Let $$ f(x)=\frac{\cos(5x^2)-1}{x^2} $$ We want to compute the $6th$ derivate of $f(x)$ at $x=0$.
Using a calculator, I found $18750$ (which is correct). But I don't understand how to find this result?
There is an additional hint, which is using the MacLaurin series for $f(x)$. I know I should use the cos Maclaurin series, but I find $0$ as a result.
I can't seem to find it.
$\endgroup$ 13 Answers
$\begingroup$Hint:
$$\cos (5x^2) - 1 = \sum_{n=1}^{\infty}(-1)^n\frac{(5x^2)^{2n}}{2n!}$$
$\endgroup$ 2 $\begingroup$Use the formula by Aaron, we have: $$f(x)=\sum_{n=1}^{\infty}(-1)^n{{(5x^2)^{2n} \cdot x^{-2}} \over {(2n)!}}$$ $$f(x)=\sum_{n=1}^{\infty}(-1)^n{{(25)^{n}} \over {(2n)!}} \cdot x^{4n-2}$$
For $f^{(6)}$ when $x \to 0$, only this term is concerned when $4n-2=6$, that is $n=2$.
$$f^{(6)}(x)=(-1)^2{{(25)^{2}} \over {(4)!}} 6!=18750$$
$\endgroup$ 1 $\begingroup$First, you define f (x) in two different ways - first you multiply by $x^2$, next you divide by $x^2$. Which one is it?
To solve the problem: Write down the Taylor polynomial for cos x. Substitute $5x^2$ for x, which turns for example $x^6$ into $5^6 x^{12}$. The sixth derivative of $x^n$ is 0 if n<6. It is some constant times $x^{n-6}$ if n > 6 which is 0 when x = 0. And it is 6! when n = 6.
Now the 18,750 that you mention seems to be the sixth derivative of $(\cos 5x - 1)x^2$ at x = 0 which is not quite what you asked.
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