Page:Calculus Made Easy.pdf/157

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 £$270$. $8$s.

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

$y_{n}=$£$100\left(1+{\tfrac {1}{1000}}\right)^{1000}$;

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

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

$y_{n}=$£$100\left(1+{\tfrac {1}{10,000}}\right)^{10,000}$;

which amounts to £$271$. $16$s. $4$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