Divergent Series and Analytical Continuation (LONG post)

[tommy1729]

Ok consider we have only 1 natural boundary.

Now let A be the interior of the boundary and B the exterior.

Then A union B = C the complex sphere without the boundary.

And A intersect B is empty.

Now we want a function L(z) such that 

1) L(z) is meromorphic on the whole complex sphere.

2) L(A) = B and L(B) = A , L(A) =/= A , L(B) =/= B

to be clear every element x of A maps to some L(x) in B , AND every element y of B maps to some L(y) in A. 

3) L© maps bijectively to C. ( f: C arrow C )

A consequence is that L(L(A)) = A , L(L(B) = B 

In fact it follows that L(L(z)) = z.

Not many functions have that property.

In fact the only possible ones are these

L(z) = (a z + b)/(c z + d)

such that one of them is true :

1) a + d = 0
2) L(z) = b - z
3) L(z) = b/z

 

Then every reflection formula equation is of the type 

L_1 ( f(z) ) = f( L_2(z) )

( And this even holds without the existance of a boundary )

(   in fact almost every reflection formula in math books is of the form L_1 ( f(z) ) = f( L_2(z) ) or f( L_1 (z) ) = f( L_2(z) )   )

***

now I mentioned ( here and else ) equations like

f(c/x) + f(x) = W

f(c/x) - f(x) = W - 2W log(x)/ln©

The first one is of the type L_1 ( f(z) ) = f( L_2(z) )

Since L_1 is entire this implies that A and B are entire regions.
So that is even better if you do not likes poles outside the boundary.


f(c/x) - f(x) = W - 2W log(x)/ln©

is a bit different.

one solution is f(x) = W * ln(x)/ln©


f(c/x) = W ln(c/x)/ln© = W (ln© - ln(x)) / ln© = W * ( 1- ln(x)/ln© )

so  f(c/x) - f(x) = W * ( 1 - ln(x)/ln© - ln(x)/ln© ) = W - 2W log(x)/ln©


However 

if f(x) = W * log(x)/ln©

f(c/x) = W ln(c/x)/ln© = W (ln© - ln(x)) / ln© = W * ( 1- ln(x)/ln© )

so  f(c/x) + f(x) = W * ( 1 - ln(x)/ln© + ln(x)/ln© ) = W 


so we solved ( at least one solution )

f(c/x) + f(x) = W

without any boundaries.

But also notice there may be other solutions.

And also notice the log© ; if c = 1 then

f(1/x) + f(x) = W

f(x) = f(1/x) = W/2 

is a solution and others ??

Does this mean we need expansions in terms of log to solve 


L_1 ( f(z) ) = f( L_2(z) )

?

What about the equation I mentioned earlier 

f(x) - f(c/x) = k ln(x)

It seems similar.



I know I only used "algebraic" ideas.
And I only used ideas based on the boundary , not even considering functions usually.

But as I show here , the options are already limited by just considering basic properties.

I did not consider other properties such as the determinant ( ad - bc ) and such ( the theory is rich ).


We use different approaches but seem to get to the same idea.
As often in math.

btw i see no reason to write f as two functions based on f(A) or f(B) but I guess that is a matter of taste.


@james : calm down

so I use only f and not 2 f's , never said it was analytic everywhere ofcourse so no contradiction just a function defined for A and B.

Also I am aware of the terminology generalized holomorphic function.

I did not say nonsense.

I feel you are not calm and rather offensive or defensive.

easy on the fu.. and !!!!!

I also want to point out I was more replying to caleb then to you.

And Im not saying you guys are fundamentally wrong or anything.

Just minor things.


Maybe I misinterpret your reaction.
Maybe I sounded hostile or disagreeing more than I intended.
( debunk was a big word )


@caleb ; what are you thoughts on my comment to your C needing to be 0 ? Or did you made a sign mistake ? I did not double check.


More investigation is needed.



regards

tommy1729