On true Compound Interest and the Law of Organic Growth

$\epsilon$ 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 $\epsilon^x$ when we make $n$ indefinitely great. \begin{align*} \epsilon^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!} · \frac{n^2x^2 - nx}{n^2} + \frac{1}{3!} · \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{align*}

But, when $n$ is made indefinitely great, this simplifies down to the following: \[ \epsilon^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \text{etc.}\dots \]

This series is called the exponential series.

For Fig. 39:

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 $\epsilon$, 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 \[ \log_\epsilon a + \log_\epsilon b = \log_\epsilon ab. \] Also the rule of powers holds good; \[ n × \log_\epsilon a = \log_\epsilon a^n. \] 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 logarithm simply by multiplying it by $0.4343$; or \begin{align*} \log_{10} x &= 0.4343 × \log_{\epsilon} x, \\ \text{ and conversely,}\; \log_{\epsilon} x &= 2.3026 × \log_{10} x. \end{align*}

A Useful Table of “Naperian Logarithms”

(Also called Natural Logarithms or Hyperbolic Logarithms)

Exponential and Logarithmic Equations.

First change to natural logarithms by multiplying by the modulus $0.4343$. This gives us \begin{align*} y &= 0.4343 \log_\epsilon x; \\ \text{whence}\; \frac{dy}{dx} &= \frac{0.4343}{x}. \end{align*}

Let us now attempt further examples.

Examples (1) $y=\epsilon^{-ax}$. Let $-ax=z$; then $y=\epsilon^z$. \[ \frac{dy}{dz} = \epsilon^z;\quad \frac{dz}{dx} = -a;\quad\text{hence}\quad \frac{dy}{dx} = -a\epsilon^{-ax}. \]

Or thus: \[ \log_\epsilon y = -ax;\quad \frac{1}{y}\, \frac{dy}{dx} = -a;\quad \frac{dy}{dx} = -ay = -a\epsilon^{-ax}. \]

(2) $y=\epsilon^{\frac{x^2}{3}}$. Let $\dfrac{x^2}{3}=z$; then $y=\epsilon^z$. \[ \frac{dy}{dz} = \epsilon^z;\quad \frac{dz}{dx} = \frac{2x}{3};\quad \frac{dy}{dx} = \frac{2x}{3}\, \epsilon^{\frac{x^2}{3}}. \]

Or thus: \[ \log_\epsilon y = \frac{x^2}{3};\quad \frac{1}{y}\, \frac{dy}{dx} = \frac{2x}{3};\quad \frac{dy}{dx} = \frac{2x}{3}\, \epsilon^{\frac{x^2}{3}}. \]

(3) $y = \epsilon^{\frac{2x}{x+1}}$. \begin{align*} \log_\epsilon y &= \frac{2x}{x+1},\quad \frac{1}{y}\, \frac{dy}{dx} = \frac{2(x+1)-2x}{(x+1)^2}; \\ hence \frac{dy}{dx} &= \frac{2}{(x+1)^2} \epsilon^{\frac{2x}{x+1}}. \end{align*}

Check by writing $\dfrac{2x}{x+1}=z$.

(4) $y=\epsilon^{\sqrt{x^2+a}}$. $\log_\epsilon y=(x^2+a)^{\frac{1}{2}}$. \[ \frac{1}{y}\, \frac{dy}{dx} = \frac{x}{(x^2+a)^{\frac{1}{2}}}\quad\text{and}\quad \frac{dy}{dx} = \frac{x × \epsilon^{\sqrt{x^2+a}}}{(x^2+a)^{\frac{1}{2}}}. \] For if $(x^2+a)^{\frac{1}{2}}=u$ and $x^2+a=v$, $u=v^{\frac{1}{2}}$, \[ \frac{du}{dv} = \frac{1}{{2v}^{\frac{1}{2}}};\quad \frac{dv}{dx} = 2x;\quad \frac{du}{dx} = \frac{x}{(x^2+a)^{\frac{1}{2}}}. \]

Check by writing $\sqrt{x^2+a}=z$.

(5) $y=\log(a+x^3)$. Let $(a+x^3)=z$; then $y=\log_\epsilon z$. \[ \frac{dy}{dz} = \frac{1}{z};\quad \frac{dz}{dx} = 3x^2;\quad\text{hence}\quad \frac{dy}{dx} = \frac{3x^2}{a+x^3}. \]

(6) $y=\log_\epsilon\{{3x^2+\sqrt{a+x^2}}\}$. Let $3x^2 + \sqrt{a+x^2}=z$; then $y=\log_\epsilon z$. \begin{align*} \frac{dy}{dz} &= \frac{1}{z};\quad \frac{dz}{dx} = 6x + \frac{x}{\sqrt{x^2+a}}; \\ \frac{dy}{dx} &= \frac{6x + \dfrac{x}{\sqrt{x^2+a}}}{3x^2 + \sqrt{a+x^2}} = \frac{x(1 + 6\sqrt{x^2+a})}{(3x^2 + \sqrt{x^2+a}) \sqrt{x^2+a}}. \end{align*}

(7) $y=(x+3)^2 \sqrt{x-2}$. \begin{align*} \log_\epsilon y &= 2 \log_\epsilon(x+3)+ \tfrac{1}{2} \log_\epsilon(x-2). \\ \frac{1}{y}\, \frac{dy}{dx} &= \frac{2}{(x+3)} + \frac{1}{2(x-2)}; \\ \frac{dy}{dx} &= (x+3)^2 \sqrt{x-2} \left\{\frac{2}{x+3} + \frac{1}{2(x-2)}\right\}. \end{align*}

(8) $y=(x^2+3)^3(x^3-2)^{\frac{2}{3}}$. \begin{align*} \log_\epsilon y &= 3 \log_\epsilon(x^2+3) + \tfrac{2}{3} \log_\epsilon(x^3-2); \\ \frac{1}{y}\, \frac{dy}{dx} &= 3 \frac{2x}{(x^2+3)} + \frac{2}{3} \frac{3x^2}{x^3-2} = \frac{6x}{x^2+3} + \frac{2x^2}{x^3-2}. \end{align*} For if $y=\log_\epsilon(x^2+3)$, let $x^2+3=z$ and $u=\log_\epsilon z$. \[ \frac{du}{dz} = \frac{1}{z};\quad \frac{dz}{dx} = 2x;\quad \frac{du}{dx} = \frac{2x}{x^2+3}. \] Similarly, if $v=\log_\epsilon(x^3-2)$, $\dfrac{dv}{dx} = \dfrac{3x^2}{x^3-2}$ and \[ \frac{dy}{dx} = (x^2+3)^3(x^3-2)^{\frac{2}{3}} \left\{ \frac{6x}{x^2+3} + \frac{2x^2}{x^3-2} \right\}. \]

(9) $y=\dfrac{\sqrt[2]{x^2+a}}{\sqrt[3]{x^3-a}}$. \begin{align*} \log_\epsilon y &= \frac{1}{2} \log_\epsilon(x^2+a) - \frac{1}{3} \log_\epsilon(x^3-a). \\ \frac{1}{y}\, \frac{dy}{dx} &= \frac{1}{2}\, \frac{2x}{x^2+a} - \frac{1}{3}\, \frac{3x^2}{x^3-a} = \frac{x}{x^2+a} - \frac{x^2}{x^3-a} \\ and \frac{dy}{dx} &= \frac{\sqrt[2]{x^2+a}}{\sqrt[3]{x^3-a}} \left\{ \frac{x}{x^2+a} - \frac{x^2}{x^3-a} \right\}. \end{align*}

(10) $y=\dfrac{1}{\log_\epsilon x}$ \[ \frac{dy}{dx} = \frac{\log_\epsilon x × 0 - 1 × \dfrac{1}{x}} {\log_\epsilon^2 x} = -\frac{1}{x \log_\epsilon^2x}. \]

(11) $y=\sqrt[3]{\log_\epsilon x} = (\log_\epsilon x)^{\frac{1}{3}}$. Let $z=\log_\epsilon x$; $y=z^{\frac{1}{3}}$. \[ \frac{dy}{dz} = \frac{1}{3} z^{-\frac{2}{3}};\quad \frac{dz}{dx} = \frac{1}{x};\quad \frac{dy}{dx} = \frac{1}{3x \sqrt[3]{\log_\epsilon^2 x}}. \]

(12) $y=\left(\dfrac{1}{a^x}\right)^{ax}$. \begin{align*} \log y &= -ax \log a^{x} = -ax^{2} \cdot \log a.\\ \frac{1}{y} \frac{dy}{dx} &= -2ax \cdot \log a\\ \frac{dy}{dx} &= -2ax\left(\frac{1}{a^{x}}\right)^{ax} \cdot \log a = -2x a^{1-ax^{2}} \cdot \log a. \end{align*}

Try now the following exercises.

Exercises XII

(1) Differentiate $y=b(\epsilon^{ax} -\epsilon^{-ax})$.

(2) Find the differential coefficient with respect to $t$ of the expression $u=at^2+2\log_\epsilon t$.

(3) If $y=n^t$, find $\dfrac{d(\log_\epsilon y)}{dt}$.

(4) Show that if $y=\dfrac{1}{b}·\dfrac{a^{bx}}{\log_\epsilon a}$,   $\dfrac{dy}{dx}=a^{bx}$.

(5) If $w=pv^n$, find $\dfrac{dw}{dv}$.

Differentiate

(6) $y=\log_\epsilon x^n$.

(7) $y=3\epsilon^{-\frac{x}{x-1}}$.

(8) $y=(3x^2+1)\epsilon^{-5x}$.

(9) $y=\log_\epsilon(x^a+a)$.

(10) $y=(3x^2-1)(\sqrt{x}+1)$.

(11) $y=\dfrac{\log_\epsilon(x+3)}{x+3}$.

(12) $y=a^x × x^a$.

(13) It was shown by Lord Kelvin that the speed of signalling through a submarine cable depends on the value of the ratio of the external diameter of the core to the diameter of the enclosed copper wire. If this ratio is called $y$, then the number of signals $s$ that can be sent per minute can be expressed by the formula \[ s=ay^2 \log_\epsilon \frac{1}{y}; \] where $a$ is a constant depending on the length and the quality of the materials. Show that if these are given, $s$ will be a maximum if $y=1 ÷ \sqrt{\epsilon}$.

(14) Find the maximum or minimum of \[ y=x^3-\log_\epsilon x. \]

(15) Differentiate $y=\log_\epsilon(ax\epsilon^x)$.

(16) Differentiate $y=(\log_\epsilon ax)^3$.

Answers

(1) $ab(\epsilon^{ax} + \epsilon^{-ax})$.

(2) $2at + \dfrac{2}{t}$.

(3) $\log_\epsilon n$.

(5) $npv^{n-1}$.

(6) $\dfrac{n}{x}$.

(7) $\dfrac{3\epsilon^{- \frac{x}{x-1}}}{(x - 1)^2}$.

(8) $6x \epsilon^{-5x} - 5(3x^2 + 1)\epsilon^{-5x}$.

(9) $\dfrac{ax^{a-1}}{x^a + a}$.

(10) $\left(\dfrac{6x}{3x^2-1} + \dfrac{1}{2\left(\sqrt x + x\right)}\right) \left(3x^2-1\right)\left(\sqrt x + 1\right)$.

(11) $\dfrac{1 - \log_\epsilon \left(x + 3\right)}{\left(x + 3\right)^2}$.

(12) $a^x\left(ax^{a-1} + x^a \log_\epsilon a\right)$.

(14) Min.: $y = 0.7$ for $x = 0.694$.

(15) $\dfrac{1 + x}{x}$.

(16) $\dfrac{3}{x} (\log_\epsilon ax)^2$.

The Logarithmic Curve.

Let us return to the curve which has its successive ordinates in geometrical progression, such as that represented by the equation $y=bp^x$.

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

Then when \[ x=1,\quad y=bp;\qquad x=2,\quad y=bp^2;\qquad x=3,\quad y=bp^3,\quad \text{etc.} \]