Sachchidanand Prasad

09-05-2023

Problem: Find all entire functions $f$ such that $f^\prime (0) = 0$ and $f^{\prime\prime} \left( 1+ \frac{1}{n} \right) = 7 - \frac{3}{n}$ for all $n\in \mathbb{N} $.

Solution: Before solving this problem, we recall the identity theorem.

Let $\Omega $ be a domain (open and connected) in $\mathbb{C} $ and $f,g:\Omega \to \mathbb{C} $ be analytic functions. If the set $E= \{z\in \Omega :f(z)=g(z)\}$ contains a limit point, then $f(z) = g(z)$ for all $z\in \Omega $.

Consider the function \[ g(z) = f^{\prime\prime} (z) - 10 + 3z. \] Note that \[ g\left( 1+\frac{1}{n} \right) = f^{\prime\prime} \left( 1+\frac{1}{n} \right) - 10 + 3\left( 1 + \frac{1}{n} \right) = f^{\prime\prime} \left( 1+\frac{1}{n} \right) -7 + \frac{3}{n} = 0 \] The set $\left\{ 1+\frac{1}{n}:n\in \mathbb{N} \right\} $ has a limit point, so $g\equiv 0$ on all of $\mathbb{C} $(by using the identity theorem)

Thus, we have \[ f^{\prime\prime} (z) = 10 - 3z. \] This implies \[ f^\prime (z) = 10z - \frac{3z^2}{2} + c. \] Using the condition that $f^\prime (0) = 0$, gives $c=0$. Thus, \[ f^\prime (z) = 10z - \frac{3z^2}{2} \implies f(z) = 5z^2 -\frac{z^3}{2} + c, \] where $c$ is an arbitrary constant.