B I N 



413 



B I N 



numbers) a quantity which, multiplied n \ times by itself, 



gives (1 + x)" ; and by the expansion of (1 + x) ", we mean 

 an algebraical series of powers of a: (positive or negative, 

 whole or fractional) which has all the algebraical properties 



of (1 + x)", and which, when it is convergent, has (1 + x)' 

 for its arithmetical limit or sum. 



Theorem 1. The well-known proof of the expansion of 

 (1 + x) m , where m is a positive whole number, giving 



(1 + a;) s = 1 + Zx + x' 

 (1 + x) 3 = 1 + 3X + 3 3 + a? &c. 



This theorem is not absolutely necessary, as we shall see. 

 Theorem 2. If there be any function of a, namely, a, 

 which satisfies the condition 



<t> (a) x (6) = (a + b) 



then $ (a) must be C* where C is any quantity independent 

 of a. 

 For the condition gives 



p (a) x <t> (b + c) = <f (a + 6 + c) 

 or $ (a) x <j> (b) x # (c) = <p (u + b + c) &c. 



which leads (supposing b, c, &c., to be severally equal to a) 

 to the equations 



(0a)L= <t>(na) ($ a')" = <t> (m a'), &c. 

 where n and m are any whole numbers, and a, a', &c., any 

 quantities whatsoever. Let us suppose m a' = n a, which 

 gives 



<t> (m a!) = <t> (n a) or (<p a') 1 " = ($ a)" 



- f \ - 



or#a' = (pa) m or ff a J= (>a) 



Again, the supposed universality of the first equation gives 



<j, (0) x <t> (a) = ? (a +0 ) = <p (a) 

 or f (0) = 1 : and also 



?(a) x <p(-a) = ?(a a) = 1 



* ( n a) = 



whence ? ( - a) = 



fa 



M> that the equation <?(na) = (p a)" is true for all values of 

 n: if a be = 1, this gives <p n = {?(!) }", and <f(\) is not a 

 function of the general symbol n : let f (1) = C, which gives 

 the theorem asserted. 



This theorem is the fundamental part of Euler's proof of 

 the binomial theorem. 



Theorem 3. If the values of '.t and b may be made as 

 near to equality as we please, then the limit of the fraction 



o" -If 



T is no."' ' 



a b 



In the case where n is a whole number, this is evident by 

 the well-known theorem 



(a 8 - 6") (a - b) = a + b 

 (a> - 6 3 ) (a - b) = a + ab + 6* 

 (a 4 6 4 ) (a - b) = a 3 + a*b + a 6* + V &c. 

 Let be a positive fraction, for instance, | ; and let a = a 3 , 

 b = /3 s . Then a* = a*, 6 1 = and 



a b a /3 3 a* + a/3 + /3 s 

 the limit of which, when a approaches to b, is 2a--3 a 2 or 

 2 -i 



-a 



2 - 

 or- 



2 3-1 

 or -a , 



In the same way any other 



case may be proved. 



Now let n be negative, say it ig t, where t is positive. 

 Then 



ar-b" a-'-b- 1 _ \ a'- &' 



a - b a b a' If a b 



of which the limit, by the two preceding cases (t being 

 positive), is 



or" x ta'~ l or - ta- 1 - 1 orna"- 1 

 Theorem 4. If (1 + x)" admit of being expanded in a 

 scries of whole powers of x, then that series must bo 



Let 



(1 +x)" = t a + f, j; + < 2 a; 8 + & c . 

 (1 + y)" = *. + <,y + ^ 2 + &c. 



+ xr - (i + vr 



' 



(1 + x) - 



which two sides being always equal, the limits to which 

 they approach, as x approaches to y, are equal ; or 

 n (1 + x) ' = t, + 2 t, x + 3 t, x 2 + &c. 

 Multiply both sides by 1 + x, which gives 

 n (1 + x)' = t, + (2 t, + < t ) x + (3 t, + 2 < 2 ) x + &c. 

 but by the original assumption 



n (I + x) " = n t + n t l x + n t^x* + &c. and 

 therefore <, = nt a 



2 < 2 



if = 





&c. &c. 



But, making x = in the original series, we find * = (!)" 

 = 1. Whence follows the theorem. 



Theorem 5. The value of (1 + x)" is in all cases the 

 series above investigated. 



Consider that series as a function of TV Or let 



f (n) = 1 + nx 



, &c. 



f (m) = 1 + m x + m ,r s + , &c. 



Actual multiplication will be found to give 



_ m + n ] 

 <fn X f m = 1 + (m + n)x + m + n -- 



X ' 



or <pnx fw = ?(ra-t- m). 



Or we may dispense with this multiplication by remem- 

 bering that since f n is (1 + x) a and $m is (1 -j- x} m , when 

 n and m are whole numbers, we must have, in that case 

 (Theorem !.) 



<fn X <?m = (1 4- 30" + " = P(m + n) 



but the result of a multiplication does not depend upon the 

 values of the letters; if therefore fm and fn give tp(m + n) 

 when m and n are any whole numbers, they give the same 

 result when m and n are fractional or negative. But we do 

 not yet know that if m in the latter cases represents (1 +x)". 

 But by theorem (2.) it follows from <f m X fn = f(m + n) 

 that <(n is {?>(!) }",or 



\ 



. V 



The greater part of the preceding proof is a concession to 

 the analytical taste of the age, which requires that synthe- 

 tical demonstration shall not appear in algebra. The theo- 

 rem is demonstrated rigorously as soon as it shall be proved 

 that from <f m X \n = f (m + n), it necessarily follows that 

 <f m is (P (1)}, and that the series above-mentioned satisfies 

 the equation just named. And in reading the objections 

 which have been made against the various proofs of tho 

 binomial theorem, the student must bear in mind that there 

 is one class of objections against the actual logic of tho pro- 

 cesses, and another arising out of the conventions already 

 alluded to. Against the demonstration of Euler, which 

 consists in theorems 2. and 5. of the preceding, one says 

 that it is ' tentative' (synthetical would have been the proper 

 word) ; another that it is not ' algebraical,' meaning ana- 

 lytical, and assuming that algebra must be analysis. To 

 all of which we should reply by another question, Is it 

 logical ? 



The last attempt to produce an unanswerable demonstra- 

 tion of the binomial theorem was made by Messrs. Swin- 

 burne and Tylecote of St. John's College, Cambridge 

 (Deighton, 1827). The details are much too complicated to 

 describe, but the general result is the expansion of (1 + x)" 

 to any number of terms, with a finite expression for the re- 

 mainder. This expression is however so complicated and 

 lon>', that it can be of no use, except as proved that the re- 

 mainder can be assigned by the ordinary operations of 



