Problem

Preamble

Newton’s Method

Let \(f\) be a differentiable function on an interval \(I\) with a root in \(I\). To approximate the value of the root, accurate to \(d\) decimal places:

1. Choose a value \(x_0\) as an initial approximation of the root. (This is often done by looking at a graph of \(f\).)

2. Create successive approximations iteractevely; given an approximation \(x_n\) , compute the next approximation \(x_{n+1}\) as

\[x_{n+1}= x_n - \frac{f(x_n)}{f^l(x_n)}\]

3. Stop the iterations when successive approximations do not differ in the first \(d\) places after the decimal point.

Solution

\(f(x) = x^2 − 2\)

curve(x^2-2, -4, 4, col = "red", ylab = 'f(x) = x^2 − 2')
abline(v = 0, col = 'black')
abline(v = sqrt(2), col = 'blue')
abline(v = -sqrt(2), col = 'blue')
abline(h = 0, col = 'black')
abline(h = -2, col = 'green')

\(f^l(x) = 2x\) <- first derivative.

In order to solve this, I will create a small function that iterates and find the solutions.

f <- function(x){return(x^2 -2)}
f1 <- function(x){return(2*x)}
res <- list('iteration' = 0, 'result' = 1.5)
for (i in 2:6){
  res$iteration[i] <- i-1
  res$result[i] <- res$result[i-1] - f(res$result[i-1])/f1(res$result[i-1])
}
res <- data.frame(res)

From the above procedure we obtain the following table:

And as we know the solution for the above will be \(x = \sqrt{2}\).

res$true_ans <- sqrt(2)
res$diff <- round(res$result - res$true_ans,6)

END