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E2Solns - Department of Mathematics Solutions for the exam...

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Department of Mathematics Solutions for the exam paper for course 157 sat in June 2001.
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Question 1 x = 12 . 74 ˜ x = (+0 . 127 , +2) RE = - 0 . 04 y = 0 . 0025 ˜ y = (+0 . 250 , - 2) RE = 0 z = - 12 . 55 ˜ z = ( - 0 . 126 , +2) RE = - 0 . 05 ˜ x - ˜ y := (+0 . 127 , +2) RE = - 0 . 0025 ˜ x + ˜ z := (+0 . 100 , 0) RE = 0 ˜ x - ˜ y ˜ x z := (+0 . 127 , +3) RE = 0 Total RE = 127 - 12 . 74 - 0 . 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 Δ f Δ 2 f Δ 3 f 1 . 0 0 . 8988 625 1 . 5 0 . 9613 - 293 332 - 8 2 . 0 0 . 9945 - 301 31 2 . 5 0 . 9976
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To interpolate at x = 1 . 5 using these values and Newton’s Forward Formula with x 0 = 1 . 0, we have x 0 + 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 0 . 0625 - 1 2 u ( u - 1) 0 . 0293 + etc. Differentiating this we get f 0 (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 0 (1 . 5) 1 0 . 5 (0 . 0625) u =1 = 0 . 125 f 0 (1 . 5) 1 0 . 5 0 . 0625 - 2 u - 1 2 0 . 0293 ¶‚ u =1 = 0 . 0957 f 0 (1 . 1) " 1 0 . 5 ˆ 0 . 0625 - 2 u - 1 2 0 . 0293 - 3 u 2 - 6 u + 2 6 0 . 0008 !# u =1 = 0 . 0960
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Question 2 Taylor series for f ( a ) = 0 0 = f ( a ) = f ( x + ( a - x )) = f ( x ) + ( a - x ) f 0 ( x ) + 1 2 ( x - a ) 2 f 0 0 ( x ) + · · · and if ( x - a ) is small 0 = f ( a ) f ( x ) + ( a - x ) f 0 ( x ) therefore a x - f ( x ) f 0 ( x ) This is made the basis of the Newton Raphson method by choosing an approximation x i to a and getting a new approximation x i +1 by assuming equality in the above expression. That is x i +1 = x
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