The function $\tan x - x$ has exactly one root $x_n$ in the interval $(n\pi, (n + \frac{1}{2})\pi)$. Show that $$x_n = n\pi + \frac{\pi}{2} - \frac{1}{n\pi} + r_n$$ where $\lim_{n\rightarrow \infty} n r_n = 0$.
I think I should try to use Taylor expansion some way, but I am not seeing how to do this.
Let $y_n = \pi/2-x_n + n\pi$, then $y_n\in(0,\pi/2)$ and $$\frac{1}{\tan y_n}=\frac{\pi}{2}-y_n+n\pi\Leftrightarrow \tan y_n=\frac{1}{\frac{\pi}{2}-y_n+n\pi}$$ Since $y\in(0,\pi/2)$, it follows that $$\frac{1}{n\pi}> \tan y_n > \frac{1}{(n+1/2)\pi}\\ \Rightarrow 1 > n\pi\tan y_n > \frac{1}{(1+\frac{1}{2n})}$$ By squeeze theorem we have that $\lim_{n\rightarrow\infty}n\pi \tan y_n=1$, but we also know that $\lim_{n\rightarrow\infty}\frac{\tan y_n}{y_n} = \lim_{y\rightarrow 0}\frac{\tan y_n}{y_n} = 1$ therefore $$\lim_{n\rightarrow +\infty} n\pi \,y_n = 1\\ \Leftrightarrow \lim_{n\rightarrow +\infty} n\pi(\pi/2 - x_n + n\pi) = 1\\ \Leftrightarrow \lim_{n\rightarrow +\infty} n\pi(x_n - \pi/2 + n\pi - \frac{1}{n\pi})=0 \Leftrightarrow \lim_{n\rightarrow +\infty} n\pi r_n=0$$
I am not sure but let $y_n=\tan[x_n-n\pi]$. Then we have $$\tan x_n - n\pi = x_n - n\pi.$$ Thus, using $\tan(x_n)=\tan(x_n-n\pi)$, $$y_n - n\pi = \arctan(y_n).$$ Now use $\arctan(y)+\arctan(1/y)=\pi/2$ to obtain $$y_n+\arctan(1/y_n)=n\pi + \pi/2.$$ Now, because $y_n \rightarrow \infty$, you can use a Taylor expansion of $\arctan$ around 0. This should give you the solution.
