+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3)
+--- Thread: fourier superfunction ? (/showthread.php?tid=460)
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
\( f(e^x) \) 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...
RE: fourier superfunction ? - tommy1729 - 06/23/2010
(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