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Unformatted text preview: Department of Mathematics Solutions for the exam paper for course 157 sat in June 2001. Question 1 x = 12 . 74 x = (+0 . 127 , +2) RE = . 04 y = 0 . 0025 y = (+0 . 250 , 2) RE = 0 z = 12 . 55 z = ( . 126 , +2) RE = . 05 x y := (+0 . 127 , +2) RE = . 0025 x + z := (+0 . 100 , 0) RE = 0 x y x + z := (+0 . 127 , +3) RE = 0 Total RE = 127 12 . 74 . 0025 12 . 74 12 . 55 60 [Note: This rounding error is relatively much larger than any individual error. It is caused by cancellation in the denominator.] When i becomes large, e i becomes so small that 1 . 5 + e i := 1 . 5 . When this happens S i is computed as zero. Thus for large i , S i = 0. x f f 2 f 3 f 1 . . 8988 625 1 . 5 . 9613 293 332 8 2 . . 9945 301 31 2 . 5 . 9976 To interpolate at x = 1 . 5 using these values and Newtons Forward Formula with x = 1 . 0, we have x + uh = 1 . 5 with h = 0 . 5 gives u = 1. Using quadratic interpolation, for example, we get f (1+0 . 5 u ) N n (1+0 . 5 u ) = 0 . 8988 + u . 0625 1 2 u ( u 1)0 . 0293+etc. Differentiating this we get f (1 . 5) " 1 h d du ( N n (1 + 0 . 5 u )) # u =1 Thus, we get the sequence of approximations based on linear, quadratic and cubic interpolation f (1 . 5) 1 . 5 (0 . 0625) u =1 = 0 . 125 f (1 . 5) 1 . 5 . 0625 2 u 1 2 . 0293 u =1 = 0 . 0957 f (1 . 1) " 1 . 5 . 0625 2 u 1 2 . 0293 3 u 2 6 u + 2 6 . 0008 !# u =1 = 0 . 0960 Question 2 Taylor series for f ( a ) = 0 0 = f ( a ) = f ( x + ( a x )) = f ( x ) + ( a x ) f ( x ) + 1...
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 Spring '09
 MarniMishna

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