Let $\displaystyle f(t)= e^{t^2/2}$

by taylor series: $\displaystyle f(t)= \sum\limits_{j = 0}^\infty \frac{(t^2/2)^j}{j!}$= $\displaystyle \sum\limits_{j = 0}^\infty \frac{t^{2j}}{2^jj!}$

So $\displaystyle f'(0)= 0$

$\displaystyle f''(0)= 1$

$\displaystyle f'''(0)=0$... where $\displaystyle f^{(n)}(0)=0$ when n is odd and $\displaystyle f^{(n)}(0)=1$ when n is even.

Differentiating $\displaystyle \frac{t^{2j}}{2^jj!}$ w.r.t (t) = $\displaystyle \frac{2jt^{2j-1}}{2^jj!}$

when n=2,t =1, give $\displaystyle \frac{2j}{2^jj!}$

when n=4, t=1, give $\displaystyle \frac{(2j)(2j-1)}{2^jj!}$

and so on

This is quite obvious but i have difficulty putting in word to show that

$\displaystyle f^{(n)}(0)= 0$ when n is odd and

$\displaystyle f^{(n)}(0)= \frac{(2j)!}{2^jj!}$ when n=2j

Any better way to show?