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Solution to HW5_S09 - HW#5 From textbook by Atkinson...

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HW#5 EGM6341 From textbook by Atkinson pp185-194 #6, The total error e(x) is the sum of the truncation error TE(x) and the round-off error R(x) ) ( ) ( ) ( x R x TE x + = ε The truncation error is given by ) )( )( ( " 2 1 ) ( 1 0 x x x x f x TE - - = ξ by for some ξ between x 0 and x i . Note: max | 4 / | ) )( ( | 2 1 0 h x x x x = - - , h= | | 0 1 x x - , and max| ) ( " ξ f | = max |sin ξ | = 1. The maximum roundoff error is Max R(x) = 5 x 10 -7 Thus we need ) ( ) ( ) ( x R x TE x + = ε < 8 / 2 h + 5 x 10 -7 < 10 -6 which gives h < 2 x 10 -3 . #11 Let z = e x . Then the polynomial becomes ) ( ) ( 0 0 z p z c e c x p n j j n j jx j n j n = = = = = …(the rest of the proof follows naturally). #18. f x D x D 2 x D 3 x D 4 x D 5 x Linear Quad Cubic 4th order 5th order 0.22389 1 2 -1.74569 0.2840 7 -0.7793 0.7648 -1.672 2.39084 2.4014 4 2.4046 5 2.40482 5 2.404825 6 0.16660 7 2.1 -1.77794 0.4152 8 -0.9487 1.2202 -2.787 2.39622 2.4038 5 2.4048 2 2.40482 6 2.404825 5 0.11036 2 2.2 -1.82407 0.5709 6 -1.2110 1.9545 -4.946 2.40131 2.4048 1 2.4048 3 2.40482 5 2.404825 7 0.05554 2.3 -1.88565 0.7632 1 -1.6159 3.2050 -9.336 2.40473 2.4048 3 2.4048 2 2.40482 6 2.404825 2 0.00250 8 2.4 -1.96497 1.0093 9 -2.2505 5.4511 -18.826 2.40493 2.4048 1 2.4048 3 2.40482 2 -0.04838 2.5 -2.06521 1.3356 2 -3.2728 9.7211 2.40008 2.4063 3 2.4041 5 2.40535 -0.0968 2.6 -2.19085 1.7828 6 -4.9830
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-0.14245 2.7 -2.34815 2.4182 4 -0.18504 2.8 -2.54612 -0.22431 2.9 D k x= kth order divided difference; I have discussed the procedure for this problem in class. The inverse interpolation can be carried out as long as the function is monotonic in that region.
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