Marx's Mathematical Manuscripts 1881

Written: August, 1881;

Source: Marx's Mathematical Manuscripts, New Park Publications, 1983;

First published: in Russian translation, in *Pod znamenem marksizma*, 1933.

Let us now develop according to d’Alembert’s method:

a) *f(u) ^{98}* or

b) *f(x)* or u = *x³ + ax²* .

*y = 3u²* , (1)

*f(u) = 3u²* . (1a)

*f(u + h) = 3(u + h)² ,*

*f(u + h) - f(u) = 3(u + h)² - 3u²*

*= 3u² + 6uh + 3h² - 3u² = 6uh + 3h²* (2)

(here is the derived function, already complete in the coefficient of h by means of the binomial theorem),

*(f(u + h) - f(u))/h = 6u + 3h .*

*f’(u) = 6u*, already given complete in (2), is freed of its factor h by means of division.

*(f(u + 0) - f(u))/0 = 6u ,*

*(y _{1} - y)/(u1 - u) ,* alias

Substituting in here the value of u from equation b) gives

*dy/du = 6(x³ + ax²) .*

Since *y* in a) is differentiated with respect to *u*, thus

*(u1 - u) = h* or *h = (u1 - u) ,*

since *u* is the independent variable.

And so:

*dy/du = 6(x³ + ax²) .*

(This is obtained from *f(u)* or *y = 3u²* .)

[We now develop b) in the same manner, so that]

b) *f(x)* or *u = x³ + ax²* ,

*f(x + h) = (x + h)³ + a(x + h)² ,*

*f(x + h) - f(x) = (x + h)³ + a(x + h)² - x³ - ax²*

*= x³ + 3x²h + 3xh² + h³ + ax² + 2axh + ah² { - x³ || - ax²*

*= (3x² + 2ax)h + (3x + a)h² + h³ ,*

*(f(x + h) - f(x))/h = 3x² + 2ax + (3x + a)h + h² .*

If we now set *h = 0*, on the second side:

*0/0* or *du/dx = 3x² + 2ax .*

The derived function is already contained complete however, in

*f(x + h) = (x + h)³ + a(x + h)² ,*

since this produces

*x³ + 3x²h + 3xh² + h³ + ax² + 2axh + ah² .*

Thus

*x³ + ax² + (3x² + 2ax)h + (3x + a)h² + h³ .*

It already appears complete as the coefficient of *h*. This derivative is therefore not obtained by means of differentiation, but rather by means of an increase from *f(x)* to *f(x + h)* and thus from *x³ + ax²* to *(x + h)³ + a(x + h)²*. It is obtained simply by virtue of the fact that when x becomes x + h we obtain a binomial in *x + h* of defined degree on the second side, a binomial whose second term, multiplied *(behaftetes)* by *h*, contains the derived function of *u, f’(u)*, ready-made *(fix und fertig)*.

The rest of the procedures serve only to liberate the *f’(x)* thus given from the very beginning from its own coefficient *h* and from all other terms.

The equation

*(f(x + h) - f(x))/h =* etc.

provides two things: first, it makes it possible to obtain the numerator on the first side as the difference of *f(x)*, presently *= Δf(x)*; on the second side, however, it provides the algebraic opportunity to extract the original function given in *x, x³ + ax²*, from the product of *(x + h)³ + a(x + h)²* etc.

*___________*

So we continue. We have obtained for a):

*dy/du = 6(x³ + ax²) ,*

and for b):

*du/dx = 3x² + 2ax .*

We multiply *dy/du* by *du/dx*, so that

*dy/du ⋅ du/dx = dy/dx ,*

which was to be found. Let us substitute in here the values found for *dy/dx* and *du/dx*; so that

*dy/dx = 6(x³ + ax²)(3x² + 2ax)*

and therefore, generally expressed, if we have:

*y = f(u); dy/du = df(u)/du, u = f(x); du/dx = df(x)/dx ,*

hence

*dy/du ⋅ du/dx or dy/dx = df(u)/du ⋅ df(x)/dx*

If we now substitute *h = u1 - u* into equation a) and *h = x _{1} - x* into equation b), things are so arranged that:

*y* or *f(u) = 3u² ,*

*f(u + (u1 - u)) = 3(u + (u1 - u))² ,*

*= 3u² + 6u(u1 - u) + 3(u1 - u)² ,*

*f(u + (u1 - u)) - f(u) = 3u² + 6u(u1 - u) + 3(u1 - u)(u1 - u) - 3u² ,*

hence:

*f(u + (u1 - u)) - f(u) = 6u(u1 - u) + 3(u1 - u)² ,*

*(f(u + (u1 - u)) - f(u))/(u1 - u) = 6u + 3(u1 - u) .*

Hence [if] *u1 - u* in the first term = 0, then

*dy/du = 6u + 0 = 6u .*

*____________*

This shows that when *f(u)* from the very beginning becomes *f(u + (u1 - u))*, then its increment appears as the positive second term of a defined binomial on the second side, and this second term, which is multiplied by *(u1 - u)* or *h* by the binomial theorem, immediately becomes the coefficient to be found. If the second term is polynomial, as it is in

*x³ + ax²,* which becomes *(x + h)³ + a(x + h)² ,*

or

*(x + (x _{1} - x))³ + a(x + (x_{1} - x))² ,*

then we have only to sum the terms multiplied by *x _{1} - x* to the first power, alias h to the first power, as the coefficient of

This result shows:

1) that when in d’Alembert’s development *x _{1} - x = h* is put in reverse

The second term which one finds in that manner attached to *h* or *(x _{1} - x)* is the complete first derived function. The problem now consists of freeing it of

The crucial procedure *(Mittelprozedur)*, however, consists of the development of the equation

*f(x + h) - f(x)* or *f(x + (x _{1} - x)) - f(x) =* [...].

The equation has the sole purpose *(Zweck)* here of making the original function vanish on the second side, since the development [of] *f(x + h)* necessarily contains *f(x)* together with its increment developed by means of the binomial. This [*f(x)*] is thus extracted from the second side.

Therefore what happens, for example, in

*(x + h)³ + a(x + h)² - x³ - ax² ,*

is, that the first terms *x³* and *ax²* are extracted from the binomial *(x + h)³ + a(x + h)²*; we thus obtain, multiplied by *h* or *(x _{1} - x)*, the already complete derived function as the first term of the equation.

The first differentiation on the second side is nothing but the simple subtraction of the original function from its increased expression, which thus gives us the increment by which it has increased and whose first term, multiplied by *h*, is already the complete derived function. The other terms can only contain *h²* etc. or *(x _{1} - x)²* etc. as coefficients; they are reduced by one power with the first division of both sides by

2) The difference from the method of *f(x _{1}) - f(x) =* etc. lies in the fact that, when we have for example

*f(x)* or *u = x³ + ax² ,*

*f(x _{1})* or

the first increment *(Anwachs)* of the variable *x* by no means provides us with *f’(x)* ready-made from the very beginning.

*f(x _{1}) - f(x)* or

Here by no means is it a matter of extracting the original function again, since *x _{1}³ + ax_{1}²* does not contain

Namely,

*= (x _{1}³ - x³) + a(x_{1}² - x²) .*

It is now clear that when we again resolve both of these two terms into factors of *x _{1} - x*, we obtain functions in

*f(x _{1}) - f(x)* or

We divide this by x_{1} - x, and the left-hand side as well, so that:

*(f(x _{1}) - f(x))/(x_{1} - x)* or

By means of this division we have obtained the preliminary derivative. Each of its parts contains terms in *x _{1}*.

Thus can finally obtain the first derived function in *x* only when we set *x _{1} = x*, so that

*x _{1}² = x², x_{1}x = x² ,*

and thus:

*x(x _{1}² + x_{1}x + x²) = 3x²* and

so that:

*a(2x) = 2xa .*

The result on the other [side]

*df(x)/dx = du/dx = 0/0 .*

Thus the derived function is here only obtained by setting *x _{1} = x*, so that

But *x _{1} = x* also leads to

We could have said from the very beginning: we have to obtain a derivative in *x*, and *x* in the end. This can only be transformed into the derivative in *x* when *x _{1}* is set

3) Even if this treatment of *x*, where an increment (*x _{1} - x = Δx*, for example, or

What distinguishes this method from Lagrange, however, is that it really differentiates, so that the differential expression also originates on the symbolic side, while with him the derivation does not represent the differentiation algebraically, but instead derives the functions algebraically directly from the binomial and simply accepts their differential form ‘by symmetry’, since it is known from differential calculus that the first derived function *= dy/dx*, the second *= d²y/dx²*;, etc.

^{97}
These notes represent the contents of sheets a to g. Sheets h to n, containing only first-draft fragments or unfinished notes the sense of which is hard to make out, are not published here; concerning them see the Description, pp.468-470 [Yanovskaya, 1968]. Sheets a to g are devoted to an analysis of d’Alembert’s method applied to the same example of a compound function which Marx considers in the manuscript ‘On the Differential’.

^{98}
The symbols *f(x)*, *f(u)* are employed here as contractions for the expressions, ‘some function in *x*’ and ‘some (other) function in *u*’. In the manuscript ‘On the Differential’ written later, Marx already designates these functions with different letters in the analysis of the same example.