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The Quartic Equation, Invariants and Euler's Solution Revealed

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The Quartic Equation, Invariants and Euler's Solution Revealed The quartic equation: invariants and Euler’s solution revealed 1 RWD Nickalls 2 The Mathematical Gazette (2009); vol. 93 (March; No. 526), pp. 66–75 http://www.nickalls.org/dick/papers/maths/quartic2009.pdf 1 Introduction The central role of the resolvent cubic in the solution of the quartic was first appreciated by Leonard Euler (1707–1783). Euler’s quartic solution first appeared as a brief section (§ 5) in a paper on roots of equations [1, 2], and was later expanded into a chapter entitled Of a new method of resolving equations of the fourth degree (§§ 773–783) in his Elements of algebra [3, 4]. Euler’s quartic solution was an important advance, in which he showed that each of the roots of√ a reduced√ quartic√ can be represented as the sum of three square roots, say ± 푟1 ± 푟2 ± 푟3, where the 푟푖 (푖 = 1, 2, 3) are the roots of a resolvent cubic. A quartic equation in 푥 is said to be reduced if the coefficient of 푥3 is zero. This can always be achieved by a simple change of variable. Motivated by the recent tercentenary of Euler’s birth, this article describes√ the geometric basis underlying both the 푟푖 and the sign of the product 푟1푟2푟3, these being two key aspects of Euler’s solution. Finally, we reveal the beautiful dynamic between Euler’s resolvent cubic and the quartic invariants 퐺, 퐻, 퐼, 퐽 [5, 6, 7], and propose a new class of algebraic object. 2 Geometric basis for the 푟푖 A significant property of the reduced quartic equation is that the four roots can be completely defined using only three parameters. For example, let 푧푗 (푗 = 1, 2, 3, 4) be the roots (see Figure 1) of a reduced quartic equation, 푍(푥) ≡ 푎푥4 + 푝푥2 + 푞푥 + 푟 = 0. (1) As the sum of the roots is zero (the coefficient of the cubic term is zero), it follows that we can define the points midway between 푧1, 푧2 and 푧3, 푧4 as ±푔. Let 푧2 − 푧1 = 2훼 and 푧4 − 푧3 = 2훽. The four roots can then be expressed as follows: {︂ 푧1, 푧2 = −푔 ± 훼, 푧3, 푧4 = +푔 ± 훽. Since specifying one pair of quartic roots necessarily defines the remaining pair, there are just three different ways of allocating the pairs of roots, each associated with its own 푔, 훼, 훽, the inter-relationship between which lies at the heart of a remarkable symmetry which underpins the solution of the quartic. 1This minor revision of the original article corrects typographic errors and incorporates some explanatory footnotes. The original published version is available from the jstor archive at http://www.jstor.org/stable/40378672. 2Department of Anaesthesia, Nottingham University Hospitals, City Hospital Campus, Nottingham, UK. email: [email protected] RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 2 gY g ... ... α...... α β....... β ..... ..... X z••1 z2 z •3 z •4 Figure 1: For example if, with no loss of generality, we let ⎧ ⎨ 푧3 + 푧4 = 2푔1, 푧3 + 푧1 = 2푔2, (2) ⎩ 푧3 + 푧2 = 2푔3, then 2(푔2 + 푔3) = 2푧3 + 푧1 + 푧2, = (푧1 + 푧2 + 푧3 + 푧4) + 푧3 − 푧4, = 푧3 − 푧4 = −2훽1, and similarly 2(푔2 − 푔3) = 푧1 − 푧2 = −2훼1, and hence {︂ 훼1 = −(푔2 − 푔3), 훽1 = −(푔2 + 푔3). Thus the 훼푘, 훽푘 (푘 = 1, 2, 3) are actually simple functions of the 푔푖 (푖 ≠ 푘) such that each of the four roots 푧푗 can be expressed as a function of the 푔푖 alone, as follows 3: ⎧ 푧1 = −푔1 − 훼1 = −푔1 + (푔2 − 푔3) = −푔1 + 푔2 − 푔3, ⎪ ⎨ 푧2 = −푔1 + 훼1 = −푔1 − (푔2 − 푔3) = −푔1 − 푔2 + 푔3, (3) 푧3 = +푔1 − 훽1 = +푔1 + (푔2 + 푔3) = +푔1 + 푔2 + 푔3, ⎪ ⎩ 푧4 = +푔1 + 훽1 = +푔1 − (푔2 + 푔3) = +푔1 − 푔2 − 푔3. 2 Thus Euler’s 푟푖 are the same as the 푔푖 . 3 Euler’s resolvent cubic Using these observations we can reconstruct a given reduced quartic equation, say Equation 1, which then leads to a resolvent cubic and hence to the solution. Let the roots of 푍(푥) = 0 be −푔 ± 훼 and 푔 ± 훽 (Figure 1). 푍(푥) ≡ {푥 − (−푔 − 훼)}{푥 − (−푔 + 훼)}{푥 − (푔 − 훽)}{푥 − (푔 + 훽)} = 0. Expanding and letting 퐴 = 푔2 − 훼2 and 퐵 = 푔2 − 훽2, gives 푥4 + (−4푔2 + 퐴 + 퐵)푥2 + (2푔)(퐵 − 퐴)푥 + 퐴퐵 = 0. 3 For a 3D version of Euler’s solution in which the ±푔푖 are associated with the mid-points of the six edges of a regular tetrahedron, see Fig 2 in Nickalls (2012), The quartic equation: alignment with an equivalent tetrahedron, Mathematical Gazette, 96, 49–55; http://www.nickalls.org/dick/papers/maths/tetrahedron2012.pdf. RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 3 We can eliminate 훼, 훽 by first equating coefficients with the monic form of Equation 1 giving ⎧ 2 ⎨ 푝/푎 = −4푔 + 퐴 + 퐵, 푞/푎 = 2푔(퐵 − 퐴), ⎩ 푟/푎 = 퐴퐵, and then eliminating 퐴 and 퐵 (using the identity 4퐴퐵 = 2퐴 × 2퐵), which gen- erates a resolvent sextic in 푔, the roots of which are the six values ±푔1, ±푔2, ±푔3. The substitution 푔2 ↦→ 푥 then generates Euler’s original resolvent cubic [1, 2, 3, 4] 푝 (︂푝2 − 4푎푟 )︂ 푞2 푅(푥) ≡ 푥3 + 푥2 + 푥 − = 0, (4) 2푎 16푎2 64푎2 푟 푔2, 푔2, 푔2 whose roots 푖 are therefore 1 2 3. The four√ roots√ of the√ reduced quartic 푍(푥) = 0 are among the eight possible values of ± 푟1 ± 푟2 ± 푟3; but in order to determine which four they are we need a way of allocating the signs correctly. 푥4 − 푙푥2 − 푚푥 − 푛 Euler, using a monic quartic of the form√ = 0, says he resolved the sign problem by noting that 푟1푟2푟3 = 푚/8, as follows [3, § 773]: √ . But it is to be observed, that the product . 푟1푟2푟3, must be 푚/ 푚/ √equal√ to √8, and that if 8 be positive, the product of the terms 푟1, 푟2, 푟3, must likewise be positive; Unfortunately Euler did not elaborate further on this, but the key to under- standing the sign problem is not difficult to find, since from Equation 2we have 8푔1푔2푔3 = (푧3 + 푧4)(푧3 + 푧1)(푧3 + 푧2), 3 2 = 푧3 + 푧3 (푧1 + 푧2 + 푧4) + 푧3(푧2푧1 + 푧2푧4 + 푧1푧4) + 푧4푧1푧2. Now 푧1 + 푧2 + 푧4 = −푧3 (since Σ푧푗 = 0), hence 8푔1푔2푔3 = 푧1푧2푧3 + 푧2푧3푧4 + 푧3푧4푧1 + 푧4푧1푧2, (5) and so 8푔1푔2푔3 is actually one of the four elementary symmetric functions of the roots 푧푗. Its value is therefore equal to −1× the coefficient of the 푥-term of the monic form of the reduced quartic equation 푍(푥) = 0, and so we have √ 8 푟1푟2푟3 = 8푔1푔2푔3 = −푞/푎, (5푎) √ which is equivalent to Euler’s 푟1푟2푟3 = 푚/8. √ 4 Geometric basis for the sign of 푟1푟2푟3 A useful way of ‘seeing’ the quartic algebra at work is to express the coefficients in terms of the key ‘visible’ parameters 휀, 푦푁푧, 푦푁푧′ shown in Figure 2, as follows: Let 퐹 (푋) be a quartic polynomial with real coefficients (푎 ≠ 0) 퐹 (푋) ≡ 푎푋4 + 푏푋3 + 푐푋2 + 푑푋 + 푒, (6) with invariants [6, p. 76] ⎧ 퐺 = 푏3 + 8푎2푑 − 4푎푏푐, ⎪ ⎨ 퐻 = 8푎푐 − 3푏2, (7) 퐼 = 12푎푒 − 3푏푑 + 푐2, ⎪ ⎩ 퐽 = 72푎푐푒 + 9푏푐푑 − 27푎푑2 − 27푒푏2 − 2푐3. RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 4 . Y . Z(x) . ......... ... .. T2 . .. .. ... .......................... .. ..... .....Nz . .. .. ....... • •.... .. .... .... .. .. I .... .. ... .. 1 .... ... .. .. .... ... ... .. ....• ε ... ... .. .... .. ... ... ..... ... ..... ..... .. .. X ................. .. .. .. .• N . .. T1 z′ .. .. •. .. .. .. .. .. ε .. I2 . .. .. .. Z′(x) . •.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ... .. ... ... .. .... .. ..... .... ... .. ... .. ... .. ... ... .... ... ................ • T3 Figure 2: The reduced quartic 푍(푥), turning points (푇1, 푇2, 푇3), points of inflection ′ (퐼1, 퐼2), and first differential 푍 (푥). The 푥-coordinates of the points of inflection are ±휀. The curves intersect the 푦-axis at points 푁푧 and 푁푧′ . Let its reduced form 푍(푥) be generated by the translation 푋 ↦→ 푥 + 푋푁푓 , where 푋푁푓 = −푏/(4푎). Using Taylor’s theorem we have 퐹 ′′ 푋 4 ( 푁푓 ) 2 ′ 푍(푥) ≡ 퐹 (푥 + 푋푁 ) = 푎푥 + 푥 + 퐹 (푋푁 )푥 + 퐹 (푋푁 ). (8) 푓 2 푓 푓 ′ If 푍(푥) and 푍 (푥) intersect the 푦-axis in points 푁푧 and 푁푧′ respectively, then Equation 8 can be expressed as 4 2 2 푍(푥) ≡ 푎푥 − 6푎휀 푥 + 푦푁푧′ 푥 + 푦푁푧 (9) where (see Equation 4 and Figures 2, 3) ⎧ (3푏2 − 8푎푐) −퐻 −푝 휀2 ≡ ≡ , ⎪ = 2 2 ⎪ 48푎 48푎 6푎 ⎪ 2 ⎪ 퐼 3퐻 ⎨ 푦 = 퐹 (푋푁 ) ≡ − ≡ 푟, 푁푧 푓 12푎 482푎3 (10) ⎪ ′ 퐺 ⎪ 푦 ′ 퐹 푋 ≡ ≡ 푞, ⎪ 푁푧 = ( 푁푓 ) 2 ⎪ 8푎 ⎩⎪ 2 ′′ −12푎휀 = 퐹 (푋푁푓 ). Expressing the reduced quartic 푍(푥) in this form (Equation 9) greatly facilitates visualisation, since we can now ‘see’ how the configuration of the curves 푍(푥) and 푍′(푥) is related to the coefficients. For example (assuming 푎 > 0), if the 푥2 term is positive then 휀 is complex (휀2 < 0), and so the quartic will have two complex points of inflection and hence only one real turning point (cf. [10]). RWD Nickalls Mathematical Gazette (2009); vol. 93, p. 66–75 5 If 푥푇푖 are the 푥-coordinates of the turning points of 푍(푥), then by differenti- ating Equation 9 we have (see Equations 5푎 and 10) −푦푁푧′ √ 4푥푇 푥푇 푥푇 = = 8 푟1푟2푟3, (11) 1 2 3 푎 √ and hence the sign of 푟1푟2푟3 is the same as that of −푦 ′ /푎 and 푥푇 푥푇 푥푇 . It 푁푧 √ 1 2 3 follows, therefore, that we can actually ‘see’ the correct sign of 푟1푟2푟3 simply by observing the signs of the abscissae of the turning points of the reduced quartic, or by noting the location of 푁푧′ in relation to the abscissa. For example (assuming 푎 > 0), if the roots 푧푗 are such that the middle turning point, 푇2, is to the left of the 푦-axis, then not only will 푦푁푧′ be negative 푥 (Figure 2) but just two of the three√ 푇푖 will be negative resulting in a positive product for 푥푇1푥푇2푥푇3, and hence 푟1푟2푟3 will also be positive (see Equation 11).
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