Are there polynomials $$R_2(x) ,R_1(x),R_0(x)$$ $$R_2(x)\text{ not the zero polynomial} $$ such as $$R_2(x)\sin^2(x) +R_1(x)\sin(x)+R_0(x)=0\quad \forall x\in[a,b]?$$


Notice that we are dealing with analytic functions here. Therefore, by the identity theorem, if the equality$$R_2(x)\sin^2(x)+R_1(x)\sin(x)+R_0(x)=0$$holds on $[a,b]$, it holds for every $x\in\mathbb R$.

So, the answer is negative, because then$$(\forall n\in\mathbb{Z}):R_0(n\pi)=-R_2(n\pi)\sin^2(n\pi)-R_1(n\pi)\sin(n\pi)=0,$$Since a non-zero polynomial cannot have infinitely many zeros, $R_0(x)\equiv0$. So, we have$$R_2(x)\sin^2(x)-R_1(x)\sin(x)=0,$$and this implies that $R_2(x)\sin(x)+R_1(x)=0$. By the same argument as above, $R_1(x)\equiv0$. So, $R_2(x)\sin(x)=0$ and this implies that $R_2(x)\equiv0$, which goes against our assumptions.

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    $\begingroup$ @TZakrevskiy I have edited my answer. $\endgroup$ – José Carlos Santos Oct 18 '17 at 22:25
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    $\begingroup$ @JoséCarlosSantos Using the fact they're entire functions was also my idea. +1 $\endgroup$ – egreg Oct 18 '17 at 22:26
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    $\begingroup$ Nice one with entire functions, +1 $\endgroup$ – TZakrevskiy Oct 18 '17 at 22:29
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    $\begingroup$ @TZakrevskiy Note that I believe that there must be a not too complicated way of solving the problem without using this approach. $\endgroup$ – José Carlos Santos Oct 18 '17 at 22:29
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    $\begingroup$ @JoséCarlosSantos I believe it, too, but I can't find it right now $\endgroup$ – TZakrevskiy Oct 18 '17 at 22:32


Let $$f(x):=p\cos t+q\sin t+r=0$$ where $p,q,r$ are polynomials.

Differentiating twice and adding the initial equation,

$$f''(t)+f(t)=p''\cos t-2p'\sin t+q''\sin t+2q'\cos t+r+r''=0.$$

In this expression, the degrees of the factors of $\cos t$ and $\sin t$ have decreased by one. We can iterate and in the end we will obtain an expression of the form

$$a\cos t+b\sin t+s=0$$ where at least one of $a,b$ is a nonzero constant. Then by further differentiations,

$$c\cos t+d\sin t=0,$$ which is not possible.

I am fairly confident that this generalizes to quadratic expressions in $\cos t$ and $\sin t$.


If you apply the differential operator $(D^4+5D^2+4)$ to an expression of the form

$$p\cos(2t)+q\sin(2t)+r\cos(t)+s\sin(t)+u$$ where at least one of $p,q,r,s$ is a non-constant polynomial, you get an expression of the same form, where the largest polynomial degree is lower.

(For instance, $x^4\cos(2x)\to48x^3\sin(2x)-228x^2\cos(2x)-192x\sin(2x)+24\cos(2x)$. It can be proven formally for any monomial, then polynomial, but the computation is a little tedious.)

By iterating, you eventually obtain a non-trivial linear combination of $$\cos(2t),\sin(2t),\cos(t),\sin(t),$$

which cannot be identically zero.

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    $\begingroup$ If we replace $\sin^2x$ with $\frac {1-\cos 2x}{2} $ and then try to apply this method directly, the extra factor $4$ in $-4p(x) \cos 2x$ (and the absence of this factor in $-q(x)\sin x $) will not allow us to go further. $\endgroup$ – TZakrevskiy Oct 19 '17 at 9:10
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    $\begingroup$ @TZakrevskiy: doesn't $g(t):=f''(t)+f(t)$ then $h(t):=g''(t)+4g(t)$ deal with that ? $\endgroup$ – Yves Daoust Oct 19 '17 at 9:22
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    $\begingroup$ We can try something easier: $g=f'+f$, this will give us a combination of $\sin 2x$, $\cos 2x$, $\sin x$, and $\cos x$ with polynomial coefficients. After that, by a sufficient number of maps $h\to h''+4h$ we can reduce this expression to a combination of just $\cos x$ and $\sin x$, and then apply your method. There will be some points where we would need to prove that along all those differentiations we don't loose some information, but that seems easy. $\endgroup$ – TZakrevskiy Oct 19 '17 at 9:45
  • $\begingroup$ Sorry guys... I have just noticed that I can vote your nice answers! $\endgroup$ – dmtri Nov 22 '17 at 12:39
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    $\begingroup$ @dmtri: you are also deemed to accept (√ sign) the answer that suits you (if any). $\endgroup$ – Yves Daoust Nov 22 '17 at 14:37

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