Re: Seeking Analytic Function


"Jannick Asmus" <jannick.newsREMOVE_ME@xxxxxx> wrote in message
news:g78u9d$j28$1@xxxxxxxxxxxx
On 05.08.2008 04:27, Bill wrote:
"Bill" <Bill_NOSPAM@xxxxxxxxxxx> wrote in message
news:MbCdnQOG5pnYKgrVnZ2dnUVZ_uOdnZ2d@xxxxxxxxxxxxxx
I have reason to belive that there exists an analytic function f defined
on the open unit disk in the complex plane with

1/4 * (z+1)^2 = f^{-1} (f(z) +1), for all z in the open unit disk.

Clearly z=1 is fixed, so f(1) = "infinity". Any suggestions on how to
formulate f?

Thank you,
Bill


By the way, in case there is any confusion, f^{-1} simply denotes the
inverse of f.

Did you try to show that f can be extended to an analytic function on an
open set containing the open unit disk _and_ 1 to get a contradiction by
evaluating the functional relation at z=1? ... I did not check the details
here - just guessing ... ;)

--
Best wishes,
J.

The function f exists. It is implicit that f(1) = "infinity", and so f(1)+1
is too--although technically 1 is not in the domain of f. Note too, that
1/4*(z+1)^2 is not 1-1 on any neighborhood of -1. The suggestion that was
given in the previous post has merit and it leads to a (complicated) closed
form expression for f involving many binomial coefficients. I will
investigate whether I can reduce it. I was working at equating Taylor
Series coefficients and, after applying the chain rule so many times
tonight, started to not be able to see the forest for the trees!

Thanks,
Bill


.

Relevant Pages

  • Complex Analysis question
    ... Show that there is an analytic function f defined on G = ann ... Here D is the open unit disk and annis the punctured open unit ... Riemann Mapping Theorem is discussed. ...
    (sci.math)
  • Re: Seeking Analytic Function
    ... Did you try to show that f can be extended to an analytic function on an open set containing the open unit disk _and_ 1 to get a contradiction by evaluating the functional relation at z=1? ... I was working at equating Taylor Series coefficients and, after applying the chain rule so many times tonight, started to not be able to see the forest for the trees! ... If 1 is a pole of order k - and f does not have other poles on the boundary of E - then you could have a closer look at ^k.f. ...
    (sci.math)
  • Re: Seeking Analytic Function
    ... I have reason to belive that there exists an analytic function f defined ...
    (sci.math)
  • Re: Two challenging analysis problems :)
    ... A N Niel wrote: ... [A uniformly convergent series of analytic ... but *why* is it an analytic function of x in the open unit disk. ...
    (sci.math)