Babylonian algorithm - square root of a number
14 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Francesco Rossi
le 21 Oct 2019
Commenté : Francesco Rossi
le 22 Oct 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
0 commentaires
Réponse acceptée
Stephan
le 21 Oct 2019
Modifié(e) : Stephan
le 21 Oct 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.
Plus de réponses (0)
Voir également
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!