# Asymptotics applied to approximate solutions of trascendental equations

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 $$|z|. Then for fixed $$n$$ it is $$\sum_{s=n}^\infty a_s z^s=O(z^n)$$ in any disk $$|z|\le\rho$$ such that $$\rho .

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 $$n and so $$\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.

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: $$\frac{1}{n\pi + O\left(\frac{1}{n}\right)} = \frac{1}{n\pi}\frac{1}{1 + O\left(\frac{1}{n^2}\right)} = \frac{1}{n\pi}\left(1+O\left(\frac{1}{n^2}\right)\right) = \frac{1}{n\pi} + O\left(\frac{1}{n^3}\right)\,.$$

• 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?
– Gwyn
Jul 25, 2022 at 14:06
• 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 $O\left(\frac{1}{n^3}\right)$. 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

$$\large\square$$

Convergence

$$(5)$$ 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 $$(1)$$.

Computation

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}