glebovg
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Compute the sine coefficients for f(x)=e^{-x^{2}} on the interval [0,2\pi]. Does this mean f(x+2\pi k)=f(x), k\in\mathbb{Z}? Can x\in[0,\infty)?
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glebovg said:Compute the sine coefficients for f(x)=e^{-x^{2}} on the interval [0,2\pi]. Does this mean f(x+2\pi k)=f(x), k\in\mathbb{Z}? Can x\in[0,\infty)?
glebovg said:It says on the interval [0,2\pi]. Does this mean f(x+2\pi k)=f(x), k\in\mathbb{Z}?
A Fourier series is an expansion of a periodic function in terms of an infinite sum of sines and cosines. A real-valued function f(x) of a real variable is called periodic of period T>0 if f(x+T) = f(x) for all x\in\mathbb{R}.
So, can x\in[0,\infty)?
LCKurtz said:I think you have answered your own question. Of course if a function is defined on ##[0,2\pi]## and periodic with period ##2\pi## it is defined for all x. But if you want to calculate the coefficients for your f(x) you need to know what periodic extension you are using. That's why I asked you if you mean to use the odd periodic extension of ##e^{-x^2}## to calculate the coefficients.
glebovg said:I need to use the the half-range sine expansion. Correct? However, the problem does not state that the function is periodic, nor that it is defined on [0,2\pi]. After all, the Gaussian function is not periodic. My instructor said that I should only consider [0,\infty), but then this would not satisfy the definition of a periodic function.
glebovg said:In this case T=\pi, but he said that I should work on the positive x-axis only. That's what bothers me. You can't do that, right? f(x) would not be periodic and it would not work because a Fourier series is an expansion of a periodic function in terms of an infinite sum of sines and cosines.