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They told me $P(1 + r)^n$ can be used to calculate money interest, example:

You invert $15,000.00, 20% interest annual, for 3 years:

$15,000(1 + 0.20)^3$ = 25,920

And that $P(2.71828)^{rn}$ can be used to calculate population growth, example:

We have 500 bacterias, growing at the rate of 25% monthly, for 16 months:

$(500)(2.71828)^{(0.25)(16)}$ = roughly 26,968

I can't tell difference on when to use one over another. I mean different than seeing the keywords money and population. How do I tell when to use one over the another?

As far as I can tell, they both are something like:

  • We have a starting value, a percent, a period of time
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Unless I am missing something major, the formula for bacteria should be $500 \cdot 1.25^{16} \approx 17,764$. –  copper.hat Jun 8 '12 at 22:41
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The first uses the plain meaning of compound interest and $n$ does not need to be an integer, while the second uses the confusing so-called "continuous compounding" –  Henry Jun 8 '12 at 22:43
    
as an aside, that number 2.71828... is an important number and it's known as $e$. 2.71828 is just an approximation—it has a decimal expansion that goes on forever without repeating, a little bit like $\pi$. So usually that formula would be written as $Pe^{rn}$. –  crf Jun 8 '12 at 23:59

2 Answers 2

up vote 3 down vote accepted

The difference is whether you try to model each discrete step or whether you choose a continuous model.

If you regard money, then you know that your interest will be calculated annually, so the discrete model will be exact.

If you have bacteria or population, you do not have any control about the number of babies or bacteria born at a particular time. You just know that you have enough people or bacteria, so that a statistical approach to the growth rate applies.

This means that you say that there are so many bacteria or people that you model it in a continuous way.

But you have to take care. In the continuous model, $r=0.25$ in your formula does not mean that you have a quarter more bacteria at the end of the month. This is a different $r$ from the one in the discrete model.

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which makes me wonder why it says "25% growth monthly" when it means something else. You could use $500\times(1.25)^n$ for real $n$ as a continuous model. –  Henry Jun 8 '12 at 22:47
    
@Henry The point is that the $r$ in question is $\log 1.25$ which amounts to the same thing as your formula, of course. –  Phira Jun 8 '12 at 22:50
    
Yes - though $\log_e (1.25) \approx 0.223$ not $0.25$ –  Henry Jun 8 '12 at 22:52
    
@Henry maybe because I translated that part wrong, I should have included the word "rate" –  ajax333221 Jun 8 '12 at 23:18

If something grows at a rate of 25% per month, then the amount after 16 months will be $ P\cdot1.25^{16}$. This is true of money, of population, or of anything else. The phrase "grows at a rate of 25% per month" means that at the end of every month the quantity is 25% bigger than it was at the beginning of the month. The factor of increase for one month is therefore 1.25, and the factor of increase for 16 months is $1.25^{16}$.

It is always possible to rewrite the expression for quantity in a base other than base 1.25. The usual choice is the natural base $e\approx2.718$. In this base, the amount is given by $P\cdot e^{16k}$, where $k=\log_e1.25\approx0.2231$. But $k$ is not the same thing as the rate. Sometimes $k$ is referred to as the "rate under continuous compounding" or "continuous rate". This terminology comes from the theory of compound interest. So, for example, if you invest \$15000 at an annual rate of 20%, compounded continuously, then the amount after three years will be $15000\cdot e^{(3)(0.2)}$, whereas without the continuous compounding it will be $15000\cdot1.2^3$. If you are not familiar with compound interest, you can simply think of $k$ as the base conversion factor - that is, the factor needed to convert from base 1.25 to base $e$.

You always need to be clear about what kind of rate you are dealing with. If the word "rate" is used without a modifier such as "continuous", you should assume that it is the ordinary sort of rate described in the first paragraph above.

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