Given the iterative definition
$$x_{n+1} = 2x_n - x_n^2y$$
Identify the origin function from the Newton's iteration.
My thoughts:
From Newton's method, $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$, so it must be that $\frac{f(x_n)}{f'(x_n)} = x_n^2y - x_n$, but this would require implicit differentiation, making it more complicated than necessary.
Any thoughts?
First, we note that since $$x_{n+1} =x_n - \frac{f(x_n)}{f'(x_n)} = 2x_2 - x_n^2y,$$ we have that $$f'(x_n) = \frac{1}{x_n^2y-x_n}f(x_n).$$ Next, we note that if $f(x)$ has the form of an exponential function, i.e., $f(x) = \exp(u(x))$, we have $$f'(x) = u'(x)\exp(u(x)) = u'(x)f(x).$$ Hence, we want some function $u(x)$, such that $$u'(x) = \frac{1}{x^2y-x}.$$ Integrating both sides, we get $$u(x) = \log\left(\left|\frac{1}{x} - y\right|\right)+C.$$ Hence, our original function is $$f(x) = \exp\left(\log\left(\left|\frac{1}{x} - y\right|\right)+C\right) = e^c\left|\frac{1}{x} - y\right|.$$
We confirm this by considering $$x -\frac{f(x)}{f'(x)} = x - \frac{e^c\left|\frac{1}{x} - y\right|}{\frac{e^c(yx-1)}{\left|1/x-y\right|x^3}} = 2x-x^2y$$
EDIT
As Dr. Israel has pointed out in the comments, any function of the form $$f(x) = a\left(\frac{1}{x}-y\right), \qquad a \ne 0$$ will work.