Compute a derivative using MacLaurin series

[kamil]

Homework Statement

Compute the 9th derivative of
$$f(x) = \frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}$$
at x=(0)

Homework Equations

$$f(x)=\sum^{\infty}_{n=0} \frac{f^{(n)}(c)}{n!}x^n$$
$$cos(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n)!}x^{2n}$$

The Attempt at a Solution

The correct answer is 1224720.
I've already read this thread(https://www.physicsforums.com/showthread.php?t=362369&highlight=Maclaurin), and I think I know how to do these kind of problems. However, I have a problem with that one since it has $${-1}}$$

What I did is write the power series for cos(x),
$$cos(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n)!}x^{2n}$$

Substitute $$3 x^{4}$$ , substract 1, and divide by $$x^{7}$$.
And getting:
$$\frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}=\sum^{\infty}_{n=0} \frac{\frac{(-1)^{n}3^{2n}}{(2n)!}x^{8n}-1}{x^7}$$
But after that I don't know.

 

[Dick]

The only term that will produce a nonzero derivative when evaluated at x=0 will have a power of x^9. What is that term and what is it's derivative?

 

[Mute]

Substitute $$3 x^{4}$$ , substract 1, and divide by $$x^{7}$$.
And getting:
$$\frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}=\sum^{\infty}_{n=0} \frac{\frac{(-1)^{n}3^{2n}}{(2n)!}x^{8n}-1}{x^7}$$
But after that I don't know.

Nope. You can't move the 1 inside the sum like that. Sum(something) -1 is not equal to sum(something - 1).

What you should do is recognize that the n = 0 term of the cosine series is 1, so this cancels with your minus 1. Your sum is then from n = 1 to infinity, and you can safely divide the x^7 through. Then I think you probably know what to do from there!

 

[HallsofIvy]

But the n= 0 term for the MacLaurin series for cosine is just "1". The "-1" term just cancels that.

$$\frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}=\sum^{\infty}_{n=1} \frac{\frac{(-1)^{n}3^{2n}}{(2n)!}x^{8n}}{x^7}$$

 

[kamil]

Thank you, it worked. I knew there was something wrong with the -1. I don't know why I thought I could do that operation on a sum. I had also simplified $$\frac{x^{8n}}{x^{7}}$$ to $${x^{n}}$$ , which is totally wrong. I don't why I did this.