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\title{$o$-Symbols}

\author{Math 166: Honors Calculus II}

\begin{document}

\maketitle

\section*{Definitions}

The symbol $o(g(x))$ stands for a function $f(x)$ that satisfies
  \[ \lim_{x\to a}\frac{f(x)}{g(x)} = 0. \]
The limit is usually taken as $x\to 0$. When it is some other number,
$a\ne0$, the number $a$ will usually be mentioned explicitly. Our primary
use of $o$-symbols is with powers of $x$. So, for example, we use the
term $o(x^3)$ to stand for an arbitrary function $f(x)$ that satisfies
$f(x)/x^3 \to 0$ as $x \to 0$. These $o$-symbols are often very useful
for calculating Taylor polynomials and finding limits.

As an extension of the notation, we write ``$f(x) = h(x) + o(g(x))$''
to mean that
  \[ \lim_{x\to a}\frac{f(x)-h(x)}{g(x)} = 0. \]
For example, the Lagrange form of the remainder term in Taylor's
formula is
  \[ E_n f(x ; a) = \frac{f^{(n+1)}(c_x)}{(n+1)!}(x-a)^{n+1}. \]
Since
  \[ \lim_{x\to a} \frac{E_nf(x;a)}{(x-a)^n} =
  \lim_{x\to a} \frac{f^{(n+1)}(c_x)}{(n+1)!}(x-a) =
  \frac{f^{(n+1)}(a)}{(n+1)!}\cdot 0 = 0 \]
we see that $E_nf(x;a) = o\big((x-a)^n\big)$ and we may express Taylor's
formula as
  \[ f(x) = \sum_{k=0}^n\frac{f^{(k)(a)}}{k!}(x-a)^k + o\big((x-a)^n\big). \]
This allows us to rewrite any expression that involves complicated functions
in terms of simple polynomials and $o$-terms. We already know how to
combine and simplify polynomials. To be able to simplify the more general
expressions we need to know how to combine and simplify $o$-terms.

\section*{Algebra of $o$-Symbols}

\begin{theorem}
As $x\to a$ we have,
\begin{enumerate}
\item $o(g(x)) \pm o(g(x)) = o(g(x))$
\item $o(cg(x)) = o(g(x))$ \quad $c\ne 0$
\item $f(x)\cdot o(g(x)) = o(f(x))o(g(x)) = o(f(x)g(x))$
\item $o(o(g(x))) = o(g(x))$
\item $\frac{1}{1-g(x)} = 1 + g(x) + o(g(x))$ if $g(x)\to 0$ as $x \to a$.
\end{enumerate}
\end{theorem}

\begin{proof}

1. The symbol $o(g(x)) \pm o(g(x))$ stands for a function of the form
$f(x) \pm h(x)$ where we know nothing about $f(x)$ or $h(x)$ other than
the limits: $f(x)/g(x) \to 0$ and $h(x)/g(x) \to 0$ as $x\to a$. That is
enough information to compute the limit
  \[ \lim_{x\to a} \frac{f(x) \pm h(x)}{g(x)} =
  \lim_{x\to a}\frac{f(x)}{g(x)} \pm \lim_{x\to a}\frac{h(x)}{g(x)} =
  0 \pm 0 = 0. \]
This shows that $f(x) \pm h(x) = o(g(x))$. Therefore, a function of
the form $o(g(x)) \pm o(g(x))$ is also of the form $o(g(x))$ and we
write this fact as $o(g(x)) \pm o(g(x)) = o(g(x))$.

2. The symbol $o(cg(x))$ stands for a arbitrary function $f(x)$
that has the property $f(x)/(cg(x))\to0$ as $x\to a$. But then,
  \[ \lim_{x\to a} \frac{f(x)}{g(x)} =
   c\cdot\lim_{x\to a}\frac{f(x)}{cg(x)} = c\cdot 0 = 0 \]
Therefore $f(x)$ is also of type $o(g(x))$.

3. The symbol $f(x)o(g(x))$ stands for a function of the form $f(x)h(x)$
where we know nothing about $f(x)$ and the only thing we know about $h(x)$
is that $h(x)/g(x)\to0$ as $x\to a$. This is enough information to
conclude that
  \[ \lim_{x\to a}\frac{f(x)h(x)}{f(x)g(x)} =
  \lim_{x\to a}\frac{h(x)}{g(x)} = 0. \]
By definition, this implies that $f(x)h(x) = o(f(x)g(x))$ and that is what
is meant by ``$f(x)o(g(x)) = o(f(x)g(x))$.'' Similarly, $o(f(x))o(g(x))$
stands for a function of the form $k(x)h(x)$ where $k(x)/f(x)\to 0$
and $h(x)/g(x)\to 0$ as $x\to a$. As above, it is easy to see that
$k(x)h(x)/(f(x)g(x)) \to 0$ so $k(x)h(x)$ is of type $o(f(x)g(x))$ as well
and ``$o(f(x))o(g(x)) = o(f(x)g(x))$.''

4. The symbol $o(o(g(x)))$ is a little harder to decode. Starting on the
inside, $o(g(x))$ stands for an arbitrary function $h(x)$ satisfying
$h(x)/g(x)\to0$ as $x\to a$. So $o(o(g(x)))$ is really $o(h(x))$
and stands for an arbitrary function $f(x)$ satisfying $f(x)/h(x)\to0$
as $x\to a$. Putting these two limits together we get:
  \[ \lim_{x\to a} \frac{f(x)}{g(x)} =
  \lim_{x\to a} \frac{f(x)}{h(x)}\frac{h(x)}{g(x)} =
  \lim_{x\to a} \frac{f(x)}{h(x)}\lim_{x\to a} \frac{h(x)}{g(x)} =
  0 \cdot 0 = 0. \]
Therefore, $f(x)$ is also of type $o(g(x))$ and that is what is meant by
``$o(o(g(x))) = o(g(x))$.''

5. Simple algebra implies that for any number $u\ne1$ we have
  \[ \frac{1}{1-u} = 1 + u + \frac{u^2}{1-u}. \]
Since $\frac{u^2}{1-u}\Big/ u = \frac{u}{1-u} \to 0$ as $u\to 0$ we may
write this as
  \[ \frac{1}{1-u} = 1 + u + o(u). \]
We are assuming that $g(x)\to 0$ as $x\to a$, so we may replace $u$ with
$g(x)$ to get
  \[ \frac{1}{1-g(x)} = 1 + g(x) + o(g(x)). \]
\end{proof}

We will usually apply the above rules to $o$-terms of the form $o(x^n)$ or
$o\big((x-a)^n\big)$, so it is helpful to see what the above statements
imply when $f(x) = x^n$ and $g(x) = x^m$. Modifications for powers of
$(x-a)$ should be obvious.

\begin{corollary}
\
\begin{enumerate}
\item $x^n = o(x^m)$ for $m<n$
\item If $f(x) = o(x^n)$ then $f(x) = o(x^m)$ for $m\le n$.
\item $x^n o(x^m) = o(x^n)o(x^m) = o(x^{n+m})$
\item $o(x^n) \pm o(x^m) = o(x^k)$ where $k = \min\{n,m\}$.
\item $o(x^n)^m = o(x^{nm})$
\end{enumerate}
\end{corollary}

\begin{proof}

1. $\lim_{x\to 0} x^n/x^m = 0$ only if $n>m$.

2. If $f(x)=o(x^n)$ then $\lim_{x\to 0}f(x)/x^n = 0$. If $m\le n$, then
$x^{n-m}\to0$ or $1$ as $x\to0$ so
  \[ \lim_{x\to 0}\frac{f(x)}{x^m} = \lim_{x\to0} x^{n-m}\frac{f(x)}{x^n}=
  \lim_{x\to 0}x^{n-m} \lim_{x\to0} \frac{f(x)}{x^n} = 0 \]
and $f(x) = o(x^m)$.

3. This is Theorem~1.3.

4. By property~2, we can replace $o(x^m)$ with $o(x^n)$ and Theorem~1.1
then implies,
  \[ o(x^n) \pm o(x^m) = o(x^n) \pm o(x^n) = o(x^n). \]

5. This can be established by applying property~3 and induction on~$m$. It
is obviously true for $m=1$. Suppose it is true for $m=k$ and consider
the case $m=k+1$:
  \[ o(x^n)^{k+1} = o(x^n)^ko(x^n) = o(x^{nk})o(x^n) = o(x^{nk}x^n) =
  o(x^{n(k+1)}). \]
By induction the formula is true for all positive integers~$m$.
\end{proof}

Note the strict inequality in Corollary~2.1. In fact, $x^n \ne o(x^n)$
since $x^n/x^n\to1\ne0$ as $x\to a$.

\section*{Simplifying $o$-Symbols}

Let $P(x)$ be a polynomial expression involving various powers of $x^n$,
$o(x^m)^k$, and products of these, $x^n o(x^m)^k$. Then $P(x)$ can be
reduced to an expression of the form
  \[ P(x) = a_0 x^{n_0} + o(x^{m_0}). \]
Furthermore,
  \[ o(P(x)) = o(x^k)\ \mbox{where}\ k = \min\{n_0,m_0\}. \]

To see why this is true, perform the following steps.  First, find the
term that has the smallest power of $x$ (only), say $a_0x^{n_0}$. The sum
of all the other terms involving only $x$ can replaced by $o(x^{n_0})$
since they all have exponents greater than $n_0$ (apply Corollary~2.1
and 2.4). Thus $P(x)$ can be replaced with an expression of the form
$a_0x^{n_0} + Q(x)$ where all the terms of $Q(x)$ involve some $o$-symbol.

Next simplify $Q(x)$ to $o(x^{m_0})$ by replacing any terms like
$c\cdot o(x^m)^k$ with $o(x^{mk})$, replacing any terms like $c\cdot
x^no(x^m)^k$ with $o(x^{n+mk})$ and replacing any sums like $o(x^m) +
o(x^n)$ with $o(x^k)$ where $k = \min\{m,n\}$ (apply Corollary~2.2--5).

The final conclusion above can be shown by considering which exponent,
$n_0$ or $m_0$, is smaller.  If $m_0 < n_0$ then $a_0x^{n_0}=o(x^{m_0})$
so $P(x)$ reduces to $P(x) = o(x^{m_0})$ and $o(P(x)) = o(o(x^{m_0}))=
o(x^{m_0})$.

If $m_0 \ge n_0$, then $o(P(x)) = o(a_0x^{n_0}+o(x^{m_0}))$ stands for a
function $f(x)$ satisfying
\[ \lim_{x\to 0}\frac{f(x)}{a_0x^{n_0}+h(x)} = 0 \]
Here $h(x)=o(x^{m_0})$, so $h(x)=o(x^{n_0})$ and $h(x)/x^{n_0}\to 0$
as $x\to 0$. Therefore,
\[ \lim_{x\to0}\frac{f(x)}{x^{n_0}}
  = \lim_{x\to0}\frac{f(x)}{a_0x^{n_0}+h(x)}\Big(a_0+\frac{h(x)}{x^{n_0}}\Big)
  = 0\cdot(a_0+0) = 0. \]
Therefore, $f(x) = o(x^{n_0})$ and we conclude $o(P(x)) = o(x^{n_0})$
as claimed.

\subsection*{Example}
Let $P(x) = (x-x^2 + o(x-x^2))^2$. We simplify the $o$-symbol first:
$o(x-x^2) = o(x+o(x)) = o(x)$. Then,
\[ P(x) = (x-x^2+o(x))^2 = (x-x^2)^2 + 2(x-x^2)o(x) + o(x)^2 \]
Let's deal with each term one at a time:
  \begin{eqnarray*}
  (x-x^2)^2 &=& x^2 - 2x^3 + x^4 = x^2 + o(x^2)\\
  2(x-x^2)o(x) &=& 2xo(x) - 2x^2o(x) = o(x^2)+o(x^3) = o(x^2)\\
  o(x)^2 &=& o(x^2)
  \end{eqnarray*}
Therefore,
\[ P(x) = x^2 + o(x^2) + o(x^2) + o(x^2) = x^2 + o(x^2) \]
and
\[ o(P(x)) = o(x^2 + o(x^2)) = o(x^2). \]

\subsection*{Example}
\begin{eqnarray*}
\lefteqn{\cos(\sin(x))
=1-\frac{1}{2}\sin^2(x)+\frac{1}{4!}\sin^4(x)+o(\sin^5(x))}\\
&=&1-\frac{1}{2}\big(x-\frac{1}{3!}x^3+o(x^4)\big)^2
   +\frac{1}{4!}\big(x-\frac{1}{3!}x^3+o(x^4)\big)^4
   +o\big(\big(x-\frac{1}{3!}x^3+o(x^4)\big)^5\big)\\
&=&1-\frac{1}{2}\big(x^2-\frac{1}{3}x^4+o(x^5)\big)
   +\frac{1}{4!}\big(x^4+o(x^5)\big)+o(x^5+o(x^6))\\
&=&1-\frac{1}{2}x^2 + \frac{5}{24}x^4 + o(x^5)
\end{eqnarray*}

\end{document}
