Approximating square root function using loops
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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 comentarios
Matt J
el 6 de Feb. de 2019
I can't seem to make this work for negative inputs of a
Why would you need to?
Respuestas (1)
AKARSH KUMAR
el 24 de Jun. de 2020
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
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