Question

Using MATLAB or FreeMat

----------------------------

Bisection Method and Accuracy of Rootfinding Consider the function f(0) = 3 cos 2r cos 4-2 cos Garcos 3r - 6 cos 2r sin 2r-5.03r +5/2. This function has exactly one root in the interval <I<1. Your assignment is to find this root accurately to 10 decimal places, if possible. Use MATLAB, which does all calculations in double precision, equivalent to about 16 decimal digits. You should use the Bisection Method as described below to determine the root, and then try to assess whether all 10 places are correct 1. Begin by applying the Bisection Method three times, using three different starting intervals (a,b). (Try to make sure they aren't thinly-disguised repetitions of one another.) Be sure to check that the Intermediate Value Theorem guarantees a solution before you start. If you are using the textbook's code, choose a tolerance TOL that will guarantee 10 correct digits. Report your three solutions, rounded to 10 digits after the decimal point Do they agree? 2. Let's try to check your computed root (or roots, if you got more than one). What is the checking error" 1()? Make a table of ll nearby numbers (including your r) and their checking errors: - 5 x 10-10 - 4 x 1p-10 +5 x 10-10 When writing down the checking errors, you only need to report a few significant digits in the right side of the table, but be sure to list the exponent Does the table change your opinion about your best guess of the root? Report your best guess for the root of f1(1) up to 10 decimal places, and underline or highlight the digits you are confident are correct 3. Repeat Step 1 for the function () that is obtained by replacing the 5/2 in the function by 7/2 Just as for f(x), the revised function f(x) has exactly one roor in the intervalo S S L 4. Repeat Step 2 with f(t). 5. Summarize any differences between the two cases, fi and f2. Do you believe you found the root to 10 correct places in both cases? In theory, the Bisection Method gives as much precision as you ask it for Can you explain any difficulties the Bisection Method is having in practice?

Answer 1

%&Matlab code for Bisection method clear all close all function for which root have to find fun @(x) 3.*cos(2.*x).*cos(4.*)-2.*cos(6.**). *COS (3. *x)-6.*cos(2.**). *(sin(2.*x)). 2-5.C xx=linspace(0,1); yy=fun (xx); fprintf('for the function f(x)=') disp(fun) figure (1) plot (xx, YY) xlabel('x') ylabel('f(x)') title('x vs. f(x) plot') (root)=bisection_method(fun, 0,1,1000); fprintf('\tRoot using Bisection method is $f.\n',root) hold on plot (root, fun(root), 'r*') grid on; box on; $table for x and fl(x) fprintf('\n\t x\t\t\tf(x)\n') xx=-5:1:5; for i=1:length(xx) rr=root+xx(i)*10^-10; fprintf('\t%f+(8d*10^-10 )= te\n',root, xx(i), fun(rr)) end function for which root have to find fun= (x) 3. *COS(2.**).*cos(4.*)-2. *cos( 6.*). *cos(3.*X)-6. *cos(2.**). *(sin(2.*x)). 2-5.C xx=linspace(0,1); yy=fun (xx); fprintf('for the function f(x)=') disp(fun) figure(2) plot (xx,yY) xlabel('x') ylabel('f(x)') title('x vs. f(x) plot') (root)=bisection_method(fun, 0,1, 1000); fprintf('\tRoot using Bisection method is $f.\n', root) hold on plot (root, fun(root), 'r*') grid on; box on; %$table for x and fl(x) fprintf('\n\t x\t\t\tf(x)\n')

xx=-5:1:5; for i=1:length(xx) rr=root+xx(i)*10^-10; fprintf('\t%f+({d*10^-10 )= te\n',root, xx(i), fun(rr)) end Matlab function for Bisection Method function [root]=bisection_method(fun, x0, x1,maxit) if fun (x0)<=0 t=x0; x0=xl; x1=t; end fprintf('\nRoot using Bisection method\n') f(xl) should be positive f (x0) should be negative k=10; count=0; while k>10^-10 count=count+l; xx (count)=(x0+x1)/2; mm=double( fun(xx (count))); if mm>=0 x0=xx (count); else xl=xx (count); end err(count) =abs(fun (xl)); k=abs(fun(x1)); if count>=maxit break end end fprintf('\tAfter $d iteration root using Bisection method is $f \n', count, xx (count)) root=xx(end); end for the function f(x)= @ (x)3.*cos(2.**). *cos (4.**)-2.*cos(6.**). *COS (3. *X)-6.*cos(2.*x).*(sin(2.*x)). 2- Root using Bisection method After 37 iteration root using Bisection method is 0.567042 Root using Bisection method is 0.56 7042. x f(x) 0.567042+(-5*10^-10)= -7.124996e-09 0.567042+(-4*10^-10)= -5.708451e-09 0.567042+(-3*10^-10)= -4.291906e-09

0.567042+(-2*10^-10)= -2.875361e-09 0.567042+(-1*10^-10)= -1.458817e-09 0.567042+(0*10^-10)= -4.22724 le-11 0.567042+(1+10^-10)= 1.374272e-09 0.567042+(2*10^-10)= 2.790817e-09 0.567042+(3*10^-10)= 4.207363e-09 0.567042+(4*10^-10)= 5.623908-09 0.567042+(5*10^-10)= 7.040452e-09 for the function f(x)= @(x)3. *cos(2.*x). *cos(4.*x)-2. *cos(6.*x). *cos(3. *x)-6. *cos(2. *x). (sin(2. *x)). 2- Root using Bisection method After 13 iteration root using Bisection method is 0.348999 Root using Bisection method is 0.348999. x f(x) 0.348999+(-5*10^-10)= -2.093170e-11 0.348999+(-4*10^-10)= -2.093126e-11 0.348999+(-3*10^-10)= -2.0931 70e-11 0.348999+(-2*10^-10)= -2.0931 70e-11 0.348999+(-1*10^-10)= -2.0931 70e-11 0.348999+(0*10^-10)= -2.093170e-11 0.348999+(1*10^-10)= -2.093170e-11 0.348999+(2*10^-10)= -2.093081e-11 0.348999+(3*10^-10)= -2.093126e-11 0.348999+(4*10^-10)= -2.093081e-11 0.348999+(5*10^-10)= -2.093126e-11

X VS. f(x) plot fo 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 9 0. 1 X VS. f(x) plot 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 9 0. 1

%%Matlab code for Bisection method
clear all
close all

%function for which root have to find
fun=@(x) 3.*cos(2.*x).*cos(4.*x)-2.*cos(6.*x).*cos(3.*x)-6.*cos(2.*x).*(sin(2.*x)).^2-5.*cos(3.*x)+5/2;
xx=linspace(0,1);
yy=fun(xx);
fprintf('for the function f(x)=')
disp(fun)
figure(1)
plot(xx,yy)
xlabel('x')
ylabel('f(x)')
title('x vs. f(x) plot')

[root]=bisection_method(fun,0,1,1000);
fprintf('\tRoot using Bisection method is %f.\n',root)
hold on
plot(root,fun(root),'r*')
grid on; box on;

%%table for x and f1(x)
fprintf('\n\t x\t\t\tf(x)\n')
xx=-5:1:5;
for i=1:length(xx)
    rr=root+xx(i)*10^-10;
    fprintf('\t%f+(%d*10^-10)= %e\n',root,xx(i),fun(rr))
end

%function for which root have to find
fun=@(x) 3.*cos(2.*x).*cos(4.*x)-2.*cos(6.*x).*cos(3.*x)-6.*cos(2.*x).*(sin(2.*x)).^2-5.*cos(3.*x)+7/2;
xx=linspace(0,1);
yy=fun(xx);
fprintf('for the function f(x)=')
disp(fun)
figure(2)
plot(xx,yy)
xlabel('x')
ylabel('f(x)')
title('x vs. f(x) plot')

[root]=bisection_method(fun,0,1,1000);
fprintf('\tRoot using Bisection method is %f.\n',root)
hold on
plot(root,fun(root),'r*')
grid on; box on;

%%table for x and f1(x)
fprintf('\n\t x\t\t\tf(x)\n')
xx=-5:1:5;
for i=1:length(xx)
    rr=root+xx(i)*10^-10;
    fprintf('\t%f+(%d*10^-10)= %e\n',root,xx(i),fun(rr))
end
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%Matlab function for Bisection Method
function [root]=bisection_method(fun,x0,x1,maxit)
if fun(x0)<=0
    t=x0;
    x0=x1;
    x1=t;
end
fprintf('\nRoot using Bisection method\n')
%f(x1) should be positive
%f(x0) should be negative
k=10; count=0;
while k>10^-10
    count=count+1;
    xx(count)=(x0+x1)/2;
    mm=double(fun(xx(count)));
    if mm>=0
        x0=xx(count);
    else
        x1=xx(count);
    end
    err(count)=abs(fun(x1));
    k=abs(fun(x1));
    if count>=maxit
        break
      
    end
    %
end
fprintf('\tAfter %d iteration root using Bisection method is %f\n',count,xx(count))
root=xx(end);
end

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