Exponential, Logarithmic, Hyperbolic Functions, and Their Derivatives

(1) At simple interest. Consider a concrete case. Let the capital at start be $$100$, and let the rate of interest be $10$ percent per annum. Then the increment to the owner of the capital will be $$10$ every year. Let him go on drawing his interest every year, and hoard it by putting it by in a stocking, or locking it up in his safe. Then, if he goes on for $10$ years, by the end of that time he will have received $10$ increments of $$10$ each, or $$100$, making, with the original $$100$, a total of $$200$ in all. His property will have doubled itself in $10$ years. If the rate of interest had been $5$ percent, he would have had to hoard for $20$ years to double his property. If it had been only $2$ percent, he would have had to hoard for $50$ years. It is easy to see that if the value of the yearly interest is ${\displaystyle \frac{1}{n}}$ of the capital, he must go on hoarding for $n$ years in order to double his property.

Or, if $y$ be the original capital, and the yearly interest is ${\displaystyle \frac{y}{n}}$, then, at the end of $n$ years, his property will be $$y+n\frac{y}{n}=2y.$$

But this mode of reckoning compound interest once a year, is really not quite fair; for even during the first year the $$100$ ought to have been growing. At the end of half a year it ought to have been at least $$105$, and it certainly would have been fairer had the interest for the second half of the year been calculated on $$105$. This would be equivalent to calling it $5$% per half-year; with $20$ operations, therefore, at each of which the capital is multiplied by $\frac{21}{20}$. If reckoned this way, by the end of ten years the capital would have grown to $$265.33$.; for $${(1+\frac{1}{20})}^{20}=2.6533.$$

But, even so, the process is still not quite fair; for, by the end of the first month, there will be some interest earned; and a half-yearly reckoning assumes that the capital remains stationary for six months at a time. Suppose we divided the year into $10$ parts, and reckon a one percent interest for each tenth of the year. We now have $100$ operations lasting over the ten years; or $$y_n=\mathrm{\$}100{(1+\frac{1}{100})}^{100};$$ which works out to $$270.481$.

Even this is not final. Let the ten years be divided into $1000$ periods, each of $\frac{1}{100}$ of a year; the interest being $\frac{1}{10}$ percent for each such period; then $$y_n=\mathrm{\$}100{(1+\frac{1}{1000})}^{1000};$$ which works out to $$271.692$.

Go even more minutely, and divide the ten years into $10,000$ parts, each $\frac{1}{1000}$ of a year, with interest at $\frac{1}{100}$ of $1$ percent. Then $$y_n=\mathrm{\$}100{(1+\frac{1}{10,000})}^{10,000};$$ which amounts to $$271.815$.

Finally, it will be seen that what we are trying to find is in reality the ultimate value of the expression ${(1+{\displaystyle \frac{1}{n}})}^n$, which, as we see, is greater than $2$; and which, as we take $n$ larger and larger, grows closer and closer to a particular limiting value. However big you make $n$, the value of this expression grows nearer and nearer to the figure $$2.71828\dots$$ a number never to be forgotten.

Let us take geometrical illustrations of these things. In the next figure, $OP$ stands for the original value. $OT$ is the whole time during which the value is growing. It is divided into $10$ periods, in each of which there is an equal step up. Here ${\displaystyle \frac{dy}{dx}}$ is a constant; and if each step up is $\frac{1}{10}$ of the original $OP$, then, by $10$ such steps, the height is doubled. If we had taken $20$ steps, each of half the height shown, at the end the height would still be just doubled. Or $n$ such steps, each of ${\displaystyle \frac{1}{n}}$ of the original height $OP$, would suffice to double the height. This is the case of simple interest. Here is $1$ growing till it becomes $2$.

We have seen that we require to know what value is reached by the expression ${(1+{\displaystyle \frac{1}{n}})}^n$, when $n$ becomes indefinitely great. Arithmetically, here are tabulated a lot of values (which anybody can calculate out by the help of a calculator) got by assuming $n=2$; $n=5$; $n=10$; and so on, up to $n=10,000$. \begin{aligned} &\left(1 + \frac{1}{2}\right)^2 &=& 2.25. \\ &\left(1 + \frac{1}{5}\right)^5 &=& 2.488. \\ &\left(1 + \frac{1}{10}\right)^{10} &=& 2.594. \\ &\left(1 + \frac{1}{20}\right)^{20} &=& 2.653. \\ &\left(1 + \frac{1}{100}\right)^{100} &=& {2.705}. \\ &\left(1 + \frac{1}{1000}\right)^{1000} &=& {2.7169}. \\ &\left(1 + \frac{1}{10,000}\right)^{10,000} &=& {2.7181}. \end{aligned}

It is, however, worth while to find another way of calculating this immensely important figure.

Accordingly, we will avail ourselves of the binomial theorem, and expand the expression ${(1+{\displaystyle \frac{1}{n}})}^n$ in that well-known way.

The binomial theorem gives the rule that \begin{aligned} (a + b)^n &= a^n + n \dfrac{a^{n-1} b}{1!} + n(n - 1) \dfrac{a^{n-2} b^2}{2!} + n(n -1)(n - 2) \dfrac{a^{n-3} b^3}{3!} + \cdots. \end{aligned} Putting $a=1$ and $b={\displaystyle \frac{1}{n}}$, we get \begin{aligned} \left(1 + \dfrac{1}{n}\right)^n &= 1 + 1 + \dfrac{1}{2!} \left(\dfrac{n - 1}{n}\right) + \dfrac{1}{3!} \dfrac{(n - 1)(n - 2)}{n^2} + \dfrac{1}{4!} \dfrac{(n - 1)(n - 2)(n - 3)}{n^3} + \cdots. \end{aligned}

Now, if we suppose $n$ to become indefinitely great, say a billion, or a billion billions, then $n-1$, $n-2$, and $n-3$, etc., will all be sensibly equal to $n$; and then the series becomes $$\boxed{e=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots .}$$

By taking this rapidly convergent series to as many terms as we please, we can work out the sum to any desired point of accuracy. Here is the working for ten terms:

$e$ is incommensurable with $1$, and resembles $\pi$ in being an interminable non-recurrent decimal.

The great reason why $e$ is regarded of importance is that $e^x$ possesses a property, not possessed by any other function of $x$, that when you differentiate it its value remains unchanged; or, in other words, its derivative is the same as itself. This can be instantly seen by differentiating it with respect to $x$, thus: \begin{aligned} \frac{d(e^x)}{dx} &= 0 + 1 + \frac{2x}{1 \cdot 2} + \frac{3x^2}{1 \cdot 2 \cdot 3} + \frac{4x^3}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{5x^4}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} + \cdots. \end{aligned} or \begin{aligned} \frac{d(e^x)}{dx}= 1 + x + \frac{x^2}{1 \cdot 2} + \frac{x^3}{1 \cdot 2 \cdot 3} + \frac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} + \cdots, \end{aligned} which is exactly the same as the original series.

$$\boxed{\frac{d(e^x)}{dx}=e^x}$$

Now we might have gone to work the other way, and said: Go to; let us find a function of $x$, such that its derivative is the same as itself. Or, is there any expression, involving only powers of $x$, which is unchanged by differentiation? Accordingly; let us assume as a general expression that $$y=A+Bx+C{x}^2+D{x}^3+E{x}^4+\cdots ,$$ (in which the coefficients $A$, $B$, $C$, etc. will have to be determined), and differentiate it. $$\frac{dy}{dx}=B+2Cx+3D{x}^2+4E{x}^3+\cdots .$$

Now, if this new expression is really to be the same as that from which it was derived, it is clear that $A$ must $=B$; that $C={\displaystyle \frac{B}{2}}={\displaystyle \frac{A}{1\cdot 2}}$; that $D={\displaystyle \frac{C}{3}}={\displaystyle \frac{A}{1\cdot 2\cdot 3}}$; that $E={\displaystyle \frac{D}{4}}={\displaystyle \frac{A}{1\cdot 2\cdot 3\cdot 4}}$, etc.

The law of change is therefore that

$$y=A(1+\frac{x}{1}+\frac{x^2}{1\cdot 2}+\frac{x^3}{1\cdot 2\cdot 3}+\frac{x^4}{1\cdot 2\cdot 3\cdot 4}+\cdots .).$$

If, now, we take $A=1$ for the sake of further simplicity, we have $$y=1+\frac{x}{1}+\frac{x^2}{1\cdot 2}+\frac{x^3}{1\cdot 2\cdot 3}+\frac{x^4}{1\cdot 2\cdot 3\cdot 4}+\cdots .$$

Differentiating it any number of times will give always the same series over again.

If, now, we take the particular case of $A=1$, and evaluate the series, we shall get simply \begin{aligned} \text{when } x &= 1,\quad & y &= 2.718281\dots; & \text{that is, } y &= e; \\ \text{when } x &= 2,\quad & y &=(2.718281\dots)^2; & \text{that is, } y &= e^2; \\ \text{when } x &= 3,\quad & y &=(2.718281\dots)^3; & \text{that is, } y &= e^3; \end{aligned} and therefore $$\text{when }x=x,y=(2.718281\dots .{)}^x;\text{that is, }y=e^x,$$ thus finally demonstrating that $$e^x=1+\frac{x}{1}+\frac{x^2}{1\cdot 2}+\frac{x^3}{1\cdot 2\cdot 3}+\frac{x^4}{1\cdot 2\cdot 3\cdot 4}+\cdots .$$

The logarithm with base $e$ is called the natural logarithm. The natural logarithm is so important that it has its own shorthand: $${\log }_{e}y\text{is often written as}\ln y.$$

The two curves plotted in Figs. 14.3 and 14.4 represent these equations.

The points calculated are:

It will be seen that, though the calculations yield different points for plotting, yet the result is identical. The two equations really mean the same thing.

As many persons who use ordinary logarithms, which are calculated to base $10$ instead of base $e$, are unfamiliar with the “natural” logarithms, it may be worth while to say a word about them. The ordinary rule that adding logarithms gives the logarithm of the product still holds good; or $$\ln a+\ln b=\ln (ab).$$ Also the rule of powers holds good; $$n\times \ln a=\ln a^n.$$ But as $10$ is no longer the basis, one cannot multiply by $100$ or $1000$ by merely adding $2$ or $3$ to the index. One can change the natural logarithm to the ordinary logarithm¹ simply by multiplying it by $\frac{1}{\ln 10}\approx 0.4343$; or \begin{aligned} \log_{10} x = \frac{1}{\ln 10}\ln x\approx0.4343 \times \ln x, \end{aligned} and conversely, \begin{aligned} \ln x = \frac{1}{\log_{10} e}\times \log_{10} x=\ln 10\times \log_{10}x\approx2.3026 \times \log_{10} x. \end{aligned}

Now this is a very curious result. It may be written $$\boxed{\frac{d(\ln x)}{dx}=x^{-1}.}$$

Note that $x^{-1}$ is a result that we could never have got by the Power Rule for differentiating powers. That rule is to multiply by the power, and reduce the power by $1$. Thus, differentiating $x^3$ gave us $3x^2$; and differentiating $x^2$ gave $2x^1$. But differentiating $x^0$ does not give us $x^{-1}$ or $0\times x^{-1}$, because $x^0$ is itself $=1$, and is a constant. We shall have to come back to this curious fact that differentiating $\ln x$ gives us ${\displaystyle \frac{1}{x}}$ when we reach the chapter on integrating.

Now, try to differentiate \begin{aligned} y = \ln(x+a), \end{aligned} that is, \begin{aligned} e^y = x+a; \end{aligned} we have ${\displaystyle \frac{d(x+a)}{dy}}=e^y$, since the differential of $e^y$ remains $e^y$.

This gives $$\frac{dx}{dy}=e^y=x+a;$$ hence, reverting to the original function (see here), we get $$\frac{dy}{dx}=\frac{1}{{\displaystyle \frac{dx}{dy}}}=\frac{1}{x+a}.$$

Next try $$y={\log }_{10}x.$$

First change to natural logarithms by multiplying by the modulus ${\log }_{10}e={\displaystyle \frac{1}{\ln 10}}\approx 0.4343$. This gives us \begin{aligned} y = \frac{1}{\ln 10} \ln x; \end{aligned} whence \begin{aligned} \frac{dy}{dx} = \frac{1}{x\ln 10}. \end{aligned}

In general, because $${\log }_{a}x=\frac{{\log }_{e}x}{{\log }_{e}a}=\frac{\ln x}{\ln a}$$ and ${\displaystyle \frac{1}{\ln a}}$ is a constant, we have $$\frac{d({\log }_{a}x)}{dx}=\frac{d({\displaystyle \frac{1}{\ln a}}\cdot \ln x)}{dx}=\frac{1}{\ln a}\cdot \frac{d(\ln x)}{dx}=\frac{1}{\ln a\cdot x}$$

The next thing is not quite so simple. Try this: $$y=a^x.$$

Taking the logarithm of both sides, we get \begin{aligned} \ln y &= x \ln a, \end{aligned} or \begin{aligned} x &= \frac{\ln y}{\ln a}\\ &= \frac{1}{\ln a} \times \ln y. \end{aligned}

Since ${\displaystyle \frac{1}{\ln a}}$ is a constant, we get $$\frac{dx}{dy}=\frac{1}{\ln a}\times \frac{1}{y}=\frac{1}{a^x\times \ln a};$$ hence, reverting to the original function. $$\frac{dy}{dx}=\frac{1}{{\displaystyle \frac{dx}{dy}}}=a^x\times \ln a.$$

We see that, since $$\frac{dx}{dy}\times \frac{dy}{dx}=1\text{and}\frac{dx}{dy}=\frac{1}{y}\times \frac{1}{\ln a},\frac{1}{y}\times \frac{dy}{dx}=\ln a.$$

We shall find that whenever we have an expression such as $\ln y=$ a function of $x$, we always have ${\displaystyle \frac{1}{y}}{\displaystyle \frac{dy}{dx}}=$ the derivative of the function of $x$, so that we could have written at once, from $\ln y=x\ln a$, $$\frac{1}{y}\frac{dy}{dx}=\ln a\text{and}\frac{dy}{dx}=a^x\ln a.$$

We can see, by putting $x=0$, that $b$ is the initial height of $y$.

Then when $$x=1,y=bp;x=2,y=b{p}^2;x=3,y=b{p}^3,\text{etc.}$$

Also, we see that $p$ is the numerical value of the ratio between the height of any ordinate and that of the next preceding it. In the next figure, we have taken $p$ as $\frac{6}{5}$; each ordinate being $\frac{6}{5}$ as high as the preceding one.

If two successive ordinates are related together thus in a constant ratio, their logarithms will have a constant difference; so that, if we should plot out a new curve, the next figure, with values of $\ln y$ as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows from the equation, that \begin{aligned} \ln y = \ln b + x \cdot \ln p, \end{aligned} whence \begin{aligned} \ln y - \ln b = x \cdot \ln p. \end{aligned}

Now, since $\ln p$ is a mere number, and may be written as $\ln p=a$, it follows that $$\ln \frac{y}{b}=ax,$$ and the equation takes the new form $$y=b{e}^{ax}.$$

The equation is still $$y=b{p}^x;$$

but since $p$ is less than one, $\ln p$ will be a negative quantity, and may be written $-a$; so that $p=e^{-a}$,² and now our equation for the curve takes the form $$y=b{e}^{-ax}.$$

The importance of this expression is that, in the case where the independent variable is time, the equation represents the course of a great many physical processes in which something is gradually dying away. Thus, the cooling of a hot body is represented (in Newton’s celebrated “law of cooling”) by the equation $${\theta }_t={\theta }_0{e}^{-at};$$ where ${\theta }_0$ is the original excess of temperature of a hot body over that of its surroundings, ${\theta }_t$ the excess of temperature at the end of time $t$, and $a$ is a constant—namely, the constant of decrement, depending on the amount of surface exposed by the body, and on its coefficients of conductivity and emissivity, etc.

A similar formula, $$Q_t=Q_0{e}^{-at},$$ is used to express the charge of an electrified body, originally having a charge $Q_0$, which is leaking away with a constant of decrement $a$; which constant depends in this case on the capacity of the body and on the resistance of the leakage-path.

Oscillations given to a flexible spring die out after a time; and the dying-out of the amplitude of the motion may be expressed in a similar way.

In fact $e^{-at}$ serves as a die-away factor for all those phenomena in which the rate of decrease is proportional to the magnitude of that which is decreasing; or where, in our usual symbols, ${\displaystyle \frac{dy}{dt}}$ is proportional at every moment to the value that $y$ has at that moment. For we have only to inspect the curve, Fig. 14.7 above, to see that, at every part of it, the slope ${\displaystyle \frac{dy}{dx}}$ is proportional to the height $y$; the curve becoming flatter as $y$ grows smaller. In symbols, thus $$y=b{e}^{-ax}$$ or $$\ln y=\ln b-ax\ln e=\ln b-ax,$$ and, differentiating, $$\frac{1}{y}\frac{dy}{dx}=-a;$$ hence $$\frac{dy}{dx}=b{e}^{-ax}\times (-a)=-ay;$$ or, in words, the slope of the curve is downward, and proportional to $y$ and to the constant $a$.

We should have got the same result if we had taken the equation in the form \begin{aligned} y &= bp^x; \end{aligned} for then \begin{aligned} \frac{dy}{dx}= bp^x \times \ln p. \end{aligned} But $$\ln p=-a;$$ giving us \begin{aligned} \frac{dy}{dx} = y \times (-a) = -ay, \end{aligned} as before.

The Time-constant. In the expression for the “die-away factor” $e^{-at}$, the quantity $a$ is the reciprocal of another quantity known as “the time-constant,” which we may denote by the symbol $T$. Then the die-away factor will be written $e^{-\frac{t}{T}}$; and it will be seen, by making $t=T$ that the meaning of $T$ $\left(\text{or of}{\displaystyle \frac{1}{a}}\right)$ is that this is the length of time which it takes for the original quantity (called ${\theta }_0$ or $Q_0$ in the preceding instances) to die away ${\displaystyle \frac{1}{e}}$th part—that is to $0.3678$—of its original value.

As an example, suppose there is a hot body cooling, and that at the beginning of the experiment (i.e. when $t=0$) it is $72{}^{\circ }$C hotter than the surrounding objects, and if the time-constant of its cooling is $20$ minutes (that is, if it takes $20$ minutes for its excess of temperature to fall to ${\displaystyle \frac{1}{e}}$ part of $72$ degrees), then we can calculate to what it will have fallen in any given time $t$. For instance, let $t$ be $60$ minutes. Then ${\displaystyle \frac{t}{T}}={\displaystyle \frac{60}{20}}=3$, and we shall have to find the value of $e^{-3}$, and then multiply the original $72$ degrees by this. Since $e^{-3}$ is $0.0498$, at the end of $60$ minutes the excess of temperature will have fallen to $72{}^{\circ }\text{C}\times 0.0498=3.586{}^{\circ }$C.