lsp-mode, Jump to Next Warning

Question: In a C# lsp-mode buffer, when something is highlighted, is there a "jump to next warning or error" function?
I don't see anything like lsp-next* or lsp-jump*.
Any pointers?
Answer: lsp-mode integrates with flycheck or flymake. Please check the variable lsp-diagnostics-provider.
For example, flycheck provides flycheck-next-error and flycheck-previous-error to help navigate across errors/warnings. You can also use flycheck-list-errors to see the errors list. Similarly, flymake provides flymake-goto-next-error and flymake-goto-prev-error.
Answer: For real-time diagnostics, lsp-mode utilizes either flycheck or a built-in package flymake (in Emacs open manual with C-h R flymake RET).
flycheck documentation on interaction with errors states:

By default Flycheck hooks into Emacs' standard error navigation on M-g
n (next-error) and M-g p (previous-error).

This is what, supposedly, you're looking for, since flycheck is a recommended package for use by lsp-mode and is configured by default.
flymake documentation in its turn also describes how to navigate between errors. To reach this section in Emacs use M-: (info "(flymake) Finding diagnostics") RET. Now the section states the following:

'flymake-goto-next-error' and 'flymake-goto-prev-error' are let you
navigate to the next/previous erroneous regions, respectively.  It
might be a good idea to map them to 'M-n' and 'M-p' in 'flymake-mode',
by adding to your init file:
 (define-key flymake-mode-map (kbd "M-n") 'flymake-goto-next-error)
 (define-key flymake-mode-map (kbd "M-p") 'flymake-goto-prev-error)

And, since we've touched the subject of diagnostics, you may find a list of related lsp-mode custom options useful. Especially note lsp-diagnostics-provider custom option on that list, which can be used for selecting -- as its name broadly hints -- one of the mentioned providers of the diagnostics.

lsp-mode, Jump to Next Warning


References in Ring Theory

Question: I am very interested in ring theory. I want to know the best textbooks about that subject, i.e about PID's, UFD's, Dedekind domain, Euclid domain. Can anyone tell me?
Comment: Dummit and Foote is a good start.
Comment: What have you read so far? There are a lot of books that cover it to some extent or another. Aluffi's *Algebra: Chapter 0* is what I'm working on myself at the moment. Mac Lane and Birkhoff's *Algebra* seems to be a classic. I used Dummit and Foote in a class once, and I couldn't stand it.
Comment: The best books are those of Lang and Bourbaki, Mac Lane's is also ok. Dummit and Foote is too elementary for you
Answer: It would be a good choice to go for Abstract algebra text by Dummit Foote for the part which you have asked for .
Simultaneously Keep doing problems (*in the same order as it is given)
http://www.math.kent.edu/~white/qual/list/ring.pdf
Do not take it much seriously but it would be a good idea if you can try looking at "Algebra : Michael Artin" mainly for irreducible polynomials part.
Simultaneously, you please keep a copy(*soft) of "CONTEMPORARY ABSTRACT ALGEBRA-Joseph Gallian". This have only small portion in chapter "Divisibility in Integral Domains" but i would say this is very useful.
Caution : My idea of suggesting three books is not to make you get confused but just for reference and one more use is something which is asked as an exercise in one book  is proved as lemma in another book :P which would be useful.
Answer: I find the book "P.M.Cohn: Introduction to Ring Theory" very useful. For noncommutative rings
there is the book "T.Y.Lam : A First Course in Noncommutative Rings". The level is a little higher than that of the first reference, but the book is not difficult. Good luck!
(I must say I am pleased that you are interested in ring theory. I find the area very rich and exciting.)

References in Ring Theory


MacBook Pro displays Apple Logo and Prohibitory Symbol

Question: My  MacBook Pro 15" displays the  Prohibitory Symbol on start-up. What can I do to fix it?
Answer: The Apple website mentions what a prohibitory symbol means:

Prohibitory symbol

  A circle with a line or slash through it means that the selected startup disk contains a Mac operating system, but it's not a macOS that your Mac can use. You should reinstall macOS on that disk.
Answer: If you have an external drive with the OS on it, boot an hold down Option (to select an alternate boot volume) -- boot with that and see if you can deal with the primary drive.
Failing that, get an external boot drive, boot holding down Command-R, and install a fresh copy of the operating system (which should be compatible with your hardware), and then boot from that to address the primary drive.
...basically getting the data you want to save off the primary drive, and then reinstalling the OS on it.


Forward Secrecy not supported with referenced browsers?

Question: I am running a scan against one of our servers from the Qualys SSL tester site and it says that my server "does not support forward secrecy with the referenced browsers". What exactly does that mean? My server is setup to use ECDHE_RSA_* ciphers which I know do support forward secrecy if negotiated and is recognized on the Qualys scan. Is this a browser issue since it says "referenced browsers".
Answer: The «reference browsers» are marked with an R in the table below. What results were you given there? The problem may be on some of them (IE?) not supporting those ECDHE_RSA_* ciphers.
Comment: Ok yes that's what I thought I just didn't know how to look. I'm on IE 10 which is among the browsers with "R". Thanks!


Is Hellinger transformation suitable for repeated (in time) site-species abundance data?

Question: I have fish survey data from four different years and several locations, where I would like to study the difference in abundance and biomass between sites and check whether time dependence influences the significance in differences found between the sites.  Because my data contains a lot of zeros and low values I believe using Hellinger transformation will be appropriate. However I am not sure if I can use this transformation on repeated site-species data. If it is possible, can anybody send me a link or info on how I can/should apply this transformation? I have not been able to find a study where they applied Hellinger transformation on similar data.
Answer: Yes, it's fine to do this. You apply it in the same way as you would for non-repeated observations.
Say you have a matrix $\mathbf{Y}$ of biomass values with dimensions $i,j$ where $j$ indexes the species and $i$ indexes the samples, the Hellinger transformation is
$$y^{\prime}_{ij} = \sqrt{\frac{y_{ij}}{y_{i+}}}$$
where $y^{\prime}_{ij}$ is the transformed biomass for species $j$ in sample $i$, $y_{ij}$ is the observed biomass, and $y_{i+}$ is the row sum over the $j$ species for the $i$th sample.
There is nothing in this transformation that relates to time or particular observations. There is nothing here that is going to get messed up by applying this to repeated observations. Just arrange each observation as a row in your response matrix $\mathbf{Y}$, arrange the species in the columns, and then have the meta data (site, time, etc) in a separate matrix in the same row order as $\mathbf{Y}$, to keep track of which samples come from which sites, times, etc.
Comment: Gavin: Would it be possible for you to help me with this question? https://stats.stackexchange.com/questions/533115/how-to-decide-what-spatial-scale-to-analyse-pca-on-general-questions-aboutpca


Linear stability analysis of a 2-cycle

Question: In a discrete $N$-dimensional Hamiltonian map $\mathbf{X}^{(n+1)}=f(\mathbf{X}^{(n)})$, we often find a 2-cycle which shows oscillation between two points in phase space. In such a Hamiltonian map we analyze the stability of a fixed point from the eigenvalues of the jacobian matrix of the corresponding linearized equation. My question is, what is the procedure for the stability of a 2-cycle in such a system?  Do I need to obtain $\mathbf{X}^{(n+2)}=f(f(\mathbf{X}^{(n)}))$, and then linearize it or anything else?
Answer: One can introduce supplementary variables:
$$
Y^{(n+1)}=f(Z^{(n)}),\\
Z^{(n+1)}=f(Y^{(n)}),
$$
and analyze the stability of this two-dimensional system. This is indeed equivalent to analyzing the stable point of mapping
$$
Y^{(n+1)}=f(f(Y^{(n)}))
$$

Linear stability analysis of a 2-cycle


How to Rotate Object around a Tilted Axis

Question: fairly new to blender! im trying to animate an atom for my friend and I have it modeled, but am trying to animate the rotations of the electrons. the problem is they dont follow a set axis, and are on a tilted axis. i have tried using a circle path and setting it as a constraint on the object, but it doesnt actually rotate the object. if anyone could help, that would be amazing <3
Comment: maybe check my answer here: https://blender.stackexchange.com/questions/129834/how-do-i-rotate-an-object-along-its-own-axis-windmill
Answer:
Create the orbit rings as torus objects
In Edit mode add a sphere to the torus to create the electron
In Object mode rotate the entire ring on one of the major axis.
Do not apply rotation.

Here is an example.

I created a torus and added a sphere in edit mode.  Then I duplicated the ring/electron and rotated it 120 degrees on the Y axis. (Shortcut Shift–DR120Enter)
I repeated this step to generate the 3rd ring/electron.
Because I did not apply the rotation, each ring still has a Local coordinate system where the Z axis is perpendicular to the ring.  If I want to rotate and ring, I select it in Object mode and rotate along the local Z axis.  (Shortcut RZZ180Enter)  The technique for rotating on a local axis rather than the global axis is to type the letter representing the axis twice: ZZ
You can tell that you are rotating on the local axis, because after you type the second Z, the blue line representing the Z axis will be tilted:
.
Answer: This approach is quite flexible: if you want to make late adjustments, there's a minimum of work.
Working in the XY plane, Transform Orientation: 'Local', Pivot: '3D Cursor' at World 0:

Create a sphere for your nucleus
ShiftD Duplicate it.. move it off to one side, and in Edit Mode, scale it down to make an electron.
Create a Curve > Circle to be an orbit.
Give the electron a Follow Path constraint, aimed at the orbit:
(If, now, or at any stage, the electron is not on the curve, select it, and hit AltG to set its translation to 0.)

In the timeline, decide on a length for the animation loop (e.g. 50 frames). At frame 1, key the 'Offset' in the constraint to 0.
At frame loop+1 (e.g. 51) key the 'Offset' to 100.
In the timeline, (just in case) hit T, and set the electron's (selected) keyframes to 'Linear' interpolation.

You should now have one electron, looping on an orbit.
For more electrons-animated-on-orbits, ShiftLeft-Arrow go to frame 0, and:

Select the curve and its electron, AltD make instances of them, and right-click to drop them in place.
(Deselect the electron,) and  RY rotate the curve to a new angle. It will take the electron with it.You can change the phase of the electron by RZ locally rotating the curve.

For more .. rinse, repeat.
Because everything is instanced, you can use the 'Bevel' in the orbit's Data tab > Geometry panel to change the thickness of the orbits all at once, or, in Edit Mode, change the size of the electrons all at once. If you revise the speed of one electron, the CtrlL menu will let you copy the animation to the others.. and so on.


What makes the speed of lightning different to the speed of light?

Question: The difference between light and lightning, their different speeds and a reason why the speeds differ
Comment: Can you please clarify what you mean by saying "speed of lightning" ?
Comment: @AgniusVasiliauskas - He means how fast does a bolt of lightning advance. See [Lightning](https://en.wikipedia.org/wiki/Lightning) in Wikipedia for a slow motion video.
Answer: Lightning is bright and looks like a streak of light in the sky. But light and lightning are two different things. After all, you can see a bolt of lightning even if it doesn't hit you. Lightning is more like electricity flowing in a wire than light.
Electricity in a wire is caused by moving electrons. Electrons repel each other. If more electrons are pushed into one end of the wire, the new electrons push on electrons already in the wire, and those push on their neighbor, and so on down the wire. None of the electrons move very far or very fast. But in a very short time, electrons come out the other end of the wire.
It is something like pulling the wire an inch. The wire doesn't move very far or very fast. But in a very short time, the far end of the wire moves an inch.
Lightning is something like this, except that it must make its own wire. Air is normally an insulator. Charges in clouds build up until the electric forces are strong enough to pull electrons off air molecules. The freed electrons can move. Electricity can flow in this region.

Electrons and light are very different. Light travels at the speed of light (at least it does in a vacuum. It travels slower in glass.)
Electrons flow at whatever speed they are pushed. However, the forces between electrons travel at the speed of light. Suppose two electrons repel each other. If one electron moves, the other doesn't notice instantly. The "news" that the electron has moved travels at the speed of light. When it arrives, the other electron feels the repulsion from a new place and begins to respond. It takes time for the pushed electron to build up speed and push differently on the next electron down the line. This slows the propagation of electricity in a wire. In copper, electricity travels at about $1/3$ the speed of light.
Lightning advances much slower than that. If a bolt of lightning advances into newly ionized channel of air, charges build up at the end of the channel. Before the lightning can advance further, more air must be ionized. The process is complex and still not totally understood. Light travels more than $10000$ times faster than lightning.
For more, see Lightning
Answer: It is rather like the distinction between firelight and fire. When you strike a match, the light from the flame travels at the speed of light, but the flame itself propagates slowly along the match because it has to pass heat energy to each neighbouring piece of wood in the match before that part will ignite. Likewise with lightening, the path of the bolt has to be heated to the point at which it will ionise, and the speed at which air will ionise to conduct the bolt is much slower than the speed of light.


VMWare Virtual Machine Ignores DCHP Lease

Question: I have a VMWare Player (Workstation 9 )virtual machine on an Ubuntu 12.10 (13.10 Kernel) host running Ubuntu 12.04 using a bridged connection and set to replicate the physical network connection. Everything usually works properly in a variety of locations. But at one location that I often frequent, the ip address of the virtual machine changes roughly every 10 minutes -rendering the vm entirely useless as it is a postgresql server and thus needs a dedicated local ip. Not only that, but when I copied a database dump into a shared folder, the file ended up getting corrupted.
I can verify that the network caused this problem, as the actual on the vm was not corrupted. I managed to temporarily solve the problem by going into a local modem and setting a DHCP Mac Address. Everything was working and files were not getting corrupted. However, it only lasted temporarily, and another random address was assigned, breaking several running processes on my machine. Between the router/gateway, there is a redundant apple router involved in the network that is likely causing the issue -but I cannot just throw it away or deactivate it, as it is not my network
Furthermore, DHCP leases work just fine for every other machine on the network; so
I believe the root issue is with vmware.
I have no clue what could possibly cause something like this to occur, as IP address assignment is one of those things that normally "just works". I am thinking about just switching to VitualBox, as I have used it in the past and never had a problem (except with properly running Windows 8. However,I have never actually seen any article suggesting VirtualVox over WMWare, as the latter supposedly performs better and has more intuitive shared folder support. Obviously though, any benefit from a shared folder is negated if it just shares corrupt garbage.
Answer: So you manually set a MAC address on your VM? In the past, I've seen VM's change MACs quite often; generally only after a reboot or cold start.  It shouldn't happen on the fly... You could install Wireshark and grab a few packet captures to see if anything in there points you in the direction of the root cause.


Covariance of two random variables with monotone transformation

Question: Suppose I know that, for two random variables $X,Y$, we have
$$Cov(X,Y)\neq 0.$$
What happens if we take a monotone transformation of $X$; will the inequality persist? That is, say $f(.)$ is a strictly increasing (not necessarily linear) function, then does it follow that
$$Cov(f(X),Y)\neq 0?$$
Since covariance depicts linear relationships, this would only follow if $f$ were linear, right? In that case, under which assumptions on the relation of $X,Y$ can we say that it does follow for nonlinear transformations $f$? (If it helps, I am currently looking at the case where $f$ is the cdf of the standard normal $\Phi$.)
Answer: (I have edited the question changing $f$ to $g$ in order to keep the symbol $f$ for probability density functions).
First, indeed, when the transformation is linear(affine actually), the covariance involving the transformation will continue to be non-zero. For non-linear transformations you cannot obtain very specific conditions. My approach is the following:
For compactness, set $Z=g\left(X\right)$. Then you require the conditions under which
$$\operatorname{Cov}(Z,Y)\neq 0? \;\Rightarrow\; E\left(ZY\right)\,-\,E\left(Z\right)E\left(Y\right)\, \neq0?\qquad [1]$$
Using Integrals to express the expected values, we require (assuming certain regularity conditions) $$\int_{D_Z} \int_{D_Y} zyf_{Z,Y}\left(z,y\right)\,dy\,dz \;-\;\left(\int_{D_Z}zf_{Z}\left(z\right)dz\right) \left(\int_{D_Y} yf_{Y}\left(y\right)\,dy\right)\neq0?$$
$$\Rightarrow \int_{D_Z} \int_{D_Y} zyf_{Z,Y}\left(z,y\right)\,dy\,dz \;-\;\int_{D_Z} \int_{D_Y} zyf_{Z}\left(z\right)f_{Y}\left(y\right)\,dy\,dz\neq0?$$
$$\Rightarrow \int_{D_Z} \int_{D_Y} zy\left[f_{Z,Y}\left(z,y\right) \;-\;f_{Z}\left(z\right)f_{Y}\left(y\right)\right]\,dy\,dz\neq0?\qquad[2]$$
where the $f()$ functions are joint or marginal probability density functions, and $D$ denotes the support. Now define $\omega \equiv(z,y)$ and note that $zy\;$,$\,z\;$,$\;y\;$, can all be derived as linear transformations of $\omega$, say $zy=\tau_0(\omega)\;$,$\,z=\tau_1(\omega)\;$,$\;y=\tau_2(\omega)\;$. For example $z=\omega[1\; 0]'$, the prime denoting the transpose. Also define $C\equiv D_Z\times D_Y$. Then we can compact even further by writing
$$\cdots\Rightarrow \int_C h\left[\omega,\tau_0(\omega),\tau_1(\omega),\tau_2(\omega)\right]d\omega\neq0?\qquad[3]$$
Why all this trouble? Because now it is clear that what you are asking is the requirements under which a double definite integral of a function of its variable does not equal zero when integrated over the whole domain of its variable. Reversely, we want to exclude the conditions under which this integral equals zero.
First since $\operatorname{Cov}(X,Y)\neq 0\;$, it implies that the random variables $X$ and $Y$ are not independent. Therefore we have $f_{Z,Y}\left(z,y\right) \;\neq\;f_X \left(x\right)f_Y \left(y\right)$ . But this does not guarantee that the double integral will be non-zero. It may still equal zero if the integrand in eq.[2] is an odd function  w.r.t. to at least one of $z$ or $y$, and if the domain w.r.t to this variable is symmetric around the origin (which will hold if for example $D=  \Bbb R$ - extended). Assume that indeed $C=\Bbb R\times\Bbb R$ extended. Therefore we must guarantee that the integrand
$zy\left[f_{Z,Y}\left(z,y\right) \;-\;f_{Z}\left(z\right)f_{Y}\left(y\right)\right]$
is not an odd function w.r.t. to either $z$ or $y$ separately.  Now if you work the integrand, substituting $-z$ for $z$ etc, you will be able to reduce the condition into the following: "The non-linear transformation should not be such as to make both $f_{Z,Y}\left(z,y\right) \;$ and $f_{Z}\left(z\right)$ even functions of $z$."
For $Y$ this could be stated as "If the pdf of $Y$ is an even function, then the non-linear transformation should not be such as to make  $f_{Z,Y}\left(z,y\right) \;$ an even function of $y$.".  Obviously this is distribution-specific (or at least distribution family- specific) and you cannot go more specific than that. If it looks disappointingly general, imagine a true jungle. Then independence is a straight clear path through this jungle, and linear dependence is an elliptical clear path approaching asymptotically the straight path of independence. The rest of the jungle is non-linear dependence.
$$ \text{}$$
Afterthought: A less "mathematical" and more "statistical" condition (although not necessarily more useful) would be "The non-linear transformation must be such as to not render the variables $Z$ and $Y$ sub-independent". In a nutshell, sub-independent random variables are dependent but uncorrelated: their dependence is "perfectly non-linear".
Comment: Thanks for your answer. I see how the integrand in [2] being an odd function will make the covariance zero if we integrate over the reals. If I rewrite [2] for the odd $z$ part: $$\int_{-\infty}^0 f_{Z,Y}(z)-f_Z(z)f_y(z) dz+\int_{0}^\infty f_{Z,Y}(z)-f_Z(z)f_Y(z) dz$$ so $$\Leftrightarrow \int_0^\infty f_{Z,Y}(-z)-f_Z(-z)f_y(-z) +f_{Z,Y}(z)-f_Z(z)f_Y(z) dz\neq0.$$ But that means the transformation should not make both $f_{Z,Y}$ and $f_Z$ an odd function of $Z$, right? You write it should not make them an even function.
Comment: Don't forget the "z" outside the brackets!
Comment: Yes, you're right, they have to be even functions. But am I right that if both $X,Y$ are positive, then the "even function condition" is not applicable?
Both of you provided sufficient conditions and I tossed a virtual coin to determine who gets the accept. You won.
Comment: Luck is always welcome, although not chased. Indeed, the property of "evenness" or "oddness" can only be defined for functions whose domain includes both positive and negative values. If $X,Y$ have positive range => the domain of the densities will be only in the positive orthant.
Answer: Assume that $\mathrm{Cov}(X,Y)\ne0$ and consider some increasing function $g$ (you say that the case of interest is when $g=\Phi$ the standard normal CDF, which is increasing). The question is to find some conditions on $(X,Y)$ implying that $\mathrm{Cov}(g(X),Y)\ne0$, presumably for every such $g$.
Here is a sufficient condition: one knows that $E[Y\mid X]=h(X)$ for some function $h$ depending only on the joint distribution of $(X,Y)$. Then:

If $h$ is increasing or if $h$ is decreasing, then $\mathrm{Cov}(g(X),Y)\ne0$.

To prove this, assume without loss of generality that $h$ is increasing, note that $\mathrm{Cov}(g(X),Y)$ is also $\mathrm{Cov}(g(X),h(X))=E[g(X)h(X)]-E[g(X)]E[h(X)]$, and that for nondecreasing functions $g$ and $h$, $E[g(X)h(X)]\geqslant E[g(X)]E[h(X)]$ with equality if and only if $g$ or $h$ is almost surely constant on $D$ such that $P[X\in D]=1$. Here $g$ and $h$ are increasing everywhere hence $E[g(X)h(X)]\gt E[g(X)]E[h(X)]$ except if $X$ is almost surely constant but then $\mathrm{Cov}(X,Y)=0$ hence this does not happen and the claim is proved.
Of course, the partial result above is only useful when one can compute $E[Y\mid X]=h(X)$ easily. Here are two remarks in this direction. First, when $(X,Y)$ has a density,
$$
h(x)=f_X(x)^{-1}\int_{-\infty}^{+\infty}yf_{(X,Y)}(x,y)\mathrm dy,
$$
hence one should be able to check the monotony of $h$ in some specific situations.
Second, when $(X,Y)$ is normal, $h(x)=ax+b$ where $a=\mathrm{Cov}(X,Y)\sqrt{\mathrm{var}(Y)/\mathrm{var}(X)}$ hence $h$ is increasing if and only if $\mathrm{Cov}(X,Y)\gt0$.


Evaluating $\int(2x^2+1)e^{x^2}dx$

Question: $$\int(2x^2+1)e^{x^2}dx$$
The answer of course:
$$\int(2x^2+1)e^{x^2}\,dx=xe^{x^2}+C$$
But what kind of techniques we should use with problem like this ?
Comment: @Babak Sorouh thanks .
Answer: You can expand the integrand, and get
$$2x^2e^{x^2}+e^{x^2}=$$
$$x\cdot 2x e^{x^2}+1\cdot e^{x^2}=$$
Note that $x'=1$ and that $(e^{x^2})'=2xe^{x^2}$ so you get
$$=x\cdot (e^{x^2})'+(x)'\cdot e^{x^2}=(xe^{x^2})'$$
Thus you integral is $xe^{x^2}+C$. Of course, the above is integration by parts in disguise, but it is good to develop some observational skills with problems of this kind.
Answer: As @Peter note above, you can integrate this by separating the integrand and integrating by parts: \begin{align} \int(2x^2+1)e^{x^2}dx &=\int2x^2e^{x^2}dx+\int e^{x^2}dx\\ &=\int x(2xe^{x^2})dx+\int e^{x^2}dx\\ &= \int x\left(\frac{d}{dx}e^{x^2}\right)dx\ + \int e^{x^2}dx\\ &= xe^{x^2}-\int e^{x^2}dx+ \int e^{x^2}dx= xe^{x^2} + C \end{align}
Comment: $\quad$:+)$\quad$
Answer: Look at the integral
$$\int 2x^2e^{x^2}\, dx.$$ Try integrating by parts as follows.
$u = x$, $dv = 2xe^{x^2}dx$, $v = e^{x^2}$, $du = dx$.
Answer: My solution is essentially based on a theorem proved by Liouville in 1835, applied to your problem: There has to be a polynomial function of linear nature that satisfies an antiderivative of the form (ax+b) times the e-power from your integral.
Left and right differentiation and "removing" the e-power gives (2x^2+1)=(2ax^2+2bx+a)
Matching coefficients according to corresponding polynomial terms yields a = 1 and b = 0
I do admit, it may not be the most beautiful solution, but Liouville's theorem is the reason why the above methods actually come out good!

Evaluating $\int(2x^2+1)e^{x^2}dx$


Help finding an alternative to the time average of the Poynting vector

Question: The time averaged Poynting vector is:
$$\langle \vec{S} \rangle = \frac{1}{T}\int_0^T\frac{1}{\mu_0}(\vec{E} \times \vec{B}) dt$$
But I am not given the total time of the event. Is there another way to find the time average of the Poynting vector here?
The context:
A plane wave hits a conducting surface at an angle $\theta_I$ and I need to find the radiation pressure given by:
$$P = \frac{2\langle S \rangle}{c}\cos ^2(\theta_I)$$
Based on an answer given that said that I should use the real parts the electric and magnetic fields, I've found $\vec{E} \times \vec{B}$ to be:
$$\vec{E} \times \vec{B} = (E_0)^2 \left [\cos(zk_I\sin(\theta)+xk_I\cos(\theta)-wt)\right ]^2 \left [\frac{\sin \theta_I}{c}\hat x + \frac{\cos \theta_I}{c}\hat z \right ]$$
Answer: You seem to have two misconceptions:

If $\vec{E}$ is defined to be the real part of the complex field $\vec{E}_c$, where $\vec{E}_c$ is a mathematical tool and not physical. The formula for the Poynting vector involves $\vec{E}\times \vec{B}$, which is very much NOT the real part of $\tilde{E}_c\times \tilde{B}_c$. That is to say, you WANT to calculate  $\mathrm{Re}(\vec{E}_c)\times \mathrm{Re}(\vec{B}_c)$, not $\mathrm{Re}(\vec{E}_c\times\vec{B}_c)$ (which is what you have calculated). If you proceeded with the cross product you have in your post, you would in fact find the time average of $\vec{E}_c\times\vec{B}_c$ to be zero!
For computing a time average, you don't need the duration of the event. The time average as $T\to \infty$ and the time average for $T$ equal to one period (which would be $T=2\pi/\omega$) are equal, so you can compute either. For example, $$\frac{1}{2\pi} \int_0^{2\pi} \sin^2(\theta)d\theta=\lim_{T\to \infty} \frac{1}{T} \int_0^{T}\sin^2(\theta)d\theta=\frac{1}{2}$$(the graph of sine squared is centered around 1/2, so this agrees with intuition)

To solve the problem, recalculate the cross product correctly as $\mathrm{Re}(\vec{E}_c)\times \mathrm{Re}(\vec{B}_c)$, and then plug in $T=2\pi/\omega$.
Comment: Ok so I've edited my question to use the real parts. But now something doesn't make sense - I can't really do the time average here because all the $\theta \to \theta_I$. These are fixed angles and therefore fixed values for $\sin \theta_I$ and $\cos \theta_I$. Does that make sense?
Comment: Furthermore, do I use $\vec{E_I}$ or $\vec{E_R}$ when computing the cross product? What about for $\vec{B}$?
Comment: @loltospoon There is no time dependence on theta and you should look at your book for what "complex fields" mean in the first place. The time average of the whole cosine squared term is just going to be 1/2. You really should try understanding why the equations are what they are and why complex numbers are used, instead of just plugging things into equations you are given!
Answer: The canonical expression, valid in the far field, is to use the (complex) phasor form $\vec E_s$ and $\vec B_s$.  In this form the Poynting vector is simply
$$
\langle\vec S\rangle =\frac{1}{2\mu}\hbox{Re}\left[\vec E_s\times \vec B_s^*\right]\, . \tag{1}
$$
You can find a derivation of this result on this wiki page.  The Poynting vector is usually calculated in terms of $\vec H$, rather than $\vec B$.

Edit: for propagation in a lossless medium the result is just
$$
\langle \vec S\rangle =\frac{1}{2}\frac{\vert \vec E\vert^2}{\eta_0}\hat k
$$
where $\eta_0\approx 120\pi$ is the impedance of vacuum, and $\hat k$ is a unit vector in the direction of propagation: $\hat k= \vec k/\vert \vec k\vert$.
Comment: I'm sorry to say this but we haven't really learned about phasors, so I don't really understand any of this.
Comment: Write your $\vec E$ and $\vec B$ in terms of exponentials and strip off the time-dependent factor.  This should give you the phasor, i.e. something like $$ \vec E\to \vec E_s: = \hat n_\perp E_{0}e^{i k (x\sin\theta_i-z\cos\theta_i)}.$$  Here, $\hat n_\perp$ is a unit vector in the direction of $\vec E$ that is perpendicular to the direction of propagation. Do same with $\vec B$.  Proceed as per (1).


Having multiple keywords separated by comma in the title of the page?

Question: I want to target multiple keywords with same piece of content (I am thinking to add multiple keywords separated by commas in title and description of the page). So, is it the right approach or I am making any mistake?
Comment: Is your only purpose in doing this to rank better for the comma-separated keywords?
Comment: Yes. I want to rank easily for long trailing (multiple [related to each other]) keywords with the same page. What I am thinking wrong?
Answer: by using multiple related keywords in the title and description separated by commas, you'll actually be creating a bad overall user experience for your readers and users.
I'd suggest not using multiple keywords (even if they are related) in your title and meta description. Instead, only focus on using your main target keyword once or twice.
Comment: Seeing as these keywords are closely related I don't think you'll be making a mistake by using all of them on the same page. Here are my suggestions:

Title: Upcoming NFT and NFT Drops - Check out our NFT calendar
Description: Stay updated on the latest NFT drops and upcoming non-fungible tokens by using our NFT calendar.
Comment: Thanks @Ruben I would love to ask for an example.
Comment: @ManishShahi well if you could share the target keyword for your page I can give you my input?
Comment: NFT Calendar, Upcoming NFT and NFT Drop
Comment: That's clean. Thank you so much.
Answer: I've always used a pipe character (|) to separate my keyword strings, but Google also recognizes commas. I use a similar approach to my SEO; I include keyword phrases that are supported by the content and try to limit it to no more than 3 very closely related terms. Example here: https://driltech.net/mud-motor-drilling.php
If I have trouble narrowing it down I look at the possibility I need to refine my keywords or I need to move some of the content to a page of its own. Each page is your elevator pitch, it should be focused and precise(ish). This is always an issue when I am trying to optimize overview-type pages because it covers a broad swath of content and you know you can't win on every term.
I found this article about writing title tags and meta descriptions, it is from 2015 but everything pretty much still rings true.
Comment: @ManishShahi I don't stress about them being the same as long as the H1 is the "most right" choice for the page and the user experience. Example 1. https://www.mavtechglobal.com/solutions/dcsnext/ Example 2. https://www.mavtechglobal.com/enterprise-integration/  Screaming Frog is brilliant for helping you track your titles, H1s, and tons more, even the free version.  Maybe think of title tags as description(s), and the H1 as the introduction.
Comment: Thanks @TexasB. I have a quick question from the link that you referred in the last paragraph. I want to ask as per article title and H1 tag shouldn't be same .i.e, repeating. Can you explain why? And what should we do to make H1 tag differentiate than the TITLE tag of the page?


Should I use the word "might" to express someone's wish/hope in a past situation?

Question: The dictionaries are very clear as to the use of "may" to express a wish or a hope - and one can easily infer that a sense of future is always implied in such use. For example: "May she rest in peace"; "May you achieve all you wish"; "May the Force be with you". However, when it comes to expressing someone's wish or hope in a past situation, it seems to me the use of "might" - which would be a "natural substitute" to the word "may" - sounds somewhat strange. Let's take the following example: "I was going through the most difficult moment of my life. Might God help me!". Is the use of "might" correct in this sentence? If not, what would be the right way to express such hope/wish in the past?
Comment: @WS2 Are you happy with 'might' rather than 'may' there?
Comment: @WS2 'May God help me!' is a fixed phrase, using a very special usage of 'may'. I don't believe 'might' is usable instead here: this is an idiom, in this case with restricted choice of tense. See  TaliesinMerlin's final bolded assertion, and Lambie's answer.
Comment: I believe that @Lambie and Edwin Ashworth have persuaded me I am wrong on this, so I am going to delete my comment and my answer below. The point is that "may" in this instance is not being used in the indicative mood - so does not have a past tense. (It does not gainsay the fact that "might" is the past of "may" when used indicatively.) I apologise if I have given you  a wrong steer.
Comment: Effectively the only way I can think of saying what is intended in the final example of your  question is *"May God help me", I thought*.
Comment: Qu'à Dieu ne plaise. On ne prête qu'aux riches, malheureusement. –
Comment: @EdwinAshworth I don't believe I would actually use it - if only for its lack of clarity. But it is another thing altogether to say it is not grammatical.
Comment: @WS I'm saying it's unacceptable on non-idiomatic grounds. [Google ngrams for may God help me / might God help me / may God help them / might God help them](https://books.google.com/ngrams/graph?content=may+God+help+me%2Cmight+God+help+me%2Cmay+God+help+them%2Cmight+God+help+them&year_start=1800&year_end=2008&corpus=15&smoothing=3&share=&direct_url=t1%3B%2Cmay%20God%20help%20me%3B%2Cc0%3B.t1%3B%2Cmay%20God%20help%20them%3B%2Cc0) (where they exist). And even if argued to be grammatical, they don't employ the mast tense 'might' of 'may', but a different 'might'.
Comment: @EdwinAshworth But presumably you accept "He asked if he might smoke" as legitimate?
Answer: I think a little grammatical clarity is needed:
First of all:
"May she rest in peace". is uttered at a present time. It is a spoken phrase.
In the unusual case that one wants to report on that (reported speech), one could say:
"She was laid gently in the grave so she might rest in peace."
In that instance, might is not really the past tense of may. As May [head word] in a sentence is not a modal verb. might rest in peace = have the possibility of resting in peace, is a modal and here, it is in the past tense.
Present tense: She may rest in peace if laid gently in the grave. Tricky wicky is what this is.
In in terms of the OP's question, let's put it in the present tense:
"I am going through the most difficult moment of my life. May God help me!".
That is grammatical. However, in the past tense, you can't just use might for may in the utterance as May [something occur] is an exhortation to express a wish or a hope and is not a modal verb as in:

He may arrive early tomorrow. [possibility]
or
He might arrive early tomorrow. [also possibility but less likely]

To express May God help me! in direct speech in the past, one would have to use something like:
"I was going through the most difficult moment of my life. I asked for God's help!".
OR
"Would that God had helped me."
In sum:

may [verb] and might [verb] can be modals in the present referring to a future.

might can sometimes be used as a past tense of may in modal form.

may [something occur] can be a head word used to express hope or a wish (a formal usage). In this sense, there is no past tense for it and a workaround must be used.
Comment: @Tuffy Please don't do this. Your grammar in those sentences simply does not work in English and Greek is completely irrelevant, pedakimou. We also use (somewhat archaic): **Would that the UK had ordered to lockdown** before it did. [That is "to will a past event not to have happened." By the way, I am not puzzled at all.
Comment: This is a very clear answer, taking this use of 'may' as a virtual imperative.  I just have one small reservation.  To the extent that the impossibility of altering the past (or time travel) is a logical but not a grammatical impossibility, you could use 'may' of the past.  "May Iphigeneia not have died at Aulis" or "May the lockdown in the UK have been ordered two weeks before it was."  Even though the though it is logical nonsense (because, if fulfilled, it must be true that "the lockdown in the UK was ordered two weeks before it was ordered" is both true and false), it is not ungrammatical.
Comment: Zi don't think so.  That would be equivalent to not the Greek imperative/subjunctive  but to the Greek optative with 'eithe' or 'ei gar') -i 'if only'.  That does not produce a logical contradiction.  It does not, in a sense have a truth value at all.  It is expresses a wish.
Comment: @Tuffy What the heck has Greek got to do with it? Your grammar is faulty in both sentences. "May the lockdown in the UK have been ordered two weeks before it was." *May* and *have been ordered* [present perfect]  do not work. **If only the lockdown *had been ordered* two weeks before it was**. is the idea.
Comment: I understand why you are puzzled.  The idea that someone can will a past event not to have happened stretches the credulity for reasons I explained.  But if someone (illogically) thought it possible, the sentence I have given is grammatically correct.  It follows the sequence of tense.  They are (now) willing that in the future it comes about that an event which has happened in the past will not have happened.
Comment: @Lambie  I believe you are confusing two entirely different senses of might. "He might arrive early tomorrow" means that there is a possibility that he will arrive. That is "might", meaning "will possibly*. However the OP here is talking about the past tense of may - he asked if he might arrive on Tuesday. That is the past tense of may. Hence I believe the OP had it exactly right to begin with.
Comment: @WS2 You clearly have not read the OP's question properly: "I was going through the most difficult moment of my life. Might God help me!". A -**past situation** states the OP. And I addressed that in my answer.
Answer: In short, making may into might doesn't make it clear when the wish was made, and so you'll need an alternative strategy (like direct speech) to make your intent clear.
May is indeed used to wish
The usage of may you're asking about expresses a hope or wish while inverting may and the subject. The Oxford English Dictionary documents this usage in def. 12 of "may, v.1":

Used (with inversion of verb and subject) in exclamatory expressions of wish (synonymous with the simple present subjunctive, which (exc. poetic and rhetorically) it has superseded).

All of the quoted examples of this usage use may and not might, e.g.,

1986   B. Gilroy Frangipani House vii. 30   May your soul never wander and may you find eternal peace.

Might was historically used to wish (with a present tense context)
You're wondering whether, nonetheless, one could treat might as an ordinary past tense version of may to represent a past wish. On its face, using might in this way is possible, but it runs into an issue - might could already represent a wish in the present. The OED classifies a wishing usage with might under definition 24:

Used (since Middle English with inversion of verb and subject) in exclamatory expressions of wish (sometimes when the realization of the wish is thought hardly possible). poetic. Perhaps Obsolete.

For instance:

1852   M. Arnold To Marguerite in Empedocles 97   Now round us spreads the watery plain—Oh might our marges meet again!

There is no sense of past tense in this particular usage of might, since the previous sentence is already in present tense.
Might is a lot less common for wishing today
Besides that, using might for wishing is a lot less common than using may in the same position. As an example when I did a search for "May_v God" in the Corpus of Contemporary American English, I had dozens of results that expressed a wish on the first page. For "Might_v God," I had one result in total, from the comments section (now scrubbed from the web, but documented in COCA's database) of an opinion piece in Washington Monthly:

You get the government you deserve.? Our forefather? s got good government. We? ve got Bush. In surrendering our rights and freedoms we are descending into cowardice and captivity. Might God help us, for we are not helping ourselves.

Again, might is used in a present tense context: we are descending. I conclude that might, when used to wish, doesn't have a strong temporal sense.
Might doesn't draw the neat temporal distinction you want
All of this helps explain how I initially read your usage. It felt odd, but I couldn't put my finger on it. I think it's because, when I read "Might God help me!" I don't read it as past tense. I read it as you - right now, narrating a past event - wishing for God to help either you or (in context) your past self. If it's you wishing this, I don't see the need for might to draw a distinction in tense.
Maybe another difficulty is a function of "May God help me!" being an interjection or exclamation; I'll read it as pertaining to whoever is immediately saying it whether you use may or might. If you really want to express someone's wish in a past situation, you could make it clear that someone in the past was saying or thinking this by using a reporting verb. For example:

I was going through the most difficult moment of my life. 'May God help me!' I thought.
Comment: I wonder why you make your answer so convoluted. Obviously, the OP's first problem lies in confusing May you [x] with: I may go. And in the sense of "I may go", one can say might instead of may, which is even less probable.
Comment: I don't follow what you think is obvious here. I didn't see OP using "I may go" or a similar form. I also don't think it's safe to assume that distinctions in usage between "I may go" and "I might go" necessarily pertain to wish-may and wish-might. I felt the need to show my work, and that includes examples.


Can a company be sued for reporting unpaid bills to lower credit score and send the bill to collections if their payment method doesn't work?

Question: I'm trying to pay a bill but I exhausted all means to pay the bill using their payment methods. Now they sent me an email threatening to send the bill to collection since I have an unpaid bill. I have evidence that I tried to pay using different methods and the email doesn't offer any other method. If the company tries to collect using a collection agency or goes to court, am I still liable for collection fees or court fees? The jurisdiction is in California.
Comment: I'm not entirely clear on this, are you saying that **all** of the payment methods offered by this company aren't working? Are they not working for technical reasons? Even paying by card over the phone? And the company has refused to accept that there is a problem? Have you actually spoken to them about their payment methods not working?
Comment: Or are you not able to use any of the offered payment options for some reason or another?
Comment: @GeoffAtkins Company only accepts payments over their payments gateway. Over the phone they told me to go and pay through their payment gateway.
Comment: @Trish There must be a glitch on their website. They are charging for something extra, so it is a payment method that doesn't seem to be very maintained.
Comment: So, does their payment gateway not work? If so, you need to speak to them and escalate. The burden of proof is on you to demonstrate that you have attempted to pay and been unable due to problems at their end.
Comment: Write them back (at their registered agent address, if necessary) and enclose a check in the amount owed.
Answer: No
You owe a contractual fee for a service or good, which is to be paid in a manner of the companies choosing. Invoices contain information about what to pay and till when. This is also known as debt and the payment date is a clear indication of when to pay. You have tried to pay your debt, but an alternative method not offered is not in a manner that satisfies your contractual obligation. So you have not paid this debt.
They informed you that they will sell the right to that outstanding amount to another company or employ another company to try and get their money back. You are liable to pay your debts in a manner the company you have the debt with chooses. You are also liable for damages your non-payment creates, such as collection or court fees.
Reporting about your unpaid debt in truth is not libel or any other tort, but might be an obligation for financial service providers. Unpaid debt impacts your credit score.
However, if their payment methods accepted don't work at all, you need to inform them of such so they can investigate and possibly decide to take a different method - but which is up to the company.
Comment: Thanks for the explanation. Still, before a civil lawsuit, aren't they obliged to show that they requested the payment? I remember having to show in court that I sent a letter requesting the payment before I could file a lawsuit.
Comment: they need to show you owe a payment and that there was a payment date, which usually is already on each invoice ("Payment due till XY")


How can I better work coffee into my morning routine

Question: I am a pretty lazy person and sometimes it takes me awhile to get around to making coffee. I have a french press and I grind my own coffee so it takes a little longer. What are some simple ways I can speed up my coffee making process or better work it into my morning routine without sacrificing quality?
Comment: You can try to get your hands on one of these http://www.joshrenoufdesign.com/new-gallery-5/
You have hot coffee ready as soon as you wake up, unless you snooze a lot :)
Comment: Can you clarify if you're looking for new brewing methods or if you're looking for streamlining your existing system?
Comment: @schvaba Ok, you just blew my mind!
Comment: You should sign up, maybe it speeds up the development :)
Answer: I had a similar problem and I ended with aeropress. To speed up the process I use pre-grinded coffee which I store in the fridge. Once you press the coffee, the cleanup is very fast.
Comment: Cleanup is very fast with the aeropress
Answer: I am pretty ineffective before my first cup, and my family doesn't want to hear anything since I'm the first one up. So I make it easier on myself and them:

grind enough for the first cup the night before. Store grounds at room temperature, not in the fridge. This is a slight tradeoff in quality. If you don't need silence, then just have the beans ready to grind.
wash out the press and leave it ready
prep the water (I use a hot water dispenser built in to my sink)
and your cup
do it exactly the same way every day so you don't have to think or decide
Answer: Though now deleted, I most prefer @EdChum's comment: brew a pot of cold-brew the night before! Pour and enjoy. Cannot get easier or tastier.
For nearly instant hot coffee, use instant coffee powder. Mix boiled water with coffee powder. Attempt to drink while you prepare a better-tasting cup of coffee. See tag instant.
For fast, automatic laziness -- yielding a quick, easy cup of passable coffee -- you can't beat pod-based automatic machines. Fill the pitcher with water, insert pod, press button... done. The only thing to mess up could be forgetting to place a coffee cup below the spigot. Clean-up is practically non-existent. I personally find the quality of this coffee to be low and unsatisfying.
Otherwise, streamline your French Press routine:

Start your water boiling. Go about other business for a few minutes (or stare at the wall if other business is impossible before coffee).
When the water has boiled, grind your coffee and put it in your сafetière (French Press). Add water; submerge coffee. Wait.
Press and drink.
Begin clean-up: dump grounds into compost and allow to dry on its own (a few hours).
Shake out dried coffee and rinse; repeat as necessary.

Commentary:

This permits really only 3 steps of brewing: boil, mix, press. This is about as simple as possible. By grinding after boiling, this minimizes the amount of time between grind and brew (crucial minutes!) and minimizes the overhead of manual interaction.
By using the multi-step clean-up process that I describe, this maximizes efficiency: get most of the coffee out by shaking when still wet (just brewed). Allow to dry on its own, and the rest of the coffee grounds will come out easily. A simple rinse is enough to get most of the remnants out without clogging a sink drain.

Long live manual coffee production methods!  See other guides from Coffee Geek, Blue Bottle, and How To Brew Coffee -- I prefer a simpler approach that I describe than all of those.
Comment: "Attempt to drink," that is really funny. I have never tried instant coffee but I have heard it is not that good.


Possible to have an RDS > 3072GB on AWS?

Question: I am a complete newb to AWS. I have a MYSQL database that is about 4TB and growing each day and would like to move it to AWS. According to this link the maximum allocated storage size is 3072 GB. How can I run my DB on AWS? (I can think of a dozen of AWS's customers that have a DB > 3TB so there must be a way, right?)
Answer: 3TB is massive for the majority of Amazon's customers
The majority of Amazon's customers consume far less than 3TB databases in fact the majority will use less than 5GB. The problem with with databases or generally those offered by services such as MySQL is that the bigger the database the more resources they use. It would be economy not viable for most to host databases even anywhere close to that because 3TB is over $300 in storage costs a month along with bandwidth costs on the RDS instance it becomes very expensive to do so unless you have very good margins on your products and services.
Multiple RDS Instances
The work around to this problem is to use multiple RDS databases, I suspect that when a database gets to a certain size it becomes better to use multiple databases unless you use some kind of horizontal scaling which I believe SQL does not unless its MongoDB or similar.
Answer: http://aws.amazon.com/rds/aurora/, a custom MySQL from Amazon, supposedly 100% compatible and faster, will scale up to 64TB when released in final. Now in preview (5/2015).


Why aren't my post showing under my custom menu?

Question: I have tried everything I've read through google search as well as through the wordpress forums and support. I have created the custom menus using the categories and adding my post I write with the proper categories in hopes they will show on my webpage but when i go to my page it's now showing anything I post anywhere and when I go through my dashboard click on a published post and hit view it sends me to my page with an error message saying the content I was trying to find isn't available. What am I doing wrong?!?!?! I want to take my page off maintence and start working on views but it's difficult when i have ZERO content wanting to show under my tabs no matter what I do. I've posted a screen shot on this question to show how I've created my custom menu
Comment: do you have the category.php template in your theme folder?
Answer: For the error message "not available": have you tried rebuilding your permalinks?
Try: Admin > Aettings > Permalinks
…then hit "save changes" to rebuild the permalinks. I hope this helps to solve the issue.


Can results be predicted?

Question: I wanted to know that, what can we assume as the result of some experiment which we have not conducted on the basis of mathematical proofs? I mean, in general, equations are created after analyzing the real things that we have studied. But can we assume the being of things in a way based on mathematical proofs?
Comment: What do you mean by "assume"?  You can assume anything, with or without support, but you may be wrong.  If you predict something using a mathematical model, you can't be sure it's correct.  You might have very good reason to expect it to be correct, but that's not the same thing.  You might assume it as a basis for further reasoning, but it still might not be correct.
Comment: Generally empiricism is used to 'prove' a theory. The theory is often mathematical in origin. But clearly the experiment is intended to prove the theory.
Answer: Of course YES.
See Dirac's prediction (1928) of the existence of the positron (observed 1929) as well as Higgs boson, predicted in 1964 and whose existence has been confirmed in 2013.
Comment: "can we assume the being of something in a way based on mathematical proofs?" The answer is YES. See Dirac mathematical prediction of the *positron* as a solution of an equation.
Comment: Thanks for the answer but i didn't want examples.
Comment: Dirac also predicted the magnetic monopole, as I remember, and nobody's found evidence of them.
Answer:
I wanted to know that can we assume the result of some experiment
  which we have not conducted on the basis of mathematical proofs?

The answer is yes, assuming that the experimenter starts with good data. My example here is the discovery of the planet Neptune (1846). Astronomers had noticed errors in their description of the orbit of the planet Uranus, at that time the farthest known.
The errors could be explained by the presence of a farther planet. Astronomers worked on a likely location for the new object.
When the calculations were finished, they put the new planet behind the sun in relation to the earth. So the location of the new planet could not be known; it could only be predicted. But when that part of the sky became visible again, the planet was there, as the mathematics had shown.
Source: Pannekoek, A. 1969. A history of astronomy (New York: Barnes and Noble, Inc.)

Can results be predicted?


What is a homographic solution in three body problem?

Question: I came across Saari's homographic conjecture in Three Body problem.
I need more information on what exactly is a homographic solution and how is it different from a homothetic solution?
Answer: A homographic solution is one in which the configuration of all the masses is preserved for all time. One example of this are the relative equilibrium solutions. Each mass rotates at a fixed distance from the center of mass at constant angular velocity. For the 3BP these are the colinear solutions found by Euler and the equilateral triangle configurations found by Lagrange.
A homothetic solution is one in which the scale size changes while retaining the configuration and no rotation is permitted. An example of this would be a total symmetric collapse: start 3 equal masses from rest at the vertices of an equilateral triangle and they will retain the configuration as they collapse to the origin.
Both of these solution types are examples of central configurations. Saari talks about these things extensively in his book: "Collisions, Rings, and other Newtonian N-Body Problems"
Comment: Thanks a lot.  That was very helpful
Answer: JEM's answer is not exactly true.  I am not an expert on the subject but I have also looked at many of the relevant papers (for eg. this one and this one) out of curiosity.
From these (also see good summary here) you can see that, in the context of central configurations and Saari's homographic conjecture, a "homographic configuration" of masses is one which is invariant under scaling and rotation.
Formally, the configuration is said to be homographic if at any time $t$ the coordinates of the masses in the configuration $q(t)$ satisfy: $$q(t)=\lambda(t)Q(t)q(t_0)$$ where $Q(t)$ is a rotation matrix, $\lambda(t)$ is a time-dependent scalar, and $t_0$ is some specific time in the orbit of the configuration.
JEM's answer refers to Saari's original conjecture and a special restrictive case of homographic solutions wherein there is rotation but no scaling--the $\lambda(t)$ drops out of the equation above and the masses rotate as a rigid body (with constant moment of inertia $I$).
In Saari's extended or homographic conjecture, the concern is with a wider range of homographic configurations that scale and rotate with time ($I$ is not constant but $UI$ is constant, where $U$ is the potential energy of the system).
Note, however, that this notion of homography is more restrictive than the usual definition of homography from projective geometry (see for eg. this, and this, and this MSE post).  It seems to me a better way of referring to n-body orbital configurations that are invariant under scaling and rotation (as one of the lead researchers Moeckel suggests), is to call them self-similar configurations.
Comment: Thank you so much ben. That gave a deeper understanding. Greatly appreciated.


How can the translators of the New World Translation say that their version is near what was originally written?

Question: The New World Translation of the Bible says much concerning the original Bible writings. As none are known to exist, how can the translators of the New World Bible, JWs, say that anything is anywhere near what was originally  written?
Comment: The answer to "how can translators know that anything is anywhere near what was originally written" can be found at [What is "Manuscript Evidence" and how is it useful?](http://christianity.stackexchange.com/questions/9934/what-is-manuscript-evidence-and-how-is-it-useful?s=1|2.0404)  That addresses all written works of antiquity, not just the Bible, but shows how the Bible has superior reason for being trusted as accurately transmitted through time compared to other works that aren't in dispute..
Comment: Please reference a quote from the translation committee that gives rise to your question
Comment: I'm not sure what you mean by the NWT saying "much concerning the original biblical writings".  It seems you're asking "how can the NWT be an accurate translation when we don't have any original manuscripts."  But that's really two questions: 1. How do we know what the original manuscripts said, and 2. How accurately does the NWT translate the original manuscripts.  For #2, see http://christianity.stackexchange.com/q/14596/32341.  #1 is almost certainly also to be found in this site.
Comment: NWT Translators say that they translated from original languages Hebrew Greek koine Greek Aramaic they make no claims to having translated from original writings like they found Moses' autographed original Pentateuch
Answer: Jehovah's Witnesses base their faith in the preservation of the Bible on the merits of scholarly studies of thousands of manuscripts, as well as on the reasoning that if the Bible is truly inspired of God, Jehovah will not allow its message to be lost (Isaiah 40:8). The New World Translation was translated using the same master texts that many other Bibles use.
Jehovah's Witnesses have published many articles about the subject. Some central resources on the subject are the Insight book and Appendix A of the NWT. The appendix states:

Not all copies of ancient Bible manuscripts contain identical wording. How, then, can we know what the original text contained?
The situation could be likened to that of a teacher who asks 100 students to copy a chapter of a book. Even if the original chapter was later lost, a comparison of the 100 copies would still reveal the original text. While each student might make some errors, it is highly unlikely that all the students would make exactly the same ones. Similarly, when scholars compare the thousands of fragments and copies of ancient Bible books available to them, they can detect copyist error and determine the original wording.
How confident can we be that the thoughts contained in the original Bible texts have been accurately transmitted to us? Commenting on the text of the Hebrew Scriptures, scholar William H. Green stated: "It may be safely said that no other work of antiquity has been so accurately transmitted." Regarding the Christian Greek Scriptures, or so-called New Testament, Bible scholar F. F. Bruce wrote: "The evidence for our New Testament writings is ever so much greater than the evidence for many writings of classical authors, the authenticity of which no one dreams of  questioning." He also said: "If the New Testament were a collection of secular writings, their authenticity would generally be regarded as beyond all doubt."

Some sources from the Insight book can be found here and here.
Comment: This article, published by the Church of Jesus Christ of Latter-day Saints, describes some of the problems and challenges of translating the Bible: https://www.lds.org/ensign/1990/01/the-bible-only-4263-languages-to-go?lang=eng  While the article is nearly thirty years old (and from a different denomination), much of the information is still very relevant to the topic of translations in general.


Lingustic relationship between plural suffixes in Turkish and -er in some Nordic/Germanic languages

Question: What's the linguistic relationship between plural suffixes "-ler/-lar" in Turkish and "-er" in some Nordic/Germanic languages?
Comment: Got any examples? Such brief questions often don't captivate the audience.
Comment: There is no relationship, it's a coincidence.
Answer: This similarity is rather coincidental. According to widely accepted theories, Turkish and Germanic languages aren't cognates. And plural endings usually don't get borrowed from one language to another, except a part of a noun.
Are you sure that the Turkish plural ending is ''-er'' and ''-ar''? Not ''-ler'' and ''-lar''? For example: kitap ("book") — kitaplar, medrese — medreseler.
The Scandinavian (Norwegian) examples could be: bok ("book") — bøker, skole ("school") — skoler. So ''-er'', ''-r'', ''-ar'' without ''l''.
Comment: No. They might be similar by mere conincidence as well. Example: Dutch "daarin" ("therein") and Persian "dar în" ("in that"). They are completely unrelated, even if they look the same and mean the same.
Comment: If you find several hundred such pairs, all showing regular sound variations, then you have the basis to argue for some kind of relationship. But without several hundred such pairs between specific languages, the odds are overwhelming it's coincidence.
Comment: If "plural endings usually don't get borrowed from one language to another", then their similarity shows an intrinsic relation between languages, doesn't it?
Comment: "plural endings usually don't get borrowed from one language to another" - does not English borrow them quite frequently?
Answer: In Nordic Germanic languages the final -r came from -s via regular sound change.
Comment: Are you sure?  Their proto-language pluralised via stem changes as much as suffixes.
Comment: @AdamBittlingmayer We can be sure because the final s is still found in Gothic.


High temperature phonon heat capacity and equipartition theorem

Question: I have seen the derivation of the high temperature heat capatcity sing integration of the density of states. The heat capacity is $C=3Nk_B$ with $N$ the number of atoms in the lattice. According to my lecturer 'this is in agreement with the equipartition theorem'. While I agree that the dependence on the temperature and the number of atoms is the same, I would think that the equiparition theorem would predict $C=\frac{3}{2}Nk_B$, i.e. $1/2 k_B T$ of energy per quadratic degree of freedom, and there are 3N modes: N atoms, 2 transverse and 1 lngitudinal vibrational mode. Where did the factor of 2 go?
Answer: There are actually six quadratic degrees of freedom per atom: the three vibrational potential energies, which are quadratic in displacement, and the three kinetic energies, which are quadratic in velocity. This gives you the proper specific heat.
Answer: Each mode is an independent harmonic oscillator. Thus, there is a quadratic kinetic energy term and a quadratic potential energy term in the Hamiltonian of each normal mode. Therefore the average total energy is $3Nk_BT$.


vb.net Is there a faster way to substract one list of string from another

Question: Public done As New List(Of String)
Public thinkingofdoing As New List(Of String)
Public todo As New List(Of String)

done.AddRange(System.IO.File.ReadAllLines("C:\Users\Work\Desktop\done.txt"))
thinkingofdoing.AddRange(System.IO.File.ReadAllLines("C:\Users\Work\Desktop\thinkingofdoing.txt"))

For i = 0 To thinkingofdoing.Count - 1
    ThreadPool.QueueUserWorkItem(AddressOf caldiff, thinkingofdoing(i))
Next

Public Sub caldiff(ByVal tobedone)
    If done.Contains(tobedone) = False Then
        todo.Add(tobedone)
    End If
End Sub

done.txt and thinkingofdoing.txt have anywhere from 5 million to 8 million lines
It's taking very long :(, even with a quad core AMD 965 overclocked to 4.2 GHZ.
Comment: You are using the thread pool with non-thread safe collections. Fixing that won't fix your performance, but it will probably prevent subtle bugs later down the line.
Answer: First off, the above code is not valid.  List(Of T) is not thread safe, so doing this from multiple threads will actually cause significant problems without synchronization, as the calls to Add and Contains aren't themselves safe to be called from multiple threads.
A better option would be to choose better collections, such as HashSet(Of T), which would cause the checks to be far faster.  I would recommend something like:
public Done as New HashSet(Of String)
public ThinkingOfDoing as IList(Of String)
public Todo as New List(Of String)

ThinkingOfDoing = System.IO.File.ReadAllLines("C:\Users\Work\Desktop\thinkingofdoing.txt")
Done.AddRange(System.IO.File.ReadAllLines("C:\Users\Work\Desktop\done.txt"))

ToDo = ThinkingOfDoing.Where(Function(i) Done.Contains(i) = False).ToList()

By using a HashSet(Of T), the Contains() check will become far faster (O(1) instead of O(n)), which will cause this to run a lot faster, even single threaded.
If you don't need to store Done, you could just keep the array, and use Enumerable.Except directly (which uses a Set internally):
ThinkingOfDoing = System.IO.File.ReadAllLines("C:\Users\Work\Desktop\thinkingofdoing.txt")
Dim done = System.IO.File.ReadAllLines("C:\Users\Work\Desktop\done.txt")

Dim Todo = ThinkingOfDoing.Except(done).ToList();

Comment: The last bit of code sorted everything. What was probably going to take 15-20 hours was completed in 15 seconds. thanks a lot
Answer: You can use Enumerable.Except which should be much more efficient because it's implemented asHashSet<T> :
IEnumerable(Of String) newLines = thinkingofdoing.Except(done)

You should also use File.ReadLines instead of File.ReadAllLines since the former uses streams whereas the latter loads all into memory at once.
I would test the performance without using ThreadPool first.
Answer: how about this ...
Public done As ISet(Of String)
Public toDo As New List(Of String)();

done = New HashSet(Of String) _
    (System.IO.File.ReadAllLine("C:\Users\Work\Desktop\done.txt")

Using reader As New StreamReader(New FileStream _
        ("C:\Users\Work\Desktop\thinkingofdoing.txt"), FileMode.Open)
    Do While reader.Peek() >= 0
        Dim line = reader.ReadLine()
        If Not done.Contains(line) Then
            toDo.Add(line)
        EndIf
    Loop
End Using

This loads all the done lines into a HashSet which has excellent lookup performance, then instead of loading the whole of the thinking of doing file into memory it gets parsed line by line and only added to todo if it is not already in done.
If VB.Net had a yield return I would have put that in a function and done ToList on the IEnumerable but hey ho.


Pressure on the walls of a balloon

Question: Is the atmospheric pressure the only pressure acting on the exterior walls of a balloon keep the air/gas inside the ballon at equilibrium? Or the walls of balloon being elastic, do exert pressure on the gas inside but its magnitude is too low compared to that of atmospheric pressure?
Answer: I vote for your second iterpretation: the elastic tension of the balloon takes part in the pressure balance.
Just consider a very strong balloon, it will hold gas inside even in a vacuum condition.

Pressure on the walls of a balloon


Are perpetual american options traded on real stock exchanges?

Question: I am looking for any information about perpetual american options from practical point of view?
Are they traded on stock exchanges?
Do investment banks deal with such products?
Comment: No, I believe they are a theoretical construct by P. Samuelson nd R. C. Merton. There are none actually traded to my knowledge.
Answer: I am not aware of perpetual call options being traded on any exchanges, but there are two very close analogues.
Perpetual warrants are issued directly by companies so trading is done OTC or through a registered B/D.
Common equity is also a very close analogue to a perpetual American call option.
The theoretical value of a perpetual warrant (or option) for a non-dividend paying stock is equal to the value of the stock itself (Samuelson, 1965; Merton, 1990).
If we have standard Black-Scholes, we can see that as $t \to \infty$, $\varPhi[d_1] \to 1$, and $\varPhi[d_2] \to 0$:
$$ V_t[S_t,K,\sigma_S,r,t] = S_t \varPhi[d_1] - K e^{-r (T-t)} \varPhi[d_2]$$
where: $d_1 = \frac{\ln\left(\frac{S_t}{K}\right)+{(r+\sigma^2/2)(T-t)} }{\sigma \sqrt{T-t}}$; $d_2 =d_1 - \sigma \sqrt{T-t}$; and $\varPhi[X]$ is cumulative distribution function (of the normal distribution).
Therefore, common stock may be viewed as a perpetual warrant on itself with an exercise price of zero. However, this results in a paradox because we would expect a lower price for a warrant with a strike price greater than zero.
The reason for this paradox is two-fold:

Actual underlying prices may not actually follow a random walk, a-la GBM;
Stocks' values are premised on future dividend payments; there is no such thing as an equity (with $S_t\gt0$) which has the perpetual expectation of zero dividend payments.


A geometric look at $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$?

Question: Is there a geometric way of looking at the relationship between the positive real numbers $a$, $b$ and $c$ if $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$?
Comment: rearranging you get $(a+b)c = ab$, this might be easier to see geometricly...
Answer: If you have two poles of length $a$ and $b$, and string wire from the top of one pole to the bottom of the other, and vice versa, the height $c$ at which the wires  intersect will be given by half the harmonic mean of the height of the two poles, $\displaystyle \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$, no matter how far apart they are.  This can be easily shown using similar triangles.
I can also think of one more place where the harmonic mean is commonly found in a geometric context:  Given a right triangle with legs $a$, $b$, the squared height of the altitude to the hypotenuse is given by half the harmonic mean of the squared lengths. Wikipedia provides a handy diagram so I'll just quickly write it out:

(Here $A$, $B$, $C$ are the angles opposite the lowercase sides, as is usual, and $P$ is the point where the altitude intersects the hypotenuse.) Then, since $\bigtriangleup ABC \; \sim \; \; \bigtriangleup CBP \;$,
$$\dfrac{a}{f} = \dfrac{c}{b}  \Rightarrow \dfrac{a^2}{f^2} = \dfrac{c^2}{b^2} = \dfrac{a^2 + b^2}{b^2}$$
So, by cross-multiplying we get,
$$\dfrac{1}{f^2} = \dfrac{a^2 + b^2}{a^2b^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$$
half the harmonic mean of the squared lengths.
Comment: (should be a comment to Billare's answer) In your pole-and-wire example, the crossing height is correctly given in the formula, but $c$ is *half* the harmonic mean of $a$ and $b$. See http://en.wikipedia.org/wiki/Harmonic_mean
Comment: Right, right.  Fixed, thank you.
Answer: Given three circles of areas $a\geq b\geq c >0$, they can be put in a position such that they are tangent to each other and to a straight line if and only if
$$
\frac{1}{c} = \frac{1}{a} + \frac{1}{b}.
$$

Note that this a special case of Descartes' theorem.
Answer: Not necessarily geometric, but if $a$ and $b$ are electrical resistances placed in parallel, then their combined resistance is $c$.
Answer: You find this kind of "addition" happening frequently. For instance, if you have lenses of focal length $a$ and $b$ and put them together, the combined focal length is $c$. What's happening in this case is that the "strength" of the first lens is $1/a$, of the second is $1/b$, and the strengths add.
Another example from optics is that if you have a lens of focal length $c$ and place something a distance of $a$ in front of it, the image forms at a distance of $b$ behind the lens.
Answer: I don't know what its origin is. I learned that from a book long time ago. And I thought it was used to calculate electrical resistances placed in parallel, which Tpofofn also mentioned.
Answer: Another non-mathematical example is Sum Frequency Generation in non-linear optics, e.g.

...generation of red light (→ red lasers), e.g. by mixing the outputs
  of a 1064-nm Nd:YAG laser and a 1535-nm fiber laser, resulting in an
  output at 628 nm

http://www.rp-photonics.com/sum_and_difference_frequency_generation.html
Here we have 1/1064 nm + 1/1525 nm = 1/628.41 nm
Answer: In a right-angled triangle (with legs a and b, and hypotenuse c) you can inscribe a square in two "obvious" ways:

The square shares the right angle with the triangle and touches c. Then 1/s = 1/a + 1/b
One side of the square lies on c, and the square touches a and b. Then 1/t = 1/c + 1/h (where h is the height on c)


Rails shows "active_admin is not yet checked out." error in production mode

Question: I'm trying to deploy Rails app in Digital Ocean. Right now unicorn's logs shows this line:
ERROR -- : git://github.com/gregbell/active_admin.git (at master) is not yet checked out. Run `bundle install` first. (Bundler::GitError)

This is the part of Gemfile that uses 'active_admin' gem:
gem 'activeadmin', github: 'gregbell/active_admin'

I'm using Rails 4.1.0 and Ruby 2.1.2p95. Does anyone know what is this error about and how it can be fixed?
Answer: Try
bundle install --no-deployment


Can I knock someone back if their speed is zero?

Question: If someone is restrained by the entangle spell making their speed zero, can I knock them back with spells like thunderwave or thunderous smite?
Comment: The creature is not trying to move, the spell effect is moving the creature. What about that interaction are you having trouble with?
Answer: Yes.
You can also shove your allies out of the area of effect (and they can opt to not resist the shove). So any forced movement form works.
The only effect of being restrained is to have move set to zero. If the spell description stated that the creature cannot be moved, then it wouldn't, but that is not the case in most situations.
The above is in line with the line of ruling from 5e where spells do what they say, and not what they don't say.
When an effect force-moves the creature, it does not spend their movement allowance, and doesn't cause Attacks of Opportunity also.
The status that set move to zero would only end if the creature is moved outside the effect.
For a grapple, it is the reach of the grappler.
Comment: You should add that forced movement doesn't expend creature's own movement. So if its speed is 30ft, and you pushed it to 10ft, it still has 30 ft in their turn.
Comment: @enkryptor be my guest to, at any moment in this site, use the rights granted to **you** by the CC-BY-SA 3.0 Licence. XD - - - -  I'll add them.
Comment: @Mindwin the OP specifically asks about entangle - it is not clear to me that in that case, the restrained condition that sets the speed to 0 is actually removed by forcing them out of the area of effect.  Shoving and thunderwaving will move them, to be sure, but might not end the restraint.  See https://rpg.stackexchange.com/questions/172570/what-happens-when-restrained-creatures-are-removed-from-the-area-of-effect-of-en
Answer: Yes
A creature's movement speed only affects its ability to move, not outside forces abilities to move it. If the spell only affects the creature's movement speed, then it only affects the movement speed. If it makes it incapable of moving or being moved, then it can't be moved.
Answer: Yes
There is a lot of good evidence of this, but the best I can find is in the description of the Grappled condition.

A grappled creature's speed becomes 0, and it can't
  benefit from any bonus to its speed.
The condition ends if the grappler is incapacitated
  (see the condition).
The condition also ends if an effect removes the
  grappled creature from the reach of the grappler or grappling effect, such as when a creature is hurled away by the thunderwave spell.-PHB p.290 (bold added)

Thus, we have RAW support for the idea that it is indeed possible to force movement on something whose speed is 0.


How to re-assign channels to different field groups?

Question: I have channels sharing the same field group and am thinking about re-assigning them to different groups because they will develop in different directions (= different fields needed). Is this possible and what would be the best approach? Thanks a lot for suggestions!
Answer: If you've already got a lot of entries published in any of these channels I'd recommend not changing the field group for those channels. The reason being as soon as you go to edit one of your existing entries you will not see the custom fields where you data was stored and as soon as you resave the entry you will lose the data that was stored in the custom fields in the old field group.
So instead just add any new fields to the existing field group to accommodate the different directions your channels are headed in. Then use Publish Layouts on a per-channel basis to hide any irrelevant fields.

If you really need to change the field groups and you have a lot of entries you can migrate the existing field data to new the new fields. You'd need to create analogous custom fields in your new field group and then use SQL to copy the data over to the new fields in exp_channel_data.
Here is some SQL to use as a starting point:
UPDATE exp_channel_data
SET field_id_X = field_id_Y
WHERE channel_id = Z

X = the new field ID number
Y = the old field ID number
Z = the channel ID for the channel at hand
I recommend using a tool like Sequel Pro to help you inspect the data and run the queries. Make sure you backup your database first!
Comment: Thank you for the fast reply, Alex! Will look into the number entries, maybe there are channels with just few entries that can easily be copied. If not, I have found the  following queries for duplicating the table: "CREATE TABLE new_table_name LIKE old_table_name;" and then "INSERT INTO new_table_name SELECT * FROM `old_table_name`;" Can you confirm that these should work (very limited MySQL knowledge here). I had thought about extending the fieldset to accommodate for all channels but different sets seemed cleaner (and maybe faster as well)?
Comment: You do not want to be creating any new tables. I will update my answer with sample SQL. Unless your fields are _very_ different between channels then sharing one field group and using publish layouts is less work to set up in my opinion.
Comment: Thanks again for your time and advice, Alex! Out of curiosity, I think I will try my luck with Sequel Pro and copying the fields contents. If it doesn´t work out, I´ll take the recommended road of extending the field group with the backed up DB.


How to pass enumerated values to a web service

Question: My dilemma is, basically, how to share an enumeration between two applications.
The users upload documents through a front-end application that is on the web. This application calls a web service of the back-end application and passes the document to it. The back-end app saves the document and inserts a row in the Document table.
The document type (7 possible document types: Invoice, Contract etc.) is passed as a parameter to the web service's UploadDocument method. The question is, what should the type (and possible values) of this parameter be?
Since you need to hardcode these values in both applications, I think it is O.K. to use a descriptive string (Invoice, Contract, WorkOrder, SignedWorkOrder).
Is it maybe a better approach to create a DocumentTypes enumeration in the first application, and to reproduce it also in the second application, and then pass the corresponding integer value to the web service between them?
Answer: I'd suggest against passing an integer between them, simply for purposes of readability and debugging.  Say you're going through your logs and you see a bunch of 500 errors for DocumentType=4.  Now you've got to go look up which DocumentType is 4.  Or if one of the applications refers to a number that doesn't exist in the other, perhaps due to mismatched versions.
It's a bit more code, and it rubs the static typing part of the brain a bit raw, but in protocols on top of HTTP the received wisdom is to side with legible strings over opaque enumerations.
Answer: I would still use enumeration internally but would expect consumers to pass me only the name, not the numeric value itself.
just some silly example to illustrate:
public enum DocumentType
{
  Invoice,
  Contract,
  WorkOrder,
  SignedWorkOrder
}

[WebMethod]
public void UploadDocument(string type, byte[] data)
{
  DocumentType docType = (DocumentType)Enum.Parse(typeof(DocumentType), type);
}

Answer: I can only speak about .net, but if you have an ASP.net Webservice, you should be able to add an enumeration directly to it.
When you then use the "Add Web Reference" in your Client Application, the resulting Class should include that enum
But this is from the top of my head, i'm pretty sure i've done it in the past, but I can't say for sure.
Comment: The 'Add Web Reference' does create enums, but only those that are referenced in some method. I can manually add enums to the automatically generated Reference.cs file and all works well. Maybe this is separate question, but is there any way to get the 'Add Web Reference' to add all enums in a referenced assembly (but not used in the code) without hacking the Reference.cs file manually?
Comment: @Dave Web Services encapsulate methods, so if there are no methods using an Enum, there is no point - from a Web Service Perspective - to have it referenced. I recommend asking a separate question with a description what you want to do/why you want an unreferenced enum in it.
Answer: In .NET, enumeration values are (by default) serialized into xml with the name. For instances where you can have multiple values (flags), then it puts a space between the values. This works because the enumeration doesn't contain spaces, so you can get the value again by splitting the string (ie. "Invoice Contract SignedWorkOrder", using lubos's example).
You can control the serialization of values of in asp.net web services using the XmlEnumAttribute, or using the EnumMember attribute when using WCF.
Answer: If you are consuming your Web service from a .NET page/application, you should be able to access the enumeration after you add your Web reference to the project that is consuming the service.
Answer: If you are not working with .NET to .NET SOAP, you can still define an enumerator provided both endpoints are using WSDL.
<s:simpleType name="MyEnum">
     <s:restriction base="s:string">
          <s:enumeration value="Wow"/>
          <s:enumeration value="This"/>
          <s:enumeration value="Is"/>
          <s:enumeration value="Really"/>
          <s:enumeration value="Simple"/>
     </s:restriction>
</s:simpleType>

Its up to the WSDL -> Proxy generator tool to parse that into a enum equivalent in the client language.
Answer: There are some fairly good reasons for not using enums on an interface boundary like that. Consider Dare's post on the subject.
Answer: I've noticed that when using "Add Service Reference" as opposed to "Add Web Reference" from VS.net, the actual enum values come across as well as the enum names.   This is really annoying as I need to support both 2.0 and 3.5 clients. I end up having to go into the 2.0 generated web service proxy code and manually adding the enum values every time I make a change!


Did Martin Luther King, Jr. say that "I will not rejoice in the death of one, not even an enemy"?

Question: Did Martin Luther King, Jr. say

I mourn the loss of thousands of precious lives, but I will not rejoice in the death of one, not even an enemy

It's been going around the tweeterwebs today in response to Osama bin Laden's death.
Answer: The source of that particular quote was a Facebook post by someone named Jessica Dovey:

@pennjillette I am the original author of the "MLK" quote. Somewhere my words got mixed with his. The post: https://i.stack.imgur.com/1XqLi.jpg

Dovey said "I mourn the loss of thousands of precious lives, but I will not rejoice in the death of one, not even an enemy", and then quoted Martin Luther King Jr.:

"Returning hate for hate multiplies hate, adding deeper darkness to a night already devoid of stars. Darkness cannot drive out darkness: only light can do that. Hate cannot drive out hate: only love can do that." - Martin Luther King, Jr.

The Facebook post itself appeared as follows:
Comment: +1. related: [*The Atlantic* article "Anatomy of a Fake Quotation" by Megan McArdle](http://www.theatlantic.com/national/archive/2011/05/anatomy-of-a-fake-quotation/238257/)
Answer:
"There is within us a moral instinct
  which forbids us to rejoice at the
  death of even an enemy."
-- Henryk Sienkiewicz, Without Dogma (1891)

Full text at Project Gutenberg
So, if Martin Luther King, Jr. did say it, he was likely paraphrasing Sienkiewicz (whether knowingly or unknowingly).
Answer: "The meaning of this story is not found in the drowning of the Egyptian soldiers, for no one should rejoice at the death or defeat of a human being. Rather, this story symbolizes the death of evil and of inhuman oppression and unjust exploitation."  Martin Luther King Jr, Strength to Love, page 79
Answer: Proverbs 24, New International Version translation:

[17] Do not gloat when your enemy falls;
  when they stumble, do not let your heart rejoice,
  [18] or the Lord will see and disapprove
  and turn his wrath away from them.

This is similar to the attributed quotation.  As a Protestant minister who championed a cause that placed himself and those around him in mortal danger, it is quite plausible that MLK saw fit to paraphrase and discuss this proverb and its application to life from time to time.
Similarly, the quotation "Darkness cannot drive out darkness" from the accepted answer can be considered a paraphrase of the Gospel of Mark, chapter 3:

[23] So Jesus called them over to him and began to speak to them in
  parables: "How can Satan drive out Satan? [24] If a kingdom is divided
  against itself, that kingdom cannot stand. [25] If a house is divided
  against itself, that house cannot stand...."

This pattern would also explain why other speakers and authors have similar quotations; perhaps they are paraphrasing the same passages, or even giving second-degree or third-degree attributions.  (Some people attribute the "house divided against itself" proverb to Abraham Lincoln, for example.)
Comment: Why did you answer a 4-year-old question, when the correct answer (MLK did not say this) is clearly given and accepted already?
Comment: This answer seems to be of the form "The bible says X, so maybe MLK said Y." Please provide an actual reference that he said it, or remove the answer.
Comment: Are you going to take down the other upvoted answer RE Sienkiewicz, that has a similar form, or only the Bible quote?


Why are hub shells such different diameters?

Question: I'm curious: why are some hub shells much larger diameter than others?  Is it an aesthetic consideration? Or are there cost/functionality tradeoffs?  Take for example this Dura-Ace hub -- pretty skinny.  Same for a classic Campy record hub.  Compare to this much fatter Tiagra hub, or a Phil Wood front hub.  I'm just talking standard hubs, not internal gears, powertaps, dynamos, etc.  [Addition after DR Hicks answer: nor I am I talking about hub flanges, but rather the barrel of the hub shell.]
Comment: Well, interestingly enough, the Dura-Ace and Campy hubs appear to be large flange (comparing flange diameter to axle diameter).  The Tiagra is small flange.  Part of the apparent difference is probably just illusion, but there clearly is some "real" difference as well.  Phil Wood hubs are by definition built like tank, so that's kind of out of the spectrum.  (But when you compare prices you see part of it -- the Dura-Ace is about 5 times the price of the Tiagra.)
Comment: Marking you as accepted for now, but interested to see if there are any other theories out there . . .
Answer: A smaller diameter hub is lighter (and likely cheaper).
A larger diameter hub has several advantages:

The bearings can be larger
Since the spoke flange has a larger diameter, it also has a larger circumference, meaning that spoke holes are farther apart.  This reduces the stress on the spoke flange with heavily-loaded bikes, and also makes higher spoke counts practical.
Related to 2, the spokes, for a given "cross", leave the hub at a more oblique angle, meaning that the force of the spoke is pulling more tangentially and less radially.  This again reduces stress on the spoke flange and also increases the rigidity of the wheel against torque.

A non-obvious disadvantage of the larger hub, when it gets MUCH larger, is that, for 3 or 4 cross, the spoke arrives at the rim at a definite angle.  Unless the rim had been drilled at a matching angle, this increases the stress on the spoke where it enters the nipple, increasing the chance of spoke failure.  But this is mostly a problem with geared hubs.
It's pretty much tradition to have larger diameter hubs on touring bikes and tandems.  Not sure to what extent this is a practical measure vs "expectations" -- the larger flange looks like its built to be rugged, while the small flange looks "light on its feet".
Comment: I got the impression he was comparing the main body of the hub axle rather than the spoke flange. So the more expensive hubs have narrower axle bodies, but the cheaper ones are wider
Comment: Correct, sorry I was asking about hub barrel, not flange.  I suppose it could be bigger bearings, but why wouldn't designers of expensive road hubs (Campy record, Dura Ace) want to have larger bearings also?
Answer: I suspect the reason may be aesthetic. As you've noticed, the cheaper/entry level hubs have large barrels and the more expensive versions have narrower barrels. It must be hard to upsell a customer to a more expensive hub... there's not much to differentiate them apart from the looks.
Personally, the narrow barrels look sleek and fast to me and I might be prepared to drop a few more $$ on them if I'm putting them on a fast road bike.


Can we Throttle SMS Sends in Marketing Cloud?

Question: Is it possible to throttle SMS sends in marketing cloud?
We would like to have the ability to specify how many sms can be send from Marketing Cloud within an hour. If the threshold is hit, then sms should be queued to be sent next hour.
Answer: Yes, its possible to throttle SMS sends in Marketing Cloud. Please log a case with Marketing Cloud support and they will enable the same for you.
once it's enabled, you can see the option to throttle SMS sends. When you configure an SMS send, in the "Schedule/Activate" screen of the wizard, you can specify the how many messages can be send during a time frame (1 min, 15 min, 30 min or 1 Hour)
Screenshot below.

you can read more about throttling function in my blog.


Export contacts with last donation date/amount (if there is a donation)

Question: I need to export a list of our contacts which is easy enough. However, I cannot figure out how to include in that export if they are a donor - and ideally the date/amount of their last donation. When building the fields to export I don't see any option for this so I'm running two reports - a contact export and a donor export and merge them in excel and then spend time sorting them all out. What am I missing? Surely there is a way to export all contacts with donation information for those who have donated. Thanks in advance!
Answer: You could get something close with SearchKit, but not quite what you are looking for. With the search below, you'll get Contacts with their most recent donation date and their total donations over whatever period you are interested in (you could also add the count of donations over that period too). I can't see a way to only show the amount of the most recent contribution in SearchKit, but maybe someone else can.
You can find:

Contacts
with (optional) Contact Contributions with Status = Completed and Financial Type and Date Received as needed
Group by Contact ID so you have one row per contact
Add the Date Received and Total Amount columns on the top right of
the table.
Under Field Transformations, select Max for Date Received (so you are
showing the largest date, i.e. the most recent donation date) and
select Sum for Total Amount (so you'll have the sum of all donations
in the time period).
If you want a count of donations too, add the Contribution ID column and then select Count under Field Transformations.

The date might not come through in a spreadsheet-usable format (though that has just been fixed so won't be a problem in the future). If you need that sorted out, you can do it by adding a table and adjusting the date format with a rewrite.
Comment: In the near future (July 2023?) you will be able to create a SK search that provides the most recent contribution date per contact ID, with a display of type "entity". Then you can create a second SK search with a join to this entity. That should get you to the finish line. See https://github.com/civicrm/civicrm-core/pull/25871. In the meantime, @petendz-fuzion's solution is probably the best fit.
Answer: If you use the Summary Fields extension or possibly Synopsis then you should be able to set up so you have a Custom Field which contains the data of their last donation, and then use the token for that Custom Field to get the outcome you want.


The connection between the rank of a matrix and its zero-mode eigenvectors

Question: I would be most thankful if you could help me prove that if an arbitrary n by n matrix has rank m < n, then the matrix has (n-m) linearly independent eigenvectors corresponding to the eigenvalue zero. Thank you!
Answer: Use the rank-nullity theorem.  What is the relationship between the nullspace of a matrix and its eigenvectors corresponding to eigenvalue zero?
Answer: if rank of $A_{n*n}$ is $m <n$ then Ax=0 will have non trivel answer as $x_1 $and so 0 will be a $eigen value$ $Ax_1=0x_1$.
Now we try to find eigen vectors of zero egien value:
 (A-0I)x=0 Ax=0 and so $X_1$ and other non trivel answersof Ax=0 will be eigen vector correspondence with zero eigen value too
since Ax=0 has n-m independence answer so number of independ eigen vectors corresponding to 0 eigen value  is  n-m too!:)