The square root of 2…

·Biagio

The square root of 2 (approximately 1.4142) is a positive real number that, when multiplied by itself, equals the number 2. 

There are a number of algorithms for approximating √2 as a ratio of integers or as a decimal. The most common algorithm for this, which is used as a basis in many computers and calculators, is the Babylonian method for computing square roots, which is one of many methods of computing square roots. It goes as follows:

First, pick a guess, a0 > 0; the value of the guess affects only how many iterations are required to reach an approximation of a certain accuracy. Then, using that guess, iterate through the following recursive computation:

a_{n+1}=\frac{a_{n}+\frac{2}{a_{n}}}{2}=\frac{a_{n}}{2}+\frac{1}{a_{n}}

In general, suppose we want to evaluate the square root of a number n. Let’s assume the n=a^2+b, where a^2 is the biggest square number that is less or equal to n.

Then, we can write it as:

n-a^2=b

or:

\left( {\sqrt n + a} \right)\left( {\sqrt n - a} \right) = b \Leftrightarrow \sqrt n - a = \frac{b}{{\sqrt n + a}}

and then:

\sqrt n = a + \frac{b}{{\sqrt n + a}}

Recursively replacing the \sqrt{n}:

\sqrt n = a + \frac{b}{{2a + \frac{b}{{\sqrt n + a}}}}

or:

For n=2 we have a=1 and b=2-1=1. So:

In general, we will write the sequence:

{a_{n + 1}} = 1 + \frac{1}{{{a_n} + 1}}

In Python:

import numpy as np
import decimal

def twoSqrt(EPS):
    decimal.getcontext().prec = 10000
    
    x = decimal.Decimal(1)
    t = x
    # while error is still big
    while abs(x*x - decimal.Decimal(2)) > EPS:
        x = decimal.Decimal(1) + decimal.Decimal(1)/(decimal.Decimal(1)+t)
        # converge the answer
        t = x
    return x

EPS = decimal.Decimal(1e-200)

r1 = twoSqrt(EPS) 
r2 = decimal.Decimal(2).sqrt()

print('%.19f' % r1)
def twoSqrt(EPS):
    decimal.getcontext().prec = 10000
    
    myList = []
    myCount = []
    
    x = decimal.Decimal(1)
    t = x
    i = 0
    myList.append(x)
    myCount.append(i)
    # while error is still big
    while abs(x*x - decimal.Decimal(2)) > EPS:
        x = decimal.Decimal(1) + decimal.Decimal(1)/(decimal.Decimal(1)+t)
        # converge the answer
        myList.append(x)
        i = i+1
        t = x
        myCount.append(i)
    plt.xlabel("iterations")
    plt.ylabel("$\sqrt{2}$ values")    
    plt.plot(myCount,myList, marker = "x")
    plt.savefig("sqrt2.jpg")
    plt.show()
    return x

With this code change, the following diagram can be obtained:

Lascia un commento