1.1. Fourier Series¶

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1.1.1. Definition¶

For simplicity, we will restrict ourselves to functions \(f(x)\) defined on the domain \(x \in [0, 1]\). Then the \(N\)th partial sum of the Fourier series of \(f(x)\) is

(1.1)¶\[ f_N(x) = \frac{a_0}{2} + \sum_{n=1}^N \left[ a_n \cos\left(2\pi nx\right) + b_n \sin\left(2\pi nx\right) \right] \]

where the so-called Fourier coefficients are

(1.2)¶\[\begin{split} a_n = 2 \int_{0}^{1} f(x) \cos\left(2\pi nx\right), \\ b_n = 2 \int_{0}^{1} f(x) \sin\left(2\pi nx\right). \end{split}\]

When \(N\) is small, \(f_N(x)\) is usually a poor approximation of \(f(x)\) because \(f_1(x)\) consists of only one frequency. As \(N\) increases the approximation improves, reaching exact convergence everywhere when \(N = \infty\).

1.1.2. Example: \(f(x) = x\)¶

The Fourier coefficients, (1.2), are

(1.3)¶\[\begin{align} a_n &= 2 \int_{0}^{1} x \cos\left(2\pi nx\right), \\ b_n &= 2 \int_{0}^{1} x \sin\left(2\pi nx\right). \end{align}\]

It’s not too hard to calculate these integrals via integration by parts, but it’s even easier to use sympy:

import sympy as sp
sp.init_printing(use_latex='mathjax')

x, a_n, b_n = sp.symbols('x a_n b_n', real=True)
n = sp.symbols('n', integer=True, nonnegative=True)
f, u = x, 2*sp.pi*n*x

sp.integrate(2*f*sp.cos(u), (x, 0, 1))

\[\begin{split}\displaystyle \begin{cases} 0 & \text{for}\: n \neq 0 \\1 & \text{otherwise} \end{cases}\end{split}\]

sp.Eq(a_n, sp.integrate(2*f*sp.cos(u), (x, 0, 1)))

\[\begin{split}\displaystyle a_{n} = \begin{cases} 0 & \text{for}\: n \neq 0 \\1 & \text{otherwise} \end{cases}\end{split}\]

sp.Eq(b_n, sp.integrate(2*f*sp.sin(u), (x, 0, 1)))

\[\begin{split}\displaystyle b_{n} = \begin{cases} - \frac{1}{\pi n} & \text{for}\: n \neq 0 \\0 & \text{otherwise} \end{cases}\end{split}\]

Direct substitution into (1.1) yields Fourier series

(1.4)¶\[ f_N(x) = \frac{1}{2} - \sum_{n=1}^N \frac{\sin\left(2\pi nx\right)}{\pi n}. \]

1.1.3. Visualization¶

Let’s plot (1.4) for various \(N\):

import matplotlib.pyplot as plt
import numpy as np

def get_fourier_sums(xs, N):
    fourier_sums = np.zeros(len(xs)) + .5
    for n in range(1, N+1):
        fourier_sums -= np.sin(2*np.pi*n*xs) / (np.pi*n)
    return fourier_sums

dx = .01
xs = np.arange(0+dx, 1, dx)
fig, ax = plt.subplots()
ax.set(xlabel='$x$', ylabel='$f_N(x)$', title='Fourier series for $f(x)=x$')
ax.plot(xs, get_fourier_sums(xs, 1), label='$N=1$')
ax.plot(xs, get_fourier_sums(xs, 10), label='$N=10$')
ax.plot(xs, get_fourier_sums(xs, 100), label='$N=100$')
ax.legend();

Clearly \(N = 1\) is a poor approximation because it consists of a single sinusoidal term. \(N = 10\) captures the linear portion but has visible oscillations. \(N = 100\) appears much smoother and matches \(f(x) = x\) closely except at the boundaries. The boundary disagreement is because the Fourier series is periodic by nature and thus approximates a sawtooth wave when the domain is extended:

xs = np.arange(0+dx, 3, dx)
fig, ax = plt.subplots()
ax.set(xlabel='$x$', ylabel='$f_N(x)$', title='Extended Fourier series for $f(x)=x$')
ax.plot(xs, get_fourier_sums(xs, 100));

Due to discontinuities at integer values of \(x\), the Fourier series actually takes the midpoint value \(\frac{1}{2}\).

Proof

When \(x\) is an integer, \(nx\) is also an integer and thus \(2\pi n x\) is an integer multiple of \(2\pi\). It follows that \(\sin(2\pi n x) = 0\) for all \(n\). Looking at the calculated Fourier series, (1.4), it follows that \(f_N(x) = \frac{1}{2}\). \(\blacksquare\)

1.1.4. Interactive Animation¶

The Fourier series plot for various \(N\) in § Visualization is helpful in distingushing different qualitative behaviors, but difficult to see how \(f_N(x)\) changes with \(N\). Of course, we could repeatedly make plots tuning \(N\) by hand, but interactive slider plots do that much better, for example:

import ipywidgets

@ipywidgets.interact(N=ipywidgets.IntSlider(min=1, max=100, step=1))
def fourier_slider(N):
    dx = .01
    xs = np.arange(0+dx, 1, dx)
    fig, ax = plt.subplots()
    ax.set(xlabel='$x$', ylabel='$f_N(x)$', title='Fourier series for $f(x)=x$')
    ax.plot(xs, get_fourier_sums(xs, N), label=f'$N={N}$')
    ax.plot(xs, xs, '--', label='$f(x) = x$')
    ax.legend()

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Exercise

Compute the Fourier series \(f_N(x)\) of the function \(f(x) = \lvert x - \frac{1}{2}\rvert\) over the interval \(x \in [0, 1]\). Plot \(f_N(x)\) for three values of \(N\) with different quailtative behavior. Finally, make an interactive slider plot of \(f_N(x)\), letting \(N\) be an adjustable parameter.

Hint

In sympy, \(\lvert x - \frac{1}{2}\rvert\) can be represented symbolically as

sympy.Abs(x - sp.Rational(1/2))

Solution

def get_fourier_sums(xs, N):
    fourier_sums = np.zeros(len(xs)) + .25
    for n in range(1, N+1, 2):  # Even terms are zero
        fourier_sums += np.cos(2*np.pi*n*xs) * 2 / (np.pi*n)**2
    return fourier_sums

@ipywidgets.interact(N=ipywidgets.IntSlider(min=1, max=50, step=1))
def fourier_slider(N):
    dx = .01
    xs = np.arange(0+dx, 1, dx)
    fig, ax = plt.subplots()
    ax.set(xlabel='$x$', ylabel='$f_N(x)$', title='Fourier series of $|x - \\frac{1}{2}|$')
    ax.plot(xs, get_fourier_sums(xs, N), label=f'$N={N}$')
    ax.plot(xs, np.abs(xs - .5), '--', label='$|x - \\frac{1}{2}|$')
    ax.legend()

# Your code goes here.

Jupyter Book Demo

1.1. Fourier Series¶

1.1.1. Definition¶

1.1.2. Example: \(f(x) = x\)¶

1.1.3. Visualization¶

1.1.4. Interactive Animation¶