FOURIER ANALYSIS ERIK LØW AND RAGNAR WINTHER 1. The best approximation onto trigonometric polynomials Before we start the discussion of Fourier series we will review some basic results on inner–product spaces and orthogonal projections mostly presented in Section 4.6 of [1]. 1.1. Inner–product spaces. Let V be an inner–product space. As usual we let hu, vi denote the inner–product of u and v. The corresponding norm is given by p kvk = hv, vi. A basic relation between the inner–product and the norm in an inner– product space is the Cauchy–Scwarz inequality. It simply states that the absolute value of the inner–product of u and v is bounded by the product of the corresponding norms, i.e. (1.1) |hu, vi| ≤ kuk kvk. An outline of a proof of this fundamental inequality, when V = Rn and k · k is the standard Eucledian norm, is given in Exercise 24 of Section 2.7 of [1]. We will give a proof in the general case at the end of this section. Let W be an n dimensional subspace of V and let P : V 7→ W be the corresponding projection operator, i.e. if v ∈ V then w ∗ = P v ∈ W is the element in W which is closest to v. In other words, kv − w ∗ k ≤ kv − wk for all w ∈ W. It follows from Theorem 12 of Chapter 4 of [1] that w ∗ is characterized by the conditions (1.2) hv − P v, wi = hv − w ∗ , wi = 0 for all w ∈ W. In other words, the error v − P v is orthogonal to all elements in W . It is a consequence of the characterization (1.2) and Cauchy–Schwarz inequality (1.1) that the norm of P v is bounded by the norm of v, i.e. (1.3) kP vk ≤ kvk for all v ∈ V. To see this simply take w = w ∗ in (1.2) to obtain kw ∗ k2 = hw ∗ , w ∗ i = hv, w ∗i ≤ kvk kw ∗k, Notes written for for Mat 120B, Fall 2001, Preliminary version. 1 or kw ∗ k ≤ kvk. Hence, since P v = w ∗ , we established the bound (1.3). Let {u1 , u2 , . . . , un } be an orthogonal basis of the subspace W . Such an orthogonal basis can be used to give an explicit representation of the projection P v of v. It follows from Theorem 13 of Chapter 4 of [1] that P v is given by (1.4) Pv = n X cj u j where the coefficients cj = j=1 hv, uj i . kuj k2 ¿From the orthogonal basis we can also derive an expression for the norm of P v. In fact, we have (1.5) 2 kP vk = n X j=1 c2j kuj k2 . This follows more or less directly from the orthogonality property of the basis {u1 , u2 , . . . , un }. We have kP vk2 = hP v, P vi n n X X cj u j , ck u k i =h = = j=1 n X n X k=1 cj ck huj , uk i j=1 k=1 n X c2j kuj k2 . j=1 The situation just described is very general. Some more concrete examples using orthogonal basises to compute projections are given Section 4.6 of [1]. Fourier analysis is another very important example which fits into the general framework described above, where V is a space of functions and W is a space of trigonometric polynomials. The Fourier series correspons to orthogonal projections of a given function onto the trigonometric polynomials, and the basic formulas of Fourier series can be derived as special examples of general discussion given above. Proof of Cauchy–Schwarz inequality (1.1). If v = 0 we have zero on both sides of (1.1). Hence, (1.1) holds in this case. Therefore, we can assume that v 6= 0 in the rest of the proof. For all t ∈ R we have ku − tvk2 ≥ 0. 2 However, ku − tvk2 = hu − tv, u − tvi = hu, ui − thu, vi − thv, ui + t2 hvv, vi = kuk2 − 2thu, vi + t2 kvk2 . Taking t = hu, vi/kvk2 we therefor obtain 0 ≤ ku − tvk2 = kuk2 − or hu, vi2 kvk2 hu, vi2 ≤ kuk2 kvk2 . By taking square roots we obtain (1.1). 1.2. Fourier series. A trigonometric polynomial of order m is a function of t of the form m X p(t) = a0 + (ak cos kt + bk sin kt), k=1 where the coefficients a0 , a1 , . . . , am , b1 , . . . , bm are real numbers. Hence, trigonometric polynomials of order zero are simply all constant functions, while first order trigonometric polynomials are functions of the form p(t) = a0 + a1 cos t + b1 sin t. A function f (t) is called periodic with period T if f (t) = f (t + T ) for all t. Such a function is uniquely determined by its values in the interval [−T /2, T /2] or any other interval of length T . The trigonometric polynomials are periodic with period 2π. Hence we can regard them as elements of the space C[−π, π]. The space of trigonometric polynomials of order m will be denoted by Tm . More precisely, m X (ak cos kt+bk sin kt), ak , bk ∈ R} Tm = {p ∈ C[−π, π] : p(t) = a0 + k=1 C[−π, π] is equipped with a natural inner product Z π hf, gi = f (t)g(t)dt −π The norm is then given by Z ||f || = ( π f 2 (t)dt)1/2 −π 2 We call this the L -norm of f on [−π, π]. Any periodic function can be regarded as a 2π-periodic function by a simple change of variable. Hence everything that follows can be applied to general periodic functions. 3 It is easy to see that the constant function 1, together with the functions sin(kt) and cos(kt), 1 ≤ k ≤ m constitute an orthogonal basis for Tm . To prove this, it is sufficient to prove that for all integers j, k the following identities hold: Z π sin(jt) sin(kt) dt = 0 j 6= k, −π Z π cos(jt) cos(kt) dt = 0 j 6= k, −π Z π cos(jt) sin(kt) dt = 0. −π Notice that setting j = 0 the cos(jt) factor becomes the constant 1. To prove the first identity, we use the trigonometric formula 1 sin(u) sin(v) = (cos(u − v) − cos(u + v)). 2 ¿From this identity we obtain for j 6= k, using the fact that sin(lπ) = 0 for all integers l, that Z Z π 1 π (cos((j − k)t) − cos(j + k)t) dt sin(jt) sin(kt) dt = 2 −π −π 1 1 = sin((j − k)t) − sin((j + k)t) |π−π 2(j − k) 2(j + k) = 0. The two other equalities follow in a similar fashion. Note that we can also compute the norm of these functions using the same equation. Clearly the norm of the constant function 1 is (2π)1/2 . Setting j = k in the integrals above yields Z π Z 1 π 2 sin (kt) dt = (1 − cos(2kt) dt 2 −π −π t 1 = − sin(2kt) |π−π 2 4k = π. (This also follows easily from the fact that sin2 t + cos2 t = 1, hence both of these functions have average value 1/2 over a whole period.) Hence the norm of sin(kt) and cos(kt) equals π 1/2 . The projection of a function f ∈ C[−π, π] onto Tm is the best approximation in L2 -norm of f by a trigonometric polynomial of degree m and is denoted by Sm (t). Notice that Sm depends on the function f , although this is suppressed in the notation. By (1.4) the coefficients 4 are given by the formulae Z π 1 f (t) dt a0 = 2π −π Z 1 π ak = f (t) cos kt dt π −π Z 1 π bk = f (t) sin kt dt π −π A central question in Fourier analysis is whether or not the approximations Sm (t) converge to f (t),i.e. if formula ∞ X f (t) = a0 + (ak cos kt + bk sin kt) k=1 holds. The series on the right is called the Fourier series of f , whether it converges to f or not.In Figure 1 we see the graphs of f (t) = t and Sm (t) for m = 1, 3, 5, 10. x x 4 4 3 3 2 2 1 1 0 0 −1 −1 −2 −2 −3 −3 −4 −4 −3 −2 −1 0 x 1 2 3 −3 −2 −1 x 0 x 1 2 3 1 2 3 x 4 4 3 3 2 2 1 1 0 0 −1 −1 −2 −2 −3 −3 −4 −4 −3 −2 −1 0 x 1 2 3 −3 −2 −1 0 x Figure 1. Fourier approximations to f (t) = t There are three things to observe from these curves. First, the functions Sm (t) do get closer to f (t) as m incresases, at least in a somewhat smaller interval. Second, at the endpoints it is not possible for Sm to converge to f , since Sm is 2π-periodic and hence has the same values at these enpoints. In this case the value is 0. Finally, close to the endpoints there are blips which do not approach zero, although they do get closer to the endpoints. This is called the Gibbs phenomenon 5 and again is related to the fact that f does not have the same values at the endpoints. We do not investigate this phenomenon further. In Figure 2 we see the corresponding curves for f (t) = t2 . This function has the same values at the endpoints. Notice that the blips have dissappeared and the convergence is faster, but it is still somewhat slower at the endpoints. x.2 x.2 12 12 10 10 8 8 6 6 4 4 2 2 0 0 −3 −2 −1 0 x 1 2 3 −3 −2 −1 x.2 0 x 1 2 3 1 2 3 x.2 12 12 10 10 8 8 6 6 4 4 2 2 0 0 −3 −2 −1 0 x 1 2 3 −3 −2 −1 0 x Figure 2. Fourier approximations to f (t) = t2 1.3. Exercises. 1.Determine the Fourier coefficients of f (t) = et 2. a) A function f is called even if f (−t) = f (t) for all t and odd if f (−t) = −f (t). Which polynomials are even or odd ? b)If f ∈ C[−π, π] is even, prove that bk = 0 for all k > 0. If f is odd, prove that ak = 0 for all k ≥ 0. c)Prove that any f may be written as a sum of an even and an odd function. (Hint : Let fe (t) = (f (t) + f (−t))/2 and fo (t) = (f (t) − f (−t))/2. d)Determine the Fourier coefficients of sinht and cosht. 3.a) If p is an odd polynomial, prove that Z π Z 1 π ′′ k+1 2 p(t)sin(kt) dt = (−1) p(π) − 2 p (t)sin(kt) dt k k −π −π b) If p is an even polynomial, prove that Z π Z 1 π ′′ k 2 ′ p(t)cos(kt) dt = (−1) 2 p (π) − 2 p (t)cos(kt) dt k k −π −π 6 c) Find the Fourier coefficients of f (t) = t3 + t2 . 4.The curves in the figures above were made by using the Matlab toolbox fourgraph. Download fourgraph from http://www.mathworks.com (for instance by writing fourgraph in the search box). Reproduce the curves above and try some other functions, for instance et or t3 + t2 . 2. Trigonometric polynomials 2.1. The complex exponential function. The real exponential function ex can be extended to complex values of the argument. Let z = s + it be a complex number. If the complex exponential function satisfies the usual product rule, we must have ez = es+it = es eit Hence we must define the complex function f (t) = eit . If f satisfies the usual chain rule for differentiation of the exponential function, we ′ must have f (t) = ieit = if (t). Hence, if g(t) and h(t) denote the real and imaginary parts of f (t), i.e. f (t) = g(t) + ih(t), we get ′ ′ g (t) + ih (t) = i(g(t) + ih(t)) and we must have ′ h (t) = g(t) ′ g (t) = −h(t) which gives ′′ ′ g (t) = −h (t) = −g(t) ′′ ′ h (t) = g (t) = −h(t) Since we also must have f (0) = 1, this gives g(0) = 1 and h(0) = 0. The solutions to these equations are g(t) = cos t and h(t) = sin t.Hence we must have eit = cos t + i sin t. We therefore define the complex exponential function ez by (2.1) es+it = es (cos t + i sin t) The complex number eit = cos t + i sin t is located on the unit circle in the complex plane at the angle t with the real axis. It follows that the complex number w = es+it in the point in the complex plane whose length is es and angle is t. It follows from (2.1) that the complex exponential function satisfies the usual addition rule for the exponent: ez1 ez2 = ez1 +z2 From this we obtain the famous DeMoivre’s formula: (2.2) cos nt + i sin nt = (cos t + i sin t)n since both sides equal eint = (eit )n . This formula contains the formulas for cos nt and sin nt as functions of cos t and sin t and is the most 7 efficient way of expressing these formulas. It is also the easiest to remember. If f is a real or complex function defined on the unit circle in the complex plane, then f (eit ), as a function of t, is 2π-periodic. Conversely, given any function f of t that is 2π-periodic, we may think of f as being defined on the unit circle. In this way we identify 2π-periodic functions with functions defined on the unit circle in the complex plane. Also, if f is a function defined on [−π, π), then f extends uniquely to a 2π-periodic function on the whole real line. This is actually true for any function defined on a (half-open) interval of length 2π. 2.2. Complex representation of trigonometric polynomials. In this section we shall write a trigonometric polynomial p ∈ Tm in the form m X 1 p(t) = a0 + (ak cos kt + bk sin kt). 2 k=1 1 2 Notice the factor in front of a0 . Hence if p(t) = 4 + cos t + sin t, then a0 = 8, a1 = 1 and b1 = 1. This convention will make formulas simpler later.( Notice also that if p is the projection of some 2π periodic functionRf onto Tm , then all the ak -coefficients are given by the formula π ak = π1 −π f (t) cos kt dt, also for k = 0.) By convention, we also set b0 = 0. For any complex number z = x + iy, the real and imaginary parts are given by 1 ℜz = (z + z̄) 2 1 ℑz = (z − z̄) 2i Applying this to z = eikt , we see that (2.3) 1 1 cos kt = (eikt + e−ikt ) , sin kt = (eikt − e−ikt ) 2 2i so m X 1 (ak cos kt + bk sin kt) p(t) = a0 + 2 k=1 m X 1 1 1 = a0 + ak (eikt + e−ikt ) + bk (eikt − e−ikt ) 2 2 2i k=1 m X 1 1 1 (ak − ibk )eikt + (ak + ibk )e−ikt = a0 + 2 2 2 k=1 Hence if we define (2.4) 1 1 ck = (ak − ibk ), c−k = (ak + ibk ) 2 2 8 we see that we can write this as (2.5) p(t) = m X ck eikt . k=−m The constants satisfy c−k = c¯k . This will be called the complex representation of the trigonometric polynomial p(t) Conversely, any function of the form (2.5) is a real trigonometric polynomial, provided the constants satisfy c−k = c¯k . The coefficients are then given by ak = ck + c−k bk = i(ck − c−k ) On the unit circle z = eit , hence if we consider p as a function on the unit circle, we have (2.6) p(z) = m X ck z k k=−m This shows the connection between real trigonometric polynomials and complex polynomials (if we admit negative powers in a polynomial). P k By P2m+1 we denote the set of complex functions of the form m k=−m ck z . We have shown that Tm is naturally isomorphic to the real subspace of P2m+1 defined by the equations c−k = c¯k for k = 0, 1, .., m.If we had allowed complex values for the coefficients ak , bk in Tm , i.e. considered the complex vector space CTm of complex trigonometric polynomials, the argument above shows that CTm and P2m+1 are isomorphic as complex vector spaces. Finally, if Sm (t) is the projection of a function f onto Tm , then the coefficients of the complex representation of Sm (t) are given by the one formula Z π 1 (2.7) ck = f (t)e−ikt dt 2π −π These coefficients are called the complex Fourier coefficients of f . 2.3. Sampling of functions in Tm . The vector space Tm is a real vector space of dimension 2m + 1. It is therefore reasonable to expect that a trigonometric polynomial p ∈ Tm is uniquely determined by its values at 2m + 1 points. We shall see that this is the case if we divide the interval [0, 2π] into 2m + 1 equally long pieces, i.e. consider the points 2π tj = j , j = 0, 1, · · · , 2m 2m + 1 Furthermore, we shall determine the precise relationship between the coefficients a0 , a1 , · · · , am , b1 , · · · , bm and the sampled values vj = p(tj ), j = 0, 1, · · · , 2m. 9 To do this, we consider the evaluation map L : Tm → R2m+1 defined by L(p) = v where the components of v are defined by evaluating p at tj , i.e. vj = p(tj ). The colums of the matrix A of L with respect to the basis {1, cos kt, sin kt} of Tm and the standard basis of R2m+1 are given by: 1 1 u0 = L(1) = ... 1 uk = L(cos kt) = cos kj 2π , j = 0, 1, · · · , 2m 2m + 1 2π vk = L(sin kt) = sin kj , j = 0, 1, · · · , 2m 2m + 1 for k = 1, 2, · · · , m. We shall prove that the vectors u0 , · · · , um , v1 , · · · , vm are orthogonal, hence form an orthogonal basis for R2m+1 . To prove this, we shall need the following formula: (2.8) N X cos jα = N X j=0 j=0 ℜ(eijα ) = ℜ N X j=0 (eiα )j = ℜ( 1 − ei(N +1)α ) 1 − eiα where we have used the formula for the sum of a (finite) geometric series. Similarly, we also have N X (2.9) j=0 sin jα = ℑ( 1 − ei(N +1)α ) 1 − eiα We shall prove that the vectors v1 , · · · , vm are orthogonal. The argument for the u-vectors is similar and also that the u’s and v’s are ′ mutually orthogonal. Therefore let k and k be two indices. The usual trig formula for the product of two sines then gives hvk , vk′ i = 2m X sin kj j=0 2π 2π ′ sin k j 2m + 1 2m + 1 2m = ′ 2m ′ 2π(k − k ) 1 X 2π(k + k ) 1X − cos j cos j 2 j=0 2m + 1 2 j=0 2m + 1 Setting N = 2m and α = ′ 2π(k+k ) 2m+1 we have ′ 1 − ei(N +1)α = 1 − ei2π(k+k ) = 0 10 , hence the second sum is zero by (2.8). Also, the first sum is zero for ′ ′ for k = k . This show that the vectors all are k 6= k and equal to 2m+1 2 orthogonal and the norms are given by ku0 k2 = 2m + 1, 2m + 1 2 Hence A is an ortogonal matrix whose inverse is given by 1 T u 2 0 uT1 .. . 2 T A−1 = u 2m + 1 vm 1T . .. T vm kuk k2 = kvk k2 = In other words, if p is a trigonometric polynomial whose values vj = p(tj ) are given by v = [vj ], then the coefficients of p are given by 1 hu0 , vi, 2m + 1 2 huk , vi, 1 ≤ k ≤ m, ak = 2m + 1 2 bk = hvk , vi, 1 ≤ k ≤ m 2m + 1 a0 = 2.4. Signal processing. The terms in a trigonometric polynomial have different frequencies. In figure 1 the top curve is sin t and the middle curve is sin 5t. We see that the middle curve oscillates five times as fast as the top curve, i.e. the frequency is five times higher. So the constant k that appears in the terms sin kt and cos kt measures the frequency of the oscillation. If we draw the graph of a trigonometric polynomial p(t), then we call the curve the time representation of p, or the representation of p in the time domain. The bottom curve is p(t) = sin t−0.5 sin 2t+sin 5t. This curve therefore is the representation of p in the time domain. A trigonometric polynomial, however, is defined by p(t) = a0 + m X (ak cos kt + bk sin kt). k=1 We call this the representation of p in the frequency domain because this representation explicitly states how much (specified by the constants ak and bk ) p contains of oscillations with frequency k. In the bottom curve, p contains the frequencies 1, 2 and 5, or more explicitly, b1 = 1, b2 = −0.5 and b5 = 1, while all the other a’s and b’s are 0. 11 sin(x) 1 0.5 0 −0.5 −1 −3 −2 −1 0 x 1 2 3 1 2 3 2 3 sin(5 x) 1 0.5 0 −0.5 −1 −3 −2 −1 0 x sin(x)−0.5 sin(2 x)+sin(5 x) 2 1.5 1 0.5 0 −0.5 −1 −1.5 −2 −3 −2 −1 0 x 1 Figure 3. Time representations of trigonometric polynomials We saw in the previous paragraph that p also was caracterized by its values at regularly spaced points. We may think of this as a discrete time representation of p. If the points are chosen very close (i.e. m is chosen big), then these points will trace out the graph. We also saw that one could pass from the frequency representation to the time representation and back. This is the central theme of Fourier analysis; that functions may be described either in the time domain (through its 12 graph) or in the frequency domain, by breaking the function down into its frequency components (i.e. components of different frequencies). It follows from the orthogonality of the basis {1, cos kt, sin kt} that the projection of Tm onto Tn for some n < m simply consists of omitting the highest frequencies, i.e. m n X X P (a0 + ak cos kt + bk sin kt) = a0 + ak cos kt + bk sin kt k=1 k=1 2 So, the best (L ) approximation of a trigonometric polynomial of degree m by a trigonometric polynomial of degree n is simply the degree n part of the trigonometric polynomial. 2.5. Convolution in Tm . If f and g are 2π-periodic functions, we define the convolution of f and g by Z π 1 f ∗ g(t) = f (s)g(t − s) ds 2π −π It follows that we can actually perform the inegration over any interval of length 2π and that f ∗ g also is 2π-periodic and f ∗ g = g ∗ f . The functions can be real or complex. If f, g ∈ Tm , the convolution is most easily described using the complex Fourier representation. Therefore, ′ assume that f (t) = eikt and g(t) = eik t . Then Z π ′ 1 f ∗ g(t) = eiks eik (t−s) ds 2π −π Z π ′ ′ 1 eik t ei(k−k )s ds = 2π −π Z ′ 1 ik′ t π i(k−k′ )s 0 if k 6= k = e ds = e ′ eikt if k = k 2π −π Pm P ′ ikt ikt c e and f = Hence, if f = m k k=−m ck e , then k=−m f ∗ g(t) = m X ′ ck ck eikt k=−m Hence the complex Fourier coefficients of the convolution of f and g equal the product of the Fourier coefficients of f and g. In other words, convolution in the time domain corresponds to pointwise products in the frequency domain. If we project a trigonometric polynomial p ∈ Tm to Tn , then we simply omit the terms of degree greater than n, i.e. we multiply the coefficients ck by 1 for |k| ≤ n and by 0 for |k| > n. By the discussion above, this means that we convolve p with the function n X eikt (2.10) Dn (t) = k=−n 13 which is called the Dirichlet kernel (of degree n). This is a finite geometric series, whose sum is ei(2n+1)t − 1 eit − 1 ei(n+1)t − e−int = eit − 1 1 1 ei(n+ 2 )t − e−i(n+ 2 )t = t t ei 2 − e−i 2 sin(n + 12 )t = sin 12 t Dn (t) = e−int by (2.3).In Figure 2 we see the Dirichlet kernel for n = 1, 3, 5, 10. sin(3.5 x)/sin(0.5 x) sin(1.5 x)/sin(0.5 x) 7 3 6 2.5 5 2 4 1.5 3 1 2 0.5 1 0 0 −0.5 −1 −1 −2 −3 −2 −1 0 x 1 2 −3 3 −2 −1 0 x 1 2 3 1 2 3 sin(10.5 x)/sin(0.5 x) sin(5.5 x)/sin(0.5 x) 20 10 8 15 6 10 4 5 2 0 0 −2 −3 −2 −1 0 x 1 2 −3 3 −2 −1 0 x Figure 4. Dirichlet kernel for n = 1, 3, 5, 10 Rπ Notice that the peaks are higher and higher. We have −π Dn (t)dt = 2π for all n. From the figures above, we see that as n increases Dn (t) more and more concentrates a mass (i.e. integral) of 2π at the origin. 14 If f is a 2π-periodic function, we have Sm (t) = m X ck eikt k=−m m X Z π 1 = ( f (s)e−iks ds)eikt 2π −π k=−m Z π m X 1 f (s)( eik(t−s) ) ds = 2π −π k=−m Z π 1 = f (s)Dm (t − s) ds 2π −π = f ∗ Dm (t) We shall see later that if f is differentiable, this converges to f . This 1 is easily understood from the fact that for large m, 2π Dm (t − s) concentrates a mass of 1 at s=t. 2.6. Exercises. 1.Show that the complex Fourier coefficients of an even/odd function are real/pure imaginary. Use exercise 3c) in Chapter 1 to find the complex Fourier coefficients of f (t) = t3 + t2 2 a)For m = 1, determine the 3x3 matrix A of the evaluation map L : T1 → R3 described in this chapter. ) = 1 and p( 2π ) = 4. b) Determine p ∈ T1 such that p(0) = p( 4π 3 3 c)Using MATLAB, compute the 5x5 matrix A for L : T2 → R5 . Use this to find a trigonometric polynomial p ∈ T2 such that p(0) = p( 6π )= 5 4π 8π ) = p( ) = p( ) = 1. (Plot your curve to check the result, 0 and p( 2π 5 5 5 for instance using fourgraph). 3. Let f be the 2π-periodic function whose values in [−π, π) is given by f (t) = t. a)Determine f ∗ f and draw the graph. (Requires some work!) b)Determine the complex Fourier coefficients of f . c)Determine Sm ∗ Sm for m = 1 and 2 and draw the graph. 4.(Parseval’s identity. Move to Chapter 1 next year!) If p(t) = a0 + Pm (a cos kt + bk sin kt), show that k=1 k Z π m 1X 2 1 2 2 |p(t)| dt = a0 + (ak + b2k ) 2π −π 2 k=1 5.Using fourgraph, plot the projection of f (t) = t6 − 13t4 + 36t2 − 2 onto T0 , T1 , T3 , T5 , T10 . 15 3. Orthogonal systems of functions The basic properties of the Fourier expansion of a function f discussed above is a consequence of the orthogonality properties of the basis functions cos(kx) and sin(kx), i.e. Z π cos(jx) cos(kx) dx = 0 j 6= k, −π Z π sin(jx) sin(kx) dx = 0 j 6= k, −π Z π cos(jx) sin(kx) dx = 0. −π In fact, as soon as we have a set of basis functions satisfying similar orthogonality properties, the a corresponding “generalized Fourier expansion” can be derived. A sequence of functions {φk (x)}∞ k=1 , defined on an interval [a, b], is referred to as an orthogonal system of functions if Z (3.1) a b φj (x)φk (x) dx = 0 j 6= k. Of course, the standard trigonometric basis functions fits into this set up if we let the interval [a, b] be [−π, π], φ2k+1 (x) = cos(kx) and φ2k (x) = sin(kx). However, there are many more examples of orthogonal systems. Example 3.1 Let the interval [a, b] be [−π, π] and φk (x) = sin(kx), k ≥ 1. Then the system is orthogonal, since we already know that Z π sin(jx) sin(kx) dx = 0 j 6= k. −π Example 3.2 Let the interval [a, b] be [0, π] and φk (x) = sin(kx), k ≥ 1. Note that we have changed the interval of definition. In order to show that the system is orthogonal we have to show that Z π sin(jx) sin(kx) dx = 0 j 6= k. 0 To show this we use the identity 1 sin(u) sin(v) = (cos(u − v) − cos(u + v)). 2 16 ¿From this identity we obtain for j 6= k, using the fact that sin(kπ) = 0 for all integers k, that Z π Z 1 π sin(jx) sin(kx) dx = (cos((j − k)x) − cos(j + k)x) dx 2 0 0 1 1 sin((j − k)π) − sin((j + k)π) = 2(j − k) 2(j + k) = 0. Hence, the system is orthogonal. Example 3.3 Let the interval [a, b] be [0, π] and φk (x) = sin((k + 21 )x), k ≥ 1. As above we have for j 6= k Z π Z π 1 1 φj (x)φk (x) dx = sin((j + )x) sin((k + )x) dx 2 2 0 0 Z π 1 = (cos((j − k)x) − cos(j + k + 1)x) dx 2 0 1 1 sin((j − k)π) − sin((j + k + 1)π) = 2(j − k) 2(j + k + 1) = 0. Therefore, the system is orthogonal. All the examples we have studied up to now are basesd on the trigonometric functions sin and cos. However, there also exists many other orthogonal systems, for example orthogonal polynomials. Example 3.4 The Legendre polynomials are orthogonal functions with respect to the interval [−1, 1]. For k ≥ 0 these polynomials are of the form d Lk (x) = αk ( )k (x2 − 1)k , dx d k ) is the derivative of order where αk is a suitable constant and ( dx k. For the discussion here we let αk = 1, even if other scalings, like αk = 1/2k k! is more standard. Note that (x2 − 1)k is a polynomial of degree 2k. By differentiating this function k times we therefore end up with a polynomial of degree k. We have therefore established that Lk (x) is a polynomial of degree k. In fact, from the definition above we can easily compute L0 (x) = 1, L1 (x) = 2x, L2 (x) = 12x2 − 4. Next we like to establish that the polynomials Lk are orthogonal, i.e. we like to show that Z 1 Lj (x)Lk (x) dx = 0 for j 6= k. −1 17 Consider first the polynomial L2 (x). The integral of this polynomial over [−1, 1] is given by Z 1 L2 (x) dx = (4x3 − 4x)|1−1 = 0. −1 Similarly, the first order moment of L2 (x) is given by Z 1 Z 1 xL2 (x) dx = (12x4 − 4x2 ) dx −1 −1 4 = (3x − 2x2 )|1−1 = 1 − 1 = 0. By combining these two results it follows that for arbritary constants a0 and a1 we have Z 1 Z 1 Z 1 (a0 + a1 x)L2 (x) dx = a0 L2 (x) dx + a1 xL2 (x) dx = 0. −1 −1 −1 In orther words, the quadratic polynomial L2 (x) is orthogonal to all linear polynomials. In particular, Z 1 Z 1 L2 (x)L1 (x) dx = L2 (x)L0 (x) dx = 0, −1 −1 i.e. L2 is orthogonal to L1 and L0 . A similar argument can be used to show that Lk is orthogonal to Lj for j < k. Since Lj is a polynomial of degree j it is enough to show that Z 1 xj Lk (x) dx = 0 for all j < k. −1 Note that the function (x2 − 1)k = (x − 1)k (x + 1)k has a root of order k at each endpoint ±1. Therefore, if i < k then d i 2 ) (x − 1)k = 0 for x = ±1. dx Integration by parts therefore gives Z 1 Z 1 d j x Lk (x) dx = xj ( )k (x2 − 1)k dx dx −1 −1 Z 1 d d j d k−1 2 k 1 = x ( ) (x − 1) )|−1 − ( xj )( )k−1 (x2 − 1)k dx dx dx −1 dx Z 1 d =− jxj−1 ( )k−1 (x2 − 1)k dx. dx −1 ( By repeating this argument k times we get Z 1 Z 1 d j k x Lk (x) dx = (−1) (( )k xj )(x2 − 1)k dx = 0, −1 −1 dx 18 where we have used that k th derivative of xj equals zero for j < k. Hence, we have shown that Lk is orthogonal to Lj for j < k, or, in other words, the sequence {Lk } is an orthogonal system of polynomials. Example 3.5 If {ψk (x)}∞ k=1 is any sequence of functions in C[a, b], then we can produce an orthogonal sequence {φk (x)}∞ k=1 by Gram-Schmidt orthogonalization : φ1 (x) = ψ1 (x) .. . Pn−1 hψn ,φk i φk (x) φn (x) = ψn (x) − k=1 kφk k2 Returning to the general situation, let {φk (x)}∞ k=1 be an orthogonal system of continuous functions defined on an interval [a, b]. For each integer n ≥ 1 we let Wn be the subspace of C[a, b] spanned by {φ1 , φ2 , . . . φn }, i.e. n X Wn = {w ∈ C[a, b] : w(x) = ak φk (x), ak ∈ R}. k=1 It is clear that W1 ⊂ W2 ⊂ . . . ⊂ Wn ⊂ Wn+1 . . . since an element in Wn corresponds to an element in Wn+1 with an+1 = 0. Let Pn : C[a, b] 7→ Wn be the orthogonal projection. It follows from (1.4) that (3.2) Pn f = n X ak φk , k=1 where the coefficients ak are given by Z b Z b 1 2 (3.3) ak = f (x)φk (x) dx and kφk k = (φk (x))2 dx. kφk k2 a a The finite series (3.2) is a generalized finite Fourier expansion of f , and the function Pn f is the best L2 -approximation of f by a function in Wn . We observe that, just as in the case of the ordinary Fourier expansion, the coefficients ak are independent of n. Hence, in order to compute Pn+1 f , when Pn f is known, all we have to do is to add the extra term an+1 φn+1 . The coefficients ak are called the generalized Fourier coefficients of f and the series ∞ X ak φk (x) k=1 is called the generalized Fourier series of f . 19 It also follows from (3.2) that 2 kPn f k = n X k=1 a2k kφk k2 . By combining this identity with (1.3) we obtain the inequality Z b n X 2 2 2 ak kφk k ≤ kf k = (f (x))2 dx, a k=1 where the coefficients ak are given by (3.3). We observe that the right hand side of this inequality is independent of n. Hence, the partial sums on the left hand side are bounded independent of n. Since all the terms in the series are positive it therefore follows that the inequality still holds with n = ∞, cf. Theorem 12.2.1 in [2]. In particular, the infinite series converges. This result is usually referred to as Bessel’s inequality. We state the result precisely as a theorem. Theorem 3.1 (Bessel’s inequality). If f ∈ C[a, b] and {φk }∞ k=1 is an orthogonal system of continuous functions on [a, b] then Z b ∞ X 2 2 ak kφk k ≤ (f (x))2 dx, (3.4) a k=1 where the generalized Fourier coefficients ak are given by (3.3). Since the infinite series (3.4) converges we must have, in particular, that the sequence {ak kφk k} converges to zero. This result will be used later. We therefore state it as a corollary. Corollary 3.1. Let f , {φk }∞ k=1 and {ak } be as in Theorem 3.1 above. Then ak kφk k −→ 0 as k → ∞. ∞ In particular, if {φk }k=1 is an orthonormal family (or more generally, kφk k > c for all k for some c > 0), then ak → 0. Example 3.6 If we consider the usual Fourier series ∞ X ak cos kt + bk sin kt a0 + k=1 of a continuous 2π-periodic function f (t) defined for all real t, then we have the following formula for the n-th partial sum : n X ak cos kt + bk sin kt Sn (t) = a0 + k=1 1 = f ∗ Dn (t) = 2π 20 Z π −π f (t − s)Dn (s) ds where Dn is the Dirichlet kernel Dn (s) = sin(n + 21 )s sin 21 s We have also shown that the integral of Dn is 2π, which gives Z π 1 f (t)Dn (s) ds f (t) = 2π −π Subtracting these two equations gives Z π 1 (f (t) − f (t − s))Dn (s) ds f (t) − Sn (t) = 2π −π Z π 1 f (t) − f (t − s) 1 = sin(n + )s ds 1 2π −π 2 sin 2 s 1 f (t)−f (t−s) The function g t (s) = 2π is continuous for s 6= 0. It is easy sin 21 s to see, using L’Hoptal’s rule, that it is also continuous for s = 0 if f is differentiable at t. Furthermore, the functions φn (s) = sin(n + 21 )s √ are orthogonal on [−π, π], by example 3.3, and kφn k = π. Hence the integral above is just the generalized Fourier coefficients of g t with respect to the orthogonal family φn , and therefore converges to zero, by Corollary 3.1. Theorem 3.2 (Convergence of Fourier series). If f is a 2π-periodic continuous function which is differentiable at t, then ∞ X ak cos kt + bk sin kt, f (t) = a0 + k=1 i.e. the Fourier series of f converges to f at t. Example 3.7 If we use the orthononal family sin nt in C[0, π] (see example 3.2), then the generalized Fourier series of f ∈ C[0, π] is called the sine series of f and is given by Z ∞ X 2 π bk sin kt and bk = (3.5) f (t) sin kt dt π 0 k=1 It will converge to f (t) for all t ∈ (0, π) where f is differentiable. Since the series is always zero at t = 0 and t = π, it will converge there if and only if f (0) = 0 (f (π) = 0). Example 3.8 It is also easy to see that the family {1, cos t, cos 2t, · · · } also is orthogonal in C[0, π] . In this case the generalized Fourier series of f ∈ C[0, π] is called the cosine series of f and is given by Rπ ∞ X aR0 = π1 0 f (t) dt ak cos kt and (3.6) a0 + π ak = π2 0 f (t) cos kt dt (k > 0) k=1 21 It will converge to f (t) for all t ∈ (0, π) where f is differentiable and at the endpoints if f has one-sided derivatives at both endpoints. ∞ 3.1. Exercises. 1. Let {φk (x)}∞ k=1 = {sin(kx)}k=1 be the orthogonal system studied in Example 3.1 above. a) Compute Pn f for f (x) = x. b) Compute Pn g for g(x) = 1. ∞ 2. Let {φk (x)}∞ k=1 = {sin(kx)}k=1 be the orthogonal system studied in Example 3.2 above, i.e. the functions are considered on the interval [0, π]. a) Compute Pn f for f (x) = x. b) Compute Pn g for g(x) = 1. 3. a) Let f (x) be the sign–function, i.e. f (x) = 1 for x > 0, f (x) = −1 for x < 0 and f (0) = 0. Compute the ordinary Fourier expansion of f and compare the result with the result of Exercise 2b above. b) Let f be a function defined on [0, π]. Show that the generalized Fourier expansion of f , with respect to the orthogonal system {sin(kx)}∞ k=1 , is exactly the ordinary Fourier expansion of the odd extension of f . 4. Let f (x) be the sign–function. Use Bessel’s inequality to show that ∞ X 1 π2 ( )2 ≤ . 2k − 1 8 k=1 5. Let {Lk (x)}∞ k=1 be the Legendre polynomials studied in Example 3.4. a) It can be shown that the polynomials Lk (x) satisfies the recurrence relation (3.7) Lk+1 (x) = 2(2k + 1)xLk (x) − 4k 2 Lk−1 (x) for k ≥ 1. Use this relation, and the fact that L0 (x) = 1 and L1 (x) = 2x, to compute the polynomials L2 (x), L3 (x) and L4 (x). b) Establish the relation (3.7). References [1] L.W. Johnson, R.D. Riess and J.T. Arnold Linear algebra, 4. edition, Addison Wesley 1998. [2] Tom Lindstrøm, Kalkulus, Universitetsforlaget 1995. 22

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