fourier functions form an orthonormal set

[[concept]]

Topics

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Proposition

The subset of functions ${\left\{\frac{e^{inx}}{\sqrt{2\pi }}\right\}}_{n\in \mathbb{Z}}$ is an orthonormal subset of $L^2([-\pi ,\pi ])$

NOTE

if we don’t like working with complex exponentials we can use $e^{ix}=\cos x+i\sin x$ And work out everything we need.

Proof

Note that

\langle e^{inx}, e^{imx} \rangle &= \int _{-\pi}^\pi e^{inx} \overline{e^{imx}} \, dx \\ &=\int _{-\pi}^\pi e^{i(n-m)x} \, dx \\ &= \begin{cases} 2\pi, & m=n \\ \frac{1}{i(n-m)} e^{i(n-m)x} \bigg{|}_{-\pi}^\pi=0 & m \neq n \end{cases} \end{align}$$ Since the exponential is periodic in $2\pi$. Normalizing by $2\pi$ yields $$\left\langle \frac{e^{inx}}{\sqrt{ 2\pi }} ,\frac{e^{imx}}{\sqrt{ 2\pi }} \right \rangle =\begin{cases} 1 & m=n \\ 0 & m \neq n \end{cases}$$ Giving us our orthonormal set.

References

References

See Also

• Fourier coefficient
• Fourier Transform
• Fourier series

Mentions

Mentions

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