# Approximations of fixed points of tangent.

This question comes from an exam, years ago.

Show that $f(x)=\tan x-x$, for every positive integer $n$, has exactly one root $x_n$ in the interval $(n\pi,n\pi+\pi/2)$. And show that $$x_n=n\pi+\frac{\pi}{2}-\frac{1}{n\pi}+\text{o}\left(\frac{1}{n}\right).$$

I can prove the claim about the existence of $x_n$, by intermidiate value-theorem, but am stuck by the second point.
Since $\tan x$ is not a contraction, we cannot apply the contraction mapping theorem here. Also, using Taylor approximation, or the Lagrange form of the remainder, I arrived at $$\tan x=(x-n\pi)+\frac{f^{(3)}(\theta)}{6}x^3$$ for some $\theta$ between $x$ and $n\pi$. It seems that, however, I can only obtain information about this $\theta$ along this direction, not about $x_n$.
Any hint or help is greatly appreciated. Thanks in advance.

• And any edit is also appreciated. :D – awllower Jul 25 '13 at 7:58

First note that $x_n-n\pi\to\frac\pi2\cdot$ This is because $0<x_n-n\pi<\frac\pi2$ and $\tan(x_n-n\pi)=\tan(x_n)=x_n\to+\infty$.

Now, put $z_n=x_n-n\pi-\frac\pi2\cdot$ Then $z_n\to 0$, so $z_n\sim \tan(z_n)$ as $n\to\infty$. But $\tan(z_n)=\tan(x_n-\frac\pi2)=-\tan(x_n)=-\frac1{x_n}$, so we get $z_n\sim -\frac1{x_n}\cdot$ Now, $x_n\sim n\pi$ since $n\pi<x_n<n\pi+\frac\pi2$; so $z_n\sim-\frac{1}{n\pi}$ and hence we may write $z_n=-\frac{1}{n\pi}+o(\frac1n).$ This is what you want by the definition of $z_n$.

• Very interestig use of the tangent properties! Thanks for this wonderful answer! – awllower Jul 25 '13 at 8:26
• You're welcome! – Etienne Jul 25 '13 at 8:40