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Full Version: [2014] tommy's theorem sexp ' (z) =/= 0 ?
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Let sexp(z) be a solution that is analytic in the entire complex plane apart from z=-2,-3,-4,...

if w is a (finite) nonreal complex number such that

sexp ' (w) = 0

then it follows that for real k>0 :

sexp ' (w+k) = 0.

Proof : chain rule

exp^[k] is analytic :

sexp(w+k) = exp^[k](sexp(w))

sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0

Hence we get a contradiction : sexp is not nonpolynomial analytic near w (or w + k).
Conclusion there is no w such that sexp'(w) = 0.

Consequences : since 0 < sexp ' (z) < oo

slog ' (z) is also 0 < slog ' (z) < oo

since exp^[k](v) = sexp(slog(v)+k)

D exp^[k](v) = sexp ' (slog(v)+k) * slog ' (v) = nonzero * nonzero = nonzero..

=> 0 < exp^[k] ' (z) < oo

Tommy's theorem

Strongly related to the TPID 4 thread and some recent conjectures of sheldon.

the analogue difference is not understood yet.

regards

tommy1729
(06/17/2014, 12:18 PM)tommy1729 Wrote: [ -> ]Let sexp(z) be a solution that is analytic in the entire complex plane apart from z=-2,-3,-4,...

if w is a (finite) nonreal complex number such that

sexp ' (w) = 0

then it follows that for real k>0 :

sexp ' (w+k) = 0.

Proof : chain rule

exp^[k] is analytic :

sexp(w+k) = exp^[k](sexp(w))

sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0
....
Hence we get a contradiction : sexp is not nonpolynomial analytic near w (or w + k).
Conclusion there is no w such that sexp'(w) = 0.

http://math.eretrandre.org/tetrationforu...452&page=2 see post#19 and post#20, for the Taylor series of an sexp(z) function with first and second derivatives, sexp'(n)=0 and sexp''(n)=0, for all integers n>-2. This is sexp(z) from the secondary fixed point, analytic in the upper and lower halves of the complex plane. Where the derivative of sexp(x)=0, the slog(z) inverse has a cuberoot(0) branch singularity.

The flaw in your proof is that k is an integer, but you state k as a real number. "sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0". For k as a fraction, the chain rule does not apply. A simple counter example to your proof is f(z), which has f'(n)=0 and f''(n)=0 for all integers>-2.

$f(z) = \text{sexp}(z - \frac{sin(2\pi z)}{2\pi})$

As a side note, $f^{-1}(z)$ has a cube root branch singularity for n>=0, at $z=\exp^{on}(0)$. This is relevant since the exp^k(z) function used in the flawed proof is $f(f^{-1}(z)+k)$