Periodic polynomial?

Question

I was thinking if it was possible to create a polynomial that would be periodic all over the reals, since polynomials can be periodic on an interval. I then I found out the following function:

$$P(x)= x \prod_{k=1}^\infty (x-k)(x+k)$$

I wonder if that function can be considered as polynomial since it's degree will be $\infty$. Then is this function periodic of period 1? Since the product is to infinity, $P(x+1) - P(x) = 0$. Moreover, how this function look like? I couldn't trace it on the Mac Grapher app nor Wolfram alpha doesn't understand my commands.

Answer 1

Well, it's not a polynomial since it has infinite degree, and also it doesn't even converge! An infinite product $\prod a_i$ can only converge if $a_i\to 1$ as $i\to \infty$. But you can see that no matter what $x$ you plug in, $(x-k)(x+k)\to-\infty$ as $k\to \infty$, so it can't converge. (Actually I guess it does converge when $x$ is an integer, since the product is $0$.)

Answer 2

The function diverges nearly everywhere, but changing it a bit to $$x \cdot \prod_{k=1}^\infty \left( 1-\frac{x^2}{k^2}\right)$$ (as $(x-k)(x+k)=x^2-k^2$) gives you the zeroes. But this one is $$x \cdot \prod_{k=1}^\infty \left( 1-\frac{x^2}{k^2}\right)=\pi \cdot \sin(\pi\cdot x)$$

Answer 3

First, this is not a polynomial. The product does not even converge on $\mathbb{R}\setminus \mathbb{Z}$.

Second, if a polynomial is periodic over $\mathbb{R}$, say $P(x+T)=P(x)$, then $P(x)-P(0)$ vanishes on $T\mathbb{Z}$, so it is constant equal to $0$. Thus $P$ is constant. Of course, every constant polynomial is periodic. So the periodic polynomials over $\mathbb{R}$ are the constant polynomials (the same is true over any infinite field).