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Math Help - Laplace transform from Taylor Series

  1. #1
    Newbie mukmar's Avatar
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    Laplace transform from Taylor Series

    Problem: Using the Taylor series for \sin t, and assuming that the Laplace transform of this series can be computed term by term, verify that \mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1}.

    Attempt at solution:
    So I start with the the taylor series of \sin t = \displaystyle\sum^{\infty}_{n=0} \frac{(-1)^n t^{2n + 1}}{(2n + 1)!},

    Then using the fact that \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, where n \in \mathbb{Z}^+, I am able to get to this step:

    \mathcal{L}\{\sin t\} = F(s) = \displaystyle\sum^{\infty}_{n=0} \frac{(-1)^n}{s^{2(n+1)}}

    However here I'm stumped, I'm guessing there's some well known summation result I should be using to get to the final step, but the ones I've looked up I haven't been able to successfully use them to get the result I want.

    Any suggestions would greatly be appreciated. Thanks in advance.
    Last edited by mukmar; December 18th 2010 at 02:23 AM.
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  2. #2
    Newbie mukmar's Avatar
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    Alright, sorry about this, I was able to solve it. I wish I could delete the thread, however here's the solution just in case anyone is wondering I guess.

    F(s) = \displaystyle\sum^{\infty}_{n=0} \frac{(-1)^n}{s^{2(n+1)}} = \displaystyle\sum^{\infty}_{n=0} \frac{1}{s^2}\left(\frac{-1}{s^2}\right)^n<br />

    We know  \displaystyle\sum^{\infty}_{n=0} ar^n = \frac{a}{1 - r}. Here a = \frac{1}{s^2} and r = \frac{-1}{s^2}. Thus \displaystyle\sum^{\infty}_{n=0} \frac{1}{s^2}\left(\frac{-1}{s^2}\right)^n = \frac{\frac{1}{s^2}}{1 + \frac{1}{s^2}} = \frac{1}{s^2 + 1} as desired.
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