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 fourier superfunction ? tommy1729 Ultimate Fellow Posts: 1,374 Threads: 337 Joined: Feb 2009 06/22/2010, 10:58 PM 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 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/23/2010, 01:20 AM $f(e^x)$ 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. $3e^{2x} \ne 3x^2$. 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... tommy1729 Ultimate Fellow Posts: 1,374 Threads: 337 Joined: Feb 2009 06/23/2010, 10:52 PM (06/23/2010, 01:20 AM)mike3 Wrote: $f(e^x)$ 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. $3e^{2x} \ne 3x^2$. 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 « Next Oldest | Next Newest »

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