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:
X_{n+1 }= ½ ( x_{n} + a/x_{n}) Eq. 1
Where: a – square root value (positive real number); x_{n} – 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 (x_{n}) |
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: x_{n} = x_{0} = initial estimated value, let say 4.5. And a = the value inside the square root, we can get the value of our first iteration (x_{n+1}):
X_{n+1 }= ½ ( x_{n} + a/x_{n})
X_{1} = ½ (4.5 + 20/4.5)
X_{1} = 4.472
Our second iteration would be:
X_{2} = ½ (x_{1} + a/x_{1})
X_{2} = ½(4.472 + 20/4.472)
X_{2} = 4.472
Since x_{1} & x_{2} 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 x_{0} = 25
Iteration (x_{n}) |
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; x_{0} = 25; we get:
X_{1} = ½(x_{0} + a/x_{0}) = ½(25 + 645/25) = 25.40
X_{2} = ½(x_{1} + a/x_{1}) = ½(25.40 + 645/25.40) = 25.39685
X_{3} = ½(x2 + a/x_{2}) = ½(25.39685 + 645/25.39685) = 25.39685
Since, x_{2} equal to x_{3}, hence, [pmath]sqrt{645}[/pmath] = 25.39685