Why does this sum of absolute values resemble a parabola?

Question

I was fooling around with a graphing calculator and I found that $\sum _{n=1} ^k |x+n|$ (where $k$ is a parameter) bore a strange resemblance to the parabola $(x+51/2)^2+5^4$. Namely, it was tangent to the parabola at exactly one point with $x$ coordinate in every interval between every integer.
I am sure that if I fiddled around with the arguments of the sum and its vertical translation, I could get them to line up at the origin, but I see no point to it. My main question is: Why do these curves resemble each other so much?

Answer 1

Consider $$\sum_{n=1}^{50}|x+n|$$

We want to look at $x=-50$ to $x=-1$ as this is where it is similar to a parabola. Outside of this range it is a straight line, because:

If $x>-1$ then $x+n>0$ for all $n$ in the sum so all the absolute values signs are positive and it becomes $50x+1275$.

Similar if $x<-50$ then $x+n<0$ for all $n$ in the sum so all the absolute values signs reverse the sign of the expression and it becomes $-50x-1275$.

for a value of $x\in[-50,-1]\cap\mathbb{Z}$ we have two (three) cases:

• for all $n<-x$ then $|x+n|<0$
• and for all $n>-x$ then $|x+n|>0$
• (trivially if $n=-x$ then $x+n|=0$)

so the sum breaks into two parts:

$$\sum_{n=1}^{-x}|x+n|+\sum_{n=-x}^{50}|x+n|$$

$$=\sum_{n=1}^{-x}(-x-n)+\sum_{n=-x}^{50}(x+n)$$

$$=\frac{1}{2}x(x+1)+\frac{1}{2}(x+50)(x+51)$$

$$=x^2+51x+1275$$

$$=\left(x+\frac{51}{2}\right)^2+\frac{2499}{4}$$

$$=\left(x+\frac{51}{2}\right)^2+625-\frac{1}{4}$$

Note: You could replace $50$ with $k$ and do similar mathematics to get a parabola of:

$$\left(x+\frac{k+1}{2}\right)^2+\frac{k^2-1}{4}$$

Answer 2

$\sum _{n=1} ^k |x+n|=$

$\sum_{n=1; n < -x}(-x-n) + \sum_{n = \max(1,\lceil -x \rceil)}^kx + n=$

$\sum_{n=1}^{\lfloor -x\rfloor}n + \sum_{n=1}^{k-\lceil -x \rceil}n=$

$\frac{(\lfloor -x\rfloor)(\lfloor -x\rfloor + 1)}2 + \frac{(k-\lceil -x \rceil)(k-\lceil -x \rceil+1)}2$

$\approx \frac 12[(x^2 - x) + (k-x)^2+(k-x)$

Which is basically a parabola.