I apologize in advance for the length.

The equation $\sin x = (\log x)^{-1}$ has exactly one solution $x_n$ in the interval $(2\pi n,2\pi n + \pi/2)$ for $n \geq 1$, and the exercise (de Bruijn, Asymptotic Methods in Analysis, ch. 2) asks me to show that

$$ x_n = 2\pi n + (\log 2\pi n)^{-1} + O((\log 2 \pi n)^{-3}). $$

To start, we have $0 < x_n - 2 \pi n < 1$ and

$$ \sin x_n = \sin (x_n - 2 \pi n) = (\log x_n)^{-1} \to 0, $$

so $x_n \to 2 \pi n$. It would make sense then to make the substitution $x_n = z + t$, where $t = 2 \pi n$, so that we're now concerned with finding the asymptotic behavior of $z$ in terms of $t$ as $t \to \infty$ in the equation

$$ \sin z = (\log (z + t))^{-1}. $$

That is, we want to show that

$$ z = (\log t)^{-1} + O((\log t)^{-3}). $$

So far I've only been able to show that $z = O((\log t)^{-1})$ through arguments which are probably not sound.

I've tried to apply the Lagrange Inversion Formula but I can't seem to get it into the right form. If we let $w = \log t$, then $w = \frac{z}{f(z)}$, where

$$ f(z) = \frac{z}{\log(e^{1/\sin z} - z)}. $$

But $f(0) = 0$ (and, probably more importantly, $f$ isn't analytic at $0$), so I can't apply Lagrange. Of course there may be a "correct" way to rearrange the equation to put it into Lagrange form.

I've also considered applying Newton's method, but I don't know if that's valid. Applying the method to $(\log (z + t))^{-1} - \sin z$ with $z_0 = 0$ I get

$$ z_1 = - (\log t)^{-1} ( 1 + O((t \log t)^{-2})), $$

which at least has the right asymptotic behavior in the first term. Trying to iterate using, for example, $x_0 = (\log t)^{-1}$ in the hopes of getting more stable terms leads me to a wall of computation, and I doubt that's the goal of the problem. More importantly, even if I did get a stable asymptotic series as I continued to iterate, I don't know whether I'm actually converging to the actual root of the equation.

Lastly I should mention that I've also tried letting $z = x_n(2 \pi n)^{-1} - 1$, but this didn't seem to lead to anywhere helpful.

Any tips?

  • $\begingroup$ You might want to somehow use the asymptotic expansion $\arcsin x = x+O(x^3)$ around $x=0$ to evaluate $x_n-2\pi n$ and the fact that $\log x_n \to \log 2\pi n$ (their difference approaches $0$). $\endgroup$ – anon Aug 11 '11 at 0:17

So... $\sin(z_n)=(\log(2\pi n+z_n))^{-1}$ and you proved that $z_n=o(1)$.

This implies that $\log(2\pi n+z_n)=\log(2\pi n)+\log(1+z_n/(2\pi n))=\log(2\pi n)+o(1/n)$ and that $\sin(z_n)=z_n+O(z_n^3)$. Hence, using the notation $\ell_n=(\log(2\pi n))^{-1}$, $$ z_n+O(z_n^3)=\frac1{1/\ell_n+o(1/n)}=\frac{\ell_n}{1+o(\ell_n/n)}=\ell_n+o(\ell_n^2/n). $$ Keeping only the terms of higher order, this proves that $z_n=\ell_n+o(\ell_n)$. Keeping all the terms, this proves that $$ z_n=\ell_n+O(\ell_n^3)+o(\ell_n^2/n). $$ Finally, since $1/n\ll\ell_n$, the term $o(\ell_n^2/n)$ is also $O(\ell_n^3)$ and you are done.

  • $\begingroup$ Hello Didier, thank you for your answer. I wanted to ask for clarification on one of the details. I see how $z_n + O(z_n^3) = \ell_n + o(\ell_n)$, but why can we disregard the $O(z_n^3)$ term to conclude that $z_n = \ell_n + o(\ell_n)$? $\endgroup$ – Antonio Vargas Aug 11 '11 at 2:37
  • 1
    $\begingroup$ Since $z_n=o(1)$, the LHS is equivalent to $z_n$. The RHS is equivalent to $\ell_n$, hence LHS=RHS implies that $z_n$ is equivalent to $\ell_n$, which means exactly that $z_n=\ell_n+o(\ell_n)$ (or that $\ell_n=z_n+o(z_n)$, by the way). $\endgroup$ – Did Aug 11 '11 at 8:04

My idea is to use Newton's formula $x_{k+1} = x_k - \frac{f(x_k)}{f'(x_{k})}$. We start with $x_0 = z_n = 2 \pi n$ and $f(x) = \log(x) \sin(x)-1$. We get $x_1 = z_n + \ell_n^{-1}$ with $\ell_n = \log(2 \pi n)$. Next

$$ x_2 = z_n + \ell_n^{-1} -\frac{\left(z_n \ell _n+1\right) \left(\sin \left(\frac{1}{\ell _n}\right) \log \left(z_n+\frac{1}{\ell _n}\right)-1\right)}{\left(z_n \ell _n+1\right) \cos \left(\frac{1}{\ell _n}\right) \log \left(z_n+\frac{1}{\ell _n}\right)+\ell _n \sin \left(\frac{1}{\ell _n}\right)} $$

The correction term is of order $(6 \ell_n^3)^{-1}$.

  • $\begingroup$ Why is the correction term of that order? $\endgroup$ – Antonio Vargas Aug 11 '11 at 17:42
  • $\begingroup$ I wrote $\log(z_n + \ell_n^{-1}) = \ell_n + \log(1 + (z_n \ell_n)^{-1})$ divided numerator and denominator by $z_n \ell_n^2$. Then, expanded to order 2 in $z_n^{-1}$ and order 3 in $\ell_n^{-1}$, i.e. $\cos(\ell_n^{-1}) = 1- \frac{1}{2} \ell_n^{-2}$, $\sin \ell_{n}^{-1} = \ell_{n}^{-1}$ , $\log(1 + (z_n \ell_n)^{-1}) = (z_n \ell_n)^{-1}$ and got $1/6 \ell_n^{-3} - z_n^{-1} \ell_n^{-3} $. $\endgroup$ – Sasha Aug 11 '11 at 18:51
  • $\begingroup$ Ah, ok. But how do we know that $x_2$ approximates the root of the equation to the desired accuracy? $\endgroup$ – Antonio Vargas Aug 11 '11 at 19:17
  • $\begingroup$ We do not, but the $\frac{x_2-x_1}{x_1-x_0}$ is small, and $\frac{x_3-x_2}{x_1-x0}$ is going to be yet smaller since terms up to $O(\ell_n^{-3} z_n^{-1})$ and $O(\ell_n^{-3})$ have already been cancelled. $\endgroup$ – Sasha Aug 11 '11 at 19:22
  • $\begingroup$ So we know it's converging to something, but it might take more work to show that something is what we want? $\endgroup$ – Antonio Vargas Aug 11 '11 at 19:30

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