TPID 4

[tommy1729]

The only weird thing I can come up with is this :

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

So exp[k] ' (sexp(w)) sexp ' (w) = exp[w] ' (sexp(k)) sexp ' (k)

So if sexp ' (w) = 0

0 = exp[w] ' (sexp(k)) sexp ' (k)

AND that is weird ...

Not sure if to relax or get excited now.
This is not satisfactional right now.

0 = exp[w] ' (sexp(k)) sexp ' (k)
Is it an argument for or against ?? Even that is not clear to me.

k = w+1 , its know that sexp ' (w+1) = 0 thus

0 = exp[w] ' (sexp(w+1)) sexp ' (w+1) is valid.

This is equivalent to sexp ' (w) = 0 => sexp ' (2w+1) = 0
Or so it appears.

But it gets stranger :

By repetition :

sexp ' (w) = 0 => sexp ' (2w+1) = 0 => sexp ' ((2w+1)^[2]) = 0 => sexp ' ((2w+1)^[oo]) = 0

The finite limit (2w+1)^[oo] is the fixpoint of 2z + 1. Thus we solve 2z+1 = z.

2z+1 = z => do -z on both sides => z + 1 = 0 => z = -1.

So if sexp ' (w) = 0 then sexp ' (-1) = 0

But if sexp ' (-1) = 0 then sexp ' (0) = 0. And sexp ' (1) = sexp ' (2) = ... = 0

However Im not finished !

(2w + 1)^[-1] also applies !

so sexp ' (2) = 0 , take a "new" w := 2w + 1 = 2.

then 2w + 1 = 2 => 2w = 1 => w = 1/2.

And then we get the result that sexp ' ( half-integer > 0) = 0.

By induction this becomes :

sexp ' ( positive real ) = 0.

So sexp(z) is of the form A z + B.

Seems like another strong case for tommy's theorem and the chain rule.

regards

tommy1729