Asymptotic expansions

Part 3 · Limits of functions and continuity · Chapter 7 · lecture notes by Fabio Furini · Chapter PDF

1. The “little \(o\)” symbol and asymptotic expansions

Definition 1: of little \(o\)

Given two functions \(f(x)\) and \(g(x)\), defined in a neighborhood of \(c \in \R^*\), we say that

\[ f(x) = o \big(g(x)\big) {\rm ~~~as~~~} x \rr c \]

(read “\(f(x)\) is little \(o\) of \(g(x)\)” as \(x \rr c\)) if and only if

\[ \frac{f(x)}{g(x)} \rr 0 {\rm ~~~as~~~} x \rr c \]

The symbol \(o(g(x))\) as \(x\) tends to \(c\) does not denote a particular function \(f(x)\), but any function \(f(x)\) such that the ratio between \(f(x)\) and \(g(x)\) tends to 0 as \(x\) tends to \(c\).

Example 1: Little \(o\)

For example:

\[ x^2 = o(x) {\rm ~~~as~~~} x \rr 0 {\rm ~~~since~~~} \frac{x^2}{x} \rr 0 {\rm ~~~as~~~} x \rr 0 \]

\[ x^3 = o(x) {\rm ~~~as~~~} x \rr 0 {\rm ~~~since~~~} \frac{x^3}{x} \rr 0 {\rm ~~~as~~~} x \rr 0 \]

\[ x^3 = o(x^2) {\rm ~~~as~~~} x \rr 0 {\rm ~~~since~~~} \frac{x^3}{x^2} \rr 0 {\rm ~~~as~~~} x \rr 0 \]

\[ e^{-1/x^2} = o(x^4) {\rm ~~~as~~~} x \rr 0 {\rm ~~~since~~~} \frac{e^{-1/x^2}}{x^4} \rr 0 {\rm ~~~as~~~} x \rr 0 \]

• From the definition of asymptotically equivalent functions, we have:

  \[ f(x) \thicksim g(x) {\rm ~~as~~} x \rr c \Longleftrightarrow \frac{f(x)}{g(x)} \rr 1 {\rm ~~as~~} x \rr c \]

  in these cases we have:

  \[ \frac{f(x)}{g(x)}-1 \rr 0 {\rm ~~as~~} x \rr c {\rm ~~~~~and~~~~~} \frac{f(x)-g(x)}{g(x)} \rr 0 {\rm ~~as~~} x \rr c \]

  from the definition of little \(o\) we can then write:

  \[ f(x) - g(x) = o\big(g(x)\big) {\rm ~~as~~} x \rr c \]

  We have the following asymptotic expansion:

  \[ f(x) \thicksim g(x) {\rm ~~as~~} x \rr c \Longleftrightarrow f(x) = g(x) + o\big(g(x)\big) {\rm ~~as~~} x \rr c \]

• Moreover, we have:

  \[ g(x) + f(x) \thicksim g(x) {\rm ~~as~~} x \rr c \Longleftrightarrow \frac{f(x)}{g(x)} \rr 0 {\rm ~~as~~} x \rr c, \]

  since:

  \[ \frac{g(x) + f(x)}{g(x)} = 1 + \frac{f(x)}{g(x)} \]

  which tends to 1 as \(x \rr c\) if and only if \(f(x) / g(x)\) tends to 0 as \(x \rr c\).

  In such situations we say that \(g(x)\) is the principal part of the sum \(g(x) + f(x)\) and that \(f(x)\) is negligible with respect to \(g(x)\) as \(x \rr c\); that is:

  \[ f(x) = o\big(g(x)\big) {\rm ~~as~~ } x \rr c \]

  We have the following asymptotic expansion:

  \[ g(x) + f(x) \thicksim g(x) {\rm ~~as~~} x \rr c \Longleftrightarrow f(x) = o\big(g(x)\big) {\rm ~~as~~} x \rr c \]

• It follows from the definition of little \(o\) that with three functions we have:

  \[ f(x) = h(x) + o\big(g(x)\big) {\rm ~~as~~} x \rr c \Longleftrightarrow \frac{f(x)-h(x)}{g(x)} \rr 0 {\rm ~~as~~} x \rr c \]

• The “little \(o\)” symbol behaves as follows with products:

  \[ f \cdot o \big(g\big) = o \big(f \cdot g\big) \]
  \[ o \big(f\big) \cdot o \big(g\big) = o \big(f \cdot g\big) \]

  Example 2: Little \(o\) and products

  For example, as \(x \rr c\):

  \[ x \cdot o\big(x^2\big) = o\big(x^3\big),~~~~~~~ \frac{o\big(x^3\big)}{x} = o\big(x^2\big),~~~~~~~ o(x) \cdot o\big(x^2\big)=o\big(x^3\big) \]

Some properties of the little \(o\) symbol, shown through examples:

\[\begin{align*} o(x) \pm o(x) &= o(x),~~~ {\rm ~~as~~} x \rr c \\[1ex] o(a \; x) &= o(x),~~~~~ \forall a \in \R,~ a \neq 0,~~~ {\rm ~~as~~} x \rr c\\[1ex] a\;o(x) &= o(x),~~~~~ \forall a \in \R,~~~ {\rm ~~as~~} x \rr c\\[1ex] o(x) + o\big(x^2\big) &= o(x),~~~ {\rm ~~as~~} x \rr 0\\[1ex] o(x) + o\big(x^2\big) &= o\big(x^2\big),~ {\rm ~~as~~} x \rr \infty \end{align*}\]

Definition 2: of \(o(1)\)

A function \(f\) is said to be infinitesimal as \(x \rr c\), and we write

\[ f(x) = o(1) {\rm ~~as~~} x \rr c {\rm ~~if~~} f(x) \rr 0 {\rm ~~as~~} x \rr c \]

Example 3: \(o(1)\)

For example:

\[ \sqrt{x} = o(1) {\rm ~~~as~~~} x \rr 0^+ {\rm ~~~since~~~} \frac{\sqrt{x}}{1} \rr 0 {\rm ~~~as~~~} x \rr 0^+ \]

\[ x^{-2} = o(1) {\rm ~~~as~~~} x \rr \ip {\rm ~~~since~~~} \frac{x^{-2}}{1} \rr 0 {\rm ~~~as~~~} x \rr \ip \]

• We have:

  \[ \lim_{ x \rr c} f(x) = \ell \Longleftrightarrow \lim_{ x \rr c} (f(x) - \ell)=0 \Longleftrightarrow \lim_{ x \rr c} |f(x) - \ell|=0 \]

  hence

  \[ \lim_{ x \rr c} f(x) = \ell \Longleftrightarrow f(x) = \ell + o(1) {\rm ~~~as~~~} x \rr c \]

  the last expression reads “\(f(x)\) is the sum of \(\ell\) and of a function that is infinitesimal as \(x \rr c\)”

Example 4: \(o(1)\)

For example:

\[ x^2 = x \big(1 + o(1)\big) {\rm ~~~as~~~} x \rr 1 {\rm ~~~since~~~} \frac{x^2}{x} = x \rr 1 {\rm ~~~as~~~} x \rr 1 \]

\[ x^2 + 3\:x = x \big(3 + o(1)\big) {\rm ~~~as~~~} x \rr 0 {\rm ~~~since~~~} \frac{x^2 + 3\:x}{x} = x + 3\rr 3 {\rm ~~~as~~~} x \rr 0 \]

\[ x^2 + 3\:x = x^2 \big(1 + o(1)\big) {\rm ~~~as~~~} x \rr \ip {\rm ~~~since~~~} \frac{x^2 + 3\:x}{x^2} = 1 + \frac{3}{x}\rr 1 {\rm ~~~as~~~} x \rr \ip \]

\[ \sqrt{x+5} = \sqrt{x} \big(1 + o(1)\big) {\rm ~~~as~~~} x \rr \ip {\rm ~~~since~~~} \frac{\sqrt{x+5}}{\sqrt{x}} =\sqrt{1 + \frac{5}{x}} \rr 1 {\rm ~~~as~~~} x \rr \ip \]

Some properties of the \(o(1)\) symbol:

\[\begin{align*} o(1) \pm o(1) &= o(1),~~~ {\rm ~~as~~} x \rr c\\[1ex] o(1) \cdot o(1) &= o(1),~~~ {\rm ~~as~~} x \rr c\\[1ex] c \cdot o(1) &= o(1),~~~ {\rm ~~as~~} x \rr c,~~ c \neq 0\\[1ex] \big(1+o(1)\big) \cdot \big(1+o(1)\big) &= 1+o(1),~~~ {\rm ~~as~~} x \rr c\\[1ex] \frac{1}{1+o(1)} &= 1+o(1),~~~ {\rm ~~as~~} x \rr c\\[1ex] \frac{c}{1+o(1)} - c &= o(1),~~~ {\rm ~~as~~} x \rr c,~~ c \neq 0 \end{align*}\]

Since

\[ \frac{c}{1+o(1)} - c = c \; \left( \frac{1}{1+o(1)} -1\right)= c \; \big(1+o(1)-1 \big)= o(1) \]

Example 5: \(o(1)\)

We have

\[ \big(x + o(1)\big)^2 = x^2 \big(1 + o(1)\big) {\rm ~~~as~~~} x \rr \ip {\rm ~~~since~~~} \frac{\big(x + o(1)\big)^2}{x^2} \rr 1 {\rm ~~~as~~~} x \rr \ip \]

2. Asymptotic expansions derived from the fundamental limits

We have seen that the following fundamental limits hold:

\[ \lim_{x \rr 0} \frac{\sin x}{x} =1,~~\lim_{x \rr 0} \frac{1 - \cos x}{x^2} = \frac{1}{2},~~ \lim_{x \rr 0} \frac{\log(1 + x)}{x} =1,~~\lim_{x \rr 0} \frac{e^x -1}{x} =1,~~\lim_{x \rr 0} \frac{ (1 + x)^{\alpha} -1}{x} = \alpha \]

and consequently the following asymptotic equivalences hold as \(x \rr 0\):

\[ \sin x \thicksim x,~~ 1 - \cos x \thicksim \frac{1}{2} x^2,~~ e^x -1 \thicksim x,~~ \log(1+x) \thicksim x,~~ (1+x)^{\alpha} -1 \thicksim \alpha \: x \]

Let us now see how to obtain the asymptotic expansions starting from the fundamental limits:

\[ \lim_{x \rr 0} \frac{\sin x}{x} =1 \]

Hence, as \(x \rr 0\):

\[ \frac{\sin x}{x} - 1 \rr 0, \qquad \frac{\sin x - x}{x} \rr 0 {\rm ~~hence~~} \sin x - x = o(x) \]

which gives the asymptotic expansion:

\[ \sin x = x + o(x) \]

\[ \lim_{x \rr 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]

Hence, as \(x \rr 0\):

\[ \frac{1 - \cos x}{x^2} - \frac{1}{2} \rr 0, \qquad \frac{2\:(1 - \cos x) - x^2}{2\:x^2} \rr 0 {\rm ~~hence~~} 2\:(1 - \cos x) - x^2 = o\big(\:2\:x^2\big) \]

which gives the asymptotic expansion:

\[ \cos x = 1 - \frac{1}{2} \: x^2 + o\big(x^2\big) \]

\[ \lim_{x \rr 0} \frac{\log(1 + x)}{x} =1 \]

Hence, as \(x \rr 0\):

\[ \frac{\log(1 + x)}{x} - 1 \rr 0, \qquad \frac{\log(1 + x) - x}{x} \rr 0 {\rm ~~hence~~} \log(1 + x) - x = o(x) \]

which gives the asymptotic expansion:

\[ \log(1 + x) = x + o(x) \]

\[ \lim_{x \rr 0} \frac{e^x -1}{x} =1 \]

Hence, as \(x \rr 0\):

\[ \frac{e^x -1}{x} - 1 \rr 0, \qquad \frac{e^x -1- x}{x} \rr 0 {\rm ~~hence~~} e^x-1 - x = o(x) \]

which gives the asymptotic expansion:

\[ e^x =1+ x + o(x) \]

\[ \lim_{x \rr 0} \frac{ (1 + x)^{\alpha} -1}{x} = \alpha {\rm ~~~~~~~~with~~~} \alpha \in \R \]

Hence, as \(x \rr 0\):

\[ \frac{ (1 + x)^{\alpha} -1}{x} - \alpha \rr 0 \qquad \frac{(1 + x)^{\alpha} -1 - \alpha \: x}{x} \rr 0 {\rm ~~hence~~} (1 + x)^{\alpha} -1 - \alpha \: x = o(x) \]

which gives the asymptotic expansion:

\[ (1 + x)^{\alpha} = 1 + \alpha \: x + o(x) \]

• These asymptotic expansions can be generalized

If \(\varepsilon(x)\) is a function that tends to zero, i.e., it is an infinitesimal (it does not matter what \(x\) tends to), we have the following asymptotic expansions derived from the fundamental limits.

As \(\varepsilon(x) \rr 0\), we have:

\[\begin{align} \sin{ \big( \varepsilon(x) \big)} &= \varepsilon(x) + o\big( \varepsilon(x)\big) \\[2ex] \cos{\big( \varepsilon(x) \big)} &= 1 - \frac{1}{2} \; \varepsilon^2(x) + o{\big(\varepsilon^2(x)\big)} \\[2ex] \log{\big(1+\varepsilon(x)\big)} &= \varepsilon(x) + o\big(\varepsilon(x)\big) \\[2ex] e^{\varepsilon(x)} &= 1 + \varepsilon(x) + o\big(\varepsilon(x)\big) \\[2ex] \big(1+\varepsilon(x)\big)^\alpha &= 1 + \alpha \; \varepsilon(x) + o\big(\varepsilon(x)\big) {\rm ~~~~~~~~with~~~} \alpha \in \R \end{align}\]