Approximation of $\frac{1}{a}$ with Newton-Raphson

Question

Approximation of $\frac{1}{a}$ with Newton-Raphson

To start the method we search an equation $F(x)$ of which $\frac{1}{a}$ is a solution.

My book uses the equation: $F(x)=\frac{1}{x}-a=0$

But why can't you use the equations:

• $F(x)=x-\frac{1}{a}=0$
• $F(x)=ax-1=0$

Answer 1

Try them! They update an arbitrary root estimate to $1/a$, presupposing knowledge of the reciprocal.

Answer 2

• For your first proposition to compute $F(x)$ you need to know $\frac{1}{a}$, but the goal of the method is precisely to know this number as the sequence write: $$x_{n+1}=x_n-\frac{x_n-1/a}{1}=x_n-x_n+\frac{1}{a}$$
• For the second one there is actually the same problem as tour sequence write: $$x_{n+1}=x_n-\frac{ax_n-1}{a}=x_n-x_n+\frac{1}{a}$$

(For linear function Newton method is simply to solve the problem in one step).

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As a side note with the function proposed: $$x_{n+1}=x_n-\frac{1/x-a}{-1/x^2}=x_n+x_n-ax_n^2$$ so you only need to use number that you actually know before.