# Finding asymptotic expansion

Let $\{ \delta_n (\varepsilon) \}$ be an asymptotic sequence as $\varepsilon \rightarrow 0$, and $a_n$ be independent of $\varepsilon$ and $\delta_n (\varepsilon)$. A series $\sum \limits_{n=0}^{\infty} a_n \delta_n (\varepsilon)$ is said to be an asymptotic expansion of a function $f(\varepsilon)$ as $\varepsilon \rightarrow 0$ if for all $N>1$, $f(\varepsilon)-\sum \limits_{n=0}^{N} a_n \delta_n (\varepsilon)=o(\delta_N (\varepsilon))$ as $\varepsilon \rightarrow 0$.

Write down an asymptotic expansion for the function $$f(\varepsilon)=\cos(2\varepsilon)$$ in the limit $\varepsilon \rightarrow 0^+$.

Following the definition, I need $$\frac{\cos(2\varepsilon)-\sum \limits_{n=0}^N a_n \delta_n (\varepsilon)}{\delta_N (\varepsilon)} \rightarrow 0$$as $\varepsilon \rightarrow 0^+$

so I need to choose my $a_n$ and $\delta_n(\varepsilon)$ so that this is satisfied, right?

But how can I do this because the numerator is always going to have the negative of the largest $N$ term as the denominator.

• WHat exactly do you mean by asymptotic expansion? One can put, for instance, Taylor expansion for $cos$ function and the higher order terms in epsilon will be small… Or do you mean something else? – GregVoit Oct 20 '15 at 12:00
• Let $\{ \delta_n (\varepsilon) \}$ be an asymptotic sequence as $\varepsilon \rightarrow 0$, and $a_n$ be independent of $\varepsilon$ and $\delta_n (\varepsilon)$. A series $\sum \limits_{n=0}^{\infty} a_n \delta_n (\varepsilon)$ is said to be an asymptotic expansion of a function $f(\varepsilon)$ as $\varepsilon \rightarrow 0$ if for all $N>1$, $f(\varepsilon)-\sum \limits_{n=0}^{N} a_n \delta_n (\varepsilon)=o(\delta_N (\varepsilon))$ as $\varepsilon \rightarrow 0$. – snowman Oct 20 '15 at 12:13
• thats the general definition – snowman Oct 20 '15 at 12:16

If $$f(x) = \sum_{n=0}^\infty a_n (x-x_0)^n$$ converges then for each $N$, $$f(x) = \sum_{n=0}^N a_n (x-x_0)^n + R_N$$ where $$R_N = (x-x_0)^{N+1}\left(\sum_{n=N+1}^\infty a_n (x-x_0)^{n-(N+1)}\right)$$ is bounded so $$R_N \in o((x-x_0)^N) \text{ as } x \to x_0$$
• like $$\sum \limits_{n=0}^{\infty} \frac{(-1)^n \varepsilon ^{2n+1}}{(2n+1)n!}$$? this is convergent... – snowman Oct 20 '15 at 12:41
• Yes, but that's not the expansion for $\cos 2\epsilon$. $$\cos(2\epsilon) = \sum_{n=0}^\infty\frac{(-4)^n \epsilon^{2n}}{(2n)!}$$ – Chris Kerridge Oct 20 '15 at 12:49
• So we can make $a_n=\frac{(-4)^n}{(2n)!}$ and $\delta_n(\varepsilon)=\varepsilon^{2n}$ right? By the way, since we have $N$ instead of infinity on the summation, we let $N$ tend to infinity right? – snowman Oct 20 '15 at 17:06
• Correct for the terms of the series, but look back to your definition of an asymptotic series to understand about the N and $\infty$: you need to ensure the asymptotic condition holds for each N. My answer shows that given the terms of a convergent power series, the asymptotic condition does hold for each N. – Chris Kerridge Oct 20 '15 at 19:56
• I see so it should hold regardless of what $N$ is. So I am a bit confused on actually showing that it tends to zero as epsilon tends to 0+. We have $$\frac{\cos(2\varepsilon)-\sum \limits_{n=0}^N \frac{(-4)^n \varepsilon^{2n}}{(2n)!}}{\varepsilon^{2N}}$$ and correct me if I am wrong but the numerator is $(1+...+N^{th}term+O(\varepsilon^{2N})-(1+...+N^{th}term)=O(\varepsilon^{2N})$ right? So $$\frac{O(\varepsilon^{2N})}{\varepsilon^{2N}} \rightarrow 0$$ as epsilon tends to 0+ right????? – snowman Oct 20 '15 at 21:07