Solution for $\arctan x = x^2$

Question

Show that the equation $\arctan x = x^2$ has at least one solution. Then explain why the equation has exactly one positive solution $r$.

So i know that $\arctan x$ is always increasing as $f'(\arctan x) > 0$. I also know that $\arctan x$ goes from negative to positive at some point and thus has one root only, but I am not sure how to apply this to the equation I'm dealing with.

I guess i could use fixed point iteration to show that the equation has at least one solution, but i don't know how i would go about doing that.

Answer 1

The function $f(x)=\arctan x-x^2$ has $$ \lim_{x\to-\infty}f(x)=-\infty, \qquad f(0)=0, \qquad \lim_{x\to\infty}f(x)=-\infty, $$ and, moreover, $$ f'(x)=\frac{1}{1+x^2}-2x=\frac{1-2x-2x^3}{1+x^2} $$ Consider $g(x)=1-2x-2x^3$; since $g'(x)=-2-6x^2<0$, the function is decreasing, so it vanishes at exactly one point $p$; note that $g(0)=1$ and $g(1)=-3$, so $0<p<1$.

Can you finish?

Since $f'(x)$ has the same sign as $g(x)$, we deduce that $f$ is increasing over $(-\infty,p]$ and decreasing (to $-\infty$) over $[p,\infty)$. Since $p>0$, we have that $f(p)>0$, so for a unique $r>p$ we have $f(r)=0$.

Answer 2

Let $f(x) = x^2$ and $g(x) = \arctan(x)$. For the first part: $x=0$.

For the second part we use 2 observations:

• $f'(x) = 2x$ and $g'(x) = \frac{1}{1+x^2}$. Thus, as $f(0) = g(0)$, but $f'(0)< g'(0)$, it follows that $f(x) < g(x)$ for small positive $x$.

• On the other hand, $\arctan(x)$ is bounded whereas $x^2$ is not, hence $f(x) > g(x)$ for large positive $x$.

Thus, by the mean value theorem (applied on $f-g$) there must be a positive $x$ for which $f(x)=g(x)$. To show that this point is unique, we can use Rolle's Theorem:

Assume $f(x) = g(x)$ and $f(y) = g(y)$ for some $x< y$. Then there must be some $\xi\in(x,y)$ such that $f'(\xi)-g'(\xi)=0$. Note that:

$$f'(x) - g'(x) = 2x - \frac{1}{1+x^2} = 0 \iff 2x^3+2x-1 = 0$$

Now if this equation has only $1$ positive solution, then also $f(x)=g(x)$ has only 1 positive solution, since we can apply Rolle for any pair of solutions. E.g. if there were 2 positive solutions $x_0<x_1$ then by Rolle $f'(x)-g'(x) = 0$ must have at least two solutions, one in $(0,x_0)$ and one in $(x_0,x_1)$.