Newton's numerical method to estimate the square root of N uses the iterative relation
[tex]x_{n+1} = \frac{1}{2} (x_{n} + \frac{N}{x_{n}}), \, for \, n=0,1,2, \, ..., [/tex]
x₀ = initial guess
We want to estimate the square root of 15, so N = 15.
We know that √{16) = 4, therefore let x₀ = 4 for the initial guess.
Calculate relative error after n iterations as
[tex]e_{n} = |1 - \frac{x_{n}}{x_{n+1}} |[/tex]
When the relative error is less than 0.01, we shall accept the approxmate answer.
Iteration #1:
x₁ = 0.5(x₀ + N/x₀) = 0.5(4 + 15/4) = 3.875
e₁ = |1 - 4/3.875| = 0.032
Iteration #2:
x₂ = 0.5(3.875 + 15/3.875) = 3.873
e₂ = |1 - 3.875/3.873| = 5.2 x 10⁻⁴
This is good enough.
Therefore the approximate value for √(15) is 3.873.
On a number line, the progression to find the square root is shown below.
Answer: 3.873
