Question

Consider the set of functions φk(x) = sin (kπx), k = 1, 2, . . . for x ∈ [0, 1]. Show that they are orthogonal according to the inner product (f, g) = Z 1 0 f(x)g(x)dx. That is, show that (φj , φk) = 0, k 6= j and that (φk, φk) = Z 1 0 sin2 (kπx)dx = 1 2 . [Hint: You will need the trigonometric identities cos (α − β) − cos (α + β) = 2 sin α sin β 1 − 2 sin2 α = cos 2α

Answer 1

Explanation:

We are going to consider the inner product

`(f,g) = \int_(0)^1 f(x) g(x) dx`

Consider the following trigonometric identity:

`\cos(\alpha-\beta) - \cos(\alpha+\beta) = 2 \sin(\alpha)\sin(\beta)`

Suppose that
`j\\eq k` consider the inner product and the identity:

`(\phi_k,\phi_j) = \int_(0)^(1)\sin(k\pi x)\sin(j\pi x)dx=(1)/(2)\int_(0)^(1)\cos((k-j)\pi x)-\cos((k+j)\pi x)dx = (1)/(2)\left.((\sin((k-j)\pi x))/((k-j)\pi)- (\sin((k+j)\pi x))/((k+j)\pi))\right|_(0)^(1)= (1)/(2)((\sin((k-j)\pi))/((k-j)\pi)-(\sin((k+j)\pi))/((k+j)\pi)`

Note that since k,j are integers then k-j and k+j are integers. Then, by properties of the sine function we have that
`sin((k-j)\pi) = \sin((k+j)\pi) = 0 = (\phi_k,\phi_j)`

Consider also the following identity:

`\sin^2(x) = (1-\cos(2x))/(2)`

Suppose that j=k. Then

`(\phi_k, \phi_k) = \int_(0)^(1)\sin^2(k\pi x) dx = \int_(0)^(1)(1-\cos(2k\pi x))/(2)dx = \left.(x)/(2)-(\sin(2k\pi x))/(4k\pi )\right|_(0)^(1) = (1)/(2)-(\sin(2k\pi))/(4k\pi )`

Since k is an integer, then
`\sin(2k\pi)=0`. So
`(\phi_k,\phi_k) =(1)/(2)`

This proves what's been asked in the question.