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Ln series

  1. Mar 29, 2007 #1
    I'm looking for a series for

    [tex]ln|cos\frac{\Theta}{2}|[/tex]

    Using r=1 for my complex variable, if that matters...

    Any ideas?
     
    Last edited: Mar 29, 2007
  2. jcsd
  3. Mar 29, 2007 #2
    u may be knowing the series of ln(1+x). Try to convert ur expression in such form.
     
  4. Mar 29, 2007 #3
    Ok, so I expand it like this:

    [tex]ln|cos\frac{\Theta}{2}|=ln|1+(cos\frac{\Theta}{2}-1)|=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(cos\frac{\Theta}{2}-1)^n[/tex]

    Ultimatly, I want to use this expansion to prove that

    [tex]\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=ln2[/tex]

    I can't quite see how this will help .... ??
     
  5. Mar 29, 2007 #4
    i think u should multiply 1/2 to ln and convert cos to cos^2(i hope i'm clear) ; then convert half angle to full angle.
     
  6. Mar 29, 2007 #5
    Clever! :)

    So now I end up with the relation:

    [tex]ln|2cos\frac{\theta}{2}|=\frac{1}{2}(ln2+ln(1+cos\theta))[/tex]

    Thanks!

    EDIT: I'm still not able to use this series to show that

    [tex]\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=ln2[/tex]

    Any sugestions?
     
    Last edited: Mar 29, 2007
  7. Mar 29, 2007 #6
    [tex]\ln(1+x) = x - \frac{x^{2}}{2} + \ldots[/tex]
    [tex]\ln(1+\cos \theta) = \cos \theta - \frac{\cos^{2}\theta}{2} + \ldots = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{\cos^{n}\theta}{n}[/tex]

    Using trig identities you have that [tex]\ln | \cos \theta/2 | = \frac{1}{2}\ln \left( \cos^{2}\frac{\theta}{2} \right) = \frac{1}{2}\ln (1+\cos \theta) - \frac{1}{2}\ln 2[/tex]

    Therefore

    [tex]\ln | \cos \theta/2 | = \frac{1}{2}\sum_{n=1}^{\infty} (-1)^{n+1}\frac{\cos^{n}\theta}{n} - \frac{1}{2}\ln 2[/tex]

    [tex]\theta = 0[/tex] gives the result.
     
    Last edited: Mar 29, 2007
  8. Mar 30, 2007 #7

    Gib Z

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    Homework Helper

    Or perhaps the most OBVIOUS way would have been to let x=1 in the taylor series for natural log, which looks just a tiny bit similar to the series you want huh?
     
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