Old Cartoon about animal powers and wristbands

Question: I am looking for an old cartoon—probably made around the '80s or '90s—that was about people who had different colored wristbands that gave them animal-like powers. I think it used to be shown on Boomerang.
Comment: Legend of the Dragon? Also, can you provide more details?
Comment: This question is very terse and could be greatly improved by going through the checklists [here](https://scifi.meta.stackexchange.com/questions/11874/how-to-ask-a-good-anime-story-id-question) and editing in any relevant info you can think to add.
Answer: Although the time period when it was made does not match what the questioner thought, I think the show he is looking for probably is, in fact, Legend of the Dragon.  The show aired from 2005–2008, and it was not actually on Boomerang either; it was broadcast in America on Disney-owned channels and on other channels elsewhere in the world.  (I looked at the long list of shows broadcast on Boomerang, and I did not see anything more promising that was older—although it is certainly conceivable that I overlooked something.)
Per Wikipedia:

The show starts when the previous Golden Dragon has passed on to the afterlife and Ang and Ling Leung, seventeen-year-old fraternal twins, born in the year of the Dragon, are next in line for the Golden Dragon power band and user of its mystical abilities to defend the mortal world from evil. Ang has no belief that he will become the Golden Dragon, as Ling is the best fighter at the Dragon dojo. Ling is positive that she will become the Golden Dragon. However, when Master Chin calls upon the power of the Golden Dragon to choose who will be the next golden Dragon, it chooses Ang. Humiliated and furious, Ling storms out of the temple and is tempted by the ultimate dark and evil power of the shadow Dragon. From then on, Ang is forced to fight his own twin sister.
...
The second and final season (which consisted of just thirteen episodes) begins with the reformed Ling finally back home as the second Golden Dragon alongside her twin brother. Now they, Chin, and the fellow Guardians of the Tiger and Monkey Temples are ready to put an end to the Zodiac Master and the Emperor of the Darkest Ying once and for all. However, the Emperor is on the move to regroup at the Temple of the Shadow Dragon, but not before he attacks and disintegrates the Temple of the Golden Dragons, leaving Ang and Ling's power-bands on the verge of powerlessness. Luckily, Chin suggests going to long-lost Temple of the Shadow Dragon and utilize its vast amounts of negative energy to rejuvenate their power-bands. As Ling was exposed to the evil energies of the Shadow Dragon for quite some time, she becomes easily tempted by the temple's negative influence on both her mind and spirit. However, with the wisdom of Master Chin, she and Ang successfully recharge their power-bands with no negative effects. Elsewhere, the Emperor manages to release his younger sister Yin Wi, the Guardian of the Shadow Rat, who had been imprisoned by one of the previous Golden Dragons for one thousand years. The two siblings concoct plans to conquer the modern world, and have their vengeance on all twelve guardians of yang; the Golden Dragon in particular. As they head to the Temple of the Shadow Wolf, the heroes get there first and encounter K-Lo, the Shadow Wolf Guardian who was one of Ling's allies during her dark times as the Shadow Dragon and who seems to have quite an interest in her, now that she has become a Golden Dragon.

The characters' dragon or animal powers are utilized through the power-bands mentioned above.  There were twelve sets of powers, based on the years of the Chinese Zodiac:  rat, ox, tiger, rabbit, dragon, snake, horse, goat, monkey, rooster, dog (interpreted as wolf on the show), pig.  The most prominent villain was the Zodiac Master.  Each band had a color, indicating its elemental affiliation,

and a gem carved with a representation of its tutelary animal.

Activating the power-bands allowed the guardians to take on animal-empowered forms.


Are the major-axis of Mars and Earth orbits on the same plane?

Question: Which means that I can draw an imaginary plane that contains the major-axis of both Earth and Mars.
Comment: Hint: the major axes intersect at the Sun.
Comment: @PM2Ring Thanks, I got it!
Answer: Yes, as PM 2Ring hinted, the major axes of the planets intersect at the Sun (more precisely, at the solar system's center of mass). Therefore, given any two planets there is a plane that that contains their major-axes.
Of course, this statement assumes the approximation that orbits are perfect ellipses.

Edit: Prompted by M. A. Golding's answer, I will attempt to find this plane and figure out how it looks like.
I found the orbital elements of Earth and Mars in the paper Numerical expressions for precession formulae and mean elements for the Moon and the planets.
At page 675, we find the value of the longitude of perihelion $\varpi$ and inclination $i$ of the planets. $\varpi$ is the angle between the vernal equinox, the focus of the orbit and the perihelion of the orbit. $i$ is the angle between the plane of the orbit and the ecliptic at J2000. I report the data here, hoping that I didn't copy anything wrong.
Earth:
$\varpi = 102.93734808° + 11612.35290'' t + 53.27577'' t^2 - 0.14095'' t^3 + 0.11440'' t^4 + 0.00478'' t^5$
$i = 469.97289'' t - 3.35035'' t^2 - 0.12374'' t^3 + 0.00027'' t^4 - 0.00001'' t^5 + 0.00001'' t^6$
Mars:
$\varpi = 336.06023395° + 15980.45908'' t - 62.32800'' t^2 + 1.86464'' t^3 - 0.04603'' t^4 - 0.00164'' t^5$
$i = 1.84972648° -293.31722'' t - 8.11830'' t^2 - 0.10326'' t^3 -0.00153'' t^4 + 0.00048'' t^5$
In these formulas, $t$ is the time in thousands of Julian days from J2000:
$$t = (\text{JD} - 2451545)/365250$$
First question: What are the values of the orbital parameters right now?
On 2021 January the first, the values where:
$$\varpi_{earth} = 103.0°, \varpi_{mars} = 336.2°$$
$$i_{earth} = 0.003°, i_{mars} = 1.848°$$
So we can say that the semi-major axes of Earth and Mars are not currently aligned, but have about 336-180-103 = 53° of separation.

This is just a sketch of the orbits to show the relative position of the semi-major axes. The ellipses are not accurate.
Are they going to align soon due to precession of the line of the apsis?

This figure shows that they are not going to be aligned for the next 20000 years or so. I have marked with red dashed lines the region where the error on the model is smaller than $1''$ (from 4000 BCE to 8000 CE). (At least, this is what I understood from the paper, I'm not 100% sure of this)
To complete the answer, let's calculate the inclination of the plane that contains both semi-major axes. The direction of the axis is defined by the vector that points from the Sun to the perihelion of the orbits:
$$\vec{v} = (\cos(\varpi), \sin(\varpi), \sin(i))$$
The vector $\vec{w} = \vec{v}_{earth} \times \vec{v}_{mars}$ will be orthogonal to the plane that contains the semi-major axes. Its inclination will just be the angle between $\vec{w}$ and the z-axis.
$$\vec{w} = (0.0314,  0.0073, -0.8002)$$
The inclination is
$$i_{plane} = \arccos \left( \frac{\hat{z} \cdot \vec{w}}{|\vec{w}|} \right) = \arccos \left( \frac{w_z}{|\vec{w}|} \right) \approx 2.31°$$
This result shows that there is indeed a plane that contains both the semi-major axes of Earth and Mars, and it is currently close to the ecliptic plane and to Mars orbital plane. I don't know how this plane could be useful, but this was the OP's question.
The last figure is the evolution in time of the inclination of the plane:
Comment: _I guess you can create a plane that contains both the semi-major axes_ - As you have stated yourself, it is possible to draw such plane. This plane does not coincide with either Earth's orbit plane or Mars' orbit plane, but the OP just asked if such plane existed. Without additional information I cannot judge how this imaginary plane will be useful for the OP
Comment: I think this answer is wrong and the planes of the orbits are tilted.  Thus there is one line where the two planes interset.  And two separate semi-major sexes can not exist in one 1-D line, unless by coincidence the longer axis passes through and beyond the shorter axis.  Which would be a big coincidence.
Comment: @MAGolding sorry, forgot to notify you
Answer: An semi-major axis of a planet is a one dimensional line contained within the two dimensional plane of the panet's orbt.  A planet's orbital plane is a 2 dimensional plane in three dimensional space.
Since the orbits of the planets are all tilted slightly with respect to each other, the orbital planes of any two planets are aligned differently in 3 demensional space.  Thus the orbital planes of two planets will intersect in a one dimensional line.
Since the semimajor axis of one planet, like Earth, is entirely within the plane of the planet's orbit, and the semi-major axis of another planet, like Mars, is entirely within the different plane of the second planet's orbit, the two semi-major axes can not be co planer -at least not within the different planes of their planetary orbits.
And since the two lines containing the semi-major axes do intersect, I guess you can create a plane that contains both the semi-major axes.  But that plane will not be identical to the orbital pane of either planet, let along both orbital planes.  And as far as I can see, that third plane should be almost at a right angle to the two very close, but different, planes of the planetary orbits.
Imagine that the planes of the orbits of Earth and Mars were parellel and separated by about a million miles.  If their orbital major axes were lined pointing in the same direction, a plane passing thorugh both the orbital major axes would be perpendicular to the two planes of the planetary orbits.
And in real life the situation is similar.
The plane of Earth's orbit is almost exactly the same as the plane of Mars's orbit.  But they are titled slightly. Thus on one side the plane of Earth's orbit will be "above" the plane of plane of Mars's orbit, and on the other side the plane of Mars's orbit will be "above" the plane of Earth's orbit.
Thus a third plane containing lines in both of those separate orbital planes will have be angled at close to a right aangle to either of those two separate orbital planes.
So if you try to draw the two planetary orbits within a plane which contrains both major axes, both planetary orbits will be very narrow ellipses, close to straight lines, since that plane will be angled at almost 90 degrees to the two planes in which the two planets orbit.  The shapes of the orbits of the two planets would be depicted very inaccurately.
It would be far more accurate to pretend the two planetary orbits are in the same plane and draw them that way.
And this discussion ignores the directions of the major axes of the orbits of the Earth and Mars.  There is little reason to suppose that they happen to be lined up at any particula moment.
In fact I believe that due to gravitational perturbations by other planets, the line of the major axis of every planet precesses.

In celestial mechanics, apsidal precession (or apsidal advance)1 is the precession (gradual rotation) of the line connecting the apsides (line of apsides) of an astronomical body's orbit. The apsides are the orbital points closest (periapsis) and farthest (apoapsis) from its primary body. The apsidal precession is the first time derivative of the argument of periapsis, one of the six main orbital elements of an orbit. Apsidal precession is considered positive when the orbit's axis rotates in the same direction as the orbital motion. An apsidal period is the time interval required for an orbit to precess through 360°.[2]

https://en.wikipedia.org/wiki/Apsidal_precession[1]
So as the major axes of the orbits of Mars and Earth precess around the Sun or the barycenter of the Solar System, the direction between them constantly changes and thus the orientation of any plane which could contain both those lines should also change drastically.
Added 08-21-21.  Prallax's revised answer show that at the present time the plane containng the major axes of the orbits of Mars and Earth is inclined only a couple of degreess from the planes of their orbit.  But thousands of years in the future, when the major axes are lined up, that plane should be close to perpendicular to the two orbital planes.
Comment: _that third plane should be almost at a right angle to the two very close_ - I do not agree. Assuming this picture is correct, I would say that the common plane of the axes intersects the ecliptic with a small angle, not close to 90 degrees. https://images.app.goo.gl/UkqYkYbLniVqiGyKA  it would be interesting to do the proper calculation and settle the matter
Comment: @Prallax  I have added several paragraphs to my answer.
Comment: @M.A.Golding I have edited my answer too. You had made me curious about this and so I have decided to do the proper calculation. Let me know what you think
Comment: @Prallax.  I guess you are right.  but I expect that the angle of thire plane containing the two major axes may change from time to time as the axes precess.  I think that when the axes are lined up the third plane should be almost perpendicular to the orbitial planes,. .
Comment: I totally agree with you on this. Indeed the question in the OP was a bit peculiar


A word to describe a situation, in which the eventual winner in the competition is hardly predictable

Question: What's the right word to describe a situation in a sports competition or in elections, in which both competitors are equal on all parameters and it is absolutely not clear which one of them will be the winner?
Comment: Your question changes when you add "Could I exclaim ....?" as you do below in a comment. With  only two candidates, "too close to call", as suggested by em in that answer, is  apt, but we could not really exclaim "What a too close to call election!".  Turning that phrase into a modifier placed before the noun like that is something most native speakers would avoid.  But *What a close election!* is idiomatic.
Comment: Almost literally a '[level playing field](https://en.wikipedia.org/wiki/Level_playing_field)'
Answer: As JeremyC mentions, there is no single word for this, but there are multiple expressions.  He already mentions "wide open", but there are others:

It's anyone's game.
It's too close to call (there is no way to confidently predict the winner).
It's a toss up (random, like tossing a coin)
It's a dead heat (as in a race, when neither runner is clearly ahead)
It's an even match/contest (neither participant has a clear advantage)
It's down to the wire (as in horse racing, with a wire across the finish line, and none of the horses clearly in the lead)
It's neck and neck (another horse racing expression, when the horses are too close together to see which is clearly in the lead)

Finally, there's the expression "photo finish", which is used after the race is over, but there is still no clear winner.  Again, with horse racing, a photo would be taken at the finish line, then developed and examined to see which horse (if any) was ahead.

Naturally this can be used for any competition:
Comment: Another informal one (in the UK, at least) is "even stevens" - see https://en.oxforddictionaries.com/definition/us/even-steven.
Comment: @alephzero  It's an idiom in the US as well, although I think it means something different, more like *"I've paid you back for the favor you did me, so now we're **even steven**."*
Answer: Not a single word, but the usual term is "wide open". See Collins Cobuild:

If you say that a competition, race, or election is wide open, you
  mean that anyone could win it, because there is no competitor who
  seems to be much better than the others. The competition has been
  thrown wide open by the absence of the world champion.
Comment: Could I say something like "What a wide-open competition!"?
Comment: Um. I can't quite put my finger on it, but something sounds a bit wrong with that. Maybe it is because a competition is wide open or not. If you say 'What a nice day!', that works because there are degrees of niceness of day, whereas with competitions it does not (to my ears).
Comment: You could say, "What a close-run competition."
Comment: +1 for **wide open**  (but it usually involves multiple candidates). I think you could exclaim something like "The competition is wide open!" or even "Such a wide-open competition!"
Comment: So, how can I put it into an exclamation? Like I am really excited about the fact that this race is wide open.
Comment: @Tᴚoɯɐuo - What do you think about "What a close-run competition!" (suggestion by chasly from UK)?
Comment: @brilliant: To my AmE ear, **close-run** doesn't sound quite right to describe an *ongoing* election, though I could easily say *The election was close run*.  Not sure why it sounds off to me for the ongoing election. FWIW, Oxford defines it as an election that was won or lost by a small margin, whereas Longman defines it as an election that is too close to call and uses an example with an upcoming election.
Comment: I agree that "The competition is wide open!" or "Such a wide-open competition" sound like much more natural exclamations.
Answer: You might call it "close":

close
  adjective
  1.4 (with reference to a competitive situation) involving only a small margin between winner and loser.
  'the race will be a close contest'
  'she finished a close second'
  (ODO)

You can also comment that it's "too close to call":

too close to call
  COMMON If a contest is too close to call, it is impossible to say who will win, because the opponents seem equally good or equally popular. The presidential race is too close to call. Exit polls in Britain say that today's parliamentary election was too close to call.
  (TFD)
Answer: Since you are looking for a single word, I would suggest crapshoot.
From the Grammarist, a crapshoot is a situation whose outcome is not predictable.
Of the five examples of usage at Grammarist, one refers to sports and another to elections:

Everybody knows that an election with a small turnout can be a crapshoot.
Talladega is a crap shoot, so you never know what's going to happen.

Notice in the second example above, it is spelled as two words, which is less common.


What is the derivative of a function of the form $u(x)^{v(x)}$?

Question: So I have a given lets say $(x+1)^{2x}$ in addition to $\frac{\mathrm dy}{\mathrm dx}a^u=a^u\log(a)u'$. I still have to multiply this by the derivative of the inside function $x+1$ correct?
Comment: To evaluate the derivative of an expression of the form $\Big[u(x)\Big]^{v(x)},~$ we must combine the two relevant formulas for the derivatives of $a^{u(x)}$ and $\Big[u(x)\Big]^n.~($ [Two](http://math.stackexchange.com/questions/534820) related [questions](http://math.stackexchange.com/questions/1017769) $).$
Answer: This is what logarithmic differentiation is for. You start with writing the function as an equation
$$y = (x + 1)^{2x},$$
then take the natural log of both sides:
$$\ln y = \ln\left[(x + 1)^{2x}\right] = 2x \ln(x+1).$$
We then implicitly differentiate both sides with respect to $x$. By chain rule (remember, $y$ is a function of $x$), the left side comes to
$$\frac{1}{y} \cdot y'.$$
The right side can be differentiated as normal:
$$\frac{2x}{x + 1} + 2\ln(x + 1).$$
So,
\begin{align*}
&\frac{1}{y} \cdot y' = \frac{2x}{x + 1} + 2\ln(x + 1) \\
\implies \, &y' = y\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right) \\
\implies \, &y' = (x + 1)^{2x}\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right).
\end{align*}
Comment: To address the comments, when working within $\Bbb{R}$, the conventional domain of $a^b$ is $a > 0$ (and $b \in \Bbb{R}$). This means that $a^b > 0$ always. While it may be observed that real values can be ascribed at certain other isolated points (e.g. when $a < 0$ but $b \in \Bbb{Z}$), many rational and all irrational values of $b$ pose a problem when $a < 0$. Therefore, talking about derivatives when $a < 0$ is tricky when the function is not even properly defined densely around such values. Thus, there is no particular problem with assuming the function is positive and taking the logarithm
Comment: Are you arguing this is a good mnemonic for what to do, or a reliably true bit of algebra that works always?
Comment: It should be noted that although logarithm is only defined for positive values, this method is applicable on general for any values inside the logarithm
Answer: Since $(x+1)^{2x}=e^{2x\ln(x+1)}=e^{u(x)}$, its derivate is $u'(x)e^{u(x)}$. Notice that if $a:I\longrightarrow\mathbb{R}^{+*}$ and $b:I\longrightarrow\mathbb{R}$, then we define $a(x)^{b(x)}$ as $e^{b(x)\ln a(x)}$ for all $x\in I$
Comment: Indeed, the definition $3^2=e^{3\ln 2}$ is compatible with the idea that $3^2=3\times 3$ but it also allows to compute much more powers such as $\pi^{\sqrt{2}}$, $\gamma^{e}$...
Comment: @EricDuminil https://math.stackexchange.com/q/1657619/79767
Comment: can you express this in terms of $a(x)^{b(x)} = e^{b(x)ln(a(x))}$?
Comment: Sure, I'm saying fold that into your answer.  Note that $e^{b(x) ln a(x)}$ can be simplified as $a(x)^{b(x)}$
Comment: What is $\mathbb R^{+*}$ ? I've seen $\mathbb R^+$ and $\mathbb R^*$ before, but not this.
Comment: It is the combination of those two : $\mathbb{R}^{+*}=]0,+\infty[$
Comment: That is what I would call $\mathbb R^+$. And $\mathbb R^*=\mathbb R\backslash \{0\}$.
Comment: Hee, $\mathbb{R}^+=[0,+\infty[$ and $\mathbb{R}^*=]-\infty,0[\cup]0,+\infty[$
Comment: @mr_e_man: As far as I know, it's a French notation.
Comment: Well I'm French, that may explain why.
Comment: It's debatable to call that a definition, since many people will take the definition of 3^2 to be 3*3, not e^(2 ln 3).
Comment: @djechlin so how would you define $3^2.5$, or $3^\pi$?
Answer: That whole $a^u$ thing works when $a$ is a constant, not another expression in terms of $x$.
To take the derivative of this, you would have to convert it to $\displaystyle e^{2x\ln(x+1)}$  and THEN use the Chain Rule.
This would be $\displaystyle e^{2x\ln(x+1)}\left(2\ln(x+1)+\frac{2x}{x+1}\right)=(x+1)^{2x}\left(2\ln(x+1)+\frac{2x}{x+1}\right)$
Answer: A nice way to do this is to use a somewhat stronger fact than the other answers: as a function of two variables, the expression $a^b$ is differentiable. Simply put, what this means is that, if $f(x)$ and $g(x)$ are functions, and $h(x)=f(x)^{g(x)}$, then $h'(x)$ is the sum of how fast this expression changes when we treat $f(x)$ as a constant plus how fast it changes when we treat $g(x)$ as a constant. You already know how to differentiate polynomials and exponentials, so this suffices - just apply both rules and add them!
Since we know that the derivative of $f^{g(x)}$ is $\log(f)\cdot f^{g(x)}\cdot g'(x)$ and the derivative of $f(x)^g$ is $g\cdot f(x)^{g-1}\cdot f'(x)$, we get
$$h'(x)=\underbrace{\log(f(x))\cdot f(x)^{g(x)}\cdot g'(x)}_{\text{Derivative treating $f$ as constant}}+\underbrace{g(x)\cdot f(x)^{g(x)-1}\cdot f'(x)}_{\text{Derivative treating $g$ as constant}}.$$
You would get the same result by using logarithmic differentiation as other answers suggest*, but I generally find this is a bit easier to remember and more generalizable - for instance, note that differentiating a product $f(x)\cdot g(x)$ can be done by this same method.

(*Logarithmic differentiation is a good way to prove this result, since you write $$\log(h(x)) = \log(f(x))\cdot g(x)$$then differentiate both sides to get $$\frac{h'(x)}{h(x)}=\frac{f'(x)\cdot g(x)}{f(x)}+\log(f(x))\cdot g'(x)$$ and moving $h(x)$ to the other side and substituting it for its formula gives the formula I claim)
Answer: Start with $$f(x) = a(x)^{b(x)}$$.
Now, we can approach this a few ways.  The one way is to go into multidimensional derivatives.  Define
$$h(x,y) = a(x)^{b(y)}$$
$$h'(x,y) = \left[ {\begin{array}{cc}
\frac{ \partial h_x }{ \partial x } & \frac{ \partial h_x }{ \partial y } \\
\frac{ \partial h_y }{ \partial x } & \frac{ \partial h_y }{ \partial y } \\
\end{array} } \right] = \left[ {\begin{array}{cc}
a'(x) b(y) a(x)^{b(y)-1} & 0 \\
0 & ln(a(x)) b'(y) a(x)^{b(y)}\\
\end{array} } \right]$$
Let $g(x) = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$.  Then $f(x) = h(g(x))$.  By the chain rule, $$f'(x) = h'(g(x)) g'(x) = h'(x,x) \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$$
or $$f'(x) =
a'(x) b(x) a(x)^{b(x)-1} + ln(a(x)) b'(x) a(x)^{b(x)}$$
$$ f'(x) = f(x) * ( b(x) \frac{a'(x)}{a(x)} + b'(x) ln(a(x)))$$
We can then do a sanity check using logarithmic derivatives.  Also we can look at what happens when $a(x) = k$:
$$ f'(x) = f(x) * ( b(x) \frac{0}{a(x)} + b'(x) ln(k))$$
$$ f'(x) = ( b'(x) ln(k) ) k^{b(x)} $$
or $b(x)=k$:
$$ f'(x) = f(x) * ( k \frac{a'(x)}{a(x)} + 0 ln(a(x)))$$
$$ f'(x) = k a'(x) a(x)^{k-1}$$
which are both right.
Conceptually, this has two terms.  You add them to find out what the instant "percent change" in $f(x)$ will be.
$b \frac{a'}{a}$ is $b$ times the instant percent change in $a$:  If $a$ is growing by $1\%$, and you raise this to the power $b$, you grow by $b\%$.
$b' ln a$ is the change in $b$ wrt $x$ scaled by the log-scale of $a$.  With exponentials, a given linear-scale unit of exponent increase causes a scale by a factor in the base.  $ln(a)$ is the ratio between the two (the linear effect in the exponent to the exponential effect in the result), and we multiply that by $f$ because we are scaling the entire value.


How to draw a curved cuboid?

Question: Sorry for boring you. After reading some answers in TEX stack exchange, it poses no problem to draw a regular cuboid. But, I am wandering someone knows how to draw a curved 'cuboid' as shown in the following fig (in red window).

Thank you very much for taking a look.
Comment: Serious 3D drawing needs something like Asymptote or a proper CAD system..
Comment: What have you tried?  Please add a [minimal working example (MWE)](https://tex.meta.stackexchange.com/q/228)
 to your question.
Comment: [Welcome to TeX.SX](//tex.meta.stackexchange.com/q/1436). Questions about how to draw specific graphics that just post an image of the desired result are really not reasonable questions to ask on the site. Please post a minimal compilable document showing that you've tried to produce the image and then people will be happy to help you with any specific problems you may have.  See [minimal working example (MWE)](//tex.meta.stackexchange.com/q/228) for what needs to go into such a document.
Comment: This should not be closed yet. Please wait at least 24 hours after asking the OP for improvements to the question before voting to close. Also, if you downvote, please don't forget to revert the vote after the question is improved.
Answer: It is fairly straightforward to draw something of this sort, but you'll have to draw additional features in the right order, or to switch to asymptote.
\documentclass[border=2mm,tikz]{standalone}
\usepackage{tikz-3dplot}

\begin{document}

\tdplotsetmaincoords{70}{50}
\begin{tikzpicture}[tdplot_main_coords,font=\sffamily]
% Uncomment these lines if you want to know where x, y and z point to
% \draw[-latex] (-7.5,0,0) -- (7.5,0,0)  node[above right]  {$x$};
% \draw[-latex] (0,-7.5,0) -- (0,7.5,0)  node[below] {$y$};
% \draw[-latex] (0,0,-7.5) -- (0,0,7.5)  node[above left]  {$z$};
\tdplotsetrotatedcoords{90}{90}{90}
\begin{scope}[tdplot_rotated_coords]
% Uncomment these lines if you want to know where x', y' and z' point to
% \draw[-latex,blue] (-7.5,0,0) -- (7.5,0,0)  node[above right]  {$x'$};
% \draw[-latex,blue] (0,-7.5,0) -- (0,7.5,0)  node[below] {$y'$};
% \draw[-latex,blue] (0,0,-7.5) -- (0,0,7.5)  node[above left]  {$z'$};
\draw[thick,fill=gray!10] plot[variable=\x,domain={-6:6},samples=120]
({\x},{4+0.25*\x-0.6*exp(\x-6)},-1) --
plot[variable=\x,domain={6:-6},samples=120]
({\x},{4+0.25*\x-0.6*exp(\x-6)},1) -- cycle;
\draw[thick,fill=gray!30] (-6,2.5,-1) -- (-6,2.5,1) -- (-5.5,0.5,1)
--(-5.5,0.5,-1) -- cycle;
\draw[thick,fill=gray!20] plot[variable=\x,domain={-6:6},samples=120]
({\x},{4+0.25*\x-0.6*exp(\x-6)},-1) --
 plot[variable=\x,domain={6:-6},samples=120]
({(5.5/6)*\x},{2+0.25*\x-0.6*exp((5.5/6)*\x-6)},-1) -- cycle;
\end{scope}
\end{tikzpicture}
\end{document}

Answer: TikZ offers support for 3D coordinates in cartesian (and per example in the manual, cylindrical) coordinates and allows to create your own coordinate system with relative ease. In this question, a spherical coordinate system was implemented, including showing ways of drawing lines and (pseudo-) 3D effects. If you intent to draw your picture, consider a proper coordinate system, set up the projection thereof onto the 2D canvas (can't find the command in the manual right now, but is possible) and then start drawing. For a free-form, it probably best to parse a series of coordinates to a plot operation (can be read from file). If you can describe your points through a mathematical function, consider drawing them with a foreach operation, give them node names and then use line to operations to join up the nodes.
Overall, it is probably better/easier to just parse your image in a svg editor like inkscape and import the svg to tikz. (Inkscape also offers limited LaTeX math support and proper scaling thereof.)
However, given that you already have an image ready, why not just use that? It is accepted practice to put pictures from other programs into articles, books and other documents. No need to spend >6h day drawing coordinates in TeX or >1h redrawing the image in an editor. (Mileage may vary)

How to draw a curved cuboid?


Why are "mobile" and "automobile" pronounced differently?

Question: I just came across the words and then I looked them both up in the dictionary app, which shows the word "mobile" pronounces as /'məʊbaɪl/, whereas the other word-"automobile", which ends with the same spelt "mobile" pronounces as /'ɔːtəməbiːl/? I'm not sure if this "mobile" word pronounces differently as shown in the app or they actually pronounce the same?
Comment: @BenjaminHarman: The pathology-pathological variation is because of English stress patterns.
Comment: They're pronounced differently as described. You'll notice that the noun "mobile" doesn't mean what "-mobile" means in the noun "automobile," like if "automobile" were really "auto mobile." Also, you'll find a lot of words that actually are forms of each other do change when adding a suffix or prefix, like "pathological" is pronounced "path-o-LO-gi-cal," whily "pathology" is pronounced "pa-THO-log-y," not "pa-tho-LO-gy."
Comment: @BenjaminHarman - actually *-mobile* in *automobile* is derived from *mobile (adjective).* https://www.etymonline.com/word/automobile
Comment: In the US I've always heard the noun, referring to a piece of art dangling from strings, pronounced the same as the ending of *automobile*.
Comment: In American English the last syllable of both is identical /bil/, but the next-to-last (i.e, first) syllable in _mobile_ is stressed, and the vowel is tense back rounded /o/, while the next-to-last syllable in _automobile_ is unstressed, and the vowel is lax central unrounded /ə/. Nobody would ever confuse them.
Answer: Because mobile entered English in the 15th century and underwent the Great Vowel Shift (GVS). The GVS was a series of changes in the pronunciation of English vowels that took place between 1400 and 1700 [Wikipedia]. It must have been pronounced with /iː/ and the GVS changed the vowel /iː/ to /aɪ/ (also see bite, which was pronounced the same as beat before the GVS).
Automobile on the other hand entered English after the GVS and preserved its original vowel.
Comment: For some people it went through the GVS; for others it's a spelling pronunciation. It's been years since I heard any English speaker pronounce _mobile_ as /mobayl/. If I heard it now, I would expect it to refer to a  named object or event, with the very special pronunciation preserved in the name, in context.


Do you need trash bins in queue lines?

Question: I know you need trash bins around your park to prevent it from becoming a garbage dump, but do you need trash bins in queue lines? Or does the sim behavior of littering or puking stop when in line?
Comment: Good question. I don't have a canonical answer, but I've never put trash bins in my queue lines and I've never seen trash dropped in there (although people will get sick in line).
Answer: Sims refuse to litter queue lines. However, they will use trash bins placed in queue lines, and if the trash bins are kept empty, nearby sims will receive a green thought about how clean the nearby area looks. Puking still happens in queue lines regardless of trash bins, requiring a janitor to patrol the queue lines every so often.
NB: For some reason, the Park Inspector does not check trash bins placed in queue lines, so you can receive "clean park" rewards despite the trash bins in your queuess looking like a set pieces for the movie Wall-E.
Comment: Perfect! Also, do you know if they use benches in queue lines, or if the queue TV monitor path extra does anything at all?

Do you need trash bins in queue lines?


May landlords evict tenants for late payment

Question: According to Halacha, if there is a contract between a tenant and landlord where the tenant will live in  the property for however long (for example one year) and the payment will be distributed over a different time period (for example every month) can the landlord evict and terminate the contract due to a late payment? So Reuven rents an apartment from Shimon and then in the third month Reuven says he can't pay that month, can Shimon evict him or must he wait till the year is over but has the right to demand compensation in the future?
Comment: Just to clarify ,is this case purely halachic without any Dina dmalchusa? Additionally, this is specifically talking in a case where the contract does not discuss defaulting on rent,correct?
Comment: http://s589827416.onlinehome.us/wp-content/uploads/2015/07/Decision5a.pdf
Comment: https://www.ou.org/life/torah/evictions_halachic_considerations/
Answer: The Be'er Heitiv in Choshen Mishpat 312:4 brings at the end a case from the Maharit about someone who rents a house for 12 dinarim (dinar per month) ,and the renter is a pauper who cannot pay what he owes or pay for future payments ,then it seems pashut that he can be removed ,since the agreement was to be payed a dinar a month.
Just to point out there are other factors which need to be considered, Dina dmalchusa,standard practice, can he eventually pay and so on ...but just bringing what is discussed .
Text of Be'er Heitev:


Confusion regarding a case of head-on elastic collision of two bodies of equal mass

Question: If two bodies collide elastically, it's common to hear that the velocities, speeds and momenta are interchanged. I thought this was obvious. But if we consider two bodies moving with equal mass and speed, moving towards each other and colliding, they would both come to rest, to conserve momentum. In this case, neither of the three aforementioned things are 'interchanged'. Where am I going wrong?
Comment: If it's an elastic collision, the two bodies don't come to rest ... they bounce off each other.  A perfectly inelastic collision is required for the two bodies to come to rest.
Answer: One of the things which you must remember in the case of elastic collisions is that with momentum, energy is also conserved.
In all e<1 collision, some energy is wasted as sound, heat, etc
Now, consider that the two objects have equal mass, and speed, and are moving towards each other. Here, there are two viable paths to conserve momentum, since initial momentum is zero.

Final velocity of both objects is 0, so net momentum will be 0, which is the one you mentioned.
The velocities are interchanged. Assume the objects are traveling along x-axis Initially, the body B-1 has speed S along the positive x-axis, and the body B-2 has speed S along negative x-axis. After colliding, the velocities are interchanged, while the momentum remains zero.
Let the mass of each body be M, speed=S
However,
since energy is scalar and has to be conserved(Elastic Collision),
<code>                                **Initial Energy=Final Energy**.
</code>
Also, let it be clear that we are only dealing with kinetic energy here.
Since initial energy= MS^2,
Final Energy should also be equal to MS^2, which is only possible in the second case.
So, the velocities are interchanged.
Answer: Prerequisites
Momentum conservation states that:
$$m_{1} \mathbf v_{1i} + m_{2} \mathbf v_{2i}=m_{1} \mathbf v_{1f} + m_{2} \mathbf v_{2f}$$
where $m_1$ and $m_2$ are the masses of the objects, $\mathbf v_{1i}$ and $\mathbf v_{2i}$ are the initial velocities and $\mathbf v_{1f}$ and $\mathbf v_{2f}$ are the initial velocities. In your specific cas the momentum equation reduces to
\begin{align}
m\mathbf v + m (-\mathbf v) & = m\mathbf v _1 + m\mathbf v _2\\
0&=m\mathbf v _1 +m \mathbf v _2\\
0&=\mathbf v _1 + \mathbf v _2 \tag{1}
\end{align}
where $m$ is the mass of the objects, $\mathbf v$ is the initial velocity of both the objects and $\mathbf v_1$ and $\mathbf v_2$ are the final velocities.
Now there are two variables of interest ($\mathbf v_1$ and $\mathbf v_2$) but there is only one equation. So as you can see, momentum conservation cannot be used alone to predict the final velocities of two colliding objects. You need to apply some other constraint/equation which will then help you in uniquely determining the final velocities.
In this case, since you are talking about elastic collisions, we can apply two equivalent constraints, one being energy conservation other being $e=1$ (where $e$ is the coefficient of restitution).
Energy conservation
Note that from $(1)$, $|\mathbf v_1| = |\mathbf v_2 |$. From now on, I will denote the magnitudes of velocities as $v_1$, $v_2$ and $v$. Thus applying energy conservation, we get
\begin{align}
\frac 1 2 m v^2+ \frac 1 2 m v^2 &= \frac 1 2 m v_1 ^2 + \frac 1 2 m v_2 ^2\\
v^2 &= v_1^2 =v_2^2\\
|\mathbf v|&=|\mathbf v_1| = |\mathbf v_2 |
\end{align}
It follows trivially that the initial velocities have reversed their directions and thus the velocities have been exchanged.
Coefficient of restitution
The coefficient of restitution is defined as the ratio of relative velocity of separation and relative velocity of approach. Thus
\begin{align}
e&=\frac{|\mathbf v_1 -\mathbf v_2|}{|\mathbf v -(-\mathbf v)|}\\
1&=\frac{|\mathbf v_1 -\mathbf v_2|}{2|\mathbf v|}\\
2|\mathbf v|&=|\mathbf v_1 -\mathbf v_2|
\end{align}
Since this is a one dimensional collision, we can convert the magnitude of difference in velocities to the difference in magnitudes of velocities (with proper sign convention):
$$2v=v_1-v_2$$
Solving this and equation $(1)$ gives us the final velocities. Again you would notice that the initial velocities have reversed and exchanged.
Fallacy in your argument
The case which you are thinking about, happens when the collision is perfectly inelastic, which implies that the coefficient of restitution is zero.
Answer: Elastic collision means Collison without loss of kinectic energy . In this collision Collison , e=1 means relative velocity of approach=. relative velocity of separation
If you know this now you will understand easily
If we take formulae for boys conservation of momentum why??  As we know both Collison follow this.let both mass m)
(mu1+mu2=mv1+mv2)
If we take equation (u1-u2=v2-v1 )
V1= (m +m)/m-m)u1+ 2mu2/m+m
Now m-m= o so this variable tend to zero
(V1= u2)
Comment: As it's currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer).


Why are my photos taking too much memory?

Question: I have a Canon 5D mark II. The weirdest thing happened today. I was shooting with a 4 GB memory in Raw and Jpeg and was only able to shoot 34 pics. I looked at my settings and couldn't figure out what I was overlooking. I usually can get close to 150 t0 200 pics.  Please Help...
Answer: Did you use the CF card in your computer and either store non-picture files on it, or delete the files without emptying the trash / recycle bin when you were done?  A bunch of hidden files can fill up the card's space pretty quickly (happened to me a couple weekends ago; luckily I always carry a spare).
It could also be a corrupted card, and some sectors have been marked as bad.
If you're sure there's nothing important on the card you could format it from the camera's menu, which will remove any undeleted or non-picture files.
Comment: I always format the card in the camera.  That way, there's just one device writing to the card, which seems to make it less likely for problems to occur - especially if the camera has any low-level ideosyncracies.
Comment: Ohhh my Gosh yes I did delete the files without emptying my trash/recycle bin.  Do you think the best thing to do from now on is to just delete pics from card via camera rather than computer?  I don't want to run into this again.  I'm going to format the card and see if that fixes it.  Thank you so much drewbenn...
Comment: I bet that it's the hidden file issues. I always format my card through the in camera menu once all the photos are safely on my computer.


Tool to help in detecting all the pages in an ASP.Net application which are performing slowly

Question: Tool to help in detecting all the pages in an ASP.Net application which are performing slowly, without having to measure performance of each individual page ?
Answer: If you're just interested in the server time, add time-taken to IIS logs (http://support.microsoft.com/kb/944884) and then use an analysis tool e.g. R, to build histograms of the response time for each page.
If you want to track the actual load time in the users browser you'll need a Real User Monitoring (RUM) tool - basically gathers timing information from the Navigation Timing API and beacons it to a central server.


If PGP is 2048 or 4098, then what is 128-bit encryption?

Question: PGP is 2048 bit encryption?
That a whole lot more than 128 bits.
But, I thought 128 bits is more "serious" and "legit"
Comment: The short version is, "good" key lengths for Asymmetric keys are different than "good" key lengths for Symmetric keys.  2048 is a reasonable minimum for Asymmetric keys, and 128 is a reasonable minimum for Symmetric keys (depending on the algorithm).
Comment: More specifically, "good" key lengths are unique to the *cipher being used*. Symmetric keys are generally unstructured random numbers and need to be resistant to brute force so 128 or 256 bit keys are common. Asymmetric schemes tend to use larger keys, as they often contain internal structure. For instance, RSA private keys consist of two large randomly-chosen primes and are significantly larger at 2,048 or 4,096 bits. For ECC keys—such as curve25519—256-bit keys are used, which provides security equivalent to 128 bits. For the McEliece cryptosystem, keys can be as large as 8,373,911 bits.
Answer: There are two different concepts at play here:
Key Size is the size of a key on disk, measured in bits. For RSA encryption you tend to see key sizes like 2048 bit, 4096 bit, etc. For Elliptic Curve Cryptography (ECC), you tend to see key sized like 224 bit, 256 bit, 384 bit etc. For AES, 128 bit, 256 bit.
Key Strength is a measure of how much computation an attacker needs to do to break your key (ie compute the private key from the public key). In crypto research papers you tend to see sentences like this:

Breaking an RSA 2048 key requires, on average, 2112 operations.

This key is said to have "112-bit security".
Symmetric block ciphers like AES tend to have the property that the key length is the same as their security level (note that this will no longer be true once quantum computers are here). Here is a table from NIST that shows relative security levels for various RSA and ECC key sizes.

Of course, as mathematical research progresses and we find more efficient ways of attacking the underlying math, the number of bits of security provided by your 2048-bit key will go down.
Why 128-bit security?
For an intuition on how long it would take you to make 2128 guesses at something, check out this youtube video:
How secure is 256 bit security?
Cryptographers like things that have >= 128-bit security because it's comfortably in the "not in my lifetime" category. When something gets below 80 bits of security we start looking for something bigger to replace it with.
Comment: Small nitpick, but 192 and 224-bit ECC is typically no longer permitted. I believe 256 is the minimum for NIST.
Comment: Is that a nitpick? According to that table, you'd need >= 256 to get 128 bit security. Is it possible to look up the NIST standards at the moment, or do we have to wait for the US government to stop throwing a temper tantrum?
Comment: I just recall reading that Bluetooth 2.1 uses P-192 and that it is lower than what is currently permitted.
Comment: p̶e̶r̶m̶i̶t̶t̶e̶d̶  recommended; for those of us who work outside of the US federal government, NIST is merely a suggestion :P
Comment: It is generally believed that symmetric encryption such as AES will be reduced to 1/2 the number of bits under a quantum attack. This 256-bit AES would degrade to 128-bit and still be secure under a quantum attack.


Did the Concorde have darkened windows?

Question: Did the Concorde and the Tu-144 have darker window glass than subsonic airliners, to protect from the Sun when flying above 9/10 of the atmosphere? Or were the crew and passengers at higher risk of eye damage if looking towards the Sun from cruise altitude?
Comment: @MichaelHall Thank you. I would upvote your comment now if I had the rights.
Answer: Concorde had tinted nose visor glass to withstand the heat and pressure produced at the cruise speed it could achieve. It was a laminate with a gold layer in the middle, thus the tint. I don't believe the passenger cabin glass was tinted.
Comment: Passenger windows were so small that passengers probably couldn't normally see the sun except when quite low in sky?
Comment: @quietflyer Would it matter if they see the entire solar disk or a part of it? Isn't it almost equally dangerous? One shouldn't look into the Sun during an annular eclipse either.
Comment: @Betternottell -- I wasn't implying that the disk would be partially hidden by the horizon, just noting that what little atmosphere is there has more of a darkening effect when the sun is quite low in the sky.  Actually I think this whole question could be grounds for a much more general question about whether viewing the sun is significantly more harmful when flying at high altitude than otherwise, and what if any measures have typically been taking to deal w/ this, other than telling crew/ passengers "don't look at the sun".  Well, actually, even including that one.
Comment: @quietflyer The higher the altitude the brighter the Sun is and the less filtered UV radiation is by the thinner atmosphere, but if I understood you correctly you mean that in Concorde windows the Sun would only be visible low enough so when looking through enough atmosphere it wouldn't be more dangerous than on the surface.
Comment: It should be noted that glass is reasonably effective at blocking UVB and UVC radiation, which cause sunburn and might reasonably be blamed for permanent damage to the eyes (visible light can cause bleaching of the optical receptors but this is reversible).

Did the Concorde have darkened windows?


Nautilus could not create the following required folders

Question: I have changed my username and when I did this,I can not log in with my new user and this error comes in the screen: nautilus could not create the following required folders: /home/"last username"/desktop and /home/'last username'/.nautilus   .Also when I press CTRL+ALT+F6 for the command line I can not login with my new username. Although I have still access to Linux by root. My Ubuntu version is 11.04 .
Answer: You're going to want to log in as root to do this
Firstly, make sure the home folder of your old username does not exist.
ls /home

If you see a folder with your old username, issue the command
mv /home/[oldusername] /home/[newusername]

Then issue the command
usermod -d /home/[newusername] [newusername]
usermod -d /home/[newusername] [oldusername] (just for good measure)

If you still can't log in after this, you'll have to edit the /etc/passwd file
nano /etc/passwd

Find the line that starts with either your old or new username
Make sure the first field is your new username
Make sure the 6th field (after the 5th colon) is your new home directory - /home/[newusername]
Take note of the UID and GID fields (the second and third respectively), this is usually 1000 on single-user systems. If it's not, write it down or remember it. I denote these fields as [UID] and [GID] below.

chown -R [UID]:[GID] /home/[newusername]
Answer: I had this problem on RHEL6.  It was because I had a gui enabled and I was trying to log in with my domain Id's.  I could not get logged in.
Then I did this step
Once the machine is on the domain open the console (not SSH) and log in as root.  Click on System>Administration>Authenication.  Change the User Account Database to Winbind.  (If you do not do this you will not be able to log in.
At this point I was able to get logged in, but this is where I started receiving the Nautilus errors.
I then went back to System>Administration>Authenication.
** Click on Advanced Options and click the Check box beside Create home directoires on the first logon.
This fixed the problem for me and my home drives were created.
Hopes this helps someone.


How can I tell IntelliJ's "Find in Files" to ignore generated files?

Question: I need to do a find in files. I want to ignore or exclude generated files, like JAX-WS artifacts or classes in target folders. How can I tell IDEA to exclude these files from the find?
Comment: Yes!  And they always put the generated results on TOP!  Sheesh!
Answer: Create a Custom Scope defining the set of files to include/exclude from your search.

CTRL+SHIFT+F for the Find
in Path dialog. (Mac users press command+shift+F)
Under Scope select Custom.

Choose a scope from the drop down list or create a Custom Scope by clicking on the ... button to the right of dropdown.
In the dialog that appears, click on the + button and select Local
On the right pane you can Include and Exclude individual files and Recursively include or exclude all files beneath a folder.

Now when you use Find in Path, use your custom scope to restrict which files are searched.
I suspect from the reference to Find in Files in your question that these instructions may not apply to your version but it is useful to know that this functionality exists and there is always the 30 day evaluation version.
Comment: That works. Thanks. I had hoped there was some setting or checkbox I had overlooked to the effect of "ignore generated files".
Comment: @ajp15243 This can be done in the settings -> project scope : )
Comment: @Freiheit You might be able to have your project ignore the generated files too, which would exclude them from searches and the like by default everywhere (unless you like seeing or need your generated files in IntelliJ, of course).
Comment: For reference, these steps work the exact same in IntelliJ Ultimate 13.
Comment: Thanks! Still works in 2016.2.2. Very helpful when web-files is copied by grunt/nodejs to a new location. Very frustrating to see duplicates.
Comment: You can use this pattern to filter out classes generated by dagger and butterknife: `!file:*InjectAdapter.java&&!file:*ModuleAdapter.java&&!file:*_ViewBinding.java`
Comment: Excellent! I was worried that I was describing steps that didn't appear in your version.
Comment: This info oudated. Please update for new Intellij Idea version
Answer: I know this is late to the party, and Rob's answer is a decent one. I'd just like to add the following alternative though: if you chose the Custom scope (as in Rob's answer), then leave the selection at Project Files, this will make IntelliJ search a bit more selectively than by default. I don't know what the exact differences are, but of particular interest is that if you mark a directory as Excluded either using the Modules tab in the Project Structure settings, or by right-clicking on a directory and selecting Mark Directory As -> Excluded.
If the files you want to exclude are in a single or relatively few directories so you can easily manually set up these exclusion rules, this is a really nice way of getting the same result without needing to configure a custom scope.
I tested this in IntelliJ Ultimate 14.1.4. I have no idea how it behaves in other versions, but I suspect most of v14 at least will behave the same.
Comment: @coding_idiot `Mark Directory As` is still visible to me when I do that (the option to choose is then `Cancel Exclusion`). If that for some reason doesn't work for you, you should be able to do the same thing by opening the `Module` settings (`Project Settings` -> `Modules` -> ``). In that dialogue all excluded folders should be listed on the right with an `x` next to it to stop the exclusion.
Comment: I marked a directory excluded by mistake, how to include it again. The context option "Mark Directory As" is no longer visible for that directory.
Comment: Related: [IDEA-87037 – Usability: Find in Path -- NOT in Whole project](https://youtrack.jetbrains.com/issue/IDEA-87037)
Answer: You can also put the search file filter starting with ! sign to exclude.
Example to search code not in Test Java files:
!*Test.java
If you have a few types of files you can separate with , sign.
Example to search in Kotlin and Groovy files only:
*.kt,*.groovy
This might be also helpful.
Comment: That's v useful.
Answer: IntelliJ Skip generated files pattern during searching
I use the next pattern to exclude generated files
!file:*intermediates*/&&!file:*generated*/&&!lib:*..*
[How to add a new pattern follow and skip tests pattern]
Answer: I use this filter to create a custom scope to exclude the target folder in play framework applications:
(src:*..*||test:*..*)&&!file:target//*
Also, this is the language reference for the scope definition language: https://www.jetbrains.com/help/idea/scope-language-syntax-reference.html


How to show post(excerpt) from specific category on wp page?

Question: I need this on 3 pages, all 3 from different categories, and then to show several post excerpts from all 3 categories in sidebar of home page.
Answer: The Easiest way to do that would be to use List category posts plugin which allows you to list posts from a category into a post/page using the [catlist] shortcode.
you can also use this shortcode in a simple text widget and add just add :
<code>add_filter('widget_text', 'do_shortcode');
</code>
to your theme's functions.php file, which will tell WordPress to render shortcodes in widgets.


Travelling to the UK from Argentina with a company computer bought in the US - will I pay VAT at UK border?

Question: I live between the UK and Argentina, but I am only a legal resident in the UK. I have a LTD company in the UK which I used to purchase a laptop (~4000USD) in New York (travelling Buenos Aires to New York return) – now I am travelling back to the UK I am starting to consider whether or not I will be asked to pay tax on this laptop to bring it into the UK.
Comment: It is an item well over the personal use free limit, can you give us reasons why you think you may bring this in free?
Comment: @a1652 From the point of view of UK customs, your company is permanently importing an item into the UK, and thus may need to pay import vat & duties. Unless there's a specific international agreement, they don't really care about other countries tax rules.
Comment: I guess the only reason I can think of is I don't understand the laws very well. If I buy it here in Argentina, pay tax on it here and use it here why can't it come back to the UK with as a used item? Is the law essential that you have to pay taxes twice?
Answer: You should have to pay tax in the UK for any personal possession you bring into the country for which you have not paid UK tax before, over a certain threshold.
A $US 4000 computer is well over that threshold.
The way not to pay double VAT (or whatever the local tax is called) is to claim it back or arrange not to pay it in the country you buy said item. Like in the tax free shops in the airport you leave from.
The UK does not get any of the foreign tax and is not concerned with you paying double, it is up to you to avoid paying outside the UK. And the UK is not special in this, all countries I know about have this tax on arrival with foreign bought goods.
We have a question at this time about traveling into the Netherlands with a used wedding dress, tax plus extra fees coming to over €1000.
The right thing for you to do upon arrival in the UK is to use the red channel, something to declare, and talk with the officers there. Have proof of the price handy, in case they evaluate the price higher than it actually was. Paying tax in the airport means that you will not run into trouble should the tax people check your company for whatever reason. (I am neither in the UK nor knowledgeable about the UK tax people but I have heard some stories how intensive those checks can be.)
If there is an exemption for business use/purchase the officers will know and implement it, on the other hand, if you have to pay more for it not being a person possession they will know that as well. I can not help you in that.
There are many people who just use the green channel and act like they did buy the computer 'at home' before their trip but as Stack Exchange site we do not tell people to break the law.
Comment: The wedding dress example is really great, thank you. I realise now the limit for the UK (£390) means suits bought abroad, for example, they would be something to declare. I will check with my accountant to see if they know if there is any exemption for business use/purchase and come back with the answer.


Is it true that it is better to get a degree with a specific option than just to get the degree in general?

Question: I'm working on my bachelors degree in computer science and the school I'm attending offers various "options" such as networking, graphics and gaming, bioinformatics and psychology. I wasn't planning on taking any of these as none of them particularly interests me but a friend was recommending I do so because a computer science degree is so broad it would be easier for an employer to know what it is I'm into if I chose one of these options. Is this true? Would it help me get a job if chose one of these options? It's not the I don't like these topics but I'm ambivalent towards them.
Comment: If you aren't particularly interested in making one of those options your career, then why would you specialize in it?
Comment: I'd find out if any have the reputation for being easy and pursue one of those to pad my GPA.  Then you have a 'specialization' which might impress some people regardless of what it actually is, and you boost your GPA(hopefully). Just a thought
Comment: Specialization is much more important at the graduate level.  An Undergraduate degree is supposed to be at least somewhat broad.
Answer: One way to find out if these help is to ask the placement office at the university. They may know what companies or government agencies are hiring which options.
If none of the options appeal to you, ask the department if you can create your own minor/specialty. Some schools allow this. Also some schools require some majors to declare a specialty or concentration.
These options can help if you know what you want to do after graduation. It can also hurt if HR can't see past the words on the paper. They are looking for a degree in A with a specialty in Y, and you have a degree in A with a specialty in ~Y, since it isn't an exact match they reject you.
After a few years these options aren't nearly as important. Employers are more interested in what you have done on the job, not in a class room.
Answer: If you want your first job out of school to be with a company in one of these sectors, it wouldn't hurt and it could help.  If I'm recruiting someone for a position in a games company, all things being equal, I'm going to prefer a candidate with the extra coursework in graphics than someone with a more general degree.  On the other hand, candidates are rarely equal and I certainly wouldn't hesitate to hire the candidate with the general degree if they were the better candidate.  If you are applying for a job that isn't in one of these specific areas, it's unlikely to matter either way.
Looking down the road, these specializations get less and less beneficial the further you are from school.  Once you've got a couple of years of practical experience, companies are going to be far more concerned with what you have experience with than what you have coursework in.
Comment: If I were hiring for a position in a games company, I would be much quicker to hire the person who actually created a game or contributed to creating one over someone who has an 'option' in graphics. The goal is to show ability and motivation toward the field you want to get in to, no? If an option helps do that, then great!


Where can I leave luggage in Frankfurt, either airport or train station?

Question: I need to find a place to leave my luggage in lockers for 2 days in Frankfurt, either the airport (Terminal 1) or at the train station. I have to go by train to Stuttgart for 2 days without luggage. The luggage is 5 bags, 28 inches.
Comment: Please clarify: it is 5 pieces of luggage, each measure 28 inches? I did edit the question, but left that as you stated it.
Answer: There are lockers at Frankfurt Main Station (Frankfurt Hbf). You can leave your bags there. A small locker costs 3 or 4 Eur per day, a big locker costs 5 or 6 Eur per day. Note, that day ends at midnight, it doesn't matter when you put your bags in the locker - at 3:00 or at 23:00.

(full picture)
You can also store your bag at the airport for up to three months. Here is the information about locations, prices and working hours.
Comment: I recently had to store a bag at Frankfurt station and it took me a while to find a locker that was both functioning and unoccupied. Bring coins. +1
Comment: Munich HBF has a note-to-coin machine - I would expect Frankfurt to have the same?


Doing squats with a medicine ball - Is it a good thing?

Question: I've been told that squats would be a good way of building up muscles in my knees as they aren't great at the moment. I have been doing squats and either use no weights at all or I do them while holding a 5Kg medicine ball. Is this safe to do or is it not recommended?
Comment: I don't recommend doing squats on the medicine ball like this.... http://www.break.com/index/crazy-exercise-ball-trick.html
Answer: If you are holding the medicine ball in front of you, what you are doing is a form of goblet squat.  A goblet squat is very good for your body, and provides a good core component to the squat as well.
The important components of any squat are:

Make sure the crease of your hip goes below the patella (knee cap).  This is going below parallel.
Make sure that your core (back, abs, upper back) is tight.

The purpose of going below parallel is to strengthen the posterior chain.  It gets your hamstrings and calf muscles involved, preventing strength imbalances between the front and back of your legs.  This of course, is what strengthens the knee.
Don't be afraid of increasing the weight you do beyond 5kg.  You can use kettlebells or dumbbells to get heavier weights on the goblet squat.  Eventually, you can graduate to back squats or front squats for even stronger knees and core.  Just make sure your increases are around 5-10lbs (2.5-5kg).
Comment: See I thought if I went below the kneecap it would cause damage to the knee and not be good for me. I can see your point and will try this tonight :)
Comment: See this for more info: http://www.exrx.net/Kinesiology/Squats.html
Answer: If you are doing goblet squats with a medicine ball (you can also use a dumbbell or kettlebell) then I would highly recommend using goblet squats to help you perfect your squat technique.
Here is a great resource on goblet squats: http://www.t-nation.com/free_online_article/most_recent/goblet_squats_101
In a nutshell, it's much harder to have bad form doing a goblet squat than a "regular" squat because the positioning of the medicine ball/kettlebell/dumbbell forces you to brace your core, keep upright, hinge from the hips, etc.
The reason why this is important is that it's not uncommon to see people in the gym attempting "regular" barbell squats but with outrageously bad form. Some of the most common ways that people butcher squats include the following:

Not going down low enough (not going to parallel or breaking parallel)
Bending forward too far at the waist
Letting the knees splay out too far (knees should not go past toes)
Not keeping head up, shoulders back, chest proud
Not "unlocking" the hips for a hip dominant hinge movement

A lot of these common errors mentioned above can be remedied by practicing proper squat form with goblet medicine ball squats.
Comment: Hi @JoelOhman, welcome to the site! Can you edit your answers and provide some more information in addition to the links you provided? That would give them more context and make them a better answer. Please see the [faq] for more information regarding how this site works. Also take note that self-promotion is frowned upon and perceived as spam which is also stated in the [faq#promotion].
Answer: Also, make sure you keep you weight on your heels during the entire squat movement, and you initiate both the up & down movements with your hips first, not your knees.
A fun variant is to extend your arms forward at the bottom of the squat, holding the medball or weight away from you at shoulder height with straight arms. Then you can bring it back in to your chest when you stand or keep it away from you as you stand. This is a great way to add more challenge without needing a lot of extra weight for lifting.


ipod touch jailbreaking on linux

Question: Santa brought me an iPod Touch (OS v3.1.2, Model MC series). I was wondering if it is possible to jailbreak it through a Linux machine (Ubuntu Karmic).
Answer: No. blackra1n works on Windows and Mac, while PwnageTool is Mac only.
Answer: Jailbreaking is not possible natively on Linux. You can setup a VM under Linux and try that option but better off doing it in Windows or a Mac

ipod touch jailbreaking on linux


Power leveling using a high level character in WoW?

Question: I have a 80 level elemental shaman and recently my friend joined WoW. I would like to help him level up, is this possible? Can i for example take him to a dungeon (low level dung) and level him up faster than leaving him doing quests etc? I am aiming to get him up to about level 30 or so, allowing him to overcome the boring start faster. What is the best way to do this, is it even possible?
Comment: Isn't the start going to be more fun with Cataclysm? Plus he could start as one of the new classes, which would definitely make it make fun ;-)
Comment: Related: http://gaming.stackexchange.com/questions/12077/in-wow-is-it-better-to-powerlevel-a-character-in-group-or-out-of-group
Answer: I'd suggest letting him play at his own pace (unless he's an experienced MMORPG player starting WoW for a specific end-game purpose such as raiding / PVP).  In particular if he's new to the genre, that first time exploring the world is such an amazing time, it would be a shame to be boosted through it.
Following the 'Shattering' the starter zones have had an overhaul and are fun to play through and also give a better step by step introduction to game mechanics and how to play your class.
Also the levelling speed now is pretty crazy - people in my guild are making jokes about being level 30 by lunchtime!  It's not quite that fast, but if he gets into it he'll be 30 before he knows it.
Comment: +1, the revamped starter experience has made 1-60 the *least* boring part of the WoW experience. Let him hit 60 at his own pace, and then, if he wants to get to endgame faster, power him through Outland with some Ramps/BF/Underbog/Mana Tombs runs.
Comment: My biggest problem is that I'm having trouble **not** leveling out of zones before I finish questing them.
Comment: Thanks, totally forgot about the Cataclysm changes. It will probably be better to let him quest on his own. He has a lot of MMO experience, but not in WoW (played Lineage 2 a lot).
Answer: Fwiw, your totem shouldn't impact his xp for mobs if you are out of the group. So if he tags them and they magma down, he should get full credit.
If you want to help someone at level 30, the best think you can do is usually get a level 30 of your own and quest together.
Comment: +1 - Questing together at the same level is not only more fun, but better all around.  He learns his toon, you learn yours... much better than grabbing a sandwich and watching while your buddy wipes the floor with 20 elite mobs for 2 hours...


can log into Minecraft at home (on Mac) but not recognised on on PC elsewhere..?

Question: My son logs into Minecraft on our computer (Mac) but since he joined the Minecraft club at school he cannot log in there. We know the login details are correct but he gets the message that they are unrecognised or the referenced account cannot be logged on to.  Other people are using Minecraft there so its not a firewall issue. Any ideas? Thank you
Comment: Did he try to login with the account's email or just the username?
Comment: Have you tried to login on Mojang website? Also have in mind, that if you fail to login like few times, Mojang servers will ban your IP for some time (an hour, or at most 24 hours), so if you try it over and over again, it won't then work at all.
Answer: From my experience, Minecraft login is really buggy. This stuff happens ALL THE TIME. Sometimes, it is because of what you're logging in with. If your son has had Minecraft for a short amount of time, then you need to log in with your username. If he's had it for a while, try logging in with the email. For more details about which to use, visit the Mojang website
Other than that, maybe you tried with failed attempts too many times. Like your phone will lock you out, maybe Mojang does the same thing-ish? I don't really know at this point.
Just remember, Minecraft login does this. Just stay calm. And if all else fails, Mojang.com or Minecraft Wiki.
Answer: It is possible that your son's minecraft account was created using the free version of minecraft. Sorry if I offended you and you did pay for minecraft.


Installing MS Office 2016 on new computer

Question: I installed MS Office 2016 Pro on a new computer and want to have it set up the same way I had it on my old computer - that is using PST files for mail, contacts, calendar, etc.  However, in the process of installing, I used some automated feature (I think) to set up the mailbox etc.  The set up then used my new MS account to setup Outlook data files which default to an email account using outlook, not gmail.  I don't want to use exchange, I just want to use the same pst file I had on my old computer with all my email, contacts, calendar, etc.  Is there a way to do this?
Answer: Go to Control Panel, Mail, and delete the profile in the Profile section (you did not want this). Then use the Email Accounts section, set up a new Account and be sure not to use any wizards. I use this approach and do not use Exchange.
Answer: If You have a copy of your "old" .PST file, then yes it is possible, and there are several ways to do that and all of them are quite similar and depend on your desired end goal.
One way:  You could just add your old PST file to outlook

Go to File -> Account Settings -> Data Files -> Add
Browse for your old PST and finish adding it.

Now You will have all data from your old PST at your disposal, You can even make it to be your primary PST file, Archive it(to make it smaller), and after archiving don't forget to compact it, or something else.... If needed, I will give You more guidance, although google knows it all ;)
By default, outlook/exchange will cashe your emails inside your local PST file, which can be seen/tweaked at this location:
File -> Account Settings -> Data Files -> Click on your PST file -> Click on Settings tab -> Switch to Advanced tab
Answer: If you want to use your gmail with old PST file on the new computer, just configure your gmail account on your Outlook as POP and import your PST file. Then remove the outlook.com account .
If you want to use your Outlook.com account with the old PST file on the new computer and would not like to use Exchange. Just re-configure your outlook.com as POP/IMAP on the new computer and import or open your PST file.(Before that, you need to enable POP/IMAP Access in Outlook.com)


How to hook into adding watchdog entries by other modules?

Question: Is it possible to write some code to intercept the process of adding Watchdog entries and filter them before adding to the log? Or is there a module that allows for watchdog filtering before adding to log? (I searched but nothing found) This would be useful if there are tons of well-known messages (like page not found from the same referrer or the same IP) that obfuscate the log.
Answer: Unfortunately hook_watchdog doesn't provide a return value which could block the insertion of the log entry in the database.
Note though that the responsible module for inserting the entry in the database is dblog. You could disable that and write your own implementation of it, although you'd have to replicate lots of functionality.
Another idea is to have a frequent cron job deleting the relevant log entries from the database table.
Comment: From looking at the dblog_cron function, I'll try to write a custom function that does db_delete('watchdog') with  conditions containing regular expressions that specify what to delete  -and it should do it
Comment: You mean the cron job as configured in Linux or in Drupal? If in Drupal, then how to specify what are the "Relevant log entries" ? I have a couple of regular expressions, where to put them?
Comment: I mean the cron mechanism of Drupal. Regarding the "relevant log entries" I meant the log entries which you consider that are worth deletion. I don't know which these are, you do. The implementation of `dblog` contains a cron function from which you could get some ideas.
Answer: Not sure if it's still relevant, but I've made a module where you can control logging per type per level: http://drupal.org/project/watchdog_filtering
Answer: I wrote a module that adds the needed D7 hook:
https://www.drupal.org/project/watchdogalter

This module adds a hook_watchdog_alter. Used in DBLog Booster.

https://www.drupal.org/project/dblogbooster


Force Anaconda to install tensorflow 1.14

Question: Now, the official TensorFlow on Anaconda is 2.0. My question is how to force Anaconda to install an earlier version of TensorFlow instead. So, for example, I would like Anaconda to install TensorFlow 1.14 as plenty of my projects are depending on this version.
Answer: You can force installing a certain version of any package found on Anaconda using simply an = operator with the package version attached to it.
So, if you want to install tensorflow 1.14, you can run the following command:
conda install -c conda-forge tensorflow=1.14

You can replace 1.14 with any other versions. To see the available versions of tensorflow on Anaconda, you can run:
conda search tensorflow

Comment: https://pastebin.com/raw/BuDBwA3F got a segmentation fault for this
Comment: I tried this command, it took **5 hours** and then failed. (error was about wrong Python version, and wrong glibc version). I then tried the process described by @sun and @brainslugs83 in the other answer (`conda create -n myenv -c conda-forge --strict-channel-priority python=3.6` ; `conda activate myenv` ; `conda install pip` ; `pip install tensorflow==1.14`) This took 15 minutes and succeeded. There was no need to install another version of Anaconda. (I had version 22.9.0)
Answer: first find the python version of tensorflow==1.14.0, then find the Anaconda version by python version.
e.g. tensorflow 1.14.0 can work well on python36, and Anaconda 3.5.1 has python36. So install the Anaconda 3.5.1, then install tensorflow==1.14.0 by pip
Comment: This was the missing link for me -- I had to create an env with Python=3.6; and then I could run `pip install --ignore-installed --upgrade tensorflow==1.14.0` -- Thanks!


Duty Cycling on FPOD

Question: I am attempting to program an FPODs software to record periodically at 1 min out of 2, but when we save the specific settings file "F_SETS.txt", the POD doesn't read it automatically (as I believe the software guide indicates) and instead keeps recording continuously. The txt file contains a line of code that reads "recording = 128; 1 min of 2". Any help is appreciated.
Comment: Hi Hannah and welcome! Please edit the main title so it is a specific question. As it stands, your title is a short phrase that doesn't let readers know what information you're looking for. [See guide for: "How do I ask a good question"](https://stackoverflow.com/help/how-to-ask)
Comment: Did you reach out to Chelonia for an answer?
Answer: Hannah, we (Chelonia) have just tested this dis-continuous logging option and it is broken in the latest version of the embedded code for one of the micro-controllers.  We are working on a fix that can be applied via code on an SD card. Apologies!
Comment: hope you provide also here link to updated version for other FPOD users
Answer: Unlike the CPODs, it is good practice to re-format your SD cards every time you deploy your FPOD to ensure all files are deleted before use. If you have other files on your SD card (like FP1 files) I think the setting files can get confused.
If you are already doing this, then reach out to Chelonia.

Duty Cycling on FPOD


How complete is support for SVG on mobile devices?

Question:
How much of the SVG specification has been introduced on the various mobile browers?
Are there many known inconsistencies between these browsers?
Are there any Javascript SVG libraries aimed at mobile browsers?
Comment: A partial answer here : http://caniuse.com/#cats=SVG
Comment: Anyone have an update on this for 2016?
Answer: Most browsers will have support that at least matches SVG Tiny 1.1, and the features of SVG Tiny 1.2 that are also in SVG Full 1.1 (yes I know, that's a bit complicated — in short: Tiny is a subset of Full, but Full 1.2 was abandoned, leaving Tiny 1.2 with some features that Full 1.1 does not have).
There are inconsistencies in some of the DOM interfaces, in referring to external content (e.g. external  or fonts), and in animation support. Most of the rest tends to work pretty well.
As for libraries, it depends on what you want to do. On smartphones I would be surprised if Raphael didn't just work. For more constrained devices there's Ikivo's Enrich framework which isn't free but works well and integrates with their IDE if you happen to like those: http://ikivo.com/tools.html (disclaimed: Ikivo is a customer of mine).
You can also check this chart for support in general, it could be updated with more mobile information: http://www.codedread.com/svg-support.php.
Answer: A quick look on caniuse.com usually answers these browser support questions
http://caniuse.com/#search=svg
Answer: You should note that the default Android browser has SVG support turned OFF up until Android 3.
Answer: Support documentation for Opera can be found here.
Like Robin wrote, Raphaël should work fine on mobile. D3.js and most other js libraries that do svg work fine too, at least the ones I've tried.


Can someone explain autocorrelation between observations in linear regression

Question: I'm new to linear regression. I created a simple linear regression model (only one feature). I am checking my assumptions, and I am having difficulty with the Independence assumption. I don't really understand what effect correlated observations has on the model and its results. Can someone explain in layman's terms. I just don't really understand why it matters or why/how we even test for this assumption. I'm just trying to learn more. I know I asked a lot of questions already, but thank you in advance if anyone can help me.
Answer: Thanks for your questions and for joining the community. The independence assumption you're referring to is the independence in the residuals.
When you apply linear regression ($\hat{f}(x) = ax + b$, where both $a,b$ are real values), you obtain an estimated value $\hat{f}(x)$ given the values $x$. But you also have the true data values (TDV) - these are the data points in your data set. The residuals (or the "noise") are the $TDV - \hat{f}(x)$.
In layman's term, if you have correlation in your noise, it's because there's information in your data that your model is not capturing. In very broad terms, you can think of your data as a source of information that it's not readily available to you. You have to "work your data" to derive meaningful information from it. So, in your case, you applied a linear regression to your data to have a model and make some informed decisions. However, because the residuals are correlated, your model is somehow not capturing all the information your data can provide. Maybe, linear regression is not a suitable approach for your data.
Once you have your sequence of residuals ($= TDV - \hat{f}(x)$), you can can calculate the autocorrelation function for them. In most computer languages, you'll find a function that will do that for you. What you want to see for correlation is a value very close to zero. Visually, you shouldn't be able to spot patterns in the residuals' (again, vaguely) behavior. You don't want residuals with a clear trend (up or down), or with cycles, for instance. These clear patterns would mean the residuals are somehow correlated. It's also beneficial to say that you'll likely find some correlation among residuals, but you have to establish a threshold to define what is acceptable or not.
If you want a more detailed explanation with graphs that will help drive the point home, this link from a Penn State class can help.
Comment: WOW! This was extremely helpful. Thank you so much.


What can I do when my parents want a divorce?

Question: My mum and dad want a divorce and I don't know what I can do. My dad likes another woman and my mom already asked three lawyers for the divorce. My BFF has heard some things about this. My life is already a dark, gloomy cloud, and I am very very unlucky! I don't know what to do. Please Help Me!
Comment: "Please help me" is unclear as a question. Do you need support? Do you need to understand what's going on and why? Do you want to know what you can do to stop it? Who to live with after the divorce? These are all specific questions. But I see no question in your post. I didn't close it as unclear because I think the duplicate might help you. If it does not, please edit your post and flag for reopening. We are here to help, but we need to know what you're asking to give you a good answer. Thanks.
Answer: I am sorry but I do not know what we can do for you. Have you read other questions and answers? Here's one: LINK
The short answer is that you can't prevent it, only they can make this decision. You can talk to them. You can tell them how you feel. You're afraid, hurt, angry, upset, and worried about what will happen. They can answer these questions for you.
My advice is to try not to choose a side as it is unlikely that you will ever know all the details. No one is perfect and I'd bet both your parents have made mistakes and hurt each other. I am not suggesting that you lie or not share how you feel, but I also think that once you've expressed how you feel (believe me they already know), then just try to support both of them and any siblings you have.
You may have to help more around the house and help your parents to remember things. They have a lot going on. "Mum, I need that permission slip signed, please." "Dad, remember I need a lunch tomorrow, thanks."
You might find that you have to grow up a little more quickly, and think before you act. You might have to take on more responsibility. This will all work out in the long run even though it is hard right now.
Oh and regarding luck: You make your own happiness. You can decide to be a happy person. That doesn't mean bad stuff doesn't happen. It means you keep going and look for things that are good.


Negative radiation pressure from negative refractive index?

Question: I see this claim being originally made by Vesselago (the discoverer of the principles of metamaterials) and indeed in contemporary papers. It means that such a metamaterial would be pulled towards its illuminating light source. Why does this not violate conservation of momentum?
Comment: Please provide a link to the papers which make this claim.
Comment: http://meetings.aps.org/Meeting/MAR09/Event/93172
Comment: @WhatRoughBeast, I've added links to both the APS paper Andrew gave in his comment, and to the original paper by Vesselago.
Answer: This does not violate conservation of momentum, because the momentum behaves in a counter-intuitive manner in a negative refractive index medium.
In most media, the refractive index is positive, and the Poynting vector $\vec{S}$ and the wave vector $\vec{k}$ point in the same direction. In a medium with a negative refractive index, the Poynting vector points in the opposite direction from the wave vector.
The wave vector of the light is what you control when you set up the light source. If you point the source towards the positive $x$ axis, for example, then $\vec{k} = |\vec{k}|\hat x$. On the other hand, the Poynting vector is what controls the light wave's momentum, and therefore the radiation pressure.
The result is (from the APS Meeting paper you cite):

a perfect mirror illuminated with a plane wave would experience a negative radiation pressure (pull) when immersed in a left-handed medium [Emphasis added]

I've emphasized that the mirror must be immersed in a left-handed medium, rather that simply made from a left-handed material, because the light wave must actually be in the negative refractive index medium for this effect to occur.


Recovery of blowfish encryption key?

Question: Is it possible to recover key used for blowfish encryption, if I have encoded and decoded string as well as initialization vector tables (bfp and ks0-3)? I don't know the length of the key.
Answer: No, not unless the input for the key small enough to be brute forced or guessed. For any secure block cipher the key cannot be retrieved even if you have the input (block of plaintext) and output (block of ciphertext). And there are no known attacks on Blowfish that break the cipher significantly.
So it really doesn't matter how the block cipher is used. There may be attacks on the mode of operation, such as CBC/PKCS#5/7 padding oracle attacks. Those may leak information about the encrypted message, but not about the key.
The small block size of Blowfish makes it harder to use in a secure mode of operation, which is probably the main reason why Bruce Schneier doesn't recommend it anymore (besides that it is an old algorithm, and there are simply better ones available by now).
Comment: @unknown123 When the answer fits your needs, please feel encouraged to push the accept button (✔) left of the answer. That way, the system and other users know that your question has been answered to your satisfaction.

Recovery of blowfish encryption key?


Using non-standard characters for accent marks

Question: I have a text of poems (psalms) that will be sung or chanted.  To indicate the word where there's a change in the music, an accent is placed above the syllable in that word.  Then the singers know that's where to change notes.  The standard acute/grave accent is too small for this.  It needs to be larger and stand out more.  I want to use the elegant-looking Bengali Currency Number One.  I don't know how to get this into LaTeX.  I'm using MikTeX2.9 with TeXMaker.  I already have some code to make any character into an accent.  However, after searching for a way to get this particular character into LaTeX, I haven't found a way to do it.  Any suggestions?
Comment: You link to a generic image for the character but to actually get a specific glyph you will need a font that covers that range. classic (pdf)TeX can not use system Unicode fonts so unless you are using xelatex or lualatex this will be more or less hard depending if it is already done by some package. But using a specific font for this rather than just using a larger accent seems to be making thing massively more complicated than need be, you can take the acute accent from any font you use in TeX already and scale it arbitrarily large then position it over the letters.
Comment: Thank you for this reply.  How do I scale an accent?  For example, If I input \'{e}, how do change the size of the accent above the e?
Comment: See also [this question for another approach to the same problem](http://tex.stackexchange.com/q/112041).
Answer: It is easy to get a large accent, eg  <code>{\Huge\'{}}</code> positioning it over a normal size letter usually means a bit of trial and error to get something that looks right but perhaps:

<code>\documentclass{article}

\newcommand\biga[1]{\leavevmode\vbox{\offinterlineskip
\halign{##\cr\hss\makebox(0,0){\Huge\'{}}\hss\cr\noalign{\vskip-3pt}#1\cr}}}
\begin{document}

aaaaaaa\biga{a}z\biga{B}zzzzz

\end{document}
</code>
Comment: @Ryan odd, try replacing `\'{}` by `^^b4` (U+00B4 is SPACING ACUTE)
Comment: Thank you (also @Caleb), the above has worked perfectly.  However, now I've changed the font from Linux Libertine to Times and instead of giving me large accents above the letters, LaTeX is simply drawing a box with an "X" around the letter in the pdf output, with no accent.  On the other hand, if I use a command like \'{e}, then it places the accent above the letter at normal size as usual.  This is happening with a lot of fonts I try.  I'm using XeLaTeX.  Any ideas?  Is there a workaround?  Thanks a lot for your assistance.
Comment: David, worked like a charm! Thank you very much!  You can disregard a comment a left on another thread with a similar question.  Appreciate it.


Private alternative to Google Docs?

Question: Is there software to install on an ubuntu server, and then use to collaboratively edit text files, sometimes adding some attachments? it's important that no data is lost if 2 people save the same doc with different contents at the same time. I've found seafile, but it seems too complicated considering it need 3 ports to work, and requires manual installation.
Answer: Try Etherpad, a nice alternative to Google Docs based on Node.js. We used it in some projects with relative success.
Answer: You can also try Firepad which is based on Firebase, and is server independent. Only using Javascript.
Answer: ownCloud is an open source, private cloud server, like dropbox or google drive, but you can install it locally. They recently added a collaborative document editing feature that you can see in action at the top of this page: http://owncloud.org/six/
I have only briefly tested the document editing aspect, but the file archival aspect has worked great across multiple machines and platforms. Just like with dropbox, you always retain a local copy of your files.
Answer: We at collatebox are creating a shared 'spreadsheet-like' offering for the private cloud / on premise installations. This will allow you cell level control with a mobile template off-the-shelf. Write to info (at) collatebox.com to get onto early release list
Answer: You could try http://www.collatebox.com/ , or simply write to them to kow if they have a solution for that.

Private alternative to Google Docs?


Is it safe to place symlinks in /bin directory ?

Question: I have many command line tools that I use frequently.
I was using the normal method to traverse till the complete path of that command through terminal and then did ./command-name BUT this is a cumbersome and time-consuming task.
Researching the topic of "How to execute a command from anywhere through terminal", I ended up in solution to include the path in the environment variable (by editing ./bashrc).
This works fine.
I even found out that all the commands that work globally(like mkdir, ifconfig, cp, etc ) are placed in "/bin" directory.
On including the "symlinks" of my commands in "/bin" directory, all is working fine and I can execute the commands from anywhere in terminal.
My Question

Is it safe to place symlinks in /bin directory directly. I am asking this because while placing the symlinks, it asked me for administrator password.
Will it create any difference if I place commands' symlinks in /bin as compared to /usr/bin directory ?

Environment (if it matters):
Ubuntu 13.10, 32-bit.
Comment: Refer [this](http://askubuntu.com/a/308048/83046) answer.
Answer: It isn't recommended to mess with /bin and with /usr/bin even though this will work it is somewhat unsafe. Also, you don't need to place your symlinks there.
I've had the same problem like you mentioned. It's cumbersome to navigate to your scripts and call them like ./myscript each and every time. Here is what I did.
Define your own bin-directory
You can create your own bin-directory and put your scripts there. Create it with
mkdir ~/bin

Now you will need to tamper with the PATH environment variable. This is a sensitive area. If you mess up the PATH variable you won't be able to execute commands with relative paths anymore. If that happens ls might not work any more /bin/ls will still do.
What I'm going to show right now will only affect the current running terminal session. So if anything goes wrong you simply log out with Ctrl+D and everything is like it was before.
echo $PATH
/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/local/games:/usr/games
export PATH=$PATH:~/bin
echo $PATH
/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/local/games:/usr/games:/home/madmike/bin

The echo-command shows what's in $PATH before and after. export allows you to change the environment variable. Your additional bin-directory is appended to the end of the variable. This means your commands have the least priority should there be more than one command with the same name. This is good, since you don't want to override essential system commands.
To make the changes permanent, edit your ~/.bashrc file.
nano ~/.bashrc

add export PATH=$PATH:~/bin to the end of that file.
Now every new terminal-session will have the $PATH variable expanded with your directory.
Comment: Just an observation: by default in Ubuntu `~/bin` is already added to the $PATH even if `~/bin` is not already created. You can check `~/.profile` file to see this.
Comment: The difference is with the multi user aspect. "/usr/local/bin" is for everyone. "~/bin" is only meant for the current user. "/usr/local/bin" will need root rights. "~/bin" won't.
Comment: nice idea @MadMike :)  But don't you think so that /usr/local/bin is meant for exhibiting same idea ?? I am asking this because I am new to Linux and not exactly sure about your idea of "~/bin" and "/usr/loca/bin".
Comment: Going to add this to my answer as soon as I get to a Ubuntu machine.
Answer: Before answering your question, let me explain "how it works" a little bit.
In your system you have Environment Variables. One of them PATH variable, that store set of directories where executable programs and scripts are located. In Linux, by default, PATH store a few directories. One of them /bin. That's why you can execute any program, which is located in /bin. You can check your PATH variable executing this:
echo $PATH

In PATH variable order of directories is important. The highest priority path names are specified first.
Now answering your questions:

Yes it is safe to create symlink in /bin. Of course if you do not rewrite other executables. But you should prefer to use /usr/local/bin. It is also must be in your PATH.
Yes it is. Depending on order in PATH. For example if you put symlinks of different files but with the same name both to /bin and /usr/bin, and /usr/bin included in PATH earlier than /bin, then it will use symlink from /usr/bin.

You can read about Filesystem Hierarchy here
About Ubuntu Environment Variables here
Answer:

Is it safe to place symlinks in /bin directory directly. I am asking this because while placing the symlinks, it asked me for administrator password.

There's a security risk. I really don't think it's wise to put your own symlinks in the /bin directory because not only your commands become global but also symbolic links can easily be broke. bash PATH environment variable is a more robust solution.

Will it create any difference if I place commands' symlinks in /bin as compared to /usr/bin directory ?

No. Although according to the Filesystem Hierarchy Standard (FHS) there's a difference between /bin and /usr/bin.

/bin Essential command binaries that need to be available in single user mode; for all users, e.g., cat, ls, cp.
/usr/bin   Non-essential command binaries (not needed in single user mode); for all users.

But the difference is not that important today. Actually there's a tendency in some distributions (esp. Fedora) to fade out /bin directory.
Answer: NO!!!
Do not touch system directories. Every script or program you wrote you should normally place in /usr/local/bin and /usr/local/sbin directories. And adding directories to PATH is not good idea, because it has security risks.
Also, the difference between those directories that 's' s for superuser programs.