Marx's Mathematical Manuscripts 1881

Appendix VI. Taylor's and MacLaurin's Theorems and Lagrange's theory of Analytic Functions in the source-books used by Marx

Written: August, 1881;
Source: Marx's Mathematical Manuscripts, New Park Publications, 1983;
First published: in Russian translation, in Pod znamenem marksizma, 1933.

1) These theorems, including Lagrange’s closely connected theory of analytic functions, attracted Marx’s particular attention, and he specifically devoted a series of longer, more important manuscripts to them (see mss 4000, 4001, 4300, 4301, 4302 [not translated]). In order to understand these manuscripts, particularly the critique to which Marx subjected the proof of Taylor’s theorem which had been introduced in the handbooks at Marx’s disposal, it is necessary to become acquainted with these proofs and with the corresponding ideas of Lagrange. Before we approach them, however, let us establish something of the history of Taylor’s and MacLaurin’s theorems.*

Taylor’s Theorem is actually included as the 7th proposition of the book Methodus incrementorum directa et inversa by the English mathematician Brook Taylor (1685-1731), published in London in 1715. Taylor had already advised his teacher John Machin by letter of this result in 1712. ‘Taylor’s Theorem’ was so called for the first time in 1784 in the article ‘Approximations’ in the French Encyclopaedia (Encycclopédie méthodique) of Condorcet. In 1786 Simon Lhuilier also used this title in the book Exposition élémentaire des calculs supérieure, honoured by an award by the Berlin Academy of Sciences (the thesis had been offered in a competition of the Academy). Since that time the theorem has entered all the handbooks of mathematical analysis and no one has called it anything else. We know nowadays, however, that the Scottish mathematician James Gregory already possessed it in the years 1671-72.

Both Gregory and Taylor approached ‘Taylor’s Theorem’ starting from finite differences. At this point Taylor addressed himself directly to the problem of considering Newton’s deliberately utterly vague explanation of his interpolation formulae. Newton had obtained his theorem by first allowing the independent variable to differ from zero by a (finite) increment and then - after a series of transformations - returning it to zero ‘by dividing it into an infinitely large number of pieces’. If we replace Taylor’s extremely cumbersome notation by more modern notation, the proof appears as follows.

Let y = f(x), where x is a variable which is varied, as he says, uniformly’, that is, obtaining the successive values x, x + Δx, x + 2Δx, ..., x + nΔx = x + h. And let the corresponding values of f(x) be y ( or y0), y1, y2, ..., yn. Let the successive differences (differences of the first order) between yk-1 and yk (k = 0, 1, ..., n - 1) be Δy, Δy1, ..., Δyn-1; the differences between these differences (differences of the second order) are Δ²y, Δ²y1, ..., Δ²yn-2; and so on. In order to visualise all this, let us write it in mathematic form:

x + Δx x + 2Δx x + 3Δx ... x + nΔx

y1 y2 y3 ... yn

Δy Δy1 Δy2 ... Δyn-1

Δ²y Δ²y1 ... Δ²yn-2

Δ³y ... Δ³yn-3

.... ..... ... .

It is then clear that:

y1 = y + Δy,

y2 = y1 + Δy1 , Δy1 = Δy + Δ²y ,

y3 = y2 + Δy2 , Δy2 = Δy1 + Δ²y1 , Δ²y1 = Δ²y + Δ³y ,

Hence we further obtain:

f(x + Δx) = y1 = y + Δy ,

f(x + 2Δx) = y2 = (y + Δy) + (Δy + Δ²y) = y + 2Δy + Δ²y ,

f(x + 3Δx) = y3 = (y + 2Δy + Δ²y) + (Δy + Δ²y) + (Δ²y + Δ³y)

= y + 3Δy + 3Δ²y + Δ³y ,

............... .. ..... .

Having observed the general regularity, Taylor concludes from this that:

f(x + nΔx) = y + nΔy + (n(n - 1)/1⋅2)⋅Δ²y + (n(n - 1) (n - 2))/1⋅2⋅3)⋅Δ³y + ... + Δny , (1)

which is Newton’s interpolation formula (for interpolation across equal intervals). Its similarity to Newton’s binomial theorem is striking - particularly the fact that the coefficients in the expansion into Δy, Δ²y, ... ,Δny are exactly the same.

Setting nΔx = h (Taylor used v instead of h), we will have:

n = h/Δx , n - 1 = (h - Δx)/Δx , n - 2 = (h - 2Δx)/Δx ,

..., n - (n - 1) = (h - (n - 1)Δx)/Δx .

Substituting these values for n, (n - 1), (n - 2), ... into formula (1), Taylor obtained (in our notation):

f(x + h) = y + h⋅(Δy/Δx) + ((h(h - Δx))/1⋅2)⋅(Δ²y/Δx²)

+ ((h(h - Δx)⋅(h - 2Δx))/1⋅2⋅3)⋅(Δ³y/Δx³) + ..., (2)

although he didn’t even write out the last term ,

(h(h - Δx)⋅(h - 2Δx)...(h - (n - 1)Δx))/1⋅2...⋅n)⋅(Δny/Δxn)

He now assumed h to be fixed, n to be actually infinitely large, and Δx to be actually infinitely small (‘zero’), inferring that this transformed Δy/Δx into the first fluxion y. (dy/dx according to Leibnitz), Δ²y/Δx² into the second fluxion y.. (d²y/dx² according to Leibnitz), and so on. This transforms formula (2) into:

f(x + h) y + y.h + y..⋅y²/1⋅2 + y...⋅h³/1⋅2⋅3 + ...,

that is, into Taylor’s series.

Thus, even beginning with finite differences and only then ‘removing’ them, Taylor still operated strictly in the style of Newton and Leibnitz, with actually infinitely large and actually infinitely small quantities and with the symbolic formulae of the calculus of fluxions, not wondering whether they had any ‘real equivalent’ and not bothering to consider, of course, the convergence of the obtained series (even so the value of f(x + h)). One must note here that, although Taylor was an ardent adherent of Newton’s in the quarrel with Leibnitz and therefore never used the latter’s notation nor ever cited him, it is nonetheless no accident that Euler presented the proof*2 in the language of Leibnitz. As D.D. Mordukai-Boltovskoi notes, in essence Taylor addressed the Newtonian fluxions from the Leibnitzian, not the Newtonian, standpoint, namely from that of finite differences (see the Kommentarii cited in Yanovskaya, 1968, p.396).

As for the history of MacLaurin’s Theorem, it must be noted first of all that is was already present in Taylor in the form of a special case of his theorem at x = 0. It is true that, unlike MacLaurin, Taylor never used the ‘MacLaurin series’ for the expansions already known at this time, for ax, sin x/a, cos x/a which are more easily obtained using this theorem.

Furthermore, with respect to the manuscripts of Marx, who specifically mentioned that he borrowed the ‘algebraic expansion’ directly from MacLaurin, it must be noted that the proofs of MacLaurin’s Theorem (by the method of indeterminate coefficients) which were presented in Boucharlat’s and Hind’s textbooks actually belonged to MacLaurin himself. Such direct borrowing from the author whose name the theorem bears may also have taken place, of course, with reference to Taylor’s Theorem. The bibliographic list which Marx compiled while preparing the historical sketch is apparent evidence that he had decided to become acquainted with Taylor’s work in the original, although he did not succeed in carrying out this intention.

2) We find the same order in which Marx criticised the proof of Taylor’s Theorem in manuscript 4302, in Boucharlat’s textbook as well (J.-L. Boucharlat, Elémens de calcul differérentiel, 5th ed., Paris, 1838; Marx apparently had an English translation done from a different edition).

Having stated the problem of successive differentiation in § 30 (pp.19-20) - where, by the way, after having obtained 6a as the third derivative of ax³ he remarks (p.20), ‘here it is no longer possible to differentiate since 6a is a constant’ - Boucharlat passes to MacLaurin’s Theorem (§31, pp.20-21), proving it by assuming the proof of Taylor’s Theorem (later proved in §§55-57, pp.34-37).

As was already mentioned, Boucharlat proves MacLaurin’s Theorem by following MacLaurin himself. He apparently did not read the latter’s work, however. In fact, with respect to the title ‘MacLaurin’s Theorem’, Boucharlat writes, ‘this theorem, as G.Peacock has noted, was discovered by G. Stirling in 1717, consequently earlier than MacLaurin used it,’ although, as we have already mentioned, MacLaurin fully acknowledged that Taylor already had the theorem.

Boucharlat’s proof - which raises not a single question about the correctness of the assumptions made, not to mention the convergence of the series under consideration - we represent below in almost literal translation.

‘Let y be a function of x; let us expand it in terms of x and assume:

y = A + Bx + Cx² + Dx³ + Ex4 + etc. ; (16)

we obtain, differentiating and dividing by dx :

dy/dx = B + 2Cx + 3Dx² + 4Ex³ + etc. ,

d²y/dx² = 2C + 2⋅3Dx + 3⋅4Ex² + etc. ,

d³y/dx³ = 2⋅3D + 2⋅3⋅4Ex + etc. ,


Let us denote by (y) that into which y is transformed when x = 0,

by (dy/dx) that into which dy/dx is transformed when x = 0,

by (d²y/dx²) that into which d²y/dx² is transformed when x = 0,

... ...

the preceeding equations give us

(y) = A, (dy/dx) = B, (d²y/dx²) = 2C, (d³y/dx³) = 2⋅3D ,

whence we extract

A = (y), B = (dy/dx), C = (1/2)⋅(d²y/dx²), D = (1/2⋅3)⋅(d³y/dx³) ;

substituting these values into (16), we will have

y = (y) + (dy/dx)⋅x + (1/2)⋅(d²y/dx²)⋅x² + (1/2⋅3)⋅(d³y/dx³)⋅x³ + ... ; (17)

and this is MacLaurin’s formula.’

In the following §§32-34 (pp.21-22) expansions are found by means of MacLaurin’s formula for

y = 1/(a + x) , y = sqrt{a² + bx}, y = (a + x)m .

By this means the binomial theorem is derived from MacLaurin’s theorem in the third example. In the first appendix to our 5th edition of Boucharlat’s textbook entitled ‘Proof of Newton’s formulae by means of differential calculus’, a direct derivation (by the same method of indeterminate coefficients) is given of Newton’s binomial theorem (for positive integer powers) by means of successive differentiation. It appears as follows.

Boucharlat begins with an expansion of (1 + z)m, from which the (?)required expansion for (a + x)m is obtained by the substitution of m⋅(x/a). Assume, he says,

(1 + z)m = A + Bz + Cz² + Dz³ + Ez4 + ... (1)

Setting z = 0 he obtains A = 1 and consequently

(1 + z)m = 1 + Bz + Cz² + Dz³ + Ez4 + ... .

Differentiating both sides of this equation with respect to z, he next finds

m(1 + z)m - 1 = B + 2Cz + 3Dz² + 4Ez³ + etc.

Referring to the fact that this equation is valid for any z, Boucharlat set z = 0 and obtains by this means m = B. Differentiating once more and again setting z = 0, he obtains

m(m - 1) = 2C,

whence he finds

C = (m(m - 1))/2 ,

after which he concludes: ‘In the same manner all remaining coefficients are determined, and upon substituting their values into equation (1) this equation is transformed to

(1 + z)m = 1 + mz + ((m(m - 1))/(1⋅2))⋅z² + ((m(m - 1)⋅(m - 2))/(1⋅2⋅3))⋅z³ + etc.’


3) Boucharlat also demonstrates Taylor’s Theorem by the method of indeterminate coefficients. In this case he not only assumes that an arbitrary function of many variables may be expanded into a series of powers of any of the variables, but he also considers this expansion unique; that is, that the coefficients of any two such expansions (in powers of one and the same variable) must be equal. This makes it possible to apply the method of indeterminate coefficients.

In order to arrive at this possibility, that is, of comparing the coefficients of two expansions of one and the same function, Boucharlat begins with a lemma which asserts that the derivatives of f(x + h) with respect to x and to h are equal. Since Marx expresses dissatisfaction in manuscript 4302 (see Yanovskaya, 1968, p.540 [not translated]) with the demonstration of this lemma in Boucharlat’s course-book, while it is impossible even to understand pp.41-42 (see Note 117 Yanovskaya, 1968 [not translated]) of manuscript 3888 without being acquainted with this proof, we present it here in full.

Devoted to this is §55 (pp.34-35), in which we read:

‘If in some function y of x the variable x changes to x + h, we then obtain one and the same differential coefficient both when x is the variable while h is constant, and when h is the variable while x is constant.

‘If in order to show this we substitute x + h = x1*3 in place of x in the equation y = f(x), we then have y1 = f(x); the differential of f(x1) will then be equal to some other function of x1, represented by φ(x1), multiplied by dx; consequently, dy1 = φ(x1)dx1 or if we replace x1 by its value x + h,

dy1 = φ(x + h)⋅d(x + h) .

But the only change which the hypothesis that x is variable while h is constant introduces into this differential refers solely to the factor d(x + h), which reduces to dx when x is variable while h is constant; consequently, in this case we have

dy1 = φ(x + h)dx ,

whence we obtain

dy1/dx = φ(x + h) . (35)

‘If on the other hand we make x constant while h is variable, the factor d(x + h) then reduces to dh and we will have

dy1 = φ(x + h)dh ,

that is,

dy1/dh = φ(x + h) ; (36)

comparing these two values for φ(x + h), we obtain

dy1/dx = dy1/dh .’

In the following §56 Boucharlat extends this lemma to derivatives of higher order and in §57 uses it to prove Taylor’s Theorem. He begins this ‘proof’ with the following words on what he considers - and as Marx calls it - his ‘starting equation’ (37), applicable to any function: ‘Let y1 be a function of x + h; let us assume that when we develop this function into powers of h we obtain

y1 = y + Ah + Bh² + Ch³ + etc. , (37)

where A, B, C, ... are unknown functions of x which are yet to be determined.’

Differentiating equation (37) with respect to h and with respect to x, and having obtained by this means

dy1/dh = A + 2Bh + 3Ch² + etc. ,

dy1/dx = dy/dx + (dA/dx)⋅h + (dB/dx)⋅h² etc. ,

Boucharlat then sets the coefficients of corresponding powers of h in the two equations equal to each other, referring to the lemma, and by this means obtains the expressions he needs for the coefficients A, B, C, ... of y and its successive derivatives. Marx gives an account of this proof on one occasion in manuscript 3888 (sheets 54-55; pp.50-51 in Marx’s enumeration), where he compares it to the proof of MacLaurin’s Theorem presented above. He criticises this proof in manuscript 4302, primarily for a lack of foundation for its initial hypothesis.

The following §§58-61 in Boucharlat’s book contain examples of expansions of f(x + h) by Taylor’s formula in the case of f(x) equals sqrt{x}, sin x, cos x, log x. Questions about the convergence of the series obtained are not even mentioned. Cases of inapplicability of the Taylor series are only considered in the very last paragraphs of the first part of the book (devoted to differential calculus) which are printed in small type.

The concluding §62 of the section on Taylor’s Theorem and its applications is devoted to a proof of MacLaurin’s Theorem from Taylor’s Theorem. Marx reproduces this proof in full in manuscript 3888 (see sheets 55-56; pp.51-52 in Marx’s enumeration).


* As sources we have used: M. Cantor, Vorlesungen über Geschichte der Mathematik, 2nd ed, Vol.3, pp.378-382; D.D. Mordukhai-Boltovskooi, ‘Kommentarii k “Metodu raznosteei”’ (Commentary on the ‘Method of Differences’) in the book İsaak Nyuton, Matematicheskie roboty, Moscow/Leningrad 1937, pp.394-396; M.V.Vygodskii, ‘Vstupitel’noe slovo k “Differentsial’nomu inschisleniya” L. Eilera’ (Introduction to L. Euler’s ‘Differential Calculus’) in the book L. Euler, Differentsial’noe ischislenie, Moscow/Leningrad, 1949, pp.10-12; G. Vileitner, Istoriya Matametiki ot Dakarta do Serediny XIX stoletiya, Moscow 1960, pp.138-140; O. Becker & J.E. Hofmann, Geschichte der Mathematik, Bonn, 1951, pp.200-201,219; G.G. Tseiten, Istoriya matematiki v XVI I XVII vekakh, Moscow/Leningrad, 1938, pp.412, 445; D.Ya. Stroik (Dirk Struik), Kratkii ocherk istorii matematiki Moscow 1964, no.153-154. For more complete coverage see the book by M. Cantor, pp.378-382.
*2 Still, Euler proved Taylor’s theorem following Taylor. See L. Euler, Differential Calculus, chapter 3, ‘On the Approximation of Finite Differences’, §§44-48.
*3 Although Boucharlat does employ Lagrange’s notation for derived function, he designates the increased x and y (i.e. (x + h) and f(x + h) ) as and . We have replaced this designation with x1, y1 .