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Given $f(a,b)=\sum_{k=1}^n(ax_k+b-y_k)^2$ where $x_k$ and $y_k$ are arbitrary real numbers such that $\exists i,k:x_i \ne x_j$.

Show that $f(a,b)$ gets minimized at exactly one point.

I've managed to show that the partial derivatives vanish at exactly one point $(a_0,b_0)$.

I then took the set $K=f^{-1}([f(a_0,b_0),f(a_0,b_0+\varepsilon)]$, and I'm trying to show that $K$ is compact which will finish the proof.

It's clear that $K$ is closed, to show that it is bound I tried to use the fact that $(a,b)\in K\implies |f(a,b)-f(a_0,b_0)| <\varepsilon$ to somehow bound $a-a_0$ and $b-b_0$, but to no avail...

Does anyone know how to bound this set, or maybe have some better idea?

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up vote 1 down vote accepted

Well, I found a better solution.

I won't go into to many details as this didn't get a lot of attention, but taking increasingly large cocentric spheres closed around $(a_0,b_0)$ compactness reveals that each sphere either attains a minimum in $(a_0,b_0)$ or on the boundary.

Assume BWoC all minima are attained on the respective boundaries, and choose a minimal point for each sphere, then this sequence is positive decreasing and thus converges.

It is easy to prove on the other hand that for any unbound sequence $x_n$ the sequence $f(x_n)$ is also unbound (a direct result of the existence of $i$ and $j$ s.t. $x_i\ne x_j$) and reach a contradiction, getting that one of said spheres attains a minimum at $(a_0,b_0)$ as needed.

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It might better to compute the Jacobian, i.e. the two by two matrix of second partials of $f(a,b)$ w.r.t. the variables $a,b$. In this case it is independent of $a,b$ and comes out $$ \sum_{k=1}^n x_k^2 - \sum_{k=1}^n x_k \cdot \sum_{k=1}^n x_k$$ which is $n^2$ times the variance of the $x_k$. This is known to be positive, so that your only critical point is a relative minimum, and all you need now is to make sure it is also the absolute minimum.

ADDED NOTE: The function $f(a,b)$ being minimized is the sum of squared errors for the linear regression of the $y_k$ as function of the $x_k$. So you could also look up "linear regression" and find various proofs that this is indeed the absoloute minimum. The values $a,b$ are slope and intercept of best fit line.

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Don't you mean Hessian? – Shai Deshe Dec 10 '12 at 1:08

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