Logistic and logarithmic functions and models

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This tutorial: Part A: Logistic functions and models

Go to Part B: Logarithmic functions and models

(This topic is also in Section 2.4 in Finite Mathematics and Applied Calculus)

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Logistic functions: Basics

The following graph shows the total population of Greenland, in thousands, from 1950 to 2050 ($t = 0$ represents 1950):

#[Data for $t \geq 75$ are projections. Sources:][los datos para $t \geq 75$ son proyecciones. Fuentes:]# https://www.worldometers.info/world-population/greenland-population/ #[and][y]# Wikipedia

During the first 20 years of the period shown, Greenland's population increased at an accelerating rate reminding us of exponential growth (see the %%ExpFuncsTut). However, after that time the population growth slowed and eventually levelled off at around 57,000. This kind of behavior—eponential growth followed by leveling off—is common in population curves as well as other situations like the spread of epidemics or the demand for a new technology, and the kinds of function we use to model it are logistic functions.

Logistic function

Logistic functions have the following form:

$f(x) = \dfrac{N}{1 + Ab^{-x}}$ \t \gap[10] $A, N,$ %%and $b$ #[positive constants with][constantes positivas con]# $b \neq 1.$

Their graphs are S-shaped as shown below; increasing (like the population graph above) when $b \gt 1$ and decreasing when $b \lt 1$:

#[Increasing when ][Aumentando cuando ]# $b \gt 1$

#[Decreasing when ][Disminuyendo cuando ]# $b \lt 1$

#[More features of the graphs][Más características de las gráficas]#

Both curves are sandwiched between the horizontal lines $y = 0$ and $y = N$ and approach those lines for large absolute values of $x.$
The curves are symmetric about the indicated inflection point which occurs when the curve is steepest, at $x = \dfrac{\ln A}{\ln b}$ #[and][y]# $y = \dfrac{N}{2}.$
The curves cross the $y$-axis when $\dfrac{N}{1 + A}.$

%%Examples

1. #[The logistic function][La función logística]# $f(x) = \dfrac{12}{1+2.3(1.5^{-x})}$ #[has][tiene]# $N= 12, A = 2.3,$ %%and $b = 1.5.$

#[As][Ya que]# $b \gt 1$, #[the graph rises with increasing $x.$][la gráfica aumenta con el aumento de $x.$]#
#[The limiting value of $f$ for large $x$ is $N = 12.$][El valor límite de $f$ para $x$ grande es $N = 12.$]#
#[$f$ is increasing most rapidly at the point of inflection,][$f$ aumenta más rápidamente en el punto de inflexión,]#
$x = \dfrac{\ln A}{\ln b} = \dfrac{\ln 2.3}{\ln 1.5} \approx 2.05.$ %%and $y = \dfrac{N}{2} = 6.$
#[The $y$-intercept occurs when][La intersección-$y$ ocurre cuando]#
$y = \dfrac{N}{1+A} = \dfrac{12}{1+2.3} \approx 3.64$

Graph:

2. #[One for you][Uno para ti]#

3. The graph showing the Greenland population data suggests an increasing logistic curve. Here we see it with a logistic curve superimposed.

$\displaystyle P(t) = \frac{57}{1+1.8(1.1^{-t})} \qquad$ $N = 57, A = 1.8, b = 1.1$

We can use the properties of logistic curves to make some inferences about the Greenland population:

According to the model,

For large values of $t$ the values of $P(t)$ approach $N = 57,$ so that, in the long term the population of Greenland can be predicted to level off at around 57,000.
The point of inflection occurs when
$t = \dfrac{\ln A}{\ln b} = \dfrac{\ln 1.8}{\ln 1.1} \approx 6.2,$
indicating that the population of Greenland was growing fastest during 1956 ($t = 6.2)$.
The intercept of the $p$-axis is
$P = \dfrac{N}{1+A} = \dfrac{57}{1+1.8} \approx 20.4,$
indicating that the population of Greenland was about 20,400 in 1950 ($t = 0$).

#[Exponential approximation of logistic functions][Aproximación exponencial de funciones logísticas]#

#[Take a look once again at the graph of a logistic function for $b \gt 1$: ][Eche un vistazo una vez más a la gráfica de una función logística para $b \gt 1$:]#

$y = \dfrac{N}{1 + Ab^{-x}}$

#[Far to the left of the point of inflection, the curve looks like an exponential curve with the same base $b$:][Lejos a la izquierda del punto de inflexión, la curva parece una curva exponencial con el mismos base $b$:]#

#[Exponential curve:][Curva exponencial:]# $y = \left(\dfrac{N}{A}\right)b^x$

Notice that the formula for the exponential approximation can be obtained from the original equation by simply ignoring the "1" in the denominator:

$y = \dfrac{N}{Ab^{-x}}$ \t \gap[20] #[Logistic formula with the "1" dropped][Fórmula logística con el "1" eliminado]# \\ \\ $=\left(\dfrac{N}{A}\right)b^x\ \ $ \t \gap[20] #[Rules of exponents][Reglas de exponentes]#

Briefly, the reason it hugs the logistic curve so closely is that, to the left of the graph (when $y$ is much smaller than the limiting value $N$), the values of $Ab^{-x}$ are a lot larger than 1, so we can ignore the 1.

#[Application: Epidemics][Aplicación: Epidemias]#

The third "gamma" wave of the COVID-19 epidemic, which peaked in the U.S. during December 2020, infected an estimated total of 45 million people in the U.S. On August 22 2020, an estimated 900,000 people had been infected with the gamma variant and this number was growing by about 3.5% each day.* Model the number of infected people with a logistic function.

Answer
Our model will have the form $P(t) = \dfrac{N}{1+Ab^{-t}}$ where $P(t)$ will be the number of people infected $t$ days after August 22 2020. We are already given the values of two of the three parameters: $N$ and $b$:

$N = 45$ \gap[10] \t Total number, in millions, eventually infected \\ $b = 1.035$ \gap[10] \t Growing by 3.5% per day \\ Intercept $\dfrac{N}{1+A} = 0.9$ \gap[10] \t $P = 0.9$ million when $t = 0$ \\ $\dfrac{45}{1+A} = 0.9$

Solving the last equation for $A$ gives $A = 49,$ so our model is

$\displaystyle N = \frac{45}{1+49(1.035)^{-t}}$ #[million people infected][millones de personas infectadas]#

* #[The estimates are based on our own modeling of the first year of the COVID-19 epidemic in the U.S., as detailed in][Las estimaciones se basan en nuestro propio modelo del primer año de la epidemia de COVID-19 en los EE. UU., como se detalla en]# Section 11.CS in Finite Mathematics and Applied Calculus (#[8th ed][8ª ed]#.)

#[Now that we have the model, we can use it to estimate when the the gamma wave peaked in the U.S.: At the point of inflection the curve is steepest, indicating the day with greatest fastest rate of growth of the epidemic; that is, the day whn the most new cases occurred. According to the model this happened when][Ahora que tenemos el modelo, podemos usarlo para estimar cuándo la onda gamma alcanzó su punto máximo en los EE. UU.: en el punto de inflexión, la curva es más pronunciada, lo que indica el día con la tasa de crecimiento más rápida de la epidemia; es decir, el día en que se presentaron más casos nuevos. Según el modelo esto sucedió cuando]#

$t =\dfrac{\ln A}{\ln b} = \dfrac{\ln 49}{\ln 1.035} \approx 113.1,$

#[corresponding to December 13, 2020. On that day, the model estimates that a total of][correspondiente al 13 de diciembre de 2020. En ese día, el modelo estima que un total de]#.

$P(113) = \dfrac{45}{1 + 49(1.035^{-113})} \approx 22.45$ #[million people][millones de personas]#

#[had been infected with the gamma variant. (If we include the alpha and beta variants, an estimated total of around 49.5 milion had in fact been infected in the U.S. by that date by one of the alpha, beta, or gamma variants acording to the Institute for Health Metrics and Evaluation (IHME).)][había sido infectado con la variante gamma. (Si incluimos las variantes alfa y beta, de hecho, un total estimado de alrededor de 49,5 millones habían sido infectados en los EE. UU. para esa fecha por una de las variantes alfa, beta o gamma según el Instituto de Métricas y Evaluación de la Salud (IHME).)]#

Now try some of the exercises in Section 2.4 in Finite Mathematics and Applied Calculus. or move ahead by pressing "Next tutorial" on the sidebar.