tutorial_wk6

tutorial_wk6 - -2 . 5)(2 . 37-3) (2-2 . 5)(2-3) ln(2) + (2...

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Tutorial week 6 1. Calculate f (0 . 001) correct to six decimal places, where f ( x ) = 1 - e - 2 x 2 x sin x . f ( x ) = 1 - ( 1 - 2 x 2 + 4 x 4 2! - 8 x 6 3! + ··· ) x ( x - x 3 3! + x 5 5! - ··· ) = 2(1 - x 2 + O ( x 4 )) 1 - x 2 ( 1 6 - x 2 120 + O ( x 4 )) = 2(1 - x 2 + O ( x 4 )) [ 1 + ± x 2 ( 1 6 - x 2 120 ) ² + O ( x 4 ) ] = 2(1 - 5 6 x 2 ) + O ( x 4 ) Thus f (0 . 001) = 2(1 - 5 6 × 10 - 6 ) = 1 . 999998 correct to 6 decimal places. 2. Find a suitable rearrangement of each of the following expressions so as to avoid serious loss of significance for values of x close to 0: 1 + x 2 - 1 - x 2 , tan x - sin x cos x sin(2 x ) . 1 + x 2 - 1 - x 2 = 2 x 2 1 + x 2 + 1 - x 2 tan x - sin x cos x sin(2 x ) = sin x/ cos x - sin x cos x 2 cos x sin x = 1 - cos 2 x 2 cos 2 x = 1 2 tan 2 x 3. The values of ln x for x = 2 . 0 , 2 . 5 , 3 . 0 are to be used to estimate ln(2 . 37) by quadratic interpolation. Calculate the quadratic approximation using the Lagrange form. Using the Lagrange form for the interpolant p 2 (2 . 37) = (2 . 37
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Unformatted text preview: -2 . 5)(2 . 37-3) (2-2 . 5)(2-3) ln(2) + (2 . 37-2)(2 . 37-3) (2 . 5-2)(2 . 5-3) ln(2 . 5) + (2 . 37-2)(2 . 37-2 . 5) (3-2)(3-2 . 5) ln(3) = 0 . 11353750818 + 0 . 85434947840-. 1056865022 = 0 . 86220048438 hence p 2 (2 . 37) = 0 . 862200484 correct to 9 decimal places. 1 Using a calculator we nd that ln(2 . 37) = 0 . 862889955 correct to 9 decimal places. ln(2 . 37)-p 2 (2 . 37) = 6 . 89471 10-4 1 Note that because of rounding errors in the calculation when I quote the nal answer I drop a couple of decimal places. MATH2051 Tutorial 1 2010-11-09...
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