Marx's Mathematical Manuscripts 1881

# Analysis of D’Alembert's Method by Means of Yet Another Example97

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 y = 3u² ;

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 ,

(y1 - y)/(u1 - u) , alias 0/0 = dy/dx = 6u .

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 = x1 - 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 + (x1 - x))³ + a(x + (x1 - x))² ,

then we have only to sum the terms multiplied by x1 - x to the first power, alias h to the first power, as the coefficient of h or x1 - x; and we have again the complete coefficient.

This result shows:

1) that when in d’Alembert’s development x1 - x = h is put in reverse h = x1 - x, thereby absolutely nothing is changed in the method itself, rather the method simply brings out more clearly how to obtain the binomial by means of f(x + h) or f(x + (x1 - x)) for the algebraic expression on the other side in place of the original function, in place of 3u² for example in the given case.

The second term which one finds in that manner attached to h or (x1 - x) is the complete first derived function. The problem now consists of freeing it of h or x1 - x, which is easily done. There the derived function is complete; it is therefore not found by setting x1 - x = 0, but rather freed of its factor (x1 - x) and accessories. Just as it is found by simple multiplication (the binomial development) as the second term [with] x1 - x, so it is finally freed of the latter by means of division of both sides by x1 - x.

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

f(x + h) - f(x) or f(x + (x1 - 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 and ax² are extracted from the binomial (x + h)³ + a(x + h)²; we thus obtain, multiplied by h or (x1 - 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 etc. or (x1 - x)² etc. as coefficients; they are reduced by one power with the first division of both sides by x1 - x, while the first term emerges without any h.

2) The difference from the method of f(x1) - f(x) = etc. lies in the fact that, when we have for example

f(x) or u = x³ + ax² ,

f(x1) or u1 = x1³ + ax1² ,

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

f(x1) - f(x) or u1 - u = x1³ + ax1² - (x³ + ax²) .

Here by no means is it a matter of extracting the original function again, since x1³ + ax1² does not contain and ax² in any form. On the contrary, this first difference equation provides us with an opportunity for development (Entwicklungsmoment), namely the transformation of both of the two original terms into differences of [power of] x1 and x.

Namely,

= (x1³ - x³) + a(x1² - x²) .

It is now clear that when we again resolve both of these two terms into factors of x1 - x, we obtain functions in x1 and x as coefficients of x1 - x, namely:

f(x1) - f(x) or u1 - u = (x1 - x) (x² +x1x + x²) + a(x1 - x) (x1 + x) .

We divide this by x1 - x, and the left-hand side as well, so that:

(f(x1) - f(x))/(x1 - x) or (u1 - u)/(x1 - x) = (x1² + x1x + x²) + a(x1 + x) .

By means of this division we have obtained the preliminary derivative. Each of its parts contains terms in x1.

Thus can finally obtain the first derived function in x only when we set x1 = x, so that x1 - x = 0, and then

x1² = x², x1x = x² ,

and thus:

x(x1² + x1x + x²) = 3x² and x1 + x = x + x = 2x ;

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 x1 = x, so that x1 - x = 0. x1 = x provides the final positive result in the real function of x.

But x1 = x also leads to x1 - x = 0 and therefore at the same time, beside this positive result, to the symbolic 0/0 or dy/dx on the other side.

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 x1 is set = x; but setting x1 = x is the same as setting x1 - x = 0, which nullification is positively expressed by the formula x1 = x which is necessary for the transformation of the derivative to a function of x, while its negative form, x1 - x = 0, must provide us with the symbol.

3) Even if this treatment of x, where an increment (x1 - x = Δx, for example, or h) is not independently introduced next to it, was already well-known, something which is very probable and of which I shall convince myself by ‘consulting J[ohn] Landen at the [British] Museum, still its essential difference cannot have been grasped.

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.

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’.
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.