# The $n$th derivative of $\frac{1}{5x^5 + 1}$ at $x=0$ is $0$ for all $n$ which is not a multiple of $5$?

How can I show this? This is part of an assigned task for which I am to find the Taylor Series expansion of $f(x) = 1 / (5x^5 + 1)$ around $0$. I have noticed by studying the $n$th derivatives of this function that for $n$ is not a multiple of $5$, the derivative evaluates to $0$.

I believe this is useful but how can I prove this? I believe it may be related to the fact that odd derivatives of even functions at $x=0$ are $0$. Is there a theorem for the function that is $1/(5x^5 + 1)$ that says its derivatives evaluate to $0$ around $0$?

Cheers!

## 3 Answers

Let $$f(x) = \frac{1}{5x^5+1}$$ $$g(x) = \frac{1}{x+1}$$ so that $f(x)=g(5x^5)$. Due to our knowledge of geometric series, we know that (when $\vert x \vert < 1$) $$g(x)=\sum_{n=0}^\infty (-1)^nx^n = 1-x+x^2-x^3+...$$ and so $$f(x) = g(5x^5) = 1-5x^5+25x^{10}-125x^{15}+...$$ So, we can see that whenever $n=5k$, we have $$f^{(n)}(0) = n!\cdot 5^kx^n$$ and whenever $n$ is not a multiple of $5$, we have $f^{(n)}(0) = 0$.

• Thanks for the speedy reply! This is very logical and concise! Exactly what I was after! :) Commented May 31, 2017 at 0:21
• Should it be $f(x)=g(5x^5)$?
– edm
Commented May 31, 2017 at 0:28
• @edm I think that's what he/she meant. :) Still very helpful. Commented May 31, 2017 at 0:31

The function depends only on $x^5$, so its series must be something like $a_0+a_1x^5+a_2x^{10}+...$, which means you are right.

In fact, just replace $5x^5$ by $y$ and do the series in $y$ (this is easy), then change back.

• Since $x\mapsto x^5$ is a bijection from $\mathbb{R}$ to $\mathbb R$, every function which only depends on $x$ can be said to only depend on $x^5$. Commented May 30, 2017 at 23:59
• You know what I mean: it is a rational function of $x^5$ Commented May 31, 2017 at 0:11

We can write $$f(x)=\frac{1}{1-(-\sqrt[5]5x)^5}=\sum_{n=0}^\infty (-1)^{5n} 5^nx^{5n}$$ by the geometric series for $|x|<1/\sqrt[5]5$. In particular the coefficients of the series are $f^{(n)}(0)/n!$ and so $f^{(n)}(0)$ is always a multiple of $5$ for all n.