Chapter 5.7 Analysis of Error

In this section, we learn about how to describe certain qualitative features of a function using limits. We introduce new notation to define the subtle differences.

Asymptotic Notation

The functions \({\rm pow}_1\), \({\rm pow}_2\), and \({\rm pow}_6\) all have the property that \[\lim_{x\to 0} {\rm pow}_1(x) = \lim_{x\to 0} {\rm pow}_2(x) = \lim_{x\to 0}{\rm pow}_6(x) = 0,\] but their behavior near 0 qualitatively appears different.

Here are sketches of these functions:

The behavior of \(\rm{pow}_1\) near \(0\) is different from \(\rm{pow}_2\) and \(\rm{pow}_6\). The behavior of \(\rm{pow}_2\) and \(\rm{pow}_6\) is similar near \(0\), but there is a difference. It appears “flatter” near \(0\) for the sketch of \(\rm{pow}_6\) compared to the sketch of \(\rm{pow}_2.\)

To continue our discussion, we first define what we mean by local behavior.

Local Behavior

For any function \(f\) and any real number \(x_0\) that is a limit point of \(\mathcal D(f)\), the local behavior of \(f\) at \(x_0\) is, roughly, the information about \(f\) that, for any open interval \(I\) that contains \(x_0\), can be recovered from the values of \(f\) on \((\mathcal D(f)\cap I)\setminus\{x_0\}\).

The Landau Notation is useful for capturing such information about a function.

The quantification of the interval \(I\) over all open intervals makes precise the idea that these are the properties of a function that depend only on the \(x\)-coordinates being “close enough” to \(x_0\).

Landau Big O Notation

For any function \(f\) and any limit point \(x_0\) of \(\mathcal D(f)\), \(f\) is \(\displaystyle \underset{x\to x_0}O (g(x))\) (reads: “big oh of \(g\) of \(x\) at \(x_0\)”) if there is

  • an open interval \(I\) that contains \(x_0\)

  • and a bounded function \(\eta\) that is defined on \((\mathcal D(f)\cap I)\setminus\{x_0\}\)

  • so that for all \(x\) in \((\mathcal D(f)\cap I)\setminus\{x_0\}\), \[f(x) = \eta(x)g(x).\]

Essentially, the function \(f\) can be “factored” into two functions, a function \(g\) that it behaves like near \(x_0\) and a function \(\eta\) that is near \(x_0\), but it may not necessarily be defined at \(x_0\).

In the next example, use the definition to identify the local behavior of the function.

Example 1

For each rational function \(f\) that is given below, classify all zeros of \(f\) by finding the largest natural number \(n\) and a real number \(x_0\) so that \(f\) is \(O((x-x_0)^n)\):

  1. \(f(x) = \frac{x^3}{x+2}\);

  2. \(f(x) = \frac{(x+3)(x-5)^2}{(x-1)}\);

  1. The zero of \(f\) is at \(x_0=0\). The function \(f\) can be rewritten like this: \[\frac{x^3}{x+2}= (x-0)^3\cdot\frac{1}{x+2}.\] The function \(\eta(x)=\tfrac{1}{x+2}\) is bounded on \(I=(-1,1)\) since it is continuous on the closed and bounded interval \([-1,1]\) (See Theorem 4 in Chapter 5.6). Hence, \(f\) is \(O(x^3)\) because
    • there is an open interval \(I=(-1,1)\) that contains \(x_0=0\)
    • a bounded function \(\eta(x)=\tfrac{1}{x+2}\) defined on \((-1,0)\cup(0,1)\)
    • so that \(f(x)=(x-0)^3\cdot\tfrac{1}{x+2}\).

  1. The zeros of \(f\) is \(x_0=-3\) and \(x_0=5\). The function \(f\) can be rewritten like this: \[\frac{(x+3)(x-5)^2}{(x-1)}= (x+3)\cdot\frac{(x-5)^2}{(x-1)}.\] The function \(\eta(x)=\tfrac{(x-5)^2}{x-1}\) is bounded on \(I=(-4,-2)\) since it is continuous on the closed and bounded interval \([-4,-2]\) (See Theorem 4 in Chapter 5.6). Hence, \(f\) is \(O(x+3)\) because
    • there is an open interval \(I=(-4,-2)\) that contains \(x_0=-3\)
    • a bounded function \(\eta(x)=\tfrac{(x-5)^2}{x-1}\) defined on \((-4,-3)\cup(-3,4)\)
    • so that \(f(x)=(x+3)\cdot\tfrac{(x-5)^2}{(x-1)}\). For \(x_0=5\), the function \(f\) can be rewritten like this: \[\frac{(x+3)(x-5)^2}{(x-1)}= (x-5)^2\cdot\frac{x+3}{x-1}.\] The function \(\eta(x)=\tfrac{x+3}{x-1}\) is bounded on \(I=(4,6)\) since it is continuous on the closed and bounded interval \([4,6]\) (See Theorem 4 in Chapter 5.6). Hence, \(f\) is \(O((x-5)^2)\) because
    • there is an open interval \(I=(4,6)\) that contains \(x_0=5\)
    • a bounded function \(\eta(x)=\tfrac{x+3}{x-1}\) defined on \((4,5)\cup(5,6)\)
    • so that \(f(x)=(x-5)^2\cdot\tfrac{x+3}{x-1}\).

A related but different way to describe local behavior is by using Landau little o notation.

Landau Little O Notation

For any function \(f\) and any limit point \(x_0\) of \(f\), a function \(f\) is \(\displaystyle \underset{x\to x_0}o (g(x))\) (reads: “little oh of \(g\) of \(x\) at \(x_0\)”) if there is

  • an open interval \(I\) that contains \(x_0\)

  • and a function \(\eta\) that is defined on \((\mathcal D(f)\cap I)\setminus\{x_0\}\) with \[\lim_{x\to x_0} \eta(x) = 0,\]

  • so that for all \(x\) in \((\mathcal D(f)\cap I)\setminus\{x_0\}\), \[f(x) = \eta(x)g(x).\]

The key difference between little o and big O notation is that for little o, we need the function \(\eta\) to not just be bounded near \(x_0\) but also have its limit at \(x_0\) go to zero. Again, it may not be defined at \(x_0\).

In the next example, we show a connection between big o and little o notation.

Example 2

Show that any function that is \(O((x-x_0)^2)\) is \(o(x-x_0)\).

Because \(f\) is \(O((x-x_0)^2)\),

  • there is an open interval \(I\) that contains \(x_0\)

  • and a bounded function \(\eta(x)\) defined on \((\mathcal{D}(f)\cap I)\setminus\{x_0\}\)

  • so that for all \(x\) in \((\mathcal{D}(f)\cap I)\setminus\{x_0\}\) \(f(x)=\eta(x)(x-x_0)^2\).

The claim is that \(f\) is \(o(x-x_0)\). First rewrite \(f\). Instead of \(f(x)=\eta(x)(x-x_0)^2\) rewrite it as \[f(x)=[\eta(x)(x-x_0)]\cdot(x-x_0)\] and define \(\eta'\) to be the function given by \[\eta'(x)=\eta(x)(x-x_0).\] To satisfy the definition of \(o(x-x_0)\), show \(\displaystyle\lim_{x\to x_0}\eta'(x)=0\) by using the squeeze theorem. First, since \(\eta\) is bounded, we have that \[\begin{align*}m\leq &\eta(x)\leq M\\ m(x-x_0)\leq&(x-x_0)\eta(x)\leq M(x-x_0) \end{align*}\] or \(m(x-x_0)\geq(x-x_0)\eta(x)\geq M(x-x_0)\) depending on if \(x>x_0\) or \(x<x_0\). However, in either case, both limits go to zero, that is \(\displaystyle\lim_{x\to x_0}m(x-x_0)=0\) and \(\displaystyle\lim_{x\to x_0}M(x-x_0)=0.\) Apply the squeeze theorem to conclude \(\lim\limits_{x\to x_0}\eta'(x)=0\).

Thus \(f\) is \(o(x-x_0).\)

In this example, practice identifying both big O and little o behavior of the following functions.

Example 3

For each function \(f\) that is given below, find the largest natural numbers \(m\) and \(n\) so that \(f\) is \(O((x-x_0)^m)\) and \(o((x-x_0)^n)\):

  1. \(x_0 = 2\) and \(f(x) = (x+3)(x-2)^3\);

  2. \(x_0 = 5\) and \(f(x) = (x+1)\sin(x-5)\);

  3. \(x_0 = 0\) and \(f(x) = (x+1)^\frac{1}{5}(1-\cos(x))\).

  1. The function \(f\) is \(O((x-2)^3)\) since \(f(x)=(x+3)(x-2)^3\) can be rewritten as \(\eta(x)(x-2)^3\) where \(\eta\) is given by \(\eta(x)=(x+3).\) This is a bounded function on the interval \(I=(1,2)\cup(2,3)\). On the other hand, \(f\) is \(o((x-2)^2)\) since \(f(x)=(x+3)(x-2)^3\) can be rewritten as \(\eta'(x)(x-2)^2\) where \(\eta'\) is given by \(\eta(x)=(x-2)(x+3).\) This is a function defined \(I=(1,2)\cup(2,3)\) with the property that \(\lim\limits_{x\to 2}\eta(x)=\lim\limits_{x\to 2}(x-2)(x+3)=0\).

  1. The function \(f\) is \(O((x-5))\) since \(f(x)=(x+1)\sin(x-5)\) can be rewritten as \[(x+1)\sin(x-5)=\tfrac{(x-5)}{(x-5)}(x+1)\sin(x-5)=\eta(x)(x-5)\] where \(\eta\) is the function given by \(\eta(x)=(x+1)\tfrac{\sin(x-5)}{x-5}.\) This is a bounded function on the interval \(I=(4,5)\cup(5,6)\). However, \(f\) is not \(o((x-5))\) since \[\displaystyle\lim_{x\to 5}\eta(x)=\displaystyle\lim_{x\to 5}(x+1)\tfrac{\sin(x-5)}{x-5}=6\cdot 1=6\] which is not equal to zero.

  1. The function \(f\) is \(O((x)^2)\) since \(f(x)=(x+1)^{1/5}(1-\cos(x))\) can be rewritten as \[(x+1)^{1/5}(1-\cos(x))=\tfrac{(x+1)^{1/5}(1-\cos(x))}{x^2}\cdot x^2=\eta(x)x^2\] where \(\eta\) is given by \(\eta(x)=\tfrac{(x+1)^2(1-\cos(x))}{x^2}.\) This is a bounded function on the interval \(I=(-1,0)\cup(0,1)\). This is because \((x+1)^{1/5}\) is bounded in \((-1,0)\cup(0,1)\) and \(\tfrac{1-\cos(x)}{x^2}\) is bounded in \((-1,0)\cup(0,1)\) since the following limit exist: \[\displaystyle\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac{1}{2}.\] To see why the limit is equal to \(\tfrac{1}{2}\), rewrite the expression like this: \[ \begin{align*} \frac{1-\cos(x)}{x^2}&=\frac{1-\cos(x)}{x^2}\cdot\frac{1+\cos(x)}{1+\cos(x)}\\ &=\frac{1-\cos^2(x)}{x^2(1+\cos(x))}\\ &=\frac{\sin^2(x)}{x^2(1+\cos(x))}\\ &=\frac{\sin(x)}{x}\cdot\frac{\sin(x)}{x}\cdot\frac{1}{1+\cos(x)}. \end{align*} \] Then take the limit to get the desired result. However \(f\) is \(o(x)\) since \(f\) can be rewritten like this: \[(x+1)^{1/5}(1-\cos(x))=\frac{(x+1)^{1/5}(1-\cos(x))}{x}\cdot x=\eta'(x)x,\] where \(\eta'\) is the function given by \(\eta'(x)= \frac{(x+1)^{1/5}(1-\cos(x))}{x}.\) This is a function defined on \((-1,0)\cup(0,1)\) with the property that \[\displaystyle\lim_{x\to 0}\eta(x)=\displaystyle\lim_{x\to 0}(x+1)^{1/5}\frac{1-\cos(x)}{x}=1\cdot 0=0.\]

Two functions can have the same behavior in the following way. Functions \(f\) and \(g\) have the same local behavior at a point \(x_0\) if \(f\) is \(\underset{x\to x_0}O(g(x))\) and \(g\) is \(\underset{x\to x_0}O(f(x)).\) In other words, we can find an open interval \(I\) and a positive real number \(C\) so that for all \(x\)-coordinate values restricted to \(I\smallsetminus\{x_0\}:\)

  • \(|f|\) lies between \(C|g|\) and the \(x\)-axis
  • \(|g|\) lies between \(C|f|\) and the \(x\)-axis.

Understand this with the following example.

Example 4

Take \(f\) and \(g\) to be the functions that are given by \[f(x)=(x+4)(x-1)^3\sin(x)\quad\text{and}\quad g(x)=\dfrac{x(x+4)^2(x-1)^3(x-5)^7}{(x+6)(x-2)^4}.\]

Identify all zeros of \(f\) and \(g\) at which the two functions have the same local behavior.

\(f\) and \(g\) have the same local behavior at \(0\) and at \(1\).

Now, understand the behavior between the following two trigonometric functions.

Example 5

Take \(f\) and \(g\) to be the functions that are given by \[f(x)=1-\cos(x)\quad\text{and}\quad g(x)=\sin^n(x).\]

Determine \(n\) so that \(f\) and \(g\) have the same local behavior at \(0\).

The function \(f\) is \(O(x^2)\) and \(\sin\) is \(O(x).\) Thus, take \(n=2\) so that \(g(x)=\sin^2(x)\) is also \(O(x^2).\)

The Big O notation can also be used to describe how the function \(f\) itself behaves without reference to another function.

Big O 1 Notation

Take \(1\) to be the constant function defined for all \(x\) by \[1(x) = 1.\]

A function \(f\) is \(\displaystyle \underset{x\to x_0} O (1(x))\) means that there is

  • an open interval \(I\) that contains \(x_0\),

  • and a bounded function \(\eta_f\),

  • so that for every \(x\) in \((\mathcal D(f)\cap I)\smallsetminus\{x_0\}\), \[f(x) = \eta_f(x)1(x) = \eta_f(x).\]

To say that a function \(f\) is \(\displaystyle \underset{x\to x_0} O (1(x))\) is just a way of saying that \(f\) is bounded in a small enough open interval containing \(x_0\), but that \(f\) might not be defined at \(x_0\).

Little o 1 Notation

A function \(f\) is \(\displaystyle \underset{x\to x_0} o (1(x))\) means that there is

  • an open interval \(I\) that contains \(x_0\),

  • a function \(\eta_f\) with \(\displaystyle\lim_{x\to 0}\eta_f(x)=0\),

  • so that for every \(x\) in \((\mathcal D(f)\cap I)\setminus\{x_0\}\), \[f(x) = \eta_f(x)1(x) = \eta_f(x).\]

Confirm your understanding with the next example.

Example 6

Show that the function \(f\) that is given by \[f(x) = \tfrac{(x+2)^2(x-4)}{(x-4)(x-6)}\] is not defined at \(4\), but is \(\displaystyle \underset{x\to 4} O (1(x))\).

The function \(f\) is undefined at \(x=4\) because of the \((x-4)\) in the denominator term. However, when \(x\not=4\), \(f\) can be rewritten like this: \[ \frac{(x+2)^2(x-4)}{(x-4)(x-6)}=\frac{(x+2)^2}{x-6}.\] Take \(\eta\) to be the function given by \(\eta_f(x)=\tfrac{(x+2)^2}{x-6}\) and \(I\) to be the interval \(I=(3,5)\).

Then \(f(x)=\eta_f(x)\) is defined and is bounded on \((3,4)\cup(4,5).\) And so \(f\) is \(\displaystyle \underset{x\to 4} O(1(x))\).

In this next example, understand the implications of big O and little o notation.

Example 7

Determine whether or not the following statements are true or false:

  1. For any function \(f\) that is \(\displaystyle \underset{x\to x_0} O(1(x))\) but not defined at \(x_0\), \(f\) has a continuous extension to \(x_0\).
  2. For any function \(f\) that is \(\displaystyle \underset{x\to x_0} o (1(x))\) but not defined at \(x_0\), \(f\) has a continuous extension to \(x_0\).

  1. Note that \(\displaystyle \underset{x\to x_0} O (1(x))\) means that there is an interval \(I\) containing \(x_0\) and a bounded function \(\eta_f(x)\) so that for every \(x\) in \((\mathcal{D}(f)\cap I)\setminus \{x_0\}\), \(f(x)=\eta_f(x)\). There is no additional information on whether \(\eta_f(x)\) has a limit at \(x_0\). So there may or may not be a continuous extension.

  2. Note \(\displaystyle \underset{x\to x_0} o (1(x))\) means that \(f(x)=\eta_f(x)\) for \(x\) near \(x_0\) but not equal to \(x_0\) and \(\displaystyle \lim_{x\to x_0}\eta_f(x)=0.\) Define a continuous extension of \(f\) to \(\mathbb{R}\) by \[ g(x)= \begin{cases} f(x)&\text{if }x\not=x_0\\ 0&\text{ if } x=x_0. \end{cases} \]

In order to advance our understanding of local behavior, we define a new function, called the finite difference operator.

Finite Difference Operator

For any function \(f\) and any \(x_0\) that is in \(\mathcal D(f)\), define the finite difference operator at \(x_0\), \(\Delta_{x_0}\), by \[(\Delta_{x_0}f)(h) = f(x_0+h) - f(x_0).\]

The \(o\)-Landau notation is useful for describing continuity at the point \(x_0\) using \(\Delta_{x_0}\).

Theorem on Continuity and Little O Notation

Given that \(x_0\) is a limit point of \(\mathcal D(f)\), the statement that \(\Delta_{x_0}f\) is \(\displaystyle \underset{h\to 0} o (1(h))\) is equivalent to the statement that \(f\) is continuous at \(x_0\).

This follows since \(\Delta_{x_0}f(h)=f(x_0+h)-f(x_0)\) is \(\displaystyle \underset{h\to 0}o(1(h))\) means that there is an open interval \(I\) containing \(x_0\) and a function \(\eta_f(x)=f(x_0+h)-f(x_0)\) defined with \(\displaystyle\lim_{h\to 0}\eta_f(h)=0\). Meaning \(\displaystyle\lim_{h\to 0}f(x_0+h)-f(x_0)=0\) so \(\displaystyle\lim_{h\to 0}f(x_0+h)=f(x_0)\). Thus \(f\) is continuous at \(x_0\).

In the next example, calculate the finite difference operator for each function.

Example 8

For each function \(f\) and each real number \(x_0\), write a formula for \(\Delta_{x_0}f(h)\) and simplify the expression as much as possible:

  1. \(x_0 = 3\) and \(f(x) = x^2\);
  2. \(x_0 = 9\) and \(f(x) = \sqrt{x}\);
  3. \(x_0 = \frac{\pi}{6}\) and \(f(x) = \sin(x)\).

  1. Expand out \(\Delta_3f(h)\) to get that \[\begin{align*}\Delta_{3}f(h)&=f(3+h)-f(3)\\&=(3+h)^2-3^2\\&=9+6h+h^2-9\\&=6h+h^2\\&=h(6+h).\end{align*}\]

  2. Expand out to get that \[\begin{align*}\Delta_{9}f(h)&=f(9+h)-f(9)\\&=\sqrt{9+h}-\sqrt{9}\\&=(\sqrt{9+h}-3)\cdot \frac{\sqrt{9+h}+3}{\sqrt{9+h}+3}&&\text{multiply and divide by conjugate}\\&=\frac{9+h-9}{\sqrt{9+h}+3}\\&=\frac{h}{\sqrt{9+h}+3}\end{align*}\]

  3. Expand out to get that \[\begin{align*}\Delta_{\tfrac{\pi}{6}}f(h)&=f(\tfrac{\pi}{6}+h)-f(\tfrac{\pi}{6})\\&=\sin(\tfrac{\pi}{6}+h)-\sin(\tfrac{\pi}{6})\\&=\sin(\tfrac{\pi}{6})\cos(h)+\sin(h)\cos(\tfrac{\pi}{6})-\sin(\tfrac{\pi}{6})&&\text{use angle addition formula}\\&=\tfrac{1}{2}\cos(h)+\tfrac{\sqrt{3}}{2}\sin(h)-\tfrac{1}{2}&\text{evaluate sine and cosine}\\&=\tfrac{1}{2}(\cos(h)-1)+\tfrac{\sqrt{3}}{2}\sin(h)&\text{rearrange terms}\end{align*}\]

What does this operator tell us? If we view a function \(f\) as describing a measuring device that probes a physical system, a given input of \(x_0\) corresponds to the output of \(f(x_0)\), then the difference \(\Delta_x0 f\) describes how \(f\) changes given certain perturbations of the input values.

Understanding how \(f\) changes under small perturbations of input values helps us to better understand how accurately the quantity that is described by \(f\) can be measured.

Sensitivity to Perturbation

For any positive real number \(\delta\) and any \(x_0\) in the domain of a function \(f\), view the set

\[ \big\{ f(x_0+h)\colon h\in (-\delta,\delta),\, x_0+h\in\mathcal{D}(f)\big\}\]

as the outputs of \(f\) given by perturbing the input \(x_0\) by \(h.\)

If \(f\) is continuous at \(x_0\), then these outputs are as close as we like to \(f(x_0)\) as long as \(\delta\) is small enough. However, even if \(f\) is continuous the sensitivity of \(f\) to perturbation may change dramatically with \(x_0.\)

That is to say that for fixed \(h\), \(|f(x_0+h)-f(x_0)|\) may be very large or very small depending on \(x_0\).

To quantify this, define the absolute finite difference operator \(|\Delta_{x_0}|\) to be the operator that takes \(f\) to the function \(|\Delta_{x_0}|f\), where \[|\Delta_{x0}|f(h)=|f(x_0+h)-f(x_0)|.\]

This function measures the sensitivity of \(f\) to perturbation at \(x_0.\)

Example 9

Take \(f\) to be the function that is given for all \(x\) in \((0,\infty)\) by \[f(x)=\frac{1}{x}.\]

  1. Identify a formula for \(|\Delta_x|f\left(\tfrac{1}{10}\right)\) for each \(x\) in \((0,\infty).\)
  2. For any natural number \(M,\) show that there is an \(x\) so that \(|\Delta_x|f\left(\tfrac{1}{10}\right)\) than \(M\).
  3. Describe qualitatively the sensitivity of \(f\) to perturbation by an amount \(\tfrac{1}{10}.\)

  1. Expand out to get that \[\begin{align*}|\Delta_x|f\left(\tfrac{1}{10}\right)&=\left|f\left(x+\tfrac{1}{10}\right)-f(x)\right|\\&=\left|\frac{1}{x+\frac{1}{10}}-\frac{1}{x}\right|\\&=\frac{1}{10}\cdot\frac{1}{x\left(x+\tfrac{1}{10}\right)}.\end{align*}\]

  2. For any natural number \(M\), if \(x\) is in \(\left(0,\tfrac{1}{10M}\right),\) then \(x+\tfrac{1}{10}\) is in \((0,1)\) and so \[|\Delta_x|f\left(\tfrac{1}{10}\right)=\frac{1}{10}\cdot\frac{1}{x\left(x+\tfrac{1}{10}\right)}>\frac{1}{10}\cdot\frac{1}{x}>\frac{1}{10}\cdot 10 M=M.\]

  3. Peturbing inputs of \(f\) by \(\frac{1}{10}\) leads to unbounded changes in outputs if \(x\) is near enough to \(0\). It means that \(f\) is highly sensitive to perturbation near \(0\).

The function above is not continuous at \(0\). In fact the function diverges to infinity as \(x\) approaches \(0\) from the right, so it is not surprising that \(f\) is highly sensitive to perturbation near \(0\).

So the question we have is that if \(f\) is continuous on some subset \(S\), under what conditions on \(S\) is the sensitivity of \(f\) to perturbation controlled by the continuity of \(f\) on \(S\)? The idea involves a modulus of continuity.

Modulus of Continuity

A modulus of continuity for a continuous function \(f\) on a subset \(S\) of \(\mathbb{R}\) is a function \(\omega_f\) on \([0,\infty)\) with the following properties:

  • \(\lim\limits_{h\to 0^+}\omega_f(h)=0\)
  • For any \(x\) and \(y\) in \(S\), \[|f(x)-f(y)|\leq \omega_f(|x-y|).\]

The second statement basically gives a global bound on the sensitivity of \(f\) to perturbation because it can be rewritten like this \[|\Delta_x|f(h)\leq \omega_f(|h|)\] where \(x\) is in \(S\) and \(h\) is any real number so that \(x+h\) is in \(S.\)

Understand this with the following example.

Example 10

For each of these choices of function \(f\) and interval \(I,\) either identify a modulus of continuity function \(\omega\) for \(f\) on \(I\), or show that \(f\) has no modulus of conintunity function on \(I\):

  1. \(f(x)=x^2\) and \(I=[0,1]\)
  2. \(f(x)=x^2\) and \(I=[0,\infty)\)
  3. \(f(x)=\tfrac{1}{x}\) and \(I=(1,\infty)\)
  4. \(f(x)=\tfrac{1}{x}\) and \(I=(0,1]\)
  5. \(f(x)=\sqrt{x}\) and \(I=[0,\infty)\)

  1. For any \(x\) and \(y\) in \([0,1],\) the sum \(|x+y|\) is bounded above by \(2\). So, we have that \[\begin{align*}|f(x)-f(y)|&=|x^2-y^2|\\&=|(x-y)(x+y)|\\&\leq 2|x-y|.\end{align*}\] So \(\omega(x)=2|x|\) is a modulus of continuity function for \(f\).

  2. In the previous part, \(|x+y|\) was bounded because we were in \([0,1].\) That is no longer the case anymore. To see why, for any positive null sequence \((h_n)\), there is a sequence \((x_n)\) so that \((|f(x_n+h_n)-f(x_n))\) is not a null sequence. For example, take \(x_n=\frac{1}{h_n}\) so that \[|f(x_n+h_n)-f(x_n)|=\left|\left(\frac{1}{h_n}+h_n\right)^2-\frac{1}{h_n^2}\right|=|2+h_n^2|\to 2\] as \(n\) goes to \(\infty\). So \(f\) does not have a modulus of continuity on \([0,\infty)\)

  3. For any \(x\) and \(y\) in \((1,\infty),\) the product \(\tfrac{1}{|xy|}\) is bounded above by \(1\). So, we have that \[\begin{align*}|f(x)-f(y)|&=\left|\frac{1}{x}-\frac{1}{y}\right|\\=\left|\frac{y-x}{xy}\right|\leq |x-y|.\end{align*}\] So \(\omega(x)=|x|\) is a modulus of continuity function for \(f\).

  4. There is no modulus of continuity because \(\tfrac{1}{|xy|}\) is unbounded in \((0,1].\)

  5. For any \(x\) in \([0,\infty)\) and any \(h\) so that \(x+h\in[0,\infty)\), we have that \[|f(x+h)-f(x)|=\left|\sqrt{x+h}-\sqrt{x}\right|=\frac{|h|}{\sqrt{x+h}+\sqrt{x}}.\] If \(x>|h|\), then \[\frac{|h|}{\sqrt{x+h}+\sqrt{x}}<\frac{|h|}{\sqrt{|h|}}=\sqrt{h}.\] If \(0\leq x\leq |h|\), then \(h>0\) or else \(x+h<0\). So we have that \[\frac{|h|}{\sqrt{x+h}+\sqrt{x}}<\frac{|h|}{\sqrt{x+h}}<\frac{|h|}{\sqrt{|h|}}<\sqrt{|h|}.\] So \(\omega(h)=\sqrt{|h}\) is a modulus of continuity function for \(f\).

This is the theorem that guarantees when we have a modulus of continuity.

Uniform Continuity Theorme

For any function \(f\) that is defined on a closed and bounded interval \([a,b]\), if \(f\) is continuous on \([a,b]\), then there is a modulus of continuity function for \(f\) on \([a,b].\)

Composite Errors

It is important to understand the local properties of functions that are sums and products of functions.

Big and Little O Theorems

For any functions \(f\) and \(g\),

  • if \(f\) is \(\displaystyle o(x-a)\) and \(g\) is \(\displaystyle o(x-a)\), then \(f+g\) is \(\displaystyle o(x-a)\) and \(f\cdot g\) is \(\displaystyle o((x-a)^2)\);
  • if \(f\) is \(\displaystyle o(x-a)\) and \(g\) is \(\displaystyle O(x-a)\), then \(f+g\) is \(\displaystyle O(x-a)\) and \(f\cdot g\) is \(\displaystyle o((x-a)^2)\);
  • if \(f\) is \(\displaystyle O(x-a)\) and \(g\) is \(\displaystyle O(x-a)\), then \(f+g\) is \(\displaystyle O(x-a)\) and \(f\cdot g\) is \(\displaystyle O((x-a)^2)\).

In this next example, we prove the first statement of the Big and Little O Theorems.

Example 11

Show that for any real valued functions \(f\) and \(g\), if \(f\) is \(\displaystyle o(x-a)\) and \(g\) is \(\displaystyle o(x-a)\), then \(f+g\) is \(\displaystyle o(x-a)\) and \(f\cdot g\) is \(\displaystyle o((x-a)^2)\)

Since \(f\) and \(g\) are \(\displaystyle o(x-a)\), there are open intervals \(I_1\) and \(I_2\) that contain \(a\) and functions \(\eta_1\) and \(\eta_2\) with \(\displaystyle\lim_{x\to a}\eta_1(x)=0\) and \(\displaystyle\lim_{x\to a}\eta_2(x)=0\) so that \[f(x)=\eta_1(x)(x-a)\] and \[g(x)=\eta_2(x)(x-a)\]. Therefore,

\[f(x)+g(x)=\eta_1(x)(x-a)+\eta_2(x)(x-a)=(\eta_1(x)+\eta_2(x))\cdot (x-a)\] and \[f(x)\cdot g(x)=\eta_1(x)(x-a)\cdot\eta_2(x)(x-a)=(\eta_1(x)\eta_2(x))\cdot (x-a)^2\]

for an open interval that is a subset of \(I_1\cap I_2\).

Since \(\eta_1\) and \(\eta_2\) are functions with \(\displaystyle\lim_{x\to a}\eta_1(x)=0\) and \(\displaystyle\lim_{x\to a}\eta_2(x)=0\), \(\displaystyle\lim_{x\to a}(\eta_1+\eta_2)=0\) and \(\displaystyle\lim_{x\to a}\eta_1\eta_2=0\).

Therefore \(f+g\) is \(\displaystyle o(x-a)\) and \(f\cdot g\) is \(\displaystyle o((x-a)^2)\)

In this next example, we prove the second statement of the Big and Little O Theorems.

Example 12

Show that for any real valued functions \(f\) and \(g\), if \(f\) is \(\displaystyle o(x-a)\) and \(g\) is \(\displaystyle O(x-a)\), then \(f+g\) is \(\displaystyle O(x-a)\) and \(f\cdot g\) is \(\displaystyle o((x-a)^2)\).

Since \(f\) is \(\displaystyle o(x-a)\), there is an open intervals \(I_1\) that contain \(a\) and a function \(\eta_1\) with \(\displaystyle\lim_{x\to a}\eta_1(x)=0\) so that \[f(x)=\eta_1(x)(x-a)\].

Since \(g\) is \(\displaystyle O(x-a)\), there is an open intervals \(I_2\) that contain \(a\) and a bounded function \(\eta_2\) so that \[g(x)=\eta_2(x)(x-a)\].

Therefore,

\[f(x)+g(x)=\eta_1(x)(x-a)+\eta_2(x)(x-a)=(\eta_1(x)+\eta_2(x))\cdot (x-a)\] and \[f(x)\cdot g(x)=\eta_1(x)(x-a)\cdot\eta_2(x)(x-a)=(\eta_1(x)\eta_2(x))\cdot (x-a)^2.\]

Since \(\eta_1\) satisfies \(\displaystyle\lim_{x\to a}\eta_1(x)=0\) and \(\eta_2\) is a bounded function, then \(\eta_1+\eta_2\) is bounded and \(\displaystyle\lim_{x\to a}\eta_1\eta_2=0\) (use the squeeze theorem).

Therefore \(f+g\) is \(\displaystyle O(x-a)\) and \(f\cdot g\) is \(\displaystyle o((x-a)^2)\)

In this next example, we prove the third statement of the Big and Little O Theorems.

Example 13

Show that for any real valued functions \(f\) and \(g\), if \(f\) is \(\displaystyle O(x-a)\) and \(g\) is \(\displaystyle O(x-a)\), then \(f+g\) is \(\displaystyle O(x-a)\) and \(f\cdot g\) is \(\displaystyle O((x-a)^2)\).

Since \(f\) and \(g\) are \(\displaystyle O(x-a)\), there are open intervals \(I_1\) and \(I_2\) that contain \(a\) and bounded functions \(\eta_1\) and \(\eta_2\) so that \[f(x)=\eta_1(x)(x-a)\] and \[g(x)=\eta_2(x)(x-a)\]. Therefore,

\[f(x)+g(x)=\eta_1(x)(x-a)+\eta_2(x)(x-a)=(\eta_1(x)+\eta_2(x))\cdot (x-a)\] and \[f(x)\cdot g(x)=\eta_1(x)(x-a)\cdot\eta_2(x)(x-a)=(\eta_1(x)\eta_2(x))\cdot (x-a)^2\]

for an open interval that is a subset of \(I_1\cap I_2\).

Since \(\eta_1\) and \(\eta_2\) are bounded functions, \(\eta_1+\eta_2\) and \(\eta_1\eta_2\) are also bounded functions.

Therefore \(f+g\) is \(\displaystyle O(x-a)\) and \(f\cdot g\) is \(\displaystyle O((x-a)^2)\)

In this example, pay attention to how the sum and products of two functions local behavior changes.

Example 14

Take \(f\) and \(g\) to be given by \[f(x)=(x+3)(x-1) \quad \text{ and} \quad g(x)=(x+2)(x-1).\] Demonstrate the following:

  1. \(f\) and \(g\) are \(\displaystyle O(x-1)\);
  2. \(f+g\) is \(\displaystyle O(x-1)\) but not \(\displaystyle O(x+3)\) and not \(\displaystyle o(x-1)\);
  3. \(f\cdot g\) is \(\displaystyle o(x-1)\) and \(\displaystyle O(x+3)\), but not \(\displaystyle o(x+2)\).

  1. Write \(f\) like this: \[f(x)=(x+3)(x-1)=\eta_f(x)(x-1),\] where \(\eta_f\) is the function given by \(\eta_f(x)=(x+3).\) This is bounded on \((0,1)\cup(1,2).\) Now rewrite \(g\) like this: \[g(x)=(x+2)(x-1)=\eta_g(x)(x-1),\] where \(\eta_g\) is given by \(\eta_g(x)=(x+2).\) This is also bounded on \((0,1)\cup(1,2).\) Thus \(f\) and \(g\) are both \(\displaystyle O(x-1)\).

  2. By the above theorem, since \(f\) and \(g\) are \(\displaystyle O(x-1)\), then \(f+g\) is \(\displaystyle O(x-1)\). However, \(f+g\) is not \(\displaystyle O(x+3)\) since \(g\) doesn’t have an \((x+3)\) factor. Also \(f+g\) is not \(\displaystyle o(x-1)\) since \(\eta\) does not have a limit equal to zero at \(x=1\).

  3. Write the product like this: \[f(x)\cdot g(x)=(x-1)(x+3)(x+2)\cdot(x-1)=\eta_{fg}(x)\cdot(x-1),\] where \(\eta_{fg}\) is the function given by \(\eta_{fg}(x)=(x-1)(x+3)(x+2)\). Since \[\displaystyle\lim_{x\to 1}\eta_{fg}(x)=0,\] we have that \(f\cdot g\) is \(\displaystyle o(x-1)\). It is \(\displaystyle O(x+3)\) since the product can be written like this: \[f(x)\cdot g(x)=[(x-1)^2(x+2)]\cdot(x+3)=\eta_{fgO}(x)\cdot (x+3)\] and the function \(\eta_{fgO}(x)\) is bounded in \((2,4)\). However, it is not \(o(x+2)\) because there is no function \(\eta\) for which the product can be written as \(f(x)\cdot g(x)=\eta(x) (x+2)\) and the function \(\eta(x)\) has a limit of zero as \(x\) approaches \(-2\).

In this next theorem, we discuss how composition can affect Big and Little O behavior.

Theorem on Big and Little O Compostion

For any real number \(x_0\) and any real valued functions \(f\) and \(g\), if there is an open interval \(I\) that contains \(x_0\) and

  • \(I\) is contained in \(\mathcal D(g)\),
  • \(g - g(x_0)\) is \(\displaystyle O(x-x_0)\),
  • \(g(x_0)\) is a limit point of \(g(I)\),
  • \(g(I)\setminus\{g(x_0)\}\) is a subset of \(\mathcal D(f)\), and \(f\) is \(\displaystyle o(x-g(x_0))\),

then the composite function \(f\circ g\) is \(\displaystyle o(x-x_0)\).

Less precisely, but more memorably, we may recall this fact by saying that if \(g\) is \(\displaystyle O(x-x_0)\) and \(f\) is \(\displaystyle o(x-g(x_0))\), then \(f\circ g\) is \(\displaystyle o(x-x_0)\).

Example 15

Take \(f\) and \(g\) to be the functions that are given by \[g(x) = x^2 +1\quad \text{ and}\quad f(x) = (x-5)^2.\] Perform the appropriate calculations in order to justify the following statements:

  1. \(g-g(2)\) is \(\displaystyle O(x-2)\);
  2. \(f\) is \(\displaystyle o(x-g(2))\);
  3. \(f\circ g\) is \(\displaystyle o(x-2)\).

  1. Write out the difference like this: \[g(x)-g(2)=x^2+1-g(2)=x^2+1-5=x^2-4=(x+2)(x-2).\] Hence \[g(x)-g(2)=(x+2)(x-2)=\eta(x)\cdot(x-2),\] where \(\eta\) is the function given by \(\eta(x)=(x+2).\) This is a bounded function in \((1,2)\cup(2,3)\). Thus \(g-g(2)\) is \(\displaystyle O(x-2)\);

  2. Notice that \(x-g(2)=x-5\) and so \[f(x)=(x-5)^2=(x-5)\cdot(x-5)=\eta(x)\cdot(x-5)\] where \(\eta(x)=(x-5)\) is a function with \(\lim\limits_{x\to 5}\eta(x)=0\). Thus \(f\) is \(\displaystyle o(x-g(2))=o(x-5)\).

  3. Since part a and part b are true, then by the above theorem we can conclude that \(f\circ g\) is \(\displaystyle o(x-2)\). Alternatively, do a direction calculation: \[\begin{align*}(f\circ g)(x)&=f(g(x))\\&=(x^2+1-5)^2\\&=(x^2-4)^2\\&=(x+2)^2(x-2)^2\\&=(x-2)(x+2)^2\cdot(x-2)\end{align*}\] and split it into \((x-2)\) and the rest is a function with a limit of zero as \(x\) approaches \(2\).

In this next example, we show the little \(o\) behavior of a composite function involving sine and cosine.

Example 16

Take \(f\) and \(g\) to be the functions that are given by \[g(x) = 3+\sin(x) \quad \text{ and}\quad f(x) = 1-\cos(x-3).\] Perform the appropriate calculations in order to justify the following statements:

  1. \(g-g(0)\) is \(\displaystyle O(x)\);
  2. \(f\) is \(\displaystyle o(x-g(0))\);
  3. \(f\circ g\) is \(\displaystyle o(x)\).

  1. Notice that \[g(x)-g(0)=3+\sin(x)-g(0)=3+\sin(x)-3-\sin(0)=\sin(x)=\tfrac{\sin(x)}{x}\cdot x.\] Hence \(g(x)-g(0)=\tfrac{\sin(x)}{x}\cdot x=\eta(x)\cdot x\), where \(\eta(x)=\tfrac{\sin(x)}{x}\) is a bounded function in \((-1,0)\cup(0,1)\) since \(\eta\) has a convergent limit at \(x=0\). Thus \(g-g(0)\) is \(\displaystyle O(x)\);

  2. Notice that \(x-g(0)=x-3\) and \[f(x)=1-\cos(x-3)=\tfrac{1-\cos(x-3)}{x-3}\cdot(x-3)=\eta(x)\cdot(x-3)\] where \(\eta(x)=\tfrac{1-\cos(x-3)}{x-3}.\) Because \(\lim\limits_{\star\to 0}\tfrac{1-\cos(\star)}{\star}=0\), it is true that \(\lim\limits_{x\to 3}\eta(x)=0.\) Thus \(f\) is \(\displaystyle o(x-g(0))=o(x-3)\).

  3. Since part a and part b are true, then by the above theorem we can conclude that \(f\circ g\) is \(\displaystyle o(x)\). Alternatively, do a direct calculation: \[\begin{align*}(f\circ g)(x)&=f(g(x))\\&=1-\cos(3+\sin(x)-3)\\&=1-\cos(\sin(x))\\&=\frac{1-\cos(\sin(x))}{\sin(x)}\cdot\frac{\sin(x)}{x}\cdot x\\&=\left[\frac{1-\cos(\sin(x))}{\sin(x)}\cdot\frac{\sin(x)}{x}\right]\cdot x\end{align*}\] and split it into \(x\) and the rest is a function with a limit of zero as \(x\) approaches \(0\).

Finally, we look at an example in which we show certain functions exhibit certain \(O\) and \(o\) behavior by perfoming the right calculations.

For example, to show a function \(f\) is \(o(x-a)\) is equivalent to showing that \(\lim\limits_{x\to a}\tfrac{f(x)}{x-a}=0.\)

Example 17

Take \(f\) and \(g\) to be the functions that are given by

\[g(x)=x^2+1 \quad \text{and} \quad f(x)=3\sin(x)(x-5)^2.\]

Perform the appropraite calculations in order to justify the following statements:

  1. \(g-g(2)\) is \(O(x-2)\)

  2. \(f\) is \(o(x-g(2))\)

  3. \(f\circ g\) is \(o(x-2)\)

  1. Rewrite \(g-g(2)\) like this: \[ g(x)-g(2)=x^2+1-5=x^2-4=(x+2)(x-2). \] The term \((x+2)\) is bounded in an open interval that contains \(2\). So \(g-g(2)\) is \(O(x-2).\)

  2. Show that \(\lim\limits_{x\to g(2)}\tfrac{f(x)}{x-g(2)}=0.\) Here is a calculation showing this: \[ \begin{align*} \lim_{x\to g(2)}\frac{f(x)}{x-g(2)}&=\lim_{x\to 5}\frac{3\sin(x)(x-5)^2}{x-5}\\ &=\lim_{x\to 5}3\sin(x)(x-5)\\ &=0. \end{align*} \] Thus, \(f\) is \(o(x-g(2)).\)

  3. Show that \(\lim\limits_{x\to 2}\tfrac{(f\circ g)(x)}{x-2}=0.\) Here is a calculation showing this: \[ \begin{align*} \lim\limits_{x\to 2}\tfrac{(f\circ g)(x)}{x-2}&=\lim_{x\to 2}\frac{3\sin(x^2+1)(x^2+1-5)^2}{x-5}\\ &=\lim_{x\to 2}\frac{3\sin(x^2+1)(x^2-4)^2}{x-2}\\ &=\lim_{x\to 2}\frac{3\sin(x^2+1)(x-2)^2(x+2)^2}{x-2}\\ &=\lim_{x\to 2}3\sin(x^2+1)(x-2)(x+2)^2\\ &=0. \end{align*} \] Thus, \(f\circ g\) is \(o(x-2).\)

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