FOURIER ANALYSIS ERIK LØW AND RAGNAR WINTHER - PDF

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FOURIER ANALYSIS ERIK LØW AND RAGNAR WINTHER. The best approimation onto trigonometric polynomials Before we start the discussion of Fourier series we will review some basic results on inner product spaces

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FOURIER ANALYSIS ERIK LØW AND RAGNAR WINTHER. The best approimation 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 []... Inner product spaces. Let V be an inner product space. As usual we let u, v denote the inner product of u and v. The corresponding norm is given by v = v, v. 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. (.) u, v u v. An outline of a proof of this fundamental inequality, when V = R n and is the standard Eucledian norm, is given in Eercise 4 of Section.7 of []. 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 W be the corresponding projection operator, i.e. if v V then w = Pv W is the element in W which is closest to v. In other words, v w v w for all w W. It follows from Theorem of Chapter 4 of [] that w is characterized by the conditions (.) v Pv, w = v w, w = for all w W. In other words, the error v Pv is orthogonal to all elements in W. It is a consequence of the characterization (.) and Cauchy Schwarz inequality (.) that the norm of Pv is bounded by the norm of v, i.e. (.3) Pv v for all v V. To see this simply take w = w in (.) to obtain w = w, w = v, w v w, Notes written for for Mat B, Fall, Preliminary version. or w v. Hence, since Pv = w, we established the bound (.3). Let {u, u,...,u n } be an orthogonal basis of the subspace W. Such an orthogonal basis can be used to give an eplicit representation of the projection Pv of v. It follows from Theorem 3 of Chapter 4 of [] that Pv is given by (.4) Pv = j= c j u j where the coefficients c j = v, u j u j. From the orthogonal basis we can also derive an epression for the norm of Pv. In fact, we have (.5) Pv = c j u j. j= This follows more or less directly from the orthogonality property of the basis {u, u,..., u n }. We have Pv = Pv, Pv = c j u j, = = j= j= c k u k c j c k u j, u k c j u j. j= The situation just described is very general. Some more concrete eamples using orthogonal basises to compute projections are given Section 4.6 of []. Fourier analysis is another very important eample 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 eamples of general discussion given above. Proof of Cauchy Schwarz inequality (.). If v = we have zero on both sides of (.). Hence, (.) holds in this case. Therefore, we can assume that v in the rest of the proof. For all t R we have u tv. However, u tv = u tv, u tv = u, u t u, v t v, u + t vv, v = u t u, v + t v. Taking t = u, v / v we therefor obtain u tv = u u, v v or u, v u v. By taking square roots we obtain (.)... Fourier series. A trigonometric polynomial of order m is a function of t of the form p(t) = a + (a k cos kt + b k sin kt), where the coefficients a, a,...,a m, b,...,b m 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) = a + a cost + b 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/, T/] or any other interval of length T. The trigonometric polynomials are periodic with period π. Hence we can regard them as elements of the space C[, π]. The space of trigonometric polynomials of order m will be denoted by T m. More precisely, T m = {p C[, π] : p(t) = a + (a k cos kt+b k sin kt), a k, b k R} C[, π] is equipped with a natural inner product The norm is then given by f, g = f = ( f(t)g(t)dt f (t)dt) / We call this the L -norm of f on [, π]. Any periodic function can be regarded as a π-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, together with the functions sin(kt) and cos(kt), k m constitute an orthogonal basis for T m. To prove this, it is sufficient to prove that for all integers j, k the following identities hold: sin(jt) sin(kt) dt = j k, cos(jt) cos(kt) dt = j k, cos(jt) sin(kt) dt =. Notice that setting j = the cos(jt) factor becomes the constant. To prove the first identity, we use the trigonometric formula sin(u) sin(v) = (cos(u v) cos(u + v)). From this identity we obtain for j k, using the fact that sin(lπ) = for all integers l, that sin(jt) sin(kt) dt = (cos((j k)t) cos(j + k)t) dt = (j k) sin((j k)t) (j + k) sin((j + k)t) π =. 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 is (π) /. Setting j = k in the integrals above yields sin (kt) dt = ( cos(kt) dt = t 4k sin(kt) π = π. (This also follows easily from the fact that sin t + cos t =, hence both of these functions have average value / over a whole period.) Hence the norm of sin(kt) and cos(kt) equals π /. The projection of a function f C[, π] onto T m is the best approimation in L -norm of f by a trigonometric polynomial of degree m and is denoted by S m (t). Notice that S m depends on the function f, although this is suppressed in the notation. By (.4) the coefficients 4 are given by the formulae a = π a k = π b k = π f(t) dt f(t) coskt dt f(t) sin kt dt A central question in Fourier analysis is whether or not the approimations S m (t) converge to f(t),i.e. if formula f(t) = a + (a k coskt + b k sin kt) holds. The series on the right is called the Fourier series of f, whether it converges to f or not.in Figure we see the graphs of f(t) = t and S m (t) for m =, 3, 5, Figure. Fourier approimations to f(t) = t There are three things to observe from these curves. First, the functions S m (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 S m to converge to f, since S m is π-periodic and hence has the same values at these enpoints. In this case the value is. 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 we see the corresponding curves for f(t) = t. 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 Figure. Fourier approimations to f(t) = t.3. Eercises..Determine the Fourier coefficients of f(t) = e t. 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] is even, prove that b k = for all k . If f is odd, prove that a k = for all k. c)prove that any f may be written as a sum of an even and an odd function. (Hint : Let f e (t) = (f(t) + f( t))/ and f o (t) = (f(t) f( t))/. d)determine the Fourier coefficients of sinht and cosht. 3.a) If p is an odd polynomial, prove that p(t)sin(kt) dt = () k+ k p(π) p (t)sin(kt) dt k b) If p is an even polynomial, prove that p(t)cos(kt) dt = () k (π) p (t)cos(kt) dt k p k 6 c) Find the Fourier coefficients of f(t) = t 3 + t. 4.The curves in the figures above were made by using the Matlab toolbo fourgraph. Download fourgraph from (for instance by writing fourgraph in the search bo). Reproduce the curves above and try some other functions, for instance e t or t 3 + t.. Trigonometric polynomials.. The comple eponential function. The real eponential function e can be etended to comple values of the argument. Let z = s + it be a comple number. If the comple eponential function satisfies the usual product rule, we must have e z = e s+it = e s e it Hence we must define the comple function f(t) = e it. If f satisfies the usual chain rule for differentiation of the eponential function, we must have f (t) = ie it = 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 and we must have which gives g (t) + ih (t) = i(g(t) + ih(t)) h (t) = g(t) g (t) = h(t) g (t) = h (t) = g(t) h (t) = g (t) = h(t) Since we also must have f() =, this gives g() = and h() =. The solutions to these equations are g(t) = cos t and h(t) = sin t.hence we must have e it = cost + i sin t. We therefore define the comple eponential function e z by (.) e s+it = e s (cost + i sin t) The comple number e it = cost + i sin t is located on the unit circle in the comple plane at the angle t with the real ais. It follows that the comple number w = e s+it in the point in the comple plane whose length is e s and angle is t. It follows from (.) that the comple eponential function satisfies the usual addition rule for the eponent: e z e z = e z +z From this we obtain the famous DeMoivre s formula: (.) cosnt + i sin nt = (cost + i sin t) n since both sides equal e int = (e it ) n. This formula contains the formulas for cosnt and sin nt as functions of cost and sin t and is the most 7 efficient way of epressing these formulas. It is also the easiest to remember. If f is a real or comple function defined on the unit circle in the comple plane, then f(e it ), as a function of t, is π-periodic. Conversely, given any function f of t that is π-periodic, we may think of f as being defined on the unit circle. In this way we identify π-periodic functions with functions defined on the unit circle in the comple plane. Also, if f is a function defined on [, π), then f etends uniquely to a π-periodic function on the whole real line. This is actually true for any function defined on a (half-open) interval of length π... Comple representation of trigonometric polynomials. In this section we shall write a trigonometric polynomial p T m in the form p(t) = a + (a k coskt + b k sin kt). Notice the factor in front of a. Hence if p(t) = 4 + cost + sin t, then a = 8, a = and b =. This convention will make formulas simpler later.( Notice also that if p is the projection of some π periodic function f onto T m, then all the a k -coefficients are given by the formula a k = π f(t) coskt dt, also for k =.) By convention, we also set π b =. For any comple number z = + iy, the real and imaginary parts are given by Rz = (z + z) Iz = (z z) i Applying this to z = e ikt, we see that (.3) coskt = (eikt + e ikt ), sin kt = i (eikt e ikt ) so p(t) = a + = a + = a + Hence if we define (a k coskt + b k sin kt) a k(e ikt + e ikt ) + i b k(e ikt e ikt ) (a k ib k )e ikt + (a k + ib k )e ikt (.4) c k = (a k ib k ), c k = (a k + ib k ) 8 we see that we can write this as (.5) p(t) = k= m c k e ikt. The constants satisfy c k = c k. This will be called the comple representation of the trigonometric polynomial p(t) Conversely, any function of the form (.5) is a real trigonometric polynomial, provided the constants satisfy c k = c k. The coefficients are then given by a k = c k + c k b k = i(c k c k ) On the unit circle z = e it, hence if we consider p as a function on the unit circle, we have (.6) p(z) = c k z k k= m This shows the connection between real trigonometric polynomials and comple polynomials (if we admit negative powers in a polynomial). By P m+ we denote the set of comple functions of the form m k= m c kz k. We have shown that T m is naturally isomorphic to the real subspace of P m+ defined by the equations c k = c k for k =,,.., m.if we had allowed comple values for the coefficients a k, b k in T m, i.e. considered the comple vector space CT m of comple trigonometric polynomials, the argument above shows that CT m and P m+ are isomorphic as comple vector spaces. Finally, if S m (t) is the projection of a function f onto T m, then the coefficients of the comple representation of S m (t) are given by the one formula (.7) c k = f(t)e ikt dt π These coefficients are called the comple Fourier coefficients of f..3. Sampling of functions in T m. The vector space T m is a real vector space of dimension m +. It is therefore reasonable to epect that a trigonometric polynomial p T m is uniquely determined by its values at m + points. We shall see that this is the case if we divide the interval [, π] into m + equally long pieces, i.e. consider the points π t j = j, j =,,, m m + Furthermore, we shall determine the precise relationship between the coefficients a, a,, a m, b,, b m and the sampled values v j = p(t j ), j =,,, m. 9 To do this, we consider the evaluation map L : T m R m+ defined by L(p) = v where the components of v are defined by evaluating p at t j, i.e. v j = p(t j ). The colums of the matri A of L with respect to the basis {, coskt, sin kt} of T m and the standard basis of R m+ are given by: u = L() =. [ π u k = L(coskt) = coskj m + [ π v k = L(sin kt) = sin kj m + ], j =,,, m ], j =,,, m for k =,,, m. We shall prove that the vectors u,, u m, v,, v m are orthogonal, hence form an orthogonal basis for R m+. To prove this, we shall need the following formula: (.8) N cosjα = j= N R(e ijα ) = R j= N j= (e iα ) j = R( ei(n+)α e iα ) where we have used the formula for the sum of a (finite) geometric series. Similarly, we also have (.9) N j= sin jα = I( ei(n+)α e iα ) We shall prove that the vectors v,, v m 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 v k, v k = = j= π sin kj m + sin π j k m + j= cosj π(k k ) m + Setting N = m and α = π(k+k ) m+ we have j= e i(n+)α = e iπ(k+k ) = cos j π(k + k ) m + , hence the second sum is zero by (.8). Also, the first sum is zero for k k and equal to m+ for k = k. This show that the vectors all are orthogonal and the norms are given by u = m +, u k = v k = m + Hence A is an ortogonal matri whose inverse is given by A = m + ut u T. u T m v T. In other words, if p is a trigonometric polynomial whose values v j = p(t j ) are given by v = [v j ], then the coefficients of p are given by v T m a = m + u, v, a k = m + u k, v, k m, b k = m + v k, v, k m.4. Signal processing. The terms in a trigonometric polynomial have different frequencies. In figure 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.5 sin t+sin 5t. This curve therefore is the representation of p in the time domain. A trigonometric polynomial, however, is defined by p(t) = a + (a k cos kt + b k sin kt). We call this the representation of p in the frequency domain because this representation eplicitly states how much (specified by the constants a k and b k ) p contains of oscillations with frequency k. In the bottom curve, p contains the frequencies, and 5, or more eplicitly, b =, b =.5 and b 5 =, while all the other a s and b s are. sin() sin(5 ) sin().5 sin( )+sin(5 ) 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 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 {, coskt, sin kt} that the projection of T m onto T n for some n m simply consists of omitting the highest frequencies, i.e. P(a + a k coskt + b k sin kt) = a + a k coskt + b k sin kt So, the best (L ) approimation of a trigonometric polynomial of degree m by a trigonometric polynomial of degree n is simply the degree n part of the trigonometric polynomial..5. Convolution in T m. If f and g are π-periodic functions, we define the convolution of f and g by f g(t) = π f(s)g(t s) ds It follows that we can actually perform the inegration over any interval of length π and that f g also is π-periodic and f g = g f. The functions can be real or comple. If f, g T m, the convolution is most easily described using the comple Fourier representation. Therefore, assume that f(t) = e ikt and g(t) = e ik t. Then f g(t) = π = π e iks e ik (t s) ds e ik t e i(k k )s ds = π t π eik e i(k k )s ds = { if k k e ikt if k = k Hence, if f = m k= m c ke ikt and f = m k= m c k eikt, then f g(t) = c k c ke ikt k= m Hence the comple 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 T m to T n, then we simply omit the terms of degree greater than n, i.e. we multiply the coefficients c k by for k n and by for k n. By the discussion above, this means that we convolve p with the function (.) D n (t) = e ikt k= n 3 which is called the Dirichlet kernel (of degree n). This is a finite geometric series, whose sum is D n (t) = e intei(n+)t e it = ei(n+)t e int e it = ei(n+ )t e i(n+ )t e i t e i t = sin(n + )t sin t by (.3).In Figure we see the Dirichlet kernel for n =, 3, 5,. sin(.5 )/sin(.5 ) sin(3.5 )/sin(.5 ) sin(5.5 )/sin(.5 ) sin(.5 )/sin(.5 ) Figure 4. Dirichlet kernel for n =, 3, 5, Notice that the peaks are higher and higher. We have D n(t)dt = π for all n. From the figures above, we see that as n increases D n (t) more and more concentrates a mass (i.e. integral) of π at the origin. 4 If f is a π-periodic function, we have S m (t) = c k e ikt = k= m ( π k= m = π = π = f D m (t) f(s)( f(s)e iks ds)e ikt k= m e ik(t s) ) ds f(s)d m (t s) ds We shall see later that if f is differentiable, this converges to f. This is easily understood from the fact that for large m, D π m(t s) concentrates a mass of at s=t..6. Eercises..Show that the comple Fourier coefficients of an even/odd function are real/pure imaginary. Use eercise 3c) in Chapter to find the comple Fourier coefficients of f(t) = t 3 + t a)for m =, determine the 33 matri A of the evaluation map L : T R 3 described in this chapter. b) Determine p T such that p() = p( 4π) = and 3 p(π) = 4. 3 c)using MATLAB, compute the 55 matri A for L : T R 5. Use this to find a trigonometric polynomial p T such that p() = p( 6π) = 5 and p( π) = 5 p(4π) = 5 p(8π ) =. (Plot your curve to check the result, 5 for instance using fourgraph). 3. Let f be the π-periodic function whose values in [, π) is given by f(t) = t. a)determine f f and draw the graph. (Requires some work!) b)determine the comple Fourier coefficients of f. c)determine S m S m for m = and and draw the graph. 4.(Parseval s identity. Move to Chapter net year!) If p(t) = a + m (a k cos kt + b k sin kt), show that π p(t) dt = a π + (a k + b k ) 5.Using fourgraph, plot the projection of f(t) = t 6 3t t onto T, T, T 3, T 5, T. 5 3. Orthogonal
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