The following question from $\textit{Statistical Inference }$by Casella &amp; Berger was given to me on an assignment last semester. On the surface, it appeared to be a straight-forward question about transformations, but I stumbled across an integral I could not understand. Here is the question:     If $X \sim \text{exponential(1)}$, show that $Y=\mu-\beta \log X$ has the $\text{Gumbel$(\mu,\beta)$}$ distribution, where $-\infty&lt;\mu&lt;\infty$ and $\beta&gt;0$.    We have that $f_X(x)=e^{-x}$ for $x \geq 0$. Now, let $Y=\mu-\beta \log X$. Then, $g(x)=\mu-\beta \log x$, which is monotone on $0&lt;x&lt;\infty$. We will use the standard theorem for transformations, with $g^{-1}(y)=e^{\frac{\mu-y}{\beta}}$ on $-\infty &lt;y&lt;\infty$. Then, the transformed pdf is given by: $$f_Y(y)=e^{-e^{\frac{\mu-y}{\beta}}}\left|\frac{d}{dy}e^{\frac{\mu-y}{\beta}}\right|=\frac{1}{\beta}e^{-e^{\frac{\mu-y}{\beta}}}e^{\frac{\mu-y}{\beta}} \text{ on $-\infty&lt;y&lt;\infty$} $$  I then attempted to find the mean of the distribution as follows. \begin{eqnarray*} \text{E}(Y)&amp;=&amp; \frac{1}{\beta}\int_{-\infty}^\infty ye^{-e^{\frac{\mu-y}{\beta}}}e^{\frac{\mu-y}{\beta}}dy \hspace{0.5in}  \text{Let $u=e^{\frac{\mu-y}{\beta}} \implies du=-\frac{1}{\beta}e^{\frac{\mu-y}{\beta}}dy$} \\ &amp;=&amp; \int_0^\infty \left(\mu e^{-u}-\beta e^{-u}\log u \right)du=\mu \int_0^\infty e^{-u}du-\beta\int_0^\infty \left(e^{-u}\log u\right) du \\ &amp;=&amp; \mu-\beta\int_0^\infty \left(e^{-u}\log u\right) du \end{eqnarray*} According to , the mean is given by $\mu+\beta\gamma$, where $\gamma\approx 0.5772$ is the Euler-Mascheroni constant. Therefore, the last integral above is equal to $-\gamma$, and indeed, that is the answer  provides. I am wondering if there is any way to solve the integral $I=\int_0^\infty (e^{-u}\log u) du$ using standard methods. I tried making every substitution I could think of and kept getting stuck. I even showed this to my instructor; he was not able to show that it equals $-\gamma$ either. I should also mention that prior to doing this question, I had never heard of the Euler-Mascheroni constant. Any help would be appreciated. Thank you!
Show  $$\int_0^\infty \frac{\ln(x)}{e^x} = -\gamma$$ (gamma is Euler-Mascheroni constant).  Can anyone please prove this result?  Also  $$ \int_0^\infty \frac{\left( \ln(x) \right)^2}{e^x}\mathrm dx. $$
My ceiling fixture has all of a sudden stopped working properly. Half the diodes are dark and the others are blinking. Oddly, the same exact ones on each strip work and don't work. I have included a picture to show which ones are blinking and which ones don't work. I am wondering if it is something that I can fix? Maybe replace that driver? And if so, is there a good guide online to walk me through that process? Thanks in advance. Given that the sequence $(v,Av,A^2v,...,A^nv)$ is linearly dependent (1), $A_{n \times n}$, over $\mathbb{F^n}$ prove that scalars $c_0,c_1,..,c_n$ exist (such that not all of them are zeroes) such that the matrix $M=c_0I+c_1A+..+c_nA^n$ is singular.    First I'm not sure how (*) can be used here exactly. We can say that the sequence $(c_0I,c_1A,..,c_nA^n)$ resembles the sequence (1) where $v=I$. But here we have scalar $c_i$ so it's not exactly the same sequence.   Maybe I could say that the sequence $(c_0I,c_1A,..,c_nA^n)$ is linearly dependent because it has $n+1$ elements while over $dim\mathbb{F^n}=n$ $\quad$(2).  Intuitively I understand that suppose we sum all the matrices $c_iA^i$ and then in the last sum operation we get to some $c_jA^j$ and if it's a linear combination of other matrices then we can have such a $c_j$ that: $$ c_oI+\sum_{i=1}^{n-1} c_iA^i+c_jA^j=0 $$ But I don't think this is enough to prove. Is their a way to move multiple vertices gradually at a time? I have moved multiple vertices gradually but a set of vertices at a time not all at a time, as shown in the animation below (First I moved the four vertices from top, then three, then two and at last the last vertex).  I wanted to do the same thing as shown in the animation below but all at a time. I've read that abelian categories can naturally be enriched in $\mathbf{Ab}.$ How does this actually work? Wikipedia defines the notion of an abelian category as follows ().     A category is abelian iff         it has a zero object,   it has all binary products and binary coproducts, and   it has all kernels and cokernels.   all monomorphisms and epimorphisms are normal.      It later explains that an abelian category can naturally be enriched in $\mathbf{Ab},$ as a result of the first three axioms above.  How does this actually work? Are Ubuntu packages security audited? I know that big O is used to describe the upper bound of a function, so doesn't this mean that if the upper bounds f and g are the same, they are the same function? If this is the case, then we can say that  f ∈ Θ(g) because if that were not the case, then the two upper bounds would not be the same and thus one would out grow the other. Is this correct? Uncaught SyntaxError: Unexpected token with JSON.parse Based on the question , I got curious. Is there a guide or resource that describes how the price of a ticket is determined?  When you search you often see curious effects like cheaper tickets for a flight from A to B with a layover compared to a direct flight. This seems counter-intuitive from an outsider's perspective since you fly twice. Also one way tickets more expensive than 2 ways.  It can be a short set of rules. Something that helps someone to get started with finding good prices. I know these can be very complex and they probably change over time and over airline interests but there is probably a simplified set of rules that can be enunciated. not working in Phonegap app on Android 2.3.3 At the moment I'm working with the . For this I read the following paper: .  But I have some mathematical problems with this formalism. I don't really see why this is a variation principle since it is opartor-valued.   The action principle is defined by: $\delta S  = G_1 -G_2$, where $S$ is the action operator $S = \int_{t_1}^{t_2} dt ~L$ and $G$ the generator operators of the transformations.  What does it mean that the action principle is operator-valued? For the classical action where $S$ is real we search for a stationary point but for operators? I also have a problem that $\delta S \neq 0$. What interpretation has this additional term $G_1 - G_2$? Why doesn't current pass through a resistance if there is another path without resistance? Why doesn't current pass through a resistance if there is another path without resistance? How does it know there is resistance on that path? Some clarification:  I understand that some current will flow through the resistance. A depiction of this phenomenon through Ohm's laws does not answer the 'why' part of the question as it does not illuminate the process behind this phenomenon. There seems to be disagreement among the answers on whether the charge repulsions are enough to cause this effect. Cited research would be appreciated to resolve this issue. To refine what I mean by this question, I think I should give an example of why I am curious as to the causal background of this phenomenon. If charge repulsions do in fact repel 'current' in the example of the short circuit, then why does the same phenomenon not happen in other parallel circuits. (Again, I know that not all circuits behave ideally) I believe the question is expressed in the most straightforward way possible. This should, however, not affect the scope of the answers. I do not wish to see or think on the level of an electron as circuit behavior always includes a cumulative effect due to the nature of electrostatic interactions. Why is there not a simpler way to calculate the standard deviation? Steps of getting standard deviation. :        Work out the average (mean value) of your set of numbers   Work out the difference between each number and the mean   Square the differences   Add up the square of all the differences   Divide this by one less than the number of numbers in your set -   this is called the variance   Take the square root of the variance and you've got the standard   deviation      Am I missing out something, or why do we need to square the differences in step 3?  Why not simply do a Abs (multiply all negative numbers by -1) in step 3?  Also, my second question is why do we need to divide by one less than the number of numbers in the set in step 5? why not simply divide by the number of numbers? Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$ How to prove the following trignometric identity? $$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$  Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$.  I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those.  Hints please!  EDIT:  What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however).  I know that $$ \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}$$  So,   $$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$ $$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$  $$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$ Value of the Hubble parameter over time There is something I don't understand about the  $H$, as it seems to clump two concepts together that I can't quite unify in my head. On the one side, we have  $$V = D H$$  which means that for a given distance $D$, there is a certain amount of new space created over time - and $H$ is simply the factor that makes this relationship work. So, for example, say we have two points 1 Mpc apart, this would mean they recede at about 70 km/s away from each other  (given our current approximation of $H$).  Now the thing I can't wrap my head around is that   $$T =\frac{1}{ H}$$  is also the age of the universe. Contrary to claims made on, say, Wikipedia this means that $H$ cannot possibly have been a constant throughout the past 13 billion years, because mathematically $1/H$ means that $H$ must be continually shrinking as the universe ages.  So if $H$ did start out as some huge value and is now shrinking over time, doesn't this mean the expansion of the universe is slowing down? Because if $H$ is shrinking, I'll get a lower value of $V$ today than I'll get tomorrow. Shouldn't the notation then be more like  $$V = DH(t)$$  So which one is it? If $1/H$ is simply the solution for $D=0$, how can we use it as the expansion-velocity-per-unit-of-distance at the same time? What's worse, how can literature say $H$ has probably been more or less constant forever and simultaneously assert that $1/H$ is the current age of the universe? What am I missing? I'm trying to continue the numbering of examples in enumerate environments in beamer. For example, I'd like one slide to feature   Foo Bar   and the next slide   Zip Yadda   My strategy so far has been to define a counter that saves the enumerate counter at the end of the first environment and can then be queried to set the enumerate counter at the start of the second enumerate environment. (This is essentially the proposal also found at )  That works wonderfully so long as I don't try to uncover elements of the slide sequentially. The problem, of course, is that as each of the pdf-pages that make up the slide is created, the counter is advanced. Here's a MWE.  \documentclass{beamer}  \setbeamercovered{highly dynamic}  \newcounter{saveenumi} \newcommand{\seti}{\setcounter{saveenumi}{\value{enumi}}} \newcommand{\conti}{\setcounter{enumi}{\value{saveenumi}}}   \begin{document}  \begin{frame}[&lt;+-&gt;]   \begin{enumerate}   \item foo   \item bar%     \seti   \end{enumerate} \end{frame}  \begin{frame}[&lt;+-&gt;]   \begin{enumerate}     \conti   \item zip   \item yadda%     \seti   \end{enumerate} \end{frame}  \end{document}   As the second slide is progressively uncovered (pages 3 and 4 of the resulting PDF), the counter at the beginning of the enumerate environment is advanced each time. So page 3 of the PDF has the right numbering, (3) and (4) respectively, while the fully revealed page has the wrong numbering. Pictures follow:   on one of the slides    on the following one How can I restrict a user on the SSH server to allow them only the privileges for ? i.e. So they cannot run commands even if they log in via SSH.  My Linux servers are Ubuntu 11.04 and OpenWrt. How to use Terminal to delete all .svn folders recursively? How to show that $\mathbb{Q}$ is not $G_\delta$ How is Leibniz's rule for the derivative of a product related to the binomial formula? Possible Duplicate:          The  describes the expansion of the $n$th power of the sum $(a+b)$:  $$(a+b)^n = \sum_{k = 0}^n {n\choose k}a^kb^{n-k}$$  In calculus, there is a generalization of the product rule called , which describes the expansion of the $n$th derivative of the product of two functions $f$ and $g$:  $$(fg)^{(n)} = \sum_{k = 0}^n {n\choose k}f^{(k)}g^{(n-k)}$$  It's not difficult to prove using the product rule and induction, what I'm really trying understand here is why the formula is true. Is there a way to prove it using the binomial formula? How are these two formulas related? Is there a combinatorial proof for Leibniz's rule like there is for the ?   I understand this question is a bit vague, but I hope you'll agree that these two formulas look so similar it just feels like there should be some relation between the two. Of course this is not necessarily true, but I'm hoping it's not just a crazy coincidence. How is MSE divided by $\sigma^2$ always a chi-square distribution?