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ece205: orthogonality and fourier
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@ -245,6 +245,37 @@ If functions $f$ and $g$ have a period $T$, then both $af+bg$ and $fg$ also must
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- odd × odd = even
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- even × odd = odd
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## Orthogonality
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$$\int^L_{-L}\cos(\frac{m\pi x}{L})\sin(\frac{n\pi x}{L})dx=0$$
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$$
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\int^L_{-L}\cos(\frac{m\pi x}{L})(\frac{n\pi x}{L})dx=\begin{cases}
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2L & \text{if }m=n=0 \\
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L & \text{if }m=n\neq 0 \\
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0 & \text{if }m\neq n
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\end{cases}
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$$
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$$
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\int^L_{-L}\sin(\frac{m\pi x}{L}\sin(\frac{n\pi x}{L})dx=\begin{cases}
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L & \text{if }m=n \\
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0 & \text{if }m\neq n
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\end{cases}
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$$
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Functions are **orthogonal** on an interval when the integral of their product is zero, and a set of functions is **mutually orthogonal** if all functions in the set are orthogonal to each other.
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If a Fourier series converges to $f(x)$:
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$$f(x)=\frac{a_0}{2} + \sum^\infty_{n=1}\left(a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})\right)$$
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The **Euler-Fourier** formulae must apply:
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$$
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\boxed{a_n=\frac 1 L\int^L_{-L}f(x)\cos(\frac{n\pi x}{L})dx} \\
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\\
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\boxed{b_n=\frac 1 L\int^L_{-L}f(x)\sin(\frac{n\pi x}{L})dx}
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$$
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## Resources
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