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Applied Mathematics

Peter Staab

Section 6.1 Properties of Functions

In Section 3.3, we reviewed inner product spaces and saw orthonormal sets of vectors (both in \(\mathbb{R}^3\) as well as polynomials). In this section, we will examine another set of functions, sines and cosines that are orthogonal. First, let’s see a short review of periodic functions.
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Subsection 6.1.1 Periodic Functions

Definition 6.1.1.

A function is periodic with period \(p\) if
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\begin{equation*} f(t) = f(t+p) \end{equation*}
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for all \(t\text{.}\) The smallest value of \(p > 0\) for which this true is called the period of the function.
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Example 6.1.2.

Show that \(f(t) = \sin t\) is periodic with period \(2\pi\text{.}\)
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Solution.
\begin{equation*} f(t+2\pi) = \sin (t + 2 \pi) = \sin t \cos 2\pi + \sin 2\pi \cos t = \sin t \end{equation*}
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where the sum of angles sine formula is used.
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Example 6.1.3.

What is the period of the function \(f(t) = \cos kt\text{?}\)
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Solution.
We know that the period of \(\cos t\) is also the same as \(\sin t\) or \(2\pi\text{.}\) If we let \(u=kt\text{,}\) then \(f(u)=\cos u\) has period \(p=2\pi\) since it is the smallest value of \(p\) such that \(f(u)=f(u+p)\) for all \(u\text{.}\) The function \(f(t)\) would then had period \(p=2\pi/k\text{,}\) since \(t=u/k\text{.}\)
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The periodic functions that we will mostly be using in this text are the sine and cosine function. We review here a few convenient identities with these functions and the complex exponential. From Euler’s formula,
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\begin{equation*} e^{i \theta} = \cos \theta + i \sin \theta \end{equation*}
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we can then write sine and cosine in terms of \(e^{i \theta}\)
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\begin{equation*} \sin \theta = \frac{e^{i \theta}- e^{-i \theta}}{2i} \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2} \end{equation*}
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Euler’s formula also leads to the following:
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Remark 6.1.4. The Most Interesting Equation in Mathematics.

\begin{equation} e^{i\pi} + 1 = 0\tag{6.1.1} \end{equation}
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and this is often called the most interesting equation in mathematics because it arguably contains the 5 most important mathematical constants: 0, 1, \(e\text{,}\) \(i\text{,}\) \(\pi\text{.}\)
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Proof.

If \(k=0\text{,}\) then the integral is of the constant function 1 over a interval of length \(2\pi\text{,}\) so the lemma holds. If \(k \neq 0\text{,}\)
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\begin{align*} \int_{x_0}^{x_0 + 2\pi} e^{i kx} \, dx \amp = \frac{e^{i k x}}{i k} \biggr \vert_{x_0} ^{x_0 + 2\pi} \\ \amp = \frac{1}{ik} \bigl( e^{ik(x_0+2\pi)} - e^{ikx_0} \bigr) \\ \amp = \frac{1}{ik} e^{ikx_0} (e^{2\pi i k} -1 ) \\ \amp = \frac{1}{ik} e^{ikx_0} ( (e^{i\pi})^{2k} -1 ) = 0 \end{align*}
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because \(e^{i\pi}=-1\) from (6.1.1), but this is raised to an even power so \((e^{i\pi})^{2k}=1\)
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Subsection 6.1.2 One-Side Limits and Derivatives; Piecewise Continuous Functions

As we will see, the notion of a piecewise continuous function is a function that is continuous on subintervals. However, there are some technical details that we need before a formal definition.
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Definition 6.1.6.

A function \(f\) has a left-hand limit at \(x=x_0\) if
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\begin{equation*} f(x_0^-) = \lim_{x \rightarrow x_0^-} f(x) \end{equation*}
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exists. In addition, a function \(f\) has a right-hand limit at \(x=x_0\) if
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\begin{equation*} f(x_0^+) = \lim_{x \rightarrow x_0^+} f(x) \end{equation*}
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exists.
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If one is talking about either a left- or right-handed limit, these are typically called one-sided limits. Also, an example of these will be shown below.
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Definition 6.1.7.

A function \(f\) has a left-hand derivative at \(x=x_0\) if
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\begin{equation*} f'(x_0^-) = \lim_{h \rightarrow 0^-} \frac{f(x_0+h)-f(x)}{h} \end{equation*}
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exists. Similarly, a function \(f\) has a right-hand derivative at \(x=x_0\) if
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\begin{equation*} f'(x_0^+) = \lim_{h \rightarrow 0^+} \frac{f(x_0+h)-f(x)}{h} \end{equation*}
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exists.
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If one is talking about either a left- or right-handed derivative, these are typically called one-sided derivatives. An example of these will be shown below.
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Definition 6.1.8.

A function \(f\) is piecewise continuous on an interval \([a,b]\) if \(f\) is continuous on all \(x \in [a,b]\) except for a finite number of points \(x_i\text{.}\) In addition for all \(x_i\text{,}\) \(f(x_i^+)\) and \(f(x_i^-)\) exist.
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Example 6.1.9.

The following function is piecewise continuous on \([-1,1]\)
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\begin{equation*} f(x) = \begin{cases} 1, \amp -1 \leq x \lt 0, \\ x, \amp 0 \leq x \lt 1/2, \\ 1-x^2, \amp 1/2 \leq x \leq 1. \end{cases} \end{equation*}
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Also, the graph of piecewise functions are helpful. These are found by finding the graphs of \(f\) on each given interval. The graph of \(f\) is shown below.
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Figure 6.1.10. Graph of a piecewise continuous function.
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In addition, we need to show that all of the one-sided limits exist. For each of the functions above, we differentiate to get
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\begin{align*} \lim_{x \rightarrow 0^-} f(x) \amp = 1 \amp \lim_{x \rightarrow 0^+} f(x) \amp = 0 \\ \lim_{x \rightarrow \frac{1}{2}^-} f(x) \amp = \frac{1}{2} \amp \lim_{x \rightarrow \frac{1}{2}^+} f(x) \amp = \frac{3}{4} \end{align*}
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And since the function is continuous at all points except at 0 and 1/2, but the one-sided limits are finite here, then the function \(f\) is piecewise continuous.
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Example 6.1.11.

Find both the left- and right-handed derivatives of the function defined in Example 6.1.9 at \(x=0\) and \(x=1/2\text{.}\)
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Solution.
First, consider the derivative of the function
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\begin{equation*} f'(x) = \begin{cases} 0, \amp -1 \lt x \lt 0, \\ 1, \amp 0 \lt x \lt 1/2, \\ -2x, \amp 1/2 \lt x \lt 1. \end{cases} \end{equation*}
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where the equality parts of the derivative have been removed (and explained later).
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Since
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\begin{align*} \lim_{x \rightarrow 0^{-}} f'(x) \amp = 0 \amp \lim_{x \rightarrow 0^+} \amp = 1 \\ \lim_{x \rightarrow \frac{1}{2}^{-}} f'(x) \amp = 1 \amp \lim_{x \rightarrow \frac{1}{2}^+} \amp = -1 \end{align*}
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then the left-handed derivative at 0 is 0, the right-handed derivative of \(f\) at 1, the left-handed derivative at 1/2 is 1 and the right-handed derivative of \(f\) at 1/2 is \(-1.\)
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Example 6.1.12.

Show that \(f(x)=1/x\) is not a piecewise continuous function on \([-1,1]\text{.}\)
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Solution.
The function on \([-1,1]\) is not continuous at \(x=0\text{.}\) Also
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\begin{align*} \lim_{x \rightarrow 0^-} f(x) \amp = -\infty \amp \lim_{x \rightarrow 0^+} f(x) \amp = \infty \end{align*}
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and since the one-sided limits are not finite, then \(f\) is not piecewise continuous on \([-1,1]\text{.}\)
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Subsection 6.1.3 Odd and Even Functions

Definition 6.1.13.

A function \(f(x)\) is an odd function if \(f(-x) = -f(x)\) for all \(x\) in its domain.
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Note: recall that an odd function is symmetric about the origin, meaning that if the graph of \(f\) is rotated a half circle about the origin, that one gets the graph back.
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Definition 6.1.14.

A function \(f(x)\) is an even function if \(f(-x)=f(x)\) for all \(x\) in its domain.
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Recall that an even function is symmetric about the \(y\)-axis. This means that if the graph is reflected over the \(y\)-axis that one gets the same graph upon the reflection.
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Example 6.1.15.

Here’s a list of a few functions that are odd or even (without showing details):
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  • The following functions are odd: \(x, x^3, x^5, \sin x, \tan x\)
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  • The following functions are even: \(1,x^2,x^4, \cos x\text{.}\)
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The following theorem is helpful for finding whether or not products of functions are odd or even.
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Proof.

Let \(f\) be an even function, then \(f(-x) = f(x)\) for all \(x\text{.}\)
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\begin{equation*} \frac{d}{dx} f(x) = \frac{d}{dx} f(-x) = -\frac{df(-x)}{dx} \end{equation*}
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by the chain rule. And the proof that the derivative of an odd function is similar.
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And as a corollary, antiderivatives work in the same way.
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Proof.

First, examine the first statement. Let \(F(x)\) be an antiderivative of \(f(x)\text{,}\) an odd function. The function \(F(x)\) can be written as \(F(x)=G(x)+C\) where \(G(x)\) is an even function.
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\begin{equation*} \int_{-a}^a f(x) \,dx = G(x) + C \biggl\vert_{-a}^a = G(a)+C - (G(-a)+C) = G(a)-G(-a)= 0 \end{equation*}
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since \(G(x)\) is even. The proof of the second statement is similar.
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Subsection 6.1.4 Tabular Integration

A very handy formula for many integrations in this section is called tabular integration, which is just a recursive version of integration by parts that works well for integrals of a certain type. Before we show this, recall that the integration by parts formula is
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\begin{equation*} \int u \, dv = uv - \int v \, du \end{equation*}
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and integration by parts is helpful for rewriting one integral (on the left) in terms of a second integral (on the right) and generally it is used to create a simpler integral. The next example shows a standard integration done with integration by parts.
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Example 6.1.20.

Find
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\begin{equation*} \int x e^x \, dx. \end{equation*}
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Solution.
In this case, we’ll let \(u= x\) and \(dv = e^x \, dx\text{,}\) finding the differential \(u\) results in \(du=dx\) and finding an antiderivative of \(dv\) results in \(v= e^x\text{,}\) so using integration by parts to get
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\begin{equation*} \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C \end{equation*}
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This example shows that in order to integrate with the by parts formula, one must replace one integral with another. In more difficult examples, this may need to be done multiple times until the resulting integral is able to be done without by parts. This is the case when tabular integration is useful.
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Remark 6.1.21.

The technique of tabular integration applied to
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\begin{equation*} \int f(x) g(x) \, dx \end{equation*}
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where there exists an \(n\) such that \(f^{(n)}(x)=0\text{,}\) that is eventually the derivative of \(f(x)\) is 0. Creates a table of three columns with
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  1. The function \(f(x)\) and its derivatives until you reach zero.
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  2. The signs \(+\) and \(-\text{,}\) starting with \(+\) and alternating signs.
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  3. The function \(g(x)\) and its antiderivatives.
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For columns 2 and 3, continue until you reach the same row as the 0 in the first column. To find the antiderivative, draw arrows from each function in the first column, to a function in the third column one row below. The result is the sum of the product of each pair of functions connected by the arrows with the sign of that above the given arrow.
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This is best seen with a couple of examples.
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Example 6.1.22.

Find
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\begin{equation*} \int x^3 e^x \, dx \end{equation*}
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using tabular integration.
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Solution.
First, we will build the table:
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Figure 6.1.23. Tabular integration of \(\int x^3 e^x \, dx\)
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Then read off the result which is the sum of the product of terms connected by the arrows with the sign above each arrow.
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\begin{equation*} \int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + 6x e^x - 6 e^x + C \end{equation*}
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and don’t forget the \(C\) for an indefinite integral.
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And the following is an example that is similar as we will see below:
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Example 6.1.24.

Find
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\begin{equation*} \int_0^{\pi} x^2 \cos 2x \, dx \end{equation*}
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using tabular integration.
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Solution.
First, we will build the table:
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Figure 6.1.25. Tabular integration of \(x^2 \cos 2x\text{.}\)
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and then read off the result which is the product of terms connected by the arrows with the sign above each arrow.
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\begin{align*} \int_0^{\pi} x^2 \cos 2x \, dx \amp = \biggl( \frac{x^2}{2} \sin 2x + \frac{x}{2} \cos 2x - \frac{1}{4} \sin 2x \biggr) \biggr \vert_{0}^{\pi} \\ \amp = \biggl(\frac{\pi^2}{2} \sin 2\pi + \frac{\pi}{2} \cos 2\pi - \frac{1}{4} \sin 2\pi \biggr) \\ \amp \qquad - \biggl(0 + 0 - \frac{1}{4} \sin 0 \biggr) \\ \amp = \frac{\pi}{2} \end{align*}
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Subsection 6.1.5 Inner Products of Functions

As we saw in Section 3.3, the inner product of two functions can be defined using the integral. In this section, we consider functions that are piecewise continuous on \([a,b]\) and
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\begin{equation*} \langle f, g \rangle = \int_a^b f(x) g(x) \, dx \end{equation*}
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In the standard way, the norm of \(f\) is defined as
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\begin{equation*} || f || = \langle f, f \rangle^{1/2} = \biggl( \int_a^b f^2(x) \, dx \biggr)^{1/2} \end{equation*}
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Definition 6.1.26.

An infinite set of continuous function \(\{f_1,f_2, \ldots\}\) is said to be orthogonal on \([a,b]\) if \(\langle f_n, f_m \rangle =0\) for all \(n,m\text{,}\) \(n \neq m\text{.}\) If in addition, \(||f_n||=1\) for all \(n\text{,}\) the set is said to be orthonormal.
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In the next couple of examples, we examine a couple of the most important orthogonal and orthonormal sets of functions.
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Example 6.1.27.

Show that \(f_n(x) = \sin nx\) for \(n=1,2,3,\ldots\) form an orthogonal set on \([-\pi,\pi]\text{.}\)
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Solution.
In this case, we will expand on Example 3.3.9 and use the handy alternative definition of the sine function. Find the inner product if \(n \neq m\) ,
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\begin{align*} \langle f_n, f_m \rangle \amp = \int_{-\pi}^{\pi} \sin nx \sin mx \, dx \\ \amp = \int_{-\pi}^{\pi} \frac{e^{i n x} - e^{-i n x}}{2i}\frac{e^{i mx} - e^{-i m x}}{2i} \, dx \\ \amp = -\frac{1}{4} \int_{-\pi}^{\pi} \bigl( e^{i(m+n)x} - e^{i(m-n) x}- e^{i(n-m) x} + e^{-i(m+n) x} \bigr) \, dx\\ \amp = 0 \end{align*}
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where Lemma 6.1.5 is used on the last step above.
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Example 6.1.28.

Show that the set of functions
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\begin{equation*} \biggl\{ \frac{1}{\sqrt{2\pi}}, \frac{1}{\sqrt{\pi}} \cos x , \frac{1}{\sqrt{\pi}} \sin x, \frac{1}{\sqrt{\pi}} \cos 2x, \frac{1}{\sqrt{\pi}} \sin 2x, \ldots \biggr\} \end{equation*}
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for an orthonormal set on \([-\pi,\pi]\text{.}\)
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Solution.
We showed above that \(\langle \sin mx, \sin nx \rangle=0\) for all \(m,n\) when \(m \neq n\text{.}\) Therefore \(\langle (\sin mx)/\sqrt{\pi}, (\sin nx)/\sqrt{\pi} \rangle\) also equals 0 for \(m \neq n\text{.}\) For simpler notation let
\begin{align*} s_n(x) \amp = \frac{1}{\sqrt{\pi}}\sin n x, \qquad \text{for $n \gt 0$} \\ c_n(x) \amp = \begin{cases} \frac{1}{\sqrt{2\pi}} \amp n = 0, \\ \frac{1}{\sqrt{\pi}} \cos nx \amp n \gt 0, \end{cases} \end{align*}
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First, we will show that \(\langle s_m (x), c_n(x) \rangle=0\) for all \(n, m \gt 0\) .
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\begin{align*} \langle s_m (x), c_n (x) \rangle \amp = \biggl\langle \frac{1}{\sqrt{\pi}}\sin mx, \frac{1}{\sqrt{\pi}}\cos nx \biggr\rangle \\ \amp = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin mx \cos nx \,dx \\ \amp = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{e^{imx} - e^{-imx}}{2i} \frac{e^{inx}+e^{-inx}}{2} \, dx \\ \amp = \frac{1}{4i\pi} \int_{-\pi}^{\pi} \bigl( e^{i(m+n)x}-e^{i(n-m)x}+e^{i(m-n)x}-e^{-i(m+n)x} \bigr) \, dx \\ \amp = 0 \end{align*}
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since each integral is zero from Lemma 6.1.5. Next, we will show that \(\langle c_m(x), c_n(x) \rangle = 0\) for all \(n,m >0\) such that \(n \neq m\text{.}\)
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\begin{align*} \langle c_m (x), c_n (x) \rangle \amp =\biggl\langle \frac{1}{\sqrt{\pi}}\cos mx, \frac{1}{\sqrt{\pi}}\cos nx \biggr\rangle\\ \amp = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos mx \cos nx \,dx\\ \amp = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{e^{imx} + e^{-imx}}{2} \frac{e^{inx}+e^{-inx}}{2} \, dx\\ \amp = \frac{1}{4\pi} \int_{-\pi}^{\pi} \bigl( e^{i(m+n)x} +e^{i(n-m)x}+e^{i(m-n)x}+e^{-i(m+n)x} \bigr) \, dx \\ \amp= 0 \end{align*}
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because again each integral is zero from Lemma Lemma 6.1.5. Also,
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\begin{align*} \langle c_m, c_0 \rangle \amp = \int_{-\pi}^{\pi} \frac{1}{\sqrt{2\pi}} \cos mx \, dx\\ \amp = -\frac{1}{m\sqrt{2\pi}} \sin mx \biggr|_{-\pi}^{\pi} = 0 \end{align*}
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Lastly, we need to show that the norm of each of the functions is 1.
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\begin{align*} \langle c_0, c_0 \rangle \amp = \int_{-\pi}^{\pi} \biggl(\frac{1}{\sqrt{2\pi}} \biggr)^2 \, dx = \int_{-\pi}^{\pi} \frac{1}{2\pi} \, dx = 1 ,\\ \langle c_n, c_n \rangle \amp =\frac{1}{\pi} \int_{-\pi}^{\pi} \cos^2 nx \, dx \\ \amp = \frac{1}{\pi} \int_{-\pi}^{\pi} \biggl( \frac{e^{inx} + e^{-inx}}{2} \biggr)^2 \, dx \\ \amp = \frac{1}{4\pi} \int_{-\pi}^{\pi} \bigl( e^{2inx} + 2 + e^{-2inx} \bigr) \, dx \end{align*}
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and the integrals of the first and third terms are zero from Lemma Lemma 6.1.5.
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\begin{align*} \amp = \frac{1}{4\pi} \int_{-\pi}^{\pi} 2 \, dx = 1 \\ \langle s_n, s_n \rangle \amp =\frac{1}{\pi} \int_{-\pi}^{\pi} \sin^2 nx \, dx \\ \amp = \frac{1}{\pi} \int_{-\pi}^{\pi} \biggl( \frac{e^{inx} - e^{-inx}}{2i} \biggr)^2 \, dx \\ \amp = \frac{1}{(2i)^2\pi} \int_{-\pi}^{\pi} \bigl( e^{2inx} - 2 + e^{-2inx} \bigr) \, dx \end{align*}
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and the integrals of the first and third terms are zero from Lemma Lemma 6.1.5.
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\begin{align*} \amp = \frac{1}{-4\pi} \int_{-\pi}^{\pi} -2 \, dx = 1 \end{align*}
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This shows that the set of functions given above is an orthonormal set of functions.
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