Asymptotic expansion (Taylor approximation)

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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
\[\lim_{x\to a}\frac{f(x)}{g(x)} = 0,\]
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 \(\lim_{x\to a}\frac{f(x)}{1} = \lim_{x\to a}f(x) = 0\).
  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 \(\lim_{x\to 0}\frac{h(x)}{x^2} = 0\).
□

Theorem (Asymptotic expansion)

Let \(f(x)\) be a \(C^n\) function defined on an open interval containing \(a\). Then
\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + o((x-a)^n) ~~ (x \to a).\]
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
\[ \begin{eqnarray*} f(x) &=& f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \cdots + \frac{f^{(n-1)}(0)}{(n-1)!}x^{n-1} + \frac{f^{(n)}(\theta x)}{n!}x^{n}\\ &=& f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \cdots + \frac{f^{(n-1)}(0)}{(n-1)!}x^{n-1} + \frac{f^{(n)}(0)}{n!}x^{n} + g(x) \end{eqnarray*} \]
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
\[\lim_{x\to 0}\frac{g(x)}{x^n} = 0,\]
 and hence \(g(x) = o(x^n) ~ (x\to 0)\). ■

Lemma

As \(x \to 0\), the following hold.
  1. \(x^mo(x^n) = o(x^{m+n})\), \(o(x^m)o(x^n) = o(x^{m+n})\).
  2. \(o(x^m) + o(x^n) = o(x^l)\) where \(l = \min\{m, n\}\).
Proof. 
  1. As \(x \to 0\), \[\frac{x^mo(x^n)}{x^{m+n}} = \frac{o(x^n)}{x^{n}},\] 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 \(m \leq n\). As \(x\to 0\), \[\frac{o(x^m) + o(x^n)}{x^m} = \frac{o(x^m)}{x^m} + x^{n-m}\frac{o(x^n)}{x^n} \to 0.\]
■

Example. Let us find the 4-th order asymptotic expansion of \((1 + x^2)e^x\) at \(x= 0\). As \(x\to 0\),
\[e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + o(x^4). \tag{eq:ex1}\]
Since \(e^x = 1 + x + \frac{1}{2}x^2 + o(x^2)\),
\[x^2e^x = x^2 + x^3 + \frac{1}{2}x^4 + o(x^4)\tag{eq:ex2}\]
where we applied the above lemma to have \(x^2o(x^2) = o(x^4)\). 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 \(\lim_{x\to 0}\frac{\sin x}{x}\) by asymptotic expansion. Since
\[\sin x = x + o(x)\]
we have
\[\lim_{x\to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{x + o(x)}{x} = \lim_{x\to 0}\left(1 + \frac{o(x)}{x}\right) = 1.\]
□


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