Here are a few summary statistics of the COVID-19 pandemic. Data is from worldometers (worldwide data) and the NYT (Arizona data). These plots will be updated daily, or close to daily, at least until the pandemic is over. Click on the images to open a higher-resolution version in a new browser tab. Last update: 2020-03-28 (statistics valid up to the day before).

The epidemic in China appears to have largely passed for now. It is uncertain if or when China will see a second wave. But the pandemic for rest of the world is currently growing exponentially, with a doubling time of about four and a half days. To increase by a factor of 10 takes about 15 days. In Arizona, the doubling time is currently about *two* days! If the Arizona rate of growth keeps up at its current pace, that’s a factor of 10 increase every 6.3 days.

The NYT maintains an informative U.S. case count map here.

**Note:** the mean time between exposure and first symptoms is 4-6 days. This means that the numbers we see today correspond to *people who were infected a week ago*. This is also, roughly, the doubling time. So the current number of cases of infection is more than double the latest numbers in the graphs. On March 22, the total number of confirmed infections, excluding China, is over 250,000. This means the real number of infected people, half of whom we don’t yet know about, is over half a million.

### Arizona

### World

### The doubling time of exponential growth

Suppose a measured quantity (such as the number of COVID-19 cases), call it $n(t)$, is growing exponentially:

\begin{equation}

n(t) = n_0 e^{c t} \label{exponential}

\end{equation}

where $c$ is a constant, the rate at which $n$ is increasing with time, and $n_0$ is the initial size of the quantity. This is a simplified—in fact, the simplest possible—description of an exponentially growing quantity. If we take the logarithm of eq. \eqref{exponential}, we have

\begin{equation}

\log{n(t)} = \log{n_0} + c t \label{logexp}

\end{equation}

We see that if we therefore plot $\log{n(t)}$ as a function of (linear) time, we should get a straight line with slope $c$. Indeed, this is a quick way to see if something is growing (or shrinking) exponentially.

In the plots above, the linear fits are least squares fits of a straight line to the indicated data. Over the time spans indicated, the increase is exponential.

A natural question is, “how long does it take to increase the quantity by a factor of 2 (or 10, or whatever)?” We can use eq. \eqref{exponential} to determine this. Suppose we want to know how long it takes to increase by a factor of $k$. We can write

\begin{eqnarray}

k \cdot n(t) &= & n_0 e^{c (t + \Delta t_k)} \\

&=& n_0 e^{c t} e^{c\,\Delta t_k} \\

&=& n(t) e^{c \Delta t_k}

\end{eqnarray}

where $\Delta t_k$ is the amount of time to increase by a factor of $k$. Thus,

\begin{equation}

\log k = c \Delta t_k

\end{equation}

or

\begin{equation}

\Delta t_k = \dfrac{\log k}{c} \label{k-period}

\end{equation}

With $k=2$, the doubling time is therefore

\begin{equation}

\Delta t_2 = \dfrac{\log 2}{c} \label{doubling-period}

\end{equation}