Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin nx n n = n. 3 n Note that all of the integrals reduce to integrals over, since fx = on,. For the a n, b n integrals, I used integration by parts. For the a n case, I took u = x, dv = cos nxdx, implying v = n sin nx. For the b n case, it was u = x and dv = sin nxdx. Noting that the cosine terms vanish for even n, the expansion is fx = 4 cos x + 3 cos 3x + cos 5x + 5 +sin x sin x + sin 3x +. 4 3.5 3.5.5-7.5-5 -.5.5 5 7.5 Figure : Problem 7.5.7
.8.6.4. -7.5-5 -.5.5 5 7.5 Figure : Problem 7.5. 7.5., 3 rd Ed. To evaluate the integrals for this problem, we will need the trig identities cos sin β = sin + β sin β and sin sin β = cos β cos + β. We have a = fxdx = sin xdx = cos x =. 5 Meanwhile, and a n = = { cos n + x n + cos nx sin xdx =, n = cos n x, n > n sin n + x sin n xdx {, n even n+ n, n odd = = { n, n even, n odd ; 6 b n = = sin x sin nxdx = Therefore, the series is given by sin n x sin n+x n sin n+x n+ cos n x cos n + xdx, n > {, n > =, n = n+, n =. 7 fx = + sin x cos x cos 4x cos 6x + + +. 8 4 6 7.6.4, 3 rd Ed. f of problem 7.5.7 is continuous at x =. Therefore, according to Dirichlet s theorem, the fourier series will converge to the value of f, namely f =. The sine terms vanish while the cosines all give a factor of, so
.5.5.75.5.5-4 - 4 Figure 3: Problem 7.9. = 4 + 3 + 5 + 8 = n N,odd n. 9 fx is discontinuous at x =, so by Dirichlet, the fourier series will converge to the midpoint of the discontinuity, which happens to be / here. Again the sine terms vanish while the cosines all contribute a factor of. Thus, we find the same result: = 4 3 5 8 = n N,odd n. Finally, f is continuous at x = /, so the fourier series will converge to the value f/ = /. Now all of the cosine terms vanish, while the sines are nonzero for odd n only and also alternate sign: = 4 + 3 + 5 7 + 4 = k k +. 7.6.5, 3 rd Ed. Evaluate the series of problem 7.5. at x =. f is continuous there, so the series converges to the value f =. All of the cosine terms give a factor of, so we find k= = + 4 + = + 4 + 6 +. 7.9., 3 rd Ed. This is an even function, with l =. Therefore, b n =, while and a = / / fxdx = 4 / xdx = 4 =, 3 3
.8.6 a.4. -3 - - 3.5-3 - - 3 -.5 b - Figure 4: Problem 7.9.5. a corresponds to f c ; b to f s a n = 4 = 4 / n x cos nxdx = 4 x sin nx n / cos nx / n / sinnxdx = cos n = n n n. 4 Note the argument of the cosines here. This is because the period is instead of the usual. Hence, we find fx = 4 cos x + 3 cos 6x + cos x +. 5 5 7.9.5, 3 rd Ed. The even function f c is the same as the problem we just did, up to a rescaling of the interval. In the previous problem we had l = /, while here we have l c =. We can obtain this series from the previous one by multiplying the coefficients by the ratio l c /l = /, ie. a cn = lc l a n, and the arguments of the cosines by l/l c = /. This follows from the following: a cn = l c lc x cos nx dx = l c l c l c l = l c l l l l x cos n x d x l In the first step, I changed variables, x = l l c x. Hence, we find for f c, x cos n x d x = l c l l a n. 6 f c x = 4 cos x + 3 cos 3x + cos 5x +. 7 5 We have f s x = x on,. The a ns are, while 4
a.8.6.4. -3 - - 3 b.5-3 - - 3 -.5 - c.8.6.4. -3 - - Figure 5: Problem 7.9.. a f c ; b f s ; c f p. Thus, b n = x cos nx x sin nxdx = + n n cos n = n + sin nx n cos nxdx = n. 8 n f s x = sin x sin x + sin 3x +. 9 3 7.9., 3 rd Ed. f c x = x on,. We have while, integrating by parts twice, a = x dx = 3, Hence, = 4 n a n = x cos nx n x x sin nx cos nxdx = n + cos nxdx = 4 n n n cos n n x sin nxdx = 4 n n. f c x = 3 + 4 cos x + cos x cos 3x +. 3 We have f s x = x x. The nonzero coefficients are 5
Thus b n = = n n n n + x sin nx n n cos nx x sin nxdx = x cos nx cos n = 4 n n = n n + n n x cos nxdx sin nxdx + 4 n 3 3 n. 3 f s x = sin x sin x + sin 3x + 3 8 sin x + 3 3 sin 3x + sin 5x +. 4 3 53 We will be able to use the previous two results to compute the fourier series for f p x. The period is now l =. We have a n = b n = where a cm, b sm are the coefficients for f c and f s. Hence, x cos nxdx = a cn, 5 x sin nxdx = b sn, 6 And so, a = /3, b n = n n a n = 4 n n = n ; 7 + 4 n 3 n = n. 8 f p x = 3 + cos x + cos x + cos 3x + 3 sin x + sin x + sin 3x +. 9 3 6
7..8, 3 rd Ed. From problem 7.9. we had fx = x on the interval /, /. The average value of fx on this interval is fx = / / fx dx / / dx = / / x dx = 3 3 8 =. 3 According to Parseval s theorem, this should equal 4 a + n= a n + b n. Hence, = + 4 n N,odd n n N,odd = 6 + n N,odd n 4 n 4 = 4 6 = 4 96. 3 7.., 3 rd Ed. The average value of fxgx on the interval, is given by fxgx = fxgxdx. 3 Now plug in the expansions and multiply out, using the linearity of the integral: = a + m= a a m cos mx + b m sin mx = 4 a a + a + m,n= + a m= + n= n= fxgxdx a n cos nx + b n sin nx dx a n cos nx + b n sin nxdx a m cos mx + b m sin mxdx a m a n cos mx cos nx + a m b n cos mx sin nx +b m a n sin mx cos nx + b m b n sin mx sin nx. 33 The cross terms, involving the sum over a single index, vanish since the integral over one period of sine or cosine is zero. For the terms in the double sum, use the orthogonality relations in formula 5. of chapter seven. Then the above reduces to 7
fxgx = 4 a a + δ mna m a n + δ mnb m b n m,n= = 4 a a + 7..3, 3 rd Ed. We have a n a n + b n b n. 34 n= Hence, g = = i e i fxe ix dx = i e i e ix dx + e ix dx = i ei i e i = cos. 35 i fx = i cos e i d. 36 This expression can be simplified by writing out e ix = cos x + i sin x. Then observe that cos x h = cos x is an odd function in. Hence, this term does not contribute. On the other hand, k = sin x is an even function, so we may write cos x fx = 7..7, 3 rd Ed. We have cos x sin xd = cos x sin xd. 37 Then, g s = f s x sin xdx = cos x = = sin xdx cos. 38 f s x = cos x sin xd = cos x sin xd. 39 This is equivalent to the result of the previous problem. 8
7..3, 3 rd Ed. We observe that f is continuous at x =, so the fourier transform expression for f should give the value f =. Thus, or = f = cos sin d, 4 = cos sin d. 4 f is discontinuous at x =, so the fourier transform will converge to the midpoint of the discontinuity, /. Hence, = 7..4, 3 rd Ed. cos sin d 4 = cos sin d. 4 a g = = e i+x i e ix e x dx = = + e i x i dxe i+x + i + + i = dxe i x +. 43 Then we have fx = e ix + d = cos x sin x + i + + d = The last step follows from evenness/oddness considerations. Hence, b Hence, g c = e x = e x cos xdx = = cos x d. 44 + cos x d. 45 + i + + i 9 e i x + e i+x dx = +. 46
f c x = which agrees with the previous result. + cos xd = cos x d, 47 + c From b, interchanging x and, we have cos x + x dx = e. 48 In other words, the cosine fourier transform of fx = is given by +x cos x g c = + x dx = e. 49 7..33, 3 rd Ed. We verify Parseval s theorem for fx and its fourier transform g, given by We have fx = e x, g = +. 5 On the other hand, fx dx = e x dx = e x dx =. 5 g d = d +. 5 To evaluate the integral, make the trig substitution = tan θ. Then + tan θ = sec θ, and d tan θ = sec θ dθ. Finally, when = ±, θ = ±/. Then Thus, d + = / / sec θ dθ = / cos θ dθ = 4 =. 53 fx dx = g d. 54