power series of arctan'x

[mathusers]

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?

[dextercioby]

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.

[mathusers]

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?

[dextercioby]

Yes, it is.

[bob1182006]

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

[D H]

That's the ratio test at work. The alternating series test also works here.

[Gib Z]

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 :(

[HallsofIvy]

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!