Week 5 The Fourier series and transformation 1. The definition of the Fourier series 2. Solving ODEs using Fourier series (forced oscillations) 3. Fourier transformation

1. The definition of Fourier series ۞ A function f(x) is said to be periodic with a period P if it’s defined for all x  ℝ and Example 1: (a) sin x is a 2π-periodic funcn; (b) sin 2x is a ? -periodic funcn. (b) sin 2x is a π-periodic funcn.

Example 2: The function f(x) is periodic with a period P = 4, and Draw the graph of f(x) for x  [− 6, 8]. Comment: If f(x) is periodic with a period P, it is also periodic with periods 2P, 3P, 4P... nP... where n is an integer.

Theorem 1: The set SP of all continuous periodic functions with a period P and the standard inner product for functions, where L = ½ P, form a Hilbert space.

Theorem 2: The (infinite) set of functions forms a basis in S2π. In other words, any continuous 2π-periodic function f(x) can be represented in the form (1) or

Expression (1) is called the Fourier series of f(x). The constants a0, a1, a2... b1, b2 ... are called the Fourier coefficients of f(x). They are, essentially, the components of f(x) with respect to the basis introduced above.

Theorem 3: The basis defined in Theorem 2 is an orthogonal one, i.e. where n, m > 1 are arbitrary integers, and where n ≠ m. Proof: This theorem can be proved by evaluating the integrals involved.

Comment: The basis defined in Theorem 2 is orthogonal but not orthonormal, e.g.

Theorem 4: The Fourier coefficients are given by (2) Proof: Formulae (2) can be derived by multiplying the Fourier series (1) by 1, cos nx, and sin nx, integrating the resulting equality w.r.t. x from –π to π, and taking into account Theorem 3.

Even and odd functions ۞ A function f(x) is said to be even if, for all x, f(–x) = f(x). ۞ A function f(x) is said to be odd if, for all x, f(–x) = – f(x).

g1(x) g2(x) is even, h1(x) h2(x) is even, g(x) h(x) is odd. Properties of even and odd functions: 1. Let g(x) and h(x) be an even and even functions. Then, g1(x) g2(x) is even, h1(x) h2(x) is even, g(x) h(x) is odd. 2.

Theorem 5: If f(x) is even, then bn = 0 for all n. If f(x) is odd, then a0 = 0 and an = 0 for all n. Proof: This theorem follows from formulae (2) and the properties of even and odd functions.

Example 2: the ‘square wave’ Consider a function defined by Sketch the graph of f(x) for x  [– 4π, 6π], and find its Fourier coefficients. f(x) is odd => a0 = an = 0.

To find bn, consider Since f(x) and sin nx are both odd, the integrand as a whole is even – hence, hence, substituting f(x),

hence, evaluating the integral,

Thus, hence, (3) Comment: Observe that our main theorem, Theorem 2, was formulated for a continuous function – whereas the ‘square wave’ is discontinuous at the points x = π × integer. So, what are the consequences?...

Consequence no. 1: the Gibbs phenomenon If a function involves ‘jumps’, the convergence of the Fourier series near these is not uniform. To illustrate this, introduce the so-called partial sums, and plot them on the same graph...

Consequence no. 2: the (possibly) wrong values at jumps Change the ‘square wave’ at a single point, x = 0, as follows (4)

Since Fourier coefficients are integral characteristics of f(x), they remain the same. Then, for x = 0, (3) yields whereas (4) yields Generally, if f(x) has a jump at x = x0, i.e. then

Theorem 5: The Fourier series of a continuous 2L-periodic function f(x) is where

Comment: Physically, Fourier series can be interpreted as an acoustic signal comprising: a constant component (the atmospheric pressure) described by a0, the main tone with the frequency ω = 2π/P and the amplitude [ (a1)2 + (b1)2 ]1/2 , an infinite series of overtones with frequencies 2ω, 3ω, etc. and the amplitudes [ (a2)2 + (b2)2 ]1/2, [ (a3)2 + (b3)2 ]1/2, etc.

For the square wave with P = 1/60 seconds (frequency 60 Hz), it is illustrated here: