1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Simple series but my answer looks wrong

  1. Oct 30, 2005 #1
    I have doing a question which I cannot get the answer to. I would like some help with it if it can be done.
    a) Derive a Taylor series about the origin for the function f(x) = log(1+x).
    b) Find the interval of convergence for the Taylor series, justifying your answer.
    c) Use the result of 'a' to deduce the Taylor series about the origin of the function [tex]\arctan h\left( x \right) = \frac{1}{2}\log \frac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}}[/tex]
    d) On what interval does your series in 'c' converge?

    I obtained:
    a) [tex]\log \left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n}} [/tex]

    b) I = (-1,1]

    c) [tex]\arctan h\left( x \right) = \frac{1}{2}\log \frac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}} = \frac{1}{2}\left( {\log \left( {1 + x} \right) - \log \left( {1 - x} \right)} \right)[/tex]

    \log \left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n}} \Rightarrow \log \left( {1 - x} \right) = - \sum\limits_{n = 1}^\infty {\frac{{x^n }}{n}}

    \arctan h\left( x \right) = \frac{1}{2}\sum\limits_{n = 1}^\infty {\left[ {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n} + \frac{{x^n }}{n}} \right]}

    = \sum\limits_{n = 1}^\infty {\left[ {\left( {\left( { - 1} \right)^{n + 1} + 1} \right)\frac{{x^n }}{{2n}}} \right]}

    That doesn't look right.

    d) I know(or rather, remember seeing somewhere) that if you have a product of series then the radius of convergence of the product is the smaller of the radii of convergence of the individual series. I don't think that applies here through since I've got a sum of series and not a product. I'm stuck on part 'c' mainly. Can someone have a look through my answer and tell me where I'm going wrong?
  2. jcsd
  3. Oct 30, 2005 #2


    User Avatar
    Homework Helper

    Good work, almost there

    Parts a and b are correct, without question.

    As for part c: The last sum is good except that [tex]\frac{{x^n }}{{2n}}}[/tex] should be [tex]\frac{{x^n }}{{n}}}[/tex]. From there, note that

    [tex]\left( {\left( { - 1} \right)^{n + 1} + 1} \right) = \left( 1 - (-1)^{n} \right)[/tex], which is 0 for even n and 2 for odd n, hence your sum reduces to

    [tex] \mbox{arctanh}(x) = \sum\limits_{n = 1}^{\infty} {\left[ {\left( {\left( { - 1} \right)^{n + 1} + 1} \right)\frac{{x^n }}{{n}}} \right]}
    = \frac{1}{2}\sum_{n=1,3,5,...}^{\infty} (2)\frac{x^{n}}{n} = \sum_{n=1,3,5,...}^{\infty} \frac{x^{n}}{n} = \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1}[/tex]

    as it must.

    As for part d, the interval of convergence of the above series is [tex]-1 < x < 1[/tex], that is [tex]x\in (-1,1) [/tex]. I obtained this by using the ratio test and testing both end points (neither of them converge).
    Last edited: Oct 30, 2005
  4. Oct 30, 2005 #3
    Thanks for the help benorin.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook