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Math Help - First three terms of hyperbolic function

  1. #1
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    First three terms of hyperbolic function

    Find the first three terms of the series
    \ln (\cosh x)

    \cosh x=\displaystyle \sum^{\infty}_{r=0} \frac{x^{2r}}{(2r)!}=\sum^{\infty}_{r=0}\frac{x^{2  r}}{(2r)!}+1

    \ln (\cosh x)=\ln\left(\displaystyle \sum^{\infty}_{r=0}\frac{x^{2r}}{(2r)!}+1\right)=\  displaystyle \sum^{\infty}_{n=1}\left(\frac{(-1)^{n+1}(\displaystyle \sum^{\infty}_{r=0}\frac{x^{2r}}{(2r)!})^n}{n}\rig  ht)
    This is as far as I got, I don't think I'm on the right track though.
    Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    The easiest way is to find the coefficients as...

    \displaystyle a_{n} = \frac{f^{(n)} (0)}{n!} (1)

    ... so that is...

    \displaystyle f(x) = \ln \cosh x = 0,  x=0

    \displaystyle f^{'} (x) = \tanh x = 0,  x=0

    \displaystyle f^{''} (x) = 1 - \tanh x = 1,  x=0

    \displaystyle f^{'''} (x) = \tanh x -1 = -1,  x=0

    ... and the series expansion required is...

    \displaystyle \ln \cosh x = \frac{x^{2}}{2} - \frac{x^{3}}{6} ... (2)

    Kind regards

    \chi \sigma
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