Till now, we have focused on Fourier series of \(2\pi\)-periodic functions \(f: \mathbb{R} \to \mathbb{R}\), and restricted our attention to the interval \([-\pi, \pi]\).
Often, we begin with a function defined only on \([-\pi,\pi]\) and form its \(2\pi\)-periodic extension to all of \(\mathbb{R}\).
Technically, this only works if \(f(-\pi) = f(\pi)\). If this is not the case, then the extension is not well-defined.
So, we should really only be considering \(2\pi\)-periodic extensions of functions defined on \((-\pi,\pi]\) or \([-\pi, \pi)\).
This all leads us to a method for computing Fourier series of functions defined on the interval \([0,\pi]\):
Provided the original function was piecewise smooth, so too will be the resulting \(2\pi\)-periodic function, and so the Fourier series will converge to the original function on \([0,\pi]\) at all points of continuity.
How do we complete step (1)? There are two natural ways to do this:
Definition
Let \(f: [0,\pi] \to \mathbb{R}\) be a function.
The even extension of \(f\) to \((-\pi, \pi]\) is the function \(f_{\text{even}}: [-\pi, \pi] \to \mathbb{R}\) defined by \[ f_{\text{even}}(\theta) = \begin{cases} f(\theta) & : \theta \in [0, \pi], \\ f(-\theta) & : \theta \in [-\pi, 0]. \end{cases} \] Notice that the even extension is in fact defined on \([-\pi, \pi]\), and that \(f_{\text{even}}(-\pi) = f_\text{even}(\pi)\).
The odd extension of \(f\) to \((-\pi, \pi]\) is the function \(f_{\text{odd}}: (-\pi, \pi] \to \mathbb{R}\) defined by \[ f_{\text{odd}}(\theta) = \begin{cases} f(\theta) & : \theta \in (0, \pi], \\ -f(-\theta) & : \theta \in (-\pi, 0), \\ 0 & : \theta = 0. \end{cases} \]
Suppose we continue with a function \(f: [0, \pi] \to \mathbb{R}\), assumed to be integrable.
The Fourier series of the even extension \(f_{\text{even}}\) has \[ a_n = \frac{2}{\pi} \int_{0}^{\pi} f(\theta) \cos{n\theta} \, d\theta \quad \text{and} \quad b_n = 0, \] since \(f_{\text{even}}\) is (of course) even. Thus its Fourier series is given by \[ \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos{n\theta}. \]
The Fourier series of the odd extension \(f_{\text{odd}}\) has \[ a_n = 0 \quad \text{and} \quad b_n = \frac{2}{\pi} \int_{0}^{\pi} f(\theta) \sin{n\theta} \, d\theta, \] since \(f_{\text{odd}}\) is (of course) odd. Thus its Fourier series is given by \[ \sum_{n=1}^{\infty} b_n \sin{n\theta}. \]
This leads us to:
Definition/Theorem
Let \(f: [0, \pi] \to \mathbb{R}\) be a piecewise smooth function. Its Fourier cosine series is given by
\[ \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos{n\theta}, \quad a_n = \frac{2}{\pi} \int_{0}^{\pi} f(\theta) \cos{n\theta} \, d\theta, \]
and its Fourier sine series is given by
\[ \sum_{n=1}^{\infty} b_n \sin{n\theta}, \quad b_n = \frac{2}{\pi} \int_{0}^{\pi} f(\theta) \sin{n\theta} \, d\theta. \]
These series both converge pointwise to \(f\) at all points of continuity.
Consider the function
\[ f:[0, \pi] \to \mathbb{R}, \quad f(\theta) = \theta. \]
Suppose we have an integrable function \(f(x)\) defined on an interval \((t, t+L]\) (or \([t, t+L)\)) of length \(L>0\). If we make the change of variables \(x = \frac{L\theta}{\pi} + t\), then the function \(g(\theta)\) is defined on the interval \((0, \pi]\) (or \([0, \pi)\)).
Thus \(g\) has its Fourier cosine and sine series \[ \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos{n\theta} \quad \text{and} \quad \sum_{n=1}^\infty b_n \sin{n\theta} \] where the coefficients are given by \[ a_n = \frac{2}{\pi} \int_0^\pi g(\theta) \cos{n\theta} \, d\theta, \quad b_n = \frac{2}{\pi} \int_0^\pi g(\theta) \sin{n\theta} \, d\theta. \]
Thus \(f\) has Fourier cosine and sine series \[ \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos{\left(\frac{n\pi(x-t)}{L}\right)} \quad \text{and} \quad \sum_{n=1}^\infty b_n \sin{\left(\frac{n\pi(x-t)}{L}\right)}, \] where the coefficients are given by \[ a_n = \frac{2}{L} \int_{t}^{t+L} f(x) \cos{\left(\frac{n\pi(x-t)}{L}\right)} \, dx, \quad b_n = \frac{2}{L} \int_{t}^{t+L} f(x) \sin{\left(\frac{n\pi(x-t)}{L}\right)} \, dx. \]