■ 
23 
he number of years, 
logarithms of £l in 
1 momently, for dif- 
ble yearly is 1.035, 
’s Logs., Table 3); 
7 V _ /407 Y 
400/ “V40oj 
•0150688358; 
7 V /807V 
00 j“ V8007 
= .0151341908. 
035 X *4342944819 
ber corresponding to 
Amount 
of i'l in 
one year. 
1.050000 
1.050025 
1.050946 
1.051271 
Logarithms of 
such amount. 
.0211892991 
.0214477308 
.0215801275 
.0217147241 
,060000 .0253058653 
,060900 .0256744494 
.061364 
1.061837 
1.070000 
1.071225 
.071859 
1.072508 
1.080000 1 
1.081600 
1.082432 
1.083287 
1.090000 
1.092025 
1.093083 
1.094175 
.0258641690 
.0260576689 
0293837777 
.0298806996 
.0301376716 
.0304006137 
.0334237555 
.0340666786 
.0344006870 
.0347435586 
.0374264979 
.0382325809 
.0386532667 
.0390865034 
1.100000 1 
1.102500 
1.103813 
1.105171 
.0413926852 
.0423785981 
.0428954616 
.0434294482 
ON THE PRESENT VALUE OF SUMS AT COMPOUND INTEREST. 
35. When money is reckoned at compound interest, the sum to be 
given in lieu of a payment at a future period, is that which laid out at 
interest until the sum is due, would just provide for the payment thereof. 
The method of finding the present value is therefore the reverse of find 
ing the amount. By Art. 19, we have the proportion as £l is to its 
amount in one year, so is any other sum to its amount in a year, which 
proportion is also true when inverted. As the amount of £l in a year 
is to the £l which produced it, so is the amount of any other sum in a 
year to the sum which produced that amount. 
Make p “ the present value, 
s = the sum due, 
n == the number of years, 
i “ the interest of ¿£l for one year. 
* = *(1 + ¿)- 1 — (present value of £l due 
\ at the end of one year, 
(1 + O -2 second year, 
(1 + i)~ s third year. 
d+0 : i :: i 
(i + 0 
(i + 0 
1 Z 
i :: (i + 0~ l : 
!::(]+ O'*: 
Generally, (1 + i) ” = present value of £l due at the end of n years, 
which multiplied by s, will give 
p s (1 + *)- " = 
(1 + it: 
By logarithms, 
log. p = — n log (1 + 0 + log s - 
Rule. Find the amount of £l in the given time, and by it divide 
the sum due. 
Example. What is the present value of ¿£350 due at the end of 
10 years, 5 per cent compound interest ? 
s = 350 n “ 10 
By Table 3, 
(1.05) 10 s= 1.628894)350 
3257788 
' 242212 
162889 
p “ .05 
(214.870 = ¿£214 17 5 
79323 
65156 
14167 
13030 
1137 
* In Algebra x - 1 means ^, x ~ 2 means &c.—{Arith. # Alg. Art, 216. ef seq.)