s08 - Ma 449: Numerical Applied Mathematics Model Solutions...

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Ma 449: Numerical Applied Mathematics Model Solutions to Homework Assignment 8 Prof. Wickerhauser Exercise 4 of Section 7.1, p.362. The integral of f0*(x-x1)/(x0-x1)+f1*(x-x0)/(x1-x0) is f0*(x-x1)^2/ 2(x0-x1)+f1*(x-x0)^2/2(x1-x0) on [x0,x1] = [f1*(x-x0)^2 - f0*(x-x1)^2] / 2(x1-x0) on [x0,x1] = [f1*(x1-x0)^2 + f0*(x0-x1)^2] / 2(x1-x0) = [f1*(x1-x0)^2 + f0*(x1-x0)^2] / 2(x1-x0) = ( [f1+f0]/2 ) * (x1-x0), which is the trapezoid rule. Exercise 5 of Section 7.1, p.363. Write I for the exact value of the integral and T for the trapezoid rule approximation. Then, for various polynomials p=p(x): p0(x)=1: I(p0) = 1*(x1-x0) T(p0) = ([1+1]/2) *(x1-x0) ==> T(p0)=I(p0) p1(x)=x: I(p1) = [x1^2-x0^2]/2 = ([x1+x0]/2)*(x1-x0) T(p1) = ([x0+x1]/2) *(x1-x0) ==> T(p1)=I(p1) p2(x)=x^2: I(p2) = [x1^3-x0^3]/3 = ([x1^2+x0*x1+x0^2]/3)*(x1-x0) T(p2) = ([x0^2+x1^2]/2) *(x1-x0) ==> T(p2)!=I(p2) Hence the degree of precision of the trapezoid rule is 1. Exercise 6 of Section 7.1, p.363. Write I for the exact value of the integral and S for the Simpson rule approximation. Then, for various polynomials p=p(x): p0(x)=1: I(p0) = 1*(x1-x0) S(p0) = ([1+4*1+1]/6) *(x1-x0) ==> S(p0)=I(p0) p1(x)=x: I(p1) = [x1^2-x0^2]/2 = ([x1+x0]/2)*(x1-x0) S(p1) = ([x0+4*(x0+x1)/2+x1]/6) *(x1-x0) = ([x0+x1]/2) *(x1-x0) ==> S(p1)=I(p1) p2(x)=x^2: I(p2) = [x1^3-x0^3]/3 = ([x1^2+x0*x1+x0^2]/3)*(x1-x0)
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This note was uploaded on 05/16/2010 for the course MATH 449 taught by Professor Wickerhauser during the Fall '09 term at Washington University in St. Louis.

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s08 - Ma 449: Numerical Applied Mathematics Model Solutions...

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