That’s one possibility, but there’s no reason it has to converge to 1 (i.e. linear). R(t) = 100/t is also not exponential.

Of course, the epidemic curve described by that function would indeed bounded below by an exponential function on part of its range, but the same is true of any function with positive derivative, and calling for example f(x) = x^2 exponential for that reason makes the term meaningless.

This is presumably what the OP meant by “as long as R>1, the curve is exponential”. But this is literally equivalent to saying “as long as f’(x) > 0, f is exponential” which is just not a useful concept.

You're nitpicking out of context, I didn't give a definition of an exponential function, I was talking about the spreading of the Covid 19 disease. For example, the R0 value of SARS-CoV-2 was estimated 5.8 in the US and "...between 3.6 and 6.1 in the eight European countries"[1] Obviously, it depends on many factors like population density and contacts of persons/day, but generally the disease will start spreading exponentially with R_t values approaching this number or staying constant.

The initial spreading will be exponential in the beginning - as every actual curve illustrates - if the disease is left unchecked as OP suggested, until R_t values go down again due to immunity. That's all I meant to say.

[1] https://www.sciencedirect.com/science/article/pii/S002251932...

Can you explain the difference between a function being "exponential in the beginning" and "having first derivative bounded away from zero at the beginning" ?

How about this:

exponential in the beginning == this part of the function can be approximated by a function ae^xb where a>0 and b>1

vs.

first derivative bounded away from zero in the beginning == any function that increases, including linear functions with constant first derivative and polynomials with linear first derivative

Or do you think all increasing functions are the same..?

Consider for example the functions f(x) = x + 1 and g(x) = e^{ln(2) * x}. Then f(0) = g(0), f(1) = g(1), and f(x) > g(x) whenever 0 < x < 1.

It is easy to show that for any function whose derivative is continuous and positive at 0, there is an exponential function (properly translated such that they agree at 0) that has similar properties.

> that has similar properties.

You should be specific about what properties you're talking about. What you're saying is that any function increasing function grows faster than some exponential function on a finite interval.

Still, you can observe f(x) on x ∈ [0, 1] and see that it is growing linearly. And you can observe g(x) on ∈ [0, 1] and see that it is growing exponentially.

I do not see the value in discussing the rate of exponential growth of f. Where as for g, there is a parameter with value ln(2).

If data looks like f, don't try to fit an exponential function to it (whether it's a least-squares fit, or any other objective f > g.

Are you trying to say anything else?