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Orthogonal Functions on [0,1]

We can construct a condition for functions to be orthogonal with respect to some inner product analogously to ordinary vectors being orthogonal being orthogonal with respect to the dot product. Many definitions are possible. Here is one.
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  
\[[0,1]\]
.
\[\begin{equation} \begin{aligned} \int^1_0 1 sin 2n \pi x dx = [-\frac{1}{2n \pi} cos 2n \pi x]^1_0 &=- \frac{1}{2n \pi} (cos 0 - cos 2n \pi) \\ &= - \frac{1}{2n \pi} (1 - 1)=0 \end{aligned} \end{equation}\]

\[\begin{equation} \begin{aligned} \int^1_0 1 cos 2n \pi x dx = [\frac{1}{2n \pi} sin 2n \pi x]^1_0 &= \frac{1}{2n \pi} (sin 0 - sin 2n \pi) \\ &= \frac{1}{2n \pi} (0 - 0)=0 \end{aligned} \end{equation}\]

\[\begin{equation} \begin{aligned} \int^1_0 sin 2n \pi x sin 2m \pi x dx &= \int^1_0 \frac{1}{2} (cos 2(n-m) \pi x \\ &- cos 2(n+m) \pi x ) dx \\ &= \frac{1}{2} [ \frac{1}{2(n-m) \pi} sin 2(n-m) \pi x \\ &- \frac{1}{2(n+m) \pi} sin 2(n+m) \pi x]^1_0 \\ &= \frac{1}{2} ((\frac{1}{2(n-m) \pi} sin 2(n-m) \pi \\ &- \frac{1}{2(n+m) \pi} sin 2(n+m) \pi ) \\ &- (\frac{1}{2(n-m) \pi} sin 0 - \frac{1}{2(n+m) \pi} sin 0 \\ &= \frac{1}{2} ((0-0)-(0-0))=0 \end{aligned} \end{equation} \]

\[\begin{equation} \begin{aligned} \int^1_0 cos 2n \pi x cos 2m \pi x dx &= \int^1_0 \frac{1}{2} (cos 2(n-m) \pi x \\ &+ cos 2(n+m) \pi x ) dx \\ &= \frac{1}{2} [ \frac{1}{2(n-m) \pi} sin 2(n-m) \pi x \\ &+ \frac{1}{2(n+m) \pi} sin 2(n+m) \pi x]^1_0 \\ &= \frac{1}{2} ((\frac{1}{2(n-m) \pi} sin 2(n-m) \pi \\ &+ \frac{1}{2(n+m) \pi} sin 2(n+m) \pi ) \\ &- (\frac{1}{2(n-m) \pi} sin 0 + \frac{1}{2(n+m) \pi} sin 0 \\ &= \frac{1}{2} ((0+0)-(0+0)) =0 \end{aligned} \end{equation} \]

\[\begin{equation} \begin{aligned} \int^1_0 sin 2n \pi x cos 2m \pi x dx &= \int^1_0 \frac{1}{2} (sin 2(n-m) \pi x \\ &+ sin 2(n+m) \pi x ) dx \\ &= \frac{1}{2} [- \frac{1}{2(n-m) \pi} cos 2(n-m) \pi x \\ &- \frac{1}{2(n+m) \pi} cos 2(n+m) \pi x]^1_0 \\ &=- \frac{1}{2} ((\frac{1}{2(n-m) \pi} cos 2(n-m) \pi \\ &+ \frac{1}{2(n+m) \pi} cos 2(n+m) \pi ) \\ &- (\frac{1}{2(n-m) \pi} cos 0 + \frac{1}{2(n+m) \pi} cos 0 \\ &= - \frac{1}{2} (( \frac{1}{2(n-m) \pi}+\frac{1}{2(n+m) \pi} )\\ &-( \frac{1}{2(n-m) \pi}+\frac{1}{2(n+m) \pi} )) =0 \end{aligned} \end{equation} \]