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+--- Thread: Coloring F(x) = F(exp(x)) (/showthread.php?tid=1000)
Coloring F(x) = F(exp(x)) - tommy1729 - 05/10/2015
Consider a 1-periodic real-analytic function of the real-analytic slog for re > 0.
That satisfies :
F(x) = F(exp(x))
However this functional equation can not hold for all nonreal x because exp is a chaotic map.
So we color the complex plane according to how many times the equation holds, more precisely :
Black x
F(x) =\= F(exp(x))
And F(x) =\= F(exp^[2](x))
Purple x
F(x) =\= F(exp(x))
And F(x) = F(exp^[2](x))
Blue x
F(x) = F(exp(x))
And F(x) =/= F(exp^[2](x))
Red x
F(x) = F(exp(x)) = F(exp^[2](x))
Tommy's 4 color conjecture
F is completely determined by its coloring and F(1).
Not to be confused with the 4 color theorem
Regards
Tommy1729
RE: Coloring F(x) = F(exp(x)) - tommy1729 - 05/10/2015
I corrected the title and first post.
Im a bit confused by this :
If F is piecewise analytic and
F(x) = F(exp(x))
Then
F(x) - F(exp(x)) = g(x)
And g(x) should be piecewise analytic too.
But g(x) = 0 locally.
However 0 has no analytic continuation apart from being 0 EVERYWHERE.
This seems like a contradiction.
I might have asked this before srr.
Regards
Tommy1729
RE: Coloring F(x) = F(exp(x)) - tommy1729 - 05/14/2015
Post 2 is simple. G has a singularity when slog has.
The functional equations holds when g = 0.
The question is
Let ln(ln(s)) = s such that ln(s) =\= s.
What is slog(s) ? And slog(exp(s)) - slog(s) ?
Maybe naive but if n > 1 and n is the smallest n such that
Ln^[n](d_n) = d_n
Then the naive conjecture is
Slog(exp^[m](d_n)) - slog(d_n) = m mod n.
Weak version : m positive integer.
Strong version : m positive real.
Variants : negative or complex.
Regards
Tommy1729
RE: Coloring F(x) = F(exp(x)) - tommy1729 - 05/14/2015
The naive conjecture seems to ignore 2 pi i.
Thinking ...
Regards
Tommy1729
RE: Coloring F(x) = F(exp(x)) - nuninho1980 - 05/14/2015
STOP due to the addiction!!!
RE: Coloring F(x) = F(exp(x)) - tommy1729 - 05/14/2015
(05/14/2015, 10:35 PM)nuninho1980 Wrote: STOP due to the addiction!!!
???
RE: Coloring F(x) = F(exp(x)) - nuninho1980 - 05/14/2015
(05/14/2015, 11:05 PM)tommy1729 Wrote: ???You should be (much) slower to reply.