Help with Series Expansion

**Lord Voldemort** · April 27th 2010, 08:01 PM

Hi - I ask for a bit of help in trying to find the series expansion of
Integral(arctan(x^3)) - I was having quite a bit of trouble, and eventually I was forced to look at Wolfram Alpha to get done with the homework, but I want to know where I messes up when I tried to do it.

Basically, this was where I started.

arctan x^3 = integral (1/(1-(-x^6)) dx
= Sum [(-1)^n*x^6n,n,0,infinity]
Then integrating both sides to get my answer, but this has been wrong.

Is my first step the bad part?

**Prove It** · April 27th 2010, 08:35 PM

Originally Posted by Lord Voldemort

Hi - I ask for a bit of help in trying to find the series expansion of
Integral(arctan(x^3)) - I was having quite a bit of trouble, and eventually I was forced to look at Wolfram Alpha to get done with the homework, but I want to know where I messes up when I tried to do it.

Basically, this was where I started.

arctan x^3 = integral (1/(1-(-x^6)) dx
= Sum [(-1)^n*x^6n,n,0,infinity]
Then integrating both sides to get my answer, but this has been wrong.

Is my first step the bad part?

$\frac{d}{dX}\left(\arctan{X}\right) = \frac{1}{1 + X^2}$ .

Notice that you can rewrite $\frac{1}{1 + X^2}$ as

$\frac{1}{1 - (-X^2)}$ .

This is the closed form of the geometric series with $a = 1, r = -x^2$ .

So you can say

$\frac{1}{1 + X^2} = \sum_{n = 0}^{\infty} (-X^2)^n$

$= \sum_{n = 0}^{\infty}(-1)^nX^{2n}$ .

So $\frac{d}{dX}(\arctan{X}) = \sum_{n = 0}^{\infty} (-1)^nX^{2n}$

which means

$\arctan{X} = \int{\sum_{n = 0}^{\infty}(-1)^nX^{2n}\,dX}$

$\arctan{X} = \sum_{n = 0}^{\infty}\int{(-1)^n X^{2n}\,dX}$

$\arctan{X} = \sum_{n = 0}^{\infty}\frac{(-1)^nX^{2n + 1}}{2n + 1}$ .

This means $\arctan{(x^3)} = \sum_{n = 0}^{\infty}\frac{(-1)^n(x^3)^{2n + 1}}{2n + 1}$

$\arctan{(x^3)} = \sum_{n = 0}^{\infty}\frac{(-1)^n x^{6n + 3}}{2n + 1}$ .

So that means:

$\int{\arctan{(x^3)}\,dx} = \int{\sum_{n = 0}^{\infty}\frac{(-1)^n x^{6n + 3}}{2n + 1}\,dx}$

$= \sum_{n = 0}^{\infty}\int{\frac{(-1)^n x^{6n + 3}}{2n + 1}\,dx}$

$= \sum_{n = 0}^{\infty}\frac{(-1)^n x^{6n + 4}}{(2n + 1)(6n + 4)}$ .

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