Math Lab 5 Assignment

Transcription

1 Math Lab 5 Assignment Dylan L. Renaud February 22nd, 2017 Introduction In this lab, we approximate the roots of various functions using the Bisection, Newton, Secant, and Regula Falsi methods. The code for all four methods is shown below %Bisection Method 4 function [r]=bisection(f,a,b,tol) 5 n = 0; 6 7 fprintf(' n a b c f(a) f(b) f(c) \n') 8 fprintf(' === ===== ===== ===== ====== ====== ====== \n') 9 while (b-a)/2 > tol 10 n = n + 1; 11 c = (a+b)/2; 12 if f(c) == 0 13 fprintf('%3i %7.4f %7.4f %7.4f %6.12f %6.12f %6.12f \n',n,a,b,c,f(a),f(b),f(c)) 14 disp('you found the root in accordance with the first condition') 15 return 16 end 17 fprintf('%3i %7.4f %7.4f %7.4f %6.12f %6.12f %6.12f \n',n,a,b,c,f(a),f(b),f(c)) 18 if f(a)*f(c) < 0 19 b = c; 20 else 21 a = c; 22 end 23 end 24 r = c; %Newton's Method 28 function [z] = newtons method(f,x 0,tol) 29 syms x 30 dfdx = diff(f,x); 31 dfdx = matlabfunction(dfdx); 32 f = matlabfunction(f); fprintf(' n f(x n) x r Err \n') 35 fprintf(' === ===== ===== === \n') Err = 1; n = 1; 38 while Err > tol; 39 if dfdx(x 0) == 0; 40 display('dfdx is equal to zero') 41 return 42 else 43 x = x 0 - (f(x 0)/dfdx(x 0)); 44 Err = abs(x-x 0); 45 if f(x) < 0 46 fprintf('%3i %15.5e %13.5e %13.5e \n', n, f(x), x,err) 47 end 48 if f(x) > 0 49 fprintf('%3i %15.5e %13.5e %13.5e \n', n, f(x), x, Err) 50 end 51 x 0 = x; n = n + 1; 54 end 55 end z = f(x 0); 1

2 %Secant Method 61 function [ roots,error ] = MySecantMethod(f,x0,x1,tol,max iter) roots(1)=x0; 64 roots(2)=x1; for i=3:max iter if (f(roots(i-1))-f(roots(i-2)))==0 69 disp('error: we are dividing by zero!'); 70 break; 71 end 72 roots(i)=roots(i-1) - f(roots(i-1))*(roots(i-1)-roots(i-2))/(f(roots(i-1))-f(roots(i-2))); 73 error(i-2)=abs(roots(i-1)-roots(i-2)); 74 if (error(i-2)<=tol) 75 fprintf('at the step i = %d, the root found is %8.7f\n',i,roots(i)) 76 return; 77 end 78 end 79 if (i == max iter) 80 disp('we have diverged!') 81 end 82 fprintf('at the step i = %d, the root found is %8.7f\n',i,roots(i)) 83 end %Regula Falsi 86 function [ r,k ] = MyRegulaFalsi( f, a, b, N, tol) if ( f(a) == 0 ) 89 r = a; 90 return; 91 elseif ( f(b) == 0 ) 92 r = b; 93 return; 94 elseif ( f(a) * f(b) > 0 ) 95 error( 'f(a) and f(b) do not have opposite signs' ); 96 end c old = Inf; for k = 1:N 101 c = (a*f(b) - b*f(a))/(f(b) - f(a)); if ( f(c) == 0 ) 104 fprintf('at the step i = %d, the root found is %8.7f\n',k,c); 105 r = c; 106 return; 107 elseif ( f(c)*f(a) < 0 ) 108 b = c; 109 else 110 a = c; 111 end if ( abs( c - c old ) < tol ) c = (a*f(b) - b*f(a))/(f(b) - f(a)); 116 r = c; 117 k = k+1; 118 fprintf('at the step i = %d, the root found is %8.7f\n',k,c); 119 return; 120 end c old = c; 123 end error( 'the method did not converge' ); end %alternative Secant Method 130 function y = secant(f,a,b,maxerr) c = (a*f(b) - b*f(a))/(f(b) - f(a)); 133 disp(' Xn-1 f(xn-1) Xn f(xn) Xn+1 f(xn+1)'); 134 disp([a f(a) b f(b) c f(c)]); 135 while abs(f(c)) > maxerr 2

3 136 a = b; 137 b = c; 138 c = (a*f(b) - b*f(a))/(f(b) - f(a)); 139 disp([a f(a) b f(b) c f(c)]); flag = flag + 1; if(flag == 100) 144 break; 145 end 146 end display(['root is x = ' num2str(c)]); 149 y = c; Problem 1 a) We are asked to find the root of 6 + cos 2 (x) = x. With an interval of [6, 7] (for Bisection, Secant, and Regula Falsi methods) and initial guess x 0 = 7 (for Newton s Method). Code 1 f 6 + cos(x)ˆ2 - x; %changed to syms x for Newton's method 2 x 0 = 7; 3 tol = 1*10ˆ(-7); 4 a = 6; 5 b = 7; 6 7 newtons method(f,x 0,tol) 8 % bisection(f,a,b,tol) 9 % MySecantMethod(f,a,b,tol,100) 10 % MyRegulaFalsi(f,a,b,100,tol) Table Newton s method e e e e e e e e e-05 Bisection method 1 n a b c f(a) f(b) f(c) == ====== ====== ====== Secant method 1 At the step i = 9, the root found is Regula Falsi method 1 At the step i = 8, the root found is

4 Comment For Newton s method, theory predicts that the convergence will be quadratic as f (r) 0 Bisection predicts approximately n = 24 terms will be required, in approximate agreement with our calculations n = ceil ln(10 7 ) ln(1/2) = 24 In other words, we expect linear convergence via the bisection method with rate S = 0.5. Secant method predicts superlinear convergence. Therefore, we expect the number of terms to be between Newton s method and Bisection method. Our calculations agree with this. Regula Falsi also behaves similarly. b) We are asked to find the root of (x 1.5) 5 = 0. With an interval of [1, 2] (for Bisection, Secant, and Regula Falsi methods) and initial guess x 0 = 2 (for Newton s Method). Code 1 f (x-1.5)ˆ5; %changed to syms x for Newton's method 2 x 0 = 2; 3 tol = 1*10ˆ(-7); 4 a = 1; 5 b = 2; 6 7 newtons method(f,x 0,tol) 8 % bisection(f,a,b,tol) 9 % MySecantMethod(f,a,b,tol,100) 10 % MyRegulaFalsi(f,a,b,100,tol) Table Newton s method e e e e e e e e e e e e e e e e e e-08 Bisection method 1 n a b c f(a) f(b) f(c) == ====== ====== ====== You found the root in accordance with the first condition Secant method 1 %Used alternative code for this entry because the MySecantMethod.m was providing incorrect results 2 Xn-1 f(xn-1) Xn f(xn) Xn+1 f(xn+1) Regula Falsi method 1 At the step i = 1, the root found is

5 Comment For Newton s method, theory predicts that the convergence will be linear as f (r) = 0 Furthermore, we find that the rate of linear convergence will be Where m is the multiplicity. In this case, m = 5 so we obtain S = m 1 m S = 4 5 Bisection again predicts approximately n = 24 terms will be required. However, our results deviate from this expectation due to the root existing at the midpoint of our initial interval. n = ceil ln(10 7 ) ln(1/2) = 24 Secant method Regula Falsi produce the same result because they are the same for the first iteration. In this first iteration, the secant line passes through the x-axis at the midpoint of the interval. Futhermore, we expect this fast convergence for these two methods due to the small derivative near the root. 0.4 f (x) x c) We are asked to find the root of x = 0. With an interval of [1, 2] (for Bisection, Secant, and Regula Falsi methods) and initial guess x 0 = 2 (for Newton s Method). Code 1 syms x 2 f (xˆ5-1.5); 3 f1 = (xˆ5-1.5); 4 x 0 = 2; 5 tol = 1*10ˆ(-7); 6 a = 1; 7 b = 2; 8 9 newtons method(f1,x 0,tol) 10 % bisection(f,a,b,tol) 11 % MySecantMethod(f,a,b,tol,100) 12 % MyRegulaFalsi(f,a,b,100,tol) 5

6 Table Newton s method e e e e e e e e e e e e e e e e e e e e e-08 Bisection method 1 n a b c f(a) f(b) f(c) == ====== ====== ====== Secant method 1 At the step i = 10, the root found is Regula Falsi method 1 At the step i = 55, the root found is Comment For Newton s method, theory predicts that the convergence will be quadratic as f (r) 0 Bisection again predicts approximately n = 24 terms will be required, in approximate agreement with our calculations n = ceil ln(10 7 ) ln(1/2) = 24 Secant method predicts superlinear convergence. Therefore, we expect the number of terms to be between Newton s method and Bisection method. Our calculations agree with this. Regula Falsi behaves differently due to the slope of our function near the root. The function has a large derivative in the interval of interest. Therefore, once the second interval is chosen for the second iteration, the root is from then on approached from only one side (the positive side). 6

7 0.4 f (x) x d) We are asked to find the root of 2x 3 15x x 23 = 0. With an interval of [0.5, 3.5] or [3.5, 0.5] (for Bisection, Secant, and Regula Falsi methods) and initial guess x 0 = 0.5 or x 0 = 3.5 (for Newton s Method). Code 1 syms x 2 f 2*xˆ3-15*xˆ2+36*x-23; 3 f1 = 2*xˆ3-15*xˆ2+36*x-23; 4 x 0 =.5; 5 tol = 1*10ˆ(-7); 6 a = 0.5; 7 b = 3.5; 8 9 newtons method(f1,x 0,tol) 10 % bisection(f,a,b,tol) 11 % MySecantMethod(f,a,b,tol,100) 12 % MyRegulaFalsi(f,a,b,100,tol) Table Newton s method at x 0 = e e e e e e e e e e e e-05 Newton s method at x 0 = e e e e e e e e e e e e e e e e e e-14 Bisection method with interval [0.5,3.5] 1 n a b c f(a) f(b) f(c) == ====== ====== ======

8 Secant method with interval [0.5,3.5] 1 At the step i = 25, the root found is Secant method with interval [3.5,0.5] 1 At the step i = 18, the root found is Regula Falsi method with interval [0.5,3.5] 1 At the step i = 17, the root found is Regula Falsi method with interval [3.5,0.5] 1 At the step i = 17, the root found is Comment For Newton s method, theory predicts that the convergence will be quadratic as f (r) 0 We observe a different number of terms between the two starting points due to the relative distances between the starting points and the root (d 0.5 =.5 vs. d 3.5 = 2.5). Bisection again predicts approximately n = 24 terms will be required, in exact agreement with our calculations n = ceil ln(10 7 ) ln(1/2) = 24 Secant method predicts superlinear convergence. Therefore, we expect the number of terms to be between Newton s method and Bisection method. Our calculations agree with this. The difference in iterations for different interval choices is likely due to the choice of the x n+1 term as either the initial or endpoint in the interval (e.g. [0.5,x n+1 ] vs [x n+1,3.5]). Regula Falsi remains the same regardless of inversion of the interval due to its root bracketing behaviour. 8