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 个评论
Matt J
2019-2-6
I think you mean
x=(x+a/x)/2
Char Jacobus
2019-2-6
Matt J
2019-2-6
Well, you have to write more than the first line of code in order to call it "help".
Char Jacobus
2019-2-6
Matt J
2019-2-6
I can't seem to make this work for negative inputs of a
Why would you need to?
回答（1 个）
AKARSH KUMAR
2020-6-24
0 个投票
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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2019-2-6
2020-6-24
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