# Asymptotics on root $x = x_n \in [2\pi n, 2\pi n + \pi/2]$ of $\sin{x}=(\log{x})^{-1}$ [duplicate]

## Problem Description

I am currently self-studying Buijn's book on asymptotic analysis "Asymptotic Methods in Analysis". I am stuck at the first exercise of chapter 2:

I have completed the first part of the question. My problem is with proving the specific form of the $$x_n$$ root.

## My Attempt

In said chapter of the book, three methods are demonstrated to tackle implicit-function problems in asymptotic analysis. Firstly, the method where one converts the equation to the form $$z/f(z) = w$$ with $$w$$ being $$z$$-independent and $$f(0) \ne 0$$. Then close enough to $$z=0$$ and for small enough $$w$$, $$z$$ can be expressed as a power series in $$w$$ which satisfies the given equation.

Secondly, the method where one derives a relation which can be used to get better and better approximations of the solution iteratively.

Thirdly, a special case of the previous method where the relation is produced using Newton's method. ie: $$x_n = x_{n-1} - f(x_{n-1})/f'(x_{n-1})$$

The first thing I noticed is that for $$x \rightarrow \infty$$, $$(\ln{x})^{-1} \rightarrow 0$$ therefore $$\sin{x} \rightarrow 0$$. This means that $$x_n \rightarrow 2 \pi n$$ since $$x_n \in [2 \pi n, 2 \pi n + \pi/2]$$. Setting $$x_n = 2 \pi n + z$$ I attempted to convert $$\sin{x}=(\ln{x})^{-1}$$ in the form of the first method descirbed above. This failed as I arrive at

$$(\sin z)^{-1} = \log{(2\pi n)} + \log {\Big(1 + \frac{z}{2\pi n}\Big)}$$

I am not able to construct a $$w$$ which decreases as $$n$$ increases and is z-independant, while keeping $$f(z)$$ $$n$$-independant (is this necessary?). I got a bit further by writting

$$\log {\Big(1 + \frac{z}{2\pi n}\Big)} = O(\frac{z}{2\pi n}) = O(\frac{1}{2\pi n}), \quad (n \rightarrow \infty, z \rightarrow 0)$$

and then $$(\sin z)^{-1} = \log (2\pi n) + O(\frac{1}{2\pi n})$$

I then tried applying the first method to $$(\sin z)^{-1} = \log (2\pi n)$$ but unfortunately the $$f(0) \ne 0$$ assumption is broken. The next thing I attempted was the third method. Using Newton's method for $$f(x) = \log{x}\sin{x} - 1$$, we arrive at the following iterative formula for recursive approximations $$\phi_k(n)$$ for $$x$$:

$$\phi_{k+1} = \phi_k - \frac{\log{\phi_k}\sin{\phi_k}-1}{\frac{\sin{\phi_k}}{\phi_k}+\log{\phi_k}\cos{\phi_k}}$$

If we start with $$\phi_0(n) = 2\pi n$$, $$\phi_1 = 2\pi n + (\log {2 \pi n})^{-1}$$ which looks promising. But then this nice pattern breaks and we do not get the result we are looking for. So, my final attempt was to use $$\phi_1$$ by writting $$x = \phi_1 + z$$ and trying to show that $$z = O((\log{2\pi n})^{-3}), \quad (n \rightarrow \infty)$$ by using $$\sin{x} \log{x} = 1$$. However this also failed

Any kind of help is very much appreciated!

• you are missing in a lot of places the $-1$ at the exponent Sep 11, 2022 at 23:12
• @Exodd thank you, I think I now fixed them all Sep 13, 2022 at 10:36

If you want to go beyond the first estimate, what you could do is a series expansion around $$x=2n\pi$$ and then write $$\sin(x)\log(x)=\log(n \pi)+ \log(2n \pi)(x-2n \pi)+ +O((x-2n\pi)^2)$$ which is the same as Newton method.
Ignoring the higher order terms, then $$x_{(n)}=2n\pi+ \frac 1{\log(2n \pi)}$$ You could even continue and use series reversion
$$x_{(n)}=2 \pi n+\frac{t}{\log (2 \pi n)}-\frac{t^2}{2 \pi n \log ^3(2 \pi n)}+O\left(t^3\right)$$ where $$t=1$$. This is equivalent to the first iteration of Halley method.