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Is there a better way to do optional function parameters in JavaScript? I've always handled optional parameters in JavaScript like this:  function myFunc(requiredArg, optionalArg){   optionalArg = optionalArg || 'defaultValue';    // Do stuff }   Is there a better way to do it?  Are there any cases where using || like that is going to fail? How can I understand and prove the "sum and difference formulas" in trigonometry? Fill the area under minimum of two functions using tikz and pgfplots It would be nice to see what users have recently been active. This is somewhat standard with many forum-style websites, showing how active the site has been in the last n-minutes or so. As a functional advantage, this also serves to let you know when is a good time to get a question in. Google Maps Location History is showing an incorrect location Solving 4 Quadratic Simultaneous Equations Consider the equations:   $$\begin{cases} ab+c+d=3\\ bc+d+a=5\\ cd+a+b=2\\ da+c+b=6 \end{cases}$$   Where $a,b,c,d \in \mathbb{R}$. How can we find $a,b,c,d$?   The furthest I've got is by adding the first two equations and the last two equations together to get: $ab+bc+c+d+d+a=(b+1)(a+c)+2d=8$ and $(d+1)(a+c)+2b=8$ We can rearrange these (assuming $b,d \neq -1$) to get $$\frac{8-2d}{b+1}=\frac{8-2b}{d+1} \implies -d^{2}+3d+4=-b^{2}+3b+4$$ Which leads us further to $$b^{2}-d^{2}-3(b-d)=0 \implies (b-d)(b+d-3)=0$$ Now since $b=d$ gives us an absurdity (equations 1 and 4 reduce to 3=6), we must have $b+d=3$. This is as far as I can get unfortunately, since trying the same approach with the other pairs doesn't work in the same way.  Help would be much appreciated; the problem is from the BMO1 2004 paper. Net work done on the body when we lift it and put it on the table is zero? Here is my candidate for  the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists. I would be interested in seeing others.  $***$ Added after some comments: I prove here by very elementary means that the limit exists. Calling the limit "$e$" names it. $***$  It only needs Bernoulli's inequality (BI) in the form $(1+x)^n \ge 1+nx$ for $x &gt; -1$ and $n$ a positive integer, with equality only if $x = 0$ or $n = 1$. This is easily proved by induction: It is true for $n=1$, and $(1+x)^{n+1} = (1+x)(1+x)^n \ge (1+x)(1+nx) = 1+(n+1)x+nx^2 \ge 1+(n+1)x $. (If $-1 &lt; x &lt; 0$, if $1+nx \ge 0$, the above proof goes through, and if $1+nx &lt; 0$, $1+mx &lt; 0$ for all $m \ge n$ so certainly $(1+x)^m &gt; 1+mx$.)  This proof originally appeared in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 and uses the arithmetic-geometric mean inequality (AGMI) in the form $\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$  (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.  Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.  We will prove that $a_n$ is an increasing sequence  and $b_n$ is an decreasing sequence.  Since $a_n &lt; b_n$, this implies,  for any positive integers $n$ and $m$ with $m &lt; n$  that $a_m &lt; a_n &lt; b_n &lt; b_m$.  For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.  Since their sum is $n+2$ and their product is $(1+1/n)^n$, by the AGMI,  $((n+2)/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $(1+1/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $a_{n+1} &gt; a_n$.  For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.  Since their sum is $n$ and their product is $(1-1/n)^n$, by the AGMI,  $(n/(n+1))^{n+1} &gt; (1-1/n)^n$  or $(1+1/n)^{n+1} &lt; (1+1/(n-1))^n$,  or $b_n &gt; b_{n+1}$.  Since $b_n-a_n  = (1+1/n)^{n+1} - (1+1/n)^n = (1+1/n)^n(1/n) =a_n/n $ and every $a_n$ is less than any $b_n$ and $b_3 = (1+1/3)^4 = 256/81 &lt; 4$, $b_n-a_n &lt; 4/n$, so $b_n$ and $a_n$ converge to a common limit.  These proofs do not seem to be really elementary,  since they use the AGMI.  However, they use a special form of the AGMI,  where all but one of the values are the same,  and this will now be shown to be implied by BI,  and thus be truly elementary.   Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.  The AGMI for these values is  $(((n-1)u+v)/n)^n \ge u^{n-1}v$  with equality if and only if $u = v$.  We will now show that this is implied by BI: $(((n-1)u+v)/n)^n \ge u^{n-1}v$  is the same as  $(u+(v-u)/n)^n \ge u^n(v/u)$.  Dividing by $u^n$,  this is equivalent to  $(1+(v/u-1)/n)^n \ge v/u$.  By BI,  since $(v/u-1)/n &gt; -1/n &gt; -1$, $(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$  with equality only if $n=1$ or $v/u-1 = 0$. Thus BI implies this version of the AGMI. What formats supporting animation are suitable for the web? What formats supporting animation are suitable for the web and what are the pros and cons of each. I have a bit of code where I've written a SOQL query that returns a list of SObject records, however I need a Set&lt;Id&gt; structure to pass to another method that I don't have control of (and thus can't change the method signature of).  How can I turn my returned List&lt;SObject&gt; into a Set&lt;Id&gt;? Is the best option just a for loop? Sum of nilpotent and element is a unit in ring? I can't tell where I went wrong, am still a beginner.    Let N be a nilpotent in a commutive ring , let X be any element.   I'll will be showing X + N is a unit .   Assume Y(X+N) =1   Then YX(N^n-1) =  N^n-1   Multiplying both side by  X+N   We get X(N^n-1) =  X(N^n-1)  Thus nilpotent summed with any element gives unit. Counting points in shapefile within polygons of another shapefile using ArcGIS Desktop? I'm creating a map with two shapefiles:   A polygon of counties Points of cities   I used the function Intersect to group the cities in each county.  I'm trying to get the number of cities in each county to show up in the legend.    For example, in county A there are 16 cities, in county B there are 7 cities and so on.  I'm trying to create a choropleth map that shows the number of cities in each county. persist port routing from 80 to 8080 I use amazon EC2 instance which works via ubuntu. By default according security restrictions I can't bin my application to port 80, so I just bind it port 8080 and then set routing redirect from port 80 to 8080 via the following command:  iptables -t nat -A PREROUTING -p tcp --dport 80 -j REDIRECT --to 8080   But I found that when I reboot the server this settings no longer active untill I invoke this command again.  So my question is how to enable port's redirect work even if system was rebooted? How to evaluate $\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta) d\theta$? For $\alpha \in \mathbb{R}$, define $\displaystyle I(\alpha):=\int_{0}^{2\pi}e^{\alpha \cos \theta}\cos(\alpha \sin \theta)\; d\theta$. Calculate $I(0)$. Hence evaluate $\displaystyle\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta)\; d\theta$.  Hint: To evaluate the integral that expresses $\displaystyle\frac{dI}{d\alpha}$, consider $\displaystyle\frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta))$.  How do I do this question? I think this might have something to do with the Fundamental Theorem of Calculus, but I'm not sure.  I computed $\displaystyle I(0)=\int_{0}^{2\pi} d\theta=2 \pi$, and $\displaystyle I(1)=\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta) d\theta$. Following the hint I get  $$\begin{align} \frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) &amp; =\alpha e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + e^{\alpha \cos \theta}\cos(\alpha \sin \theta) \alpha \cos \theta \\ &amp; =  \alpha e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + \frac{dI}{d \alpha} \cos \theta. \\ \end{align}$$  Is this correct so far?  The answers in the question referred as a duplicate does not help. I'm in a course dealing with real values, not complex. If I know for two numbers a and b that    $${a &lt; b }$$  Then is it correct to say that    $$ a \leq b $$  I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information. Conditional distributions of the multivariate normal What exactly are the Python scoping rules?  If I have some code:  code1 class Foo:    code2    def spam.....       code3       for code4..:        code5        x()   Where is x found?  Some possible choices include the list below:   In the enclosing source file In the class namespace In the function definition In the for loop index variable Inside the for loop   Also there is the context during execution, when the function spam is passed somewhere else. And maybe  pass a bit differently?  There must be a simple reference or algorithm somewhere.  It's a confusing world for intermediate Python programmers. Mod of numbers with large exponents [modular order reduction] I've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:   $13^{100} \bmod 7$ $7^{100} \bmod 13$   I've also heard of this formula:  $$a \equiv b\pmod n \Rightarrow a^k \equiv b^k \pmod n $$  But I don't see how exactly to use that here, because from $13^1 \bmod 7$ I get 6, and $13^2 \bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)  Any hints or pointers in the right direction would be great. Exploiting JavaScript's eval() method Shorter way to say "conversational partner" I'm looking for a proper English word to say "conversational partner".   Is "interlocutor" a word an average person would understand?  Are there any other words or shorter expressions which could describe a person you are having a conversation with, but who is not your friend and is basically unknown to you?