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 \(x=t_{50}\).

No vote yet

4 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestLet \(x=m \in \mathbb{R}\) 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 · 4 years ago

Log in to reply

– Mridul Sachdeva · 4 years ago

How is the assumption \( t_1 < m < t_{99} \) obvious? I mean I know its correct but can we assume that?Log in to reply

– Slice A Lot · 4 years 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.Log in to reply

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 · 4 years ago

Log in to reply

– Mridul Sachdeva · 4 years ago

Nice use of the triangle inequality. Thanks!Log in to reply

Intuitively it seems correct but I have not been able to prove it. – Mridul Sachdeva · 4 years ago

Log in to reply