Fourier Series with General Period

The Domain [-p,p]

The key idea here is that you can compute the Fourier coefficients for a function using any period of the $\sin$ and $\cos$ functions. To derive the coefficients for a function f(x) represented by a Fourier series

$$f(x)=a_0+\sum_{k=1}^{\infty}a_k\cos(k\frac{\pi}{p}x) + \sum_{k=1}^{\infty}b_k\sin(k\frac{\pi}{p}x)$$ (1)

defined on an interval [-p,p] we use relations will gotten by adjusting the period of the $\sin$ and $\cos$ by multiplying x by $\frac{\pi}{p}$. These are
$$\begin{align} \nonumber \int_{-p}^{p} \cos(\frac{\pi}{p}x)dw=0& &\displaystyle\... ...ses} p&\text{m=n}\ 0&\text{m$\neq$\space n}\ \end{cases} \ \end{align}$$

$$\int_{-p}^{p} \cos(m\frac{\pi}{p}x)\sin(n\frac{\pi}{p}x)dx =0\,\mathrm{for\;any\;}m\neq n.$$ (2)

We can then compute the integrals

$$\begin{align} \nonumber \int_{-p}^{p} f(x)dx=\int_{-p}^{p}( a_0+\sum_{j=1}^{\in... ...{\infty}b_j\sin(j\frac{\pi}{p}x))\sin(k\frac{\pi}{p}x)dx =p b_k.\ \end{align}$$
which leads to the same identities as in the period $2\pi$ case only dilated to a period 2p domain. So the coefficients of the Fourier series for a function f(x) of period 2p defined on the interval [-p,p] are

$$a_0 = \frac{1}{2p} \int_{-p}^{p} f(x)dx$$ (3)

$$\begin{align} a_k=\frac{1}{p}\int_{-p}^{p} f(x)\cos(k\frac{\pi}{p}x)dx& &\displaystyle b_k=\frac{1}{p}\int_{-p}^{p} f(x)\sin(k\frac{\pi}{p}x) \end{align}$$

The Domain [0,p]

A similar argument will compute the identities for the coefficients of a Fourier series for a function defined on the interval [0,p] of period p. Using

$$f(x)=a_0+\sum_{k=1}^{\infty}a_k\cos(k\frac{2\pi}{p}x) + \sum_{k=1}^{\infty}b_k\sin(k\frac{2\pi}{p}x)$$ (4)

defined on the interval [0,p] instead we will have that by adjusting the period of the $\sin$ and $\cos$ with the period shifting term $\frac{2\pi}{p}$,
$$\begin{align} \nonumber \int_{0}^{p} \cos(\frac{2\pi}{p}x)dw=0& &\displaystyle\... ...ac{p}{2}&\text{m=n}\\ 0&\text{m$\neq$\space n}\\ \end{cases} \\ \end{align}$$

$$\int_{0}^{p} \cos(m\frac{2\pi}{p}x)\sin(n\frac{2\pi}{p}x)dx =0\,\mathrm{for\;any\;}m\neq n.$$ (5)

and so in the integrals we compute that
$$\begin{align} \nonumber \int_{0}^{p} f(x)dx=\int_{0}^{p}( a_0+\sum_{j=1}^{\inft... ...sin(j\frac{2\pi}{p}x))\sin(k\frac{2\pi}{p}x)dx =\frac{p}{2} b_k.\\ \end{align}$$
which leads to the following identities for the coefficients of the Fourier series of a function f(x) of period p defined on the interval [0,p]

$$a_0 = \frac{1}{p} \int_{0}^{p} f(x)dx$$ (6)

$$\begin{align} a_k=\frac{2}{p}\int_{0}^{p} f(x)\cos(k\frac{2\pi}{p}x)dx& &\displaystyle b_k=\frac{2}{p}\int_{0}^{p} f(x)\sin(k\frac{2\pi}{p}x) \end{align}$$
You should experiment with different values of p to get a feeling for how these general cases reduce to the cases you have seen already, e.g. $p=\{1,2,\pi,2\pi\}$.

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