# Y 1 x 1 x 2 1 x 4 1 3 x 6 1 3 5 1 n 1 1 n x 2 n 1 3 5

• 33

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y 1 ( x ) = 1 x 2 1 + x 4 1 3 x 6 1 3 5 + = 1 + n = 1 ( 1) n x 2 n 1 3 5 (2 n 1) , y 2 ( x ) = x x 3 2 + x 5 2 4 x 7 2 4 6 + = x + n = 1 ( 1) n x 2 n + 1 2 4 6 (2 n ) 8. a. a n + 2 = − (( n + 1) 2 a n + 1 + a n + a n 1 ) (( n + 1)( n + 2)) , n = 1 , 2 , , a 2 = − ( a 0 + a 1 ) 2 b. y 1 ( x ) = 1 1 2 ( x 1) 2 + 1 6 ( x 1) 3 1 12 ( x 1) 4 + , y 2 ( x ) = ( x 1) 1 2 ( x 1) 2 + 1 6 ( x 1) 3 1 6 ( x 1) 4 + 9. a. 3( n + 2) a n + 2 ( n + 1) a n = 0 , n = 0 , 1 , 2 , b, d. y 1 ( x ) = 1 + x 2 6 + x 4 24 + 5 432 x 6 + + 3 5 (2 n 1) 3 n 2 4 (2 n ) x 2 n + , y 2 ( x ) = x + 2 9 x 3 + 8 135 x 5 + 16 945 x 7 + + 2 4 (2 n ) 3 n 3 5 (2 n + 1) x 2 n + 1 + 10. a. 2( n + 2)( n + 1) a n + 2 + ( n + 3) a n = 0 , n = 0 , 1 , 2 , b, d. y 1 ( x ) = 1 3 4 x 2 + 5 32 x 4 7 384 x 6 + + ( 1) n 3 5 (2 n + 1) 2 n (2 n ) ! x 2 n + , y 2 ( x ) = x x 3 3 + x 5 20 x 7 210 + + ( 1) n 4 6 (2 n + 2) 2 n (2 n + 1) ! x 2 n + 1 + 11. a. 2( n + 2)( n + 1) a n + 2 + 3( n + 1) a n + 1 + ( n + 3) a n = 0 , n = 0 , 1 , 2 , b. y 1 ( x ) = 1 3 4 ( x 2) 2 + 3 8 ( x 2) 3 + 1 64 ( x 2) 4 + , y 2 ( x ) = ( x 2) 3 4 ( x 2) 2 + 1 24 ( x 2) 3 + 9 64 ( x 2) 4 + 12. a. y = 2 + x + x 2 + 1 3 x 3 + 1 4 x 4 + c. about | x | < 0 . 7 13. a. y = 4 x 4 x 2 + 1 2 x 3 + 4 3 x 4 + c. about | x | < 0 . 5 14. a. y = − 3 + 2 x 3 2 x 2 1 2 x 3 1 8 x 4 + c. about | x | < 0 . 9 15. a. y 1 ( x ) = 1 1 3 ( x 1) 3 1 12 ( x 1) 4 + 1 18 ( x 1) 6 + , y 2 ( x ) = ( x 1) 1 4 ( x 1) 4 1 20 ( x 1) 5 + 1 28 ( x 1) 7 + 16. Hint: Consider using induction. 18. a. y 1 ( x ) = 1 𝜆 2 ! x 2 + 𝜆 ( 𝜆 4) 4 ! x 4 𝜆 ( 𝜆 4)( 𝜆 8) 6 ! x 6 + , y 2 ( x ) = x 𝜆 2 3 ! x 3 + ( 𝜆 2)( 𝜆 6) 5 ! x 5 ( 𝜆 2)( 𝜆 6)( 𝜆 10) 7 ! x 7 + b. 1, x , 1 2 x 2 , x 2 3 x 3 , 1 4 x 2 + 4 3 x 4 , x 4 3 x 3 + 4 15 x 5 c. 1, 2 x , 4 x 2 2, 8 x 3 12 x , 16 x 4 48 x 2 + 12, 32 x 5 160 x 3 + 120 x 19. b. y = x x 3 6 + Section 5.3, page 209 1. 𝜙 ′′ (0) = − 1 , 𝜙 ′′′ (0) = 0 , 𝜙 (4) (0) = 3 2. 𝜙 ′′ (1) = 0 , 𝜙 ′′′ (1) = − 6 , 𝜙 (4) (1) = 42 3. 𝜙 ′′ (0) = 0 , 𝜙 ′′′ (0) = − a 0 , 𝜙 (4) (0) = − 4 a 1 4. 𝜌 = ∞ , 𝜌 = ∞ 5. 𝜌 = 1 , 𝜌 = 3 , 𝜌 = 1 6. 𝜌 = 1 , 𝜌 = 3 7. a. 𝜌 = ∞ b. 𝜌 = ∞ c. 𝜌 = ∞ d. 𝜌 = ∞ e. 𝜌 = 1 f. 𝜌 = 2 g. 𝜌 = ∞ h. 𝜌 = 1 i. 𝜌 = 1 j. 𝜌 = 2 k. 𝜌 = 3

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Boyce 9131 BMAnswersToProblems 2 March 11, 2017 15:55 583 Answers to Problems 583 l. 𝜌 = 1 m. 𝜌 = ∞ n. 𝜌 = ∞ 8. a. y 1 ( x ) = 1 𝛼 2 2 ! x 2 (2 2 𝛼 2 ) 𝛼 2 4 ! x 4 (4 2 𝛼 2 )(2 2 𝛼 2 ) 𝛼 2 6 ! x 6 ((2 m 2) 2 𝛼 2 ) (2 2 𝛼 2 ) 𝛼 2 (2 m ) ! x 2 m , y 2 ( x ) = x + 1 𝛼 2 3 ! x 3 + (3 2 𝛼 2 )(1 𝛼 2 ) 5 ! x 5 + + ((2 m 1) 2 𝛼 2 ) (1 𝛼 2 ) (2 m + 1) ! x 2 m + 1 + b. y 1 ( x ) or y 2 ( x ) terminates with x n as 𝛼 = n is even or odd c. n = 0, y = 1; n = 1, y = x ; n = 2, y = 1 2 x 2 ; n = 3, y = x 4 3 x 3 9. y 1 ( x ) = 1 1 6 x 3 + 1 120 x 5 + 1 180 x 6 + , y 2 ( x ) = x 1 12 x 4 + 1 180 x 6 + 1 504 x 7 + , 𝜌 = ∞ 10.
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• Spring '16
• Anhaouy
• Districts of Vienna, Boyce, e2t, 3y, = min, + c2 sin x

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