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

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

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

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

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