MATH 4461 Numerical Analysis

MATH 4461 Numerical Analysis

<p> MATH 4461 Numerical Analysis Project 4 Due October 27, 2011</p><p>1. For a given function f (x)</p><p> a) Derive the best approximation for using the points f ''(x) f (x0  2h) , f (x0  h) , and . f (x0 ) f (x0  h)</p><p> x 2 sin(e x ) b) Use your approximation to estimate f ''(1) , if f (x)  for 1  x 4 h  0.1, h  0.05, h  0.01.</p><p> c) Use your error from part (a) to find an upper bound on the error for your approximation with h  0.01.</p><p> d) Find the absolute error for your approximation in part (b) for h  0.01.</p><p>3 2. Determine the derivative of f (x)  tan x ln(7  e 4x ) at x  0.5 . Use the forward difference method, the 3-point formula and the 5-point formula to approximate this derivative as closely as possible. You want to find your approximation for smaller and smaller h vales and see what happens. Keep going until round-off error ruins your approximation. (Use Maple to help you evaluate). </p><p>5 1 3. Consider  dx 3 x ln x</p><p> a) Code Composite Trapezoidal rule and use it to compute with the given integral with h  0.25 .</p><p> b) Determine what h would have to be for Trapezoidal Rule to have an error10 6 .</p><p> c) Code Composite Simpson rule and use it to compute with the given integral with h  0.25 .</p><p> d) Determine what h would have to be for Simpson’s Rule to have an error10 6 .</p><p> e) Use Gaussian quadrature with n=2,3, and 4 approximate the integral.</p><p> f) Calculate the exact answer and compare your approximations in part a) and c). 4. A car laps a race track in 84 seconds. The speed of the car at each 6-second interval is determined by using a radar gun and is given from the beginning of the lap, in feet/second, by the entries in the table. How long is the track?</p><p>Time 0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 Speed 124 134 148 156 147 133 121 109 99 85 78 89 104 116 123</p><p>5. Determine constants a, b, c, d, e and f that will produce a quadrature formula of degree 5 </p><p>2 where  f (x)dx  af (0)  bf (1)  cf (2)  df '(0)  ef '(1)  ff '(2) . 0</p>

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