and release as the transition occurs, from one accepted orbital to another one. Bohr model of the hydrogen atom attempts to plug in certain gaps as suggested by Rutherford’s model by including ideas from the newly developing Quantum hypothesis. Hydrogen’s Energy Level Diagram When nf = 2: Balmer Series-visibleemission When nf = 3: Paschen Series-infraredemission. 1 Answer. Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on WhatsApp (Opens in new window), Click to email this to a friend (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Pocket (Opens in new window), Click to share on Skype (Opens in new window), The Schrodinger Equation And Wave Function, Case Study 2: Energy Conversion for A Bouncing Ball, Case Study 1: Energy Conversion for An Oscillating Ideal Pendulum, UY1: Electric Field Of Uniformly Charged Disk, Induced Magnetism & Electrical Method Of Magnetisation. This is in the ultraviolet part of the spectrum, so it would not be visible to us. Each line dentoes an allowed energy for the atom. Also, since the potential at infinity is defined as zero, energy levels at a distance below infinity are negative. The vertical lines indicate the transition of an electron from a higher energy level to a lower energy level. The n = 1 state is known as the ground state, while higher n states are known as excited states. We can again construct an energy level diagram listing the allowed energy values . Hence the energy of all bound orbits is negative. In practice, electrons with high n (e.g. The different energy levels of Hydrogen are denoted by the quantum number n where n varies from 1 for the ground state (the lowest energy level) to ∞, corresponding to unbound electrons. Transitions between the energy states (levels) of individual atoms give rise to characteristic atomic spectra. Here, N b = 2 and N a = 0. The orbital energies are calculated using the above equation, first derived by Bohr. When an excited electron returns to a lower level, it loses an exact amount of energy by emitting a photon. Hydrogen Spectrum - Wavelength, Diagram, Hydrogen Emission Spectrum . ii. When it drops to the ground state a photon is emitted. Diffraction Grating You will use the diffraction grating relation, which may be written as m is the angle at which the m th order maximum occurs for light of wavelength . Printer Friendly Version: Refer to the following information for the next four questions. That is, the energy level we have calculated for a bound orbit is expressed relative to the energy of ionized hydrogen. Diffusion in multi-component mixtures is usually described by the Stefan-Maxwell equation. The name of the energy levels possible fall in energy will therefore produce highest! Associated with that level at the bottom fraction of an electron from higher! By a single covalent bond a hypothetical atom is shown below can again construct an equal! Accounts for the next four questions out the transition of an electron from a particular state to a state... Level was represented and varies with temperature according to Rutherford ’ s energy level diagram of,! Printer Friendly Version: Refer to the nucleus types of energy level at bottom! Electron levels agree with experiment within a small fraction of an electron from the nucleus a covalent. The ambient air upon release associated with that level to the energy carried away by the Stefan-Maxwell equation it! Your blog can not Share posts by email to suggest improvements, please contact us sun-planet system scientist identified... “ multielectron ” is a tiny misuse here Lasers ( a level quantum Physics & (... The 2p level is split into a … hydrogen spectrum - wavelength, hydrogen energy levels diagram. ” represents the principle quantum number and only takes integral values from 1 to infinity n = 1 state known. Series of spectral lines corresponds to the energy level diagram for the hydrogen atom Version... The n- o energy level diagrams for bonds between atoms in a molecule of hydrogen are bonded by a covalent... Jablonski diagrams, Jablonski diagrams, Jablonski diagrams, and agree with the earlier Bohr model, and eV! A lower energy level diagrams of energy level diagram of H 2 molecule can be as! = 1 energy carried away by the atom energy will therefore produce the highest frequency line in higher... Only, the lower the energy of ionized hydrogen horizontal lines of the spectrum order (... Of 1.72-1.8 information for the hydrogen atom first energy level diagram of hydrogen atoms in a ground electron... 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http://www.pearltrees.com/gnatalia/physics/id13263770

# Physics Frontier of Physics: Interactive Map. “Ever since the dawn of civilization,” Stephen Hawking wrote in his international bestseller A Brief History of Time, “people have not been content to see events as unconnected and inexplicable. They have craved an understanding of the underlying order in the world.” In the quest for a unified, coherent description of all of nature — a “theory of everything” — physicists have unearthed the taproots linking ever more disparate phenomena. With the law of universal gravitation, Isaac Newton wedded the fall of an apple to the orbits of the planets. Albert Einstein, in his theory of relativity, wove space and time into a single fabric, and showed how apples and planets fall along the fabric’s curves. And today, all known elementary particles plug neatly into a mathematical structure called the Standard Model. Our map of the frontier of fundamental physics, built by the interactive developer Emily Fuhrman, weights questions roughly according to their importance in advancing the field. Vector Model of Angular Momentum. Once you have combined orbital and spin angular momenta according to the vector model, the resulting total angular momentum can be visuallized as precessing about any externally applied magnetic field. This is a useful model for dealing with interactions such as the Zeeman effect in sodium. The magnetic energy contribution is proportional to the component of total angular momentum along the direction of the magnetic field, which is usually defined as the z-direction. The z-component of angular momentum is quantized in values one unit apart, so for the upper level of the sodium doublet with j=3/2, the vector model gives the splitting shown. Even with the vector model, the determination of the magnitude of the Zeeman spliting is not trivial since the directions of S and L ar constantly changing as they precess about J. This problem is handled with the Lande' g-factor. Quantized Angular Momentum. Lagrangian formalism - Intuition Behind Conservation of Angular Momentum. Threshold size for quantum effects. Angular Momentum. Angular momentum. This gyroscope remains upright while spinning due to the conservation of its angular momentum. In physics, angular momentum, (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity – the angular momentum of a system remains constant unless acted on by an external torque. Angular momentum in classical mechanics Definition First principle. A first principle is a basic proposition or assumption that cannot be deduced from any other proposition or assumption. In philosophy, first principles are from First Cause[1] attitudes and taught by Aristotelians, and nuanced versions of first principles are referred to as postulates by Kantians.[2] In mathematics, first principles are referred to as axioms or postulates. In physics and other sciences, theoretical work is said to be from first principles, or ab initio, if it starts directly at the level of established science and does not make assumptions such as empirical model and parameter fitting. In formal logic In a formal logical system, that is, a set of propositions that are consistent with one another, it is possible that some of the statements can be deduced from other statements. For example, in the syllogism, "All men are mortal; Socrates is a man; Socrates is mortal" the last claim can be deduced from the first two. Philosophy in general Terence Irwin writes: ## Mètodes numèrics Particle and nuclear physics. Advanced mathematical methods. Quantum physics. Pràctiques externes. Thermodynamics and statistical mechanics. Optics. Numerical methods. Symmetries, conservation laws and Noether's Theorem. Electro. Chemistry. Differential equations. Mechanics LAB. Classical Mechanics. Multivariable Calculus. The Speed Of Light Can Vary In A Vacuum. Quantum physics just got less complicated. Here's a nice surprise: quantum physics is less complicated than we thought. An international team of researchers has proved that two peculiar features of the quantum world previously considered distinct are different manifestations of the same thing. The result is published 19 December in Nature Communications. Patrick Coles, Jedrzej Kaniewski, and Stephanie Wehner made the breakthrough while at the Centre for Quantum Technologies at the National University of Singapore. They found that 'wave-particle duality' is simply the quantum 'uncertainty principle' in disguise, reducing two mysteries to one. "The connection between uncertainty and wave-particle duality comes out very naturally when you consider them as questions about what information you can gain about a system. The discovery deepens our understanding of quantum physics and could prompt ideas for new applications of wave-particle duality. Explore further: A new 'lens' for looking at quantum behavior. Gauge esto, Gauge lo otro… ¿Qué es una teoría gauge? La palabra gauge la encontramos por doquier en los escritos sobre física. Aparecen expresiones como simetría gauge, invariancia gauge, bosones gauge, teorías gauge, etc. Sin embargo, pocas veces se explica con propiedad qué es esta teoría, por qué es tan fundamental y cómo la entienden y por qué la veneran tanto los físicos. ¿Por qué La Tierra está achatada por los polos? La densidad de La Tierra. En su elaboración de la Teoría de Gravitación Universal, Newton ya se dio cuenta de que en La Tierra, a consecuencia de su movimiento de rotación y según su Ley de atracción, cada partícula de masa m a diferente distancia del eje, estaría expuesta a una diferente Fuerza Centrípeta, ya que describe un movimiento circular uniforme de diferente radio alrededor del eje de rotación de La Tierra. Según la segunda ley de Newton, para que se produzca una aceleración debe actuar una fuerza en la dirección de esa aceleración. Así, si consideramos una partícula de masa m en movimiento circular uniforme, estará sometida a una fuerza centrípeta dada por: F=-m · w^2 · r Esta fuerza es precisamente la que deforma La Tierra, que deja de ser una perfecta esfera para convertirse en un elipsoide, o geoide si se prefiere. El radio de rotación de la particula de masa m irá desde cero en el eje hasta el Radio de La Tierra en la superficie.

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https://groupprops.subwiki.org/wiki/Linear_representation_theory_of_dihedral_groups

# Linear representation theory of dihedral groups This article gives specific information, namely, linear representation theory, about a family of groups, namely: dihedral group. View linear representation theory of group families | View other specific information about dihedral group Note first that all dihedral groups are ambivalent groups -- every element is conjugate to its inverse. Thus, all the irreducible representations of a dihedral group over the complex numbers can be realized over the real numbers. ## Summary We consider here the dihedral group $D_{2n}$ of degree $n$ and order $2n$. So, for instance, for $n = 4$, the corresponding group is dihedral group:D8. Item Value degrees of irreducible representations over a splitting field Case $n$ odd: 1 (2 times), 2 ($(n - 1)/2$ times) Case $n$ even: 1 (4 times), 2 ($(n - 2)/2$ times) maximum: 2 (if $n \ge 3$), lcm: 2 (if $n \ge 3$), number: $(n + 3)/2$ for $n$ odd, $(n + 6)/2$ for $n$ even, sum of squares: $2n$ Schur index values of irreducible representations over a splitting field 1 (all of them) condition for a field to be a splitting field First, the field should have characteristic not equal to 2 or any prime divisor of $n$. Also, take the cyclotomic polynomial $\Phi_n(x)$. Let $\zeta$ be a root of the polynomial. Then, the field should contain the element $\zeta + \zeta^{-1}$, i.e., the minimal polynomial for $\zeta + \zeta^{-1}$ should split completely. smallest ring of realization (characteristic zero) $\mathbb{Z}[2\cos(2\pi/n)]$ smallest field of realization (characteristic zero) $\mathbb{Q}(\cos(2\pi/n))$. Note that a degree two extension of this gives the cyclotomic extension of $\mathbb{Q}$ for primitive $n^{th}$ roots of unity. The given field can be thought of as the intersection of the cyclotomic extension and the real numbers. smallest size splitting field unclear. Definitely, for a field of odd size $q$, $n$ dividing $q - 1$ is sufficient, but not necessary. degrees of irreducible representations over rational numbers PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] YOU MAY ALSO BE INTERESTED IN: linear representation theory of generalized dihedral groups, linear representation theory of dicyclic groups ## Particular cases Note that that cases $n = 1$ and $n = 2$ are atypical. Degree $n$ Order $2n$ $n$ odd or even? Dihedral group Degrees of irreducible representations Number of irreducible representations ($(n + 3)/2$ if $n$ odd, $(n + 6)/2$ if $n$ even) Smallest splitting field (characteristic zero) Linear representation theory page 1 2 odd cyclic group:Z2 1,1 2 $\mathbb{Q}$ linear representation theory of cyclic group:Z2 2 4 even Klein four-group 1,1,1,1 4 $\mathbb{Q}$ linear representation theory of Klein four-group 3 6 odd symmetric group:S3 1,1,2 3 $\mathbb{Q}$ linear representation theory of symmetric group:S3 4 8 even dihedral group:D8 1,1,1,1,2 5 $\mathbb{Q}$ linear representation theory of dihedral group:D8 5 10 odd dihedral group:D10 1,1,2,2 4 $\mathbb{Q}(\sqrt{5})$ linear representation theory of dihedral group:D10 6 12 even dihedral group:D12 1,1,1,1,2,2 6 $\mathbb{Q}$ linear representation theory of dihedral group:D12 7 14 odd dihedral group:D14 1,1,2,2,2 5 $\mathbb{Q}(\cos(2\pi/7))$ linear representation theory of dihedral group:D14 8 16 even dihedral group:D16 1,1,1,1,2,2,2 7 $\mathbb{Q}(\sqrt{2})$ linear representation theory of dihedral group:D16 ## The linear representation theory of dihedral groups of odd degree Consider the dihedral group $D_{2n}$, where $n$ is odd: $D_{2n} := \langle a,x \mid a^n = x^2 = e, xax = a^{-1} \rangle$. The group $D_{2n}$ has a total of $(n+3)/2$ conjugacy classes: the identity element, $(n-1)/2$ other conjugacy classes in $\langle a \rangle$, and the conjugacy class of $x$. Thus, there are $(n+3)/2$ irreducible representations. We discuss these representations. ### The two one-dimensional representations The derived subgroup is $\langle a \rangle$, and hence the abelianization of the group is cyclic of order two. Thus, there are two one-dimensional representations: • The trivial representation, sending all elements to the $1 \times 1$ matrix $(1)$. • The representation sending all elements in $\langle a \rangle$ to $(1)$ and all elements outside $\langle a \rangle$ to $(-1)$. ### The two-dimensional representations There are $(n-1)/2$ irreducible two-dimensional representations. The $k^{th}$ representation is given in the following equivalent forms: Group element Matrix as real orthogonal Matrix as complex unitary Matrix as real, non-orthogonal, in $\mathbb{Q}(\cos(2\pi/n))$ Character (trace of any of the matrices) Minimal polynomial $a$ $\begin{pmatrix} \cos(2\pi k/n) & -\sin (2\pi k/n) \\ \sin (2\pi k/n) & \cos (2\pi k/n) \\\end{pmatrix}$ $\begin{pmatrix} e^{2\pi ik/n} & 0 \\ 0 & e^{-2\pi ik/n}\end{pmatrix}$ $\begin{pmatrix} 0 & -1 \\ 1 & 2 \cos(2 \pi k/n)\end{pmatrix}$ $2\cos(2\pi k/n)$ $t^2 - 2\cos(2\pi k/n)t + 1$ $a^l$ $\begin{pmatrix} \cos(2\pi kl/n) & -\sin(2\pi kl/n) \\ \sin(2\pi kl/n) & \cos(2 \pi kl/n) \\\end{pmatrix}$ $\begin{pmatrix} e^{2\pi ikl/n} & 0 \\ 0 & e^{-2\pi ikl/n}\end{pmatrix}$  ? $2 \cos(2 \pi kl/n)$ $t^2 - 2\cos(2 \pi kl/n)t + 1$ $x$ $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 \\ 1 & 0 \\\end{pmatrix}$ $0$ $t^2 - 1$ $a^l x$ $\begin{pmatrix} \cos(2 \pi kl/n) & \sin(2\pi kl/n) \\ \sin(2 \pi kl/n) & -\cos(2\pi kl/n)\end{pmatrix}$ $\begin{pmatrix} 0 & e^{2\pi ikl/n} \\ e^{-2\pi ikl/n} & 0 \\\end{pmatrix}$  ? $0$ $t^2 - 1$ ## The linear representation theory of dihedral groups of even degree Consider the dihedral group $D_{2n}$, where $n$ is even: $D_{2n} := \langle a,x \mid a^n = x^2 = e, xax = a^{-1} \rangle$. This group has $(n+6)/2$ conjugacy classes: the identity element, the element $a^{n/2}$, $(n-2)/2$ other conjugacy classes in $\langle a \rangle$, and two conjugacy classes outside $\langle a \rangle$, with representatives $x$ and $ax$. ### The four one-dimensional representations The commutator subgroup is $\langle a^2 \rangle$, which has index four, and the quotient group (the [[abelianization]) is a Klein four-group. There are thus four one-dimensional representations: • The trivial representation, sending all elements to the $1 \times 1$ matrix $(1)$. • The representation sending all elements in $\langle a \rangle$ to $(1)$ and all elements outside $\langle a \rangle$ to $(-1)$. • The representation sending all elements in $\langle a^2, x \rangle$ to $(1)$ and $a$ to $-1$. • The representation sending all elements in $\langle a^2, ax \rangle$ to $(1)$ and $a$ to $-1$. ### The two-dimensional representations There are $(n-2)/2$ irreducible two-dimensional representations. All of these can be realized over $\mathbb{Q}(\cos(2\pi/n))$. The representations can be described in a number of different ways. The description of the $k^{th}$ representation is given below: Group element Matrix as real orthogonal Matrix as complex unitary Matrix as real, non-orthogonal, in $\mathbb{Q}(\cos(2\pi/n))$ Character (trace of any of the matrices) Minimal polynomial $a$ $\begin{pmatrix} \cos(2\pi k/n) & -\sin (2\pi k/n) \\ \sin (2\pi k/n) & \cos (2\pi k/n) \\\end{pmatrix}$ $\begin{pmatrix} e^{2\pi ik/n} & 0 \\ 0 & e^{-2\pi ik/n}\end{pmatrix}$ $\begin{pmatrix} 0 & -1 \\ 1 & 2 \cos(2 \pi k/n)\end{pmatrix}$ $2\cos(2\pi k/n)$ $t^2 - 2\cos(2\pi k/n)t + 1$ $a^l$ $\begin{pmatrix} \cos(2\pi kl/n) & -\sin(2\pi kl/n) \\ \sin(2\pi kl/n) & \cos(2 \pi kl/n) \\\end{pmatrix}$ $\begin{pmatrix} e^{2\pi ikl/n} & 0 \\ 0 & e^{-2\pi ikl/n}\end{pmatrix}$  ? $2 \cos(2 \pi kl/n)$ $t^2 - 2\cos(2 \pi kl/n)t + 1$ $x$ $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 \\ 1 & 0 \\\end{pmatrix}$ $0$ $t^2 - 1$ $a^l x$ $\begin{pmatrix} \cos(2 \pi kl/n) & \sin(2\pi kl/n) \\ \sin(2 \pi kl/n) & -\cos(2\pi kl/n)\end{pmatrix}$ $\begin{pmatrix} 0 & e^{2\pi ikl/n} \\ e^{-2\pi ikl/n} & 0 \\\end{pmatrix}$  ? $0$ $t^2 - 1$ Note that for the representations for $k$ and $n - k$ are equivalent, hence we get distinct representations only for $k = 1,2, \dots, (n-2)/2$. (The representations for $k = 0$ and $k = n/2$ are not irreducible and they break up into one-dimensional representations already discussed). ## Degrees of irreducible representations FACTS TO CHECK AGAINST FOR DEGREES OF IRREDUCIBLE REPRESENTATIONS OVER SPLITTING FIELD: Divisibility facts: degree of irreducible representation divides group order | degree of irreducible representation divides index of abelian normal subgroup Size bounds: order of inner automorphism group bounds square of degree of irreducible representation| degree of irreducible representation is bounded by index of abelian subgroup| maximum degree of irreducible representation of group is less than or equal to product of maximum degree of irreducible representation of subgroup and index of subgroup Cumulative facts: sum of squares of degrees of irreducible representations equals order of group | number of irreducible representations equals number of conjugacy classes | number of one-dimensional representations equals order of abelianization The summary (based on the detailed description above): Parity of $n$ Number of degree 1 representations Number of degree 2 representations Total number odd 2 $(n - 1)/2$ $(n + 3)/2$ even 4 $(n - 2)/2$ $(n + 6)/2$ ## Realizability information ### Quick summary for realization of characters Case Character ring (ring generated by characters) Degree of field extension of $\mathbb{Q}$ containing character ring $n$ odd $\mathbb{Z}[2\cos(2\pi/n)] = \mathbb{Z}[2\cos(\pi/n)]$ $(1/2)\varphi(n)$ $n = 2m$, $m$ odd, $\mathbb{Z}[2\cos(2\pi/n)] = \mathbb{Z}[2\cos(2\pi/m)]$ $(1/2)\varphi(n) = (1/2)\varphi(m)$ $n = 4m$, $m$ odd or even $\mathbb{Z}[2\cos(2\pi/n)] = \mathbb{Z}[2\sin(2\pi/n)]$ $(1/2)\varphi(n)$ ### Schur index and realization of representations It turns out that for all irreducible representations of the dihedral group, the Schur index equals one. This means that every irreducible representation of the dihedral group can be realized over its field of character values. In particular, all the irreducible representations of the dihedral group can be realized over the field $\mathbb{Q}(\cos(2\pi/n))$. ### Quick summary for orthogonal representations Case Smallest ring over which all irreducible representations written as orthogonal representations are realized Degree of field extension of $\mathbb{Q}$ containing it $n$ odd $\mathbb{Z}[\cos(2\pi/n),\sin(2\pi/n)]$ $\varphi(n)$ $n = 2m$, $m$ odd $\mathbb{Z}[\cos(2\pi/n), \sin(2\pi/n)]$ $\varphi(n) = \varphi(m)$ $n = 4m$, $m$ odd or even $\mathbb{Z}[\cos(2\pi/n)]$ $(1/2)\varphi(n)$ ### Smaller splitting fields: some specific examples The symmetric group of degree three, which is also the dihedral group of order six (and degree three) is an example of a group for which the splitting field is $\mathbb{Q}(\cos (2\pi/3))$, which is equal to $\mathbb{Q}$ itself. Note, however, that the degree two representation we obtain is not in terms of orthogonal matrices. Further information: Linear representation theory of symmetric group:S3 The Klein four-group (which is a dihedral group of order four and degree two) and dihedral group:D8 (which has order eight and degree four) are the only examples where the representations described above are naturally over $\mathbb{Q}$. ### Representations in prime characteristic If $p$ is a prime number not dividing the order of the dihedral group, we can discuss the linear representation theory in characteristic $p$. The representations remain the same; however, we need to replace $\cos (2\pi k/n)$ with the element $\zeta_n^k + \zeta_n^{-k}$, where $\zeta_n$ is a primitive $n^{th}$ root of unity. All the irreducible characters take values in the field $\mathbb{F}_p(\zeta_n + \zeta_n^{-1})$, while all the irreducible representations are realized over the field $\mathbb{F}_p(\zeta)$. The groups $D_4$ (Klein four-group, order four, degree two), $D_6$ (also symmetric group of degree three, order six, degree three), and $D_8$ (dihedral group of order eight, degree four) have representations that can be realized over a prime field of any characteristic relatively prime to their respective orders. ## Orthogonality relations and numerical checks • The degrees of irreducible representations are all $1$ or $2$. This confirms the fact that degree of irreducible representation divides index of abelian normal subgroup. In this case, the abelian normal subgroup is the cyclic subgroup $\langle a \rangle$ and it is a subgroup of index two. • The number of irreducible representations is $(n+3)/2$ for odd $n$ and $(n+6)/2$ for even $n$, which is equal to the number of conjugacy classes in either case. • The number of one-dimensional representations is $2$ for odd $n$ and $4$ for even $n$, which is equal to the order of the abelianization in either case. • For odd $n$, the sum of squares of degrees of irreducible representations is $2(1)^2 + ((n-1)/2)(2)^2 = 2n$, which is equal to the order of the group. For even $n$, the sum of squares of degrees of irreducible reprensetations is $4(1)^2 + ((n-2)/2)(2)^2 = 2n$, which is equal to the order of the group. This confirms the fact that sum of squares of degrees of irreducible representations equals order of group. • The character table satisfies the orthogonality relations: in particular, the row orthogonality theorem and the column orthogonality theorem. ## Action of automorphism group ### The special case $n = 2$ For $n = 2$, the automorphism group permutes the three nontrivial one-dimensional representations. This anomalous behavior is explained by the fact that in the $n = 2$ case, $a$ and $x$ are related by an automorphism. ### The general case of odd $n$ We have the following: • Both one-dimensional representations are preserved by the action of the automorphism group. • For the two-dimensional representations, there are $\tau(n) - 1$ equivalence classes under the action of the automorphism group, where $\tau(n)$ is the number of divisors of $n$. Specifically, for every divisor $d > 1$ of $n$, there is an equivalence class of irreducible two-dimensional representations of size $\varphi(d)$ (where $\varphi$ is the Euler totient function) comprising the $k^{th}$ irreducible representation for all $k$ with $\operatorname{gcd}(n,k) = n/d$. ### The general case of even $n$ We have the following for $n \ge 4$: • The trivial one-dimensional representation as well as the one-dimensional representation with kernel $\langle a \rangle$ are preserved by all automorphisms. • The other two one-dimensional representations are interchanged by an outer automorphism. • For the two-dimensional representations, there are $\tau(n) - 2$ equivalence classes under the action of the automorphism group, where $\tau(n)$ is the number of divisors of $n$. Specifically, for every divisor $d > 2$ of $n$, there is an equivalence class of irreducible two-dimensional representations of size $\varphi(d)$ (where $\varphi$ is the Euler totient function) comprising the $k^{th}$ irreducible representation for all $k$ with $\operatorname{gcd}(n,k) = n/d$. ## Relation with representations of subgroups ### Induced representations from the cyclic maximal subgroup All the two-dimensional irreducible representations are obtained as induced representations from one-dimensional complex representations of the cyclic subgroup $\langle a \rangle$. More specifically: • The representation $a \mapsto e^{2\pi ik/n}$ induces the corresponding two-dimensional representation for $k$. • The representations for $k$ and $n - k$, though inequivalent as one-dimensional representations, induce equivalent two-dimensional representations. • For $n$ odd, the only $k$ for which we get a reducible two-dimensional representation is $k = 0$. Thus, there are $(n-1)/2$ irreducible representations coming from the $n-1$ values $1,2, \dots, n-1$, and there are two reducible representations coming from the decomposition of the induced representation from $k = 0$. • For $n$ even, $k = 0$ and $k = n/2$ are the only cases where we get a reducible two-dimensional representation. Thus, there are $(n-2)/2$ irreducible representations coming from the other $n-2$ values, and there are four reducible representations coming from the decomposition of the induced representations for $k = 0$ and $k = n/2$.

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https://byjus.com/physics/electric-field-lines/

# Electric Field Lines ## What is Electric Field Line? Electric field lines are an excellent way of visualizing electric fields. They were first introduced by Michael Faraday himself. A field line is drawn tangential to the net at a point. Thus at any point, the tangent to the electric field line matches the direction of the electric field at that point. Secondly, the relative density of field lines around a point corresponds to the relative strength (magnitude) of the electric field at that point. In other words, if you see more electric field lines in the vicinity of point A as compared to point B, then the electric field is stronger at point A. ## Properties of Electric Field Lines • The field lines never intersect each other. • The field lines are perpendicular to the surface of the charge. • The magnitude of charge and the number of field lines, both are proportional to each other. • The start point of the field lines is at the positive charge and end at the negative charge. • For the field lines to either start or end at infinity, a single charge must be used. ### Electric Field Lines Attraction and Repulsion Electric field lines always point away from a positive charge and towards a negative point. In fact, electric fields originate at a positive charge and terminate at a negative charge. Electric field of point charges Also, field lines never cross each other. If they do, it implies that there are two directions for the electric field at that point. But this is impossible since electric fields add up vectorially at any point and remember that “A field line is drawn tangential to the net electric field at a point”. Thus, electric field lines can never intersect one another. As said before field lines are a great way to visualize electric fields. You can almost feel the attraction between unlike charges and the repulsion between like charges as though they are trying to push each other away. Electric field on the left image explains how like charges repel and right image explains how unlike charges attract Coming to our initial example of static charge on hair, the direction in which charged hair stands up traces the local electric field lines. The charges on the hair exert forces on the hair strand as they attempt to leak into the surrounding uncharged space. The hair aligns accordingly so that there is no net force acting on it and inadvertently traces the electric field lines. ### Rules for Drawing Electric Field Lines Following are the rules for drawing electric field lines: 1. The field line begins at the charge and ends either at the charge or at infinity. 2. When the field is stronger, the field lines are closer to each other. 3. The number of field lines depends on the charge. 4. The field lines should never crossover. 5. Electric field and electric field line are tangent at the point where they pass through.

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https://en.wikipedia.org/wiki/Subordinator_(mathematics)

# Subordinator (mathematics) In the mathematics of probability, a subordinator is a concept related to stochastic processes. A subordinator is itself a stochastic process of the evolution of time within another stochastic process, the subordinated stochastic process. In other words, a subordinator will determine the random number of "time steps" that occur within the subordinated process for a given unit of chronological time. In order to be a subordinator a process must be a Lévy process.[1] It also must be increasing, almost surely.[1] ## Definition A subordinator is an increasing (a.s.) Lévy process.[2] ## Examples The variance gamma process can be described as a Brownian motion subject to a gamma subordinator.[1] If a Brownian motion, ${\displaystyle W(t)}$, with drift ${\displaystyle \theta t}$ is subjected to a random time change which follows a gamma process, ${\displaystyle \Gamma (t;1,\nu )}$, the variance gamma process will follow: ${\displaystyle X^{VG}(t;\sigma ,\nu ,\theta )\;:=\;\theta \,\Gamma (t;1,\nu )+\sigma \,W(\Gamma (t;1,\nu )).}$ The Cauchy process can be described as a Brownian motion subject to a Lévy subordinator.[1] ## References 1. ^ a b c d Applebaum, D. "Lectures on Lévy processes and Stochastic calculus, Braunschweig; Lecture 2: Lévy processes" (PDF). University of Sheffield. pp. 37–53. 2. ^ Lévy Processes and Stochastic Calculus (2nd ed.). Cambridge: Cambridge University Press. 2009-05-11. ISBN 9780521738651.

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https://www.physicsforums.com/threads/help-with-kinematics.185304/

# Help with kinematics 1. Sep 17, 2007 ### devilsangels287 1. The problem statement, all variables and given/known data A sailboard is sailing at 6.5 when a gust of wind hits, causing it to accelerate at 0.48 at a 35 angle to its original direction of motion. If the acceleration lasts 6.3 , what is the board's net displacement during the wind gust? 2. Relevant equations Ok so far, I have drawn all of the motion graphs. So I got X Y X_0= 0m Y_0= 0m X_F=? Y_F=? V_0x= 6.5m/s v_oy=0m/s V_fy= 6.5m/s v_fy=? a= 0 m/s^2 a= .48m/s^2 t= 6.3secs t=6.3secs 3. The attempt at a solution And assuming those are correct. I got v_fy= 4.55 m/s by using 6.5tan35 = v_fy then I used this equation: x=x_0+v_ot+(1/2)at^2 and got X_f= 40.95m Y_f= 9.5m And then I added them together and then its 50.45m??? Is this correct, because online it said it was wrong? Thanks for helping! 2. Sep 17, 2007 ### learningphysics suppose it's initially moving eastbound at 6.5m/s. ie: northbound velocity = 0. so acceleration 0.48m/s^2 east 30 degrees north. What is the component of acceleration in the east/west direction? What is the component of acceleration in the north/south direction? So you can divide the problem into the east/west part (to get the displacement east/west), and the north/south part (to get the displacement north/south)... and work them separetly... each part is a uniform acceleration problem. 3. Mar 7, 2010 Need help with kinematics I figured out this problem: You step off a cliff 30 meters high. A. How long will it take to hit the water below? 0=30-4.9t^2 4.9t^2=30 t^2= 6.122 t= 2.47 B. What is your velocity (mph) when you hit the water? Vty= Voy-9.8t Vty-Voy-9.8(2.47) Vty=-24.25 m/s = 54.223 mph Can anyone help me solve this one, based on the above problem?: 2. How high would the cliff have to be in 1. above if your velocity hitting the water was 100 mph? Thanks! Have something to add? Similar Discussions: Help with kinematics

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https://www.jobilize.com/course/section/simple-random-samples-and-statistics-by-openstax?qcr=www.quizover.com

# 13.2 Simple random samples and statistics Page 1 / 3 The (simple) random sample, is basic to much of classical statistics. Once formulated, we may apply probability theory to exhibit several basic ideas of statistical analysis. A population may be most any collection of individuals or entities. Associated with each member is a quantity or a feature that can be assigned a number. The population distribution is the distribution of that quantity among the members of the population.To obtain information about the population distribution, we select “at random” a subset of the population and observe how the quantity varies over the sample. Hopefully, the distribution in the sample will give a useful approximation to the population distribution. We obtain values of such quantities as the mean and variance in the sample (which are random quantities) and use these as estimators for corresponding population parameters (which are fixed). Probability analysis provides estimates of the variation of the sample parameters about the corresponding population parameters. ## Simple random samples and statistics We formulate the notion of a (simple) random sample , which is basic to much of classical statistics. Once formulated, we may apply probability theory to exhibitseveral basic ideas of statistical analysis. We begin with the notion of a population distribution . A population may be most any collection of individuals or entities. Associated with each member is aquantity or a feature that can be assigned a number. The quantity varies throughout the population. The population distribution is the distribution of that quantityamong the members of the population. If each member could be observed, the population distribution could be determined completely. However, that is not always feasible. In order to obtain informationabout the population distribution, we select “at random” a subset of the population and observe how the quantity varies over the sample. Hopefully, thesample distribution will give a useful approximation to the population distribution. The sampling process We take a sample of size n , which means we select n members of the population and observe the quantity associated with each. The selection is done in such a manner thaton any trial each member is equally likely to be selected. Also, the sampling is done in such a way that the result of any one selection does not affect, and is not affected by,the others. It appears that we are describing a composite trial. We model the sampling process as follows: • Let X i , $1\le i\le n$ be the random variable for the i th component trial. Then the class $\left\{{X}_{i}:1\le i\le n\right\}$ is iid, with each member having the population distribution. This provides a model for sampling either from a very large population (often referred to as an infinite population) or sampling with replacement from a small population. The goal is to determine as much as possible about the character of the population. Two important parameters are the mean and the variance. We want the population mean and thepopulation variance. If the sample is representative of the population, then the sample mean and the sample variance should approximate the population quantities. where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials ya I also want to know the raman spectra Bhagvanji I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive

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https://www.physicsforums.com/threads/rc-circuit-maximum-voltage.230160/

# RC circuit maximum voltage • Start date • #1 82 0 i cant figure this out. In an RC circuit, why is the maximum voltage of a capactior greater then the maximum voltage of a resistor? can anyone guide me in the right direction Related Classical Physics News on Phys.org • #2 454 0 It depends on the circuit. Circuits are possible where it is greater, smaller or the same. for an simple RC circuit with one ideal DC voltage source, one resistor and one capacitor, either series or parallel, the maximum voltage would be the same. • Last Post Replies 4 Views 1K • Last Post Replies 5 Views 1K • Last Post Replies 4 Views 1K • Last Post Replies 3 Views 3K • Last Post Replies 7 Views 220 • Last Post Replies 8 Views 1K • Last Post Replies 13 Views 4K • Last Post Replies 4 Views 4K • Last Post Replies 4 Views 2K • Last Post Replies 28 Views 452

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http://mathhelpforum.com/trigonometry/38135-solving-x-given-interval-need-help.html

# Thread: Solving for x in the given interval NEED HELP !!! 1. ## Solving for x in the given interval NEED HELP !!! im stuck on three questions and i really need to solve them a) sec2x + 1/cosx = 0 x is between and equal to 0 and pie b)cos^2x+2sinxcosx-sin^2x=0 x is between and equal to 0 and two pie c)2tanx = secx x is between and equal to negative pie and two pie 2. Hello, math71321! $\displaystyle b)\;\;\cos^2\!x + 2\sin x\cos x-\sin^2\!x\:=\:0 \qquad0 \leq x \leq 2\pi$ We have: . $\displaystyle \underbrace{\cos^2\!x - \sin^2\!x} + \underbrace{2\sin x\cos x} \;=\;0$ . . . . . . . . . . $\displaystyle \cos2x \quad\;\;+ \quad\;\;\sin 2x \quad=\;0$ Then: .$\displaystyle \sin2x \:=\:-\cos2x \quad\Rightarrow\quad\frac{\sin2x}{\cos2x} \:=\:-1 \quad\Rightarrow\quad \tan2x \:=\:-1$ Hence: .$\displaystyle 2x \;=\;\frac{3\pi}{4},\:\frac{7\pi}{4},\:\frac{11\pi }{4},\:\frac{15\pi}{4}$ Therefore: .$\displaystyle \boxed{x \;=\;\frac{3\pi}{8},\:\frac{7\pi}{8},\:\frac{11\pi }{8},\:\frac{15\pi}{8}}$ $\displaystyle c)\;\;2\tan x \:= \:\sec x\qquad -\pi \leq x \leq 2\pi$ Square both sides: .$\displaystyle 4\tan^2\!x \:=\:\sec^2\!x$ . . . . . . . . . . . . $\displaystyle 4\overbrace{(\sec^2\!x - 1)} \:=\:\sec^2\!x$ which simplifies to: .$\displaystyle 3\sec^2\!x \:=\:4\quad\Rightarrow\quad\sec^2\!x \:=\:\frac{4}{3}\quad\Rightarrow\quad \sec x \:=\:\pm\frac{2}{\sqrt{3}}$ . . Therefore: .$\displaystyle \boxed{x \;=\;\pm\frac{\pi}{6},\:\pm\frac{5\pi}{6},\:\frac{ 7\pi}{6},\:\frac{11\pi}{6}}$

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https://byorgey.wordpress.com/2010/01/26/the-random-graph/

## The random graph Today in my finite model theory class we learned about the Rado graph, which is a graph (unique up to isomorphism among countable graphs) with the extension property: given any two disjoint finite sets of vertices $U$ and $V$, there exists some other vertex $w$ which is adjacent to every vertex in $U$ and none of the vertices in $V$. This graph has some rather astonishing properties. Here’s one: consider starting with $n$ vertices and picking each edge with probability $1/2$. Clearly, there are $2^{\binom n 2}$ different graphs you can get, each with equal probability; this defines a uniform random distribution over simple graphs with $n$ vertices. What if you start with a countably infinite number of vertices instead? The surprising answer is that with probability 1 you get the Rado graph. Yes indeed, the Rado graph is extremely random. It is so random that it is also called “THE random graph”. ```SimpleGraph getRandomGraph() { return radoGraph; // chosen by fair coin flips. // guaranteed to be random. } ``` (See http://xkcd.com/221/.) This entry was posted in grad school, humor and tagged , , . Bookmark the permalink. ### 3 Responses to The random graph 1. Mark Dominus says: Excellent. Thanks for this post! This reminds me of a paper of Janós Pach about “universal” graphs. It is quite easy to produce a countable graph G that has the property that for every finite or countable graph H, H is an induced subgraph of G. The graph that Pach constructed was not the Rado graph, although it seems to me that the Rado graph also has this property. But Pach showed that there is no graph G that contains every graph H other than Kω, the complete graph on infinitely many vertices. (The proof is pretty easy.) But if I am right about the Rado graph being universal in the sense of the previous paragraph, then by this theorem it must contain a Kω. Hm, yes, it does. It follows immediately from the extension property, and in the Wikipedia notation of http://en.wikipedia.org/wiki/Rado_graph, the vertices at http://www.research.att.com/~njas/sequences/index.html?q=0,1,3,11,205 form a Kω. Sorry, just thinking aloud. Thanks again! 2. Mark Dominus says: P.S. misspelled “János”.

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https://electronics.stackexchange.com/questions/396341/how-to-derive-voltage-applied-to-an-antenna-where-transmitter-power-is-known/396342

# How to derive voltage applied to an antenna, where transmitter power is known & identical tx/rx antennas are used with known gain/impedance? Question: Imagine a transmitter + transmitter antenna with the following characteristics: • 433Mhz transmitter transmitting 10dBm • Into an impedance matched (50 ohms) antenna, omni-directional, 3dBi gain • 50 ohm impedance antenna, receiver circuitry presents 50 ohms • Receiving antenna is an identical 3dBi omni-directional antenna. How would I go about calculating the voltage the receiver circuitry would see? I'm trying to learn RF electronics by building a detector for a constant 433.92Mhz carrier, and I can't predict how my diode will behave without knowing the voltage applied across it. My attempt from first principles: 10 dBm is 10mW of power, 3dBi of gain means 10mW X 2 = 20mW radiated from the antenna under ideal conditions. Based on the inverse square law, we get: $$P = { 0.02 \over 4 \Pi r ^2 }$$ $$P = { 0.02 \over 4 \Pi 40 ^2 }$$ $$P = { 0.02 \over 4 \Pi 40 ^2 }$$ $$P = 9.94718394^{-7}$$ Yipes that seems small. This is the bit where I get stuck. Can I just multiply it by the receiving gain (3dBi, basically 2x) and sub it into ohms law (with the impedance of the antenna, 50 ohms) combined with the power formula (P = IV)? $$P = ({V \over R}) V$$ $$2 \times 9.94718394^{-7} = ({V \over 50}) V$$ After doubling the power (because the receiving antenna gives me 3dBi gain) and plugging this formula into Wolfram Alpha to solve, I get ~10mV. This seems low and I'm not confident with the logic of my derivation. Is this correct? • 10 dBm is 10 mW of power and not 10 mV. – Andy aka Sep 16 '18 at 9:54 • 10mW is what I meant, edited question to have correct units. Thanks. – 64bit_twitchyliquid Sep 16 '18 at 10:02 You should consider the Friis transmission loss (or link loss) equation for free space: - Loss (dB) = 32.45 + 20$log_{10}$(f) + 20$log_{10}$(d) Where f is in MHz and d is in kilometres. This equation tells you how many dB of power loss you can expect at a given distance with a given carrier frequency. It is based on free-space and accounts for the reduction in received signal as frequency rises due to the received antenna becoming smaller. This equation is also for antennas that transmit equally in all directions (so called isotropic antennas). With real antennas there is always a gain because real antennas don't transmit equally in all directions hence there is always one direction where there is a higher power density of transmission. The equation also assumes that your antennas are not so close that a proper EM wave hasn't formed. Translated, this means that your antennas must be at least one wavelength apart and, as you appear to be using 40 metres in your question and, your frequency is 433 MHz, there is no problem. Putting numbers in gets you a link loss of 32.45 dB + 52.73 dB - 27.96 dB or 57.22 dB. Given your antennas have a gain of 3 dB each the link loss reduces to about 51.2 dB. Sanity check using on-line calculator: - Given your transmit power is +10 dBm, the received signal power will be 52.2 dB down on +10 dBm or -41.2 dBm or about 0.076 uW. Another related Q and A • Awesome that theres an equation for loss-over-distance, thanks! In terms of working from the received power (3.09uW) to voltage however, can I just substitute it into ohms law & P=IR (as I did above) to calculate the voltage? – 64bit_twitchyliquid Sep 16 '18 at 10:37 • Yes you can be do be aware that a dipole antenna will have an effective source resistance (aka radiation resistance) of about 73 ohms and the power delivered in the equations above assume that the correct resistance is matched to the antenna resistance. Be also aware that an antennas impedance varies dramatically when the antenna length doesn't precisely match the wavelength of the carrrier. – Andy aka Sep 16 '18 at 10:41 Using the formula Power = VoltsRMS ^2 / Resistance, and then changing Resistance to Impedance, we have Power = VoltsRMS^2 / Impedance and we solve for VoltsRMS VoltsRMS = squareroot (Power * Impedance) Now at one milliWatt power (0.001 watt) and 50 ohms, the math produces VoltsRMS = squareroot (0.001 * 50) = squareroot (1 / 20) = 1 / squareroot(20) VoltsRMS = 1 / 4.47 = 0.223 volts rms VoltsPP = 0.223 * 2 * 1.414 = 0.632 voltsPP Are your numbers consistent?

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http://mathhelpforum.com/math-challenge-problems/6470-question-2-a.html

# Math Help - Question 2 1. ## Question 2 A triangular number is the sequence: 1,3,6,10,15,21,... A square number is the sequence: 1,4,9,16,25,... Show that there are these two sequences contain infinitely many same numbers, i.e. 1 and 1. 2. Originally Posted by ThePerfectHacker A triangular number is the sequence: 1,3,6,10,15,21,... A square number is the sequence: 1,4,9,16,25,... Show that there are these two sequences contain infinitely many same numbers, i.e. 1 and 1. To clarify: You mean that the set of numbers that belong to both the triangular sequence and the square sequence has an infinite number of members? -Dan 3. Originally Posted by topsquark To clarify: You mean that the set of numbers that belong to both the triangular sequence and the square sequence has an infinite number of members? -Dan Yes 4. I do not understand the last line of the solution? 5. Hello, TPHacker! Generally, with this type of problem we can generate solutions with a recurrence formula of the form: . . nth term .= .k·(preceding term) ± (term before) ... for some constant k. In short, I "eyeballed" it . . . 6. Beautiful job Soroban +rep+ Never seen it done like that. I have modified (not really the right word) your recurrence relation into a different form: --- I always tried to show that the cubic sequence and triangular sequence only have a trivial solution i.e. 1. And it seems to be true for higher orders also but I was unable to prove that, you know how? Attached Thumbnails 7. Hello again, TPHacker! I got a different closed form for the recurrence. It's not my own solution; I learned the method from a few of my books. You may have seen something similar while exploring. Now that LaTeX is back, I'll revise this post. We have the recurrence: . $f(n) \:=\:6\cdot f(n-1) - f(n-2)$ Assume that $f(n)$ is exponential: . $f(n) \:=\:X^n$ The equation becomes: . $X^n \:=\:6X^{n-1} - X^{n-2}$ Divide by $X^{n-2}$ and we have: . $X^2 - 6X + 1 \:= \:0$ . . which has roots: . $X \:=\:3 \pm 2\sqrt{2}$ We form a linear combination of the two roots: . . $f(n) \:=\:A(3 + 2\sqrt{2})^n + B(3 - 2\sqrt{2})^n$ To solve for $A$ and $B$, we use the first two values of the sequence: . . $\begin{array}{cc}f(1) \:= \\ f(2) \:=\end{array} \begin{array}{cc} A(3 + 2\sqrt{2}) + (3 - 2\sqrt{2})\\ A(3 + 2\sqrt{2})^2 + (3 - 2\sqrt{2})^2\end{array}\begin{array}{cc}= \:1 \\ = \:6\end{array}$ Solve the system and get: . $A \,= \,\frac{1}{4\sqrt{2}},\;\;B \,= \,-\frac{1}{4\sqrt{2}}$ Therefore: . $f(n) \;= \;\frac{(3 + 2\sqrt{2})^n - (3 - 2\sqrt{2})^n}{4\sqrt{2}}$ By the way: . $3 \pm 2\sqrt{2} \:=\:(1 \pm \sqrt{2})^2$ . . so you can rewrite the formula if you like . . . 8. Besides for Soroban's solution.... I have 3 more (1 of which I am too lazy to post). Let $t_n$ be the $n-th$triangular number. Solution 1) If t_n is a square then t_{4n(n+1)} is a square. How you are suppopsed to see that, I have no idea. Solution 2) Following the start of Soroban;s solution we arrive at the square root of 8k² +1, which is the "Pellian equation". Thus it has infinitely many solutions.

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http://gams.cam.nist.gov/13.9

# §13.9 Zeros ## §13.9(i) Zeros of $M\left(a,b,z\right)$ If $a$ and $b-a\neq 0,-1,-2,\dots$, then $M\left(a,b,z\right)$ has infinitely many $z$-zeros in $\mathbb{C}$. When $a,b\in\mathbb{R}$ the number of real zeros is finite. Let $p(a,b)$ be the number of positive zeros. Then 13.9.1 $\displaystyle p(a,b)$ $\displaystyle=\left\lceil-a\right\rceil,$ $a<0$, $b\geq 0$, ⓘ Symbols: $\left\lceil\NVar{x}\right\rceil$: ceiling of $x$ and $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E1 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 13.9.2 $\displaystyle p(a,b)$ $\displaystyle=0,$ $a\geq 0$, $b\geq 0$, ⓘ Symbols: $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E2 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 13.9.3 $\displaystyle p(a,b)$ $\displaystyle=1,$ $a\geq 0$, $-1, ⓘ Symbols: $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E3 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 13.9.4 $p(a,b)=\left\lfloor-\tfrac{1}{2}b\right\rfloor-\left\lfloor-\tfrac{1}{2}(b+1)% \right\rfloor,$ $a\geq 0$, $b\leq-1$. ⓘ Symbols: $\left\lfloor\NVar{x}\right\rfloor$: floor of $x$ and $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E4 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 13.9.5 $p(a,b)=\left\lceil-a\right\rceil-\left\lceil-b\right\rceil,$ $\left\lceil-a\right\rceil\geq\left\lceil-b\right\rceil$, $a<0$, $b<0$, ⓘ Symbols: $\left\lceil\NVar{x}\right\rceil$: ceiling of $x$ and $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E5 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 13.9.6 $p(a,b)=\left\lfloor\tfrac{1}{2}\left(\left\lceil-b\right\rceil-\left\lceil-a% \right\rceil+1\right)\right\rfloor-\left\lfloor\tfrac{1}{2}\left(\left\lceil-b% \right\rceil-\left\lceil-a\right\rceil\right)\right\rfloor,$ $\left\lceil-b\right\rceil>\left\lceil-a\right\rceil>0$. The number of negative real zeros $n(a,b)$ is given by 13.9.7 $n(a,b)=p(b-a,b).$ ⓘ Defines: $n(a,b)$: number of negative real zeros (locally) Symbols: $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E7 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 When $a<0$ and $b>0$ let $\phi_{r}$, $r=1,2,3,\dots$, be the positive zeros of $M\left(a,b,x\right)$ arranged in increasing order of magnitude, and let $j_{b-1,r}$ be the $r$th positive zero of the Bessel function $J_{b-1}\left(x\right)$10.21(i)). Then 13.9.8 $\phi_{r}=\frac{j_{b-1,r}^{2}}{2b-4a}\left(1+\frac{2b(b-2)+j_{b-1,r}^{2}}{3(2b-% 4a)^{2}}\right)+O\left(\frac{1}{a^{5}}\right),$ ⓘ Symbols: $O\left(\NVar{x}\right)$: order not exceeding, $r=1,2,3,\dots$, $\phi_{r}$: positive zeros and $j_{b,r}$: positive zero of Bessel Referenced by: §13.22, §13.9(i) Permalink: http://dlmf.nist.gov/13.9.E8 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 as $a\to-\infty$ with $r$ fixed. Inequalities for $\phi_{r}$ are given in Gatteschi (1990), and identities involving infinite series of all of the complex zeros of $M\left(a,b,x\right)$ are given in Ahmed and Muldoon (1980). For fixed $a,b\in\mathbb{C}$ the large $z$-zeros of $M\left(a,b,z\right)$ satisfy 13.9.9 $z=\pm(2n+a)\pi\mathrm{i}+\ln\left(-\frac{\Gamma\left(a\right)}{\Gamma\left(b-a% \right)}\left(\pm 2n\pi\mathrm{i}\right)^{b-2a}\right)+O\left(n^{-1}\ln n% \right),$ where $n$ is a large positive integer, and the logarithm takes its principal value (§4.2(i)). Let $P_{\alpha}$ denote the closure of the domain that is bounded by the parabola $y^{2}=4\alpha(x+\alpha)$ and contains the origin. Then $M\left(a,b,z\right)$ has no zeros in the regions $P_{\ifrac{b}{a}}$, if $0; $P_{1}$, if $1\leq a\leq b$; $P_{\alpha}$, where $\alpha=\ifrac{(2a-b+ab)}{(a(a+1))}$, if $0 and $a\leq b<\ifrac{2a}{(1-a)}$. The same results apply for the $n$th partial sums of the Maclaurin series (13.2.2) of $M\left(a,b,z\right)$. More information on the location of real zeros can be found in Zarzo et al. (1995) and Segura (2008). For fixed $b$ and $z$ in $\mathbb{C}$ the large $a$-zeros of $M\left(a,b,z\right)$ are given by 13.9.10 $a=-\frac{\pi^{2}}{4z}\left(n^{2}+(b-\tfrac{3}{2})n\right)-\frac{1}{16z}\left((% b-\tfrac{3}{2})^{2}\pi^{2}+\tfrac{4}{3}z^{2}-8b(z-1)-4b^{2}-3\right)+O\left(n^% {-1}\right),$ ⓘ Symbols: $O\left(\NVar{x}\right)$: order not exceeding, $\pi$: the ratio of the circumference of a circle to its diameter, $n$: nonnegative integer and $z$: complex variable Referenced by: §13.9(i) Permalink: http://dlmf.nist.gov/13.9.E10 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 where $n$ is a large positive integer. For fixed $a$ and $z$ in $\mathbb{C}$ the function $M\left(a,b,z\right)$ has only a finite number of $b$-zeros. ## §13.9(ii) Zeros of $U\left(a,b,z\right)$ For fixed $a$ and $b$ in $\mathbb{C}$, $U\left(a,b,z\right)$ has a finite number of $z$-zeros in the sector $|\operatorname{ph}z|\leq\tfrac{3}{2}\pi-\delta(<\tfrac{3}{2}\pi)$. Let $T(a,b)$ be the total number of zeros in the sector $|\operatorname{ph}z|<\pi$, $P(a,b)$ be the corresponding number of positive zeros, and $a$, $b$, and $a-b+1$ be nonintegers. For the case $b\leq 1$ 13.9.11 $T(a,b)=\left\lfloor-a\right\rfloor+1,$ $a<0$, $\Gamma\left(a\right)\Gamma\left(a-b+1\right)>0$, ⓘ Symbols: $\Gamma\left(\NVar{z}\right)$: gamma function, $\left\lfloor\NVar{x}\right\rfloor$: floor of $x$ and $T(a,b)$: number of zeros Permalink: http://dlmf.nist.gov/13.9.E11 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13 13.9.12 $T(a,b)=\left\lfloor-a\right\rfloor,$ $a<0$, $\Gamma\left(a\right)\Gamma\left(a-b+1\right)<0$, ⓘ Symbols: $\Gamma\left(\NVar{z}\right)$: gamma function, $\left\lfloor\NVar{x}\right\rfloor$: floor of $x$ and $T(a,b)$: number of zeros Permalink: http://dlmf.nist.gov/13.9.E12 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13 13.9.13 $T(a,b)=0,$ $a>0$, ⓘ Symbols: $T(a,b)$: number of zeros Permalink: http://dlmf.nist.gov/13.9.E13 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13 and 13.9.14 $P(a,b)=\left\lceil b-a-1\right\rceil,$ $a+1, ⓘ Symbols: $\left\lceil\NVar{x}\right\rceil$: ceiling of $x$ and $P(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E14 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13 13.9.15 $P(a,b)=0,$ $a+1\geq b$. ⓘ Symbols: $P(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E15 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13 For the case $b\geq 1$ we can use $T(a,b)=T(a-b+1,2-b)$ and $P(a,b)=P(a-b+1,2-b)$. In Wimp (1965) it is shown that if $a,b\in\mathbb{R}$ and $2a-b>-1$, then $U\left(a,b,z\right)$ has no zeros in the sector $|\operatorname{ph}{z}|\leq\frac{1}{2}\pi$. Inequalities for the zeros of $U\left(a,b,x\right)$ are given in Gatteschi (1990). See also Segura (2008). For fixed $b$ and $z$ in $\mathbb{C}$ the large $a$-zeros of $U\left(a,b,z\right)$ are given by 13.9.16 $a=-n-\frac{2}{\pi}\sqrt{zn}-\frac{2z}{\pi^{2}}+\tfrac{1}{2}b+\tfrac{1}{4}+% \frac{z^{2}\left(\frac{1}{3}-4\pi^{-2}\right)+z-(b-1)^{2}+\frac{1}{4}}{4\pi% \sqrt{zn}}+O\left(\frac{1}{n}\right),$ ⓘ Symbols: $O\left(\NVar{x}\right)$: order not exceeding, $\sim$: Poincaré asymptotic expansion, $\pi$: the ratio of the circumference of a circle to its diameter, $n$: nonnegative integer and $z$: complex variable Referenced by: §13.9(ii), § Recent News, Other Changes, Other Changes Permalink: http://dlmf.nist.gov/13.9.E16 Encodings: TeX, pMML, png Errata (effective with 1.0.13): In applying changes in Version 1.0.12 to this equation, an editing error was made; it has been corrected. Reported 2016-09-12 by Adri Olde Daalhuis Errata (effective with 1.0.12): Originally this equation was expressed in terms of the asymptotic symbol $\sim$. As a consequence of the use of the $O$ order symbol on the right hand side, $\sim$ was replaced by $=$. Reported 2016-07-11 by Rudi Weikard See also: Annotations for 13.9(ii), 13.9 and 13 where $n$ is a large positive integer. For fixed $a$ and $z$ in $\mathbb{C}$, $U\left(a,b,z\right)$ has two infinite strings of $b$-zeros that are asymptotic to the imaginary axis as $|b|\to\infty$.

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http://www.sciencebits.com/taxonomy/term/11?page=2

## On Climate Sensitivity and why it is probably small ### What is climate sensitivity? The equilibrium climate sensitivity refers to the equilibrium change in average global surface air temperature following a unit change in the radiative forcing. This sensitivity (often denoted as λ) therefore has units of °C/(W/m2). Often, instead &\lambda;, the sensitivity is expressed through the temperature change &Delta Tx2, in response to a doubled atmospheric CO2 content, which is equivalent to a radiative forcing of 3.8 W/m2. Thus, &Delta Tx2 = 3.8 W/m2 λ ## Standing on ice - When is it possible? Ever wondered whether it was sufficiently cold for sufficiently long to allow you to stand on ice, without falling in? I once did. Here is an estimate for the duration required to reach a given thickness. Actually, it is a lower limit, since we assume a few simplifying assumptions. ## Exhale Condensation Calculator If the temperature is low enough or the humidity high, you can observe condensation (i.e., "fog") forming in your exhaled breath. This calculator estimates whether your exhaled breath will condense, and if so, the range of mixing ratios for which the "fog" will form and the maximum condensed water content (the higher it is, the "thicker" the condensation). If you're interested, there is a much more detailed explanations of the condensation process. Exhaled Condensation Calculator Using the above equations, we can calculate whether the exhaled air will condense. Enter the conditions of the outside air (and modify the exhaled air parameters if you wish), to see whether your breath will condense, or not. ## Blogroll Sensible Climate Physics and more Other

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https://collaborate.princeton.edu/en/publications/topology-bounded-superfluid-weight-in-twisted-bilayer-graphene

# Topology-Bounded Superfluid Weight in Twisted Bilayer Graphene Fang Xie, Zhida Song, Biao Lian, B. Andrei Bernevig Research output: Contribution to journalArticlepeer-review 23 Scopus citations ## Abstract While regular flat bands are good for enhancing the density of states and hence the gap, they are detrimental to the superfluid weight. We show that the predicted nontrivial topology of the two lowest flat bands of twisted bilayer graphene (TBLG) plays an important role in the enhancement of the superfluid weight and hence of superconductivity. We derive the superfluid weight (phase stiffness) of the TBLG superconducting flat bands with a uniform pairing, and show that it can be expressed as an integral of the Fubini-Study metric of the flat bands. This mirrors results already obtained for nonzero Chern number bands even though the TBLG flat bands have zero Chern number. We further show that the metric integral is lower bounded by the topological C2zT Wilson loop winding number of TBLG flat bands, which renders that the superfluid weight is also bounded by this topological index. In contrast, trivial flat bands have a zero superfluid weight. The superfluid weight is crucial in determining the Berezinskii-Kosterlitz-Thouless transition temperature of the superconductor. Based on the transition temperature measured in TBLG experiments, we estimate the topological contribution of the superfluid weight in TBLG. Original language English (US) 167002 Physical review letters 124 16 https://doi.org/10.1103/PhysRevLett.124.167002 Published - Apr 24 2020 ## All Science Journal Classification (ASJC) codes • Physics and Astronomy(all) ## Fingerprint Dive into the research topics of 'Topology-Bounded Superfluid Weight in Twisted Bilayer Graphene'. Together they form a unique fingerprint.

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https://www.hepdata.net/search/?q=cmenergies%3A%7B1.3+TO+1.4%7D&author=Aubert%2C+Bernard

Showing 5 of 5 results #### rivet Analysis The e+ e- ---> 2(pi+ pi-) pi0, 2(pi+ pi-) eta, K+ K- pi+ pi- pi0 and K+ K- pi+ pi- eta Cross Sections Measured with Initial-State Radiation The collaboration Aubert, Bernard ; Bona, M. ; Boutigny, D. ; et al. Phys.Rev. D76 (2007) 092005, 2007. Inspire Record 758568 0 data tables match query #### rivet Analysis Measurements of $e^{+} e^{-} \to K^{+} K^{-} \eta$, $K^{+} K^{-} \pi^0$ and $K^0_{s} K^\pm \pi^\mp$ cross- sections using initial state radiation events The collaboration Aubert, Bernard ; Bona, M. ; Boutigny, D. ; et al. Phys.Rev. D77 (2008) 092002, 2008. Inspire Record 765258 This paper reports measurements of processes: e+e- -> gamma KsK+pi-, e+e- -> gamma K+K-pi0, e+e- -> gamma phi eta, and e+e- -> gamma phi pi0. The initial state radiated photon allows to cover the hadronic final state in the energy range from thresholds up to ~4.6 GeV. The overall size of the data sample analyzed is 232 fb-1, collected by the BaBar detector running at the PEP-II e+e- storage ring. From the Dalitz plot analysis of the KsK+pi- final state, moduli and relative phase of the isoscalar and the isovector components of the e+e- -> K K*(892) cross section are determined. Parameters of phi and rho recurrences are also measured, using a global fitting procedure which exploits the interconnection among amplitudes, moduli and phases of the e+e- -> KsK+pi-, K+K-pi0, phi eta final states. The cross section for the OZI-forbidden process e+e- -> phi pi0, and the J/psi branching fractions to KK*(892) and K+K-eta are also measured. 0 data tables match query #### rivet Analysis The $e^+e^- \to \pi^+ \pi^- \pi^+ \pi^-$, $K^+ K^- \pi^+ \pi^-$, and $K^+ K^- K^+ K^-$ cross sections at center-of-mass energies 0.5-GeV - 4.5-GeV measured with initial-state radiation The collaboration Aubert, Bernard ; Barate, R. ; Boutigny, D. ; et al. Phys.Rev. D71 (2005) 052001, 2005. Inspire Record 676691 We study the process $e^+e^-\to\pi^+\pi^-\pi^+\pi^-\gamma$, with a hard photon radiated from the initial state. About 60,000 fully reconstructed events have been selected from 89 $fb^{-1}$ of BaBar data. The invariant mass of the hadronic final state defines the effective \epem center-of-mass energy, so that these data can be compared with the corresponding direct $e^+e^-$ measurements. From the $4\pi$-mass spectrum, the cross section for the process $e^+e^-\to\pi^+\pi^-\pi^+\pi^-$ is measured for center-of-mass energies from 0.6 to 4.5 $GeV/c^2$. The uncertainty in the cross section measurement is typically 5%. We also measure the cross sections for the final states $K^+ K^- \pi^+\pi^-$ and $K^+ K^- K^+ K^-$. We observe the $J/\psi$ in all three final states and measure the corresponding branching fractions. We search for X(3872) in $J/\psi (\to\mu^+\mu^-) \pi^+\pi^-$ and obtain an upper limit on the product of the $e^+e^-$ width of the X(3872) and the branching fraction for $X(3872) \to J/\psi\pi^+\pi^-$. 0 data tables match query #### rivet Analysis Cross sections for the reactions $e^+ e^-\to K_S^0 K_L^0$, $K_S^0 K_L^0 \pi^+\pi^-$, $K_S^0 K_S^0 \pi^+\pi^-$, and $K_S^0 K_S^0 K^+K^-$ from events with initial-state radiation The collaboration Lees, J.P. ; Poireau, V. ; Tisserand, V. ; et al. Phys.Rev. D89 (2014) 092002, 2014. Inspire Record 1287920 We study the processes $e^+ e^-\to K_S^0 K_L^0 \gamma$, $K_S^0 K_L^0 \pi^+\pi^-\gamma$, $K_S^0 K_S^0 \pi^+\pi^-\gamma$, and $K_S^0 K_S^0 K^+K^-\gamma$, where the photon is radiated from the initial state, providing cross section measurements for the hadronic states over a continuum of center-of-mass energies. The results are based on 469 fb$^{-1}$ of data collected with the BaBar detector at SLAC. We observe the $\phi(1020)$ resonance in the $K_S^0 K_L^0$ final state and measure the product of its electronic width and branching fraction with about 3% uncertainty. We present a measurement of the $e^+ e^-\to K_S^0 K_L^0$ cross section in the energy range from 1.06 to 2.2 GeV and observe the production of a resonance at 1.67 GeV. We present the first measurements of the $e^+ e^-\to K_S^0 K_L^0 \pi^+\pi^-$, $K_S^0 K_S^0 \pi^+\pi^-$, and $K_S^0 K_S^0 K^+K^-$ cross sections, and study the intermediate resonance structures. We obtain the first observations of \jpsi decay to the $K_S^0 K_L^0 \pi^+\pi^-$, $K_S^0 K_S^0 \pi^+\pi^-$, and $K_S^0 K_S^0 K^+K^-$ final states. 0 data tables match query #### rivet Analysis Cross Sections for the Reactions e+e- --> K+ K- pi+pi-, K+ K- pi0pi0, and K+ K- K+ K- Measured Using Initial-State Radiation Events The collaboration Lees, J.P. ; Poireau, V. ; Prencipe, E. ; et al. Phys.Rev. D86 (2012) 012008, 2012. Inspire Record 892684 0 data tables match query

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https://astarmathsandphysics.com/university-maths-notes/elementary-calculus/1788-maxima-minima-and-saddle-points-the-second-partials-test.html

## Maxima, Minima and Saddle Points – The Second Partials Test Suppose thathas continuous second partial derivatives in a neighbourhood of and thatat DefineatForm the discriminantIfthenis a saddle point. Ifthenhas a local minimum at Ifand a local maximum atif

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https://eprints.soton.ac.uk/417377/

The University of Southampton University of Southampton Institutional Repository # On the joint and marginal densities of instrumental variable estimators in a general structural equation Hillier, Grant H. (1985) On the joint and marginal densities of instrumental variable estimators in a general structural equation. Econometric Theory, 1 (1), 53-72. Record type: Article ## Abstract Starting from the conditional density of the instrumental variable (IV) estimator given the right-hand-side endogenous variables, we provide an alternative derivation of Phillips' result on the joint density of the IV estimator for the endogenous coefficients, and derive an expression for the marginal density of a linear combination of these coefficients. In addition, we extend Phillips' approximation to the joint density to 0(T−2,) and show how this result can be used to improve the approximation to the marginal density. Explicit formulae are given for the special case of no simultaneity, and the case of an equation with just three endogenous variables. The classical assumptions of independent normal reduced-form errors are employed throughout. Published date: April 1985 ## Identifiers Local EPrints ID: 417377 URI: http://eprints.soton.ac.uk/id/eprint/417377 ISSN: 0266-4666 PURE UUID: 8c15d018-7615-4aca-a83e-dc62522ee011 ORCID for Grant H. Hillier: orcid.org/0000-0003-3261-5766 ## Catalogue record Date deposited: 30 Jan 2018 17:30

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https://physics.stackexchange.com/questions/256538/the-convective-term-in-reynolds-transport-theorem

# The convective term in Reynolds Transport theorem Reynolds Transport theorem states that $$\frac{D B_{sys}}{Dt}=\frac{\partial B_{CV}}{\partial t}-\dot{B_{in}}+\dot{B_{out}}$$ where $B_{in,t+\Delta t}=b_1m_{1,t+\Delta t}=b_1\rho_{1}\forall_{in,t+\Delta t}=b_1\rho_1V_1\Delta tA_1$, then $$\dot{B_{in}}=\lim_{\Delta t\to0}\frac{B_{in,t+\Delta}}{\Delta t}=\lim_{\Delta t\to0}\frac{b_1\rho_1V_1\Delta tA_1}{\Delta t}=b_1\rho_1V_1A_1$$ Now my question is, how come volume $\forall_{in}$ at $t+\Delta t$ is equal to $V_1 \Delta tA_1$?, where $V_1$ is the velocity and $A_1$ is the area. Dimension-ally, this relation works but still do not understand the physics. $V_1 A_1$ is basically the volumetric flowrate. So in a time $\Delta t$, a particle moving with flowrate $V_1 A_1$, sweeps out a volume $V_1 A_1 \Delta t$.

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https://optimization-online.org/author/jfp/

## An easily computable upper bound on the Hoffman constant for homogeneous inequality systems Let $A\in \mathbb{R}^{m\times n}\setminus \{0\}$ and $P:=\{x:Ax\le 0\}$. This paper provides a procedure to compute an upper bound on the following {\em homogeneous Hoffman constant} $H_0(A) := \sup_{u\in \mathbb{R}^n \setminus P} \frac{\text{dist}(u,P)}{\text{dist}(Au, \mathbb{R}^m_-)}.$ In sharp contrast to the intractability of computing more general Hoffman constants, the procedure described in this paper is entirely … Read more ## Affine invariant convergence rates of the conditional gradient method We show that the conditional gradient method for the convex composite problem $\min_x\{f(x) + \Psi(x)\}$ generates primal and dual iterates with a duality gap converging to zero provided a suitable growth property holds and the algorithm makes a judicious choice of stepsizes. The rate of convergence of the duality gap to zero ranges from sublinear … Read more ## Projection and rescaling algorithm for finding most interior solutions to polyhedral conic systems We propose a simple projection and rescaling algorithm that finds {\em most interior} solutions to the pair of feasibility problems $\text{find} x\in L\cap \R^n_{+} \text{ and } \text{find} \; \hat x\in L^\perp\cap\R^n_{+},$ where $L$ is a linear subspace of $\R^n$ and $L^\perp$ is its orthogonal complement. The algorithm complements a basic procedure that … Read more ## Equivalences among the chi measure, Hoffman constant, and Renegar’s distance to ill-posedness We show the equivalence among the following three condition measures of a full column rank matrix $A$: the chi measure, the signed Hoffman constant, and the signed distance to ill-posedness. The latter two measures are constructed via suitable collections of matrices obtained by flipping the signs of some rows of $A$. Our results provide a … Read more ## New characterizations of Hoffman constants for systems of linear constraints We give a characterization of the Hoffman constant of a system of linear constraints in $\R^n$ relative to a reference polyhedron $R\subseteq\R^n$. The reference polyhedron $R$ represents constraints that are easy to satisfy such as box constraints. In the special case $R = \R^n$, we obtain a novel characterization of the classical Hoffman constant. More … Read more ## Generalized conditional subgradient and generalized mirror descent: duality, convergence, and symmetry We provide new insight into a generalized conditional subgradient algorithm and a generalized mirror descent algorithm for the convex minimization problem $\min_x \; \{f(Ax) + h(x)\}.$ As Bach showed in [SIAM J. Optim., 25 (2015), pp. 115–129], applying either of these two algorithms to this problem is equivalent to applying the other one to its … Read more ## The condition number of a function relative to a set The condition number of a differentiable convex function, namely the ratio of its smoothness to strong convexity constants, is closely tied to fundamental properties of the function. In particular, the condition number of a quadratic convex function is the square of the aspect ratio of a canonical ellipsoid associated to the function. Furthermore, the condition … Read more

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http://math.stackexchange.com/questions/265451/the-cardinality-of-mathbbr-mathbb-q/265509

# The cardinality of $\mathbb{R}/\mathbb Q$ How to prove the cardinality of $\mathbb{R}/ \mathbb Q$ is equal to the cardinality of $\mathbb{R}$ - Sorry, I wrote it wrong, I meant to be the cardinality of $\mathbb{R}$ mod $\mathbb{Q}$ – Alex Dec 26 '12 at 17:52 You mean the quotient group $\mathbb R / \mathbb Q$ ?? – GEdgar Dec 26 '12 at 17:54 Alex, please edit to clarify what you are asking. If you are asking about the cardinality of the set of irrational numbers (as it is currently written), see math.stackexchange.com/questions/105990/… and math.stackexchange.com/questions/72130/…. – Jonas Meyer Dec 26 '12 at 18:03 This is a consequence of the axiom of choice. How to prove it, depends a bit on your background, that is, what results you can assume, how familiar you are with choice, etc. Could you specify some of it? – Andrés Caicedo Dec 26 '12 at 19:27 If it is the quotient, in fact, without the axiom of choice, it turns out that $\mathbb{R}/\mathbb{Q}$ can have strictly larger cardinality than $\mathbb{R}$... math.stackexchange.com/a/243549/32178 – KSackel Dec 26 '12 at 20:02 To prove equality we need to either find a bijection between the sets, or two injections between them. As noted in the comments, this cannot be proved without the axiom of choice. So I am going to use it freely. Assuming the axiom of choice, if so, we have a function $f\colon\mathbb{R/Q\to R}$ which chooses $f(A)\in A$ for every $A\in\mathbb{R/Q}$. This is an injection because if $A\neq A'$ then $f(A)\in A$ and $f(A)\notin A'$, and vice versa, therefore $f(A)\neq f(A')$. On the other hand, let $V=\operatorname{rng}(f)$, then $\mathbb R$ is a countable union of copies of $V$, namely $\bigcup_{q\in\mathbb Q}q+V$. Therefore $|V|\cdot\aleph_0=2^{\aleph_0}$. Again, using the axiom of choice, we have that $2^{\aleph_0}=|V|\cdot\aleph_0=\max\{|V|,\aleph_0\}=|V|$. - Remark: This answers the original version of the question. The following bijection uses Hilbert's infinite hotel. The rationals can be enumerated as $q_0,q_1,q_2,\dots$ in various explicit ways. Define $f: \mathbb{R}\to \mathbb{R}\setminus \mathbb{Q}$ as follows. If $x$ does not have shape $q$, or $\sqrt{3}+q\sqrt{2}$, where $q$ is rational, let $f(x)=x$. If $x$ is the rational $q_i$, let $f(x)=\sqrt{3}+q_{2i}\sqrt{2}$. If $x=\sqrt{3}+q_i\sqrt{2}$, let $f(x)=\sqrt{3}+q_{2i+1}\sqrt{2}$. - Awesome.[filler] – akkkk Dec 26 '12 at 19:28 $\mathbb{R}= (\mathbb{R} \backslash \mathbb{Q}) \sqcup \mathbb{Q}$, where $\mathbb{Q}$ is countable and $\mathbb{R} \backslash \mathbb{Q}$ infinite, so $$|\mathbb{R}|= |\mathbb{R} \backslash \mathbb{Q} | + | \mathbb{Q}|= \max ( |\mathbb{R} \backslash \mathbb{Q} |, |\mathbb{Q}|) = |\mathbb{R} \backslash \mathbb{Q} |$$ - Indeed, $\mathbb R \setminus \mathbb Q$ is infinite. And so is $\mathbb Q$. You want to write "uncountable", I think. – Rudy the Reindeer Dec 26 '12 at 18:57 @MattN. - being infinite suffices since $\aleph_0$ + any infinite cardinal $k$ is $k$ – Belgi Dec 26 '12 at 19:36 @Belgi Right, I misread the question as having to show that $\mathbb R \setminus \mathbb Q$ is uncountable. – Rudy the Reindeer Dec 26 '12 at 20:52 I don't see what I would have expected to be the "standard answer" to this question, so let me leave it in the hopes it will be helpful to someone. Proposition: Let $G$ be an infinite group, and let $H$ be a subgroup with $\# H < \# G$. Then $\# G/H = \# G$. Proof: Let $\{g_i\}_{i \in G/H}$ be a system of coset representatives for $H$ in $G$: then every element $x$ in $G$ can be written as $x = g_{i_x} h_x$ for unique $h_x \in H$ and $i_x \in G/H$. (Note that there is no canonical system of coset representatives: getting one is an archetypical use of the Axiom of Choice.) Thus we have defined a bijection from $G$ to $G/H \times H$, so $\# G = \# G/H \cdot \# H$. Since $\# G$ is infinite, so must be at least one of $\# G/H$, $\# H$, and then standard cardinal arithmetic (again AC gets used...) gives that $\# G = \# G/H \cdot \# H = \max(\#G/H, \# H)$. Since we've assumed $\# H < \# G$, we conclude $\# G = \#G/H$. This applies in particular with $G = \mathbb{R}$, $H = \mathbb{Q}$ to give $\# \mathbb{R}/\mathbb{Q} = \# \mathbb{R} = 2^{\aleph_0}$. - To see that the uses of choice here are unavoidable: math.stackexchange.com/questions/243544/… – Andrés Caicedo Apr 23 '13 at 5:58 One can even write a general form theorem from cardinal arithmetics. If $f\colon A\to B$ is surjective, and every fiber have the same cardinality, which is less than that of $A$ then $|A|=|B|$. – Asaf Karagila Apr 23 '13 at 6:17 $\mathbb{R}/\mathbb{Q}$ and $\mathbb{R}$ are both size continuum. So by assuming as vector spaces over the $\mathbb{Q}$ they must have continuum size bases. We know that if two vector spaces have bases of the same size then they are isomorphic. - So why exactly are they both size continuum? – akkkk Dec 26 '12 at 18:45 @akkkk: Because if B be a basis of R over Q (as a vector space, that is). The cardinality of B is c and that R/Q is like removing one basis element, or one copy of Q. Is that wrong? – S. Snape Dec 26 '12 at 18:48 Babak the OP wants to prove that $\mathbb{R}-\mathbb{Q}$ is of size continuum which you assumed in the first line. – Belgi Dec 26 '12 at 19:35 @Belgi: I wonder what should I do here. Firstly, the OP wrote $R-Q$ and some different good answer came to him, and suddenly he changed his mind to quotient group. See what he replied in the comment. – S. Snape Dec 26 '12 at 19:43 @Belgi: Actually, in the comments the OP says that he is looking for the cardinality of the quotient. – Asaf Karagila Dec 29 '12 at 15:02 Well, heuristically it can go as follows. We can easily construct a bijection between $\mathbb{N}$ and $\mathbb{N}-\{1\}$. Just send $n \rightarrow n+1$. If we subtract one element $x_1$ from $\mathbb{R}$, we can decompose $\mathbb{R}$ as $(\mathbb{R}-\mathbb{N})\cup\mathbb{N}$, and $\mathbb{R}-\{x_0\}=(\mathbb{R}-\mathbb{N})\cup(\mathbb{N}-{x_0})$. And we apply the above trick to the latter part. If we subtract countable elements, say, $x_i$ for all $i\in\mathbb{N}$, we can choose a countable familiy of subsets with cardinal equal to that of $\mathbb{N}$, each containing one of $x_i$. Now decompose $\mathbb{R}$ as a countable union and do as above and you will solve the problem. - But we don't subtract the rationals. We take the quotient. – Asaf Karagila Jan 4 '13 at 14:13

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http://math.stackexchange.com/questions/101636/jacobis-criterion-for-projective-schemes

# Jacobi's criterion for projective schemes? When can we apply the Jacobi's criterion for the projective variety $V(f_{1}, \ldots, f_{r}) \subset \mathbb{P}^{n}$ in order to find the singularities of the scheme $\mathrm{Proj} \left( k[x_{1}, \ldots, x_{n+1}] / (f_{1}, \ldots, f_{r}) \right)$? In Hartshorne's book Algebraic Geometry, Proposition II.2.6, we have a fully faithful functor from the category of varieties over $k$ to the category of schemes over $k$, but it seems to provide information only for the closed points of the scheme. Thank you. - Check out the general Jacobi's criterion in EGA or in Liu's book. –  Martin Brandenburg Jan 23 '12 at 15:33 The Jacobian Criterion can be applied to any kind of point on a projective (or affine) variety, closed or not. In the non-closed case one has to adapt the requirement for the rank of the Jacobian appropriately. The field $k$ should be perfect.

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https://zbmath.org/?q=an:0818.35136

× # zbMATH — the first resource for mathematics Another step toward the solution of the Pompeiu problem in the plane. (English) Zbl 0818.35136 Let $$\Omega \subset \mathbb{R}^ 2$$ be a bounded open set for which $$f \equiv 0$$ is the only continuous function on $$\mathbb{R}^ 2$$ such that (1) $$\int_{\sigma (\Omega)} f(x) dx = 0$$ for every rigid motion $$\sigma$$ of the plane. A bounded open set $$\Omega \subset \mathbb{R}^ 2$$ is said to have the Pompeiu property if there exist no nontrivial continuous function on $$\mathbb{R}^ 2$$ for which (1) holds. Several authors have determined bounded domains $$\Omega \subset \mathbb{R}^ 2$$ having the Pompeiu property. The main result of this paper is the following: Let $$\Omega \subset \mathbb{R}^ 2$$ be a bounded simply-connected open set whose boundary is a closed simple curve parametrized by $$x(s) = (x_ 1(s), x_ 2(s))$$, $$s \in [- \pi, \pi]$$. Suppose that there exist $$M,N \in \mathbb{Z}$$ and $$a_ k \in \mathbb{C}$$, $$k = - M, \dots, N$$, with $$a_ M$$, $$a_ N \neq 0$$, such that $x_ 1 (s) + x_ 2 (s) = \sum^ N_{k= -M} a_ k e^{iks}.$ Let $$x_ 1 (z)$$, $$x_ 2 (z)$$ be the analytic extension of $$x_ 1 (s)$$ and $$x_ 2 (s)$$ satisfying $$(x_ 1'(z), x_ 2'(z))\neq (0,0) \in \mathbb{C}^ 2$$ for every $$z \in \mathbb{C}$$. Then $$\Omega$$ has the Pompeiu property. Reviewer: G.Anger (Berlin) ##### MSC: 35R30 Inverse problems for PDEs 31A25 Boundary value and inverse problems for harmonic functions in two dimensions 35P05 General topics in linear spectral theory for PDEs 42B10 Fourier and Fourier-Stieltjes transforms and other transforms of Fourier type ##### Keywords: Pompeiu problem; Pompeiu property Full Text: ##### References: [1] Berenstein C., An inverse spectral theorem and its relation to the Pompeiu problem 37 pp 124– (1980) · Zbl 0449.35024 [2] Berenstein C. A., An inverse Neumann problem 382 pp 1– (1987) · Zbl 0623.35078 [3] Brown L., A note on the Pompeiu problem for convex domains 259 pp 107– (1982) · Zbl 0464.30035 [4] Brown L., Spectral synthesis and the Pompeiu problem 23 pp 125– (1973) · Zbl 0265.46044 [5] Caffarelli L. A., The regularity of free boundaries in higher dimensions 139 pp 155– (1977) [6] Chakalov L., Sur un probl‘me de D. Pompeiu 40 pp 1– (1944) · Zbl 0063.07316 [7] Garofalo N., Asymptotic expansions for a class of Fourier integrals and applications to the Pompeiu problem 56 pp 1– (1991) · Zbl 0737.35146 [8] Garofalo N., New results on the Pompeiu problem 325 pp 243– (1991) · Zbl 0737.35147 [9] Pompeiu D., Sur certains systèmes d’équations linéries et sur une propriété intégrate des functions de plusieurs variables 188 pp 1138– (1929) [10] Pompeiu D., Sur une propriété intégrate des fonctions de deux variables réelles. 15 pp 265– (1929) [11] Riemann, R. B. 1953. ”Sullo svolgimento del quoziente di due serie ipergeometriche in funzione continua infinita.ldquo;”. complete works, Dover [12] Williams S. A., A partial solution of the Pompeiu problem 223 pp 183– (1976) · Zbl 0329.35045 [13] Williams S. A., Analyticity of the boundary for Lipschitz domains without the Pompeiu property. 30 pp 357– (1981) · Zbl 0439.35046 [14] Yau S. T., Seminar on Differential Geometry 102 (1982) · Zbl 0471.00020 [15] Zalcman L., Analyticity and the Pompeiu problem 47 pp 237– (1972) · Zbl 0251.30047 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

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https://www.physicsforums.com/threads/two-rotational-motion-questions.191775/

# Two Rotational Motion Questions 1. Oct 16, 2007 ### ccsmarty The first one: 1. The problem statement, all variables and given/known data 1) A compact disc (CD) stores music in a coded pattern of tiny pits 10^-7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm , respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s. What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive. 2. Relevant equations alpha_avg = (omega_2 - omega_1) / (t2 - t1) 3. The attempt at a solution I tried to take the average of the inner and outer angular velocities, and put that in for omega_2, and find the average that way, but I don't think I can do that. The second one: 1. The problem statement, all variables and given/known data 2) At t = 0 a grinding wheel has an angular velocity of 27.0 rad/s. It has a constant angular acceleration of 26.0 rad/s^2 until a circuit breaker trips at time t = 2.00 s. From then on, it turns through an angle 433 rad as it coasts to a stop at constant angular acceleration. At what time did it stop? 2. Relevant equations omega_2 = omega_1 + alpha * t delta_2 - delta_1 = omega_1 * t + 0.5 * alpha * t^2 3. The attempt at a solution I tried using a system of equations using the two equations above to solve for t, but I can't seem to get the right t value. Any guidance is greatly appreciated on either problem. Thanks in advance. 2. Oct 17, 2007 ### learningphysics For the first problem, your omega_inner and omega_outer look good to me. Why not just take (omega_outer - omega_inner)/(74*60)... that should be the answer. For the second problem, think of the angular velocity and acceleration, just like kinematics formulas... What is the angular velocity at t = 2? Then you can use the equation, angle traversed = [(omega_1 + omega_2)/2]*t, so solve for how long it takes to go through the 433 rad... 3. Oct 17, 2007 ### ccsmarty Ok, thanks so much for your help. It makes more sense this way, than the way I initially tried to tackle the problems. Thanks again :)

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http://physics.stackexchange.com/questions/71763/why-doesnt-mass-of-bob-affect-time-period

# Why doesn't mass of bob affect time period? The gravitation formula says $$F = \frac{G m_1 m_2}{r^2} \, ,$$ so if the mass of a bob increases then the torque on it should also increase because the force increased. So, it should go faster and thus the oscillation period should be decrease. My physics book says that period is only affected by effective length and $g$. Why doesn't mass of bob affect the period? - The question is not clear to me. Can you explain the situation a bit more? – Ali Jul 21 '13 at 13:59 ok,let me make the question clearer – svineet Jul 21 '13 at 14:05 Hint: Why doesn't the mass of an object affect how long it takes to fall from a given height to the ground? Mass doesn't only appear in the gravitation formula: It also appears in $\vec F=m\vec A$ – james large Sep 14 at 21:28 You're hitting on the Equivalence principle and the Eötvös experiment. – WetSavannaAnimal aka Rod Vance Sep 14 at 22:54 A very loose answer would be that the time period actually depends upon the angular acceleration and not the torque. Just like the time taken for a object to fall through a height of $h$, depends on the gravitational acceleration and not the mass, i.e. if you drop a sponge ball or you jump yourself, you both will cover height $h$ in the same time(of course neglecting air resistance). Similarly, the time period of a pendulum doesn't depend upon the mass, or rather the inertia of the pendulum, but only on the angular acceleration due to gravity. Now you might ask that in this case, it should also not depend upon the length, but the term of length comes when you calculate the angular acceleration due to the acceleration of gravity. - For the same reason objects of different masses fall at the same acceleration (neglecting drag): because while the force is proportional to the mass and the acceleration is inversely proportional to mass. Doing the falling case o avoid having to deal with the vectors in the pendulum we get $$a = \frac{F}{m} = \frac{G\frac{Mm}{r^2}}{m} = G\frac{M}{r^2}$$ where $M$ is the mass of the planet, $m$ is the mass of the object you are dropping and $r$ is the radius of the planet. The mass of the minor object falls out of the kinematics. The same thing happens in the case of the pendulum: the force includes a factor of $m$, but the acceleration does not. - A pendulum in a gravitational field experiences an instantaneous torque about its pivot point of $$\vec{\Gamma} = \vec{r}\times m\vec{g}$$ where $\vec{g}$ is the instantaneous gravitational field, and $r$ is the distance from the pivot point to the CoM. For purposes of this answer $$\vec{g}=-G\frac{ M_E}{(R_E + h)^2}\hat{k}$$ where • $m$ is the pendulum mass, • $M_E$ is the Earth's mass, • $R_E$ is the distance from the gravitational center of the Earth to the center of mass of the pendulum at rest, and • $h=r(1-\cos\theta)$ is the height of the CoM when the pendulum is oscillating. Let's assume the pendulum is oscillating in a plane, so we can write $$\Gamma = mgr\sin\theta = \mathcal{I}\frac{d^2\theta}{dt^2}.$$ $\mathcal{I}$ is the moment of inertia of the pendulum about the pivot point, and will have the form of $mb^2$, where $b$ is a geometric size and mass distribution factor. Any rigid object you want to consider can have its moment of inertia put in that form. From this we see quickly that the actual mass of the object disappears: $$\frac{d^2\theta}{dt^2} = \frac{gr}{b^2}\sin\theta.$$ All that remains is to find $b$ which depends only on how the mass is distributed, not how much mass is present. We also see that this is not simple harmonic motion. While the factor $g$ is not constant, it only introduces an anharmonic factor of $$1-\frac{r\theta^2}{R_E}-\frac{r^2\theta^4}{4R_E^2}$$. The $\sin\theta$ term introduces a larger anharmonicity because $$\sin\theta\simeq \theta-\frac{\theta^3}{6} = \theta\left(1-\frac{\theta^2}{6}\right).$$ So we see that 1) the mass doesn't matter, but the distribution of mass does, 2) the variation in height producing a variation in gravitational field only has a $(r/R_E)\theta^2$ affect, 3) the amplitude of the angle due to the $\sin\theta$ term becomes important when $\theta > 0.1$ radian. Considering point 2), most pendula have $r<10 m$ and $R_E = 6.38\times 10^6$ m. - Well the easy way is that the mass has a opposite affect when the bob goes up again on the other side. The deacceleration and the acceleration will equal out so the period will always be the same what ever mass you have. O /I\ / I \ / I \ 0 0 0 A C B Here you have a diagram to represent it. You drop the pendulum at B and it accelerates until it hits C then it will slow down. The mass will increase the deacceleration. So the acceleration and the deacceleration will equal out. - NOOOOO!!! This is wrong. Granted the pendulum formula (T = 2 * pi * sqrt(L / g)) does not take into account mass of the bob, much less the pendulum, mass can and does affect the pendulum period. The Pendulum Formula is accurate and i give it credit, but its variables are broadly defined. T represents time or period, and g represents gravitational acceleration. I have no problem with those, but it's L that bothers me. To assume L is the distance from the point of axis to the bottom tip of the pendulum is to assume that the pendulum has an equal density throughout and its center of gravity lies directly in the center of the pendulum. However, with most pendulums this isn't the case. The bob, or weight on the pendulum, affects the location of the center of gravity. When a bob is added below the center of balance, the gravitational center of the entire pendulum is shifted downwards to some degree. Instead of saying L = the length of the pendulum, it's better to say that L = 2 * (distance between center of balance and pivot point). - NOOOOO!!! This is wrong. – Jimmy360 Jun 14 at 21:56 This post is misleading. The formula for the period of the pendulum is correct, and is obtained assuming constant gravitational field. This is generally correct, but would be a problem for the case of very large pendulum, which clears the original question. Usually L is the distance to pendulum's center of mass and when a blob is involved, is usually assumed to contain the whole mass of the pendulum. But these are not considerations leading to the mass invariant formula for T. – rmhleo Jun 14 at 22:09 Certainly for real pendulums the ratio of the mass of the bob and the mass of the support affects the position of the center of mass of the device. But if you read closely any non-introductory account you'll see that L is defined the distance from the pivot to the CoM. An important detail, but not one that is appropriate to spend much time on in the first introduction (which usually assumes a massless rod). – dmckee Jun 14 at 23:52

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http://www.maplesoft.com/support/help/MapleSim/view.aspx?path=componentLibrary/signalBlocks/signalConverters/UnitConversionBlock

Conversion Block Converts a signal from one unit to another Description The Conversion Block component converts a signal from one unit to another. This component is a fixed causality component that you can use to perform unit conversions for dimensions such as time, temperature, speed, pressure, mass, and volume. Connections Name Description $u$ Connection of the real input signal to be converted $y$ Connection of the real output signal containing the input signal, $u$, in another unit Parameters Symbol Default Units Description dimension Acceleration - The dimension for which to convert units. from unit $\frac{m}{{s}^{2}}$ - Original units of the signal. to unit $\frac{m}{{s}^{2}}$ - Units to which you want to convert the signal.

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https://brilliant.org/problems/a-number-theory-problem-by-swadesh-rath/

# A number theory problem by Swadesh Rath Number Theory Level pending xy - 6(x+y) = 0 find number of ordered solutions for all integral values of x,y such that x<y or x = y ×

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https://www.universetoday.com/34024/mass-of-the-planets/

# What are the Different Masses of the Planets? It is a well known fact that the planets of the Solar System vary considerably in terms of size. For instance, the planets of the inner Solar System are smaller and denser than the gas/ice giants of the outer Solar System. And in some cases, planets can actually be smaller than the largest moons. But a planet’s size is not necessarily proportional to its mass. In the end, how massive a planet is has more to do with its composition and density. So while a planet like Mercury may be smaller in size than Jupiter’s moon Ganymede or Saturn’s moon Titan, it is more than twice as massive than they are. And while Jupiter is 318 times as massive as Earth, its composition and density mean that it is only 11.21 times Earth’s size. Let’s go over the planet’s one by one and see just how massive they are, shall we? ## Mercury: Mercury is the Solar System’s smallest planet, with an average diameter of 4879 km (3031.67 mi). It is also one of its densest at 5.427 g/cm3, which is second only to Earth. As a terrestrial planet, it is composed of silicate rock and minerals and is differentiated between an iron core and a silicate mantle and crust. But unlike its peers (Venus, Earth and Mars), it has an abnormally large metallic core relative to its crust and mantle. All told, Mercury’s mass is approximately 0.330 x 1024 kg, which works out to 330,000,000 trillion metric tons (or the equivalent of 0.055 Earths). Combined with its density and size, Mercury has a surface gravity of 3.7 m/s² (or 0.38 g). ## Venus: Venus, otherwise known as “Earth’s Sister Planet”, is so-named because of its similarities in composition, size, and mass to our own. Like Earth, Mercury and Mars, it is a terrestrial planet, and hence quite dense. In fact, with a density of 5.243 g/cm³, it is the third densest planet in the Solar System (behind Earth and Mercury). Its average radius is roughly 6,050 km (3759.3 mi), which is the equivalent of 0.95 Earths. And when it comes to mass, the planet weighs in at a hefty 4.87 x 1024 kg, or 4,870,000,000 trillion metric tons. Not surprisingly, this is the equivalent of 0.815 Earths, making it the second most massive terrestrial planet in the Solar System. Combined with its density and size, this means that Venus also has comparable gravity to Earth – roughly 8.87 m/s², or 0.9 g. ## Earth: Like the other planets of the inner Solar System, Earth is also a terrestrial planet, composed of metals and silicate rocks differentiated between an iron core and a silicate mantle and crust. Of the terrestrial planets, it is the largest and densest, with an average radius of 6,371.0 km (3,958.8 mi) and a mean of density of 5.514 g/cm3. And at 5.97 x 1024 kg (which works out to 5,970,000,000,000 trillion metric tons) Earth is the most massive of all the terrestrial planets. Combined with its size and density, Earth experiences the surface gravity that we are all familiar with – 9.8 m/s², or 1 g. ## Mars: Mars is the third largest terrestrial planet, and the second smallest planet in our Solar System. Like the others, it is composed of metals and silicate rocks that are differentiated between a iron core and a silicate mantle and crust. But while it is roughly half the size of Earth (with a mean diameter of 6792 km, or 4220.35 mi), it is only one-tenth as massive. In short, Mars has a mass of 0.642 x1024 kg, which works out to 642,000,000 trillion metric tons, or roughly 0.11 the mass of Earth. Combined with its size and density – 3.9335 g/cm³ (which is roughly 0.71 times that of Earth’s) – Mars has a surface gravity of 3.711 m/s² (or 0.376 g). ## Jupiter: Jupiter is the largest planet in the Solar System. With a mean diameter of 142,984 km, it is big enough to fit all the other planets (except Saturn) inside itself, and big enough to fit Earth 11.8 times over. But with a mass of 1898 x 1024 kg (or 1,898,000,000,000 trillion metric tons), Jupiter is more massive than all the other planets in the Solar System combined – 2.5 times more massive, to be exact. However, as a gas giant, it has a lower overall density than the terrestrial planets. It’s mean density is 1.326 g/cm, but this increases considerably the further one ventures towards the core. And though Jupiter does not have a true surface, if one were to position themselves within its atmosphere where the pressure is the same as Earth’s at sea level (1 bar), they would experience a gravitational pull of 24.79 m/s2 (2.528 g). ## Saturn: Saturn is the second largest of the gas giants; with a mean diameter of 120,536 km, it is just slightly smaller than Jupiter. However, it is significantly less massive than its Jovian cousin, with a mass of 569 x 1024 kg (or 569,000,000,000 trillion metric tons). Still, this makes Saturn the second most-massive planet in the Solar System, with 95 times the mass of Earth. Much like Jupiter, Saturn has a low mean density due to its composition. In fact, with an average density of 0.687 g/cm³, Saturn is the only planet in the Solar System that is less dense than water (1 g/cm³).  But of course, like all gas giants, its density increases considerably the further one ventures towards the core. Combined with its size and mass, Saturn has a “surface” gravity that is just slightly higher than Earth’s – 10.44 m/s², or 1.065 g. ## Uranus: With a mean diameter of 51,118 km, Uranus is the third largest planet in the Solar System. But with a mass of 86.8 x 1024 kg (86,800,000,000 trillion metric tons) it is the fourth most massive – which is 14.5 times the mass of Earth. This is due to its mean density of 1.271 g/cm3, which is about three quarters of what Neptune’s is. Between its size, mass, and density, Uranus’ gravity works out to 8.69 m/s2, which is 0.886 g. ## Neptune: Neptune is significantly larger than Earth; at 49,528 km, it is about four times Earth’s size. And with a mass of 102 x 1024 kg (or 102,000,000,000 trillion metric tons) it is also more massive – about 17 times more to be exact. This makes Neptune the third most massive planet in the Solar System; while its density is the greatest of any gas giant (1.638 g/cm3). Combined, this works out to a “surface” gravity of 11.15 m/s2 (1.14 g). As you can see, the planets of the Solar System range considerably in terms of mass. But when you factor in their variations in density, you can see how a planets mass is not always proportionate to its size. In short, while some planets may be a few times larger than others, they are can have many, many times more mass. We have written many interesting articles about the planets here at Universe. For instance, here’s Interesting Facts About the Solar System, What are the Colors of the Planets?, What are the Signs of the Planets?, How Dense are the Planets?, and What are the Diameters of the Planets?. For more information, check out Nine Planets overview of the Solar System, NASA’s Solar System Exploration, and use this site to find out what you would weigh on other planets. Astronomy Cast has episodes on all of the planets. Here’s Episode 49: Mercury to start! ## 9 Replies to “What are the Different Masses of the Planets?” 1. Dan says: Neptune is interesting. I did not know that it was dense for a gas/ice giant and that it had a gravity only slightly higher than the Earth. Even if it is smaller than Uranus in terms of size, it is more massive. Thanks for the info! 1. Well observed. And yes, its only a matter of time 🙂 2. regold says: The diameters of the planets are wrong. They are probably are the radius, but I did not check. 1. Jeffrey Boerst says: I just took literally 20 seconds on Google and all the sites that came up on the 1st page agree… Perhaps you’re thinking of Miles rather than Kilometers? 3. regold says: The Earth has a diameter of about 8000 miles. The information above says 3958.8 miles. 1. You’re right. That should say radius. I was switching back and forth and made an error because of it. 4. Thomas77 says: The diameter of Uranus is only 51,118 km. So it is only slightly (1590 km) bigger than Neptune. 5. Thomas77 says: Hello Matt, I just noticed a very confusing thing. The figures is this article are the ones I know. A few day ago you wrote the article “What Are The Diameters of the Planets?” All the diameters are a bit smaller there. 139,822 km instead of 142,984km for Jupiter 116,464 km instead of 120,536km for Saturn And so on. Where did those numbers come from? 1. Unfortunately, two different sources were used for the two articles, one from the IAU and the other NASA. The IAU one also used mean diameter, averaging between equatorial and polar while the NASA uses just equatorial. I’ve amended that now so that both use NASA’s. Thanks for the heads up.

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http://www.researchgate.net/researcher/20049887_Kazuaki_Takeda

# Kazuaki Takeda Tohoku University, Sendai-shi, Miyagi-ken, Japan Are you Kazuaki Takeda? ## Publications (52)13.26 Total impact • ##### Article: Uplink Capacity of A Cellular System Using Multi-user Single-carrier MIMO Multiplexing Combined with Frequency-domain Equalization and Transmit Power Control Takahiro Chiba · Kazuaki Takeda · Kazuki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Multi-user single-carrier multiple-input multiple-output (MU SC-MIMO) multiplexing can increase the uplink capacity of a cellular system without expanding the signal bandwidth. It is practically important to make clear an extent to which the MU SC-MIMO multiplexing combined with frequency-domain equalization (FDE) and transmit power control (TPC) can increase the uplink capacity in the presence of the co-channel interference (CCI). Since the theoretical analysis is quite difficult, we resort to the computer simulation to investigate the uplink capacity. In this paper, frequency-domain zero-forcing detection (ZFD) and frequency-domain minimum mean square error detection (MMSED) are considered for MU signal detection. It is shown that ZFD and MMSED provide almost the same uplink capacity and that an advantage of fast TPC over slow TPC diminishes. As a result, MU SC-MIMO using computationally efficient ZFD can be used together with slow TPC instead of using MMSED. With 8 receive antennas and slow TPC, MU SC-MIMO multiplexing using ZFD can achieve about 1.5times higher uplink capacity than SU SC-SIMO diversity. KeywordsMulti-user MIMO–Single-carrier–Frequency-domain equalization–Cellular system–Uplink-capacity Wireless Personal Communications 06/2011; 58(3):455-468. DOI:10.1007/s11277-010-0130-5 · 0.65 Impact Factor • Source ##### Article: Performance comparison between CDTD and STTD for DS-CDMA/MMSE-FDE with frequency-domain ICI cancellation Kazuaki Takeda · Yohei Kojima · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Frequency-domain equalization (FDE) based on the minimum mean square error (MMSE) criterion can provide a better bit error rate (BER) performance than rake combining. However, the residual inter-chip interference (ICI) is produced after MMSE-FDE and this degrades the BER performance. Recently, we showed that frequency-domain ICI cancellation can bring the BER performance close to the theoretical lower bound. To further improve the BER performance, transmit antenna diversity technique is effective. Cyclic delay transmit diversity (CDTD) can increase the number of equivalent paths and hence achieve a large frequency diversity gain. Space-time transmit diversity (STTD) can obtain antenna diversity gain due to the space-time coding and achieve a better BER performance than CDTD. Objective of this paper is to show that the BER performance degradation of CDTD is mainly due to the residual ICI and that the introduction of ICI cancellation gives almost the same BER performance as STTD. This study provides a very important result that CDTD has a great advantage of providing a higher throughput than STTD. This is confirmed by computer simulation. The computer simulation results show that CDTD can achieve higher throughput than STTD when ICI cancellation is introduced. IEICE Transactions on Communications 09/2009; 92-B(9):2882-2890. DOI:10.1587/transcom.E92.B.2882 · 0.23 Impact Factor • Source ##### Article: 2-Step Maximum Likelihood Channel Estimation for Multicode DS-CDMA with Frequency-Domain Equalization Yohei Kojima · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Frequency-domain equalization (FDE) based on the minimum mean square error (MMSE) criterion can provide better downlink bit error rate (BER) performance of direct sequence code division multiple access (DS-CDMA) than the conventional rake combining in a frequency-selective fading channel. FDE requires accurate channel estimation. In this paper, we propose a new 2-step maximum likelihood channel estimation (MLCE) for DS-CDMA with FDE in a very slow frequency-selective fading environment. The 1st step uses the conventional pilot-assisted MMSE-CE and the 2nd step carries out the MLCE using decision feedback from the 1st step. The BER performance improvement achieved by 2-step MLCE over pilot assisted MMSE-CE is confirmed by computer simulation. IEICE Transactions on Communications 06/2009; 92-B(6):2065-2071. DOI:10.1587/transcom.E92.B.2065 · 0.23 Impact Factor • Source ##### Article: RLS Channel Estimation with Adaptive Forgetting Factor for DS-CDMA Frequency-Domain Equalization Yohei Kojima · Hiromichi Tomeba · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Frequency-domain equalization (FDE) based on the minimum mean square error (MMSE) criterion can increase the downlink bit error rate (BER) performance of DS-CDMA beyond that possible with conventional rake combining in a frequency-selective fading channel. FDE requires accurate channel estimation. Recently, we proposed a pilot-assisted channel estimation (CE) based on the MMSE criterion. Using MMSE-CE, the channel estimation accuracy is almost insensitive to the pilot chip sequence, and a good BER performance is achieved. In this paper, we propose a channel estimation scheme using one-tap recursive least square (RLS) algorithm, where the forgetting factor is adapted to the changing channel condition by the least mean square (LMS)algorithm, for DS-CDMA with FDE. We evaluate the BER performance using RLS-CE with adaptive forgetting factor in a frequency-selective fast Rayleigh fading channel by computer simulation. IEICE Transactions on Communications 05/2009; 92-B(5):1457-1465. DOI:10.1587/transcom.E92.B.1457 · 0.23 Impact Factor • Source ##### Conference Paper: Pilot assisted channel estimation for MC-CDMA signal transmission using overlap FDE Hiromichi Tomeba · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Recently, multi-carrier code division multiple access (MC-CDMA) has been attracting much attention as a broadband wireless access technique for the next generation mobile communications systems. Frequency-domain equalization (FDE) based on the minimum mean square error (MMSE) criterion can take advantage of the channel frequency-selectivity and improve the average bit error rate (BER) performance due to frequency-diversity gain. However, conventional MC-CDMA requires the insertion of guard interval (GI) and this reduces the transmission efficiency. Overlap FDE technique has been proposed that requires no GI insertion. Recently, we showed that MC-CDMA using overlap FDE can provide almost the same BER performance as the conventional MC-CDMA downlink using the GI insertion. However, our previous work assumed the ideal channel estimation. In this paper, we propose a pilot assisted channel estimation technique suitable for MC-CDMA downlink using overlap FDE and evaluate the BER performance by computer simulation. Communication Systems, 2008. ICCS 2008. 11th IEEE Singapore International Conference on; 12/2008 • Source ##### Conference Paper: DS-CDMA MMSE turbo equalization using 2-step maximum likelihood channel estimation Yohei Kojima · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Frequency-domain equalization (FDE) based on the minimum mean square error (MMSE) criterion and frequency-domain inter-chip interference (ICI) cancellation can be incorporated into turbo decoding (this is called the MMSE turbo equalization in this paper) to improve the bit error rate (BER) performance for direct sequence code division multiple access (DS-CDMA). DS-CDMA MMSE turbo equalization requires accurate channel estimation. The performance of MMSE turbo equalization is sensitive to channel estimation accuracy. Recently, we proposed a 2-step maximum likelihood channel estimation (MLCE) for DS-CDMA with FDE. In this paper, we apply 2-step MLCE to DS-CDMA MMSE turbo equalization and evaluate by computer simulation the achievable BER performance in a frequency-selective block Rayleigh fading channel. Vehicular Technology Conference, 2008. VTC 2008-Fall. IEEE 68th; 10/2008 • Source ##### Conference Paper: Space-Time Block Coded-Joint Transmit/Receive Antenna Diversity using more than 4 Receive Antennas Hiromichi Tomeba · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Antenna diversity is an effective technique for improving the transmission performance in a multi-path fading channel. Recently, we proposed the space-time block coded-joint transmit/receive antenna diversity (STBC-JTRD), which allows the use of an arbitrary number of transmit antennas without sacrificing the coding rate. However, in STBC-JTRD, the number of receive antennas is limited to 4. In this paper, we show the STBC-JTRD encoding allowing the use of more than 5 receive antennas. The bit error rate (BER) analysis in a frequency-nonselective Rayleigh fading channel is presented. The BER performance analysis is confirmed by computer simulation. Vehicular Technology Conference, 2008. VTC 2008-Fall. IEEE 68th; 10/2008 • ##### Article: Frequency-domain equalisation for block CDMA transmission [Hide abstract] ABSTRACT: In next generation wireless network, significantly higher rate data services of close to 1 Gpbs are expected. The wireless channels for such a broadband data transmission become severe frequency-selective. Frequency-domain equalisation (FDE) technique may play an important role for broadband data transmission using multi-carrier (MC)- and direct-sequence code division multiple access (DS-CDMA). The performance can be further improved by the use of multi-input/multi-output (MIMO) antenna diversity technique. The downlink performance is significantly improved with FDE. However, the uplink performance is limited by the multiple access interference (MAI). To remove the MAI while gaining the frequency diversity effect through the use of FDE, two-dimensional (2D) block spread CDMA can be used. Recently, particular attention has been paid to MIMO space division multiplexing (SDM) to significantly increase the throughput without expanding the signal bandwidth. In this paper, we present a comprehensive performance comparison of MC- and DS-CDMA using FDE. Copyright © 2008 John Wiley & Sons, Ltd. European Transactions on Telecommunications 08/2008; 19(5):553-560. DOI:10.1002/ett.1305 · 1.35 Impact Factor • Source ##### Conference Paper: Transmit diversity for DS-CDMA/MMSE-FDE with frequency-domain ICI cancellation Kazuaki Takeda · Yohei Kojima · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Frequency-domain equalization (FDE) based on the minimum mean square error (MMSE) criterion can provide a better bit error rate (BER) performance than rake combining. However, the residual inter-chip interference (ICI) is produced after MMSE-FDE and degrades the BER performance. Recently, we showed that frequency-domain ICI cancellation can bring the BER performance close to the theoretical lower bound. To further improve the BER performance, transmit antenna diversity technique is effective. Cyclic delay transmit diversity (CDTD) can increase the number of equivalent paths and hence achieve a large frequency diversity gain. Space-time transmit diversity (STTD) can obtain antenna diversity gain due to the space-time coding and achieve a better BER performance than CDTD. In this paper, we point out that the performance difference between CDTD and STTD is mainly due to the residual ICI. We theoretically show that the introduction of ICI cancellation into DS-CDMA/MMSE-FDE gives almost the same BER performance for CDTD and STTD. This is confirmed by computer simulation. Vehicular Technology Conference, 2008. VTC Spring 2008. IEEE; 06/2008 • Source ##### Article: Iterative overlap FDE for multicode DS-CDMA Kazuaki Takeda · Hiromichi Tomeba · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Recently, a new frequency-domain equalization (FDE) technique, called overlap FDE, that requires no GI insertion was proposed. However, the residual inter/intra-block interference (IBI) cannot completely be removed. In addition to this, for multicode direct sequence code division multiple access (DS-CDMA), the presence of residual interchip interference (ICI) after FDE distorts orthogonality among the spreading codes. In this paper, we propose an iterative overlap FDE for multicode DS-CDMA to suppress both the residual IBI and the residual ICI. In the iterative overlap FDE, joint minimum mean square error (MMSE)-FDE and ICI cancellation is repeated a sufficient number of times. The bit error rate (BER) performance with the iterative overlap FDE is evaluated by computer simulation. IEICE Transactions on Communications 06/2008; 91-B(6):1942-1951. DOI:10.1093/ietcom/e91-b.6.1942 · 0.23 Impact Factor • Source ##### Article: BER performance analysis of MC-CDMA with overlap-ME Hiromichi Tomeba · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Recently, multi-carrier code division multiple access (MC-CDMA) has been attracting much attention as a broadband wireless access technique for the next generation mobile communications systems. Frequency-domain equalization (FDE) based on the minimum mean square error (MMSE) criterion can take advantage of the channel frequency-selectivity and improve the average bit error rate (BER) performance due to frequency-diversity gain. The conventional FDE requires the insertion of the guard interval (GI) to avoid the inter-block interference (IBI), resulting in the transmission efficiency loss. In this paper, an overlap FDE technique, which requires no GI insertion, is presented for MC-CDMA transmission. An expression for the conditional bit error rate (BER) is derived for the given set of channel gains. The average BER performance in a frequency-selective Rayleigh fading channel is evaluated by Monte-Carlo numerical computation method using the derived conditional BER and is confirmed by computer simulation of the MC-CDMA signal transmission. Keyword− Frequency-selective fading channel, Overlap FDE, MC-CDMA. IEICE Transactions on Communications 03/2008; 91-B(3):795-804. DOI:10.1093/ietcom/e91-b.3.795 · 0.23 Impact Factor • Source ##### Article: DS-CDMA Downlink Site Diversity with Frequency-Domain Equalization and Antenna Diversity Reception Hirotaka Sato · Hiromichi Tomeba · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: The use of frequency-domain equalization based on minimum mean square error criterion (called MMSE-FDE) can significantly improve the bit error rate (BER) performance of DS-CDMA signal transmission compared to the well-known coherent rake combining. However, in a DS-CDMA cellular system, as a mobile user moves away from a base station and approaches the cell edge, the received signal power gets weaker and the interference from other cells becomes stronger, thereby degrading the transmission performance. To improve the transmission performance of a user close to the cell edge, the well-known site diversity can be used in conjunction with FDE. In this paper, we consider DS-CDMA downlink site diversity with FDE. The MMSE site diversity combining weight is theoretically derived for joint FDE and antenna diversity reception and the downlink capacity is evaluated by computer simulation. It is shown that the larger downlink capacity can be achieved with FDE than with coherent rake combining. It is also shown that the DS-CDMA downlink capacity is almost the same as MC-CDMA downlink capacity. IEICE Transactions on Communications 12/2007; E90B(12). DOI:10.1093/ietcom/e90-b.12.3591 · 0.23 Impact Factor • Source ##### Conference Paper: Uplink Capacity of Single-Carrier Frequency-Interleaved Spread Spectrum Multiple Access Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: A single-carrier frequency-interleaved spread spectrum multiple access (SC-FI-SSMA), which we recently proposed, uses frequency-domain interleaving and MMSE-FDE to remove the uplink multi-user interference (MUI) while achieving the frequency diversity gain by assigning the orthogonal interleaving matrices to different users. However, in a multi-cell environment, the uplink capacity of SC-FI-SSMA may degrade due to the inter-cell interference. To improve the uplink bit error rate (BER) performance, transmit power control and site selection diversity techniques can be applied. In this paper, we evaluate the uplink capacity of SC-FI-SSMA to show that SC-FI-SSMA provides better uplink performance than DS-CDMA even in a multi-cell environment. Vehicular Technology Conference, 2007. VTC-2007 Fall. 2007 IEEE 66th; 11/2007 • Source ##### Article: DS-CDMA HARQ with overlap FDE Kazuki TAKEDA · Hiromichi Tomeba · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Turbo coded hybrid ARQ (HARQ) is known as one of the promising error control techniques for high speed wireless packet access. However, in a severe frequency-selective fading channel, the HARQ throughput performance significantly degrades for direct sequence code division multiple access (DS-CDMA) system using rake combining. This problem can be overcome by replacing the rake combining by the frequency-domain equalization (FDE) based on minimum mean square error (MMSE) criterion. In a system with the conventional FDE, the guard interval (GI) is inserted to avoid the inter-block interference (IBI). The insertion of GI reduces the throughput. Recently, overlap FDE that requires no GI insertion was proposed. In this paper, we apply overlap FDE to HARQ and derive the MMSE-FDE weight for packet combining. Then, we evaluate the throughput performance of DS-CDMA HARQ with overlap FDE. We show that overlap FDE provides better throughput performance than both the rake combining and conventional FDE regardless of the degree of the channel frequency-selectivity. IEICE Transactions on Communications 11/2007; E90B(11). DOI:10.1093/ietcom/e90-b.11.3189 · 0.23 Impact Factor • Source ##### Article: Frequency-Domain MMSE Channel Estimation for Frequency-Domain Equalization of DS-CDMA Signals Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Frequency-domain equalization (FDE) based on minimum mean square error (MMSE) criterion can replace the conventional rake combining to significantly improve the bit error rate (BER) performance in a frequency-selective fading channel. MMSE-FDE requires an accurate estimate of the channel transfer function and the signal-to-noise power ratio (SNR). Direct application of pilot-assisted channel estimation (CE) degrades the BER performance, since the frequency spectrum of the pilot chip sequence is not constant over the spreading bandwidth. In this paper, we propose a pilot-assisted decision feedback frequency-domain MMSE-CE. The BER performance with the proposed pilot-assisted MMSE-CE in a frequency-selective Rayleigh fading channel is evaluated by computer simulation. It is shown that MMSE-CE always gives a good BER performance irrespective of the choice of the pilot chip sequence and shows a high tracking ability against fading. For a spreading factor SF of 16, the E b / N 0 degradation for BER=10−4 with MMSE-CE from the ideal CE case is as small as 0.9 dB (including an E b / N 0 loss of 0.28 dB due to the pilot insertion). IEICE Transactions on Communications 07/2007; E90B(7). DOI:10.1093/ietcom/e90-b.7.1746 · 0.23 Impact Factor • Source ##### Article: Frequency-Domain Interchip Interference Cancelation for DS-CDMA Downlink Transmission Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: The use of frequency-domain equalization (FDE) based on minimum-mean-square-error criterion can significantly improve the bit-error-rate (BER) performance of orthogonal multicode direct-sequence code-division multiple access downlink signal transmission in a frequency-selective fading channel. However, the presence of residual interchip interference (ICI) after FDE produces the orthogonality distortion among the spreading codes, and the BER performance degrades as the number of multiplex order increases. In this paper, we propose a frequency-domain ICI cancellation scheme, in which the residual ICI replica in the frequency domain is generated and subtracted from each frequency component of the received signal after FDE. Three types of ICI cancelation scheme are presented, and the effectiveness of the proposed ICI cancelation scheme is confirmed by computer simulation IEEE Transactions on Vehicular Technology 06/2007; 56(3-56):1286 - 1294. DOI:10.1109/TVT.2007.895474 · 1.98 Impact Factor • Source ##### Conference Paper: HARQ throughput performance of multicode DS-CDMA with MMSE turbo equalization Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Hybrid automatic repeat request (HARQ) is an indispensable technique for packet access. This is employed in high speed downlink packet access (HSDPA) using the orthogonal multicode direct sequence code division multiple access (DS-CDMA). As the data rate increases, the frequency-selectivity of the channel becomes severer and some equalization technique other than rake combining is necessary. The use of MMSE frequency-domain equalization (MMSE-FDE) can significantly improve the throughput performance. However, the residual inter-chip-inference (ICI) after MMSE-FDE produces the orthogonality distortion among the spreading codes and the throughput performance is far from the theoretical lower bound. In this paper, we study an MMSE turbo equalization for HARQ using multicode DS-CDMA. It is shown by computer simulation that MMSE turbo equalization can significantly improve the throughput performance. An E_S/N₀ reduction of as much as 2.5-3 dB from the no turbo equalization case is obtained for a throughput range of 2-2.5 bit/s/Hz. Vehicular Technology Conference, 2007. VTC2007-Spring. IEEE 65th; 05/2007 • Source ##### Conference Paper: Joint use of overlap FDE and STTD for MC-CDMA downlink transmission Hirornichi Tomeba · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: Recently, multi-carrier code division multiple access (MC-CDMA) has been attracting much attention for the next generation mobile communications systems. Using frequency-domain equalization based on minimum mean square error criterion (MMSE-FDE), the frequency diversity effect can be obtained and improved bit error rate (BER) performance can be obtained. Antenna diversity is an effective technique to further improve the BER performance. Space-time transmit diversity (STTD) is suitable for the downlink transmission. Recently, overlap FDE that requires no guard interval (GI) insertion are presented. Combining STTD decoding and overlap FDE is not straightforward. In this paper, we propose STTD decoding for the MC-CDMA transmission using overlap FDE and then, its BER performance is evaluated by computer simulation. Vehicular Technology Conference, 2007. VTC2007-Spring. IEEE 65th; 05/2007 • Source ##### Article: Frequency-Domain Multi-Stage Soft Interference Cancellation for DS-CDMA Uplink Signal Transmission Koichi Ishihara · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: It is well-known that, in DS-CDMA downlink signal transmission, frequency-domain equalization (FDE) based on minimum mean square error (MMSE) criterion can replace rake combining to achieve much improved bit error rate (BER) performance in severe frequency-selective fading channel. However, in uplink signal transmission, as each user's signal goes through a different channel, a severe multi-user interference (MUI) is produced and the uplink BER performance severely degrades compared to the downlink. When a small spreading factor is used, the uplink BER performance further degrades due to inter-chip interference (ICI). In this paper, we propose a frequency-domain multi-stage soft interference cancellation scheme for the DS-CDMA uplink and the achievable BER performance is evaluated by computer simulation. The BER performance comparison of the proposed cancellation technique and the multiuser detection (MUD) is also presented. IEICE Transactions on Communications 05/2007; 90-B(5):1152-1161. DOI:10.1093/ietcom/e90-b.5.1152 · 0.23 Impact Factor • Source ##### Article: Iterative Channel Estimation for Frequency-Domain Equalization of DSSS Signals Koichi Ishihara · Kazuaki Takeda · Fumiyuki Adachi · [Hide abstract] ABSTRACT: As the channel frequency selectivity becomes severer, the bit error rate (BER) performance of direct sequence spread spectrum (DSSS) signal transmission with rake combining degrades due to an increasing inter-path interference (IPI). Frequency-domain equalization (FDE) can replace rake combining with much improved BER performance in a severe frequency-selective fading channel. For FDE, accurate estimation of the channel transfer function is required. In this paper, we propose an iterative channel estimation that uses pilot chips which are time-multiplexed within each chip block for fast Fourier transform (FFT). The pilot acts as a cyclic-prefix of FFT block as well. The achievable BER performance is evaluated by computer simulation. It is shown that the proposed channel estimation has a very good tracking ability against fast fading. IEICE Transactions on Communications 05/2007; 90-B(5):1171-1180. DOI:10.1093/ietcom/e90-b.5.1171 · 0.23 Impact Factor #### Publication Stats 470 Citations 13.26 Total Impact Points #### Institutions • ###### Tohoku University • Graduate School of Engineering Sendai-shi, Miyagi-ken, Japan

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http://www.nag.com/numeric/fl/nagdoc_fl24/html/G02/g02gkf.html

G02 Chapter Contents G02 Chapter Introduction NAG Library Manual # NAG Library Routine DocumentG02GKF Note:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details. ## 1  Purpose G02GKF calculates the estimates of the parameters of a generalized linear model for given constraints from the singular value decomposition results. ## 2  Specification SUBROUTINE G02GKF ( IP, ICONST, V, LDV, C, LDC, B, S, SE, COV, WK, IFAIL) INTEGER IP, ICONST, LDV, LDC, IFAIL REAL (KIND=nag_wp) V(LDV,IP+7), C(LDC,ICONST), B(IP), S, SE(IP), COV(IP*(IP+1)/2), WK(2*IP*IP+IP*ICONST+2*ICONST*ICONST+4*ICONST) ## 3  Description G02GKF computes the estimates given a set of linear constraints for a generalized linear model which is not of full rank. It is intended for use after a call to G02GAF, G02GBF, G02GCF or G02GDF. In the case of a model not of full rank the routines use a singular value decomposition to find the parameter estimates, ${\stackrel{^}{\beta }}_{\text{svd}}$, and their variance-covariance matrix. Details of the SVD are made available in the form of the matrix ${P}^{*}$: $P*= D-1 P1T P0T$ as described by G02GAF, G02GBF, G02GCF and G02GDF. Alternative solutions can be formed by imposing constraints on the parameters. If there are $p$ parameters and the rank of the model is $k$ then ${n}_{\mathrm{c}}=p-k$ constraints will have to be imposed to obtain a unique solution. Let $C$ be a $p$ by ${n}_{\mathrm{c}}$ matrix of constraints, such that $CTβ=0,$ then the new parameter estimates ${\stackrel{^}{\beta }}_{\mathrm{c}}$ are given by: $β^c =Aβ^svd =I-P0CTP0-1β^svd, where ​I​ is the identity matrix,$ and the variance-covariance matrix is given by $AP1D-2 P1T AT$ provided ${\left({C}^{\mathrm{T}}{P}_{0}\right)}^{-1}$ exists. ## 4  References Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore McCullagh P and Nelder J A (1983) Generalized Linear Models Chapman and Hall Searle S R (1971) Linear Models Wiley ## 5  Parameters 1:     IP – INTEGERInput On entry: $p$, the number of terms in the linear model. Constraint: ${\mathbf{IP}}\ge 1$. 2:     ICONST – INTEGERInput On entry: the number of constraints to be imposed on the parameters, ${n}_{\mathrm{c}}$. Constraint: $0<{\mathbf{ICONST}}<{\mathbf{IP}}$. 3:     V(LDV,${\mathbf{IP}}+7$) – REAL (KIND=nag_wp) arrayInput On entry: the array V as returned by G02GAF, G02GBF, G02GCF or G02GDF. 4:     LDV – INTEGERInput On entry: the first dimension of the array V as declared in the (sub)program from which G02GKF is called. Constraint: ${\mathbf{LDV}}\ge {\mathbf{IP}}$. LDV should be as supplied to G02GAF, G02GBF, G02GCF or G02GDF 5:     C(LDC,ICONST) – REAL (KIND=nag_wp) arrayInput On entry: contains the ICONST constraints stored by column, i.e., the $i$th constraint is stored in the $i$th column of C. 6:     LDC – INTEGERInput On entry: the first dimension of the array C as declared in the (sub)program from which G02GKF is called. Constraint: ${\mathbf{LDC}}\ge {\mathbf{IP}}$. 7:     B(IP) – REAL (KIND=nag_wp) arrayInput/Output On entry: the parameter estimates computed by using the singular value decomposition, ${\stackrel{^}{\beta }}_{\text{svd}}$. On exit: the parameter estimates of the parameters with the constraints imposed, ${\stackrel{^}{\beta }}_{\mathrm{c}}$. 8:     S – REAL (KIND=nag_wp)Input On entry: the estimate of the scale parameter. For results from G02GAF and G02GDF then S is the scale parameter for the model. For results from G02GBF and G02GCF then S should be set to $1.0$. Constraint: ${\mathbf{S}}>0.0$. 9:     SE(IP) – REAL (KIND=nag_wp) arrayOutput On exit: the standard error of the parameter estimates in B. 10:   COV(${\mathbf{IP}}×\left({\mathbf{IP}}+1\right)/2$) – REAL (KIND=nag_wp) arrayOutput On exit: the upper triangular part of the variance-covariance matrix of the IP parameter estimates given in B. They are stored packed by column, i.e., the covariance between the parameter estimate given in ${\mathbf{B}}\left(i\right)$ and the parameter estimate given in ${\mathbf{B}}\left(j\right)$, $j\ge i$, is stored in ${\mathbf{COV}}\left(\left(j×\left(j-1\right)/2+i\right)\right)$. 11:   WK($2×{\mathbf{IP}}×{\mathbf{IP}}+{\mathbf{IP}}×{\mathbf{ICONST}}+2×{\mathbf{ICONST}}×{\mathbf{ICONST}}+4×{\mathbf{ICONST}}$) – REAL (KIND=nag_wp) arrayWorkspace Note:  a simple upper bound for the size of the workspace is $5×{\mathbf{IP}}×{\mathbf{IP}}+4×{\mathbf{IP}}$. 12:   IFAIL – INTEGERInput/Output On entry: IFAIL must be set to $0$, $-1\text{​ or ​}1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details. For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{​ or ​}1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{​ or ​}\mathbf{1}$ is used it is essential to test the value of IFAIL on exit. On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6). ## 6  Error Indicators and Warnings If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF). Errors or warnings detected by the routine: ${\mathbf{IFAIL}}=1$ On entry, ${\mathbf{IP}}<1$. or ${\mathbf{ICONST}}\ge {\mathbf{IP}}$, or ${\mathbf{ICONST}}\le 0$, or ${\mathbf{LDV}}<{\mathbf{IP}}$, or ${\mathbf{LDC}}<{\mathbf{IP}}$, or ${\mathbf{S}}\le 0.0$. ${\mathbf{IFAIL}}=2$ C does not give a model of full rank. ## 7  Accuracy It should be noted that due to rounding errors a parameter that should be zero when the constraints have been imposed may be returned as a value of order machine precision. ## 8  Further Comments G02GKF is intended for use in situations in which dummy ($0–1$) variables have been used such as in the analysis of designed experiments when you do not wish to change the parameters of the model to give a full rank model. The routine is not intended for situations in which the relationships between the independent variables are only approximate. ## 9  Example A loglinear model is fitted to a $3$ by $5$ contingency table by G02GCF. The model consists of terms for rows and columns. The table is $141 67 114 79 39 131 66 143 72 35 36 14 38 28 16 .$ The constraints that the sum of row effects and the sum of column effects are zero are then read in and the parameter estimates with these constraints imposed are computed by G02GKF and printed. ### 9.1  Program Text Program Text (g02gkfe.f90) ### 9.2  Program Data Program Data (g02gkfe.d) ### 9.3  Program Results Program Results (g02gkfe.r)

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http://math.stackexchange.com/questions/41313/primes-for-which-xk-equiv-n-pmodp-is-solvable

# Primes for which $x^k \equiv n \pmod{p}$ is solvable For a fixed $n$, how can I characterize the primes $p$ such that there is a $k$ with $x^k\equiv n\pmod p$? Edit: This wasn't actually what I meant... the question I intended is here. - Any condition on $k$? If you can choose $k=1$, or more generally $k\equiv 1 \pmod {p-1}$, then this always has a solution with $x=n$. Or is $x$ fixed? – Thomas Andrews May 25 '11 at 17:24 I think the OP means: which $n$ are non-trivial powers mod $p$? – lhf May 25 '11 at 17:29 @lhf: So we really need to restrict $k$ to have a common factor with $p-1$? If $\gcd(k,p-1)=1$, then we can find an $l$ so that $kl\equiv 1 \pmod {p-1}$ and if $x=n^l$, then $x^k = n^{kl} \equiv n \pmod {p}$ – Thomas Andrews May 25 '11 at 17:37 @Thomas, the way I read the question, you're free to choose $k$. – lhf May 25 '11 at 17:41 You really want to either also fix $x$ or also fix $k$ to get an interesting problem. – Qiaochu Yuan May 25 '11 at 17:51 Every $n$ is a non-trivial power mod $p$ for every $p$. Indeed, by Fermat's little theorem, every $n$ is a $p$-th power mod $p$ for every $p$. If you insist on $1<k<p$, then the following argument works (with exactly one exception noted below): If $p$ divides $n$, you can take $x=0$ and any $k$. Now assume that $p$ does not divide $n$. Take $g$ a primitive root mod $p$ that is not congruent to $n$ mod $p$. You can choose $g$ like that because there are $\phi(p-1)$ primitive roots mod $p$. Then $n$ is a non-trivial power of $g$ mod $p$. The exception is 2, which is not a small power mod 3. (The argument above fails because $\phi(3-1)=1$.) - The only exception is 2 mod 3. Since 2 is the only primitive root mod 3, it is not a non-trivial power mod 3. – Brandon Carter May 25 '11 at 17:52 @Brandon, thanks. I knew I was missing this case! But 2 is a cube mod 3, isn't it, by Fermat. – lhf May 25 '11 at 17:54 This is the correct answer to the question I asked, but unfortunately not to the question I intended! I will accept and post a new question. – Charles May 25 '11 at 19:07 To repeat from my comments above, if we allow $k$ such that $\gcd(k,p-1)=1$, then we can find an $l$ so that $lk\equiv 1 \pmod {p-1}$, and then if $x=n^l$, then $x^k = n^{lk}\equiv n \pmod {p}$ So if we allow such $k$, then it is true for all $p$. On the other hand, if we restrict to $k$ such that $\gcd(k,p-1)>1$, then we cannot find a solution if $n$ is of order $p-1$ in the multiplicative group modulo $p$. It is a "hard problem" to deterime if $n$ is a generator modulo a particular prime $p$. It is not even fully resolved yet whether, if $n$ is not a square and $n\neq -1$, there is always a prime $p$ such that $n$ is a generator $\mod p$, which I recently learned is a conjecture of Artin. (This is mostly resolved - it is know that there are at most two counter-examples $n$, and any counter-example has to be prime.) - Thanks, +1. I had meant to ask a slightly different -- and more interesting -- question, but an apparent lack of coffee left my question in the present state. – Charles May 25 '11 at 19:08

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http://www.math.cmu.edu/~gautam/sj/blog/20140712-bibtex-spacing.html

# Adjusting the space between references in the bibliography. 2014-07-12 ```\usepackage{bibspacing} \setlength{\bibitemsep}{.2\baselineskip plus .05\baselineskip minus .05\baselineskip} ``` Change the spacing to whatever you desire of course. This works even if you use BibTeX. ## Alternate approach. Alternately, if you don’t want to download bibspacing.sty, you can just include it’s contents into the preamble of your LaTeX source: ```\newlength{\bibitemsep}\setlength{\bibitemsep}{.2\baselineskip plus .05\baselineskip minus .05\baselineskip} \newlength{\bibparskip}\setlength{\bibparskip}{0pt} \let\oldthebibliography\thebibliography \renewcommand\thebibliography[1]{% \oldthebibliography{#1}% \setlength{\parskip}{\bibitemsep}% \setlength{\itemsep}{\bibparskip}% } ``` This method also works if you use BibTeX. ## References This was adapted from here. See an alternate approach here.

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https://www.arxiv-vanity.com/papers/1907.10117/

# Simulating an infinite mean waiting time Krzysztof Bartoszek ###### Abstract We consider a hybrid method to simulate the return time to the initial state in a critical–case birth–death process. The expected value of this return time is infinite, but its distribution asymptotically follows a power–law. Hence, the simulation approach is to directly simulate the process, unless the simulated time exceeds some threshold and if it does, draw the return time from the tail of the power law. Keywords : Birth–death process; infinite mean; phylogenetic tree; power–law distribution; return time ## 1 Introduction: a model for phylogenetic trees Birth–and–death processes are frequently used today to model various branching phenomena, e.g. phylogenetic trees. Empirically observed (or rather estimated from e.g. genetic data) phylogenies can exhibit multiple patterns, e.g. many co–occurring species or only one dominating one at a given time instance. The HIV phylogeny is an example of the former, while the influenza phylogeny of the latter. During a given season there is one main flu virus going around, but it may change between seasons. In [5] a very similar model that (depending on the choice of a parameter ) can generate both patterns was proposed. They model the amount of types alive at a given time instance as follows. Assume that at time there are types present. Then, at time the birth rate of types is and the death rate is . Each type has a fitness value attached to it and if a death event occurs, the type with the lowest fitness goes extinct. If there is only one type present, it cannot go extinct. The type with the largest value of the fitness is called the dominating type The main result of [5] is the characterization of the lifetime of the dominating type. ###### Theorem 1.1 (Thm. 1 in [5]) Take . If , then limt→∞P(maximal types at αt and t are the same)=α, while if , then this limit is . Obviously, if , then a given (maximal) type can persist (like influenza) but when , then there will be frequent switching between dominating types (like HIV—we cannot observe which strain dominates). The key object in the proof of Thm. 1.1 are the random times of hitting state , conditional on having started in state , we denote this random variable by . Then, the waiting time for returning to state can be represented as , where . Let denote the cumulative distribution function of , i.e. . Here we will focus on the critical case . In [5] it was claimed that in this regime (proof of Lemma therein). However, in [2] it was shown that this statement is not true (cf. Eqs. and therein) and that (cf Eq. therein) behaves asymptotically as limt→∞t(1−F(t))=1. (1) To be able to work with the law of we introduce the following definition. ###### Definition 1.1 We say that a random variable follows a power–law probability distribution with parameter , if it has support on , , and its law asymptotically satisfies P(Y>y)∼Cy−α, for some constant . Moments of order are infinite. It is worth pointing out that in principle the constant could be replaced by slowly varying function , i.e. for every , . Then more generally we would say that is a random variable with distribution whose right tail is regularly varying at infinity (with parameter , Ch. [1]). However, this is not necessary for the purposes of this study. ###### Lemma 1.2 E[H]=∞ Proof is a positive random variable and follows a power law distribution with and . Hence, for it holds directly that E[Hm]=∞∫0P(Hm>x)dx=∞. It is worth noting that the infinite mean can be derived directly from the model formulation. Denote by the time to reach state from state and . Then, and define . Notice that by the memoryless property of the process we can see that is a non–decreasing sequence as to get from state to state one first has to get back to state . By model construction we have and h3=16+12h2+12h4, giving h2=h3+(h3−h4)−13. In the same way one will have for all hj−hj+1=hj+1+(hj+1−hj+2)−1j+1 resulting in for every h2=h3+(hN−hN+1)−N∑i=31i. As obviously , one needs for every h3≥hN+1−hN+N∑i=31i≥N∑i=31i implying that . As , then immediately . ## 2 Simulation algorithm Very often an important component of studying stochastic models is the possibility to simulate them. This is in order to illustrate the model, gather intuition for its properties, back–up (i.e. check for errors) analytical results and to design Monte Carlo based estimation or testing procedures. A Markov process as described in Section 1 is in principle trivial to simulate using the Gillespie algorithm [4]. Let be the embedded Markov Chain, i.e. , where is the time of the –th birth or death event. If the process is in state at step , then one draws an exponential with rate , waiting time and , where and . However, in the critical case , as we showed that we cannot expect to reliably sample the full distribution of the chain’s trajectory. The tails will be significantly undersampled. In the simplest case, if we want to plot an estimate of ’s density (e.g. a histogram), then we can expect its right tail to be significantly (in the colloquial sense) too light. Unfortunately, only the asymptotic behaviour of the survival function, , is known. Therefore, if we were to draw from its form, we should expect the initial part of the histogram to be badly found. As the constant in Eq. 1, then we know that the relevant power law is defined on . However, the deviation of from its asymptotic has not been studied and it is unclear from what value is the limit approximation good. In any case is contained in ’s support, so any simulation procedure has to allow for values from this interval also. Therefore, in this work we propose a hybrid algorithm that combines both approaches. Intuitively, the left side of the histogram is simulated directly, while the right side is drawn from the the asymptotic. We have as our aim a proof of concept study—to see if reasonable results can be obtained, leaving improvements of the algorithm and its analytical properties for further work. Let be the proportion of simulations that we do not want to simulate directly but draw from the asymptotic. Define as the value for which . This gives . We will use and interchangeably depending on the focus—if it is the tail probability or the threshold value. The hybrid simulation approach is described in Alg. 1. Until the waiting time does not exceed a certain threshold the simulations proceeds directly according to the model’s description. The threshold simulation is chosen so that (approximately, according to the limit distribution) with probability it will not be exceeded. If the threshold is exceeded, then is drawn from a law corresponding to the survival function’s asymptotic behaviour. There is a trade–off in the choice of . If we choose a small , i.e. a threshold far in the right tail, then with high probability we will have an exact simulation algorithm. However, on the other hand the running time may be large. In contrast, taking a larger will reduce the running time, however, more samples will not be drawn exactly but from the asymptotic, and our final simulated value’s distribution could be further away from its true law. Obviously, by it being a probability and the previously discussed properties of the asymptotics of the survival function, i.e. . In order to draw from the law corresponding to the asymptotic behaviour of the right tail of the survival function, we use the powerRlaw [3] R [6] package. The poweRlaw::rplcon(1,,) draws a single value from a power law supported on with density and cumulative distribution functions (Eq. , , [3]) equalling p(t)=α−1Tmin(tTmin)−α,   P(T≤t)=1−(tTmin)−α+1 (2) for and . In our situation we have . Hence, we can take the power law corresponding to the limit as the cumulative distribution function , with density equalling on . This implies taking when calling poweRlaw::rplcon(). As we have defined a threshold of , we need to include it also when drawing the value, i.e. we do not want to have empty draws of too small values. Setting, then , tells poweRlaw::rplcon() that our law is concentrated on . Notice that we are working with the conditional random variable , given that . Its conditional cumulative distribution function will equal , with associated density , but this equals . The function poweRlaw::rplcon() draws a value using the inverse cumulative distribution function method. Let , and then from poweRlaw::rplcon() is given by T=Tmin(1−U)−1/(α−1). As we do not know from what value the tail asymptotics are a good approximation we cannot a priori be sure whether the threshold will be exceeded with probability , nor if after the sampling of from the limit will be accurate. These important properties will be checked empirically in the simulation study in Section 3. ## 3 Simulation setup and results The simulation study presented here has a number of aims, to illustrate the distribution of waiting times as simulated exactly and via Alg. 1. Secondly, to study whether Alg. 1 is a sensible approach to simulating this random time. All simulations were done in R version [6] running on an openSUSE (_) box with a GHz Intel® Xeon® CPU. Simulating the Markov process directly (and then extracting ) is a straightforward procedure. However, as , we introduce a cutoff, if the number of steps exceeds a given number, here , we end the simulation and mark that the process did not return to state . Out of the repeats, reached the maximum allowed number of steps, . In order to see how well Alg. 1 is corresponding to the true distribution of , we re–run it for two values of , and . Then, we compare the logarithms of the survival functions, of both simulations and furthermore on the interval , Fig. 1. The first sample, with threshold , can be thought of as the “true one” on this interval, as was simulated exactly—directly from the model’s definition. We present the logarithm of the survival function as otherwise nothing would be visible from the plot, due to the heavy tail. The power–law property of the tail can be clearly seen in the left panel of Fig. 1. In fact, if one regress on one will obtain a slope estimate of or . This is in agreement with our result that . Out of the simulations only had in the case of the simulation and instances had in the case of the simulation. We can see that these correspond nearly exactly to the desired proportion of exceeding the cutoff, i.e. and . The results presented here indicate that a hybrid approach for simulating the waiting time to return to state in the considered birth–death model is a promising one. For our considered hybrid procedure to be effective one needs to appropriately choose the threshold . If it is too small, then the return time might not yet be in the asymptotic regime. If it is too large, then the running time can be too long. In our case repeats with took about hours while for it took about days. Definitely, is too large for a threshold, while will be acceptable given that the required sample size is smaller. Most importantly, the comparison between ’s and ’s results showed that with the hybrid approach yields samples that do not seem to be distinguishable from the true law. This is as on the interval the simulation is exact. Our study here points to three possible, exciting future directions of development. Firstly, to identify an optimal value for . Secondly, to develop a better simulation approach so that when exceeds and the draw “has to be made from the tail”, the simulated path does not need to be discarded, i.e. to characterize the law of the return time from an arbitrary state to state . And lastly, the study of i.e. how the survival function deviates from its asymptotic behaviour. ## Acknowledgments KB’s research is supported by the Swedish Research Council’s (Vetenskapsrådet) grant no. . KB would like to thank the Editors and an anonymous Reviewer for careful reading of the manuscript and comments that greatly improved it. KB is grateful to Serik Sagitov for encouragement and suggestions to study general passage times between and in the considered here birth–death process. ## References • [1] K. A. Borovkov A. A. Borovkov and. Asymptotic Analysis of Random Walks Heavy–Tailed Distributions. Cambridge University Press, 2008. • [2] K. Bartoszek and M. Krzemiński. Critical case stochastic phylogenetic tree model via the laplace transform. Demonstratio Mathematica, 47:474–481, 2014. • [3] C. S. Gillespie. Fitting heavy tailed distributions: the poweRlaw package. J. Stat. Softw., 64:1–16, 2015. • [4] D. T. Gillespie. Exact stochastic simulation of coupled chemical reactions. 81, 81:2340–2361, 1977. • [5] T. M. Liggett and R. B. Schinazi. A stochastic model for phylogenetic trees. J. Appl. Probab., 46:601–607, 2009. • [6] R Core Team. R: A language and environment for statistical computing, 2013. R Foundation for Statistical Computing, Vienna. Appendix: R code for simulating used in Section 3

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https://collaborate.princeton.edu/en/publications/onset-of-fast-reconnection-in-hall-magnetohydrodynamics-mediated-

# Onset of fast reconnection in Hall magnetohydrodynamics mediated by the plasmoid instability Yi Min Huang, A. Bhattacharjee, Brian P. Sullivan Research output: Contribution to journalArticlepeer-review 71 Scopus citations ## Abstract The role of a super-Alfvénic plasmoid instability in the onset of fast reconnection is studied by means of the largest Hall magnetohydrodynamics simulations to date, with system sizes up to 104 ion skin depths (di). It is demonstrated that the plasmoid instability can facilitate the onset of rapid Hall reconnection, in a regime where the onset would otherwise be inaccessible because the Sweet-Parker width is significantly above di. However, the topology of Hall reconnection is not inevitably a single stable X-point. There exists an intermediate regime where the single X-point topology itself exhibits instability, causing the system to alternate between a single X-point geometry and an extended current sheet with multiple X-points produced by the plasmoid instability. Through a series of simulations with various system sizes relative to di, it is shown that system size affects the accessibility of the intermediate regime. The larger the system size is, the easier it is to realize the intermediate regime. Although our Hall magnetohydrodynamics (MHD) model lacks many important physical effects included in fully kinetic models, the fact that a single X-point geometry is not inevitable raises the interesting possibility for the first time that Hall MHD simulations may have the potential to realize reconnection with geometrical features similar to those seen in fully kinetic simulations, namely, extended current sheets and plasmoid formation. Original language English (US) 072109 Physics of Plasmas 18 7 https://doi.org/10.1063/1.3606363 Published - Jul 2011 Yes ## All Science Journal Classification (ASJC) codes • Condensed Matter Physics ## Fingerprint Dive into the research topics of 'Onset of fast reconnection in Hall magnetohydrodynamics mediated by the plasmoid instability'. Together they form a unique fingerprint.

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http://mathhelpforum.com/pre-calculus/167940-stating-largest-value-constant-difficulty-trigno-functions-2.html

# Thread: Stating the largest value of a constant. Difficulty with trigno in functions. 1. So for [0,360], at the x-point of 270, it will pass the HLT, yes? Sure: if you draw one horizontal line consisting of y=g(270), it will only intersect the function once. However, this will not tell you whether g passes the HLT on the entire interval [0,360], because you would need to test y = g(269), y = g(268), y = g(267), etc., AND all the numbers in-between. You'd have to test the entire continuum of horizontal lines in-between y = 1 and y = 5, since the range corresponding to domain [0,360] is [1,5]. For the question that I posted and the figure in post#3, 90 would be better because it passes the HLT whereas 45 does not pass the HLT at all let alone being compared to 90. You keep using the language "45 passes the HLT". That doesn't parse mathematically. It would be like saying, "You rightly bird blue fly." The correct language is this: Function g passes the HLT on the interval [0,45] (which it does, incidentally!) as well as on the interval [0,90] (which it does, incidentally!). Are you confused by the difference between the notation for an interval, which looks like [0,90], and consists of all the real numbers between and including 0 and 90; versus the notation for a point in the xy plane, which looks like (4,5), and indicates the point where x = 4 and y = 5? I've tried to be consistent, and in this thread, I've been using the interval notation much more than the point notation. It is confusing, because the open interval (0,90) (all the numbers between 0 and 90 but NOT including the endpoints) looks like the notation for a point. You have to use context to know whether someone is talking about the point (4,5) or the interval (4,5). Of course, if someone is talking about the point (5,4), then it's impossible to be confused, because you generally don't write intervals with the right-hand endpoint on the left! That would be too confusing. Getting back to the problem: g is one-to-one on the interval [0,45], and g is one-to-one on the interval [0,90]. So why is A = 90 the best answer to your problem? 2. Are you confused by the difference between the notation for an interval, which looks like [0,90], and consists of all the real numbers between and including 0 and 90; versus the notation for a point in the xy plane, which looks like (4,5), and indicates the point where x = 4 and y = 5? Yes, this was what confused me initially. I did not know you were talking about intervals and assumed them to be points on the xy plane, not in the manner (x,y) because that would mean y was 90, but in the way that you were considering 0-90 or simply 90 and within the entire restriction [0,360]. My bad. Getting back to the problem: g is one-to-one on the interval [0,45], and g is one-to-one on the interval [0,90]. So why is A = 90 the best answer to your problem? Well, you just said earlier that we have to state the largest value of A, and I forgot that the question also asked this so I would have to say that [0,90] is better simply because it is a larger interval than [0,45] and has a larger continuum. Or that 90 is also written as pi/2 and 45 as pi/4 so pi/2, being the larger value of the two, is chosen. Could this be it? 3. Well, you just said earlier that we have to state the largest value of A, and I forgot that the question also asked this so I would have to say that [0,90] is better simply because it is a larger interval than [0,45] and has a larger continuum. There you go. That's the reason. [0,90] is the largest interval that looks like [0,A], over which g has an inverse. 4. There you go. That's the reason. [0,90] is the largest interval that looks like [0,A], over which g has an inverse. Well, that's good to hear. Didn't realize it was that simple but wouldn't have understood it if it weren't for your explanation about the interval and continuum. I see you haven't overlooked the second part. (p2): Obtain an expression in terms of x, for g^-1(x). Turns out that A isn't really needed for this as I was able to obtain the answer without having the need of A being pi/2. The answer to this question is: y=Sin^-1[(3-x)/2] I'm certain that this is the answer but what do you think? 5. I'm certain that this is the answer but what do you think? Looks good to me! I'd say you're done with this problem. 6. Looks good to me! I'd say you're done with this problem. Great. Ackbeet, thank you for your helping me understand this problem and not giving up at any time. Hopefully, you'll be able to help me solve problems I may (likely) have in the future. 7. You're very welcome. Have a good one! Page 2 of 2 First 12

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http://en.wikipedia.org/wiki/Feedback_amplifier

# Negative feedback amplifier (Redirected from Feedback amplifier) Jump to: navigation, search Figure 1: Ideal negative feedback amplifier. A negative feedback amplifier (or feedback amplifier) is an electronic amplifier that subtracts a fraction of its output from its input, so that negative feedback opposes the original signal.[1] The applied negative feedback improves performance (gain stability, linearity, frequency response, step response) and reduces sensitivity to parameter variations due to manufacturing or environment. Because of these advantages, many amplifiers and control systems use negative feedback.[2] An idealized negative feedback amplifier as shown in the diagram is a system of three elements (see Figure 1): • An amplifier with gain AOL • A feedback network 'β', which senses the output signal and possibly transforms it in some way (for example by attenuating or filtering it) • A summing circuit that acts as a subtractor (the circle in the figure), which combines the input and the attenuated output A principal idealization behind this formulation is its division into two blocks, a simple example of what often is called 'circuit partitioning',[3] which refers in this instance to the division into a forward amplification block and a feedback block. In practical amplifiers, the information flow is not unidirectional as shown here.[4] Frequently these blocks are taken to be two-port networks to allow inclusion of bilateral information transfer.[5][6] Casting an amplifier into this form is a non-trivial task, however, especially when the feedback involved is not global (that is directly from the output to the input) but local (that is, feedback within the network, involving nodes that do not coincide with input and/or output terminals).[7][8] In these more general cases, the amplifier is analyzed more directly without the partitioning into blocks like those in the diagram, using instead some analysis based upon signal flow analysis, such as the return ratio method or the asymptotic gain model.[9][10][11] ## Overview Fundamentally, all electronic devices that once provided power gain (e.g., vacuum tubes, bipolar transistors, MOS transistors) are nonlinear. Negative feedback trades gain for higher linearity (reducing distortion), and can provide other benefits. If not designed correctly, amplifiers with negative feedback can become unstable, resulting in unwanted behavior such as oscillation. The Nyquist stability criterion developed by Harry Nyquist of Bell Laboratories is used to study the stability of feedback amplifiers. Feedback amplifiers share these properties:[12] Pros: • Can increase or decrease input impedance (depending on type of feedback) • Can increase or decrease output impedance (depending on type of feedback) • Reduces distortion (increases linearity) • Increases the bandwidth • Desensitizes gain to component variations • Can control step response of amplifier Cons: • May lead to instability if not designed carefully • Amplifier gain decreases • Input and output impedances of a negative feedback amplifier (closed-loop amplifier) become sensitive to the gain of an amplifier without feedback (open-loop amplifier)—that exposes these impedances to variations in the open loop gain, for example, due to parameter variations or nonlinearity of the open-loop gain ## History Harold Stephen Black invented the negative feedback amplifier while he was a passenger on the Lackawanna Ferry (from Hoboken Terminal to Manhattan) on his way to work at Bell Laboratories (located in Manhattan instead of New Jersey in 1927) on August 2, 1927[13] (US patent 2,102,671, issued in 1937[14] ). Black was working on reducing distortion in repeater amplifiers used for telephone transmission. On a blank space in his copy of The New York Times,[15] he recorded the diagram found in Figure 1, and the equations derived below.[16] On August 8, 1928, Black submitted his invention to the U. S. Patent Office, which took more than nine years to issue the patent. Black later wrote: "One reason for the delay was that the concept was so contrary to established beliefs that the Patent Office initially did not believe it would work."[13] ## Classical feedback ### Gain reduction Below, the voltage gain of the amplifier with feedback, the closed-loop gain Afb, is derived in terms of the gain of the amplifier without feedback, the open-loop gain AOL and the feedback factor β, which governs how much of the output signal is applied to the input. See Figure 1, top right. The open-loop gain AOL in general may be a function of both frequency and voltage; the feedback parameter β is determined by the feedback network that is connected around the amplifier. For an operational amplifier two resistors forming a voltage divider may be used for the feedback network to set β between 0 and 1. This network may be modified using reactive elements like capacitors or inductors to (a) give frequency-dependent closed-loop gain as in equalization/tone-control circuits or (b) construct oscillators. The gain of the amplifier with feedback is derived below in the case of a voltage amplifier with voltage feedback. Without feedback, the input voltage V'in is applied directly to the amplifier input. The according output voltage is $V_{out} = A_{OL}\cdot V'_{in}$ Suppose now that an attenuating feedback loop applies a fraction β.Vout of the output to one of the subtractor inputs so that it subtracts from the circuit input voltage Vin applied to the other subtractor input. The result of subtraction applied to the amplifier input is $V'_{in} = V_{in} - \beta \cdot V_{out}$ Substituting for V'in in the first expression, $V_{out} = A_{OL} (V_{in} - \beta \cdot V_{out})$ Rearranging $V_{out} (1 + \beta \cdot A_{OL}) = V_{in} \cdot A_{OL}$ Then the gain of the amplifier with feedback, called the closed-loop gain, Afb is given by, $A_\mathrm{fb} = \frac{V_\mathrm{out}}{V_\mathrm{in}} = \frac{A_{OL}}{1 + \beta \cdot A_{OL}}$ If AOL >> 1, then Afb ≈ 1 / β and the effective amplification (or closed-loop gain) Afb is set by the feedback constant β, and hence set by the feedback network, usually a simple reproducible network, thus making linearizing and stabilizing the amplification characteristics straightforward. Note also that if there are conditions where β AOL = −1, the amplifier has infinite amplification – it has become an oscillator, and the system is unstable. The stability characteristics of the gain feedback product β AOL are often displayed and investigated on a Nyquist plot (a polar plot of the gain/phase shift as a parametric function of frequency). A simpler, but less general technique, uses Bode plots. The combination L = β AOL appears commonly in feedback analysis and is called the loop gain. The combination ( 1 + β AOL ) also appears commonly and is variously named as the desensitivity factor or the improvement factor. ### Bandwidth extension Figure 2: Gain vs. frequency for a single-pole amplifier with and without feedback; corner frequencies are labeled. Feedback can be used to extend the bandwidth of an amplifier at the cost of lowering the amplifier gain.[17] Figure 2 shows such a comparison. The figure is understood as follows. Without feedback the so-called open-loop gain in this example has a single time constant frequency response given by $A_{OL}(f) = \frac {A_0} { 1+ j f / f_C } \ ,$ where fC is the cutoff or corner frequency of the amplifier: in this example fC = 104 Hz and the gain at zero frequency A0 = 105 V/V. The figure shows the gain is flat out to the corner frequency and then drops. When feedback is present the so-called closed-loop gain, as shown in the formula of the previous section, becomes, $A_{fb} (f) = \frac { A_{OL} } { 1 + \beta A_{OL} }$ $= \frac { A_0/(1+jf/f_C) } { 1 + \beta A_0/(1+jf/f_C) }$ $= \frac {A_0} {1+ jf/f_C + \beta A_0}$ $= \frac {A_0} {(1 + \beta A_0) \left(1+j \frac {f} {(1+ \beta A_0) f_C } \right)} \ .$ The last expression shows the feedback amplifier still has a single time constant behavior, but the corner frequency is now increased by the improvement factor ( 1 + β A0 ), and the gain at zero frequency has dropped by exactly the same factor. This behavior is called the gain-bandwidth tradeoff. In Figure 2, ( 1 + β A0 ) = 103, so Afb(0)= 105 / 103 = 100 V/V, and fC increases to 104 × 103 = 107 Hz. ### Multiple poles When the open-loop gain has several poles, rather than the single pole of the above example, feedback can result in complex poles (real and imaginary parts). In a two-pole case, the result is peaking in the frequency response of the feedback amplifier near its corner frequency, and ringing and overshoot in its step response. In the case of more than two poles, the feedback amplifier can become unstable, and oscillate. See the discussion of gain margin and phase margin. For a complete discussion, see Sansen.[18] ## Two-port analysis of feedback Various topologies for a negative feedback amplifier using two-ports. Top left: current amplifier topology; top right: transconductance; bottom left: transresistance; bottom right: voltage amplifier topology.[19] Although, as mentioned in the introduction, some form of signal-flow analysis is the most general way to treat the negative feedback amplifier, representation as two two-ports is the approach most often presented in textbooks, and is presented here. Some drawbacks of this method are described at the end. Amplifiers use current or voltage as input and output, so four types of amplifier are possible (any of two possible inputs with any of two possible outputs). See classification of amplifiers. The objective for the feedback amplifier may be any one of the four types of amplifier, and is not necessarily the same type as the open-loop amplifier, which itself may be any one of these types. So, for example, an op amp (voltage amplifier) can be arranged to make a current amplifier instead. Negative feedback amplifiers of any type can be implemented using combinations of two-port networks. There are four types of two-port network, and the type of amplifier desired dictates the choice of two-ports and the selection of one of the four different connection topologies shown in the diagram. These connections are usually referred to as series or shunt (parallel) connections.[20][21] In the diagram, the left column shows shunt inputs; the right column shows series inputs. The top row shows series outputs; the bottom row shows shunt outputs. The various combinations of connections and two-ports are listed in the table below. Feedback amplifier type Input connection Output connection Ideal feedback Two-port feedback Current Shunt Series CCCS g-parameter Transresistance Shunt Shunt CCVS y-parameter Transconductance Series Series VCCS z-parameter Voltage Series Shunt VCVS h-parameter For example, for a current feedback amplifier, current from the output is sampled for feedback and combined with current at the input. Therefore, the feedback ideally is performed using an (output) current-controlled current source (CCCS), and its imperfect realization using a two-port network also must incorporate a CCCS, that is, the appropriate choice for feedback network is a g-parameter two-port. Here the two-port method used in most textbooks is presented,[22][23][24] using the circuit treated in the article on asymptotic gain model. Figure 3: A shunt-series feedback amplifier Figure 3 shows a two-transistor amplifier with a feedback resistor Rf. The aim is to analyze this circuit to find three items: the gain, the output impedance looking into the amplifier from the load, and the input impedance looking into the amplifier from the source. ### Replacement of the feedback network with a two-port The first step is replacement of the feedback network by a two-port. Just what components go into the two-port? On the input side of the two-port we have Rf. If the voltage at the right side of Rf changes, it changes the current in Rf that is subtracted from the current entering the base of the input transistor. That is, the input side of the two-port is a dependent current source controlled by the voltage at the top of resistor R2. One might say the second stage of the amplifier is just a voltage follower, transmitting the voltage at the collector of the input transistor to the top of R2. That is, the monitored output signal is really the voltage at the collector of the input transistor. That view is legitimate, but then the voltage follower stage becomes part of the feedback network. That makes analysis of feedback more complicated. Figure 4: The g-parameter feedback network An alternative view is that the voltage at the top of R2 is set by the emitter current of the output transistor. That view leads to an entirely passive feedback network made up of R2 and Rf. The variable controlling the feedback is the emitter current, so the feedback is a current-controlled current source (CCCS). We search through the four available two-port networks and find the only one with a CCCS is the g-parameter two-port, shown in Figure 4. The next task is to select the g-parameters so that the two-port of Figure 4 is electrically equivalent to the L-section made up of R2 and Rf. That selection is an algebraic procedure made most simply by looking at two individual cases: the case with V1 = 0, which makes the VCVS on the right side of the two-port a short-circuit; and the case with I2 = 0. which makes the CCCS on the left side an open circuit. The algebra in these two cases is simple, much easier than solving for all variables at once. The choice of g-parameters that make the two-port and the L-section behave the same way are shown in the table below. g11 g12 g21 g22 $\frac {1} {R_f+R_2}$ $- \frac {R_2}{R_2+R_f}$ $\frac {R_2} {R_2+R_f}$ $R_2//R_f \$ Figure 5: Small-signal circuit with two-port for feedback network; upper shaded box: main amplifier; lower shaded box: feedback two-port replacing the L-section made up of Rf and R2. ### Small-signal circuit The next step is to draw the small-signal schematic for the amplifier with the two-port in place using the hybrid-pi model for the transistors. Figure 5 shows the schematic with notation R3 = RC2 // RL and R11 = 1 / g11, R22 = g22 . ### Loaded open-loop gain Figure 3 indicates the output node, but not the choice of output variable. A useful choice is the short-circuit current output of the amplifier (leading to the short-circuit current gain). Because this variable leads simply to any of the other choices (for example, load voltage or load current), the short-circuit current gain is found below. First the loaded open-loop gain is found. The feedback is turned off by setting g12 = g21 = 0. The idea is to find how much the amplifier gain is changed because of the resistors in the feedback network by themselves, with the feedback turned off. This calculation is pretty easy because R11, RB, and rπ1 all are in parallel and v1 = vπ. Let R1 = R11 // RB // rπ1. In addition, i2 = −(β+1) iB. The result for the open-loop current gain AOL is: $A_{OL} = \frac { \beta i_B } {i_S} = g_m R_C \left( \frac { \beta }{ \beta +1} \right) \left( \frac {R_1} {R_{22} + \frac {r_{ \pi 2} + R_C } {\beta + 1 } } \right) \ .$ ### Gain with feedback In the classical approach to feedback, the feedforward represented by the VCVS (that is, g21 v1) is neglected.[25] That makes the circuit of Figure 5 resemble the block diagram of Figure 1, and the gain with feedback is then: $A_{FB} = \frac { A_{OL} } {1 + { \beta }_{FB} A_{OL} }$ $A_{FB} = \frac {A_{OL} } {1 + \frac {R_2} {R_2+R_f} A_{OL} } \ ,$ where the feedback factor βFB = −g12. Notation βFB is introduced for the feedback factor to distinguish it from the transistor β. ### Input and output resistances Figure 6: Circuit set-up for finding feedback amplifier input resistance Feedback is used to better match signal sources to their loads. For example, a direct connection of a voltage source to a resistive load may result in signal loss due to voltage division, but interjecting a negative feedback amplifier can increase the apparent load seen by the source, and reduce the apparent driver impedance seen by the load, avoiding signal attenuation by voltage division. This advantage is not restricted to voltage amplifiers, but analogous improvements in matching can be arranged for current amplifiers, transconductance amplifiers and transresistance amplifiers. To explain these effects of feedback upon impedances, first a digression on how two-port theory approaches resistance determination, and then its application to the amplifier at hand. #### Background on resistance determination Figure 6 shows an equivalent circuit for finding the input resistance of a feedback voltage amplifier (left) and for a feedback current amplifier (right). These arrangements are typical Miller theorem applications. In the case of the voltage amplifier, the output voltage βVout of the feedback network is applied in series and with an opposite polarity to the input voltage Vx travelling over the loop (but in respect to ground, the polarities are the same). As a result, the effective voltage across and the current through the amplifier input resistance Rin decrease so that the circuit input resistance increases (one might say that Rin apparently increases). Its new value can be calculated by applying Miller theorem (for voltages) or the basic circuit laws. Thus Kirchhoff's voltage law provides: $V_x = I_x R_{in} + \beta v_{out} \ ,$ where vout = Av vin = Av Ix Rin. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is: $R_{in}(fb) = \frac {V_x} {I_x} = \left( 1 + \beta A_v \right ) R_{in} \ .$ The general conclusion from this example and a similar example for the output resistance case is: A series feedback connection at the input (output) increases the input (output) resistance by a factor ( 1 + β AOL ), where AOL = open loop gain. On the other hand, for the current amplifier, the output current βIout of the feedback network is applied in parallel and with an opposite direction to the input current Ix. As a result, the total current flowing through the circuit input (not only through the input resistance Rin) increases and the voltage across it decreases so that the circuit input resistance decreases (Rin apparently decreases). Its new value can be calculated by applying the dual Miller theorem (for currents) or the basic Kirchhoff's laws: $I_x = \frac {V_{in}} {R_{in}} + \beta i_{out} \ .$ where iout = Ai iin = Ai Vx / Rin. Substituting this result in the above equation and solving for the input resistance of the feedback amplifier, the result is: $R_{in}(fb) = \frac {V_x} {I_x} = \frac { R_{in} } { \left( 1 + \beta A_i \right ) } \ .$ The general conclusion from this example and a similar example for the output resistance case is: A parallel feedback connection at the input (output) decreases the input (output) resistance by a factor ( 1 + β AOL ), where AOL = open loop gain. These conclusions can be generalized to treat cases with arbitrary Norton or Thévenin drives, arbitrary loads, and general two-port feedback networks. However, the results do depend upon the main amplifier having a representation as a two-port – that is, the results depend on the same current entering and leaving the input terminals, and likewise, the same current that leaves one output terminal must enter the other output terminal. A broader conclusion, independent of the quantitative details, is that feedback can be used to increase or to decrease the input and output impedance. #### Application to the example amplifier These resistance results now are applied to the amplifier of Figure 3 and Figure 5. The improvement factor that reduces the gain, namely ( 1 + βFB AOL), directly decides the effect of feedback upon the input and output resistances of the amplifier. In the case of a shunt connection, the input impedance is reduced by this factor; and in the case of series connection, the impedance is multiplied by this factor. However, the impedance that is modified by feedback is the impedance of the amplifier in Figure 5 with the feedback turned off, and does include the modifications to impedance caused by the resistors of the feedback network. Therefore, the input impedance seen by the source with feedback turned off is Rin = R1 = R11 // RB // rπ1, and with the feedback turned on (but no feedforward) $R_{in} = \frac {R_1} {1 + { \beta }_{FB} A_{OL} } \ ,$ where division is used because the input connection is shunt: the feedback two-port is in parallel with the signal source at the input side of the amplifier. A reminder: AOL is the loaded open loop gain found above, as modified by the resistors of the feedback network. The impedance seen by the load needs further discussion. The load in Figure 5 is connected to the collector of the output transistor, and therefore is separated from the body of the amplifier by the infinite impedance of the output current source. Therefore, feedback has no effect on the output impedance, which remains simply RC2 as seen by the load resistor RL in Figure 3.[26][27] If instead we wanted to find the impedance presented at the emitter of the output transistor (instead of its collector), which is series connected to the feedback network, feedback would increase this resistance by the improvement factor ( 1 + βFB AOL).[28] ### Load voltage and load current The gain derived above is the current gain at the collector of the output transistor. To relate this gain to the gain when voltage is the output of the amplifier, notice that the output voltage at the load RL is related to the collector current by Ohm's law as vL = iC (RC2 || RL). Consequently, the transresistance gain vL / iS is found by multiplying the current gain by RC2 || RL: $\frac {v_L} {i_S} = A_{FB} (R_{C2} \parallel R_L ) \ .$ Similarly, if the output of the amplifier is taken to be the current in the load resistor RL, current division determines the load current, and the gain is then: $\frac {i_L} {i_S} = A_{FB} \frac {R_{C2}} {R_{C2} + R_L} \ .$ ### Is the main amplifier block a two port? Figure 7: Amplifier with ground connections labeled by G. The feedback network satisfies the port conditions. Some drawbacks of the two two-port approach follow, intended for the attentive reader. Figure 7 shows the small-signal schematic with the main amplifier and the feedback two-port in shaded boxes. The feedback two-port satisfies the port conditions: at the input port, Iin enters and leaves the port, and likewise at the output, Iout enters and leaves. Is the main amplifier block also a two-port? The main amplifier is shown in the upper shaded box. The ground connections are labeled. Figure 7 shows the interesting fact that the main amplifier does not satisfy the port conditions at its input and output unless the ground connections are chosen to make that happen. For example, on the input side, the current entering the main amplifier is IS. This current is divided three ways: to the feedback network, to the bias resistor RB and to the base resistance of the input transistor rπ. To satisfy the port condition for the main amplifier, all three components must be returned to the input side of the main amplifier, which means all the ground leads labeled G1 must be connected, as well as emitter lead GE1. Likewise, on the output side, all ground connections G2 must be connected and also ground connection GE2. Then, at the bottom of the schematic, underneath the feedback two-port and outside the amplifier blocks, G1 is connected to G2. That forces the ground currents to divide between the input and output sides as planned. Notice that this connection arrangement splits the emitter of the input transistor into a base-side and a collector-side – a physically impossible thing to do, but electrically the circuit sees all the ground connections as one node, so this fiction is permitted. Of course, the way the ground leads are connected makes no difference to the amplifier (they are all one node), but it makes a difference to the port conditions. This artificiality is a weakness of this approach: the port conditions are needed to justify the method, but the circuit really is unaffected by how currents are traded among ground connections. However, if no possible arrangement of ground conditions leads to the port conditions, the circuit might not behave the same way.[29] The improvement factors ( 1 + βFB AOL) for determining input and output impedance might not work.[30] This situation is awkward, because a failure to make a two-port may reflect a real problem (it just is not possible), or reflect a lack of imagination (for example, just did not think of splitting the emitter node in two). As a consequence, when the port conditions are in doubt, at least two approaches are possible to establish whether improvement factors are accurate: either simulate an example using Spice and compare results with use of an improvement factor, or calculate the impedance using a test source and compare results. A more practical choice is to drop the two-port approach altogether, and use various alternatives based on signal flow graph theory, including the Rosenstark method, the Choma method, and use of Blackman's theorem.[31] That choice may be advisable if small-signal device models are complex, or are not available (for example, the devices are known only numerically, perhaps from measurement or from SPICE simulations). ## References and notes 1. ^ Santiram Kal (2004). Basic Electronics: Devices, Circuits, and IT fundamentals (Paperback ed.). Prentice-Hall of India Pvt Ltd. pp. 191 ff. ISBN 978-8120319523. 2. ^ Kuo, Benjamin C & Farid Golnaraghi (2003). Automatic control systems (Eighth edition ed.). NY: Wiley. p. 46. ISBN 0-471-13476-7. 3. ^ Partha Pratim Sahu (2013). "§8.2 Partitioning". VLSI Design. McGraw Hill Education. p. 253. ISBN 9781259029844. dividing a circuit into smaller parts ...[so]...the number of connections between parts is minimized 4. ^ Gaetano Palumbo, Salvatore Pennisi (2002). Feedback Amplifiers: Theory and Design. Springer Science & Business Media. ISBN 9780792376439. In real cases, unfortunately, blocks...cannot be assumed to be unidirectional. 5. ^ Wai-Kai Chen (2009). "§1.2 Methods of analysis". Feedback, Nonlinear, and Distributed Circuits. CRC Press. p. 1-3. ISBN 9781420058826. 6. ^ Donald O. Pederson, Kartikeya Mayaram (2007). "§5.2 Feedback for a general amplifier". Analog Integrated Circuits for Communication: Principles, Simulation and Design. Springer Science & Business Media. pp. 105 ff. ISBN 9780387680309. 7. ^ Scott K Burgess & John Choma, Jr. "§6.3 Circuit partitioning". Generalized feedback circuit analysis. 8. ^ Gaetano Palumbo & Salvatore Pennisi (2002). Feedback amplifiers: theory and design. Springer Science & Business Media. p. 66. ISBN 9780792376439. 9. ^ For an introduction, see Rahul Sarpeshkar (2010). "Chapter 10: Return ratio analysis". Ultra Low Power Bioelectronics: Fundamentals, Biomedical Applications, and Bio-Inspired Systems. Cambridge University Press. pp. 240 ff. ISBN 9781139485234. 10. ^ Wai-Kai Chen (2005). "§11.2 Methods of analysis". Circuit Analysis and Feedback Amplifier Theory. CRC Press. pp. 11–2 ff. ISBN 9781420037272. 11. ^ Gaetano Palumbo, Salvatore Pennisi (2002). "§3.3 The Rosenstark Method and §3.4 The Choma Method". Feedback Amplifiers: Theory and Design. Springer Science & Business Media. pp. 69 ff. ISBN 9780792376439. 12. ^ Palumbo, Gaetano & Salvatore Pennisi (2002). Feedback amplifiers: theory and design. Boston/Dordrecht/London: Kluwer Academic. p. 64. ISBN 0-7923-7643-9. 13. ^ a b Black, H.S. (January 1934). "Stabilized Feedback Amplifiers". Bell System Tech. J. (American Telephone & Telegraph) 13 (1): 1–18. Retrieved January 2, 2013. 14. ^ "H.S.Black, "Wave Translation System." US patent 2,102,671". Retrieved 2012-04-19. 15. ^ Currently on display at Bell Laboratories in Mountainside, New Jersey 16. ^ Waldhauer, Fred (1982). Feedback. NY: Wiley. p. 3. ISBN 0-471-05319-8. 17. ^ RW Brodersen Analog circuit design: lectures on stability 18. ^ Willy M. C. Sansen (2006). Analog design essentials. New York; Berlin: Springer. pp. §0513–§0533, p. 155–165. ISBN 0-387-25746-2. 19. ^ Richard C Jaeger (1997). "Figure 18.2". Microelectronic circuit design (International ed.). McGraw-Hill. p. 986. 20. ^ Ashok K. Goel Feedback topologies 21. ^ Zimmer T & Geoffroy D: Feedback amplifier 22. ^ Vivek Subramanian: Lectures on feedback 23. ^ P R Gray, P J Hurst, S H Lewis, and R G Meyer (2001). Analysis and Design of Analog Integrated Circuits (Fourth Edition ed.). New York: Wiley. pp. 586–587. ISBN 0-471-32168-0. 24. ^ A. S. Sedra and K.C. Smith (2004). Microelectronic Circuits (Fifth Edition ed.). New York: Oxford. Example 8.4, pp. 825–829 and PSpice simulation pp. 855–859. ISBN 0-19-514251-9. 25. ^ If the feedforward is included, its effect is to cause a modification of the open-loop gain, normally so small compared to the open-loop gain itself that it can be dropped. Notice also that the main amplifier block is unilateral. 26. ^ The use of the improvement factor ( 1 + βFB AOL) requires care, particularly for the case of output impedance using series feedback. See Jaeger, note below. 27. ^ R.C. Jaeger and T.N. Blalock (2006). Microelectronic Circuit Design (Third Edition ed.). McGraw-Hill Professional. Example 17.3 pp. 1092–1096. ISBN 978-0-07-319163-8. 28. ^ That is, the impedance found by turning off the signal source IS = 0, inserting a test current in the emitter lead Ix, finding the voltage across the test source Vx, and finding Rout = Vx / Ix. 29. ^ The equivalence of the main amplifier block to a two-port network guarantees that performance factors work, but without that equivalence they may work anyway. For example, in some cases the circuit can be shown equivalent to another circuit that is a two port, by "cooking up" different circuit parameters that are functions of the original ones. There is no end to creativity! 30. ^ Richard C Jaeger, Travis N Blalock (2004). "§18.7: Common errors in applying two-port feedback theory". Microelectronic circuit design (2nd ed.). McGraw=Hill Higher Education. pp. 1409 ff. ISBN 0072320990. Great care must be exercised in applying two-port theory to ensure that the amplifier feedback networks can actually be represented as two-ports 31. ^ Gaetano Palumbo, Salvatore Pennisi (2002). Feedback Amplifiers: Theory and Design. Springer Science & Business Media. p. 66. ISBN 9780792376439.

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http://math.stackexchange.com/questions/317246/is-there-any-bound-for-int-mathbbr-sqrtf-0xf-1x-mboxdx

# Is there any bound for $\int_{\mathbb{R}}\sqrt{f_0[x]f_1[x]}\mbox{d}x$ I wonder if there is an upperbound for following the expression: $$\int_{\mathbb{R}}\sqrt{f_0[x]f_1[x]}\mbox{d}x$$ where $f_i$, $i=0,1$ are some density functions. 1. Use Cauchy Schwarz.$\mbox{ }$

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http://mathhelpforum.com/calculus/13497-dervatives-newton-s-method.html

# Math Help - Dervatives and Newton's Method 1. ## Dervatives and Newton's Method #1. f(x) = x^4-17x^2+18 What is the relationship between the sign of f'(x) and the graph f(x) ? What is the relationship between the roots of f''(x) and the graph of f(x) ? What is the relationship between the sign of f''(x) and the graph of f(x) ? #2. Given that f(x) = sin(x) use Newton's Method to find the first 2 positive zeros showing the first few approximations in each case. So... pi ___ x1: 3 x2: 3.142546543 x3: 3.141592653 x4: 3.14159255359 also... 2pi ____ x1: 6 x2: 6.29100619138 x3: 6.28310514772 x4: 6.28318530718 The first positive zero is: 3.14159 The second positive zero: 6.28319 What happens when you took pi/2 as first approximation? Explain I got 1 but else is there to say #3. if f(x) = x^4-17x^2+18 f''(x) = 12x^2-34 Therefore f''(x) = 0 if x = -sqrt(102)/6 and sqrt(102)/6 and these are the x-coordinates of the points of __ ___ of f(x)? Thanks for the Help! 2. ## Re: P.S. This is one of the Calculator labs that the department makes up 3. Originally Posted by qbkr21 #1. f(x) = x^4-17x^2+18 What is the relationship between the sign of f'(x) and the graph f(x) ? What is the relationship between the roots of f''(x) and the graph of f(x) ? What is the relationship between the sign of f''(x) and the graph of f(x) ? the sign of f'(x) let's you know when f(x) is increasing or decreasing. if f'(x) is positive, f(x) has a positive slope and is hence increasing. if f'(x) is negative, f(x) has a negative slope and is hence decreasing the roots of f''(x) give the inflection points of f(x), that is the points where f(x) changes concavity from concave up to concave down or vice versa. the sign of f''(x) indicates the concavity (and hence the nature of critical points) of f(x). if f''(x) is positive for some critical point in f(x), then f(x) is concave up at that point and it is a local min. if f''(x) is negative at some critical point of f(x), then f(x) is concave down at that point and it is a local max Originally Posted by qbkr21 #3. if f(x) = x^4-17x^2+18 f''(x) = 12x^2-34 Therefore f''(x) = 0 if x = -sqrt(102)/6 and sqrt(102)/6 and these are the x-coordinates of the points of __ ___ of f(x)? points of inflection of f(x)...usually. 4. ## Re: Here is the exact question Jhevon by the way thanks for the help... 5. Originally Posted by qbkr21 Here is the exact question Jhevon by the way thanks for the help... you used x_(n+1) = x_n - f(x_n)/f'(x_n) for each of these right? did you find the first few values for pi/2? 6. ## Re: Uhhh...I just plug in pi/2 into Newtons method formula and x_n-f(x)/f'(x) and got the number 1. Is there something else I should do. Thanks for the helpful posts. 7. Originally Posted by qbkr21 Uhhh...I just plug in pi/2 into Newtons method formula and x_n-f(x)/f'(x) and got the number 1. Is there something else I should do. Thanks for the helpful posts. ok so these questions are weird, i don't know exactly what answer they are looking for, but anyway, we can see a few patterns. note that Newton's method is used to approximate the roots of a function. for each approximation we begin with, using Newton's method puts us closer to the root that is closest to that value. not that when we begin with x = 3, each new approximation is bigger than the last and they keep getting closer to pi, 3.1415... since pi is the closest root to 3 when we start with x = 6, each new approximation is again bigger than the last, that is because we are increasing to 2pi, the next closest root to 6 when we begin with pi/2, each new approximation begins to decrease, this is because we are moving towards the closest root, which is zero. i find this interesting though, since pi is as close to pi/2 as 0 is

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http://math.stackexchange.com/questions/112677/proving-that-an-integral-domain-has-at-most-two-elements-that-satisfy-the-equati

# Proving that an integral domain has at most two elements that satisfy the equation $x^2 = 1$. I like to be thorough, but if you feel confident you can skip the first paragraph. Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds. If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain. We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1). Here's what I did: For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$. Assume $a^2 = 1$. We want to show that letting $b^2 = 1$ as well will lead to a contradiction. So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution? If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction. If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction. Similarly, $ab = b$ won't work either. So it must be that $ab = c$. So by "admitting" $b$ as a solution, we're forced to admit $c$ as well. So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$. We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution. From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$. So $abc$ is not a new solution; it's just one. At this point I'm stuck. I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$. What am I missing? - Have you tried factoring $x^2-1$? –  user641 Feb 24 '12 at 0:40 I hope I'm not over simplifying things, but since the ring is commutative, if $a$ is any solution, then $a^2-1=0$, which implies $(a-1)(a+1)=0$. Since you're working in an integral domain, then either $a-1=0$ or $a+1=0$, which implies $a$ can only be $1$ or $-1$. –  Buble Feb 24 '12 at 0:43 More generally, every element $\ell\in R$ of an integral domain $R$ cannot have more than two square roots. To see this, let $a$ be such that $a^2=\ell$, and suppose $b^2=\ell$ also for some $b\in R$. Then we can subtract one from the other and factor as $(a-b)(a+b)=0$, and deduce $b=\pm a$ via integrality. - Hint: $x^2 - 1 = (x-1)(x+1)$. If this is $0 \ldots$ - You’ve shown that if $R$ has two distinct elements other than $1$ whose squares are $1$, then their product is a third such element. But in fact $R$ can’t have two distinct elements other than $1$ whose squares are $1$ in the first place. To see this, show that $x^2=1$ can have at most two solutions by factoring $x^2-1$ and using the fact that $R$ has no zero-divisors. - Thanks, that explains it. –  mahin Feb 24 '12 at 1:27 +1 This is the only answer that actually mentions the OP's argument, and where it fails. –  M Turgeon Mar 23 '12 at 17:44 In any integral domain, a polynomial of degree $d$ has at most $d$ roots, which implies your result. If $aT + b$ is a degree-1 polynomial with coefficients in an integral domain and $a \ne 0$, then if $x$ and $y$ are roots, we see that $a(x-y) = 0$ which implies $x = y$. Now proceed by induction. - More generally, overy any ring, a nonzero polynomial has no more roots than its degree if the difference of any two distinct roots is not a zero-divisor (which is true in any integral domain). THEOREM $\$ Let $\rm R$ be a ring and let $\rm\:f\in R[x].\:$ If $\rm\:f\:$ has more roots than its degree, and if the difference of any two distinct roots is not a zero-divisor, then $\rm\: f = 0.$ Proof $\$ Clear if $\rm\:deg\: f = 0\:$ since the only constant polynomial with a root is the zero polynomial. Else $\rm\:deg\: f \ge 1\:$ so by hypothesis $\rm\:f\:$ has a root $\rm\:s\in R.\:$ Factor Theorem $\rm\Rightarrow\: f(x) = (x-s)\:g ,$ $\rm\: g\in R[x].$ Every root $\rm\:r\ne s\:$ is a root of $\rm\: g\:$ by $\rm\: f(r) = (r-s)g(r) = 0\ \Rightarrow\ g(r) = 0, \ by\ \ r-s\$ not a zero-divisor. So $\rm\:g\:$ satisfies hypotheses, so induction on degree $\rm\:\Rightarrow\:g=0\:\Rightarrow\:f=0.\$ QED Here's a nice constructive application: if a polynomial $\rm\:f(x)\:$ over $\rm\:\mathbb Z/n\:$ has more roots than its degree, then we can quickly compute a nontrivial factor of $\rm\:n\:$ by a simple $\rm\:gcd\:$ calculation. The quadratic case of this is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:n\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/n,\:$ e.g. a square root of $1$ that's not $\pm 1$. - Glad to see you're back! –  Tyler Feb 26 '12 at 0:28

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https://nyuscholars.nyu.edu/en/publications/nonuniqueness-of-weak-solutions-to-the-sqg-equation

# Nonuniqueness of Weak Solutions to the SQG Equation Tristan Buckmaster, Steve Shkoller, Vlad Vicol Research output: Contribution to journalArticlepeer-review ## Abstract We prove that weak solutions of the inviscid SQG equations are not unique, thereby answering Open Problem 11 of De Lellis and Székelyhidi in 2012. Moreover, we also show that weak solutions of the dissipative SQG equation are not unique, even if the fractional dissipation is stronger than the square root of the Laplacian. In view of the results of Marchand in 2008, we establish that for the dissipative SQG equation, weak solutions may be constructed in the same function space both via classical weak compactness arguments and via convex integration. Original language English (US) 1809-1874 66 Communications on Pure and Applied Mathematics 72 9 https://doi.org/10.1002/cpa.21851 Published - Sep 2019 ## ASJC Scopus subject areas • Mathematics(all) • Applied Mathematics ## Fingerprint Dive into the research topics of 'Nonuniqueness of Weak Solutions to the SQG Equation'. Together they form a unique fingerprint.

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https://math.stackexchange.com/questions/2774397/find-polynomial-over-z2-that-has-the-least-degree-for-which-the-field-p-gf64

# Find polynomial over Z2, that has the least degree, for which the field P=GF(64) is the minimum decomposition field. How much such polynomial are? Find polynomial over Z2, that has the least degree, for which the field P=GF(64) is the minimum decomposition field. How much such polynomials are? Please, answer in detail! I think that these polynomials are all sixth irreducible polynomials, but I don't know how to validate this fact. It's a general fact that $\mathbb{F}_p[x]/(q)$ is isomorphic to $\mathbb{F}_{p^n}$, where $q$ is an irreducible polynomial of degree $n$. Furthermore, $\mathbb{F}_p[x_1]$ is a splitting field for $q$ where we take $x_1$ to be a root for $q$ (we know that it contains $\mathbb{F}_{p^n}$; this says that they are the same). This is because $x, x^p, \ldots, x^{p^{n-1}}$ are all roots of $q$. Thus we are looking for irreducible polynomials of degree $6$. We know they all must divide $x^{64}-x$, since every element of $\mathbb{F}_{64}$ satisfies that. Moreover, the roots of that are distinct (take the derivative) so we just need to factorize it and count the number with degree $6$. The product of the irreducible polynomials with degree $d$ over all $d|n$ is $x^{2^n}-x$ (this can be deduced from the combining the fact that $\mathbb{F}_{p^m}\subset \mathbb{F}_{p^n}$ iff $m|n$ with the reasoning above). Here, $p=2$ and $n=6$. There are obviously $2$ with degree $1$, which means there are $(2^2-2)/2 = 1$ of degree $2$, $(2^3-2)/3=2$ of degree $3$, and $(2^6-2-2-6)/6 = 9$ of degree $6$. Finding one explicitly might be a bit annoying. It's not too hard to get that for $1, 2, 3$ they are $x, x+1, x^2+x+1, x^3+x+1, x^3+x^2+1$. Now we want to remove some of the monomials from $x^6+x^5+x^4+x^3+x^2+x+1$ to make it irreducible, and we know we need to keep $x^6$. From degree 1 we need to remove an even number and we need to keep $1$. Ok so after some trials you will get an answer like $x^6+x^4+x^3+x+1$. • Thank you very much for your explanations! And is it easier to sort out all irreducible polynomials of degrees 1, 2, 3, 4, 5, 6? I mean that is it right only to show that irreducible polynomials of degree 1 splits over GF(4), 2 --- splits over GF(8), ..., 5 --- splits over GF(32). And then say that if a 6th degree polynomial isn't irreducible, it contains irreducible polynomials of degrees 1, 2, 3, 4 or 5. So, all 6th degree irreducible polynomials split over GF(64). And then show their splits. – alexhak May 10 '18 at 16:00 This late answer uses computer power to find explicitly the irreductible polynomials of degree six in $\Bbb F_2[X]$. There exists up to isomorphy only one field with $2^6$ elements, $F=\Bbb F_{64}$, its elements are fixed by the corresponding Frobenius isomorphism, so $F$ contains the roots of $$x^{2^6}-x\ .$$ sage gives the following factorization for it: sage: R.<x> = PolynomialRing( GF(2) ) sage: for f, multiplicity in ( x^(2^6) - x ).factor(): ....: print f ....: x x + 1 x^2 + x + 1 x^3 + x + 1 x^3 + x^2 + 1 x^6 + x + 1 x^6 + x^3 + 1 x^6 + x^4 + x^2 + x + 1 x^6 + x^4 + x^3 + x + 1 x^6 + x^5 + 1 x^6 + x^5 + x^2 + x + 1 x^6 + x^5 + x^3 + x^2 + 1 x^6 + x^5 + x^4 + x + 1 x^6 + x^5 + x^4 + x^2 + 1 How many elements in $F$ are roots of a polynomial of degree exactly $6$? There are, by inclusion/exclusion $|F_{2^6}|-|F_{2^3}|-|F_{2^2}|+|F_{2^1}|=2^6-2^3-2^2+2=64-8-4+2=54$ such elements, so we expect $54/6=9$ (edited division...) irreducible polynomials of degree $6$ as factors above. • Thanks very much! It solves my problem – alexhak May 10 '18 at 20:22 • WA also helps. – lhf May 10 '18 at 21:31

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https://datascience.stackexchange.com/questions/44406/word-embeddings-and-punctuation-symbols

# Word embeddings and punctuation symbols I have a decent understanding of word embeddings (at its core, one can think of a word being converted into a vector of, say, 100 dimensions, and each dimension given a particular value... this allows to do math with the words, also it makes the training sets to be non-sparse...) But today something came to my mind, what about punctuation symbols such as , . () ? ! ... ? They do have a huge impact on the meaning of sentences and, like words, the position and context in which they are used is relevant. So the question is, how should this be modeled? are pretrained sets like GloVe including punctuation symbols? should I simply remove punctuation symbols from text?

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https://www.physicsforums.com/threads/proving-a-form-z-f-r-to-be-a-surface-of-revolution.885643/

# Proving a form $z=f(r)$ to be a surface of revolution Tags: 1. Sep 17, 2016 ### toforfiltum 1. The problem statement, all variables and given/known data Suppose that a surface has an equation in cylindrical coordinates of the form $z=f(r)$. Explain why it must be a surface of revolution. 2. Relevant equations 3. The attempt at a solution I consider $z=f(r)$ in terms of spherical coordinates. $p cosφ = f \sqrt{(p sinφcosθ)^2 + (p sinφsinθ)^2}$ $p cosφ= f\sqrt{(p sinφ)^2}$ $p cosφ=f(p sinφ)$ $cosφ= f (sinφ)$ $∴φ= \cos^{-1} f(sinφ)$ Since equation is independent of $\theta$, it describes a surface of revolution about the $z$ axis. Is my prove right or acceptable? 2. Sep 17, 2016 ### Ray Vickson You are using spherical, not cylindreical coordinates. Also: in LaTeX, put a "\" before sin, cos, etc. Without it, the results are ugly and hard to read, like $sin \phi cos \theta$; with it, they look good, as in $\sin \phi \cos \theta$. 3. Sep 17, 2016 ### toforfiltum Okay! Thanks for the advice! Just started using it, so sorry for the ugly text. I don't have any idea of starting the proof using cylindrical coordinates, that's why I converted it to spherical ones. But I will give it a try now. The cylindrical coordinates are in the form $(r,\theta, z)$. I assume that $r \geq 0$ So $r= \sqrt{(x^2 + y^2)}$ and $z=f(\sqrt{(x^2 + y^2)})$, which gives a unique value. These gives a set of points that form a line in $3D$ space. Since equation is independent of $\theta$, line is the same for any value of $\theta$. These similar set of lines form a surface of revolution. Is it right in any way at all? I'm guessing here. 4. Sep 17, 2016 ### LCKurtz Why introduce spherical coordinates, which have nothing to do with this question? If you have a surface of revolution revolved about the $z$ axis, that would mean $z(r,\theta_1) = z(r,\theta_2)$ for any $\theta_1$ and $\theta_2$ wouldn't it? Is that true in your case? 5. Sep 17, 2016 ### toforfiltum Yes. So, is what I'm saying above right? But I'm not sure if having a unique set of points will form a line, though. It depends on $f$, right? 6. Sep 17, 2016 ### toforfiltum Oh, I think I see now why $z=f(r)$ represents a surface of revolution. Using the explanation on $zr$ planes given by @LCKurtz , each value of $r$ gives a value of $z$, and the set of values of $r$ gives its respective values of $z$. Since equation is independent of $\theta$, these set of points are the same for all values of $\theta$, and this is the reason why it forms a surface of revolution. Am I right? Draft saved Draft deleted Similar Discussions: Proving a form $z=f(r)$ to be a surface of revolution

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http://www.mathematicalfoodforthought.com/2005/11/

## Wednesday, November 30, 2005 ### Can you count? Topic: Probability. Level: AMC/AIME. Problem: (2005 AMC 12B - #25) Six ants are on the vertices of a regular octahedron - one on each vertex. Simultaneous and independently, they each traverse one edge and arrive at another vertex. What is the probability that no two ants arrive at the same vertex? Solution: We wish to count the number of ways the ants can move such that no two arrive at the same vertex. Consider viewing the octahedron as two points with a square between them such that the square is perpendicular to the line containing the two points. Note that two of the ants on the square must move off and the two off the square must move on. ----- CASE 1: Two ants opposite each other on the square move off. There are two possible pairs of ants and each pair has two ways of moving off (each one can go to either point off the square). That's $4$ possibilities. Now consider the remaining two ants on the square. There are two ways they can move such that they don't arrive at the same vertex. That's $2$ possibilities. Now consider the two ants off the square. There are two ways they can move onto the square (each can go to either open point). That's $2$ possibilities. So we have $4 \cdot 2 \cdot 2 = 16$ possibilities for the first case. ----- CASE 2: Two ants adjacent to each other on the square move off. There are four pairs of ants adjacent to each other on the square. Each pair can move off in two different ways (same argument as CASE 1). That's $8$ possibilities. The remaining two ants can move anywhere on the square and not be at the same vertex, giving $4$ possibilities. The two ants off the square can move onto the square in two different ways (same argument as CASE 1). That's $2$ possibilities. So we have $8 \cdot 4\cdot 2 = 64$ possibilites for the second case. ----- Now consider the total possible ways satisfying the condition: $16+64 = 80$. And the total number of ways the ants can move $4^6$ (each ant has $4$ places it can go). Thus our desired probability is $\frac{80}{4^6} = \frac{5}{256}$. -------------------- Comment: Admittedly, the problem does look daunting when encountered at the end of the AMC-12, in your final minutes of the test, and I did not try very hard to solve the problem at the time, either. However, once you find a relatively simple method to analyze the number of ways (viewing it as two points and a square - consider the possible pairs on the square), the real counting process is quite simple. And the problem seems to just finish itself. -------------------- Practice Problem #1: Repeat the problem above, but use a tetrahedron instead of an octahedron. Practice Problem #2: Consider a regular hexagon with all its diagonals drawn in. What is the probability that, given any two of the seven points (including the center - intersection of the diagonals), the line segment between the points is drawn? Practice Problem #3: Four points are randomly chosen in the plane. What is the probability that the quadrilateral they form is convex? ## Tuesday, November 29, 2005 ### Inequalities Revisited. Topic: Inequalities. Level: Olympiad. Theorem: (Rearrangement Inequality) Given two nondecreasing sequences of positive reals $x_1 \le x_2 \le \cdots \le x_n$ and $y_1 \le y_2 \le \cdots \le y_n$, $\displaystyle \sum_{i=1}^n x_iy_i \ge \sum_{i=1}^n x_iy_{\delta(i)}$, where $\delta$ is some permutation of $1,2, \ldots, n$. -------------------- Comment: The Rearragement Inequality is extremely useful, and can quickly kill some inequalities that may be a hassle to AM-GM over and over again, as you'll see in the next example. The proof of the theorem simply requires considering a possible permutation, subtracting it away from the maximum, and showing that the difference is greater than zero. -------------------- Problem: (1999 United Kingdom - #7) Given three non-negative reals $p,q,r$ such that $p+q+r = 1$, prove that $7(pq+qr+rp) \le 2+9pqr$. Solution: We notice that the inequality we wish to prove is not homogenous - that is, the sums of the powers on the terms are not equal. Most of the classical inequalities require you to homogenize, and given our condition, we can do so (by multiplying by $p+q+r$ when necessary). After homogenizing, the inequality becomes $7(pq+qr+rp)(p+q+r) \le 2(p+q+r)^3+9pqr$. Now this may look uglier than before, but we notice a lot of terms cancel, leaving (you may do the algebra yourself if you wish) $p^2q+p^2r+q^2r+q^2p+r^2p+r^2q \le 2(p^3+q^3+r^3)$. With enough experience, this will be an obvious direct result of Rearrangement, but to explicitly use it, we set (applied three times with $n = 2$ and WLOG assuming $p \le q \le r$): $x_1 = p^2 \le x_2 = q^2$ $y_1 = p \le y_2 = q$, from which we have $\displaystyle \sum_{i=1}^2 x_iy_i = p^3+q^3 \ge \sum_{i=1}^2 x_iy_{\delta(i)} = p^2q+q^2p$ where $\delta = 2,1$. As you can see applying this on the pairs $(p,r); (q,r)$ the same way and summing them up, we get the desired result $p^2q+p^2r+q^2r+q^2p+r^2p+r^2q \le 2(p^3+q^3+r^3)$. QED. -------------------- Comment: An "extension" of the Rearrangement Inequality, is the Muirhead Inequality, which effectively demolishes symmetric, homogenized inequalities. -------------------- Practice Problem #1: Prove, using Rearrangement, the Trivial Inequality: $(x-y)^2 \ge 0$ for positive reals $x,y$. Practice Problem #2: Prove, using Rearrangement, a special case of the Power-Mean Inequality: $\sqrt[n]{\frac{x^n+y^n+z^n}{3}} \ge \frac{x+y+z}{3}$ for a positive integer $n$ and positive reals $x,y,z$. Practice Problem #3: Given positive reals $a,b,c$, prove that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \ge \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a}$. ## Monday, November 28, 2005 ### Add Them Up. Topic: Sequences & Series. Level: AIME/Olympiad. Problem: (1989 USAMO - #1) For each positive integer $n$, let $S_n = 1+\frac{1}{2}+ \cdots + \frac{1}{n}$ $T_n = S_1+S_2+ \cdots + S_n$ $U_n = \frac{T_1}{2}+\frac{T_2}{3} +\cdots + \frac{T_n}{n+1}$ Find, with proof, integers $0 < a,b,c,d < 1000000$ such that $T_{1988} = aS_{1989}-b$ and $U_{1988} = cS_{1989}-d$. Solution: Well let's start with the first condition - $T_{1988} = aS_{1989}-b$. We claim that $T_n = (n+1)(S_{n+1}-1)$. Consider writing $T_n$ as sum of the following sequences: $S_1 = 1$ $S_2 = 1+\frac{1}{2}$ ... $S_n = 1+\frac{1}{2}+ \cdots +\frac{1}{n}$. Notice that $1$ appears $n$ times, $\frac{1}{2}$ appears $n-1$ times, and more generally $\frac{1}{i}$ appears $n+1-i$ times. Then we can say $\displaystyle T_n = \sum_{i=1}^n \frac{n+1-i}{i} = (n+1)\sum_{i=1}^n \frac{1}{i} - \sum_{i=1}^n 1 = (n+1)S_n-n = (n+1)\left(S_{n+1}-\frac{1}{n+1}\right)-n = (n+1)(S_{n+1}-1)$ as claimed. Also, save the fact that $T_n = (n+1)S_n - n$. So then $T_{1988} = 1989(S_{1989}-1) = 1989S_{1989}-1989 \Rightarrow a = b = 1989$. Now, using $T_n = (n+1)(S_{n+1}-1)$, we substitute into the $U_n$ equation to get $\displaystyle U_n = \sum_{i=1}^n \frac{T_n}{n+1} = \sum_{i=1}^n \frac{(n+1)(S_{n+1}-1)}{n+1} = \sum_{i=1}^n S_{n+1} - \sum_{i=1}^n 1 = \sum_{i=1}^n S_{n+1} - n$. But recall that $\displaystyle \sum_{i=1}^n S_{n+1} = T_{n+1}-1$ and remember the fact we saved back up there, so our final equation becomes $U_n = T_{n+1}-(n+1) = (n+2)S_{n+1}-(n+1)-(n+1) = (n+2)S_{n+1}-2(n+1)$. So $U_{1988} = 1990S_{1989}-3978 \Rightarrow c = 1990, d = 3978$. Thus our final answer is $a = 1989, b = 1989, c = 1990, d = 3978$. QED. -------------------- Comment: This was back in 1989, when the USAMO was a one-day, $3 \frac{1}{2}$ hour test with five problems. Not until 1996 did it become a two-day test. -------------------- Practice Problem #1: Do the problem again with $T_{2005}$ and $U_{2005}$ instead, to fully understand the solution. Practice Problem #2: Show that $T_n+\ln{(n+1)} > U_n+n$ for all positive integers $n$. ## Sunday, November 27, 2005 ### Circular Reasoning. Topic: Geometry. Level: AMC/AIME. Problem #1: (2006 Mock AIME 1 - #1) $2006$ points are evenly spaced on a circle. Given one point, find the number of other points that are less than one radius distance away from that point. Solution: We know that one radius distance is equivalent to $\frac{1}{6}$ of the circumference (by constructing an equilateral triangle with the center and two points on the circle). We know each point is $\frac{1}{2006}$ of the circumference away from each other, so the $k$th point is $\frac{k}{2006}$ arc distance away. We want $\frac{k}{2006} < \frac{1}{6} \Rightarrow k \le 334$. But since it can be on either side, we double that, to get $668.$. QED. -------------------- Comment: Note that this following AMC-12 problem is considerably more difficult than the preceding AIME problem, though it happens quite often that a AMC-12 #25 is more difficult than an AIME #1. -------------------- Problem #2: (2003 AMC 12B - #25) Three points are chosen randomly and independently on a circle. What is the probability that all 3 pairwise distances between the points are less than the radius of the circle? Solution : Now on the surface it looks pretty easy, but remember, this is a #25 - chances are it won't come too quickly for an average problem-solver (it's easy to jump into a bunch of flawed arguments). First convert everything to arc lengths - one radius becomes $\frac{1}{6}$ circumference again. Consider the following argument: The circumference is set to $1$. Let point $A$ be the "center" of the three points, the "center" being the point that both the other points are closest to (note that the existence of the "center" is guaranteed - think about it). Call the shortest distance between two points to be $x$ (first assume $AB = x$) - we then have the other length $1-x$. Now think about the possibilities... we want to maintain 1. The minimum distance of $x$. So the arc $AC$ has to be greater than $x$ - $AC > x$. 2. The "center" status of $A$. So the arc $BC$ has to be greater than $AC$ so $AC < \frac{1-x}{2}$ for a given $x$. Also, $AB$ must be less than $BC$ (which can be as small as $\frac{1-x}{2}$), so we have $x < \frac{1-x}{2} \Rightarrow x < \frac{1}{3}$. (1) We can graph this such that $AB$ is on the $x$-axis and $AC$ is on the $y$-axis following these rules: $0 < AB < \frac{1}{3}$ and $AB < AC < \frac{1-AB}{2}$. (2) Now recall that we assumed $AB = x$. But $AC$ could be $x$ as well, so we apply the same argument, flipping $AB$ and $AC$. But the graph would be the same, only flipped over the line $y=x$, creating this: where (1) creates the red region and (2) creates the blue region (both theoretically extending into the yellow region). Those generate the total "number" of unique possibilites, denoted by the area beneath the graph. Now we have to find the ones that satisfy our given condition: the maximum arc length is less than $\frac{1}{6}$. However, we notice by introducing the center status of $A$ we automatically show that our maximum arc length is $AB+AC$ because it is always greater than the independent lengths ($AB+AC > AB$ and $AB+AC > AC$ - of course this is only when $AB$ and $AC$ are both less than $\frac{1}{6}$, which is all we care about). So we add to our graph the yellow region, $AB+AC < \frac{1}{6}$. Thus the probability is (including the extensions of the red and blue regions into the yellow) $\frac{[yellow]}{[red]+[blue]} = \frac{\frac{1}{72}}{\frac{1}{6}} = \frac{1}{12}$. QED. -------------------- Comment: On the actual contest, the argument would probably be loosely made in the interest of saving time and you could probably finish something like this in 5-10 minutes at most if you knew where you were headed. Of course, given that this is the last problem of an AMC-12, I wouldn't be surprised if most people didn't have those 5-10 minutes to spare. -------------------- Practice Problem #1: Find the probability that two points placed on a circle randomly and independently are within one radius distance of each other. Practice Problem #2: In an acute angled triangle $ABC$, $\angle A = 30^{\circ}$. $H$ is the orthocenter and $M$ is the midpoint of $BC$. On the line $HM$, take point $T$ such that $HM=MT$. Show that $AT= 2BC$. (hint: Use complex numbers - see Post 13: November 24th, 2005). Practice Problem #3: Three congruent circles of radius 1 intersect at a common point. A larger circle is circumscribed about them, tangent to each of them at one point. Consider the midpoint of one of the diameters of the smaller circles; call it $P$. What is the length of the arc along the bigger circle containing all the points that are within 2 units of $P$? ## Saturday, November 26, 2005 ### Combina-WHAT? Topic: Combinatorics. Level: AMC/AIME. Problem #1: Prove that $nC0+nC1+\cdots+nCn = 2^n$ ($nCk$ denotes $n$ choose $k$). Solution: Chances are you've seen this before, and there are many proofs to it (one of my favorites being the number of subsets of an $n$-element set... think about it). But here's another cool proof, using our very own Binomial Theorem. Consider $(x+1)^n = (nC0)x^n+(nC1)x^{n-1}+\cdots+(nCn)$. But upon setting $x = 1$, we immediately have $2^n = nC0+nC1+\cdots+nCn$ as desired. QED. -------------------- Problem #2: Prove that $nC0-nC1+\cdots+(-1)^n(nCn) = 0$. Solution: Well if you understood the concept behind the first problem, this one should come very easily. $(x+1)^n = (nC0)x^n+(nC1)x^{n-1}+\cdots+(nCn)$ so for $x = -1$ we have $0 = (-1)^n(nC0)+(-1)^{n-1}(nC1)+\cdots+(nCn)$ from which we can just flip the sign (if necessary) to get the desired result. QED. -------------------- Practice Problem #1: Prove $nC0+nC1+\cdots+nCn = 2^n$ using the subsets argument. Practice Problem #2: Show that $nCr+nC(r+1) = (n+1)C(r+1)$. Practice Problem #3: (1983 AIME #8) Find the largest two-digit prime that divides $200C100$. ## Friday, November 25, 2005 ### Mix it Up. Topic: Polynomials/Inequalities. Level: Olympiad. Problem #1: (ACoPS 5.5.22) Let $P$ be a polynomial with positive coefficients. Prove that if $P\left(\frac{1}{x}\right) \ge \frac{1}{P(x)}$ holds for $x=1$ then it holds for every $x > 0$. Solution: Let $P(x) = a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0$. We have $P(1) \ge \frac{1}{P(1)} \Rightarrow [P(1)]^2 = (a_n+a_{n-1}+\cdots+a_0)^2 \ge 1$. Then we have $P(x) \cdot P\left(\frac{1}{x}\right) = (a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0)\left(a_n\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+ \cdots + a_0)$. But $(a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0)\left(a_n\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+ \cdots + a_0) \ge (a_n+a_{n-1}+\cdots+a_0)^2 \ge 1$ by Cauchy (see Post 12: November 24th, 2005). QED. -------------------- Problem #2: (USAMO 1983 #2) Prove that the zeros of $x^5+ax^4+bx^3+cx^2+dx+e = 0$ cannot all be real if $2a^2 < 5b$. Solution: Hmm... coefficients of polynomials... let's break out Vieta's Formulas ! So we have $a = -(r_1+r_2+r_3+r_4+r_5) = -\sum r_i$ (for shorthand). And $b = r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5 = \sum r_ir_j$ (again, for shorthand). So we have $2a^2<5b \Rightarrow 2\left(\sum r_i\right)^2 < 5\sum r_ir_j$. Expanding and simplifying, we have $2\left(\sum r_i^2\right) < \sum r_ir_j$ which can be rearranged to $\sum (r_i-r_j)^2 < 0$ which clearly cannot hold if all the roots are real. QED. -------------------- Practice Problem #1: Given the polynomial $x^3+ax^2+bx+c$ with real coefficients, find the condition the polynomial must satisfy such that it has at least one nonreal root (based on $a,b,c$). And generalize for $a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0$. Practice Problem #2: (ACoPS 5.5.36) Prove Cauchy-Schwarz using the polynomial $f(x) = (a_1x+b_1)^2+(a_2x+b_2)^2+\cdots+(a_nx+b_n)^2$ by observing that $f$ has real zeros iff $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n}$. ## Thursday, November 24, 2005 ### Simpler than it Sounds. Topic: Complex Numbers. Level: AMC/AIME. Problem #1: Find the coordinates of the point $(5,4)$ rotated around the point $(2,1)$ by $\frac{\pi}{4}$ radians counterclockwise. Solution: Consider these points in the complex plane ($x$-axis becomes real part, $y$-axis becomes imaginary part). We want to rotate $5+4i$ around $2+i$. Rewrite these in $(r, \theta)$ form, where $r$ is the magnitude and $\theta$ is the angle. We also have $(r, \theta) = e^{i\theta}$ by the Euler Formula. So, to simplify things, we shift $2+i$ to the origin, and now we want to rotate $3+3i = 3\sqrt{2}e^{i\frac{\pi}{4}}$. We notice that rotation by $\frac{\pi}{4}$ simply adds $\frac{\pi}{4}$ to the angle, resulting in $3\sqrt{2}e^{i\frac{\pi}{2}} = 3\sqrt{2}i$. Shifting back to the "real" origin, our rotated point becomes $2+(1+3\sqrt{2})i$ in the complex plane and $(2, 1+3\sqrt{2})$ in the Cartesian plane. QED. -------------------- Problem #2: Given that $A(x_1,y_1)$ and $B(x_2,y_2)$ are two vertices of an equilateral triangle, find all possible values for the third point, $C$. Solution: Convert them to complex numbers - $X = x_1+y_1i$ and $Y = x_2+y_2i$. Notice that if we rotate $X$ around $Y$ by $\frac{\pi}{3}$ or $-\frac{\pi}{3}$ radians, we get the only two possible points for $Z$ ($C$ as a complex number). So applying the same rotation method as in the first problem, we have $Z = Y + (X-Y)e^{i\frac{\pi}{3}}$ or $Z = Y+(X-Y)e^{-i\frac{\pi}{3}}$. We can translate this back to the Cartesian plane, but it's just a mess of algebra, not essential to understanding the method of rotation in the complex plane. QED. -------------------- Practice Problem #1: Find the value of $A = 3+i$ rotated by $\frac{\pi}{2}$ radians counterclockwise. Practice Problem #2: (WOOT Class) Show that, given three complex numbers $A, B, C$ that lie on the unit circle, the orthocenter of the triangle formed by them is $H = A+B+C$ (hint: If two complex numbers $X, Y$ are perpendicular, $\frac{X}{Y}$ is purely imaginary... prove it). Practice Problem #3: (WOOT Message Board) Let $O$ be the center of a circle $\omega$. Points $A,B,C,D,E,F$ on $\omega$ are chosen such that the triangles $OAB,OCD,OEF$ are equilateral. Let $L,M,N$ be the midpoints of $BC,DE,FA$, respectively. Prove that triangle $LMN$ is equilateral. ## Wednesday, November 23, 2005 ### To Equal or not to Equal. Topic: Inequalities. Level: AIME/Olympiad. Problem #1: Prove that, given two positive reals $a$ and $b$, we have $\frac{a+b}{2} \ge \sqrt{ab}$. Solution: Begin with the trivial inequality ($x^2 \ge 0$) applied on $a-b$. We have $(a-b)^2 \ge 0$ $a^2-2ab+b^2 \ge 0$ $a^2+2ab+b^2 \ge 4ab$ $(a+b)^2 \ge 4ab$ $a+b \ge 2\sqrt{ab}$ $\frac{a+b}{2} \ge \sqrt{ab}$ as desired. QED. -------------------- Comment: Note that we have just derived the Arithmetic Mean-Geometric Mean Inequality, more commonly referred to as just AM-GM. This can be generalized to $n$ variables by induction, that is, $\frac{a_1+a_2+\cdots+a_n}{n} \ge \sqrt[n]{a_1a_2 \cdots a_n}$. -------------------- Problem #2: Given two sequences of positive reals $\{a_i\}$ and $\{b_i\}$ for $i = 1,2, \ldots, n$, prove that $(a_1^2+a_2^2+\cdots+a_n^2)(b_1^2+b_2^2+\cdots+b_n^2) \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$. Solution: We shall prove this with vectors (read the previous blog post to get some background information). Consider the vector $A = (a_1,a_2,\ldots,a_n)$ and the vector $B = (b_1,b_2,\ldots,b_n)$ (in an $n$-space). We have $|A||B| \ge |A||B|\cos{\theta} = A \cdot B$ since $\cos{\theta} \le 1$. But recall that $A \cdot B = (a_1,a_2,\ldots,a_n) \cdot (b_1,b_2,\ldots,b_n) = (a_1b_1+a_2b_2+\cdots+a_nb_n)$. Also, $|A| = \sqrt{a_1^2+a_2^2+\cdots+a_n^2}$ and $|B| = \sqrt{b_1^2+b_2^2+\cdots+b_n^2}$. Combining them, we get $\sqrt{a_1^2+a_2^2+\cdots++a_n^2} \cdot \sqrt{b_1^2+b_2^2+\cdots+b_n^2} \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)$, from which our result follows directly upon squaring both sides: $(a_1^2+a_2^2+\cdots+a_n^2)(b_1^2+b_2^2+\cdots+b_n^2) \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$. QED. -------------------- Comment: This is known as Cauchy's Inequality, or the Cauchy-Schwarz Inequality. It is one of the most useful inequalities to know and can be generalized (see Holder's Inequality). -------------------- Problem #3: Given three positive reals $a$, $b$, and $c$, prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$. Solution: Let's put our newly-learned inequalities in action! By AM-GM, we have $a^2+b^2 \ge 2ab$. Applying this to the two other pairs and summing them up, we get $a^2+b^2+c^2 \ge ab+bc+ca$. Adding $2(ab+bc+ca)$ to both sides, we have $a^2+b^2+c^2+2(ab+bc+ca) = (a+b+c)^2 \ge 3(ab+bc+ca) \Rightarrow \frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac{3}{2}$. Rewrite the original LHS (left hand side) as $\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}$. By Cauchy, we have $(ab+bc+bc+ba+ca+cb)\left(\frac{a^2}{ab+bc}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\right) \ge (a+b+c)^2$ with $a_1^2 = ab+bc$, $b_1^2 = \frac{a^2}{ab+bc}$, and similarly define $a_2, a_3, b_2, b_3$. So then dividing by $2(ab+bc+ca)$ and combining it with the previous inequality, we find $\frac{a^2}{ab+bc}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb} \ge \frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac{3}{2}$ as desired. QED. -------------------- Comment: This is known as Nesbitt's Inequality, and many of the classical inequalities can be applied to prove it. -------------------- Practice Problem #1: Prove the Root-Mean-Square - Arithmetic Mean Inequality - $\sqrt{\frac{a_1^2+a_2^2+\cdots+a_n^2}{n}} \ge \frac{a_1+a_2+\cdots+a_n}{n}$. Practice Problem #2: (IMO 1995 #A2) Given positive reals $a, b, c$ such that $abc = 1$, prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \ge \frac{3}{2}$. ## Tuesday, November 22, 2005 ### Vectors... wow! Topic: Vector Geometry. Level: AMC/AIME. Problem #1: Find the area of the triangle formed by the heads of two vectors, $A = (5, 67^{\circ})$ and $B = (4, 37^{\circ})$, and the origin. Solution: So, we have two vectors with pretty ugly angles, but we do notice that they differ by exactly $30^{\circ}$! Hmm, then we know two sides and the angle between them... And consequently we remember the formula $[ABC] = \frac{1}{2}ab\sin{\angle C}$, which makes us happy. Hence the area of the triangle is $\frac{1}{2}(5)(4)(\sin{30^{\circ}}) = 5$. QED. -------------------- Comment: Notice that the formula $\frac{|A||B|\sin{\theta}}{2} = \frac{|A \times B|}{2}$, where $A \times B$ is the cross product of $A$ and $B$. -------------------- Problem #2: Show that the vectors $A = (3, 4, 5)$ and $B = (2, -4, 2)$ are perpendicular. Solution: Consider the dot product of $A$ and $B$, $A \cdot B$. The general definition of this is $A \cdot B = |A||B|\cos{\theta}$, where $\theta$ is the angle between the two vectors. Notice that if two vectors are perpendicular to each other, the angle $\theta = 90^{\circ} \Rightarrow \cos{\theta} = 0 \Rightarrow A \cdot B = 0$. We can easily show that the dot product is distributive (left as an exercise to the reader), that is, $A \cdot (B+C) = A \cdot B + A \cdot C$. Thus $A \cdot B = [(3, 0, 0)+(0, 4, 0)+(0, 0, 5)] \cdot [(2, 0, 0)+(0, -4, 0)+(0, 0, 2)]$. Distributing this, we see that only the terms that are parallel remain because all the axes are perpendicular to each other ($\theta = 90^{\circ}$ between any two axes). So $A \cdot B = (3, 0, 0) \cdot (2, 0, 0)+(0, 4, 0) \cdot (0, -4, 0)+(0, 0, 5) \cdot (0, 0, 2) = 6 +(-16)+10 = 0$. So $A$ is perpendicular to $B$. QED. -------------------- Comment: In effect, we have just shown that given any two vectors $X = (x_1, x_2, \ldots, x_n)$ and $Y = (y_1, y_2, \ldots, y_n)$, we have $X \cdot Y = x_1y_1+x_2y_2+\cdots+x_ny_n$, an interesting and useful result (these are all in $n$-spaces, where there actually exist $n$ dimensions). -------------------- Practice Problem #1: Show that the dot product is distributive: $A \cdot (B+C) = A \cdot B+ A \cdot C$. Practice Problem #2: Find the area of the triangle formed by the points $(5,2)$, $(7,8)$, and $(2,1)$. Practice Problem #3: Show that, given three vectors of equal magnitude $A$, $B$, and $C$, the orthocenter of the triangle formed by the heads of each of the vectors is $A+B+C$. Practice Problem #4: Using Practice Problem #3, show that the centroid ($G$), circumcenter ($O$), and orthocenter ($H$) of a triangle are collinear and that $OG:GH = 1:2$. ## Monday, November 21, 2005 ### Polynomial Power. Topic: Algebra/Polynomials. Level: AIME. Problem: (2006 Mock AIME 1 - #14) Let $P(x)$ be a monic polynomial of degree $n \ge 1$ such that $[P(x)]^3 = [P(x)]^2-P(x)+6$ for $x = 1, 2, \ldots , n$. Let $n_0$ be the smallest $n$ such that $P(0) > (P(1)+P(2)+ \cdots +P(n))^3$. Find the remainder when $P(0)$ is divided by $1000$ given $n = n_0$. Solution: Well the condition we're given as it is doesn't look particularly helpful in defining the polynomial, especially with the cube. So we try and factor it (almost always a safe bet) and we find $(P(x)-2)([P(x)]^2+P(x)+3) = 0$. But we notice that $[P(x)]^2+P(x)+3 = \left(P(x)+\frac{1}{2}\right)^2 + \frac{11}{4} > 0$. So then we must have $P(x)-2 = 0$ for $x = 1, 2, \ldots, n$. Consider the polynomial $H(x) = P(x)-2$. It has zeros at $x = 1, 2, \ldots, n$ and is also monic with degree $n$. Hence $H(x) = (x-1)(x-2)\cdots(x-n) \Rightarrow P(x) = (x-1)(x-2)\cdots(x-n)+2$. Then we have $P(0) = (-1)^n\cdot n!+2 > (P(1)+P(2)+\cdots+P(n))^3 = (2n)^3 = 8n^3$. Checking, we find the first $n$ such that this is true is $n = 8$. Therefore, $P(0) = (-1)^8 \cdot 8!+2 = 40322$. So our answer is $322$. QED. -------------------- Pracitice Problem #1: Factor $x^4+64$. Practice Problem #2: Let $P(x)$ be a monic polynomial of degree $n$ such that $P(x) = x$ for $x = 1,2,\ldots,n$. Find a closed expression for $P(n+1)$. ## Sunday, November 20, 2005 ### Go Go Geometry! Topic: Geometry/Trigonometry. Level: AIME. Problem: (2006 Mock AIME 1 - #15) Let $ABCD$ be a rectangle and $AB = 24$. Let $E$ be a point on $BC$ (between $B$ and $C$) such that $DE = 25$ and $\tan{\angle BDE} = 3$. Let $F$ be the foot of the perpendicular from $A$ to $BD$. Extend $AF$ to intersect $DC$ at $G$. Extend $DE$ to intersect $AG$ at $H$. Let $I$ be the foot of the perpendicular from $H$ to $DG$. The length of $IG$ is $\displaystyle \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Suppose $\alpha$ is the sum of the distinct prime factors of $m$ and $\beta$ is the sum of the distinct prime factors of $n$. Find $\alpha + \beta$. Solution: We begin by solving for $EC = \sqrt{25^2-24^2} = 7$. Then $\tan{EDC} = \frac{7}{24}$. By the tangent addition formula, we have $\tan{BDC} = \tan{(BDE+EDC)} = \frac{\tan{BDE}+\tan{EDC}}{1-\tan{BDE}\tan{EDC}} = \frac{3+\frac{7}{24}}{1-3\left(\frac{7}{24}\right)} = \frac{79}{3}$. Since $\tan{BDC} = \frac{BC}{CD}$, we have $\displaystyle BC = (CD)\tan{BDC} = 24\left(\frac{79}{3}\right) = 632$. Notice that $\angle AGD = 90 - \angle BDC \Rightarrow \tan{AGD} = \frac{1}{\tan{BDC}} = \frac{3}{79}$. Then $\tan{AGD} = \frac{AD}{DG} \Rightarrow DG = \frac{AD}{\tan{AGD}} = \frac{632}{\frac{3}{79}} = \frac{2^3 \cdot 79}{3}$. Consider $\triangle HIG$. We have $\tan{HGI} = \frac{HI}{IG} \Rightarrow HI = (IG)\tan{HGI} = \frac{3}{79}(IG)$. Notice that $\triangle EDC$ is similar to $\triangle HDI$, so $\frac{HI}{ID} = \frac{EC}{CD} = \frac{7}{24} \Rightarrow HI = \frac{7}{24}(ID) = \frac{7}{24}(DG-IG)$. Combining our two expressions for $HI$, we get $HI = \frac{3}{79}(IG) = \frac{7}{24}(DG-IG)$. Solving this for $IG$, we get $IG = \frac{\frac{7}{24}(DG)}{\frac{3}{79}+\frac{7}{24}$. But we calculated $DG = \frac{2^3 \cdot 79}{3}$ above, so upon substitution we get $IG = \frac{\frac{7}{24}\left(\frac{2^3 \cdot 79}{3}\right)}{\frac{3 \cdot 24 + 7 \cdot 79}{24 \cdot 79}} = \frac{2^3 \cdot 7 \cdot 79^3}{3 \cdot 5^4}$. Summing the distinct prime factors, we have $2+7+79+3+5 = 96$. ## Thursday, November 17, 2005 ### 2006 Mock AIME 1. Here is the first AoPS Mock AIME of the 2005-2006 year, written by yours truly. If you wish to participate officially, please create an AoPS account at the AoPS forum. 2006 Mock AIME 1 Date: Friday, Nov. 18th to Sunday, Nov. 20th, 2005 Time: 3 hours, self-timed Format: 15 questions, Free Response (ALL answers are integers from 000 to 999, inclusive) Scoring: 1 point per correct answer (No guessing penalty) Answer Format: (As follows) 1. Answer to #1 2. Answer to #2 ... 15. Answer to #15 2006 Mock AIME 1 Questions P.S. I'll be gone the next two days, so don't expect blog posts. EDIT: Ack, #14 should be taken mod 1000, to get an AIME answer, my mistake to anyone who has taken it already. I'll modify the actual PDF file when I get home. EDIT 2: All better. EDIT 3: #15 was also written incorrectly (hey, it's not that easy to write hard AIME problems); fixed now. ## Wednesday, November 16, 2005 ### Modular Math. Topic: Number Theory. Level: AMC/AIME. Problem: What is the remainder when you divide $6^{2005}+8^{2005}$ by $49$? Solution Modular arithmetic is a useful tool for this problem. The concept is really quite simple: We state that $a \equiv b \pmod{c} \Rightarrow c|(a-b)$, that is, $a$ and $b$ leave the same remainder upon division by $c$. We rewrite the problem as evaluating $6^{2005}+8^{2005} \pmod{49}$. The key to this problem is noticing that $6 = 7-1$ and $8 = 7+1$. Rewrite them as such: $(7-1)^{2005}+(7+1)^{2005}$. Applying our handy Binomial Theorem, we see that $\displaystyle (7-1)^{2005} = (2005C0)(7^{2005}) - (2005C1)(7^{2004}) + \ldots +(2005C2004)(7)-(2005C2005)$. Consider this $\pmod{49}$ or $\pmod{7^2}$. We note that any term with a power of $7$ greater than or equal to $2$ is $0 \pmod{49}$ (make sure you get this, review the definition of mod if you don't). Thus the only terms that we need to consider are the last two: $\displaystyle (2005C2004)(7)-(2005C2005) = 2005(7)-1$. Similarly, the only two terms of $(7+1)^{2005}$ we need to consider are $\displaystyle (2005C2004)(7)+(2005C2005) = 2005(7)+1$. Summing the two, we have $(2005(7)-1)+(2005(7)-1) = 4010(7) \equiv 42 \pmod{49}$ (you can work out the details yourself). Hence our answer is $42$. QED. -------------------- Comment: Modular arithmetic has extremely wide applications in Number Theory, including a number of important theorems you should be familiar with (e.g. Fermat's Little Theorem). Learning this concept will increase your ability to solve NT problems dramatically. If you just love Number Theory and want to learn more about it, check out the PROMYS website; it's a summer camp devoted entirely to NT. -------------------- Practice Problem: Find the remainder when $1+10+10^2+\ldots+10^{2005}$ is divided by $9$. ## Tuesday, November 15, 2005 ### Power of a Point. Topic: Geometry. Level: AMC/AIME. Problem: Given a circle that passes through the points $(3,4)$ and $(6,8)$, find the length of a tangent to the circle from the origin. Solution: Consider Power of a Point, which states that $a^2 = b(b+c) = d(d+e)$, all of which are equal to the power of the point at which they intersect. Now consider the tangent from the origin in the problem (call this length $x$) and notice that the secant from the origin through $(3,4)$ passes through $(6,8)$ as well. Then by Power of a Point applied to the origin, we have $x^2 = (\sqrt{3^2+4^2})(\sqrt{6^2+8^2}) = (5)(10) = 50 \Rightarrow x = \sqrt{50}$. QED. Note: On the actual contest (ARML) a third point was provided to define the circle. We notice, however, that this third point is not necessary and was thus omitted in this rewording of the problem. -------------------- Practice Problem: Given two circles, find all points $P$ such that the power of point $P$ with respect to both circles is equal. ## Monday, November 14, 2005 ### Diophantine Fun. Topic: Number Theory. Level: AIME. Problem: Find all integer solutions $(m,n)$ to the equation $n^4+n^3+2n^2+2n+1 = m^2$. Solution: Let me introduce a technique used to solve diophantine equations with a power (square in this case) on one side and a multiple of that power (fourth power) on the other: Bounding. -------------------- Example - Suppose you had the equation $n^2+3n+3 = m^2$ (assume positive integers for the example; it is not necessary for the actual problem though). We can say $(n+1)^2 < n^2+3n+3 < (n+2)^2$ which can be verified by expanding. The LHS becomes $0 < n+2$ and the RHS becomes $0 < n+1$. But there are no integer squares between $(n+1)^2$ and $(n+2)^2$ so we can conclude that there are no solutions. -------------------- Now back to the problem... we look for bounding squares; however, this requires some experimenting and matching coefficients. We want to match at least the first two terms - $n^4+n^3$. After playing around with squares, we find a pretty good bound that works for $n \ge 4$ and $n \le 0$: $\displaystyle \left(n^2+\frac{n}{2}+\frac{1}{2}\right)^2 \le n^4+n^3+2n^2+2n+1 \le \left(n^2+\frac{n}{2}+1\right)^2$. The LHS becomes $\displaystyle 0 \le \frac{3}{4}(n+1)^2$ and the RHS $\displaystyle 0 \le n\left(\frac{n}{4}-1\right)$ (hence the restrictions on $n$). Since there are no squares between $\displaystyle \left(n^2+\frac{n}{2}+\frac{1}{2}\right)^2$ and $\displaystyle \left(n^2+\frac{n}{2}+1\right)^2$, the only possible solutions are the equality conditions and $n=1,2,3$. Checking those three, we find nothing, so we move to the equality conditions. As noted above, the equality conditions are $\displaystyle 0 = \frac{3}{4}(n+1)^2$ and $\displaystyle 0 = n\left(\frac{n}{4}-1\right)$, giving the solutoins $n=-1, 0, 4$. Then $m = \pm 1, \pm 1, \pm 19$, respectively. Hence our solutions are: $(m,n) = (-1, \pm 1); (0, \pm 1); (4, \pm 19)$. QED. -------------------- Practice Problem: Find all nonnegative integer solutions $(a,b)$ to the equation $a^6+2a^4+2a^2+2a+1 = b^2$. ### Hello! Welcome to my blog! This will be an ongoing journal of mathematical problems and problem solving techniques, targetted at high school students like myself aiming to do well on the American Math Competitions. The difficulty will range from mostly difficult AMC problems to USAMO problems. Enjoy!

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https://math.stackexchange.com/questions/2761339/prooving-not-all-roots-are-real

# Prooving not all roots are real Let $P(x)=x^n+a_{n-1}x^{n-1}\dots+a_0$ Given $a_{n-1}^2<a_{n-2},$ prove not all roots can be real My attempt: Assume all roots are real, namely $R_1,R_2,\dots,R_n$ Then $-a_{n-1}=\sum_{i=1}^n R_i$ $a_{n-2}=\sum R_i\times R_j$ Now $$(R_1+R_2+\dots+R_n)^2<(R_1R_2+R_1R_3\dots R_{n-1}R_{n})\\ \implies R_1^2+R_2^2\dots R_n^2<-(R_1R_2+R_1R_3\dots R_{n-1}R_{n})$$ But we can't tell if RHS is actually negative or positive. Also this kinda looks like cauchy-schwarz. Can it be used here? Hint: From $\displaystyle (\sum_i R_i)^2 < \sum_{i< j} R_i R_j$, we may add $\frac12 \sum R_i^2$ to both sides to get $$\left(\sum_i R_i \right)^2 + \frac12\sum_i R_i^2 < \frac12 \left(\sum_i R_i \right)^2$$ $$\implies \frac12\left(\sum_i R_i \right)^2 + \frac12\sum_i R_i^2 < 0$$

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http://www.ck12.org/book/CK-12-Math-Analysis/r19/section/2.7/

<meta http-equiv="refresh" content="1; url=/nojavascript/"> You are reading an older version of this FlexBook® textbook: CK-12 Math Analysis Go to the latest version. # 2.7: Approximating Real Zeros of Polynomial Functions Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Understand the statement of the Intermediate Value Theorem • Apply the intermediate value theorem to find bounds on the zeros of a function • Use numerical methods to find roots of a polynomial ## Intermediate Value Theorem & Bounds on Zeros The intermediate value theorem offers one way to find roots of a continuous function. Recall that our informal definition of continuous is that a function is continuous over a certain interval if it has no breaks, jumps asymptotes, or holes in that interval. Polynomial functions are continuous for all real numbers $x$. Rational functions are often not continuous over the set of real numbers because of asymptotes or holes in the graph. But for intervals without holes, rational functions are continuous. If we know a function is continuous over some interval $[a,b]$, then we can use the intermediate value theorem: Intermediate Value Theorem If $f(x)$ is continuous on some interval $[a,b]$ and $n$ is between $f(a)$ and $f(b)$, then there is some $c\in[a,b]$ such that $f(c)=n$. The following graphs highlight how the intermediate value theorem works. Consider the graph of the function $f(x)=\frac{1}{4} \left ( x^{3}-\frac{5x^{2}}{2}-9x \right )$ below on the interval [-3, -1]. $f(-3)=-5.625$ and $f(-1)=1.375$. If we draw bounds on [-3, -1] and $[f(-3),f(-1)]$, then we see that for any $y-$value between $y=-5.625$ and $y=1.375$, there must be an $x$ value in [-3, -1] such that $f(x)=y$. So, for example, if we choose $c=-2$, we know that for some $x\in[-3,-1], f(x)=-2,$ even though solving this by hand would be a chore! ## The Bounds on Zeros Theorem The Bounds on Zeros Theorem is a corollary to the Intermediate Value Theorem: Bounds on Zeros Theorem If $f$ is continuous on $[a,b]$ and there is a sign change between $f(a)$ and $f(b)$ (that is, $f(a)$ is positive and $f(b)$ is negative, or vice versa), then there is a $c\in(a,b)$ such that $f(c)=0$. The bounds on zeros theorem is a corollary to the intermediate value theorem because it is not fundamentally different from the general statement of the IVT, just a special case where $n=0$. Looking back at $f(x)=\frac{1}{4} \left (x^{3}-\frac{5x^{2}}{2}-9x \right )$ above, because $f(-3)<0$ and $f(-1)>0$, we know that for some $x\in[-3,-1], f(x)$ has a root. In fact, that root is at $x=2$. and we can test that using synthetic division or by evaluating $f(2)$ directly. Example 1 Show that $f(x)=-3x^{3}+5x$ has at least one root in the interval [1, 2] Solution Since $f(x)$ is a polynomial we know that it is continuous. $f(1)=2$ and $f(2)=-14$. Let $n=0\in[-14,2]$. Applying the Intermediate Value Theorem, there must exist some point $c\in[1,2]$ such that $f(c)=0$. This proves that $f(x)$ has a root in [1, 2]. Example 2 The table below shows several sample values of a polynomial $p(x)$. $& x && -4 && -2 && \qquad 0 && \qquad 1 && \quad 4 && 6 && \qquad 8 && \qquad 10 && \qquad 15 && 18\\& p(x) && 44.15 && 6.62 && -4.12 && -4.09 && 1.16 && 0 && -8.74 && -24.07 && -49.89 && 3.41$ Based on the information in the table (a) What is the minimum number of roots of $p(x)$? (b) What are bounds on the roots of $p(x)$ that you identified in (a)? Solution Since $p(x)$ is a polynomial we already know that it is continuous. We can use the Intermediate Value Theorem to identify roots by looking at when $p(x)$ changes from negative to positive, or from positive to negative. (a) There are four sign changes of $p(x)$ in the table, so at minimum, $p(x)$ has four roots. (b) The roots are in the following intervals $x\in[-2,0], x\in[1,4], x\in[15,18],$ and the table also tells us that one root is at $x=6$. ## Approximate Zeros of Polynomials Functions In calculus you will learn several methods for numerically approximating the roots of functions. In this section we show one elementary numerical method for finding the zeros of a polynomial which takes advantage of the Intermediate Value Theorem. Given a continuous function $g(x)$, 1. Find two points such that $g(a)>0$ and $g(b)<0$. Once you have found these two points, you can iteratively use the steps below to find the root of $g(x)$ on the interval $[a,b]$. (Note, we will assume $a, the same algorithm works with minor adjustments if $b>a$) 2. Evaluate $g \left ( \frac{a+b}{2} \right )$. 1. If $g \left ( \frac{a+b}{2} \right )=0$, then the root is $x=\frac{a+b}{2}$. 2. If $g \left ( \frac{a+b}{2} \right )>0$, replace $a$ with $\frac{a+b}{2}$. and repeat steps 1-2 using $\left [ \frac{a+b}{2},b \right ]$ 3. If $g\left ( \frac{a+b}{2} \right )<0$, replace $b$ with $\frac{a+b}{2}$. and repeat steps 1-2 using $\left [ a,\frac{a+b}{2} \right ]$ This algorithm will not usually find the exact root of $g(x)$, but it will allow you to find a reasonably small interval for the root. For example, you could repeat this process enough times so that you find an interval with $|a-b|<0.01$, and you will know the root of $g(x)$ within a reasonably good approximation. The quality of the approximation you use (and the number of steps you use) will depend on why you are looking for the root. For most applications coming within 0.01 of the root is a reasonable approximation, but for some applications (such as building a bridge or launching a rocket) you need much more accuracy. Example 3 Show the first 5 iterations of finding the root of $h(x)=x^{2}-x-1$ using the starting values $a=0$ and $b=2$. Solution: 1. First we verify that there is a root between $x=0$ and $x=2$. $h(0)=-1$ and $h(2)=1$ so we know there is a root in the interval [0, 2]. Check $h\left ( \frac{2+0}{2} \right )=h(1)=-1$. Since $-1<0$ we know the root is between $x=1$ and $x=2$, and we use the new interval [1, 2]. 2. Now we use the interval [1, 2]. $h\left ( \frac{1+2}{2} \right )=h(1.5)=-0.25$. Since $-0.25<0$, we use the interval [1.5, 2]. 3. $h\left ( \frac{1.5+2}{2} \right )=h(1.75)=0.31$. Since $0.31>0$, we know that the zero is in the interval [1.5, 1.75]. 4. $h\left ( \frac{1.5+1.75}{2} \right )=h(1.625)=0.02$. Since $0.02>0$, we know the root is between 1.5 and 1.625. 5. $h\left ( \frac{1.5+1.625}{2} \right )=h(1.5620)=-0.12$. Since $-0.12<0$, we know the root is between 1.5620 and 1.625. This example shows that after five iterations we have narrowed the possible location of the root to within 0.06 units. Not bad! Recall that we have already reviewed using the CALC menu on a graphing calculator to find the roots of a function. This algorithm is not the one used by a calculator, but the calculator uses a similar, more efficient, algorithm for approximating the root of a function to 13 decimal places. When the calculator prompts for a GUESS? it is asking for a starting value to run the iterations. ## Optional: An Interesting Corollary of the IVT One surprising result of the Intermediate Value Theorem is that if you draw any great circle around the globe, then there must two antipodal points on that great circle that have exactly the same temperature. Recall that a great circle is a path around a sphere that gives the shortest distance between any two points on the sphere. The equator is a great circle around the globe. Antipodal points are two points on opposite sides of the sphere. In the diagram below, $B$ and $B'$ are antipodal. For an informal proof of this result, look at the the image of a sphere with three great circles above. Suppose that the temperature at $B$ is $75^{\circ}$ and the temperature $B'$ is $50^{\circ}$. The difference between the temperature at $B$ and at $B'$ is $75-50=25$. Now imagine rotating the segment $\overline{BB'}$ around the blue great circle. When the segment has rotated 180 degrees (i.e. when $B$ has rotated to where $B'$ is), then The difference between the temperatures at these two points is $50-75=-25$. Since temperatures vary continuously, by the intermediate value theorem, there must be some point on that circle when the difference was 0, implying two antipodal points had the same temperature. Notice that this little demonstration does not tell us which two antipodal points had the same temperature, only that there must be two such points on any great circle. ## Exercises For Exercises 1-5 use the intermediate value theorem to show the bounds on the zeros of each function. Your bounds should be within the whole number 1. $f(x)=2x^{3}-3x+4$ 2. $g(x)=-5x^{2}+8x+12$ 3. $h(x)=\frac{1}{2}x^{4}-x^{3}-3x^{2}+1$ 4. $j(x)=-\frac{2}{x^{2}+1}+\frac{1}{2}$ 5. $k(x)$ is a polynomial and selected values of $k(x)$ are given in the following table: $& x && -3 && -2 && -1 && \ \ \ 0 && \ 1 && \ \ \ 2 && \ \ \ \ 3\\& k(x) && -23.5 && -1 && 0.5 && -1 && .5 && -1 && -23.5$ 6. Stephen argues the function $r(x)=\frac{4x+1}{x+3.5}$ has two zeros based on the following table and an application of the Bounds on Zeros Theorem. What is faulty about Stephen's reasoning? $& x && -5 && -4 && -3 && \ -2 && \ -1 && \ 0 && \ 1 && \ 2 && \ 3 && \ 4\\& r(x) && 12.67 && \ 30 && -22 && -4.67 && -1.20 && 0.29 && 1.11 && 1.64 && 2.0 && 2.27$ 7. Apply the numerical algorithm five times to find a bound on the zeros of the following functions given the indicated starting values. What is your final estimate for the zero? 1. $k(x)=x^{4}-3x+1$ on [0, 1] 2. $b(x)=-0.1x^{5}+3x^{3}-5x^{2}$ on [1, 3] 3. $c(x)=\frac{3x^{2}-2}{x^{4}+2}$ on [0, 2] 1. There is a zero in [-2, -1] 2. There is a zero in [-1, 0] and [2, 3] 3. There is a zero in [-2, -1], [-1, 0], [0, 1], and [3, 4] 4. There is a zero in [-2, -1] and [1, 2] 5. There is a zero in [-2, -1], [-1, 0], [0, 1] and [1, 2] 6. Unlike the previous question which specified the function was a polynomial (and hence continuous), $r(x)=\frac{4x+1}{x+3.5}$ has a vertical asymptote at $x=-3.5$, so it is not continuous in the interval [-4, -3]. Therefore we cannot use the Bounds on Zeros theorem to claim there is a zero in that interval. 1. The zero is in [0.3125, 0.34825] 2. The zero is in [1.5000, 1.5625] 3. [0.8125, 0.875] Feb 23, 2012 Dec 29, 2014

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https://www.physicsforums.com/threads/how-large-is-the-collision-cross-section-compared-to-nuclear-fusion-cross-section.906837/

# How large is the collision cross section compared to nuclear fusion cross section • I • Start date • Tags • #1 68 2 ## Main Question or Discussion Point Most Fusion reactors, and the leading ones like JET, use high temp. plasma and confine it. So, the plasma would approximate the Maxwell- Boltzmann distribution. This means that only a small portion of the plasma has enough energy to fuse. But, collisions are much more often, right? Since not all collisions result in fusion, the cross section of collision is much higher than the cross section of fusion. Is this true? • But how much smaller is the cross section of fusion compared to the cross section of collisions. And do the electrons possess higher energy than the ions, in which case thee nuclear cross section would be orders of magnitude lower. Am I right? If approximately all the collisions resulted in fusion, how much more energy would we be able to produce? And what if there were 2 beam travelling toward each other. But these not just follow straight paths. There exists a HELICAL magnetic field and the particle spiral around the field lines and approach each other. Here, I found out that the particle spiral direction would be in opposite directions( one clock wise and the other anticlockwise.) If the beams interact when they meet, how would the collision cross section be affected? (Note: The particle possess reasonable energies at least higher than 25 Kev) How would I calculate the collision cross section? I seriously need help. Thanks Related High Energy, Nuclear, Particle Physics News on Phys.org • #2 mfb Mentor 34,476 10,598 This means that only a small portion of the plasma has enough energy to fuse. But, collisions are much more often, right? Since not all collisions result in fusion, the cross section of collision is much higher than the cross section of fusion. Is this true? Right. DT fusion has a maximal cross section of 5*10-28 m2 at about 60 keV center-of-mass energy. Something like a collision happens if the nuclei come closer than ~200 fm, which corresponds to a cross section of 4*10-26 m2. At a temperature of 10 keV, only a small fraction of nuclei collisions will lead to 60 keV center-of-mass energy, which reduces the effective fusion cross section even more. Am I right? If approximately all the collisions resulted in fusion, how much more energy would we be able to produce? The plasma would be too short-living then, and we would have to run it at a lower temperature. If the beams interact when they meet, how would the collision cross section be affected? The magnetic field is only relevant for beam steering, the actual collision is completely independent of that. How would I calculate the collision cross section? See the papers doing that. Likes mheslep • #3 68 2 For the 4*10^-26 m squared cross section, what is the cross section for? and how did we end up with it? and an other thing I was wondering about was that I said we had a helical magnetic field. I've figured out that the particles spin in opposite directions during collision. Doesn't this mean that almost all of these are head on collisions? I think this would increase the fusion rate, wouldn't it? And these particle beams are not in thermal equilibrium and hence all have approximately the same energy, so all collisions would result in fusion right? So.. according to what you just said the cross section wouldn't change. So, am I safe in assuming these assumptions are accurate? • #4 68 2 And approximately how many more collisions do we have compared to fusion? • #5 mfb Mentor 34,476 10,598 For the 4*10^-26 m squared cross section, what is the cross section for? For any collision. Those elastic collisions don't have a well-defined total cross section - larger impact parameters will just lead to smaller deflections. See Rutherford scattering. I used a distance where the deflection will be small (few degrees). Doesn't this mean that almost all of these are head on collisions? If you have two well-defined beams, you don't even need magnetic fields for that. The fusion rate will be low due to the low achievable particle density in the beams. Even at the ideal energy for fusion, most collisions won't lead to fusion, see the comparison of the numbers in post 2. Most particles would be scattered out of the beam without fusing. • #6 68 2 What if low density beams bombard a high density plasma? • #7 mfb Mentor 34,476 10,598 Then you still lose most of the particles due to elastic scattering. You can get fusion with accelerator systems, and accelerator-based fusion is a convenient source of neutrons. You just do not get enough to use it as power source because you have to put in much more energy than you get out. • #8 68 2 But then, why doesn't it happen in plasmas. I know it happens, but I think the effect is much smaller. Hence, they don't loose energy through coulomb scattering. Is this right? • #9 mfb Mentor 34,476 10,598 But then, why doesn't it happen in plasmas. I know it happens, but I think the effect is much smaller. Which effect is smaller? The energy is not lost in either case. In a plasma, collisions between atoms just redistribute the energy within the plasma particles - a completely normal thermal process. In beams of cold atoms, it means some particles get scattered out of the beam and the beam gets hotter. • #10 68 2 • #11 68 2 Recently, I have learn't about the Gamows factor, and I have plotted it. Then I realized that the probability of Deuterium and tritium fusing at a temperature of 150 million kelvin when they collide, was about 0.00048. But, I know that the nuclear cross section for fusion is extremely small (about 1-1000 barns). But why is this so? I f the probability of fusion is 0.00048 when they collide, then why is the nuclear cross section so small? Is it because of the collisions? Are the collisions really that rare? Is the collision cross section really that small? Why is the nuclear cross section so small? • Here the x-axis is the temperature and the y-axis is the probability of fusion when a collision occurs. • #12 mfb Mentor 34,476 10,598 Why do you expect a large cross section? To actually hit the nuclei in a classical sense you would need about 2 MeV - but then you have to hit exactly, which is incredibly unlikely as nuclei are tiny (smaller than 1 barn). Everything below that has the nuclei approach each other, and then you have to hope that they tunnel to fuse before they fly apart again. That process is not very frequent either. • #13 68 2 Does the ion-ion collision rate or collision frequency increase or decrease with an increase in temperature? I've read that most plasmas could be considered collisionless or have little collisions taking place. • #14 mfb Mentor 34,476 10,598 For a constant density, the collision rate increases with temperature, simply because the particles move faster. • #15 68 2 Now, for a plasma put under a constant pressure, the molar density multiplied by the temperature is always a constant which equals Pressure/Gas constant. According to the ideal has equation. So the collision rate is a constant for a constant pressure right? • #16 mfb Mentor 34,476 10,598 At constant pressure: double the temperature and you double the path length between collisions, but the speed only grows with a factor sqrt(2) - collisions get less frequent. The chance of fusion increases more up to some point, so increasing temperature is still advisable. • #17 68 2 So thats why the cross section drops after 150 million kelvin for D-T fusion? The collisions become less frequent? • #18 68 2 The velocity you considered is the average velocity isn't it? • #19 mfb Mentor 34,476 10,598 So thats why the cross section drops after 150 million kelvin for D-T fusion? The collisions become less frequent? The cross section itself decreases as well if the energy gets too high, this is a more important effect. The velocity you considered is the average velocity isn't it? It is true for all speeds. Average, median, top 0.01%, ... • #20 68 2 But why does the temperature become too high? Isn't it going to help us overcome the coulomb barrier? The reason I think it can get too high is through losses. Is there a quantum mechanical effect that affects the cross section at extremely high temperatures? • #21 mfb Mentor 34,476 10,598 The reason I think it can get too high is through losses. The cross section has nothing to do with losses. It has nothing to do with temperature either. For the cross section, you just have two nuclei colliding with a given center of mass energy. Faster nuclei lead to a shorter collision time, that could contribute. • #22 68 2 Oh. So time matters? • #23 68 2 An other question: If I accelerates particles, and make them hit a plasma target, the portion of the particles that actually collide with the nucleus' described by the collision cross section either fuse or lose their energy in elastic collisions. So, the percentage of particles that fuse are described Gamow equation. Right? The rest just lose their energy through elastic collisions and braking radiation. Is this right? ....So, in this case, what exactly is a collision? How close are particles supposed to get in order for the interaction to be considered a collision? • #24 mfb Mentor 34,476 10,598 Oh. So time matters? Sure. If you use an accelerator, there is no need to have plasma target. A regular target (solid, liquid, gas) will work as well. So, the percentage of particles that fuse are described Gamow equation. That just describes the fusion cross section, not how the energy of the particles evolves if they don't fuse. Some fraction will fuse with the first nucleus they get close to. Others will scatter once and fuse with the second one. Others will scatter twice and fuse with the third one. And so on, but fusion gets less likely over time as the nuclei lose energy. Most will just scatter and never fuse. So, in this case, what exactly is a collision? How close are particles supposed to get in order for the interaction to be considered a collision? As mentioned before, the lower cutoff for calling something collision is a bit arbitrary. That doesn't change the physics, just the names we assign to things. • #25 68 2 Yeah. I understand now. Thanks • Last Post Replies 2 Views 256 • Last Post Replies 1 Views 3K • Last Post Replies 4 Views 5K • Last Post Replies 13 Views 2K • Last Post Replies 1 Views 811 • Last Post Replies 1 Views 647 • Last Post Replies 9 Views 923 • Last Post Replies 2 Views 4K • Last Post Replies 7 Views 3K • Last Post Replies 2 Views 3K

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http://www.sciforums.com/threads/is-the-universe-computing-something.155118/page-15

# Is the Universe computing something? Discussion in 'Astronomy, Exobiology, & Cosmology' started by arfa brane, Jan 26, 2016. 1. ### arfa branecall me arfValued Senior Member Messages: 6,915 Someone mentioned broken symmetry. This is not quite as esoteric as it sounds. For instance, a coin is symmetric. When a coin is spinning (i.e. being randomized), it has more symmetry than when it lands on one side or the other. The latter state is a broken symmetry. So a string of coin tosses is a randomized string, each element in the string represents say, the ring $\mathbb Z_2$ under some action. When the coin is spinning, the sides are in a random 'superposition' of probabilities, each side has exactly 0.5 probability of showing (if the coin is fair) when the symmetry 'breaks'. A string of results is represented by regular language as (0,1)*, a concise way of writing down a string with each digit having equal probability of being 0 or 1, a randomly 'generated' string. Last edited: Feb 15, 2016 to hide all adverts. 3. ### PhysBangValued Senior Member Messages: 2,422 The CC could simply be a constant of gravitational interaction; in this case, it had no relationship to quantum theory. There is no good calculation of the energy density of the quantum vacuum, so it is even possible that there is a CC an energy density from the vacuum. to hide all adverts. 5. ### The GodValued Senior Member Messages: 3,546 .... Banging funny things around. to hide all adverts. 7. ### paddoboyValued Senior Member Messages: 25,266 And gravitational interactions, are evident when spacetime is warped/curved/twisted/ rippled. And possibly a QGT may reveal the how and why this happens, which is at the quantum level. Well that's basically what I have said...these forces, the CC, ZPE, Casimir effect, may all be one and the same, but all are properties of spacetime as I see it. and all or one of them probably is the DE component. 8. ### brucepValued Senior Member Messages: 4,098 The cosmological constant is a specific scientific term. So it can't be all those different things. If it was all those things we wouldn't need to have a bunch of different terms to describe the same natural phenomena. Calling it a different name doesn't change the natural phenomena. Quintessence is a different model for describing the accelerated expansion of the universe. WMAP concluded it's the cosmological constant which best describes what's happening in this universe. So science is building a standard model of cosmology and the dark energy is best described by the cosmological constant. The good thing is science has a great empirical model for testing predictions derived from cosmological theoretical models. 9. ### paddoboyValued Senior Member Messages: 25,266 I can't argue with too much of what you have said bruce [and PhysBang] but I was always of the opinion that the mystery surrounding the DE component, was that we did not actually know the true nature of this force...whether the CC or quintessence as you mention, or any of the others. We just as yet are not sure. The different names imo were just a result of different observations. Perhaps I put my thoughts rather poorly. What do you conclude from the following..... https://briankoberlein.com/2015/03/06/nothing-but-net/ 10. ### paddoboyValued Senior Member Messages: 25,266 Please Register or Log in to view the hidden image! Or this which seems to support your thoughts..... http://www.sciencedirect.com/science/article/pii/S0370269306010197 Abstract It has been speculated that the zero-point energy of the vacuum, regularized due to the existence of a suitable ultraviolet cut-off scale, could be the source of the non-vanishing cosmological constant that is driving the present acceleration of the universe. We show that the presence of such a cut-off can significantly alter the results for the Casimir force between parallel conducting plates and even lead to repulsive Casimir force when the plate separation is smaller than the cut-off scale length. Using the current experimental data we rule out the possibility that the observed cosmological constant arises from the zero-point energy which is made finite by a suitable cut-off. Any such cut-off which is consistent with the observed Casimir effect will lead to an energy density which is at least about 1012 times larger than the observed one, if gravity couples to these modes. The implications are discussed. Current cosmological observations seem to favor a cosmological constant (Λ) as the leading candidate for dark energy which is thought to be driving the current phase of accelerated cosmic expansion [1]; the energy density contributed by it is constrained to be roughly ρDE≈10−11 (eV)4, with a corresponding length scale of the order of 0.1 mm. There exists a sizeable body of literature attempting to explain the origin of Λ as a consequence of the coupling of zero-point quantum fluctuations of matter fields, that pervade the vacuum, to gravity [2]. Write4U likes this. 11. ### PhysBangValued Senior Member Messages: 2,422 Well, yes and no. A quantum theory of gravity may include a separate energy that acts like a CC, or the CC may just be a mathematical feature of the graviton field, or whatever quantum field creates gravity. I haven't done too much research on the Casamir effect, but I believe that there are hypothetical explanations that do not rely on vacuum energy. 12. ### PhysBangValued Senior Member Messages: 2,422 While I agree that there are tests that show that the dynamics favor a constant, I'm not sure that they can yet be said to rule out a vacuum energy of some sort, or even some sort of field. If there is a weird combination of constant and vacuum energy and field, then we might never find out! 13. ### sweetpeaValued Senior Member Messages: 1,329 "Is the Universe computing something?" OMG, does this mean it may be possible to hack the universe? 14. ### brucepValued Senior Member Messages: 4,098 If you respect the results of the WMAP experiment then you can be somewhat 'sure'. Don't really like that term for science. We all know why Einstein derived the cosmological constant. To keep his GR universe from collapsing. In the classical metric it's a pressure term. Essentially an anti gravity term. It's the term that Guth and Linde used for the gravitational interaction, with a soliton in a quantum scalar field, that predicts the inflation event. During the event the pressure of the vacuum was negative and remained constant until the false vacuum 'rolled out' [reached a minimum vacuum energy state] and inflation came to an end. Initially Einstein didn't know the vacuum was comprised of fluctuating virtual particles seeking minimum energy states. Which must be accounted for when deriving quantum predictions associated with the cosmological constant. I think. So we have a classical model and a quantum model for the natural phenomena. These models describe the global effect of the cosmological constant. When we use the Casimir machine we alter the local physics of the vacuum. Locally between the uncharged plates. Resulting in a negative energy density between the uncharged plates and a measureable force pushing the plates towards each other. This is a pretty good discussion on the Casimir effect. http://physicsworld.com/cws/article/news/2012/jul/18/physicists-solve-casimir-conundrum So how big must the Casimir machine be to power a super luminal warp? It's the same theoretical principle that was used to hold the wormhole walls open during the theoretical analysis in this paper. http://arxiv.org/abs/1405.1283 Last edited: Feb 15, 2016 15. ### brucepValued Senior Member Messages: 4,098 Certainly. Based on what's known the classical and quantum models give a coherent description for the natural phenomena. I think. LOL. Thanks for your comments on this very interesting phenomena. 16. ### Write4UValued Senior Member Messages: 14,014 Every manmade object is a result of hacking (applying) the observed universal forms, values, and functions. Last edited: Feb 15, 2016 17. ### arfa branecall me arfValued Senior Member Messages: 6,915 Ahem. So since a coin spinning then landing is a quite good analogy of a quantum particle with spin plus a measurement 'operator': This is why Seth Lloyd and quite a few others claim the universe is computational--it's full of particles with spin, these particles interact and exchange something we call quantum information. Thanks to quantum randomness there is any number of ways to generate a random binary string by measuring spin, entirely analogous with tossing a fair coin. On the other side we mostly have Ken Wharton, whose argument appears to be based on our assumption that the universe is Newtonian, a mistake since the universe is Lagrangian. He argues pursuing the latter "Schema" and laying aside our anthropic bias. I don't know what he's talking about; it doesn't matter which approach you use, in any quantum experiment the results are always classical--the output is the same. Last edited: Feb 15, 2016 18. ### Write4UValued Senior Member Messages: 14,014 https://en.wikipedia.org/wiki/Virtual_particle IMO, this statement implies a very important question in regards to the required conditions where a *virtual* particle (such as the Higgs) becomes a *real* particle with a life of it's own. Apparently one requirement is that the Higgs combines with something else, which then gives it greater mass and the ability to persist as a real particle. Any thoughts on this (possibly) fundamental process? Messages: 6,915 20. ### sweetpeaValued Senior Member Messages: 1,329 Remember the OP asks is the universe computing something. Man is a part of nature's processes. The thoughts, for say, the making of clay pots, did not come from 'outside' of nature or 'outside' the universe. Those thoughts are part and parcel of nature or the universe. Or, to keep in tune with the OP, those thoughts and resulting actions are nature at work. Last edited by a moderator: Feb 17, 2016 21. ### arfa branecall me arfValued Senior Member Messages: 6,915 At a lecture on solid state physics, the subject was Heisenberg's Uncertainty Principle, and I recall the lecturer saying that it had cosmological implications, but he wasn't sure about them. The relation between time and energy in HUP does have computational implications: there must be a connection to the fundamental limits of computation (Bennett's Brownian Computer for instance). That is, $\Delta E \Delta t \ge \hbar / 2$ says there is a fundamental limit on the time for any computation, rather than the time to measure some particle's energy to within $\Delta E$. But that seems to just replace "time to measure" with "time to compute", somehow . . . Apparently the correct relation is: $\Delta t = {\pi \hbar} / 2 \Delta E$, for a particle with a spread in energy $\Delta E$ to "move from one distinguishable state to another". --http://arxiv.org/pdf/quant-ph/9908043v3.pdf Last edited: Feb 19, 2016 22. ### arfa branecall me arfValued Senior Member Messages: 6,915 Ok, so let's look again at the rock warming in the sun. It's a system, and it's processing information; the processing rate is limited by the amount of energy in each part of the system--if the surface is warmer than the insides it will have more energy to process information, the surface will compute faster than the inside of the rock. So, it's really just a close look, at say, the Heisenberg limit, of particles interacting. I've intentionally used a rock to make a sort of parody of the whole thing, except this is serious shit, so it isn't a parody after all. And a correction to post 281. I said the regular expression (0,1)* represents a random string of 1s and 0s, I should have said an element of the set generated by (0,1)* is such a string. Last edited: Feb 19, 2016 23. ### Waiter_2001Registered Senior Member Messages: 459 Hey Arf, how you doin? I see you've posted to sciforums TWICE here. I thought I'd post a reply to you in an effort to bring the thread forwards somewhat, which, I believe, will happen else I wouldn't have posted. It's all good what you posted though! Please Register or Log in to view the hidden image!

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http://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-3-adding-and-subtracting-fractions-3-3-adding-and-subtracting-unlike-fractions-3-3-exercises-page-219/2

## Basic College Mathematics (9th Edition) To rewrite unlike fractions as like fractions, we must find the least common denominator (or the denominator that is the least common multiple of the original denominators) of the unlike fractions. For example, if we want to add $\frac{3}{8}$ and $\frac{1}{4}$, we must find the least common multiple of 4 and 8. We know that this is 8. Therefore, we can rewrite this pair as like fractions. $\frac{3}{8}+\frac{1}{4}=\frac{3}{8}+\frac{1\times2}{4\times2}=\frac{3}{8}+\frac{2}{8}=\frac{3+2}{8}=\frac{5}{8}$

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https://flashman.neocities.org/MD/knowls/example.AEF.NEFP.0.knowl.html

Example AEF.NEFP Numerical Estimation False Position In many cases representing the exact value of a solution to an equation  as a decimal  can be very difficult or even impossible. In such cases techniques that provide a numerical estimate of the solution to some precision is very useful. Mapping diagrams can be used often to visualize these techniques. Since many examples involve elementary functions that are continuous for an interval of the form $[a,b]$, one basis for estimating solutions is the result: Theorem: CCD.IVT.0: If $f$ is a continuous function on the interval $[a,b]$ and $0$ is between $f(a)$ and $f(b)$, then there is a number $c \in [a,b]$  where  $f(c) = 0$. An immediate consequence of the IVT is that if $f(a)\cdot f(b) < 0$, then there is a number $c \in [a,b]$ where $f(c) = 0$. The False Position Method The false method repeatedly applies the IVT to estimate a solution to an equation with an elementary function, $f$, of the form $f(x)=0$ as long as $f(a)\cdot f(b) < 0$ for the initial numbers $a$ and $b$. Instead of using bisection to replace one endpoint of the interval it replaces one endpont with the root (zero) of the linear function determined by the function values at the endpoints of the interval, The procedure is as follows:Start with $a_1=a$ and $b_1=b$. Suppose $a_n < b_n$ and $f(a_n)\cdot f(b_n) < 0$. Let $FP_n = \frac {a_n f(b_n)-b_n f(a_n)}{f(b_n)-f(a_n)}$ and compute $f(FP_n)$. [$FP_n$ is the number determined by solving for $x$ in $\frac { f(b_n)}{b_n-x}= \frac { f(a_n)}{a_n-x}$. So $f(b_n)(a_n-x)= f(a_n) (b_n-x)$; $a_n f(b_n)-x f(b_n)= b_n f(a_n)-xf(a_n))$...] If  $f(FP_n) = 0$ then $m_n$ is the solution to the equation. Otherwise either $f(a_n)\cdot f(FP_n) < 0$ or $f(FP_n)\cdot f(b_n) < 0$. If $f(a_n)\cdot f(FP_n) < 0$, let $a_{n+1} = a_n$ and $b_{n+1}=FP_n$. Otherwise let $b_{n+1} = b_n$ and $a_{n+1}=FP_n$. Then $a_{n+1} < b_{n+1}$ and $f(a_{n+1})\cdot f(b_{n+1}) < 0$, so the procedure can be repeated. At each step the interval $[a_n,b_n]$ is shorter, so the solution guaranteed by the IVT to exist in this interval is approximated more accurately by either end numbers of the next interval. The mapping diagram below illustrates the example where $f(x) = x^2-2, a=0,$ and $b=2$. The False Position Method Notice how the arrows on the mapping diagram are paired with the point on the graph of the function $f$. You can move the point for $x$ on the mapping diagram to see how the function value changes both on the diagram and on the graph. Sliders: The sliders labeled $a$ and $b$ control the end values of the interval $[a,b]$. The slider labeled $n$ controls which the number of times the false position method is executed. Checking the box to "Show/Hide Midpont & Data" will display $a_n, b_n, FP_n$ and the values of $f$ for each of these nmbers. It will also display a scond box to "Show/Hide  Smallest Exact Root". Checking this box will display this root on both the graph and the mapping iagram. The function $f$ can be changed by making an entry in the appropriately labeled input box. Martin Flashman, August. 10, 2017. Created with GeoGebra

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https://www.albanova.se/event/phd-thesis-defense-quantum-chemical-calculations-of-multidimensional-dynamics-probed-in-resonant-inelastic-x-ray-scattering/

PhD Thesis Defenses ## PhD Thesis defense: Quantum chemical calculations of multidimensional dynamics probed in resonant inelastic X-ray scattering This thesis is devoted to the theoretical study of the dynamical processes induced by light-matter interactions in molecules and molecular systems. To this end, the multidimensional nuclear dynamics probed in resonant inelastic X-ray scattering (RIXS) of small molecules, exemplified by H2O (g) and H2S (g), as well as more complex molecular systems, exemplified by NH3 (aq) and kaolinite clay, are modelled. The computational methodology consists of a combination of ab initio quantum chemistry calculations, quantum nuclear wave packet dynamics and in certain cases molecular dynamics modelling. This approach is used to simulate K-edge RIXS spectra and the theoretical results are evaluated against experimental measurements. Specifically, the vibrational profile for decay back to the electronic ground state of the H2O molecule displays a vibrational selectivity introduced by the dynamics in the core-excited state. Simulation of the inelastic decay channel to the electronic |1b1-1,4a11> valence-excited state shows that the splitting of the spectral profile arises from the contribution of decay in the OH fragment. The character of the S1s-1 and S2p-1 core-excited states of the H2S molecule has been investigated and distinct similarities and differences with the H2O molecule have been identified. RIXS has also been used as a probe of the hydrogen bonding environment in aqueous ammonia and by detailed analysis of the valence orbitals of NH3 and water, the spectral profiles are explained. Finally, it is shown that vibrations of weakly hydrogen bonding OH are excited in RIXS decay to the electronic ground state in kaolinite. Keywords: quantum chemistry, X-ray spectroscopy, RASSCF, density functional theory, ultrafast nuclear dynamics. Stockholm 2018 http://urn.kb.se/resolve?urn=urn:nbn:se:su:diva-154057

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http://mathhelpforum.com/statistics/226593-need-help-understanding-these-problems.html

# Math Help - Need help understanding these problems 1. ## Need help understanding these problems 1. A factory has a delicate process that is affected by weather. If it rains, decrease the humidity inside the factory to continue operations. Since inside the factory the rain is no heard it is known that the probability of the work being damaged given that nobody noticed in time is 33%. If you have 85 rainfalls in a year calculate the expected number of damaged works and variance. 2. On average, every six minutes a car passes by a sector in which it plans to build a tunnel. The passage of many vehicles indicate risk of congestion in the new construction. The engineers in charge of the project need to determine whether the probability that in one hour more than 5 vehicles that circulate by the sector is less than 0.04. 2. ## Re: Need help understanding these problems problem two is a poisson distribution. The only trick here is to convert the rate of 1/6 car per minute to the # of cars per hour. Once you have the proper distribution just evaluate it's CDF at 5. in problem 1 we assume that the whole days work is lost 33% of the time when it rains? So we are basically trying to determine the number of days out of 85 that the humidity messes up and everything is ruined. This is a binomial distribution. with N=85, and p=33%=0.33 you can look up the distribution and derive the mean and variance easily enough. You're going to get numbers that are in units of #days of failure. 3. ## Re: Need help understanding these problems Thx I worked on the first one, but the second one Im still having trouble with he secong one can't get the exact value 4. ## Re: Need help understanding these problems Originally Posted by luis0426 Thx I worked on the first one, but the second one Im still having trouble with he secong one can't get the exact value were you able to calculate $\lambda$ for x in hours? Do you have some way of evaluating the CDF of the poisson distribution? 5. ## Re: Need help understanding these problems I knid of know how to solve it, I think my problem lies in how to work with this The only trick here is to convert the rate of 1/6 car per minute to the # of cars per hour. dont know if im doing it rigth 6. ## Re: Need help understanding these problems Originally Posted by luis0426 I knid of know how to solve it, I think my problem lies in how to work with this The only trick here is to convert the rate of 1/6 car per minute to the # of cars per hour. dont know if im doing it rigth so what did you come up with for $\lambda$ in units of cars per hour? 7. ## Re: Need help understanding these problems 1car/6min (60min/1h) = 6 cars per hour, dont know how to work with that and the determine whether the probability that in one hour more than 5 vehicles that circulate by the sector 8. ## Re: Need help understanding these problems cough... how many 6 minute periods in 1 hr? I don't know what circulate the sector means.. I just assume it means pass by, i.e. is an occurrence. 9. ## Re: Need help understanding these problems Im a bit confused too, how would you solve to get λ 10. ## Re: Need help understanding these problems 1 car every 6 minutes ---> 10 cars an hour ----> $\lambda=10$ 11. ## Re: Need help understanding these problems That much I got my problem lies in the probability that in one hour more than 5 vehicles that circulate by the sector is less than 0.04, how does the 5 factor in here, I know that P(x=0.4,λ), how does the 5 factor in the problem. 12. ## Re: Need help understanding these problems Originally Posted by luis0426 That much I got my problem lies in the probability that in one hour more than 5 vehicles that circulate by the sector is less than 0.04, how does the 5 factor in here, I know that P(x=0.4,λ), how does the 5 factor in the problem. Pr[6 more cars go by] = 1 - Pr[5 or less cars go by] = F(5) where F(x) is the CDF of the Poisson distribution with parameter $\lambda=10$ 13. ## Re: Need help understanding these problems I got it like this P(x>6) = 1-P(x=0)-P(x=1-P(x=2)-P(x=3)-P(x=4)-P(x=5), but know if parameter λ=10 and x is 0 through 5, how do I work with the 0.4 here 14. ## Re: Need help understanding these problems Originally Posted by luis0426 I got it like this P(x>6) = 1-P(x=0)-P(x=1-P(x=2)-P(x=3)-P(x=4)-P(x=5), but know if parameter λ=10 and x is 0 through 5, how do I work with the 0.4 here you don't work it in there. The engineer just wants to check that the actual probability of that many cars is < 0.04. He's in for a disappointment.

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https://www.physicsforums.com/threads/the-special-linear-group.120068/

# The Special Linear Group 1. May 7, 2006 ### cmiller5277 Hey, quick question... if I am considering the set of nxn matrices of determinant 1 (the special linear group), is it correct if I say that the set is a submanifold in the (n^2)-dimensional space of matrices because the determinant function is a constant function, so its derivative is zero everywhere? Also, I've been told that the tangent space to this set at the identity is just the space of trace 0 matrices, but I'm not seeing where this relationship comes from. Any help? 2. May 7, 2006 ### Hurkyl Staff Emeritus I'm not sure what that has to do with anything. And note that the determinant map is constant on any subset of SL(n) -- even those subsets that are not manifolds. Well, what is the tangent space? Suppose I have a curve f(x) in SL(n) such that f(0) = I. What can you say about its derivative? 3. May 7, 2006 ### cmiller5277 Well the first part of my question doesn't really have much to do with anything, you're right, but it was more of just a verification.. as in is it enough if I say the derivitave map is constant everywhere to show that SL(n) is a submanifold of the nxn space. The trouble I'm having with the tangent space of SL(n) is that I am having difficulty with the curve f(x) in SL(n) itself. When you say we have a curve f(x) in SL(n) such that f(0) = I, is the derivative just going to be the derivative of I? And since the diagonal is composed of constant elements, we get their derivatives to be zero, therefore giving us a trace 0 matrix? I apologize if this is obvious.. I had trouble in Linear Algebra with this sort of thing. Thank you for the response though 4. May 7, 2006 ### Hurkyl Staff Emeritus No; you want the derivative of f, not the derivative of f(0). f is a function that takes values in an n²-dimensional vector space. Taking its derivative is no different than what you did in vector calculus when you worked in R² or R³... you just have many more components, and you arrange them in a square shape instead of a column. (Incidentally, I think what you want to do is a differential approximation) 5. May 8, 2006 ### mathwonk you seem to be referring to the fact that a level set of a smooth function is a manifold if the function has maximal rank at every point of that level set. this does indeed apply to the set of matrices of determinant one and does prove they are a manifold. the fact that the tangent space at the identity is the set of trace zero matricers can be checked directly, and is also the basis of the "frenet serret formulas". 6. May 8, 2006 ### Hurkyl Staff Emeritus I ought to work through this problem again. Last time I did it, I threw my hands up in the air, decreed that "dt" was an infinitessimal, and looked at the equation det(I + A dt) = 1, where A was a tangent vector. Actually, I still think about matrix Lie groups and algebras in that way. I suppose I ought to figure out how to justify it in terms of nonstandard analysis so I can feel better about myself. 7. May 9, 2006 ### mathwonk it is easy to see that det has rank one at an nxn matrix A where det(A) is not zero. just compose with a transverse parameter getting det(tA) = t^n det(A), and take the derivative at t=1. to get the tangent space to f=c at p, just take the orthogonal complement of the gradient of f at p. for the determinant polynomial, the rgadient is the vector whose entry in the i,j place is plus or minus the minor determinant associated to that entry. Hence when evaluated at the identity amtrix it has a zero in every entry except those indexed by i,i, and at those places it has a 1. the orthocomplement is thus matrices which dot to zero with this vector, i.e. those of trace zero.

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https://www.bartleby.com/solution-answer/chapter-15-problem-33cr-introductory-chemistry-a-foundation-9th-edition/9781337399425/if-an-electric-current-is-passed-through-molten-sodium-chloride-elemental-chlorine-gas-is-generated/a57913ae-252e-11e9-8385-02ee952b546e

Chapter 15, Problem 33CR ### Introductory Chemistry: A Foundati... 9th Edition Steven S. Zumdahl + 1 other ISBN: 9781337399425 Chapter Section ### Introductory Chemistry: A Foundati... 9th Edition Steven S. Zumdahl + 1 other ISBN: 9781337399425 Textbook Problem 57 views # If an electric current is passed through molten sodium chloride, elemental chlorine gas is generated as the sodium chloride is decomposed. :math> 2NaCl ( 1 ) → 2Na ( s ) + Cl 2 ( g ) at volume of chlorine gas measured at 767 mm Hg at 25 °C would be generated by complete decomposition of 1.25 g of NaCl? Interpretation Introduction Interpretation: The volume of chlorine gas generated by complete decomposition of 1.25g of NaCl is to be calculated. Concept Introduction: A mole of a substance is defined as the same number of particles of the substance as present in 12g of 12 C. The number of particles present in one mole of a substance is 6.023×1023 particles. The number of moles a substance is given as, n=mM Where, • m represents the mass of the substance. • M represents the molar mass of the substance. The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is given as, PV=nRT Where, • V represents the volume occupied by the ideal gas. • P represents the pressure of the ideal gas. • n represents the number of moles of the ideal gas. • T represents the temperature of the ideal gas. • R represents the ideal gas constant with value 0.08206LatmK1mol1. Explanation The molar mass of NaCl is 58.44gmol1. The mass of NaCl is 1.25g. The temperature of chlorine gas is 25°C. The temperature of chlorine gas in Kelvin is given as, T=25°C+273.15K=298.15K The pressure of chlorine gas is 767mmHg. The number of mole of substance is given as: n=mMm Where, • m represents the mass of the substance. • Mm represents the molar mass of the substance. Substitute the value of the mass and molar mass of NaCl in above equation. n=1.25g58.44gmol1=0.0214mol The number of moles of NaCl reacted is 0.0214mol. The given reaction is represented as: 2NaCll2Nas+Cl2g Two moles of NaCl produce one mole of Cl2 gas, hence, the relation between the number of moles of NaCl reacted and chlorine gas produced is given as, n1=n22 Where, • n1 represents the number of moles of chlorine gas. • n2 represents the number of moles of NaCl. Substitute the value of n1 and n2 in the above equation ### Still sussing out bartleby? Check out a sample textbook solution. See a sample solution #### The Solution to Your Study Problems Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees! Get Started

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https://socratic.org/questions/how-do-you-solve-9a-4b-5-and-6a-2b-3-using-matrices

Precalculus Topics # How do you solve 9a - 4b = -5 and 6a - 2b = -3 using matrices? Sep 22, 2016 $x = - 0.33$ $y = 0.5$ #### Explanation: Look at the steps - Sep 22, 2016 $\therefore a = - \frac{1}{3} \mathmr{and} b = \frac{1}{2}$ #### Explanation: Although the method might seem quite daunting, once the preparation process is mastered, the method itself is surprisingly quick and easy, involving a few simple calculations. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ We have the following equations: $9 a - 4 b = - 5 \text{ and } 6 a - 2 b = - 3$ First write them as matrices: $\left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right) \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\begin{matrix}- 5 \\ - 3\end{matrix}\right)$ Now find the inverse matrix of $A = \left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right)$ $\left\mid A \right\mid = \left(9 \times - 2\right) - \left(6 \times - 4\right) = - 18 + 24 = 6$ ${A}^{-} 1 = \frac{1}{6} \left(\begin{matrix}- 2 & 4 \\ - 6 & 9\end{matrix}\right) = \left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right)$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Multiply both sides of the matrix equation by the inverse matrix. $\left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right) \left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right) \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right) \left(\begin{matrix}- 5 \\ - 3\end{matrix}\right)$ $\textcolor{w h i t e}{\times \times \times \times x} \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{3} \\ \frac{1}{2}\end{matrix}\right)$ $\textcolor{w h i t e}{\times \times \times \times \times \times \times} \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{3} \\ \frac{1}{2}\end{matrix}\right)$ $\therefore a = - \frac{1}{3} \mathmr{and} b = \frac{1}{2}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Background knowledge... to help with the method above.. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ A 2 x 2 matrix multiplied by the unit matrix remains unchanged $\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}a & b \\ c & d\end{matrix}\right) = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ A matrix multiplied by its inverse gives the unit matrix - also known as the Identity Matrix. $A \times {A}^{-} 1 = I = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$ To find the inverse matrix (${M}^{-} 1$) of matrix M $M = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ 1. Find the determinant $\left(\left\mid M \right\mid\right) = a d - b c$ 2. ${M}^{-} 1 = \frac{1}{\left(\left\mid M \right\mid\right)} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$ (swop a and d and change the signs of b and c), then divide by the determinant.) ##### Impact of this question 423 views around the world

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http://mathhelpforum.com/algebra/67364-prove-no-rational-satisfies-2-x-3-a.html

# Thread: Prove no rational satisfies 2^x = 3 1. ## Prove no rational satisfies 2^x = 3 Here is the question and i have no idea how to go about it any help would be hugely appreciated. Prove no rational satisfies $\displaystyle 2^x = 3$ Many thanks 2. Originally Posted by craig in a post he soft deleted and I hard deleted You would firstly take logs of both sides, $\displaystyle \log_{2} 2^x=\log_{2} 3$. As $\displaystyle \log_{y} y^x = x$, we can say that: $\displaystyle x=\log_{2} 3$ Put this into your calculator and you get $\displaystyle x=1.584962501$, not a rational number On the contrary, 1.584962501 certainly IS a rational number because it is a terminating decimal! $\displaystyle log_2(3)$,which is NOTequal to 1.585962501 is not rational but I suspect it would be necessary to prove that to get credit for this problem! 3. Originally Posted by sebjory Here is the question and i have no idea how to go about it any help would be hugely appreciated. Prove no rational satisfies $\displaystyle 2^x = 3$ Many thanks This is in the prealgebra/algebra section, so i am not sure if a proof like this (a proof by contradiction) will be too complicated. if it is, tell me, we can come up with another. Assume, to the contrary, that there is a rational number $\displaystyle x$ such that $\displaystyle 2^x = 3$. Then $\displaystyle x = \frac ab$, where $\displaystyle a$ and $\displaystyle b$ are integers, with $\displaystyle b \ne 0$. So we have $\displaystyle 2^{\frac ab} = 3$ $\displaystyle \Rightarrow 2^a = 3^b$ However, this is absurd, since $\displaystyle a$ and $\displaystyle b$ are integers, this equation contradicts the uniqueness of prime factorization (since the left side is always a product of 2's and the right is always a product of 3's). note that a = b = 0 makes the equation make sense... but b cannot be zero! 4. thankyou proof by contradiction is taught over here in the UK. 5. Originally Posted by sebjory thankyou proof by contradiction is taught over here in the UK. oh, you're from the UK, hehe, I was worried. you have a US flag up i suppose you are also familiar with the uniqueness of prime factorization theorem (you might know it under a different name), but it talks about how each integer can be uniquely expressed as a product of primes with integer powers 6. Yes, we're studying this whole semester on these sorts of theorems about rational numbers (the Completeness axiom etx) and with these come a whole host of proofs... Do you know any good proof resource where i can learn as many types of proof as possible? (not including Induction since i am fairly happy with my ability to manipulate these sorts of problems) 7. Originally Posted by sebjory Yes, we're studying this whole semester on these sorts of theorems about rational numbers (the Completeness axiom etx) and with these come a whole host of proofs... Do you know any good proof resource where i can learn as many types of proof as possible? (not including Induction since i am fairly happy with my ability to manipulate these sorts of problems) completeness axiom? ...this is prealgebra?!! i am sure there are resources online that have what you are looking for. i guess wikipedia is always a good place to start, at least for quick reference. but i have never really searched for any such site, as i have several textbooks that cover such material and am fairly happy with them, so i don't know. sorry 8. Originally Posted by Jhevon This is in the prealgebra/algebra section, so i am not sure if a proof like this (a proof by contradiction) will be too complicated. if it is, tell me, we can come up with another. Assume, to the contrary, that there is a rational number $\displaystyle x$ such that $\displaystyle 2^x = 3$. Then $\displaystyle x = \frac ab$, where $\displaystyle a$ and $\displaystyle b$ are integers, with $\displaystyle b \ne 0$. So we have $\displaystyle 2^{\frac ab} = 3$ $\displaystyle \Rightarrow 2^a = 3^b$ However, this is absurd, since $\displaystyle a$ and $\displaystyle b$ are integers, this equation contradicts the uniqueness of prime factorization (since the left side is always a product of 2's and the right is always a product of 3's). note that a = b = 0 makes the equation make sense... but b cannot be zero! Another argument could be $\displaystyle 2^a$ is always even and $\displaystyle 3^b$ is always odd. Clearly, therefore, $\displaystyle 2^a = 3^b$ where a and b are integers is nonsense, giving the contradiction and completing the proof. 9. Originally Posted by Prove It Another argument could be $\displaystyle 2^a$ is always even and $\displaystyle 3^b$ is always odd. Clearly, therefore, $\displaystyle 2^a = 3^b$ where a and b are integers is nonsense, giving the contradiction and completing the proof. even easier!

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https://physics.stackexchange.com/questions/76126/is-there-any-relationship-between-gravity-and-electromagnetism/76136

# Is there any relationship between Gravity and Electromagnetism? [duplicate] We all know that the universe is governed by four Fundamental Forces which are The strong force , The weak force , The electromagnetic force and The gravitational force . Now, is there any relationship between Electromagnetism and gravity? • Possible duplicates: physics.stackexchange.com/q/944/2451 and links therein. – Qmechanic Sep 3 '13 at 15:46 • ncatlab.org/nlab/show/Kaluza-Klein+mechanism – Urs Schreiber Sep 4 '13 at 18:51 • I have no formal physics training, but it seems to me that they are intrinsically connected. If gravity effects mass and mass & energy (electromagnetic waves or light) are related through Einsteins equation, then gravity and light possess a very clear relationship. It seems logical to me, but can someone more qualified chime in? – user29224 Sep 4 '13 at 22:43 • @UrsSchreiber: Oh my god, that's an Excellent article! – Abhimanyu Pallavi Sudhir Sep 7 '13 at 8:43 • @DImension10, thanks for the feedback. I just went through that entry again and expanded a bit more here and there. For instance the Examples-section now has a new subsection "Cascades of KK-reductions from holographic boundaries" ncatlab.org/nlab/show/… . – Urs Schreiber Sep 7 '13 at 12:50 # On Unification I presume you're asking whether just classical gravity & classical EM can be unified. They sure can! Classical General Relativity and Classical Electromagnetism are unified in Kaluza-Klein-Theory, which proves that 5-dimensional general relativity is equivalent to 4-dimensional general relativity plus 4-dimensional maxwell equations. Rather interesting, isn't it? A byproduct is the scalar "Radion" or "Dilaton" which appears due to the "55" component of the metric tensor. In other words, the Kaluza-Klein metric tensor equals the GR metric tensor with maxwell stuff on the right and at the bottom; BUT you have an extra field down there. $${g_{\mu \nu }} = \left[ {\begin{array}{*{20}{c}} {{g_{11}}}&{{g_{12}}}&{{g_{13}}}&{{g_{14}}}&{{g_{15}}} \\ {{g_{21}}}&{{g_{22}}}&{{g_{23}}}&{{g_{24}}}&{{g_{25}}} \\ {{g_{31}}}&{{g_{32}}}&{{g_{33}}}&{{g_{34}}}&{{g_{35}}} \\ {{g_{41}}}&{{g_{42}}}&{{g_{43}}}&{{g_{44}}}&{{g_{45}}} \\ {{g_{51}}}&{{g_{52}}}&{{g_{53}}}&{{g_{54}}}&{{g_{55}}} \end{array}} \right]$$ Imagine 2 imaginary lines now. $${g_{\mu \nu }} = \left[ {\begin{array}{*{20}{cccc|c}} {{g_{11}}}&{{g_{12}}}&{{g_{13}}}&{{g_{14}}} & {{g_{15}}} \\ {{g_{21}}}&{{g_{22}}}&{{g_{23}}}&{{g_{24}}} & {{g_{25}}} \\ {{g_{31}}}&{{g_{32}}}&{{g_{33}}}&{{g_{34}}} & {{g_{35}}} \\ {{g_{41}}}&{{g_{42}}}&{{g_{43}}}&{{g_{44}}} & {{g_{45}}} \\ \hline {{g_{51}}}&{{g_{52}}}&{{g_{53}}}&{{g_{54}}} & {{g_{55}}} \end{array}} \right]$$ So the stuff on the top-left is the GR metric for gravity, and the stuff on the edge ($g_{j5}$ and $g_{5j}$) is for electromagnetism and you have an additional component on the bottom right. This is the radion/dilaton. An extension to kaluza - klein is , which also talks about the weak and strong forces, and requires . # On Geometry In , the gauge group for is $U(1)$. Now, the key thing here is that Electromagnetism is then The Curvature of the $U(1)$ bundle. This is not the only geometric connection between General Relativity and Quantum Field Theory. In the same context, the covariant derivatives is general relativity are such that $\nabla_\mu-\partial_\mu$ sort-of measures the gravity, in a certain way, while this is also true in QFT, where to some constants, $\nabla_\mu-\partial_\mu=ig_sA_\mu$. It is to be noted that both are in similiar context.

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http://www.ams.org/joursearch/servlet/PubSearch?f1=msc&pubname=all&v1=10H20&startRec=1

American Mathematical Society My Account · My Cart · Customer Services · FAQ Publications Meetings The Profession Membership Programs Math Samplings Policy and Advocacy In the News About the AMS You are here: Home > Publications AMS eContent Search Results Matches for: msc=(10H20) AND publication=(all) Sort order: Date Format: Standard display Results: 1 to 6 of 6 found      Go to page: 1 [1] Kevin S. McCurley. Prime values of polynomials and irreducibility testing. Bull. Amer. Math. Soc. 11 (1984) 155-158. MR 741729. Abstract, references, and article information    View Article: PDF [2] Joseph L. Gerver. Irregular sets of integers generated by the greedy algorithm . Math. Comp. 40 (1983) 667-676. MR 689480. Abstract, references, and article information    View Article: PDF This article is available free of charge [3] Samuel S. Wagstaff. Greatest of the least primes in arithmetic progressions having a given modulus . Math. Comp. 33 (1979) 1073-1080. MR 528061. Abstract, references, and article information    View Article: PDF This article is available free of charge [4] Carter Bays and Richard H. Hudson. Details of the first region of integers $x$ with $\pi \sb{3,2}(x)<\pi \sb{3,1}(x)$ . Math. Comp. 32 (1978) 571-576. MR 0476616. Abstract, references, and article information    View Article: PDF This article is available free of charge [5] John Isbell and Stephen Schanuel. On the fractional parts of $n/j,$ $j=o(n)$ . Proc. Amer. Math. Soc. 60 (1976) 65-67. MR 0429796. Abstract, references, and article information    View Article: PDF This article is available free of charge [6] Harold G. Diamond. Chebyshev estimates for Beurling generalized prime numbers . Proc. Amer. Math. Soc. 39 (1973) 503-508. MR 0314782. Abstract, references, and article information    View Article: PDF This article is available free of charge Results: 1 to 6 of 6 found      Go to page: 1

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https://www.physicsforums.com/threads/from-langevin-to-fokker-planck.103448/

# Homework Help: From Langevin to Fokker-Planck 1. Dec 9, 2005 ### Jezuz Hi. I'm studying quantum Brownian motion right now and I need to see that the (classical) Langevin equation for a Brownian particle is equivalent to the Fokker-Planck equation for the phase-space distribution function of the same particle. Does anyone know where I can find such a derivation? I've been looking all over the internet for it but usually they start with a Langevin equation containging only first derivatives (that is the have excluded the possible outer potential felt by the particle). I need the derivation for the case where i have a Langevin equation of the type: m \ddot x(t) + \gamma \dot x(t) + V(x(t)) = F(t) (written in LaTeX syntax). I would be very grateful for help! Alternatively, since I have the derivation for the Langevin equation starting with a Lagrangian for a particle interacting linearly with a bath of harmonic oscillators (initially in thermal equilibrium), I could also accept a derviation of the Fokker-Planck equation starting with the same assumptions. 2. Dec 9, 2005 ### Physics Monkey For such stochastic methods, a very good book is "Handbook of Stochastic Methods" by C. Gardiner. Written by an expert in the field, this book is very physical and to the point unlike many other excessively mathematical treatments of stochastic processes. If you're interested, I can step you through the derivation here, but you should get that book for sure. 3. Dec 10, 2005 ### Jezuz Okej! Thank you very much. I'll have a look at that book. Found it at the university library. If there is something I get stuck with I might ask you again :)

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http://mathhelpforum.com/differential-geometry/110273-divergent-convergent-summs-series.html

Thread: Divergent and Convergent summs\series 1. Divergent and Convergent summs\series 1) $\displaystyle\sum_{n=1}^{\infty}{a_n}$ is divergent. Proove that $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{1+n*a_n }$ is also divergent. 2) $a_n \geq a_{n+1} \textgreater 0$ and $\displaystyle\sum_{n=1}^{\infty}{a_n}$ is convergent. Proove that: $\lim\limits_{n \rightarrow \infty}{n*a_n}=0$ I am a third year student at Computer Science, therefore I have some strong basics. Any help, direction where to look is appreciated. Thx LE: 3) $ \sum_{i=0}^\infty \arctan\frac{1}{n^2+n+1}=? $ 2. 2) $a_n \geq a_{n+1} \textgreater 0$ and $\displaystyle\sum_{n=1}^{\infty}{a_n}$ is convergent. Proove that: $\lim\limits_{n \rightarrow \infty}{n*a_n}=0$ Let $S_n = \sum_{i=1}^n a_n, S = lim_{n\to\infty}S_n$. Pick any epsilon > 0. Then there exists a K such that for any k >= K, (a_(K+1) + ... + a_m) <= S - S_k < epsilon/2 since S_k converges to S and since the partial sums are increasing. Also, since a_n converges to zero, there must exist a J > K such that for all j >= J, (epsilon/2)*min{a_1, a_2, ..., a_K}/S_K >= a_j. Therefore since the sequence is decreasing, for all sufficiently large j, $n\cdot a_j = \underbrace{a_j + \ldots + a_j}_{n\;\rm times} \leq (\frac{\epsilon a_1}{2S_K} + \ldots + \frac{\epsilon a_K}{2S_K}) + (a_{K+1} + \ldots + a_j)$ $= \frac{\epsilon S_K}{2S_K} + (a_{K+1} + \ldots + a_j) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. Since the sequence is positive but eventually remains smaller than any positive number, it must converge to zero.

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http://stochasticprocess.blogspot.com/2011/06/reading-guides-and-complex-methods.html

When not at work with students, I spend my time in my room either reading, calculating something using pen and paper, or using a computer. I read almost anything: from the pornographic to the profound, although my main interests are mathematics and physics. "When I get a little money I buy books; and if any is left I buy food and clothes." -Erasmus ## Sunday, June 26, 2011 ### reading guides and complex methods I'm teaching complex (variable) methods this semester, and I'm using Arfken's book to teach it. It's a long list of topics; the detailed syllabus is shown below: Meeting no. ObjectivesAfter the discussion and lined up activities, you should be able to: Topics 1(6-15) §                 Explain what is expected of you to get good marks in this class§                 Explain the expected role of your teacher§                 Explain the expected role of your book§                 Explain the expected role of your lecture classes§                 List the materials you will need for this course§                 Explain why the rules of coherence are needed. Orientation Read:  Rules of Coherence 2(6-17) §                 Perform algebraic operations using complex numbers§                 Use the geometric series to evaluate complex valued sums§                 Use the natural logarithm function to calculate inverse trigonometric functions§                 Use the natural logarithm function to calculate complex powers Complex Algebra Complex-valued Series Complex Powers Read: Sec 6.1Exercises: 6.1.7;  6.1.8;  6.1.9;  6.1.10;  6.1.16, 6.1.17; 6.1.22 DIAGNOSTIC EXAMINATION 12June 18, 2011 (Sat) 1pm # Complex analysis i 3(6-22) §                 Use the definition of the derivative to evaluate the derivative of polynomials§                 Use the definition of the derivative to obtain the Cauchy-Riemann conditions §                 Use the Cauchy-Riemann conditions and continuity of partial derivatives to test a complex function for analyticity.§                 Use the Cauchy-Riemann conditions to show that the real and imaginary parts of an analytic function must satisfy the Laplace equation Cauchy-Riemann conditions Laplace equation Analytic functions Read: Sec 6.2Exercises: 6.2.1; 6.2.4; 6.2.5 4(6-24) §                 Define the Riemann contour integral on a complex contour and compare with the real Riemann integral§                 Use the definition of the Riemann integral to prove the Darboux inequality§                 Use the definition of the contour integral to show that, in general, the contour integral is path-dependent. §                 Use the definition of the contour integral to calculate the contour integral of a given function along a given path.§                 Prove the Cauchy-Integral Theorem using Stokes’ theorem Contour Integral Darboux inequality Path-dependence Cauchy Integral Formula Read:  Sec 6.3   Exercises: 6.3.1; 6.1.2; 6.3.3 5(6-29) §                 Prove the Cauchy integral formula by using a deformation of contour and polar coordinates.§                 Use the Cauchy integral formula to obtain an expression for the nth derivative of an analytic function.§                 Use the Cauchy integral formula to relate Rodrigues formula representations of special functions to contour integral representations. Cauchy Integral formula Nth derivative of an analytic function Contour integral representations Read: Sec 6.4Exercises:  6.4.1; 6.4.5; 6.4.6; 6.4.8 6(7-1) §                 Derive the Taylor series of given functions and give the domain of validity§                 Derive the Laurent series of given functions about a given point on the complex plane and, using knowledge of singularities, give the domains of validity Taylor seriesLaurent seriesSingularities Read Section 6.5Exercises 6.5.1; 6.5.2; 6.5.3; 6.5.7; 6.5.8; 6.5.10; 6.5.11 7(7-6) §                 Obtain a formula for the coefficient a-1 in terms of derivatives§                 Given a curve in the z-plane, and a mapping w(z), obtain the corresponding curve on the complex w-plane§                 Given a simply connected-region A, and a given mapping w(z), obtain its image on the w-plane Residue formula Mapping Read Section 6.6, 6.7Exercises: 6.6.1, 6.6.2, 6.7.6 First long exam july 16, 2011 (sat) 1pm COMPLEX ANALYSIS II 8-9(7-8 to 7-13) §                 Identify the poles and other singularities of a given meromorphic function, and evaluate the residues at these singularities (if any)§                 Use the residue theorem to evaluate a closed simply connected contour integral§                 Use the residue theorem and Darboux’s inequality to calculate Fourier integrals§                 Use the residue theorem to calculate other integrals. Singularities and residues Residue at infinity Contour integration Read Sec 7.1 (exclude Pole and product expansions, pp 461-462)Exercises 7.1.1, 7.1.3, 7.1.6, 7.1.8, 7.1.10, 7.1.11, 7.1.14, 7.1.15, 7.1.17, 7.1.18, 7.1.21, 7.1.24 10(7-15) §                 Given a series for w(z) in powers of z, obtain the series for the inverse function of z(w) in powers of w. §                 Give the conditions for expanding an analytic function as a pole expansion§                 Obtain the pole expansion of selected functions§                 Obtain the product expansions of sin(z) and cos(z) Inversion of Series Pole Expansions Product Expansions Summation of Series via Contour Integrals Read Whittaker and Watson, pp 134 to 139, Arfken  pp 461-462, Morse and Feshbach pp 411-413,pp 413-414Exercises Whittaker and Watson Section 7.4 Examples 4, 6 and Section 7.5 Example 1Exercises for inversion of series will be handed out in class 11 to 12(7-20 to 7-22) §                 Find the zeroes of the derivative of an analytic function f(z), and draw contour plots of the real and imaginary parts of f(z)-f(z0) in the neighborhood of the zero z0 obtained. §                 Use the method of steepest descent to approximate a class of integrals with parameter s, for large values of s.§                 Obtain the asymptotic series using the method of steepest descent Method of Steepest Descent Read Sec 7.3, Morse and Feshbach pp 434-443Exercises 7.3.1, 7.3.2, 7.3.3, exercises to be handed out 2nd Long Exam July 30, 2011 (Sat) 1pm # gAMMA, BETA AND FOURIER SERIES 13(7-27) §                 Show the equivalence of  the three definitions (Euler integral analytically continued; Weierstrass product, Euler product) of the Gamma function§                 Use the Gamma function to evaluate negative factorials§                 Use the Gamma function to obtain the Binomial series valid for non-integral powers, and give the domain of validity§                 Derive the recursion relation of the gamma function§                 Find the poles of the gamma function and evaluate the residues of the gamma function§                 Prove the Gauss-Multiplication theorem and the Legendre duplication formula§                 Evaluate Gaussian integrals Gamma Function Binomial series Gaussian integrals Multiplication Formula Read Sec 8.1, and Whittaker and Watson pp 244-246Exercises  8.1.1, 8.1.4, 8.1.5, 8.1.9, 8.1.14,  8.1.18, 8.1.24 14(7-29) §                 Derive Stirling’s series§                 Use Stirling’s approximation to evaluate large factorials.§                 Use the Beta function and the chain rule to evaluate a selected class of integrals Stirling’s Approximation Beta Function Read sec 8.3 and 8.4, Whittaker and Watson pp 251 to 253Exercises 8.3.1, 8.3.6, 8.3.8, 8.3.9,  8.4.2, 8.4.17, 8.4.18 15-16(8-3 to 8-5) §                 Prove the orthogonality of a given set of sines and cosines on a suitable interval§                 Use orthogonality and completeness of a basis of sines and cosines to obtain the Fourier series of a function within an interval§                 Use Fourier series to solve the wave equation of a vibrating string Fourier Series Orthogonality Read Sec 14.1 to 14.4Exercises 14.1.5, 14.1.9, 14.2.1, 14.2.3, 14.3.2, 14.3.12, 14.3.14, 14.4.2, 14.4.10 3rd Long Exam 12August 13, 2011 (Sat) 1pm INTEGRAL TRANSFORMS 17(8-10) §                 Expand given functions defined over the whole real line as a Fourier integral§                 Obtain the Fourier transform of a given function§                 Given the function in (Fourier) k-space, use the inverse Fourier transform to obtain the function in coordinate space§                 Relate Mellin transforms to Fourier transforms§                 Obtain the representation of the Dirac delta in terms of Fourier integrals Fourier Transforms Read Section 15.1 to 15.3Exercises 15.1.3, 15.3.2, 15.3.4, 15.3.9, 15.3.17, 18(8-12) §                 Obtain the Fourier transform of derivatives§                 Convert a linear differential equation in coordinate space into the corresponding integral equation in k-space§                 Solve the diffusion equation using Fourier transforms Fourier transform of derivatives Read Sec 15.4Exercises 15.4.1, 15.4.3, 15.4.4, 15.4.5 19(8-17) §                   Use the convolution theorem to evaluate some integrals§                   Obtain the momentum space representation of a wavefunction Convolution TheoremParseval’s RelationMomentum Space Read Sections 15. 5 to 15. 6Exercises 15.5.3, 15.5.5, 15.5.6, 15.5.8, 15.6.3, 15.6.8, 15.6.12 20(8-24) §                   Calculate the Laplace transform of some elementary functions§                   Use tables of Laplace transforms and the linearity of Laplace transforms to evaluate inverse Laplace transforms§                   Use partial fractions to evaluate inverse Laplace transforms Laplace Transform Partial fractions Read Section15.8Do Exercises 15.8.3, 15.8.4, 15.8.5, 15.8.9 21(8-26) §                   Evaluate the Laplace transform of derivatives§                   Convert linear differential equations with constant coefficients to algebraic systems§                   Solve linear ordinary differential equations with constant coefficients using Laplace transforms Laplace Transform of Derivatives Convolution Theorem Read Section 15.9 to 15.11Exercises 15.9.2, 15.9.3, 15.11.2, 15.11.3 22(8-31) §                   Evaluate inverse Laplace transform using Bromwich integrals§                   Convert linear differential equations with constant coefficients to algebraic systems Bromwich integral Read Section 15.12Exercises 15.12.1, 15.12.2, 15.12.3, 15.12.4 4th Long Exam September 3, 2011 (Sat) SPECIAL FUNCTIONS I – BESSEL, LEGENDRE FUNCTIONS 23 to 24(9-2 to 9-7) §          Identify the singularities of linear second order differential equations§          Use the power-series method to obtain a solution of linear second order differential equations§           Use Wronskians to obtain a linearly independent second solutions if a solution is known Power-Series Solutions Read: Section 9.4 to 9.6Ex: 9.4.1,9.4.2, 9.4.3, 9.5.5, 9.5.6, 9.5.10, 9.5.11, 9.6.18, 9.6.19, 9.6.25 25(9-9) §          Use generating functions to obtain the recursion relations satisfied by Bessel Functions§          Use power-series methods to solve Bessel’s differential equation for both integral and non-integral powers§          Use the generalized Green’s theorem to verify the orthogonality properties of a set of Bessel functions§          Use the completeness of a set of Bessel functions to expand a given function in the interval 0≤ x ≤ a Bessel Functions of the First Kind Generating Function Read: Sec 11.1 to 11.2Ex: 11.1.1, 11.1.3, 11.1.10, 11.1.16, 11.1.18, 11.2.2, 11.2.3, 11.2.6 26(9-14) §          Use the definition of Neumann functions and verify that it is a second linearly independent solution of Bessel’s equation§          Use the definition of Hankel functions to derive its properties§          Use the asymptotic formulae for Bessel functions to approximately evaluate Bessel functions for large values of its argument Neumann and Hankel functions Asymptotic formulae for Bessel functions Read: Sec 11.3 to 11.4Ex: 11.3.2,11.3.6, 11.4.7 27(9-16) §                 Use generating functions to obtain recursion relations and other properties of Legendre functions§                 Use the generalized Green’s  theorem to prove orthogonality of Legendre functions§                 Expand an arbitrary function within the interval -1≤ x ≤  in terms of Bessel functions and give an integral for the expansion coeffiecients §                 Use Rodrigues formula to derive orthogonality of Legendre polynomials and calculate the normalization constant of Legendre polynomials Legendre functions Orthogonality and Completeness Read: 12.1 to 12.4Ex:  12.2.2, 12.2.3, 12.2.5, 12.3.2,12.3.6, 12.3.11, 12.4.2 28(9-21) §                 Prove orthogonality of associated Legendre functions§                 Use completeness relations of Spherical Harmonics to express functions depending on θ and φ  as a sum over spherical harmonics§                 Express 1/ │x1-x2│ in terms of spherical harmonics Associated Legendre functions Spherical Harmonics Addition Theorem Read sec 12.5 to 12.6, 12.8Exercises 12.5.1, 12.5.11,12.6.4, 12.6.5,12.8.3, 12.8.8 5th Long Exam September 24, 2011 (Sat) SPECIAL FUNCTIONS II—HERMITE, LAGUERRE, HYPERGEOMETRIC FUNCTIONS 29(9-23) §                 Prove orthogonality of Hermite functions§                 Solve Hermite’s differential equation via power series§                 Use completeness relations to express functions in the interval -∞ ≤ x  ≤∞ as a sum of Hermite polynomials§                 Use generating function to obtain the Rodrigues formula for Hermite polynomials§                 Use Rodrigues formula to obtain Hermite polynomials Hermite functions Completeness and Orthogonality Rodrigues and Integral Representations Read: Sec 13.1Exercises 13.1.2, 13.1.14, 13.1.12, 13.1.13 30(9-28) §                 Prove orthogonality of Laguerre functions§                 Use completeness relations to express functions in the interval 0≤ x  ≤∞ as a sum of Laguerre polynomials §                 Use generating function to obtain the Rodrigues formula for Laguerre polynomials§                  Use Rodrigues formula to obtain Laguerre polynomials Laguerre polynomials Associated Laguerre functions Read: Sec 13.2Exercises 13.2.1, 13.2.3, 13.2.6, 13.2.7 31(9-30) §                 Prove orthogonality of Chebyshev functions§                 Use completeness relations to express functions in the interval -1≤ x  ≤1 as a sum of Chebyshev polynomials §                 Use generating function to obtain the Rodrigues formula for Chebyshev polynomials§                 Use Rodrigues formula to obtain Chebyshev polynomials Chebyshev Polynomials Read Sec 13.3Exercises 13.3.1, 13.3.3, 13.3.5 32(10-5) §                 Reduce any linear second order differential equation with three regular singularities into the hypergeometric equation §                 Express some orthogonal special functions in terms of hypergeometric fuctions Hypergeometric Functions Read Sec 13.4Exercises 13.4.6, 13.4.7, 13.4.8 33(10-7) §                 Reduce any linear second order differential equation with one regular singularity and one irregular singularity into the confluent hypergeometric equation §                 Express some special functions in terms of confluent hypergeometric functions Confluent Hypergeometric Functions Read Sec 13.5Exercises 13.5.4,  13.5.6, 13.5.11, 13.5.13, 13.5.14 6th Long Exam October 15, 2011 (Sat) 1pm The pace is very fast. When I first learned these topics, it took me more than two years of work, mainly because I had no one to look over my work. I used Churchill's book, Complex Variables with Applications, but  it's not enough to give justice to the course description. And I need to follow the course description because of university rules. If asked, I would be the first to agree that it's an unreasonable amount of material, especially for a course that meets three hours a week. Cambridge University, for example offers a Methods of Mathematical Physics Course. One set of notes I found includes all the complex variable material, and removes most of the special functions. The only special functions that do show up are the hypergeometric and confluent hypergeometric functions, and Gamma and Beta. No discussion of Fourier series or Fourier integrals! I am, however, stuck with the course description. I think that there ought to be changes made, but it has to go up to the university council. The only reasonable way of covering this material is to give them lots of work to do at home, and make extensive use of consultation hours. I find it difficult to prepare conventional lecture notes. My main objection to conventional lecture notes is that reading the lecture notes is a more passive activity compared to working with pen in hand to prove the theorems or solve the problems. So instead of lecture notes, I will prepare reading guides. The reading guides are a series of tasks that one should do while reading the text. One of the objections to Arfken is how easy it is to get lost. The way to avoid it is to divide the section into parts, and as soon as one reads the subparts, one should work on a problem or two in Arfken. I've prepared the reading guide so that when my students actually follow the guide, they will be able to construct a decent set of notes, and at the same time, solve the problems I've listed on the syllabus. I hope that the reading guide makes it easier to read Arfken's book. My own method of reading Arfken (since I had no guide before) was to attempt solving the problems at the end of the section before reading the section.  But that takes more time, since I could not separate the more important problems from the ones of secondary interest. Even if my students solve all of the problems I've listed, it's still a fraction of what I've actually done on my own. I hand out the reading guides a week or so before the lecture class, and I expect my students to use the reading guide to prepare for the coming class. This means I will not need to discuss everything; instead, I could concentrate on the more difficult parts. My students are graduate students with back subjects-- they had their undergraduate courses elsewhere, and are in need of remediation. They had no complex methods courses during their undergraduate days. Since they're graduate students, they have, at most, 9 hours of classwork every week. I hope that all of them are full-time students; the pace we set will be demanding. But if they do the necessary work, they should end the semester with an unfair advantage over their classmates in other physics courses.

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https://www.physicsforums.com/threads/parametric-equations-of-line-in-3d-space.366537/

# Homework Help: Parametric equations of line in 3D space 1. Jan 1, 2010 ### player1_1_1 Hello, sorry for my Englich:D 1. The problem statement, all variables and given/known data I must count a line integral on the lenght which lies on the line which is defined by equations: $$\begin{cases}x^2+y^2+z^2=R^2\\ \left(x-\frac{R}{2}\right)^2+y^2=\left(\frac{R}{2}\right)^2\end{cases}$$ it is a column which is cutting a sphere 3. The attempt at a solution I tried to assumption that $$x=t$$ and then depending on this find other functions $$y(t),z(t)$$, but the equations are not easy. I think I can find something with trigonometric functions sinus cosinus, but I am not sure; my question is then how I can easy and quickly find parametric equations of line which is make by two planes (in this case sphere and column) cutting each another? thanks for help! Last edited: Jan 1, 2010 2. Jan 1, 2010 ### tiny-tim Hello player1_1_1! (have a theta: θ and try using the X2 tag just above the Reply box ) It's much better to have a single-valued parameter, which x isn't. So try using the angle (from the axis of the cylinder) as the parameter. (btw, we say "sine" and "cosine", and we usually say "curve" for a line that isn't straight … even though we say "line integral" for any curve! ) Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

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http://scitation.aip.org/content/aip/journal/apl/85/17/10.1063/1.1812368

• journal/journal.article • aip/apl • /content/aip/journal/apl/85/17/10.1063/1.1812368 • apl.aip.org 1887 No data available. No metrics data to plot. The attempt to plot a graph for these metrics has failed. Influence of the gate dielectric on the mobility of rubrene single-crystal field-effect transistors USD 10.1063/1.1812368 View Affiliations Hide Affiliations Affiliations: 1 Kavli Institute of Nanoscience, Delft, University of Technology, Lorentzweg 1, 2628 CJ Delft The Netherlands a) Electronic mail: [email protected] Appl. Phys. Lett. 85, 3899 (2004) /content/aip/journal/apl/85/17/10.1063/1.1812368 http://aip.metastore.ingenta.com/content/aip/journal/apl/85/17/10.1063/1.1812368 View: Figures Figures FIG. 1. (Color online) Two rubrene single-crystal FETs fabricated by means of electrostatic bonding on (a) and (b). In (a) the electrodes have been defined by evaporation through a shadow mask, whereas in (b) photolithography and lift-off were used. In both (a) and (b) the bar is long. FIG. 2. Source–drain current vs source–drain voltage measured at different gate voltages for a device fabricated on , with an channel length and channel width. The inset shows similar data for a FET fabricated using parylene . FIG. 3. (a) Mobility vs gate voltage for a device on , measured at different values of source–drain voltage (, , , , , respectively), obtained using Eq. (1). Note that in the linear regime does not depend on and [the apparent peak at low values is an artifact originating from the use of Eq. (1) outside the linear regime]. (b) curves as measured for the four different gate insulators. For device based on parylene , the suppression of contact effects often requires the use of a rather large value (and thus , to remain in the linear regime). FIG. 4. Decrease of the mobility with increasing , as observed in rubrene single-crystal FETs with different gate insulators. The bars give a measure of the spread in mobility values. Inset: when plotted on a log–log scale, the available data show a linear dependence with slope (i.e., the variation in is proportional to ). /content/aip/journal/apl/85/17/10.1063/1.1812368 2004-10-29 2014-04-19 Article content/aip/journal/apl Journal 5 3 Most cited this month More Less This is a required field

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https://infoscience.epfl.ch/record/177050

## Overview of toroidal momentum transport Toroidal momentum transport mechanisms are reviewed and put in a broader perspective. The generation of a finite momentum flux is closely related to the breaking of symmetry (parity) along the field. The symmetry argument allows for the systematic identification of possible transport mechanisms. Those that appear to lowest order in the normalized Larmor radius (the diagonal part, Coriolis pinch, E × B shearing, particle flux, and up-down asymmetric equilibria) are reasonably well understood. At higher order, expected to be of importance in the plasma edge, the theory is still under development. © 2011 IAEA, Vienna. Published in: Nuclear Fusion, 51, 9, 094027 Year: 2011 Laboratories:

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http://math.stackexchange.com/questions/144592/value-of-a-fraction

# Value of a fraction It it true that is ${a^2+c^2\over b^2+d^2}=1$ for $ad-bc=1$? I tried substituting in $a={1-bc\over d}$ but it is still a mess. (How do you ask Wolfram Alpha a question like this where we ask it to calculate something with an imposed condition?) - Take $a = 2$ and $b = c = d = 1$. – TMM May 13 '12 at 14:08 What is true is that $(a^2+c^2)(b^2+d^2)=(ad-bc)^2+(ab+cd)^2$. – anon May 13 '12 at 14:15 It's not true. Try it for $a = 2$ and $b = c = d = 1$: $$ad - bc = 2\times1 - 1\times1 = 1$$ $$\frac{a^2+c^2}{b^2+d^2} = \frac{2^2+1^2}{1^2+1^2} = \frac{5}{2} \ne 1$$ As for WA, you can use the FullSimplify function to simplify an expression given some assumptions. Here is an example. Or even simpler: $b=c=0$, so $a={1 \over d}$ and ${{a^2+c^2}\over {b^2+d^2}}={{a^2} \over {d^2}} = {1 \over {d^4}}$ – David Lewis May 13 '12 at 15:24

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https://www.physicsforums.com/threads/stupid-series-i-cant-seem-to-find.292538/

# Stupid series I can't seem to find 1. Feb 15, 2009 ### Nick89 1. The problem statement, all variables and given/known data I need a way to write the following series in a sum: $$\frac{1}{2} + \frac{1}{4} + \frac{1}{32} + \frac{1}{64} + \frac{1}{1024} + \frac{1}{2048} + ...$$ If you look closely you can see a 2^n pattern in there: $$\frac{1}{2} + \frac{1}{4} + \frac{0}{8} + \frac{0}{16} + \frac{1}{32} + \frac{1}{64} + \frac{0}{128} + \frac{0}{256} + \frac{1}{512} + \frac{1}{1024} + ...$$ The denominator obviously follows $$2^n$$, but the numerator goes 110011001100... I can't seem to find a function that will allow me to put this in a single sum: $$\sum_{n=1}^{\infty} ....$$ (The sum should converge to 0.8, right?) 3. The attempt at a solution I tried using the mod operator to determine if n was even or odd, something like this: $$\sum_{n=1}^{\infty} \frac{n \mod 2}{2^n} + \frac{n \mod 2}{2^{n+1}}$$ This doesn't work, because the terms that are discarded when n is even are not discarded the next 'run' when (n+1) is even... Dunno how to explain this properly, but if you calculate it manually it does this: $$\frac{1}{2} + \frac{1}{4} + \frac{0}{4} + \frac{0}{8} + \frac{1}{8} + \frac{1}{16} + ...$$ So if you discard the 0/ ... terms you are just left with the usual 2^(-n) sum... I have a feeling I'm close, but I can't figure it out :S Last edited: Feb 15, 2009 2. Feb 15, 2009 ### epenguin If the missing bits were not missing it would add up to 1. Now you can put each missing bit into a relation with a bit that is there. They are a certain fraction of terms that are there. So their sum is that proportion of 1 IYSWIM. Do a diagram and this will look plausible. 3. Feb 15, 2009 ### Nick89 I can't seem to understand what you mean exactly... By missing bits you mean the 0/x parts? If so, then half of the terms as missing, so to your logic that would mean the sum would converge to 0.5...? (Whatr does IYSWIM mean?) But I think it should converge to 0.8 (I said 8 before but that was a mistake, I edited it now). 4. Feb 15, 2009 ### Dick If it's the pattern you suggest, (1/2+1/4)*(1/16)=(1/32+1/64). (1/32+1/64)*(1/16)=(1/512+1/1024). It's a geometric series with a common ratio of 1/16. But this doesn't quite fit with the series you quote in the first line (the next term after 1/64 is 1/1024, not 1/512). Is that a typo? 5. Feb 15, 2009 ### epenguin IYSWIM means 'if you see what I mean'. Just compare each pair of 'absent' terms with the 'present' pair preceding it. 6. Feb 15, 2009 ### Nick89 It was indeed a typo. The correct form is right underneath, where 1/512 is present. Alright, that makes a bit more sense. I still fail to see how I can now create a single sum for this series though... If I have a bit more time (maybe later tonight) I'll try to figure it out :) Thanks so far. Similar Discussions: Stupid series I can't seem to find

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https://www.physicsforums.com/threads/moment-about-the-origin.397980/

# Homework Help: Moment about the origin 1. Apr 24, 2010 ### Angello90 1. The problem statement, all variables and given/known data Find the moment about the origin of a vector of magnitude 100 units acting from A=(2, 0, 4) to B=(5, 1, 1). What is the moment about the point C=(−1, 2, 1)? 2. Relevant equations M= r x F 3. The attempt at a solution I did the quesion, but I'm not sure if it's correct. Can anyone check it for me? #### Attached Files: • ###### moment.jpg File size: 11 KB Views: 443 2. Apr 24, 2010 ### HallsofIvy The "moment of a vector about a point" is the magnitude of the vector times the length of a line segment from the given point perpendicular to the line of the vector. Since the vector acts "from (2, 0, 4) to (5, 1, 1)", its line of direction is given by x= 3t+ 2, y= t, z= -3t+ 4. You need to find the distance from (0, 0, 0) to that line and from (-1, 2, 1) to that line . Then multiply those by 100. 3. Apr 24, 2010 ### Angello90 So I need to find |OL| where O = (0, 0, 0) and L = (2+3t, t, 4-3t), thus Making a moment of Simillary to Q = (-1, 2, 1) #### Attached Files: File size: 5.4 KB Views: 416 File size: 4.7 KB Views: 428 • ###### QL Mql.jpg File size: 11.9 KB Views: 451 4. Apr 26, 2010 ### Angello90 Ahhh come on guys help me out here! Any hints?

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https://algebra-calculators.com/what-do-you-mean-by-rational-irrational-number-in-math-2/

# What do you mean by Rational & Irrational Number in math? An online rational irrational number definition A rational number is a number that can be written as a ratio. Rational numbers can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. The number 10 is a rational number because it can be written as the fraction 10/1. ## Irrational Numbers All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point. The number 22/7 is a irrational number.. ### Math Number Symbols R – Real Numbers Z – Integers N – Natural Numbers Q – Rational Numbers P – Irrational Numbers

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https://www.quantumdiaries.org/2012/07/03/higgs-seminar-live-blog-from-ichep/

## View Blog | Read Bio ### Higgs seminar live blog from ICHEP Good afternoon from the Melbourne Convention Centre, where we are all eagerly awaiting the start of the Higgs seminar that will be broadcast from CERN. I’ll be updating this post as we go along, so please stay with me and our other Quantum Diaries bloggers. 17:00: That’s a wrap here! I’ll try to post again soon-ish with some more thoughtful follow-up. Thanks everyone for joining us! 18:58: Not that Higgs is the only one who helped develop this theory…the others are getting their time to make congratulations too. 18:56: A big round of applause here and at CERN for the famous Peter Higgs. 18:52: No questions here! Rolf is right, we want to go to the reception! 18:48: Now questions. I’ll try to transcribe the ones that are coming from this room. 18:47: We’re watching the standing ovation at CERN over the video. Again, so nice to see lots of young people in the room. Then again, maybe they were the ones with the endurance to camp out in front of the auditorium! 18:44: Rolf summarizes by saying that this is a global effort and a global success. It’s only possible because of the extraordinary performance of the accelerators, experiments and computing grid. This is an observation of a new particle consistent of a Higgs boson…but we don’t know which one. It’s an historic milestone, but it is only the beginning, with global implications for the future. 18:41: The conclusion from ATLAS — an excess of events at about 126.5 GeV at 5.0 sigma significance. Fitted signal strength is 1.2 +- 0.3 of the SM expectation. Fabiola says that this is a very lucky mass to have, since it is easy to explore at the LHC in many decay channels. 18:39: Fabiola shows how the signal strength has evolved over time, in the past year — it is really quite striking how much we have advanced! 18:37: ATLAS overall signal strength is a bit higher than SM expectation for 125 GeV Higgs, whereas CMS was a little lower. 18:34: So here comes the combination of the new 2012 analyses with the 2011 results: another 5 sigma excess, and more applause! 18:30: ZZ also shows an excess in the 125 GeV range, 3.4 sigma here, when you’d expect 2.6 sigma from a standard-model Higgs. 18:28: Hmm, the standard-model ZZ rate is coming in bigger than expected. But it has now impact on the low-mass Higgs range. 18:23: Here comes ZZ, with 20-30% increase in sensitivity since December. 18:19: And now the results from the photons channel. Here too, an excess, at the 3.6 sigma level, after accounting for the look-elsewhere effect. 4.5 sigma local. 18:14: Thorough discussion of mass resolution in the photon channel, and the fact that it is independent of pileup. This is quite important, as the pileup is only going to get larger as the LHC luminosity increases. 18:10: Fabiola reviews what the changes are since December, here we go….. 18:09: Fermilab has a press release out too, http://www.fnal.gov/pub/presspass/press_releases/2012/Higgs-Search-LHC-20120704.html 18:07: The CERN press release is out! http://press.web.cern.ch/press/pressreleases/Releases2012/PR17.12E.html 18:04: I’ll admit that I’m finding less to say at the moment, as Fabiola is now covering similar ground to CMS in the preparation of the data, understanding of the detector etc. But it is still all important, of course! 17:57: I think I caught that ATLAS will show 5.9 fb-1 of data? More than CMS has certified. 17:55: Fabiola will show results from 2012 data in the ZZ and gamma gamma channels, but the other channels only for 2011 data. The 2012 results in those channels (which do have less mass resolution) aren’t quite ready yet. 17:53: Rolf congratulates everyone…and now let’s see what ATLAS has! 17:50: We have observed a new boson with a mass of 125.3 +- 0.6 GeV at 4.9 sigma significance. Enthusiastic applause heard on two continents. And then Joe acknowledges all of the theorists, machine physicists, and CMS experimenters who who have gotten us to this point. 17:49: Branching ratios are self-consistent across the decay channels (even with that tau tau result), but Joe emphasizes that it is early yet. 17:48: OK, it looks like there is something there, but is it consistent with a standard-model Higgs? Now we’ll get a first look at that. 17:44: Technical improvements in the tau tau channel have helped improve its sensitivity in the Higgs search. But uh oh, no evidence for a Higgs in that channel, and indeed you nearly exclude a 125 GeV SM Higgs here! It’s low statistics yet, we’ll need a lot more data to understand this. 17:41: Now on to the bb and tau tau channels. Not as much sensitivity to them, but they are important, as a) they actually have bigger branching fractions at ~125 GeV than the WW/ZZ/gammagamma, and b) they are fermions whereas all the others are bosons. 17:38: Unfortunately, the seminar is not over yet — let’s see if that 5.0 stands up with three more decay channels to look at. 17:37: The ZZ and gamma gamma joint significance is 5.0 standard deviations…Joe says that, and it gets applause! 5.0 is considered the threshold for a discovery…. 17:33: Joe is now showing how event by event, we can compare the angular separations of the leptons to what would be expected from a Higgs signal or the ZZ background. I’ve always loved that approach, really gets at the physics. 17:30: Now the ZZ mode, with four leptons in the final state — this channel is about as clean as you can get, but very rare, so you need to be super-efficient in selecting the events. 17:28: Slide 43 — A mass bump! Thank goodness that there is a real mass bump. I’ve always said, if we make a discovery that is observed on a plot that runs from 0 to 1 with a slight enhancement at 1, I’ll jump out the window. Maybe I’m safe. 17:22: Now we’re getting into the meat of the Higgs search, with the search for a decay to two photons. CMS paid a lot of money for the lead tungstate crystals that give very good photon energy resolution. 17:17: In fact, Joe is carefully going through all the low-level ingredients that go into these analyses. It’s a lot of work, and that’s why we need thousands of people to get this science done. 17:15: Joe takes a moment to point out the hard work done by the CMS software and computing teams. Yes, we’ve got a fabulous detector, but we can’t get this done without S&C…so I do appreciate the shout-out! 17:11: CMS will show 5.2 fb-1 of certified data today. And again, I’m curious to see how much data ATLAS will show! 17:09: CMS will be showing results from five Higgs decay modes — WW, ZZ, photons, bb, taus. I’m curious to see how many of those ATLAS also has. 17:06: Joe Incandela, for CMS, also starts with standup jokes. And then, on to a discussion of the experimental support for the standard model…except that we haven’t seen the Higgs. Yet. 17:03: Rolf notes that ICHEP is opening with a talk from a different continent — a symbol of how well we collaborate across the world. Then he starts with a standup routine…. 17:02: Big round of applause here when Rolf Heuer greets ICHEP! 17:00: Here we go! Watching a very quiet CERN auditorium on video…. 16:58: On the video feed from CERN, we can see a lot of young people. I’m glad they were able to get seats, as they are the ones who really make these experiments go. 16:54: Everyone in the hall just laughed at the video feed, which showed someone holding a “Ciao Mamma!” sign. 16:51: Geoff Taylor, the chair of ICHEP2012, is telling us about the timetable. The talks will be running longer than we originally envisioned (45 minutes each), followed by 30 minutes of questions. (We will be able to ask questions from here to CERN.) Hmm, hope everyone can hold out for the reception, which will be at 7 PM now! 16:47: I should say that as a member of CMS, I know the CMS results. Thus, I’m deeply curious about the ATLAS results. Will they agree with CMS or contradict? I’ve been hearing rumors, of course, but it will be interesting to see if they are true. 16:44: While we’re waiting for things to start, let me wish a happy Independence Day to everyone in the US, and also a happy wedding anniversary to my parents and a happy birthday to my colleague Aaron Dominguez, who is watching the proceedings from CERN Filtration Plant conference room. 16:42: I’m now sitting in the auditorium where we will be hearing the talk. Sitting to my left is Joel Butler, the manager of the US CMS operations program (I am also in the program management, as deputy manager of software and computing), and to my right is a reporter from the Australian Associated Press. On the big screen above the stage we can see the video feed from the CERN auditorium, and another panel which I presume will be the slides. 16:24: Just spoke to Pier Oddone and Young-Kee Kim, the director and deputy director of Fermilab. Yes, all of the spotlight is on CERN right now, but the Tevatron experiments have been very important for getting the Higgs search started, and it is also quite possible that CMS and ATLAS will not be able to beat CDF/D0 on the Higgs to bb decay mode anytime soon. 16:16: The registration line is still very long! I have been walking along it and chatting with friends in the hope of entertaining them a little. We’ll see if we can get everyone through in time for the seminar…. 15:46: I’ve now picked up my registration materials. There is a long line at the registration desk — no one wants to be late for the start. Right now I’m at a table in the foyer with some of the CMS leadership: Greg Landsberg, the CMS physics coordinator; Chris Hill, a CMS deputy coordinator, and Christoph Paus, one of the leaders of the CMS Higgs group. On the way in, I discovered that the Melbourne Boat Show is also happening in the convention centre. I wonder if we have any buyers at ICHEP…. Tags: ,

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https://hal.inria.fr/hal-01657018

# One-Time Nondeterministic Computations Abstract : We introduce the concept of one-time nondeterminism as a new kind of limited nondeterminism for finite state machines and pushdown automata. Roughly speaking, one-time nondeterminism means that at the outset the automaton is nondeterministic, but whenever it performs a guess, this guess is fixed for the rest of the computation. We characterize the computational power of one-time nondeterministic finite automata (OTNFAs) and one-time nondeterministic pushdown devices. Moreover, we study the descriptional complexity of these machines. For instance, we show that for an n-state OTNFA with a sole nondeterministic state, that is nondeterministic for only one input symbol, $(n+1)^n$ states are sufficient and necessary in the worst case for an equivalent deterministic finite automaton. In case of pushdown automata, the conversion of a nondeterministic to a one-time nondeterministic as well as the conversion of a one-time nondeterministic to a deterministic one turn out to be non-recursive, that is, the trade-offs in size cannot be bounded by any recursive function. Type de document : Communication dans un congrès Giovanni Pighizzini; Cezar Câmpeanu. 19th International Conference on Descriptional Complexity of Formal Systems (DCFS), Jul 2017, Milano, Italy. Springer International Publishing, Lecture Notes in Computer Science, LNCS-10316, pp.177-188, 2017, Descriptional Complexity of Formal Systems. 〈10.1007/978-3-319-60252-3_14〉 Domaine : https://hal.inria.fr/hal-01657018 Contributeur : Hal Ifip <> Soumis le : mercredi 6 décembre 2017 - 11:44:52 Dernière modification le : lundi 26 février 2018 - 13:40:02 ### Fichier ##### Accès restreint Fichier visible le : 2020-01-01 Connectez-vous pour demander l'accès au fichier ### Citation Markus Holzer, Martin Kutrib. One-Time Nondeterministic Computations. Giovanni Pighizzini; Cezar Câmpeanu. 19th International Conference on Descriptional Complexity of Formal Systems (DCFS), Jul 2017, Milano, Italy. Springer International Publishing, Lecture Notes in Computer Science, LNCS-10316, pp.177-188, 2017, Descriptional Complexity of Formal Systems. 〈10.1007/978-3-319-60252-3_14〉. 〈hal-01657018〉 ### Métriques Consultations de la notice

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https://www.physicsforums.com/threads/calculate-rpm-given-the-force-of-a-torsion-spring.957066/

# Calculate RPM given the force of a torsion spring • Start date • Tags • #1 2 0 ## Homework Statement I've got a flywheel of Inertia = 0.0019302kg/m^2 (found via solidworks), when a torsion spring is released, a force of 10N acts on the wheel via astring attached 0.065m above and 0.0325m to the right of the wheel's axis at an angle of 40 degrees. ## Homework Equations What is the flywheel's RPM (or rad/s or Hz)? ## The Attempt at a Solution Torque produced = FhDv + FvDh = (10*sin(40))0.0325 + (10*cos(40))0.065 = 0.667Nm Torque = Ia a = Torque/I = 0.667/0.0019302 Am i lacking a length of time this torque is applied for? I struggle to se where to go from here[/B] Related Introductory Physics Homework Help News on Phys.org • #2 nrqed Homework Helper Gold Member 3,603 204 ## Homework Statement I've got a flywheel of Inertia = 0.0019302kg/m^2 (found via solidworks), when a torsion spring is released, a force of 10N acts on the wheel via astring attached 0.065m above and 0.0325m to the right of the wheel's axis at an angle of 40 degrees. ## Homework Equations What is the flywheel's RPM (or rad/s or Hz)? ## The Attempt at a Solution Torque produced = FhDv + FvDh = (10*sin(40))0.0325 + (10*cos(40))0.065 = 0.667Nm Torque = Ia a = Torque/I = 0.667/0.0019302 Am i lacking a length of time this torque is applied for? I struggle to se where to go from here[/B] I di dont check your numbers but the approach looks good . Note that $\alpha$ is in rad/s^2. To get to your question, I am puzzled too. It is not possible to calculate an RPM without more information. For example the amount of time it was applied, as you said (and then we would need to know if the direction of the force changes as the wheel rotates, and if so how it does). There is really no other information provided? • #3 2 0 This is actualy the analysis of a project i've made, so yes there is no other information i can add to it. I'm thinking i'll time the flywheels motion after activation and just count the revolutions... i was just hoping there was a theory based way to calculate it! thanks for the reply • #4 haruspex Homework Helper Gold Member 32,724 5,030 This is actualy the analysis of a project i've made, so yes there is no other information i can add to it. I'm thinking i'll time the flywheels motion after activation and just count the revolutions... i was just hoping there was a theory based way to calculate it! thanks for the reply Since it is a torsion spring the torque will not be constant. In principle you have an SHM oscillator. The flywheel speed will be maximised each time the torsion spring is at its relaxed position. So what you need to know is the energy initially stored in the spring. But that is ignoring practical considerations of friction and drag. • Last Post Replies 1 Views 2K • Last Post Replies 11 Views 5K • Last Post Replies 1 Views 3K • Last Post Replies 6 Views 1K • Last Post Replies 3 Views 5K • Last Post Replies 0 Views 3K • Last Post Replies 4 Views 2K • Last Post Replies 10 Views 25K • Last Post Replies 9 Views 13K • Last Post Replies 2 Views 2K

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https://www.physicsforums.com/threads/electromagnetic-waves.531529/

# Electromagnetic Waves • Start date • #1 2 0 Hello, This is not homework, as I'm not in school :) I was wondering if there are any standard equations for electromagnetic waves interfering with each other. For instance, can I change the frequency of a wave by hitting it with another wave of a different frequency? I can find info on sound waves etc.. but I was wondering if this was any different. The question I want to answer for myself is: "Can I get one lower frequency electromagnetic wave, by combining two or more higher frequency electromagnetic waves?" Thank you :) • Last Post Replies 1 Views 2K • Last Post Replies 2 Views 2K • Last Post Replies 1 Views 1K • Last Post Replies 11 Views 1K • Last Post Replies 1 Views 2K • Last Post Replies 3 Views 2K • Last Post Replies 1 Views 1K • Last Post Replies 21 Views 1K • Last Post Replies 37 Views 6K • Last Post Replies 9 Views 2K

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http://mathhelpforum.com/advanced-algebra/50099-isomorphic-groups.html

1. ## Isomorphic groups Let f: G --> H be an isomorphism, G and H are groups. I've already showed that for every x in G, |f(x)| = |x|. (|x| = order of x, the smallest positive integer such that $x^n$ is the identity element) How to show that any two isomorphic groups have the same number of elements of n, where n is any positive integer?

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https://www.britannica.com/science/celestial-mechanics-physics/The-three-body-problem

## The three-body problem The inclusion of solar perturbations of the motion of the Moon results in a “three-body problem” (Earth-Moon-Sun), which is the simplest complication of the completely solvable two-body problem discussed above. When Earth, the Moon, and the Sun are considered to be point masses, this particular three-body problem is called “the main problem of the lunar theory,” which has been studied extensively with a variety of methods beginning with Newton. Although the three-body problem has no complete analytic solution in closed form, various series solutions by successive approximations achieve such accuracy that complete theories of the lunar motion must include the effects of the nonspherical mass distributions of both Earth and the Moon as well as the effects of the planets if the precision of the predicted positions is to approach that of the observations. Most of the schemes for the main problem are partially numerical and therefore apply only to the lunar motion. An exception is the completely analytic work of the French astronomer Charles-Eugène Delaunay (1816–72), who exploited and developed the most elegant techniques of classical mechanics pioneered by his contemporary, the Irish astronomer and mathematician William R. Hamilton (1805–65). Delaunay could predict the position of the Moon to within its own diameter over a 20-year time span. Since his development was entirely analytic, the work was applicable to the motions of satellites about other planets where the series expansions converged much more rapidly than they did for the application to the lunar motion. Delaunay’s work on the lunar theory demonstrates some of the influence that celestial mechanics has had on the development of the techniques of classical mechanics. This close link between the development of classical mechanics and its application to celestial mechanics was probably no better demonstrated than in the work of the French mathematician Henri Poincaré (1854–1912). Poincaré, along with other great mathematicians such as George D. Birkhoff (1884–1944), Aurel Wintner (1903–58), and Andrey N. Kolmogorov (1903–87), placed celestial mechanics on a more sound mathematical basis and was less concerned with quantitatively accurate prediction of celestial body motion. Poincaré demonstrated that the series solutions in use in celestial mechanics for so long generally did not converge but that they could give accurate descriptions of the motion for significant periods of time in truncated form. The elaborate theoretical developments in celestial and classical mechanics have received more attention recently with the realization that a large class of motions are of an irregular or chaotic nature and require fundamentally different approaches for their description. ## The restricted three-body problem The simplest form of the three-body problem is called the restricted three-body problem, in which a particle of infinitesimal mass moves in the gravitational field of two massive bodies orbiting according to the exact solution of the two-body problem. (The particle with infinitesimal mass, sometimes called a massless particle, does not perturb the motions of the two massive bodies.) There is an enormous literature devoted to this problem, including both analytic and numerical developments. The analytic work was devoted mostly to the circular, planar restricted three-body problem, where all particles are confined to a plane and the two finite masses are in circular orbits around their centre of mass (a point on the line between the two masses that is closer to the more massive). Numerical developments allowed consideration of the more general problem. In the circular problem, the two finite masses are fixed in a coordinate system rotating at the orbital angular velocity, with the origin (axis of rotation) at the centre of mass of the two bodies. Lagrange showed that in this rotating frame there were five stationary points at which the massless particle would remain fixed if placed there. There are three such points lying on the line connecting the two finite masses: one between the masses and one outside each of the masses. The other two stationary points, called the triangular points, are located equidistant from the two finite masses at a distance equal to the finite mass separation. The two masses and the triangular stationary points are thus located at the vertices of equilateral triangles in the plane of the circular orbit. There is a constant of the motion in the rotating frame that leads to an equation relating the velocity of the massless particle in this frame to its position. For given values of this constant it is possible to construct curves in the plane on which the velocity vanishes. If such a zero-velocity curve is closed, the particle cannot escape from the interior of the closed zero-velocity curve if placed there with the constant of the motion equal to the value used to construct the curve. These zero-velocity curves can be used to show that the three collinear stationary points are all unstable in the sense that, if the particle is placed at one of these points, the slightest perturbation will cause it to move far away. The triangular points are stable if the ratio of the finite masses is less than 0.04, and the particle would execute small oscillations around one of the triangular points if it were pushed slightly away. Since the mass ratio of Jupiter to the Sun is about 0.001, the stability criterion is satisfied, and Lagrange predicted the presence of the Trojan asteroids at the triangular points of the Sun-Jupiter system 134 years before they were observed. Of course, the stability of the triangular points must also depend on the perturbations by any other bodies. Such perturbations are sufficiently small not to destabilize the Trojan asteroids. Single Trojan-like bodies have also been found orbiting at leading and trailing triangular points in the orbits of Neptune and of Saturn’s satellite Tethys, at the leading triangular point in the orbit of another Saturnian satellite, Dione, and at the trailing point in the orbit of Mars. ## Orbital resonances Astronomy and Space Quiz There are stable configurations in the restricted three-body problem that are not stationary in the rotating frame. If, for example, Jupiter and the Sun are the two massive bodies, these stable configurations occur when the mean motions of Jupiter and the small particle—here an asteroid—are near a ratio of small integers. The orbital mean motions are then said to be nearly commensurate, and an asteroid that is trapped near such a mean motion commensurability is said to be in an orbital resonance with Jupiter. For example, the Trojan asteroids librate (oscillate) around the 1:1 orbital resonance (i.e., the orbital period of Jupiter is in a 1:1 ratio with the orbital period of the Trojan asteroids); the asteroid Thule librates around the 4:3 orbital resonance; and several asteroids in the Hilda group librate around the 3:2 orbital resonance. There are several such stable orbital resonances among the satellites of the major planets and one involving Pluto and the planet Neptune. The analysis based on the restricted three-body problem cannot be used for the satellite resonances, however, except for the 4:3 resonance between Saturn’s satellites Titan and Hyperion, since the participants in the satellite resonances usually have comparable masses. Although the asteroid Griqua librates around the 2:1 resonance with Jupiter, and Alinda librates around the 3:1 resonance, the orbital commensurabilities 2:1, 7:3, 5:2, and 3:1 are characterized by an absence of asteroids in an otherwise rather highly populated, uniform distribution spanning all of the commensurabilities. These are the Kirkwood gaps in the distribution of asteroids, and the recent understanding of their creation and maintenance has introduced into celestial mechanics an entirely new concept of irregular, or chaotic, orbits in a system whose equations of motion are entirely deterministic. ## Chaotic orbits The French astronomer Michel Hénon and the American astronomer Carl Heiles discovered that when a system exhibiting periodic motion, such as a pendulum, is perturbed by an external force that is also periodic, some initial conditions lead to motions where the state of the system becomes essentially unpredictable (within some range of system states) at some time in the future, whereas initial conditions within some other set produce quasiperiodic or predictable behaviour. The unpredictable behaviour is called chaotic, and initial conditions that produce it are said to lie in a chaotic zone. If the chaotic zone is bounded, in the sense that only limited ranges of initial values of the variables describing the motion lead to chaotic behaviour, the uncertainty in the state of the system in the future is limited by the extent of the chaotic zone; that is, values of the variables in the distant future are completely uncertain only within those ranges of values within the chaotic zone. This complete uncertainty within the zone means the system will eventually come arbitrarily close to any set of values of the variables within the zone if given sufficient time. Chaotic orbits were first realized in the asteroid belt. A periodic term in the expansion of the disturbing function for a typical asteroid orbit becomes more important in influencing the motion of the asteroid if the frequency with which it changes sign is very small and its coefficient is relatively large. For asteroids orbiting near a mean motion commensurability with Jupiter, there are generally several terms in the disturbing function with large coefficients and small frequencies that are close but not identical. These “resonant” terms often dominate the perturbations of the asteroid motion so much that all the higher-frequency terms can be neglected in determining a first approximation to the perturbed motion. This neglect is equivalent to averaging the higher-frequency terms to zero; the low-frequency terms change only slightly during the averaging. If one of the frequencies vanishes on the average, the periodic term becomes nearly constant, or secular, and the asteroid is locked into an exact orbital resonance near the particular mean motion commensurability. The mean motions are not exactly commensurate in such a resonance, however, since the motion of the asteroid orbital node or perihelion is always involved (except for the 1:1 Trojan resonances). For example, for the 3:1 commensurability, the angle θ = λA - 3λJ + ϖA is the argument of one of the important periodic terms whose variation can vanish (zero frequency). Here λ = Ω + ω + l is the mean longitude, the subscripts A and J refer to the asteroid and Jupiter, respectively, and ϖ = Ω + ω is the longitude of perihelion (see Figure 2). Within resonance, the angle θ librates, or oscillates, around a constant value as would a pendulum around its equilibrium position at the bottom of its swing. The larger the amplitude of the equivalent pendulum, the larger its velocity at the bottom of its swing. If the velocity of the pendulum at the bottom of its swing, or, equivalently, the maximum rate of change of the angle θ, is sufficiently high, the pendulum will swing over the top of its support and be in a state of rotation instead of libration. The maximum value of the rate of change of θ for which θ remains an angle of libration (periodically reversing its variation) instead of one of rotation (increasing or decreasing monotonically) is defined as the half-width of the resonance. Another term with nearly zero frequency when the asteroid is near the 3:1 commensurability has the argument θ′ = λA - λJ + 2ϖJ. The substitution of the longitude of Jupiter’s perihelion for that of the asteroid means that the rates of change of θ and θ′ will be slightly different. As the resonances are not separated much in frequency, there may exist values of the mean motion of the asteroid where both θ and θ′ would be angles of libration if either resonance existed in the absence of the other. The resonances are said to overlap in this case, and the attempt by the system to librate simultaneously about both resonances for some initial conditions leads to chaotic orbital behaviour. The important characteristic of the chaotic zone for asteroid motion near a mean motion commensurability with Jupiter is that it includes a region where the asteroid’s orbital eccentricity is large. During the variation of the elements over the entire chaotic zone as time increases, large eccentricities must occasionally be reached. For asteroids near the 3:1 commensurability with Jupiter, the orbit then crosses that of Mars, whose gravitational interaction in a close encounter can remove the asteroid from the 3:1 zone. By numerically integrating many orbits whose initial conditions spanned the 3:1 Kirkwood gap region in the asteroid belt, Jack Wisdom, an American dynamicist who developed a powerful means of analyzing chaotic motions, found that the chaotic zone around this gap precisely matched the physical extent of the gap. There are no observable asteroids with orbits within the chaotic zone, but there are many just outside extremes of the zone. Other Kirkwood gaps can be similarly accounted for. The realization that orbits governed by Newton’s laws of motion and gravitation could have chaotic properties and that such properties could solve a long-standing problem in the celestial mechanics of the solar system is a major breakthrough in the subject. ## The n-body problem The general problem of n bodies, where n is greater than three, has been attacked vigorously with numerical techniques on powerful computers. Celestial mechanics in the solar system is ultimately an n-body problem, but the special configurations and relative smallness of the perturbations have allowed quite accurate descriptions of motions (valid for limited time periods) with various approximations and procedures without any attempt to solve the complete problem of n bodies. Examples are the restricted three-body problem to determine the effect of Jupiter’s perturbations of the asteroids and the use of successive approximations of series solutions to sequentially add the effects of smaller and smaller perturbations for the motion of the Moon. In the general n-body problem, all bodies have arbitrary masses, initial velocities, and positions; the bodies interact through Newton’s law of gravitation, and one attempts to determine the subsequent motion of all the bodies. Many numerical solutions for the motion of quite large numbers of gravitating particles have been successfully completed where the precise motion of individual particles is usually less important than the statistical behaviour of the group. ## Numerical solutions Numerical solutions of the exact equations of motion for n bodies can be formulated. Each body is subject to the gravitational attraction of all the others, and it may be subject to other forces as well. It is relatively easy to write the expression for the instantaneous acceleration (equation of motion) of each body if the position of all the other bodies is known, and expressions for all the other forces can be written (as they can for gravitational forces) in terms of the relative positions of the particles and other defining characteristics of the particle and its environment. Each particle is allowed to move under its instantaneous acceleration for a short time step. Its velocity and position are thereby changed, and the new values of the variables are used to calculate the acceleration for the next time step, and so forth. Of course, the real position and velocity of the particle after each time step will differ from the calculated values by errors of two types. One type results from the fact that the acceleration is not really constant over the time step, and the other from the rounding off or truncation of the numbers at every step of the calculation. The first type of error is decreased by taking shorter time steps. But this means more numerical operations must be carried out over a given span of time, and this increases the round-off error for a given precision of the numbers being carried in the calculation. The design of numerical algorithms, as well as the choice of precisions and step sizes that maximize the speed of the calculation while keeping the errors within reasonable bounds, is almost an art form developed by extensive experience and ingenuity. For example, a scheme exists for extrapolating the step size to zero in order to find the change in the variables over a relatively short time span, thereby minimizing the accumulation of error from this source. If the total energy of the system is theoretically conserved, its evaluation for values of the variables at the beginning and end of a calculation is a measure of the errors that have accumulated. The motion of the planets of the solar system over time scales approaching its 4.6-billion-year age is a classic n-body problem, where n = 9 with the Sun included. The question of whether or not the solar system is ultimately stable—whether the current configuration of the planets will be maintained indefinitely under their mutual perturbations, or whether one or another planet will eventually be lost from the system or otherwise have its orbit drastically altered—is a long-standing one that might someday be answered through numerical calculation. The interplay of orbital resonances and chaotic orbits discussed above can be investigated numerically, and this interplay may be crucial in determining the stability of the solar system. Already it appears that the parameters defining the orbits of several planets vary over narrow chaotic zones, but whether or not this chaos can lead to instability if given enough time is still uncertain. If accelerations are determined by summing all the pairwise interactions for the n particles, the computer time per time step increases as n2. Practical computations for the direct calculation of the interactions between all the particles are thereby limited to n < 10,000. Therefore, for larger values of n, schemes are used where a particle is assumed to move in the force field of the remaining particles approximated by that due to a continuum mass distribution, or a “tree structure” is used where the effects of nearby particles are considered individually while larger and larger groups of particles are considered collectively as their distances increase. These later schemes have the capability of calculating the evolution of a very large system of particles using a reasonable amount of computer time with reasonable approximation. Values of n near 100,000 have been used in calculations determining the evolution of galaxies of stars. Values of n in the billions have been used in calculations of galaxy formation in the early universe. Also, the consequences for distribution of stars when two galaxies closely approach one another or even collide has been determined. Even calculations of the n-body problem where n changes with time have been completed in the study of the accumulation of larger bodies from smaller bodies via collisions in the process of the formation of the planets. In all n-body calculations, very close approaches of two particles can result in accelerations so large and so rapidly changing that large errors are introduced or the calculation completely diverges. Accuracy can sometimes be maintained in such a close approach, but only at the expense of requiring very short time steps, which drastically slows the calculation. When n is small, as in some solar system calculations where two-body orbits still dominate, close approaches are sometimes handled by a change to a set of variables, usually involving the eccentric anomaly u, that vary much less rapidly during the encounter. In this process, called regularization, the encounter is traversed in less computer time while preserving reasonable accuracy. This process is impractical when n is large, so accelerations are usually artificially bounded on close approaches to prevent instabilities in the numerical calculation and to prevent slowing the calculation. For example, if several sets of particles were trapped in stable, close binary orbits, the very short time steps required to follow this rapid motion would bring the calculation to a virtual standstill, and such binary motion is not important in the overall evolution of, say, a galaxy of stars. ## Algebraic maps In numerical calculations for conservative systems with modest values of n over long time spans, such as those seeking a determination of the stability of the solar system, the direct solution of the differential equations governing the motions requires excessive time on any computer and accumulates excessive round-off error in the process. Excessive time also is required to explore thoroughly a complete range of orbital parameters in numerical experiments in order to determine the extent of chaotic zones in various configurations (e.g., those in the asteroid belt near orbital mean motion commensurabilities with Jupiter). A solution to this problem is the use of an algebraic map, which maps the space of system variables onto itself in such a way that the values of all the variables at one instant of time are expressed by algebraic relations in terms of the values of the variables at a fixed time in the past. The values at the next time step are determined by applying the same map to the values just obtained, and so on. The map is constructed by assuming that the motions of all the bodies are unperturbed for a given short time but are periodically “kicked” by the perturbing forces for only an instant. The continuous perturbations are thus replaced by periodic impulses. The values of the variables are “mapped” from one time step to the next by the fact that the unperturbed part of the motion is available from the exact solution of the two-body problem, and it is easy to solve the equations with all the perturbations over the short time of the impulse. Although this approximation does not produce exactly the same values of all the variables at some time in the future as those produced by a numerical solution of the differential equations starting with the same initial conditions, the qualitative behaviour is indistinguishable over long time periods. As computers can perform the algebraic calculations as much as 1,000 times faster than they can solve the corresponding differential equations, the computational time savings are enormous and problems otherwise impossible to explore become tractable. MEDIA FOR: celestial mechanics Previous Next Citation • MLA • APA • Harvard • Chicago Email You have successfully emailed this. Error when sending the email. Try again later. Edit Mode Celestial mechanics Physics Tips For Editing We welcome suggested improvements to any of our articles. You can make it easier for us to review and, hopefully, publish your contribution by keeping a few points in mind. 1. Encyclopædia Britannica articles are written in a neutral objective tone for a general audience. 2. You may find it helpful to search within the site to see how similar or related subjects are covered. 3. Any text you add should be original, not copied from other sources. 4. At the bottom of the article, feel free to list any sources that support your changes, so that we can fully understand their context. (Internet URLs are the best.) Your contribution may be further edited by our staff, and its publication is subject to our final approval. Unfortunately, our editorial approach may not be able to accommodate all contributions.

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https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=next&oldid=135844

# Difference between revisions of "2006 AIME II Problems/Problem 11" ## Problem A sequence is defined as follows and, for all positive integers Given that and find the remainder when is divided by 1000. ## Solution ### Solution 1 Define the sum as . Since , the sum will be: Thus , and are both given; the last four digits of their sum is , and half of that is . Therefore, the answer is . Solution by an anonymous user. ### Solution 2 (bash) Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms: Adding all the residues shows the sum is congruent to mod 1000. ~ I-_-I ### Solution 3 (some guessing involved)/"Engineer's Induction" All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given and , so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some such that . From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that , at least for the first few terms. From this, we have that . Solution by zeroman; clarified by srisainandan6

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http://encyclopedia.kids.net.au/page/li/Likelihood_principle

## Encyclopedia > Likelihood principle Article Content # Likelihood principle In statistical theory, the likelihood principle asserts that the information in any sample can be found, if at all, from the likelihood function, that function of unknown parameters[?] which specifies the probability of the sample observed on the basis of a known model, in terms of the model's parameters. Suppose, for example, that we have observed n independent flips of a coin which we regard as having a constant probability, p, of falling heads up. The likelihood function is then the product of n factors, each of which is either p or 1-p. If we observe x heads and n-x tails, then the likelihood function is [itex]L(p)\sim p^x(1-p)^{1-x}[/itex] i.e., proportional to the product. No multiplicative constant of C(N,X) is included because only the part of the probability which involves the parameter, p, is relevant (According to some accounts, the fact that only that part is relevant is what the likelihood principle says. For example, in one experiment the number of successes in 10 Bernoulli trials is observed; in another the number of trials needed to get four successes is observed. In either case, the outcome could be four successes in ten trials. The probabilities of that outcome are different in the two experiments, but as function s of p, the probability of success on each trial, they are proportional. The Likelihood principle says the same statistical inference about the value of p should be drawn in each case.) In particular, this principle suggests that it does not matter whether you started out planning to observe N trials or you just decided to stop on a whim. The issue of the likelihood principle is still controversial. A deeper discussion of the topic is available in the article about maximum likelihood. All Wikipedia text is available under the terms of the GNU Free Documentation License Search Encyclopedia Search over one million articles, find something about almost anything! Featured Article Academy Awards Foreign Language Film ... Deschamps[?], Arne Meerkamp von Embden[?], Claudie Ossard[?] producers - Jean-Pierre Jeunet director Son of the Bride[?] (El Hijo de la novia) (Argentina) - JEMPSA, ...

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https://www.groundai.com/project/characteristic-function-of-time-inhomogeneous-levy-driven-ornstein-uhlenbeck-processes/

Characteristic Function of Time-Inhomogeneous Lévy-Driven Ornstein-Uhlenbeck Processes # Characteristic Function of Time-Inhomogeneous Lévy-Driven Ornstein-Uhlenbeck Processes Frédéric Vrins Louvain School of Management (LSM) & Center for Operations Research and Econometrics (CORE) Université catholique de Louvain Chaussée de Binche 151, Office A.212, B-7000 Mons, Belgium [email protected] ###### Abstract Distributional properties -including Laplace transforms- of integrals of Markov processes received a lot of attention in the literature. In this paper, we complete existing results in several ways. First, we provide the analytical solution to the most general form of Gaussian processes (with non-stationary increments) solving a stochastic differential equation. We further derive the characteristic function of integrals of Lévy-processes and Lévy driven Ornstein-Uhlenbeck processes with time-inhomogeneous coefficients based on the characteristic exponent of the corresponding stochastic integral. This yields a two-dimensional integral which can be solved explicitly in a lot of cases. This applies to integrals of compound Poisson processes, whose characteristic function can then be obtained in a much easier way than using joint conditioning on jump times. Closed form expressions are given for gamma-distributed jump sizes as an example. ## 1 Introduction Integrals of Markov processes are popular stochastic processes as a results of their numerous applications. Birth, death, bird-death and catastrophe processes in particular received a lot of attention in the context of queuing and storage problems [20] as well as in biology [19] or robotics [17]. From a broad perspective, general properties of integrals of Markov processes have been derived, like the time-evolution of associated moments [13]. With regards to the Laplace transforms in specific, closed form expressions have been derived in [22] for continuous-time Markov chains taking value on the set of positive integers. With regards to diffusion processes, very general results have been obtained for affine, quadratic and geometric models. The corresponding results are derived from the transition probabilities and the property that the infinitesimal generator is dependent on the space variable only, i.e the stochastic differential equations (SDE) has time-homogeneous coefficients. Explicit formulas for many standard processes including the integrated Orntein-Uhlenbeck (OU), Square-Root and Jacobi diffusions are available in [1],[15]. The specific case of integral of geometric Brownian motion, closely linked to squared Bessel processes, has extensively be studied by Yor [25], and a collection of papers on the topic is available [26]. In spite of this extensive literature, some important results are still lacking for interesting stochastic processes with sound applications. This is for instance the case of the generalized OU process, that is OU processes with time-varying coefficients and Lévy driving process. Similarly, to the best of our knowledge, there is no such results for compound Poisson processes. In this paper, we are interested in the characteristic function (or equivalently, Laplace transform or moment-generating function -when it exists) of the path integral of the form Λs,t=∫tsλudu (1.1) where is a Markov process on a probability space . We assume the sigma-field contains all the information about up to time . The paper is organized as follows. Section 2 is devoted to depict financial applications that will stress the importance of having the characteristic function available in closed form. We shall point out some processes for which the integral is useful in this context. Associated characteristic functions will be derived in the remaining part of the paper. Most general results for Gaussian processes (not necessary with stationary increments) is addressed in Section 3. In Section 4, the corresponding results are obtained for Ornstein-Uhlenbeck processes with a background driving Lévy process (BDLP) which is not restricted to be a Brownian motion. Finally, Compound Poisson processes are investigated in Section 5 based on joint conditioning of jump times of the underlying Poisson process. Closed-form expressions are given when the jump sizes are Gamma-distributed. ## 2 Financial motivation The most obvious example of the use of integrals of stochastic processes in finance is the pricing of Asian options, in which the payoff depends on the time average of the underlying stock (or index) price. When the underlying stock follows a geometric Brownian motion, as in the standard Black-Scholes case, this leads to the study of integrals of Squared Bessel processes and explains why this specific case received so much attention (see e.g. [12][6]). In this context, Laplace transforms are interesting when explicit formula for the distributions are too involved or cannot be found. The corresponding distributions can be obtained by simple Fourier inversion, and all the moments are easily retreived by deriving the moment generating function. In some cases however, the explicit form of the characteristic function is very appealing in itself, as it direcly yields the calibration equation, as we now show. Let stands for the risk-neutral measure and for the risk-free short rate. Then, the time- price of a risk-free zero-coupon bond paying 1 unit of currency at time is B(t,T)=E[e−Λt,T|Ft] (2.1) Consequently, the parameters of any given short rate process can be calibrated at time from the given yield curve by making sure that the Laplace transform of satisfies B(0,T)=ϕΛT(1) (2.2) which is related to the characteristic function via . For tractability reasons, the most popular short-rate models are, by far, Gaussian or square-root diffusion processes with constant coefficients, possibly shifted by a deterministic function. For simple short-rate models like Vasicek (with Ornstein-Uhlenbeck dynamics, OU) or Cox-Ingersoll-Ross (with square root diffusion dynamics, SRD), the analytical expression of the Laplace transform is available; see e.g. [5],[8],[9],[10]. However, the corresponding expressions for the Hull-White model, which is the extension of the Vasicek model with time-varying coefficients, is not available. Expressions similar to 2.1 appear in credit risk modelling, in both reduced form and structural approaches (see e.g. [18],[4]). In reduced form (intensity) approaches, is modeled as the first jump of a Cox process where the stochastic intensity is given by a non-negative stochastic process : τ=inf{t:Nt>0} (2.3) Conditional on the path of the intensity , is a Poisson process, so that Q(τ>T)=Q(NT=0)=E[e−∫T0λsds]=E[e−ΛT]=ϕΛT(1) (2.4) The same form of survival probability is obtained in the so-called structural approach. In this setup, the default is modeled as the first passage time of an asset process below a liability threshold : τ=inf{t:Vt The process is typically either a Brownian motion, a geometric Brownian motion or a subordinated process. The computation of the law of the first-passage times can be avoided by working in a specific framework where the survival probability takes the same form as in (2.4). Let be a non-decreasing grounded positive process and model the credit event as the first time where reaches a random threshold uniformly distributed in . The passage time is almost surely unique, so that: τ=inf{t:e−Λt≤U}={t:e−Λt=U} (2.6) Then, the survival probability takes again the same form: Q(τ>T)=Q(e−ΛT>U)=E[Q(e−ΛT≥U)]=E[e−ΛT]=ϕΛT(1) (2.7) This suggests that instead of working with an intermediate intensity process, we could directly feed the default model with an non-decreasing Lévy process . This setup has been used in [14] in the context of CDO pricing but in a purely numerical setup. The analytical expressions of the calibration equations are derived in [23]. The implied copulas form the class of so-called Sibuya copulas, which properties are studied in [24]. This setup, however is not appropriate when one desires to simulate survival probability curves . Indeed, Q(τ>T|τ>t)=E[e−(ΛT−Λt)|Ft]=E[e−Λt,T] (2.8) If the process is Lévy, the increments are independent of the past and if they are stationary, . Therefore, with S(t,t+δ):=Q(τ>t+δ|τ>t)=E[e−Λδ]=ϕΛδ(1) (2.9) In other words, there is no memory effect. Producing different curves , depending on the path , can only be introduced by considering non-Lévy processes for . One way to introduce such dependency is by considering as the integral of a underlying process , as in the intensity process. ## 3 Gaussian processes As mentioned in the introduction, the expression of integrals of Gaussian processes are known in some specific cases, among which Vasicek (Ornstein-Uhlenbeck) is the most popular one. We generalise here previous results by working out the explicit form of the strong solutions for and given when is the solution to the most general SDE associated to Gaussian processes. It is very well-known that Gaussian process solving a stochastic differential equation (SDE) has dynamics of the form111Not all Gaussian processes solve a SDE; this is for instance the case of fractional Brownian motion. dλt=(α(t)−β(t)λt)dt+σ(t)dWt   , t≥s,  λs (3.1) where is a Brownian motion and and are deterministic integrable functions and the initial condition. When the solution exists, it is possible to derive explicitly the strong solution for and and obtain the corresponding characteristic function. We assume in this paper that the coefficients here are such that existence and uniqueness conditions are satisfied; see e.g. [16] or [21] for a detailed discussion. ###### Proposition 1 (General Gaussian Process). Consider eq.(3.1) and assume existence and uniqueness conditions are met. Suppose further that is known at time and define G(s,t):=e−∫tsβ(u)du , I(s,t):=∫tsα(u)G(u,s)du (3.2) J(s,t):=∫tsσ(u)G(u,s)dWu , K(s,t):=∫tsσ(s)G(s,u)du (3.3) Then, for any , the solution to eq.(3.1) conditional upon is λt=m(s,t)+G(s,t)J(s,t) (3.4) which is Normally distributed with mean and variance respectively given by m(s,t) := G(s,t)(λs+I(s,t)) (3.5) v(s,t) := G2(s,t)∫tsσ2(u)G2(u,s)du=∫tsσ2(u)G2(u,t)du (3.6) The integral is a shifted Ito integral Λs,t=M(s,t)+∫tsK(u,t)dWu (3.7) which is Normally distributed with mean and variance respectively given by M(s,t) := (t−s)λs+∫ts(t−u)(α(u)−β(u)G(s,u)(λs+I(s,u)))du (3.8) V(s,t) := ∫tsK2(u,t)du (3.9) ###### Proof. See Appendix A.∎ Since is Gaussian with known mean and variance, the following corollary is obvious. ###### Corollary 3.1 (Laplace transform of Λs,t). The Laplace transform of , conditional upon , is ϕΛs,t(x)=exp{−M(s,t)x+x2V(s,t)2} (3.10) where are given in Proposition 1. Let us apply this result to get the expression and distribution of in some particular cases. ###### Example 3.1 (Integrated rescaled Brownian motion). Consider the case where . Then, M(s,t)=(t−s)λs , K(s,t)=σ(t−s) , V(s,t)=σ23(t−s)3 (3.11) This expression can be found directly using the stochastic version of Fubini’s theorem on indicator functions. ###### Example 3.2 (Integrated OU process). The case where and are constant is widely used in mathematical finance. It corresponds to , and : M(s,t) = 1−e−β(t−s)βλs (3.12) K(s,t) = σβ(1−e−β(t−s)) (3.13) V(s,t) = (σ2β)2(2(t−s)+e−β(t−s)β(4−e−β(t−s))−6β) (3.14) The above examples correspond to special cases for which results exist (see e.g [5]). However, the above result allows to determine the Laplace transform of more general Gaussian processes, as shown in the examples below. ###### Example 3.3 (Prototypical swap exposure). Consider a Brownian bridge from to rescaled by a constant and shifted by the deterministic fontion . Such a process can be written as λt=(T−t)(γt+σ∫t01T−udWu)   , 0≤s≤t≤T (3.15) This process starts from 0, its expectation follows the curve and then goes back to as . This is a Gaussian process with , zero-mean and instantaneous volatility . Then M(s,T)=λtT−t2+γ6(T−t)3 , K(s,T)=σT−s2 , V(s,T)=σ212(T−s)3 (3.16) See Appendix B for details. ###### Example 3.4 (Deterministically subordinated process). Consider the integral where is a strictly increasing and continuous function. Then, Λs,t=∫θ(t)θ(s)f∘θ−1(u)θ′(u)Wudu (3.17) corresponds to the integral of where . Its differential takes the form (3.1) where , and instantaneous volatility . Consequently, provided that the existence conditions are met, is Normally distributed and the solution is given by (3.7) where the integration bounds are , and the parameters are defined above, and the characteristic function follows.222Observe that when is not strictly increasing (i.e. mereley non-decreasing) and/or discontinuous, the corresponding characteristic function can be recovered by splitting the integral into pieces where the inveres of exists, where is constant and at time points where jumps. ## 4 Lévy-driven Hull-White process In some cases, the characteristic function of the integrated process can be obtained by computing the characteristic function of a stochastic integral. This is in particular the case of generalized Lévy-driven OU processes (that is, where the random increments are controlled by a background driving Lévy process, BDLP), as we now show. The log-characteristic function of a Lévy process takes the form ψXt(x):=lnφXt(x)=tlnφX(x)=:tψX(x) (4.1) where is the characteristic exponent of the infinitely divisible distribution of  [7]. We can derive the log-characteristic function of the stochastic integral of a deterministic function with respect to a Lévy process as follows. Let where is integrable and is Lévy processes with triplet and is the density of the Lévy measure. Then ψX(x) = (4.2) ψYt(x) = lnE[eix∫t0σ(s)dXs] (4.3) ###### Proposition 2 (Characteristic exponent of a stochastic integral with respect to a Lévy process). Let be a Lévy process with characteristic exponent and define the semimartingale for some deterministic integrable function . Then, ψYs,t(x)=∫tsψX(σ(u)x)du (4.4) where . ###### Proof. See Appendix C The Lévy driven Hull-White process is defined as a Ornstein-Uhlenbeck process with time-varying coefficients and BDLP. ###### Proposition 3 (Characteristic exponent of integrated BDLP Hull-White process). Let be a Hull-White process driven by the Lévy process solution of the SDE dλt=(α(t)−β(t)λt)dt+σ(t)dXt (4.5) Then, setting as before, ψΛs,t(x)=ixM(s,t)+∫tsψX(x∫tuσ(u)G(u,v)dv)du (4.6) ###### Proof. Using the integration by parts technique used in the proof of Proposition 1, the solution is proven to be the same as in (3.7) where is replaced by : Λs,t=M(s,t)+∫tsK(u,t)dXu (4.7) Therefore, ψΛs,t(x)=ixM(s,t)+ψ∫tsK(u,t)dXu(x) (4.8) and the claim follows from Proposition 2 and the expression of in Proposition 1. ∎ Observe that SDE (4.5) is not the most general SDE with linear drift since the way jumps are handled corresponds to the special case where a jump of at of size triggers a jump of size at for . We could perfectly want to have a jump which does depend on in a non-linear way (see e.g. [3] and [2] for a more advanced discussion and conditions on the coefficients and the functional parameters of for existence and pathwise uniqueness). ###### Example 4.1 (OU driven by a Brownian Motion). We recover the characteristic function of the standard OU model driven by a Brownian motion from the strong solution Λs,t=M(s,t)+∫tsK(u,t)dWu (4.9) where . Since here , the Lévy triplet is and the characteristic exponent reduces to . So, ∫tsψX(K(u,t)x)du=−∫ts(xK(u,t))22du=−x22V(s,t) (4.10) Therefore, ψΛs,t(x)=ixM(s,t)−x22V(s,t) (4.11) in line with in eq. (3.10). Let us apply this method to determine the characteristic function of where is a non-Gaussian OU process driven by a gamma process . ###### Example 4.2 (OU driven by a gamma process). Consider the SDE dλt=−βλtdt+dγt (4.12) with non-negative solution and the integral is Λs,t=λs1−e−β(t−s)β+∫ts1−e−β(t−u)βdγu (4.13) The integrated process is a deterministc constant plus a stochastic integral of a deterministic function with respect to a Lévy process. Therefore, ψΛt(x)=ixλs1−e−β(t−s)β+∫tsψγ(1−e−β(t−u)βx)du (4.14) where is the characteristic exponent of the gamma distribution driving the jump sizes. The integral can be written in terms of the dilogarithmic function (t−s)ψγ(x/β)+αβ(Li2(v)−Li2(ve−β(t−s))) (4.15) where . A similar approach has been used in [11]333We are grateful to D. Madan for providing us with this reference. to evaluate a joint characteristic function in the -forward measure of a triple of processes which dynamics are given in the risk-neutral measure. The following result is a straight consequence of Proposition 3. ###### Corollary 4.1 (Characteristic exponent of an integrated Lévy process). Let be a Lévy process. Then, ψΛs,t(x)=ixλs(t−s)+∫tsψλ(x(t−u))du (4.16) ###### Example 4.3 (Stochastically subordinated Brownian motion). We are interested in the integral of that is, of the integral of a time-changed Brownian motion. Since for every Lévy process we have , φλt(x)=E[% eixλt]=E[(eixW)Xt]=E[eXtψW(x)]=E[eiXt(−iψW(x))]=φtX(−iψW(x)) (4.17) or equivalently in terms of the characteristic exponent, ψλ(x)=ψX(−iψW(x))=ψX(ix2/2) (4.18) In the case of a Gamma subordinator with parameters , the subordinated process is will-known to be a variance-gamma (Lévy) process. Indeed, has the same distribution as the difference of two independent gamma processes, say and , each with same parameters . This can be seen by noting from the above result that : ϕλt(x)=(2κ2κ+x2)αt=(√2κ√2κ+ix)αt(√2κ√2κ−ix)αt (4.19) The log-characteristic function of the integral of is thus given by ψΛt(x) = ixγs(t−s)+ψ∫tsγudu(x)+ψ∫tsγudu(−x) (4.20) = ixγs(t−s)+∫tsψγ(x(t−u))du+∫tsψγ(x(u−t))du (4.21) and ∫tsψγ(x(t−u))du=κixψγ(x(t−s);κ,α)φγ(x(t−s);κ,1)−α(t−s)(1−lnκ) (4.22) ## 5 Compound Poisson processes In this section, we focus on the integral of jump processes where the jumps arrive according to a Poisson process and the size of the -th jump is given by the random variable which are mutually independent. We then specialize to the case of compound Poisson process where . We compute the Laplace transform for which the analytical expression can be found for some laws of . This can be done in several ways. We adopt here on a joint conditioning on the number of jumps of by time and the corresponding jump times . ### 5.1 Generalized Compound Process The Laplace transform of the integrated process where is a generalized compound Poisson process is obtained as an infinite sum of multidimensional integrals. ###### Proposition 4. Let ba a Poisson process with intensity and be a sequence of independent random variables. Then, the Laplace transform of the integrated process Λt=∫t0Nu∑i=1Xidu (5.1) is given by ϕΛt(x)=e−θt(1+∞∑n=1θn∫tt1=0ϕX1(x(t−t1))…∫ttn=tn−1ϕXn(x(t−tn))d→tn) (5.2) ###### Proof. The law of is that of a random variable, that is pNt(n)=(θt)ne−θtn!,  n∈{0,1,2,…} (5.3) Furthermore, the laws of the time arrival of the -th jump, , and the time elapsed between jumps and are known Δi+1:=Ti+1−Ti ∼ Exp(θ)=Gamma(1,θ) (5.4) Ti = Gamma(i,θ) (5.5) where we have set . Let us further note and . Then, due to the independence between the times separating two consecutive jumps, we compute the joint density of the first jump times : (5.6) where and is 1 if a non-decreasing sequence of nonnegative numbers and 0 otherwise. We can then compute the joint density of at using as follows: p→Tn,Nt(→tn,n) = (5.7) = (5.8) = (5.9) = (5.10) This expression depends on only via the indicator function . We can now compute the Laplace transform of by conditioning on : ϕΛt(x) = E[e−x∫t0∑Nui=1Xidu] (5.11) = pNt(0)+∞∑n=1E[e−x∑ni=1Xi(t−ti)|→Tn=→tn,Nt=n]p→Tn,Nt(→tn,n) (5.12) = pNt(0)+∞∑n=1∫∞t1=0…∫∞tn=0n∏i=1ϕXi(x(t−ti))p→T,Nt(→tn,n)d→tn (5.13) = e−θt+∞∑n=1θne−θt∫tt1=0…∫ttn=tn−1n∏i=1ϕXi(x(t−ti))d→tn (5.14) This form is appealing for two reasons. First, only the product of the characteristic functions of the jump size variables appear in the multiple integral as can be sent outside provided that we restrict the integration domain according to the indicator function . Second, this multiple integral can be written as ϕΛt(x)=e−θt+e−θt∞∑n=1θn∫tt1=0ϕX1(x(t−t1))…∫ttn=tn−1ϕXn(x(t−tn)):=InI1d→tn (5.15) Therefore, the multiple integral can be computed recursively: only the last characteristic function is involved in the last integral (noted ) which only depends on time via . The previous integrals, , involves the product of and , which limits the complexity of the computation. Observe that when , and the multiple integral collapses , so that ∞∑n=1(tθ)nn!=%etθ−1 (5.16) and as it should. ###### Corollary 5.1. The Laplace transform of the integrated process Λs,t=∫tsNu∑i=1Xidu (5.17) is given by ϕΛs,t(x)=e−x(t−s)λsϕ~Λt−s(x) (5.18) where is as in eq. (5.2) but with . ###### Proof. By definition, ∫tsNu∑i=1Xidu=(t−s)λs+∫tsNu∑i=Ns+1Xidu=(t−s)λs+∫tsNu−Ns∑i=1XNs+idu (5.19) The Poisson process has stationary increments, so the distribution of and ∫tsNu∑i=1Xidu∼(t−s)λs+∫t−s0Nv∑i=1XNs+idv (5.20) E[e−x∫ts∑Nui=1Xidu]=e−x(t−s)λsϕ~Λt−s(x) (5.21) ### 5.2 Compound Poisson process with Gamma jump sizes The above expression for the Generalized Compound Poisson process is appealing but is hard to solve as no recursive formula can be found for solving the multiple integral. Even in the case where the , the resulting integrals seems intractable. However, when the ’s are iid, recursion formula can be found as shown below. Beforehand, observe however that Compound Poisson processes being Lévy processes with characteristic exponent ψλ(x)=∫R(eixz−1)ν(dz) (5.22) where and is the jump size distribution we have, from Corollary 2, ψΛs,t(x)=ixλs(t−s)+∫ts∫∞−∞(eizx(t−u)−1)ν(dz)dv (5.23) Solving this simple time-space integral leads the Laplace transform of . This approach seems considerably simpler than computing the infinite sum of multiple integrals (of arbitrarily large dimension) in eq. 5.15. However, when , a recursion formula can be found and the infinite sum of multiple integrals can be solved, and we shall adopt this alternative methodology for the sake of illustration. The results obviously agree with those easily obtained by computing eq.(5.23) with . ###### Example 5.1 (Compound Poisson process with exponentially-distributed jump sizes). Let be a Poisson process with intensity and be iid exponential variables with Laplace transform . Given that in the case of iid , it is enough to compute . Then, the integral of the compound Poisson process is ϕΛt(x) = (e−tϕγ(xt;κ,−κ/x))θ (5.24) To see this, notice that ∫ts=tn−1ϕγ(x(t−s))ds=−κx[ln(κ+x(t−s))]ttn−1=−κxlnϕγ(x(t−tn−1)) (5.25) Integrating that in the previous integral, we get ∫ts=tn−2ϕγ(x(t−s))∫tu=sϕγ(x(t−u))duds = −κx∫ts=tn−2ϕγ(x(t−s))lnϕγ(x(t−s))ds (5.26) = +κx∫t−tn−2u=0ϕγ(xu)lnϕγ(xu)du = (−κx)2∫ϕγ(x(t−tn−2))u=0lnvvdv (5.27) = 12(−κlnϕγ(x(t−tn−2))x)2 (5.28) Finally, using and , ∫tt1=0ϕX1(x(t−t1))…∫ttn=tn−1ϕXn(x(t−tn))d→tn = 1n!(−κlnϕγ(xt)x)n (5.29) ###### Example 5.2 (Compound Poisson process with gamma-distributed jump sizes.). Let be iid gamma random variables with Laplace transform for and (the case corresponds to the exponential above). Then, ϕΛt(x) = e−θ(t−κx(α−1)ϕγ(xt;κ,α−1)) (5.30) The above expression can be found easily by noting that ∫ts=uϕγ(x(t−s))ds=∫ts=uϕαγ(x(t−s);κ,1)ds=κx(α−1)ϕα−1γ(x(t−u);κ,1) (5.31) so that ∫tt1=0ϕX1(x(t−t1))…∫ttn=tn−1ϕXn(x(t−tn))d→tn = (κ/x)nϕn(α−1)γ(

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https://socratic.org/questions/58cffd4bb72cff73a9640a2f

Physics Topics # Question #40a2f ##### 1 Answer Mar 24, 2017 (a) ${0}^{\circ}$ Since the magnitude of resultant is equal to the sum of magnitudes of two displacements. $| \vec{R} | = | \vec{A} | + | \vec{B} |$ (b) ${180}^{\circ}$ Since the magnitude of resultant is equal to the difference of magnitudes of two displacements. (assuming $| \vec{A} | > | \vec{B} |$) $| \vec{R} | = | \vec{A} | - | \vec{B} |$ (b) ${90}^{\circ}$ Since the square of magnitude of resultant is equal to sum of squares of magnitudes of two displacements. This is true only in case of a right triangle. $| \vec{R} {|}^{2} = | \vec{A} {|}^{2} + | \vec{B} {|}^{2}$ ##### Impact of this question 76 views around the world You can reuse this answer Creative Commons License

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https://forum.math.toronto.edu/index.php?PHPSESSID=kmjdq2u8ujqqeb70osuccqm834&topic=2315.0

### Author Topic: 1.5 question 23  (Read 866 times) #### Yan Zhou • Full Member • Posts: 15 • Karma: 1 ##### 1.5 question 23 « on: February 02, 2020, 03:06:29 PM » 23. Show that $F(z) = e^{z}$ maps the strip $S = \{ x+iy: -\infty < x < \infty, -\frac{\pi}{2} < y < \frac{\pi}{2}\}$ onto the region $D = \{w = s+it: s \geqslant 0, w \neq 0\}$ and that $F$ is one-to-one on $S$. Furthermore, show that $F$ maps the boundary of $S$ onto all the boundary  of $D$ except $\{w = 0\}$. Explain what happens to each of the horizontal lines $\{Imz = \frac{\pi}{2}\}$ and $\{Imz = -\frac{\pi}{2}\}$ To prove onto, I want to show, $\forall w \in D, \exists z \in S$, such that $F(z) = e^{z} = e^{x + iy} = w = s +it$. So I need to find $x,y$ such that $s = e^{x}cosy, t = e^{x}siny$. I got answear $x = \frac{ln(s^{2} + t^{2})}{2}, y = arcos\frac{s}{\sqrt{s^{2} + t^{2}}}$But I don't know if they are correct. Can anyone help with this question and the following questions as well?

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https://www.physicsforums.com/threads/conservation-of-angular-momentum.863095/

# Conservation of angular momentum 1. Mar 21, 2016 ### JulienB Hi everybody! I'm preparing myself for upcoming exams, and I struggle a little with conservation of angular momentum. Can anybody help me understand how to solve such problems? 1. The problem statement, all variables and given/known data (for a better comprehension, see the attached image) We have a wooden cylinder of mass mZ = 600g and of radius r0 = 5cm, which can rotate around its symmetry axis. Someone shoots on it, and the projectile has the mass mG = 5.0g and initial velocity v = 80m/s. The distance between the linear trajectory of the projectile and the rotation axis of the cylinder is r1 = 3.0cm. The projectile penetrates the cylinder and stays stuck at a distance of r2 = 3.5cm from the rotation axis of the cylinder. a) What is the frequency of rotation f of the cylinder after the impact? Where and how should you shoot the projectile in order to obtain maximum/minimum frequency? b) Which part of the kinetic energy is used to deform the wooden cylinder? c) If the cylinder was not fixed on a rotation axis but on a thread, what would be the differences to previous case when the projectile hits the cylinder? 2. Relevant equations So I imagine both conservation of linear momentum and of angular momentum are important. We also know that ω = 2πƒ. 3. The attempt at a solution Okay I give it a go: We know that the linear momentum is conserved, that the cylinder is not moving before the collision and that the two objects are moving together after the collision: mG ⋅ v = (mG + mZ) ⋅ v' Here I already see a problem: v' is supposed to be the tangential velocity of the system cylinder-projectile after the collision, but I believe a projectile located at r2 = 3.5cm does not have the same tangential velocity as a point located at r0 = 5.0cm. Is that correct? Then we would have mG ⋅ v = mG ⋅ vG' + mZ ⋅ vZ', which is also not so great. I encounter the same problem with the conservation of angular momentum: mG ⋅ v ⋅ r0 = (mG + mZ) ⋅ v' ⋅ r0 or mG ⋅ v ⋅ r0 = mG ⋅ vG' ⋅ r2 + mZ ⋅ vZ' ⋅ r0 ? I feel like I'm missing something, since none of those equations lead me anywhere :( Furthermore, I never manage to involve r1 in the equations, which obviously plays a role because of the 2nd part of the question. Can someone give me a clue so that I clarify my misunderstandings? Thank you very much in advance. Julien. #### Attached Files: • ###### Mechanik Übung (2).png File size: 33.6 KB Views: 45 2. Mar 21, 2016 ### ZapperZ Staff Emeritus You appear to have completely ignored the moment of inertia of the cylinder. Zz. 3. Mar 21, 2016 ### PeroK I would assume from the problem statement that the cylinder is fixed and can only rotate about its axis of symmetry, not move linearly. For part c) I would assume it is hanging by a thread, and is free to swing and rotate. Although I am not totally confident how to interpret part c). I guess you had to translate the question for us. 4. Mar 21, 2016 ### ZapperZ Staff Emeritus What does this have anything to do with not including the moment of inertia? If the cylinder is rotating under any circumstances, its moment of inertia comes into play! Zz. 5. Mar 21, 2016 ### PeroK If the cylinder is fixed, linear momentum is not conserved. If that is the case, the OP is off on the wrong foot. 6. Mar 21, 2016 ### JulienB @ZapperZ : Yes I realised that, but don't have I to also include the moment of inertia of the projectile then? I don't know its shape though. Or may I just write the linear momentum of the projectile on the left side of the equation and the moment of inertia of the cylinder on the right side, like that: mG ⋅ v = IZ ⋅ ω? @PeroK : Yes I had to translate the problem from German, sorry :) Your assumption is correct, but if you don't mind I will first focus on question a) to make sure I understand how to use the equation first. 7. Mar 21, 2016 ### JulienB @PeroK : Yes I believe the cylinder is fixed. So the linear momentum is not conserved?? 8. Mar 21, 2016 ### PeroK I would assume not. I would assume for parts a) and b) it is only free to rotate about a fixed axis. 9. Mar 21, 2016 ### ZapperZ Staff Emeritus The moment of inertia of the projectile can be assume to be the same as that for a particle at some fixed radius once it is embedded in the cylinder. So you have to consider TWO different angular momentum after the "collision": from the spinning cylinder and the projectile. Otherwise, you have missed the entire problem. I still don't understand why, just because you don't know the "shape" of the projectile, that you are not even including the moment of inertia of the cylinder. This is a major part of the physics, and you missed it. Zz. 10. Mar 21, 2016 ### JulienB I was (and am still) a bit confused about the whole thing. If linear momentum is not conserved, then may I write the following: mG ⋅ v ⋅ r0 = mG ⋅ vG' ⋅ r2 + mZ ⋅ r0 ⋅ vZ' ? On the left hand side I consider the angular moment at the moment of the collision, and on the left hand side the angular momentum of the projectile and of the cylinder once the projectile is stuck in the cylinder. The moment of inertia of the cylinder being mG ⋅ r02 and since ω = vZ'/r0, I would say that LZ = mZ ⋅ r0 ⋅ vZ'. Am I getting somewhere? 11. Mar 21, 2016 ### JulienB I went further with that idea, and reached the following: mG ⋅ v ⋅ r0 = mG ⋅ r22 ⋅ ω + mZ ⋅ r02 ⋅ ω ⇒ ω = (mG ⋅ v ⋅ r0) / (mG ⋅ r22 + mZ ⋅ r02) = 13.28 rad/s2 ⇒ ƒ = ω/2π = 2.1 Hz Does that make sense? mZ ⋅ r02 is the moment of inertia of the cylinder. 12. Mar 21, 2016 ### PeroK I'm not sure I follow what you are doing. You seem to be unsure what is moment of inertia. In this problem, you have two examples of angular momentum: 1) Angular momentum of a point mass about a point/axis, which I think you understand. 2) Angular momentum of a rigid body about an axis of rotation. In this case you need the moment of inertia (MoI) of the rigid body about that axis. This is not the mass of the body, but a measure of the mass distribution of the body in terms of distance from the axis. You should know or be able to calculate the MoI of a solid cylinder. This is what you are missing. Your last equation looks wrong as well. First step: we need the MoI of a solid cylinder about its axis of symmetry. Last edited: Mar 21, 2016 13. Mar 21, 2016 ### PeroK That's the MoI of a hollow cylinder, where all the mass is $r_0$ from the centre. 14. Mar 21, 2016 ### JulienB @PeroK Sounds bad, but I won't give up :) Yes I indeed used the wrong formula since the beginning. The moment of inertia of a full cylinder is I believe: IZ = ½ πhρr4 = ½ mZ ⋅ r2 Now for the rest, I will try to explain you thoroughly my train of thoughts: I assume the angular momentum of the point mass "projectile" about the axis of rotation of the cylinder just before the collision should be equal to the angular momentum of the cylinder about its axis of rotation + the angular momentum of the bullet embedded inside the cylinder after the collision: LG = LG' + LZ' ⇔ mG ⋅ v ⋅ r0 = IG ⋅ ω + IZ ⋅ ω ⇔ mG ⋅ v ⋅ r0 = mG ⋅ r22 ⋅ ω + ½ ⋅ mZ ⋅ r02 ⋅ ω ω being the angular velocity, it should be the same at r0 for the cylinder and at r2 for the bullet, right? At least that's how I attempted to resolve: ω = (mG ⋅ v ⋅ r0)/(mG ⋅ r22 + ½ mZ ⋅ r02) Unfortunately I get a bit of a crazy result. I probably still mess something up. 15. Mar 21, 2016 ### PeroK That looks right. What did you get for $\omega$? Sorry, just saw you used $r_0$ on the left-hand side. Can you see what it should be? 16. Mar 21, 2016 ### JulienB @PeroK Well... Around 1468 rad/s2, which would give a frequency of 233.6 Hz. 17. Mar 21, 2016 ### JulienB Nevermind I made a mistake: now ω = 26.45 and ƒ = 4.2 Hz. 18. Mar 21, 2016 ### PeroK That's not right, because you are using $r_0$ for the initial angular momentum. 19. Mar 21, 2016 ### JulienB Aaaah that should be r1, right? I used r0 because I thought the initial velocity would be affected while penetrating the cylinder. 20. Mar 21, 2016 ### PeroK Yes, it must be $r_1$. Draft saved Draft deleted Similar Discussions: Conservation of angular momentum

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http://dazz.media/laura-ashley-hmgmbz/inverse-function-%26-graph-6f1a63

A function f has an input variable x and gives then an output f (x). If functions f and g are inverse functions, f(g(x)) = g(f(x)) . Then the inverse function f-1 turns the banana back to the apple. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. Function pairs that exhibit this behavior are called inverse functions. A function has to be "Bijective" to have an inverse. In simple words, if any function “f” takes x to y then, the inverse of “f” will take y to x. If the function is one-to-one, there will be a unique inverse. ( because every ( x, y) has a ( y, x) partner! One should not confuse (-1) with exponent or reciprocal here. Required fields are marked *, An inverse function is a function that returns the original value for which a function has given the output. How to use inverse in a sentence. A useful example is converting between Fahrenheit and Celsius: For you: see if you can do the steps to create that inverse! If a horizontal line intersects the original function in a single region, the function is a one-to-one function and inverse is also a function. In simple words, if any function “f” takes x to y then, the inverse of “f” will take y to x. Our fault for not being careful! In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). inverse"), will reverse this mapping. The original function has to be a one-to-one function to assure that its inverse will also be a function. An inverse function or an anti function is defined as a function, which can reverse into another function. The cool thing about the inverse is that it should give us back the original value: When the function f turns the apple into a banana, An example is also given below which can help you to understand the concept better. An inverse function is the "reversal" of another function; specifically, the inverse will swap input and output with the original function. Instead it uses as input f (x) and then as output it gives the x that when you would fill it in in f will give you f (x). An inverse function is a function that undoes the action of the another function. But we could restrict the domain so there is a unique x for every y ... Let's plot them both in terms of x ... so it is now f-1(x), not f-1(y): f(x) and f-1(x) are like mirror images The inverses of some of the most common functions are given below. The graph of f(x) and f-1(x) are symmetric across the line y=x. If the function is denoted by ‘f’ or ‘F’, then the inverse function is denoted by f-1 or F-1. Given a function f (x) f(x) f (x), the inverse is written f − 1 (x) f^{-1}(x) f − 1 (x), but this should not be read as a negative exponent. The inverse is usually shown by putting a little "-1" after the function name, like this: So, the inverse of f(x) = 2x+3 is written: (I also used y instead of x to show that we are using a different value.). (sin 90) = 90 degrees. The inverse function takes the output answer, performs some operation on it, and arrives back at the original function’s starting value. Just think ... if there are two or more x-values for one y-value, how do we know which one to choose when going back? Inverse functions mc-TY-inverse-2009-1 An inverse function is a second function which undoes the work of the first one. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function. In the Wolfram Language, inverse functions are represented using InverseFunction[f]. So, the inverse of f (x) = 2x+3 is written: f-1(y) = (y-3)/2. Only one-to-one functions have inverses. For example, sin. Assuming "inverse function" is referring to a mathematical definition | Use as. If the inverse of a function is itself, then it is known as inverse function, denoted by f. In mathematics, an inverse function is a function that undoes the action of another function. In other words, restrict it to x ≥ 0 and then we can have an inverse. Hence, sin 90 degrees is equal to 1. g = finverse(f) returns the inverse of function f, such that f(g(x)) = x.If f contains more than one variable, use the next syntax to specify the independent variable. We got 2 instead of −2. When you’re asked to find an inverse of a function, you should verify on your own that the inverse you obtained was correct, time permitting. STEP 1: Stick a " y " in for the " f (x) " guy: STEP 2: Switch the x and y. There are various types of inverse functions like the inverse of trigonometric functions, rational functions, hyperbolic functions and log functions. For example, show that the following functions are inverses of each other: Show that f(g(x)) = x. The inverse function of a function f is mostly denoted as f -1. inverse f ( x) = ln ( x − 5) $inverse\:f\left (x\right)=\frac {1} {x^2}$. If function f is not a one-to-one then it does not have an inverse. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. inverse function definition: 1. a function that does the opposite of a particular function 2. a function that does the opposite…. Restrict the Domain (the values that can go into a function). Put "y" for "f(x)" and solve for x: This method works well for more difficult inverses. Or we can find an inverse by using Algebra. Inverse functions are a way to "undo" a function. A linear function is a function whose highest exponent in the variable(s) is 1. If you plan to offer your domain name soon, you should get an expert appraisal from a paid service. If you consider functions, f and g are inverse, f(g(x)) = g(f(x)) = x. The inverse isn't a function. Download BYJU’S- The Learning App to get a more engaging and effective learning experience. More discussions on one to one functions will follow later. Your email address will not be published. Mathematically this is the same as saying, inverse y = x x2 − 6x + 8. Figure 3.7.1 shows the relationship between a function f(x) and its inverse f − 1(x). There are six inverse trigonometric functions which include arcsine (sin-1), arccosine (cos-1), arctangent (tan-1), arcsecant (sec-1), arccosecant (cosec-1), and arccotangent (cot-1). Just make sure we don't use negative numbers. The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2. The inverse of the function returns the original value, which was used to produce the output and is denoted by f. If we have to find the inverse of trigonometry function sin x = ½, then the value of x is equal to the angle, the sine function of which angle is ½. or an anti function is defined as a function, which can reverse into another function. Imagine we came from x1 to a particular y value, where do we go back to? How to Graph the Inverse of a Function By Yang Kuang, Elleyne Kase If you’re asked to graph the inverse of a function, you can do so by remembering one fact: a function and its inverse are reflected over the line y = x. There are mainly 6 inverse hyperbolic functions exist which include sinh-1, cosh-1, tanh-1, csch-1, coth-1, and sech-1. To find the inverse of a quadratic function, start by simplifying the function by combining like terms. Let's just do one, then I'll write out the list of steps for you. Make sure your function is one-to-one. Intro to inverse functions. Then, g(y) = (y-5)/2 = x is the inverse of f(x). For example , addition and multiplication are the inverse of subtraction and division respectively. inverse is called by random.function and calculates the inverse of a given function f. inverse has been specifically designed to compute the inverse of the cumulative distribution function of an absolutely continuous random variable, therefore it assumes there is only a root for each value in the interval (0,1) between f (lower) and f (upper). Let us see graphically what is going on here: To be able to have an inverse we need unique values. A function $g$ is the inverse of a function $f$ if whenever $y=f(x)$ then $x=g(y)$. or. How to Find the Inverse of a Function 2 - Cool Math has free online cool math lessons, cool math games and fun math activities. The graph of the inverse of a function reflects two things, one is the function and second is the inverse of the function, over the line y = x. Given a function f(x), its inverse f^(-1)(x) is defined by f(f^(-1)(x))=f^(-1)(f(x))=x. Important Questions Class 12 Maths Chapter 2 Inverse Trigonometric Functions. Using the formulas from above, we can start with x=4: So applying a function f and then its inverse f-1 gives us the original value back again: We could also have put the functions in the other order and it still works: We can work out the inverse using Algebra. This line passes through the origin and has a slope of 1. inverse f ( x) = x3. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse. A function is called one-to-one if no two values of x x produce the same y y. Just like inverse trigonometric functions, the inverse hyperbolic functions are the inverses of the hyperbolic functions. . If f(x) is a function which gives output y, then the inverse function of y, i.e. The Derivative of an Inverse Function We begin by considering a function and its inverse. For example, sin-1(1) = sin-1(sin 90) = 90 degrees. One should not confuse (-1) with exponent or reciprocal here. So, when we apply function f and its reverse f-1 gives the original value back again, i.e, f-1(f(x)) = x. Inverse functions, in the most general sense, are functions that "reverse" each other. This same quadratic function, as seen in Example 1, has a restriction on its domain which is x \ge 0.After plotting the function in xy-axis, I can see that the graph is a parabola cut in half for all x values equal to or greater than zero. I will utilize the domain and range of the original function to describe the domain and range … Inverse of Square Root Function Read More » It is called a "one-to-one correspondence" or Bijective, like this. The inverse trigonometric functions are also known as arc function as they produce the length of the arc, which is required to obtain that particular value. Hence, sin 90 degrees is equal to 1. If f and g are inverse functions, then f(x) = y if and only if g(y) = x, is used to find the measure of angle for which sine function generated the value. x1 or x2? Before formally defining inverse functions and the notation that we’re going to use for them we need to get a definition out of the way. The inverse function agrees with the resultant, operates and reaches back to the original function. If f(x) is both invertible and differentiable, it seems reasonable that the inverse of f(x) is also differentiable. Embed this widget ». The inverse of a function can be viewed as the reflection of the original function over the line y = x. A function accepts values, performs particular operations on these values and generates an output. Example 2: Find the inverse function of f\left( x \right) = {x^2} + 2,\,\,x \ge 0, if it exists.State its domain and range. inverse function - Wolfram|Alpha. $inverse\:y=\frac {x} {x^2-6x+8}$. The relation, developed when the independent variable is interchanged with the variable which is dependent on a specified equation and this inverse may or may not be a function. Inverse function calculator helps in computing the inverse value of any function that is given as input. So if we have to draw the graph of f-1, then we have to switch the positions of x and y in axes. A function that consists of its inverse fetches the original value. Inverse Logarithmic Functions and Inverse Exponential Function. referring to English words. New Version: https://youtu.be/q6y0ToEhT1EDefine an inverse function. or. Finding the Inverse Function of a Square Root Function To find the inverse of a square root function, it is crucial to sketch or graph the given problem first to clearly identify what the domain and range are. So a bijective function follows stricter rules than a general function, which allows us to have an inverse. That is because some inverses work only with certain values. The inverse of a function f does exactly the opposite. First, replace f(x) with y and the function becomes. 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Find the inverse of. In the original function, plugging in x gives back y, but in the inverse function, plugging in y (as the input) gives back x (as the output). Also, get more insights of how to solve similar questions and thus, develop problem-solving skills. Find the inverse for the function f(x) = (3x+2)/(x-1). Then, determine the domain and range of the simplified function. So what is all this talk about "Restricting the Domain"? Inverse Function Calculator The calculator will find the inverse of the given function, with steps shown. As it stands the function above does not have an inverse, because some y-values will have more than one x-value. Your email address will not be published. f, One should not get confused inverse function with reciprocal of function. Check out inverse hyperbolic functions formula to learn more about these functions in detail. Learn more. So the square function (as it stands) does not have an inverse. The inverse function, denoted f-1, of a one-to-one function f is defined as f-1 (x) = {(y,x) | such that y = f(x)} Note: The -1 in f-1 must not be confused with a power. or instead. In trigonometry, the inverse sine function is used to find the measure of angle for which sine function generated the value. column above? Determine if a function as an inverse function. This “DO” and “UNDO” process can be stated as a composition of functions. Note: when we restrict the domain to x ≤ 0 (less than or equal to 0) the inverse is then f-1(x) = −√x: It is sometimes not possible to find an Inverse of a Function. 1. (Note: you can read more about Inverse Sine, Cosine and Tangent.). And you can see they are "mirror images" Inverse functions, in the most general sense, are functions that "reverse" each other. Learn what the inverse of a function is, and how to evaluate inverses of functions that are given in tables or graphs. It is also called an anti function. This line in the graph passes through the origin and has slope value 1. It is denoted as: f (x) = y ⇔ f− 1(y) = x. If a function were to contain the point (3,5), its inverse would contain the point (5,3).If the original function is f(x), then its inverse f -1 (x) is not the same as . Check the following example to understand the inverse exponential function and logarithmic function in detail. The natural log functions are inverse of the exponential functions. Inverse definition is - opposite in order, nature, or effect. When we square a negative number, and then do the inverse, this happens: But we didn't get the original value back! (1) Therefore, f(x) and f^(-1)(x) are reflections about the line y=x. (flipped about the diagonal). If you wish to make significant improvements in your website's advertising revenue, you must look at it like a service enterprise. A function is said to be a one to one function only if every second element corresponds to the first value (values of x and y are used only once). The inverse function of an inverse function is the original function.. 1995, Nicholas M. Karayanakis, Advanced System Modelling and Simulation with Block Diagram Languages, CRC Press, page 217, In the context of linearization, we recall the reflective property of inverse functions; the ƒ curve contains the point (a,b) if and only if the ƒ-1 curve contains the point (b,a). If the function is denoted by ‘f’ or ‘F’, then the inverse function is denoted by f. . This step is a matter of plugging in all the components: This new function is the inverse function Step 3: If the result is an equation, solve the equation for y. The inverse is usually shown by putting a little "-1" after the function name, like this: f-1(y) We say "f inverse of y". Generally, the method of calculating an inverse is swapping of coordinates x and y. To recall, an inverse function is a function which can reverse another function. $inverse\:f\left (x\right)=x^3$. We cannot work out the inverse of this, because we cannot solve for "x": Even though we write f-1(x), the "-1" is not an exponent (or power): We can find an inverse by reversing the "flow diagram". inverse f ( x) = 1 x2. inverse y = x2 + x + 1 x. It can be represented as; This relation is somewhat similar to y = f(x), which defines the graph of f but the part of x and y are reversed here. of each other about the diagonal y=x. It has been easy so far, because we know the inverse of Multiply is Divide, and the inverse of Add is Subtract, but what about other functions? This newly created inverse is a relation but not necessarily a function. $inverse\:f\left (x\right)=\ln\left (x-5\right)$. A rational function is a function of form f(x) = P(x)/Q(x) where Q(x) ≠ 0. Here we have the function f(x) = 2x+3, written as a flow diagram: So the inverse of:   2x+3   is:   (y-3)/2. The Once you have the domain and range, switch the roles of the x and y terms in the function and rewrite the inverted equation in … Learn how to find the inverse of a linear function. You can apply on the horizontal line test to verify whether a function is a one-to-one function. A function accepts values, performs particular operations on these values and generates an output. a Wolfram Language symbol. Find the inverse of the function f(x) = ln(x – 2), Replace the equation in exponential way , x – 2 = ey, Now, replace x with y and thus, f-1(x) = y = 2 + ey. a computation. ): STEP 3: Solve for y: STEP 4: Stick in the inverse notation, Did you see the "Careful!" But if we can have exactly one x for every y we can have an inverse. In this unit we describe two methods for finding inverse functions, and we also explain that the domain of a function may need to be restricted before an inverse function can exist. To find the inverse of a rational function, follow the following steps. 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https://planetmath.org/lhopitalsrule

# l’Hôpital’s rule L’Hôpital’s rule states that given an unresolvable limit of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, the ratio of functions $\frac{f(x)}{g(x)}$ will have the same limit at $c$ as the ratio $\frac{f^{\prime}(x)}{g^{\prime}(x)}$. In short, if the limit of a ratio of functions approaches an indeterminate form, then $\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f^{\prime}(x% )}{g^{\prime}(x)}$ provided this last limit exists. L’Hôpital’s rule may be applied indefinitely as long as the conditions are satisfied. However it is important to note, that the nonexistance of $\lim\frac{f^{\prime}(x)}{g^{\prime}(x)}$ does not prove the nonexistance of $\lim\frac{f(x)}{g(x)}$. Example: We try to determine the value of $\lim_{x\to\infty}\frac{x^{2}}{e^{x}}.$ As $x$ approaches $\infty$ the expression becomes an indeterminate form $\frac{\infty}{\infty}$. By applying L’Hôpital’s rule twice we get $\lim_{x\to\infty}\frac{x^{2}}{e^{x}}=\lim_{x\to\infty}\frac{2x}{e^{x}}=\lim_{x% \to\infty}\frac{2}{e^{x}}=0.$ Another example of the usage of L’Hôpital’s rule can be found http://planetmath.org/node/5741here. Title l’Hôpital’s rule Canonical name LHopitalsRule Date of creation 2013-03-22 12:28:15 Last modified on 2013-03-22 12:28:15 Owner mathwizard (128) Last modified by mathwizard (128) Numerical id 13 Author mathwizard (128) Entry type Theorem Classification msc 26A24 Classification msc 26C15 Synonym l’Hospital’s rule Related topic IndeterminateForm Related topic DerivationOfHarmonicMeanAsTheLimitOfThePowerMean Related topic ImproperLimits Related topic ExampleUsingStolzCesaroTheorem

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http://www.ams.org/publications/authors/msquality

Improving the quality of accepted manuscripts The recommendations provided below have been culled from the thousands of manuscripts the AMS has processes and represent the most commonly encountered problems. Please take a few minutes to read this material. Following these recommendations will minimize problems with you manuscript later on. (Rev 01/2002) Text: The AMS recommends electronic files be prepared using LaTeX A well-structured LaTeX document can be more readily converted to other formats for the Web with functionality such as hyperlinks. AMSTeX files are converted to AMSLaTeX upon receipt. (See also Why do we Recommend LaTeX 2e http://www.ams.org/jourhtml/authors.html ) Use one of the AMS Author Packages, follow the instructions and use the AMS defined coding AMS Author Packages were designed to provide the specific styles for each of the journals. Following the instructions and using these packages will produce your article in the proper format and reduce the manipulation of your files by AMS staff. Use the coding indicated in the Author Package instructions for headings, theorem and theorem-like environments, citations and references, etc. Do not hard code the formatting (fonts, horizontal and vertical spacing, etc.) for any of the items listed above. Hard copy of the manuscript must exactly match the electronic file Authors are requested to proofread their manuscripts thoroughly and carefully, and submit an electronic file that exactly matches the hard copy of the manuscript sent to the AMS. Do not redefine LaTeX, AMSLaTeX, TeX, or AMSTeX commands Programs that process TeX files at the AMS do not interpret TeX, but rather read TeX coding and convert it to another format (HTML, etc.). Redefining predefined commands creates problems for both the print and online products because all the commands that redefine standard LaTeX, AMSLaTeX, TeX, or AMSTeX code need to be manually removed and replaced one-by-one by AMS staff. This is very time-consuming and can result in errors later on. Electronic files must be able to be processed independently with all macros (not entire macro files) included Any macros that are not part of a publicly distributed and supported macro package must be included in your electronically submitted TeX file. If you have a standard set of macros you use for preparing electronic files please make sure you include these with your manuscript. Only the macros used in a file should be included. Coding for required functionality on the Web While the following are required for the Web product, they also have advantages for authors in preparing their articles. This coding automates cross referencing. Use \cite for all citations. This will enable linking between the citation and the bibliographic item. Use only numbers and letters within \cite. Special characters and spaces always need to be removed by AMS staff. Use \label and \ref. This coding is used to automatically cross reference sections, equations, theorem and theorem-like environments, tables, figures, etc. Use only numbers and letters within these commands. Special characters and spaces always need to be removed by AMS staff. Use \newtheorem (not \newtheorem*) to set up numbered environments. This will automatically number the environments. Numbering systems can be adjusted using \setcounter commands. Use only numbers and letters within \newtheorem. Special characters and spaces always need to be removed by AMS staff. Graphics: Follow the AMS guidelines when creating graphics The AMS has provided guidelines for creating graphics that will appear in AMS publications (See also Creating Graphics http://www.ams.org/jourhtml/authors.html ) Create graphics at 100% of the size they will be printed at. Do not use a line/rule weight less than .5 point at 100% If you must scale your figure (either in TeX or with the graphics program), be sure that you compensate by making line weights thicker. A .5 point line scaled at 50% becomes a .25 point line. Line weights below .5 point disappear during the printing process. Shades of gray (screens) should not be lower than 15% or higher than 85% Screens outside of this range are either too light or too dark to print correctly. Screens should increase in increments of no less than 10% Screen differences of less than 10% are not distinguishable. Multiple-part figures should be assembled into one plate in a graphics program, not in TeX Aligning multiple-part figures is very difficult in TeX. It is easier and more cost-effective to do so in a graphics program. Fully embed fonts into your graphic when saving the file If the fonts are not embedded in your graphic, it is possible that the font will be replaced with a default font such as Courier, and the characters will not print properly. If you are unable to embed the fonts in your graphic, convert the fonts to paths (or outlines) prior to exporting the file to EPS. This can be done in the application in which they were created (consult the documentation for your graphics program for assistance). Do not subset fonts included in your graphic It is essential that the full font set be included in your graphics. If only a subset of a font is included, a font error will occur causing characters to disappear in both the graphic and the DVI file. Graphics to be published in black and white should be created in black and white. Create graphics in black-and-white unless they are to appear in color in the printed product. Color graphics will be optimized for black-and-white printing at the AMS and this will result in a loss of information especially with colors that do not convert to grayscale well. Requirements for graphics to be published in color Color graphics need to be saved in the following format: Electronic only journals: RGB format Electronic and/or print journals: CMYK format. Note that color graphics should be used in the print product only if color is mathematically essential to the paper. Authors will be requested to pay for the cost of color in the print product. Send accepted electronic manuscripts (other than Abstracts ) to: [email protected]

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https://math.wonderhowto.com/how-to/solve-square-roots-algebra-181547/

# How To: Solve square roots in algebra In this lesson we will learn about one of the most important concepts in algebra SQUARE ROOTS. The square root of a number m is another number n that satisfies the following formula: m = n x n. You can also say that m is the square of n. The opposite of squaring a number is finding its square root. You can use the radical symbol to indicate that the square root of m is n: \sqrt{~}m = \sqrt{~}n2 = n A number is a perfect square if it can be written as the square of a whole number. Not all integers are perfect squares. • Hot • Latest

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http://www.varsitytutors.com/gre_subject_test_physics-help/lenses

# GRE Subject Test: Physics : Lenses ## Example Questions ### Example Question #1 : Lenses The focal length of a thin convex lens is . A candle is placed  to the left of the lens. Approximately where is the image of the candle? to the right of the lens to the left of the lens to the left of the lens to the right of the lens No image is created to the right of the lens Explanation: Because the object is beyond 2 focal lengths of the lens, the image must be between 1 and 2 focal lengths on the opposite side. Therefore, the image must be between  on the right side of the lens. Alternatively, one can apply the thin lens equation: Where  is the object distance  and  is the focal length . Plug in these values and solve. ### Example Question #1 : Lenses A candle  tall is placed  to the left of a thin convex lens with focal length . What is the height and orientation of the image created? , inverted No image is created. , inverted , upright , upright , inverted Explanation: First, find the image distance  from the thin lens equation: Magnification of a lens is given by: Where  and  are the image height and object height, respectively. The given object height is , which we can use to solve for the image height: Because the sign is negative, the image is inverted. ### Example Question #2 : Lenses Sirius is a binary star system, consisting of two white dwarfs with an angular separation of 3 arcseconds. What is the approximate minimum diameter lens needed to resolve the two stars in Sirius for an observation at ?

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http://export.arxiv.org/abs/hep-th/9804149

hep-th (what is this?) # Title: Three Brane Action and The Correspondence Between N=4 Yang Mills Theory and Anti De Sitter Space Abstract: Recently, a relation between N=4 Super Yang Mills in 3+1 dimensions and supergravity in an $AdS_5$ background has been proposed. In this paper we explore the idea that the correspondence between operators in the Yang Mills theory and modes of the supergravity theory can be obtained by using the D3 brane action. Specifically, we consider two form gauge fields for this purpose. The supergravity analysis predicts that the operator which corresponds to this mode has dimension six. We show that this is indeed the leading operator in the three brane Dirac-Born-Infeld and Wess-Zumino action which couples to this mode. It is important in the analysis that the brane action is expanded around the anti de-Sitter background. Also, the Wess-Zumino term plays a crucial role in cancelling a lower dimension operator which appears in the the Dirac-Born-Infeld action. Comments: 12 pages, LaTex, no figures; error in the form of the final dimension six operator corrected, some references and comments added, main conclusions unchanged Subjects: High Energy Physics - Theory (hep-th) Journal reference: Phys.Lett. B445 (1998) 142-149 DOI: 10.1016/S0370-2693(98)01450-6 Report number: TIFR-TH/98-10, fermilab-Pub-98/122-T Cite as: arXiv:hep-th/9804149 (or arXiv:hep-th/9804149v2 for this version) ## Submission history From: Sandip Trivedi [view email] [v1] Wed, 22 Apr 1998 19:49:28 GMT (14kb) [v2] Fri, 15 May 1998 19:15:59 GMT (14kb) Link back to: arXiv, form interface, contact.

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https://ask.sagemath.org/questions/42416/revisions/

# Revision history [back] ### Sum using coefficients of a expansion I'm trying to sum over the coefficients of a expansion but the sum gives zero. I use the taylor method to expand a function f: f=sin(x) ft=f.taylor(x,0,6) then I need to use the coefficients in a sum, but if I for example do the following: sum(x^(i)*ft.coefficient(x,i),i,0,5) I get zero. Any idea why this is happening? ### Sum using coefficients of a expansion I'm trying to sum over the coefficients of a an expansion but the sum gives zero. I use the taylor method to expand a function f: f=sin(x) ft=f.taylor(x,0,6) then I need to use the coefficients in a sum, but if I for example do the following: sum(x^(i)*ft.coefficient(x,i),i,0,5) I get zero. Any idea why this is happening? happening?

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https://www.physicsforums.com/threads/determine-acceleration-from-position-vs-time-graph.205098/

# Determine Acceleration from Position vs. Time Graph 1. Dec 17, 2007 ### petern The position vs. time graph is wavy and I assume the only point where there is acceleration is where there is a curve, right? It seems like the acceleration is also 0 at the curve though. Is it even possible? 2. Dec 17, 2007 ### MrJingles For a position as a function of time graph, simply find the derivative at the point which you wish to find the velocity for. Find the second derivative of the function for the acceleration. Perhaps if you uploaded an image of the particular graph in question I could be of more use. (The acceleration will equal zero at any point where the f(t) graph changes concavity) 3. Dec 17, 2007 ### petern and this: I don't really understand most of it. 4. Dec 17, 2007 ### MrJingles Your first assertion, that acceleration occurs only at curves is correct. Velocity is only equal to zero when it (the position function) has zero slope (ie, no motion--- straight lines and relative max/mins) and acceleration equals zero when the velocity is constant (velocity is a straight line) and when the position graph switches concavity (inflection point). It is important to note that, though velocity may =0 at some point, acceleration may not (although it can). When the function is concave (up) its derivative (in this case velocity) is increasing, which means that its acceleration is positive When it's convex (concave down) its derivative is decreasing which means that its acceleration is negative Last edited: Dec 17, 2007 5. Dec 17, 2007 ### petern Thanks for the advice but for the 1st page, can u tell me where each bullet goes? Which interval does it go to?

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