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
Related questions
Question
100%
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)
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4810b3f8-7174-4e11-ad5b-d1003066dd42%2F9d44e24a-97f9-4327-a1cf-2e01c8f95dcc%2Fvsqniq_processed.png&w=3840&q=75)
Transcribed Image Text: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
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