practice exam b

2 m m1 agreed 8 tm t 1t xm line 23590 id

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Unformatted text preview: 1 − tx √√ t 1−t = √ 1 √ t 1−t 2 m − 3 !! ∞ − 2 m !! m=1 Agreed. 8 √ tm √ t 1−t xm . Line #: 23590. ID # : Name : ([V] continued) (3) The Taylor series expression for √ 1 − tx √√ t 1−t 1 = Ex t=0 dt is found as follows: Integrate the both sides of the identity in (2) with respect to t, over 1 Ex = √ t=0 1 √ t 1−t 0, 1 : dt π ∞ − m=1 2 m − 3 !! 1 2 m !! t=0 √ tm √ t 1−t 2m − 1 2m !! dt xm π !! = . 9 Line #: 23590. Name : ID # : ([V] continued) (4) Prove that E x in (3) has the following alternative expression: Ex = π 2 2 u=0 Hint : 1− sin u 2 x du. In the above E x , do the following substitution: t= sin u Proof. 10 2 . Line #: 23590. Name : ID # : ([V] continued) (5) Evaluate 2 ∞ 2 m − 3 !! 2 m=1 2 m !! 2m − 1 = . 11...
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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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