Why does this differentation output correct result? I noticed something funny. If you differentiate $x^x$ treating the exponent as a constant, you get $xx^{x-1}=x^x$. If you treat the base as a constant, you get $x^x \ln{x}$. If you add these two bizzare and incorrect derivatives of $x^x$, you get $x^x(1+\ln{x})$, which is correct!  Is it merely a weird and funny coincidence or is it a part of a deeper result?  (I'm checking it for $(x^x)^x$, but it will take a while :) )
Proof of this fairly obscure differentiation trick? Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$  But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:  $$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$  which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.  It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with $$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$ which is the correct expression for $f'(x)$.  This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.  These observations have led me to the following conjecture:     Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).$$   (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)  This formulation also seems to lend itself to following generalisation:     Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x_1,x_2,\ldots,x_N$. For $n=1,2,\ldots,N$ define $f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum_n f'_n(T).$$   Thus my question is:   Is this true? How can it be proven? (Specifically in the case $N=2$ but also in the general case.)
Adjective for "terrified yet calm and capable" What is an adjective to describe the duality of a first responder (such as an EMT responding to a terrorist attack) who is terrified for the wounded, yet simultaneously calm, detached, and capable enough to give aid? Ruby block and unparenthesized arguments I extracted simple example:  require 'pp' x = 1..3 pp x.map do |i| {:value =&gt; i, :double =&gt; (i*2)} end pp x.map { |i| {:value =&gt; i, :double =&gt; (i*2)} }  pp(x.map do |i| {:value =&gt; i, :double =&gt; (i*2)} end) pp(x.map { |i| {:value =&gt; i, :double =&gt; (i*2)} })   I am wondering why first pp produces:  [1, 2, 3]   While all the oders are giving:  [{:value=&gt;1, :double=&gt;2}, {:value=&gt;2, :double=&gt;4}, {:value=&gt;3, :double=&gt;6}]   I assume it has something to do with operator precedence. Where can I find good explanation? Why does shrinkage work? I've been informed by those who know much more about Harry Potter than I that Hogwarts is protected with a Disillusionment Charm so all muggles will see is ruins with a "Danger: Keep out!" sign around them.  (Apparently this was in one of the first three books, but I'm not sure where it was.)  My first thought was, "Well, a danger sign would encourage some muggles to investigate!"  Which leads to what I think is an obvious question: Will a disillusionment charm on a place like that still work when a person is real close to the location?  Will it work from inside the location?  For example, if a curious muggle made it to the gates of Hogwarts and started walking around inside, would they still see it as abandon and in ruins, or once they entered, would they see it as it is? $_SERVER['QUERY_STRING']; not returning full query string Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ I have 2 graphics cards (Tesla k10, K40) in my supermicro SuperServer (7047GR-TPRF). I installed Ubuntu Server 14.04, but when I install cuda or Nvidia graphics driver (version 311 or 340), my Ubuntu desktop hangs on login. I have a table that I want to insert on a page, but at least one (perhaps both) of the following conditions are met:   The table is too wide to fit within the text block or page. That is, I'm exceeding some horizontal restriction. The table is too tall to fit within the text block or page. That is, I'm exceeding some vertical restriction.   What are my options to make this table fit? If it doesn't fit, regardless of my attempts, what other options exist? I want to describe my business and want to refer to it in a sentence.  I am trying to say:      "In 2015 we established a business (that is) named X".   Intuitively, the "that is" part is redundant, and makes the sentence more cumbersome. But I cannot think of a reason why we should be able to drop the verb.  Is the version without the "that is" part grammatically correct? How can it be explained? I'm trying to find a way of building a new power supply for my workshop, in order to keep it small and not so hot I will be using a spare xformer to step down mains to 24v and then a variable buck converter to adjust it (and CC mode, displays and blinky stuff).   I thought it would be a good idea to use a linear regulator on the output stage to smooth things out, but I can't seem to think of a control scheme that allows me to change the voltage using only one potentiometer. The buck stage is this topology but with a external PNP transistor.(It has already been prototyped, it works alright, I know the mc34063A is old)   The only idea I got that ""works"" is using the buck converter output as a reference for a linear regulator (IC or discrete parts), but using a SMPS output as reference for anything is a bad idea, due to the amount of noise and fluctuations.   This is the sort of output stage I can and have built (or bought) over time. Just a big discrete regulator.  In short: How to control a MC34063A step down (buck) converter so it follows the output stage +1-3 V of headroom. I am looking to identify a story that I read around 2001 (it is probably much older).  Someone showed me the story on pages printed from the internet, and it was attributed to Isaac Asimov (although I cannot state that is the true author).    The story is told in glimpses at different points of time in the future.  In each glimpse, humans are asking a more and more "intelligent" computer how to prevent the universe from collapsing into a singularity in the future, and the computer states it needs more computational power and resources to solve this problem.  At some point, man becomes one with the computer, and the computer exists in hyperspace (or subspace?).    In the final glimpse, it is the end of time and the universe is collapsing into a singularity.  At the instant the universe reaches the singularity, the computer figures out how to prevent it and says "Let there be light" and the universe begins to expand again.  Is anyone familiar with this story? How many elements does the ring $ℤ[X]/(X^2-3,2X-4)$ have? Describe the structure of this ring. I'm making exercises to prepare for my ring theory exam:     How many elements does the ring $ℤ[X]/(X^2-3,2X-4)$ have? Describe the   structure of the this ring.   I find it always difficult if the ideal is generated by two elements.   I thought about someting like this. $X=2$ and $X^2=3$. Therefore $4=3$ therefore $1=0$. As $2=1$ then $X=0$. And then everyting seems to become $0$.   I'm not sure if I'm allowed to see it this way, but this is the first thing that comes to my mind. Converting GeoPDF to vector format like CAD or shapefile? I want to convert a GeoPDF that has vectors entity in it to a vector format and preserve the real coordinates of these vectors. The GeoPDF has been exported from ArcMap and has layers in it.  I have try GDAL, but it rasterize the PDF before exporting it. Which is not good for me. Also I don't want a solution that would convert this image back to vectors, too complicated for the data I have.  Is someone aware of any tool that can do that? The resulting format can be a CAD file or a shapefile or anything that is a vector format. how to prove that $\ln(1+x)< x$ On page 19 of the textbook  (by James, Witten, Hastie and Tibshirani--it is freely downloadable on the web, and very good), the following is stated:  Consider a given estimate $$\hat{Y} = \hat{f}(x)$$ Assume for a moment that both $$\hat{f}, X$$ are fixed. Then, it is easy to show that: $$\mathrm{E}(Y - \hat{Y})^2 = \mathrm{E}[f(X) + \epsilon - \hat{f}(X)]^2$$ $$ = [f(X) - \hat{f}(X)]^2 + \mathrm{Var}(\epsilon)$$  It is further explained that the first term represents the reducible error, and the second term represents the irreducible error. I am not fully understanding how the authors arrive at this answer. I worked through the calculations as follows: $$\mathrm{E}(Y - \hat{Y})^2 = \mathrm{E}[f(X) + \epsilon - \hat{f}(X)]^2$$ This simplifies to $[f(X) - \hat{f}(X) + \mathrm{E}[\epsilon]]^2 = [f(X) - \hat{f}(X)]^2$ assuming that $\mathrm{E}[\epsilon] = 0$. Where is the $\mathrm{Var}(x)$ indicated in the text coming from? Any suggestions would be greatly appreciated. I am trying to use the Directory.GetFiles() method to retrieve a list of files of multiple types, such as mp3's and jpg's.  I have tried both of the following with no luck:  Directory.GetFiles("C:\\path", "*.mp3|*.jpg", SearchOption.AllDirectories); Directory.GetFiles("C:\\path", "*.mp3;*.jpg", SearchOption.AllDirectories);   Is there a way to do this in one call? Bluetooth headset never gives me sound options From the  article on Wikipedia, the following with regard to the Coriolis Effect on a rotating sphere:     By setting vn = 0, it can be seen immediately that (for positive φ and ω) a movement due east results in an acceleration due south. Similarly, setting ve = 0, it is seen that a movement due north results in an acceleration due east. In general, observed horizontally, looking along the direction of the movement causing the acceleration, the acceleration always is turned 90° to the right and of the same size regardless of the horizontal orientation.   My intuitive (but possibly incorrect) understanding is that if there are two points, Point A and Point B, at different latitudes in the Northern hemisphere, the Eastward velocities of these points are different because they are at different distances from the Earth's axis of rotation, and this causes the Coriolis effect for a rotating sphere.  If a projectile is fired due North from Point A near the Equator towards Point B near the North Pole, the projectile will start off with the higher Eastward velocity of Point A, and will land to the East of Point B, which is moving East at a slower velocity than the projectile.  Firstly, is that correct?  If that IS correct, that brings me to the quote off Wikipedia. The quote implies that, if a projectile is fired due East, it will experience a Coriolis force to the South. My intuitive explanation based on differences in velocities between origin and destination does not account for a Southwards movement at all, since the Point A and Point B velocities are identical if the points are on the same latitude.  What am I missing? Would a projectile fired due East or due West experience any North or South drift caused by the Coriolis effect (or anything else for that matter)?  Why? Can't quantum teleportation be superluminal some percentage of times? I apologize if this is a really silly question.   In the (textbook) quantum teleportation algorithm, in the step right after Alice has measured her system but before she has sent her classical information to Bob, she is about to send one of the following values: 00,01,10,11.   What if Bob doesn't want to wait and simply takes a guess? Wouldn't there then be superluminal communication 25% of the time? \documentclass[11pt]{article}  \begin{document}   The above equations on application of Laplace Transform become:  $$sL_1\hat{i}_{L1}(s) = D\cdot \hat{v}_{C1}(s) + D\cdot \hat{v}_{C2}(s) - D\prime \cdot \hat{v}_g(s) + (C_{C1} + V_{C2} - V_g + V_D)\cdot \hat{d}(s)$$ $$sL_2\hat{i}_{L2}(s) = -D\prime \cdot \hat{v}_{C1}(s) + D\cdot \hat{v}+{C2}(s) + D\prime \cdot \hat{v}_g(s) + (C_{C1} + V_{C2} - V_g + V_D)\cdot \hat{d}(s)$$ $$sC_1\hat{v}_{C1}(s) = -D\cdot \hat{i}_{L1}(s)+ D\prime \cdot \hat{i}_{L2}(s) + (-I_{L1} -I_{L2} - I_1)\cdot \hat{d}(s) -D\prime \cdot \hat{i}_l(s)$$ $$sC_1\hat{v}_{C1}(s) = D\prime \cdot \hat{i}_{L1}(s)- D\cdot \hat{i}_{L2}(s) + (-I_{L1} -I_{L2} - I_1)\cdot \hat{d}(s) -D\prime \cdot \hat{i}_l(s)$$ $$sL_l\hat{i}_l(s) = D\prime \cdot \hat{c}_{C1}(s) + D\prime \cot \hat{v}_{C2}(s) - D\prime \cdot \hat{v}_g(s) +(V_g - V_D - V_{C1} - V_{C2})\cdot \hat{d}(s) - R_l \cdot \hat{i}_l(s)$$  \end{document}   Error:   181 Underfull \hbox (badness 10000) in paragraph at lines 181--181 181 Overfull \hbox (22.63293pt too wide) detected at line 181