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 f(f(x)) = exp(x) + x tommy1729 Ultimate Fellow Posts: 1,359 Threads: 331 Joined: Feb 2009 12/13/2009, 06:26 PM (12/12/2009, 12:25 PM)bo198214 Wrote: Indeed $F(x)=e^x+x$ (which I will call here added exponential) is a very interesting function as it has no complex fixpoints. If there was some complex fixpoint $z$ then $e^z+z=z$, means $e^z=0$ which is never true. So one could think that one can not apply regular iteration. However as Tommy already suggested there is a fixed point at (complex) infinity. Well, the function is not analytic there, but for regular iteration it suffices that the function has an asymptotic powerseries development (approaching the fixed point in some sector) or even merely that the function is asymptotically real differentiable. (((snip part of quote))) Another way is to consider the analytic conjugate $M = \exp \circ F \circ \log$. $M(x)=\exp(x+\log(x))=e^x x$ is the (what I call) multiplied exponential, which moves the fixed point at $-\infty$ of $F$ to the fixed point 0 of $M$. This function is well regularly iterable at 0 (with powerseries and limit formula) at then $F^{[t]}=\log\circ M^{[t]} \circ\exp$. The conjugation formula shows also how to compute the inverse of $F(x)=x+e^x$ which is needed e.g. in the limit formula. The inverse is $F^{[-1]} = \log\circ M^{[-1]}\circ \exp$, i.e. $F^{[-1]}(x) = \log(W(e^x))$, where $W$ is the Lambert function. that is intresting , though i dont know why you call it 'analytic conjugate' ? funny thing is i considered the above things in reverse order :p back to business , i wanted to point out that perhaps the above approach might also work for real-iterates of exp(z) ?!? basicly bo replaced the fixpoint from - oo to 0 , and i consider the possibility of doing the same with the 2 complex fixpoints of exp(z) towards 0. ( if succesfull i bet we arrive at kouznetsov's solution but with more proven properties ) so we look for strictly increasing functions T(x) resp f(x) with T(x) = f(x) ° exp(x) ° f°-1(x) with a fixpoint at 0. ( i didnt check but ) f(x) might just be simple like f(x) = (x - fp1)^a (x - fp2)^a realpoly(x) note fp1 and fp2 are the solutions of exp(z) = z and are eachother conjugate ( which is very important here ) of course also the correct branches of f°-1(x) resp f(x) are important. regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread f(f(x)) = exp(x) + x - by tommy1729 - 12/12/2009, 12:54 AM RE: f(f(x)) = exp(x) + x - by bo198214 - 12/14/2009, 09:52 AM RE: f(f(x)) = exp(x) + x - by tommy1729 - 12/14/2009, 09:47 PM

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