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The Exponential Function

Suppose the newspaper headlines read, ``The cost of living went up 10% this year.'' Can we translate this information into an equation? Let ``V'' denote the value of a dollar, in terms of the ``real goods'' it can buy - whatever economists mean by that. Let the elapsed time  t  be measured in years (y). Then suppose that  V  is a function of  t,  V(t),  which function we would like to know explicitly. Call now ``t = 0'' and let the initial value of the dollar (now) be  V₀,  which we could take to be $1.00 if we disregard inflation at earlier times.¹

Then our news item can be written

$$V(0) = V_0, \qquad \hbox{\rm whereas} \qquad V(1 \, {\rm y}) \; = \; (1 - 0.1) \, V_0 \; = \; 0.9 \, V_0 .$$

This formula can be rewritten in terms of the changes in the dependent and independent variables,   $\Delta V = V(1 \, {\rm y}) - V(0)$  and   $\Delta t = 1 \, {\rm y}$:

$${\Delta V \over \Delta t} = - 0.1 \, V_0 ,$$ (1)

where it is now to be understood that  V  is measured in ``1998 dollars'' and  t  is measured in years. That is, the average time rate of change of V is proportional to the value of V at the beginning of the time interval, and the constant of proportionality is -0.1 y<sup>-1</sup>. (By y<sup>-1</sup> or ``inverse years'' we mean the per year rate of change.) This is almost like a derivative. If only $\Delta t$ were infinitesimally small, it would be a derivative. Since we're just trying to describe the qualitative behaviour, let's make an approximation: assume that $\Delta t = 1$ year is ``close enough'' to an infinitesimal time interval, and that the above formula (1) for the inflation rate can be turned into an instantaneous rate of change:²

$${dV \over dt} = - 0.1 \, V .$$ (2)

This means that the dollar in your pocket right now will be worth only $0.99999996829 in one second. Well, this is interesting, but we cannot go any further with it until we ask a crucial question: ``What will happen if this goes on?'' That is, suppose we assume that equation (2) is not just a temporary situation, but represents a consistent and ubiquitous property of the function V(t), the ``real value'' of your dollar bill as a function of time.³

Applying the d/dt ``operator'' to both sides of Eq. (2) gives

$${d \over dt}\left({dV \over dt}\right) = {d \over dt}(- 0.1 \... ...x{\rm or} \qquad {d^2V \over dt^2} = - 0.1 \, {dV \over dt} .$$ (3)

But dV/dt is given by (2). If we substitute that formula into the above equation (3), we get

$${d^2V \over dt^2} = (- 0.1)^2 \, V = 0.01 \, V .$$ (4)

That is, the rate of change of the rate of change is always positive, or the (negative) rate of change is getting less negative all the time.⁴ In general, whenever we have a positive second derivative of a function (as is the case here), the curve is concave upwards. Similarly, if the second derivative were negative, the curve would be concave downwards.

So by noting the initial value of V, which is formally written V₀ but in this case equals $1.00, and by applying our understanding of the ``graphical meaning'' of the first derivative (slope) and the second derivative (curvature), we can visualize the function V(t) pretty well. It starts out with a maximum downward slope and then starts to level off as time increases. This general trend continues indefinitely. Note that while the function always decreases, it never reaches zero. This is because, the closer it gets to zero, the slower it decreases [see Eq. (2)]. This is a very ``cute'' feature that makes this function especially fun to imagine over long times.

We can also apply our analytical understanding to the formulas (2) and (4) for the derivatives: every time we take still another derivative, the result is still proportional to V - the constant of proportionality just picks up another factor of (- 0.1). This is a really neat feature of this function, namely that we can write down all its derivatives with almost no effort:

 

$${dV \over dt} - 0.1 \, V$$ = (5)

$${d^2V \over dt^2} (-0.1)^2 \, V = + 0.01 \, V$$ = (6)

$${d^3V \over dt^3} (-0.1)^3 \, V = - 0.001 \, V$$ = (7)

$${d^4V \over dt^4} (-0.1)^4 \, V = + 0.0001 \, V$$ = (8)

$$\vdots$$

$${d^nV \over dt^n} (-0.1)^n \, V \qquad \hbox{\rm for {\sl any\/} } n .$$ = (9)

This is a pretty nifty function. What is it? That is, can we write it down in terms of familiar things like t, t², t³, and so on? First, note that Eq. (9) can be written in the form

$${d^nV \over dt^n} = k^n \, V , \qquad \hbox{\rm where} \qquad k = - 0.1$$ (10)

A simpler version would be where k = 1, giving

$${d^nW \over dt^n} = W,$$ (11)

W(t) being the function satisfying this criterion. We should perhaps try figuring out this simpler problem first, and then come back to V(t). Let's try expressing W(t), then, as a linear combination⁵ of such terms. For starters we will try a ``third order polynomial'' (i.e. we allow terms up to t³):

$$\begin{array}[c]{ccrcrcrcl} W(t) &=& a_0 &+& a_1 t &+& a_2 t... ...W \over dt}} &=& a_1 &+& 2 a_2 t &+& 3 a_3 t^2 & & \end{array}$$

follows by simple ``differentiation'' [a single word for ``taking the derivative'']. Now, these two equations have similar-looking right-hand sides, provided that we pretend not to notice that term in t³ in the first one, and provided the constants a_n obey the rule a_n-1 = n a_n [i.e. a₀ = a₁, a₁ = 2 a₂ and a₂ = 3 a₃]. But we can't really neglect that t³ term! To be sure, its ``coefficient'' a₃ is smaller than any of the rest, so if we had to neglect anything it might be the best choice; but we're trying to be precise, right? How precise? Well, precise enough. In that case, would we be precise enough if we added a term a₄ t⁴, preserving the rule about coefficients [ a₃ = 4 a₄]?  No?  Then how about a₅ t⁵? And so on. No matter how precise an agreement with Eq. (11) we demand, we can always take enough terms, using this procedure, to achieve the desired precision. Even if you demand infinite precision, we just [just?] take an infinite number of terms:

$$W(t) = \sum_{n=0}^{\infty} a_n \, t^n, \qquad \hbox{\rm whe... ... \qquad \hbox{\rm or} \qquad a_n \, = \, {a_{n-1} \over n} .$$ (12)

Now, suppose we give W(t) the initial value 1. [If we want a different initial value we can just multiply the whole series by that value, without affecting Eq. (11).]  Well, W(0) = 1 tells us that a₀ = 1. In that case, a₁ = 1 also, and $a_2 = {1\over2}$, and $a_3 = {1\over2} \times {1\over3}$, and $a_4 = {1\over2} \times {1\over3} \times {1\over4}$, and so on. If we define the factorial notation,

$$n! \; \equiv \; n \times (n-1) \times (n-2) \times (n-3) \times \dots \times 3 \times 2 \times 1$$ (13)

(read, ``n factorial'') and define   $0! \equiv 1$,  we can express our function W(t) very simply:

$$W(t) = \sum_{n=0}^{\infty} {t^n \over n!}$$ (14)

We could also write a more abstract version of this function in terms of a generalized variable ``x'':

$$W(x) = \sum_{n=0}^{\infty} {x^n \over n!}$$ (15)

Let's do this, and then define   $x \equiv k \, t$  and set   $V(t) = V_0 \; W(x)$.  Then, by the CHAIN RULE for derivatives,⁶

$${dV \over dt} = \, V_0 \, {dW \over dx} \, {dx \over dt}$$ (16)

and since   ${d \over dt}(k \, t) = k$,  we have

$${dV \over dt} \; = \; k \, V_0 \, W \; = \; k \, V.$$ (17)

By repeating this we obtain Eq. (10). Thus

$$V(t) \; = \; V_0 \; W(kt) \; = \; V_0 \, \sum_{n=0}^{\infty} \, {(kt)^n \over n!}$$ (18)

where  k = - 0.1  in the present case. This is a nice description; we can always calculate the value of this function to any desired degree of accuracy by including as many terms as we need until the change produced by adding the next term is too small to worry us.⁷ But it is a little clumsy to keep writing down such an unwieldy formula every time you want to refer to this function, especially if it is going to be as popular as we claim. After all, mathematics is the art of precise abbreviation. So we give W(x) [from Eq. (15)] a special name, the ``exponential'' function, which we write as either⁸

$$\exp(x) \qquad \hbox{\rm or} \qquad e^x .$$ (19)

In FORTRAN it is represented as EXP(X). It is equal to the number

$$e \; = \; 2.71828182845904509\cdots$$ (20)

raised to the   $x^{\rm th}$  power. In our case we have   $x \equiv - 0.1 \, t$,  so that our ``answer'' is

$$V(t) \; = \; V_0 \; e^{-0.1 \, t}$$ (21)

which is a lot easier to write down than Eq. (18). Now, the choice of notation  e^x  is not arbitrary. There are a lot of rules we know how to use on a number raised to a power. One is that

$$e^{-x} \; \equiv \; {1 \over e^x}$$ (22)

You can easily determine that this rule also works for the definition in Eq. (15).

The ``inverse'' of this function (the power to which one must raise  e  to obtain a specified number) is called the ``natural logarithm'' or ``$\ln$'' function. We write

$$\hbox{\rm if} \qquad W = e^x , \qquad \hbox{\rm then} \qquad x = \ln(W)$$

or

$$x \; = \; \ln(e^x)$$ (23)

A handy application of this definition is the rule

$$y^x = e^{x \ln(y)} \qquad \hbox{\rm or} \qquad y^x = \exp[x \ln(y)].$$ (24)

Before we return to our original function, is there anything more interesting about the ``natural logarithm'' than that it is the inverse of the ``exponential'' function? And what is so all-fired special about  e,  the ``base'' of the natural log? Well, it can easily be shown⁹ that the derivative of  $\ln(x)$  is a very simple and familiar function:

$${d[\ln(x)] \over dx} \; = \; {1 \over x} .$$ (25)

This is perhaps the most useful feature of  $\ln(x)$,  because it gives us a direct connection between the exponential function and a function whose derivative is  1/x. [The handy and versatile rule   ${d(x^r) \over dx} = r x^{r-1}$  is valid for any value of  r, including  r=0, but it doesn't help us with this task. Why?]  It also explains what is so special about the number  e.