Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 6 pps

We use the definition of the Fourier coefficients to evaluate the integrals in the last sum ∞ π (f (x)) dx − πa2 ∞ a2 n − 2π −π + b2 n n=1 πa2 a2 + b = + +π n n n=1 ∞ π a2 + a2 + b = n n π n=1 f (x)2 dx −π We determine the Fourier coefficients for f (x) = x Since f (x) is odd, all of the an are zero π x sin(nx) dx π −π π 1 + = − x cos(nx) π n −π b0 = π −π cos(nx) dx n n+1 = The Fourier series is 2(−1) n ∞ x= n=1 2(−1)n+1 sin(nx) for x ∈ (−π π) n We apply Parseval’s theorem for this series to find the value of ∞ n=1 = n2 π ∞ n=1 ∞ n=1 π x2 dx −π 2π = n2 π2 = n2 1374 ∞ n=1 n2 Consider f (x) = x2 Since the function is even, there are no sine terms in the Fourier series The coefficients in the cosine series are π x dx π 2π = π x cos(nx) dx an = π 4(−1)n = n2 a0 = Thus the Fourier series is ∞ π2 (−1)n x = +4 cos(nx) for x ∈ (−π π) n2 n=1 ∞ n=1 n4 We apply Parseval’s theorem for this series to find the value of ∞ 2π 1 + 16 = n4 π n=1 ∞ π x4 dx −π 2π 2π + 16 = n4 n=1 ∞ n=1 π4 = n4 90 1375 Now we integrate the series for f (x) = x2 x ξ2 − ∞ π2 3 dξ = n=1 ∞ (−1)n n2 x cos(nξ) dξ x π (−1)n − x=4 sin(nx) 3 n3 n=1 We apply Parseval’s theorem for this series to find the value of ∞ 16 n=1 π 1 = n6 π −π ∞ 16 n=1 ∞ n=1 x3 π − x 3 16π = n6 945 π6 = n6 945 1376 ∞ n=1 n6 dx Solution 28.2 We differentiate the partial sum of the Fourier series and evaluate the sum N SN = n=1 N 2(−1)n+1 sin(nx) n (−1)n+1 cos(nx) SN = n=1 N (−1)n+1 eınx SN = n=1 − (−1)N +2 eı(N +1)x + eıx + e−ıx −(−1)N eı(N +1)x −(−1)N eıN x SN = + cos(x) cos((N + 1)x) + cos(N x) SN = − (−1)N + cos(x) cos N + x cos x SN = − (−1)N cos2 x SN = cos N + dSN = − (−1)N dx cos x 1377 x We integrate SN x SN (x) − SN (0) = x − (−1)N cos x − SN = sin (ξ ξ−π N+ sin ξ ξ cos x N+ − π) dξ We find the extrema of the overshoot E = x − SN with the first derivative test sin E = N+ sin (x x−π − π) =0 We look for extrema in the range (−π π) N+ x=π 1− (x − π) = −nπ n N + 1/2 n ∈ [1 2N ] , The closest of these extrema to x = π is x=π 1− N + 1/2 Let E0 be the overshoot at this point We approximate E0 for large N sin π(1−1/(N +1/2)) E0 = sin (ξ ξ−π N+ N+ − π) We shift the limits of integration π sin E0 = π/(N +1/2) 1378 sin ξ ξ dξ dξ dξ We add and subtract an integral over [0 π/(N + 1/2)] π E0 = sin N+ sin ξ 2 ξ π/(N +1/2) dξ − sin N+ sin ξ ξ dξ We can evaluate the first integral with contour integration on the unit circle C π sin N+ sin ξ 2 ξ π dξ = π sin ((2N + 1) ξ) dξ sin (ξ) −π z 2N +1 dz = − C (z − 1/z)/(ı2) ız z 2N +1 = − dz C (z − 1) z 2N +1 = ıπ Res , + ıπ Res (z + 1)(z − 1) 12N +1 (−1)2N +1 =π + −2 =π = sin ((2N + 1) ξ) dξ sin (ξ) 1379 z 2N +1 , −1 (z + 1)(z − 1) We approximate the second integral π/(N +1/2) sin N+ ξ ξ sin dξ = π 2N + sin(x) dx sin 2Nx+1 π sin(x) dx x π x ≈2 =2 ∞ ∞ n=0 π =2 n=0 ∞ =2 n=0 (−1)n x2n dx (2n + 1)! (−1)n π 2n+1 dx (2n + 1)(2n + 1)! ≈ 3.70387 In the limit as N → ∞, the overshoot is |π − 3.70387| ≈ 0.56 Solution 28.3 The eigenfunctions of the self-adjoint problem −y = λy, y(0) = y(1) = 0, are φn = sin(nπx), 1380 (−1)n x2n+1 dx (2n + 1)! n ∈ Z+ We find the series expansion of the inhomogeneity f (x) = ∞ fn sin(nπx) 1= n=1 fn = sin(nπx) dx fn = − cos(nπx) nπ fn = (1 − (−1)n ) nπ for odd n nπ fn = for even n We expand the solution in a series of the eigenfunctions ∞ an sin(nπx) y= n=1 We substitute the series into the differential equation y + 2y = ∞ ∞ 2 − an π n sin(nπx) + n=1 ∞ an sin(nπx) = n=1 odd n n=1 an = nπ(2−π n2 ) for odd n for even n ∞ y= n=1 odd n sin(nπx) nπ(2 − π n2 ) 1381 sin(nπx) nπ Now we solve the boundary value problem directly y + 2y = y(0) = y(1) = The general solution of the differential equation is √ √ y = c1 cos 2x + c2 sin 2x + We apply the boundary conditions to find the solution √ √ 1 c1 + = 0, c1 cos + c2 sin + =0 2 √ cos − 1 √ c = − , c2 = 2 sin √ √ √ cos − 1 √ y= − cos 2x + sin 2x sin We find the Fourier sine series of the solution ∞ y= an sin(nπx) n=1 an = y(x) sin(nπx) dx √ √ √ cos − √ 2x + sin 2x − cos sin an = an = an = 2(1 − (−1)2 nπ(2 − π n2 ) nπ(2−π n2 ) for odd n for even n 1382 sin(nπx) dx which has the solutions, λn = nπ , b−a nπ(x − a) b−a φn = sin , n ∈ N We expand the solution and the inhomogeneity in the eigenfunctions ∞ y(x) = yn sin n=1 ∞ f (x) = nπ(x − a) b−a fn sin n=1 , nπ(x − a) b−a b fn = b−a f (x) sin a nπ(x − a) b−a dx Since the solution y(x) satisfies the same homogeneous boundary conditions as the eigenfunctions, we can differentiate the series We substitute the series expansions into the differential equation y + αy = f (x) ∞ ∞ yn −λ2 n + α sin (λn x) = n=1 fn sin (λn x) n=1 yn = fn α − λ2 n Thus the solution of the problem has the series representation, ∞ α − λ2 sin n y(x) = n=1 nπ(x − a) b−a Solution 28.13 The eigenfunction problem associated with this problem is φ + λ2 φ = 0, φ(a) = φ(b) = 0, 1398 which has the solutions, nπ nπ(x − a) , φn = sin b−a b−a We expand the solution and the inhomogeneity in the eigenfunctions λn = ∞ yn sin y(x) = n=1 ∞ fn sin f (x) = n=1 nπ(x − a) b−a , fn = n ∈ N , nπ(x − a) b−a b−a b f (x) sin a nπ(x − a) b−a dx Since the solution y(x) does not satisfy the same homogeneous boundary conditions as the eigenfunctions, we can differentiate the series We multiply the differential equation by an eigenfunction and integrate from a to b We use integration by parts to move derivatives from y to the eigenfunction y + αy = f (x) b b b a a a f (x) sin(λm x) dx y(x) sin(λm x) dx = y (x) sin(λm x) dx + α b b−a b−a ym = fm 2 a b b−a b−a ym = fm − [yλm cos(λm x)]b − yλ2 sin(λm x) dx + α m a 2 a b−a b−a −Bλm (−1)m + Aλm (−1)m+1 − λ2 ym + α ym = fm m 2 fm + (−1)m λm (A + B) ym = α − λ2 m Thus the solution of the problem has the series representation, [y b sin(λm x)]a − ∞ y(x) = n=1 y λm cos(λm x) dx + α fm + (−1)m λm (A + B) sin α − λ2 m 1399 nπ(x − a) b−a Solution 28.14 A + ıB = 1 − z2 ∞ z 2n = n=0 ∞ r2n eı2nx = n=0 ∞ ∞ r = 2n n=1 n=0 ∞ ∞ r2n cos(2nx), A= r2n sin(2nx) cos(2nx) + ı r2n sin(2nx) B= n=1 n=0 1 − z2 = − r2 eı2x A + ıB = 1 − cos(2x) − ır2 sin(2x) − r2 cos(2x) + ır2 sin(2x) = (1 − r2 cos(2x))2 + (r2 sin(2x))2 = A= r2 − r2 cos(2x) , − 2r2 cos(2x) + r4 1400 B= r2 sin(2x) − 2r2 cos(2x) + r4 We consider the principal branch of the logarithm A + ıB = log(1 + z) ∞ = n=1 ∞ = n=1 ∞ = n=1 ∞ A= n=1 (−1)n+1 n z n (−1)n+1 n ınx r e n (−1)n+1 n r cos(nx) + ı sin(nx) n (−1)n+1 n r cos(nx), n ∞ B= n=1 (−1)n+1 n r sin(nx) n A + ıB = log(1 + z) = log (1 + r eıx ) = log (1 + r cos x + ır sin x) = log |1 + r cos x + ır sin x| + ı arg (1 + r cos x + ır sin x) = log A= (1 + r cos x)2 + (r sin x)2 + ı arctan (1 + r cos x, r sin x) log + 2r cos x + r2 , 1401 B = arctan (1 + r cos x, r sin x) n zk An + ıBn = k=1 − z n+1 1−z − rn+1 eı(n+1)x = − r eıx − r e−ıx −rn+1 eı(n+1)x +rn+2 eınx = − 2r cos x + r2 = An = − r cos x − rn+1 cos((n + 1)x) + rn+2 cos(nx) − 2r cos x + r2 Bn = r sin x − rn+1 sin((n + 1)x) + rn+2 sin(nx) − 2r cos x + r2 n zk An + ıBn = k=1 n rk eıkx = k=1 n n k An = r cos(kx), k=1 rk sin(kx) Bn = k=1 Solution 28.15 π · sin x dx = [− cos x]π = 0 1402 Thus the system is not orthogonal on the interval [0, π] Consider the interval [a, a + π] a+π · sin x dx = [− cos x]a+π = cos a a a a+π · cos x dx = [sin x]a+π = −2 sin a a a Since there is no value of a for which both cos a and sin a vanish, the system is not orthogonal for any interval of length π First note that π cos nx dx = for n ∈ N If n = m, n ≥ and m ≥ then π cos nx cos mx dx = π cos((n − m)x) + cos((n + m)x) dx = 0 Thus the set {1, cos x, cos 2x, } is orthogonal on [0, π] Since π dx = π π cos2 (nx) dx = π , the set , π cos x, π is orthonormal on [0, π] 1403 cos 2x, π If n = m, n ≥ and m ≥ then π sin nx sin mx dx = π cos((n − m)x) − cos((n + m)x) dx = 0 Thus the set {sin x, sin 2x, } is orthogonal on [0, π] Since π sin2 (nx) dx = π , the set sin x, π sin 2x, π is orthonormal on [0, π] Solution 28.16 Since the periodic extension of |x| in [−π, π] is an even function its Fourier series is a cosine series Because of the anti-symmetry about x = π/2 we see that except for the constant term, there will only be odd cosine terms Since the periodic extension is a continuous function, but has a discontinuous first derivative, the Fourier coefficients will decay as 1/n2 ∞ |x| = an cos(nx), for x ∈ [−π, π] n=0 a0 = π π x2 x dx = π 1404 π = π 2 π x cos(nx) dx π π sin(nx) = x − π n π π cos(nx) =− π n2 = − (cos(nπ) − 1) πn 2(1 − (−1)n ) = πn2 an = |x| = π + π ∞ n=1 odd n π sin(nx) dx n cos(nx) for x ∈ [−π, π] n2 Define RN (x) = f (x) − SN (x) We seek an upper bound on |RN (x)| |RN (x)| = π ≤ π = π Since ∞ cos(nx) n2 n=N +1 odd n ∞ n2 n=N +1 odd n ∞ n=1 odd n ∞ n=1 odd n − n π π2 = n2 1405 N n=1 odd n n2 We can bound the error with, π |RN (x)| ≤ − π N n=1 odd n n2 N = is the smallest number for which our error bound is less than 10−1 N ≥ is sufficient to make the error less that 0.1 π 1 |R7 (x)| ≤ − 1+ + + ≈ 0.079 π 25 49 N ≥ is also necessary because ∞ |RN (0)| = π n=N +1 n2 odd n Solution 28.17 ∞ 1∼ an sin(nx), 0≤x≤π n=1 Since the odd periodic extension of the function is discontinuous, the Fourier coefficients will decay as 1/n Because of the symmetry about x = π/2, there will be only odd sine terms π · sin(nx) dx π = (− cos(nπ) + cos(0)) nπ = (1 − (−1)n ) nπ an = 1∼ π ∞ n=1 odd n 1406 sin(nx) n It’s always OK to integrate a Fourier series term by term We integrate the series in part (a) x a dx ∼ π x−a∼ π ∞ x n=1 odd n ∞ n=1 odd n a sin(nξ) dx n cos(na) − cos(nx) n2 Since the series converges uniformly, we can replace the ∼ with = x−a= π ∞ n=1 odd n cos(na) − n2 π ∞ n=1 odd n cos(nx) n2 Now we have a Fourier cosine series The first sum on the right is the constant term If we choose a = π/2 this sum vanishes since cos(nπ/2) = for odd integer n π x= − π ∞ n=1 odd n cos(nx) n2 If f (x) has the Fourier series ∞ f (x) ∼ a0 + (an cos(nx) + bn sin(nx)), n=1 then Parseval’s theorem states that ∞ π π (a2 + b2 ) f (x) dx = a2 + π n n −π n=1 1407 We apply this to the Fourier sine series from part (a) ∞ π f (x) dx = π −π n=1 odd n π ∞ n=1 ∞ (−1) dx + −π 16 (1) dx = π πn n=1 (2n − 1)2 π2 = (2n − 1)2 We substitute x = π in the series from part (b) to corroborate the result π x= − π π= π − π ∞ n=1 ∞ n=1 ∞ n=1 cos((2n − 1)x) (2n − 1)2 cos((2n − 1)π) (2n − 1)2 π2 = (2n − 1)2 Solution 28.18 ∞ f (x) ∼ a0 + an cos(nx) n=1 Since the periodic extension of the function is discontinuous, the Fourier coefficients will decay like 1/n Because of the anti-symmetry about x = π/2, there will be only odd cosine terms a0 = π π f (x) dx = 1408 2 π f (x) cos(nx) dx π π/2 = cos(nx) dx π = sin(nπ/2) πn (−1)(n−1)/2 , for odd n πn = for even n an = The Fourier cosine series of f (x) is f (x) ∼ + π ∞ n=0 (−1)n cos((2n + 1)x) 2n + The N th partial sum is SN (x) = + π N n=0 (−1)n cos((2n + 1)x) 2n + We wish to evaluate the sum from part (a) First we make the change of variables y = x − π/2 to get rid of the 1409 (−1)n factor ∞ n=0 (−1)n cos((2n + 1)x) 2n + N = n=0 N = (−1)n cos((2n + 1)(y + π/2)) 2n + (−1)n (−1)n+1 sin((2n + 1)y) 2n + n=0 N =− n=0 sin((2n + 1)y) 2n + 1410 We write the summand as an integral and interchange the order of summation and integration to get rid of the 1/(2n + 1) factor N y =− cos((2n + 1)t) dt n=0 y N =− cos((2n + 1)t) dt y n=0 2N +1 N cos(nt) − =− y n=1 2N +1 eınt − =− y n=1 eıt cos(2nt) n=1 N ı2nt e dt dt n=1 ı(2N +2)t e eı2t − eı2(N +1)t − − dt − eıt − eı2t y (eıt − eı2(N +1)t )(1 − eı2t ) − (eı2t − eı2(N +1)t )(1 − eıt ) =− (1 − eıt )(1 − eı2t ) y eıt − eı2t + eı(2N +4)t − eı(2N +3)t =− dt (1 − eıt )(1 − eı2t ) y eıt − eı(2N +3)t dt =− − eı2t y eı(2N +2)t −1 =− dt eıt − e−ıt y −ı eı2(N +1)t +ı =− dt sin t y sin(2(N + 1)t) =− dt sin t x−π/2 sin(2(N + 1)t) =− dt sin t 1411 =− dt Now we have a tidy representation of the partial sum SN (x) = We solve dSN (x) dx x−π/2 1 − π sin(2(N + 1)t) dt sin t = to find the relative extrema of SN (x) SN (x) = sin(2(N + 1)(x − π/2)) =0 − π sin(x − π/2) (−1)N +1 sin(2(N + 1)x) =0 − cos(x) sin(2(N + 1)x) =0 cos(x) nπ , n = 0, 1, , N, N + 2, , 2N + x = xn = 2(N + 1) Note that xN +1 = π/2 is not a solution as the denominator vanishes there The function has a removable singularity at x = π/2 with limiting value (−1)N πN −π/2 2(N +1) 1 SN (xN ) = − π We note that the integrand is even πN −π/2 2(N +1) π − 2(N +1) = SN (xN ) = 1 + π sin(2(N + 1)t) dt sin t π 2(N +1) =− π 2(N +1) 1412 sin(2(N + 1)t) dt sin t ... Parseval’s theorem for this series to find the value of ∞ 16 n=1 π 1 = n6 π −π ∞ 16 n=1 ∞ n=1 x3 π − x 3 16? ? = n6 945 ? ?6 = n6 945 13 76 ∞ n=1 n6 dx Solution 28.2 We differentiate the partial sum of... 16 = n4 π n=1 ∞ π x4 dx −π 2π 2π + 16 = n4 n=1 ∞ n=1 ? ?4 = n4 90 1375 Now we integrate the series for f (x) = x2 x ξ2 − ∞ π2 3 dξ = n=1 ∞ (−1)n n2 x cos(nξ) dξ x π (−1)n − x =4 sin(nx) 3 n3 n=1... = π 4( −1)n = n2 a0 = Thus the Fourier series is ∞ π2 (−1)n x = +4 cos(nx) for x ∈ (−π π) n2 n=1 ∞ n=1 n4 We apply Parseval’s theorem for this series to find the value of ∞ 2π 1 + 16 = n4 π