# is there any analytic function f on open unit disk $f(0)=1/2$ , $f(1/2)=1/3$ and $f(1/3)=1/4$

is the following statement is true?
there exists an analytic function $f$ on open unit disk $f(0)=1/2$ , $f(1/2)=1/3$ and $f(1/3)=1/4$

can anyone help me please.I have no idea how to solve this problem

## 2 Answers

We can fit a polynomial through any $n$ points, for instance, look at Lagrange interpolation. Clearly, the polynomials are analytic.

Hint: start with $3(x-1/2)(x-1/3)$