Ex. 1. To express an+b" in terms of powers of ab and a+b. From the identity (1 − ax) (1 − bx) = 1 − (a + b) x + abx2 Equate the coefficients of x" on the two sides of the last equation. [This is allowable since the series can clearly be made convergent by taking a sufficiently small.] Then the coefficient of x" on the left is n (a+b). On the right we have to pick out the coefficient of xn from the terms (beginning at the highest in which it can appear) arbr (a+b)n−2r + .. 10 (a2 + b2 + c2) = 7 (a2 + b2+c2) (a5 + b5+c3). Put p for be+ca+ab, and q for abc; then we have the identity (1-ax) (1 - bx) (1 − cx) = 1 -- px2 – qx3. Now take logarithms, and equate the coefficients of the different powers of x in the two expansions. This gives — (a2 + b2 + c2) in r terms of p and q, and the required result follows at once. [See also Art. 129.] Ex. 3. To express an+b+cn in terms of abc and bc + ca+ab, when a+b+c=0. Put p= bc + ca + ab, and q=abc; then we have the identity (1 − ax) (1 − bx) (1 − cx) = 1 − px2 — qx3. Hence, by taking logarithms, and equating the coefficients of like powers of x, we have n 1 (an+bn+cn) = coefficient of x" in Σ = x2 (p+qx)", Now by inspection we see that the coefficient of x6m-1 in each of the above terms in which it occurs contains pq as a factor; and also that the coefficient of x6m+1 in each of the terms in which it occurs contains p2q as a factor. Hence, when a+b+c=0, an+b+c" is algebraically divisible by abc (bc+ca + ab) when n is of the form 6m - 1, and an + b + cn is algebraically divisible by abc (bc + ca + ab)2 when n is of the form 6m+1. If we put c (a + b), bc+ca + ab becomes · (a2+ab+b2), and we have Cauchy's Theorem, namely that an+b2 - (a+b)n is divisible by ab (a+b) (a2 + ab + b2) when n is of the form 6m 1, and by ab (a+b) (a2+ab+b2)2 when n is of the form 6m+1. [See papers on Cauchy's Theorem by Mr J. W. L. Glaisher and Mr T. Muir in the Quarterly Journal, Vol. xvI., and in the Messenger of Mathematics, Vol. VIII.] 305. In order to diminish the labour of finding the approximate value of the logarithm of any number, more rapidly converging series are obtained from the fundamental logarithmic series. Changing the sign of y in the logarithmic series from which it is easy to obtain the value log, 2 = ·693147..... Having found log, 2, we have from (iv) Hence log, 3693147 +405465 = 1.09861. Proceeding in this way, the logarithm to base e of any number can be found to any requisite degree of approximation. 306. Logarithms to base e are called Napierian or natural logarithms. The logarithms used in all theoretical investigations are Napierian logarithms; but when approximate numerical calculations are made by means of logarithms, the logarithms used are always those to base 10, for reasons which will shortly appear on this account logarithms to base 10 are called Common logarithms. We have shewn how logarithms to base e can be found; and having found logarithms to base e, the logarithms to base 10 are obtained by multiplying by the constant factor loge, or by 1/log 10. [Art. 303, V.] This constant factor is called the Modulus: its value is 43429... 11. Shew that, if log. (1+x+x) be expanded in powers 12. If log (1-x+x) be expanded in ascending powers of x in the form a,x+α ̧x2 +α ̧ï3 + ................, then will a ̧ +a+a +... 2 3 log 2. 13. Expand loge 1 + x + x2 in ascending powers of x. From the identity 2 log (1-x)= log (1 - 2x + x2), prove 15. 1 16. If loge integral powers of x, the coefficient of x" will be ing as n is even or odd. n be expanded in a series of positive n |