CHAP, xxiv] PROBLEM IN THE THEORY OF LOGARITHMS. 715 



series for log (1 - x/a) from that for log (1+x/a) . [If in the second formula 

 we take d=x 2, we obtain Borda's 56 result. If in the first we take 

 d = xl, we obtain the example m = x 2 , n = x z l given before the report 

 on Delambre. 56 H 



W. Allman 60 gave the result quoted under Delambre 56 and the first two 

 results cited under Lavernede. 



T. Knight 61 started with x=(x+ri) {xf(x+ri)}, changed x into x+ri 

 in the fraction and multiplied by such a fraction as will restore equality: 



x+n ' 



x+n+rix+n)(x+ri)' 



In the final fraction change x into x-\-n" and restore equality by annexing 

 the new factor 



x(x+n+n')(x+n+n")(x+n'+n") 



(x+ri)(x+ri)(x+n")(x+n+n f +n"Y 



The expanded numerator has its first three terms the same as the corre- 

 sponding terms of the expanded denominator, and also the fourth terms 

 alike if n" n-\-ri. The rest of the paper is on the case n = n f = n" = 

 = 1, and gives the general factor explicitly. 

 Secretan 62 noted that 



E. B. Escott 63 spoke of a x n +aiX n ~ l -\ ---- and a' x n -{ ---- as having 

 exactly their first r terms alike if a = a' , , a r _i = a' r _i, a r 4= a' r . He readily 

 proved theorem (I) : If / and g are two polynomials in x having exactly 

 their first r terms alike, then f(x} -g(x+d) and g(x)-f(x+d) have exactly 

 their first r+1 terms alike. Starting with f=xa, g = x, and taking 

 d= b, we see that (x a) (x b} and x(x a 6) have two terms alike. 

 Taking the latter as / and g, and d = c, we see that (Knight) 



(x a)(x b}(x c)(x a b c), x(x a b}(x a c)(x b c) 



have three terms alike. Proceeding similarly, we obtain theorem (II): 

 If we form the equation whose roots are the sums of a\ t , a n taken 1, 3, 

 5, at a time, and that whose roots are the sums of the a's taken 2, 4, 6, 

 at a tune, we obtain two functions of degree 2 n ~ l having exactly their 

 first n terms alike. For special a's common factors occur and may be 

 removed. Thus, if w = 4 and if the a's are a, b, a+b, a+2b, four of the eight 

 roots will be common and the remaining ones are 0, a +36, 2a+6, 3a+46, 

 and a, b, 2a+46, 3a+36. If in (I) we take g = P(x}=x(x+d)(x+2d) 

 {x+(n l}d} and/=P+c, and remove the common factor Pfx, we obtain 

 two functions (P+c)(x+nd) and (x-\-nd)P+cx of degree n+1 with exactly 

 their first n-\-\ terms alike. Again, taking g = P(x}-P(x-\-a) and/=0+c 

 in (I), and removing the common factor g[{x(x-\-a) }, we get 



(x+nd)(x+a+nd)(g+c}, (x+nd)(x+a+nd)g+cx(x+a}, 



60 Trans. Roy. Irish Acad., 6, 1797, 391-434. 



61 Phil. Trans. Roy. Soc. London, 1817, 217-33. 



62 Comptes Rendus Paris, 44, 1857, 1276-9. 

 83 Quar. Jour. Math., 41, 1910, 141-167. 



