Functions
\[f(x), \: g(x) \in C[0,1]\]
are orthogonal on \[[0,1]\]
if \[\int^1_0 f(x) g(x) dx\]
The set of functions
\[\left\{1, \: cos 2n \pi x, sin 2n \pi x \colon \: n \in \mathbb{N} \right\}\]
is orthogonal on \[[0,1]\]
.\[\begin{equation} \begin{aligned} \int^1_0 1 sin 2n \pi x dx = [-\frac{1}{2n \pi} cos 2n \pi x]^1_0 &=- \frac{1}{2n \pi} (cos 0 - cos 2n \pi) \\ &= - \frac{1}{2n \pi} (1 - 1)=0 \end{aligned} \end{equation}\]
\[\begin{equation} \begin{aligned} \int^1_0 1 cos 2n \pi x dx = [\frac{1}{2n \pi} sin 2n \pi x]^1_0 &= \frac{1}{2n \pi} (sin 0 - sin 2n \pi) \\ &= \frac{1}{2n \pi} (0 - 0)=0 \end{aligned} \end{equation}\]
\[\begin{equation} \begin{aligned} \int^1_0 sin 2n \pi x sin 2m \pi x dx &= \int^1_0 \frac{1}{2} (cos 2(n-m) \pi x \\ &- cos 2(n+m) \pi x ) dx \\ &= \frac{1}{2} [ \frac{1}{2(n-m) \pi} sin 2(n-m) \pi x \\ &- \frac{1}{2(n+m) \pi} sin 2(n+m) \pi x]^1_0 \\ &= \frac{1}{2} ((\frac{1}{2(n-m) \pi} sin 2(n-m) \pi \\ &- \frac{1}{2(n+m) \pi} sin 2(n+m) \pi ) \\ &- (\frac{1}{2(n-m) \pi} sin 0 - \frac{1}{2(n+m) \pi} sin 0 \\ &= \frac{1}{2} ((0-0)-(0-0))=0 \end{aligned} \end{equation} \]
\[\begin{equation} \begin{aligned} \int^1_0 cos 2n \pi x cos 2m \pi x dx &= \int^1_0 \frac{1}{2} (cos 2(n-m) \pi x \\ &+ cos 2(n+m) \pi x ) dx \\ &= \frac{1}{2} [ \frac{1}{2(n-m) \pi} sin 2(n-m) \pi x \\ &+ \frac{1}{2(n+m) \pi} sin 2(n+m) \pi x]^1_0 \\ &= \frac{1}{2} ((\frac{1}{2(n-m) \pi} sin 2(n-m) \pi \\ &+ \frac{1}{2(n+m) \pi} sin 2(n+m) \pi ) \\ &- (\frac{1}{2(n-m) \pi} sin 0 + \frac{1}{2(n+m) \pi} sin 0 \\ &= \frac{1}{2} ((0+0)-(0+0)) =0 \end{aligned} \end{equation} \]
\[\begin{equation} \begin{aligned} \int^1_0 sin 2n \pi x cos 2m \pi x dx &= \int^1_0 \frac{1}{2} (sin 2(n-m) \pi x \\ &+ sin 2(n+m) \pi x ) dx \\ &= \frac{1}{2} [- \frac{1}{2(n-m) \pi} cos 2(n-m) \pi x \\ &- \frac{1}{2(n+m) \pi} cos 2(n+m) \pi x]^1_0 \\ &=- \frac{1}{2} ((\frac{1}{2(n-m) \pi} cos 2(n-m) \pi \\ &+ \frac{1}{2(n+m) \pi} cos 2(n+m) \pi ) \\ &- (\frac{1}{2(n-m) \pi} cos 0 + \frac{1}{2(n+m) \pi} cos 0 \\ &= - \frac{1}{2} (( \frac{1}{2(n-m) \pi}+\frac{1}{2(n+m) \pi} )\\ &-( \frac{1}{2(n-m) \pi}+\frac{1}{2(n+m) \pi} )) =0 \end{aligned} \end{equation} \]