1. ## Complex variables

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

2. 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}
$
, 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)?)

3. 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

$\chi$ $\sigma$

4. 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.