Find all functions $\displaystyle f (z ) $ satisfying the following two conditions:
(1) $\displaystyle f (z ) $ is analytic in the disk $\displaystyle |z - 1| < 1 $ .
(2) $\displaystyle f ( \frac{n}{n + 1} ) = 1 - \frac{1}{2n^2 + 2n + 1} $.
Find all functions $\displaystyle f (z ) $ satisfying the following two conditions:
(1) $\displaystyle f (z ) $ is analytic in the disk $\displaystyle |z - 1| < 1 $ .
(2) $\displaystyle f ( \frac{n}{n + 1} ) = 1 - \frac{1}{2n^2 + 2n + 1} $.
$\displaystyle \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 $\displaystyle f(z) = 1 - \frac{(1-z)^2}{1+z^2}$. (But could there be any other analytic functions taking those values at the points n/(n+1)?)
In a problem i'm working about one important step is to demonstrate this lemma...
Let be $\displaystyle f(*)$ an analytic function whose value $\displaystyle f_{n}$ are known for $\displaystyle z=0,1,...,n, ...$. In this case, under certain conditions, there is only one analytic $\displaystyle f(*)$ for which is $\displaystyle f(n)= f_{n}$.
Does Opalg think that is an interesting question to be discussed in MHF?... if yes, in which section?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$