Fourier Series Introduction One important application of linear algebra is in the representation and approximation of functions. For example, suppose E(t) is a function that represents the amount of acoustical energy received by a microphone at time t. Since sound energy is in the form of sound waves of varying frequencies, it seems plausible to represent E(t) as a linear combination of simple waves. In mathematics, simple waves are represented using sine and cosine functions of the form sin(kt) and cos(kt), where k is a nonnegative integer, and the greater the k value, the higher the wave frequency. Then E(t) = c0 + c1 cos(t) + ··· + cn cos(nt) + d1 sin(t) + ··· + dn sin(nt); where c0; : : : ; cn; d1; : : : ; dn are constant coefficients, and n is a nonnegative integer tht is sufficiently large enough to capture all of the frequencies that comprise E(t). The expression on the right is called a trigonometric polynomial of order n (assuming cn 6= 0 and dn 6= 0). A trigonometric- polynomial representation of E(t) seems quite useful, since it alllows one to determine the dominant acoustical frequencies that comprise the energy function. These frequencies will correspond with values of k for which either ck or dk seems substantial. But how to determine the values of these coefficients? Recall that the span of the set of wave functions f1; sin(t); cos(t); sin(2t); cos(2t);:::; sin(nt); cos(nt)g forms a subspace Wn of the inner product space C[0; 2π], the set of continuous functions over [0; 2π]. Recall also that C[0; 2π] admits the inner product Z 2π <f; g> = f(t)g(t)dt: 0 Moreover, the wave functions are pairwise orthogonal, and can be normalized to form an orthonormal basis. 1 Example 1. Let jj cos(kt)jj denote the length of cos(kt) as vector in C[0; 2π]. Show that p jj cos(kt)jj = jj sin(kt)jj = π; p while jj1jj = 2π. From the lecture on inner product spaces, to obtain the best approximation of E(t) as a linear com- bination of wave functions (i.e. as an element of Wn), one must project E(t) on to Wn. Furthermore, the coefficient ck can be computed as 1 1 1 Z 2π ck = p <E(t); p cos(kt)> = E(t) cos(kt)dt: π π π 0 p1 p1 Note that the π outside the inner product arises from the fact that the basis vector is ek = π cos(kt). Moreover, when projecting E(t) on to Wn, the inner product <E(t); ek> must be multiplied with ek, p1 which accounts for the extra π . Similarly, 1 Z 2π dk = E(t) sin(kt)dt: π 0 Each ck and dk is called a Fourier coefficient of E(t), and the series 1 1 X X c0 + ck cos(kt) + dk sin(kt) k=1 k=1 is called the Fourier series of E(t). Notice that the above expression actually represents two series (one of cosines and one of sines), but we will think of it as a single series. In summary, the projection theorem states that the best approximation of E(t) as an element of Wn is proj(E(t); Wn). But how accurate is this approximation? In other words, what will be the size ? of the vector wn = E(t) − proj(E(t); Wn)? It turns out that we can make some fairly conservative assumptions about E(t) that guarantee that ? limitjjwn jj = 0; n!1 2 meaning that, as n increases, proj(E(t); Wn) becomes an increasingly better approximation of E(t). The following theorems are stated without proof. Theorem 1. If E(t) is continuous on [0; 2π], then limitjjE(t) − proj(E(t); Wn)jj = 0: n!1 Theorem 2. If E(t) is continuous on [0; 2π], then the Fourier series 1 1 X X c0 + ck cos(kt) + dk sin(kt) k=1 k=1 converges uniformly to E(t). In other words, for any > 0, there exists a sufficiently large n, independent of t, for which jjE(t) − proj(E(t); Wn)jj < . In general, whenever a vector v of some inner product space is approximated by its projection on to a subspace W, then jjw?jj2 = jjv − proj(v; W)jj2 is called the sum squared error (sse) of the approximation. In particular, if the inner product space is C[a; b], then the sse is computed as Z b sse = (f(x) − proj(f; W))2dx: a Example 2. Find the best approximation of f(x) = x over the subspace spanned by f1; exg and compute the sum squared error. 3 At this point the reader may wonder why both sines and cosines are used in the Fourier series. After all, sin(t) = cos(t − π=2). The reason is that, when E(t) is measured at time t = 0, the simple waves that comprise E(t) will likely be in different phases of their cycles. For example, at t = 0, cos(t) is in the phase of being at its peak, where as cos(t + π=3) is n the phase of being 1/2 the distance to its highest peak. Therefore, a simple wave should have the general form A cos(kt + φ), where π is called the phase angle of the wave, and A is called its amplitude. Now, using the identity, cos(x + y) = cos(x) cos(y) − sin(x) sin(y) yields A cos(kt + φ) = A cos(kt) cos(φ) − A sin(kt) sin(φ): Therefore, A cos(kt+φ) can be written as a linear combination of cos(kt) and sin(kt), which explains why both sines and cosines are needed in the Fourier series. Of course, the advantage in using both sines and cosines lies in the fact that all the simple waves have the same phase angle of φ = 0. Example 3. Write 4 cos(3t + π=4) as a linear combination of simple waves, each with phase angle φ = 0. 2π Another valid concern is the assumption that the simple waves of E(t) have periods of k , where k is a positive integer. This is a somewhat limited set of periods. Suppose in practice that there are a finite number n of significant (in the sense of having a sufficiently large amplitude) simple waves of E(t), and that their respective periods are T1;:::;Tn, where each Ti is a rational number. Let T be the least common multiple of T1;:::;Tn, meaning that T is the least rational number for which T=Ti is a positive integer, for every i = 1; : : : ; n. Note that T is called the fundamental period. Then measuring E(t) over the interval [0;T ] will completely capture at least one period of each of its simple waves, and 2πt 2πt 2πt 2πt E(t) = c0 + c1 cos( ) + ··· + cn cos( ) + d1 sin( ) + ··· + dn sin( ) T1 Tn T1 Tn over the interval [0;T ]. To analyze E(t) over the preferred period of [0; 2π], we define a linear function s(t) of time so that 0 maps to 0, and T maps to 2π. Thus s(t) = 2πt=T . For example, if T = 10, and for t = 5 we measure 4 E(5) = 7, then on the s time scale this would correspond to s = (2π5)=10 = π, and we would write E(π) = 7. Thus, when measuring E on the s time scale, the fundamental period changes from T to 2π. Moreover, if 2πt 2πt 2πt 2πt E(t) = c0 + c1 cos( ) + ··· + cn cos( ) + d1 sin( ) + ··· + dn sin( ); T1 Tn T1 Tn T s then the formula for E on the s time scale can be obtained by using the fact that t = 2π . Thus T s substituting 2π for t in the above formula for E(t) yields E(s) = c0 + c1 cos(k1s) + ··· + cn cos(kns) + d1 sin(k1s) + ··· + dn sin(kns); where ki = T=Ti is an integer. Therefore, for given E(t), we may first derive a Fourier series for E(s) on the s time scale, with fundamental period 2π, and then derive E(t) by substituting s with 2πt=T to yield a Fourier series with fundamental period T . Example 4. If −3 cos(πt=4) is a term of a Fourier series with fundamental frequency T = 24, then re-write this term if using the s time scale with fundamental period 2π. Conversely, if 1:5 sin(10s) is a term of a Fourier series over an s time scale with fundamental period 2π, re-write this term using the t time scale with a fundamental period 30. The remainder of the lecture will assume a fundamental period of 2π so that we may exclusively use the orthonormal basis 1 1 1 1 1 fp ; p cos(t); p cos(2t);:::; p sin(t); p sin(2t)g: 2π π π π π Given function E(t), the n th-order Fourier approximation of E(t) is defined as F (E; n; t) = c0 + c1 cos(t) + ··· + cn cos(nt) + d1 sin(t) + ··· + dn sin(nt); where 1 Z 2π c0 = E(t)dt; 2π 0 5 and for k = 1; : : : ; n 1 Z 2π ck = E(t) cos(kt)dt; π 0 and 1 Z 2π dk = E(t) sin(kt)dt: π 0 Example 5. Find the 2nd-order Fourier approximation of E(t) = t. Provide a general formula for the n th-order Fourier approximation of E(t) = t.