# Tetration Forum

Full Version: fourier superfunction ?
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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
$f(e^x)$ 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...
(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