0 Motivating Example: Fourier Series

0.1  Fourier series: basic notions

Before proceed with an abstract theory we consider a motivating example: Fourier series.

0.1.1 2π-periodic functions

In this part of the course we deal with functions (as above) that are periodic.

We say a function f:ℝ→ℂ is periodic with period T>0 if f(x+T)= f(x) for all x∈ ℝ. For example, sinx, cosx, e^ix(=cos x+i sinx) are periodic with period 2π. For k∈ R∖{0}, sinkx, coskx, and e^ikx are periodic with period 2π/|k|. Constant functions are periodic with period T, for any T>0. We shall specialize to periodic functions with period 2π: we call them 2π-periodic functions, for short. Note that cosnx, sinnx and e^inx are 2π-periodic for n∈ℤ. (Of course these are also 2π/|n|-periodic.)

Any half-open interval of length T is a fundamental domain of a periodic function f of period T. Once you know the values of f on the fundamental domain, you know them everywhere, because any point x in ℝ can be written uniquely as x=w+nT where n∈ ℤ and w is in the fundamental domain. Thus f(x) = f(w+(n−1)T +T)=⋯ =f(w+T) =f(w).

For 2π-periodic functions, we shall usually take the fundamental domain to be ]−π, π]. By abuse of language, we shall sometimes refer to [−π, π] as the fundamental domain. We then have to be aware that f(π)=f(−π).

0.1.2 Integrating the complex exponential function

We shall need to calculate ∫_a^b e^ikx d x, for k∈ℝ. Note first that when k=0, the integrand is the constant function 1, so the result is b−a. For non-zero k, ∫_a^b e^ikx d x= ∫_a^b (coskx+isinkx) d x = (1/k)[ (sinkx − icoskx)]_a^b = (1/ik)[(coskx+isinkx)]_a^b = (1/ik)[e^ikx]_a^b = (1/ik)(e^ikb−e^ika). Note that this is exactly the result you would have got by treating i as a real constant and using the usual formula for integrating e^ax. Note also that the cases k=0 and k≠0 have to be treated separately: this is typical.

This follows immediately from Ex. 2 and Prop.4.

For real-valued functions, the introduction of complex exponentials seems artificial: indeed they can be avoided as follows. We work with (1) in the case of a finite sum: then we can rearrange the sum as

Here

for n>0 and

for n>0. a₀ = 1/π∫_−π^πf(x) d x, the constant chosen for consistency.

The a_n and b_n are also called Fourier coefficients: if it is necessary to distinguish them, we may call them Fourier cosine and sine coefficients, respectively.

We note that if f is real-valued, then the a_n and b_n are real numbers and so ℜ f(n) = ℜ f(−n), ℑ f(−n) = −ℑf(n): thus f(−n) is the complex conjugate of f(n). Further, if f is an even function then all the sine coefficients are 0 and if f is an odd function, all the cosine coefficients are zero. We note further that the sine and cosine coefficients of the functions coskx and sinkx themselves have a particularly simple form: a_k=1 in the first case and b_k=1 in the second. All the rest are zero.

For example, we should expect the 2π-periodic function whose value on ]−π,π] is x to have just sine coefficients: indeed this is the case: a_n=0 and b_n=i(f(n)−f(−n)) = (−1)ⁿ⁺¹2/n for n>0.

The above question can then be reformulated as “to what extent is f(x) represented by the Fourier series a₀/2 + ∑_n>0(a_ncosx + b_nsinx)?” For instance how well does ∑(−1)ⁿ⁺¹(2/n)sinnx represent the 2π-periodic sawtooth function f whose value on ]−π, π] is given by f(x) = x. The easy points are x=0, x=π, where the terms are identically zero. This gives the ‘wrong’ value for x=π, but, if we look at the periodic function near π, we see that it jumps from π to −π, so perhaps the mean of those values isn’t a bad value for the series to converge to. We could conclude that we had defined the function incorrectly to begin with and that its value at the points (2n+1)π should have been zero anyway. In fact one can show (ref. ) that the Fourier series converges at all other points to the given values of f, but I shan’t include the proof in this course. The convergence is not at all uniform (it can’t be, because the partial sums are continuous functions, but the limit is discontinuous.) In particular we get the expansion

which can also be deduced from the Taylor series for tan⁻¹.

0.2 The vibrating string

In this subsection we shall discuss the formal solutions of the wave equation in a special case which Fourier dealt with in his work.

We discuss the wave equation

subject to the boundary conditions

for all t≥0, and the initial conditions

This is a mathematical model of a string on a musical instrument (guitar, harp, violin) which is of length π and is plucked, i.e. held in the shape F(x) and released at time t=0. The constant K depends on the length, density and tension of the string. We shall derive the formal solution (that is, a solution which assumes existence and ignores questions of convergence or of domain of definition).

0.2.1 Separation of variables

We first look (as Fourier and others before him did) for solutions of the form y(x,t) = f(x)g(t). Feeding this into the wave equation (2) we get

and so, dividing by f(x)g(t), we have

The left-hand side is an expression in x alone, the right-hand side in t alone. The conclusion must be that they are both identically equal to the same constant C, say.

We have f^′′(x) −Cf(x) =0 subject to the condition f(0) = f(π) =0. Working through the method of solving linear second order differential equations tells you that the only solutions occur when C = −n² for some positive integer n and the corresponding solutions, up to constant multiples, are f(x) = sinnx.

Returning to equation (4) gives the equation g^′′(t)+K²n²g(t) =0 which has the general solution g(t) = a_ncosKnt + b_nsinKnt. Thus the solution we get through separation of variables, using the boundary conditions but ignoring the initial conditions, are

for n≥ 1.

0.2.2 Principle of Superposition

To get the general solution we just add together all the solutions we have got so far, thus

ignoring questions of convergence. (We can do this for a finite sum without difficulty because we are dealing with a linear differential equation: the iffy bit is to extend to an infinite sum.)

We now apply the initial condition y(x,0) = F(x) (note F has F(0) =F(π) =0). This gives

We apply the reflection trick: the right-hand side is a series of odd functions so if we extend F to a function G by reflection in the origin, giving

we have

for −π≤ x ≤ π.

If we multiply through by sinrx and integrate term by term, we get

so, assuming that this operation is valid, we find that the a_n are precisely the sine coefficients of G. (Those of you who took Real Analysis 2 last year may remember that a sufficient condition for integrating term-by -term is that the series which is integrated is itself uniformly convergent.)

If we now assume, further, that the right-hand side of (5) is differentiable (term by term) we differentiate with respect to t, and set t=0, to get

This equation is solved by the choice b_n=0 for all n, so we have the following result

0.3 Historic: Joseph Fourier

Joseph Fourier, Civil Servant, Egyptologist, and mathematician, was born in 1768 in Auxerre, France, son of a tailor. Debarred by birth from a career in the artillery, he was preparing to become a Benedictine monk (in order to be a teacher) when the French Revolution violently altered the course of history and Fourier’s life. He became president of the local revolutionary committee, was arrested during the Terror, but released at the fall of Robespierre.

Fourier then became a pupil at the Ecole Normale (the teachers’ academy) in Paris, studying under such great French mathematicians as Laplace and Lagrange. He became a teacher at the Ecole Polytechnique (the military academy).

He was ordered to serve as a scientist under Napoleon in Egypt. In 1801, Fourier returned to France to become Prefect of the Grenoble region. Among his most notable achievements in that office were the draining of some 20 thousand acres of swamps and the building of a new road across the alps.

During that time he wrote an important survey of Egyptian history (“a masterpiece and a turning point in the subject”).

In 1804 Fourier started the study of the theory of heat conduction, in the course of which he systematically used the sine-and-cosine series which are named after him. At the end of 1807, he submitted a memoir on this work to the Academy of Science. The memoir proved controversial both in terms of his use of Fourier series and of his derivation of the heat equation and was not accepted at that stage. He was able to resubmit a revised version in 1811: this had several important new features, including the introduction of the Fourier transform. With this version of his memoir, he won the Academy’s prize in mathematics. In 1817, Fourier was finally elected to the Academy of Sciences and in 1822 his 1811 memoir was published as “Théorie de la Chaleur”.

For more details see Fourier Analysis by T.W. Körner, 475-480 and for even more, see the biography by J. Herivel Joseph Fourier: the man and the physicist.

What is Fourier analysis. The idea is to analyse functions (into sine and cosines or, equivalently, complex exponentials) to find the underlying frequencies, their strengths (and phases) and, where possible, to see if they can be recombined (synthesis) into the original function. The answers will depend on the original properties of the functions, which often come from physics (heat, electronic or sound waves). This course will give basically a mathematical treatment and so will be interested in mathematical classes of functions (continuity, differentiability properties).