# s10 - Ma 449: Numerical Applied Mathematics Model Solutions...

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Unformatted text preview: Ma 449: Numerical Applied Mathematics Model Solutions to Homework Assignment 10 Prof. Wickerhauser Exercise 1* of Section 8.1, pp.427--428. (a) f'(x) = 6x^2-18x+12 = 6(x-1)(x-2), so f(x) increases on (-infty, 1) where f'(x) > 0; decreases on (1, 2) where f'(x) < 0; increases on (2, infty) where f'(x) > 0. (b) f'(x)=1/(x+1)^2 >0, so f(x) increases on R - {-1} (c) This is the reciprocal of the function in part (b). It decreases where the part (b) function increases, so f(x) decreases on R - {-1}. (d) f(x)=x^x = exp(x log x), so f'(x) = (1+log x) exp(x log x), so f(x) decreases on (0, 1/e), where f'(x)<0 increases on (1/e, infty), where f'(x)>0. Exercise 2(b) of Section 8.1, pp.427--428. Observe that cos(x) is decreasing on [0, pi] and increasing on [pi, 2*pi], and that 0 < pi < 4 < 2*pi. Conclude that f is unimodal on [0,4] with a unique minimum at pi. Exercise 7(d*) of Section 8.1, pp.427--428. C PROGRAM: /* Golden ratio search with N iterations for the minimum of a function f(x) unimodal on [a0,b0] */ #include <assert.h> #include <math.h> #include <stdio.h> double f(double x) { return -sin(x)-x+x*x/2; } double a0=0.8, b0=1.6; int N=20; /* number of iterations */ main() { double a, b, c, d, fa, fb, fc, fd; const double r=(sqrt(5.0)-1)/2; /* golden ratio */ int k; a=a0; b=b0; fa=f(a); fb=f(b); /* initialize */ c=r*a+(1-r)*b; d=r*b+(1-r)*a; fc=f(c); fd=f(d); /* a < c < d < b */ for(k=0; k<N; k++) { printf("k=%d: a[%d]=%.4f, b[%d]=%.4f\n", k, k,a, k,b); if( fc > fd ) { /* ==> min lies in [c,b] */ a=c; c=d; d=r*b+(1-r)*a; /* b is still the same */ fa=fc; fc=fd; fd=f(d); /* f(b) is still the same */ } else { /* min lies in [a,d] */ b=d; d=c; c=r*a+(1-r)*b; /* a is still the same */ fb=fd; fd=fc; fc=f(c); /* f(a) is still the same */ } } return 0; } OUTPUT: k=0: a[0]=0.8000, b[0]=1.6000 k=1: a[1]=1.1056, b[1]=1.6000 k=2: a[2]=1.1056, b[2]=1.4111 k=3: a[3]=1.2223, b[3]=1.4111 k=4: a[4]=1.2223, b[4]=1.3390 k=5: a[5]=1.2669, b[5]=1.3390 k=6: a[6]=1.2669, b[6]=1.3115 k=7: a[7]=1.2669, b[7]=1.2944 k=8: a[8]=1.2774, b[8]=1.2944 k=9: a[9]=1.2774, b[9]=1.2879 k=10: a[10]=1.2814, b[10]=1.2879 k=11: a[11]=1.2814, b[11]=1.2854 k=12: a[12]=1.2830, b[12]=1.2854 k=13: a[13]=1.2830, b[13]=1.2845 k=14: a[14]=1.2830, b[14]=1.2839 k=15: a[15]=1.2833, b[15]=1.2839 k=16: a[16]=1.2833, b[16]=1.2837 k=17: a[17]=1.2833, b[17]=1.2835 k=18: a[18]=1.2834, b[18]=1.2835 k=19: a[19]=1.2834, b[19]=1.2835 Exercise 8(d) of Section 8.1, pp.427--428. Modify the Fibonacci method Matlab function from the textbook's web site: fib.m from http://math.fullerton.edu/mathews/software/NumericalMethods7.1.zip function y=fib(n) %Generates fibonacci numbers for the program fibonacci fz(1)=1;fz(2)=1; for k=3:n fz(k)=fz(k-1)+fz(k-2); end y=fz(n); fibonacci10.m (modified from fibonacci.m) from http://math.fullerton.edu/mathews/software/NumericalMethods7.1.zip function X=fibonacci10(f,a,b,tol,e) %Input - f, the object function % - a, the left endpoint of the interval % - b, the right endpoint of the interval...
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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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s10 - Ma 449: Numerical Applied Mathematics Model Solutions...

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