Simple question about exponential growth and the use of the number e

Question

I'm in the process of reading about and beginning to study the number e and how to work with it. I'm also reading Growth by Vaclav Smil. He describes the growth function as: $${N_t}={N_0}(1+r)^t$$ He then says that a "trivial multiplicative unit-of-timekeeping adjustment" can be made to express this as: $${N_t}={N_0}e{^r}{^t}$$ This suggests that you can replace (1+r) with $e^r$, but I'm not sure it's that simple, and if it is, why it is OK to do that. I'm also curious about why the latter is preferable to the former. I've dug around through books and on the web but haven't found anything that answers my questions. I'd appreciate any light anyone can shed on this. And any recommendations on good tutorials on working with e would also be appreciated!

Answer 1

I think the confusion here comes from the bad notational habits some analysts have. The symbol $r$ denotes different quantities here, otherwise we should have $$1+r=e^r,$$ or in other words that $$1+r=\log r.$$ But this equality is not satisfied for any $r$ as any polynomial grows faster than the logarithm, and at zero, $1+r$ is already far above $\log r,$ and the former increases, etc.

Anyway, my point is that the $r$'s mean differently, otherwise we'd be forced to deduce that the man does not know what he's talking about, which might be absurd given the circumstances.

So, given this, what can we say? Well, what's happened is that the scalar $1+r$ has been expressed as a power of $e,$ namely as $e^s$ (this is what he calls $e^r$ again, frustratingly!). This means that $s=\log (1+r).$ This is well done provided $1+r$ is positive (which it will be if $r>-1$), since otherwise a real logarithm does not exist; of course a complex logarithm may then exist, but it's not definite or unique.

You also ask why it's preferable to have a number as a power of $e,$ the base of the natural logarithm. Well, the reason is that performing calculations that have to do with the infinitesimal calculus is easiest with the base $e$ than with any other, so that reexpressing the base $1+r$ as some power of $e$ makes the later calculations easier and more familiar. One example is that the derivative of $e^x$ is again just $e^x$ (or, in other words, that differentiation does not change the exponential function -- even though there are infinitely many bases, mathematicians have chosen this as their fixed base, and study other bases in terms of it). Thus any time one has something like $a^y,$ it's always preferable to make the substitution $a\mapsto e^{\ell}$ for some appropriate $\ell,$ to have $a^y=e^{y\ell}.$ (This $\ell$ is what's called the logarithm of the number $a.$)

Answer 2

$${N_t}={N_0}(1+r_1)^t=N_0e^{\log(1+r_1)t}=N_0 e^{r_2t}$$