Deducing the original function from Newton's method

Question

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?

So you have a separable differential equation for $f$, try and solve it — Joshhh, May 08 '17 at 16:39
Note that the Newton iteration for $x \mapsto f(x)$ and $x \mapsto a f(x)$ ($ a \neq 0$) are the the same, so you can, at best, determine the function up to a multiplicative constant. — copper.hat, May 08 '17 at 16:50

David · Accepted Answer · 2017-05-08T18:10:41.847

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.

Actually you don't need the absolute values: $f(x) = a \left(\frac{1}{x} - y \right)$ would work for any constant $a \ne 0$. This has the advantage that $f'(1/y)$ is well-defined, whereas in your version it is not. — Robert Israel, May 08 '17 at 17:57

Deducing the original function from Newton's method

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