# Periodic polynomial?

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.

• Close relative to Euler's product formula $\dfrac{\sin x}{x} = \prod_{k \ge 1}\left( 1 - \frac{x^2}{k^2 \pi^2}\right)$... – vonbrand Mar 25 '13 at 18:20
• That function, assumed it is well defined, is far from being a polynomial. To begin with, it has infinite zeros or, if you will, roots: this can't happen with polynomials. Second, I can't see how some definition of "degree" could make some sense as to make this thing "polynomialish". Third, polynomials are everywhere infinitely differentiable: here I am not sure... – DonAntonio Mar 25 '13 at 18:22
• All other commentary aside, what do you mean by "polynomials can be periodic on an interval"? I'm pretty sure the only way that could happen is if the polynomial were constant. – alex.jordan Mar 25 '13 at 18:25
• As I know, the Taylor series approaches any function (then also periodic ones) on an interval. So aren't they periodic on the interval that the function approaches another periodic one? – moray95 Mar 25 '13 at 18:51
• @moray95, approaching a periodic function is not the same as being a periodic function. The partial products of your expression (suitably scaled) get closer to $\sin \pi x$, but aren't really periodic. – vonbrand Mar 25 '13 at 20:10

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 $0$.)
• Look up infinite product on wikipedia. There it is shown that if $a_i>0$, then the product converges only if $a_i\to 1$. If all $a_i\neq 0$, but could be positive or negative, you can argue that if the product converges, then the absolute value of the product converges... – Cheerful Parsnip Mar 25 '13 at 19:46
• This is wrong -- consider $\prod (1/n)$. – Glen Wheeler Apr 12 '13 at 3:33
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 $$x \cdot \prod_{k=1}^\infty \left( 1-\frac{x^2}{k^2}\right)=\pi \cdot \sin(\pi\cdot x)$$
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).