Viète's equations are used in some proofs of the Basel problem, which was allegedly solved by Euler.

Viète's equations include the following: given a polynomial,

$$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$ with roots $\alpha_1, \alpha_2,...\alpha_n,$

$$a_{n-1}= -a_n\,(\alpha_1+\alpha_2\,+...+\,\alpha_n).$$

This can be applied to a polynomial of the form:

$$b_0 - b_1\, x^2 +b_2\,x^4-\cdots+(-1)^n\,b_n\,x^{2n},$$

in which case we find that,

$$b_0 - b_1\, x^2 +b_2\,x^4-\cdots+(-1)^n\,b_n\,x^{2n}=b_0\left(1-\frac{x^2}{\beta_1^2}\right)\,\left(1-\frac{x^2}{\beta_2^2}\right)\, \cdots\,\left(1-\frac{x^2}{\beta_n^2}\right).$$


$$b_1 = b_0\,\left(\frac{1}{\beta_1^2}\right)+\,\left(\frac{1}{\beta_2^2}\right)\,+\cdots+\,\left(\frac{1}{\beta_n^2}\right).$$

This form allows expressing the second coefficient in:

$$\frac{sin (x)}{x}= 1 - \frac{x^2}{2\cdot 3}+\frac{x^4}{2\cdot 3\cdot 4\cdot 5}-\frac{x^6}{2\cdot3\cdots7}+\cdots$$


$$\frac{1}{2\cdot3}= \frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+\cdots.$$

The question is whether Euler truly resorted to Viète's equations, and how big of a mathematical figure Viète name truly is in his own right.


1 Answer 1


According to Varadarajan's Euler and his Work on Infinite Series (p.518)

"Euler’s idea was based on an audacious generalization of Newton’s formula for the sums of powers of the roots of a polynomial to the case when the polynomial was replaced by a power series".

The "formula" is the one relating power sums to elementary symmetric polynomials, and Girard is usually credited with deriving it in full generality in 1629 (Newton independently rederived it in 1666). Simpler formulas relating elementary symmetric polynomials of the roots to coefficients of the polynomial were given by Viète (Vieta in Latin) for positive real roots, and in general again by Girard. While Newton may have been unaware of Girard

"...to some extent Newton cut his mathematical teeth on Viète, under the keen eye of Isaac Barrow, and Vol. 1 of D.T. Whiteside's monumental athematical Papers of Isaac Newton contains some relevant material from Newton's notes: if nothing else, these demonstrate the calibre of the young Newton, who had transcribed some of the theorems of this work of Viète and others to aid his own understanding."

"This work" is Ad Angulares Sectiones, which contained polynomial expansions for sine and cosine of multiple angles, among other things. It was also the basis of Viète's trigonometric solution to the irreducible case of the cubic, with three real roots, none of which is given by the Cardano formula (it involves extracting cube root of a complex number, which as Bombelli showed earlier, requires solving another cubic). Viète showed that the problem reduces to angle trisection, so in modern terms all three roots can be found by inverting cosines.

As for Viète's significance, his Isagoge (1591) did for algebra what Descartes's La Géométrie (1637) did for geometry, in other words, he was the father of symbolic algebra. Modern style symbolic notation, especially systematic use of letter notation for both variables and parameters, enabled wide use of algebraic manipulation rules (Viète still uses words for powers, these were symbolized by Descartes, but they are attached to variables). Familiar approach of converting problems to equations, and then solving them symbolically, was also coined by Viète. Hartshorne gives a nice description of Viète applying it to classical geometric problems in Supplementum Geometriae (1593). Bos in Redefining Geometrical Exactness gives a detailed historical analysis of Viète's contributions:

"It was Viète who first introduced and promoted the idea that algebra was proper method for analysis of problems both in geometry, and in the theory of numbers... Viète did not see algebra as a technique concerning numbers... but as a method of symbolic calculation concerning abstract magnitudes."

For more on his influence see Esteve's paper.

  • 1
    $\begingroup$ Great entry. I didn't forget to "accept" it, but I want to see if there are additional answers before I do. Ty $\endgroup$ Jun 3, 2015 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.