02/11/2009, 10:56 PM
I dont really see what you mean.
And I gave you exactly those two distinct analytic functions:
\( f_1(x)=F(\frac{1}{2}+F^{-1}(x)) \)
and
\( f_2(x)=F_\theta(\frac{1}{2}+F^{-1}_\theta(x)) \)
Let one solution be \( f(x)=\exp(x) \) and let the other solution be some other solution be \( f_2=\exp\left(x+\frac{1}{2\pi}\sin(2\pi x)\right) \). Then
\( f(1+f^{-1}(x))=ex=f_2(1+f_2^{-1}(x)) \) but
\( f\left(\frac{1}{2}+f^{-1}(x)\right)=e^{\frac{1}{2}}x\neq f_2\left(\frac{1}{2}+f_2^{-1}(x)\right) \)
tommy1729 Wrote:taking that into account , its probably clear that i will only accept examples of 2 distinct analytic solutions f(x) that map all reals to a subset of reals and satisfy f(f(x)) = exp(x).
And I gave you exactly those two distinct analytic functions:
\( f_1(x)=F(\frac{1}{2}+F^{-1}(x)) \)
and
\( f_2(x)=F_\theta(\frac{1}{2}+F^{-1}_\theta(x)) \)
Quote:( as an analogue : there are multiple functions that satisfy f(x+1) = e*f(x) but that doesnt mean there are multiple functions that are a solution to exp( log(x) + 1 ) ( being e*x ) )
Let one solution be \( f(x)=\exp(x) \) and let the other solution be some other solution be \( f_2=\exp\left(x+\frac{1}{2\pi}\sin(2\pi x)\right) \). Then
\( f(1+f^{-1}(x))=ex=f_2(1+f_2^{-1}(x)) \) but
\( f\left(\frac{1}{2}+f^{-1}(x)\right)=e^{\frac{1}{2}}x\neq f_2\left(\frac{1}{2}+f_2^{-1}(x)\right) \)