Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

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

    Gib Z

    User Avatar
    Homework Helper

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Power series of arctan'x
  1. Power series (Replies: 1)

  2. Power series (Replies: 2)

  3. Power Series (Replies: 1)

  4. Power Series (Replies: 2)

Loading...