Math WW09 Solutions November 24, 2008

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1 Math 352- WW09 Solutions November 24, 2008 Assigned problems: 8.7 0, 6, ww 4; , ww 5, ww 6 Always read through the solution sets even if your answer was correct. Note that like many of the integrals in this course, there is frequently more than one way to determine convergence or divergence of a series. Your solution may be correct even if you used a different method than what I use here.. (8.7 #0) (x 5) k k (k+) : First we ll notice that the series converges (trivially) for x 5 since each term is identically 0. For x 5, we ll try the generalized ratio test: a k+ a k (x 5) k / (k + 2) (x 5) k / (k + ) (k + )(x 5) k+ (k + 2)(x 5) k (k + ) x 5 (k + 2) (k + 2) x 5 (L hopital) (k + ) (k + 2) x 5 (k + ) x 5 L According to the generalized ratio test, the series converges for L < and diverges for L >. So this series converges when x 5 <, i.e., when < (x 5) < which gives us 4 < x < 6. Similar, the series converges when L >, so x 5 > giving x > 6 or x < 4. The generalized ratio test is inconclusive when L x 5, so we need to check the endpoints x 4 and x 6 separately. When x 4, we have k (x 5) k (k + ) k ( ) k (k + ) This is an alternating series, so we ll test with the alternating series test. (k + ) 0 (k + 2) > (k + ) < (k + 2) (k + ) a k+ < a k

2 Math 352- WW09 Solutions November 24, 2008 Therefore the series converges at x 4. When x 6, k (x 5) k (k + ) k (k + ) We ll do a it comparison test with b k k : a k / (k + ) b k /k k (k + ) (k + ) (L hopital) (k + ) This it goes to infinity and b k diverges, so by the 0 it comparison test the series diverges, too. So the interval of convergence is 4 x < (8.7 #6) (2k)!x k k (3k)! : First we ll notice that the series converges (trivially) for x 0 since each term is identically 0. For x 0, we ll try the generalized ratio test: a k+ a k (2k + 2)!x k+ /(3k + (2k)!x k /(3k)! (3k)!(2k + 2)!x k+ (3k + 3)!(2k)!x k (Note (2k + 2)! (2k + 2)(2k + )(2k)! and (3k + 3)! (3k + 3)(3k + 2)(3k + )(3k)!) (2k + 2)(2k + ) (3k + 3)(3k + 2)(3k + ) x 4k 2 + 6k k k 2 + 2k + 3 x 4/k + 6/k 2 + 2/k /k + 2/k 2 + 3/k 3 x 0 The it is 0 for any value of x. Therefore the series converges absolutely for any x, < x <. (Since the interval is the whole real line, there are no endpoints to check.) 2

3 Math 352- WW09 Solutions November 24, (8.8 #32) First 4 coefficients of Taylor series for f(x) (x) at c 3. We ll need the first 3 derivatives of f evaluated at c 3: f(x) (x) f(3) (3) f (x) x f (3) 3 f (x) x f (3) 2 9 f (x) 2 x f (3) So the Taylor series at c 3 for f(x) (x) is: where f(x) f(3) + f (3)(x 3) + f (3)(x 3) 2 2! (3) + ( ( /9)(x 3)2 )(x 3) (x 3) (x 3)2 (x 3)3 (3) c 0 (3) c /3 + f (c)(x 3) 3 (2/27)(x 3)3 6 c 0 + c (x 3) + c 2 (x 3) 2 + c 3 (x 3) 3 c 2 /8 c 3 /8 4. (Note that webwork creates random, but similar, problems for each student on this question. So the solutions here are for a sample of problems to give you some examples. If you have questions on any particular problem you were given, please ask. (4.) n ( ) n 4n+3 : This is an alternating series, so we can try the alternating series test. We need to check that a k 0 and a k+ a k. And, k 4n n + 7 > 4n + 3 4(n + ) + 3 > 4n + 3 < 4(n + ) + 3 4n + 3 a n+ < a n So the the series converges. 3

4 Math 352- WW09 Solutions November 24, 2008 Now we ll check if it converges absolutely. ( ) n 4n + 3 4n + 3 This looks almost like a divergent p-series. We can confirm that it diverges with a it comparison test with bk k. a k /(4k + 3 b k /k k 4k + 3 (4.2) 4 The it is positive and finite, so the two series converge or diverge together. Therefore a k diverges. So the original series converges conditionally. n ( ) n n n+2 : This is an alternating series, so we can try the alternating series test. We need to check that a k 0 and a k+ a k. k a k k + 2 And, 0 (/2)k /2 2 n a k+ a k? k + k k + 3 k + 2? k + (k + 2)? k(k + 3) (L hopital) (k + )(k + 2) 2 k(k + 3) 2? k 3 + 5k 2 + 8k + 4 k 3 + 6k 2 + 9k? k 3 + 5k 2 + 8k + 4 k 3 + 6k 2 + 9k? 4 k 2 + k? This is true for all x 2. The conditions of the alternating series test are met, so the series converges. 4

5 Math 352- WW09 Solutions November 24, 2008 (4.3) Now we ll check if the series converges absolutely. Consider ( ) n n n + 2 n n + 2 We ll do a direct comparison test with b k /(n+2) (which diverges by a it comparison with /n). n n + 2 n So the series is term by term a divergent series. So by the direct comparison test, the series diverges. Therefore, since the original series converges but the absolute value series diverges, the original series converges conditionally. n (n+)(6 2 ) n 6 2n : Because of the powers of n, I ll try the generalized root test. n a n (n + ) /n (6 2 ) (6 2n ) /n (n + ) /n (6 2 ) 35 (n + )/n 36 (This is 0 form) L 35 (n + )/n L (n + )/n ( 35 L (n + )/n) ( 35 L (n + ) /n) 35 L (n + ) n 35 L (n + ) n 35 L n + (L hopital) 35 L 0 36 L 35 (Take exponential of both sides.) L The it is <, therefore the series converges absolutely

6 Math 352- WW09 Solutions November 24, 2008 (4.4) (4.5) n ( 2) n n 2 : We ll try the generalized ratio test. a n+ a n ( 2) n+ /(n + ) 2 ( 2) n /n 2 n 2 ( 2) n+ (n + ) 2 ( 2) n 2n 2 (n + ) 2 2 2n 2 n 2 + 2n /n + /n 2 The it is >, so the series diverges. n sin(5n) n : 2 This series can be negative, but does not alternate in a regular +, pattern, so it s not an alternating series. I ll try testing for absolute convergence on n sin(5n) n 2 We can use a direct comparison test. Since sin(x) for any x, we have 0 sin(5n) n 2 n 2 Therefore, the absolute value series is dominated by the convergent p-series n 2 absolutely convergent. (p 2). Therefore the series is 5. Here we are matching a set of series with the Maclaurin series for different functions. I ll derive the Maclaurin series (Taylor series at c 0) for the given functions. Each time, we ll get one of the given series. f(x) cos(x) f(x) cos(x) f(0) f (x) sin(x) f (0) 0 f (x) cos(x) f (0) f (x) sin(x) f (0) 0 f (4) (x) cos(x) f (4) (0) 6

7 Math 352- WW09 Solutions November 24, 2008 f(x) e x cos(x) + 0 x2 2! x4 4! + 0 x6 6! x2 2! + x4 4! x6 6! ( ) k x 2k (2k)! k0 f(x) e x f(0) f (x) e x f (0) f (x) e x f (0) e x + x! x k k! k0 f(x) arctan(x) + x2 2! + x3 + x4 4! f(x) arctan(x) f(0) 0 f (x) +x f (0) 2 f (x) (2x)( + x 2 ) 2 f (0) 0 f (x) (8x 2 )( + x 2 ) 3 2( + x 2 ) 2 f (0) 2 f (4) (x) (6x)( + x 2 ) 3 (48x 3 )( + x 2 ) 4 + (8x)( + x 2 ) 3 f (4) (0) 0 f(x) sin(x) arctan(x) 0 + x + 0 2x3 x x3 3 ( ) k x 2k+ 2k + k0 f(x) sin(x) f(0) 0 f (x) cos(x) f (0) f (x) sin(x) f (0) 0 f (x) cos(x) f (0) f (4) (x) sin(x) f (4) (0) 0 7

8 Math 352- WW09 Solutions November 24, 2008 sin(x) 0 + x! + 0 x3 x x3 + x5 5! ( ) k x 2k+ (2k + )! k0 6. Taylor series for f(x) cos(x) at c π/ x5 5! f(x) cos(x) f(π/2) 0 f (x) sin(x) f (π/2) f (x) cos(x) f (π/2) 0 f (x) sin(x) f (π/2) f (4) (x) cos(x) f (4) (π/2) 0 f (5) (x) sin(x) f (5) (π/2) cos(x) (x π/2) ! c k (x π/2) k k0 Where the first 5 coefficients are: c 0 0 c c 2 0 c 3 /6 c 4 0 (x π/2)3 + 0 (x π/2)5 5! 8