Question

1. 10 points Find the positive solution of the equation cosx=0.3x (ie. the positive root of cosx-03x=0) using the bisection method with [0, 2] chosen as the initial interval. (a) Perform the first 3 iterations by hand and calculate the relative error for the last iteration; (b) Write a MATLAB code to find the root. The solution will be considered satisfactory if its relative error is smaller than 1%. Plot the relative error vs the number of iterations (must be computer generated plot!). c) Attach your MATLAB code with your submission.

Answer 1

(a) The function is, f(x)=cos x -0.3x The interval of x is [0,2]. f(0)=cos(0) -0.3x0)=1 f(2)= cos(2)-0.3x2 =-1.016 Find the solution using bisection method. Step1: Calculate the first estimate of the numerical solution. a+b 0+2 22 Step 2: Determine whether the true solution lies betw determining the value of the function at x equal to 1. f(1) = cos(1)-0.3x1=0.24 As f (1) is positive, the true solution lies between [1,2]. X NS1 = - = 1 Repeat Step1: Calculate the second estimate of the numerical solution. a+b 1+2 -=1.5 X NS2 = 2 2 Repeat Step 2: Determine whether the true solution lies between by determining the value of the function at x equal to 1.5. f(1.5) = cos(1.5)-0.3x1.5=-0.379 As f (1.5) is negative, the true solution lies between [1,1.5]. Again Repeat Step1: Calculate the third estimate of the numerical solution. a+b 1+1.5 =1.25 XnS3 2 2 Again Repeat Step 2: Determine whether the true solution lies between [1,1.25] or determining the value of the function at x equal to 1.25. F (1.25)=cos(1.25)– 0.3x1.25 = -0.0596 As f (1.25) is negative, the true solution lies between [1,1.25]. The value of x after three iterations is x =1.25.

Again Repeat Step 2: Determine whether the true solution lies between [1,1.25] or [1.25,1.5] by determining the value of the function at x equal to 1.25. f(1.25)=cos(1.25)-0.3x1.25 =-0.0596 As f (1.25) is negative, the true solution lies between [1,1.25]. The value of x after three iterations is x =1.25. Calculate the relative error for the last iteration. 1.25-1 %relative error = — -x 100 %relative error = 25% Determine the exact solution of the function using MATLAB. fx=@ (x) cos (x) -0.3*x; fzero (fx, 1.25) ans = 1.2019 (b) Write the MATLAB code to find the root using bisection method. % define the anonymous function f f=@ (x) cos(x) -0.3*x; f define the interval a b , iterations, tolerance a=1;b=2;imax=20;re=1; % determine the values of the function at x=a, x=b fa=f(a) ;fb=f(b); if fa*fb>0 disp('error: function has the same sign at points a and b') else disp('iteration a relative') for i=1:imax & determine un xn= (a+b)/2; rei (i)=abs((b-a) *100/a); fn=f(xn); printf("%3i $11.6f $11.6f $11.6f $11.6f %11.6f\n',i,a,b, xn, fn, rei(i)) if fn==0 fprintf('exact solution of x=$11.6f is found',xn) break f(x) end if rei(i) <re nai; break end if i=simax fprintf('solution cannot be determined in max iterations \n') break end

ſ check for the condition if fa* fn<0 b=xn; else a=xn; end end end it=1:1:n; plot(it,rei) xlabel ('Number of iterations') ylabel ('Percentage tolerance') iteration Can w 1.000000 1.000000 1.000000 1.125000 1.187500 1.187500 1.187500 1.195313 2.000000 1.500000 1.250000 1.250000 1.250000 1.218750 1.203125 1.203125 1.500000 1.250000 1. 125000 1.187500 1.218750 1.203125 1.195313 11.199219 f (x) -0.379263 -0.059678 0.093677 0.017730 -0.020806 -0.001494 0.008129 0.003320 relative 100.000000 50.000000 25.000000 11.llllll 5.263158 2.631579 1.315789 0.653595 Obtain the MATLAB output for the plot of the relative error versus the number of iterations. Percentage tolerance 1 2 7 8 3 4 5 6 Number of iterations (C) MATLAB code attached.

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% MATLAB code to find the zero value of the function

fx=@(x) cos(x)-0.3*x;

fzero(fx,1.25)

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

% define the anonymous function f

f=@(x) cos(x)-0.3*x;

% define the interval a b , iterations, tolerance

a=1;b=2;imax=20;re=1;

% determine the values of the function at x=a,x=b

fa=f(a);fb=f(b);

if fa*fb>0

    disp('error: function has the same sign at points a and b')

else

    disp('iteration     a           b             x              f(x)           relative')

    for i=1:imax

% determine xn

        xn=(a+b)/2;

        rei(i)=abs((b-a)*100/a);

        fn=f(xn);

        fprintf('%3i    %11.6f   %11.6f    %11.6f    %11.6f     %11.6f\n',i,a,b,xn,fn,rei(i))

        if fn==0

            fprintf('exact solution of x=%11.6f is found',xn)

            break

        end

       

        if rei(i)<re

            n=i;

            break

        end

       

        if i==imax

            fprintf('solution cannot be determined in max iterations \n')

            break

        end

% check for the condition

if fa*fn<0

            b=xn;

        else

            a=xn;

        end

    end

end

it=1:1:n;

plot(it,rei)

xlabel ('Number of iterations')

ylabel (' Percentage tolerance')