[MATPY00] The Exponential Function

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The exponential function is a special kind of function because it is the solution to the following differential equation: 

\[\begin{equation}\label{eqn:diffeqn}\frac{dx}{dt} = rx\end{equation}\]

i.e. the derivative (here shown as a rate of change) in the value of \(x\) is proportional to itself. The constant \(r\) is the exponential rate: if \(r > 0\), we will find an exponential growth, while if \(r < 0\), we will have an exponential decay. 

Before solving Eqn (\(\ref{eqn:diffeqn}\)), let us discuss why this equation is important. Many processes in the real world starts out with trends comparable to the one modeled by this equation. For example, consider the case of the spread of infectious diseases, just like the one we are having right now. Let \(x(t)\) represent the number of infected individuals at a particular time \(t\). These infected individuals are the ones who can transmit the infection to the other individuals; more other people can be infected by, say, 100 infected individuals than, say, 10. And thus, contagions usually start out with mechanisms similar to Eqn (\(\ref{eqn:diffeqn}\)). 

And now, for the math. We can rewrite the differential equation above as follows (by cross-multiplication): 

\[\begin{equation}\label{eqn:solving01}\frac{dx}{x} = rdt\end{equation}\]

Notice that we placed all quantities with \(x\) on the left-hand side and everything else, particularly the term \(dt\), on the right. 

This is where our previous lesson comes in. To solve Eqn (\(\ref{eqn:solving01}\)), we need to do definite integrations on both sides. Why a definite integration? Because the quantities \(x\) and \(t\) refer to real-world quantities, so they must have definite values at the limits. Additionally, the lower and upper limits of integration for both sides have to be related. 

Here, let us consider first the right hand side. It is customary to integrate time from a \(t = 0\) mark; i.e. we usually start our physical observations with the timer set to zero. As such, when we proceed to the left hand side, the lower limit of \(x\) should be its value corresponding to the lower limit of \(t\), i.e. \(x(t = 0)\). For exponential functions, this lower limit is usually denoted by \(x_0\), and is oftentimes nonzero. Think about the analogy with the spread of infection: \(x_0\) denotes the original number of people infected in the population, so it cannot be zero (if it is zero, then no infection will spread in the first place). 

How about the upper limit? Here, let us consider the upper limits to be the arbitrary values \(t\) for the right side and \(x(t)\) for the left. This means that we are looking at the value at some unspecified later time after \(t = 0\). Because we use this variables for the upper limits, it is customary to make the substitutions inside the integral:

\[\begin{equation}dx \rightarrow dx'\end{equation}\]

\[\begin{equation}x \rightarrow x'\end{equation}\]

\[\begin{equation}dt \rightarrow dt'\end{equation}\]

i.e. the terms inside the integral sign become dummy variables (formality). 

Now, we can proceed! The next step becomes: 

\[\begin{equation}\label{eqn:solving02}\int_{x_0}^{x(t)}\frac{dx'}{x'} = r \int_{0}^{t}dt'\end{equation}\]

where \(r\), the rate, is taken outside the integral because it is a constant. Solving further, the antiderivative of the left hand side is a natural logarithmic function, while it becomes a simple linear power for the right. 

\[\begin{equation}\label{eqn:solving03}\left[\ln |x'| \right]_{x_0}^{x(t)} = r \left[t'\right]_{0}^{t}\end{equation}\]

Consider again the analogy of the spread of infection: we know that the quantity \(x\) refers to the number of infected individuals, which can't be negative, so we can take out the absolute value symbol. Evaluating at the limits, we get:

\[\begin{equation}\label{eqn:solving04}\left[\ln x(t) - \ln x_0 \right] = r \left[t - 0\right]\end{equation}\]

And since we have the property of logarithms that the difference of logarithms is equal to the logarithm of the quotient, we will have:

\[\begin{equation}\label{eqn:solving05}\left[\ln \frac{x(t)}{x_0} \right] = rt\end{equation}\] 

At this point, let us go back to an even older lesson on inverse functions. We want to isolate \(x(t)\) in Eqn (\(\ref{eqn:solving05}\)) to get the trend in time. But \(x(t)\) is an argument of the natural logarithm function. For this, recall that the composition of the inverse function \(f^{-1}(x)\) and the function \(f(x)\) gives:

\[\begin{equation}\left(f^{-1} \circ f\right)(x) = f^{-1}\left(f(x)\right) = x\end{equation}\]

What is the inverse of the natural logarithm function? The natural exponential function. [By now you should recognize that \(\exp(x) = e^{x}\)].

\[\begin{equation}\label{eqn:solving06}\exp\left(\left[\ln \frac{x(t)}{x_0} \right]\right) = \exp(rt)\end{equation}\] 

\[\begin{equation}\label{eqn:solving07}\frac{x(t)}{x_0} = \exp(rt)\end{equation}\] 

\[\begin{equation}\label{eqn:solving08} \boxed{x(t) = x_0 \exp(rt) = x_0e^{rt}} \end{equation}\] 

Now, going back to contagions: The exponential growth trend, \(r > 0\), is valid at the onset of contagious diseases in a population when the number of infected individuals \(x\) is very small relative to the total population \(N\), \(x << N\). A quick look at Eqn (\(\ref{eqn:diffeqn}\)) shows that the rate, i.e. the number of new individuals to be infected, is proportional to the number of individuals already having the disease at the previous time step. This has serious real-world implications.

Try this yourself: Plot the curve given by Eqn (\(\ref{eqn:solving08}\)) for the simplest case of \(x_0 = 1\) (one infected individual at the initial time) and \(r = 1\) (every individual infects one other individual). Comment on the shape of the plot, specifically on how fast the number of infected individuals grow.

Because of the finite number of individuals in the population, \(N\), the exponential curve will not continue after some time. The trend to model the case when a significant number of individuals are infected will be discussed for our next meeting.