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Unformatted text preview: CS 257/MATH 257 Numerical Methods  Homework 2 CS257 1. [2pt] Section 1.2 #1 Solution: f ( x ) = f (0) + xf (0) + x 2 2! f 00 (0) + Â·Â·Â· f ( x ) = (1 + x ) n , f (0) = 1 f ( x ) = n (1 + x ) n 1 , f (0) = n f 00 ( x ) = n ( n 1)(1 + x ) n 2 , f 00 (0) = n ( n 1) f 000 ( x ) = n ( n 1)( n 2)(1 + x ) n 3 , f 000 (0) = n ( n 1)( n 2) Thus, (1 + x ) n = 1 + nx + n ( n 1) 2 x 2 + n ( n 1)( n 2) 3! x 3 + Â·Â·Â· = âˆ‘ âˆž i =0 f ( i ) (0) i ! x i n = 2 : (1 + x ) 2 = 1 + 2 x + x 2 n = 3 : (1 + x ) 3 = 1 + 3 x + 3 x 2 + x 3 n = 1 2 : (1 + x ) 1 2 = 1 + 1 2 x 1 2 3 x 2 + 1 2 4 x 3 5 2 7 x 4 + 7 2 8 x 5 21 2 9 x 6 + Â·Â·Â· = 1 + âˆ‘ âˆž i =1 Q k = i 1 k =0 ( 1 2 k ) i ! x i (1+0 . 0001) 1 2 = 1+0 . 00005 . 00000000125+0 . 0000000000000625 = 1 . 0000499987500624 Â·Â·Â· = 1 . 000049998750062 2. [1pt] Why does the function f ( x ) =  x 1  not posses a Taylor series at x = 1? Solution: The function does not have a continuous first derivative at 0....
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 Fall '05
 ThomasKerkhoven
 Taylor Series, 1 k, Continuous function, 1pt, Bisection Method

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