Math 5 HW #5 Solutions From 29 4 Find the power series representation for the function and determine the interval of convergence Answer: Using the geometric series formula, f(x) = 3 x 4 3 x 4 = 3(x 4 ) n = 3x 4n Using the Ratio Test, 3x 4(n+) 3x 4n = x 4, which is less than when x < Checking the endpoints, we see that when x =, the series is 3 () 4n = 3, which diverges When x =, the series is 3 ( ) 4n = 3, which diverges the interval of convergence is (, ) 0 Find a power series representation for the function f(x) = and determine the interval of convergence x2 x 3 Answer: Re-writing f as ( ) ( ) f(x) = x 2 x 3 = x2, x3 we can use the geometric series to see that f(x) = x2 ( ) x 3 n = x2 x 3n n = x 3n+2 n+3
Using the Ratio Test, x 3(n+)+2 (n+)+3 x 3n+2 n+3 = x 3, which is less than when x < a Checking the endpoints, when x = a, the series is which diverges When x = a, the series is n+2 n+3 = a, ( a) 3n+2 n+3 = ( ) 3n+2 a, which also diverges the interval of convergence is ( a, a) 6 Find a power series representation for the function f(x) = and determine the radius of convergence Answer: Write f(x) as f(x) = x 2 n= x 2 ( 2x) 2 ( ) f(x) = x 2 ( 2x) 2 n(2x) n = x 2 n= 2 n nx n = 2 n nx n+ n= Using the Ratio Test, 2 n (n + )x n+2 2 n nx n+ = 2 x n + n n + = 2 x = 2 x, n which is less than when x < /2 the radius of convergence is /2 24 Evaluate the indefinite integral ln( t) dt t as a power series What is the radius of convergence? Answer: From Example 6, we know that the power series for ln( t) is t t2 2 t3 3 = t n n 2 n=
the series for ln( t) t t ln( t) dt = t is ( t t22 t33 ) = t 2 t2 3 = t n n + ( t2 t23 ) dt = C t t2 4 t3 9 = C n= t n n 2 Using the Ratio Test, t n+ (n+) 2 t n n 2 = t n 2 n 2 + 2n + = t, so the radius of convergence is From 20 8 Find the Maclaurin series for f(x) = cos 3x using the definition of a Maclaurin series Also find the associated radius of convergence Answer: We compute the first few derivatives: So, by the definition of the Maclaurin series, f (x) = 3 sin 3x f (x) = 9 cos 3x f (x) = 27 sin 3x f (4) (x) = 8 cos 3x f(0) = f (0) = 0 f (0) = 9 f (0) = 0 f (4) (0) = 8 f(x) = f(0) + f (0)x + f (0) x 2 = 9 2! 2 x2 + 8 4 x4 = 3 ( ) n 32n (2n)! x2n
Using the Ratio Test, ( ) n+ 32n+2 (2n+2)! x2n+2 ( ) n 3 2n (2n)! x2n = 3 2 (2n + 2)(2n + ) x 2 = x 2 so this series always converges the radius of convergence is 6 Find the Taylor series for centered at a = 3 Answer: Note that f(x) = x 9 4n 2 + 6n + 2 = 0, f (x) = x 2 f (x) = 2 x 3 f (x) = 6 x 4 f (4) (x) = 24 x 5 f (n) (x) = ( ) n n! x n+ f( 3) = 3 f ( 3) = 9 f ( 3) = 2 27 Hence, by the definition of the Taylor series, f ( 3) = 6 8 f (4) ( 3) = 24 243 f (n) (3) = n! 3 n+ f(x) = (x + 3) n 3 n+ is the Taylor series for f(x) = x centered at 3 4
34 Use a Maclaurin series in Table to obtain the Maclaurin series for the function f(x) = x 2 tan ( x 3) Answer: From Table, we know that the Maclaurin series for tan x is tan x = the series for tan (x 3 ) is ( ) n x2n+ 2n + = x x3 3 + x5 5 x7 7 tan (x 3 ) = ( ) n (x3 ) 2n+ 2n + = ( ) n x6n+3 2n + = x3 x9 3 + x5 5 x2 7 In turn, this means that the series for f is x 2 tan (x 3 ) = x 2 ( ) n x6n+3 2n + = ( ) n x6n+5 2n + = x5 x 3 + x7 5 x23 7 48 Evaluate the indefinite integral e x dx x as an infinite series Answer: The Maclaurin series for e x is e x = x n n! = + x + x2 2! + x3 3!, so the series for the numerator is ) e x = ( + x + x2 2! + x3 3! = x + x2 In turn, this means the series for the integrand is 2! + x3 3! = n= x n n! x n= x n n! = n= x n n! = + x 2! + x2 3! we can integrate term-by-term to get e x dx = C + x + x x2 2 2! + x3 3 3! = C + n= x n n n! 56 Use series to evaluate the it x 0 cos x + x e x 5
Answer: Using the Maclaurin series for cos x we can write the numerator as the series ) cos x = ( x2 2! + x4 4! x6 6! = x2 2! x4 4! + x6 6! Using the Maclaurin series for e x, we can write the denominator as ) + x e x = + x ( + x + x2 2! + x3 3! = x2 2! x3 3! x 0 cos x + x e x = x 0 x 2 2! x4 4! x2 2! x3 3! Dividing both numerator and denominator by x 2, this is equal to x 0 2! x2 4! 2! x 3! = 2! 2! = 68 Find the sum of the series ln 2 + (ln 2)2 2! (ln 2)3 3! Answer: Notice that e ln 2 = ln 2 + (ln 2)2 2! is just the given series, so the sum of the series is (ln 2)3 3! e ln 2 = e ln 2 = 2 From 2 6 (a) Approximate f(x) = sin x by a Taylor polynomial with degree 4 at the number π/6 Answer: The first four derivatives of f are f (x) = cos x f (x) = sin x f (x) = cos x f (4) (x) = sin x 6
so we have f(π/6) = /2 f (π/6) = 3/2 f (π/6) = /2 f (π/6) = 3/2 f (4) (π/6) = /2 the degree 4 Taylor polynomial for f at π/6 is 3 2 + 3 (x π/6) 2 2 2! (x π/6)2 2 3! (x π/6)3 + 2 4! (x π/6)4 (b) Use Taylor s Inequality to estimate the accuracy of the approximation f(x) = T 4 (x) when x lies in the interval 0 x π/3 Answer: When 0 x π/3, Taylor s Inequality says that the remainder R 4 (x) is bounded by R 4 (x) M x π/6 5 5! where M is an upper bound on f (5) in this interval Since f (5) (x) = cos x and since 2 cos x for all x between 0 and π/3, we can pick M = R 4 (x) 5! x π/6 5 = 20 x π/6 5 When 0 x π/3, the quantity x π/6 π/6, so 240 x π/6 5 20 (π/6)5 = is the worst the error could possibly be on this interval π 5 20 7776 0000328 7
(c) Check your result in (b) by graphing R 4 (x) Answer: 4 0-4 32 0-4 24 0-4 6 0-4 8 0-5 0 025 05 075 35 If a water wave with length moves with velocity v across a body of water with depth d, as in the figure, then v 2 = g 2πd tanh 2π (a) If the water is deep, show that v g/(2π) Answer: As u, tanh u, so, when the water is very deep, 8 v 2 g 2π,
meaning that as desired v g 2π, (b) If the water is shallow, use the Maclaurin series for tanh to show that v gd (Thus in shallow ater the velocity of a wave tends to be independent of the length of the wave) Answer: The first few derivatives of f(x) = tanh x are f (x) = sech 2 x f (x) = 2 sech x sech x tanh x = 2 sech 2 x tanh x f (x) = 4 sech x sech x tanh x tanh x 2 sech 2 x sech 2 x = 4 sech 2 x tanh 2 x 2 sech 4 x Hence, f(0) = tanh 0 = 0 f (0) = sech 2 0 = This gives a Maclaurin series for v 2 : f (0) = 2 sech 2 0 tanh 0 = 0 f (0) = 4 sech 2 0 tanh 2 0 2 sech 4 0 = 2 v 2 = g 2π tanh x = x x3 3 ( 2πd = gd g 6π ( 2πd ( 2πd ) 3 ) 3 ) 3 When d is small, 2πd is also small and higher powers of it are even smaller the first term above gives a good approximation for v 2 when d is small Thus, v gd (c) Use the Alternating Series Estimation Theorem to show that if > 0d, then the estimate v 2 gd is accurate to within 004g Answer: The error is no bigger than the first unused term in the series: error g 6π ( ) 2πd 3 When > 0d, the right hand side (and, thus, the error) is smaller than ( ) g 2πd 3 = g ( π ) 3 π 3 π2 = g = g 003g 6π 0d 6π 5 6π 53 750 9