## power series of arctan'x

how could i expand something such as arctan'x into a power series. also how would you be able to find the power series 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?

 Blog Entries: 9 Recognitions: Homework Help Science Advisor Integrate the series you wrote term by term. Watch out for the first line you wrote. You're missing a derivative operator acting on the "arctan" function.
 just to clear confusion.. i did mean the derivate of arctanx .. i.e. d/dx arctan x , hence arctan'x.... how would i show the radius of convergence as |x|<1 though please? to work it out 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?

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## power series of arctan'x

Yes, it is.

 the way my book (Stewart) does it is they say that since it's a geometric series the series will be convergent when |-x^2n|<1 = x^2<1=|x|<1
 Mentor That's the ratio test at work. The alternating series test also works here.
 Recognitions: Homework Help Another way to check would have been to see where the expression $\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +...+ (- 1)^n x^{2n}$ is valid, since that is the basis for the new power series. We can see that the expression fails for values of x larger than 1. Really, its just a tiny variation of what DH and dex said :(
 Recognitions: Gold Member Science Advisor Staff Emeritus In general, a power series will converge as long as has no reason not too! $$\frac{1}{1+x^2}$$ is defined for all complex x except i or -i. The radius of the "disk" of convergence in the complex numbers is 1 so, restricting to the real numbers, the radius of the interval of convergence is also 1. Of course, you can look at it as a geometric series: it is of the form arn with a= 1, r= -x2: its sum is $$\frac{1}{1+x^2}$$ and it converges as long as |-x2|< 1 or |x|< 1. Similarly, the ratio test gives the same result: |x|< 1. Oh, and the root test: $^n\sqrt{a_n}= |x|< 1$ as well. I think we have determined that the radius of convergence is 1!