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You are not logged in. #1 20110810 20:59:17
Laplace Transform of sin^2tHi all, Here's my working: Using integration by parts we get then I noticed that is the Laplace transform of since by the doubleangle rule sin(2t) = 2sintcost, and we can separate the 1/2 since it's a constant. So if that bit is the Laplace transform of , then we can use our identity for Laplace transform of sin(at) to rewrite the integral as; So; But when I evaluate the RHS at t = and t = , I end up with y(s) = 0... is that really correct? I haven't checked the answer online anywhere because I don't want the answer spoiled for me... I also tried a different method using integration by parts 3 times before noticing a repeat. Not sure what to do though. Would appreciate your help. Thanks. #2 20110810 21:08:08
Re: Laplace Transform of sin^2tHi; That I am not getting? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20110810 21:18:08
Re: Laplace Transform of sin^2tThanks for the quick response, I don't understand why I'm not getting the same answer as you... Using the formula uv  ∫vdu for integration by parts, that's how I got my answer... And using the identity for sin(at) (where a = 2), I am getting why is this incorrect? Thanks. #4 20110810 21:39:37
Re: Laplace Transform of sin^2tSorry, I am being disconnected. I will check your IBP please hold. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #5 20110810 22:03:07
Re: Laplace Transform of sin^2tHi, #6 20110810 22:11:21
Re: Laplace Transform of sin^2tHi; The LHS has to be evaluated at infinity and then you subtract the evaluation of it at 0. The RHS is untouched. How are you getting 0 for the LHS? If s is very small then the LHS is not zero. Were you given some interval for s? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #7 20110810 22:18:46
Re: Laplace Transform of sin^2tHi, Then evaluate RHS at 0 and subtract that from the evaluation at infinity. I got 0... so then we have Therefore assuming s > 0. I also tried the Laplace transformation for sin^{a}(t) and got . #8 20110810 22:20:29
Re: Laplace Transform of sin^2tI wasn't given an interval for s, sorry. I am just waiting for my GCSE results (I turn 16 in August) and I'm just trying to extend my knowledge of calculus. I want to learn about Fourier transforms too hopefully but I need some practice with that. #9 20110810 22:21:32
Re: Laplace Transform of sin^2tHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #10 20110810 22:38:37
Re: Laplace Transform of sin^2tSorry if it's a bother but do you know how to compute Fourier transforms? I'm trying to learn how, I've seen the Wikipedia article and saw this: Does this mean that if I put in some function of x, such as sin(x), I'll get f(ξ) where ξ is a real number? Not sure, I'll post my working in a second. Sorry if I sound stupid... #11 20110810 22:43:40
Re: Laplace Transform of sin^2tAlso what is the notation for a Fourier transform? For Laplace it's a fancy L, is it a fancy F for Fourier transforms? #12 20110810 22:43:50
Re: Laplace Transform of sin^2tHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #13 20110810 22:50:01
Re: Laplace Transform of sin^2tThanks... #14 20110810 22:52:21
Re: Laplace Transform of sin^2tWolfram is going to put it in terms of the Dirac delta function, which I think is a step function. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #17 20121026 15:38:46
Re: Laplace Transform of sin^2tHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #18 20130513 08:41:54
Re: Laplace Transform of sin^2tShouldn't this be #19 20130513 08:58:08
Re: Laplace Transform of sin^2tHi;
That is not correct. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 