# Thread: power series

1. ## power series

I have a problem that i can't figure out.
a.) Consider the function f(x) = 1/(1+x^2). Write this functio as power series centered at 0.

b.) Now, use the first part find a power series for g(x) = arctan(x) also centered at x.

E is a summation.
I think i have found the first part but the second part is giving me problems i came out with this answer E(-x^(2n+2))/ (2n+2) but the answer is apparently E(-x^2n)/ (2n+2)

if someone could explain to me why that would be great!!

2. Originally Posted by snorris103
I have a problem that i can't figure out.
a.) Consider the function f(x) = 1/(1+x^2). Write this functio as power series centered at 0.

b.) Now, use the first part find a power series for g(x) = arctan(x) also centered at x.

E is a summation.
I think i have found the first part but the second part is giving me problems i came out with this answer E(-x^(2n+2))/ (2n+2) but the answer is apparently E(-x^2n)/ (2n+2)

if someone could explain to me why that would be great!!
I don't see how anyone can tell you why your answer is wrong if you don't tell us how you got that.
In particular what did you get for (a)?

Here'es a guess: What is the anti-derivative of $\displaystyle x^{2n}$?

3. (a) Use the fact that: $\displaystyle \frac{1}{1-y} = \sum_{n=0}^{\infty} y^n \qquad |y| < 1$

Here, imagine: $\displaystyle y = x^2$

So: $\displaystyle f(x) = \frac{1}{1+x^2} = \frac{1}{1 - \left(- x^2\right)} = \sum_{n=0}^{\infty} \left(-x^2\right)^n = \sum_{n=0}^{\infty}(-1)x^{2n}$

____________________

(b) Recall that: $\displaystyle \arctan (x) = \int \frac{1}{1+x^2} dx$

So: $\displaystyle \arctan (x) = \int \left(\sum_{n=0}^{\infty} (-1)x^{2n}\right) dx = \sum_{n=0}^{\infty} \left( (-1)^n \int x^{2n} dx\right)$

The power series of $\displaystyle \arctan x$ should be: $\displaystyle \arctan (x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} \qquad |x| < 1$

4. Originally Posted by snorris103
I have a problem that i can't figure out.
a.) Consider the function f(x) = 1/(1+x^2). Write this functio as power series centered at 0.

b.) Now, use the first part find a power series for g(x) = arctan(x) also centered at x.

E is a summation.
I think i have found the first part but the second part is giving me problems i came out with this answer E(-x^(2n+2))/ (2n+2) but the answer is apparently E(-x^2n)/ (2n+2)

if someone could explain to me why that would be great!!
Part a is the sum of the geometric series with $\displaystyle r=-x^2$

so its series is $\displaystyle \sum_{n=0}^\infty (-1)^nx^{2n}$

This series is valid on (-1,1)

Note that the derivative of $\displaystyle g(x)=\tan^{-1}(x)$

$\displaystyle g'(x)=\frac{1}{1+x^2}$

Now from part a we know that

$\displaystyle g'(x)=\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$

Now we can integrate the equation(There are some technical points that you may need to justify to exchange a sum and an integral) to get

$\displaystyle g(x)=\tan^{-1}(x)=C+\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}$

Since $\displaystyle g(0)=0$ we get

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