Is there a quick way to prove that $\sin(x)-x^2$ has exactly $2$ real roots?

Question

I found a proof of this fact, but it is not concise as I would like. Would anyone have any advise on how to shorten it? Or an alternative shorter proof? Thank you very much in advance.

$\textbf{Proof}$

Notice that $\sin(x)-x^2$ is $0$ at $0$. Since $(\sin(x)-x^2)''=-\sin(x)-x^2\leq 0$ and is $0$ exactly when $x=0$ it follows that on $(-\infty,0)\cup(0,\infty)$ that $(\sin(x)-x^2)'$ is strictly decreasing. This implies that $(\sin(x)-x^2)'$ has at most one root (its positive at $0$). Since $$\cos(0)-2(0)>0>\cos(1)-2(1)$$ it has a root on $(0,1)$ by Darboux Theorem. Say $y$ is the unique root of $(\sin(x)-x^2)'$.

It follows that $\sin(x)-x^2$ can have most one root on $(-\infty,y)$ and at most one root on $(y,\infty)$. Moreover, by continuity $y$ is not a root. It remains to show that $\sin(x)-x^2$ has a root on $(y,\infty)$. Since $\sin(x)-x^2$ is $0$ at $0$, and on $[0,y)$ it is strictly increasing by continuity it follows that $\sin(y)-y^2>0$. Since $$\sin(2)-(2)^2<0$$ it follows from I.V.T. that $\sin(x)-x^2$ has a root on $(y,2)$. This completes the proof.$\square$

Answer 1

You already know that $0$ is a root of $\sin x - x^2$. Therefore, you only need to find a range for the other root.

It is easy to see that $\sin x - x^2$ does not have roots for $|x| \geq 1$, so you only need to check for $|x| < 1$.

In this case, we may try simple values of $x$ for which $\sin x$ is known, and see if things work out. For example, $\sin \frac{\pi}{6} = \frac 12$, so we can check if $\frac 12 - \frac{\pi^2}{36} > 0$. Thankfully, it is, since $\pi^2 < 10$. Hence, by IVT there is a root of $\sin x - x^2$ between $\frac{\pi}{6}$ and $1$, since $\sin 1 - 1^2 < 0$. Therefore, there are at least two roots of $\sin x - x^2$.

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That there cannot be more follows from Rolle's theorem : if there were three or more distinct roots of $\sin x - x^2$, this would give rise to two distinct roots of $\cos x - 2x$, which would then give rise to a root of $\sin x -2$, which is impossible, as $|\sin x| \leq 1$. Therefore,there are at most two distinct roots of $\sin x - x^2$. Combining with the previous frame gives the desired result.

This proof avoids the use of the unknown $y$, and instead makes use of the fact that a nice value when plugged into $\sin x - x^2$ gives a positive value, which can be used to produce a root.

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Let us call the second root as $l$. So $l$ satisfies $l^2 = \sin l$. The important thing is that $0$ does not satisfy the derivative which is $\cos x - 2x$. In fact, neither does $l$, since $\cos l = 2l$ would imply that $l^4 + 4l^2 = \sin^2 l + \cos^2 l = 1$, and therefore $l$ gives four values, which can be checked to not satisfy $l^2 = \sin l$. Consequently, up to multiplicity as well, $\sin x - x^2$ has only two real roots.

Answer 2

Note that $x=0$ is a solution by inspection.

Now, $x^2 > \sin x$ for all $x<0$ and all $x>1$, so we need only consider the interval $x\in(0,1)$. Since $\sin x = x - \mathcal{O}(x^3)$, we see that $\max(x^2,\sin x)=\sin x$ for small enough $x$, say $x=0.1.$ Because both functions monotonically increase on the interval, there must be exactly one intersection.

Therefore, we have two real roots.

Answer 3

Obviously, any solution must satisfy $x\in[-1,1]$, since $|sin(x)|\leq1$. $sin(x)=sin(x+2\pi)$, which could mean infinite solutions, but if $x\in[-1,1]$ then $x+2n\pi\notin[-1,1]$, so there are at most two solutions. Zero is a solution, and $sin(x)\leq x^2$ if $sin(x)\leq \cfrac{\pi}{6}$, which you can see from a graph or by looking at their concavity and using the fact that they both pass through the origin, and also $x^2$ is strictly increasing with no asymptote, while $sin(x)$ is bounded, so at some point $x^2\geq sin(x)$, meaning they intersect at some point, since they are both finite in that interval. That means they have at least two solutions, and the fact that it is essentially a quadratic means that there are no more solutions. You can also check this, because $sin(x)$ is negative inside $[-2\pi,0]$, and when it is positive again $x^2$ is too big

Answer 4

Well, consider $f(x) =\sin x-x^2$ and $$f'(x) =\cos x- 2x=1-2\sin^2(x/2)-2x$$ Since $|\sin x|\leq 1$ we should observe that any root of $f(x) $ must lie in $[-1,1]$ and moreover since $x^2\geq 0$ any root must lie in $[0,1]$. Since $0$ is a root of $f(x) $ we consider $x\in(0,1]$ and then $$f'(x) > 1-2(x/2)^2-2x=\frac{6-(x+2)^2}{2}$$ and the rightmost expression is positive for all $x$ with $0<x<\sqrt{6}-2$ and hence $f$ is strictly increasing to the right of $0$ and therefore positive in some interval of type $(0,h)$. And $f(1)=\sin 1 - 1<0$ and hence by intermediate value property $f$ has a root in $(0,1)$. Thus we have established that $f$ has two roots and it can't have more than two because $f''(x) =-\sin x - 2<0$ everywhere. The above approach avoids knowing any specific value of $x\in(0,1)$ for which $f(x) >0$.