Approximating square root function using loops

Please help me solve this:
the "divide and average" method, an old-time method for approximating the square root of any positive number a, can be formulated as x=(x+a/x)/2. write a well-structured function to implement this alogrithm.
here is my first line:
function estSqrt= ApproxSqrt(a,tol)
a is the number whos square root i want to find
tol is the tolerance that must be greater than abs(xOld-xNew)/abs(xOld)

5 commentaires

I think you mean
x=(x+a/x)/2
yeah sorry
Matt J
Matt J le 6 Fév 2019
Well, you have to write more than the first line of code in order to call it "help".
I can't seem to make this work for negative inputs of a
function estSqrt= ApproxSqrt(a,tol)
% function estSqrt= ApproxSqrt(a,tol)
e = 1;
x = a/2;
estSqrt = 1;
if a == 0
estSqrt = 0;
end
while e > tol
xOld = x;
x = (x+a/x)/2;
e=abs((x - xOld)/x);
estSqrt = x;
if a < 0
a = abs(a);
estSqrt = x*1i;
end
end
end
Matt J
Matt J le 6 Fév 2019
I can't seem to make this work for negative inputs of a
Why would you need to?

Connectez-vous pour commenter.

Réponses (1)

AKARSH KUMAR
AKARSH KUMAR le 24 Juin 2020

0 votes

function estSqrt= ApproxSqrt(a,tol)
if a<0
msg='Can't calculate square root of negative number';
error(msg);
else if a==0
estSqrt=0;
else
e = 1;
x = a/2;
estSqrt = 1;
while e > tol
xOld = x;
x = (x+a/x)/2;
e=abs((x - xOld)/x);
estSqrt = x;
if a < 0
a = abs(a);
estSqrt = x*1i;
end
end
end
end

Catégories

En savoir plus sur Mathematics dans Centre d'aide et File Exchange

Produits

Version

R2018b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by