Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 5 pdf

We integrate Equation 28.1 from −π to π to determine a 0 .  π −π f(x) dx = 1 2 a 0  π −π dx +  π −π ∞  n=1 a n cos(nx) + b n sin(nx) dx  π −π f(x) dx = πa 0 + ∞  n=1  a n  π −π cos(nx) dx + b n  π −π sin(nx) dx   π −π f(x) dx = πa 0 a 0 = 1 π  π −π f(x) dx Multiplying by cos(mx) and integrating will enable us to solve for a m .  π −π f(x) cos(mx) dx = 1 2 a 0  π −π cos(mx) dx + ∞  n=1  a n  π −π cos(nx) cos(mx) dx + b n  π −π sin(nx) cos(mx) dx  All but one of the terms on the right side vanishes due to the orthogonality of the eigenfunctions.  π −π f(x) cos(mx) dx = a m  π −π cos(mx) cos(mx) dx  π −π f(x) cos(mx) dx = a m  π −π  1 2 + cos(2mx)  dx  π −π f(x) cos(mx) dx = πa m a m = 1 π  π −π f(x) cos(mx) dx. 1334 Note that this formula is valid for m = 0, 1, 2, . . Similarly, we can multiply by sin(mx) and integrate to solve for b m . The result is b m = 1 π  π −π f(x) sin(mx) dx. a n and b n are called Fourier coefficients. Although we will not show it, Fourier series converge for a fairly general class of functions. Let f(x − ) denote the left limit of f(x) and f(x + ) denote the right limit. Example 28.2.1 For the function defined f(x) =  0 for x < 0, x + 1 for x ≥ 0, the left and right limits at x = 0 are f(0 − ) = 0, f(0 + ) = 1. Result 28.2.1 Let f(x) b e a 2π-periodic function for which  π −π |f(x)|dx exists. Define the Fourier coefficients a n = 1 π  π −π f(x) cos(nx) dx, b n = 1 π  π −π f(x) sin(nx) dx. If x is an interior point of an interval on which f(x) has limited total fluctuation, then the Fourier series of f(x) a 0 2 + ∞  n=1  a n cos(nx) + b n sin(nx)  , converges to 1 2 (f(x − ) + f(x + )). If f is continuous at x, then the series converges to f(x). 1335 Periodic Extension of a Function. Let g(x) be a function that is arbitrarily defined on −π ≤ x < π. The Fourier series of g(x) will represent the periodic extension of g(x). The periodic extension, f(x), is defined by the two conditions: f(x) = g(x) for − π ≤ x < π, f(x + 2π) = f(x). The period ic e xtension of g(x) = x 2 is shown in Figure 28.1. -5 5 10 -2 2 4 6 8 10 Figure 28.1: The Periodic Extension of g(x) = x 2 . Limited Fluctuation. A function that has limited total fluctuation can be written f(x) = ψ + (x) − ψ − (x), where ψ + and ψ − are bounded, nondecreasing functions. An example of a function that does not have limited total fluctuation 1336 is sin(1/x), whose fluctuation is unlimited at the point x = 0. Functions with Jump Discontinuities. Let f(x) be a discontinuous function that has a convergent Fourier series. Note that the series does not necessarily converge to f(x). Instead it converges to ˆ f(x) = 1 2 (f(x − ) + f(x + )). Example 28.2.2 Consider the function defined by f(x) =  −x for − π ≤ x < 0 π −2x for 0 ≤ x < π. The Fourier series converges to the function defined by ˆ f(x) =          0 for x = −π −x for − π < x < 0 π/2 for x = 0 π −2x for 0 < x < π. The function ˆ f(x) is plotted in Figure 28.2. 28.3 Least Squares Fit Approximating a function with a Fourier series. Suppose we want to approximate a 2π-periodic function f(x) with a finite Fourier series. f(x) ≈ a 0 2 + N  n=1 (a n cos(nx) + b n sin(nx)) Here the coefficients are computed with the familiar formulas. Is this the best approximation to the function? That is, is it possible to choose coefficients α n and β n such that f(x) ≈ α 0 2 + N  n=1 (α n cos(nx) + β n sin(nx)) 1337 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 Figure 28.2: Graph of ˆ f(x). would give a better approximation? Least squared error fit. The most common criterion for finding the best fit to a function is the least squares fit. The best approximation to a function is defined as the one that minimizes the integral of the square of the deviation. Thus if f(x) is to be approximated on the interval a ≤ x ≤ b by a series f(x) ≈ N  n=1 c n φ n (x), (28.2) 1338 the best approximation is found by choosing values of c n that minimize the error E. E ≡  b a      f(x) − N  n=1 c n φ n (x)      2 dx Generalized Fourier coefficients. We consider the case that the φ n are orthogonal. For simplicity, we also assume that the φ n are real-valued. Then most of the terms will vanis h when we interchange the order of integration and summation. E =  b a  f 2 − 2f N  n=1 c n φ n + N  n=1 c n φ n N  m=1 c m φ m  dx E =  b a f 2 dx − 2 N  n=1 c n  b a fφ n dx + N  n=1 N  m=1 c n c m  b a φ n φ m dx E =  b a f 2 dx − 2 N  n=1 c n  b a fφ n dx + N  n=1 c 2 n  b a φ 2 n dx E =  b a f 2 dx + N  n=1  c 2 n  b a φ 2 n dx − 2c n  b a fφ n dx  We complete the square for each term. E =  b a f 2 dx + N  n=1    b a φ 2 n dx  c n −  b a fφ n dx  b a φ 2 n dx  2 −   b a fφ n dx  b a φ 2 n dx  2   Each term involving c n is non-negative, and is minimized for c n =  b a fφ n dx  b a φ 2 n dx . (28.3) 1339 We call these the generalized Fourier coefficients. For such a choice of the c n , the error is E =  b a f 2 dx − N  n=1 c 2 n  b a φ 2 n dx. Since the error is non-negative, we have  b a f 2 dx ≥ N  n=1 c 2 n  b a φ 2 n dx. This is known as Bessel’s Inequality. If the series in Equation 28.2 converges in the mean to f(x), lim N → ∞E = 0, then we have equality as N → ∞.  b a f 2 dx = ∞  n=1 c 2 n  b a φ 2 n dx. This is Parseval’s equality. Fourier coefficients. Previously we showed that if the series, f(x) = a 0 2 + ∞  n=1 (a n cos(nx) + b n sin(nx), converges uniformly then the coefficients in the series are the Fourier coefficients, a n = 1 π  π −π f(x) cos(nx) dx, b n = 1 π  π −π f(x) sin(nx) dx. 1340 Now we show that by choosing the coefficients to minimize the squared error, we obtain the same result. We apply Equation 28.3 to the Fourier eigenfunctions. a 0 =  π −π f 1 2 dx  π −π 1 4 dx = 1 π  π −π f(x) dx a n =  π −π f cos(nx) dx  π −π cos 2 (nx) dx = 1 π  π −π f(x) cos(nx) dx b n =  π −π f sin(nx) dx  π −π sin 2 (nx) dx = 1 π  π −π f(x) sin(nx) dx 28.4 Fourier Series for Functions Defined on Arbitrary Ranges If f(x) is defined on c − d ≤ x < c + d and f(x + 2d) = f(x), then f(x) has a Fourier series of the form f(x) ∼ a 0 2 + ∞  n=1 a n cos  nπ(x + c) d  + b n sin  nπ(x + c) d  . Since  c+d c−d cos 2  nπ(x + c) d  dx =  c+d c−d sin 2  nπ(x + c) d  dx = d, the Fourier coefficients are given by the formulas a n = 1 d  c+d c−d f(x) cos  nπ(x + c) d  dx b n = 1 d  c+d c−d f(x) sin  nπ(x + c) d  dx. 1341 Example 28.4.1 Consider the function defined by f(x) =      x + 1 for − 1 ≤ x < 0 x for 0 ≤ x < 1 3 − 2x for 1 ≤ x < 2. This function is graphed in Figure 28.3. The Fourier series converges to ˆ f(x) = (f(x − ) + f(x + ))/2, ˆ f(x) =                − 1 2 for x = −1 x + 1 for − 1 < x < 0 1 2 for x = 0 x for 0 < x < 1 3 − 2x for 1 ≤ x < 2. ˆ f(x) is also graphed in Figure 28.3. The Fourier coefficients are a n = 1 3/2  2 −1 f(x) cos  2nπ(x + 1/2) 3  dx = 2 3  5/2 −1/2 f(x − 1/2) cos  2nπx 3  dx = 2 3  1/2 −1/2 (x + 1/2) cos  2nπx 3  dx + 2 3  3/2 1/2 (x − 1/2) cos  2nπx 3  dx + 2 3  5/2 3/2 (4 − 2x) cos  2nπx 3  dx = − 1 (nπ) 2 sin  2nπ 3   2(−1) n nπ + 9 sin  nπ 3  1342 [...]... 0 .5 -1 1 0 .5 -0 .5 1 0 .5 2 1 .5 -1 -0 .5 1 0 .5 -0 .5 2 -0 .5 -1 1 .5 -1 Figure 28.3: A Function Defined on the range −1 ≤ x < 2 and the Function to which the Fourier Series Converges bn = = = 1 3/2 2 3 2 3 2 f (x) sin −1 5/ 2 f (x − 1/2) sin −1/2 1/2 dx dx + (4 − 2x) sin 2nπx 3 −1/2 2 3 2nπx 3 dx 2nπx 3 (x + 1/2) sin + =− 2nπ(x + 1/2) 3 5/ 2 3/2 2 nπ sin2 2 (nπ) 3 2 3 3/2 (x − 1/2) sin 1/2 dx 2(−1)n nπ + 4nπ... n for even n The function and the sum of the first three terms in the expansion are plotted, in dashed and solid lines respectively, in Figure 28.7 Although the three term sum follows the general shape of the function, it is clearly not a good approximation 13 54 1 0 .5 -1 0 .4 0.2 -0 .5 0 .5 1 -1 -0 .5 0 .5 -0 .5 -0.2 -1 1 -0 .4 Figure 28.7: Three Term Approximation for a Function with Jump Discontinuities and. .. oddn ∞ 4 n sin(nt) dt −π 4 1 − cos(nt) n n x −π 4 (− cos(nx) + (−1)n ) n2 ∞ =4 n=1 oddn −1 cos(nx) 4 2 n n2 n=1 oddn Since this series converges uniformly, ∞ 4 n=1 oddn ∞ cos(nx) −1 4 = F (x) = 2 n n2 n=1 −x − π x−π oddn The value of the constant term is ∞ 4 n=1 oddn −1 2 = 2 n π π 0 1 F (x) dx = − π 1360 for − π ≤ x < 0 for 0 ≤ x < π Thus ∞ − −x − π x−π 1 cos(nx) 4 = π n2 n=1 for − π ≤ x < 0 for. .. Continuous Function A Continuous Function Consider the function defined by  −x − 1  f2 (x) = x   −x + 1 for − 1 < x < −1/2 for − 1/2 < x < 1/2 for 1/2 < x < 1 1 355 0 .5 1 0.2 1 0.1 -1 -0 .5 1 0 .5 0. 25 1 0.1 -0.1 1 -0.2 0 0.1 Figure 28.8: Three Term Approximation for a Function with Continuous First Derivative and Comparison of the Rates of Convergence Since this function is continuous, the Fourier coefficients... sin(nx) and cos(nx) in terms of eınx and e−ınx we can obtain the complex form for a Fourier series ∞ ∞ a0 a0 1 1 + an cos(nx) + bn sin(nx) = + an (eınx + e−ınx ) + bn (eınx − e−ınx ) 2 2 2 ı2 n=1 n=1 ∞ = a0 + 2 n=1 1 1 (an − ıbn ) eınx + (an + ıbn ) e−ınx 2 2 ∞ cn eınx = n=−∞ where cn =  1  2 (an − ıbn )  a0  12  (a 2 −n + ıb−n ) 1 346 for n ≥ 1 for n = 0 for n ≤ −1 1 .5 1 0 .5 -2 -3 -1 1 2 3 -0 .5 -1... π f (x) cos(nx) dx 0 Example 28 .5. 1 Consider the function defined on [0, π) by f (x) = for 0 ≤ x < π/2 for π/2 ≤ x < π x π−x The Fourier cosine coefficients for this function are an = = π/2 2 π x cos(nx) dx + 0 2 π π 4 8 πn2 cos nπ 2 sin 2 nπ 4 π (π − x) cos(nx) dx π/2 for n = 0, for n ≥ 1 In Figure 28 .4 the even periodic extension of f (x) is plotted in a dashed line and the sum of the first five nonzero... is f (x) ∼ 4 π ∞ n=1 1 sin(nπx) n 1 For any fixed x, the series converges to 2 (f (x− ) + f (x+ )) For any δ > 0, the convergence is uniform in the intervals −1 + δ ≤ x ≤ −δ and δ ≤ x ≤ 1 − δ How will the nonuniform convergence at integral values of x affect the Fourier series? Finite Fourier series are plotted in Figure 28.9 for 5, 10, 50 and 100 terms (The plot for 100 terms is closeup of the behavior... 1 344 1 .5 1. 25 1 0. 75 0 .5 0. 25 -2 -3 -1 1 2 3 Figure 28 .4: Fourier Cosine Series 28.6 Fourier Sine Series If f (x) is an odd function, (f (−x) = −f (x)), then there will not be any cosine terms in the Fourier series Since f (x) cos(nx) is an odd function, the cosine coefficients will be zero Since f (x) sin(nx) is an even function,we can rewrite the sine coefficients bn = 2 π π f (x) sin(nx) dx 0 1 3 45 Example... 28.10.1 Consider the step function f (x) = π −π for 0 ≤ x < π for − π ≤ x < 0 1 358 1 1 1 1 1.2 1 1 0.1 0.8 Figure 28.9: Since this is an odd function, there are no cosine terms in the Fourier series bn = 2 π π π sin(nx) dx 0 π 1 = 2 − cos(nx) n 0 2 = (1 − (−1)n ) n 4 for odd n = n 0 for even n 1 359 ∞ f (x) ∼ n=1 oddn 4 sin nx n Integrating this relation, x x ∞ 4 sin(nt) dt n f (t) dt ∼ −π n=1 oddn ∞ x −π... by f (x) = for 0 ≤ x < π/2 for π/2 ≤ x < π x π−x The Fourier sine coefficients for this function are 2 π/2 2 π x sin(nx) dx + (π − x) sin(nx) dx π 0 π π/2 16 nπ nπ = cos sin3 2 πn 4 4 In Figure 28 .5 the odd periodic extension of f (x) is plotted in a dashed line and the sum of the first five nonzero terms in the Fourier sine series are plotted in a solid line bn = 28.7 Complex Fourier Series and Parseval’s . − 1 (nπ) 2 sin  2nπ 3   2(−1) n nπ + 9 sin  nπ 3  1 342 -1 -0 .5 0 .5 1 1 .5 2 -1 -0 .5 0 .5 1 -1 -0 .5 0 .5 1 1 .5 2 -1 -0 .5 0 .5 1 Figure 28.3: A Function Defined on the range −1 ≤ x < 2 and the Function to which the Fourier. ıb n ) e −ınx  = ∞  n=−∞ c n e ınx where c n =      1 2 (a n − ıb n ) for n ≥ 1 a 0 2 for n = 0 1 2 (a −n + ıb −n ) for n ≤ −1. 1 346 -3 -2 -1 1 2 3 -1 .5 -1 -0 .5 0 .5 1 1 .5 Figure 28 .5: Fourier Sine Series. The functions { nonzero terms in the Fourier cosine series are plotted in a solid line. 1 344 -3 -2 -1 1 2 3 0. 25 0 .5 0. 75 1 1. 25 1 .5 Figure 28 .4: Fourier Cosine Series. 28.6 Fourier Sine Series If f(x) is an odd