proof of 2 equations of Standard deviation

Question

i have following question:

i can not prove why the following equations are the same: $$ \begin{align} A=(a_j),\ 1 \leqslant j \leqslant N. \\\\ \mu = mean(A). \\\\ S_1 = \sqrt{ \frac{1}{N-1} \sum_{i=1}^{N} (a_i - \mu)^2 }. = \sqrt{ \frac{1}{N-1} \sum_{i=1}^{N} (a_i^2 - 2 a_i \mu + \mu^2) } ] \\\\ S_2 = \sqrt{ \frac{1}{N-1} \left[ \sum_{i=1}^{N} (a_i^2) - \frac{\left(\sum_{i=1}^{N} (a_i) \right)^2}{N} \right] } \\ \end{align} $$

why does $S_1 = S_2$?

Thanks.

Answer 1

\begin{align} S_1^2&=\frac{1}{N-1} \sum_{i=1}^{N} (a_i - \mu)^2 =\frac{1}{N-1} \sum_{i=1}^{N} (a_i^2-2\mu a_i +\mu^2)\\ &=\frac{1}{N-1} \left[\sum_{i=1}^{N} a_i^2 +\sum_{i=1}^{N} \mu^2-2\mu\sum_{i=1}^{N} a_i\right]=\frac{1}{N-1}\left[\sum_{i=1}^{N} a_i^2 +N \mu^2-2\mu N\left(\frac{1}{N}\sum_{i=1}^{N} a_i\right)\right]\\ &=\frac{1}{N-1}\left[\sum_{i=1}^{N} a_i^2 +N \mu^2-2N\mu^2 \right]=\frac{1}{N-1}\left[\sum_{i=1}^{N} a_i^2 -N\frac{\left(\sum_{i=1}^{N} a_i\right)^2}{N^2} \right]\\ &=\frac{1}{N-1}\left[\sum_{i=1}^{N} (a_i^2) -\frac{\left(\sum_{i=1}^{N} a_i\right)^2}{N} \right]=S^2_2 \end{align}