Find all polynomials $P(x)$ with real coefficients, such that,

$$ |P(x)| \leq \sum_{r=0}^{\infty} |\sin (x)|^{r} \quad \forall x\in\mathbb{R}$$

This question looks daunting. Please help.


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If $x$ is a multiple of $\pi$, the series converges to $1$.

If $\deg P\ge 1$, we know that $|P(x)|>1$ for all $|x|\gg 0$, contradicting the above conditon.

Remains the case that $P$ is constant, $P(x)=c$. Then the the remark about $x=k\pi$ leads to $|c|\le 1$. As already the first summand in the series equals $1$ for all $x$, this conditio0n is also sufficient.

Hence the solution consists precisely of the constant polynomials with constant of absolute value $\le 1$.

  • $\begingroup$ So if I understand correctly, this is because $\sin(\pi)=0$, but wouldn't the infinite sum be undefined because at $r=0$ you would end up with $0^0$? $\endgroup$ – JAD Jun 18 '17 at 8:08
  • 3
    $\begingroup$ The issue of the value of $0^0$ has been discussed at length in other questions at this site. Suffice it to say that when the zero in the exponent is limited to nonnegative integers, as for the index of a summation, we define $0^0=1$. Most programming languages follow that convention for $x^y$ when $y$ has integer type. So this answer is correct. $\endgroup$ – Rory Daulton Jun 18 '17 at 9:03
  • $\begingroup$ It'd be trivial to change the answer for $0^0=0$ anyway, you'd just end up with $P(x)=0$ $\endgroup$ – Jam Jun 22 '17 at 14:04

In addition to the good and correct answer by Hagen von Eitzen, let me point out that the series is a geometric one and can be evaluated easily: $$ \sum_{r=0}^{\infty} |\sin (x)|^{r} = \begin{cases} \frac1{1-|\sin(x)|},&\text{when }|\sin(x)|<1\\ \infty,&\text{when }|\sin(x)|=1. \end{cases} $$ The sum of the series is $1$ when $x$ is an integer multiple of $\pi$, and that simple case is enough to draw the conclusion, but even more information is available if you want to use it.


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