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:

sin⁡mxcos⁡nx=12[sin⁡(m+n)x+sin⁡(m−n)x]cos⁡mxcos⁡nx=12[cos⁡(m+n)x+cos⁡(m−n)x]sin⁡mxsin⁡nx=12[cos⁡(m−n)x−cos⁡(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 i≠jc>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) m≠nm \neq n. For the case where m=nm=n, Eq. (1.10) becomes:

∫−ππsin⁡2(nx)dx=2∫0πsin⁡2(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 sin⁡2(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)=1−sin⁡2(nx)\cos (2 n x)=1-\sin ^{2}(n x). This gives:

2∫0πsin⁡2(nx)dx=2∫0π1−cos⁡(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 m≠nm \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⁡(m−n)x−cos⁡(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⁡(m−n)x−cos⁡(m+n)x]dx=sin⁡(m−n)xm−n−sin⁡(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 m≠nm \neq n and therefore the denominator is well-defined. Secondly, observe that mm and nn are integers and hence m−nm-n and m+nm+n are also integers. It follows that sin⁡[(m−n)x]\sin [(m-n) x] and sin⁡[(n−m)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 m≠n=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,cos⁡nx1, \cos n x and sin⁡nx\sin n x are mutually orthogonal on [−π,π][-\pi, \pi] giving rise to the following:

  1. ∫−ππcos⁡mxdx=∫−ππsin⁡mxdx=0\int_{-\pi}^{\pi} \cos m x d x=\int_{-\pi}^{\pi} \sin m x d x=0 if mm is integral and m≠0m \neq 0;
  2. ∫−ππsin⁡nxcos⁡mxdx=0\int_{-\pi}^{\pi} \sin n x \cos m x d x=0 for any integers nn and mm;
  3. ∫−ππcos⁡nxcos⁡mxdx=∫−ππsin⁡nxsin⁡mxdx=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 n≠mn \neq m;
  4. ∫−ππcos⁡2mxdx=∫−ππsin⁡2mxdx=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 m≠0m \neq 0.

These relations are useful in determining the Fourier coefficients.