RECHERCHÉS. P. S. Laplace

Transcription

1 RECHERCHÉS Sr le calcl intégral ax différences infiniment petìtes, & ax différences finies. P. S. Laplace Mélanges de philosophie et mathématiqe de la Société royale de Trin, 4, pp , I. Among the great nmber of differential eqations which we enconter in the resoltion of the problems where the concern is to apply the calcls to natre, the most ordinary, & the most remarkable are comprehended nder the general form. X = y + H dy + H d2 y 2 + H d3 y 3 + Hn 1 dn y n X, H, H, H, being some fnctions any whatever of the variable x, of which the difference is spposed constant. It wold be therefore very important to have a general method to resolve them, & there is no dobt that mechanics, & more particlarly yet physical astronomy may derive great advantages from it. Messrs. d Alembert, & Eler have resolved a long time ago the case where we have H, H, H constants; the first, by his fine method of ndetermined coefficients which is srely one of the most ingenios, & of the most fertile of analysis, the second, by a very fine consideration, & which is of the greatest sage in the integral calcls. In the third volme of his memoirs, we find some profond researches of Mr. de la Grange, in which this great Geometer integrates the case where we have, H = Ax + h H = A x + h 2 H = A x + h 3 &c, Translator s note: The notation sed by Laplace is very awkward, bt has been retained as far as possible. As is sal with papers of this period, the typesetting was poorly done. Characters and accents are omitted, pt in the wrong place, or inverted. Consistent se of pper and lower case is lacking. A raised dot, often sed to represent mltiplication, also seems to be an artifact of setting space into formlas. Symbols of groping are not matched. It is often not clear if Laplace intended to begin a new sentence or not. Therefore, in the interest of making the paper more readable, I have taken the liberty to display eqations which were in-line, correct what appear to be obvios errors of langage and notation, and break long seqences of material into sentences. With very few exceptions, there will be no indication that a correction has been made. Translated by Richard J. Plskamp, Department of Mathematics & Compter Science, Xavier University, Cincinnati, OH. Agst 14,

2 A, A, & h, being some constants. He shows moreover that this eqation X = y + H dy + H d2 y 2 + is generally integrable in the same case as this 0 = y + H dy + H d2 y 2 + This beatifl theorem of which Mr. d Alembert has given in the same volme a qite simple demonstration, is a very important step towards the general resoltion of this kind of eqations. Mr. Bezot has made long since on these eqations some analogos remarks which he has since given in the forth volme of his corse of mathematics. Here is presently a method which has led me not only to the demonstration of this theorem, bt moreover to find immediately the integral of the first of these eqations when we have that of the second. This method is not limited moreover to the infinitely small differences, we will see, hereafter that it is applicable eqally well to the finite differences. Remark. By a particlar integral of a differential eqation between x, & y, we can nderstand, either a fnction of, x, which sbstitted for, y, in this eqation makes vanish all the terms or else a parallel fnction of, x, which moreover is comprehended in the general integral of this eqation by determining in a certain manner the arbitrary constants that the integration introdces there 1 ; becase Mr. Eler has shown that the first of these two properties very well sbsist in the second. It is in the first sense that I will take for gold by having the particlar integral of a differential eqation. PROBLEM I. Let it be proposed to integrate the differential eqation X = y + H dy ddy + H 2 + Hn 1 dn y n A X, H, H being some fnctions any whatever of, x, being spposed constant. Let there be, 2 Soltion. w dy + y = T 1 Translator s note:... qe l integration y introdit; I have read this rather as qe l integration y introdit. 2 Eqation B is actally written as œ dy + y = T. Bt the symbol œ seems to have been sed in error. 2 B

3 w, & T, being some fnctions of, x; this eqation differentiated sccessively becomes w ddy dw dy = dt w d3 y 2dw ddy 2 + ddw 2 dy = ddt 2 w d4 y 3dw d3 y 3 + 3ddw 2 ddy 2 + d3 w 3 dy = d3 T 3 w d5 y 4dw d4 y 4 + 6ddw 2 d3 y 3 + 4d3 w 3 d2 y 2 + d4 w 4 dy = d4 T 4 w dn y n 1 n + 1 dw + 1 dn 1 y n 1 + n 1 n 2 ddw dn 1 w dy n 1 = dn 1 T n 1 2 dn 2 y + n 2 I mltiply the first of these eqations by, w, the second by w. the third by w, I add them with the eqation B, this which gives T + w dt + w d2 T 2 + w d3 T 3 + wn 1 dn 1 T n 1 = y + dy w + w + w dw + w ddw 2 + w d3 w 3 + w n 1 dn 1 w n 1 + ddy 2 ww + w + 2w dw + 3w d2 w 2 + 4wiv d3 w 3 n 1 w n 1 dn 2 w 1 n 2 ww + w + 3w dw + 6wiv ddw wv d3 w + n 1 n dn 2 y n 2 w n 1 ddw 2 + dn y n wwn 1. w n 1 dn 3 w n 3 ww n 3 + w n 2 + n d3 y 3 3 dw + n 2 n dn 1 y n 1 ww n 2 + w n 1 + n 1 w n 1 dw 1 by comparing this eqation with eqation A, we will have 1 this. C X = T + w dt + w d2 T 2 + wn 1 dn 1 T n 1 3

4 2 the following. ww n 1 = H n 1 ww n 2 + w n 1 + n 1 1 ww n 3 + w n 2 + n 2 1 w n 1 dw = Hn 2 whence we will conclde the following hence w n 1 = Hn 1 w w n 2 = Hn 2 w w n 3 = Hn 3 w [ 1 + n 1 dw 1 w = H w w n 2 dw + n 1 n 2 w n 1 ddw = Hn n n 2 1 ] Hn 1 w 3 dw dw 1 + 2dw H ww H 3ddw 2 w = H w w = H w w 3 Hn 2 ww H n 2 ww n 1 n 2 ddw Hn 1 ww 1 + 3dw If we sbstitte these vales into the eqation 1 + 3dw [ ] H w 3 H w 3 & into this one H = w + w + w dw ddw + w 2 + X = T + w dt + 4

5 we will form the following two H w 1 + 2dw [ ] H ww 0 = w H dw 1 + 3dw [ ] H w 3 3ddw [ ] H 2 w 3 + ddw H 2 w 1 + 3dw [ ] H w 3 + d3 w H 3 w + dn 1 w Hn 1 n 1 w H w 1 + 2dw [ H ww 1 + 3dw [ ] H w 3 3ddw [ ] H 2 w 2 + ddt H 2 w 1 + 3dw [ ] H w 3 + d3 T H 3 w X = T + dt + dn 1 T Hn 1 n 1 w III. D E Eqations D & E are of a degree inferior to the proposed, ths eqation A can always be redced to two others of a degree less by one nit. It is not even necessary to resolve generally these eqations, it sffices to find in eqation D a nmber, n, of particlar vales for, w, which satisfy this differential eqation. Let, β, β, β, be these vales or these particlar integrals, we will sbstitte them sccessively into eqation E, & we will form from them by this manner a nmber, n, of eqations of which it will sffice to find a particlar integral for each; becase β, β, β, being the particlar vales of, w; let T, T, T, be the particlar 5

6 and corresponding vales of, T. We will have the nmber, n, of eqations following, β dy + y = T β dy + y = T dy β + y = T β n 1 dy + y = T n 1 Nonetheless the complete integral of eqation A, will be y =e β T C + β e β + e β T C + e β + + e β n 1 C n 1 + β T n 1 β n 1 e β n 1, ω e being the nmber of which the hyperbolic logarithm, is nity. If we sbstitte into eqation E, β, in the place of, w, it will become X = T + dt [ H β 1 + 2dβ H ββ + ddt H 2 β IV. + dn 1 T Hn 1 n 1 β If we sppose that T = Z, is the complete integral of this eqation, Z, then it will contain a nmber, n 1, of arbitrary constants. Hence the complete integral of eqation A will be 3 y = e β Z C + β e β since this integral contains a nmber, n, of arbitrary constants. 3 Translator s note: The eqation label is lacking and has been added. ω 6

7 V. If in eqation A, X = 0, we can sppose then in the expression ω of, y, of article III., T = 0, T = 0, T = 0, & we will have y = Ce β + C e β + C n 1 e β n 1 & if we sppose that in the eqation 0 = T + dt H β 1 + 2dβ [ ] H ββ + ddt H 2 β we have for the complete integral + dn 1 T Hn 1 n 1 β T = AR + A R + A R + A n 2 R n 2 A, A, A being some arbitrary constants, & R, R, R being some particlar integrals of the preceding eqation, the expression gives y =Ce β + e β + + e β + e β A R β ω AR β e β e A n 2 R n 2 By comparing this expression of, y, with this one y = Ce β + C e β we will form the following eqations e β AR β e e β A R β e β β of, y, of the preceding article e β + C e β + C n 1 e β n 1 β β = C e β = C e β 7

8 whence we will have AR = C β β β A R = C β β β e β e β A n 2 R n 2 = C n 1 βn 1 β β n 1 e β n 1 We sppose now that we know how to integrate eqation A in y by making X = 0, & let, C, C, C be the particlar vales of y, which satisfy this eqation, so that its complete integral is By comparing it with this one we will have y = Ce β Whence we will conclde y = C + C + C + C n 1 n 1 + C e β C = Ce β, β = d, + C e β + C n 1 e β n 1 β = d, C = e β, β = d these vales of β, β, β will satisfy conseqently for, w, in eqation D, & we will conclde from it, AR, A R, A R, & conseqently if we know how to integrate the eqation 0 = y + H dy ddy + H 2 + Hn 1 dn y n we will know 1 a nmber, n, of vales which satisfy for, w, in eqation D, 2 a nmber, n 1, of particlar vales for, T, in the eqation 0 =T + dt H β + dn 1 T Hn 1 n 1 β & hence we will know the integral completely. 8

9 VI. Now if we know how to integrate the eqation X =T + dt H β + dn 1 T Hn 1 n 1 β by spposing, Z, its complete integral, we will have by article IV. y = e β Z C + β e β Therefore the difficlty of integrating the eqation X = y + H dy + Hn 1 dn y n when we know how to integrate 0 = y + H dy + Hn 1 dn y n is redced to integrating this one X =T + dt H β + dn 1 T Hn 1 n 1 β of degree, n 1, & if we know how to integrate, when we sppose X = 0, we will make in the same manner, & by the same method the resoltion of that one to depend on another of degree, n 2, & ths in seqence ntil we arrive to an eqation of degree, n n, or prely algebraic; whence there reslts that the eqation X = y + H dy + Hn 1 dn y n is integrable in the same case as this one 0 = y + H dy + Hn 1 dn y n this which is the beatifl theorem of Mr. de la Grange. If we knew only a nmber, n 1, of particlar vales for, y, in this last eqation, or that which comes to the same of a particlar vale for, w, in eqation D, the integration wold have no more difficlty, becase instead of arriving to a prely algebraic eqation, we wold arrive to an eqation of the first degree of this form X = S + Q ds 9

10 an eqation that we know how to resolve generally, X, & Q, being some fnctions of x. VII. The preceding method frnishes s not only the demonstration of the theorem of Mr. de la Grange, it leads s yet to find immediately the expression of, y, in the eqation X = y + H dy + Hn 1 dn y n when we know how to resolve this one 0 = y + H dy + Hn 1 dn y n becase let, as previosly,,,,, be the particlar vales of, y, in this same eqation, so that its complete integral is y = C + C + C + C n 1 n 1 The complete integral of eqation A will be by the preceding article Zd y = C Z, being the complete integral of, T, in eqation. If we name,,,, the particlar vales of, T, in this eqation by spposing in it, X = 0, we will have in the same manner Z Z = C d We will form likewise Z Z = C d Z Z = C d ntil finally we arrive to this eqation, Z n 1 = X. We seek now the vales of,,, now we have by article V. R = = β β β e β by sbstitting in place of, β, & β their vales, = d d, where,,= d d in the same manner = d d d = d d d A d. 10

11 we will form likewise = d d d = d d d This pt by sbstitting, we will have [ [ d y = C C [ C [n 1 d n 1 = d d d = d d d [ [ d d C n 2 [ C n 1 n 2 n 1 Xd n 1 we can give to this expression a simpler form by considering that d = d d = d ; d in the same manner d = d, the formla Z will become conseqently [ [ [ [ y = C + d C + d [ n 2 [ + d C n 1 n 2 & we will have [ C n 1 Xd n 1 d = d d d d d d d = d d d d n 1 n 1 If we wold know only a nmber, n 1, of particlar vales for, y, in the eqation 0 = y + H dy + Hn 1 dn y n 11 Z Y

12 or that which reverts to the same, a nmber, n 1, of vales for, w, in eqation D, then we cold arrive, as we can be assred by the preceding article to the following eqation X = Z n 2 + dz n 2 H n 1 d d by means of which we will have easily, Z n 2, & we will have n 2 n 2 d y = [ C [ C n 2 [ d C n 2 Z n 2 d n 2 Remark. n 2 n 2 n 3 d n 2 n 2 X We will observe here that if we know a nmber, n 1, of particlar vales for, w, in eqation D, we will know how to integrate it completely; becase by means of these vales we will have by the preceding articles the complete integral of this one 0 = y + H dy + Hn 1 dn y n We sppose that this complete integral of, y, is as that above y = C + C + C n 1 n 1 & that the complete vale of, w, is γ; γ, will contain conseqently a nmber, n 1, of arbitrary constants; bt we will have γ dy + y = 0 hence y = Ae γ & this expression will be the complete integral of, y, since it contains a nmber, n, of arbitrary constants; therefore Ae γ = C + C + C n 1 n 1 Whence we will conclde γ = + C C + C C + Cn 1 C d + C C d + Cn 1 C an expression which contains, n 1, arbitrary constants. n 1 d n 1 12

13 VIII. We will resme now the formla [ [ [ [ [ y = C + d C + d C [ n 2 [ + d C n 1 Xd 1 n 2 n 1 if we divide by,, & if we differentiate we will have [ y d = d C [ [ + d C [ n 2 [ + d C n 1 Xd 1 n 2 n 1 by dividing by d, & differentiating again we will have } { d y d d = d [C + ] by contining to operate ths, we will arrive to a differential eqation of this form C n 1 + Xd 1 = γy + γ dy n 1 + γn 1 dn 1 y n 1 1 γ, γ, being some fnctions of,,, & of their differences. We will n 1 form,, by article VII., & for this we have considered in this article the vales of,,, in this order,,,,... n 1 ; bt if instead of giving to,, the second rank, we had pt it in the first, & in the second in the following order,,,,... n 1, then we will be arrived to this eqation C n 1 + Xd 1 n 1 = γ y + γ dy + γn 1 dn 1 y n 1 2 n 1 n 1, γ, being that which,, γ, become when we change,, into, & reciprocally,, into, ; now if we sppose, X = 0, the two eqations 1 13

14 & 2 become hence, C n 1 = γy + γ dy + γn 1 dn 1 y n 1 C n 1 = γ y + γ dy + γn 1 dn 1 y n 1 γy + γ dy + γn 1 dn 1 y n 1 = γ y + γ dy + γn 1 dn 1 y n 1 3 an eqation which mst be identical, becase withot this, as we have y = C + C + C n 1 n 1 the integral of eqation 3, whatever it be of order, n 1, wold contain a nmber, n, of arbitrary constants, that which is absrd. We will have therefore by comparing the eqations 1 & 2 C n 1 + Xd 1 = C n 1 + Xd n 1 1 n 1 n 1 n 1 n 1 hence we will have, =. Ths the expression, will remain the same, whether we change or not,, into, &, into. We will prove in the same manner that it will remain constantly the same, be it that we change,, into,, n 1 &, into ;, into, &, into,, & that in general, by forming, we can withot changing its vale give to,,,, sch order as we will wish, n 1 provided that we consider,, as the last of these qantities. d 1 n 1 Let now =, Z n 1 ; let, Z n 2, that which Z n 1 becomes, when we change, n 1, into, n 2, &, n 2, into, n 1 ; we will have by the same method which has made s arrive to the eqation 1, C n 2 + XZ n 2 = γy + γ dy + γn 1 dn 1 y n 1 γ, γ, γ being that which γ, γ become when we change, n 1, into n 2, & reciprocally; we will have likewise by treating sccessively, n 3, n 4,, as the last of the qantities,,, C n 3 + Xz n 3 = γy + γ dy + γn 1 dn 1 y n 1 14

15 by arranging all these eqations in the following order C n 1 + z n 1 X = γy + γ dy + γn 1 dn 1 y n 1 C n 2 + z n 2 X = γy + γ dy + γn 1 dn 1 y n 1 C n 3 + z n 3 X = γy + γ dy + γn 1 dn 1 y n 1 C + zx = γ + γ n 1y dy n 1 + γ n 1 dn 1 y n 1 n 1 & adding them together after having mltiplied the first by, n 1, the second by, n 2, & ths in seqence. We will have an eqation of this form λy + λ dy + λn 1 dn 1 y = C + zx n 1 + C + z X + C + z X + C n 1 n 1 + z n 1 X If we sppose, X = 0, we will have bt we have, therefore λy + λ dy + λn 1 dn 1 y n 1 = C + C + C + C n 1 n 1 y = C + c + c n 1 n 1. y = λy + λ dy + λn 1 dn 1 y n 1 an eqation which mst be identical, becase withot this, whatever be the order, n 1, its integral wold contain a nmber, n, of arbitrary constants, that which is absrd. We will have therefore y = C + ZX + C + Z X + C n 1 n 1 + z n 1 X 15

16 thence reslts this very simple rle in order to have the complete integral of the eqation when we know how to integrate this one let be the integral of this last, & let s make X = y + H dy ddy + H 2 + Hn 1 dn y n 0 = y + H dy ddy + H 2 + Hn 1 dn y n y = C + C + C + C n 1 n 1 = d d d = d d = d d d d = d d d = d d d = d d d 1 n 1 d n 1 & let s arrive to form ths,, let then = Z n 1. If in the expression of, Z n 1, we change, n 1, into, n 2, & reciprocally, we will form, Z n 2 ; if in the same expression of, Z n 1, we change, n 1, into n 3, & reciprocally we will form, Z n 3, & ths in seqence, I say that the complete integral of the eqation will be X = y + H dy + Hn 1 dn y n y = C + + C + ZX Z X A + C n 1 n 1 + Z n 1 X IX. If we sppose now in eqation A, H, H, H constants, we see easily that in order to have a nmber, n, of particlar vales which satisfy for, w, in the eqation D it sffices to sppose, w, constant, & then this eqation will be changed into this one 0 = w H + H w H w 2 + H w 3 ± Hn 1 w n 1 16

17 By resolving this last eqation we will have a nmber, n, of vales for, w, by means of which we will integrate easily eqation A. We apply to this case the rle given in the preceding article. Let for this, 1 p 1 p 1 p be the roots of the eqation we will have hence 0 = w H + H w, β = 1 p = d = e px, = e px, = e px, whence we will form = e p x p p p p p = e px p p p p = e p x p p p p p p p p p = e p x p p p = e p x p p p p p p = e p x p p p therefore hence n 1 = e pn 1x Z n 1 = whence we will conclde p p n 1 p p p n 1 p n 2 p n 1 p p n 2 p p p p n 1 p p n 1 p p n 1 p p n 1 p n 2 p n 1 e pn 1 x Z n 2 = Z = p p p p n 1 p p n 2 p p n 2 p n 1 p n 2 e pn 2 x p p p p n 1 p p p p p n 1 p e px 17

18 hence p p p p n 1 e px y = p p p p p n 1 p + p p p p n 1 e p x p p p p + A A X e p x X we sppose frther in eqation A, H = Ax, H = A x 2, H = A x 3 it is easy to perceive that by spposing in eqation D, w = mx, we will satisfy this eqation, & we will have by dividing by, x, e px 0 = m A m A m 1 + m 1 + 2m A mm m1 + 2m1 + 3m A m 3 By resolving this eqation, we will have a nmber, n, of vales for, m, & conseqently a nmber, n, of particlar integrals of eqation D. This case is known enogh for s to dispense of entering into any detail in its regard. This method of integrating the eqations of the form X = y + H dy ddy + H 2 + Hn 1 dn y n by seeking a nmber n, or n 1, of particlar integrals of eqation D embraces therefore all the known cases where the integration has scceeded to the present, & there is no dobt that we can by its means discover new, & mch more extensive of them. X. If we sppose that eqation A ascends only to the second degree, eqation D will become 0 = H Hw + ww + H dw o of which it will sffice to find a single particlar integral, in order to integrate the eqation X = y + H dy ddy + H 2. for let, β, be this integral by sbstitting it into eqation E, we will have whence we will conclde T = e β H X = T + dt H β βx A + H e β H 18

19 therefore we will have by article IV. T = e β C + e [ βx H β ] β βx A + H e β H if instead of knowing a vale of, w, we wold know a particlar vale of, y, in the eqation 0 = y + H dy ddy + H 2. let, y = A, be this vale, we will have w = β = d, & we will integrate as, above, X = y + H dy ddy + H 2. thence, & from the remark which we have made article VII. reslts the following two theorems. The eqation, 0 = H Hw + ww + H dw o is completely integrable, when we know a particlar integral of it. The eqation of Riccati will be therefore integrable yet when we will know a particlar integral, since this eqation is contained in the preceding, & more generally still the eqation 0 = P + Qy + Ryy + Sdy P, Q, R, S, being any fnctions whatever of, x, is integrable, when we know a single particlar vale which satisfies for, y, in this eqation; becase by dividing by, R, we will have 0 = P R + Q R y + yy + S R dy In order to restore this eqation to formla o I mltiply it by, P S, & I make, P S = dz, this which gives 0 = P R dz + Q R ydz + yydz + P R dy P, Q, R, becoming then some fnctions of z. Let there be P R = H, & Q R = H, we will have, 0 = dzh Hy + yy + H dy; an eqation which is the same as eqation o. The eqation X = y + H dy + H ddy 2 7 is integrable yet when H = w + H w 1 + dw. If we sppose, w = m + nx + px 2 + hx r q, 19

20 the eqation X = y + dy [m + nx + px2 + hx r q + Q1 + qm + nx + hx r q 1 [n + 2px hrx r 1 ] + ddy 2 [Qm + nx + px2 + hx r q ] is integrable whatever be, Q. We can therefore by means of the preceding theorem, find in an infinite nmber of cases the integral of eqation 7 whatever be, Q, namely by giving to w, an infinity of vales. XI. Eqation D will become for the third order [ 0 = w H dw ] H w 1 + 2dw H + ddw ww 2 H w P of which it sffices to find two particlar integrals, in order to have the complete integral of this one X = y + H dy ddy + H 2 + H d3 y 3 it will be easy to find ths an infinity of eqations of third order, & of sperior orders integrable, & to form ths for each order of differentials, a class of eqations very general, & integrable, bt I content myself to have given the method. XII. Application of the preceding method to the integral calcls in finite differences. Althogh the integral calcls in the finite differences is the fondation of the entire theory of series, however this interesting branch of analysis is yet qite far from the point of perfection where we have carried the others. Mr. Eler has given in trth in his instittions many very good, & very ingenios methods in order to integrate a differential fnction in finite differences & in a single variable, bt the integration of the differential eqations is an absoltely new part, if we except from it one or two cases which the theory of recrrent series contains, & the excellent essay which Mr. the Marqis de Condorcet has given on this material in his integral calcls. 4 I myself am therefore here proposed to stdy it thoroghly, by applying the method of which I have made sage, above for the infinitely small differences; it has led me to find the general term of a qite extensive class of series, & of which the known series are only some particlar cases, at the same time to many other remarks which have appeared to me 4 When I wrote this in the month of March 1771 there had appeared nothing more on this matter; since this time I have had occasion to see a qite beatifl memoir of Mr. the Marqis de Condorcet on the finite differences which will appear in the volme of memoirs of the Académie des Sciences de France, for the year 1770 bt the researches of this illstrios geometer have nothing in common with mine. Except that he observes in the same manner that the theorem of Mr. de la Grange holds eqally for the finite differences. 20

21 important to demonstrate, for example, that the beatifl theorem of Mr. de la Grange according to which the eqation X = y + H dy + H ddy 2 + X, H, H being some fnctions whatever of, x, is integrable in the same cases as this 0 = y + H dy + H ddy 2 + holds eqally for the finite differences, & to determine in a qite simple manner the integral of the first of these eqations, when we know the integral of the second. XIII. In order for s to form a precise idea of the differential eqations in the finite differences, we imagine a series y + y + y + y x formed according to a law sch as we have constantly X x = M x y x + V x y x + p x x y x + S x n y x A X x, M x, V x being some fnctions any whatever of the index, x, of which the difference is spposed constant & eqal to 1, & the characteristic designating in the manner of Mr. Eler the finite difference of a qantity. The preceding eqation will be a differential eqation in the finite differences which can generally represent the eqations of this kind where the variable y x, & its differences are nder a linear form. Althogh we can easily form other differential eqations in which for example y x & its differences wold be mltiplied by themselves, or the ones by the others, however those which are contained in eqation A are the only ones which it is veritably interesting to know well, becase they alone can serve in the theory of series; ths we will adhere to examine them with care. XIV. Many principles of the integral calcls in the infinitely small differences, hold eqally for the finite differences, ths every fnction of x, for example which will satisfy for y x, in eqation A & which will contain a nmber n, of arbitrary constants, will be the complete integral of it. This principle which is of greatest sage in the integral calcls in the infinitely small differences, is not of a less extensive sage, in the calcls in the finite differences. By a particlar integral of a differential eqation, I nderstand as previosly every fnction of x, which sbstitted for y x, in this eqation makes vanish all the terms of it. I will cation here that for the convenience of the calcls I will sppose that H, H, H express some different qantities, & which can have no relation 21

22 among themselves, instead as this, H, H, H express as ordinarily that which, H, becomes when the index increases sccessively by 1, 2, 3,, & the following, H, H, H, express that which this same qantity, H, becomes when the index diminishes sccessively by 1, 2, 3 this pt. As we have XV. y x = y x+1 y x 2 y x = y x+2 2y x+1 + y x 3 y x = y x+3 3y x+2 + 3y x+1 y x we can pt eqation A nder this form X x =y x M x N x + P x + y x+1 N x 2p x y x+n S x we can conseqently give to it this form X x = y x + H x y x+1 + H x y x+2 + H x y x+3 + n 1 H x y x+2 B & this eqation represents generally every eqation linear in the finite differences. The eqation X x = y x + H x y x+1 is of the first order; this one X x = y x + H x y x+1 + H x y x+2 is of the second order, & ths in seqence. We are going presently to resolve the following problems. XVI. PROBLEM II. Let be proposed to integrate the eqation of the first order X x = y x + H x y x+1 22

23 I pt this eqation nder this form, Soltion. y x+1 = 1 H x yx + Xx H x let 1 H + R x, &, Xx x H = Z x, & we will have, x y x+1 = R x y x + Z x an eqation which it is necessary to integrate; now it gives y = Ry + Z y = R y + Z by sbstitting into this second eqation, in the place of, y, its vale dedced from the first, we will have y = R Ry + R Z + Z therefore, y = R R y + R Z + Z By sbstitting always in the place of, y, its vale, we will have & However we will have in the same manner y = R R y + R R Z + R Z + Z y iv = R R R y + R R Z + R Z + Z y iv = R R R Ry + R R R Z + R R Z + R Z + Z y v = R iv R R R Ry + R iv R R R Z + R iv R R Z + R iv R Z + R iv Z + R iv 23

24 & generally y x = R x 1 R x 2 R x 3 R x 4 Ry + R x 1 R x 2 R Z + R x 1 R x 2 + R Z + R x 1 R x 2 + R Z + R x 1 Or by spposing, y = A y x = R R R R x 1 A + Z R + Z RR + Z x 1 R R R x 1 Therefore, y x = R R R R x 1 A + Z x R R R x the characteristic, Σ, designating the integral in the finite differences. For more simplicity, I will denote by, R x 1, the prodct R R R x 1. This which gives, y x = R x 1 A + Z x R x Whence we will conclde by sbstitting in the place of, Z x, & R x 1, their vales y x = 1 H x 1 A + [ X x H x 1 ] if, H x, is constant, & eqal to 1 p we will have y x = p A x 1 [ ] X x p x 1 XVII. PROBLEM III. Let it be proposed to integrate the differentio-differential eqation X x = y x + H x y x+1 + H x y x+2 + H x y x+3 + n 1 H x y x+n B I sppose that we have, Soltion. w x y x+1 + y x = T x C 24

25 & we will form the following eqations w x+1 y x+2 + y x+1 = T x+1 w x+2 y x+3 + y x+2 = T x+2 w x+3 y x+4 + y x+3 = T x+3 w x+n 1 y x+n + y x+n 1 = T x+n 1 I mltiply the first of these eqations by, β, the second by, β, the third by, β,, & I add them with eqation C this which gives T x + β T x+1 + β T x+2 + β T x+3 + n 1 β T x+n 1 =y x + w x + βy x+1 + βw x+1 + βy x+2 + βw x+2 + βy x+3 + βw x+3 + iv βy x+4 + n 1 βw x+n 1 y x+n by comparing this eqation with eqation B, we will have 1 X x = T x + β T x+1 = β T x+2 n 1 β T x+n 1 2 The following w x + β = H x β w x+1 + β = H x β w x+2 + β = H x n 1 βw x+n 1 = n 1 H x whence we will conclde β =H x w x β = H x H x w x+1 + w x w x+1 β = H x H x w x+2 + H x w x+1 w x+2 w x w x+1 w x+2 n 1 β = ± w x w x+1 w x+n 2 H x w x+1 w x+n 2 n 1 H x + H x w x+2 w x+n 2 = w x + w x+n 1 the sign, +, having place if, n, is odd, & the sign,, if it is even. We will have therefore in order to resolve the problem, the following two eqations. X x = T x + T x+1 H x w x + T x+2 H x Hw x+1 + w x w x+1 n 1 H x + + T x+n 1 w x + w x+n 1 D 25

26 0 = 1 Hx w x + H x H x n 1 H x w x w x+1 w x w x+1 w x+2 ± w x w x+n 1 E XVIII. Eqations D & E are of a degree inferior to the proposed, & eqation D is of the same form; now it is not necessary to resolve generally these eqations. It sffices to know a nmber, n, of vales which satisfy for, w x, in eqation E, becase by sbstitting these vales into eqation D, we will form in it a nmber, n, of eqations of which it will sffice to find for each a particlar integral. Let V x, V x, V x be the particlar vales of, w x, & L x, L x, L x the corresponding vales of, T x, we will have V x y x+1 + y x = L x V x y x+1 + y x = L x V x y x+1 + y x = L x & the complete integral of eqation B will be y x 1 = V x 1 [A L x V x 1 ] 1 + V x 1 [ A L x V x 1 ] 1 + n 1 V x 1 [ n 1 A n 1 L x n 1 V x 1 ] since this integral contains a nmber, n, of arbitrary constants. will be XIX. I sppose that X = 0, & the complete integral of the eqation 0 = y x + H x y x+1 + H x y x+2 + n 1 H x y x+n 2 y x A = V x 1 + A n 1 V x y x 1 A n 1 V x y x 1 Presently if we sbstitte into eqation D, V x, in the place of, w x, & if we sppose next that the complete integral of this new eqation which I call D, is when we sppose, X x = 0, T x = CR x + C R x + C R x + n 2 C n 2 R x 26

27 It is easy to see that since we have, V x y x+1 + y x = T x the complete integral of the eqation 0 = y x + H x y x+1 + n 1 H x y x+2 will be y x = 1 V x 1 A C [R x V x 1 ] C [ R x V x 1 ] n 2 C [ n 2 R x V x 1 ] By comparing this last integral with the preceding, we will have Therefore 1 V x 1 [R x V x 1 ] = 1 V x 1 1 V x 1 [ R x V x 1 1 ] = V x 1 R x = V x 1 V x 1 V x 1 R x = V x 1 V x 1 V x 1 Therefore if we know how to resolve eqation B by spposing X x = 0 in it, we will know how to resolve eqation D by spposing in the same manner X x = 0 in it. Let then,,,, be the particlar vales of, y x, in eqation B, so that its complete integral is y x = A + A + A + + n 1 A n 1 1 we will have, = V x 1, and the integral of eqation D by spposing Xx = 0 in it will be T x = C + C + C + n 2 C n 1 Now if we know how to integrate eqation D by spposing in it whatever X x, we will know in the same manner how to integrate nder the same spposition eqation B. Becase let then, Z x, be the complete integral of, T x, in eqation D, we will have for the complete integral of, y x, in eqation B y x = 1 V x 1 [A Z x V x 1 ] 27

28 since this integral contains a nmber, n, of arbitrary constants. Therefore the difficlty to integrate X x = y x + H x y x+1 + n 1 H x y x+n when we know how to integrate this one is redced to integrating the eqation 0 = y x + H x y x+1 + n 1 H x y x+n X x =T x + T x+1 H x V x + + T x+n 1 n 1 H x V x+n 1 which is of degree, n 1, & since we know how to resolve by spposing in it X x = 0; we will make in the same manner the resoltion of this one to depend, on the resoltion of an eqation of degree, n 2, & ths in seqence; whence there reslts that the eqation X x = y x + H x y x+1 + n 1 H x y x+n is integrable in the same cases as this one 0 = y x + H x y x+1 + n 1 H x y x+n This which is the beatifl theorem that Mr. de la Grange has fond for the infinitely small differences, & that the preceding method gives s ths means to extend to the finite differences. XX. This method leads s yet frther, namely to find all at once the general expression of, y x. Becase, Z x, being the complete integral of T x in eqation D, being as above,,, the particlar vales of, y x, in eqation B when we sppose X x 1 = 0 becase of V x 1 =, we will have y x = A Z for the complete integral of eqation B, X x, being any whatever. If we name,,, the particlar vales of, T x, in eqation D when we sppose in it X x = 0, we will have in the same manner Z x = A Z x D 28

29 We will have similarly Z x = A Z x ntil finally we arrive to this algebraic eqation n 1 Z x = X x We will have therefore by sbstitting y x =A [ A A n 1 n 2 A n 1 A X x n 1 It is necessary presently to determine,, Now we have by article XVIII. =, =, = We will have likewise =, & formla K will become [ y x = A [ =, = [ A [ = A [ n 2 n 2 [ n 1 A X x n 1 K o If we wold know only a nmber, n 1, of particlar vales in the eqation 0 = y x + H x y x+1 + the integration wold no longer have difficlty; becase instead of arriving as previosly to the algebraic eqation, n 1 z x = X x, we wold arrive to an eqation of this form, X x = n 2 Z x + S x n 2 Z x+1 29

30 an eqation which we know how to integrate by the second problem. If instead of knowing how to resolve the eqation, 0 = y x + H x y x+1 +, we wold know a nmber, n, or, n 1, of vales for, w x, in eqation E, the preceding formlas wold serve eqally, becase we have, = 1 V x 1 = XXI. 1 V x 1 We can frther simplify formla o, by ptting it nder this form y x =A + A + A + n 1 A n 1 ± X x n 1 the sign, +, having place, if, n, is even, & the sign,, if it is odd: now we imagine first that the differential eqation B is only of the second order, & we will have Now Therefore Bt we have, y x = A + A + ± X x = X x X x 1 y x = A X x 1 + A + X x 1 = By mltiplying by, we will have = =,. X x 1 σ 30

31 Hence if we make, = Z & if we call, Z, that which becomes, when we change,, into, & into, we will have y x = A + X x 1 Z + A + X x 1 & Z = Z = I will observe however that by integrating differently, we wold have a vale of y x, which wold appear different, bt of which it is easy to recognize the identity with the preceding. Becase we will have whence we will conclde X x = Z X x X x y x = A + X x Z + A + X x Z. Now it is easy to see that this expression of y x is the same as the preceding, that is to say, that we have A + X x Z A + X x 1 Z + A + X x = Z + A + X x 1 or that which reverts to the same, that Z = Z ; Z or that or finally that, =, = + 31

32 that which is clear. If we sppose presently that eqation B is of the third order, we will have, y x =A + A + A [ [ X x now Hence [ [ [ Therefore Xx = Xx 1 = Xx X x 1 X x = X x 2 + X x 2 y x = A [ X x 2 + A X x 2 A X x 2 + Xx 1 ] X x 2 by observing the same process, we will have the vale of y x, by spposing the differential eqation B of sch order as we will wish. XXII. Bt here is a qite simple method in order to conclde this vale of y x. We resme for this the formla [ y x = A [ A [ [ A [ n 2 n 2 [ n 1 A X x n 1 32

33 By differentiating, we will have y x [ = A [ Whence we will conclde by dividing by, y x = & differentiating [ A We will have therefore by contining to differentiate ths. n 1 A X x = γ x y x + γ x y x+1 + n 1 γ x y x+n 1 n 1 γ x, γ x being some fnctions of,, If instead of, we had considered,, as the first of the vales of y x, of eqation B when we sppose in it X x = 0, &, as the second we wold have had n 1 A X x n 1 = γ x y x + γ x y x+1 + n 1 γ x y x+n 1 n 1 n 1, & γ x being that which, γ x become when we sppose in it,, & reciprocally: now if we sppose X x = 0, we will have We will have therefore n 1 A = γ x y x + γ x y x+1 + n 1 γ x y x+n 1 n 1 A = γ x y x + γ x y x+1 + n 1 γ x y x+n 1 γ x y x + γ x y x+1 + n 1 γ x y x+n 1 = γ x y x + n 1 γ x y x+n 1 an eqation which mst be identical, becase if it were not, then this eqation being differential of the order, n 1, wold have however for complete integral y x = A + + n 1 A n 1 which contains, n, arbitrary constants, that which is absrd. It is necessary therefore that n 1 A X x = n 1 A X x n 1 n 1 33

34 n 1 n 1 n 1 hence that, = Ths the expression, will remain always the same, granted we change,, into,, &, into. It wold be easy to establish n 1 in the same manner that if in,, we change,, into, &, into, or,, into, & into, or,, into, &, into,, & generally, k, into i, n 1 & i, into k, k, & i, being less than, n 1, the expression, will remain always the same, & that ths, whatever order that we give to the vales,, n 1 in order to form, this expression will remain constantly the same, provided that, n 1, is always considered as the last of these vales. n 1 Let now, = n 1 z, & we imagine that after having treated, n 1, as the last of the vales,, we regard, n 2, as this last, let, n 2 z, be that which n 1 z becomes, when we change, n 2, into n 1, & reciprocally we will have n 2 A X x n 1 z = γx y x + n 1 γ x y x+n 1 γ x, being that which, γ x, become, when we change, n 1, into n 2, & reciprocally. I give to, n 2 A, the sign,, becase formla o gives, by spposing in it X x = 0 y x = A A n 2 A n 2 + n 1 A n 1 the sign, +, having place if, n, is odd, & the sign,, if it is even. We will have likewise n 3 A X x n 3 z = γx y x + n 1 γ x y x+n 1 & ths in seqence; by arranging all these eqations in the following order, n 1 A X x n 2 A X x n 3 A X x n 1 z = γx y x + n 1 γ x y x+n 1 4 n 2 z = γx y x + n 1 γ x y x+n 1 10 n 3 z = γx y x + n 1 γ x y x+n 1 7 ±A X x z = γ x y x + n 1 γ x y x+n 1 15 n 1 n 1 & adding them together after having mltiplied the first by n 1, the second by n 2, 34

35 , & the last by, we will have an eqation of this form λ x y x n 1 λ x y x+n 1 = A ± X x z + A ± X x this which gives by spposing in it, X x = 0, + n 1 z n 1 A ± X x n 1 z λ x y x n 1 λ x y x+n 1 = A A n 1 A n 1 Bt we have nder this same spposition Hence y x = A A + A + n 1 A n 1 y x = λ x y x n 1 λ x y x+n 1 Now this eqation mst be identical, otherwise althogh it is of order, n 1, as we have y x = A + n 1 A n 1 its integral wold contain a nmber, n, of arbitrary constants, that which is absrd. We will have therefore for the complete integral of eqation B, by changing of sign, as this is permitted the arbitrary negative constants y x = A ± X x z + A ± X x z n + n 1 n 1 A ± X x n 1 z the sign, +, having place if, n, is odd, & the sign,, if it is even; we resme now the eqations 4, 10, 7, 15; they give n 1 A X x 1 ± A X x 1 n 1 z = γx 1 + n 1 γ x 1 y x+n 2 z = γ n 1 x 1 y x 1 + n 1 x 1 γ y x+n 2 n 1 35

36 mltiplying the first of these eqations by, n 1, the second by n 2,, & ths in seqence, we will have by adding them an eqation of this form λ x y x 1 n 2 λ x y x+n 2 = A ± X x 1 z + A ± X x 1 z We will have therefore by spposing, X x = 0 Therefore + λ x y x 1 n 1 λ x y x+n 2 = A A + λ x y x 1 n 1 λ x y x+n 1 = y x an eqation which mst be identical. Hence we will have, by changing of sign the negative constants y x = A ± X x 1 z + A ± X x 1 z + We will find in the same manner y x = A ± X x 2 z + A ± X x 2 z + and ths in seqence, ntil we arrive to this last expression inclsively y x = A ± X x n+1 z n 1 + A ± X x n+1 z n 1 + & all these expressions of, y x, mst be the same, as we have remarked above that this holds for the eqations of the second order; by comparing together these expressions, 36

37 we will form the following eqations z + z + z + z + z + + z n 1 + z n 1 + We resme now eqation E, which is XXIII. n 1 n 1 z = 0 n 1 n 1 z = 0 n 1 n 1 z n 1 = 0 We sppose 0 = 1 Hx w x + H x H x n 1 H x w x w x+1 w x w x+1 w x+2 + w x w x+n 1 H x = C φ x H x = C φ x φ x+1 E H x = C φ x φ x+1 φ x+2 C, C, C being some constants any whatever, & φ x, a fnction any whatever of, x, then the eqation X x = y x + C φ x y x+1 + C φ x φ x+1 y x+2 + C φ x φ x+1 φ x+2 y x n 1 C φ x φ x+n 1 y x+n F will be integrable; becase if we sppose in eqation E, a, being constant, it will become w x = a φ x, 0 = 1 C a + C aa C n 1 a 3 ± C a n whence we will have a nmber, n, of vales for, a, & conseqently for, w x, becase w x = a φ x. We can pt formla F nder this form y x+n = + n 2 C n 1 C y x+n 1 n 3 φ x+n 1 C y x+n 2 n 1 C φ x+n 1 + φx+n 2 X x n 1 C φ x φ x+n 1 37

38 & conseqently nder this one which is simpler y x =A φ x y x 1 + A φ x φ x 1 y x 2 + A φ x φ x 1 φ x 2 y x 3 + H + X x this formla of which we can find the general term by the preceding article is mch more extensive than any of those which the geometers have examined to the present; becase if, φ x, = 1, that which is the simplest case, then it is changed into this one y x = Ay x 1 + A y x 2 + A y x 3 + X x this which is, as we know, the general expression of recrrent series. XXIV. Among the infinite nmber of series which formla H offers s, we will choose in first place those, in which we have φ x = x, & we will consider the series formed according to this law y x = A y x 1 + A x x 1 y x 2 + X x by spposing in it first, X x = 0; we will see in the series that the researches which we are going to make on this formla extend easily to the case, where we will have φ x, eqal to a fnction any whatever of x. We examine presently this eqation when it rises only to the second order. It becomes then y x = Ax y x 1 + A x x 1 y x 2 in order to give an example of a series formed according to this law, we sppose A = 2, & A = 3, & we will have, whence we will form the seqence, y x = 2x y x 1 + 3x x x 1 y x 2, 1, 4, 42, 480, 7320, , we seek now the vale of, y x, in the differential eqation I pt it nder this form y x = Ax y x 1 + A x x 1 y x 2 0 = y x + A A x + 1 yx+1 y x+2 A x + 1 x + 2. by comparing it with eqation B of problem II; we will have X x = 0, H x = A A x + 1, H x = 1 A x + 1 x

39 & eqation E will give, 0 = 1 A A x + 1 w 1 x A x + 1 x + 2 w x w x+1 I sppose w x = 1 Aa, & we will have, 0 = 1 ax+1 a2 or A = Aa + a 2. Hence A A a = 1 2 A + A A2 are expressed by, p, & p these two vales of a; we will have therefore by that which precedes, y x = xbp x + B B x this is the general term of this new kind of series, when the differential eqation does not exceed the second order. In order to determine B & B, it is necessary to sppose that the first two terms of the series are given; let M & M, be these two terms, & we will have Therefore Hence B = M = Bp + B p M = 2 B p B p 2 M 2M p 2pp p ; B = y x M 2M p = x 2pp p p x + M 2Mp 2 p p p M 2Mp 2 p p p px λ if we have p = p, that is to say, if the two roots of the eqation A = Aa + a 2, are eqals, we will make p = p + dp, whence we will conclde easily y x = x p x 1 [ 4Mp M 2p M 2Mp + 2 p ] x in order to apply the preceding formlas to an example, let as above A = 1, A = 3, we will have A = 1 + 2, therefore p = 1, & p = 3. We sppose that the first two terms of the series are 1, & 4, so that M = 1 & M = 4. We will have therefore B = 1 4, B = 1 4. Hence formla λ will give y x = x 1 4 3x ± 1, the sign, +, having place if, x, is odd, & the sign,, if it is even. If we wish to have for example the fifth term of the series, we will have, as previosly. y x = x = 7320 λ 39

40 XXV. We sppose presently that we have to resolve the differential eqation of the third order. y x = Ax y x 1 + A x x 1 y x 2 + A x x 1 x 2 y x 3 I pt it nder this form 0 = y x + A A x + 1 yx+1 + y x+3 A x + 1 x + 2 x + 3 A A x + 1 x + 2 yx+2 by comparing with eqation B of problem II we will have, X x = 0, H x = H x = & eqation E will give I sppose & we will have, A A x + 1 x + 2, H x = 0 = 1 + A A x + 1, 1 A x + 1 x + 2 x + 3 A A x + 1 w x + A A x + 1 x + 2 w x w x+1 1 A x + 1 x + 2 x + 3 w x w x+1 w x+2 w x = 1 a x + 1, 0 = A Aa + Aa 2 + a 3 Let, p, p, p, be the three vales of, a, & we will have y x = xbp x + B p x + B p x Let M, M, M, be the three terms of the series, we will have in order to determine the constant B, B, B,the three eqations M = Bp + B p + B p M 1 2 = Bp2 + B p 2 + B p 2 M = Bp3 + B p 3 + B p 3 40

41 Frther we can raise orselves to some more general considerations, becase if we examine attentively the process of the preceding article, it is easy to see that if we have generally y x =A x y x 1 + A x x 1 y x 2 + we will have + n 1 Ax x n + 1 y x n y x = xbp x + B p x + n 1 B n 1 p x γ p, p, p being the roots of, a, in the eqation 0 = n 1 A n 2 Aa + n 3 Aa 3 + Aa n 1 + a n the sign, +, having place if, n, is odd, & the sign,, if it is even, or if M, M, M, are the first terms of the series, we will have in order to determine the constants, B, B, B the nmber, n, of the following eqations M = Bp + B p + B p + n 1 B n 1 p M 1 2 = Bp2 + B p 2 + B p 2 + n 1 B n 1 p 2 M = Bp3 + B p 3 + B p 3 + n 1 B n 1 p 3 n 1 M n = Bpn + B p n + B p n + n 1 B n 1 p n in order to resolve these eqations we can avail orselves of the ordinary methods of elimination, bt here is one which seems to me more convenient & simpler. I mltiply the first eqation by, n 1 p, & I sbtract it from the second, I mltiply in the same manner the second by, n 1 p, & I sbtract it from the third, & ths in seqence, this which prodces the following eqations M 1 2 M n 1 p =Bp p n 1 p + B p p n 1 p + n 2 B n 2 p n 2 p n 1 p M M 1 2 n 1 p =Bp 2 p n 1 p + B p 2 p n 1 p + n 2 B n 2 p 2 n 2 p n 1 p n 1 M 1 2 n n 2 M 1 2 n 1 n 1 p =Bp n 1 p n 1 p + n 2 B n 2 p n 1 n 2 p n 1 p 41

42 I mltiply again the first of these eqations by n 2 p, & I sbtract it from the second. I mltiply similarly the second by n 2 p, & I sbtract it from the third, & ths in seqence, this which gives M M 1 2 n 1 p + n 2 p + M n 1 p n 2 p = Bp p n 2 p p n 1 p + B p p n 2 p p n 1 p + n 3 B n 3 p n 3 p n 2 p n 3 p n 1 p M M n 1 p + n 2 M p n 1 p n 2 p = Bp 2 p n 2 p p n 1 p + B p 2 p n 2 p p n 1 p + n 3 B n 3 p n 3 p n 2 p n 3 p n 1 p by operating on these last eqations as on the preceding, we will have M M n 1 p + n 2 p + n 3 p M [ n 2 p + n 1 p n 3 p + n 1 p n 2 p] M n 1 p n 2 p n 3 p =Bpp n 1 p p n 2 p p n 3 p + whence it is easy to see that if we name, f, the sm of all the roots, p, p with the exception of p, h, the sm of their prodcts two by two, l, the sm of their prodcts three by three, q, the sm of their prodcts for by for, f, the sm of all the roots p, p with the exception of p, h, the sm of their prodcts two by two, f, the sm of all the roots, p, p, p with the exception of p, 42

43 we will have n 1 M nf n 2 M + n n 1h n 3 M n n 1 n 2 l n 4 M + B = n p p pp pp p n 1 M n f n 2 M + n n 1 h n 3 M n n 1 n 2 l n 4 M + B = n p p p p p p p Let for brevity, n 1 M nf n 2 M + n n 1h n 3 M = N n. n 1 M n f n 2 M + = N n. We will have therefore N p pp pp p px 1 y x = x N + p p p p p px 1 p + the qantities f, h, l, q, can be determined qite easily in the following manner. Let the eqation of this article be resmed 0 = a n + A a n 1 A a n 2 + A a n 3 ± n 1 A I divide it by a + p that which gives whence it is easy to conclde 0 = a n 1 + a n 2 A p a n 3 A + Ap pp + a n 4 A + Ap + Ap 2 p 3 f = A p, h = A + Ap pp l = A + Ap + Ap 2 p 3 f = A p ω h = A + A p p 2 43

44 XXVII. If we have p = p we will have N = N; let then p = p + dp, & 1 p pp p = Q, we will have N p p p p = NQ + Ndp dq dp Now it is easy to see by the formation of N, & N, that N = N + d N dp N = N dn dp dp. Therefore & N NQ p p p p = dp dp + NdQ dp QdN, dp p x 1 N p p p p = dp NQ dp + NdQ dp QdN dp Hence becase p p = dp, & p p = dp, we will have p x 1 N p x 1 p pp p + N p p p p NdQ = p x 1 dp QdN dp + NQ x 1 p = N 2 p x 2 x 1 R + p dr dp dp. Hence + x 1 NQ p x 1 p by spposing R = Q N. We will have N 2 p x 2 x 1 R + pdr dp y x = x N + p p 2 p p p x 1 + ω If moreover we have in this last formla p = p, we will make yet p = p + dp, & ths in seqence it will be sperflos to pase orselves frther in these different cases. 44

45 XXVIII. If we consider with attention the process of the preceding articles, we mst notice that it gives s the general term of the series formed according to this eqation y x =A φ x y x 1 + A φ x φ x 1 y x 2 + A φ x φ x 1 φ x 2 y x 3 + becase the whole sbsisting as in these articles with this sole difference that we mst make here n 1 M nf n 2 M + N = φ φ φ n the formlas ω, & ω N = will become n 1 M n f n 2 M + φ φ φ n N p p 2 p p px 1 y x = φ φ φ φ x N + p p p px 1 p + y x = φ φ φ φ x + + N 2 p x 2 x 1 R + p dr dp N p p 2 p p p x 1 ω If φ x = 1, the preceding series are changed into recrrent series, & we will have ths in a very direct manner, the general term of recrrent series: thence reslts this theorem. If we name T x, the general term of a recrrent series sch that y x = A y x 1 + A y x 2 + A y x 3 + n 1 A y x n, the general term of a series sch that we have y x = Aφ x y x 1 + A φ x φ x 1 y x 2 + n 1 A φ x φ x+1 n y x n & in which the first n terms which are arbitraries, are the same as in the preceding, will be y x = φ n+1 φ n+2 φ x T x ω 45

46 This is that of which it is easy to assre orselves besides; becase if we sbstitte this vale of y x, into the eqation we will have or y x = Aφ x y x 1 n 1 A φ x φ x+1 n y x n φ n+1 φ x T x =A φ n+1 φ n+2 φ x T x 1 + n 1 A φ n+1 φ x y x n T x = A T x 1 n 1 A T x n Now this last eqation holds, since T x, is the general term of the recrrent series formed by the eqation y x = A y x 1 + n 1 A y x n XXIX. We have spposed ntil here in formla H of article XXII. X x = 0 & this was indispensable before integrating it. By spposing X x whatever, we are going presently to examine this formla nder this hypothesis. Now if in the eqation y x = Aφ x y x 1 n 1 A φ x φ x n+1 y x n + X x, we sppose X x = 0, we will have as we have seen previosly y x = φ φ φ x Bp x + B p x + n 1 B n 1 p x whence we will form by the articles XVIII & following = φ φ φ x p x = φ φ φ x p x n 1 = φ φ φ x n 1 p x We form presently the expressions of n 1 z, n 2 z in order to sbstitte them into formla n of article XXI. We will have = = φ φ φ x p x p x p x = φ φ φ x p p p x = φ φ φ x p p p = φ φ φ x p p p p p x p x 46

47 Hence = = φ φ φ x p p p p p = φ φ φ x p p p = p = φ φ φ x p p p p p p p p p x p x p p p p & ths in seqence, we will have therefore p z = φ φ φ x p x p p p p p p n 1 p z = φ φ φ x+n 1 p x p p p p p p p p whence we will conclde by sbstitting these vales into formla n y x = φ φ φ x p x p p p A ± X x p φ φ φ x p x p p + φ φ φ x p x p p p p p + p p n 1 p A ± X x φ φ φ x p x XXX. If p = p; we will make p = p + dp. Let p p n 1 p N = p pp p we will have therefore y x = N dp p φ φ φ x p x A ± X x φ φ φ x p x Np + dp + pdn φ φ φ x p + dp x dp A ± X x φ φ φ x p + dp + 47