Complex variables

**vincisonfire** · March 25th 2009, 03:02 PM

Find all functions $f (z )$ satisfying the following two conditions:
(1) $f (z )$ is analytic in the disk $|z - 1| < 1$ .
(2) $f ( \frac{n}{n + 1} ) = 1 - \frac{1}{2n^2 + 2n + 1}$ .

**Opalg** · March 26th 2009, 12:25 AM

Originally Posted by vincisonfire

Find all functions $f (z )$ satisfying the following two conditions:
(1) $f (z )$ is analytic in the disk $|z - 1| < 1$ .
(2) $f ( \frac{n}{n + 1} ) = 1 - \frac{1}{2n^2 + 2n + 1}$ .

$\frac1{2n^2+2n+1} = \frac1{(n+1)^2+n^2} = \frac{\frac1{(n+1)^2}}{1+\bigl(\frac n{n+1}\bigr)^2} = \frac{\bigl(1-\frac n{n+1}\bigr)^2}{1+\bigl(\frac n{n+1}\bigr)^2}<br />$ , so you can take $f(z) = 1 - \frac{(1-z)^2}{1+z^2}$ . (But could there be any other analytic functions taking those values at the points n/(n+1)?)

**chisigma** · March 26th 2009, 03:30 AM

Originally Posted by Opalg

... but could there be any other analytic functions taking those values at the points n/(n+1)?...

In a problem i'm working about one important step is to demonstrate this lemma...

Let be $f(*)$ an analytic function whose value $f_{n}$ are known for $z=0,1,...,n, ...$ . In this case, under certain conditions, there is only one analytic $f(*)$ for which is $f(n)= f_{n}$ .

Does Opalg think that is an interesting question to be discussed in MHF?... if yes, in which section?...

Kind regards

$\chi$ $\sigma$

**Opalg** · March 26th 2009, 01:06 PM

Originally Posted by chisigma

In a problem i'm working about one important step is to demonstrate this lemma...

Let be $f(*)$ an analytic function whose value $f_{n}$ are known for $z=0,1,...,n, ...$ . In this case, under certain conditions, there is only one analytic $f(*)$ for which is $f(n)= f_{n}$ .

Does Opalg think that is an interesting question to be discussed in MHF?... if yes, in which section?...

Yes, that's a nice question. You could ask it as a new thread in this section.

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