Question

d/dt arctan t= 1/(1+t^2) the integral of arctan x=1/(1=t^2) from 0 to xsub in power series (-1)^nt^(n+1) and integrate to get a power series expansion...

Answer 1

integrate term by term, but not sure how to go about this one

Answer 2

\[\sum_{n=0}^{\infty}(n+1)x ^{n}\] for x>0, for the original equation.

Answer 3

Now to substitute \[\sum_{n=0}^{\infty}(-1)^{n-1}t ^{2n}\]

Answer 4

\[\arctan x=\int\limits_{0}^{x}1/1+t ^{2} dt \] was the original integral

Answer 5

And so they're asking for the power series representation of that?

Answer 6

with the second power series substituted in I think...

Answer 7

Honestly i'm quite confused then. :/

Answer 8

Me too, but wait, looks like there is more...

Answer 9

looks to be a geometric series, the convergence is going to be (-1, 1), but the end point is x=1.

Answer 10

This is "cheating" a bit but might offer some help: http://www.wolframalpha.com/input/?i=power+series+of+arctan

I need to head to bed though. Good luck!

Answer 11

Thanks for your help