Math128_hw8

# Math128_hw8 - (0) = 3 Y 2 + 4 Y 2 + 13 Y 1 = 40 cos( x ) Y...

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Math 128A, Spring 2007 Homework 8 Solution 5.4.14 We approximate f 00 (0 . 5) as 0 . 0800 and 0 . 0775 for h = 0 . 1 and h = 0 . 2, respectively. To understand the roundoﬀ error, note that we know all of the data to within 0 . 00005. So, f 00 ( x ) f ( x + h ) - 2 f ( x ) + f ( x - h ) h 2 = ˆ f ( x + h ) - 2 ˆ f ( x ) + ˆ f ( x - h ) h 2 + ± 1 - 2 ± 2 + ± 3 h 2 where each | ± i | < 0 . 00005 and ˆ f ( x ) is the value in the table. The roundoﬀ error term can be bounded ± ± ± ± ± 1 - 2 ± 2 + ± 3 h 2 ± ± ± ± < 4 × 0 . 00005 h 2 yeilding 0 . 0200 and 0 . 0050 for h = 0 . 1 and h = 0 . 2, respectively. 8.7.3 (a) If we let Y 1 = Y , Y 2 = Y 0 and Y 3 = Y 00 , then the system Y 0 1 = Y 2 Y 1 (0) = 1 Y 0 2 = Y 3 Y 2 (0) = - 1 Y 0 3 + 4 Y 3 + 5 Y 2 + 2 Y 1 = 2 x 2 + 10 x + 8 Y 3 (0) = 3 is equivalent to the original one. (b) Letting Y 1 = Y and Y 2 = Y 0 , Y 0 1 = Y 2 Y 1
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Unformatted text preview: (0) = 3 Y 2 + 4 Y 2 + 13 Y 1 = 40 cos( x ) Y 2 (0) = 4 is equivalent to the original system. 8.7.4 If we let X 1 = x , X 2 = x , Y 1 = y , Y 2 = y , Z 1 = z , Z 2 = z , and R = ( X 2 1 + Y 2 1 + Z 2 1 ) 1 / 2 , then X 1 = X 2 X 2 =-cX 1 /R 3 Y 1 = Y 2 Y 2 =-cY 1 /R 3 Z 1 = Z 2 Z 2 =-cZ 1 /R 3 is equivalent to the original system. Euler We get data i x_i y_i 0 0.00 1.0000 1 0.05 0.9500 2 0.10 0.9050 3 0.15 0.8645 4 0.20 0.8283 1...
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## This note was uploaded on 05/07/2008 for the course MATH 128A taught by Professor Rieffel during the Spring '08 term at Berkeley.

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