# 18128 - 10.1#2 Pn(x = C0 C1 x C2 x2 Cn xn where Cj =...

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P n ( x ) = C 0 + C 1 x + C 2 x 2 + ··· + C n x n , where C j = f ( j ) (0) /j !. Now f (0) ( x ) = (1 + x ) 1 / 2 , f (1) ( x ) = (1 / 2)(1 + x ) - 1 / 2 , f (2) ( x ) = - (1 / 2)(1 / 2)(1 + x ) - 3 / 2 , f (3) ( x ) = (3 / 2)(1 / 2)(1 / 2)(1 + x ) - 5 / 2 , f (4) ( x ) = - (5 / 2)(3 / 2)(1 / 2)(1 / 2)(1 + x ) - 7 / 2 , f (0) (0) = 1 , f (1) (0) = 1 / 2 , f (2) (0) = - 1 / 4 , f (3) (0) = 3 / 8 , f (4) (0) = - 15 / 16 . Thus P 2 ( x ) = 1 + (1 / 2) x - (1 / 8) x 2 , P 3 ( x ) = 1 + (1 / 2) x - (1 / 8) x 2 + (1 / 16) x 3 , P 4 ( x ) = 1 + (1 / 2) x - (1 / 8) x 2 + (1 / 16) x 3 - (5 / 128) x 4 . 10.1, #4. f (0) ( x ) = (1 + x ) - 1 , f (1) ( x ) = ( - 1)(1 + x ) - 2 , f (2) ( x ) = ( - 1) 2 2!(1 + x ) - 3 , f (3) ( x ) = ( - 1) 3 3!(1 + x ) - 4 , ··· f ( n ) ( x ) = ( - 1) n n !(1 + x ) - n - 1 , f (0) (0) = 1 , f (1) (0) = - 1 , f (2) (0) = 2! , f (3) (0) = - 3! , ··· f ( n ) (0) = ( - 1) n n ! . Thus

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## This note was uploaded on 07/06/2011 for the course MATH 180 taught by Professor Tan during the Fall '08 term at Ill. Chicago.

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18128 - 10.1#2 Pn(x = C0 C1 x C2 x2 Cn xn where Cj =...

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