1. 1 Review of Fourier series Let S P be the space of all continuous periodic functions of a period P. We’ll also define an inner (scalar) product by where L = P/2. 1. The definition and basics 2. Solving PDEs using Fourier series 1. The definition and basics Week 3

2. 2 Theorem 1: The (infinite) set of functions forms a basis in S 2π, i.e. any continuous 2π -periodic function f(x) can be represented as (1) is called the Fourier series of f(x). The constants a 0, a 1, a 2... b 1, b 2... are called the Fourier coefficients of f(x). They are, essentially, the components of f(x) with respect to the above basis. (1)

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

4. 4 Comment: The basis defined in Theorem 2 is orthogonal but not orthonormal, e.g.

5. 5 Theorem 3: The Fourier coefficients are given by (2) Proof: These three formulae 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 2.

6. 6 1.If f(x) is even, i.e. f(–x) = f(x), then b n = 0 for all n. 2.If f(x) is odd, i.e. f(–x) = –f(x), then a 0 = 0 and a n = 0 for all n. The proof: This theorem follows from formulae (2). Theorem 4:

7. 7 Example 1: the ‘square wave’ Consider a function defined by Theorem 4: Draw the graph of f(x) for –4π < x < 6π, and find its Fourier coefficients.

8. 8 f(x) is odd => a 0 = a n = 0. Since f(x) and sin nx are both odd, the integrand as a whole is even – hence, hence, substituting f(x), To find b n, consider

9. 9 hence, hence, evaluating the integral,

10. 10 Thus, hence, Comment: (3) Observe that our main theorem, Theorem 1, was formulated for a continuous function – whereas the ‘square wave’ is discontinuous at the point such that x = π × integer. So, what are the consequences?...

11. 11 Consequence no. 1: the Gibbs phenomenon. If a function has ‘jumps’, the convergence of the Fourier series near them is not uniform. and plot them on the same graph. To illustrate this, introduce the so-called partial sums,

12. 12

13. 13 Consequence no. 2:the (possibly) wrong values at jumps. Change the ‘square wave’ at a single point, x = 0, as follows Since Fourier coefficients are integral characteristics of f(x), they remain the same, and so does the whole FS (3). Then, for x = 0, (3) yields (4) but (4) yields

14. 14 Generally, if f(x) has a jump at x = x 0, i.e. then

15. 15 Theorem 5: The Fourier series of a continuous 2L -periodic function f(x) is where the Fourier coefficients are

16. 16 Comment: Physically, Fourier series can be interpreted as an acoustic signal comprising:  a constant component (atmospheric pressure) described by a 0 ;  the main tone with the frequency ω = 2π/P and the amplitude [ (a 1 ) 2 + (b 1 ) 2 ] 1/2 ;  an infinite series of overtones with frequencies 2ω, 3ω, etc. and the amplitudes [ (a 2 ) 2 + (b 2 ) 2 ] 1/2, [ (a 3 ) 2 + (b 3 ) 2 ] 1/2, etc.

17. 17 For the square wave with P = 1 /60 s (frequency 60 Hz ), it is illustrated here: