Estimating SQRT of a non-negative number the long way

Estimate the square root of a non-negative number, then displaying the original number, the answer, and the amount of iterations it took to calculate. Cannot accept a negative or a 0. Using the divide and average method. Estimate with a tolerance less than 1e-6.

I don't know how to actually do the math of it. and I can't figure out how to get the input prompt to show up after the error messges.

clear
clc
num = input("Enter any number greater than 0 to calculate it's square root: ");
if num < 0
    msg = 'Cannot calculate the square root of a negative number. Try again.';
    error (msg); %display error message
else
    if num == 0
        msg = 'Its not very hard to calculate the square root of 0. Try again.';
        error (msg); %display error message
        
    else
        n = 0;
        k = 1:100; %initialize the counter?
        e = 1e-6; %error tolerance <.000001
        
        s = num/2; %the first approximation, n= )
        while e > tol
            sOld = s;
            s = (s + (n/s)) / 2;
            e = abs((s - sOld)/s);
        end
    end
end
%disp ("The square root of" num "is" s "and it took" __ "iterations to calculate.")

Answer 1

[John D'Errico]

If you generate an error message, MATLAB stops execution. An error is a failure, and nothing gets past that. However, a warning message does not cause termination.

But here is a thought for you: Use a while loop. So you might do this.

num = input("Enter any number greater than 0 to calculate it's square root: ");
while num < 0
  warning("Aw come on. I need it to be non-negative. Please try harder.")
  num = input("Enter any number greater than 0 to calculate it's square root: ");
end

When the loop finally terminates, num will be at least zero. But until the while loop is willing to let num pass, it will just keep on asking until you make the test in the while statement fail. Actually, I might even make the warning message complain more severely, the more times it gets a negative number. You could be creative in this of course.

Even better code would worry about the user giving you a complex number, or a NaN.

[Dylan]

John,

Thanks for helping with that. I'm still in need of assistance with the actual math of the problem. I have to write code to approximate/calculate the squareroot of the number the user input. I'm being advised to use the divide and average method. I have a tolerance of 1e-6. So it has to iterate/calculate until it gets within that tolerance. then I have to display how many iterations/calculations it took.