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 f(f(x)) = exp(x) + x bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 12/12/2009, 12:25 PM (This post was last modified: 12/12/2009, 12:39 PM by bo198214.) 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. This condition is met, the asymptotic derivatives for $z\to -\infty$ of $F$ are: $F'(-\infty)=\lim_{z\to -\infty} e^z+1=1$ and all higher derivatives are $F^{(n)}(-\infty) = 0$. I.e. $F$ has asymptotically the same derivatives as the identity function. So we can apply the limit formula for regular iteration with multiplier 1 (derived from Lévy's formula for the Abel function): (*) $F^{[t]}(z) = F^{[n]}( t F^{[-n+1]}(z) + (1-t) F^{[-n]}(z) )$ This formula converges only very slowly, but it suffices to get 2 or 3 digits to plot a graph of $f=F^{[0.5]}$ (blue):     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. « 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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