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 $x=t_{50}$ .

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Comments

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 $m=t_{50}$ .

Tim Vermeulen - 7 years, 10 months ago

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

Mridul Sachdeva - 7 years, 10 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.

Slice A Lot - 7 years, 10 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 $x = t_{50}$ .

Jimmy Kariznov - 7 years, 10 months ago

Nice use of the triangle inequality. Thanks!

Mridul Sachdeva - 7 years, 10 months ago

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

Mridul Sachdeva - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$