From: edgar@math.ohio-state.edu (G. A. Edgar)
Newsgroups: sci.math.symbolic
Subject: Re: The correct definition of a strange function ...
Date: Mon, 27 Jul 1998 09:35:21 -0400

In article <6peunq$dac$1@news00.btx.dtag.de>, tobi.kamke@t-online.de
(Tobias Kamke) wrote:

> The function f is defined as follows:
> 
>   f(x) = p * f(2x)                   0 < x <= 0.5
>   f(x) = p + (1 - p) * f(2x - 1)     0.5 < x < 1
>   f(1) = 1
>   f(0) = 0
> 
>   0 <= p <= 1
> 
> How can I calculate f((1/3)), f((2/5)), f((1/1984)), ... ?
> 
> What can you say about this function? Is it monotone? What's the
> derivation of f(x)?
> 
> 
> Thanks,
> 
>   Tobi

There is a unique continuous
monotone function f with these properties.  It is "singular" in the sense
that its derivative is zero almost everywhere.

Reference: Billingsley, PROBABILITY AND MEASURE (first edition) Example 31.1.

This example goes back to Cesàro (1906).

For f(1/3) you get an equation:  f(1/3) = p f(2/3) = p (p + (1-p)f(1/3))
and thus f(1/3) = p^2/(1-p+p^2) .

Similarly for any rational: the base 2 expansion is eventually periodic,
so you get an equation to solve.  The process may be computerized.
If I have not made a mistake, f(2/5) = p^2(2-p)/(1-p^2+2p^3-p^4) ;
f(1/1984) = p^11/(1-p^4+p^5) .
-- 
Gerald A. Edgar                   edgar@math.ohio-state.edu