Good closed form approximation for iterates of $x^2+(1-x^2)x$

Question

Let $f(x) := x^2+(1-x^2)x$. Is there a nice nontrivial closed form approximation $g_n(x)$ over $[0,1]$ for the $n$-fold composition $f^{\circ n}(x)$? Obviously near $0$ we have that $f^{\circ n}(x) = x+nx^2+...$ but this is not much use to me. Rather than try to pin down what "nice" ought to mean, I'll channel Potter Stewart and just say I (and I'm sure also a respondent) would know it upon sight.

One might be tempted to mumble "solve Schroder's equation" but I don't see how that helps. Nor do I see how computing the Carleman matrix of $f$ helps (but for what it's worth, I believe the matrix elements are $M_{jk} := \sum_{r=0}^j \binom{j}{r} (-1)^{j-r} \binom{r}{k-3j-2r}$). Such tactics are suggested in How would I go about finding a closed form solution for $g(x,n) = f(f(f(...(x))))$, $n$ times?

Answer 1

We have that $f(x) - x = x^2 - x^3$, so if we pretend that $n$ is a continuous variable, and treat the iterates $x(n)$ as functions of this continuous variable, then we are led to the differential equation:

$$\frac{dx}{dn} = x^2 - x^3$$

Which has the solution:

$$\log\left[\frac{x(n)}{1-x(n)}\right]-\frac{1}{x(n)} = \log\left[\frac{x(0)}{1-x(0)}\right]-\frac{1}{x(0)} + n$$

You can then approximately solve for $x(n)$, what is then clear is that for $x(n)$ near zero the $\frac{1}{x(n)}$ term dominates while for $x(n)$ near $1 $ the logarithmic term dominates.

Answer 2

In blue: $g(x,25)$. In orange: $0.5\tanh(x(1+x)^{25})+0.5$. Your function does stay at $y=1$ thereafter.

The behaviour near zero is entirely influenced by $u(x)=x(1+x)$, neglecting the $-x^3$ term. The obvious approximate composition that I mentioned above, $u^n(x)\approx x(1+x)^n$ simply doesn't increase fast enough. Perhaps your Carlemann matrix aproach could be used on this simpler function to yield a more meaningful approximation near zero?

The $\tanh$ motivation should also be obvious. I hope this helps.

Answer 3

FYI, here is a picture illustrating Count's answer at https://math.stackexchange.com/a/1872168/241. Blue curves are the zeroth through tenth iterates of $f$; red curves are obtained by numerically inverting $F(f^{\circ N}(x)) \approx N + F(x)$ for $F(x) := \log \frac{x}{1-x} + \frac{1}{x}$. I will mention that this is an improvement on the exact fourth-order approximation $f^{\circ N}(x) \approx x+Nx^2+(N^2-2N)x^3+([N-1]^3-2[N-1]^2-3[N-1])x^4$ near $0$, and I imagine an improvement on all exact fixed-order approximations.

On a separate note, I will also mention the Carleman approach I originally mentioned can be pursued along the lines of http://www.sciencedirect.com/science/article/pii/S0022247X98959868.

Answer 4

You have worked out that $f^{\circ n}(x)\approx a(x)=x+nx^2$ near $x=0$. You can do something similar near $x=1$ and then "glue" the two ends together.

$$\begin{align} f(x)&=x^2+(1-x^2)x=x+x^2-x^3=1-2(1-x)^2+(1-x)^3\\ p(x)&:=1-f(1-x)=2x^2-x^3 \end{align}$$

Approximating $f^{\circ n}(x)$ near $x=1$ is essentially the same as approximating $p^{\circ n}(x)$ near $x=0$. $p^{\circ n}(x)\approx 2^{2^n-1}x^{2^n}$, and near $x=1$, you have $$f^{\circ n}(x)\approx b(x)=1-2^{2^n-1}(1-x)^{2^n}\qquad{x\text{ near }1}$$

If you could find some way to glue these, you'd have an approximation. Shooting from the hip, if you could follow each curve ($a$ forward from $0$, $b$ backward from $1$) until they no longer worked well as approximations, then connect the curves with a cubic spline, that's one idea. [It would require figuring out where to make the transitions though].

But there may be better ways to glue them.