# How to show an aymptotic expansion is uniformly valid?

I have an equation $$nt = u - \epsilon \sin(u)$$ which asks for the first four terms in the asymptotic solution.
Hence if the solution is $u_0 + \epsilon u_1 + \cdots.$, expand $\sin(u)$ around $u_0$, $$\sin(u) = \sin(u_0) + \cos(u_0)(\epsilon u_1 + \cdots) - \frac{\sin(u_0)(\epsilon u_1 + \cdots)^2 }2- \frac{\cos(u_0)(\epsilon u_1 + \cdots)^3}6 + \cdots.$$
By collecting terms in different orders of $\epsilon$, I got the following equations $$u_0 = nt$$ $$u_1 = \sin(u_0) = \sin(nt)$$ $$u_2 = \sin(nt)\cos(nt)$$ $$u_3 = \sin(nt)\cos^2(nt) - \frac{1}{2}\sin^3(nt)$$ The question is after getting these terms, how do I show "the four term expansion is uniformly valid for all $nt$". I am thinking the calculation itself already shows, but there is a separate question asking for that. I am not sure what needs to be done here.

• The usual tool to compute approximate solutions to Kepler's equation is Newton's method. Aug 18 '14 at 22:02

A series $\sum_{m=0}^{\infty} x^m u_m(t)$ is asymptotic to $u(t)$ as $x\rightarrow 0$ uniformly in $t$ if for any $\epsilon>0$ there exists $\delta$ independent of $x$ such that \begin{align} |x|<& \delta& \Rightarrow&& \left| u(t) - \sum_{m=0}^M x^m u_m(t)\right| <& \epsilon |x^M| & \text{ for all } M=0,1,2,\dots. \end{align} This is probably equivalent to that \begin{align} \left|\frac{u_{m+1}(t)}{u_m(t)}\right| <& \infty & \text{ for all } t, && \text{ for all } m=&0,1,2,\dots. \end{align} I think you can check it as you have the expressions for $\{u_m(t)\}$.
• $x$ in this answer corresponds to $\epsilon$ in the question, and $\epsilon$ in this answer is an arbitrary positive real number conventionally used in the $\epsilon$-$\delta$ argument. Jun 24 '18 at 9:15