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Homework Help: Laplace Transform of cos(kt) using Power Series expansion

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem just states to find the Laplace Transform of cos(kt) from its power series expansion, instead of using the formula for the transform of a periodic function.

    2. Relevant equations
    Equation for Laplace transform of a function f(t) ->[tex]\int(e^{-st}f(t))dt[/tex]
    Power Series Expansion for cos(x)-> [tex]\sum\frac{(-1)^{n}}{(2n)!}x^{2n}[/tex]

    3. The attempt at a solution
    I've been trying to apply the formula for the Laplace Transform directly to the expansion of cos, but I get stuck in the integration.. Once you apply the formula, I figured you can bring the e[tex]^{-st}[/tex] inside the sum since it doesn't depend on n, and therefore you treat it like a constant wrt the sum. Then interchange the order of the sum and the integral, and end up with [tex]\sum\frac{(-1)^{n}}{(2n)!}k^{2n}\int(e^{-st}t^{2n})dt[/tex]..
    This is what I cant figure out how to integrate, if you try it by parts you just get t to the 2n-1, then 2n-2.... etc.
    Any ideas?
  2. jcsd
  3. Apr 5, 2009 #2


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    Science Advisor
    Homework Helper

    You've got it down to the laplace transform of a power t^(2n). If you don't have a formula that you can use for it, then you derive it just as you say. Use integration by parts and induction to find the formula. Start by doing t, t^2, t^3... It should be pretty obvious what the formula for t^(2n) looks like. Hint: the answer will have a factorial in it.
  4. Apr 5, 2009 #3
    You can also compute the integral of exp(-s t) and then differentiate that 2n times w.r.t. s to bring down a factor t^(2n) in the integrand.
  5. Apr 16, 2009 #4
    oh..ok i see, i was forgetting to input the bounds, this causes each term to go to zero except for the last one, where the exponent of t is 0. Thanks.
  6. Apr 9, 2011 #5
    Hi i'm a physics student studying theoretical physics 2, we haven't learned math induction. Is there another way around it?

    What about applying the formula

    | exp(- sigma t) f(t) | = M

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