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Wildcard not working in subprocess call using shlex On the proof that the product of diagonal matrices is a diagonal matrix? Let $A [a_{i,j}]$  and $B =[b_{ij}]$   be two diagonal matrices in   $M_n(F)$. Let $i \ne j \in \left\{1, \ldots, n \right\}$. Using the   product rule, we obtain      $$(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$$      We claim that $a_{ik}b_{kj} =0$ for all $k \in \left\{1, \ldots, n \right\}$ thus $(AB)_{ij}=0$ for all $i \ne j$ and $AB$ is diagonal. To    prove the claim, note that since $i \ne j$ , we have $i \ne k$ or $    k\ne j$. Thus either $a_{ik} = 0$ (since $A$ is diagonal) or $b_{kj} =  0$ (since $B$ is diagonal), thus in all cases $a_{ik}b_{kj}=0$ and the   claim is proved.   Where does the bit "we have $i \ne k$ or $k\ne j$" come from? I understand that since $i \ne j$, and $k$ and $j$ run over the same set of natural numbers, we also have $i \ne k$. But the "or $k \ne j$" baffles me. I understand that $k \ne j$ implies that we can have $i=k$. But that's from the other direction ($\Longleftarrow).$  Also, when it says $a_{ik}=0$ or $b_{kj}=0$ it's talking about when $i \ne k$ or $k \ne j$ right? Many thanks. How to evaluate the product: How to evaluate the product:  $$\prod _{n=2}^\infty\left(1+\dfrac{1}{n^2}+\dfrac{1}{n^4}+\dfrac{1}{n^6}+\dfrac{1}{n^8}+\ldots\right)$$  $$\left(1+\dfrac{1}{n^2}+\dfrac{1}{n^4}+\dfrac{1}{n^6}+\dfrac{1}{n^8}+\ldots\right)=\dfrac{1}{1-\dfrac{1}{n^2}}$$  How to proceed now? Automatically change é to \'e – most convenient setup for accented letters? I assume this is really basic question, I'm new to LaTeX, I read a first step tutorial to get ready to write some documents but I couldn't answer my own question at this time and I really don't know where to search.  When I try to write é and compile my document, this character is not shown. I know the sequence is \'e to make it happen in the document but that is really disturbing to write.  Is there someway to write a paragraph "normally" and convert all the special chars into the LaTeX syntax? Or is there a way to allow é, à, ... writing so the compilation includes them into the document? Homemade coconut milk separates into different layers after a day in the fridge I made some coconut milk. After a day in the fridge there was two layers of liquid in the jar, coconut water and solid coconut which I found out afterwards that it is called coconut butter! When I stir them together to  have my coconut milk back, it didn't work. Is there a way to prevent this? If not, is there a way to have coconut milk again? How do I summon a named mob? In Java Edition, I can name a mob by using NBT tags like this: /summon zombie ~ ~ ~ {CustomName:'{&quot;text&quot;:&quot;TheUndead&quot;}'}  But in Bedrock Edition, I've heard NBT is not accessible from commands. What can I do to summon a mob with a custom name tag using the /summon command? Adjective for someone who exaggerates emotions to get sympathy Does mathematics become circular at the bottom? What is at the bottom of mathematics? I am trying to understand what mathematics is really built up of. I thought mathematical logic was the foundation of everything. But from reading a book in mathematical logic, they use "="(equals-sign), functions and relations.  Now is the "=" taken as undefined? I have seen it been defined in terms of the identity relation.  But in order to talk about functions and relations you need set theory.  However,  seems to be a part of mathematical logic.  Does this mean that (naive) set theory comes before sentential and predicate logic? Is (naive)set-theory at the absolute bottom, where we can define relations and functions and the eqality relation. And then comes sentential logic, and then predicate logic?  I am a little confused because when I took an introductory course, we had a little logic before set-theory. But now I see in another book on introduction to proofs that set-theory is in a chapter before logic. So what is at the bottom/start of mathematics, logic or set theory?, or is it circular at the bottom?  Can this be how it is at the bottom?  naive set-theory $\rightarrow$ sentential logic $\rightarrow $ predicate logic $\rightarrow$ axiomatic set-theory(ZFC) $\rightarrow$ mathematics  (But the problem with this explanation is that it seems that some naive-set theory proofs use logic...)  (The arrows are of course not "logical" arrows.)  simple explanation of the problem:  a book on logic uses at the start: functions, relations, sets, ordered pairs, "="  a book on set theory uses at the start: logical deductions like this: "$B \subseteq A$", means every element in B is in A, so if $C \subseteq B, B \subseteq A$, a proof can be "since every element in C is in B, and every element in B is in A, every element of C is in A: $C \subseteq A$". But this is first order logic? ($(c \rightarrow b \wedge b \rightarrow a)\rightarrow (c\rightarrow a)$).  Hence, both started from each other? Does anyone know where I can find cheat codes for Goldeneye 007 Wii? I know that the game hasn't even been out for a week, but does anyone know if any cheat codes have been released by Activision yet?  I know that they must exist since there is a cheat codes menu within the game. Evaluate $\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$ I tried l'Hospital but that will require a lot (and I mean A LOT!!!) of differentiating  Is there a shortcut? $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$  Thanks in advance How to minimize the ink-to-data ratio for pgfplots Considering the irreducible (by condition) polynomial $f=X^4-2aX^2+b\in\mathbb Q[X]$ and the field extension $K=\mathbb Q(\sqrt{a^2-b})$ of $\mathbb Q$, I want to show the following result:    For $[L:\mathbb Q]=4$ we have $\sqrt b\in K$ where $L$ is the splitting field of $f$.   My thoughts so far: The four roots of $f$ are $\pm\sqrt{a\pm\sqrt{a^2-b}}$. If $L$ is of degree $4$ it must be primitive since $f$ is irreducible, so $L=\mathbb Q\left(\sqrt{a+\sqrt{a^2-b}}\right)$. Then, by multiplying the two roots $\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}$ we have that $\sqrt{b}\in L$. Now I want to look at $$\mathbb Q\subset K\subset K(\sqrt b)\subset L$$ and differentiate some cases. The case where $[L:K]=1$ is trivial and the case $[L:K]=4$ can't happen because if we square any root and subtract $a$, we get $\pm\sqrt{a^2-b}\in K$ which means $[L:K]\leq 2$. So let's look at the case where $[L:K]=2$. As I see it, we get two subcases:   For $[L:K(\sqrt b)]=2$ we have $K(\sqrt b)=K$ which finishes the proof. For $[L:K(\sqrt b)]=1$ we have $K(\sqrt b)=L$. This means that $L$ is also the splitting field of $(x^2-b)(x^2-(a^2-b))$. Is this a contradiction? How do I finish this case? What is the most efficient way to concatenate strings in SQL Server? I have this code:  DECLARE @MyTable AS TABLE (    [Month] INT,    Salary INT );  INSERT INTO @MyTable VALUES (1,2000), (1,3100);  SELECT [Month], Salary FROM @MyTable;   Output:    I want to concat the Salary (grouping by month) so that it will be NVARCHAR like this: '2000,3100'  How would I do this efficiently? Representing every positive rational number in the form of $(a^n+b^n)/(c^n+d^n)$ About a month ago, I got the following :      For every positive rational number $r$, there exists a set of four positive integers $(a,b,c,d)$ such that    $$r=\frac{a^\color{red}{3}+b^\color{red}{3}}{c^\color{red}{3}+d^\color{red}{3}}.$$      For $r=p/q$ where $p,q$ are positive integers, we can take    $$(a,b,c,d)=(3ps^3t+9qt^4,\ 3ps^3t-9qt^4,\ 9qst^3+ps^4,\ 9qst^3-ps^4)$$   where $s,t$ are positive integers such that $3\lt r\cdot(s/t)^3\lt 9$.      For $r=2014/89$, for example, since we have $(2014/89)\cdot(2/3)^3\approx 6.7$, taking $(p,q,s,t)=(2014,89,2,3)$ gives us $$\frac{2014}{89}=\frac{209889^3+80127^3}{75478^3+11030^3}.$$   Then, I began to try to find every positive integer $n$ such that the following proposition is true :   Proposition : For every positive rational number $r$, there exists a set of four positive integers $(a,b,c,d)$ such that $$r=\frac{a^\color{red}{n}+b^\color{red}{n}}{c^\color{red}{n}+d^\color{red}{n}}.$$  The followings are what I've got. Let $r=p/q$ where $p,q$ are positive integers.   For $n=1$, the proposition is true. We can take $(a,b,c,d)=(p,p,q,q)$. For $n=2$, the proposition is false. For example, no such sets exist for $r=7/3$. For even $n$, the proposition is false because the proposition is false for $n=2$.   However, I've been facing difficulty in the case of odd $n\ge 5$. I've tried to get a similar set of four positive integers $(a,b,c,d)$ as the set for $n=3$, but I have not been able to get any such set. So, here is my question.     Question : How can we find every odd number $n\color{red}{\ge 5}$ such that the following proposition is true?      Proposition : For every positive rational number $r$, there exists a set of four positive integers $(a,b,c,d)$ such that $$r=\frac{a^n+b^n}{c^n+d^n}.$$   Update : I posted this question on .  Added :  asks the case $n=3$. Query to collect the recent TimeSampled row using inner joins in MS SQL In the following setup, does method B run in a (new) transaction?  An EJB, having two methods, method A and method B  public class MyEJB implements SessionBean     public void methodA() {        doImportantStuff();        methodB();        doMoreImportantStuff();     }      public void methodB() {        doDatabaseThing();     } }   The EJB is container managed, with methodB in requires_new transaction, and method A in required transaction. thus:  &lt;container-transaction id="MethodTransaction_1178709616940"&gt;   &lt;method id="MethodElement_1178709616955"&gt;     &lt;ejb-name&gt;MyName&lt;/ejb-name&gt;     &lt;method-name&gt;*&lt;/method-name&gt;   &lt;trans-attribute&gt;Required&lt;/trans-attribute&gt;   &lt;/method&gt;   &lt;method id="MethodElement_1178709616971"&gt;     &lt;ejb-name&gt;MyName&lt;/ejb-name&gt;     &lt;method-name&gt;methodB&lt;/method-name&gt;   &lt;/method&gt;   &lt;trans-attribute&gt;RequiresNew&lt;/trans-attribute&gt; &lt;/container-transaction&gt;   Now let another EJB call methodA with an EJB method call. methodA now runs in an transaction. Will the subsequent call to methodB from methodA run in the same transaction, or does it run in a new transaction? (mind, it's the actual code here. There is no explicit ejb-call to method B) calculator's angle answer for trig ratios that can work in more than 1 quadrant on the unit circle Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,  and a clockwise rotation for negative sine &amp; tan instead of cc and a counterclockwise rotation for negative cos ratios instead of a clockwise   ie. in degree mode  $\cos^{-1}(-5/12)=114.62$  $\sin^{-1}(-5/12)=-24.62$  $\tan^{-1}(-5/12)=-22.61$  Is it maybe picking the value that involves the least amount of computing power? or is it a matter of convention? or am I overlooking something? Prove that there is no continuous functions $f:[0,1]\to R$, such that   $$ f(x)+f(x^2)=x. $$   My try. Assume that there is a continuous function with this property. Thus, for any $n\ge 1$ and all $x\in [0,1]$, \begin{align*} f(x)&amp;=x-f(x^2)=x-\big(x^2-f(x^4)\big)=x-x^2+\big(x^4-f(x^8)\big)=\cdots\\ &amp;=x-x^2+x^4-\cdots+(-1)^n\left(x^{2^n}-f\big(x^{2^{n+1}}\big)\right) \end{align*}  since $f(0)=0$ and $\displaystyle\lim_{n\to \infty}x^{2^{n+1}}=0$ for any $x\in(0,1)$,it follows by the continuity of $f$ that $\displaystyle\lim_{n\to \infty}f\big(x^{2^{n+1}}\big)=0$, hence $$f(x)=x-x^2+x^4-x^8+\cdots+(-1)^nx^{2^n}+\cdots$$ for any $x\in (0,1)$  Why do I have prove that there is exists such a continuous functions $f$? Maybe  my example is wrong? Why?  if my   method is wrong,then How prove this problem ? Thank you. LaTeX template for typesetting a novel Draw power from a portable mobile charger to power circuits