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François Viète

François Viète, seigneur de la Bigotière (Franciscus Vieta)

A course in two strands: the career of a Poitevin lawyer who served three kings and broke Spanish ciphers in the Wars of Religion, and the “analytic art” in which he first wrote letters for known quantities and so made algebra a language of general formulas.

Born 1540, Fontenay-le-Comte Died 1603, Paris Signature work In artem analyticem isagoge, Tours, 1591 Level advanced undergraduate
Viète’s product for π (1593). The areas of the inscribed square, octagon, 16-gon, … approach the circle, and each ratio of successive areas is a nested square root: \(\tfrac{2}{\pi}=\sqrt{\tfrac12}\cdot\sqrt{\tfrac12+\tfrac12\sqrt{\tfrac12}}\cdots\)
Lesson 00

How to use this course

François Viète was never a professional mathematician. He was a lawyer and royal official whose working life ran through the French Wars of Religion. In the gaps of that career he wrote the books that turned algebra from a collection of recipes for particular numbers into a symbolic method. His key idea was to denote given quantities by letters, as well as unknowns. With it he could write a solution as a formula valid for all values, study how the coefficients of an equation depend on its roots, and treat geometry, trigonometry and arithmetic in one language.

Prerequisites for Part II
  • High-school algebra and trigonometry, including addition formulas.
  • Complex numbers in polar form, for Lessons 08 and 09 (Viète did not use them; we do, for proofs).
  • Some Euclidean geometry: similar triangles, tangent circles.

Each technical lesson opens with a paragraph for the general reader, then translates Viète’s statements into modern symbols, sketches his argument or a modern proof, and ends with exercises with folded solutions. Viète wrote in Latin and in his own notation. When this course writes \(x^3+px=q\), Viète wrote something like A cubus + B planum in A aequatur Z solido, and Lesson 04 explains how to read it.

Reference

Timeline

1540Born in Fontenay-le-Comte, Bas-Poitou, son of Étienne Viète, a lawyer.
1558–59Studies law at the University of Poitiers; bachelor of laws in 1559 (MacTutor gives a law degree in 1560).
1560Begins practice as an advocate in Fontenay.
1564Enters the service of Antoinette d’Aubeterre, lady of Soubise, as secretary and tutor to her daughter Catherine de Parthenay.
1571Advocate in Paris.
1573–74Appointed councillor to the Parlement of Brittany at Rennes.
1579Canon mathematicus, trigonometric tables and their theory, printed in Paris.
1580Maître des requêtes in Paris and royal privy councillor under Henry III.
1584–85Removed from office under pressure from the Catholic League; withdraws to Poitou (Beauvoir-sur-Mer, then the Soubise house at Le Parc).
1589Recalled by Henry III to the royal court at Tours. Deciphers letters of the League and of Spain.
1590Publishes the decipherment of a letter from the Spanish commander Moreo.
1591In artem analyticem isagoge and Zeteticorum libri quinque, Tours.
1593Supplementum geometriae; Variorum de rebus mathematicis responsorum liber VIII, with the infinite product for π.
1594–95Solves Adriaan van Roomen’s equation of degree 45; publishes the Responsum in 1595.
1600De numerosa potestatum resolutione; Apollonius Gallus. Begins attacking Clavius over the Gregorian calendar.
1602December: leaves royal service.
1603Dies in Paris (most sources: 23 February; MacTutor: 13 December).
1615Alexander Anderson publishes De aequationum recognitione et emendatione and Ad angularium sectionum theorems.
1646Frans van Schooten’s collected Opera mathematica, Leiden.
Part I · Formation
Lesson 01

Fontenay and Poitiers, 1540–1564

François Viète was born in 1540 at Fontenay-le-Comte, a market town of Bas-Poitou in what is now the Vendée. His father, Étienne Viète, was an attorney in Fontenay and a notary. His mother was an aunt of Barnabé Brisson, later a leading Paris magistrate, and his grandfather had been a merchant of La Rochelle. The family belonged to the provincial legal class, comfortable and well connected.

He attended a Franciscan school in Fontenay and then studied law at the University of Poitiers from 1558, taking his bachelor’s degree in 1559. From 1560 he practised as an advocate in his home town. Parish records of the time already call him seigneur de la Bigotière, after a small farm property at Foussais inherited from his father. No formal training in mathematics is recorded. He appears to have taught himself from printed books, the Greek geometers and the Italian algebraists among them.

A divided province

Poitou was one of the regions where the Protestant Reformation was strongest, and Viète grew up among families on both sides. Whether he himself was Protestant is disputed. His opponents in the Catholic League called him a Huguenot, and MacTutor describes him as one. Other biographers point out that he made a Catholic profession of faith on entering the Parlement of Brittany in 1574, and describe him as a politique, a moderate who put the stability of the state before religious allegiance. All agree that his closest patrons were Protestant.

Lesson 02

Tutor, councillor, courtier, 1564–1584

In 1564 Viète entered the household of Antoinette d’Aubeterre, lady of Soubise, a leading Protestant family of Poitou, to manage legal affairs and to tutor her daughter Catherine de Parthenay. Catherine became an accomplished scholar, and Viète later counted her as his first pupil. For her he wrote lessons in cosmography and astronomy, which survive in manuscript; one strand later became the unpublished Harmonicon coeleste on Ptolemaic and Copernican planetary theory. After the death of Catherine’s father in 1566 the household moved to La Rochelle.

Paris and Rennes

By 1571 Viète was enrolled as an advocate in Paris. He was in the city in August 1572 during the St Bartholomew’s Day massacre of Protestants, a dangerous time for a man with his connections. In 1573 Charles IX named him a councillor at the Parlement of Brittany in Rennes, and he took up the post in 1574.

His first major mathematical work was printed in Paris in 1579, though parts may have been ready years earlier: the Canon mathematicus seu ad triangula, with tables of the trigonometric functions and a systematic treatment of plane and spherical triangles (Lesson 09). In 1580 Henry III appointed him maître des requêtes, an officer reviewing petitions to the king, and royal privy councillor. He was now a member of the king’s administration in Paris.

Lesson 03

Disgrace and the new algebra, 1584–1589

In 1584 the death of the king’s brother made the Protestant Henry of Navarre heir to the throne. The Catholic League, backed by Spain, mobilized against him and against officials suspected of sympathy with Protestants. Viète was removed from his posts, in 1584 according to MacTutor, in 1585 according to French biographers. Letters from Henry of Navarre in 1585 failed to restore him.

He spent the next four or five years in Poitou. He stayed at Beauvoir-sur-Mer on the coast, where he had a house and where Françoise de Rohan, lady of La Garnache, gave him shelter, and later at the Soubise house of Le Parc near Mouchamps, with Catherine de Parthenay. These years without office were his most productive. The ideas of the Isagoge, the Zetetica and the theory of equations that appeared from 1591 onward were worked out in this retreat.

What he was reading

Three sources shaped the new algebra. Greek geometrical analysis, as described by Pappus of Alexandria in Book VII of his Collection (Latin translation by Federico Commandino, 1588), gave him the idea of assuming a problem solved and reasoning back to what is known. Diophantus’s Arithmetica, available in Wilhelm Xylander’s Latin of 1575, gave him a stock of problems and techniques. The Italian algebraists gave him the solution of cubic and quartic equations. Viète’s ambition was to fuse them into a single art of analysis that could, as he put it at the end of the Isagoge, leave no problem unsolved.

Exercise 3.1Pappus distinguishes “theoretical” analysis, which seeks the truth of a statement, from “problematical” analysis, which seeks a construction. Viète divides his art into three parts: zetetic, poristic and exegetic (Lesson 04). Match them as well as you can, and say which part is new.

Show solution
Poristics, which checks the truth of a stated theorem by the analytic art, corresponds to theoretical analysis. Exegetics, which produces the actual solution, numerical or by construction, corresponds to the synthetic completion of problematical analysis. Zetetics, which sets up an equation among letters from the conditions of a problem, has no Greek equivalent: it is the new symbolic step.
Part II · The mathematics
Lesson 04

The analytic art

For the general reader. Before Viète, an algebra book would say “find a number which, multiplied by itself and added to 10 times itself, makes 39.” The numbers 10 and 39 were fixed, and readers had to guess the general rule from examples. Viète introduced letters for the given numbers too. Now the problem and its solution could be written once, for every case.

Goals
  • Read an equation in Viète’s notation.
  • Understand the law of homogeneity and its geometric origin.
  • Distinguish logistica speciosa from logistica numerosa.

Letters for species

In the Isagoge (1591) Viète distinguishes two kinds of calculation. Logistica numerosa operates on numbers. Logistica speciosa operates on “species”, forms of quantities, denoted by letters. Unknowns are written with vowels (A, E, …) and given quantities with consonants (B, C, D, …). René Descartes later replaced this with the convention, still used, of late letters for unknowns and early letters for knowns.

Powers are named, not written as exponents: A quadratum (\(A^2\)), A cubus (\(A^3\)), A quadrato-quadratum (\(A^4\)), and so on. Viète used + and − and wrote “in” for multiplication and a fraction bar for division. For “equals” he wrote the word aequatur.

The law of homogeneity

Every quantity in Viète’s algebra has a dimension: lengths (longitudo), planes, solids, plano-planes, and so on. Only homogeneous quantities can be added or compared. So a given coefficient multiplying \(A\) in a cubic equation must be a “plane”, and the constant term a “solid”:

\[ \textit{A cubus} + \textit{B plano in A aequatur Z solido}\quad\longleftrightarrow\quad A^3+B\,A=Z,\quad \dim B=2,\ \dim Z=3. \]
Principle (Viète, Isagoge, 1591)

Homogeneous quantities may be added, subtracted and compared; heterogeneous ones may not. Multiplying magnitudes adds their dimensions; dividing subtracts them. Every equation must be homogeneous in all its terms.

The rule kept algebra tied to geometry, where it came from: a product of two lengths is an area, so it cannot be added to a length. It also acts as a check on calculations, like dimensional analysis in physics. Descartes removed it in 1637 by fixing a unit length, so that the product of two lengths could again be represented as a length.

Zetetic, poristic, exegetic

Viète divided analysis into three stages. Zetetics: translate the conditions of a problem into an equation among species. Poristics: test the truth of theorems by the same art. Exegetics: extract from the equation the unknown itself, either as a number (by logistica numerosa, Lesson 10) or by geometric construction (Lesson 07).

Exercise 4.1Translate into modern notation and check homogeneity: A quadratum + B in A bis aequatur Z plano. Then solve for \(A\).

Show solution
\(A^2+2BA=Z\), with \(B\) a length and \(Z\) a plane: all terms have dimension 2. Completing the square, \((A+B)^2=Z+B^2\), so \(A=\sqrt{Z+B^2}-B\), again a length.

Exercise 4.2In \(A^3-3B^2A=2C\), what dimensions must \(B\) and \(C\) have? Rewrite the equation in Viète’s style.

Show solution
\(B^2A\) must be a solid, so \(B\) is a length (and \(B^2\) a plane). \(2C\) must be a solid, so \(C\) is a solid. In Viète’s style: A cubus − B quadrato in A ter aequatur C solido bis, where ter means “three times” and bis “twice”.
Lesson 05

The Zetetica: Diophantus in letters

For the general reader. To show what the new notation could do, Viète took problems from Diophantus, who always worked with particular numbers, and solved them once and for all with letters. The result was a short book of general formulas, and a demonstration that algebra could prove things, not just compute them.

Goals
  • Solve sum-and-difference problems in general form.
  • Reproduce Viète’s general solution of the two-square problem.
  • Compare a numerical solution with a general one.

The first zetetic

Book I of the Zetetica begins: given the difference and sum of two sides, find the sides. Let \(B\) be the difference, \(D\) the sum, and \(A\) the smaller side. Then the larger side is \(A+B\), and \(2A+B=D\), so

\[ A=\frac{D-B}{2},\qquad A+B=\frac{D+B}{2}. \]

Viète then states the result in words, as a theorem, and gives a numerical example (difference 40, sum 100, sides 30 and 70). The proof is the calculation itself. Nothing like this is possible in Diophantus, where the given numbers vanish into the computation.

Two squares in general

Diophantus II.8 splits 16 into two rational squares using a line of slope 2 (see the Diophantus course). Viète redoes it in general: to split a given square \(B^2\) into two squares, take any two numbers \(S\) and \(R\) and set the sides of the parts proportional to \(2SR\) and \(S^2-R^2\):

General solution (modern form) \[ B^2=\Big(\frac{2SR}{S^2+R^2}B\Big)^2+\Big(\frac{S^2-R^2}{S^2+R^2}B\Big)^2. \]

With \(B=4\), \(S=2\), \(R=1\) this gives \(\frac{16}{5}\) and \(\frac{12}{5}\), Diophantus’s answer. The general formula makes visible what Diophantus’s method could only suggest: every right triangle with rational sides yields a solution.

Exercise 5.1Given the sum \(D\) and the product \(P\) of two quantities, find them in Viète’s style. What condition on \(D,P\) makes the answer rational?

Show solution
Let the quantities be \(\frac D2+E\) and \(\frac D2-E\) (Diophantus’s trick, now with letters). Then \(\frac{D^2}{4}-E^2=P\), so \(E=\sqrt{\frac{D^2}{4}-P}\). The answer is rational exactly when \(D^2-4P\) is the square of a rational. Note the homogeneity: \(P\) is a plane, \(D^2\) a plane.

Exercise 5.2Use Viète’s formula with \(B=5\), \(S=3\), \(R=2\) to split 25 into two rational squares, and check.

Show solution
\(S^2+R^2=13\), \(2SR=12\), \(S^2-R^2=5\). The sides are \(\frac{60}{13}\) and \(\frac{25}{13}\). Then \(\frac{3600+625}{169}=\frac{4225}{169}=25\).
Lesson 06

Roots and coefficients

For the general reader. Multiply out \((x-2)(x-3)\) and you get \(x^2-5x+6\): the 5 is the sum of the roots and the 6 their product. Viète was the first to state this kind of relation for equations of higher degree. The relations are now called Vieta’s formulas, and they are the start of the theory of equations that led to Galois.

Goals
  • State and prove Vieta’s formulas.
  • See Viète’s own formulation, restricted to positive roots.
  • Use transformations of equations, including removal of the second term.

Viète’s formulation

In De aequationum recognitione et emendatione, written before 1603 and published by Alexander Anderson in 1615, Viète gives statements of the following kind. If \(A(B+D)-A^2=BD\), then \(A\) can be \(B\) or \(D\). For the cubic: if \(A^3-(B+D+G)A^2+(BD+BG+DG)A=BDG\), then \(A\) can be \(B\), \(D\) or \(G\). He continues the pattern to higher degrees. Because he admitted only positive roots, he states it only when all roots are positive. The general form, with negative and complex roots allowed, was given by Albert Girard in 1629.

Theorem (Vieta’s formulas)

Let \(x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=\prod_{i=1}^n(x-r_i)\) over a field, with roots \(r_1,\dots,r_n\) in an extension. Then for \(1\le k\le n\),

\[ a_{n-k}=(-1)^k\,e_k(r_1,\dots,r_n),\qquad e_k=\sum_{i_1<\cdots<i_k}r_{i_1}\cdots r_{i_k}. \]

Proof. Expand the product. A term of degree \(n-k\) arises by choosing \(-r_i\) from exactly \(k\) of the factors and \(x\) from the rest; summing over all choices gives \((-1)^ke_k\,x^{n-k}\).

Transforming equations

The second half of the same work, the emendatio, collects transformations: multiply all roots by a constant, increase them by a constant, take reciprocals. The most important is the removal of the second term. In \(x^n+a_{n-1}x^{n-1}+\cdots\), the substitution \(x=y-\frac{a_{n-1}}{n}\) shifts the roots so that their sum is zero, so the \(y^{n-1}\) term disappears. Every cubic thus reduces to \(y^3+py=q\), where Cardano’s formula or Viète’s methods (Lesson 07) apply. For the reduced cubic Viète also used the substitution \(y=z-\frac{p}{3z}\), now called Vieta’s substitution, which turns \(y^3+py=q\) into the quadratic \(z^6-qz^3-\frac{p^3}{27}=0\) in \(z^3\).

Exercise 6.1Write the cubic with roots 1, 2, 3 and check Viète’s statement for it.

Show solution
\(B+D+G=6\), \(BD+BG+DG=2+3+6=11\), \(BDG=6\). So \(A^3-6A^2+11A=6\), that is \(x^3-6x^2+11x-6=0\). Substituting \(x=1,2,3\) gives \(0\) each time.

Exercise 6.2Remove the second term from \(x^3-6x^2+11x-6=0\). Then verify Vieta’s substitution on the result.

Show solution
Put \(x=y+2\): \((y+2)^3-6(y+2)^2+11(y+2)-6=y^3-y\). So \(y^3-y=0\), with roots \(-1,0,1\) as expected. Here \(p=-1\), \(q=0\); Vieta’s substitution gives \(z^6-0\cdot z^3+\frac1{27}=0\), whose roots \(z^3=\pm\frac{i}{3\sqrt3}\) are complex although all three roots of the cubic are real. This is the irreducible case of Lesson 07.

Exercise 6.3The roots of \(x^3-7x+6=0\) are \(r_1,r_2,r_3\). Without solving, compute \(r_1^2+r_2^2+r_3^2\).

Show solution
\(e_1=0\), \(e_2=-7\). Then \(\sum r_i^2=e_1^2-2e_2=14\). (The roots are \(1,2,-3\): \(1+4+9=14\).)
Lesson 07

Cubics, trisection and the neusis

For the general reader. Cardano’s formula for the cubic has a strange flaw: when all three roots are real, the formula expresses them through square roots of negative numbers. Viète found a way around this using trigonometry. A cubic with three real roots, he showed, is the same problem as dividing an angle into three equal parts.

Goals
  • Derive the triple-angle identity and use it to solve a cubic.
  • State Viète’s reduction of the irreducible case to trisection.
  • Understand the role of the neusis postulate in the Supplementum geometriae.

The trigonometric solution

The identity \(\cos3\theta=4\cos^3\theta-3\cos\theta\) says that \(c=2\cos\theta\) satisfies \(c^3-3c=2\cos3\theta\). Scaling turns any reduced cubic with three real roots into this form.

Theorem (Viète)

Let \(p>0\) and \(|q|\le 2\big(\frac p3\big)^{3/2}\). Then the three roots of \(y^3-py=q\) are real and are given by

\[ y_k=2\sqrt{\frac p3}\,\cos\Big(\frac{\varphi+2\pi k}{3}\Big),\quad k=0,1,2,\qquad\text{where}\quad \cos\varphi=\frac{q}{2}\Big(\frac{3}{p}\Big)^{3/2}. \]

Proof. Put \(y=2\sqrt{p/3}\,\cos\theta\). Then \(y^3-py=2\big(\frac p3\big)^{3/2}(4\cos^3\theta-3\cos\theta)=2\big(\frac p3\big)^{3/2}\cos3\theta\). Setting this equal to \(q\) gives \(\cos3\theta=\cos\varphi\), so \(3\theta=\varphi+2\pi k\). The hypothesis on \(q\) is the condition \(|\cos\varphi|\le1\), equivalently that the discriminant \(4p^3-27q^2\) be non-negative.

Viète expressed this geometrically: solving such a cubic is equivalent to trisecting a given angle. The cases where Cardano’s formula required imaginary quantities were precisely the cases of trisection.

The Supplementum geometriae (1593)

Euclid’s constructions use only ruler and compass, and they cannot trisect a general angle or double a cube (as was proved in 1837 by Pierre Wantzel). Viète proposed to supplement geometry with one further postulate: given two lines and a point, one may draw through the point a line on which the two lines cut off a segment of given length. This is the Greek neusis (verging) construction. With it he showed that every cubic and quartic equation can be solved geometrically, by reducing it either to the trisection of an angle or to the insertion of two mean proportionals between two lengths.

Exercise 7.1Solve \(y^3-3y=1\) by Viète’s method.

Show solution
Here \(p=3\), \(q=1\), so \(y=2\cos\theta\) with \(\cos3\theta=\frac12\), \(3\theta=60^\circ+360^\circ k\). The roots are \(2\cos20^\circ\approx1.879\), \(2\cos140^\circ\approx-1.532\), \(2\cos260^\circ\approx-0.347\). They sum to 0, as Vieta’s formulas require.

Exercise 7.2Show that the two-mean-proportional problem, finding \(x,y\) with \(a:x=x:y=y:b\), reduces to a pure cubic.

Show solution
From \(x^2=ay\) and \(y^2=xb\): \(y=\frac{x^2}{a}\), so \(\frac{x^4}{a^2}=xb\) and \(x^3=a^2b\). With \(b=2a\) this is \(x^3=2a^3\), the doubling of the cube.
Lesson 08

Angular sections and van Roomen’s challenge

For the general reader. If you know the chord of an angle in a circle, how do you find the chord of twice the angle, three times, \(n\) times? Viète worked out the general rule. It let him solve, almost at once, an equation of degree 45 that a Flemish mathematician had posed as a challenge to all the mathematicians of Europe.

Goals
  • State Viète’s formulas for \(\cos n\theta\) and \(\sin n\theta\).
  • Understand how van Roomen’s equation encodes \(\sin45\theta\).
  • See why Viète found 23 positive roots.

Multiple angles

In Ad angularium sectionum analyticen theoremata, published by Anderson in 1615, Viète gives general expressions for the chords of multiple angles. In modern form they follow from expanding \((\cos\theta+i\sin\theta)^n\), a route opened only by de Moivre and Euler in the next century:

Theorem (Viète’s multiple-angle formulas, modern form) \[ \begin{aligned} \cos n\theta&=\sum_{j\ge0}(-1)^j\binom{n}{2j}\cos^{n-2j}\theta\,\sin^{2j}\theta,\\ \sin n\theta&=\sum_{j\ge0}(-1)^j\binom{n}{2j+1}\cos^{n-2j-1}\theta\,\sin^{2j+1}\theta. \end{aligned} \]

Viète found the pattern of binomial coefficients with alternating signs and verified it by repeated use of the addition formulas.

The challenge

In his Ideae mathematicae (Louvain, 1593), Adriaan van Roomen (Adrianus Romanus) proposed the equation

\[ x^{45}-45x^{43}+945x^{41}-12300x^{39}+\cdots+95634x^5-3795x^3+45x=C \]

and gave its solution for a few values of \(C\). The French tradition, recorded by Tallemant des Réaux, says that a Dutch envoy at the court of Henry IV boasted that France had no mathematician able to solve it, and that Viète, summoned, found a solution almost at once. His answer, the Responsum ad problema quod omnibus mathematicis totius orbis construendum proposuit Adrianus Romanus, appeared in 1595.

Viète’s key observation

If \(x=2\sin\theta\), the left side equals \(2\sin45\theta\). Since \(45=3\cdot3\cdot5\), solving the equation amounts to dividing a given angle into three, then three again, then five equal parts: two cubics and a quintic of the same type.

For van Roomen’s main example the constant was \(C=\sqrt{\tfrac74-\sqrt{\tfrac5{16}}-\sqrt{\tfrac{15}{8}-\sqrt{\tfrac{45}{64}}}}\), which is \(2\sin12^\circ\). A root is \(2\sin\frac{12^\circ}{45}\). Viète gave 23 positive roots, \(2\sin\big(\frac{12^\circ+360^\circ k}{45}\big)\) for those \(k\) that make the value positive. He ignored the 22 negative roots, which he did not count as solutions. Van Roomen was impressed, visited Viète in Fontenay, and the two became friends.

Exercise 8.1Find the polynomial \(P_3\) with \(2\sin3\theta=P_3(2\sin\theta)\), and compare it with the lowest-degree terms of van Roomen’s equation.

Show solution
\(\sin3\theta=3\sin\theta-4\sin^3\theta\). With \(x=2\sin\theta\): \(2\sin3\theta=3x-x^3\). The pattern \(nx-\cdots\) recurs: van Roomen’s equation begins \(45x-3795x^3\), and \(45\) is the multiple of the angle.

Exercise 8.2Show that among \(2\sin\big(\frac{12^\circ+360^\circ k}{45}\big)\), \(k=0,\dots,44\), exactly 23 are positive.

Show solution
The angles are \(\frac{12^\circ}{45}+8^\circ k\), from about \(0.27^\circ\) to \(352.27^\circ\). The sine is positive for angles in \((0^\circ,180^\circ)\), which requires \(8k<179.73\), so \(k=0,\dots,22\): 23 values. The other 22 lie in \((180^\circ,360^\circ)\) and give negative roots. All 45 are distinct, so they are all the roots.
Lesson 09

Trigonometric tables and the product for π

For the general reader. Archimedes trapped π between the perimeters of polygons with 96 sides. Viète pushed the same idea much further, to polygons with nearly 400,000 sides, and then saw that the process could go on forever. He wrote π as an infinite product, the first infinite product in the history of mathematics.

Goals
  • Know the scope of the Canon mathematicus.
  • Derive Viète’s product from areas of inscribed polygons.
  • Estimate its rate of convergence.

The Canon mathematicus (1579)

The Canon contains tables of all six trigonometric functions, and the accompanying Universalium inspectionum collects the formulas for solving plane and spherical triangles. Viète worked systematically with all six functions and used decimal fractions some years before Simon Stevin’s pamphlet on decimals of 1585.

π to ten places and to infinity

In the Variorum responsorum liber VIII (1593) Viète computed π to nine or ten decimal places (sources differ) from a polygon of \(6\cdot2^{16}=393{,}216\) sides. In the same book he gave the formula

\[ \frac{2}{\pi}=\sqrt{\tfrac12}\cdot\sqrt{\tfrac12+\tfrac12\sqrt{\tfrac12}}\cdot\sqrt{\tfrac12+\tfrac12\sqrt{\tfrac12+\tfrac12\sqrt{\tfrac12}}}\cdots \]
Theorem (Viète, 1593)

Let \(c_1=\sqrt{1/2}\) and \(c_{k+1}=\sqrt{\tfrac12(1+c_k)}\). Then \(c_k=\cos\frac{\pi}{2^{k+1}}\) and \(\displaystyle\prod_{k=1}^{\infty}c_k=\frac{2}{\pi}\).

Proof. The half-angle formula \(\cos\frac\alpha2=\sqrt{\frac{1+\cos\alpha}{2}}\) gives \(c_k=\cos\frac{\pi}{2^{k+1}}\) by induction. Repeated use of \(\sin2\alpha=2\sin\alpha\cos\alpha\) gives

\[ \sin x=2^n\sin\frac{x}{2^n}\prod_{k=1}^{n}\cos\frac{x}{2^k}. \]

Take \(x=\frac\pi2\): \(1=2^n\sin\frac{\pi}{2^{n+1}}\prod_{k=1}^n c_k\). As \(n\to\infty\), \(2^n\sin\frac{\pi}{2^{n+1}}\to\frac\pi2\), so the product tends to \(\frac2\pi\). Viète’s own argument used areas: the area of the regular \(2^{k}\)-gon inscribed in a circle divided by that of the \(2^{k+1}\)-gon is \(c_{k-1}\) (Exercise 9.2), and the product telescopes from the square to the circle. A rigorous proof of convergence was first published by Ferdinand Rudio in 1891.

Exercise 9.1Compute \(2/(c_1c_2c_3)\) to four decimals. How many terms give roughly ten correct digits?

Show solution
\(c_1\approx0.70711\), \(c_2\approx0.92388\), \(c_3\approx0.98079\); the product is \(\approx0.64073\) and \(2/0.64073\approx3.1214\). From the proof, \(2/\prod_{k\le n}c_k=2^{n+1}\sin\frac{\pi}{2^{n+1}}\approx\pi-\frac{\pi^3}{6}4^{-(n+1)}\). For \(n=3\) the error is \(\approx0.020\), as found. Each term divides the error by 4, gaining about 0.6 digits, so about 18 terms give ten correct decimals.

Exercise 9.2Show that the area of a regular \(m\)-gon inscribed in a unit circle is \(\frac m2\sin\frac{2\pi}{m}\), and that the ratio of the \(2^k\)-gon to the \(2^{k+1}\)-gon is \(\cos\frac{\pi}{2^k}\).

Show solution
The polygon is \(m\) isosceles triangles with two sides 1 and apex angle \(\frac{2\pi}{m}\), each of area \(\frac12\sin\frac{2\pi}m\). The ratio is \(\frac{2^{k-1}\sin(2\pi/2^k)}{2^{k}\sin(2\pi/2^{k+1})}=\frac{\sin2\beta}{2\sin\beta}=\cos\beta\) with \(\beta=\frac{\pi}{2^k}\). For \(k=2\) (square to octagon) this is \(\cos\frac\pi4=\sqrt{1/2}=c_1\).
Lesson 10

Apollonius Gallus and numerical roots

For the general reader. Two works of 1600 show the two sides of Viète’s exegetics. One solves a classical geometry problem with ruler and compass: draw a circle touching three given circles. The other gives a digit-by-digit method for finding the roots of equations as numbers, a forerunner of the numerical methods of Newton.

Goals
  • State the problem of Apollonius and count its solutions.
  • Understand Viète’s strategy of reducing the problem to simpler cases.
  • Carry out a step of numerical root extraction.

The problem of Apollonius

Apollonius of Perga’s lost treatise Tangencies, known from a summary by Pappus, asked for a circle tangent to three given objects, each a point, a line or a circle. The hardest case is three circles. Van Roomen had published a solution in 1596 using the intersection of hyperbolas, which is not a ruler-and-compass construction. Viète answered in Apollonius Gallus (1600), styling himself the French Apollonius, with a construction by ruler and compass for all ten combinations of points, lines and circles.

Problem of Apollonius

Given three circles in general position, there are at most eight circles tangent to all three, and the tangent circles can be constructed with ruler and compass.

Viète’s strategy is reduction. Shrink every given circle by the same amount \(r\) and change the radius of the sought circle by \(r\) in the appropriate direction: tangency is preserved. Choosing \(r\) equal to the smallest radius turns that circle into a point, so three circles reduce to two circles and a point. Viète works through the ten cases in order of difficulty, solving each with ruler and compass and using the centre of similitude of two circles. The up to eight solutions correspond to the choices of internal or external tangency with each circle.

De numerosa potestatum resolutione (1600)

For numerical solution Viète generalized the school method for extracting square and cube roots. For an equation such as \(x^2+7x=60750\) the root is found one decimal digit at a time: the leading digit from the dominant term, then each further digit by dividing the remainder by a trial divisor, which in modern terms approximates the derivative of the left side. The method works for any degree, and Viète carried it to equations of the sixth. It is laborious, and it was replaced by Newton’s method in the next century, but it was the first general procedure in Europe for approximating roots of arbitrary polynomial equations.

Exercise 10.1Solve \(x^2+7x=60750\) digit by digit, assuming a three-digit root.

Show solution
Hundreds digit: \(200^2+7\cdot200=41400\le60750<300^2+2100\), so \(x=2\ldots\). Write \(x=200+y\): \(y^2+407y=60750-41400=19350\). Tens: \(40^2+407\cdot40=17880\le19350<50^2+407\cdot50\), so \(y=40+z\): \(z^2+487z=19350-17880=1470\). Units: \(1470/487\approx3\), and \(9+1461=1470\). So \(x=243\).

Exercise 10.2Explain why there are at most \(2^3=8\) circles tangent to three given circles, and give a configuration with none.

Show solution
Let the circles have centres \((x_i,y_i)\) and radii \(r_i\). A circle \((x,y,r)\) tangent to all three satisfies \((x-x_i)^2+(y-y_i)^2=(r+s_ir_i)^2\) with signs \(s_i=\pm1\). Subtracting the equations pairwise gives two linear equations, so \(x,y\) are linear in \(r\), and the first equation becomes a quadratic in \(r\): at most two solutions per sign choice. Changing all three signs gives the same circles (with \(r\) replaced by \(-r\)), so there are 4 essentially different sign choices and at most \(4\times2=8\) circles. For a configuration with none, put \(C_1\) strictly inside \(C_2\) and \(C_3\) strictly outside \(C_2\). A circle touching \(C_1\) and \(C_3\) has points strictly inside and strictly outside \(C_2\), so it crosses \(C_2\) and cannot be tangent to it.
Part III · Life and legacy
Lesson 11

Code-breaker to Henry IV, 1589–1603

In May 1588 the League drove Henry III out of Paris. The king moved his court and parlement to Tours, and in 1589 he recalled Viète. On 1 August 1589 Henry III was assassinated, and Henry of Navarre became Henry IV. Viète served him as councillor at Tours and later as privy councillor in Paris.

Cryptanalysis

Viète’s most valued service was breaking codes. In 1589 he deciphered letters of the League. In 1590, after the battle of Ivry, Henry IV allowed him to publish his decipherment of an intercepted letter from the Spanish commander Moreo to Philip II, which showed that the Duke of Mayenne was aiming at the crown himself. The cipher used more than 500 symbols. According to a story repeated by MacTutor and others, Philip II, finding his ciphers read, complained to the pope that the French were using sorcery. From 1594 Viète was assigned to deciphering enemy correspondence, and late in life he wrote a memoir on cryptanalysis.

In 1593 Henry IV converted to Catholicism, and in 1594 he entered Paris. Viète followed the court. In these years he published the Supplementum, the π product, the reply to van Roomen and the two works of 1600.

The calendar quarrel

Pope Gregory XIII had reformed the calendar in 1582, using calculations of Aloysius Lilius as defended by the Jesuit astronomer Christopher Clavius. From 1600 Viète attacked Clavius in pamphlets, proposing his own corrections. He was wrong on several points, and Clavius’s rebuttal appeared in 1603, after Viète’s death. Historians generally count the episode against him.

Death

Viète left royal service in December 1602 with a payment of 20,000 écus, said to have been found at his bedside when he died. He died in Paris in 1603. The day is uncertain: most modern references give 23 February 1603, while MacTutor gives 13 December 1603. He left two daughters.

Exercise 11.1A simple substitution cipher can be attacked by letter frequencies. Explain why a nomenclator with more than 500 symbols, including codes for whole words and “nulls” that mean nothing, defeats naive frequency analysis, and what a cryptanalyst like Viète could still exploit.

Show solution
With many symbols for each letter (homophones) and whole-word codes, no single symbol is frequent and letter statistics flatten. Nulls add noise. A cryptanalyst can still exploit repeated symbol groups, the predictable structure of diplomatic letters (salutations, names, dates, titles), known or guessed plaintext, and errors by clerks who reuse symbols carelessly. Viète worked from partial decipherments sent in stages, consistent with such an iterative attack.
Lesson 12

Editions and inheritors

Viète printed his works at his own expense in small runs and gave copies away, so his ideas spread slowly. Several works remained in manuscript at his death. The Scottish mathematician Alexander Anderson, working in Paris, published a supplement to Viète’s Apollonian work in 1612, then in 1615 the treatises on equations and on angular sections, adding his own demonstrations. Marino Ghetaldi of Ragusa, a pupil of Viète, also completed his work on Apollonius. In 1631 a second edition of the Isagoge appeared in Paris with the posthumous Ad logisticen speciosam notae priores. In 1646 Frans van Schooten collected the Opera mathematica at Leiden, with help from Marin Mersenne and Jacob Golius; this is the edition through which most later readers knew him.

Who learned from him

  • Thomas Harriot in England developed a symbolic algebra close to Viète’s, with lowercase letters and products written by juxtaposition. Nathaniel Torporley, who had worked as Viète’s secretary, became Harriot’s friend.
  • Albert Girard (1629) stated the relations between roots and coefficients in full generality, admitting negative and imaginary roots.
  • Pierre de Fermat wrote all his mathematics in Viète’s notation and used his theory of equations in the method of maxima and minima.
  • René Descartes simplified the notation, dropped the law of homogeneity, and in La Géométrie (1637) made algebra the language of geometry. He played down the debt, writing that he began where Viète had stopped; historians are sceptical.
  • Isaac Newton studied van Schooten’s Viète as a student; his numerical method replaced Viète’s.

Assessment

Viète did not invent algebra. Diophantus, the Arabic algebraists and the Italians of the sixteenth century came before him. His contribution was the idea of a calculus on general quantities, which allowed formulas, identities and proofs about equations as objects. Everything that followed, from the theory of symmetric functions to Galois theory, presupposes it. The relations between roots and coefficients still bear his name.

Final projectTake one problem from Diophantus and one from the Italian algebraists (a cubic). Solve each first with fixed numbers, in the style of the original, and then with letters for all given quantities, in the style of Viète. Write two pages explaining what the general solution shows that the numerical one hides.

Reference

Connections

People linked to François Viète across this collection. A name without a link has no page here yet. See François Viète in the network of all 337 lives →

Students

  • Alexander Anderson
  • Catherine de Parthenay 1564-1570
  • Marino Ghetaldi

Influenced by

Influenced

Rivals

  • Christopher Clavius 1600-1603

Friends

  • Adriaan van Roomen 1594-1603

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, François Viète, MacTutor History of Mathematics.
  2. François Viète, Wikipedia; François Viète, Wikipédia (French, with detail from local archives).
  3. F. Ritter, François Viète, inventeur de l’algèbre moderne, Revue occidentale, 1895.
  4. J. Grisard, François Viète, mathématicien de la fin du seizième siècle, thesis, EHESS, Paris, 1968.
  5. H. L. L. Busard, Viète, François, in Dictionary of Scientific Biography, vol. 14, Scribner, 1976.

Texts and mathematics

  1. F. van Schooten (ed.), Francisci Vietae Opera mathematica, Leiden, 1646.
  2. F. Viète, The Analytic Art, translated by T. R. Witmer, Kent State University Press, 1983.
  3. J. Klein, Greek Mathematical Thought and the Origin of Algebra, MIT Press, 1968.
  4. Viète’s formula and Vieta’s formulas, Wikipedia.
  5. H. J. M. Bos, Redefining Geometrical Exactness: Descartes’ Transformation of the Early Modern Concept of Construction, Springer, 2001 (chapters on Viète).
  6. J. Stedall, From Cardano’s Great Art to Lagrange’s Reflections, European Mathematical Society, 2011.
  7. P. Beckmann, A History of π, St. Martin’s Press, 1971.