UnkleRhaukus
  • UnkleRhaukus
laplace transform of a periodic functions
Differential Equations
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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UnkleRhaukus
  • UnkleRhaukus
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anonymous
  • anonymous
What is this you're doing lol
UnkleRhaukus
  • UnkleRhaukus
i have taken the Laplace transform of some periodic functions i dont know how to check if i have done it right

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UnkleRhaukus
  • UnkleRhaukus
i have used \[\mathcal L\{f(x)\}=\int\limits_0^\infty f(x)e^{-px}\text dx\] \[=\int\limits_0^\tau f(x)e^{-px}\text dx+\int\limits_\tau^{2\tau} f(x)e^{-px}\text dx+\int\limits_{2\tau}^{3\tau} f(x)e^{-px}\text dx+\dots\]\[=(1+e^{-p\tau}+e^{-2p\tau}+\dots)\int\limits_0^\tau f(z)e^{-pz}\text dz\]\[=\frac{1}{1-e^{-p\tau}}\int\limits_0^\tau f(z)e^{-pz}\text dz\]
anonymous
  • anonymous
how 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
  • UnkleRhaukus
i have been using a program called TeXShop
tkhunny
  • tkhunny
How sure are you that it converges?
UnkleRhaukus
  • UnkleRhaukus
i dont know
anonymous
  • anonymous
As long as the function grows slower than Me^pt then it converges.
UnkleRhaukus
  • UnkleRhaukus
ok it converges because the slope is only positive or negative 1 , which is less slope than any exponential
anonymous
  • anonymous
Yeah, 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.
UnkleRhaukus
  • UnkleRhaukus
geometric series?

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