The Fourier Series

Definition

For a function $f(x)$ what is periodic and integrable over one period of length $T$, the Fourier series expansion is,

$$f(x)=\frac{a_0}{2}+\sum _{n=1}^{\infty }(a_n\cos (\frac{2\pi nx}{T})+b_n\sin (\frac{2\pi nx}{T}))$$

where,

$$a_0=\frac{1}{T}{\int }_0^{T}f(x)dx$$

and,

$$a_n=\frac{2}{T}{\int }_0^{T}f(x)\cos (\frac{2\pi nx}{T})dx,$$ $$b_n=\frac{2}{T}{\int }_0^{T}f(x)\sin (\frac{2\pi nx}{T})dx$$

To decompose an aperiodic function in this way, the Fourier transform must be used. It should also be noted that the bounds of integration, here $0$ to $T$, can be any bounds so long as it is the length of a period, e.g. from $-\frac{T}{2}$ to $\frac{T}{2}$, or from $4T$ to $5T$, etc.

Derivation

The trigonometric functions sine and cosine are orthogonal to each other, and since they are orthogonal, they can be used as basis functions to define a new function space. We can then transform any periodic function within a normal space of $n$-dimensions, ${\mathbb{R}}^n$, into that new function space using a change of basis vectors.

Intuitively, we will transform a function of normal variables, e.g. $x$ and $y$, into a new space where the basis vectors are $sin$ and $cos$.

Orthogonality

See Orthogonality.

Theorem:

If there are two integers, $j$ and $k$, and for the natural period of $-\pi$ to $\pi$, then,

$${\int }_{-\pi }^{\pi }\cos (jx)\cos (kx)dx={\int }_{-\pi }^{\pi }\sin (jx)\sin (kx)dx=\{\begin{array}{ll}\pi & \text{if}j=k\\ 0& \text{otherwise}\end{array}$$

and

$${\int }_{-\pi }^{\pi }\sin (jx)\cos (kx)dx=0$$

Proof:

For some integer $n$, $n\in \mathbb{Z}$, observe that,

$${\int }_{-\pi }^{\pi }\sin (nx)dx=0\text{and}{\int }_{-\pi }^{\pi }\cos (nx)dx=\{\begin{array}{ll}2\pi & \text{if}n=0\\ 0& \text{otherwise}\end{array}$$

If $j$ and $k$ are likewise integers, $j,k\in \mathbb{N}$, we can use the product-to-sum identities to see that,

$${\int }_{-\pi }^{\pi }\cos (jx)\cos (kx)dx={\int }_{-\pi }^{\pi }\frac{\cos ([j-k]x)+\cos ([j+k]x)}{2}dx=\{\begin{array}{ll}\pi & \text{if}j=k\\ 0& \text{otherwise}\end{array}$$

and

$${\int }_{-\pi }^{\pi }\sin (jx)\sin (kx)dx={\int }_{-\pi }^{\pi }\frac{\cos ([j-k]x)-\cos ([j+k]x)}{2}dx=\{\begin{array}{ll}\pi & \text{if}j=k\\ 0& \text{otherwise}\end{array}$$

and

$${\int }_{-\pi }^{\pi }\sin (jx)\cos (kx)dx={\int }_{-\pi }^{\pi }\frac{\sin ([j-k]x)-\sin ([j+k]x)}{2}dx=0$$

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Because of this conclusion, we can see that for two integers $j$ and $k$, $\sin (jx)$ and $\cos (kx)$ are orthogonal, since their inner product is equal to zero everywhere.

References

1. The Fourier Series from Cornell
2. An Introduction to Fourier Series and Transforms by Justin Tarquino, U of Chicago
3. Fourier Series Derivation from University of Michigan