function [outputs] function_name (inputs) % Name % Date % Function description %

Question

function [outputs] function_name (inputs) % Name % Date % Function description %Description of inputs, including units % Description of outputs,including units % Any test cases you've used to test your code. % These are here mainly for convenience so that you don't need to remember what you % were testing from session to session. Be sure to include not only the inputs but % the expected outputs. % Input testing and default values % Main body code, properly documented end % while the final end is not required by MATLAB, % I have found that including it does help with debugging

Explanation / Answer

Solution:-

The false-position method in Matlab is as follows.

we will assume a file f.m with contents

function y = f(x)

If y = x3- 2; exists. Then:

format long

epsabs = 1e-5;

epsstep = 1e-5;

a = 0.0;

b = 2.0;

step_size = Inf;

while (step_size >= epsstep || ( abs( f(a) ) >= epsabs && abs( f(b) ) >= epsabs ) )

c = (f(a)*b - f(b)*a)/(f(a) - f(b));

if ( f(c) == 0 )

break;

elseif ( f(a)*f(c) < 0 )

step_size = b - c;

b = c;

else

step_size = c - a;

a = c;

end

end

[a b]

Ans = 1.259915864579067 2

abs(f(a))

Ans = 0.0000246934256663

abs(f(b))

ans = 6

Thus, we will choose 1.259915864579067 as our approximation to the cube-root of 2, which has an actual value of 1.259921049894873.

//// Now function will be implemented like that ///////

function [ r ] = false_position( f, a, b, N, epsstep, epsabs )

% Check that that neither end-point is a root

% and if f(a) and f(b) have the same sign, throw an exception.

if ( f(a) == 0 )

r = a;

return;

elseif ( f(b) == 0 )

r = b;

return;

elseif ( f(a) * f(b) > 0 )

error( 'f(a) and f(b) do not have opposite signs' );

end

% We will iterate N times and if a root was not

% found after N iterations, an exception will be thrown.

cold = Inf;

for k = 1:N

% % Find the false position%%

c = (a*f(b) + b*f(a))/(f(b) - f(a));

% % Check if we found a root or whether or not %%

% we should continue with:

% [a, c] if f(a) and f(c) have opposite signs, or

% [c, b] if f(c) and f(b) have opposite signs.

if ( f(c) == 0 )

r = c;

return;

elseif ( f(c)*f(a) < 0 )

b = c;

else

a = c;

end

% If |b - a| < epsstep, check whether or not

% |f(a)| < |f(b)| and |f(a)| < epsabs and return 'a', or

% |f(b)| < epsabs and return 'b'.

if ( abs( c - cold ) < epsstep )

if ( abs( f(a) ) < abs( f(b) ) && abs( f(a) ) < epsabs )

r = a;

return;

elseif ( abs( f(b) ) < epsabs )

r = b;

return;

end

end

cold = c;

end

error( 'the method did not converge' );

end

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