# Math Adventures/Square Roots using Newton’s Method

This activity introduces the student to a simple method for computing square roots as an example of using Newton’s Method for calculating solutions to a general class of equations.

We will begin by finding a number that when multiplied by itself equals 2. This is called the square root of 2, and is written as ${\displaystyle {\sqrt {2}}}$.

The square root of any number (called the operand, or radicand) is a number that when multiplied by itself (e.g. “squared”) equals the operand. Every number has two square roots, one is a positive number and the other is the corresponding negative number.

Geometrically the square root of a number is equal to the length of a square having an area equal to that number.

Square roots are often irrational numbers and cannot be represented exactly by any (finite) number of digits.

## Activity

To compute the square root of 2 (the operand):

1. Begin with a guess (1 will always work) This is the current estimate
2. Divide the operand by current estimate (in this case, 2/1 = 2) to get the balance.
3. Average the current estimate and the balance to get the next estimate. ((1+2)/2 = 1.5)
4. Repeat steps 2 and 3 as often as you need to obtain the required accuracy.

Here is the resulting sequence (for the operand=2):

Computing the Square Root of Two
Guess (Current Estimate) Balance
1 2
1.5 1.333333333
1.416666666667 1.411764705882
1.414215686275 1.414211438475
1.414213562375 1.414213562372
1.414213562373 1.414213562373
1.414213562373 1.414213562373

The difference in the two columns gives an indication of the accuracy of the current estimate to the true square root. Notice how quickly the estimates converge.

The method can be interpreted geometrically. Each row in the above table describes the two sides of a rectangle having an area equal to 2, the operand in this case. For example, the first row describes a rectangle with sides of length 1 and 2, and an area of 2. By averaging the length of the two sides to get the next estimate (appearing in the first column of the next row) the next rectangle more closely approximates a square. The method continues by “squaring up” each subsequent rectangle.

The accepted value of the square root of 2 is approximately: 1.4142135623730950488016887242096980785696718753769480731766797379…

Repeat the steps to compute the square roots of 3, 4, 5, 10, 100, 1,000, and 10,000 or other number you have chosen.

The accepted values are given in this table:

Square Roots
X Square Root (X)
3 1.7320508075688772935274463415058723669428052538103806280558069794…
4 2
5 2.2360679774997896964091736687312762354406183596115257242708972454…
10 3.1622776601683793319988935444327185337195551393252168268575048527…
100 10
1,000 31.622776601683793319988935444327185337195551393252168268575048527…
10,000 100

How can this method be extended to compute cube roots?

Here is the answer, illustrated by computing the cube root of 10.

Begin by extending the geometric interpretation to compute a series of rectangular cuboids, converging toward a cube. The steps are as follows:

To compute the cube root of 10 (the operand):

1. Begin with a guess (1 will always work). This is the current estimate.
2. Divide the operand by the square of the current estimate (in this case 10 / (1*1) = 10) to get the balance.
3. Average the current estimate and the balance to get the next estimate. ((1+10)/2 = 5.5)
4. Repeat steps 2 and 3 as often as you need to obtain the required accuracy.

Here is the resulting sequence (for the operand=10):

Computing the cube root of 10
uess (Current Estimate) Balance
1 10
5.5 0.330578512
2.915289256 1.176621272
2.045955264 2.388953702
2.217454483 2.033717249
2.125585866 2.213312276
2.169449071 2.124716945
2.147083008 2.169213658
2.158148333 2.147026564
2.152587449 2.15813393
2.155360689 2.152583885

This converges slowly toward the accepted value for the cube root of 10= 2.1544346900318837217592935665193504952593449421921085824892355063…

Although this “geometric approach” does work for cube roots, a stricter use of newton’s method converges much more quickly. The formula for each iteration is given by:[1]

${\displaystyle x_{(n+1)}=x_{n}-((x_{n})^{3}-10)/(3*(x_{n})^{2})}$

Where 10 is the operand and 3 is a constant.

The resulting sequence converges quickly and is shown here:

Estimating the cube root of 10
Successive Estimates
1
4
2.875
2.319943
2.165962
2.154496
2.154435
2.154435

## Notes:

1. Socratic Q&A, Calculus, How do you use Newton's Method to approximate the value of cube root?