arXiv:1804.00698v2 [math.CV] 29 Apr 2018 eea oml htrpeet l the all represents that formula general a xmlsg nado.Ide,asxi qaincnb ovdi ter These in solved be solutions. can possible sextic of a functio Indeed, set (Bessel) on. the and special on in go the included examples functions, and elementary introduced in were solved be not n h eyrcn aesb od[]adNs [10]. Nash eve and interest [2] Mora Boyd Teo of by of is papers monograph topic recent current The t very comprehensive the for results. a and [7] or these e.g., [4] of see, e.g., details today; consult, arithmet the can the and reader of history The means radicals. by and coefficients operations its through degree oraihei prtosadrdcl,wiefrno for while superposition radicals, as and coefficients operations its arithmetic through four explicitly expressed be can adpt fetnigteeitn olo.Frisac,teequ the instance, For stan- toolbox. the existing [8 followed x the have book extending mathematicians of famous path quintics, his dard the published solving Klein mod in elliptic later Thus, through century quintic general quarter a functions; g of indep more Kronecker roots and and the Hermite, quintics expressed Brioschi, dently the 1858, of in roots Already the . represent to means other for xedteaalbenmest h mgnr ubr;t td th study to numbers; imaginary equation the differential to ordinary numbers available the extend ot sitgaso lmnayfunctions. elementary of integrals as roots 2 h blGli eutddnt fcus,so cetssfo sear from scientists stop course, of not, did result Abel-Galois The 00MteaisSbetClassification Subject Mathematics 2010 words: Key les or four degree of equation algebraic univariate any of roots The = LSDFR OUINO POLYNOMIAL OF SOLUTION CLOSED-FORM − in fteroso nvraeplnma faydge sinte- as degree any of polynomial functions. univariate elementary a of of grals roots the of sions Abstract. antb ovdi elnmes u tcnb oei we if done be can it but numbers, real in solved be cannot 1 nvraeplnmas xlctfrua xrsigthe expressing formulas Explicit ; Univariate ope nlssi sdt n lsdfr expres- closed-form find to used is analysis Complex LXNE KHEYFITS ALEXANDER 1. EQUATIONS Introduction x 2 y ′′ 1 + xy ′ n ( + ope ot fevery of roots complex 21;1E2 97H30. 12E12; 12D10; : x 2 − ν 2 ) y n ,wihcan- which 0, = ≥ hr exists there 5 eneral n ation − ular of s en- ms [9] he ch ns th ic n ]. e s 2 ALEXANDERKHEYFITS of Kamp´e de F´eriet functions, and a septic in terms of the hyperelliptic functions and associated theta functions of degree 3. To the best of the author’s knowledge, no information regarding algebraic equations of higher degree is currently available. In this note we solve any in quadratures, i.e., we find explicit representations of the roots of univariate polynomials of any degree as integrals of elementary functions along the positive real axis. To derive these integral formulas, we apply the method from [1, 5, 6].

2. Integral Representations of Roots of Polynomials with Simple Roots It is convenient to us to represent a generic polynomial with any real or complex coefficients as n n−1 n−2 2 P (z)= z + a z + a z + + a − z + a − z + a . 1 2 ··· n 2 n 1 n If P (z) has at least one pair of multiple roots, then the of P is zero, and the multiplicities of all the roots can be determined algebraically, e.g., by Horner’s method. Now, by making use of the Euclidean algorithm, we can reduce P to a polynomial of smaller de- gree without multiple roots. Thus from now on, we suppose that the polynomials under consideration have only simple roots. Denote the roots of P (z)=0as ξi, i = 1, 2,...,n, where all the n roots ξi are different complex (or maybe real) numbers. We can also assume that all ξ = 0, since otherwise the problem is again reduced to i 6 a polynomial of lesser degree. Set M = 1 + max aj , 1 j n , and let A be any number such that A > M. Introduce{| another| ≤ polynomial≤ } n n−1 n−2 f(z)= P (z + A)= z + b z + b z + + b − z + b . 1 2 ··· n 1 n The equations P (z) = 0 and f(z) = 0 are equivalent, however, all the roots of f(z) = 0 have negative real parts; the goal of the shift z z + A is to clear the positive semi-axis x = z 0 from the zeros → ℜ ≥ of f. In particular, bn = f(0) = 0. We also mention that if A is large enough, then b An(1 + a /A6 ). n ≈ 1 The equation f(z) = 0 can be rewritten as

n n−1 n−2 (1) z = (b1z + b2z + + bn−1z + bn). p− ··· To work witth single-valued holomorphic functions, we cut the complex z plane along the positive x axis, and in the slit plane consider a − − CLOSED-FORM SOLUTION OF POLYNOMIAL EQUATIONS 3 closed contour Ω, consisting of two horizontal segments (2) z = x ıǫ, 0 x < R, ± ≤ whose left ends are connected with a small semi-circle z = ǫ, z < 0, and their right end-points R ıǫ are connected with a| large| arcℜ of the circle z = R′; if R is big enough,± the contour contains all the roots of the polynomial| | f(z).

In the domain D bounded by the contour Ω = ∂D, all arguments satisfy 0 < arg w< 2π. Consider n holomorphic functions n n−1 n−2 fk(z)= z (k) (b1z + b2z + + bn−1z + bn) − p− ··· k = 0, 1, 2,...,n 1, z D, where (k) √n is the k th branch of the − ∈ − radical in (1), i.e.,

n n−1 n−2 1/n (k) √ = b1z + b2z + + bn−1z + bn · ··· ×

ı − − 2kπ n 1 n 2 n ı (3) exp arg b1z + b2z + + bn−1z + bn e . × n−n ··· o In particular, n−1 n−2 1/n f (z)= z b z + b z + + b − z + b 0 −| 1 2 ··· n 1 n| × ı n−1 n−2 exp arg b1z + b2z + + bn−1z + bn . × n−n ··· o We also consider a linear h(z)= αz + β with positive coefficients α > 0 and β > 0, which has one root in the left half-plane. We will show that the coefficients α and β can be chosen so that f (z) < h(z) on the contour Ω for every k = 0, 1,...,n 1, hence | k | | − we can apply Rouch´e’s theorem to each pair h and fk for every k = 0, 1,...,n 1, and conclude that every function fk(z) has the unique root in the− plane.

The contour Ω consists of three parts; a) small semi-circle of ra- dius ǫ in the left half-plane; b) two parallel horizontal segments z = x ıǫ, 0

a) If z = ǫ, ǫ> 0 and z < 0, or 0 z 1, then | | ℜ ≤ℜn ≤ f (z) ǫ + √2B, | k | ≤ 4 ALEXANDERKHEYFITS where B = max bj , 1 j n. On the other hand, here h(z) β ǫα, hence we| must| choose≤ ≤ the parameters to satisfy the inequality| | ≥ − n (4) √2B +(α + 1)ǫ β. ≤ b) In this case f (z) √n nBx, while | k | ≤ n h(z) α(x ǫ)+ β √nBx. | | ≥ − ≥ Hence, in this case the parameters have to satisfy two inequalities α √n nB and β αǫ. ≥ ≥ n c) Finally, if z = R, then fk(z) √nB, while n | | | | ≤ h(z) = α2R2 +2αβR cos θ + β2, leading to inequality | | α β > (pnB)2/nR. − We collect the inequalities together, as α β > (nB)2/nR − αǫ<β (α + 1)ǫ + (2B)1/n < β α (nB)1/n. ≥ This system can be solved if we first choose big R to embrace all the roots of f(z), then α to satisfy the last equation, then β, and finally ǫ as small as we need.

Since the linear function h(z) has exactly one root in the slit domain D, by Rouch´e’s theorem we conclude that every function fk(z), k = 0, 1,...,n 1, has exactly one root, call it ξk, in the complex plane. We have proved− the main claim of this note. Proposition 1. Every function f (z), k =0, 1, 2,...,n 1, has exactly k − one root ξk in the complex plane. 

To derive explicit representations for the roots ξk, we consider con- tour integrals

z ξk (5) Ik = 2 − dz, k =0, 1, ..., n 1, IΩ (z + 1)fk(z) − and compute each of them twice, first by the residue theorem, and then by distorting the contour of integration. For every k, fk(ξk) = 0, thus, the denominator of (5) has a simple pole. However, due to the numerator, at z = ξk the integrand of (5) has a removable singularity. CLOSED-FORM SOLUTION OF POLYNOMIAL EQUATIONS 5

In addition to that, the integrand has two simple poles at z = ı. Computing residues at these points, we get ± π I = (ı (f (ı)+ f ( ı)) + ξ (f (ı) f ( ı))) . k f (ı)f ( ı) k k − k k − k − k k − On the other hand, letting in (5) ǫ 0 and R , the contour Ω becomes the positive x axis, traversed→ twice, thus,→∞ ∞ − ∞ (x ξk)dx (x ξk)dx Ik = 2 − 2 − 2πı . Z0 (x + 1)fk(x) − Z0 (x + 1)fk(xe )

Equating the two expressions for Ik, we get a for

ξk. Solving this equation, we derive the explicit formulas for the roots ξ , k =0, 1,...,n 1, k − 2πı x(fk(xe ) fk(x)) f (ı)+f ( ı) ∞ − k k − 0 (x2+1)f (x)f (xe2πı)dx πı f (ı)f ( ı) k k − k k − (6) ξk = 2πı . R fk(xe ) fk(x) fk(ı) fk( ı) ∞ 2 − 2πı dx + π − − 0 (x +1)fk(x)fk(xe ) fk(ı)fk( ı) R − Remark 1. These formulas are unsuitable for numerical computations, since the integrands decay very slowly. However, there are simple sub- stitutions, which make such computations feasible even on a good cal- culator like TI-89, not to mention computers; see [6].

References

[1] Anastasselou E. G., Ioakimidis N. I., Application of the Cauchy theorem to the location of zeros of sectionally analytic functions, Z. Angew. Math. Phys., Vol. 35(1984), No. 5, 705-711. [2] Boyd J. P., Finding the Zeros of a Univariate Equation: Proxy Rootfind- ers, Chebyshev Interpolation, and the Companion Matrix, SIAM Rev., Vol. 55(2013), No. 2, 375-396. [3] Burniston E. E., Sievert C. E., Exact analytical solution of the transcendental equation α sin ξ = ξ, SIAM J. Appl. Math., Vol. 24(1973), No. 4, 460-466. [4] Dickson L. E. Algebraic Theories, Dover, NY, 1959. [5] Kheyfits A. I., Eigenvalues of a boundary-value problem for a plane layer, Soviet Physics. Acoustics, Vol. 35(1989), No. 3, 330-332. [6] Kheyfits A. I., Closed-form representations of the Lambert W function, Frac- tional Calculus and Applied Analysis, Vol. 7(2004), No. 2, 177-190 [7] King R. B., Beyond the Quintic Equation, Birkh¨auser, Boston, 1996. [8] Klein F., Vorlesungen Uber¨ Das Ikosaeder Und Die Aufl¨osung Der Gleichungen Vom F¨uften Grade, Leipzig, 1884. [9] Mora T., Solving Polynomial Equation Systems I, Cambridge Univ. Press, 2003. 6 ALEXANDERKHEYFITS

[10] Nash O., On Klein?s icosahedral solution of the quintic. Expo. Mathematicae, Vol. 32(2014), 99-120.

Bronx Community College of The City University of New York USA E-mail: alexander.kheyfi[email protected]