Theory

Fourier series

How a periodic function is expanded in a series of sines and cosines, how the coefficients are computed, at which points and in what sense the series converges, and how Parseval's identity links the function to its coefficients.

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Periodic functions and trigonometric polynomials
Orthogonality of the trigonometric system

Periodic functions and trigonometric polynomials

1. Periodic function

$f : R \to R$ is periodic with period $T > 0$ if $f (x + T) = f (x)$ for every $x \in R$ . A function defined only on $[- π, π)$ extends uniquely to a $2 π$ -periodic function on $R$ , the periodic extension, by repeating its values on each interval $[π, 3 π)$ , $[- 3 π, - π)$ and so on. For an integrable $T$ -periodic function the integral over one period does not depend on the starting point:

\int_{a}^{a + T} f (x) d x = \int_{0}^{T} f (x) d x for every a \in R .

Periodic functions describe phenomena that repeat, such as oscillations, waves and signals. The theory is developed for the period $2 π$ , the common period of $sin x$ and $cos x$ , and is carried over to an arbitrary period by a change of variable.

2 π

-periodic extension of

f (x) = x

x \in [- π, π)

: the piece on

[- π, π)

repeats in every period, with a jump at the points

π + 2 k π

2. Trigonometric polynomial

A trigonometric polynomial of degree $N$ is a function of the form

s_{N} (x) = \frac{a _{0}}{2} + n = 1 \sum N (a_{n} cos (n x) + b_{n} sin (n x)),

with $a_{0}, a_{1}, \dots, a_{N}, b_{1}, \dots, b_{N} \in R$ . Every trigonometric polynomial is $2 π$ -periodic and infinitely differentiable, since it is a finite sum of sines and cosines. Trigonometric polynomials are the building blocks of the theory: the Fourier series represents a periodic function as a limit of trigonometric polynomials, just as the Taylor series represents a smooth function as a limit of algebraic polynomials.

Orthogonality of the trigonometric system

For every pair of integers $n, m \geq 1$ the following relations hold:

\int_{- π}^{π} cos (n x) cos (m x) d x = {0 π if n \neq = m if n = m \int_{- π}^{π} sin (n x) sin (m x) d x = {0 π if n \neq = m if n = m \int_{- π}^{π} sin (n x) cos (m x) d x = 0, \int_{- π}^{π} cos (n x) d x = \int_{- π}^{π} sin (n x) d x = 0.

On $[- π, π]$ the product of two different functions of the system $1, cos x, sin x, cos (2 x), sin (2 x), \dots$ has zero integral. This property makes it possible to isolate one coefficient at a time in the expansion of a function, and it is the foundation of the formulas for the Fourier coefficients.

y = \sin x \sin \left(2 x\right)

y = sin x sin (2 x)

[- π, π]

: the regions above and below the

x

-axis have the same area, therefore the integral is zero

Proof

For every integer $k \neq = 0$ we have $\int_{- π}^{π} cos (k x) d x = [\frac{sin ( k x )}{k}]_{- π}^{π} = 0$ , since $sin (k π) = 0$ ; the integral of $sin (k x)$ is zero because the function is odd on the symmetric interval. For the products we apply the product-to-sum formulas:

cos (n x) cos (m x) = \frac{cos ( ( n - m ) x ) + cos ( ( n + m ) x )}{2} .

If $n \neq = m$ both terms are cosines of nonzero integer multiples of $x$ , therefore their integrals are zero. If $n = m$ the first term is $cos 0 = 1$ , hence

\int_{- π}^{π} cos^{2} (n x) d x = \int_{- π}^{π} \frac{1 + cos ( 2 n x )}{2} d x = π .

For the sines we use $sin (n x) sin (m x) = \frac{cos ( ( n - m ) x ) - cos ( ( n + m ) x )}{2}$ and argue in the same way. Finally, $sin (n x) cos (m x)$ is an odd function, being the product of an odd and an even function, therefore its integral over $[- π, π]$ is zero.

The theory continues with 6 more sections (coefficients and fourier series, even functions and odd functions, pointwise convergence…), plus all worked examples, the solutions and the PDF.

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