Periodic polynomial?

3,410

Solution 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 $0$.)

Solution 2

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).

Solution 3

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 $$x \cdot \prod_{k=1}^\infty \left( 1-\frac{x^2}{k^2}\right)=\pi \cdot \sin(\pi\cdot x)$$

Share:
3,410

Related videos on Youtube

moray95
Author by

moray95

Updated on August 01, 2022

Comments

  • moray95
    moray95 over 1 year

    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.

    • vonbrand
      vonbrand over 10 years
      Close relative to Euler's product formula $\dfrac{\sin x}{x} = \prod_{k \ge 1}\left( 1 - \frac{x^2}{k^2 \pi^2}\right)$...
    • DonAntonio
      DonAntonio over 10 years
      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...
    • alex.jordan
      alex.jordan over 10 years
      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.
    • moray95
      moray95 over 10 years
      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?
    • vonbrand
      vonbrand over 10 years
      @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.
  • moray95
    moray95 over 10 years
    Okay, got your purpose but can just explain briefly why An infinite product ∏ai can only converge if ai→1 as i→∞?
  • Cheerful Parsnip
    Cheerful Parsnip over 10 years
    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...
  • Glen Wheeler
    Glen Wheeler over 10 years
    This is wrong -- consider $\prod (1/n)$.
  • Cheerful Parsnip
    Cheerful Parsnip over 10 years
    @Glen: thanks; I wasn't careful.