[math-fun] "new" kind of fourier series
However, it actually looks like any real-valued series in Warren's form can be converted to an equivalent real-valued usual Fourier series, if you include both positive and negative frequencies (technically, positive and negative for every coordinate except the first - more on that later) with the same number of coefficients as the new kind of Fourier series.
--aha. Neil Bickford has redeemed himself! :) So, originally, I thought the "new" kind of Fourier series had about 2^D or 2^(D-1) or whatever (depends if real or complex, and whether I count right...) times more coefficients than the "old" kind -- and that is definitely true if the old kind happened to consist of exactly one nonzero term, as you can show by expressing exp as a sum of two trigs and expanding out a product. However, Bickford points out that kinda also works in reverse. Namely, if the "new" series happens to consist of exactly one nonzero term, then its "old"-form equivalent, will consist of about 2^D terms! ("About" means up to a constant factor, since I'm not carefully counting.) And this you can also show, albeit the demonstration is somewhat messier. His idea is basically to write a trig as a sum of two exps, then expand out product. Oh. So then, you realize the "new" and "old" Fourier series actually are equivalent. However, the equivalence can cause either the new or the old thing to have up to about 2^D times more nonzero terms than the other, in the right circumstances. So, given that 2^D can be pretty large, that could under the right circumstances be a serious savings. Also, in practice one would truncate the series at some point, using only a finite set of terms, and that finite number might actually be comparable or even much smaller than 2^D... for example if D=40... in which case the two kinds of series could exhibit considerably different practical behaviors. So, that was interesting, and I thank Neil Bickford for figuring that out. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Warren D Smith