06/17/2014, 12:18 PM

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.

(posted that already)

regards

tommy1729

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.

(posted that already)

regards

tommy1729