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f(f(x)) = exp(x) + x
#8
(12/14/2009, 09:52 AM)bo198214 Wrote:
(12/13/2009, 06:26 PM)tommy1729 Wrote: 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)

Yes, I was also playing with this idea. However when you map a simply connected region with two fixed points on its boundary biholomorphically and real-analytically such that both fixed points go to 0, then you probably have a branch cut on the real axis (consider a sickle with the two fixed points at its ends, if you bent the fixed points to 0, there will be a slit or overlapping on the real axis up to the point where one part of the boundary intersects the real axis).

(12/14/2009, 07:18 AM)BenStandeven Wrote: Using the simplest conjugation function produces something of a mess; to simplify, I'll consider the conjugate F of exp(pi/2 x) instead; the fixed points are +/-I, so we can conjugate with w=z^2 + 1. The inverse function is z=sqrt(w-1); then I get: F(x) = exp(pi/2 * sqrt(x-1))^2 + 1 = exp(pi * sqrt(x-1)) + 1.

Very good example, it also illustrates my point above. The function F is non-real on (0,1) it has a cut from 1 to -oo (as you described)

still i believe in the basic idea ...

didnt say it would be easy Wink

regards

tommy1729
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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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