solutions for quiz xxix

# solutions for quiz xxix - Math 116 CALCULUS II SOLUTION FOR...

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Math 116 CALCULUS II SOLUTION FOR QUIZ – XXIX (11/19) November 19 (Tue), 2013 Instructor: Yasuyuki Kachi Line #: 23590. [I] (10pts) Let J ( x ) be defned over the interval p − ∞ , + P . Assume J ( 0 ) = 1 and moreover that For an arbitrary positive integer J (2 1) ( 0 ) = 0, J (2 ) ( 0 ) = p 1 P p 2 1 P ! ! p 2 P ! ! . (1) J (1) ( 0 ) = 0, J (2) ( 0 ) = 1 !! 2 !! , J (3) ( 0 ) = 0, J (4) ( 0 ) = 3 !! 4 !! , J (5) ( 0 ) = 0, J (6) ( 0 ) = 5 !! 6 !! , J (7) ( 0 ) = 0, J (8) ( 0 ) = 7 !! 8 !! , · · · · · · 1 (2a) J ( x ) = s =0 p 1 P ± p 2 P !! ² 2 · x 2 . (2b) J ( x ) = 1 1 p 2 !! P 2 x 2 + 1 p 4 !! P 2 x 4 1 p 6 !! P 2 x 6 + 1 p 8 !! P 2 x 8 − ··· . 1

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b Work for (2a), (2b) B : J ( x ) = J ( 0 ) + J (1) ( 0 ) 1 ! x + J (2) ( 0 ) 2 ! x 2 + J (3) ( 0 ) 3 ! x 3 + J (4) ( 0 ) 4 ! x 4 + J (5) ( 0 ) 5 ! x 5 + J (6) ( 0 ) 6 ! x 6 + J (7) ( 0 ) 7 ! x 7 + J (8) ( 0 ) 8 ! x 8 + J (9) ( 0 ) 9 ! x 9 + J (10) ( 0 ) 10 ! x 10 + ··· 1 !! 2 !! 3 !! 4 !! = 1 + 0 1 ! x + 2 ! x 2 + 0 3 ! x 3 + 4 ! x 4 5 !! 6 !! 7 !! 8 !! + 0 5 ! x 5 + 6 ! x 6 + 0 7 ! x 7 + 8 ! x 8 9 !! 10 !! + 0 9 ! x 9 + 10 ! x 10 + = 1 1 2 · 2 · 1 x 2 + 3 · 1 4 · 2 · 4 · 3 · 2 · 1 x 4 5 · 3 · 1 6 · 4 · 2 · 6 · 5 · 4 · 3 · 2 · 1 x 6 + 7 · 5 · 3 · 1 8 · 6 · 4 · 2 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 x 8 9 · 7 · 5 · 3 · 1 10 · 8 · 6 · 4 · 2 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 x 10 + 2
= 1 1 2 · 2 x 2 + 1 4 · 2 · 4 · 2 x 4 1 6 · 4 · 2 · 6 · 4 · 2 x 6 + 1 8 · 6 · 4 · 2 · 8 · 6 · 4 · 2 x 8 1 10 · 8 · 6 · 4 · 2 · 10 · 8 · 6 · 4 · 2 x 10 + ··· = 1 1 p 2 !! P 2 x 2 + 1 p 4 !! P 2 x 4 1 p 6 !! P 2 x 6 + 1 p 8 !! P 2 x 8 − ··· 1 = s =0 p 1 P ± p 2 P !! ² 2 x 2 . b Alternative Answers for (2a), (2b) B : Since ± p 2 P !! ² 2 = p ! P 2 · 2 2 , you may write your answers alternatively as 1 (2a) J ( x ) = s =0 p 1 P p ! P 2 · ± 1 2 ² 2 · x 2 . (2b) J ( x ) = 1 1 p 1! P 2 p x 2 P 2 + 1 p 2! P 2 p x 2 P 4 1 p 3! P 2 p x 2 P 6 + 1 p 4! P 2 p x 2 P 8 . 3

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1 (3) The radius of convergence for s =0 p 1 P ± p 2 P !! ² 2 · x 2 is R = + . b Work B : Consider the series with variable u : 1 s =0 p 1 P ± p 2 P !! ² 2 · u . 1 Set a = p 1 P ± p 2 P !! ² 2 . Then 1 v v a v v = ± p 2 P !! ² 2 , 1 v v a +1 v v = ± p 2 + 2 P !! ² 2 . 1 ± p 2 P !! ² 2 = lim −→ v v a v v v v a +1 v v = lim 1 ± p 2 + 2 P !! ² 2 4
= lim −→ p 2 + 2 P !!

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