PythonWe are trying to find the square root using RECURSION. Look at the instructions. The following code implements Newton's algorithm forfinding the square root of a number using repetition:n17# we want the square root of 17for exampleg = 4# our guess is 4 initiallyerror0.0000000001# we want to stop when weare this||closewhile abs (n(g**2)) >error:( (g**2 - n)/(2g))# g holds the square root of n at this pointImplement this algorithm in a function sqRoot (n, g, e),that returns the square root using recursion.Assume that the function will be called with positivenumbers only.

Input : n = Number g = inital guess of square root e = permissible error Output : Tuned value of…

The following code implements Newton's algorithm for finding the square root of a number using repetition: 17 # we want the square root of 17 for example # our guess is 4 initially # we want to stop when we g = 4 error 0.0000000001 are thi close while abs (n (g**2)) > error: ( (g**2 n)/(2 g)) g = g # g holds the square root ofn at this point Implement this algorithm in a function sqRoot (n, g, e), that returns the square root using recursion. Assume that the function will be called with positive numbers only.

The following code implements Newton's algorithm for finding the square root of a number using repetition: 17 # we want the square root of 17 for example # our guess is 4 initially # we want to stop when we g = 4 error 0.0000000001 are thi close while abs (n (g2)) > error: ( (g2 n)/(2 g)) g = g # g holds the square root ofn at this point Implement this algorithm in a function sqRoot (n, g, e), that returns the square root using recursion. Assume that the function will be called with positive numbers only.

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter4: Selection Structures

Section: Chapter Questions

Problem 14PP

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le.com/forms/d/e/1FAlpQLSc6PlhZGOLJ4LOHo5cCGEf9HDChfQ-tT1bES-BKgkKu44eEnw/formResponse The following iterative sequence is defined for the set of positive integers: Sn/2 3n +1 ifn is odd if n is even Un = Using the rule above and starting with 13, we generate the following sequence: 13 u13 = 40 u40 =20 u20 = 10→ u10 =5 u5 = 16 u16 = 8 ug = 4 → Us =2 u2 =1. It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. The below function takes as input an integer n and returns the number of terms generated by the sequence starting at n. function i-Seq (n) u=n; i=%3; while u =1 if statement 1 u=u/2; else statement 2 end i=i+1; end statement 1 and statement 2 should be replaced by: None of the choices statement 1 is "mod(u,2)=D%3D0" and statement 2 is "u = 3*u+1;" statement 1 is "u%2" and statement 2 is "u = 3*u+1;" O statement 1 is "mod(n,2)=30" and statement 2 is "u = 3*n+1;"
The following code implements Newton's algorithm for finding the square root of a number using repetition: n = 17 # we want the square root of 17 for exampleg = 4 # our guess is 4 initiallyerror = 0.0000000001 # we want to stop when we are this closewhile 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. Assume that the function will be called with positive numbers only. Note: Your initial guess cannot be zero!
The following code implements Newton's algorithm for finding the square root of a number using repetition: n = 17 # we want the square root of 17 for example # our guess is 4 initially # we want to stop when we are this close g = 4 error = 0.0000000001 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. Assume that the function will be called with positive numbers only. Note: Your initial guess cannot be zero!
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