$E(x)= |x|-|x+1|+|x+2|-|x+3|+\dots+|x+2016|$. 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. 
 A: If you look at the graph of $E(x)= |x|-|x+1|+|x+2|-|x+3|+\dots+|x+n|$, you can notice that if $n$ is even (our case), then the graph of $E(x)$ is symmetric to the line $x=-\frac{n}{2}$. In fact here the graph of $E(x)= |x|-|x+1|+|x+2|-|x+3|+|x+4|$:

To find the minimum value I have to set $x\geq0$, so: $$E(x)= |x|-|x+1|+|x+2|-|x+3|+\dots+|x+2016|=x-x-1+x+2-x-3+\dots+x+2016$$ Removing absolute values, I obtain: $2+4+\cdots+2016-(1+3+5+\cdots+2015)=A-B$. I have now two aritmetic series; their values are:$A=\frac{2016}{2}\cdot(4+1007\cdot2)=1008\cdot2018$ and $B=\frac{2016}{2}\cdot(2+1007\cdot2)=1008\cdot2016$. 
Now I obtain: $A-B=1008\cdot2018-1008\cdot2016=1008\cdot2=2016$.
I hope it has helped you...
A: 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.
A: Calling $f(x) = |x|-|x+1|$ we have
$$
E_n(x) = \sum_{k=0}^{k=n}f(x+2k)
$$
and
$$
-(n+1) \le E_n(x) \le n+1
$$
