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  2. Newton's Method (simple code) This is the method suggested in Think Python, 2nd edition, page 67, and doesn't need any library. Newton's method takes a number a and returns its square root as follows: y = (x + a/x) / 2. where x is an arbitrary estimation and y is a better estimation of a. def sqr_root(a):

  3. Is there a short-hand for nth root of x in Python?

    stackoverflow.com/questions/19255120

    Any nth root is an exponentiation by 1/n, so to get the square root of 9, you use 9**(1/2) (or 9**0.5) to get the cube root, you use 9 ** (1/3) (which we can't write with a simpler fraction), and to get the nth root, 9 ** (1/n). Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary.

  4. This is simply a question of how the underlying code actually works. What is the theory of how Python code works? I sent Guido van Rossum an email cause I really wanted to know the differences in these methods. My email: There are at least 3 ways to do a square root in Python: math.sqrt, the '**' operator and pow(x,.5).

  5. math - Integer square root in python - Stack Overflow

    stackoverflow.com/questions/15390807

    Square Roots in Binary. Longhand Square Roots. And here's an implementation in Python: def exact_sqrt(x): """Calculate the square root of an arbitrarily large integer. The result of exact_sqrt(x) is a tuple (a, r) such that a**2 + r = x, where. a is the largest integer such that a**2 <= x, and r is the "remainder". If.

  6. How to perform square root without using math module?

    stackoverflow.com/questions/3047012

    the below code is to find the square root of a number without using built in methods using python.the code is very simple to understand because i wrote the code using mathematics simple solution. x=float(input()) min1=0. max1=x. for i in range(10):

  7. Square root of complex numbers in python. Ask Question Asked 8 years, 6 months ago. Modified 8 years, 6 ...

  8. Writing your own square root function - Stack Overflow

    stackoverflow.com/questions/1623375

    This square root is the first digit of your final answer. Lets denote the digits we have already found of our final square root as B. So at the moment B = 2. 2) Next compute the difference between {5} and B^2: 5 - 4 = 1. 3) For all subsequent 2 digit groups do the following: Multiply the remainder by 100, then add it to the second group: 100 ...

  9. This can be solved using the decimal module to get arbitrary precision square roots and easy checks for "exactness": import math. from decimal import localcontext, Context, Inexact. def is_perfect_square(x): # If you want to allow negative squares, then set x = abs(x) instead. if x < 0: return False.

  10. Python handles ridiculously large integer numbers without problems. For floating point numbers, it uses standard (C) conventions, and limits itself to 64 bit precision. You almost always get floating point numbers when taking a square root of something. 1e309 means 10**309, and 3.1415e209 means 3.1415 * 10**209. This is a standard programming ...

  11. Python- Square Root of list - Stack Overflow

    stackoverflow.com/questions/71581215

    Taking the square root of each number in a list. For this problem in the sqrt_list function: Take the square root of each item in the list, Store each squared item in another list, and. Return this list. alist = [11,22,33] def sqrt_list(alist): ret = [] for i in alist: