Derive the Newton's method for finding 1/N, N > 0. Hence find 1/17 using initial approximation as 0.05.

Steps to Solve

1. Understanding the Problem and Deriving the Method

Newton's method is an iterative root-finding algorithm. To find the reciprocal of a number $N$, consider the equation:

$$ f(x) = \frac{1}{x} - N = 0 $$

We want to find $x = \frac{1}{N}$.

2. Finding the Derivative

Next, we need the derivative of $f(x)$:

$$ f'(x) = -\frac{1}{x^2} $$

3. Applying Newton's Method Formula

Newton's method is given by the formula:

$$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$

Substituting $f(x)$ and $f'(x)$ into the formula:

$$ x_{n+1} = x_n - \frac{\frac{1}{x_n} - N}{-\frac{1}{x_n^2}} $$

4. Simplifying the Expression

Simplifying the expression gives us:

$$ x_{n+1} = x_n + x_n^2 \left( \frac{1}{x_n} - N \right) $$

This simplifies further to:

$$ x_{n+1} = x_n + (1 - Nx_n) $$

So we have:

$$ x_{n+1} = x_n(2 - Nx_n) $$

5. Finding $ \frac{1}{17} $ Using Initial Approximation

Start with an initial approximation $x_0 = 0.05$ and $N = 17$. We now compute:

• First Iteration:

$$ x_1 = 0.05(2 - 17 \times 0.05) = 0.05(2 - 0.85) = 0.05 \times 1.15 = 0.0575 $$

• Second Iteration:

$$ x_2 = 0.0575(2 - 17 \times 0.0575) = 0.0575(2 - 0.9775) = 0.0575 \times 1.0225 \approx 0.0588 $$

• Third Iteration:

$$ x_3 = 0.0588(2 - 17 \times 0.0588) = 0.0588(2 - 0.9996) = 0.0588 \times 1.0004 \approx 0.0588 $$

Continuing this iteration until the result stabilizes.