370

MISCELLANEOUS

TRACT 38.

‘ l /(x—x r —x' + ‘ &c) = x^x~ xf-^x &c. Now tbe exponentsof the first terms made equal, give r — 1 — tlieref.r — 1 + A = £; and those of the 2 d terms made equal,'give r + s — 1 = r— l, theref. s— 1 = — a, and s=l — i=i >conseq. the whole assumed series of exponents r, r + •?,r-{-2s, &c, become |, a, &.c, as assumed above in pa. 368.

Again, for the 2d equation mk or k = (12-{-z)V(x—z)x= (a + z) V{x — z)x ; assuming Z = x r +x r +‘ &c as before,then k—x r ~ T x-{-x r + s ~ I x 8 tc, and V(x — z)x = x^x — x r — r x&c, both as above; this mult, by a -f Z or a -f x r + x r + s &c,gives ax*x — ax r ~^x See : then equating the first exponentsgives r— 1 = 4 . or r=\, and r+s— lr=r — 4 or s= 1 —4 = 4 ;hence the series of exponents is A, 4 , i, &e, the same as theformer, and as assumed in pa. 366.

Lastly, assuming the same form of series for z and k as inthe above two cases, for the 1st fluxional equation also, viz,mk = ( 2x +z) V(x — z)x : then \f(x — z)x =x T x— x'—^x &c,which mult, by 2x-f-Z, gives 2x^x—x' +ix 8tc : here equat-ing the first exponents gives r — 1 = 4 , or r = 4 ; and equat-ing the 2 d exponents gives r + s- 1 = r + 4 > or s = 4 ;hence the series of exponents in this case is 4 , y, &c, asused for this case in pa. 365. Then, in every case, the ge-neral coefficients a, b, c, 8tc, are joined to the assumedterms x r , x r + s ,&c, and the whole process conducted as in thethree pages just referred to.

Such then is the regular and legitimate way of proceeding,to obtain the form of the series with respect to the expo-nents of the terms. But, in many cases we may perceive atsight, without that formal process, what the law of the ex-ponents will be, as I indeed did in the solutions in the pagesabove referred to ; and any person with a little practice mayeasily do the same.