How do I calculate square root in Python

Question

Why does Python give the  quot wrong quot  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

User · Answer

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)

User · Answer

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

User · Answer

You can use NumPy to calculate square roots of arrays    import numpy as np  np sqrt  1  4  9

User · Answer

performs an integer division in Python 2    gt  gt  gt  1 2 0   If one of the numbers is a float  it works as expected    gt  gt  gt  1 0 2 0 5  gt  gt  gt  16   1 0 2  4 0

User · Answer

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

User · Answer

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

User · Answer

sqrt x   1 2  is doing integer division  1 2    0   So you re computing x 1 2  in the first instance  x 0  in the second   So it s not wrong  it s the right answer to a different question

User · Answer

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

User · Answer

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

User · Answer

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

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

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