TPID 4

[tommy1729]

More about that :

Let f(z) be a holomorphic function in all relevant domains.
At complex infinity f(z) is either undefined or it has a singularity.
( now clearly f(z) is not a polynomial )
Let g(z) be a holomorphic function in all relevant domains apart for z=0 and ignoring branches.
Let h(z) be a meromorphic function in all relevant domains.
Also g ' (z) =/= h(z).

To prove under these conditions :
f ( g(z) ) is not analytic at z = 0.

Proof :

If f ( g(z) ) is analytic at z = 0 then so is D f ( g(z) ) = f ' ( g(z) ) g ' (z).
( use the chain rule )


Since f ' ( g(z) ) is either analytic or not , and g ' (z) is not ( g ' (z) =/= h(z) ) then if f ' ( g(z) ) g ' (z) needs to be analytic then there are only a few options :

A) f ' ( g(z) ) is analytic :

A1) h is a small number , g ' (h) is bounded ==>

A1*) if g ' (h) =/= close to 0 => f ' ( g(h) ) is bounded.

A1**) if g ' (h) = close to 0 => f ' ( g(z) ) * g'(z) = 0.

THUS a product of an analytic function with a singularity = a singularity OR 0.

Hence we require f ' ( g(z) ) * g'(z) = 0.

But also f ' ( g(h) ) * g ' (h) = 0 , thus one of f ' ( g(z) ) or g'(z) must be identical 0.

Since f is not a polynomial this cannot be !

Hence if f ' ( g(z) ) is analytic near z = 0 then f ( g(z) ) is not.
This is 1/3 of the proof.

A2) g ' (h) is unbounded => f ' ( g(h) ) is close to 0.

(notice : If g ' (h) is unbounded then so is g(h).)

This requires f ' ( g(0) ) = f ' ( complex oo ) = 0.
But f ' ( complex oo ) has a singularty because f ( complex oo) does (from def).


So 2) is not possible if f ' ( g(z) ) must be analytic.

B) f ' ( g(z) ) is not analytic :

The product of 2 nonanalytic functions is nonanalytic ??

Not that simple.

We reduced the problem

f(g(z)) is not analytic near z = 0

=>

f ' (g(z)) is nonanalytic near z = 0 -> f ' (g(z)) g'(z) is not analytic near z = 0.

Lemma ?? : if f ' (g(z)) is not analytic then f ' ( g(z) ) is well approximated by truncated_Taylor_f( g(z) ).

By this dubious lemma we get :

a0 g'(z) + a1 g(z) g'(z) + a2 g(z)^2 g'(z) + ...

By integration ( if integral G dx is analytic then so is G , if integral G dx is not analytic then neither is G )

t(z) = C0 + c1 g(z) + c2 g(z)^2 + c3 g(z)^3 + ...

Now if g(z) has no algebraic singularities then this t(z) is not analytic.

A step closer to a proof perhaps.

...

A(z) * g ' (z) = analytic.
show A(z) =/= f ' ( g(z) )

Assume A(z) = f ' ( g(z) )

Let g(Q(z)) = z.

A(Q(z)) = f ' (z).

=>
C) Q(z) is not analytic =>

C1) A(Q(z)) * g ' (Q(z)) is not analytic.

...

C2) A(Q(z)) * g ' (Q(z)) is analytic.

...

D) Q(z) is analytic

...

Im getting tired ...

Maybe this is in the textbooks.
I have not even considered the special case of sexp or 1-periodic and the question is if that is the right strategy or the wrong strategy.


regards

tommy1729