By integrating a power series for 1/1+z^2 term-by-term, find a power series for arctan(z) centered at z0=0. Compute its radius of convergence.

Any help would be great!!

Printable View

- May 4th 2011, 12:06 PMshaheen7complex analysis2
By integrating a power series for 1/1+z^2 term-by-term, find a power series for arctan(z) centered at z0=0. Compute its radius of convergence.

Any help would be great!! - May 4th 2011, 12:31 PMgirdav
So, what is the power series of $\displaystyle \frac 1{1+z^2}$ ?

- May 6th 2011, 04:29 AMshaheen7
i know it equals (1/2i)[1/(z-i)-1/(z+i)]? where do I go from there?

- May 6th 2011, 11:41 AMgirdav
If you know the power series of $\displaystyle \frac 1{1-Z}$ you can find the power series of $\displaystyle \frac 1{1+z^2}$ by a substitution.

- May 6th 2011, 12:26 PMshaheen7
would it be similar to 1/1-x which is 1 + x + x^2 + ... so 1/1-z = 1 + z + z^2 + z^3 onwards? and then how would one go about the substitution?

- May 6th 2011, 12:59 PMgirdav
Put $\displaystyle Z:=-z^2$.

- May 7th 2011, 12:25 PMshaheen7
okay so when you put that in then it becomes 1 -z^2 + z^4 - z^6 ... but then what is the next step from that in order to find the power series for arctan(z). do you simply integrate the series we have found through substitution??

and how would you get the radius of convergence??

sorry to be a pain!! - May 7th 2011, 12:32 PMgirdav
You have to integrate term by term the series $\displaystyle \sum_{n=0}^{+\infty}(-1)^nz^{2n}$.

Use the fact that the radius of convergence is the same as the radius of convergence of its derivative. - May 7th 2011, 12:53 PMshaheen7
okay thank you i think ill be able to handle the integration part! im just confused about the radius of convergence. i know the interval of convergence but how do i go about getting the radius of convergence for this series?

- May 7th 2011, 12:57 PMgirdav
You can use the ratio test.