NOTE: in 37, the last term in the given integral should be
changed to (kf^(3))(c).
- Please Use Matlab Note In 37 The Last Term In The Given Integral Should Be Changed To Kf 3 C 1 (161.49 KiB) Viewed 26 times
Please use matlab. NOTE: in 37, the last term in the given integral should be changed to (kf^(3))(c).
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Please use matlab. NOTE: in 37, the last term in the given integral should be changed to (kf^(3))(c).
Please use matlab.
NOTE: in 37, the last term in the given integral should be
changed to (kf^(3))(c).
36. Consider the integral Vare dx = 1.2563 0825 51861. (a) Evaluate using the midpoint rule with a sequence of doubling n's, and find the errors. Take the ratios of errors. Are they four as expected? (b) Can you explain why the errors are not behaving as expected? Hint: remember that 1 f"(e)=s"(d;) where each di € [11–1,17). T2 n i=1 (c) Apply the substitution = uz (or u = ) to the integral. Then try the midpoint rule again, listing errors. Why do you think it is working well this time? 37. (a) Explain why you can't use Trapezoidal or Simpson's rules to approximate / in af (2) ds. (b) Derive an integral formula of the form ( in a (a) dr x a50) +651/2) + cf(1)+43"(c), In f( + by making it exact for f(x) = 1,1,12 and finding k by setting f(x) = z'. It will be 1 useful to know that " Inc dr = Hint: you should get k = -1/288. (n + 1)2 (c) Use this formula (without the f(1) term) to approximate In r cos r d -0.91608 30704. What is the error? Remember that in this case f(x) = cos r. (d) Using the midpoint rule with f(x) = In z cos r, how big does n need to be to get the same accuracy as the previous formula? Clearly building integration rules that take into account the form of the integrand is clearly a good idea! -
NOTE: in 37, the last term in the given integral should be
changed to (kf^(3))(c).
36. Consider the integral Vare dx = 1.2563 0825 51861. (a) Evaluate using the midpoint rule with a sequence of doubling n's, and find the errors. Take the ratios of errors. Are they four as expected? (b) Can you explain why the errors are not behaving as expected? Hint: remember that 1 f"(e)=s"(d;) where each di € [11–1,17). T2 n i=1 (c) Apply the substitution = uz (or u = ) to the integral. Then try the midpoint rule again, listing errors. Why do you think it is working well this time? 37. (a) Explain why you can't use Trapezoidal or Simpson's rules to approximate / in af (2) ds. (b) Derive an integral formula of the form ( in a (a) dr x a50) +651/2) + cf(1)+43"(c), In f( + by making it exact for f(x) = 1,1,12 and finding k by setting f(x) = z'. It will be 1 useful to know that " Inc dr = Hint: you should get k = -1/288. (n + 1)2 (c) Use this formula (without the f(1) term) to approximate In r cos r d -0.91608 30704. What is the error? Remember that in this case f(x) = cos r. (d) Using the midpoint rule with f(x) = In z cos r, how big does n need to be to get the same accuracy as the previous formula? Clearly building integration rules that take into account the form of the integrand is clearly a good idea! -
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