Finite solution for x

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#1 h Apr 21, 2025, 4:19 AM

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$P(t)$ be a non constant polynomial with real coefficients. Prove that the system of simultaneous equations —
$\int_{0}^{x} P(t)sin t dt =0$ $\int_{0}^{x}P(t) cos t dt=0$ has finitely many solutions $x$ .

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#2 h Apr 21, 2025, 10:41 AM

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One can show by induction on the degree of $P$ that $\begin{align*} \int P(t) \sin t \mathrm{d}t &=F'(t)\sin t - F(t)\cos t + \mathcal{C}, \\ \int P(t) \cos t \mathrm{d}t &= F'(t) \cos t + F(t)\sin(t) + \mathcal{C}, \end{align*}$ where $F(t)=P(t)-P''(t)+P''''(t)- \ldots.$

Thus, our system is equivalent to $\begin{cases}F'(x)\sin x - F(x)\cos x + F(0) = 0 \\ F'(x) \cos x + F(x)\sin x - F'(0) = 0 \end{cases}.$
Multiplying the first equation by $-\cos x$ and the second one by $\sin x,$ and then adding them, we get $F(x)-F(0) \cos x - F'(0) \sin x=0.$
It would therefore be enough to show that an equation of the form $F(x)+a \sin x + b \cos x=0$ with $F$ a nonconstant polynomial has finitely many solutions. If $a=b=0$ this is clear, so let's assume $a^2+b^2 \neq 0.$ Then the equation can be rewritten as $\frac{F(x)}{\sqrt{a^2+b^2}}+ \frac{a}{\sqrt{a^2+b^2}} \sin(x+ \varphi) = 0,$ where $\varphi$ satisfies $\cos \varphi = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \varphi = \frac{b}{\sqrt{a^2+b^2}}.$

So now it is enough to show that an equation of the form $G(x) + a \sin (x+b)=0$ has finitely many solutions, where $G$ is a nonconstant polinomial. We will prove this by induction on the degree of $G.$

Let $h(x) = G(x) + a \sin (x+b).$ If $\deg G = 1,$ then $h'(x)=c+a \cos(x+b),$ where $c$ is some constant. The zero set of $h'$ is either the empty set or a discrete set. If it is the empty set, then $h$ is strictly monotonic, so it has at most one zero. If it is discrete, then so is the zero set of $h$ (as a corollary of Rolle's theorem, between any two consecutive zeroes of $h'$ there is at most one zero of $h$ ). On the other hand, $\lim_{x \to \infty} |h(x)|= \infty,$ so there is some $C$ such that $|h(x)| \ge 1$ for any $x \ge C.$ Thus, the zero set of $h$ is contained in the interval $(-C,C).$ Since the zero set of $h$ is bounded and discrete, it is finite.

Now assume the claim is true for $\deg G = k.$ Then $h'(x)=0$ has finitely many solutions by the induction hypothesis (notice that the equation $G'(x) + a \cos (x+b)$ is still of the form $A(x) + \alpha \sin ( x+ \beta)=0$ because $\cos(x+b)=\sin(x+b + \pi/2)$ ). With the same corollary of Rolle's theorem we conclude that $h$ has finitely many zeroes, which concludes the proof.

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