Does there exist an analytic function $f:D\to\mathbb{C}$ such that $f(1/n)=f(-1/n)=1/n^3$ "Does there exist an analytic function $f:D\to\mathbb{C}$ such that $f(1/n)=f(-1/n)=1/n^3$?"
This is one of the past qualifying exam problems that I am working on and I found that 
$f(0)=0$, $f^{(n)}(0)=0,n=1,2$, $f^{(3)}(0)=1$ using the definition of derivative of a function. I am trying to use a Taylor expansion at z=0 since f is analytic in $D=\{z\in \mathbb{C}||z|=1\}$. However I do not know how to use $f(1/n)=f(-1/n)=1/n^3$ to prove or disprove the existence of such function $f$. 
Any help would be appreciated. 
Thank you in advance. 

No. Use identity theorem to show that if $g(z)=z^3$ and $h(z)=-z^3$ then $f=g=h↯.$
