Lecture 13 Fermat's discoveries on Integration and Tangents.

Pierre de Fermat (1601--1665) was one of a group of 6 French mathematicians, who worked on similar problems and communicated their results, leading to rapid advances. The others were:

  1. Rene Descartes (1596--1650)
  2. Gilles Personne de Roberval (1602--1675)
  3. Gerard Desargues (1591--1661) and
  4. Blaise Pascal (1623--1662)

The communication was by letters circulated by Friar Marin Mersenne.

Fermat studied classical mathematics at the Univesity of Toulouse, and during his Law studies he read Viete, with his new ideas for notation, discovery and elucidation, and also the works of Apollonius and Diophantos. Using the comments of Pappus, he tried to restore the lost works of Apollonius and express them in the algebraic notation of Viete. This marks the real beginning of analytic geometry. For example, he generalized the theorem of Galileo: if sraight lines are drawn from a variable point P to any number of points such that the sum of the squares of the distances is constant, the locus of P is a circle.

He developed the idea that whenever you have an equation relating two unknown quantities, you have a curve, and vice versa. Whereas Descartes used this idea to solve geometric problems, Fermat developed analytic geometry systematically, beginning with linear equations, then going to quadratics. Since Fermat's work was published posthumously, Descartes gets the credit. Fermat suggests his ideas could be used for quadratics in 3 variables, ie curves in space.

He developed methods of finding points at which a function reaches max and min values, and finding tangents to curves.


The text: Integration


Archimedes did not employ geometric progressions except for the quadrature of the parabola; in comparing various quantities he restricted himself to arithmetic progressions. Was this because he found that the geometric progression was less suitable for the quadrature? Was it because the particular device that he used to square the parabola by this progression can only with difficulty be applied to other cases? Whatever the reason may be, I have recognized and proved that this progression is very useful for quadratures, and I am willing to present to modern mathematicians my invention which permits us to square, by a method absolutely similar, parabolas as well as hyperbolas.

The entire-method is based on a well-known property of the geometric progression, namely the following theorem: Given a geometric progression the terms of which decrease indefinitely, the difference between two consecutive terms of this progression is to the smaller of them as the greater one is to the sum of all following terms.

This established, let us discuss first the quadrature of hyperbolas: . I define hyperbolas as curves going to infinity, which, like DSEF [Fig. 1], have the following property.

Let RA and AC be asymptotes which may be extended indefinitely; let us draw parallel to the asymptotes any lines EG, HI, NO, MP, RS, etc.

We shall then always have the same ratio between a given power of AH and the same power of AG on one side, and a power of EC (the same as or different from the preceding) and the same power of Hi on-the other. I mean by powers not only squares, cubes, fourth powers, etc., the exponents of which are 2, 3, 4, etc., but also simple roots the exponent of which is un ity.

I say that all these infinite hyperbolas except the one of Apollonius, or the first, may be squared by the method of geometric progression according to a uniform and general procedure. Let us consider, for example, the hyperbolas the property of which is defined by the relations (AH)2:(AG)2 = EG:HI and (AO)2:(AH)2= HI:NO, etc. I say that the indefinite area which has for base EG and which is bounded on the one side by the curve ES and on the other side by the infinite asymptote GOR is equal to a certain rectilinear area.

Let us consider the terms of an indefinitely decreasing geometric progression; let AG be the first term, AH the second, AO the third, etc. Let us suppose that those terms are close enough to each other that following the method of Archimedes we could adequate according to Diophantos, that is, equate approximately the rectilinear parallelogram GE x GH and the general quadrilateral GHIE; in addition we shall suppose that the first intervals GH, HO, OM, etc. of the consecutive terms are sufficiently equal that we can easily employ Archimedes' method of exhaustion by circumscribed and inscribed polygons. It is enough to make this remark once and we do not need to repeat it and insist constantly upon a device well known to mathematicians.

Now, since AG:AH = AH:AO = AO:AM, we have also AG:AH = GH:HO = HO:OM, for the intervals. But for the parallelograms,

(EG x GH)/( HI x HO) = (HI x H0)/(ON x OM)

Indeed, the ratio (EG x GH):(HI x HO) of the parallelograms consists of the ratios EG/HI and GH/HO; but, as indicated, GH/HO = AG/AH; therefore, the ratio (EG x CH)/(HI x HO) can be decomposed into the ratios EG/HI and AG/AH.

On the other hand, by construction, EG/HI = (AH)2 /(AG)2 or AO/AG, because of the proportionality of the terms; therefore, the ratio (GE x GH)/(HI x HO) is decomposed into the ratios AO/AG and AG/GH; now AO/AH is decomposed into the same ratios; we find consequently for the ratio of the parallelograms: (EG x GH)/(HI x HO) = AO/AH = AH/AG.

Similarly we prove that (HI x HO)/(NO x MO) = AO/AH. But the lines AO, AH, AG, which form the ratios of the parallelograms, define by their construction a geometric progression; hence the infinitely many parallelograms EG x GH, HI x HO, NO x OM, etc., will form a geometric progression, the ratio of which will be AH/AG.

Consequently, according to the basic theorem of our method, GH, the difference of two consecutive terms, will be to the smaller term AG as the first term of the progression, namely, the parallelogram GE x GH, to the sum of all the other parallelograms in infinite number. According to the adequation of Archimedes, this sum is the infinite figure bounded by HI, the asymptote HR, and the infinitely extended curve IND.

Now if we multiply the two terms by EG we obtain GH/AG = (EG x GH)/(EG x AG); here EG x GH is to the infinite area the base of which is HI as EG x GH is to EG x AG. Therefore, the parallelogram EG x AG, which is a given rectilinear area, is adequated to the said figure; if we add on both sides the parallelogram EG x GH, which, because of infinite subdivisions will vanish and will be reduced to nothing, we reach a conclusion that would be easy to confirm by a more lengthy proof carried out in the manner of Archimedes, namely, that for this kind of hyperbola the parallelogram AE is equivalent to the area bounded by the base EG, the asymptote GR, and the curve ED infinitely extended.

It is not difficult to extend this idea to all the hyperbolas defined above except the one that has been indicated.


The text: Maxima and minima

The whole theory of evaluation of maxima and minima presupposes two unknown quantities and the following rule:

Let a be any unknown of the problem (which is in one, two, or three dimensions, depending on the formulation of the problem). Let us indicate the maximum or minimum by a in terms which could be of any degree. We shall now replace the original unknown a by a + e and we shall express thus the maximum or minimum quantity in terms of a and e involving any degree. We shall adequate , to use Diophantus' term, the two expressions of the maximum or minimum quantity and we shall take out their common terms. Now it turns out that both sides will contain terms in e or its powers. We shall divide all terms by e, or by a higher power of e, so that e will be completely removed from at least one of the terms. We suppress then all the terms in which e or one of its powers will still appear, and we shall equate the others; or, if one of the expressions vanishes, we shall equate, which is the same thing, the positive and negative terms. The solution of this last equation will yield the value of a, which will lead to the maximum or minimum, by using again the original expression.

Here is an example: To divide the segment AC at E so that AE x EC may be a maximum.

We write AC = b; let a be one of the segments, so that the other will be b - a, and the product, the maximum of which is to be found, will be ba - a2.

Let now a + e be the first segment of b; the second will be b - a - e, and the product of the segments, ba - a2 + be - 2ae - e2; this must be adequated with the preceding: ba - a2.

Suppressing common terms: be adequal 2ae + e2.

Suppressing e: b = 2a. To solve the problem we must consequently take the half of b.

We can hardly expect a more general method.



Fermat's Integration: The text concerns integration in the sense of computing area under a certain curve. Curves of the form y = xn, where n is a rational not -1, he calls "higher order parabolas" if n > 0 and " hyperbolas" if n < 0 .

Cavalieri, a student of Galileo, had found the integral for n a positive integer. His methods were dubious, while Fermat's where rigorous.

To approximate the area by the sum of areas under inscribed rectangles, the problem is to find a formula for this area involving the width w of the rectangles and find the limit as w -> 0 . For example, for y=xn between x = 1 and x = m, n a positive integer, Fermat and others used 1n + 2n +...+ mn > mn+1/n + 1 > 1n + 2n... +(m-1)n, and known formulae for the sum of the first m powers of n. But this does not work for n negative or rational.

Fermat's solution was to divide the domain not into finitely many equal intervals, but into infinitely many subintervals whose widths form a G.P. Take points whose abscissae are G, rG, rnG,... where r < 1. Erect ordinates at these points with lengths Gn, rnGn, r2nGn, ... Then the areas of the rectangles have sum 2G + rnGn(rnG - r3nG) +....

This is a G.P.with initial term Gn+1(1-r) and ratio rn+1

so sum = Gn+1}(1-r)/(1-rn+1. If n is an integer, this =

Gn+1/1+r+r2+...+r n

Then as r -> 1 , the rectangles become narrower and more equal and the area approaches Gn+1/n+1 .

Fermat shows that this remains true when n is replaced by a rational.


Method of evaluating maxima and minima, and of tangents.

Fermat was lead to his idea on computing maxima and tangents by his study of Kepler's work on the largest parallelipiped which can be inscribed ina sphere and Viete's work on the relationship between coefficients and roots of a polynomial.

Kepler noted that if if you inscribe symmetric blocks of decreasing heights, the volume increases till you get a cube, then decreases, and the rate of change is slow when you are near the maximum. Viete had noted that if the roots of Bx - x2 = C are x1 and x2, then B x1 - x12 = C = Bx2 - x22, so B(x1 - x2) = x12 - x22 and hence B = x1 + x2. Now what happens if x1 = x2 ? We still have x1 = B/2 = x2, but how can you divide by x1 - x2? Note that the equation Bx - x2 = C arises from the geometric problem of dividing a line of length B into two parts whose product is C. It was known since Euclid that the maximum possible value of C is B2/4 , obtained when B is divided into half, and that for any C < B2/4 there are two values of x giving the same product.

So Fermat's insight is to see that to find the maximum value of any polynomial p(x) , you set p(x1) = p(x2) and divide by x1 - x2 to find the relationship between the coefficients and two roots x1 and x2. When you have such a relationship, set x1 = x2 and solve. For example, to maximise Bx2=x3, let Bx12 - x13 = Bx23 - x23, so B(x1-2x2 2) = x13 - x2 3.

Divide by (x1 - x2) to get Bx1 + Bx2 = x12 + x1x2 + x2 2. Set x1 = x2, 2Bx = 3x2, so x = 2B/3 gives the maximum.

The problems: Fermat ignores the possibility x = 0 or negative as solutions, because it is a problem in Geometry. How can you divide by x1 - x2 when they are equal?

The answer to the latter is that Fermat never considered he was dividing by zero. He just assumed that the relationship between roots and coefficients which holds when x1 not equal to x2 continues to hold when x1 = x2.

In fact he calles the roots x and x +e , so he divides by e, and he never assumes p(x) = p(x+e) but p(x) adequal p(x+e) . (Sufficiently equal? becoming equal?)

A similar method is used to determine tangents. In the example the curve is a parabola, given by its geometric description. To find the tangent at B, Fermat needs the point E where the tangent meets the axis, the distance CE is called the subtangent. Pick an arbitrary point A on the tangent and drop perpendiculars BC and AI.

Let t be the subtangent and e = CI. Then f(x+e):f(x) adequal (t+e):t so tf(x+e) adequal (t+e)f(x). By calculating both sides, dividing by e and then discarding the remaining terms in e, Fermat can find t.

For example , say the curve is y = x^2. Then t(x 2 - 2ex +e2) ~ tx2 + ex2 so x2 ~ 2xt + te and hence t~ x/2

Fermat also modified his method to deal with curves we would write as f(x,y) = 0. For example he found the tangent to the curve x3 + y3 = pxy proposed by Descartes.


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 Author: Phill Schultz, schultz@maths.uwa.edu.au