TITLE: Find $f^{(1001)}(0)$
QUESTION [6 upvotes]: I am to find the value in 0 of 1001th derivative of the function $$f(x) = \frac{1}{2+3x^2}$$
How should I approach this kind of problem? I tried something like : $$\frac{1}{2+3x^2} = \frac{1}{2}\cdot\frac{1}{1-(-\frac{3}{2}x^2)}= \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{3x^2}{2}\right)^n$$ and compare whats next to $x^{1001}$ in this sum and in MacLaurin series but damn, we've got only even powers of x here. How should I do that?

REPLY [3 votes]: You’re missing a minus sign:
$$f(x)=\frac12\sum_{n\ge 0}(-1)^n\left(\frac32\right)^nx^{2n}\;.$$
Thus, 
$$f^{(1001)}(x)=\frac12\sum_{n\ge 0}(-1)^n\left(\frac32\right)^n(2n)^{\underline{1001}}x^{2n-1001}\;,$$
where $$(2n)^{\underline{1001}}=(2n)(2n-1)\ldots(2n-1001+1)=(2n)(2n-1)\ldots(2n-1000)$$ is a falling factorial. This is zero if $n\le 500$, so
$$\begin{align*}
f^{(1001)}(x)&=\frac12\sum_{n\ge 501}(-1)^n\left(\frac32\right)^n(2n)^{\underline{1001}}x^{2n-1001}\\\\
&=\frac12\left(-\left(\frac32\right)^{501}x+\left(\frac32\right)^{502}x^3-\ldots\right)\;,
\end{align*}$$
and $f^{(1001)}(0)=0$.
Of course you don’t actually have to do all of this calculation: the fact that the original series has only even powers means that any odd-order derivative must have only odd powers and hence a constant term of $0$.