I'm self studying "Olver - Asymptotics and Special Functions": I'm new to asymptotics and so there are some things I am not getting. First, the author states the following theorem:

Theorem 3.1. Let $\sum_{s=0}^\infty a_s z^s$ converge when $|z|<r$. Then for fixed $n$ it is $\sum_{s=n}^\infty a_s z^s=O(z^n)$ in any disk $|z|\le\rho$ such that $\rho <r$.

And proceeds to study the asymptotic behavior of the roots of trascendental equations: for instance, he considers the equation $x \tan x=1$ for large $x>0$. Inverting, it is $x=n\pi+\arctan(1/x)$ for $n\in\mathbb{Z}$ for the principal value of arctan. Since arctan is bounded, it is $x \sim n\pi$ as $n \to \infty$.

Since for $x>1$ the Taylor's series of $\arctan(1/x)$ converges, he then writes: $$x=n\pi+\frac{1}{x}-\frac{1}{3x^3}+\frac{1}{5x^5}-\frac{1}{7x^7}+\dots$$ Finally, he claims that hence $x=n\pi+O(x^{-1})=n\pi+O(n^{-1})$ and that next two substitutions produce $$x=n\pi+\frac{1}{n\pi}+O\left(\frac{1}{n^3}\right) \tag{1}$$ $$x=n\pi+\frac{1}{n\pi}-\frac{4}{3(n\pi)^3}+O\left(\frac{1}{n^5}\right) \tag{2}$$

First question: why is it $n\pi+O(x^{-1})=n\pi+O(n^{-1})$? I know that $f(x)=O(g(x))$ as $x \to x_0$ means that $|f(x)|\le K|g(x)|$ for some constant $K$ in a neighborhood of $x_0$, maybe this comes from the Archimedean property and I can always find a natural $n$ such that $n<x$ and so $\frac{1}{x}<\frac{1}{n}$, so I can say (for $x>0$) that $f(x)=O(x^{-1})\implies f(x)\le \frac{K}{x}<\frac{K}{n}\implies f(x) \le \frac{K}{n} \implies f(x)=O(n^{-1})$? Or this is wrong and he is using another kind of reasoning?

Second question: I am not getting the calculations to obtain equations $(1)$ and $(2)$. What I tried so far is the following. From theorem 3.1 with $n=1$, I know that $$x=n\pi+\frac{1}{x}+O\left(\frac{1}{x^3}\right)$$ Substituting $x=n\pi+O(n^{-1})$ and using the properties $O(cf)=O(f)$ for $c>0$, $fO(g)=O(fg)$ and $O(f)O(g)=O(fg)$, it is $$x=n\pi+\frac{1}{n\pi+O\left(\frac{1}{n\pi}\right)}+O\left(\frac{1}{\left(n\pi+O\left(\frac{1}{n\pi}\right)\right)^3}\right)=n\pi+\frac{1}{n\pi+O\left(\frac{1}{n\pi}\right)}+O\left(\frac{1}{n^3\pi^3+O(n)+O\left(\frac{1}{n}\right)+O\left(\frac{1}{n^3}\right)}\right)$$ But I don't know how to manipulate the big-Os in the denominators. I have similar problems for $(2)$, putting $n=2$ in theorem 3.1 and have no clue how to manipulate the big-Os.


2 Answers 2


For your first question, the author first proved that $x\sim n\pi$, and this implies that $x^{-1} \sim \frac{1}{\pi}n^{-1} = O(n^{-1})$, thus a $O(x^{-1})$ is a $O(n^{-1})$ (I'll let you work out the details yourself).

For your second question, you can use a Taylor expansion by factoring the dominant term in the denominator. For instance: $$\frac{1}{n\pi + O\left(\frac{1}{n}\right)} = \frac{1}{n\pi}\frac{1}{1 + O\left(\frac{1}{n^2}\right)} = \frac{1}{n\pi}\left(1+O\left(\frac{1}{n^2}\right)\right) = \frac{1}{n\pi} + O\left(\frac{1}{n^3}\right)\,.$$

  • $\begingroup$ Thank you for the help Astyx. So I can use Taylor expansion for quantities that are big-O provided that there is convergence? For instance, I can say that $\log[1+O(z)]=O(z)-\frac{[O(z)]^2}{2}+...$ when $O(z) \to 0$ as $z \to z_0$? If this is the case, I assume that you used $\frac{1}{1+O(z)}=1-O(z)+[O(z)]^2-...=1+O(z)$ in your second equality of the last line? $\endgroup$
    – Gwyn
    Commented Jul 25, 2022 at 14:06
  • $\begingroup$ You are correct, just be careful that $O(z)\to 0$ can only be obtained if $z\to 0$. In this case, as $n\to \infty$, $\frac{1}{n^3}\to0$, therefore $O\left(\frac{1}{n^3}\right)$ is a $o(1)$ (meaning it goes to $0$) and only then can you use Taylor expansion. Also, I am using Taylor expansions with remainders, not Taylor series, so the sum is finite, and the last term should be something like $O\left(O\left(\frac{1}{n^3}\right)\right)$, which you can prove is just $O\left(\frac{1}{n^3}\right)$. $\endgroup$
    – Astyx
    Commented Jul 26, 2022 at 6:21

The Recursion

Rather than use $$ x=n\pi+\tan^{-1}\left(\frac1x\right)\tag1 $$ it might be easier to see what is going on if we set $u=\frac1x$. Equation $(1)$ then becomes $$ u=\frac1{n\pi+\tan^{-1}(u)}\tag2 $$ In light of $(2)$, define $$ f(u)=\frac1{n\pi+\tan^{-1}(u)}\tag3 $$ Then we have $$ f'(u)=-\frac1{\left(1+u^2\right)\left(n\pi+\tan^{-1}(u)\right)^2}\tag4 $$ and $$ |f'(u)|\le\frac1{n^2\pi^2}\tag5 $$ Equation $(4)$ says that $f$ is decreasing, and therefore, $$ f:\left[0,\frac1{n\pi}\right]\to\left[\frac1{(n+1/2)\pi},\frac1{n\pi}\right]\subset\left[0,\frac1{n\pi}\right]\tag6 $$ Proposition: Define the sequence $u_0=0$ and $u_{k+1}=f(u_k)$. Then we have $$ |u_{k+1}-u_k|\le\frac1{(n\pi)^{2k+1}}\tag7 $$ Proof: $u_0=0$ and $u_1=\frac1{n\pi}$. Therefore, $(7)$ is satisfied for $k=0$.

Assume that $|u_k-u_{k-1}|\le\frac1{(n\pi)^{2k-1}}$. Then, the Mean Value Theorem says that for some $c$ between $u_{k-1}$ and $u_k$, $$ \begin{align} |u_{k+1}-u_k| &=|f(u_k)-f(u_{k-1})|\tag{8a}\\[9pt] &=|u_k-u_{k-1}|\,|f'(c)|\tag{8b}\\[3pt] &\le\frac1{(n\pi)^{2k-1}}\cdot\frac1{n^2\pi^2}\tag{8c}\\ &=\frac1{(n\pi)^{2k+1}}\tag{8d} \end{align} $$ Explanation:
$\text{(8a):}$ definition
$\text{(8b):}$ Mean Value Theorem
$\text{(8c):}$ assumption and $(5)$
$\text{(8d):}$ simplification



$(5)$ and $(6)$ show that $f$ is a contraction map on $\left[0,\frac1{n\pi}\right]$. Therefore, the sequence $u_k$ converges to a solution of $(2)$ in $\left[\frac1{(n+1/2)\pi},\frac1{n\pi}\right]$.

$(7)$ says that $u_k$ is a solution to $(2)$ mod $O\!\left(\frac1{n^{2k+1}}\right)$.

For $k\ge1$, $u_k\sim\frac1{n\pi}$. Thus, $$ u_k=\frac1{n\pi}\left(1+\sum_{j=1}^{k-1}\frac{a_j}{(n\pi)^{2j}}+O\!\left(\frac1{n^{2k}}\right)\right)\tag9 $$ Taking the reciprocal of $(9)$, we get $$ \frac1{u_k}=n\pi\left(1+\sum_{j=1}^{k-1}\frac{b_j}{(n\pi)^{2j}}+O\!\left(\frac1{n^{2k}}\right)\right)\tag{10} $$ we have $\frac1{u_k}+O\!\left(\frac1{n^{2k-1}}\right)$ is a solution to $(1)$.


Let's compute $u_5$ mod $O\!\left(\frac1{n^{11}}\right)$: $$ \begin{align} u_0&=0\\[6pt] u_1&=\color{#090}{\frac1{n\pi}}\\ u_2&=\color{#090}{\frac1{n\pi}-\frac1{n^3\pi^3}}+\frac4{3n^5\pi^5}-\frac{28}{15n^7\pi^7}+\frac{836}{315n^9\pi^9}+O\!\left(\frac1{n^{11}}\right)\\ u_3&=\color{#090}{\frac1{n\pi}-\frac1{n^3\pi^3}+\frac7{3n^5\pi^5}}-\frac{31}{5n^7\pi^7}+\frac{5414}{315n^9\pi^9}+O\!\left(\frac1{n^{11}}\right)\\ u_4&=\color{#090}{\frac1{n\pi}-\frac1{n^3\pi^3}+\frac7{3n^5\pi^5}-\frac{36}{5n^7\pi^7}}+\frac{7724}{315n^9\pi^9}+O\!\left(\frac1{n^{11}}\right)\\ u_5&=\color{#090}{\frac1{n\pi}-\frac1{n^3\pi^3}+\frac7{3n^5\pi^5}-\frac{36}{5n^7\pi^7}+\frac{8039}{315n^9\pi^9}}+O\!\left(\frac1{n^{11}}\right) \end{align} $$ Due to $(7)$, the terms in green are stable as $k$ increases. Thus, $$ \begin{align} x &=1/u_5+O\!\left(\frac1{n^9}\right)\\ &=n\pi+\frac1{n\pi}-\frac4{3n^3\pi^3}+\frac{53}{15n^5\pi^5}-\frac{1226}{105n^7\pi^7}+O\!\left(\frac1{n^9}\right)\\ \end{align} $$


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