Orthogonality relations

3 Orthogonality relations

In general two functions $f (x), g (x)$ are said to be orthogonal to each other over an interval $a \leq x \leq b$ if

$\int_{a}^{b} f (x) g (x) d x = 0$

It follows from the previous Task that $sin x$ and $cos x$ are orthogonal to each other over the interval $0 \leq x \leq 2 π$ . This is also true over any interval $α \leq x \leq α + 2 π$ (e.g. $π ∕ 2 \leq x \leq 5 π$ , or $- π \leq x \leq π$ ).

More generally there is a whole set of orthogonality relations involving these trigonometric functions on intervals of length $2 π$ (i.e. over one period of both $sin x$ and $cos x$ ). These relations are useful in connection with a widely used technique in engineering, known as Fourier analysis where we represent periodic functions in terms of an infinite series of sines and cosines called a Fourier series. (This subject is covered in HELM booklet 23.)

We shall demonstrate the orthogonality property

$I_{m n} = \int_{0}^{2 π} sin m x sin n x d x = 0$

where $m$ and $n$ are integers such that $m \neq n$ .

The secret is to use a trigonometric identity to convert the integrand into a form that can be readily integrated.

You may recall the identity

$sin A sin B \equiv \frac{1}{2} (cos (A - B) - cos (A + B))$

It follows, putting $A = m x$ and $B = n x$ that provided $m \neq n$

\begin{array}{rcl} I_{m n} & = & \frac{1}{2} \int_{0}^{2 π} [cos (m - n) x - cos (m + n) x] d x \\ = & \frac{1}{2} {[\frac{sin (m - n) x}{(m - n)} - \frac{sin (m + n) x}{(m + n)}]}_{0}^{2 π} \\ = & 0 \end{array}

because $(m - n)$ and $(m + n)$ will be integers and $sin (integer \times 2 π) = 0$ . Of course $sin 0 = 0$ .

Why does the case $m = n$ have to be excluded from the analysis? (left to the reader to figure out!)

The corresponding orthogonality relation for cosines

$J_{m n} = \int_{0}^{2 π} cos m x cos n x d x = 0$

follows by use of a similar identity to that just used. Here again $m$ and $n$ are integers such that $m \neq n$ .

Example 23

Use the identity $sin A cos B \equiv \frac{1}{2} (sin (A + B) + sin (A - B))$ to show that

$K_{m n} = \int_{0}^{2 π} sin m x cos n x d x = 0$ $m$ and $n$ integers, $m \neq n$ .

Solution

\begin{array}{rcl} K_{m n} & = & \frac{1}{2} \int_{0}^{2 π} [sin (m + n) x + sin (m - n) x] d x \\ = & \frac{1}{2} {[- \frac{cos (m + n) x}{(m + n)} - \frac{cos (m - n) x}{(m - n)}]}_{0}^{2 π} \\ = & - \frac{1}{2} [\frac{cos (m + n) 2 π - 1}{(m + n)} + \frac{cos (m - n) 2 π - 1}{(m - n)}] = 0 \end{array}

(recalling that $cos (integer \times 2 π) = 1$ ).

Task!

Derive the orthogonality relation

$K_{m n} = \int_{0}^{2 π} sin m x cos n x d x = 0$ $m$ and $n$ integers, $m = n$

Hint: You will need to use a different trigonometric identity to that used in Example 23.

$K_{m n} = \int_{0}^{2 π} sin m x cos m x d x$

Putting $m = n \neq 0$ , and then using the identity $sin 2 A \equiv 2 sin A cos A$ we get

\begin{array}{rcl} K_{m m} & = & \int_{0}^{2 π} sin m x cos m x d x \\ = & \frac{1}{2} \int_{0}^{2 π} sin 2 m x d x \\ = & \frac{1}{2} {[- \frac{cos 2 m x}{2 m}]}_{0}^{2 π} = - \frac{1}{4 m} (cos 4 m π - cos 0) = - \frac{1}{4 m} (1 - 1) = 0 \end{array}

Putting $m = n = 0$ gives $K_{00} = \frac{1}{2} \int_{0}^{2 π} sin 0 cos 0 d x = 0$ .

Note that the particular case $m = n = 1$ was considered earlier in this Section.

3 Orthogonality relations

Example 23

Solution

Task!

Answer