Suppose X is a random variable. Is there a way to find $\min\{E|X-c|, c \in \mathbb{R}\}$?  What have I attempted to do so far:$\DeclareMathOperator{\Pr}{Pr}$ $$\Pr(|X - c| \leq t) = \begin{cases} 0 &amp;  \text{if } \;t \leq 0\newline\Pr(X \leq t + c) - \Pr(X \leq c - t)&amp; \text{if } t &gt; 0 \end{cases}$$ That results in the probability density function of $|X - c|$ being  $$P_{|X - c|}(t) =  \begin{cases} 0 &amp;  \text{if } t \leq 0,\cr P_{X}(t + c) + P_{X}(c - t)&amp;   \text{if } t &gt; 0,\end{cases}$$ where $P_X$ is the probability density function of $X$. That means, that $$E|X-c| = \int_0^\infty t(P(t + c) + P(c - t))\,dt.$$ However this gives me nothing, as finding $\min\{\int_0^\infty t(P_X(t + c) + P_X(c - t))\,dt,\; c \in \mathbb{R}\}$ does not seem to be any easier.  Any help will be appreciated.
Suppose $X$ is a real-valued random variable and let $P_X$ denote the distribution of $X$. Then $$ E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x). $$  of $X$ are defined as any number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$.  Why do the medians solve $$ \min_{c \in \mathbb{R}} E(|X-c|) \, ? $$
Is it possible to install an earlier version of Skype on Windows 8? Windows 8 skype is terrible.  They've removed many buttons, most of the preferences, and have in general made it impossibly oversimplified and difficult to use.  Old Skype was fine when I had windows 7.  Is there any way to get it back on Windows 8? Using the method of characteristics, find a solution to Burgers' equation    \begin{cases} u_t+\left(\frac{u^2}{2} \right)_x =0 &amp; \text{in }\mathbb{R}\times(0,\infty) \\ \qquad \qquad \, \, u=g &amp; \text{on } \mathbb{R} \times\{t=0\} \end{cases}   with the initial conditions    $$g(x)=\begin{cases} 0 &amp; \text{if }x &lt; 0 \\ 1 &amp; \text{if }0 \le x \le 1 \\ 0 &amp; \text{if }x &gt; 1 \end{cases}$$   First, I realized that the equation $u_t+\left(\frac{u^2}{2} \right)_x =0$ is equivalent to this form: $$u_t+uu_x =0$$  Then should I generally follow the method of solution as outlined in the answer of ?  Note that this is not a duplicate question of that page. Rather I want to know if that page can be used for my problem, even though my ICs are different.  By the way, the solution printed in my book (PDE Evans, 2nd edition, page 142) is      $$u(x,t) =  \begin{cases} 0 &amp; \text{if } x &lt; 0 \\  \frac xt &amp; \text{if } 0 &lt; x &lt; t  \\ 1 &amp; \text{if } t &lt; x &lt; 1 + \frac t2  \\ 0 &amp; \text{if } x &gt; 1 + \frac t2 \tag{$0 \le t \le 2$}  \end{cases}$$ Do lens hoods need to be matched to focal length? Is it true that not all generic lens hoods work on all lenses (provided the filter diameter matches, of course)? I read that the focal length plays a role in it - but I cannot find such readings on any lens hoods I see available online. Usage and example of the word “litotes” I've come across the word litotes, which means a rhetorical understatement. However, I’m having trouble understanding how to use it in colloquial English. Could someone please give an example? Proving $\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$ I am being asked to prove that $$\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$$ I have some progress made, but I am stuck and could use some help. What I did: It holds that $$\sum\limits_{k=0}^{n}\cos(kx)=\sum\limits_{k=0}^{n}Re(\cos(kx))=\sum\limits_{k=0}^{n}Re(\cos(x)^{k})=Re(\sum\limits_{k=0}^{n}\cos(x)^{k})=Re\left(\cos(0)\cdot\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right) $$ For any $z_{1},z_{2}\in\mathbb{C}$ we have it that if $z_{1}=a+bi,z_{2}=c+di$ then $$\frac{z_{1}}{z_{2}}=\frac{z_{1}\overline{z2}}{|z_{2}|^{2}}=\frac{(a+bi)(c-di)}{|z_{2}|^{2}}=\frac{ac-bd+i(bc-ad)}{|z_{2}|^{2}}$$ hence $$Re\left(\frac{z_{1}}{z_{2}}\right)=\frac{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}{|z_{2}|^{2}}$$ Thus, $$Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=\frac{(\cos(nx)-1)(\cos(x)-1)-\sin(nx)\sin(x)}{(\cos(x)-1)^{2}+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{\cos^{2}(x)-2\cos(x)+1+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2\cos(x)+2}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(\cos(x)-1)}= \frac{=\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(-2\cdot\sin^{2}(x/2))}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(x(n+1))-\cos(nx)-\cos(x)+1}{4\sin^{2}(x/2)} $$ This is the part where I am stuck, I would appriciate any help or hint on how to continue. Edit: Given the corrections by André I get: $$(\cos(nx+x)-1)(\cos(x)-1)+\sin(nx+x)\sin(x)=\cos(nx+x)\cos(x)-\cos(nx)-\cos(x)+1+\sin(nx+x)\sin(x)$$ so $$\cos(nx+x)\cos(x)+\sin(nx+x)\sin(x)=\cos(xn+x-x)-\cos(nx)=0$$ Edit 2: I found anoter mistake in the above, I will try to correct Edit 3: When multiplying correctly the above it works out :-) How to check whether a string contains a substring in JavaScript? $\max(\{n: n+5 | n^5 + 5\})$ What is the maximum value of $n$ if $n + 5$ divides $n^5 + 5$, $n$ being a natural number? I tried to solve by binomial theorem but it failed. Removal of lines with no more or fewer than 'N' fields? Since the version of the word "that" might matter to the correct answer, I'll provide the actual title in question:  Title:  Existing Solutions that Didn't Work  If someone could explain the why of the answer, I would be appreciative.  EDIT:  It has been suggested that this is a duplicate, and that may be, in a sense.  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