ython implements Newton's algorithm for finding the square root of a number using recursion NOTE:  n = 17                # we want the square root of 17 for example g = 4                 # our guess is 4 initially error = 0.0000000001  # we want to stop when we are this close while abs(n - (g**2)) > error:   g = g - ((g**2 - n)/(2 * g)) # g holds the square root of n at this point   Implement this algorithm in a function sqRoot(n), that returns the square root using recursion your code to return the square root of n to an accuracy within the error e, using Newton's algorithm with an initial guess of g def sqRoot(n, g, e):

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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Python implements Newton's algorithm for finding the square root of a number using recursion

NOTE: 

n = 17                # we want the square root of 17 for example
g = 4                 # our guess is 4 initially
error = 0.0000000001  # we want to stop when we are this close
while abs(n - (g**2)) > error:
  g = g - ((g**2 - n)/(2 * g))
# g holds the square root of n at this point  

Implement this algorithm in a function sqRoot(n), that returns the square root using recursion

your code to return the square root of n to an accuracy within the error e,
using Newton's algorithm with an initial guess of g

def sqRoot(n, g, e):

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