0
$\begingroup$

Basic exponential growth is

$$x(t) =ab^{t/\tau} $$

where $b$ is the growth rate or factor. Now, as Wikipedia describes with a bacteria example, the growth rate is $2$. This comes from starting with one bacteria (initial condition), then "doubling," which means $100\%$, or 1.0, i.e., $1 + 1.0$ means the growth rate is $2$.

This is very confusing to a beginner like me. Investigation led me to this alternate formula for exponential growth, which seems to break down the growth rate:

$$y = a(1 + r)^x $$

that is, $(1 + r)$ is $b$, although I'm guessing the initial starting point has only one single item. So, e.g., if there were two initial bacteria, we would have

$$ y = a(2 + r)^x$$

Is this true? Basically, I'd like to see how we can say $(1 + r)$ is derived from just plain old $b$. I'm not finding a source that explains this very well. The Wikipedia certainly doesn't.

$\endgroup$
1
$\begingroup$

The form $ab^{t/\tau}$ has more free parameters than it really needs. You can decide, once and for all, what your base $b$ is, and then it is not part of the model anymore. Or you can decide, once and for all, what the coefficient $1/\tau$ in the exponent is, and then it is not a part of the model anymore. People use both approaches. Generally if they fix the base, they fix it to be either $2$ (for "doubling" or "halving" problems), $e$ (for calculus convenience), or sometimes $(1+r)$ (usually just for compound interest). Generally if they fix the coefficient, they fix it to be $1$. In all cases, this decision is a matter of convenience, because an exponential model is really determined by two parameters, not three.

The initial quantity in an exponential problem appears as the coefficient on the outside; for example if a population of bacteria double every $\tau$ time units, and they start out at a population of $10$, then their population can be modeled as $10 \cdot 2^{t/\tau}$.

$\endgroup$
  • $\begingroup$ I thought $(1 + r)$ in the bacteria meant we start with one bacteria, and then double it, i.e., 2% growth in decimal notation is $0.02$ and 100% growth in decimal notation is therefore $1.0$; hence, $1 + 1.0 = 2$. But now I'm confused. If we start with three bacteria, then $3 + 1.0 = 4$, which is bogus. So, e.g., 2% growth is simply $1.02$, then times the initial population $P_0$ to give the next round of $P_1$. That means 100% is simply $2.0$. But I'd still like to see some proof, or something deeper explaining the seemingly arbitrary way we got 100% means $2$ in this formula. $\endgroup$ – 147pm Oct 10 '17 at 3:10
  • $\begingroup$ @147pm As I said, the starting amount is on the outside, not in the base. So if you have 3 bacteria growing by 2% each time unit then that's described by $3 \cdot 1.02^t$. And yes, an increase of 100% is 2.0...simply because it means you add the starting amount to itself, so you're left with twice that. $\endgroup$ – Ian Oct 10 '17 at 4:11
  • $\begingroup$ Yes, this is simply one of those arithmetic tricks, such as figuring out how to add tax, i.e., multiply by $1.09$ if the tax is $9\%$. There is no "proof" per se, it's just a trick. Still, Wikipedia doesn't explain this, which is confusing for beginners. So I'm someone who "gets to the bottom" of this issue. How many others simply give up? Alas. . . . $\endgroup$ – 147pm Oct 10 '17 at 12:15
1
$\begingroup$

The point is that the number of new bacteria in a period is proportional to the number at the start of the period. If $1$ bacterium becomes $1+r$, two bacteria will become $2(1+r)=2+2r$. Thus we can replace $1+r$ by $b$. The initial quantity can be factored out and becomes a multiplier.

$\endgroup$
  • $\begingroup$ Yes, agreed. One text called the increase per round a "constant ratio," e.g., a $2^x$ sequence increases per round by doubling the previous number of the sequence. Thanks. $\endgroup$ – 147pm Oct 10 '17 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.