sols_wk6 - √ 10 3-5 √ 10 ± 3-5 = a-5 a ±-5 = a ± 5-a...

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Homework 6 due Thursday November 18 before 13:30 in corre- sponding group folder on door of office CM110 1. Prove that ( 10 + 3) - 5 = 1405 10 - 4443 . Estimate the errors involved in evaluating each of these expressions, given an ap- proximation for 10 with a small error ± . { Hint: You may want to use Taylor series expansions of a - 5 and ( a + ± ) - 5 , where a = ( 10 + 3). } ——————————————————————————————— Multiplying the top and bottom by ( 10 - 3) 5 yields ( 10 + 3) - 5 = ( 10 - 3) 5 ( 10 + 3) 5 ( 10 - 3) 5 = ( 10) 5 + 5( 10) 4 3 + 10( 10) 3 3 2 + 10( 10) 2 3 3 + 5( 10)3 4 + 3 5 = 1405 10 - 4443 The error in the second approximation is: [1405( 10 + ± ) - 4443] - [1405( 10) - 4443] = 1405 ± while in the first approximation, using a Taylor series expansion, is ( 10 + 3) - 5 - ( 10 + ± + 3) - 5 5 ± ( 10 + 3) 6 This can be show if we take a = ( 10 + 3) (
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Unformatted text preview: √ 10 + 3)-5-( √ 10 + ± + 3)-5 = ( a )-5-( a + ± )-5 = ( a + ± ) 5-a 5 a 5 ( a + ± ) 5 Using binomial expansion leads to ( a + ± ) 5-a 5 a 5 ( a + ± ) 5 ≈ 5 a 4 ± + O ( ± 2 ) a 10 + O ( ± ) ≈ 5 ± ( √ 10 + 3) 6 2. Use cubic Lagrange interpolation to estimate f (3), given the function values f (0) = 1, f (2) = 1 . 2, f (4) = 11 . 8 and f (5) = 24 . 75. ——————————————————————————————— f (3) ≈ p 3 (3) = 3 ∑ i =0 f ( x i ) L i ( x ) = 1 × (3-2)(3-4)(3-5) (0-2)(0-4)(0-5) +1 . 2 × (3-0)(3-4)(3-5) (2-0)(2-4)(2-5) +11 . 8 × (3-0)(3-2)(3-5) (4-0)(4-2)(4-5) +24 . 75 × (3-0)(3-2)(3-4) (5-0)(5-2)(5-4) = 4 . 45 MATH2051 Homework 1 2010-11-09...
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This note was uploaded on 02/09/2011 for the course MATH 2051 taught by Professor Dr.a.wacher during the Fall '11 term at Durham.

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