C++ Programming Help Please!
NOTE: DO the *Challenging* Version!
READ THE ENTIRE PROGRAM DESCRIPTION CAREFULLY PLEASE!
The Babylonian algorithm to compute the square root of a number n is as follows:
1. Make a guess at the answer (you can pick n/2 as your initial guess).
2. Compute r = n / guess
3. Set guess = (guess + r) / 2
4. Go back to step 2 for as many iterations as necessary. The more that steps 2 and 3 are repeated, the closer guess will become to the square root of n.
Write a program that inputs a double for n and iterates through the Babylonian algorithm 100 times. For a more challenging version, iterate until guess is within 1% of the previous guess, and outputs the answer as a double.
More Important Info:
Babylonian Algorithm for square root:
Do the Challenging version!!! (I would do the simple version first, then add the challenging loop condition)
This is a math problem:
The challenge is to find the condition that will stop the loop when guess is within 1% of the previous guess. This is not so hard to do on paper. It’s a simple inequality: | guess – previous_guess | < 1% * previous_guess
You will NOT use absolute value function (if you know what that is)
You will have to convert this expression to a simple boolean expression, removing the absolute value.