Some well-known mathematician was reported to ahve announced at the start of a course in complex analysis that he would set 2 pi i = 1 . History does not relate the outcome. WFL On 12/27/15, Dan Asimov <asimov@msri.org> wrote:
Which all suggests this question, that I've wondered before in any case:
Suppose we are given a complex Fourier series of form
f(t) = Sum_{-oo < n < oo} c_n exp(nit)
(with each c_n in C), defining a function that by abuse of notation we call
f: R/(2pi)Z —> C
. Then:
How can you tell from the coefficients {c_n} when f is one-to-one (i.e., self-avoiding)?
—Dan ___________________________________________
P.S. Random thought: Isn't it a bit vexing that if the basis functions are exp(it), then the period is 2pi; and if the period is 1, then the basis functions are {exp(2pi*it)} — always with the obnoxious 2pi in the way.
Maybe it would be best if the basis functions were {exp(sqrt(2pi)*it} and so the period would be sqrt(2pi), so at least there's some symmetry. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun