TPID 4

[tommy1729]

Ok so

f ' (g(z)) is nonanalytic.

f := sexp.

g(z) = z + theta(z).

Whenever Re(g(z)) > 0 and g(z) is bounded then

f ' (g(z)) is bounded as is f(g(z)).

And in that case we have another bounded sexp yet that is not an analytic one.
But the conjecture was bounded and analytic.

Notice that if g(z) is not bounded then f(g(z)) is not bounded and hence the boundedness uniqueness is not broken.

so g(z) must be bounded.

We continue the quest for this hardcore edition of TPID 4 :


1) f ' ( g(z) ) is nonanalytic.

2) g ' (z) is nonanalytic and 1periodic.

3) g(z) is bounded.

to do : proof f(g(z)) is nonanalytic.

From 3) => f ' ( g(z) ) is bounded.

SO if D f(g(z)) is bounded then so is g ' (z).

IF ...
D f(g(z)) is bounded/analytic => f(g(z)) is bounded/analytic => g'(z) is bounded.

CASE ALPHA : f(g(z)) is not bounded near 0 => f(g(z)) not analytic.
Case closed.

CASE BETA : f(g(z)) is bounded => g'(z) is bounded.

HENCE

I) f ' ( g(z) ) is nonanalytic and bounded.

II) g ' (z) is nonanalytic and 1periodic and bounded.

III) g(z) is nonanalytic and bounded.

to do : proof f(g(z)) is nonanalytic.

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sidenote :

assume f(g(z)) is not analytic :
If ln(f(g(z))) =/= log(0)
then
f ' (g(z)) g ' (z)/ f(g(z)) is not analytic.

hence IF f ' / f (g(z)) is analytic then f(g(z)) is not analytic.

now f = sexp

f ' = sexp '

thus f ' / f = sexp'(g(z))/ sexp(g(z)) and I could bring out the continuum product again but that would not make it trivial ...

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It seems natural to consider G( f ' (x) ) = f (x) and hoping that G is analytic.

Then we get G(sexp ' (g(z))) = sexp(g(z)) QED.

But G = f( f ' ^[-1](x)).

So it comes down to sexp ' ^[-1](x) being analytic ?

Now if sexp ' (T) =/= 0 and Re(T) > -2 then it seems

sexp ' ^[-1] (z) is analytic.

But what if sexp ' (T) = 0 ? Then D sexp ' ^[-1](T) = oo.

Now solve for z : (T* such T that lead to D sexp ' ^[-1](T) = oo) :

T* = sexp ' ( g(z) )

then those values z are the only ones where the singularity MIGHT be cancelled If z + theta(z) = g(z) has a singularity there.

partial QED.

Finally !

regards

tommy1729