COVID-19/Mathematical Modelling/Exponential Growth

We start the learning resource with a verbal description of a growth process and then go into the mathematical modelling by the exponential function. Then we discuss the limits of epidemiological modeling by exponential growth in the context of logistic growth.

Graph of the exponential function $f(x)=e^{x}$ (red) with the tangent (light blue dotted line) through the point 0/1

The number of new infections doubles every four days.

Task for Learners

• Explain the difference between
  • Number of infected and
  • the number of newly infected

mathematically. What is the different meaning of function $f(x)$ and derivation of the function $f'(x)$ in general? What is the special property of $f(x)=e^{x}$?

• Compare exponential growth with logistic growth and explain the role of population immunization on epidemioligical growth. Compare the derivation of the exponential function and the function of the logistical growth!

Table of values

First we create a table of values:

Day $t$	Number of infected	Exponential representation	New infections after 4 days
0	1	$2^{0}$	-
4	2	$1\cdot 2=2^{1}$	$1=2^{0}$
8	4	$2\cdot 2=2^{2}$	$2=2^{1}$
12	8	$4\cdot 2=2^{3}$	$4=2^{2}$
16	16	$16\cdot 2=2^{4}$	$8=2^{3}$
20	32	$32\cdot 2=2^{5}$	$16=2^{4}$
24	64	$64\cdot 2=2^{6}$	$32=2^{5}$

If you now want to display the number of infected persons as a function depending on the tag, you can use this function $f:\mathbb {R_{0}^{+}} \to \mathbb {R} ^{+}$ as follows. $f(t):=2^{{\frac {1}{4}}\cdot t}=e^{{\frac {2}{4}}\cdot t}$ with$t\in \mathbb {R_{0}^{+}}$, where $t\in \mathbb {R_{0}^{+}}$ describes the time period from the day $0$ of the initial infection. The function $f:\mathbb {R_{0}^{+}} \to \mathbb {R} ^{+}$ interpolates the points.

Start population

Data collection may start with a starting population (e.g. 1000 persons), then the table changes as follows.

Day $t$	Number of infected	Exponential representation	New infections after 4 days
0	1000	$2^{0}$	-
4	2000	$1000\cdot 2=1000\cdot 2^{1}$	$1000$
8	4000	$2000\cdot 2=1000\cdot 2^{2}$	$2000=1000\cdot 2^{1}$
12	8000	$4000\cdot 2=1000\cdot 2^{3}$	$4000=1000\cdot 2^{2}$
16	16000	$16000\cdot 2=1000\cdot 2^{4}$	$8000=1000\cdot 2^{3}$
20	32000	$32000\cdot 2=1000\cdot 2^{5}$	$16000=1000\cdot 2^{4}$
24	64000	$64000\cdot 2=1000\cdot 2^{6}$	$32000=1000\cdot 2^{5}$

If you now want to integrate into the function depending on the star resolution, you can use this function $f:\mathbb {R} \to \mathbb {R} ^{+}$ as follows. $f(t):=1000\cdot 2^{{\frac {1}{4}}\cdot t}=1000\cdot e^{{\frac {2}{4}}\cdot t}$ with$t\in \mathbb {R}$, where $t\in \mathbb {R}$ denotes the period of time since the day $0$ of the mathematical modelling with the selected starting population $s_{0}\in \mathbb {R} ^{+}$. In this context, it also makes sense to allow negative numbers for the time $t\in \mathbb {R} ^{-}$. E.g. how many infected people were there 4 days before the reference day $t=0$, the calculation can be deduced from the content, because the number of infected people doubles every 4 days and then the population must have had 500 infected people 4 days before with a starting population of $s_{0}=1000$. Functionally this can be described as follows: $f(-4):=1000\cdot 2^{{\frac {-4}{4}}\cdot t}=1000\cdot 2^{-1}=500$ with $t\in \mathbb {R}$. Overall, exponential growth can thus be described for epidemiological spread as follows: $f(t):=s_{0}\cdot 2^{{\frac {1}{4}}\cdot t}=s_{0}\cdot e^{{\frac {2)}{4}}\cdot t}$ with $t\in \mathbb {R}$,

For a general presentation of the textual description The number of new infections increases every math day. record the following representation by an exponential function,

$f(t):=s_{0}\cdot n^{{\frac {1}{d}}\cdot t}=s_{0}\cdot e^{{\frac {\ln(n)}{d}}\cdot t}$

• with $t\in \mathbb {R}$,

$d\in \mathbb {R} ^{+}/math>definesthetimeintervalinwhichthenumberofinfectedpersonsmultiplies<math>n\in \mathbb {R} ^{+}$ describes the factor by which the number of infected persons per time interval $d$ multiplies.

New infection and number of infected

However, the term "new infection" does not refer to the absolute number of COVID-19 patients identified in the country, but the detected new infections give an indication of how the increase in function at the time $t\in \mathbb {R}$ can be calculated. Therefore, one considers the derivation of the function terms for the number of infected persons in a population.

Comparison exponential growth - logistic growth

At school you may have contact with exponential growth. Epidemiologically, the exponential growth model assumes that there are no limits to growth. In fact, for epidemiology, the maximum number of infected people is given by the total population. If one considers the other biological growth processes (e.g. cell division), there are also limits to growth (e.g. limits of resources, limits of space, ...). The logistic growth includes a capacity in the modelling. A logistic function or logistic curve is a common "S" form (sigmoid curve), with the equation:

$f(x)={\frac {M}{1+e^{-k(x-x_{0})}}}$

where

• $e$ = the basis of the exponential function, also known as Euler's number)
• $x_{0}$ = the $x$ value of the inflection point, which in epidemiology is the point in time with the maximum growth rate (maximum value of the derivative)
• $M$ = the smallest upper bound of the function values (supremum of the quantity $M:=\sup _{x\in \mathbb {R} }f(x)$. capacity of the logistic growth of the curve (the capacity of growth), and
• $k$ = the logistic growth rate or steepness of the curve.