1. ## proving question?

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?

2. I see in your post a statement of a function and a statement of the Taylor polynomial, along with some conclusions about derivatives.

But what are you supposed to be proving?