From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: solve f(x) + a = f( x + a*sqrt(x) )
Date: 4 May 1995 16:44:51 GMT

In article <3o80k8$f9@larry.rice.edu>,
Owen Ernest Kelly <oekelly@rice.edu> wrote:
>Is there a function satisfying
>
>f(x) + a = f( x + a*sqrt(x) )
>
>for positive real values of x ?

The answer is yes. But the real question is, where on earth 
did this problem come from?

Let  g(x) = x+a, h(x) = x+a*sqrt(x). Then you are asking if there's a
function  f  with  g o f = f o h. 

Let's assume first that  a>0.
Note that  h'(x) > 1, so  h  is invertible. For all integers  n  let
a_n = h^n( 1 ). These  a_n  form an ascending sequence of real numbers
partitioning  (0, oo)  into an infinite union of intervals; h  simply
stretches each interval into the next one to the right.

Likewise  g  is invertible  ( on the set of  _all_ reals); we create a
similar sequence  b_n = g^n( 0 ).

Now just let  f  be, say, a linear map carrying the interval
(a_0, a_1) to the interval  (b_0, b_1)  and define  f  on the rest of
(0, oo)  by  f( h^n( x ) ) = g^n( f(x) )  for  x  in this first interval.

This  f  is continuous, although you didn't require it. I suppose we may
even make  f  smooth by adjusting our choice of  f  near the endpoints of the
initial interval.

Now, if  a<0,  h is no longer  one-to-one: it's two-to-one on  [0, a^2].
Here what we can do is define  f  arbitrarily on  [0, a^2/4]; extend
to the rest of [0, a^2] using  f(x) = f([-a-sqrt(x)]^2); then for any
other  x, choose  n  so that  h^n(x) lies in [0, a^2]  (there is a 
unique such  n)  and there define  f(x) = f( h^n(x) ) - n a. 

Incidentally, it seems to me difficult to write a "formula" for  f; 
you would need to be able to write a succint description of  h^n (1),
for example, which seems unlikely.

dave