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