Laplace Transform of cos(kt) using Power Series expansion

eyenkay
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Homework Statement


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.

Homework Equations


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

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^{-st} 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 \sum\frac{(-1)^{n}}{(2n)!}k^{2n}\int(e^{-st}t^{2n})dt..
This is what I can't 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?
 
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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.
 
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.
 
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.
 
Dick said:
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.
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) | = Mthanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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