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.
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 $\epsilon&gt;0$ there exists $\delta$ independent of $x$ such that
\begin{align}
|x|&lt;&amp; \delta&amp; \Rightarrow&amp;&amp;
\left| u(t) - \sum_{m=0}^M x^m u_m(t)\right| &lt;&amp; \epsilon |x^M| &amp;
\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| &lt;&amp; \infty &amp;
\text{ for all } t, &amp;&amp;
\text{ for all } m=&amp;0,1,2,\dots. 
\end{align}
I think you can check it as you have the expressions for $\{u_m(t)\}$.
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 $\epsilon&gt;0$ there exists $\delta$ independent of $x$ such that
\begin{align}
|x|&lt;&amp; \delta&amp; \Rightarrow&amp;&amp;
\left| u(t) - \sum_{m=0}^M x^m u_m(t)\right| &lt;&amp; \epsilon |x^M| &amp;
\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| &lt;&amp; \infty &amp;
\text{ for all } t, &amp;&amp;
\text{ for all } m=&amp;0,1,2,\dots. 
\end{align}
I think you can check it as you have the expressions for $\{u_m(t)\}$.