Artificial Intelligence 🤖
Function Basics
Orthogonality

Orthogonality

In order to derive some of the results included in this section, we make use of the product-to-sum trigonometric identities, given as follows:

sinmxcosnx=12[sin(m+n)x+sin(mn)x]cosmxcosnx=12[cos(m+n)x+cos(mn)x]sinmxsinnx=12[cos(mn)xcos(m+n)x]\begin{aligned} \sin m x \cos n x & =\frac{1}{2}[\sin (m+n) x+\sin (m-n) x] \\ \cos m x \cos n x & =\frac{1}{2}[\cos (m+n) x+\cos (m-n) x] \\ \sin m x \sin n x & =\frac{1}{2}[\cos (m-n) x-\cos (m+n) x] \end{aligned}

Theorem: Orthogonality

Two nonzero functions, f(x)f(x) and g(x)g(x) are said to be orthogonal on [a,b][a, b] if their inner product is given by

<f(x),g(x)>=abf(x)g(x)dx=0.<f(x), g(x)>=\int_{a}^{b} f(x) g(x) d x=0 .

We are more interested in the so-called orthogonality relations a set of nonzero functions have: a sequence of functions {fi(x)}\left\{f_{i}(x)\right\} is said to be mum u tually orthogonal on [a,b][a, b] if,

abfi(x)fj(x)dx={0 for ijc>0 for i=j\int_{a}^{b} f_{i}(x) f_{j}(x) d x= \begin{cases}0 & \text { for } i \neq j \\ c>0 & \text { for } i=j\end{cases}

Next, we prove the orthogonality relations given by Theorem 1.1 for the set of functions {sin(nx),sin(mx)}\{\sin (n x), \sin (m x)\}. We want to prove that they are mutually orthogonal on [π,π][-\pi, \pi]; to do that we will make use of the appropriate identity from (1.7a)-(1.7c) and relevant double angle formula relations. We start with the integral on the LHS of Eq. (1.9) where, here, fi(x)=sin(nx)f_{i}(x)=\sin (n x) and fj(x)=f_{j}(x)= sin(mx)\sin (m x) and a=π,b=πa=-\pi, b=\pi :

I=ππsin(nx)sin(mx)dx\mathcal{I}=\int_{-\pi}^{\pi} \sin (n x) \sin (m x) d x

We need to consider two cases separately for (i) m=nm=n and (ii) mnm \neq n. For the case where m=nm=n, Eq. (1.10) becomes:

ππsin2(nx)dx=20πsin2(nx)dx;\int_{-\pi}^{\pi} \sin ^{2}(n x) d x=2 \int_{0}^{\pi} \sin ^{2}(n x) d x ;

note that in going from left to right in Eq. (1.11), we made use of Eq. (1.5) since the function sin2(nx)\sin ^{2}(n x) is an even function. To integrate the RHS of Eq. (1.11), we make use of the double angle formula cos(2nx)=1sin2(nx)\cos (2 n x)=1-\sin ^{2}(n x). This gives:

20πsin2(nx)dx=20π1cos(2nx)2dx=π\begin{aligned} 2 \int_{0}^{\pi} \sin ^{2}(n x) d x & =2 \int_{0}^{\pi} \frac{1-\cos (2 n x)}{2} d x \\ & =\pi \end{aligned}

Next, for the case mnm \neq n, we need to make use of the sum-to-product identity given by Eq. (1.7c).

ππsin(nx)sin(mx)dx=12[cos(mn)xcos(m+n)x].\int_{-\pi}^{\pi} \sin (n x) \sin (m x) d x=\frac{1}{2}[\cos (m-n) x-\cos (m+n) x] .

Note that the integrand is the product of two odd functions, sin(nx)sin(mx)\sin (n x) \sin (m x) which gives an even function. Integrating Eq. (1.14) with respect to xx, gives

0π[cos(mn)xcos(m+n)x]dx=sin(mn)xmnsin(m+n)xm+n.\int_{0}^{\pi}[\cos (m-n) x-\cos (m+n) x] d x=\frac{\sin (m-n) x}{m-n}-\frac{\sin (m+n) x}{m+n} .

Notice that in Eq. (1.15), we made use of Eq. (1.5) since we are integrating an even function from π-\pi to π\pi. Firstly, we know that mnm \neq n and therefore the denominator is well-defined. Secondly, observe that mm and nn are integers and hence mnm-n and m+nm+n are also integers. It follows that sin[(mn)x]\sin [(m-n) x] and sin[(nm)x]\sin [(n-m) x] are zero-valued and the integral in Eq. (1.15) is zero. We have therefore shown that:

ππsin(nx)sin(mx)dx={=0, if mn=π if m=n\int_{-\pi}^{\pi} \sin (n x) \sin (m x) d x=\left\{\begin{array}{ccc} =0, & \text { if } & m \neq n \\ =\pi & \text { if } & m=n \end{array}\right.

Equation (1.50) implies that the set of functions given by {sin(nx),sin(mx)}\{\sin (n x), \sin (m x)\} are mutually orthogonal on [π,π][-\pi, \pi].

Similarly, we can show that the functions 1,cosnx1, \cos n x and sinnx\sin n x are mutually orthogonal on [π,π][-\pi, \pi] giving rise to the following:

  1. ππcosmxdx=ππsinmxdx=0\int_{-\pi}^{\pi} \cos m x d x=\int_{-\pi}^{\pi} \sin m x d x=0 if mm is integral and m0m \neq 0;
  2. ππsinnxcosmxdx=0\int_{-\pi}^{\pi} \sin n x \cos m x d x=0 for any integers nn and mm;
  3. ππcosnxcosmxdx=ππsinnxsinmxdx=0\int_{-\pi}^{\pi} \cos n x \cos m x d x=\int_{-\pi}^{\pi} \sin n x \sin m x d x=0 if nmn \neq m;
  4. ππcos2mxdx=ππsin2mxdx=π\int_{-\pi}^{\pi} \cos ^{2} m x d x=\int_{-\pi}^{\pi} \sin ^{2} m x d x=\pi if mm is integral and m0m \neq 0.

These relations are useful in determining the Fourier coefficients.