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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.

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polynomial cannot take any value infinitely many times, unless it is a constant function –  leshik Jan 28 '13 at 18:04
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up vote 5 down vote accepted

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.

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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.

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Maybe a nitpick: The main point of FT of Algebra is existence of a root of a nonconstant polynomial in $\mathbb C$, whereas here we need only finiteness of the set of the roots, which is not particular to $\mathbb C$ and would hold over any other field (or integral domain). It is enough to consider the factor theorem. –  Jonas Meyer Jan 28 '13 at 18:21
    
According to Wikipedia, the FToA is also stated as follows: "Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ roots." The equivalence of the two statements can be proven through the use of successive polynomial division. Using the FToA (or indeed a lemma) is perfectly valid and there are no nits to be picked. Perhaps a simpler proof exists, but I didn't say one didn't. –  Fly by Night Jan 28 '13 at 18:29
    
Yes it is possible. The polynomial constant equal to this value. You need to use the value at odd integers to get a contradiction. –  1015 Jan 28 '13 at 18:42
    
@julien I'm sorry, I don't follow. It is made explicit that $P(n) = (-1)^n$ for all $n \in \mathbb{Z}$ and so $P$ cannot possibly be constant because $P(1) \neq P(2)$. –  Fly by Night Jan 28 '13 at 18:48
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@julien I mentioned the non-constant condition in my original post. It is clear that $P$ is non-constant. –  Fly by Night Jan 28 '13 at 19:05
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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.

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Yeah, should be easy as pie. –  Hagen von Eitzen Jan 28 '13 at 18:39
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