# Power series of arctan'x

1. Sep 9, 2007

### 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?

Last edited: Sep 9, 2007
2. Sep 9, 2007

### 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.

Last edited: Sep 9, 2007
3. Sep 9, 2007

### 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?

4. Sep 9, 2007

Yes, it is.

5. Sep 9, 2007

### 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

6. Sep 9, 2007

### D H

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

7. Sep 10, 2007

### 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 :(

8. Sep 10, 2007

### 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!