Exponentially Varying FunctionsDate: 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/ |
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