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Power series of arctan'x

  1. Sep 9, 2007 #1
    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 = [itex] \frac{1}{1 + x^2} [/itex]

    [itex]\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +...+ (- 1)^n x^{2n}[/itex]

    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. jcsd
  3. Sep 9, 2007 #2


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    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
  4. Sep 9, 2007 #3
    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
    [itex](-1)^n x^{2n}[/itex]
    i ended up with

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

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

    is this working out correct?
  5. Sep 9, 2007 #4


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    Yes, it is.
  6. Sep 9, 2007 #5
    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
  7. Sep 9, 2007 #6

    D H

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    That's the ratio test at work. The alternating series test also works here.
  8. Sep 10, 2007 #7

    Gib Z

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    Another way to check would have been to see where the expression [itex]\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +...+ (- 1)^n x^{2n}[/itex] 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 :(
  9. Sep 10, 2007 #8


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    In general, a power series will converge as long as has no reason not too!

    [tex]\frac{1}{1+x^2}[/tex] 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 [tex]\frac{1}{1+x^2}[/tex] 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: [itex]^n\sqrt{a_n}= |x|< 1[/itex] as well.

    I think we have determined that the radius of convergence is 1!
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