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Fourier Series When the French mathematician Joseph Fourier (1768–1830) was trying to solve a problem in heat conduction, he needed to express a function f as an infinite series of sine and cosine functions: ϱ ͚ ͑a f ͑x͒ a ϩ n cos nx ϩ bn sin nx͒ n1 a ϩ a1 cos x ϩ a2 cos 2x ϩ a3 cos 3x ϩ и и и ϩ b1 sin x ϩ b2 sin 2x ϩ b3 sin 3x ϩ и и и Earlier, Daniel Bernoulli and Leonard Euler had used such series while investigating problems concerning vibrating strings and astronomy The series in Equation is called a trigonometric series or Fourier series and it turns out that expressing a function as a Fourier series is sometimes more advantageous than expanding it as a power series In particular, astronomical phenomena are usually periodic, as are heartbeats, tides, and vibrating strings, so it makes sense to express them in terms of periodic functions We start by assuming that the trigonometric series converges and has a continuous function f ͑x͒ as its sum on the interval ͓Ϫ, ͔, that is, f ͑x͒ a ϩ ϱ ͚ ͑a n Ϫ ഛ x ഛ cos nx ϩ bn sin nx͒ n1 Our aim is to find formulas for the coefficients a n and bn in terms of f Recall that for a power series f ͑x͒ cn͑x Ϫ a͒ n we found a formula for the coefficients in terms of deriv͑n͒ atives: cn f ͑a͒͞n! Here we use integrals If we integrate both sides of Equation and assume that it’s permissible to integrate the series term-by-term, we get f ͑x͒ dx y y Ϫ Ϫ a dx ϩ y ϱ Ϫ cos nx ϩ bn sin nx͒ dx ϱ n But y n ͚ a y cos nx dx ϩ ͚ b y sin nx dx ͬ sin nx n n Ϫ n1 cos nx dx ͚ ͑a Ϫ n1 2 a ϩ ϱ n1 Ϫ Ϫ ͓sin n Ϫ sin͑Ϫn͔͒ n because n is an integer Similarly, xϪ sin nx dx So y Ϫ f ͑x͒ dx 2 a0 ■ FOURIER SERIES and solving for a0 gives ■ ■ Notice that a is the average value of f over the interval ͓Ϫ, ͔ a0 2 f ͑x͒ dx y Ϫ To determine an for n ജ we multiply both sides of Equation by cos mx (where m is an integer and m ജ 1) and integrate term-by-term from Ϫ to : y Ϫ f ͑x͒ cos mx dx a0 y Ϫ y Ϫ ͫ a0 ϩ ϱ ͚ ͑a n ͬ cos nx ϩ bn sin nx͒ cos mx dx n1 ϱ ϱ ͚ a y cos nx cos mx dx ϩ ͚ b y sin nx cos mx dx cos mx dx ϩ n n1 n Ϫ n1 Ϫ We’ve seen that the first integral is With the help of Formulas 81, 80, and 64 in the Table of Integrals, it’s not hard to show that y sin nx cos mx dx for all n and m Ϫ y cos nx cos mx dx Ϫ ͭ for n m for n m So the only nonzero term in (4) is am and we get y Ϫ f ͑x͒ cos mx dx am Solving for am , and then replacing m by n, we have an y Ϫ f ͑x͒ cos nx dx n 1, 2, 3, Similarly, if we multiply both sides of Equation by sin mx and integrate from Ϫ to , we get bn y Ϫ f ͑x͒ sin nx dx n 1, 2, 3, We have derived Formulas 3, 5, and assuming f is a continuous function such that Equation holds and for which the term-by-term integration is legitimate But we can still consider the Fourier series of a wider class of functions: A piecewise continuous function on ͓a, b͔ is continuous except perhaps for a finite number of removable or jump discontinuities (In other words, the function has no infinite discontinuities See Section 2.4 for a discussion of the different types of discontinuities.) FOURIER SERIES ■ Definition Let f be a piecewise continuous function on ͓Ϫ, ͔ Then the Fourier series of f is the series a0 ϩ ϱ ͚ ͑a n cos nx ϩ bn sin nx͒ n1 where the coefficients an and bn in this series are defined by 2 a0 an y Ϫ y Ϫ f ͑x͒ cos nx dx f ͑x͒ dx bn y Ϫ f ͑x͒ sin nx dx and are called the Fourier coefficients of f Notice in Definition that we are not saying f ͑x͒ is equal to its Fourier series Later we will discuss conditions under which that is actually true For now we are just saying that associated with any piecewise continuous function f on ͓Ϫ, ͔ is a certain series called a Fourier series EXAMPLE Find the Fourier coefficients and Fourier series of the square-wave function f defined by f ͑x͒ ͭ if Ϫ ഛ x Ͻ if ഛ x Ͻ f ͑x ϩ 2͒ f ͑x͒ and So f is periodic with period 2 and its graph is shown in Figure y ■ ■ Engineers use the square-wave function in describing forces acting on a mechanical system and electromotive forces in an electric circuit (when a switch is turned on and off repeatedly) Strictly speaking, the graph of f is as shown in Figure 1(a), but it’s often represented as in Figure 1(b), where you can see why it’s called a square wave _π π 2π x π 2π x (a) y _π FIGURE Square-wave function (b) SOLUTION Using the formulas for the Fourier coefficients in Definition 7, we have a0 2 y Ϫ f ͑x͒ dx 2 y dx ϩ 2 y Ϫ dx ϩ 1 ͑͒ 2 ■ FOURIER SERIES and, for n ജ 1, an y Ϫ y Ϫ ͭ n ͬ Ϫ y dx ϩ y Ϫ cos nx dx ͑sin n Ϫ sin 0͒ n f ͑x͒ sin nx dx cos nx Ϫ n ■ ■ Note that cos n equals if n is even and Ϫ1 if n is odd ͬ sin nx n 0ϩ bn f ͑x͒ cos nx dx y dx ϩ y Ϫ sin x dx ͑cos n Ϫ cos 0͒ n if n is even if n is odd The Fourier series of f is therefore a ϩ a1 cos x ϩ a2 cos 2x ϩ a3 cos 3x ϩ и и и ϩ b1 sin x ϩ b2 sin 2x ϩ b3 sin 3x ϩ и и и ϩ ϩ ϩ ϩ иии ϩ 2 sin x ϩ sin 2x ϩ sin 3x ϩ sin 4x ϩ sin 5x ϩ и и и 3 5 2 2 ϩ sin x ϩ sin 3x ϩ sin 5x ϩ sin 7x ϩ и и и 3 5 7 Since odd integers can be written as n 2k Ϫ 1, where k is an integer, we can write the Fourier series in sigma notation as ϱ ϩ͚ sin͑2k Ϫ 1͒x k1 ͑2k Ϫ 1͒ In Example we found the Fourier series of the square-wave function, but we don’t know yet whether this function is equal to its Fourier series Let’s investigate this question graphically Figure shows the graphs of some of the partial sums Sn͑x͒ 2 ϩ sin x ϩ sin 3x ϩ и и и ϩ sin nx 3 n when n is odd, together with the graph of the square-wave function FOURIER SERIES ■ y y 1 y S£ S¡ _π π x _π y x π _π y 1 x x π x S¡∞ S¡¡ π π y S¶ _π S∞ _π x π _π FIGURE Partial sums of the Fourier series for the square-wave function We see that, as n increases, Sn͑x͒ becomes a better approximation to the square-wave function It appears that the graph of Sn͑x͒ is approaching the graph of f ͑x͒, except where x or x is an integer multiple of In other words, it looks as if f is equal to the sum of its Fourier series except at the points where f is discontinuous The following theorem, which we state without proof, says that this is typical of the Fourier series of piecewise continuous functions Recall that a piecewise continuous function has only a finite number of jump discontinuities on ͓Ϫ, ͔ At a number a where f has a jump discontinuity, the one-sided limits exist and we use the notation f ͑aϩ͒ limϩ f ͑x͒ f ͑aϪ͒ lim f ͑x͒ x l aϪ xla Fourier Convergence Theorem If f is a periodic function with period 2 and f and f Ј are piecewise continuous on ͓Ϫ, ͔, then the Fourier series (7) is convergent The sum of the Fourier series is equal to f ͑x͒ at all numbers x where f is continuous At the numbers x where f is discontinuous, the sum of the Fourier series is the average of the right and left limits, that is ͓ f ͑xϩ͒ ϩ f ͑xϪ͔͒ If we apply the Fourier Convergence Theorem to the square-wave function f in Example 1, we get what we guessed from the graphs Observe that f ͑0ϩ͒ limϩ f ͑x͒ xl0 and f ͑0Ϫ͒ lim f ͑x͒ x l 0Ϫ and similarly for the other points at which f is discontinuous The average of these left and right limits is 12 , so for any integer n the Fourier Convergence Theorem says that ϱ ϩ͚ sin͑2k Ϫ 1͒x k1 ͑2k Ϫ 1͒ (Of course, this equation is obvious for x n.) ͭ f ͑x͒ if n n if x n ■ FOURIER SERIES Functions with Period 2L If a function f has period other than 2, we can find its Fourier series by making a change of variable Suppose f ͑x͒ has period 2L, that is f ͑x ϩ 2L͒ f ͑x͒ for all x If we let t x͞L and t͑t͒ f ͑x͒ f ͑Lt͒͞ then, as you can verify, t has period 2 and x ϮL corresponds to t Ϯ The Fourier series of t is ϱ ͚ ͑a a0 ϩ n cos nt ϩ bn sin nt͒ n1 where 2 a0 an y t͑t͒ dt Ϫ bn y t͑t͒ cos nt dt Ϫ y t͑t͒ sin nt dt Ϫ If we now use the Substitution Rule with x Lt͞, then t x͞L, dt ͑͞L͒ dx, and we have the following If f is a piecewise continuous function on ͓ϪL, L͔, its Fourier series is ϱ ͚ a0 ϩ n x L a n cos n1 Notice that when L these formulas are the same as those in (7) ͫ ͩ ͪ ͩ ͪͬ ϩ bn sin n x L where ■ ■ 2L a0 and, for n ജ 1, an L y L ϪL ͩ ͪ f ͑x͒ cos n x L y L f ͑x͒ dx ϪL bn dx L y L ϪL ͩ ͪ f ͑x͒ sin n x L dx Of course, the Fourier Convergence Theorem (8) is also valid for functions with period 2L Խ Խ EXAMPLE Find the Fourier series of the triangular wave function defined by f ͑x͒ x for Ϫ1 ഛ x ഛ and f ͑x ϩ 2͒ f ͑x͒ for all x (The graph of f is shown in Figure 3.) For which values of x is f ͑x͒ equal to the sum of its Fourier series? y FIGURE The triangular wave function _1 x FOURIER SERIES ■ SOLUTION We find the Fourier coefficients by putting L in (9): a 12 y Ϫ1 ■ ■ Notice that a is more easily calculated as an area Խ x Խ dx y ] Ϫ1 Ϫ 14 x ϩ 14 x Ϫ1 ] ͑Ϫx͒ dx ϩ 12 y x dx 12 and for n ജ 1, an y Ϫ1 Խ x Խ cos͑n x͒ dx y x cos͑n x͒ dx Խ Խ because y x cos͑n x͒ is an even function Here we integrate by parts with u x and dv cos͑n x͒ dx Thus, ͫ ͬ x an sin͑n x͒ n 0Ϫ n ͫ Ϫ n y ͬ cos͑n x͒ n Ϫ 1 sin͑n x͒ dx ͑cos n Ϫ 1͒ n2 Խ Խ Since y x sin͑n x͒ is an odd function, we see that bn y Ϫ1 Խ x Խ sin͑n x͒ dx We could therefore write the series as ϱ 2͑cos n Ϫ 1͒ ϩ͚ cos͑n x͒ n1 n2 But cos n if n is even and cos n Ϫ1 if n is odd, so an ͭ ͑cos n Ϫ 1͒ n2 Ϫ 2 n if n is even if n is odd Therefore, the Fourier series is 4 Ϫ cos͑ x͒ Ϫ cos͑3 x͒ Ϫ cos͑5 x͒ Ϫ и и и 9 25 ϱ Ϫ͚ cos͑͑2k Ϫ 1͒ x͒ n1 ͑2k Ϫ 1͒2 The triangular wave function is continuous everywhere and so, according to the Fourier Convergence Theorem, we have f ͑x͒ ϱ Ϫ͚ cos͑͑2k Ϫ 1͒ x͒ n1 ͑2k Ϫ 1͒2 for all x ■ FOURIER SERIES In particular, Խ Խ x ϱ Ϫ͚ cos͑͑2k Ϫ 1͒ x͒ k1 ͑2k Ϫ 1͒2 for Ϫ1 ഛ x ഛ Fourier Series and Music One of the main uses of Fourier series is in solving some of the differential equations that arise in mathematical physics, such as the wave equation and the heat equation (This is covered in more advanced courses.) Here we explain briefly how Fourier series play a role in the analysis and synthesis of musical sounds We hear a sound when our eardrums vibrate because of variations in air pressure If a guitar string is plucked, or a bow is drawn across a violin string, or a piano string is struck, the string starts to vibrate These vibrations are amplified and transmitted to the air The resulting air pressure fluctuations arrive at our eardrums and are converted into electrical impulses that are processed by the brain How is it, then, that we can distinguish between a note of a given pitch produced by two different musical instruments? The graphs in Figure show these fluctuations (deviations from average air pressure) for a flute and a violin playing the same sustained note D (294 vibrations per second) as functions of time Such graphs are called waveforms and we see that the variations in air pressure are quite different from each other In particular, the violin waveform is more complex than that of the flute t FIGURE Waveforms t (a) Flute (b) Violin We gain insight into the differences between waveforms if we express them as sums of Fourier series: ͩ ͪ P͑t͒ a ϩ a1 cos t L ͩ ͪ ϩ b1 sin t L ͩ ͪ ϩ a2 cos 2 t L ͩ ͪ ϩ b2 sin 2 t L ϩ иии In doing so, we are expressing the sound as a sum of simple pure sounds The difference in sounds between two instruments can be attributed to the relative sizes of the Fourier coefficients of the respective waveforms The n th term of the Fourier series, that is, ͩ ͪ ͩ ͪ a n cos n t L ϩ bn n t L is called the nth harmonic of P The amplitude of the n th harmonic is A n sa 2n ϩ b2n and its square, A2n a 2n ϩ b2n , is sometimes called energy of the n th harmonic (Notice that FOURIER SERIES ■ Խ Խ for a Fourier series with only sine terms, as in Example 1, the amplitude is A n bn and the energy is A2n b 2n.) The graph of the sequence ͕A2n ͖ is called the energy spectrum of P and shows at a glance the relative sizes of the harmonics Figure shows the energy spectra for the flute and violin waveforms in Figure Notice that, for the flute, A2n tends to diminish rapidly as n increases whereas, for the violin, the higher harmonics are fairly strong This accounts for the relative simplicity of the flute waveform in Figure and the fact that the flute produces relatively pure sounds when compared with the more complex violin tones A@n A@n FIGURE Energy spectra n 10 (a) Flute n 10 (b) Violin In addition to analyzing the sounds of conventional musical instruments, Fourier series enable us to synthesize sounds The idea behind music synthesizers is that we can combine various pure tones (harmonics) to create a richer sound through emphasizing certain harmonics by assigning larger Fourier coefficients (and therefore higher corresponding energies) Exercises 7–11 Click here for solutions S Find the Fourier series of the function ͭ A function f is given on the interval ͓Ϫ, ͔ and f is periodic with period 2 (a) Find the Fourier coefficients of f (b) Find the Fourier series of f For what values of x is f ͑x͒ equal to its Fourier series? ; (c) Graph f and the partial sums S2, S4, and S6 of the Fourier series if x Ͻ if ഛ x Ͻ f ͑x ϩ 4͒ f ͑x͒ f ͑x͒ if Ϫ2 ഛ x Ͻ if ഛ x Ͻ if ഛ x Ͻ f ͑x ϩ 4͒ f ͑x͒ ͭ 1–6 ͭ ͭ ■ f ͑x͒ x ͭ ͭ f ͑x͒ ■ ■ ■ Ϫ1 ■ ■ ■ f ͑x ϩ 2͒ f ͑x͒ Ϫ1 ഛ t ഛ ■ ■ ■ ■ ■ ■ ■ ■ so-called half-wave rectifier that clips the negative part of the wave Find the Fourier series of the resulting periodic function if Ϫ ഛ x Ͻ Ϫ͞2 if Ϫ͞2 ഛ x Ͻ if ഛ x Ͻ ■ ■ Ϫ1 ഛ x ഛ f ͑x ϩ 8͒ f ͑x͒ ■ 12 A voltage E sin t, where t represents time, is passed through a if Ϫ ഛ x Ͻ if ഛ x Ͻ ■ ■ Խ Խ if Ϫ4 ഛ x Ͻ if ഛ x Ͻ 11 f ͑t͒ sin͑3 t͒, f ͑x͒ x f ͑x͒ cos x Ϫx 10 f ͑x͒ Ϫ x, if Ϫ ഛ x Ͻ if ഛ x Ͻ x f ͑x͒ ͭ f ͑x͒ if Ϫ ഛ x Ͻ if ഛ x Ͻ 1 f ͑x͒ Ϫ1 Խ Խ f ͑x͒ ■ ͭ f ͑t͒ ■ ■ ■ ■ ■ E sin t ഛtϽ0 if ഛ t Ͻ if Ϫ f ͑t ϩ 2͒͞ f ͑t͒ 10 ■ FOURIER SERIES 18 Use the result of Example to show that 13–16 Sketch the graph of the sum of the Fourier series of f without actually calculating the Fourier series ͭ ͭ 1ϩ Ϫ1 if Ϫ4 ഛ x Ͻ if ഛ x Ͻ 13 f ͑x͒ 19 Use the result of Example to show that x if Ϫ1 ഛ x Ͻ 14 f ͑x͒ Ϫ x if ഛ x Ͻ 15 f ͑x͒ x 3, Ϫ1 ഛ x ഛ 16 f ͑x͒ e x, Ϫ2 ഛ x ഛ ■ ■ ■ ■ ■ ■ 1Ϫ 1 ϩ Ϫ ϩ иии 20 Use the given graph of f and Simpson’s Rule with n to ■ ■ ■ ■ 17 (a) Show that, if Ϫ1 ഛ x ഛ 1, then x2 1 2 ϩ ϩ ϩ и и и 32 52 72 ■ ■ ■ estimate the Fourier coefficients a 0, a1, a 2, b1, and b2 Then use them to graph the second partial sum of the Fourier series and compare with the graph of f y ϱ ϩ ͚ ͑Ϫ1͒ n 2 cos͑n x͒ n n1 (b) By substituting a specific value of x, show that ϱ ͚ n1 2 n 0.25 x FOURIER SERIES ■ 11 SOLUTIONS (a) a0 = an = 2π π −π π π −π π π −π f (x) dx = π −π 2π −π dx − f(x) cos nx dx = π f (x) sin nx dx = π = − [1 − cos(−nπ)] = − nπ nπ bn = (b) f (x) = ∞ k=0 π −π − π dx = 0 −π cos nx dx − π π cos nx dx = [since cos nx is even] −π sin nx dx − π π sin nx dx = π −π sin nx dx [since sin nx is odd] if n even if n odd sin(2k + 1)x when −π < x < and < x < π (2k + 1)π (c) y 0.5 -2.5 -1.25 1.25 2.5 x -0.5 -1 (a) a0 = 2π π −π f (x) dx = 2π π −π x dx = an = π π −π f (x) cos nx dx = π π −π x cos nx dx = [because x cos nx is odd] bn = π π −π f (x) sin nx dx = π π −π x sin nx dx = =− (b) f (x) = cos nπ n ∞ n=1 [using integration by parts] = (−1)n+1 sin nx n π π x sin nx dx [since x sin nx is odd] −(2/n) if n even (2/n) if n odd y (c) 2.5 when −π < x < π 1.25 -2.5 -1.25 1.25 2.5 x -1.25 -2.5 12 ■ FOURIER SERIES (a) a0 = an = bn = = 2π π −π π π (b) f (x) = 2π f (x) dx = π cos x dx = π −π f (x) cos nx dx = π π cos x cos nx dx = π −π f (x) sin nx dx = π π cos x sin nx dx 2n π(n2 − 1) if n even if n odd cos x + if n = if n = [by symmetry about x = π ] using an integral table, and simplified using the addition formula for cos(a + b) 4k sin(2k) when −π < x < 0, < x < π − 1) π (4k k=1 ∞ y (c) 0.5 -2.5 -1.25 1.25 2.5 x -0.5 -1 Use f (x) = a0 = 2L L −L an = L L −L L L −L bn = if − ≤ x ≤ −1 if − < x < 1, L = if ≤ x ≤ f (x) dx = −1 dx = nπx dx = f (x) cos L nπx dx = L f (x) sin Fourier Series: + + −1 −1 π nπx 2/nπ dx = sin n = cos nπ −2/nπ sin k=1 if n = 4n + if n = 4n + nπx dx = 2 πx 3πx cos − cos π 3π ∞ if n even + 5πx cos 5π −··· π π sin (4k + 1) − sin (4k + 3) (4k + 1) π (4k + 3) π y 0.75 0.5 0.25 -5 -2.5 2.5 x FOURIER SERIES ■ 13 Use f (x) = a0 = 2L an = L L −L L −L −x if − ≤ x < 0 if ≤ x ≤ f (x) dx = f (x) cos −4 , L = −x dx = nπx dx = L −4 nπx dx = (cos (nπ) − 1) = (nπ)2 −x cos if n is even −8/(nπ) bn = L L −L if n is odd f(x) sin nπx dx = L −4 −x sin nπx dx = cos (nπ) = nπ 4/nπ if n is even −4/nπ if n is odd Fourier Series: 1+ ∞ k=1 − π π π cos sin (2k − 1)x − (2k − 1) x + sin (2k)x (2k − 1)π (2k − 1)2 π2 (2k)π y -5 -2.5 2.5 x 11 Use f (x) = {sin(3πt) if − ≤ t ≤ , L = Note: This can be done instantly if one observes that the period of sin(3πt) is 23 , and the period of f (x) = which is an integer multiple of 23 Therefore f(x) is the same as sin(3πt) for all t, and its Fourier series is therefore sin(3πt) We can get this result using the standard coefficient formulas: a0 = 2L L −L f (x) dx = −1 sin(3πx) dx = 1 nπx f (x) cos dx = −1 sin(3πx) cos(nπx) dx L −1 L = [applying change of variables to a formula in the section] an = L nπx dx = −1 sin(3πx) sin(nπx) dx f (x) sin L −L L sin nπ 6 if n = π (−9 + n2 ) = [using integral table and addition formula = if n = bn = if n = if n = 14 ■ FOURIER SERIES Fourier Series: sin(3πx) y 0.5 -5 -2.5 2.5 x -0.5 -1 13 −1 −1 if −5 ≤ x < −4 y if −4 ≤ x < if ≤ x < if ≤ x < -5 -2.5 2.5 x -1 15 y 0.5 -1.25 1.25 2.5 3.75 x -0.5 -1 17 (a) We find the Fourier series for f (x) = {x2 a0 = 2L L −L f (x) dx = −1 x2 dx = if −1 ≤ x ≤ 1, L = 1 (nπ)2 1 nπx an = f (x) cos cos nπ = dx = −1 x2 cos(nπx) dx = L −1 L (nπ)2 − (nπ)2 bn = L L −L f (x) sin So we have x2 = + nπx dx = L ∞ n=1 (−1)n −1 x2 sin(nπx) dx = because x2 sin(nπx) is odd cos(nπx) for −1 ≤ x ≤ (nπ)2 (b) We let x = in the above to obtain 1= + ∞ n=1 (−1)n cos(nπ) (nπ)2 = π2 n n=1 ∞ ∞ π2 = n=1 n if n even if n odd FOURIER SERIES ■ 15 19 Example says that, for ≤ x < π, = Let x = π 2 + + ∞ k=1 sin((2k − 1)x) (2k − 1)π to obtain 1= ∞ k=1 π sin (2k − 1) (2k − 1)π ∞ π = sin((2k − 1)) k=1 (2k − 1) π =1− + − + ··· −1.25(−1) + 4(0.25) − √12 + 2(−2.25)(0) + 4(−1.25) √12 a1 = −1 f (x) cos(πx) dx ≈ 12 + 2(3.5)(1) + 4(3.5) √1 + 2(0)(0) + 4(−2.75) − √1 − 1.25(−1) 2 √ = 19 1+ 24 a2 = = b1 = ≈ −1 f (x) cos(2πx) dx ≈ −1 12 f (x) sin(πx) dx −1.25(0) + 4(0.25) − √12 + 2(−2.25)(−1) + (−1.25) − √12 + 2(3.5)(0) + 4(3.5) + 4(−2.75) √ 9+7 = 24 b2 = = −1.25(1) + 4(0.25)(0) + 2(−2.25)(−1) + 4(−1.25)(0) 12 + 2(3.5)(1) + 4(3.5)(0) + 2(0)(−1) + 4(−2.75)(0) − 1.25(1) −1 √1 f (x) sin(2πx) dx ≈ − 1.25(0) + 4(3.5) 12 √ + 2(0)(1) + 4(−2.75) √1 − 1.25(0) −1.25(0) + 4(0.25)(1) + 2(−2.25)(0) + 4(−1.25)(−1) + 2(3.5)(0) + 4(3.5)(1) + 2(0)(0) + 4(−2.75)(−1) − 1.25(0) 31 12 f (x) ≈ − 12 + 19 24 √ + cos(πx) + √ 9+7 24 sin(πx) + cos(2πx) + y -1 -0.5 0.5 x -2 31 12 sin(2πx) √1 + 2(0)(1)