Newton's method to find an update rule to compute $\frac{1}{y}$ given $f(x)=\frac{1}{x}-y$ I have to use Newton's method to derive an update rule for finding a root of the form $\frac{1}{y}$ given a specific $f(x)$, where $f(x)=\frac{1}{x}-y$.
From the given, $\frac{1}{y}$ is a valid root of $f(x)$, since $f(x)=\frac{1}{\frac{1}{y}} -y = 0$.
But when I use Newton's method:
$$x_{i+1} = {x_i} - \frac{f(x)}{f'(x)}$$
I get the following with $f'(x)=-\frac{1}{x^2}$:
$$x_{i+1} = x_{i} - \frac{\frac{1}{x_{i}}-y}{-\frac{1}{x_{i}^2}} = 2x_{i}-x_{i}^2y$$
To eliminate the $y$, I can use the best current estimate I have of it which is $y=\frac{1}{x_i}$. But then I get the following equation after some algebra:
$$x_{i+1}=x_{i}$$
This equation only works if the initial value of $x_i$ is the root of $f(x)$ itself or that $x_0=\frac{1}{y}$.
Was there something that I missed here? Or is it that the form of $f(x)$ is just not 'friendly' for a Newton's method.. I know that if I use another form $f(x)'= 1-yx$, then I get an update rule that converges and despite various initial values.
 A: You don't have to eliminate $y$ it's just a constant. It has to be used in order to get the result. The problem is probably that you think of $y$ as a variable but it's not. So instead take a different name for it, say $b$.
For instance, consider that you need to find the value of $\frac{1}{2}$ hence $b=2$, so the equation becomes:
$$x_{i+1} = x_{i} - \frac{\frac{1}{x_{i}}-b}{-\frac{1}{x_{i}^2}} = x_{i}(2-bx_{i})$$
$$\implies x_{i+1}= x_{i}\left(2-(2)x_{i}\right)$$
Now you can just iterate over using this equation for increasing values of $i$ with an already given $x_0$
Edit: You need to be careful with your guess when choosing the initial value because the method may also lead to divergence from the root.
Look at the graph of $\displaystyle f(x)=\frac{1}{x}-3$ below :

Newton-Raphson works by using the tangent from the initial point $x_0$ and then uses the root $x_1$ where the tangent crosses the $x$-axis and this process continues. If you look closely, at $x_0=1$ the tangent intersects the negative $x$-axis and then draws a perpendicular to the function below the negative $x$-axis and with every iteration, it keeps on moving further negative and hence, it keeps on diverging from the root. Take a look at the below gif. 

Source: Wikipedia
