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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]

    [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}}
    [/tex]

    [tex]
    \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]}
    [/tex]

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

    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

    benorin

    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.
     
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