proving question?

**noob mathematician** · April 6th 2009, 07:43 AM

Let $f(t)= e^{t^2/2}$
by taylor series: $f(t)= \sum\limits_{j = 0}^\infty \frac{(t^2/2)^j}{j!}$ = $\sum\limits_{j = 0}^\infty \frac{t^{2j}}{2^jj!}$
So $f'(0)= 0$
$f''(0)= 1$
$f'''(0)=0$ ... where $f^{(n)}(0)=0$ when n is odd and $f^{(n)}(0)=1$ when n is even.
Differentiating $\frac{t^{2j}}{2^jj!}$ w.r.t (t) = $\frac{2jt^{2j-1}}{2^jj!}$
when n=2,t =1, give $\frac{2j}{2^jj!}$
when n=4, t=1, give $\frac{(2j)(2j-1)}{2^jj!}$
and so on
This is quite obvious but i have difficulty putting in word to show that
$f^{(n)}(0)= 0$ when n is odd and
$f^{(n)}(0)= \frac{(2j)!}{2^jj!}$ when n=2j
Any better way to show?

**stapel** · April 6th 2009, 07:51 AM

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?

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