$E(x)= |x|-|x+1|+|x+2|-|x+3|+\dots+|x+2016|$. Find the minimum value of $E$. Can any one help me finding the minimum value of the following expression:
$$
E(x)= |x|-|x+1|+|x+2|-|x+3|+\dots+|x+2016|
$$
where $x$ is a real number. 

Note that $$\begin{align}E(x+2)&=|x+2|-|x+3|
\mp\cdots+|x+2016|-|x+2017|+|x+2018|\\&=E(x)-|x|+|x+1|-|x+2017|+|x+2018|\end{align} $$
and $$|x+1|-|x|=\begin{cases}1&x\ge 0\\-1&x\le -1\\2x+1&-1\le x\le0\end{cases} $$
$$|x+2017|-|x+2018|=\begin{cases}1&x\ge -2017\\-1&x\le -2018\\2(x+2017)+1&-2018\le x\le2017\end{cases} $$
From this, $$\tag1E(x+2)\le E(x) \qquad \text{for }x\le -1$$
$$\tag2E(x+2)\ge E(x) \qquad \text{for }x\ge -2017$$
so that the minimum of $E$ on the interval $[-2017,-1]$ (which exists because $E$ is continuous and the interval is compact) is also the global minimum of $E$. Combining $(1)$ and $(2)$, 
$$\tag3E(x+2)= E(x) \qquad \text{for }-2017\le x\le -1$$
so that we can look for the minimum on any interval of length $2$ within $[-2017,-1]$, for example on $[-3,-1]$. There,
$$ E(x)=|x|-|x+1|+|x+2|-(x+3)+(x+4)\mp\cdots+(x+2016)=|x|-|x+1|+|x+2|+const$$
and the rest is easy.
