does there exist an entire function with this property?

Question

We need to show that there is no polynomial $P$ with complex coefficients such that $P(n)=(-1)^n\forall n\in \mathbb{Z}$. Does there exist an entire function with this property? Hints only, please.

Answer 1

If a polynomial $P(z)$ satisfies your condition, then $P+1$ has infinitely many zeroes (the odd integers) so $P+1=0$, hence $P=-1$. But then it cannot satisfy the condition on the even integers. Contradiction.

For the existence of an entire function satisfying your condition, let's look at $\cos (\pi z)$. This is an entire function. We have $\cos (\pi 2n)= 1$ and $\cos(\pi (2n+1))=-1$ for all $n\in\mathbb{Z}$ . So it works.

Answer 2

HINT:

Regarding the polynomial. Assume that there is a polynomial $P(z) \in \mathbb{C}[z]$ such that $P(n)=(-1)^n$ for all $n \in \mathbb{Z}$. Clearly $P$ is not constant: $P(1)=-1$ while $P(2)=1$. Consider the polynomial restricted to the even integers. We see that $P(2k) = 1$ for all $k \in \mathbb{Z}$.

If such a polynomial were to exist then you would need a non-constant polynomial to take a given value infinitely many times. Is that possible?

Think about the Fundamental Theorem of Algebra. Any non-constant degree-$n$ polynomial has exactly, when counted with multiplicities, exactly $n$-roots.

Answer 3

If there were such a polynomial $P$, then $Q(z):=P(z)-1$ is a non-constant polynomial with infinitely-many zeros. (Why?) Can this happen?

As for finding an entire function with the desired property, try basic transformations of a familiar periodic function.