History of Mathematics

Transcription

1 History of Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Summer B: Solution of cubic and uartic euations

2 Cardano s solution of the cubic euation In his book Ars Magna, Geronimo Cardano ( ) admitted that he was not the original discoverer of the solution of the cubic euation, but rather obtained hints from Niccolo Tartaglia ( ). In Ars Magna, the solution of the cubic euation is discussed in 20 different cases: x 3 + px =, x 3 = px +, etc. x 3 + px + = 0. 1

3 Cardano, Ars Magna, Chapter XI On the cube and first power eual to the number x 3 + px = Cube one-third the coefficient of x; ( ) p 3 3 add to it the suare of one-half the constant of the euation; ( p 3 ( 3) + 2 (p 3 ( and take the suare root of the whole. 3) + ) 2 2 You will duplicate this (Take two copies of this uantity), and to one of the two you add one-half the number you have already suared (p 3 ( 3) + ) (p 3 ( 3) + ) and from the other you subtract one-half the same. You will then have a binomium and its apotome. Take this to mean a + b (a binomium) and changing the sign of exactly one term to yield an analogous positive uantity (apotome). Then subtracting the cube root of the apotome from the cube root of the binomium, (p 3 3 ( 3) + ) (p ) 3 3 ( + ) the remainder [or] which is left is the value of x. ) 2 2

4 Examples (1) x 3 + 6x = 20. Answer: x = (ii) x 3 + 6x = 2. Answer: x =

5 Solution of cubic euation Cubic euation: x 3 + px + = 0. Key idea: Find a solution of the form x = u + v for suitable choice of u and v. Note that x 3 + px + = (u + v) 3 + p(u + v) + = u 3 + v 3 + 3uv(u + v) + p(u + v) + = u 3 + v (u + v)(3uv + p) If we can find u and v satisfying 3uv + p = 0 and u 3 + v 3 + = 0, then we would obtain a solution x = u + v of the cubic euation. Now, u 3 + v 3 = and u 3 v 3 = (uv) 3 = ( p 3 )3 = p3 27. It follows that u 3 and v 3 are the roots of the uadratic euation t 2 + t p3 27 = 0, 4

6 Solution of cubic euation (continued) Now, u 3 + v 3 = and u 3 v 3 = (uv) 3 = ( p 3 )3 = p3 27. It follows that u 3 and v 3 are the roots of the uadratic euation and we may take t 2 + t p3 27 = 0, and obtain u 3 = p3 27, v3 = p3 27, x = p p

7 Complete solutions of cubic euation Let ω be a complex cube root of unity. This is a complex number satisfying ω 2 + ω + 1 = 0. Explicitly, ω = 1 2 ( 1 + 3), or 1 2 (1 + 3). If x 1 = u + v is one root of the cubic euation x 3 + px + = 0 obtained above, then x 2 = ωu + ω 2 v and x 3 = ω 2 u + ωv are the remaining two roots of the euation. The expression = p3 27 is called the discriminant of the cubic euation. If > 0, the cubic euation has one real root and two complex roots. If = 0, it has three real roots, at least two of which are eual. If < 0, the cubic euation has three distinct real roots. 6

8 Ferrari s solution of the uartic euation Ludovico Ferrari ( ) of the uartic euation x 4 + px 2 + x + r = 0. Key idea: Find a uantity y so that (x 2 + y) 2 is also a suare of the form (Ax + B) 2 : (x 2 + y) 2 = x 4 + 2yx 2 + y 2 = (2y p)x 2 x + (y 2 r). This last expression is a suare of the form (Ax + B) 2 if and only if ( ) 2 4(2y p)(y 2 r) = 0. This latter is a cubic euation in y and can be solved by Cardano s method. 7

9 Transformation of an euation A polynomial euation of degree n x n + a 1 x n 1 + a 2 x n a n 1 x + a n = 0 can be transformed into one without second highest order term by a subsitution x = y a 1 n : y n + b 2 y n b n 1 y + b n = 0. 8

10 Example To solve the uartic euation x 4 + 6x 2 60x + 36 = 0, we follow Ferarri s method by finding y such that (x 2 + y) 2 = x 4 + 2yx 2 + y 2 = (2y 6)x y + (y 2 36) has its right hand side eual to the suare of a linear form. This reuires or (2y 6)(y 2 36) = 0, y 3 3y 2 36y 342 = 0. 9

11 Example To solve the uartic euation x 4 + 6x 2 60x + 36 = 0, we follow Ferarri s method by finding y such that (x 2 + y) 2 = x 4 + 2yx 2 + y 2 = (2y 6)x y + (y 2 36) has its right hand side eual to the suare of a linear form. This reuires or By putting y = z + 1, we transform this cubic euation into z 3 39z 380 = (2y 6)(y 2 36) = 0, y 3 3y 2 36y 342 = 0. This cubic euation can be solved by Cardano s method. It has a root

12 Euler: Factorization of uartics Every euation of the fourth degree, as x 4 + Ax 3 + Bx 2 + Cx + D = 0, can always be decomposed into two real factors of the second degree. Proof. It is known that by setting x = y 1 4 A, we can change this euation into another one of the same degree without the second term, and, since this transformation can always be performed, let us suppose that in the proposed euation the second term is already missing, and that we have to resolve this euation: x 4 + Bx 2 + Cx + D = 0, into two real factors of the second degree. It is clear that these factors will be of the form (xx + ux + α)(xx ux + β) = 0. 11

13 Euler: Factorization of uartics... to resolve x 4 + Bx 2 + Cx + D = 0, into two real factors of the second degree: (xx + ux + α)(xx ux + β) = 0. If now we compare this product with the proposed euation, we shall find from which we derive hence B = α + β = uu, C = (β α)u, D = αβ, α + β = B + uu, β α = C u, 2β = uu + B + C u, and 2α = uu + B C u, and since we have 4αβ = 4D we obtain the euation u 4 + 2Buu + BB CC uu = 4D 12

14 Euler: Factorization of uartics... to resolve x 4 + Bx 2 + Cx + D = 0, into two real factors of the second degree: (xx + ux + α)(xx ux + β) = u 4 + 2Buu + BB CC uu = 4D, or u 6 + 2Bu 4 + (BB 4D)uu CC = 0, from which the value of u must be found. And since the absolute term CC is essentially negative, we have shown that this euation (of degree 6) has at least two real roots. When we take one of them as u, then the values of α and β will also be real, and hence the two supposed factors of the second degree xx + ux + α and xx ux + β will be real. 13

15 Euler also established the following for a polynomial euation with real coefficients. (i) When a root x+y 1 exists there is also one of the form x y 1. (ii) An euation of odd degree has at least one root. (iii) An euation of even degree has either no real roots or pairs of such roots. Euler continued to give a proof on the following theorem. Every euation of which the degree is a power of the binary, as 2 n (n being a number greater than 1), can be resolved into two factors of degree 2 n 1. 14

16 The Fundamental Theorem of Algebra Every nonconstant polynomial with complex coefficient has a complex root. Origin of the theorem. In order to reduce in general a differential rational function to the uadrature of the hyperbola or to that of the circle, it is necessary, according to the method of M.Bernoulli, to show that every rational polynomial, without a divisor composed of a variable x and of constants, can always be divided, when it is of even degree, into trinomial factors xx + fx + g, xx + hx + i, etc., of which all coefficients f, g, h, i, are real. 1 Euler ( ) proved the theorem for uartic euations. Gauss ( ) gave 4 different proofs of the Fundamental Theorem of Algebra. 1 J.L.R. D Alembert ( ), Recherches sur le calcul intégral (1746). 15

17 The Fundamental Theorem of Algebra Every nonconstant polynomial with complex coefficient has a complex root. Up to the present, there is no purely algebraic proof of this theorem; it remains a theorem in topology, and its validity depends on the fact that the punctured plane (the plane with one point removed) is not the same topological object as the plane itself. 16

18 Unsolvability of general polynomial euations of degree 5 N. H. Abel ( ): For n 5, there is no formula that expresses the roots of a general polynomial euation of degree n in terms of its coefficients using on the arithmetic operations and radicals. 17