Code A function []=find_root_A() myfun = @f; x0 = 0.5; x = fzero(myfun,x0) end function y f(x) y = cos(x).^3-x; end Code B function []=find_root_B() myfun @(x) cos(x).^3 - x; x_vec [0.5:0.1:4]; A=myfun_4(x_vec)'* myfun_4(x_vec); dummy=sum(eig(A)); x=sqrt(dummy) end Code C function []=find_root_C() myfun @(x) cos(x). ^3 - x; h=1e-2; derivative=@(x,h) (myfun(x+h)-myfun(x))/h; dummy=1; accuracy 1e-15; Code D function []=find_root_D() myfun @(x) cos(x).^3 - x; x_vec=[0.5:0.1:4]; B=myfun(x_vec)'*myfun(x_vec); dummy-pinv(B)*x_vec'; x=max(abs(dummy)) end x=0.5; while (abs(dummy-x)>accuracy) dummy=x; if (abs(derivative(x,h))>2*eps) x=x-myfun(x)/derivative(x,h); else disp(derivative(x,h)) sprintf('derivative too close to zero!') break end end disp(x) end
Code A function []=find_root_A() myfun = @f; x0 = 0.5; x = fzero(myfun,x0) end function y f(x) y = cos(x).^3-x; end Code B function []=find_root_B() myfun @(x) cos(x).^3 - x; x_vec [0.5:0.1:4]; A=myfun_4(x_vec)'* myfun_4(x_vec); dummy=sum(eig(A)); x=sqrt(dummy) end Code C function []=find_root_C() myfun @(x) cos(x). ^3 - x; h=1e-2; derivative=@(x,h) (myfun(x+h)-myfun(x))/h; dummy=1; accuracy 1e-15; Code D function []=find_root_D() myfun @(x) cos(x).^3 - x; x_vec=[0.5:0.1:4]; B=myfun(x_vec)'*myfun(x_vec); dummy-pinv(B)*x_vec'; x=max(abs(dummy)) end x=0.5; while (abs(dummy-x)>accuracy) dummy=x; if (abs(derivative(x,h))>2*eps) x=x-myfun(x)/derivative(x,h); else disp(derivative(x,h)) sprintf('derivative too close to zero!') break end end disp(x) end
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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You want to use MATLAB in order to find an approximate solution to the equation cos3 (x) - x = 0. With which of the presented MATLAB codes is it possible to find a solution to the equation above? (There might be
more than one correct answer)
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