10/28/2013, 11:11 PM
Ok some guts is needed to admit I do not fully understand and to say I find some things not perfectly well explained.
In particular because it sounds stupid and ungrateful , which I am not.
But it needs to be done.
For Kneser's solution alot of attention is going to the Riemann mapping but the " a priori " is not clear to me.
Maybe Im getting old and I am asking question that I have asked before or understood before so plz forgive me if so.
From my experience it is best to ask very specific questions so I will point to the post I find most confusing.
But first a silly question probably , what I will probably know right after I asked :p
About the schroeder equation
F( f(x) ) = q F(x).
Let c be the fixpoint of f(x).
Now it appears to me that F© must be either 0 or infinite.
What usefull stuff can be said if F© = oo ? Or is that completely useless ?
Second , Why do we prefer f ' © = 1 ?
I assume it is ONLY for the easy solvability of the Taylor series or limit formula for the " principal " schroeder function.
Having probably answered those questions somewhat myself , let continue with the MAIN question(s) and the principal schroeder function.
Since we have f © = 0 we thus have a Taylor series expanded at c.
However there is a limit radius of convergence.
And the real line is not included in the radius , so the trouble begins.
I was only able to find ONE POST adressing how to continue before the Riemann mapping ( all the others seemed to be copied or linked to that " mother post " )
It is still not clear to me what exactlty is mapped , what happened to the singularities and if all that does not loose the property of analyticity.
Also Im not sure what kind of " solution " we are suppose to end up with ?? A sexp that has no singularities for Re > 0 ?
What properties are claimed for the Kneser solution ?
Is it only the property of sexp being analytic near the real line for Re > 0 ?
The post that failed to enlighten me was this one :
- with respect to the poster of course -
http://math.eretrandre.org/tetrationforu...hp?tid=213
POST NR 3
It is said : we analyticly continue to ...
HOW ?
What happened to the singularities and limited radius ?
If you use Taylor series you CANNOT have a function that converges on the entire upper plane ; the Taylor series ALWAYS converges in a circle !!?
It is not said how the continuation is done , how we know it is possible , what series expansion we end up with etc etc
So what to make of that ?
I note that mapping a singularity or pole with exp or ln remains a problem ?
Then there follows a claim of simply connected which I find a bit handwaving ?? And what if it contains singularities ??
How is this different from Gottfriend's brown curve ?
I also note that the Riemann mapping may not change the functional equation.
Those 3 pics do not explain all that and perhaps a longer post should have been made.
With respect to the efforts though.
I hope I have sketched what I believe confuses most people about Kneser's method.
regards
tommy1729
In particular because it sounds stupid and ungrateful , which I am not.
But it needs to be done.
For Kneser's solution alot of attention is going to the Riemann mapping but the " a priori " is not clear to me.
Maybe Im getting old and I am asking question that I have asked before or understood before so plz forgive me if so.
From my experience it is best to ask very specific questions so I will point to the post I find most confusing.
But first a silly question probably , what I will probably know right after I asked :p
About the schroeder equation
F( f(x) ) = q F(x).
Let c be the fixpoint of f(x).
Now it appears to me that F© must be either 0 or infinite.
What usefull stuff can be said if F© = oo ? Or is that completely useless ?
Second , Why do we prefer f ' © = 1 ?
I assume it is ONLY for the easy solvability of the Taylor series or limit formula for the " principal " schroeder function.
Having probably answered those questions somewhat myself , let continue with the MAIN question(s) and the principal schroeder function.
Since we have f © = 0 we thus have a Taylor series expanded at c.
However there is a limit radius of convergence.
And the real line is not included in the radius , so the trouble begins.
I was only able to find ONE POST adressing how to continue before the Riemann mapping ( all the others seemed to be copied or linked to that " mother post " )
It is still not clear to me what exactlty is mapped , what happened to the singularities and if all that does not loose the property of analyticity.
Also Im not sure what kind of " solution " we are suppose to end up with ?? A sexp that has no singularities for Re > 0 ?
What properties are claimed for the Kneser solution ?
Is it only the property of sexp being analytic near the real line for Re > 0 ?
The post that failed to enlighten me was this one :
- with respect to the poster of course -
http://math.eretrandre.org/tetrationforu...hp?tid=213
POST NR 3
It is said : we analyticly continue to ...
HOW ?
What happened to the singularities and limited radius ?
If you use Taylor series you CANNOT have a function that converges on the entire upper plane ; the Taylor series ALWAYS converges in a circle !!?
It is not said how the continuation is done , how we know it is possible , what series expansion we end up with etc etc
So what to make of that ?
I note that mapping a singularity or pole with exp or ln remains a problem ?
Then there follows a claim of simply connected which I find a bit handwaving ?? And what if it contains singularities ??
How is this different from Gottfriend's brown curve ?
I also note that the Riemann mapping may not change the functional equation.
Those 3 pics do not explain all that and perhaps a longer post should have been made.
With respect to the efforts though.
I hope I have sketched what I believe confuses most people about Kneser's method.
regards
tommy1729
