The minimum of $\sum_{i=1}^m(c_ix-b_i)^2$

Question

We know the minimum of $(c_ix-b_i)^2$ is $$x_i^*=\frac{b_i}{c_i}$$

How to show that the minimum of $\sum_{i=1}^m(c_ix-b_i)^2$ with $c_i \neq0$ is
$$x^*=\frac{\sum_{i=1}^mc_ib_i}{\sum_{i=1}^mc_i^2}$$

I know $x^* \in [\text{min}_i \ \ x_i^*, \text{max}_i\ \ x_i^*]$

I have no idea to manipulate the summation.

Answer 1

$$\sum_{i=1}^{m}(c_ix-b_i)^2=Ax^2-2Bx+C$$ where $$A=c_1^2+c_2^2+\cdots+c_m^2$$ $$B=b_1c_1+b_2c_2+\cdots+b_mc_m$$ $$C=b_1^2+b_2^2+\cdots+b_m^2.$$ Now, note that $$\begin{align}Ax^2-2Bx+C&=A\left(x^2-\frac{2B}{A}x\right)+C\\&=A\left(\left(x-\frac BA\right)^2-\frac{B^2}{A^2}\right)+C\\&=A\left(x-\frac BA\right)^2-\frac{B^2}{A}+C\end{align}$$

So, the minimum is attained when $x=B/A$.

Answer 2

$$0=\frac{d}{dx}\sum_{i=1}^{m}(c_ix-b_i)^2 = \sum_{i=1}^{m}\frac{d}{dx}(c_ix-b_i)^2 = \sum_{i=1}^{m}2c_i(c_ix-b_i)= 2x\sum_{i=1}^{m}c_i^2 - 2\sum_{i=1}^{m}b_ic_i$$

Now just isolate $x$ and you're done

Answer 3

Let $f(x)=\sum_{i=1}^m(c_ix-b_i)^2$.

Then it is clear, that $f(x)\in \textrm{D}^1(\mathbb{R}).$

It is well-known that in this case $f'(x^*)=0$. If you compute $f’(x)$ you will see, that equation $f’(x)=0$ has only one root…