224. Except when x is very small the series for log, (1 + x) is of little use for numerical calculations. We can, however, deduce from it other series by the aid of which Tables of Logarithms may be constructed. 1 + :) .(3). + From this formula by putting n 3 we obtain log, 4 – log, 2, that is log, 2; and by effecting the calculation we find that the value of log, 269314718...; whence log,8 is known. Again by putting n = 9 we obtain log, 10-log, 8; whence we find log, 10 = 2.30258509.... To convert Napierian logarithms into logarithms to base 10 1 we multiply by which is the modulus [Art. 216] of the loge 10' common system, and its value is we shall denote this modulus by μ. In the Proceedings of the Royal Society of London, Vol. XXVII. page 88, Professor J. C. Adams has given the values of e, μ, log, 2, log, 3, log, 5 to more than 260 places of decimals. 225. If we multiply the above series throughout by μ, we obtain formulæ adapted to the calculation of common logarithms. Thus from (1), μlog.(n + 1) - μ log,n =! μ + 2n2 3n3 -- ... ; From either of the above results we see that if the logarithm of one of two consecutive numbers be known, the logarithm of the other may be found, and thus a table of logarithms can be constructed. It should be remarked that the above formulæ are only needed to calculate the logarithms of prime numbers, for the logarithm of a composite number may be obtained by adding together the logarithms of its component factors. In order to calculate the logarithm of any one of the smaller prime numbers, we do not usually substitute the number in either of the formulæ (1) or (2), but we endeavour to find some value of n by which division may be easily performed, and such that either n + 1 or n-1 contains the given number as a factor. We then find log (n + 1) or log (n-1) and deduce the logarithm of the given number. Example. Calculate log 2 and log 3, given μ=43429448. By putting n=10 in (2), we have the value of log 10- log 9; thus 1-2 log 3=043429448+002171472+000144765 +000010857 +000000868+000000072 +000000006; 1-2 log 3=045757488, log 3=477121256. Putting n=80 in (1), we obtain log 81 - log 80; thus 4 log 3-3 log 2-1-005428681-000033929+000000283 –·000000003; 3 log 2.908485024-005395032, log 2=301029997. In the next article we shall give another series for log, (n+1) - log, n which is often useful in the construction of Logarithmic Tables. For further information on the subject the reader is referred to Mr Glaisher's article on Logarithms in the Encyclopædia Britannica. Н. Н. А. 13 log (n+1) - log, n = 2x + so that x= 2 1 + ..). 1 2n + 1 1 ; we thus obtain 1 {(2n + 1 + 3 (2n+1)2 + 5 (2n + 1)2 + ...} · NOTE. This series converges very rapidly, but in practice is not always so convenient as the series in Art. 224. 227. The following examples illustrate the subject of the chapter. that Example 1. If a, ẞ are the roots of the equation ax2+ bx+c=0, shew a2+82 a3+ß3 log (a - bx + cx2) = log a+ (a+ẞ) x · x2+ 2 a- · bx + cx2= a {1+ (a + ẞ) x+aßx2} =a (1+ax) (1+ßx). 3 .. log (a - bx + cx2) = log a +log (1 + ax) + log (1+ßx) Example 2. log (1+x+x2) is Prove that the coefficient of x in the expansion of log (1+x+x2)=log -X x6 according as n is or is not a multiple of 3. = log (1 − x3) – log (1 − x) x9 x3r - 1 If n is a multiple of 3, denote it by 3r; then the coefficient of 2" is 1 from the first series, together with from the second series; that is, the 3r If n is not a multiple of 3, xn does not occur in the first series, therefore 1 that is, less than -; hence an integer is equal to an integer plus n a fraction, which is absurd; therefore e is incommensurable. |