Page:Calculus Made Easy.pdf/156

 will have grown to £$$259$$. $$7$$s. $$6$$d. In fact, we see that at the end of each year, each pound will have earned 110 of a pound, and therefore, if this is always added on, each year multiplies the capital by $$\tfrac {11}{10}$$; and if continued for ten years (which will multiply by this factor ten times over) will multiply the original capital by $$2.59374$$. Let us put this into symbols. Put $$y_0$$ for the original capital; $$\frac{1}{n}$$ for the fraction added on at each of the $$n$$ operations; and $$y_n$$ for the value of the capital at the end of the $$n^{th}$$ operation. Then

But this mode of reckoning compound interest once a year, is really not quite fair; for even during the first year the £$$100$$ ought to have been growing. At the end of half a year it ought to have been at least £$$105$$, and it certainly would have been fairer had the interest for the second half of the year been calculated on £$$105$$. This would be equivalent to calling it $${5}{\%}$$ per half-year; with $$20$$ operations, therefore, at each of which the capital is multiplied by $$\tfrac{21}{20}$$. If reckoned this way, by the end of ten years the capital would have grown to £$$265$$. $$8$$s.; for

But, even so, the process is still not quite fair; for, by the end of the first month, there will be some interest earned; and a half-yearly reckoning assumes that the capital remains stationary for six months at