Let $f_0,f_1,\ldots$ be a sequence of functions $f_n : [0,1] \rightarrow R$ defined as follows:

$$f_0(x) =1+2x$$

$$f_{n}(x) := \left\{\frac{5+t}{2} : \text{ where t solves } f_{n-1}\left(\frac{x}{t}\right) = \frac{5+t}{2}\right \}$$

In other words, $f_n(x)$ is equal to the intercept (with respect to $y$) of the functions $f_{n-1}(x/t)$ and $\frac{5+t}{2}$.

It isn't hard to prove by induction that each $f_n$ (for $n>0$) will be a continuous increasing function such that $f_n(0)=5/2$ and $f_n(1)=3$. If we fix $0<x<1$ and consider the function $f_{n-1}(x/t)$ as a function of $t$, we see that it is decreasing as $t$ increases and $f_{n-1}(x/t)=3$ for $t=x$, and $f_{n-1}(x/t) <3$ for $y=1$. From this it follows that there is a unique intercept and that $f_n$ is well defined.

One can show that: $$f_1(x)= \frac{1}{4}(\sqrt{16x+9}+7),$$ $$f_2(x) = \{\frac{5+t}{2} : \text{ where t solves } x = (t^3+3t^2)/4 \}.$$

Is there a method for finding a good approximation (upper and lower bounds) by elementary functions of the $n$-th iterate $f_n(x)$ in a neighborhood of $x=1$?

I'm interested in both a solution to this particular problem, as well as understanding methods that work in similar situations.

  • $\begingroup$ I edited the LaTeX so that it matches your description. Previously it said $f_{n-1}(x/y=(3+y)/2)$. Hopefully I didn't add any errors... $\endgroup$ Jan 3 '14 at 5:59
  • $\begingroup$ It might be clearer to replace $y$ by $t$. Isn't $f_1(1)=\frac{1+\sqrt{17}}{4}\approx 2.28$? $\endgroup$ Jan 3 '14 at 6:52
  • $\begingroup$ @Aaron, thanks, I have corrected the question. $\endgroup$
    – Mark Lewko
    Jan 3 '14 at 7:07
  • 1
    $\begingroup$ so $$f_2(x)=2+\frac{\sqrt [3]{-1+2\,x+2\,\sqrt {-x+{x}^{2}}}}{2}+\frac {1}{2\sqrt [3]{-1+2\,x+2\,\sqrt {-x+{x}^{2}}}}$$ but $f_3$ is not going to be pretty. $\endgroup$ Jan 3 '14 at 8:27

From your implicit iteration, $f_n(x)=f_{n-1}\big(\frac{x}{2f_n(x)-5}\big)$, it follows that $f_n$ is the inverse function to the function $g_n:[5/2,3]\to[0,1]$ defined by $g_n(y):=\frac{1}{2}(y-1)(2y-5)^n$.

If we put $z=z_n(x):=(2f_n(x)-5)/3$ the above relation writes $z(1+z)^{1/n}=4^{\frac{1}{n}} 3^{-\frac{n+1}{n}} x^{\frac{1}{n} }$.

The local inverse at $0$ of $z\mapsto z(1+z)^{1/n}$ is an analytic function whose expansion at $z=0$ is
$$F_n(z):=\sum_{k=1}^\infty\frac{1}{k}{-\frac{k}{n}\choose k-1}z^k\; ,$$

so we have $$f_n(x)=\frac{5}{2}+\frac{3}{2}F_n\big( 4^{\frac{1}{n}} 3^{-\frac{n+1}{n}} x^{\frac{1}{n} }\big) \; .$$

It should be easy (I did not try) to compute the radius of convergence of the above power series expansion, to check whether it also cover $x=1$. For $n=1$ the radius of convergence is $1/4$, so it does not cover $F_n(4/9)$, needed for $f_n(1)$, but for larger $n$ it could be better.

  • $\begingroup$ It would be even better to get an expansion of $f_n$ at $x=1$ by the Lagrange Inversion Formula, but that leads to a residue which is not easy at all to compute, as it is for the above expansion at $x=0$. $\endgroup$ Jan 3 '14 at 22:34
  • $\begingroup$ Thanks, this is very helpful! Although, for my application, I'm going to need an expansion at the point $x=1$. $\endgroup$
    – Mark Lewko
    Jan 3 '14 at 22:42
  • $\begingroup$ For an expansion at $x=1$, you can also write $f_n(x)=3+\sum_{k=1}c_{n,k}(x-1)^k$ and find the coefficients recursively, by equating the series $$\frac{1}{2}(f_n-1)(2f_n-5)^n-1=x-1$$. If you only need few terms that should be fine. $\endgroup$ Jan 3 '14 at 22:57

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