Computing $n$-th derivative $f(x)= \frac{\cos(5x^4)-1}{x^7}$ using MacLaurin Series.

Question

The prompt is to find the $9$-th derivative of the function f(x) defined as, $$f(x)= \frac{\cos(5x^4)-1}{x^7}$$ at $x = 0$.

We are suggested to use the MacLaurin Series for $f(x)$.

Answer 1

$$f(x) = \frac{\cos(5x^4)-1}{x^7} = \sum_{n=1}^{+\infty} \frac{(-1)^n \cdot 5^{2n} \cdot x^{8n-7}}{(2n)!} $$ Therefore if $k \not\equiv -7 \mod 8$ we have $f^{(k)}(0)=0$.

Otherwise $\displaystyle f^{(8n-7)}(0) = \frac{(-25)^n \cdot (8n-7)!}{(2n)!}$

So $\displaystyle f^{(9)}(0) = \frac{625 \cdot 9!}{24}$.

I used $\displaystyle \cos(x) = \sum\limits_{n=0}^{+\infty} (-1)^n \cdot \frac{x^{2n}}{(2n)!}$ and $\displaystyle f(x) = \sum\limits_{n=0}^{+\infty} f^{(n)}(0) \cdot \frac{x^n}{n!}$

Answer 2

The coefficient $a_9$ of $x^9$ in the Taylor series of $f(x)$ centered at $x = 0$ is the ninth derivative of $f(x)$ evaluated at $0$, so we have $f^{(9)}(0) = 9!\cdot a_9$.

Using the series for $\cos$, we have \begin{align*} f(x) &= \frac{1}{x^7}\cdot\bigg\{\bigg(1-\frac{(5x^4)^2}{2!} + \frac{(5x^4)^4}{4!}-\dotsb\bigg)-1\bigg\} \\ &= \frac{1}{x^7}\bigg(\frac{-(5x^4)^2}{2!} + \frac{(5x^4)^4}{4!}-\dotsb\bigg) \\ &= \frac{-25x}{2!} + \frac{5^4x^9}{4!}-\dotsb, \end{align*} and the rest is algebra.