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Re:Germany locks down unvaccinated people, as leaders plan t... by 13415 on Thursday December 02, 2021 @05:08PM (#29423438)
That's a bit misleading. As long as the R value is above 1 the spreading is exponential. That follows directly from how the R-value is defined. What you mean is that the curve will take a sigmoidal shape for a while once the number of immune people has reached a certain threshold. In that case, the R-value is slowly approaching 1 again, and finally will go below 1 (which finally results in a curve that is not sigmoidal, of course). Other measures like lockdowns do the same to the curve, as is easy to observe by overlaying the measures over the curve, just not with an initial increase of cases that leads to hundreds of thousands of deaths. Estimated Infection Fatality Rates during the first Covid wave - where there was full hospital care available - ranged from 0.5% to 2%. So for Germany's 80M people that would be in the ballpark of 400,000 to 1.6M deaths - under medical care. Even if you take the original estimates of around 0.3% the results wouldn't have been flattering.

In a nutshell, the German government and every other country on earth is doing what you propose, except that they are vaccinating at the same time and try to keep the curves flat (but your use of "slow" also suggests this, so it's not clear what else you're suggesting).

Re:Germany locks down unvaccinated people, as leaders plan t... by umanwizard on Thursday December 02, 2021 @08:27PM (#29425519)
> As long as the R value is above 1 the spreading is exponential

No, this is only true if the R value is constant in time. Not all functions with a positive first derivative are the exponential function.

Re:Germany locks down unvaccinated people, as leaders plan t... by gugagore on Friday December 03, 2021 @12:04AM (#29426804)
Are you talking about a situation where the R value approaches 1 from above? e.g. R(t) = 1 + 1/t?

Because otherwise the growth is indeed exponential, yeah?

Re:Germany locks down unvaccinated people, as leaders plan t... by umanwizard on Friday December 03, 2021 @01:00AM (#29427077)
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.

Re:Germany locks down unvaccinated people, as leaders plan t... by 13415 on Friday December 03, 2021 @01:12PM (#29433127)
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...

Re:Germany locks down unvaccinated people, as leaders plan t... by umanwizard on Friday December 03, 2021 @01:21PM (#29433253)
Can you explain the difference between a function being "exponential in the beginning" and "having first derivative bounded away from zero at the beginning" ?
Re:Germany locks down unvaccinated people, as leaders plan t... by 13415 on Saturday December 04, 2021 @08:32AM (#29440718)
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..?

Re:Germany locks down unvaccinated people, as leaders plan t... by umanwizard on Tuesday December 07, 2021 @11:38AM (#29474615)
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.

Re:Germany locks down unvaccinated people, as leaders plan t... by gugagore on Tuesday December 07, 2021 @04:56PM (#29478713)
> 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?

Re:Germany locks down unvaccinated people, as leaders plan t... by ipaddr on Thursday December 02, 2021 @06:32PM (#29424499)
Exponential implies a doubling of R value. Starting with an R value of 1.1 the length of time required for doubling of the infected population is longer than other factors which will lower the R value to under 1.