What value of $\alpha$ makes $\sum_{i=0}^n (x_i-\alpha)^2$ minimum?

Question

I know that $\overline{x}$ makes $\sum_{i=0}^n (x_i-\alpha)$ minimum. In fact it makes it zero. But how to find what value for $\alpha$ makes $\sum_{i=0}^n (x_i-\alpha)^2$ minimum ? what is the best approach?

Answer 1

If you like a physical interpretation, the parallel axis theorem clearly gives that the minimum is attained by the centroid $\alpha=\frac{x_0+x_1+\ldots+x_n}{n+1}$. That also follows from the fact that:

$$\frac{d}{d\alpha}\sum_{i=0}^{n}(x_i-\alpha)^2 = -2\sum_{i=0}^{n}(x_i-\alpha). $$

Answer 2

hint: write it as $f(\alpha) = (n+1)\alpha^2-2A\alpha + B$. And use property of parabola where the minimum attains at the "vertex"...