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The Pamphlet Collection of Sir Robert Stout: Volume 8

Appendix

page 31

Appendix.

To construct a table of the logarithms of the present value of £1 due at the end of any number of years.

The present value of £1 due a year hence is Mathematical equation, the logarithm of which is the arithmetical complement of log (1+i). When the rate is 3 per cent, log (l+i)=log 1.03=0.01283,72247, the arithmetical complement of which is 1.98716,27753. By taking the first nine multiples of this logarithm may be constructed a table of the logarithms of the present value of £1 due at the end of any number of years from 1 to 100 which will be true to the last figure to six places of decimals, thus:

Logarithm of the present value of £1.

Years.

11.987102 775
2.974325 550
3.961488 325
4.948651 100
5.935813 875

* *

To construct columns D and N.

Since Dxlxvx;

therefore log Dx=log lx+log vx.

Log Dx is formed in reverse order to facilitate the formation of column N, thus: Mathematical equation

page 32

To construct columns C and M.

Since Cx=dxvx+1;

Mathematical equation

To construct columns C′and M′.

Since C′x=s′xwDx, where s′x denotes the average number of weeks' sickness to each person in the year following the age x, and w the present value of £1 due half a year hence;

therefore log C′x=log s′x+log w+log Dx.

The present value of £1 due half a year hence at 3 per cent. is Mathematical equation the logarithm of which is the arithmetical complement of log 1.015. Log 1.015=0.00646,60422, the arithmetical complement of which is 1.99353,39578. Writing this logarithm to six places of decimals at the bottom of a card to be added to the other two logarithms at each age, the formation is as follows: Mathematical equation