<s>[INST] Hundredth Derivative From Taylor's polynomial for $\frac{x^2}{1+x^4}$ I'm trying to solve this problem:

Find the hundredth derivative (at $x=0$) from Taylor's polynomial for $\dfrac{x^2}{1+x^4}$.

I keep getting the wrong answer; can someone help?

I have tried to solve this by extracting the coefficient of $x^{100}$ in the sum
$$x^2\cdot\sum_{i=0}^{\infty}(-x^4)^i.$$
I see that the coefficient is $1$ and since, by Taylor's formula, this should be $1=\frac{f^{(100)}(0)}{100!}$ I get that the hundredth derivative is $100!$, but this doesn't appear to be correct.
 [/INST] Notice that $x^4+1=\Phi_8(x)$, i.e. the roots of $x^4+1$ are the primitive eighth roots of unity $\frac{\pm 1\pm i}{\sqrt{2}}$.
The residue theorem or equivalent techniques give:
$$\begin{eqnarray*} f(x)&=&\frac{x^2}{x^4+1}\\ 
&=& \frac{1-i}{4\sqrt{2}}\cdot\frac{1}{x-\frac{1+i}{\sqrt{2}}}+\frac{1+i}{4\sqrt{2}}\cdot\frac{1}{x-\frac{1-i}{\sqrt{2}}}-\frac{1+i}{4\sqrt{2}}\cdot\frac{1}{x-\frac{-1+i}{\sqrt{2}}}-\frac{1-i}{4\sqrt{2}}\cdot\frac{1}{x-\frac{-1-i}{\sqrt{2}}} \end{eqnarray*}$$
hence $f^{(100)}(x)$ equals:
$$-100!\cdot\left[\frac{1-i}{4\sqrt{2}}\cdot\frac{1}{\left(x-\frac{1+i}{\sqrt{2}}\right)^{101}}+\frac{1+i}{4\sqrt{2}}\cdot\frac{1}{\left(x-\frac{1-i}{\sqrt{2}}\right)^{101}}-\frac{1+i}{4\sqrt{2}}\cdot\frac{1}{\left(x-\frac{-1+i}{\sqrt{2}}\right)^{101}}-\frac{1-i}{4\sqrt{2}}\cdot\frac{1}{\left(x-\frac{-1-i}{\sqrt{2}}\right)^{101}}\right].$$
Since $f(x)$ is an analytic function and the radius of convergence of its Taylor series at $x=0$ is one,
$$ f(x) = \frac{x^2}{1+x^4} = x^2-x^6+x^{10}-x^{14}+\ldots = \sum_{n\geq 0}(-1)^{n} x^{4n+2} $$
for any $x$ such that $|x|<1$. That also gives:
$$ f^{(100)}(x) = 100!\cdot\sum_{n\geq 25}(-1)^n\binom{4n+2}{100}  x^{4n+2-100} $$
for any $|x|<1$.
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