Why is $\nabla f(p) = \sum_{i=1}^n 2(p - p_i) = 0$?

Question

Given $n$ points in space, $p_1,p_2, \cdots , p_n$, find the point $p$ for which $f(p) = \sum_{i=1}^n |p-p_j|^2$.

I have the following in my notes:

(1) $f(p) = \sum_{i=1}^n |p-p_i|^2 = \sum_{i=1}^n (p-p_i)\cdot(p-p_i)$

=$\sum_{i=1}^n p \cdot p - 2p \cdot p_i + p_{i} \cdot p_{i}$

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(2) Find $p$ to minimize $f(p)$.

$\nabla f(p) = \sum_{i=1}^n 2(p - p_i) = 0$

$np = \sum_{i=1}^n p_i \rightarrow \frac{1}{n} \sum_{i=1}^n p_i = \bar{p}$

$\bar{p}$ is a critical point for $f(p)$

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(3) $D^2 f = 2I_n$

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(4) $f(p)-f(\bar{p}) = \sum_{i=1}^n |p-\bar{p_i}|^2 - |\bar{p} - p_i|^2$

= $\sum_{i=1}^n |p|^2 -2p \cdot p_i + |p_i|^2 - |\bar{p}|^2 +2\bar{p} \cdot p_i - |p_i|^2$

= $\sum_{i=1}^n |p|^2 - |\bar{p}|^2 + 2(\bar{p}-p) \cdot p_i$

=$2\bar{p} \cdot p_i = \bar{p} \cdot (\sum p_i) \cdot 2 = \bar{p} \cdot n\bar{p} \cdot 2 = 2n|\bar{p}|^2$

= $n(|p|^2 + |\bar{p}|^2 -2np\cdot \bar{p} = n(|p-\bar{p}|^2) \ge 0$

I find that part (1) is self-explanatory.

For part (2), why do we do $\nabla f(p) = \sum_{i=1}^n 2(p \cdot p_i) = 0$?

Answer 1

Take

$$ f({\bf p}) = \sum_i \left({\bf p}\cdot {\bf p} - 2{\bf p}\cdot {\bf p}_i + {\bf p}_i\cdot {\bf p}_i\right) $$

The gradient is just

\begin{eqnarray} \nabla f({\bf p}) &=& \frac{\partial}{\partial {\bf p}}\sum_i \left({\bf p}\cdot {\bf p} - 2{\bf p}\cdot {\bf p}_i + {\bf p}_i\cdot {\bf p}_i \right) \\ &=& \sum_i \left( 2{\bf p} - 2{\bf p}_i \right) = 2\sum_i {\bf p} - 2\sum_i{\bf p}_i \\ &=& 2n {\bf p} - 2\sum_i {\bf p}_i = 2n \left({\bf p}-\frac{1}{n}\sum_i{\bf p}_i\right) \\ &=& 2n ({\bf p} - \overline{{\bf p}}) \end{eqnarray}