Exponentially Varying Functions

Date: 11/11/2001 at 19:42:27
From: Eric Burgess
Subject: Are there functions that behave this way?

I'm imagining a function that doubles after a certain interval, then 
doubles again in half that time, doubles again in a quarter of the 
original interval, etc.  Like this:

f(0)      =   1
f(8)      =   2
f(12)     =   4
f(14)     =   8
f(15)     =  16
f(15.5)   =  32
f(15.75)  =  64
f(15.875) = 128

I can tell that this function has a vertical asymptote at x = 16, and 
will not be defined for x >= 16.  Are there any elementary functions 
that behave in this way? I mean, is there a type of formula rather 
than an algorithm that will produce these numbers?  

Thanks,
Eric Burgess

Date: 11/11/2001 at 23:24:30
From: Doctor Peterson
Subject: Re: Are there functions that behave this way?

Hi, Eric.

My first impression is that you are putting two exponential functions 
"back to back," in the sense that both the intervals you take on the 
left, and the results on the right, are varying exponentially. Let's 
see what we can do with that idea.

Let's express your x and y values in terms of n, the number in your 
sequence:

     n          x                f(x)
    ---  -------------------  ---------
     0    0     = 16 - 16       1 = 2^0
     1    8     = 16 - 8        2 = 2^1
     2   12     = 16 - 4        4 = 2^2
     3   14     = 16 - 2        8 = 2^3
     4   15     = 16 - 1       16 = 2^4
     5   15.5   = 16 - 0.5     32 = 2^5
     6   15.75  = 16 - 0.25    64 = 2^6
     7   15.875 = 16 - 0.125  128 = 2^7

If you look at this, you realize that

    f(16 - 2^(4-n)) = 2^n

In other words,

    x = 16 - 2^(4-n) = 16 - 16*2^-n = 16(1 - 2^-n)
    y = 2^n

(Do you see what I meant by two exponential functions back to back?)

Eliminating n from these equations, we get

    x = 16(1 - 1/y)

Now you just have to solve this for y:

    y = 1/(1 - x/16) = 16/(16 - x)

As you can see, the exponential functions disappear, and we are left 
with nothing but a reciprocal function. That doesn't surprise me much; 
I could have solved it more directly, but this approach makes it very 
clear that our function does exactly what you want. 

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/