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fourier superfunction ? - tommy1729 - 06/22/2010
the idea of slog(e^x) = slog(x) + 1 in terms of taylor series has been around for a while. i propose to consider f(e^x) = f(x) + exp(x). since exp(x) =/= 0 , f(e^x) =/= f(x). thus f has no "fixed point" of e^x. now instead of trying taylor on both sides of the equation , how about using fourier series on one side ? afterall we have a periodic function ! define f(0) = some real A with A as you wish ! (whatever makes the equations the easiest ) thus we could set the values of the fourier coefficients equal to those of the taylor series : keeping in mind that we have 'expressions' for n'th derivate ( taylor ) and fourier coefficients. ( in fourier ) f(exp(x)) - exp(x) = ( in taylor ) f(x). nth fourier coef = nth taylor coef. it follows that f(-oo) = f(0) ( and = A ) thus i would expand f(x) at x >= 0 and consider it valid only for real z >= 0. i guess A can be 0.... maybe this is related to kouznetsov and/or andrew slog ? it certainly seems to have similar properties at first sight ... not everything is totally clear to me though ... hoping to be correct ... regards tommy1729 RE: fourier superfunction ? - mike3 - 06/23/2010
can be expressed as a Fourier series, but then how do you express the right hand side? If it's a Taylor series, you can't just equate coefficients, since what they're multiplying is different! E.g. . Just because the coefficients are equal doesn't mean the terms are equal. Equating coefficients of the two distinct series types will not produce a solution of the equation. I don't see what you're trying to get at here... RE: fourier superfunction ? - tommy1729 - 06/23/2010
(06/23/2010, 01:20 AM)mike3 Wrote: can be expressed as a Fourier series, but then how do you express the right hand side? If it's a Taylor series, you can't just equate coefficients, since what they're multiplying is different! E.g. . Just because the coefficients are equal doesn't mean the terms are equal. Equating coefficients of the two distinct series types will not produce a solution of the equation. I don't see what you're trying to get at here... yes , but the thing is this : we compute the fourier coefficients with the taylor expression and the n'th derivate of the four series. this crossed self-reference might be solvable directly or recursively. regards tommy1729 |