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 PART II.

109. We now proceed to treat the subject of Annuities Algebraically.

110. Then, reasoning as in the first number of this article, it will be found that $$ m = s (1 + r)^n$$. And by No. 2. it appears, that the present value of $$s$$ pounds to be received certainly at the expiration of $$n$$ years, is $$s \frac{1}{(1+r)^n}$$, or $$s(1+r)^{-n}$$.

111. The amount of L. 1 in $$n$$ years being $$(1+r)^n$$, its increase in that time will be $$(1+r)^n-1$$, and when it is considered that this increase arises entirely from the simple interest ($$r$$) of L. 1 being laid up at the end of each year, and improved at compound interest during the remainder of the term; it must be obvious that $$(1+r)^n-1$$ is the amount of an annuity of $$r$$ pounds in that time, but $$r : a :: (1+r)^n-1 : \frac{a}{r} \left [ (1+r)^n-1 \right ]$$, which, therefore, is equal to $$\text{M}$$, the amount of an annuity of $$a$$ pounds in $$n$$ years.

112. Reasoning as in No. 8. it will be found, that since $$r : 1 :: a : \frac{a}{r}$$, the present value of a perpetual annuity of $$a$$ pounds is $$\frac{a}{r}$$.

113. If two persons, A and B, purchase a perpetuity of $$a$$ pounds between them, which A and his heirs or assigns are to enjoy during the first $$n$$ years, and B, and his heirs and assigns, for ever after. Since the value of the perpetuity to be entered upon immediately, has just been shown to be $$\frac{a}{r}$$, the present value of B’s share, that is, the present value of the same perpetuity when the entrance thereon is deferred until the expiration of $$n$$ years, will be $$\frac{a}{r}(r+1)^{-n}$$, (110); and the value of the share of A will be thus much less than that of the whole perpetuity (21), and therefore equal to $$\frac{a}{r} \left [ 1-(1+r)^{-n} \right ] = \text{V}$$, the present value of an annuity of $$a$$ pounds for the term of $$n$$ years certain.

114. If the annuity is not to be entered upon until the expiration of $$d$$ years, but is then to continue $$n$$ years, its value at the time of entering upon it will be $$\frac{a}{r} \left [ 1-(1+r)^{-n} \right ]$$, as has just been shown; therefore its present value will be $$\frac{a}{r} \left [ (1+r)^{-d} - (1+r)^{-(d+n)} \right ] = \text{V}$$, (110.)

115. In the same manner, it appears that, when the entrance on a perpetuity of $$a$$ pounds is deferred $$d$$ years, its present value will be $$\frac{a}{r} (1+r)^{-d}$$ (110 and 112.)

116. $$q$$ being any number whatever, whole, fractional, or mixed, let $$\lambda q$$ denote its logarithm, and $$\varkappa q$$ the arithmetical complement of that logarithm; so that these equations may obtain, $$\lambda \frac{1}{q} = -\lambda q = \varkappa q$$.

Then, for the resolution of the principal questions of this kind by logarithms, we shall have these formulæ.

By means of each of these equations, it is manifest that any one of the quantities involved in it may be found, when the rest are given.

117. If the interest be convertible into principal $$\nu$$ times in the year, at $$\nu$$ equal intervals, since the interest of L. 1 for one of these intervals will be $$\frac{r}{\nu}$$, (109), and the number of conversions of interest into principal in $$n$$ years $$\nu n$$; to adapt the formula in No. 110. to this case, we have only to substitute $$\frac{r}{\nu}$$ for $$r$$, and $$\nu n$$ for $$n$$, in the equation $$m = s(1 + r)^n$$ there given, whereby it will be transformed to this, $$m = s\left ( 1 + \frac{r}{\nu} \right) ^ {\nu n}$$.

118. According as $$\nu$$ is equal to 1, 2, 4, or is infinite; that is, according as the interest is convertible into principal yearly, half-yearly, quarterly, or continually, let $$m$$ be equal to $$\text{Y}$$, $$\text{H}$$, $$\text{Q}$$, or $$\text{C}$$; so shall Rh