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

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