solutions for b

Xm xm proof in the above e x do substitution t

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Unformatted text preview: identity in (2) with respect to t, over 1 Ex = √ t=0 1 √ t 1−t 0, 1 : dt π − ∞ 2 m − 3 !! 1 m=1 2 m !! t=0 √ tm √ t 1−t 2m − 1 2m !! dt π !! 2 = (4) π − 2 m − 3 !! ∞ π 2m − 1 2 m=1 2 m !! Prove that the integral in (3): 1 = Ex t=0 √ 1 − tx √√ t 1−t dt has the following alternative expression: π 2 u=0 2 1− sin u 14 2 x du. xm . xm Proof. √ In the above E x , do substitution √ t = sin u, 1−t = cos u, and t= sin u dt = 2 sin u 2 . cos u Then d u. Hence Ex 1− π 2 = π 2 = 2 u=0 (5) 1− x 2 sin u u=0 2 sin u sin u cos u cos u 2 sin u x du. We may evaluate the inifinite series 2 2 m − 3 !! ∞ 2m − 1 2 m=1 2 m !! as follows. By (3), 2 = Ex Substitute π−π x=1 ∞ 2 m − 3 !! 2m − 1 2 m=1 2 m !! xm . into this result: 2 ( ∗) E1 = π−π 2 m − 3 !! ∞ 2 m=1 2 m !! 15 2m − 1 . du We may easily calculate the left-hand side of (∗) as follows: √ 1 − t ·1 √√ t 1−t 1 E1 = t=0 1 = √ t=0 1 t=0 = 2 √ √ 1−t √ t 1−t 1 √ t = dt dt dt 1 t t=0 =2 In short, E1 = 2. √ 1 −2 √ 0 = 2. Substitute this back into (∗): 2 2 = π−π 2 m − 3 !! ∞ 2 m=1 2 m !! From this we conclude that the value of the infinite series 2 ∞ 2 m − 3 !! 2 m=1 2 m !! equals 1− 16 2 . π 2m − 1 2m − 1 ....
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This note was uploaded on 01/12/2014 for the course MATH 116 taught by Professor Scholle,minho during the Fall '08 term at Kansas.

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