Finding the big-$O$ estimate of the solutions to $\cot x = x$

Question

So a problem for my class I am asked to find the big-$O$ estimate (up to 3 terms) of the solutions to $\cot x = x$. We label the solutions to this equation in increasing order $0<x_1<x_2<\cdots$ and we wish to find the expansion of $x_n$ as $n \rightarrow\infty$.

I've computed the expansion $$\cot x = x^{-1} - \frac{1}{3}x - \frac{1}{45}x^3 + O(x^5)$$

I observe graphically that for integer $m\ge0$ there exists exactly one solution between $m\pi$ and $(m+1/2)\pi$ ($x_{m+1}$ by the previous labelling). So we know that the leading order of $x_n$ should be $(n-1)\pi$. I'm not sure where to proceed from here. I tried iterating $x$ in the series expansion, but the increasing powers of $x$ gives larger errors with each iteration.

Answer 1

You already find the first term $n\pi$ for $x_n$ (with $0=x_0<x_1<...$). So, let's denote $x_n=n\pi+y_n$ with $y_n=\mathcal{o}(n)$ and $0<y_n<\frac{\pi}{2}$.

$$\begin{align} &\Longrightarrow \cot(y_n)=n\pi+y_n \xrightarrow{n\to+\infty}+\infty \\ &\Longrightarrow y_n \xrightarrow{n\to+\infty} 0 \end{align}$$

So $y_n=\mathcal{o}(1)$ and we can compute the series expansion of $\cot(y_n)$ when $y_n \to 0$, we have

$$\begin{align} n\pi+y_n=\cot(y_n) = \frac{1}{y_n}+\mathcal{O}(y_n) &\Longleftrightarrow n\pi = \frac{1}{y_n}+\mathcal{O}(y_n)\\ &\Longleftrightarrow y_n = \frac{1}{n\pi + \mathcal{O}(y_n)}\\\ &\Longleftrightarrow y_n = \frac{1}{n\pi } \left(1 + \mathcal{O}(\frac{y_n}{n}) \right)\\\ &\Longleftrightarrow y_n = \frac{1}{\pi}\cdot\frac{1}{n}+\mathcal{o}\left(\frac{1}{n} \right)\\ \end{align}$$

We obtain then the second term $\frac{1}{\pi}\cdot\frac{1}{n}$.

Let's denote $x_n=n\pi+ \frac{1}{\pi}\cdot\frac{1}{n}+z_n$ with $z_n=\mathcal{o}\left(\frac{1}{n} \right)$. We have

$$\begin{align} n\pi+\frac{1}{\pi}\cdot\frac{1}{n}+z_n &=\cot\left(\frac{1}{\pi}\cdot\frac{1}{n}+z_n \right) \\&=\frac{1}{\frac{1}{\pi}\cdot\frac{1}{n}+z_n} -\frac{1}{3}\left( \frac{1}{\pi}\cdot\frac{1}{n}+z_n \right)+\mathcal{o}\left( \frac{1}{n} \right)\\ &=\pi n \left(1+ \pi n z_n \right)^{-1}-\frac{1}{3}\left( \frac{1}{\pi}\cdot\frac{1}{n}+z_n \right)+\mathcal{o}\left( \frac{1}{n} \right)\\ &=\pi n \left(1- \pi n z_n +\mathcal{o}(n^2z_n^2) \right)-\frac{1}{3\pi}\cdot\frac{1}{n}+\mathcal{o}\left( \frac{1}{n} \right)\\ &=\pi n - \pi^2 n^2 z_n +\mathcal{o}(n^2z_n)-\frac{1}{3\pi}\cdot\frac{1}{n}+\mathcal{o}\left( \frac{1}{n} \right)\tag{1}\\ \end{align}$$ Then $$\begin{align} (1) &\Longleftrightarrow \frac{4}{3\pi}\cdot\frac{1}{n} = -\pi^2 n^2 z_n++\mathcal{o}(n^2z_n)+\mathcal{o}\left( \frac{1}{n} \right)\\ &\Longleftrightarrow z_n = -\frac{4}{3\pi^3}\cdot\frac{1}{n^3} +\mathcal{o}\left( \frac{1}{n^3} \right)\\\ \end{align}$$

Conclusion: $$x_n = n\pi+\frac{1}{\pi}\cdot\frac{1}{n}-\frac{4}{3\pi^3}\cdot\frac{1}{n^3} +\mathcal{o}\left( \frac{1}{n^3} \right)$$

Answer 2

It is better to consider that you look for the zeros of function $$f(x)=\cos(x)-x\sin(x)$$ which shows that the solutions are closer and closer to $n\pi$.

Expanding as a series around $x=n \pi$ gives $$f(x)=(-1)^n+(-1)^n \sum_{k=1}^\infty \frac{(k+1) \cos \left(\frac{\pi k}{2}\right)-\pi n \sin \left(\frac{\pi k}{2}\right)}{k!} (x-n\pi)^k$$ Truncate to some order and use series reversion to obtain $$x_n=n\pi+\frac{1}{\pi n}-\frac{4}{3 \pi ^3 n^3}+\frac{53}{15 \pi ^5 n^5}-\frac{1226}{105 \pi ^7 n^7}+O\left(\frac{1}{n^9}\right)$$

If you use the above truncated series $$f(x_n)=(-1)^n \frac{13597}{315 \pi ^8 n^8}+O\left(\frac{1}{n^{10}}\right)$$ which is not much as soon as $n\geq 2$.