A mismatch in quantum mechanics

Question: One of the basic rules of Quantum mechanics is that  after a measurement of an observable, the wavefunction is an eigenstate and any subsequent measurement will give the same result. This is not so in case of coordinate (x) measurement, because of the Heisenberg uncertainty or because the interaction used to specify (x) brings uncontrollable impulse to the particle. What is this mismatch due to?
Answer:
[...] and any subsequent measurement will give the same result.

This is only the case if the operator commutes with the Hamiltonian of the system. Otherwise, time evolution will take the system out of the eigenstate and subsequent measurements will not give the same answer.
Edit
This can be seen easily from the Schrödinger equation
$$ i\frac{\partial}{\partial t} |\psi\rangle = \hat{H} |\psi\rangle$$
which has the solution
$$ |\psi(t)\rangle = e^{-i\hat{H}t} |\psi(0)\rangle.$$
So if at $t=0$ the system is in the eigenstate $|\psi(0)\rangle =|a\rangle$ of the operator $\hat{A}$ with eigenvalue $a$, it will for $t > 0$ be
$$ |\psi(t)\rangle = e^{-i\hat{H}t} |a\rangle.$$
Applying the operator $\hat{A}$ again at time $t$ gives
$$ \hat{A}|\psi(t)\rangle = \hat{A} e^{-i\hat{H}t} |a\rangle = e^{-i\hat{H}t} \hat{A} |a\rangle + [e^{i\hat{H}t},\hat{A}]|a\rangle = a |\psi(t)\rangle + [e^{i\hat{H}t}, \hat{A}]|a\rangle.$$
Clearly, if $[\hat{H},\hat{A}]$ = 0, the later application of $\hat{A}$ will give the same answer. But if not, this will generally not be the case.
Comment: Yes, a measurement of $a$ at time $t$ means the system is in the eigenstate of $\hat{A}$ corresponding to the eigenvalue $a$ at time $t$. But that does not mean it will *stay* in this eigenstate for times $t_1 >t$. This is only the case if $\hat{A}$ commutes with the Hamiltonian.
Comment: I do not have a reference at hand, but you should be able to find one yourself. I also edited the answer, so it should be clear from that.
How did you get $[\hat{p},\hat{H}] \neq 0$? The free particle Hamiltonian is $\hat{p}^2/2m$, which clearly commutes with $\hat{p}$. If you have a potential $V(\hat{x})$, this will not commute with $\hat{p}$, but is also equivalent to a force acting on the particle, thus changing the momentum.
Comment: The rule can be read here http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html and it reads: Postulate 3. In any measurement of the observable associated with operator A, the only values that will ever be observed are the eigenvalues a, which satisfy the eigenvalue equation Aw=aw (w = wavefunction). If the system is in an eigenstate of A with eigenvalue a, then any measurement of the quantity A will yield a. There is nothing said about your proposal?

If the system is in an eigenstate of $\hat{A}$ with eigenvalue $a$, then any measurement of the quantity $A$ will yield $a$
Comment: Thank you very much Noah! But would you be so kind to cite a book or a site where in the 3rd principle is explicitly stated that it is true only when [A,H]not=0.
Comment: By the way [p,H] not=0 also. Does it mean that p changes for a single body (e.g. 1 Newton principle not true in QM)?
Answer: There is no such mismatch. If you ever manage to measure the position to an infinite precision, it will project it onto a position eigenstate $|x\rangle$; this is required by the uncertainty principle to have infinite momentum uncertainty, which among other things will cause the position uncertainty to begin growing immediately after the measurement, but does not introduce any inconsistencies into the theory.
Of course, in physical measurements we can only ever perform experiments with finite precision, which need to be modelled by the application of a suitable projector such as
$$
\hat \Pi_{(x_0-\Delta x/2, x_0+\Delta x/2)} = \int_{x_0-\Delta x/2}^{x_0+\Delta x/2} |x\rangle \langle x| \: \mathrm dx
$$
to the state. This will produce a state with a small-but-finite position uncertainty, and therefore with a large-but-finite momentum uncertainty, and all traces of unphysicality are removed from the situation.

It's also worth emphasizing that your understanding of the measurement postulate,

the wavefunction is an eigenstate and any subsequent measurement will give the same result

is rather incomplete. If one performs a measurement of an observable $A$ and obtains the result $A=a$, then subsequent measurements of $A$ performed after a negligible time interval will also yield $A=a$. However, if the second measurement happens after some finite time interval $\Delta t$ and $A$ is not preserved by the time evolution (or, in technical language, $A$ does not commute with (is incompatible with) the hamiltonian) then that ceases to be true.
For the case of the position of a massive particle, the observable $A=x$ is incompatible with the hamiltonian $H=p^2/2m$, so if you let $H$ act for any length of time at all (i.e. if you propagate for any finite time) on a position eigenstate, or indeed after projecting with finite precision with $\hat \Pi_{(x_0-\Delta x/2, x_0+\Delta x/2)} $ as above, then the "uncontrollable impulse" (or, with the finite-precision measurement, the finite-but-large impulse) brought on by the measurement will start to act and the state will spread out.
However, as above, if the two measurements are performed in quick succession, i.e. if there is not enough time for the hamiltonian to act, then the first measurement result will be preserved.
Comment: I cant understand you. You say there is no mismatch but in fact you repeat exactly what the mismatch is: the postulate in the case of x measurement is not true!
Answer: The postulate as stated is correct, and you are also correct. Here's what you're saying:

If I measure $x$ and get into an eigenstate of $\hat{X}$, then there is huge momentum uncertainty. So if I come back ten seconds later and measure $x$ again, it will probably be in a totally different place!

Here is what the postulate is saying:

If I measure $x$ and get into an eigenstate of $\hat{X}$, and then immediately (no waiting) measure $x$ again, I'll get the same result.

There is no contradiction here. It's true that a particle with a well-defined position has a huge momentum uncertainty, but that only means that in the future the wavefunction will spread out. The postulate isn't saying anything about what happens in the future, it's talking about what would happen if you measure two times in a row with no elapsed time in between.

A mismatch in quantum mechanics


How do I derive a surjective transformation if annihilator of range T is zero

Question: In Linear Algebra Done Right 3rd ed. Axler tries to prove a transformation $T\in \mathcal{L}(V,W)$ is surjective iff the dual map $T^*$ is injective. His reasoning starts with the following equation:$$\operatorname{range}(T)=W\iff (\operatorname{range}(T))^0=\{0\}.$$ How do I derive this? Is there something special about the fact that the only transformation that takes $\operatorname{range}(T)$ to the zero vector is the zero transformation?
Answer: Let $\Bbbk$ be the field over which you are working. If $U$ is a subspace of $W$, then you have two possibilities:

$U=W$: then the only linear form $\alpha\in W^*$ such that $\alpha(W)=\{0\}$ is the null form.
$U\varsubsetneq W$: then, if $v\in W\setminus U$, if $U'$ is a subspace of $V$ such that $U\subset U'$ and that $U'\oplus\Bbbk v=W$, then if $\alpha\colon W\longrightarrow\Bbb k$ is defined by $\alpha(u+\lambda v)=\lambda$ ($u\in U'$ and $\lambda\in\Bbb k$), then $\alpha\in W^*$, $\alpha\neq0$, and $\alpha\in U^0$.

So, this proves that$$U=W\quad\text{if and only if}\quad U^0=\{0\}.$$


Finding the spectrum of a polynomial of the creation and annihilation operators

Question: Is there a general algorithm to find the spectrum of $S S^\dagger$, where $S$ is a homogenous polynomial (of degree $n$) of the annihilation operators (of $d$ variables)?
Answer: The supspaces $V_n = Span \{ (a_1^{\dagger})^{n_1}, . . . (a_d^{\dagger})^{n_d} |0>\}$, $n_i \ge 0$,  $ n_1 + . . . n_d = n$, constitute of invariant subspaces of the operator $ S S^{\dagger}$ action. The dimension of $V_n$ is $ \frac{(d+n-1)!}{(d-1)! n!}$. Thus the operator can be represented on each of these subspaces as a square matrix of size $ \frac{(d+n-1)!}{(d-1)! n!}$ for which the spectrum can be found by elementary linear algebra. The spectrum on the whole of the Fock space is the union of the spectra over $V_n$, $ n = 0, 1, . . .$
Comment: @Moshe, did you lose your old account?
Comment: @Moshe, thank you a lot. Pushing in further - for a polynomial of degree $k$ does it suffice to know eigenvalues in the first $i$ subspaces (i.e. $V_0,\ldots, V_i$) to predict all the other? Like for $n=1$ it suffices to know the eigenvalue for $i=1$ (all others are their multipilicites).
Comment: @David Bar Moshe: The formula (v1) for $\dim V_n$ must contain an error/typo(?), which can e.g. be seen by putting $d=1$.
Comment: @Piotr - sorry for the error, of course the dimension is equal to the number of ordered partitions of $n$ into at most $d$ pieces, or equivalently the dimension of the fully symmetric $n$-tensorial representation of $SU(d)$, which can be calculated for example by using the hook length formula. By the way I am trying to think occasionally on your interesting suggestion in your last comment, but I haven't reached an answer yet.
Answer: One can always reorder the operators in your polynomial to make it a polynomial of individual particle number operators. E.g. $a^+_k a^+_k a_k a_k = \pm n_k^2+n_k$ (the sign depends on the statistics of your particles). Since the particle number operators for different modes commute, the calculation of the spectrum is straightforward.
Comment: I am afraid it is not that simple. Even for the simplest non-trivial case $S=\frac{\alpha}{\sqrt{2}} a_1^2+\beta a_1 a_2+\frac{\gamma}{\sqrt{2}} a_2^2$ you get cross-terms in $S S^\dagger$, e.g. $\frac{\alpha \beta^*}{\sqrt{2}} a_1^2 a_1^\dagger a_2^\dagger$. Do you know a general algorithm to 'diagonalize' it, so there are not any cross-terms?


Closest cross-platform alternative to Direct2D/DirectWrite?

Question: What are some good cross-platform alternatives to Windows-specific 2D APIs? I'm looking for an API for rendering simple geometric objects as well as an API for rendering text with subpixel antialiasing (ClearType) and OpenType support, specifically support for hinting.
Answer: OpenGL is the defacto cross-platform graphics API. OpenGL supports the FreeType and OpenType libraries. Newer versions of FreeType (2.7+) have support for sub-pixel hinting similar to what DirectWrite offers.
Comment: Apparently, OpenGL does not guarantee identical pixel output across implementations.
Answer: Have a look at the Simple DirectMedia Layer (SDL), it contains exactly what you need. SDL_ttf is an extension to libsdl that provides functionality for rendering of text.


Silverlight compatible Telnet/SSH library

Question: Is there a Silverlight compatible Telnet or SSH library.
Answer: Our SecureBlackbox product includes SSH client components, which can be used in Silverlight (subject to network access restrictions, imposed by Microsoft on Silverlight code).
Telnet is "low-level" TCP access + handling of some control sequences. While SecureBlackbox doesn't include control sequence interpretation (neither it does for SSH), you can use TElSocket class for simple synchronous socket access (also in Silverlight) and add control sequence handling.
Answer: There are a number of open source projects out there that implement SSH, mostly deriving from SharpSSH, such as SharpSSH2 which is "updated for the v3.5 framework", and a rewrite for the v4 framework is SSH.Net (although it also claims to support ".NET 3.5, Silverlight and Windows Phone"). There are other projects such as Granados as well, so depending on your preference of open source licence, you may find one of these will suit your needs.


Can a galactic center be my sun?

Question: Suppose there is a rogue planet. I was imaging what could be its energy source for life, and I remembered that photos of the center of a Galaxy look very bright. So my question is, can a galactic center be the sun of a rogue planet?
Comment: We cannot see the galatic centre in the visible spectrum from earth, so I want to know where you got those pictures from. Other than that, please read: https://en.wikipedia.org/wiki/Galactic_Center - not because I think you premise is flawed but because you use the wrong words for stuff
Comment: Picking on one singular aspect of "sun of a rogue planet". I know you're mostly interested in the energy part, but hear me out o the orbit part: Isn't the definition of Rogue Planet that it is _not_ in orbit around a single object? Yes, in the end, everything close to a galaxy is directly or indirectly in orbit around the galactic center, but I suppose we're talking about the Hill sphere based definition of "in orbit" here. I'd be very confused if the story I was reading were to have a Rogue Planet in orbit around something.
Comment: Those photos are likely extremely over-exposed, which is why it looks bright. If you look up into the night-sky, you can't actually see the galactic center, but those photos were taken from Earth. (Or near enough, anyway)
Comment: The view towards the center of the galaxy is brighter because there are more stars in that direction than in other directions.  You probably want to be close enough for at least one to be a sun instead of a star.
Answer: Probably Not
As far as we can tell, the center of each galaxy contains one, or sometimes more, supermassive black hole hundreds of thousands to billions of times more massive than our Sun. I'll use our own, Sagittarius A*, for the rest of this answer. It has a mass of roughly 4 million Suns packed into a space roughly the size of the orbit of Mercury.
While the planet could orbit the supermassive black hole, plenty of stars do, it's not a pleasant place for life. It would have an enormous accretion disk of material spiraling into the black hole, compressing, and radiating massive amounts of energy.
It's this accretion disk which glows and could provide the energy necessary to sustain life on a rogue planet. Since the amount of energy radiating from the accretion disk will change as more or less material swirls around, it's unlikely the habitable zone around the disk, if one even exists, would remain stable.
In addition, your rogue planet would have to deal with powerful and unstable magnetic fields, X-Ray bursts, and jets of material. For example, in 2015 a massive X-Ray flare was detected from Sagittarius A*, possibly from an asteroid falling into the black hole, or tangled magnetic field lines.
Comment: They could make a rule:  No wusses allowed on the rogue planet.  Probably that already is a rule.
Comment: https://www.space.com/20303-black-hole-star-speed-record.html describes a star circling a black hole very fast.
Comment: @Will I think over time they simplified that rule. "You die, you lose"
Answer: Interesting concept, but probably a bad idea.
Even if the galactic core could be used as a power source, it is thought that life as we know it could not evolve close to the center due to the high concentration of high-mass stars that are found there.  Dangerous cosmic objects like neutron stars, pulsars, and black holes roam wild near the center, and supernovas or gamma ray bursts would sterilize a planet every few million years, causing evolution to start over.  Complex life would find it hard to evolve in such a hostile environment.
Comment: What if there was a healthy later of water/ice? If there are "enough" meters of water, ionizing radiation will have no effect. On longer time-scales, the problem might be if large portions are vaporized and removed from the planet. The strange chemical compounds might actually provide an energy source if they seep into an underground ocean
Answer: Signs point to no. As the answer to this Physics Stack Exchange discussion mentions, even a hundred times higher stellar density of the night sky as expected closer to the galactic center would brighten up the sky less than the full moon.
While 100 stars per cubic parsec, which seems to be the common estimate for the galactic core, is a lot (we're at less than one star per P3) it's still not very dense as I understand it. And as others have pointed out, you want to be a bit off from the supermassive black hole in the middle so at best you'd get a very exciting night sky during your eternal nights.
Answer: It's unlikely
Galaxy Photo Clarification
You stated

I remembered that photos of the center of a Galaxy look very bright

Many photos of extra-galactic galaxies use filters, false-coloring or are enhanced to emphasize certain features of the galaxy or combat disruptive effects such as light pollution. Examples: M42, Crab Nebula, Horsehead Nebula, and M31: Andromeda [4] [5]. If you look at software used in [5] , you'll see that the astrophotographer used stacking, masking, and photoshop to develop the final photo from a set of 44-291 photos (44 + 86 +71 + 90).
Nevertheless raw photos are remarkable.
Sol & The Milky Way
From Earth, if you travel far enough from light polluted regions, you can see the milky way galaxy center with the naked eye.
(Paranal Observatory)
(wallpaper)
(space.com gallery).
Sol is 27,000 LY from the galactic center in the Orion–Cygnus Arm. The bulgy core of the Milky Way extends to about 10,000 LY and the 1 parsec galactic center contains 10 million stars. That being said, while it is relatively dim compared to nearby light sources with an absolute visual magnitude of −20.9* it is still a very bright object. By comparison the Sun has an absolute magnitude of +4.83 which is 20 billion times dimmer.
Note: Absolute magnitude is defined as the apparent magnitude of the object at 10 parsecs.
We Need To Go Deeper Morty!
We've established that the Milky Way Galaxy is MUCH brighter than the Sun in terms of its absolute magnitude. It then follows that there must be some distance at which the Milky Way looks as bright as the sun (apparent magnitude).
We're interested in whether there's a distance ($d$) at which the apparent magnitude is equivalent to the apparent magnitude of The Sun from Earth. This is a fairly simple calculation given the conversion formula between absolute magnitude ($M$) and apparent magnitude ($m$):
$$M = m - 5 (log_{10} d + 1 )$$
$$m = M + 5 (log_{10} d + 1 )$$
Calculating the apparent magnitude of The Sun from earth using $M=+4.83$ and $d = 1/206265$ (1 AU in Parsecs) returns us an apparent magnitude of -26.74. Restructure the formula in terms of the distance (in parsecs):
$$d = 10^{\frac{m-M}{5}+1}$$
Then plug and chug with $M = -20.9$ and $m = -26.74$ to get 0.67 parsecs (2.2 LY). **This suggests that at 2 lightyears the rogue planet will see the milky way as brightly as we can see the Sun on Earth; however, it ignores a key premise:
The absolute magnitude of a galaxy is calculated differently compared to stars since galaxies are larger than 10 par secs.**

A galaxy's magnitude is defined by measuring all the light radiated over the entire object, treating that integrated brightness as the brightness of a single point-like or star-like source, and computing the magnitude of that point-like source as it would appear if observed at the standard 10 parsecs distance.
source

Therefore, this may not be very reliable for the Milky Way. That being said, it is worthwhile to note that within 1 parsec of the galactic center there are 10 million stars, and that this region would be very bright since it does sit within the galactic core.
My Conclusion: it can go either way.
(I'll confer with an astrophysics buddy of mine for clarification and update this answer.)
But we're not done. We've only considered the Milky Way Galaxy. What of the brightest objects in the universe?
Quasars
Quasars, or quasi-stellar objects, are active galactic nuclei of obscenely high luminosity. The brightest Quasar, 3C 273 has an apparent visual magnitude of +12.9 and an absolute magnitude of −26.7 making it 4 trillion times more bright than the sun. Using the same calculation above, we find that it'll have the same brightness as the sun at 9.8 parsecs or 32 LY. The same arguments as before applies, but less strictly so.
Side note: if a civilization were to develop on a rogue planet orbiting a quasar, they'd see some fantastic jets. I'd also expect them to use the jets as a time-keeping mechanism (much as we can use the moon to keep time in Lunar Calendars).
Other Considerations
Since your planet is rogue, you'll need to consider the following as well:

Electromagnetic Protection from solar wind. (Magnetosphere)
Goldilocks Zone (The planet needs to be warm and pressurized enough to sustain liquid water.)
Heating: Earth's surface heat is maintained primarily through solar radiation; but, 0.03% of the energy budget is originates from the radioactive decay of uranium-238 (238U), uranium-235 (235U), thorium-232 (232Th), and potassium-40 (40K). (source). In contrast, this radiogenic heating is what's responsible for our geologic activity from the soft-iron dynamo outer core that drives our geomagnetic shield to plate tectonics.
Comment: A magnetosphere does *nothing* to protect against X-rays, and not much to protect against cosmic rays. On Earth, those are blocked by our thick atmosphere. The magnetosphere only protects against relatively low energy charged particle radiation, like solar wind.
Comment: Woops. I meant solar wind. Updated.
Answer: From a visual standpoint, there  is a possibility of the accretion disk of the supermassive black hole at the centre of the galaxy releasing enough radiation to make a rogue planet orbiting sufficiently far enough away be lit up like the Sun.

The accretion disk as portrayed in the movie "Interstellar"
Of course, as many other answers point out, visible light isn't the only sort of radiation you will be getting, and it seems very likely that visible radiation is not going to be more than a small percentage of the energy output of the accretion disk. I'm not able to find a good estimate as to how much energy is coming off the disk as light, but consider if we can see the supermassive black hole at the centre of the Milky way in frequencies from radio to x-ray from 30,000 light years away, then the radiation environment is probably sufficient to fry a planet orbiting a black hole, and possibly boil it away as it's orbit takes it closer to the centre.
The ever helpful Atomic Rockets "Boom Table" tells us that it would take:
3.2 × 10^26J    77 Pt   Energy required blow off Terra's atmosphere into space
6.6 × 10^26J    158 Pt  Energy required to heat all the oceans of Terra to boiling
4.5 × 10^27J    1 Et    Energy required to vaporize all the oceans of Terra into the atmosphere
7.0 × 10^27J    2 Et    Energy required to vaporize all the oceans of Terra and dehydrate the crust
1.0 × 10^29J    24 Et   Energy required blow off Terra's oceans into space
1.5 × 10^30J    359 Et  Energy required blow off Terra's crust into space
2.9 × 10^31J    7 Zt    Energy required to blow up Terra (reduce to gravel orbiting the sun)
5.9 × 10^31J    14 Zt   Energy required to blow up Terra (reduce to gravel flying out of former orbit)
So making some assumptions about the energy output of the central black hole gives you an idea of how far away the planet could be before the energetic output of the accretion disk destroys the planet, which in turn would tell you how many light years away it would have to be to be in the "safe" zone.
This calculation does not take into account things like the presence of other stars caught in the orbit of the black hole, or take into account if the black hole is currently "active" (i.e. irregularly consuming larger amounts of gas, planets or stars), or if you have entered in a "starburst" phase (the radiative pressure of the black hole's accretion disk has triggered a massive burst of star formation in the gas clouds that have been drawn into close proximity (several light years) of the black hole.
Answer: There are also dwarf galaxies out there, like this one. They don't seem to have a black hole in the middle, which makes an inhabitable planet in that region more likely. It is theorized that such galaxies may be formed around an agglomeration of dark matter instead of a supermassive black hole present in regular galaxies like Milky Way.
However, unless you're willing to imagine an unknown way in which energy required to support life could be extracted from dark matter directly, your best bet is to have a regular star in the middle of a dwarf galaxy with your rogue planet orbiting it.
Comment: Welcome to WorldBuilding! If you have a moment please take the [tour] and visit the [help] to learn more about the site. Have fun!
Answer: There is a brilliant video called "Civilizations at the End of Time: Black Hole Farming" by Isaac Arthur: https://www.youtube.com/watch?v=Qam5BkXIEhQ
It describes civilizations that could live off of energy of a supermassive black hole, but definitely not in a sense you would expect. Worth checking out.
Potentially, a sufficiently advanced civilization could also live on a planet orbiting black hole in old enough galaxy (where most of the orbits stabilized and material does not fall uncontrollably into a BH). By throwing material into accretion disk at constant rate it is probably possible to sustain consistent habitable zone. Although I don't really see the point in it for a society this advanced, they can just use artificial fusion instead.

Can a galactic center be my sun?


When is it permissible for a women to go to the Mikvah during the 7th day?

Question: M'Dioraysa a zavah gedolah can go to the mikvah on the morning of Day 7 as Miktzas Hayom K'Kulo. However, the rabanan were gozer that a women must wait until nightfall to go. The concern is: If she were to go during the day, and come home and have relations with her husband during the day, and afterwards would see blood before nightfall, that would invalidate her Shivah Nikiyyim.
My question is: Is it ever permitted for a women to go during day 7? I believe I once saw a heter to go on Erev a 3-day Yom Tov if otherwise mikvah would be pushed off 3 days. I have heard of heterim in cases where the couple wouldn't see each other until nightfall. I was unsuccessful at locating any sources.
Comment: I have also heard of one specific Rav giving one specific woman a heter to go Friday late afternoon when the couple wouldn't see each other until the night, in a very specific set of circumstances (I think the mikve was too far from a hotel they were at for Shabbat). I think the reason these don't get written up is that they are specific and come with instructions to avoid issues. Writing them up and "generalizing" them defeats the purpose
Comment: https://www.yeshiva.org.il/midrash/23369
Answer: Rav Moshe Feinstein is lenient on Erev Shabbos and Yom Tov, bshas hadchak (pressing situations) where she does not return home until after nightfall or the husband will not return home until after nightfall.  Igros Moshe, Vol. 6, Yoreh Deah, Siman 60.
In the United States, in places where there is no Mikvah in walking distance, it is not uncommon for rabbanim to be matir on Friday evenings for the woman to drive to the mikvah and be tovel shortly before Shabbos and only return home after their husband has gone to shul.
Consult with a competent Rabbi for any practical situations.
Answer: the ערוה"ש סי' קצ"ג סט"ו talks about a kallah, and he says that בשעת הדחק in any case where the husband and wife cannot meet before night (i.e. one of them are out of town/half a day away)she can go on day 7.


What's the most recent version of OS X I can run on a 1999 iBook and where can I buy it?

Question: I've got a 1999 Blueberry clamshell iBook an I want to get it up and running agin. I'm wondering if I can get Mac OS 9 or OS X on it and more importantly where I could buy the software.
Answer: The maximum OS version is 9.2.2 for Classic Mac OS or OS X 10.3.9 (Panther). I'm not sure which one to recommend. OS 9 might be faster, but OS X is certainly more powerful and will be far better supported by software (though Panther is so old that you'll have to use old versions of many apps).
You'll probably have to go digging on Craigslist or eBay for install discs that old.
Comment: I had an iBook SE which had a 66MHz CPU boost vs. the one @GregGuida mentions. With that, 10.3.9 was significantly more responsive for most tasks, particularly web-browsing. I can confirm 10.3.9.
Answer: You may want to start by checking Low End Mac's "Best Deals" pages. They have one for Classic Mac OS and another for OS X 10.0-10.3.
It's a long-shot, but you could also check with independent local Mac shops to see if they still have old copies of OS 9 or OS X for sale (Apple Retail almost certainly will not). If nothing else, you could probably pay them to get it up and running again, which would include reinstalling the OS if necessary. However, checking it in for service would most likely cost much more than finding an old copy you can purchase and install yourself.
Other than that, you'll probably need to follow CajunLuke's recommendation of checking Craigslist and eBay.


Why is it "what is that supposed to mean?" And not "...to means?"

Question: I know it looks stupid but I really couldn't understand why it's "mean" and not "means". Is it because it's on the past?
Answer: It's because it is in the infinitive, which always uses the uninflected form.  Likewise:

It is supposed to do that.
It is not supposed to jump like that.
It is supposed to mean nothing.
Answer: The answer is due to basic English infinitives.

She has {to take} her medication.
Not - She has to takes her medication.

He is supposed {to attend} to his own wounds.
Not - He is supposed to attends to his own wounds.

Infinitives are a basic feature of the English language, and many other languages including Hebrew. They should be among the first features to in which {to gain} expediency when learning English.


Printing an Array as an Unordered List

Question: I'm using Bill Erickson Custom Meta Boxes and Fields for WordPress and was wondering how I'd go about printing an array of checkboxes as an unordered list.
This is what I've got in my theme's functions.php file:
array(
'name' => 'Other Features',
'desc' => 'field description (optional)',
'id' => $prefix . 'other_features',
'type' => 'multicheck',
'options' => array(
    'language' => 'English Speaking',
    'furniture' => 'Modern & Classic Furniture',
    'fridge' => 'Fridge/Freezer',
    'stove' => 'Wood Burning Stove',
    'oven' => 'Electric Oven',
    'internet' => 'Free Wi-Fi',
    'tv' => 'Television/DVD Player',
    'plugs' => 'UK Power Plugs',
)
),

And this is how it appears in my Add/Edit page panel:

What I'd like to know how to do is print only the selected checkboxes as an unordered list with their ID, like so:

English Speaking

Modern & Classic Furniture

Wood Burning Stove

Electric Oven

Free Wi-Fi

Television/DVD Player

Any help at all to point me in the right direction would be appreciated. Thanks in advance.
Answer: Here's what I ended up using, courtesy of Bill Erickson himself:
<?php
global $post;
$other_features = get_post_meta( $post->ID, 'other_features' );
echo '
';
foreach( $other_features as $id => $value )
    echo '
' . $value . '
';
echo '
';
?>

Printing an Array as an Unordered List


Bank of wall outlets not working

Question: I have a bank of wall outlets that are not working. I have checked the fuses and found all of them check ok at 120 v ac. I then replaced all three outlets with new ones and still have no power. I did check the power on all of the lines to each outlet and found no power to every socket. (There are three). I then went to the attic where the line goes into the wall and traced the line to the power source and found no breaks that I could see or identify. So now I am stuck????
Comment: Sounds similar to an issue I had - http://diy.stackexchange.com/questions/56359/what-is-causing-my-electric-problems
Answer: If the line from the wall to the power source is good, and the power source is good, then the problem will be in the wall.  It's hard to imagine how something could go wrong between the top of the wall and the outlet, but I had this very problem once and traced it to a mystery, unclosed box with the two unterminated wires in it in the middle of the wall for who-knows-what reason, completely hidden by drywall.
Is there a switch on this wall as well?  If so, the outlets are probably connected there to the power source.  If not, your best bet may be to start at the outlet and trace the wire back to the source through the wall.  This will involve removing wall covering, and the repairs and cleanup that that entails.

Bank of wall outlets not working


Spotlight doesn't show Finder

Question: I typed "Finder" in Spotlight. Finder isn't shown. Why?
I use Mac OS X 10.9 (Mavericks)

On Mac OS X 10.10 (Yosemite) typing "Finder" in Spotlight does display the finder app.
Answer: What are you trying to accomplish? Finder.app is hidden. If you wish to quickly open finder, command-tab is easy.
If that's your goal, some of these suggestions may be helpful. Personally I use finder windows accessed via command-tab. I dragged my Downloads directory into the sidebar (it should be there already). So: command-tab to finder, click Downloads in the sidebar and viola. Alternatively, you could drag the downloads folder to the dock (it should be there already). Or, swipe 4 fingers down to reveal desktop (an option in system preferences -> trackpad), then click desktop to select finder.
In Automator you could build a workflow and save it as a service. In Automator use open finder and save as a service. This will put it in your Library to be available across applications. Then enable a quick key assigning a keyboard shortcut in system preferences -> keyboard -> keyboard shortcuts -> services.
Comment: Thanks. I was surprised as on Yosemite typing "Finder" does display the finder app.
Comment: cheers. that is an interesting change, perhaps another can say why...
Answer: Because Finder is located inside /System/Library. (Specifically, at /System/Library/CoreServices/Finder.app.) Everything inside any of the Library folders is considered a "system file", and not normally visible.
Interestingly, Yosemite seems to have made an exception for /System/Library/CoreServices, treating it more like an applications folder than a library folder. Spotlight will show Finder in Yosemite.
In a Finder search window (which is basically Spotlight with more focus on finding files but little or none on finding non-files), you need to add a [System files][are included] search term to see system files. If you don't see System files as an option in that first menu, select "Other..." and find "System files". (Note that you can search for it!) For future convenience, put a checkmark on it to add it to the menu.

Spotlight doesn't show Finder


How do I write a multi-line message to the quickfix or location list?

Question: I've got a program that spits out multi-line messages separated with something other than a newline. I want these messages to show in the quickfix and location lists spanning multiple lines as intended. What character can I replace the separator with to get that result? For example, setqflist([{'text': "1\r2"}]) prints 1^M2 on a single line. What character can I use to get two lines?
Answer: It can depend on your system. Some characters may be right on one system and not work at all on another. Between Macintosh, Windows, and Unix/Linux, there are three different types of carriage returns.
\n may work for you. You could also try them both together: \n\r (I think that's the right way, but you could also try it reversed)
Alternately, you can instruct Vim to use a particular kind of carriage return, no matter what system you are on, by playing with the fileformat (abbreviated as ff) option:
set ff=dos for example switches to the Windows formatting when it reads the file or buffer. You can re-read a file you already have open in a different format by using this command:
:e ++ff=unix

You may have to play around with it and see what works best.
For far more detail, see this page on the Vim wiki.


Cache-control in Tomcat

Question: We have run into a problem with Internet Explorer where it is unable to handle the Cache-control header set to no-store for SSL downloads. See http://support.microsoft.com/kb/323308 for more details.
Unfortunately when Tomcat authentication is enabled it appears to automatically add the Cache-control header with the value no-store.
I know that there is a workaround for when using the BasicAuthenticator in Tomcat. See http://daveharris.wordpress.com/2007/07/09/how-to-configure-cache-control-in-tomcat/. This solution describes using the disableProxyCaching attribute on the authenticator.
However, we are using SSO authentication using org.apache.catalina.ha.authenticator.ClusterSingleSignOn which does not appear to have the disableProxyCaching attribute. Is there a way to change the cache-control headers when using SSO?
Answer: You can use a Filter that will run over the cache-control header that tomcat imposes.
response.setHeader("cache-control", );
chain.doFilter(request, response);

Cache-control in Tomcat


QGIS 3.4.4 georeferencing issues

Question: I have upgraded to QGIS 3.4.4 and using the georeferencer and there appears to be shift when zooming in and out of a georeferenced image (Transformation type is Helmert, Resampling Method is Linear and CRS is 27700) which I haven't had on 2.18. Also getting colour issues when zooming out as well. Not sure if anyone has had the issue or knows If I'm doing anything wrong.
Answer: I had a similar issue with raster data when I displayed layers with different coordinate systems within one QGIS-project (using spontaneous projection/ on the fly projection). So maybe it might help to disable the on-the-fly-projection an to use only layers with a same coordinate system that is the same as those of the georeferenced image.
Comment: The project is set to the same projection.

QGIS 3.4.4 georeferencing issues


Component configuration file in xcode?

Question: Lets say i want to develop an RSS Reader reusable component for iOS with a configuration file which can me updated by user. So my wish list is as follow:

Add the component to a new project in single click. Just like adding framework.
I also want that my source files are NOT visible to other user (developer).
And yes if there are resources associated with it should also seamlessly be added to new project.
And most importantly I want the developer to be able to adjust few parameters for my component eg specifying the URL for my RSS Reader.

In short after i am done all i want the developer to see are my header files and configuration file. Any ideas ?
Answer: You should use a fake framework as described here.

Component configuration file in xcode?


Can you host WordPress using Amazon Web Services such as EC2, RDS, EBS etc?

Question: I've been looking into the feasibility of using something like Amazon Web Services for hosting WordPress sites. I was wondering if it is possible and if so what people's experiences are of having done so. What aspects of AWS did you use?
Answer: You explicitly mention three services in your question title: EC2, RDS and EBS.
If they're the three services you're interested in, then yes, very easily.
EC2 + EBS are pretty much the same thing, EBS is simply a persistent storage extension to EC2, and RDS is a full MySQL database - you should be able to get those three services running together very easily.
If you were looking to extend to some of the other AWS products, S3 is very easy to implement on WordPress for file uploads - check some of these plugins.
SimpleDB, however, has very limited usefulness to us without a rewrite of how WordPress reads and writes to the database. There is an article by Amazon on how to create a plugin which adds tagging functionality that uses SimpleDB as the storage, but that is as much as I could find regarding the use of SimpleDB with WordPress.
Comment: Thanks for your response, it seems like it could be a pretty good solution. I'll have to schedule some time to try setting up an instance and see how it performs.
Comment: what EC2 instance would i need for my personal private weblog on wp 3.1 mu (and 15 domains on subblogs) with low traffic overall? micro?
Comment: nevermind i just found the "free tier" program FREE is good! signed up.
Answer: Thomas McDonald is pretty much right on track.  Just wanted to add that if you're going to use EC2 to scale your sites as you need more servers, then you will need to contend with the fact that you will have multiple instances of your site out there with no common shared volumes.
In other words, let's say you have 2 EC2 instances up -- each one is completely separate from the other, so what happens when a user uploads an image?  That image only exists on one server and not the other.
To combat this, you'll want to use something like S3 with Cloudfront or a some other type of CDN.  Then something like W3 Total Cache, which will propagate your uploads to the CDN.  Then when an image is shown in your theme the image URL will be automatically rewritten to the CDN source URL by W3 Total Cache and, despite that image not existing on the local server, it will still be visible to visitors.
Comment: Thanks Gabrielk, I'll keep that in mind although I don't think I would need multiple instances initally. It is one of the attractions of AWS though, the potential of scaling up for projects that get bigger but without having to fork out a lot of cash for big servers up front.
Answer: Yes you can, performance wise things should be just as fast, if not faster than a standard dedicated box, you just need to figure out if the varying costs are worth it or not.
If you use something like Scalr, you can do some neat auto scaling stuff as well.


Investment Clubs Net Worth

Question: If a company is asking for investors with a net worth of $1M, can that net worth be collectively, i.e., investment club members?
Comment: @RonJohn I think the question is can the club be considered "an investor".
Comment: @DStanley maybe.  OP would have to clarify.
Comment: The club's assets are split across multiple members, so this would be like claiming your NW is a trillion dollars because you own some shares of Apple stock.
Answer: When you state net worth of $1 Million, it sounds like the limit set in the US by the Securities and Exchange Commission for a Accredited Investor which is set at $1M net worth excluding primary residence for Natural Persons.
Assuming the Investment Club qualifies by not being formed just to purchase a specific security, such a club would need assets of at least $5M.
From the above link:

Who is an accredited investor?
An accredited investor, in the context of a natural person, includes anyone who:

earned income that exceeded $200,000 (or $300,000 together with a spouse) in each of the prior two years, and reasonably expects the same for the current year, or
has a net worth over $1 million, either alone or together with a spouse (excluding the value of the person's primary residence).

There are other categories of accredited investors, including the following, which may be relevant to you:

any trust, with total assets in excess of $5 million, not formed specifically to purchase the subject securities, whose purchase is directed by a sophisticated person, or
any entity in which all of the equity owners are accredited investors.

In this context, a sophisticated person means the person must have, or the company or private fund offering the securities reasonably believes that this person has, sufficient knowledge and experience in financial and business matters to evaluate the merits and risks of the prospective investment.
Answer: Probably not initially, but it will depend on how desperate they are later. Consider the probable reason for the $1M requirement. They want low-hassle investors who will follow through.
They don't want someone who initially thinks that they can invest $100k, but later decide they really can't afford to. They also want someone with reserves, who could invest more later in a crunch, or invest more later to enable more rapid growth.
Your investment group will be high-hassle, one person can't make the decisions. The paperwork will be more complicated, and your group will need to vote on everything.

Investment Clubs Net Worth


Shared hosting vs VPS with CiviCRM & Wordpress

Question: I would like to start a new civi installation on WP to keep around 1000 contact records. What is the best option, Shared hosting vs VPS?
Pros & cons?
Any recommended hosting provider?
Thanks
Answer: Shared hosting can work well, but use one that specifically supports CiviCRM.
Typically VPS offers more resources (RAM, CPU etc) but at a higher price.  VPS's are not all the same: on unmanaged ones you need to do more sysadmin work but have more flexibility; on managed ones there are more restrictions but more of the work is done for you.
Good shared hosting gives you flexibility while taking care of all the basics.
Have you considered CiviCRM's own Spark offering?
I've had good experiences with shared hosting on CiviHosting


From Deore MTB to GRX Gravel

Question: I want to convert my MTB to gravel, and the main issue for me is the transmission.
I plan to completely replace the transmission with 10x2, because it is pretty worn out. However, for me the main question is this: is it possible to use drop bars designed for 10 and 2 speeds respectively on a deore transmission?
Cassette M4100 11-42T
Rear Derailleur RD-M4120-SGS
Chain CN-HG54
Front Derailleur FD-M4100
Bottom Bracket SM-BB52
Comment: Is your frame size really suitable for a conversion to drop bars? Keep in mind that drop bars make the bike much "longer" so it usually only works with a frame which is one or two sizes too small and a short stem.
Comment: It doesn't really answer your question, but if you want to use MTB components with brifters, this guide is comprehensive: https://bikepacking.com/gear/guide-to-mullet-drivetrains/
Answer: No, you won't be able to find compatible Shimano drop bar shifters 10-speed Shimano MTB derailleurs. The MTB and road groupsets are not compatible for 10 speed and higher.
I suggest this question for more details about your options for shifters compatible with 10-speed Deore. You also have to think about the compatibility with your brakes, which is not given.
You also won't be able to put Shimano road and GRX cranks on 73 mm bottom brackets and you will have issues with proper fit of your frame. If it fits you with flat bars, it may fit poorly with drop bars.

From Deore MTB to GRX Gravel


ArcPy polyline shape object to numpy array

Question: I am learning to work with the geometry objects included in ArcPy.
I'm really interested in using them since I don't want to create files at all since most of the data I'm going to use is just mere intermediaries.
Reading the documentation I was able to discover that according to them they say an ArcPy Geometry object can be used together with the geoprocessing tools. I have tried this and it is really very interesting to be able to use these objects instead of writing files, the main problem is that not all the tools work (or at least I can't make them work). More specifically the FeatureClassToNumpyArray() function, which when passing Polyline ArcPy object returns me the following error:
'in_table' is not a table or a featureclass.
I get the same error when I try to use searchCursor(), and I would like to know if it is possible to use some of the previous two functions (or if there is another efficient way) to obtain a numpy array of points through a Polyline () object.
Comment: You could try creating an in-memory dataset with `mem_fc = arcpy.CopyFeatures_management([polyline0, polyline1], 'in_memory/test')` then feeding that in.  However since you won't have any attributes anyway, you could also just loop over the vertices of the geometry objects themselves to manually construct your numpy array
Comment: @mikewatt really what I want to store is the output of "contourLists" (which by the way, it would be great if it accepted a tin as input, since I have to previously convert the tin to raster, maybe it is not the best approach), the first thing I tried as you Well it indicates it was to use the "in_memory" space, however, when I stored the contour list there, the generated contours do not have the "Shape_Lenght" field and it is precisely one of the fields that I need to extract when I pass it to "contourList". For this reason I got rid of that idea.
Answer: You can see the official docs for how to extract Point() objects from geometries.  Using that to create numpy arrays would look something like this:
polylines = arcpy.CopyFeatures_management('polyline_fc', arcpy.Geometry())

for geom in polylines:
    coords = []
    for point in geom.getPart(0):
        coords.append((point.X, point.Y))
    array = np.array(coords)
    print(array)

For giggles lets try another approach that parses the well-known binary representation:
import numpy as np

polylines = arcpy.CopyFeatures_management('polyline_fc', arcpy.Geometry())

for geom in polylines:
    buff = geom.WKB

    dtype_struct = np.dtype([('byteOrder', np.byte), ('wkbType', np.uint32), ('numPoints', np.uint32)])
    array_struct = np.frombuffer(buff[:dtype_struct.itemsize], dtype_struct)

    assert array_struct['byteOrder'] == 1
    assert array_struct['wkbType'] == 2

    array_points = np.frombuffer(buff[dtype_struct.itemsize:], np.float64).reshape((-1, 2))

    print(array_points)

If you do away with the validation it boils down to:
for geom in polylines:
    array = np.frombuffer(geom.WKB[9:], np.float64).reshape((-1, 2))

WKB reference: https://www.ibm.com/docs/en/ias?topic=formats-well-known-binary-wkb-representation
Comment: Thank you very much for the demonstration. I knew that I could extract the coordinates one by one, however, I don't know if by having to construct the coordinates one by one I lose the speed improvement that I gained by storing the output of "contourList" in a python object instead of in a file . I really don't understand how they don't provide an easier way to pass geometry objects to numpy and I would like to know why it's not possible to use "FeatureClassToNumpyArray" since I can use other tools, which is a shame.

Thanks again for answering me
Comment: No problem.  I think only the ESRI devs can answer that one, but perhaps because `FeatureClassToNumpyArray` is geared towards converting tabular data into numpy format they didn't bother to accept straight geometry objects.  However my 2nd example shows how to bypass the one-by-one point unpacking if that's your goal
Comment: There's also no way to store the coordinates from multiple polyline geometries in a single numpy array, because they likely have different numbers of vertices.  And in numpy arrays all of the fields must be the same size.  So there's no avoiding having to unpack each polyline into an individual numpy array
Comment: I have been testing your suggestions and they work really well, however I am facing the following problem:
Since the output of contourLists () is a collection of polylines that I must filter, until now what I did was group the lines based on the "Contour" field that is generated by default when using this tool and then keep only the polyline that present the longest length for each group and then I pass them to points.
Comment: The problem is that while the "Shape_Length" field can be easily obtained using geom.length, the "Contour" field that is generated when I write the output of contourList () in a file is not generated (or at least not I know how to access it) when I use arcpy objects. Is there a way to maintain that field? If you think it is the reason for a new question, please say so without problems.
Thanks, as always. @mikewatt
Comment: To my knowledge, there's no way to retain field values when creating geometry objects directly.  You'd probably have to create an `in_memory` feature class instead, then use a search cursor to extract both the geometry and field values


method for creating a unique validation key/number

Question: I'm using django for a web-magazine with subscriber-content.  when a user purchases a subscription, the site will create a validation key, and send it to the user email address.
The validation key would be added to a list of "valid keys" until it is used.
What is the best method for creating a simple yet unique key?  Can someone suggest a standard python library for key-creation/validation/ect?
This might be a very simple question, but I'm very new. ;)
Comment: Amazing - Thanks guys.  I accepted Gardner, since he also included the tip on django extensions.
Answer: I'd recommend using a GUID. They are quickly becoming industry standard for this kind of thing.
See how to create them here: How to create a GUID/UUID in Python
Answer: As other posters mentioned, you are looking for a GUID, of which the most popular implemntation UUID (see here) . Django extensions (see here) offer a UUID field just for this purpose.
Answer: Well, you can always use a GUID. As you said it would be stored as a valid key.


Will a 3x10 or 3x9 front derailleur work on a 6/7 speed chain?

Question: Will a 3x10 or 3x9 front derailleur work on a 6/7 speed chain? Can't seem to find a good MTB front derailleur for 6/7 speed chain. Tourneys won't do it for me
Comment: @Criggie yeah. You are right. But i love my current bike too much and also i don't want to spend too much money on a new bike
Comment: You've asked three different questions about using more modern parts on a 6/7 speed setup.  At some point you would be better to move completely to the newer designs, which means a new shifter.  But parts availability will be easier and you won't have as many compatibility problems.
Answer: Generally yes, the front mech can work acceptably with other sizes of chain.
Since you will have a thicker chain than expected, the front derailleur cage may need to be widened a little to reduce chain rub.
The actuation ratio is between your shifter and the chainrings, so a friction shifter will work perfectly, and an indexed shifter should be acceptable.  If you're mixing shifter/chainrings then things could get odd but that's more likely above 10speed.  Below 10 speed, the chainring spacing generally was the same.
Remember the front shifter is more of a "brute force" mechanism having to move the chain under tension, compared to the rear shifter whose chain is mostly slack.  So the front shifter has to deal with higher pressures and tensions.

I have an old MTB that was 15 speed, (3x5) which now has a 9 speed rear cassette/derailleur/chain.  The front mech is still the original triple, on the original chainrings.  All I had to do was close the cage a bit to improve shifting.  I do use friction shifter on the left hand which makes trim very flexible.
Answer: Might I suggest looking at new-old-stock or used items from say the 1990s? Something like this (example only: https://www.ebay.com/p/1300427514 )


How can I replace Xubuntu with Lubuntu on a Windows XP laptop?

Question: I installed Xubuntu 11.10 on an old Windows XP in a separate partition. Later, I realized that I should have installed Lubuntu 11.10. Is there an easy way I can delete Xubuntu and replace it with a clean install of Lubuntu without deleting the partition? If not, what's the best way to delete Xubuntu and start over? By the way, this is my very first Linux install so I'm a definite newbie and would really like advice that does not involve using command lines. Thanks.
Comment: You know you could just install LXDE log out select it as your session log in, then remove XFCE and that's it.. no Command Line, no partitioning, no reinstalling,  no anything... and definitively not hard to do.
Answer: Well I know you want advice that does not involve command lines but you only have to copy and paste one line into the terminal. (although you'll need a root terminal)
Open a terminal (Ctrl+Alt+T usually works)
type
sudo -i
enter your password
copy and paste (paste in a terminal is CTRL+SHIFT+V) in:
apt-get install lubuntu-desktop -y && apt-get remove xubuntu-desktop -y
That's it
the "apt-get install" bit installs the ldxe desktop environment and all the programs that go with it, the "apt-get remove" bit removes all of xubuntu. You could also run apt-get autoremove to make sure everything has gone. Or you can just reinstall by downloading lubuntu and selecting remove xubuntu and install lubuntu instead during the install process.
Comment: If you install, just install into the Xubuntu partition. That will remove xfce by over writing your data. If you want to keep your data, install lubuntu-desktop as suggested. You will have to remove more then xubuntu-desktop to remove all the xfce packages. See http://www.psychocats.net/ubuntu/puregnome
Comment: Thanks for your suggestion. Since I have Lubuntu installed on a USB flash drive can I boot from the USB drive and just click on the "Install Lubuntu" icon on the desktop? Would doing so give me the option to completely remove Xubuntu during the install process?
Comment: I second bodhi.zazen's comment. If you recently installed Xubuntu, I'm sure you don't have a ton of data. A clean install into the Xubuntu partition is absolutely the way to go. If you don't know how to do that, just ask and we'll give some pointers...
Comment: Oh... and welcome to Lubuntu!
Answer: I don't know how to answer the first question but the second I can handle:
Best way to delete and start over is to format a usb key so it's FAT32 and then use a program like unetbootin to create a bootable disk.
Then all you need to do is set your bios so your computer boots from USB. You can do this at start-up by pressing one of the F keys (f2 or f4 or f8 something, I can't remember and it's different on each computer anyway - I just press them all and it usually works out ;-)) then you go into the page of the bios and change the order so USB is it's the first on the boot priority list.
Once you did this you need to save and exit it (be careful in the Bios too if you are not sure about anything I have said it is best to ask before fiddling about in there). Then reboot your computer and the lubuntu installer should prompt you for what to do next.
After that your computer will be all better.
Answer: Yes, you can. You need a liveCD/liveusb Lubuntu. Insert it in your PC, boot in live mode. Open GParted and select your Xubuntu partition. Right-click and choose format/reformat to ext. Do not delete that partition. Just reformat it. Apply changes in GParted and close it. Click the Install Lubuntu icon. Install Lubuntu to the partition you just reformatted.
Answer: Just use the software center or Synaptic to add the Lubuntu package, and boot into Lubuntu instead of Xubuntu on boot. Simple as that.
The answer from   Uri Herrera Jan 13 '12 at 22:51 is correct.
Comment: Not sure why this was voted down. Not only is this a VERY old post I commented on, but the answer was useful. I doubt the original poster even read it.Adding Lubuntu and logging in at boot in session manager screen is what I'd do, unless of course you have a low powered system without a lot of memory and/or small drive (almost impossible nowadays.) Then, you'd want to back up all your data and just install Lubuntu from USB stick and restore your data from the backup later.


Multinomial logistic regression assumptions

Question: What are the proper assumptions of Multinomial Logistic Regression? And what are the best tests to satisfy these assumptions using SPSS 18?
Answer: The key assumption in the MNL is that the errors are independently and identically distributed with a Gumbel extreme value distribution. The problem with testing this assumption is that it is made a priori. In standard regression you fit the least-squares curve, and measure the residual error. In a logit model, you assume that the error is already in the measurement of the point, and compute a likelihood function from that assumption.
An important assumption is that the sample be exogenous. If it is choice-based, there are corrections that need to be employed.
As far as assumptions on the model itself, Train describes three:

Systematic, and non-random, taste variation.
Proportional substitution among alternatives (a consequence of the IIA property).
No serial correlation in the error term (panel data).

The first assumption you mostly just have to defend in the context of your problem. The third is largely the same, because the error terms are purely random.
The second is testable to a certain extent, however. If you specify a nested logit model, and it turns out that the inter-nest substitution pattern is entirely flexible ($\lambda = 1$) then you could have used the MNL model, and the IIA assumption is valid. But remember that the log-likelihood function for the nested logit model has local maxima, so you should make sure that you get $\lambda =1$ consistently.
As far as doing any of this in SPSS, I can't help you other than suggest you use the mlogit package in R instead. Sorry.
Comment: Also, the multinomial probit model gives comparable output with a different set of assumptions. So an MNP/MNL comparison can be valuable as well.
Answer: Assumptions:

Outcome follows a categorical distribution (http://en.wikipedia.org/wiki/Categorical_distribution), which is linked to the covariates via a link function as in ordinary logistic regression
Independence of observational units
Linear relation between covariates and (link-transformed) expectation of the outcome

For assumption 1 to be fulfilled, the categories of your outcome need to be exclusive (non-overlapping) and exhaustive (covering all possible forms the outcome can take).
I don't really know if there are any proper statistical tests for assumption 2. For time-series data there is a test of autocorrelation called Durbin-Watson test. For other forms of correlated data, I think you would rather make that decision based on theoretical considerations (e.g., if your data come are derived from a cluster-sampling procedure, you would expect the data within clusters to be correlated).
As for assumption 3, in binary logistic regression you can plot binned residuals against estimated probabilities to see if the average residual is around 0 over the entire range of estimated probabilities. I suppose this can be generalized to multinomial regression by making (k-1) such plots instead for the different categories of an outcome with k categories.
EDIT:
Concerning alternative models: Assumption 1 is fairly straightforward to fulfill. You might run into trouble because you have to estimate a large number of parameters (k-1 different sets of intercepts and slope parameters). In such a case, you could for example collapse the outcome into a binary outcome and do a simple logistic regression.
If assumption 2 is violated, you could use a mixed model, which allows you to specify a dependence structure-
As for assumption 3, you could transform variables of which you suspect that they have a non-linear effect. A common transformation is for example to include age squared in health-related outcomes.
Comment: To follow up on the first reply, you can use the Hausman-McFadden test for independence.
Answer: One of the most important practical assumptions of multinomial logistic is that the number of observations in the smallest frequency category of $Y$ is large, for example 10 times the number of parameters from the right hand side of the model.
Comment: Sounds like a Bayesian prior is being called for - can't disagree.  But without external information the unconstrained multinomial logistic has an awful lot of parameters to estimate.
Comment: It turns out that this is not always completely true. There has been some (very) recent work showing that consistent parameters can be estimated for alternatives that you never observe, provided you have some exogenous information on what the actual population frequency is. But this requires a different estimator, so in general you are correct.
Answer: gmacfarlane has been very clear. But to be more precise, and I assume you perform a cross section analysis, the core assumption is the IIA (independence of irrelevant alternatives).
You can not force your data fit into the IIA assumption, you should test it and hope for it to be satisfied. SPSS could not handle the test until 2010 for sure. R of course does it, but it might me easier for you to migrate to Stata and implement the IIA tests provided by the mlogit postestimation commands.
If the IIA does not holds, mixed multinomial logit or nested logit are reasonable alternatives. The first one can be estimated within the gllamm, the second with the far more parsimonious nlogit command.
Answer: @h_bauer has provided a good answer.  Let me add a small complementary point:  You can also test for a curvilinear relationship by adding curvilinear terms and performing a nested model test.  For example, imagine you have $X_1$ as an explanatory variable, but you aren't sure whether the relationship between it and the link transformed expectation is a straight line.  You could form a new model by adding $X_1^2$, and $X_1^3$, and then test to see if your new model fits better than your original model.
Another assumption of generalized linear models, like the multinomial logistic, is that the link function is correct.  Strictly speaking, multinomial logistic regression uses only the logit link, but there are other multinomial model possibilities, such as the multinomial probit.  Many people (somewhat sloppily) refer to any such model as "logistic" meaning only that the response variable is categorical, but the term really only properly refers to the logit link.  For more on links, it may help you to read my answer here:  Difference between logit and probit models.
Regarding addressing violations of these assumptions, it is mostly self-explanatory.  If the observations are not independent, you can add the relevant fixed or random effects to make them so.  If the relationship with a predictor is not linear, you can add transformed variables so that it is linear in the augmented predictor space.  If the link is not appropriate, you can change it, etc.  In general, multinomial logistic regression does not make very constraining assumptions.


Change the color of an empty?

Question: I know what empties are for, but I'd like to change the colors of empties as a visual aid (for example, red for camera controls, green for object controls etc.) while animating. Is this possible?
Answer: They can be changed from User Preferences > Themes.

Edit:
As you commented, per object viewport color is unsupported officially. Such idea has been discussed before, and decision is to allow per type color via Themes setting (see above), instead. So the answer is no (or not yet).
However, you may consider to use another trick in case you have to:

Create a simple mesh object, assign material with specific Diffuse color (Viewport Color for Cycles);
Parent it to the empty (or other objects);
Toggle off the renderability (or CtrlH), and also selectbility in Outliner if necessary.

Then, objects like empties and cameras can be indentified as you expect (except in Wireframe or Bounding Box view though). It might be the only solution so far, if you really have to.
Edit 2:
A specific feature to set per-object vewport color is available in Gooseberry branch build, I believe it will be soon merged into master trunk.
Comment: I've updated the answer. :)
Comment: Is there a way to change the color per empty (and not globally with a theme)?
Comment: I've decided to just use meshes with properties; empties weren't flexible enough (but meshes are too flexible, so I depend on some properties I set instead of the type of empty). Thanks anyway!
Comment: some Idea about how to do this in blender 2.8?

Change the color of an empty?


Change the icon of all the folders on mac

Question: I want to change the default folder icon to something nicer.
please help
Answer: Most standard system wide icons for the macOS are stored here:
/System/Library/CoreServices/CoreTypes.bundle/Contents/Resources
The folder icon in which you're interested is named GenericFolderIcon.icns.
If you're running Mac OS X El Capitan (10.11) or later, due to System Integrity Protection, you won't be able to modify these files.
However, if you turn off System Integrity Protection, you'll gain the ability to edit the icon files. Here're instructions:

Boot into recovery mode (boot while holding down ⌘R)
Open the terminal
Type csrutil disable.
Reboot normally.
Now edit the icon file(s).
Repeat steps 1–4 to re-enable System Integrity protection, except that the Terminal command you'll use is csrutil enable.

Needless to say, be very careful when turning off System Integrity Protection and editing system files. There's a lot of potential to royally screw things up when doing so.
For step 5, you'll need some sort of program that can edit or generate ICNS files. Your best bet is probably to search around this site or with Google to find recommendations. You may have to try several options before you find something that fits your needs.
Comment: I've updated to Mac OS Catalina. I have successfully turned off SIP via csrutil disable, however I still can not modify the folder icons. Is it not possible to do in Catalina or is there an additional step now?
Comment: @user906357 Changes to the security of Catalina and later make modification of the system impossible.
Answer: Liteicon automates all of this and works on the latest update (High Sierra). CandyBar is an older program for versions below 10.9.
http://freemacsoft.net/liteicon/
I believe you might still have to disable system integrity first though
Comment: An update to this answer: The linked page states it's unfortunately not possible for Big Sur, so they've stopped development, though they apparently still offer downloads for Mojave and Catalina.
Answer: jefe2000 provides an excellent step-by-step overview of the process you will need to follow.
However, if you already have an icon file you want to use, or you don't wish to use other software to edit/create your own, for Step 5 in jefe2000's instructions you may find it easier to do as follows:

Browse the web for an already existing icon file (.icns file). There are millions of .icns files that you can download.
Now when you're at Step 5 of jefe2000's instructions, rename the existing GenericFolderIcon.icns file to something else (e.g. OldGenericFolderIcon.icns)
Add the newly downloaded icon file you want to use into the same directory and rename that file to GenericFolderIcon.icns.
Continue on to Step 6 of jefe2000's instructions

Finally, you may want to test how an icon will look (or you may only want to change the icon of a specific file or folder. To do so you can follow these instructions.


How to derive fixed-end moment of a beam?

Question: Here is my attempt but I am not getting the right answer. Can someone tell me what I am doing wrong?
Answer: your calculations of reaction forces $F_A\ and\ F_B \  $ are wrong. This beam is indeterminate and you are handling it as a simply supported determinate beam.
One way of finding the fixed end reactions is to select the two fixed end moments as two degrees of redundancy and solve the beam. we will refer to this diagram at the bottom of the page.
$$R_A = \frac{Pb}{L}+ \frac{M_a}{L} -\frac{M_b}{L}\\\ RB = \frac{Pa}{L}-\frac{M_A}{L}+\frac{M_S}{L}$$
Force displacement relations:
$ (\theta A)_1 = \frac{Pab(L + b) }{6LEI}\quad$and $\ (\theta B)_1=\frac{Pab(L + a) }{6LEI}  $
$ (\theta A)_2=\frac{M_AL}{3EI}( \theta B )_2=\frac{M_AL }{6EI}  $
$( \theta A )_3=\frac{M_BL}{6EI}(\theta B  )_3=\frac{M_BL }{3EI}  $
Compatibilty equations:
$\theta A = (\theta A)_1 - (\theta A)_2 - (\theta A)_3 = 0  $
$ \theta B = (\theta B)_1 - (\theta B)_2 - (\theta B)_3 = 0 $
$ \therefore \frac{M_AL}{3EI} +\frac{M_BL}{6EI}= \frac{Pab(L + b) }{6LEI}\\ \frac{M_AL}{6EI }+\frac{M_BL}{3EI }= \frac{Pab(L + a) }{6LEI} $
Solving these equations, we obtain
$ M_A =\frac{Pab^2}{L^2} \quad M_B= \frac{Pa^2
b}{L^2} $
and the reactions:
$ R_A =\frac{Pb^2 }{L^3}(L + 2a ) \quad R_B=\frac{Pa^2}{L^3}(L + 2b ) $


Is the potential at each point on a circuit same?

Question: Is the potential at each point on a circuit same,if so why?
I have read that in order for current to flow through any kind of resistance,the potential of charges reaching resistance is higher than those exiting it. ie. all the charges before it are at same potential and the charges exiting it are at some low but same potential.this means that there are only two potential levels in a circuit. for better understanding of my doubt please check out this video
https://youtu.be/-Rb9guSEeVE
Comment: Does this answer your question? [How electric currents can flow between 2 points at the same potential?](https://physics.stackexchange.com/q/45040/)
Answer: Kirchoff's law says that $\Delta V_{\rm loop} = 0$, i.e. that the voltage summed around the closed circuit must add to zero.
If you have a circuit with one battery and one resistive element, like a resistor or a lightbulb, like this:

Circuits are typically modeled with ideal wires, which have no resistance. That means that in this situation, if the battery is 3V, then every point in the wire between the (+) terminal of the battery and the "upper" terminal of the resistor is 3V. At the resistor, you see a voltage drop of -3V. Then, from the bottom of the resistor back to the (-) terminal of the battery, every point is 0 volts. So in this very simple circuit, the wires are held at two distinct potentials.
In reality, wires do have resistance and dissipate energy through heat, but this is very small, so it is a good approximation in many applications to model the circuits with ideal wires.
Comment: There are circuits where the resistance of the wire matters and must be modeled, but they tend to be rare.  You find them in "power circuits," that are moving large numbers of watts through low resistance loads.  For a point of reference, 18ga wire typically has a resistance of about 6 ohm per 1000ft.  In most "reasonable" circuits, this loss is far less than the uncertainty you have about the resistor's actual resistance, and lots of other fun fiddly details, like how the resistance of the resistor changes with temperature.
Comment: Can I conclude that at each point the potential changes in real life circuits
Comment: Yes, but only by a very small amount.
Comment: Now if the potential changes even with small amount..then as I go from positive terminal of battery to starting end of resistance,then the potential would have changed significantly, isn't??
Comment: No, when I said a very small amount, I meant a very small change in potential from the terminal of the battery to the starting end of the resistor. For most practical purposes, consider the potential across the wire to be constant. In reality it is never really more than 2-5% I believe.


How do I Change my old iPod Apple ID to my new Current ID to match the one on my iPhone?

Question: About 2 almost 3 years ago I purchased an iPod touch 5. My phone and my iPod both have 2 different Apple ID's (same as 2 different email addresses). The email address used for my iPod is an email I no longer use and I'm not sure if I deactivated it or not. Anyway when I go to Settings> iCloud> it still tells me that same old email address.
Answer: Sign in with your old Apple ID at iCloud.com, go to Settings->Manage (Apple ID)->Click on your iPod and remove it from the list of devices. And finally use your new Apple ID on your iPod.
Answer: Contact apple directly, only they can authorise Apple ID merges or the transfer of products from one ID to another.
Use the link below to get the details. I would have my passport/ID at hand, you may also have to send them a scan of proof of purchase or proof of identity.
https://support.apple.com/en-us/HT204169


Recursive Sequence Proof

Question: Take the function defined by $a_n= 3n + 1$, for all $n \in \mathbb{N} \geq 0$. Show that this sequence satisfies the recurrence relation $ a_k = a_{k-1} + 3$ $\forall k \in \mathbb{Z}, k \geq 1$. My main problem is finding the basis because no conditions were assigned to $a_{k-1}$.
Comment: By definition, $a_k=3k+1$ and $a_{k-1}=3(k-1)+1$. You are supposed to show that $(3k+1)-(3(k-1)+1)=3$. That is not hard.
Comment: I just got confused because we're supposed to be doing this via proof by induction and the basis was eluding me. Thank you.
Answer: Since $a_n=3n+1$ for all $n\geq0$, we can see that $a_0=1$ and $a_1=4$. Using the recurrence relation $a_k=a_{k-1}+3$ for all $k\geq1$ we see that $a_1=4$ when $a_0=1$. Assume that $a_l=3l+1$ for some arbitrary positive integer $l$. We must show that $a_{l+1}=a_l+3=3l+4$. Consider $a_l+3$, using our induction hypothesis we obtain $a_l+3=(3l+1)+3=3l+4$. Thus by the Principle of Mathematical Induction $a_n=3n+1$ for all $n\geq0$ is a solution to the recurrence relation $a_k=a_{k-1}+3$ for all $k\geq 1$.

Recursive Sequence Proof


Leaf stains on fresh concrete driveway not sealed yet

Question: My driveway was poured when I was on vacation and leaves fell on it and they made a black mark I believe it's called tar leaf in about 20 places in the driveway and before they come and seal it I wanted to know how to get it off before the winter comes and the Salt destroys it
Comment: Possible duplicate of [What is the best way to remove and prevent leaf stains on concrete pavers?](http://diy.stackexchange.com/questions/1876/what-is-the-best-way-to-remove-and-prevent-leaf-stains-on-concrete-pavers)
Comment: The pavers are little bit different this is a california-style driveway with a brush stroke so on the rough spots the leaves have made marks that really need to come off before he seals it I understand the sun will fade in time the winter is less than 2 months away and he's thinking we shouldn't seal it until the leaves marks are gone. and he didn't see like at the beginning once he finished like within 2 hours because he was doing three of my neighbors houses all in the same month... we all just figured to get it done at the very end
Comment: Hello, and welcome to Stack Exchange. Rather than put lots of details in a comment, it's best to edit the information right into your question.
Answer: Tar leaf, simply known  as tar, is caused from the organic pigments that give leaves and pine cones their color get embedded in the micro-pores and small voids on the surface of the concrete. They can be hard to remove without a good detergent and some scrubbing action. The best detergent to use is one with bleaching action or ammonia. Typically detergents designed to remove organic stains (blood, food stains, etc.) work best. OxiClean is one example. Granular detergents offer an added benefit, because they provide additional friction when you scrub the surface.
Here are the basic steps to follow:
Use a pressure washer to remove all solid leaf matter and debris from the concrete.
Apply a household detergent (as recommended above) to damp concrete. Let it sit and wet out for a few minutes.
Scrub vigorously with a stiff broom or scrub brush.
Rinse the concrete with water until you remove all the soap.
If this procedure doesn't completely remove the stains, repeat the process.
Comment: I don't think I would chose to use a pressure washer on fresh concrete. A weak Muriatic acid and water solution would etch the stain off. but the detergent may be the best method but requires more physical labor.
Comment: I have different nozzles for the pressure washer and the weaker one has been fine.buy I do understand your point thank you..I have the dry detergent and bleach liquid and crystals and I will try all if I have to before they come and seal it..thx again
Answer: Tannin , get used to it. Unless you are going to paint it , the concrete will weather. Tannin stains will come and go. I use a 50:50 bleach/ water and a garden sprayer every coupe years to make my wife happy.


Why use the `+` (plus) or `!` (logical not) operator in an IIFE?

Question: I know the typical form of an IIFE is:
(function(){/* code */})()

However, recently I found a new one:
!function(){/* code */}()
+function(){/* code */}()

which also work as IIFEs.
I think in the first ! makes the value a boolean, so the code is equivalent to true(), and in the second + makes the value a number, so the code is the same as NaN()?  But true() and NaN() cannot possibly work, while the above two forms work well.
How do those two forms of IIFE work?
Answer: The function keyword has a duel purpose. Depending on the context you use it in, it either starts a function declaration or a function expression.
Function declarations declare a (hoisted) local variable in the scope of the current function and assign the function to that variable.
Function expressions don't, they require that you do something with the function immediately. Common examples include assigning it to a property of an object, passing it as an argument to a function (i.e. making it a callback) or using it as an IIFE.
If you try to follow a function declaration with () then it will error.

function myFunction() {
    console.log("Example");
}();

Only function expressions can be immediately invoked function expressions. (Well, arrow functions can too, but IIFEs are much older than them).
So to use an IIFE you have to do something to change the context the function keyword appears in so it is used to create an expression instead of a declaration.
The classic way is to wrap the function in parentheses.

(function myFunction() {
    console.log("Example");
})();

But anything that makes it an expression will work.
Putting an operator (such as ! or +) in front of the function keyword will do that.
While ! will take the return value of calling the IIFE and turn it into a negated boolean and + will turn it into a number that isn't the point. There's nothing to the left hand side of the operator so that value is discarded.
It is just a way to make the function keyword create a function expression that can be turned into an IIFE.

Note that the key benefits of IIFEs are also provided by block scoped variables (let and const) and JavaScript modules, so there isn't much reason to use an IIFE at all in 2022.
Answer: In the cases of !function() {}() and +function() {}() JavaScript will first evaluate the right handside of the given statement/IIFE and then cast the return value of the IIFE to a boolean in case of ! or to type number in case of +.
If the IIFE returns a boolean ! will negate the return value.
Example

console.log(
  !function() {
    return "something";
  }()
);

/**
!function() {
  return "something";
}()

evaluates to

!"something" and not true() or false()

lastly !"something" is equivalent to false
*/
Answer: There are many ways to get a function to execute immediately upon its definition. The common thing is that you must make sure the function is not interpreted as a function statement, but as a function expression.
By prefixing the function keyword with a unary operator like ! or + or any other unary operator (~, -, void, ...etc) you avoid that the function is interpreted as a statement.
Also assignment will work:
var x = function(){/* code */}()

Or array literal:
[function(){/* code */}()]

Or passing it as argument to a (cheap) function:
Object(function(){/* code */}())

Or applying a binary operator, provided the function expression is the second operand and not the first. Here with the comma operator:
0,function(){/* code */}()


Ubuntu on samsung 9 series thin laptop

Question: Has anyone tried loading ubuntu 11.04/10.10 on new Samsung Series 9 (NP900X3A). It looks pretty thinner and similar to Macbook Air. Does it has good support with ubuntu ?.
Comment: if you could go to the store you can try the livecd on it.
Comment: There is no CD drive on this laptop. He'll need a USB stick.
Answer: I've started using my new Series 9 laptop by immediately installing Ubuntu/Kubuntu on it.
My experience so far (other than having a fancy machine) is mixed.
Pros:
 - almost everything worked out of the box
 - amazing speeds
 - best quality laptop display I have ever encountered
Cons:
 - had to make a research on how to make the wireless driver work
 - had to make a research on how to get the right click functionality
 - had unidentifiable crashes almost every day, usually couple of times a day (8-10 hrs of usage daily).
 - had a lock up once that even made it impossible to do a hard reset holding down the power button. Given that it has built in battery and 6..7hrs of battery life, it was a huge annoyance.
I am unsure if the crashes are somewhat Linux-specific, did not use Windows. Because of them I am thinking of returning the laptop.
You may find some other guy's similar crash experience here.
Other issues are covered here.
Answer: Since the original answer by Greg, support seems to have improved considerably.  I am running 12.04 (beta 2) on a NP900X3B model.  There are workarounds for features that don't work out of the box (right-click, keyboard backlight, etc.).  See here.
I occasionally experience minor glitches but nothing like Greg described.  For example, once after waking from suspend the trackpad did not function properly.  But the overall experience has been quite good.
Comment: The only thing I can't get working on the np900x3b is the backlit keyboard controls... Can you advise me on your experiences with this? (By the way I'm not the same Greg as the other answer...)
Comment: There is an excellent set of instructions at: http://jablonskis.org/2012/linux-and-samsung-series-laptop-9-fn-keys/.  The instructions are for Fedora, but also work for Ubuntu.  For 900X3{B,C} models you may need to change the scancodes slightly.  For example, use ATTR{[dmi/id]product_name}=="900X3B*" for the 900X3B model.
Answer: Update, using Ubuntu 12.10, Kernel 3.5.
My Samsung 900x3cA02 runs completely fine. Upgraded the BIOS under Win7 to P04AAC then installed Ubuntu 12.10 from USB. Out of the box things are running quite well. No problems with lid closure detection, keyboard backlight, battery status, function keys. Battery life is OK (5-6 h with wireless on, using office software, browser, thunderbird). Screen brightness can be set to very low values (15% or so) indoors.

Ubuntu on samsung 9 series thin laptop


What is mast camera? And what does it do?

Question: The mast camera is on curiosity but I don't really know what its about . it is know that it's official on curiosity but I don't know what it's used for.
Answer: The mast on Curiosity is used to place cameras and other instruments far higher than the top deck of the rover body. Other rovers use a similar mast. This gives these instruments a better field of view.

Mastcam is used to provide scientific images (for geology), context images for the other instruments, and to help with navigation.

The Mast Camera, or Mastcam for short, takes color images and color video footage of the Martian terrain. The images can be stitched together to create panoramas of the landscape around the rover.

The mast contains more cameras: Navcams for navigation (4x) and Chemcam, which is used in combination with a powerful laser for spectroscopic analysis of the soil.
Answer: The Mastcams (there are 2) on the Mars Science Laboratory aka Curiosity were built by Malin Space Science Systems and use similar components to the other imagers on Curiosity such as MAHLI and MARDI. They are developments and upgrades of similar cameras that were flown on Spirit and Opportunity, the previous "generation" of Mars rovers.
There are more details on the Mastcams on the specific page at MSSS but basically there is a short focal length (34mm) Mastcam, giving a wider field of view (15$^\circ$), and a longer 100m focal length Mastcam for more close-up images (5.1$^\circ$ field of view). Both are equipped with several filters in the visible and near-infrared part of the spectrum (the filters span 440 to 1035nm) to allow the mineralogy of rocks and the terrain to be determined but also to allow natural color images to be taken. In addition both Mastcams are capable of 720p video (at 10 frames/second) and in theory can be used for making 3d stereo pairs (although this is apparently not often used due to the different focal lengths and fields of view of the 2 cameras).


Direct descendants of Moses today

Question: I'm just curious to know if there are any direct descendants of Moses today. We know that Cohen is the name for the descendants of Aaron. This is because it seems Moses generation ended with him? Their significance or influence was nothing compared to Aaron.
Comment: This question is off-topic on this site. We deal strictly with the hermeneutical analysis of the text of scripture. Your question is not about scripture.
Answer: This question will probably be closed soon since it is not a hermeneutical question. However, you may be interested in this:
Judges 18:30-31 might indicate Moses' line: Jonathan son of Gershom son of Moses/Manasseh. In many Hebrew manuscripts the letter nun (נ) is suspended above the name in question between the first two letters (מ^נ^שה) suggesting the name Manasseh. But other manuscripts as well as the LXX indicate Moses. If the latter reading, then the Shilohite priests appear to have been descendants of Moses. From this priesthood came the chief priest Abiathar who was exiled to Anathoth by King Solomon when he came to power (1 Kgs 2:26-27). Jeremiah appears to have been of this same priesthood (Jer 1:1) and thus potentially a descendant of Moses as well. I am not aware of those who we might be able to trace to Moses after the destruction of the first Jerusalem temple and certainly not after the destruction of the second Jerusalem temple. Today, even tribal distinctions have disappeared from memory among the Jewish people except for Levi.

Direct descendants of Moses today


Is the circuit a voltage doubler or not?

Question: I'm not able to actually understand how the circuit behaves as a voltage doubler.From my understanding the circuit would behave as a voltage doubler if during the positive half cycle only D1 would conduct and only C2 capacitor would charge to positive max of input and for negative cycle only D2 would conduct and only C1 would charge to negative max of input and hence the voltage drop across resistor would be 2*max input voltage. However my doubt is during positive half cycle apart from cap C2 wouldn't C1 as well get charged through the resistor similarly during negative half cycle wouldn't C2 also be charged through the resistor. If that is the case would the circuit still act as voltage doubler?
Answer:
wouldn't C1 as well get charged through the resistor...[?]

You are correct that current would flow through the resistor and then through C1.  C1 would actually be discharged by this action, but by being in the circuit, its voltage (on the - side of the resistor) is added to the voltage of the power supply (D1 to the + side), resulting in double the voltage across the resistor (which is the load). C1 should not discharge all the way; if it does, then there is too much load current and the voltage ripple would be nearing 100%.  And so you would want to use a bigger capacitor.  The same action applies to C2 during the negative cycle.
I've never seen a full-wave doubler drawn this way, but of course it is correct. I slapped my head when I saw this.  And for your benefit, you should search the Web for a full-wave doubler, and look at the images, and see how it's usually drawn. It should make more sense that way.
Comment: How one usually draws a voltage [doubler](http://www.augustica.com/ElectronicsTutorialsImages/Tubes/Full%20Wave%20Voltage%20Doubler.jpg), [tripler](http://www.augustica.com/ElectronicsTutorialsImages/Tubes/Full%20Wave%20Voltage%20Tripler.jpg) and [quadrupler](http://www.augustica.com/ElectronicsTutorialsImages/Tubes/Full%20Wave%20Voltage%20Quadrupler.jpg). Note how the same approach can be used to go even further. [(source)](http://www.augustica.com/full-wave-voltage-doubler-tripler-and-quadrupler-ezp-36)
Comment: Feel free to include any and all of the above comment to improve your answer. Re-directing to "you should search the web" is less helpful than actually showing it. Thought I'd save you some time.
Comment: Your response is appropriate, but in this case I wanted the OP to do their own search so they could appreciate the breadth of circuits, see the ones that are most popular, and the fact that the one represented here is really quite unusual.


How to Add Name of Different Category in Attribute Table

Question:
I'm stuck in a problem when I started calculating area in arcmap 10.1. In the picture, I categorize my polygon into four unique labels - i.e. vegetation, sand fill, water body, built-up area. Now I want to calculate the area of these four categories. But when I opened the attribute table and try to calculate the area of each category, there are many polygons having the same gridcode. Also I tried to add another field called Name but can't. Can anyone please tell me how I can do this ?
Comment: It's probable you can't add a field because you're editing. How are you trying to calculate your category? There are (at least) two ways to do this, one is an advanced field calculation, the other is select the GRIDCODE using select by attributes then calculate the Name accordingly for each unique GRIDCODE value; calculate field applies to selected records but will calculate ALL records if there isn't a selection.
Comment: this is really 2 questions. In gis stackexchange it is important to view the help center how to ask good questions.  the first question 1. calculate area?  if you are trying to get a total area report for each category, you want to summarize on the gridcode field. question 2. you need to clarify because your picture shows you already have a field with name as the name.
Comment: I have done in the second way, you described @MichaelMiles-Stimson
Comment: I managed to add the `Name` field but I can't edit this. I thought, if i, somehow, manage to edit the field, I can calculate area of each category. @BradNesom
Answer: If you already have the Name field, as your images shows, then right-click on the field header and select "Field Calculator...". In the field calculator window, enter an expression in either VB Script or Python which checks the value of GRIDCODE and returns the appropriate name for each of the four possible values.

To calculate the total area for each type of polygon, use the Summary Statistics tool (ArcToolbox > Analysis Tools > Statistics).

The Input Table will be your existing polygon dataset ('polygon' in your example).
Set your Output Table as desired.
The Statistics Field will be the field which contains the area. Once you have selected the field, you will also need to enter the Statistics Type of sum in the table.
The Case Field will be the attribute which differntiates between the polygon type (GRIDCODE in your example)

Note that when you first enter the Statistics Field, a red X will appear next the that part of the tool until you enter the statistic type into the table
Comment: Thank you very much. It worked ! I calculated area under each GRIDCODE successfully but I can't understand how to enter an expression in VB Script or Python  as I have very poor knowledge in Python.
Comment: I think that really could be a second question. If you post it I'm sure it'll get a quick response.


Windows Phone apps crash on startup

Question: Sometimes apps on my Lumia 650 will crash when started. Restarting the phone does not help, neither is it a problem with installing it on the SD card etc. Most of the time a reinstall cycle will work, but this will wipe the app's data. Is there any other way to fix this?
Answer: Try a complete soft or a hard reset of the phone (that mostly helps here on my side, and these problems normally occur after an update like the Creators Update), but be sure to backup your files etc. before!
Comment: Even I had the same problem. The whole system was in a messed up state. But once after a hard reset, phone is working fine. As John Archer said, make sure to backup all your files and applications before a hard reset.
Comment: The problem will persist after a hard reset after some time. A hard reset is basically a reinstall cycle.
Answer: Navigate to the store page of said app, and click on "Launch" or "Install" from there. The app will then start normally.

Windows Phone apps crash on startup