human
Babylonian method for square root | Returns the square root of n . Note that the function.
We are using n itself as initial approximation This can definitely be improved.
e decides the accuracy level.
Driver program to test above function"
Solution in Python:
 What is a code to babylonian method for square root and the other steps required?
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#Here is code to babylonian method for square root and the other steps required.
 def squareRoot ( n ) :
	x = n
	y = 1
	e = 0.000001
	while ( x - y > e ) :
		x = ( x + y ) / 2
		y = n / x
	return x
n = 50
print ( " Square ▁ root ▁ of " , n , " is " , round ( squareRoot ( n ) , 6 ) )