I have a task where I need to write the power series:
$\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^n$
differentiated 2 times where $x=0$.
This is what I have done so far:
$f'(x) = \sum\limits_{n=1}^\infty n\frac{(-1)^n}{(2n)!}x^{n-1} \\
f''(x) = \sum\limits_{n=1}^\infty n(n-1)\frac{(-1)^n}{(2n)!}x^{n-2} = \sum\limits_{n=1}^\infty\frac{((-1)^n (n – 1) n x^{n – 2})}{((2 n)!)}$
However I'm not quite sure how to proceed.
Update:
So I found the solution by replacing x with 0 which meant that the only term that wouldn't be undefined was 2 which resulted in the sollution:
$ f''(0) = \frac{(-1)^2(2-1)2}{(2*2)!} = \frac{2}{4!} = \frac{2}{4*3*2*1} = \frac{1}{4*3} = \frac{1}{12}$
Best Answer
It's easiest to see this when you write out the series.
For an arbitrary power series: $$ \begin{align} f(x) =& a_0 + a_1x + a_2x^2 + a_3x^3 + \dots \\ \\ f'(x) =& a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + \dots \\ \\ f''(x) =& 2a_2 + 6a_3x + 12a_4x^2 + 20a_5x^3 + \dots \\ \\ f''(0) =& 2a_2 \end{align} $$