MATH 124B Solution Key HW 01

Transcription

1 5.1 THE COEFFICIENTS MATH 14B Soution Key HW THE COEFFICIENTS 1. In the expansion 1 = n odd (4/nπ) sin nx, vaid for < x < π, put x = π/4 to cacuate the sum = Hint: Since each of the series converges, then can be combined as indicated. However, they cannot be arbitrariy rearranged because they are ony conditionay, not absoutey, convergent. SOLUTION. Putting x = π/4 makes it pain that 1 = 4 1 π n sin nπ 4 n odd sin π sin 3π sin 5π sin 7π4 + = 4 π = 4 π Hence; π 4 = 4 π = π = 4 π = = Let φ(x) x for x 1 =. (a) Cacuate its Fourier sine series (b) Cacuate its Fourier cosine series. SOLUTION. 1 Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13

2 5.1 THE COEFFICIENTS (a) For = 1, the Fourier coefficients are given by A m = x sin(x). Integrating by parts twice makes it pain that A m = x cos x cos x x cos sin x 1 = + 4x () 4 = ( 1)m+1 x 1 + 4cos () 3 = ( 1)m+1 + 4( 1)m 1 () 3. sin x () (b) For = 1, the Fourier cosine coefficients are When m = it s easy to see that A m = x cos(x). A = /3. Otherwise, we integrate by parts twice and discover A m = x sin(x) 1 4 cos x 1 = 4x () 4 = 4 ( 1)m (). sin x x cos x () 5. Given the Fourier sine series of φ(x) x on (, ). Assume the series can be integrated term by term, a fact that wi be shown ater. 1 (a) Find the Fourier cosine series of the function x /. Find the constant of integration that wi be the first term in the cosine series. (b) Then by setting x = in your resut, find the sum of the series ( 1) n+1 1 This resut is highy non-trivia. The proof invoves advanced techniques from Math 118B (rea anaysis). n. Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13

3 5. EVEN, ODD, PERIODIC, AND COMPLEX FUNCTIONS SOLUTION. (a) Armed with the resuts from Exampe 3 on page 19 in our text we see that Fourier sine series of φ(x) x on (, ) is given by φ(x) x = sin πx π 1 πx sin + 1 3πx sin 3 x m+1 = ( 1) sin. m=1 Integrating both sides our equaity in the variabe x aong with the assumption vaidating term-by-term integration, we indeed obtain the Fourier cosine series x = ( 1) m x cos () m=1 + C, where C = A = 1 x = 6. (b) Putting x = in the resut of part (a) we see that = 6 + ( 1) m (), m=1 or equivaenty, π 1 = m=1 ( 1) m+1 m. 5. EVEN, ODD, PERIODIC, AND COMPLEX FUNCTIONS 1. For each of the foowing functions, state whether it is even or odd or periodic. If periodic, what is its smaest period? (a) sin ax (a > ) (b) e ax (a > ) (c) x m (d) tan x (m = integer) (e) sin(x/b) (b > ) (f) x cos ax (a > ) 3 Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13

4 5. EVEN, ODD, PERIODIC, AND COMPLEX FUNCTIONS SOLUTION. (a) Odd, period π/a. (b) Neither even nor odd nor periodic. (c) Even when m is even, whie odd when m is odd, not periodic. (d) Even, not periodic. (e) Even, period πb. (f) Odd, not periodic. 5. Show that the Fourier sine series on (, ) can be derived form the fu Fourier series on (, ) as foows. Let φ(x) be any (continuous) function on (, ). Let φ(x) be its odd extension. Write the fu series for φ(x) on (, ). Assume that its sum is φ(x). By Exercise 4, this series has ony sine terms. Simpy restrict your attention to < x < to get the sine series for φ(x). SOLUTION. Reca that definite integras on symmetric intervas of odd and even functions have the foowing usefu property (odd) = and (even) = (even). Now, given a function φ(x) on (, ), the odd extension is given by φ odd (x) = φ(x) φ(x), < x <, = φ( x), < x <,, x =. Each of the terms in the Fourier sine series for φ(x), sin nπx, is odd. As with the fu Fourier series, each of these terms aso has period. So we may think of the Fourier series as the expansion of an odd function with period defined on the entire ine which coincides with φ(x) on (, ). In fact, the fu Fourier seres of φ odd (x) is precisey the same as the Fourier series of φ(x). To see this, et 1 A + A n cos nπx + B n sin nπx be the Fourier series for φ odd (x), with the usua coefficients. Then A n = 1 φ odd (x) cos nπx =, 4 Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13

5 5. EVEN, ODD, PERIODIC, AND COMPLEX FUNCTIONS inasmuch, φ odd is odd and cos is even, so their product is again odd. Aso, both φ odd and sin are odd, so their product is even, hence; B n = 1 = = φ odd (x) sin nπx φ odd (x) sin nπx φ(x) sin nπx which are the Fourier sine coefficients of φ(x). Therefore, as the Fourier sine series of φ(x) is the fu Fourier series of φ odd, the -periodic odd function that the Fourier sine series expands is just the periodic extension of φ odd. 6. Show that the cosine series on (, ) can be derived from the fu series on (, ) by using the even extension of a function., SOLUTION. The argument is neary identica as that presented in Exercise (a) Let φ(x) be a continuous function on (, ). Under what conditions is its odd extension aso a continuous function? (b) Let φ(x) be a differentiabe function on (, ). Under what conditions is its odd extension aso a differentiabe function? (c) Same as part (a) for the even extension. (d) Same as part (b) for the even extension. 15. Without any computation, prediction which of the Fourier coefficients of sin x on the interva ( π, π) must vanish. SOLUTION. Since sin x is odd it foows that sin x defines an even function. By the resut estabished in Exercise 4 we note that the fu Fourier series on ( π, π) has ony cosine terms. That is, a of the sine coefficients must necessariy vanish. 16. Use the De Moivre formuas sin θ = eiθ e iθ, and cos θ = eiθ + e iθ. (5.11) i to derive the standard formuas for cos(θ + φ) and sin(θ + φ). 5 Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13

6 5. EVEN, ODD, PERIODIC, AND COMPLEX FUNCTIONS SOLUTION. A direct cacuation makes it pain that cos θ + i sin θ = eiθ + e iθ = eiθ + e iθ = eiθ = e iθ. + i e iθ e iθ i + eiθ e iθ Armed with this resut we now assume we have two compex numbers which we write as: and e iθ = cos θ + i sin θ e iφ = cos φ + i sin φ. We mutipy these compex numbers together. Mutipying the eft hand sides: We can write this answer as: Mutipying the right hand sides: e iθ e iφ = e i(θ+φ). e i(θ+φ) = cos(θ + φ) + i sin(θ + φ). (5.1) (cos θ + i sin θ) cos φ + i sin φ = cos θ cos φ sin θ sin φ + i(cos θ sin φ + sin θ sin φ), (5.) inasmuch i = 1. Now, equating (5.1) and (5.): cos(θ + φ) + i sin(θ + φ) = cos θ cos φ sin θ sin φ + i(cos θ sin φ + sin θ cos φ). Equating the rea parts gives: Equating the imaginary parts gives: cos(θ + φ) = cos θ cos φ sin θ sin φ. sin(θ + φ) = sin θ cos φ + cos θ sin φ. 6 Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13

7 5.3 ORTHOGONALITY AND GENERAL FOURIER SERIES 5.3 ORTHOGONALITY AND GENERAL FOURIER SERIES. (a) On the interva [ 1, 1], show that the function x is orthogona to the constant functions. (b) Find a quadratic poynomia that is orthogona to both 1 an. (c) Find a cubic poynomia that is orthogona to a quadratics. (These are the first few Legendre poynomias.) SOLUTION. (a) A direct cacuation makes it pain that 1 x C = C x 1 1 =. (b) Since x and 1 are even functions it must be the case that they are orthogona to x, which is odd. This suggest that we produce some constant C such that x + C is orthogona to 1. Hence, 1 (x + C) 1 = x C x = + C =. 1 3 We have a soution for C = 1/3, so the desired quadratic is any non-zero mutipe of x 1/3. (c) In a simiar fashion we ook for x 3 + C x orthogona to x. We find 1 (x 3 + C x) x = x C x = 5 + C 3 =. Taking C = 3/5 we get that any non-zero mutipe of x 3 3x/5 is orthogona to x by construction and orthogona to both 1 an by an even-odd argument. 3. Consider u t t = c u x x for < x <, with boundary conditions u(, t) =, u x (, t) = and the initia conditions u(x, ) = x, u t (x, ) =. Find the soution expicity in series form. SOLUTION. Reca that a separated soution is a soution of the form u(x, t) = X (x)t(t). Pugging this form into the wave equation we get X (x)t (t) = c X (x)t(t), 7 Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13

8 5.3 ORTHOGONALITY AND GENERAL FOURIER SERIES or equivaenty, T (t) c T(t) = X (x) X (x) = λ for some constant λ. Indeed, since the eft-hand-side is independent of x and the righthand-side is independent of t, respectivey, it foows that λ is independent of both t and x and must be constant. Now, et λ = β, where β >, then we obtain two separate ODE s. Namey, X + β X = and T + (cβ) T =. Next, we sove for the function X by imposing our given boundary conditions. That is, u(, t) = X ()T(t) = X () = X () =, u x (, t) = X ()T(t) = and we obtain the Sturm-Liouvie eigenvaue probem X + λx = X () = X () =. Since d defines a sef-adjoint operator on the space of functions satisfying the Dirichet boundary conditions, it foows that a of its eigenvaues are rea. More specificay, λ. The cases λ =, λ < and λ > must be addressed separatey. λ = : We get λ < : We get X = hence, X (x) = Ax + B. The conditions X () = X () = makes it pain that X () = B = an () = A = so X and we have a trivia soution. and λ > : We get X β X = X (x) = Ae β x + Be β x Again, X () = A + B = and X () = βae β βbe β = reveas that A = B = and our soution is again trivia. and X + β X = X (x) = Acos β x + B sin β x. The conditions X () = A = and X () = βb cos β x = make it pain that This is a standard resut in upper-division inear agebra. β = nπ + π/ = (n + 1/)π/. 8 Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13

9 5.3 ORTHOGONALITY AND GENERAL FOURIER SERIES Indeed, the soution is non-trivia when λ > and by the superposition principe we obtain the famiy X n (x) = B n sin(n + 1/)πx/, where B n denotes the usua Fourier cosine coefficient. Furthermore, we sove for T in a simiar fashion, that is, u(x, ) = X (x)t() = x T () =, u t (x, ) = X (x)t () = Since, it foows that T + (cβ) T =, T(t) = C cos(cβ t) + D sin(cβ t), thus T (t) = cβc sin(cβ t) + cβ D cos(cβ t). Imposing the condition T () = makes it pain that D =, hence; T(t) = C cos(cβ t) and by the superposition principe we get the famiy T n (t) = C n cos(cβ n t) where C n denotes the usua Fourier cosine coefficient with β n = (n + 1/)π/. Putting a of this together we see that u(x, t) = T n (t)x n (x). The condition u(x, ) = x now gives that x = T n ()X n (x) = C n X n (x) = C n B n sin(n+1/)πx/ = where K n = C n B n is given by the inner product K n = (n + 1/)πx x sin 4 [π(n + 1) sin(πn) + cos(πn)] = (πn + π) = 8( 1)n (πn + π), since cos(nπ) = ( 1) n and sin(nπ) = for a n = 1,, 3,... Therefore, 8( 1) n u(x, t) = T n (t)x n (x) = (πn + π) cos(cβ nt) sin(β n x). K n sin(n+1/)πx/, 9 Soutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 14B, Winter 13