View Full Version : Exponential Functions and Population
05-11-2003, 05:43 PM
As part of my math homework I have to discuss why exponenetial functions are often useful for represnting the size of a population over time. Could someone help me UNDERSTAND why they are useful. As far as I know, they dont have a constant rate of change, just like population. Also, the more people you have, the more they multiply so with exponents, the bigger the number is, the more it will get bigger. Uhm..help?
05-11-2003, 05:49 PM
What math class are you in? The best answer to this involves Calculus, but that's not very helpful if your class is Algebra.
05-11-2003, 05:53 PM
I am in this course called IMP math III. I think it just scratches the surface of calculus if even that.
05-11-2003, 07:24 PM
Exponential functions are useful for modeling population because they obey the recurrence relation
ƒ(t) = c * ƒ(t - 1),
where c is some constant. To give a concrete example, say ƒ(t) = et, where e = 2.71828.... Then ƒ obeys the recurrence relation
ƒ(t) = e * ƒ(t - 1).
Now, if we take t to be time, and take ƒ(t) to be the number of individuals in a population at time t, then the recurrence relation above says that the number of individuals present at time t is always c times the number of individuals that were present one time-unit prior to time t.
And this matches how we'd expect populations to behave, at least under ideal conditions. Why? Because we expect that, on average, each individual has the same number of offspring each time-unit. Call that number n. So say we're at time t - 1, and each individual then has its allotment of n offspring. How many individuals do we have at time t? Well, there are the ƒ(t - 1) individuals we had one time-unit ago, plus the n*ƒ(t - 1) individuals that were born this last time-unit. So there are a total of (n + 1)*ƒ(t - 1) individuals in the population at time t. In equations, we've just shown that
ƒ(t) = (n + 1)*ƒ(t - 1).
If we write c for (n + 1) we have exactly the recurrence relation above.
05-11-2003, 08:22 PM
Here's one reason why exponential functions are necessary: they ensure you end up with a number that makes sense.
Day in, day out, I'm dealing with population changes in the course of my job. Frequently what I'm citing is the change in population between the 1990 and 2000 U.S. Census. For example, let's say "TownVille" had a population of 4,952 in 1990, and 7,413 in 2000. Any fool can calculate that there was a 49.7% increase in population between 1990 and 2000.
Unfortunately, the more common fool then figures "hey, that means there was a 4.97% annual increase between 1990 and 2000!" Nope. If you apply 4.97% to 1990, and then 4.97% to that figure for 1991, and so on, you do not end up with 7,413 in 2000. Of course, if you apply 4.97% to 4,952, and then only apply it to that 1990 count 10 times and add those numbers to 4,952, then you do come to the correct figure, but that's hardly how one would interpret "annual rate of change", now is it?
The correct answer is [(7,413/4,952)^0.1]-1, or 4.12%. This is the "annual growth rate" of TownVille as one would normally interpret it, and fundamentally it's calculated the same way one would calculate a compound interest rate. Nonetheless, I find myself explaining this concept at least weekly to very well-paid bankers that are convinced that I flunked high school algrebra...
So, the short answer is, exponential functions are useful because we tend to think that a population increases, say 4.12% in 1990, and then that larger population, that is 4.12% larger, then increases 4.12% of its new, larger size. This is how we typically think of population growth, and its the same way we traditionally calculate interest rates. At the end of the year, you want 3% interest on the amount you had in your bank account that year, not the amount you first put in 10 years ago, right?
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