Let there be a quantity growing in such a way that the increment of its growth, during a given time, shall always be proportional to its own magnitude. This resembles the process of reckoning interest on money at some fixed rate; for the bigger the capital, the bigger the amount of interest on it in a given time.

Now we must distinguish clearly between two cases, in our calculation, according as the calculation is made by what the arithmetic books call “simple interest,” or by what they call “compound interest.” For in the former case the capital remains fixed, while in the latter the interest is added to the capital, which therefore increases by successive additions.

(1) at simple interest.

Consider a concrete case. Let the capital at start be £$$100$$, and let the rate of interest be $$10$$ per cent. 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 $$5$$ per cent., he would have had to hoard for $$20$$ years to double his property. If it had been only $$2$$ per cent., he would have had to hoard for $$50$$ years. It is easy to see that if the value of the yearly interest is $$\dfrac{1}{n}$$ 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 $$\dfrac{y}{n}$$, then, at the end of $$n$$ years, his property will be $y + n\dfrac{y}{n} = 2y.$

(2) at compound interest.

As before, let the owner begin with a capital of £$$100$$, earning interest at the rate of $$10$$ per cent. per annum; but, instead of hoarding the interest, let it be added to the capital each year, so that the capital grows year by year. Then, at the end of one year, the capital will have grown to £$$110$$; and in the second year (still at $$10$$%) this will earn £$$11$$ interest. He will start the third year with £$$121$$, and the interest on that will be £$$12$$$$2$$s.; so that he starts the fourth year with £$$133$$$$2$$s., and so on. It is easy to work it out, and find that at the end of the ten years the total capital will have grown to £$$259$$$$7$$s$$6$$d. In fact, we see that at the end of each year, each pound will have earned $$\tfrac{1}{10}$$ of a pound, and therefore, if this is always added on, each year multiplies the capital by $$\tfrac{11}{10}$$; and if continued for ten years (which will multiply by this factor ten times over) will multiply the original capital by $${2.59374}$$. Let us put this into symbols. Put $$y_0$$ for the original capital; $$\dfrac{1}{n}$$ for the fraction added on at each of the $$n$$ operations; and $$y_n$$ for the value of the capital at the end of the $$n^{\text th}$$ operation. Then $y_n = y_0\left(1 + \frac{1}{n}\right)^n.$

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 $$\tfrac{21}{20}$$. If reckoned this way, by the end of ten years the capital would have grown to £$$265$$$$6$$s$$7$$d.; for $(1 + \tfrac{1}{20})^{20} = {2.653}.$

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-per-cent. interest for each tenth of the year. We now have $$100$$ operations lasting over the ten years; or $y_n = £100 \left( 1 + \tfrac{1}{100} \right)^{100};$ which works out to £$$270$$$$9$$s$$7\frac{1}{2}$$d.

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 $$\frac{1}{10}$$ per cent. for each such period; then $y_n = £100 \left( 1 + \tfrac{1}{1000} \right)^{1000};$ which works out to £$$271$$$$13$$s$$10$$d.

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 $$1$$ per cent. Then $y_n = £100 \left( 1 + \tfrac{1}{10,000} \right)^{10,000};$ which amounts to £$$271$$$$16$$s$$3\frac{1}{2}$$d.

Finally, it will be seen that what we are trying to find is in reality the ultimate value of the expression $$\left(1 + \dfrac{1}{n}\right)^n$$, 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 $2.71828\ldots$ a number never to be forgotten.

Let us take geometrical illustrations of these things. In Fig. 36, $$OP$$ 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 $$\dfrac{dy}{dx}$$ 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 $$\dfrac{1}{n}$$ 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 $$2$$.

In Fig. 37, we have the corresponding illustration of the geometrical progression. Each of the successive ordinates is to be $$1 + \dfrac{1}{n}$$, that is, $$\dfrac{n+1}{n}$$ times as high as its predecessor. The steps up are not equal, because each step up is now $$\dfrac{1}{n}$$ of the ordinate at that part of the curve. If we had literally $$10$$ steps, with $$\left(1 + \frac{1}{10} \right)$$ for the multiplying factor, the final total would be $$(1 + \tfrac{1}{10})^{10}$$ or $${2.594}$$ times the original $$1$$. But if only we take $$n$$ sufficiently large (and the corresponding $$\dfrac{1}{n}$$ sufficiently small), then the final value $$\left(1 + \dfrac{1}{n}\right)^n$$ to which unity will grow will be $$2.71828$$.

The number $e$.[1]

To this mysterious number (2.7182818) etc., the mathematicians have assigned as a symbol the letter $e$ (called Euler’s number). All schoolboys know that the Greek letter $$\pi$$ (called pi) stands for $$3.141592$$ etc.; but how many of them know that e means $$2.71828$$? Yet it is an even more important number than $$\pi$$!

What, then, is the number e?

Suppose we were to let $$1$$ grow at simple interest till it became $$2$$; then, if at the same nominal rate of interest, and for the same time, we were to let $$1$$ grow at true compound interest, instead of simple, it would grow to the value epsilon.

This process of growing proportionately, at every instant, to the magnitude at that instant, some people call a logarithmic rate of growing. Unit logarithmic rate of growth is that rate which in unit time will cause $$1$$ to grow to $$2.718281$$. It might also be called the organic rate of growing: because it is characteristic of organic growth (in certain circumstances) that the increment of the organism in a given time is proportional to the magnitude of the organism itself.

If we take $$100$$ per cent. as the unit of rate, and any fixed period as the unit of time, then the result of letting $$1$$ grow arithmetically at unit rate, for unit time, will be $$2$$, while the result of letting $$1$$ grow logarithmically at unit rate, for the same time, will be $$2.71828\ldots$$ .

A little bit more about e.

We have seen that we require to know what value is reached by the expression $$\left(1 + \dfrac{1}{n}\right)^n$$, when $$n$$ becomes indefinitely great. Arithmetically, here are tabulated a lot of values (which anybody can calculate out by the help of an ordinary table of logarithms) got by assuming $$n = 2$$; $$n = 5$$; $$n = 10$$; and so on, up to $$n = 10,000$$. \begin{aligned} {2} &(1 + \tfrac{1}{2})^2 &&= 2.25. \\ &(1 + \tfrac{1}{5})^5 &&= {2.488}. \\ &(1 + \tfrac{1}{10})^{10} &&= 2.594. \\ &(1 + \tfrac{1}{20})^{20} &&= 2.653. \\ &(1 + \tfrac{1}{100})^{100} &&= {2.705}. \\ &(1 + \tfrac{1}{1000})^{1000} &&= {2.7169}. \\ &(1 + \tfrac{1}{10,000})^{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 $$\left(1 + \dfrac{1}{n}\right)^n$$ 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!} \\ & \phantom{= a^n\ } + n(n -1)(n – 2) \dfrac{a^{n-3} b^3}{3!} + \text{etc}. \end{aligned} Putting $a = 1$ and $b = \dfrac{1}{n}$, 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} \\ &\phantom{= 1 + 1\ } + \dfrac{1}{4!} \dfrac{(n – 1)(n – 2)(n – 3)}{n^3} + \text{etc}.\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 $e = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \text{etc}.\ldots$

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:

 $$1.000000$$ dividing by $$1$$ $$1.000000$$ dividing by $$2$$ $$0.500000$$ dividing by $$3$$ $$0.166667$$ dividing by $$4$$ $$0.041667$$ dividing by $$5$$ $$0.008333$$ dividing by $$6$$ $$0.001389$$ dividing by $$7$$ $$0.000198$$ dividing by $$8$$ $$0.000025$$ dividing by $$9$$ $$0.000002$$ Total $$2.718281$$

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

The Exponential Series

We shall have need of yet another series.

Let us, again making use of the binomial theorem, expand the expression $$\left(1 + \dfrac{1}{n}\right)^{nx}$$, which is the same as $$e^x$$ when we make $$n$$ indefinitely great. \begin{aligned} e^x &= 1^{nx} + nx \frac{1^{nx-1} \left(\dfrac{1}{n}\right)}{1!} + nx(nx – 1) \frac{1^{nx – 2} \left(\dfrac{1}{n}\right)^2}{2!} \\ & \phantom{= 1^{nx}\ } + nx(nx – 1)(nx – 2) \frac{1^{nx-3} \left(\dfrac{1}{n}\right)^3}{3!} + \text{etc}.\\ &= 1 + x + \frac{1}{2!} \cdot \frac{n^2x^2 – nx}{n^2} + \frac{1}{3!} \cdot \frac{n^3x^3 – 3n^2x^2 + 2nx}{n^3} + \text{etc}. \\ &= 1 + x + \frac{x^2 -\dfrac{x}{n}}{2!} + \frac{x^3 – \dfrac{3x^2}{n} + \dfrac{2x}{n^2}}{3!} + \text{etc}.\end{aligned}

But, when $$n$$ is made indefinitely great, this simplifies down to the following: $\bbox[#F2F2F2,5px,border:2px solid black]{e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \text{etc.}\dots}$

This series is called the exponential series.

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 differential coefficient 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} \\ &\phantom{= 0 + 1 + \frac{2x}{1 \cdot 2} + \frac{3x^2}{1 \cdot 2 \cdot 3}\ } + \frac{5x^4}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} + \text{etc}. \\ &= 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} + \text{etc}., \end{aligned}

which is exactly the same as the original series.

$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx} e^x=e^x}$

Now we might have gone to work the other way, and said: Go to; let us find a function of $$x$$, such that its differential coefficient 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 \begin{aligned} y &= A + Bx + Cx^2 + Dx^3 + Ex^4 + \text{etc}.,\end{aligned} (in which the coefficients $A$, $B$, $C$, etc. will have to be determined), and differentiate it.. \begin{aligned}\dfrac{dy}{dx} &= B + 2Cx + 3Dx^2 + 4Ex^3 + \text{etc}.\end{aligned}

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 $$C=\dfrac{B}{2}=\dfrac{A}{1\cdot 2}$$; that $$D = \dfrac{C}{3} = \dfrac{A}{1 \cdot 2 \cdot 3}$$; that $$E = \dfrac{D}{4} = \dfrac{A}{1 \cdot 2 \cdot 3 \cdot 4}$$, etc.

The law of change is therefore that $y = A\left(1 + \dfrac{x}{1} + \dfrac{x^2}{1 \cdot 2} + \dfrac{x^3}{1 \cdot 2 \cdot 3} + \dfrac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} + \text{etc}.\right).$

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

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 \text{ etc.}; & \text{that is, } y &= e; \\ \text{when } x &= 2,\quad & y &=(2.718281 \text{ etc.})^2; & \text{that is, } y &= e^2; \\ \text{when } x &= 3,\quad & y &=(2.718281 \text{ etc.})^3; & \text{that is, } y &= e^3;\end{aligned} and therefore $\text{when } x=x,\quad y=(2.718281 \text{ etc}.)^x;\quad\text{that is, } y=e^x,$ thus finally demonstrating that $e^x = 1 + \dfrac{x}{1} + \dfrac{x^2}{1\cdot 2} + \dfrac{x^3}{1 \cdot 2 \cdot 3} + \dfrac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} + \text{etc}.$

Of course it follows that $$e^y$$ remains unchanged if differentiated with respect to $$y$$. Also $$e^{ax}$$, which is equal to $$(e^a)^x$$, will, when differentiated with respect to $$x$$, be $$ae^{ax}$$, because $$a$$ is a constant.

Natural or Naperian Logarithms

Another reason why $$e$$ is important is because it was made by Napier, the inventor of logarithms, the basis of his system. If $$y$$ is the value of $$e^x$$, then $$x$$ is the logarithm, to the base $$e$$, of $$y$$. Or, if

$y=e^x,$

then

$x=\log_e y$

The two curves plotted in Figs. 38 and 39  represent these equations.

The points calculated are:

For Fig. 38  \begin{array}{|c||*{5}{c|}} \hline {x} & {0} & {0.5} & {1} & {1.5} & {2} \\ \hline {y} & {1} & {1.65} & {2.71} & {4.50} & {7.39} \\ \hline \end{array} For Fig. 39  \begin{array}{|c||*{5}{c|}} \hline {y} & {1} & {2} & {3} & {4} & {8} \\ \hline {x} & {0} & {0.69} & {1.10} & {1.39} & {2.08} \\ \hline \end{array}

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.