On true Compound Interest and the Law of Organic Growth

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.

* 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 \[ y + n\dfrac{y}{n} = 2y. \]

Let us take geometrical illustrations of these things. In Figure 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 Figure 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 $2.71828$.

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

The binomial theorem gives the rule that \begin{align*} (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{align*} Putting $a = 1$ and $b = \dfrac{1}{n}$, we get \begin{align*} \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{align*}

$1.000000$ | |

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 | $2.718281$ |

$\epsilon$ 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*.

The great reason why $\epsilon$ is regarded of importance
is that $\epsilon^x$ 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{align*}
\frac{d(\epsilon^x)}{dx}
&= 0 + 1 + \frac{2x}{1 · 2} + \frac{3x^2}{1 · 2 · 3} + \frac{4x^3}{1 · 2 · 3 · 4} \\
&\phantom{= 0 + 1 + \frac{2x}{1 · 2} + \frac{3x^2}{1 · 2 · 3}\ } + \frac{5x^4}{1 · 2 · 3 · 4 · 5} + \text{etc}. \\
or
&= 1 + x + \frac{x^2}{1 · 2} + \frac{x^3}{1 · 2 · 3} + \frac{x^4}{1 · 2 · 3 · 4} + \text{etc}.,
\end{align*}
which is exactly the same as the original series.

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

*Natural or Naperian Logarithms.*

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

The points calculated are:

For Fig. 38:

$x$ | $0$ | $0.5$ | $1$ | $1.5$ | $2$ |

$y$ | $1$ | $1.65$ | $2.71$ | $4.50$ | $7.39$ |

For Fig. 39:

xy$ | $1$ | $2$ | $3$ | $4$ | $8$ |

xx$ | $0$ | $0.69$ | $1.10$ | $1.39$ | $2.08$ |

*(Also called Natural Logarithms or Hyperbolic Logarithms)*

Number | $\log_{\epsilon}$ | Number | $\log_{\epsilon}$ | |
---|---|---|---|---|

$1 $ | $0.0000$ | $6$ | $1.7918$ | |

$1.1$ | $0.0953$ | $7$ | $1.9459$ | |

$1.2$ | $0.1823$ | $8$ | $2.0794$ | |

$1.5$ | $0.4055$ | $9$ | $2.1972$ | |

$1.7$ | $0.5306$ | $10$ | $2.3026$ | |

$2.0$ | $0.6931$ | $20$ | $2.9957$ | |

$2.2$ | $0.7885$ | $50$ | $3.9120$ | |

$2.5$ | $0.9163$ | $100$ | $4.6052$ | |

$2.7$ | $0.9933$ | $200$ | $5.2983$ | |

$2.8$ | $1.0296$ | $500$ | $6.2146$ | |

$3.0$ | $1.0986$ | $1000$ | $6.9078$ | |

$3.5$ | $1.2528$ | $2000$ | $7.6009$ | |

$4.0$ | $1.3863$ | $5000$ | $8.5172$ | |

$4.5$ | $1.5041$ | $10 000$ | $9.2103$ | |

$5.0$ | $1.6094$ | $20 000$ | $9.9035$ |

*Exponential and Logarithmic Equations.*

Now let us try our hands at differentiating certain expressions that contain logarithms or exponentials.

Take the equation: \[ y = \log_\epsilon x. \] First transform this into \[ \epsilon^y = x, \] whence, since the differential of $\epsilon^y$ with regard to $y$ is the original function unchanged (see here), \[ \frac{dx}{dy} = \epsilon^y, \] and, reverting from the inverse to the original function, \[ \frac{dy}{dx} = \frac{1}{\ \dfrac{dx}{dy}\ } = \frac{1}{\epsilon^y} = \frac{1}{x}. \]

Now this is a very curious result. It may be written \[ \frac{d(\log_\epsilon x)}{dx} = x^{-1}. \]

Note that $x^{-1}$ is a result that we could never have got by the 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 $0 × x^{-1}$, because $x^0$ is itself $= 1$, and is a constant. We shall have to come back to this curious fact that differentiating $\log_\epsilon x$ gives us $\dfrac{1}{x}$ when we reach the chapter on integrating.

Now, try to differentiate \begin{align*} y &= \log_\epsilon(x+a),\\ \text{that is}\; \epsilon^y &= x+a; \end{align*} we have $\dfrac{d(x+a)}{dy} = \epsilon^y$, since the differential of $\epsilon^y$ remains $\epsilon^y$. This gives \begin{align*} \frac{dx}{dy} &= \epsilon^y = x+a; \\ \end{align*} hence, reverting to the original function, we get \begin{align*} \frac{dy}{dx} &= \frac{1}{\;\dfrac{dx}{dy}\;} = \frac{1}{x+a}. \end{align*}

Next try \begin{align*} y &= \log_{10} x. \end{align*}

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*}

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

Let us now attempt further examples.

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

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_\epsilon y &= ax(\log_\epsilon 1 - \log_\epsilon a^x) = -ax \log_\epsilon a^x. \\ \frac{1}{y}\, \frac{dy}{dx} &= -ax × a^x \log_\epsilon a - a \log_\epsilon a^x. \\ and \frac{dy}{dx} &= -\left(\frac{1}{a^x}\right)^{ax} (x × a^{x+1} \log_\epsilon a + a \log_\epsilon a^x). \end{align*}

Try now the following exercises.

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

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

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

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 Figure 40, 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, Figure 41, with values of $\log_\epsilon y$ as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows from the equation, that \begin{align*} \log_\epsilon y &= \log_\epsilon b + x · \log_\epsilon p, \\ \text{whence }\; \log_\epsilon y &- \log_\epsilon b = x · \log_\epsilon p. \end{align*}

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

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