$$\int_0^{\pi/2} \frac 1 {1+\tan^{2017} x} \, dx$$ I have tried substitution method, multiplying numerator and denominator with $\sec^2x$, breaking the numerator in terms of linear combination of the denominator and the derivative of it. None of these methods work. Some hints please?
Evaluate the integral $$\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\, \mathrm dx.$$  How can i evaluate this one? Didn't find any clever substitute and integration by parts doesn't lead anywhere (I think).  Any guidelines please?
I've been working through a bit of Ireland and Rosen's Number Theory for fun. Problem 4.11 of Ireland and Rosen asks     Prove that $1^k+2^k+\cdots+(p-1)^k\equiv 0\pmod{p}$ if $p-1\nmid k$, and $-1\pmod{p}$ if $p-1\mid k$.    My work is leading where I didn't expect to go. Here I assume $p$ id odd. First, if $p-1\mid k$, then we can wrike $k=(p-1)j$ for some $j$. But then $$ 1^k+\cdots+(p-1)^k\equiv 1^{(p-1)j}+\cdots+(p-1)^{(p-1)j}\equiv \underbrace{1+\cdots+1}_{p-1\text{ times}}=\frac{(p-1)p}{2}\equiv 0\pmod{p} $$ since $p-1$ is even.   Secondly, I suppose $p-1\nmid k$. WLOG, I can assume $0&lt;k&lt;p-1$, Since if $k=(p-1)j+r$ with $0&lt;r&lt;p-1$, then  $$ a^k\equiv a^{(p-1)j+r}\equiv a^r\pmod{p}. $$  Then if $g$ is a primitive root, $$ 1^k+\cdots+(p-1)^k=\sum_{i=1}^{p-1}(g^k)^i=\frac{g^k(1-g^{k(p-1)})}{1-g^k}. $$ Since $g$ is primitive, $p\nmid 1-g^k$, but $p\mid 1-g^{k(p-1)}$, so the above sum is again congruent to $0\pmod{p}$.  So in all cases, I get the sum is congruent to $0$ regardless. Have I messed up, or is there a typo? Thanks. Smallest Width for Galaxy S and Galaxy S2 I have to do some manual reordering or rows in Excel and cut/paste is too hard to use. I'm looking for something that would enable me to use drag'n'drop or to add some buttons to move-up/down move-top/bottom. Generated $\sigma$-algebras with cylinder set doesn't contain the space of continuous functions Product Notation for Multiplication in Reverse Order Is there a standard notation for multiplication in reverse order? For example consider the problem  $$x_{k+1} = A_k x_k$$  where $x_i \in \mathbb{R}^n$ and $A_i \in M_n(\mathbb{R})$, ($i=0,1,2,\dots$) without any further assumptions on $A_i$. The solution to this problem is  $$x_k = A_{k-1} \dots A_1 A_0 x_0$$  Obviously, writing  $$x_k = \prod_{i=0}^{k-1} A_i x_0$$  is wrong because of the multiplication order. But I'm also unconfortable to use  $$x_k = \prod_{i=k-1}^{0} A_i x_0$$  as the notation suggests that we need to increment $i$, not decrement it. Also, one may need to write $\prod_{i=k}^{m} A_i$. Then, we need to explicitly state $m &lt; k$ to imply ordering.  I can always write the open form but as similar products frequently occur in calculations in my work I want to express them in a more compact way. The question is, is there a standard and nice notation to do so? There are some suggestions in , but I think none of them are "nice enough" to work with.  I came up with the notation  $$x_k = \coprod_{i=k-1}^{0} A_i x_0$$  to imply decrementing $i$ rather than increment it, but I don't know if it is used in somewhere else or there is a standard notation. What is the need of exponential generating functions on combinatorial problems? I've been introduced in my last lectures. And for the following problem:     Having 3 different types of books $a,b,c$ in how many ways can we take four different books putting them in a shelf such that the book $a$ can only be taken at most $1$ time, $b$ can be taken at most $3$ times and $c$ can only be taken at most $2$ times?   The ordinary generating function for this problem is:  $\quad \quad \quad \quad \quad \quad \quad $  Looking at the exponential generating function of the same problem, the only part of interest is the coefficient of $x^4$, which is:  $$\left(\frac{ab^3}{1!3!}+\frac{b^3c}{3!1!}+\frac{ab^2c}{1!2!1!}+\frac{b^2c^2}{2!2!}+\frac{abc^2}{1!1!2!}\right)$$ And then multiply it by $4!/4!$. But why do I need an exponential generating function? Isn't only needed to take the coefficient of $x^4$, divide each term of it by $a_1!a_2!\dots a_n!$ with $a_n=\text{power of a,b,c,}\dots$ and then multiply it by $4!4!$? It's not clear why such complication is needed.  By complication I mean to expand that generating function with the factorials, I could expand the ordinary generating function and then artificially add the factorials. cp - cannot create regular file : permission denied I want copy to /lib/udev but gives error  cp -f /tmp/ok_pcscd_hotplug.sh /lib/udev  cp:cannot create regular file /lib/udev/ok_pcscd_hotplug : Permission denied   I am root , but I can't understand . what can I do ? Number of vertices in the Klein bottle triangulation given in "Topology and geometry for physicists" by Nash and Sen I stumbled upon the following triangulation of the Klein Bottle in page 76 of the book Topology and geometry for physicists by Charles Nash and Siddhartha Sen.    Below the figure, it says "the triangulation $K$ consists of eight $0$-simplexes[...]".  I think this is not an eight-vertex triangulation, but rather a nine-vertex triangulation of the Klein bottle. I also know, thanks to the great comments under  question, that an $8$-vertex triangulation of the Klein bottle is a minimal one, hence the one portrayed in the book isn't minimal.  Am I correct? Meaning of convolution? Why do all Winforms programs require the [STAThread] attribute? Can I transfer saved data from my old PS3 hard drive to my new PS3? A while ago, my PS3 completely broke. Every time I turn it on, it just completely shuts down a few seconds later. I decided to buy a new ps3 to play my old games. The problem is that I don't have my saved data, but I still have the original hard drive. Is it possible to somehow transfer the saved data from the old hard drive to my new PS3? I was thinking maybe I could clone the hard drive, but I don't have a spare hard drive to do it on. I do however have a 2 TB external hard drive. Would it be possible to clone it on the external? Are there any alternative methods? How to enlarge thickbox media library iframe? The iframe in thickbox media library is not big enough to display my custom tab. I tried to add "width=900&amp;height=600" at the end of the link that pointing to the library, but no luck. How can I modify the inframe size? CPT Template Not Showing - Getting 404 I have a sword that currently rotates around a pivot point thats off of the actual object. Im new to blender and im not sure how i would go about changing it so that the sword rotates around the handle. Any help would be appreciated! It is easy to find the minimum of $\|Ax-b\|_2$, when $A$ has full column rank. But how is the case when we add the constraint $\|x\|_2=1$? Or, to be explicit,  $$\min_{\|x\|_2=1}\|Ax-b\|_2=?$$  My idea is to construct the corresponding Lagrange function  $$L(x,\lambda) = (Ax-b)^T(Ax-b) + \lambda(x^T x - 1)$$  Differentiate the function with respect to $x,\lambda$, we have  $$A^T A x - A^T b + \lambda x = 0$$  $$x^T x=1$$  However, I find it hard to move forward to get the answer. Could anyone help me? Thanks a lot! On many occasions, I've found questions that should be moved to AskUbuntu, or Unix/Linux, but when I click the "flag" link, I don't have those as options. I hate to leave something flagged as off-topic with no suggestion of where to put it, so I just cancel.   What determines the sites that show up on that list? Is it best to flag them off topic even if the ideal site doesn't appear? How many connected components does $\mathrm{GL}_n(\mathbb R)$ have? How can the following inequation be proven?  $$a^2 + b^2 + c^2 \ge ab + bc + ca$$ How do  methods compare in terms of model variance and bias?   My question is partly motivated by this thread: . The answer there suggests that models learned with leave-one-out cross-validation have higher variance than those learned with regular $K$-fold cross-validation, making leave-one-out CV a worse choice.  However, my intuition tells me that in leave-one-out CV one should see relatively lower variance between models than in the $K$-fold CV, since we are only shifting one data point across folds and therefore the training sets between folds overlap substantially.  Or going in the other direction, if $K$ is low in the $K$-fold CV, the training sets would be quite different across folds, and the resulting models are more likely to be different (hence higher variance).  If the above argument is right, why would models learned with leave-one-out CV have higher variance? What is the purpose of the Character Table in style files? What is the reason to insert the code below at the beginning of a style file? For example, the code below can be found on lipsum.sty.   %% \CharacterTable %%  {Upper-case    \A\B\C\D\E\F\G\H\I\J\K\L\M\N\O\P\Q\R\S\T\U\V\W\X\Y\Z %%   Lower-case    \a\b\c\d\e\f\g\h\i\j\k\l\m\n\o\p\q\r\s\t\u\v\w\x\y\z %%   Digits        \0\1\2\3\4\5\6\7\8\9 %%   Exclamation   \!     Double quote  \"     Hash (number) \# %%   Dollar        \$     Percent       \%     Ampersand     \&amp; %%   Acute accent  \'     Left paren    \(     Right paren   \) %%   Asterisk      \*     Plus          \+     Comma         \, %%   Minus         \-     Point         \.     Solidus       \/ %%   Colon         \:     Semicolon     \;     Less than     \&lt; %%   Equals        \=     Greater than  \&gt;     Question mark \? %%   Commercial at \@     Left bracket  \[     Backslash     \\ %%   Right bracket \]     Circumflex    \^     Underscore    \_ %%   Grave accent  \`     Left brace    \{     Vertical bar  \| %%   Right brace   \}     Tilde         \~} %%   What is it for?