I need to compute $\frac{1}{x+\Delta x}$ avoiding divisions, is this possible through an iterative method starting from $\frac{1}{x}$? I thought of two possibilities that could be different from the classical computation of the reciprocal using the Newton-Raphson algorithm:
1) find $\frac{x}{x+\Delta x}$ such that $\frac{1}{x}\cdot\frac{x}{x+\Delta x}=\frac{1}{x+\Delta x}$
2) find $\frac{\Delta x}{x(x+\Delta x)}$ such that $\frac{1}{x}-\frac{1}{x}+\frac{1}{x+\Delta x}=\frac{1}{x}-\frac{\Delta x}{x(x+\Delta x)}=\frac{1}{x+\Delta x}$
Is one of these viable or am I stuck with the classical way of finding the reciprocal? Can you help me find if there are any equations whose roots are the values I'm looking for?
One slick method (assuming $|\Delta x| < x$) is to use the geometric series. We have $$ \frac 1{x + \Delta x} = \frac 1x \cdot \frac{1}{1 + \frac{\Delta x}{x}} = \frac 1x \sum_{n=0}^\infty (-1)^n\left(\frac{\Delta x}{x}\right)^n $$ for an approximation, simply take the sum out to $N$ steps rather than infinitely many.
This approach gives you a linear convergence rate, which is very much outdone by Newton-Raphson.