f(x) = x2-10x + 23 Write a MATLAB code with the following requirements: Set the

Question

f(x) = x2-10x + 23 Write a MATLAB code with the following requirements: Set the maximum number of iterations to 30 and set the accuracy on the root to 0.01. . Solve for the root with the specified accuracy. Show the number of iterations. Include a break statement or use a while loop so when the solution reaches the specified accuracy it will stop. a) Solve using the secant method and xo-1,xi-2. Plot the function and the root on the same figure. b) Solve using the Newton-Raphson method and xo 2.5 c) Comment on the results. How many iterations did it take for the specified accuracy for each method?

Explanation / Answer

%%%%%%%%%% Solution using secant method %%%%%%%%%%%%%%%%%%%

tolerance=0.01;

f=@(x) x^2-10*x+23;

x0=1;

x1=2;

iterations=1;  

maximumIterations=30;

c=(x0*f(x1)-x1*f(x0))/(f(x1)-f(x0));

while abs(f(c))>tolerance

x0=x1;

x1=c;

c=(x0*f(x1)-x1*f(x0))/(f(x1)-f(x0));

iterations=iterations+1;

if(iterations==maximumIterations)

break;

end

end

if iterations<maximumIterations

disp(['Root of this Equation = ' num2str(c)]);

fprintf('Number of iterations taken: %d ', iterations);

else

disp('Error! Cannot find Root of this equation.');

end

% plotting

x = -10:0.01:10;

f = x.^2-10.*x+23;

figure;

plot(x,f)

hold on;

plot(c,'r');

grid on

%%%%%%%%%%%%%%%%%%%%%%%%%% Solution using Newton Raphson %%%%%%%%%%%%%%%%%%%

tolerance=0.01;

f=@(x) x^2-10*x+23;

iterations=1;  

maximumIterations=30;

d=@(x) 2*x-10;

x(1)=2.5;

error=100;

while error>tolerance

x(iterations+1)=x(iterations)-((f(x(iterations))/d(x(iterations))));

error=abs((x(iterations+1)-x(iterations))/x(iterations));

  

iterations=iterations+1;

if(iterations==maximumIterations)

break;

end

end

root=x(iterations);

disp(root);

fprintf('Number of iterations taken: %d ', iterations);

%%%%%%%%%%%%% Question c %%%%%%%%%%%%%%%%

% Secant method took 3 iterations whereas newton raphson took 4 iterations.

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