# Finding $f^{(n)}(0)$ if $f(x)= \frac{x^5}{ 1 + x^2}$

In an exercise I'm asked the following:

Let $$f: \Bbb R \to \Bbb R$$ such that: $$f(x)= \frac{x^5}{ 1 + x^2}$$ Determine the value of: $$f^{(n)}(0),\ \ \forall n \in \mathbb N$$

Because we have a fraction, then $$f^{(n)}(0)$$ will take of the following form, because of the quotient rule for derivatives:

$$f^{(n)}(x)=\frac{P_n(x)}{(1-x^2)^{2n}}$$

So if follows that: $$f^{(n)}(0) = P_n(0)$$

So to solve this we basically need to find out what the numerator of that quotient will be for any $$n \in \mathbb N$$. That is precisely the part that I'm struggling with. How can I solve this?

You have that $$\frac{1}{1+x^2}=\sum_{k=0}^\infty (-1)^nx^{2n}.$$ Therefore $$\frac{x^5}{1+x^2}=\sum_{k=0}^\infty (-1)^nx^{2n+5}=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^{k}.$$ I let you find $$f^{(n)}(0)$$ for all $$n$$.
• Your infinite sum is true when $|x|<1$. Commented Mar 23, 2021 at 19:49
• @hamam_Abdallah: And so ? as far as I know, the OP want $f^{(n)}(0)$, so no need to have $|x|\geq 1$. Commented Mar 23, 2021 at 20:04
prove that the numerator is $$x \cdot Q_n(x)$$ for every $$n$$
• you can prove it by induction on $n$ :) Commented Mar 23, 2021 at 19:37