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Coloring F(x) = F(exp(x))
#1
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
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#2
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
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#3
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
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#4
The naive conjecture seems to ignore 2 pi i.

Thinking ...

Regards

Tommy1729
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#5
STOP due to the addiction!!! Huh
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#6
(05/14/2015, 10:35 PM)nuninho1980 Wrote: STOP due to the addiction!!! Huh

???
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#7
(05/14/2015, 11:05 PM)tommy1729 Wrote: ???
You should be (much) slower to reply.
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