Extracting Square Roots by Babylonian Method

Extracting Square Roots by Babylonian Method

I’ve came across an interesting way of extracting square roots using just pencil and paper. As far as I could remember, I haven’t encountered this method (nor any other method) except thru a calculator in solving square roots.

The Babylonian method also known as Heron’s method involves making an estimate, dividing and averaging, over and over, to find a more accurate output. In this Babylonian Method, we start with an arbitrary positive number (closer to the actual square root), and then apply the following iterative process:

Xn+1 = ½ ( xn + a/xn)     Eq. 1

Where:  a – square root value (positive real number);  xn – arbitrary positive number

Babylonian Method of extracting Square Roots uses an iterative step, as follows:

Step 1: Make a guess (make a close guess to the actual square root, using perfect squares)

Step 2: Divide your original number by your guess

20 / 4.5  =  4.444

Step 3: Find the average

(4.444  +  4.5)  /  2  =  4.472

Step 4: Use the resulting average as your next guess

Example:

[pmath]sqrt{20}[/pmath]

We know that the [pmath]sqrt{16}[/pmath] = 4 and the [pmath ]sqrt{25}[/pmath] = 5, hence, we could initially estimate that [pmath]sqrt{20}[/pmath] is somewhere 4.5 (if your initial estimate is close to the actual extracted square root , the shorter the iteration)

Iteration (xn)

Guess

Divide

Average

0

4.5

20 / 4.5 = 4.444

(4.444 + 4.5) / 2 = 4.472

1

4.472

20 / 4.472 = 4.472

(4.472 + 4.472) / 2 = 4.472

Since the second guess (i.e., 4.472) is the same as the first iteration of the average, we can say that:

[pmath]sqrt{20}[/pmath] = 4.472

Using, Eq. 1, substituting the values:  xn = x0 = initial estimated value, let say 4.5.  And a = the value inside the square root, we can get the value of our first iteration (xn+1):

Xn+1 = ½ ( xn + a/xn)

X1  = ½ (4.5 + 20/4.5)

X1 = 4.472

Our second iteration would be:

X2  =  ½ (x1  +  a/x1)

X2  =  ½(4.472 + 20/4.472)

X2  =  4.472

Since x1 & x2 are the same, hence, [pmath size=18]sqrt{20}[/pmath]  = 4.472

 Example 2:

Let’s try:

[pmath size=24]sqrt{645}[/pmath]

Our initial estimate would be:

[pmath]sqrt{625}[/pmath] = 25 and [pmath]sqrt{676}[/pmath] = 26, we can set x0 = 25

Iteration (xn)

Guess

Divide

Average

0

25

645 / 25 = 25.80

(25.80 + 25) / 2 = 25.40

1

25.40

645 / 25.40 = 25.3937

(25.3937 + 25.40) / 2 = 25.39685

2

25.39685

645 / 25.39685 = 25.39685

(25.39685 + 25.39685) / 2 = 25.39685

Therefore, [pmath]sqrt{645}[/pmath] = 25.39685

Alternatively, using Eq. 1, setting a = 645; x0 = 25; we get:

X1  =  ½(x0 + a/x0)  =  ½(25 + 645/25)  =  25.40

X2  =  ½(x1 + a/x1)  =  ½(25.40 + 645/25.40)  = 25.39685

X3 = ½(x2 + a/x2) = ½(25.39685 + 645/25.39685)  =  25.39685

Since, x2 equal to x3, hence, [pmath]sqrt{645}[/pmath] = 25.39685

Leave a Reply

Your email address will not be published. Required fields are marked *