What is the family of functions that satisfies

Question

Let $f$ be a continuous function in $\Re$, differentiable (at least one time). Then, which functions can satisfy:

$$ \frac {f(x)} {f'(x)}= - \frac {f(x-\frac {f(x)} {f'(x)})}{f'(x-\frac {f(x)} {f'(x)})}$$

I solved it with a change of variable

$$h(x)=\frac{f(x)}{f'(x)}$$

So I imposed that f has no maximum or minimum in $\Re$ as $h$ is well behaved. If it is imposed that h is odd and bijective, then there is an easy solution. But I couldn´t get anything better. Can you help me?

PD: All this comes from finding a function for which it is impossible to use the Newton-Raphson method for finding roots of a function, giving any first approach of the root.

Answer 1

$\newcommand\fix{\mathrm{fix}}$ Assume that $h$ is defined and continuous on all of $ℝ$ and define $g(x) := x - h(x)$. Then the equation becomes $$x=g(g(x)).$$ So $g$ is a bijection $ℝ→ℝ$, in particular strictly monotonous.

1. If $g$ is monotonously increasing $g(x) > x ⇒ x = g(g(x)) > g(x) > x$ (analogous for $<$), so the only possible solution is $g(x) = x$. That corresponds to $h(x) = x-g(x) = 0$, but there is no appropriate $f$.

2. If $g$ is monotonously decreasing there are lots of solutions. For any $g$ there is exactly one $x_\fix \in ℝ$ such that we have a fixed point $g(x_\fix) = x_\fix$ (the intersection of its graph with the diagonal in $ℝ^2$). Then $g$ maps $[x_\fix, ∞)$ bijectively onto $(-∞, x_\fix]$ and vice-versa.

   On the other hand taking any $x_\fix\in ℝ$ and any continuous, strict monotonously decreasing, unbounded function $\hat{g}$ on $[0, ∞)$ with $\hat{g}(0) = 0$, we get a solution to our problem as $$g(x) := \begin{cases}x_\fix+\hat{g}^{-1}(x)&x≤x_\fix\\x_\fix+\hat{g}(x)&x>x_\fix\end{cases}$$

   Each of those in turn defines an $h: h(x) = x-g(x)$. As $h$ has exactly one zero (at $x_\fix$), you get two families of solutions for $f$ depending on which side you do the integration $$f(x) = \exp{\int_{x_0}^x\frac{dt}{t-g(t)}}$$

   Regularity of $g$ at $x_\fix$ is not important because it would not help the convergence of the integral at that point.

Of course when you relax the conditions on the domain of $h$ and $g$ you will get more solutions. E.g. in 2. you could take functions $\hat{g}$ bounded below. Then $g$ will be defined on $(x_\fix+\inf{\hat{g}}, ∞)$ and of course there are additional solutions for an analogous construction on left-open intervals.