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?
Power series for the sawtooth wave
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0while 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
1 Answers
If the sawtooth wave is defined by, say
$f(x)=x, 0
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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0@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