# power series of arctan'x

• Sep 9th 2007, 12:04 PM
joanne_q
power series of arctan'x
how could i expand something such as arctan'x (derivate of arctanx ... i.e. d/dx arctanx) into a power series. also how would you be able to find the radius of convergence for it?

so far i have managed to work out that:

arctan'x = $\frac{1}{1 + x^2}$

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

how do you work out the "radius of convergence" though: i know it is : |x|< 1.. but how do you work it out please?

i tried it on $(-1)^n x^{2n}$
i ended up with

$a_{n+1} / a_{n} = \frac{|x|^{2n + 2}}{|x|^{2n}} = |x|^2/1$ as n tends to infinity... ...

so radius of convergence is |x|< 1...

is this working out correct?
• Sep 9th 2007, 01:22 PM
ThePerfectHacker
If $|x|<1$ then,
$1-x^2+x^4-x^6 + ... = \frac{1}{1+x^2}$

Integrate both side from $0\mbox{ to }t$ where $|t|<1$ to get,
$\tan^{-1}t = t - \frac{t^3}{3}+...$

Now this hold if $-1 < t < 1$.
But it also hold when $t=1$.
That is a little more difficult to show.