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I am having a problem with the following exercise.

I need to show the $x^2 = \cos x $ has two solutions.

Thank you in advance.

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In what interval? – DonAntonio Nov 14 '12 at 12:45
The interval is not specified – user43418 Nov 14 '12 at 12:49
up vote 2 down vote accepted

$f(x) = x^2 - \cos x$ is a continuous function. Since $f(0) = -1$ and $f(\frac{\pi}{2}) = \frac{\pi^2}{4}$, $f$ has at least one zero in the interval $(0, \pi/2)$. Its derivative $f'(x) = 2x + \sin x$ is strictly positive in the interval $(0,\pi/2)$ so $f$ is strictly increasing and we conclude that $f$ has exactly one root $f(x_0) = 0$ in the interval $(0,\pi/2)$.

For $x \geq \pi/2$, $f(x) \geq \frac{\pi^2}{4} - 1 > 0$, so $f$ has no roots in $[\pi/2, +\infty)$.

Since $f$ is symmetric with respect to the $y$-axis, meaning $f(x) = f(-x)$, $f$ has only two roots -- $x_0$ and $-x_0$.

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I was asked as well to solve this equation using Taylor polynomials (I did that) and found that $x_0=± \sqrt{\frac{2}{3}}$. But what are the bounds of the error? – user43418 Nov 14 '12 at 13:55
@user43418: The bound for the error in $y$ is the error in the Taylor series, $\frac {x^4}{4!}$. You can use the slope of the function at this point to approximate the error in $x$. – Ross Millikan Nov 14 '12 at 14:02
I don't understand... – user43418 Nov 14 '12 at 14:05

Let $f(x)=x^2-\cos x$. Note that the curve $y=f(x)$ is symmetric about the $y$-axis. It will thus be enough to show that $f(x)=0$ has a unique positive solution. That there is a unique negative solution follows by symmetry.

There is a positive solution, since $f(0)\lt 0$ and $f(100)\gt 0$. (Then use the Intermediate Value Theorem.)

For uniqueness of the positive solution, note that $f'(x)=2x+\sin x$. In the interval $(0,\pi/2)$, $f'(x)$ is positive because both terms are positive. And for $x\ge \pi/2$, we have $f'(x)\ge \pi-1$.

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Assuming you want to show that $f(x) = \cos x - x^2$ has two solutions over $\mathbb R$, you could try the following approach: Since both $\cos x$ and $x^2$ are even functions, if $f(x)$ has a solution $x > 0$, $-x$ is also a solution. Therefore, it is sufficient to prove that $f(0) \neq 0$ and there is exactly one solution $x \in ]0,\infty[$.

The first part is easy, since $f(0) = \cos 0 - 0^2 = 1$. So, split up the interval $]0,\infty[$ into two parts: $]0,1]$ and $]1,\infty[$. In the following, we will use three well-known properties of $\cos x$, namely:

  1. $\cos x$ is strictly decreasing on $[0,\pi]$.
  2. $|\cos x| \le 1$ for all $x \in \mathbb R$.
  3. $\cos$ is continuous.

In the first interval, we have that $1 = \cos 0 > \cos 1$. On the other hand, $0^2 = 0 < 1$ and $1^2 = 1 > \cos 1$, so $f(0) = 1$ and $f(1) < 0$. By continuity of $\cos$ and $x^2$, $f$ is continuous. Hence by the Intermediate Value Theorem, the equation $f(x) = 0$ has a solution in $[0,1]$, and that solution is not $0$.

Furthermore, let $x > 1$. Then $f(x) \le 1- x^2 < 0$ by property (2), so there are no solutions for $x > 1$ (and, by $f(x)$ even, $x < -1$).

It remains to show that there is no further solution in $[0,1]$. But this follows from the fact that $f(x)$ is strictly decreasing on $[0,1]$.

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Here is a different approach. Consider the function $$ F(x)=\frac{x^3}{3}-\sin x. $$ Since $DF(x)=x^2-\cos x$, any critical point of $F$ solves your equation. Now, $F(x) \to -\infty$ as $x \to -\infty$, and $F(x) \to +\infty$ as $x \to +\infty$. Since $DF(0)=-1$ and $F(0)=0$, there are points $x_1<0$ and $x_2>0$ where $F(x_1)>0$ and $F(x_2)<0$. Therefore $F$ must have a local maximum point and a local minimum point: two critical points.

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