Newton: born 1642, 1727 (85 years old). Philosophiae naturalis principia mathematica (first published 1687; c.f. Lemma I and Lemma XI, Schol.) Then in particular: Analysis per quantitatum series fluxiones etc., first published 1711, but composed in 1665, while Leibnitz first made the same discovery in 1676. Leibnitz: born 1646, 1716 (70 years old). Lagrange: born 1736, during the Empire (Napolean I); he is the discoverer of the method of variations. Th orie des fonctions analytiques (1797 and 1813). D Alembert: born 1717, 1783 (66 years old). Trait des fluides, 1744. 1) Newton. The velocities or fluxions, of for example the variables x, y etc. are denoted by x.,y. Etc. For example if u and x are connected quantities (fluents) generated by continuous movement, then u. and x. denote their rates of increase, and therefore u./x. the ratio of the rates at which their increments are generated. Since the numerical quantities of all possible magnitudes may be represented by straight lines, and the moments or infinitely small portions of the quantities generated = products of their velocities and the infinitely small time intervals during which these velocities exist,56 so then [we have] denoting these infinitely small time intervals, and the moments of x and y represented by x. and y., respectively. For example: y = uz; [with] y., z., u. denoting the velocities at which y, z, u respectively [are] increasing, then the moments of y., z., u. are y., z., u., and we obtain y = uz , y + y. = (u + u.) (z + z.) = uz + u z. + z u. + u.z. ; y. = u z. + z u. + u.z. . Since is infinitesimal, it disappears by itself and even more as the product u.z. altogether, since it is not that of the infinitely small period of time , but rather its 2nd power. We thus obtain y. = u.z + z.u , 2) Leibnitz. The differential of uz is to be found. u becomes u + du, z becomes z + dz; so that uz + d(uz) = (u + du) (z + dz) = uz + udz +zdu + dudz . If from this the given quantity uz is subtracted, then there remains udz + zdu + dudz as the increment; dudz, the product d un infiniment petit du par un autre infiniment petit dz, (of an infinitely small du times another infinitely small dz)* is an infinitesimal of the second order and disappears before the infinitesimal udz and zdu of the first order; therefore d(uz) udz + zdu .58 [3)] D Alembert. Puts the problem is general terms thus: y = f(x) , y1 = f(x + h) ; ____________ Newton and Leibnitz, like the majority of the successors from the beginning performed operations on the ground of the differential calculus, and therefore valued differential expressions from the beginning as operational formulae whose real equivalent is to be found. All of their intelligence was concentrated on that. If the independent variable x goes to x1, then the dependent variable goes to y1 x1 - x, however, is necessarily equal to some difference, let us say, = h. This is contained in the very concept of variables. In no way, however, does it follow from this that this difference, which = dx, is a vanished [quantity], so that in fact it = 0. It may represent a finite difference as well. If, however, we suppose from the very beginning that x, when it increases, goes to x + x. (the which Newton uses serves no purpose in his analysis of the fundamental functions and so may be suppressed60), or, with Leibnitz, goes to x + dx, then differential expressions immediately become operational symbols (Operationssymbole) without their algebraic origin being evident. To 15*2 (Newton). Let us take Newton s beginning equation for the product uz that is to be differentiated; then: y = uz , y + y. = (u + u. ) (z + z. ) . If we toss out the , as he does himself if you please, after he develops the first differential equation, we then obtain: y + y. = (u + u.) (z + z.) , y + y. = uz + u.z + z.u + z.u. , y + y. - uz = u.z + z.u + u.z. . So that, since uz = y, y. = u.z + z.u + u.z. . Now, whence arises the term to be forcibly suppressed, u.z.? Quite simply from the fact that the differentials y. of y, u. of u, and z. of z have from the very beginning been imparted by definition*3 a separate, independent existence from the variable quantities from which they arose, without having been derived in any mathematical way at all. On the one hand one sees what usefulness this presumed existence of dy, dx or y., x. has, since from the very beginning, as soon as the variables increase I have only to substitute in the algebraic function the binomials y + y., x + x. etc. and then may must manipulate (man vrieren) these themselves as ordinary algebraic quantities. I obtain, for example, if I have y = ax: y + y. = ax + ax. ; y - ax + y. = ax. ; y. = ax. . I have therewith immediately obtained the result: the differential of the dependent variable is equal to the increment of ax, namely ax.; it is equal to the real value a derived from ax*4 (that this is a constant quantity here is accidental and does nothing to alter the generality of the result, since it is due to the circumstance that the variable x appears here to the first power). If I generalise this result,61 then I know y = f(x), for this means that y is the variable dependent on x. If I call the quantity derived from f(x), i.e. the real element of the increment, f (x), then the general result is: y. = f (x)x. . I thus know from the very beginning that the equivalent of the differential of the dependent variable y is equal to the first derived function of the independent variable, multiplied by its differential, that is dx or x. . So then, generally expressed, if y = f(x) dy = f (x)dx But y. = ax. Gives me immediately y./x. = a, and in general: y./x. = f (x) . I have thus found for the differential and the differential coefficients two fully-developed operational formulae which form the basis of all of differential calculus. And furthermore, put in general terms, I have obtained, by means of assuming dx, dy etc. or x., y. etc. to be independent, insulated increments of x and y, the enormous advantage, distinctive to the differential calculus, that all functions of the variables are expressed from the very beginning in differential forms. Were I thus to develop the essential functions of the variables in this manner, such as ax, ax b, xy, x/y, xn, ax, log x, as well as the elementary trigonometric functions then the determination of dy, dy/dx would thus become completely tamed, like the multiplication table in arithmetic. If we now look, however, on the reverse side we find immediately that the entire original operation is mathematically false. Let us take a perfectly simple example: y = x . If x increases then it contains an indeterminate increment h, and the variable y dependent on it has an indeterminate increment k, and we obtain y + k = (x + h) = x + 2hx + h , y + k - x or y + k - y = 2hx + h ; (y + k) - y or k = 2hx + h ; k/h = 2x + h . 2x + h = 2x + 0 = 2x . On the other side, however, k/h goes to k/0 . Since, however, y only went to y + k because x went to x + h, and then y + k goes back to y when h goes to 0, therefore when x + h goes back to x + 0, to x. So then k also goes to 0 and k/0 = 0/0, which may be expressed as dy/dx or y./x.. We thus obtain: 0/0 or y./x. = 2x . If on the other hand we [substitute h = 0] in y + k - x = 2hx + h or (y + k) - y = 2xh + h In no way do we obtain what Newton makes of it: k = 2xdx + dxdx y. = 2xx. + x.x. ; Since Newton, however, does not immediately determine the increments of the variables x, y, etc by means of mathematical derivation, but instead immediately stamps x., y., etc on to the differentials, they cannot be set = 0; for otherwise, were the result 0, which is algebraically expressed as setting this increment from the very beginning = 0, it would follow from that, just as above in the equation (y + k) - y = 2xh + h , (y1 - y)/h = 2x + h . dy/dx = 2x dy = 2xdx Therefore nothing more remains than to imagine the increments h of the variable to be infinitely small increments and to give them as such independent existence, in the symbols x., y. etc. or dx, dy [etc.] for example. But infinitely small quantities are quantities just like those which are infinitely large (the word infinitely (unendlich) [small] only means in fact indefinitely (unbestimmt) small); the dy, dx etc. or y., x. [etc.] therefore also take part in the calculation just like ordinary algebraic quantities, and in the equation above (y + k) - y or k = 2xdx + dxdx But (Oder), if in y. = u.z + z.u + u.z. But then in walks the still greater miracle that by this method you don t obtain an approximate value at all, but rather the unique exact value (even when as above it is only symbolically correct) of the derived function, such as in the example y. = 2xx. +x.x. . y. = 2xx. y./x. = 2x , But the miracle is no miracle. It would only be a miracle if no exact result emerged through the forcible suppression of x.x.. That is to say, one suppresses merely a computational mistake which nevertheless is an unavoidable consequence of a method which brings in the undefined increment of the variable, i.e. h, immediately as the differential dx or x., a completed operational symbol, and thereby also produces from the very beginning in the differential calculus a characteristic manner of calculation different from the usual algebra. ___________ The general direction of the algebraic method which we have applied may be expressed as follows: Given f(x), first develop the preliminary derivative , which we would like to call f1(x): 1) f1(x) = y/ x or y/ x = f1(x) . From this equation it follows y = f1(x) x . f(x) = f1(x) x (since y = f(x), [thus] y = f(x) ) . By means of setting x1 - x = 0, so that y1 - y = 0 as well, we obtain [2)] dy/dx = f (x) . dy = f (x)dx ; df(x) = f (x)dx When we have once developed 1) f(x) = f1(x)dx 2) df(x) = f (x)dx [____________] 1) If we have x going to x1, then A) x1 - x = x ; Aa) x = x1 - x ; a) x1 - x = x ; The difference may therefore be expressed in two ways: directly as the difference between the increased variable and its state before the increase, and this is its negative expression ; positively as the increment,*5 as a result : as the increment of x to the state in which it has not yet grown, and this is the positive expression. We shall see how this double formulation plays a role in the history of differential calculus. b) x1 = x + x . x1 is the increased x itself; its growth is not separated from it; x1 is the completely indeterminate form of its growth. This formula distinguishes the increased x, namely x1, from its original form prior to the increase, from x, but it does not distinguish x from its own increment. The relationship between x1 and x may therefore only be expressed negatively, as a difference, as x1 - x. In contrast, in x1 = x + x 1) The difference is expressed positively as an increment of x. 2) Its increase is therefore not expressed as a difference, but instead as the sum of itself in its original state + its increment. 3) Technically x is expelled from its monomial into a binomial, and wherever x appears to any power in the original function a binomial compose of itself and its increment enter for the increased x; the binomial (x+h)m in general for xm. The development of the increase of x is therefore in fact a simple application of the binomial theorem. Since x enter as the first and x as the second term of this binomial - which is given by their very relationship, since x must be [there] before the formation of its increment x - by means of the binomial, in the event only the functions of x will be derived, while x figures next to it as a factor raised to increasing powers; indeed, x to the first power must [appear], so that x1 is a factor of the second term of the resulting series, of the first function, that is, of x1 derived, using the binomial theorem. This shows up perfectly when x is given to the second power. x goes to (x + h) , which is nothing more than the multiplication of x + x by itself, [and which] leads to x + 2x x + x : that is, the first term must be the original function of x and the first derived function of x , namely x here, comprises the second term together with the factor x1, which entered into the first term only as the factor x = 1. So then, the derivative is not found by means of differentiation but rather by means of the application of the binomial theorem, therefore multiplication; and this because the increased variable x1 takes part from the very beginning as a binomial, x + x. 4) Although x in x + x is just as indefinite, so far as its magnitude goes, as the indefinite variable x itself, x is defined as a distinct quantity separate from x like a fruit beside the mother who had previously borne her (als Frucht neben ihrer Mutter, bevor diese geschwangert war). x + x not only expresses in an indefinite way the fact that x has increased as a variable; rather, it [also] expresses by how much it has grown, namely, by x. 5) x never appears as x1; the whole development centres around the increment x as soon as the derivative has been found by means of the binomial theorem, by means, that is, of substituting x + x for x in a definite way (in bestimmten Grad). On the left-hand side, however, if in (y1 - y)/ x, the x becomes = 0, it finally appears as x1 - x again, so that: (y1 - y)/ x = (y1 - y)/(x1 - x) .*6 The positive side, where x1 - x = 0 takes place, namely x1 becoming = x, can therefore never enter into the development, since x1 as such never enters into the side of the resultant series (Entwicklungsreihe); the real mystery of the differential calculus makes itself evident as never before. 6) If y = f(x) and y1 = f(x + x), then we can say that in using this method the development of y1 solves the problem of finding the derivative. c) x + x = x1 (so that y + y = y1 as well). x here may only appear in the form x = x1 - x, therefore in the negative form of the difference between x1 and x, and not in the positive form of the increment of x, as in x1 = x + x. 1) Here the increased x is distinguished as x1 from itself, before it grows, namely from x; but x1 does not appear as an x increased by x, so x1 therefore remains just exactly as indefinite as x is. 2) Furthermore: however x enters into any original function, so x1 does as the increased variable in the original function now altered by the increase. For example, if x takes part in the function x , so does x1 in the function x1 . Whereas previously, by means of substituting (x + x) wherever x appeared in the original function, the derivative had been provided ready-made by the use of the binomial, leaving it burdened with the factor x and the first of other terms in x burdened with x etc., so no there is just as little which can be derived directly from the immediate form of the monomial - x1 - as could be got from x . It does provide, however, the difference x1 - x . We know from algebra that all differences of the form x - a are divisible by x - a ; the given case, therefore, is divisible by x1 - x. In therefore dividing x1 - x by x1 - x (instead of, [as] previously, multiplying the term (x + x) by itself to the degree specified by the function), we obtain an expression of the form (x1 - x)P, wherein nothing is affected whether the original function of x contains many terms (and so contains x to various powers) or as in our example is of a single term. This x1 - x passes by division to denominator of y1 - y on the left-hand side and thus produces (y1 - y)/(x1 - x) there, the ratio of the difference of the function to the difference of the independent variable x in its abstract difference-formula (Differenzform). The development of the difference between the function expressed in x1 and that expressed in x into terms, all of which have x1 - x as a factor, may well require algebraic manipulation (Man ver) to a greater or lesser degree, and thus may not always shed as much light as the form x1 - x . This has no effect on the method. When by is nature the original function does not allow the direct development into (x1 - x)P, as was the case with f(x) = uz (two variables both dependent on x), (x1 - x) appears [in] the factor 1/(x1 - x). Furthermore, after the removal of x1 - x to the left hand side by means of dividing both sides by it, x1 - x still continues to exist in P itself (as, for example, in the derivation from y = ax, where we find (y1 - y)/(x1 - x) = (a^{x}){(a - 1) + ((x1 - x) - 1)/1 2 + etc.}, = (a^{x}){(a - 1) - (a - 1) /2 + (a - 1) /3 - etc.} ; ___________ The first finite difference, x1 - x , where y = x and y1 = x1 , has therefore been evolved to y1 - y = (x1 - x)P , (y1 - y)/(x1 - x) = P . ____________
Marx's Mathematical Manuscripts 1881