[python] How do I calculate square root in Python?

Why does Python give the "wrong" answer?

x = 16

sqrt = x**(.5)  #returns 4
sqrt = x**(1/2) #returns 1

Yes, I know import math and use sqrt. But I'm looking for an answer to the above.

If you want to do it the way the calculator actually does it, use the Babylonian technique. It is explained here and here.

Suppose you want to calculate the square root of 2:

a=2

a1 = (a/2)+1
b1 = a/a1
aminus1 = a1
bminus1 = b1


while (aminus1-bminus1 > 0):
    an = 0.5 * (aminus1 + bminus1)
    bn = a / an
    aminus1 = an
    bminus1 = bn
    print(an,bn,an-bn)

You can use NumPy to calculate square roots of arrays:

 import numpy as np
 np.sqrt([1, 4, 9])

I hope the below mentioned code will answer your question.

def root(x,a):
    y = 1 / a
    y = float(y)
    print y
    z = x ** y
    print z

base = input("Please input the base value:")
power = float(input("Please input the root value:"))


root(base,power) 

/ performs an integer division in Python 2:

>>> 1/2
0

If one of the numbers is a float, it works as expected:

>>> 1.0/2
0.5
>>> 16**(1.0/2)
4.0

What you're seeing is integer division. To get floating point division by default,

from __future__ import division

Or, you could convert 1 or 2 of 1/2 into a floating point value.

sqrt = x**(1.0/2)

This might be a little late to answer but most simple and accurate way to compute square root is newton's method.

You have a number which you want to compute its square root (num) and you have a guess of its square root (estimate). Estimate can be any number bigger than 0, but a number that makes sense shortens the recursive call depth significantly.

new_estimate = (estimate + num / estimate) / 2

This line computes a more accurate estimate with those 2 parameters. You can pass new_estimate value to the function and compute another new_estimate which is more accurate than the previous one or you can make a recursive function definition like this.

def newtons_method(num, estimate):
    # Computing a new_estimate
    new_estimate = (estimate + num / estimate) / 2
    print(new_estimate)
    # Base Case: Comparing our estimate with built-in functions value
    if new_estimate == math.sqrt(num):
        return True
    else:
        return newtons_method(num, new_estimate)

For example we need to find 30's square root. We know that the result is between 5 and 6.

newtons_method(30,5)

number is 30 and estimate is 5. The result from each recursive calls are:

5.5
5.477272727272727
5.4772255752546215
5.477225575051661

The last result is the most accurate computation of the square root of number. It is the same value as the built-in function math.sqrt().

Perhaps a simple way to remember: add a dot after the numerator (or denominator)

16 ** (1. / 2)   # 4
289 ** (1. / 2)  # 17
27 ** (1. / 3)   # 3

You have to write: sqrt = x**(1/2.0), otherwise an integer division is performed and the expression 1/2 returns 0.

This behavior is "normal" in Python 2.x, whereas in Python 3.x 1/2 evaluates to 0.5. If you want your Python 2.x code to behave like 3.x w.r.t. division write from __future__ import division - then 1/2 will evaluate to 0.5 and for backwards compatibility, 1//2 will evaluate to 0.

And for the record, the preferred way to calculate a square root is this:

import math
math.sqrt(x)

import math
math.sqrt( x )

It is a trivial addition to the answer chain. However since the Subject is very common google hit, this deserves to be added, I believe.