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:

enter image description here

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!

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

1 Answer 1


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.


Not the answer you're looking for? Browse other questions tagged .