4 ms·

A constant 2nd-derivative means quadratic growth, not exponential. I'm bit confused by the graphs, because they show the 2nd-derivative going to 0. But to stop

A constant 2nd-derivative means quadratic growth, not exponential. I'm bit confused by the graphs, because they show the 2nd-derivative going to 0. But to stop this thing, we actually need to decelerate, eg going negative on the 2nd-derivative.

dpwm 7y ago
I think they're talking about the 2nd-derivative of the cumulative deaths. As they can (presumably) only increase, the first derivative is >= 0. The second derivative tends to zero as the cumulative deaths flattens off. Edit: Now that I think about this some more, you're right. There has to be a decrease in the daily cases, which implies a negative 2nd-derivative. They mention "relative 2nd-derivative." It seems to be defined in the paper in a way that leaves me more confused: > Daily fatality rates from the included countries were then used to calculate estimates of the relative second derivative of total deaths, N, 1/N d²N/dt², for a period of at least ten days. Does this mean they are taking the second derivative of the reciprocal of the cumulative deaths?
- 9wzYQbTYsAIc 7y ago
  They are taking the 2nd derivative of total deaths and dividing that by the number of total deaths and calling it the relative second derivative (rate of increase of rate of increase of total deaths, relative to total deaths).
9wzYQbTYsAIc 7y ago
The only way the velocity (number of cases) would go negative is if they discovered a disproportionate number of false positives or (number of deaths) if deaths were discovered to be due to something else [or people came back to life]. As time goes on, the velocity will reach a constant of 0, presumably. The acceleration at a constant velocity is 0.
- zeroimpl 7y ago
  Eventually acceleration would go to 0, but it needs to go negative first. Right now acceleration is positive and velocity is positive, hopefully soon acceleration will become negative, then eventually velocity will approach 0 from above and acceleration will approach 0 from below.
  - 9wzYQbTYsAIc 7y ago
    I believe that you are correct in your calculation, but incorrect in your assumption. You are treating the curve of daily deaths as the thing to be differentiated. They are treating the curve of total deaths as the thing to be differentiated. While the curve of daily deaths certainly would have a negative acceleration, the curve of total deaths is monotonically non-decreasing and therefore the second derivative is always positive but trending towards zero. Like I mentioned above, the only way for total deaths to decrease (making the function not monotonically increasing and therefore capable of having a negative acceleration) would be through errors of accounting.
dralley 7y ago
The goal was never to "stop" Coronavirus. Just slow it down enough that it doesn't crush the medical system.
9wzYQbTYsAIc 7y ago
> A constant 2nd-derivative means quadratic growth, not exponential. A constant relative second derivative of total deaths is what they are talking about, meaning that they are flattening the 2nd derivative of the exponential by dividing by the exponential, making it constant. Of course, this is all numerical methods, not analytical methods, so it doesn’t necessarily make pure analytical sense that they are even talking about an exponential.

dpwm 7y ago

I think they're talking about the 2nd-derivative of the cumulative deaths. As they can (presumably) only increase, the first derivative is >= 0. The second derivative tends to zero as the cumulative deaths flattens off. Edit: Now that I think about this some more, you're right. There has to be a decrease in the daily cases, which implies a negative 2nd-derivative. They mention "relative 2nd-derivative." It seems to be defined in the paper in a way that leaves me more confused: > Daily fatality rates from the included countries were then used to calculate estimates of the relative second derivative of total deaths, N, 1/N d²N/dt², for a period of at least ten days. Does this mean they are taking the second derivative of the reciprocal of the cumulative deaths?

9wzYQbTYsAIc 7y ago
They are taking the 2nd derivative of total deaths and dividing that by the number of total deaths and calling it the relative second derivative (rate of increase of rate of increase of total deaths, relative to total deaths).

9wzYQbTYsAIc 7y ago

The only way the velocity (number of cases) would go negative is if they discovered a disproportionate number of false positives or (number of deaths) if deaths were discovered to be due to something else [or people came back to life]. As time goes on, the velocity will reach a constant of 0, presumably. The acceleration at a constant velocity is 0.

zeroimpl 7y ago
Eventually acceleration would go to 0, but it needs to go negative first. Right now acceleration is positive and velocity is positive, hopefully soon acceleration will become negative, then eventually velocity will approach 0 from above and acceleration will approach 0 from below.
- 9wzYQbTYsAIc 7y ago
  I believe that you are correct in your calculation, but incorrect in your assumption. You are treating the curve of daily deaths as the thing to be differentiated. They are treating the curve of total deaths as the thing to be differentiated. While the curve of daily deaths certainly would have a negative acceleration, the curve of total deaths is monotonically non-decreasing and therefore the second derivative is always positive but trending towards zero. Like I mentioned above, the only way for total deaths to decrease (making the function not monotonically increasing and therefore capable of having a negative acceleration) would be through errors of accounting.

dralley 7y ago

The goal was never to "stop" Coronavirus. Just slow it down enough that it doesn't crush the medical system.

9wzYQbTYsAIc 7y ago

> A constant 2nd-derivative means quadratic growth, not exponential. A constant relative second derivative of total deaths is what they are talking about, meaning that they are flattening the 2nd derivative of the exponential by dividing by the exponential, making it constant. Of course, this is all numerical methods, not analytical methods, so it doesn’t necessarily make pure analytical sense that they are even talking about an exponential.