Orthogonal and Orthonormal Systems of Functions Examples 1

# Orthogonal and Orthonormal Systems of Functions Examples 1

Recall from the Orthogonal and Orthonormal Systems of Functions page that if $I$ is any interval in $\mathbb{R}$, then a collection of functions $\{ \varphi_0, \varphi_1, \varphi_2, ... \}$ is said to be an orthogonal system of functions on $I$ if:

(1)
\begin{align} \quad (\varphi_i, \varphi_j) = \left\{\begin{matrix} 0 & \mathrm{if} \: i \neq j \end{matrix}\right. \end{align}

Furthermore, said that $\{ \varphi_0, \varphi_1, \varphi_2, ... \}$ is an orthonormal system of functions on $I$ if:

(2)
\begin{align} \quad (\varphi_i, \varphi_j) = \left\{\begin{matrix} 0 & \mathrm{if} \: i \neq j\\ 1 & \mathrm{if} \: i = j \end{matrix}\right. \end{align}

We then described some common examples of orthonormal systems of functions. For any interval $I$ of length $2 \pi$ we defined the trigonometric system of functions on $I$ as the collection:

(3)
\begin{align} \quad \left \{ \frac{1}{\sqrt{2\pi}} , \frac{\cos x}{\sqrt{\pi}}, \frac{\cos x}{\sqrt{\pi}}, \frac{\cos 2x}{\sqrt{\pi}}, \frac{\sin 2x}{\sqrt{\pi}}, ... \right \} \end{align}

And we defined another trigonometric system for all $n \in \{ 0, 1, 2, ... \}$ by:

(4)
\begin{align} \quad \left \{ \frac{e^{inx}}{\sqrt{2\pi}} \right \} = \left \{ \frac{1}{\sqrt{2\pi}}, \frac{\cos x + i \sin x}{\sqrt{2\pi}}, \frac{\cos 2x + i \sin 2x}{\sqrt{2\pi}}, ... \right \} \end{align}

We will now look at some example problems regarding orthogonal and orthonormal systems of functions.

## Example 1

Consider the interval $I = [0, 2\pi]$. Prove that indeed the system of functions $\left \{ \frac{e^{inx}}{\sqrt{2\pi}} \right \}$ ($n \in \{0, 1, 2, ... \}$) is orthonormal on $I$.

Let $n \in \{ 0, 1, 2, ... \}$. Then:

(5)
\begin{align} \quad \left ( \frac{e^{inx}}{\sqrt{2\pi}}, \frac{e^{inx}}{\sqrt{2\pi}} \right ) &= \int_0^{2\pi} \frac{e^{inx}}{\sqrt{2\pi}} \cdot \overline{\frac{e^{inx}}{\sqrt{2\pi}}} \: dx \\ &= \int_0^{2\pi} \frac{\cos nx + i \sin nx}{\sqrt{2\pi}} \cdot \frac{\cos nx - i \sin nx}{\sqrt{2\pi}} \: dx \\ &= \int_0^{2\pi} \frac{\cos^2 nx - i \sin nx \cos nx + i \sin nx \cos nx - i^2 \sin^2 nx}{2\pi} \: dx \\ &= \int_0^{2\pi} \frac{\cos^2 nx + \sin^2 nx}{2\pi} \: dx \\ &= \int_0^{2\pi} \frac{1}{2\pi} \: dx \\ &= \frac{1}{2\pi} \int_0^{2\pi} 1 \: dx \\ &= \frac{2\pi}{2\pi} \\ &= 1 \end{align}

Now let $m, n \in \{0, 1, 2, ... \}$ with $m \neq n$ and assume without loss of generality that $m > n$. Then:

(6)
\begin{align} \quad \left ( \frac{e^{inx}}{\sqrt{2\pi}}, \frac{e^{inx}}{\sqrt{2\pi}} \right ) &= \int_0^{2\pi} \frac{e^{imx}}{\sqrt{2\pi}} \cdot \overline{\frac{e^{inx}}{\sqrt{2\pi}}} \: dx \\ &= \int_0^{2\pi} \frac{\cos mx + i \sin mx}{\sqrt{2\pi}} \cdot \frac{\cos nx - i \sin nx}{\sqrt{2\pi}} \: dx \\ &= \int_0^{2\pi} \frac{\cos mx \cos nx - i \sin nx \cos mx + i \sin mx \cos nx - i^2 \sin mx \sin nx}{2\pi} \: dx \\ &= \int_0^{2\pi} \frac{\cos mx \cos nx + \sin mx \sin nx}{2\pi} \: dx + i \int_0^{2\pi} \frac{\sin mx \cos nx - \sin nx \cos mx}{2\pi} \: dx \end{align}

For any $m \neq n$ the integrals above equal $0$ and so for $m \neq n$, $\displaystyle{\left ( \frac{e^{inx}}{\sqrt{2\pi}}, \frac{e^{inx}}{\sqrt{2\pi}} \right ) = 0}$.