# Power series involving arctan(x)

MeMoses

## Homework Statement

The function f(x) = 8x*arctan(6x) is represented as a power series f(x) = sum from n=0 to infinity of Cn * x^n
Find the first few coefficients in the power series

## The Attempt at a Solution

I deduced that
8xarctan(6x) = sum from n=0 to infinity of 8(-1)^n * 6^(2n+1) * x^(2n+2) / (2n+1)
but I'm not sure how to go about to get it into the from Cn*x^n
Any help will be great. Thanks

## Answers and Replies

Gold Member
Do you know the general form of the power series?
$$\sum^{\infty}_{n=0}c_n (x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+...+c_n (x-a)^n+...$$
Make an effort to use LaTeX for clearer equations: $$f(x) = 8x\arctan(6x) \\a=0$$

Last edited:
MeMoses
Yes but how do i go about getting what I have into the from Cn * x^n? I have the right series representation but how to I get that to a power series?

Gold Member
$\sum^{\infty}_{n=0}c_n (x)^n$ is a general form of the power series, but it's not specifically applicable to your function, f(x). Instead of giving you the exact power series to be used, you were given the general form. You are expected to know the appropriate power series to use based on the type of function, f(x). The question does not ask you to represent the function, f(x) in the given power series form -- you are asked to give "the first few coefficients" as your answer to this problem.
$$\sum^{\infty}_{n=0}\frac{(-1)^n(6x)^{2n}48x^2}{2n+1}=48x^2-576x^4+\frac{62208}{5}x^6+...$$
The first few coefficients are: $c_2=48,\,c_4=-576,\,c_6=\frac{62208}{5},\,...$
The radius of convergence is, $R=\frac{1}{6}$.

Last edited:
MeMoses
Yes that is equivalent to what I have, but that does not answer my question. I need the first few coefficients(Cn), but all I can get up to is a series representation, but it does not match the form of the power series. Perhaps you don't understand my question or I'm not getting what your trying to say.

Gold Member
I was still editing my post #4 when you replied. Refresh the page in your browser to see the edited post.

MeMoses
Well thats a lot simpler than I thought. Thanks