Page:Calculus Made Easy.pdf/157

a time. Suppose we divided the year into ${\displaystyle 10}$ parts, and reckon a one-per-cent. interest for each tenth of the year. We now have ${\displaystyle 100}$ operations lasting over the ten years; or

which works out to £${\displaystyle 270}$. ${\displaystyle 8}$s.

Even this is not final. Let the ten years be divided into ${\displaystyle 1000}$ periods, each of ${\displaystyle {\tfrac {1}{100}}}$ of a year; the interest being ${\displaystyle {\tfrac {1}{10}}}$ per cent. for each such period; then

which works out to £${\displaystyle 271}$. ${\displaystyle 14}$s. ${\displaystyle 2{\tfrac {1}{2}}}$d.

Go even more minutely, and divide the ten years into ${\displaystyle 10,000}$ parts, each ${\displaystyle {\tfrac {1}{1000}}}$ of a year, with interest at ${\displaystyle {\tfrac {1}{100}}}$ of ${\displaystyle 1}$ per cent. Then

which amounts to £${\displaystyle 271}$. ${\displaystyle 16}$s. ${\displaystyle 4}$d.

Finally, it will be seen that what we are trying to find is in reality the ultimate value of the expression ${\displaystyle \left(1+{\dfrac {1}{n}}\right)^{n}}$, which, as we see, is greater than ${\displaystyle 2}$; and which, as we take ${\displaystyle n}$ larger and larger, grows closer and closer to a particular limiting value. However big you make ${\displaystyle n}$, the value of this expression grows nearer and nearer to the figure

a number never to be forgotten.

Let us take geometrical illustrations of these things. In Fig. 36, ${\displaystyle OP}$ stands for the original value. ${\displaystyle OT}$ is