Asymptotic expansion (Taylor approximation)

In many situations, the remainder term in the finite Taylor (Maclaurin) expansion is unimportant. To denote that some terms are not as important as others, we introduce a new notation of great convenience called the "little o." Using this little o notation, we define the asymptotic expansion, which is (almost) the same as the finite Taylor expansion except that the remainder term is replaced by the little o term.

See also: Taylor's theorem

Definition (Landau's asymptotic (``little $o$'') notation)

Let $f(x)$ and $g(x)$ be functions defined in the neighbor of $x=a$. If

$$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=0,$$

then, we write

$$f(x)=o(g(x))(x\to a).$$

This ``$o$'' is called Landau's symbol} (or ``little o''), and this notation is called Landau's notation (or little-o notation).

Remark. When an equation involves Landau's symbol, it does not represent exact equality. □

Example. 

1. $f(x)=o(1)(x\to a)$ means $\underset{x\to a}{lim}\frac{f(x)}{1}=\underset{x\to a}{lim}f(x)=0$.
2. $f(x)=2x^2+o(x^2)(x\to 0)$ means that there exists some function $h(x)$ such that $f(x)=2x^2+h(x)$ and $\underset{x\to 0}{lim}\frac{h(x)}{x^2}=0$.

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Theorem (Asymptotic expansion)

Let $f(x)$ be a $C^n$ function defined on an open interval containing $a$. Then

$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2}(x-a{)}^2+\cdots +\frac{f^{(n)}(a)}{n!}(x-a{)}^n+o((x-a{)}^n)(x\to a).$$

The above expansion is called the asymptotic expansion of $f(x)$ at $x=a$.

Proof. For simplicity, we prove the case when $a=0$.

The finite Taylor expansion gives

$$\begin{array}{rcl}f(x)& =& f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2}{x}^2+\cdots +\frac{f^{(n-1)}(0)}{(n-1)!}{x}^{n-1}+\frac{f^{(n)}(\theta x)}{n!}{x}^n\\ & =& f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2}{x}^2+\cdots +\frac{f^{(n-1)}(0)}{(n-1)!}{x}^{n-1}+\frac{f^{(n)}(0)}{n!}{x}^n+g(x)\end{array}$$

where $\theta \in (0,1)$ is a real number dependent on $x$, and

$$g(x)=\frac{f^{(n)}(\theta x)}{n!}{x}^n-\frac{f^{(n)}(0)}{n!}{x}^n=\frac{f^{(n)}(\theta x)-f^{(n)}(0)}{n!}{x}^n.$$

Since $f(x)$ is of class $C^n$, $f^{(n)}(x)$ is continuous, which implies that $f^{(n)}(\theta x)-f^{(n)}(0)\to 0$ as $x\to 0$. Therefore

$$\underset{x\to 0}{lim}\frac{g(x)}{x^n}=0,$$

 and hence $g(x)=o(x^n)(x\to 0)$. ■

Lemma

As $x\to 0$, the following hold.

1. $x^{m}o(x^n)=o(x^{m+n})$, $o(x^m)o(x^n)=o(x^{m+n})$.
2. $o(x^m)+o(x^n)=o(x^l)$ where $l=min\{m,n\}$.

Proof. 

1. As $x\to 0$, $$\frac{x^{m}o(x^n)}{x^{m+n}}=\frac{o(x^n)}{x^n},$$ and $$\frac{o(x^m)o(x^n)}{x^{m+n}}=\frac{o(x^m)}{x^m}\frac{o(x^n)}{x^n}\to 0.$$
2. Without losing generality, we may assume $m\le n$. As $x\to 0$, $$\frac{o(x^m)+o(x^n)}{x^m}=\frac{o(x^m)}{x^m}+x^{n-m}\frac{o(x^n)}{x^n}\to 0.$$

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Example. Let us find the 4-th order asymptotic expansion of $(1+x^2)e^x$ at $x=0$. As $x\to 0$,

$$\begin{array}{}\text{(eq:ex1)}& e^x=1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\frac{1}{24}{x}^4+o(x^4).\end{array}$$

Since $e^x=1+x+\frac{1}{2}{x}^2+o(x^2)$,

$$\begin{array}{}\text{(eq:ex2)}& x^2{e}^x=x^2+x^3+\frac{1}{2}{x}^4+o(x^4)\end{array}$$

where we applied the above lemma to have $x^{2}o(x^2)=o(x^4)$. Adding both sides of Eqs. (eq:ex1) and (eq:ex2) yields the asymptotic expansion of $(1+x^2)e^x$

$$(1+x^2)e^x=1+x+\frac{3}{2}{x}^2+\frac{7}{6}{x}^3+\frac{13}{24}{x}^4+o(x^4)$$

where we applied the above lemma to have $o(x^4)+o(x^4)=o(x^4)$. □

Example. We can find the limit $\underset{x\to 0}{lim}\frac{\sin x}{x}$ by asymptotic expansion. Since

$$\sin x=x+o(x)$$

we have

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=\underset{x\to 0}{lim}\frac{x+o(x)}{x}=\underset{x\to 0}{lim}(1+\frac{o(x)}{x})=1.$$

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