352 EEPORT— 18S0. 



Section 7. — Sijmmetrical Biffcrcnces. 



If the successive values of a function are written down in a column 

 and diflPerenced, the successive differences belonffinar to a sriven value run 

 across the sclieme m a diagonal line, down or up accordingly as the 

 process is begun from top or bottom. Thus, beginning from the top, the 

 series u^ . . . . u^ gives the following scheme : — 



u 







«i . . . A^Ho 



A«i . . . l\^Uq 

 U2 . . . A-a(i . . . A*Uq 



A21.2 . . . A3?<i . . . A^Ho 

 ■M3 . . . A^Mg . . . A-*i<i 



A?f3 . . . A3?«2 

 «4 . . . A^Us 



An. 



•!(. 



This process is essentially unsymmetrical, as is evidenced by its diagonal 

 character. But any horizontal line has symmetry as regards the general 

 scheme, and accordingly the line n.y, A^h,, A^Mq is said to be a set of 

 symmetrical differences, that is to say, symmetrical to a value or ordinate. 

 So again the set A^^2, ^^Uy, A^w^, is said to be symmetrical to an interval. 

 It Avill be observed that this process is an alternate one — that it is 

 not possible to pass from one column to the next, but always to the next 

 but one. The operative symbol at each actual step, as, for instance, from 

 A^(ti to A^mq) is always A^ : (1 + A) and the direct problems of interpo- 

 lation and quadrature by symmetrical differences are to express (1 + A)-^ 



aiid tuclx in a series of ascending powers of A'' : 1 + A = Z^. It so 



happens that (1 + A)-^ can be expanded by ascending powers of Zj but 

 not by powers of Z-. This introduces terms of the form A : s/ (1 + A) 

 which cannot be interpreted so long as terms used are confined to one 

 horizontal line, thus implying that the expansion must be a double series. 

 The series itself was given by Newton,* but without proof. The connec- 

 tion between the two parts of the series is a difEerential one. This is 

 perhaps best shown as follows, using the notation 



Z = A2 : ( 1 -f A) = (B - 1)2 : E = E + E-i - 2, 



M= 1 (E - E->) = A (a + ^) = M2^±_^). 

 2 '^ ^ 2 V ^ eJ 2 (1 + A) 



Solving the first equation with respect to E, gives 



E = l + A = l + |z±^(z-flz^) 



E-i =i + |zT^/(z-flz2) 



The form of these values shows that E and its powers involve, in their 



* See his Mctliodust DiffercntidUs, ah-eady quoted, Prop. III. ; also Stirling, 

 Metlwdia Differentialis, Prop. XX. pp. 104-8; De Morgan, Calculus, pp. 644-7, 

 Lacroix, 2nd ed. of his Calculus, vol. iii. pp. 26-31 and 327-330, 



