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This wikipedia article described a Fourier expansion of the sawtooth wave. Does this wave have a power series expansion (around any point)? If so, what is it? Does every function with a Fourier expansion also has a power series expansion?

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    while I'm very flattered that you would accept my fairly sketchy answer, I suggest leaving the question open for a couple days. More qualified people than me may skip the question thinking you've received an answer you find fully satisfactory, when you'll definitely want someone with more expertise covering the interesting issues you're finding.2012-08-30

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If the sawtooth wave is defined by, say

$$f(x)=x, 0 $$f(x)=f(x+1)$$

Then yes, it has a power series expansion around any point that isn't a discontinuity, but it's not particularly exciting:

$$x$$

Since what we're dealing with on any interval around a non-jump point is merely a line, and so has constant derivative and $0$ higher ones. Power series are only affected by local behaviour, and since areas within some positive radius of convergence are locally straight lines, that's what you get. The radius of convergence is just the distance to the nearest discontinuity.

I'm not quite sure about the last result (so I suppose this is an incomplete answer), but I can say that the converse isn't true. Just take any unbounded function, like $e^x$, which is entire but has no fourier transform.

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    Since every sine in the Fourier expansion has a power series expansion, can't we just sum all of the individual series, or is it an ill defined operation?2012-08-30
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    isn't it a little more complicated? Agreed the power series expansion is at most linear, but the details depend on the point. Even within the initial window: f(x) = a+(x-a).2012-08-30
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    @zxc That's what I would expect, but that's also the snag I don't know enough analysis to solve. My assumption is that each coefficient of the resulting power series would converge and be equal to the corresponding coefficient in the power series of the function it represents, but I'm not sure how to prove it. I know convergence results have been tricky with fourier series, but I don't know if assuming the fourier series already converges removes them in this instance.2012-08-30
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    @RobertMastragostino: So you would expect that the coeffs converge to $f(x)=x$, or that such summation is ill defined?2012-08-30
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    @JamesS.Cook I'm not quite clear on why that would cause issues. My definition of the power series would be that it *is* linear within a certain interval, and then periodic, which would directly imply a linear power series outside of discontinuities, no? If I took the derivative of the sawtooth wave it would be constant except at discontinuities, and then higher derivatives would then be zero except at discontinuities. I know the argument sounds hand-wavy, but I don't see the hole in it.2012-08-30