# Interesting Modulus problem. Kindly Help!

Let $$t_1 < t_2 < t_3 < \dots < t_{99}$$ be real numbers. Consider a function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ given by $$f(x) = |x-t_1|+|x-t_2|+ \dots +|x-t_{99}|$$. Show that $$f(x)$$ attains minimum value at $$x=t_{50}$$.

Note by Mridul Sachdeva
4 years, 11 months ago

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Let $$x=m \in \mathbb{R}$$ be the value such that $$f(x)$$ is minimized. Obviously, $$t_1<m<t_{99}$$. Then

$\lvert m - t_1 \rvert + \lvert m - t_{99} \rvert = (m-t_1) + (t_{99} - m)=t_{99} - t_1,$

which is independant of $$m$$, so $$t_1$$ and $$t_{99}$$ can be disregarded. This process of omitting the smallest and largest value continues, until only the middle one is left, which is $$t_{50}$$, so $$m=t_{50}$$.

- 4 years, 11 months ago

How is the assumption $$t_1 < m < t_{99}$$ obvious? I mean I know its correct but can we assume that?

- 4 years, 11 months ago

$$f(x)$$ is the sum of distances from $$x$$ to $$t_i$$. If for example $$x>t_{99}$$, then moving $$x$$ to the left decreases all distances.

- 4 years, 11 months ago

This is similar to what Tim V. did, but perhaps a bit more rigorous.

By triangle inequality, $$|x-t_{k}| + |x - t_{100 - k}| \ge t_{100-k} - t_{k}$$ for $$k \in \overline{1,49}$$ with equality iff $$t_k \le x \le t_{100-k}$$.

Trivially, $$|x - t_{50}| \ge 0$$ with equality iff $$x = t_{50}$$.

Adding these up gives $$f(x) = |x - t_{50}| + \displaystyle\sum_{k = 1}^{49} \left(|x-t_{k}| + |x - t_{100 - k}|\right) \ge \displaystyle\sum_{k = 1}^{49} \left( t_{100-k} - t_{k} \right)$$.

Equality occurs iff $$t_k \le x \le t_{100-k}$$ for all $$k \in \overline{1,49}$$ and $$x = t_{50}$$.

Clearly, $$x = t_{50}$$ satisfies those conditions, and is the only real number to do so.

Therefore, $$f(x)$$ attains its minimum value at $$x = t_{50}$$.

- 4 years, 11 months ago

Nice use of the triangle inequality. Thanks!

- 4 years, 11 months ago

Intuitively it seems correct but I have not been able to prove it.

- 4 years, 11 months ago