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

$y_{n}=y_{0}\left(1+{\frac {1}{n}}\right)^{n}$.

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

$(1+{\tfrac {1}{20}})^{20}=2.653$.

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