Babylonian algorithm - square root of a number
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Francesco Rossi
el 21 de Oct. de 2019
Comentada: Francesco Rossi
el 22 de Oct. de 2019
Dear all,
I am trying to bulid a function that should calculate the square root o a positive number. Unfortunatly I cannot run the code as expected.
Does anyone spot the error? Thank you in advance.
function y=sqrtB(x)
% It returns a row vector containing the approximation of the square roots of the elements of x.
% Using the Babylonian method.
% with precision 10^-10
%
% INPUT x ... 1xn vector of positive numbers
%
% OUTPUT y ... 1xn vector of square roots of x
%format long
xn=x./2; %starting number
err=abs(xn-x./xn); % the absolute error
while any(err > 1e-10) % upper bundary for the absolute error (vector compatible)
xn = 0.5 .* (xn + x./xn);
err=abs(xn-x./xn);
%disp(err);
end
y=xn;
disp(x);
disp(y);
end
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Respuesta aceptada
Stephan
el 21 de Oct. de 2019
Editada: Stephan
el 21 de Oct. de 2019
What is the problem - works for me:
y = sqrtB([16 4 9]);
function y=sqrtB(x)
% It returns a row vector containing the approximation of the square roots of the elements of x.
% Using the Babylonian method.
% with precision 10^-10
%
% INPUT x ... 1xn vector of positive numbers
%
% OUTPUT y ... 1xn vector of square roots of x
%format long
xn=x./2; %starting number
err=abs(xn-x./xn); % the absolute error
while any(err > 1e-10) % upper bundary for the absolute error (vector compatible)
xn = 0.5 .* (xn + x./xn);
err=abs(xn-x./xn);
%disp(err);
end
y=xn;
disp(x);
disp(y);
end
i get correct results by calling the function properly.
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