Tutorial_1_Pbms 2 and 3_Answers.pdf - Answers to problems 2 and 3 Tutorial 01 2 z f x y ofx 2data y xyerror x 3.0 y 2.0 x y 0.1 2 Propagation 2 f f 2 xy

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Unformatted text preview: Answers to problems 2 and 3: Tutorial 01 2 z f ( x, y ) ofx 2data y xyerror ; x 3.0, y 2.0, x y 0.1 2. Propagation 2 f f 2 xy y 2 8 2y 4 x x 2 f 2 f x 2 2 xy 3 2 x 6 y y 2 2 z f f ( x, y ) x 2 y xy 2 ; x 3.0, y 2.0, x y 0.1 2x 2 y 2 xf y 2 f 2 xy y 2 8 2y 4 x x 2 fz f 2( x x, y y ) 2 f ( x, y ) x 2 xy 3 2 x 6 2 y f f 1y 2 f 2 f 2 f 2 2 x y x y 2 x y R 2 2 2 y 2! x y xy f x 2x 2 y 2 xyorder error analysis (i) First z z x y f f 1 2 f 2 f 2 f 2 2 x y 2 x 2 y 2 x y R x by second y order 2! error xy (ii) Similarly, x analysis y f ( xxx, y yy) 0.8 f (x0.3 , y ) 1.1 z 1 ≈ 1.1 + (0.04+0.06+0.04)=1.17 2 1 3. Round-off error f ( x) ( x 1) x (i) Condition number of the problem: For large values of x ( x 1) 1/ x x 1 fx( x) x 11 1/ ( x 1) 1/ x x 1 x 1 Cp 2 ( x f1)( x)x 22 2 ( x 1) x 2 ( x 1) x ( x 1) x The problem is well-conditioned (ii) Relative error (for double precision calculate upto twice the given precision; 12 significant digits) x 208208 f ( x) 0.456299 103 0.456298 103 0.110 2 f ( x) 0.1095774 102 er f ( x) f ( x) 100 8.74% f ( x) (iii) Modified algorithm Multiply and divide f ( x) by f ( x) ( x 1) x 1 ( x 1) x 1 2 f ( 1x) 2 3 0.109577 10 3 f ( x) 10 0.456299 10 0.109577 0.456298 10 0.456299 103 0.456298 103 f ( x) f ( x) f ( x) f ( x) er 100 0.4 105% 5 er 100 0.4 f 10 % ( x) f ( x) 2 ...
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