6
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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, 2014 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, 2014 at 6:52
  • $\begingroup$ @Aaron, thanks, I have corrected the question. $\endgroup$
    – Mark Lewko
    Jan 3, 2014 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, 2014 at 8:27

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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, 2014 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, 2014 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, 2014 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.