Find solution set of $200x=\sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}$

Question

Let $f(x)=\sum\nolimits_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}$. How can I find all the roots when $f(x)=200x$?

I first tried to use the Triangle Inequality as $|a|+|b|>|a+b|$ but this doesn't make sense. I then used a graphing device and found the solution set is $\{x:x\geq 100\}$.

My question: is there any shortcut to attack such a problem ?

Answer 1

If you are looking for

$$200x = \sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}$$

then you can notice that $|x+k| \geq 0$ and $|x-k| \geq$ $0$ for all $k$ when $x \geq 100$ so you have your equality.

When $x \leq 0$ instead you can't have equality because lhs is negative or zero while rhs is positive. When $0 < x < 1$, instead you have the rhs greater than the lhs.

Remains to investigate $1 < x< 100$, say you have $q \leq x < q+1$ with $q \in \{1,2,...,99\}$, then you can rewrite the rhs as

$$\sum_{k=1}^{q}{\left( \left| x-k \right|+\left| x+k \right| \right)} + \sum_{k=q+1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)} = 2qx +100\cdot101-q(q+1)$$

but

$$200x < 2qx +100\cdot101-q(q+1) \iff x < \frac{q+101}{2}$$

and this is always the case when $q \leq x < q+1$.

So it's proven that the solution set is $\{x:x\geq 100\}$.

Answer 2

Clearly, $x\geq 0.$ Rewrite the equation $\;200x = \sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}$ as $$\sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right|-2x \right)}=0$$ Consider two possible cases:

1. If $\;x\geq k\;$ for some $k,\;$ then $\;| x-k |+| x+k |-2x=0.$
2. If $\;0 \leq x < k\;$ for some $k,\;$ then $\;| x-k |+| x+k |-2x=2k-2x>0.$

From 1. and 2. it follows that
$$\sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right|-2x \right)}=0 \iff \big(x\geq k \;\;\text{for all} \;\; k\big),$$ that is, if $x\geq 100.$