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UnkleRhaukus
 4 years ago
laplace transform of a periodic functions
UnkleRhaukus
 4 years ago
laplace transform of a periodic functions

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is this you're doing lol

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2i have taken the Laplace transform of some periodic functions i dont know how to check if i have done it right

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2i have used \[\mathcal L\{f(x)\}=\int\limits_0^\infty f(x)e^{px}\text dx\] \[=\int\limits_0^\tau f(x)e^{px}\text dx+\int\limits_\tau^{2\tau} f(x)e^{px}\text dx+\int\limits_{2\tau}^{3\tau} f(x)e^{px}\text dx+\dots\]\[=(1+e^{p\tau}+e^{2p\tau}+\dots)\int\limits_0^\tau f(z)e^{pz}\text dz\]\[=\frac{1}{1e^{p\tau}}\int\limits_0^\tau f(z)e^{pz}\text dz\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how do you type a document like that is it LaTex ,it looks great csry i cant help with the content but the presentation is fine hw do you create such a presentation

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2i have been using a program called TeXShop

tkhunny
 4 years ago
Best ResponseYou've already chosen the best response.0How sure are you that it converges?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0As long as the function grows slower than Me^pt then it converges.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2ok it converges because the slope is only positive or negative 1 , which is less slope than any exponential

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, I'm not sure of the name of the theorem but I just went over that in one of my classes not too long ago.
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