On an Improvement of the Numerical Application for Cardano’s Formula in Mathematica Software Alicja Wróbel∗ , Edyta Hetmaniok∗ , Mariusz Pleszczyński∗ , and Roman Wituła∗ , ∗ Institute of Mathematics, Silesian University of Technology, ul.Kaszubska 23, 44-100 Gliwice, Poland Abstract—The aim of this paper is to develop a program The program, developed by us, gave the following form of the for determining, by symbolic description, the roots of each real solutions of equation (1): cubic polynomial on the basis of the well known Mathematica √ ! √ ! software. We have obtained a program completely satisfying our 1 87 π 1 87 expectations and even more. For example, for many tested cases −4 cos arctan , 4 sin ± arctan . of the cubic polynomials, on the way of comparing the description 3 13 6 3 13 of the roots of these polynomials received by using our program The possibly excessive optimism about the effectiveness of with their trigonometric form obtained as a result of geometric discussion or respective trigonometric transformations, we have symbolic computations realized by our program has been got some new attractive relations of algebraic and trigonometric suppressed by other examples. For instance, for the equation nature. By applying our elaborated program we can also decide, x3 − 12x + 11 = 0, symbolically, whether the given cubic polynomial is a Ramanujan cubic polynomial (one of two kinds). Moreover, in case of Mathematica answer are the following solutions these polynomials we have got many new additional pieces of 1 √  information essentially completing the facts discovered by now. 1, −1 ± 3 5 . Additionally, we have solved some, posed ad hoc, theoretical 2 problem. To the contrast our program produces the following trigono- Index Terms—cubic polynomials, Ramanujan cubic polynomi- metric form of the same solutions als, Cardano’s formula. √ ! √ ! 1 3 15 π 1 3 15 −4 cos arctan , 4 sin ± arctan , I. I NTRODUCTION 3 11 6 3 11 While preparing papers [12], [14] a part of computations but only numerically Mathematica verified the equality has been done with the aid of Mathematica software. To √ ! π 1 3 15 our surprise, the program did not manage to derive, in the 1 = 4 sin − arctan way as we expected, the symbolic transformations needed for 6 3 11 √ ! determining the zeros of polynomials, especially the real cubic π 1 3 15 and quartic polynomials. We had to make by hand a part of = 4 cos + arctan . 3 3 11 the final transformations. For example, for the equation In the third section of this paper we present a number of x3 − 12x + 13 = 0 (1) such completely unexpected trigonometric relations obtained by comparing the ”model” decompositions of some cubic Mathematica program gave us the following set of solutions polynomials with the decompositions obtained by us by ap- √  r  4 3 1 plying the introduced Cardano’s type formulae (implemented q √ + −13 + i 87 , for our computer procedure). 3 1  2 2 −13 + i 87 It should be also emphasized that the mentioned model √  √  3 1 √  polynomials and their decompositions were usually obtained r 2 1+i 3 1 −q √ −  2 1 − i 3 −13 + i 87 , in result of some trigonometric transformations of the known 3 1 2 2 −13 + i 87 trigonometric relations (see [5], [17], [18], [19], [28], [33], √  √  3 1 √  [34]). In consequence, these two different ways of decom- r 2 1−i 3 1 −q √  2 − 1 + i 3 −13 + i 87 . posing the cubic polynomials resulted in many interesting 3 1 −13 + i 87 2 2 equalities of trigonometric nature shedding a new light on many quantities, mysterious till now. For example, we dis- We decided to change this situation which resulted in elaborat- k covered that the values of cos 2 7π ,√k ∈ N, are strictly ing the appropriate computer procedure calculating the roots of all real cubic polynomials starting from the Cardano’s formula. connected   with  the k+2  of arctan(3 3), whereas values of values 2k π 2 π cos 7 / cos 7 , k ∈ N, are strictly connected with 1 Copyright c 2016 held by the authors. the values of arctan 3 3 – see Section 3. √ 71 Finally, in the fourth section we describe the recently popu- which implies (the formulae below are satisfied by all three lar subject matter concerning the Ramanujan cubic polynomi- real zeros of p(z)): als. Discussion executed for the goal of preparing this paper r    a 0 1 ∗ resulted in creating a new paper devoted to these polynomials z = u + v = 2 − cos + (π − arctan ∆ ) [2] – the obtained there original result is announced at the end 3 ±2π 3 r    3  of Section 4. a 1 ±π = 2 − cos − arctan ∆∗ (2) As the Authors, we want to emphasize that our aim now is 3 3 3π ∗  p a 1  to initiate the investigations on the continuation of this work p − −2 3 cos 3 arctan ∆ , = (3) 2 − a3 cos 13 (±π + arctan ∆∗ ) ,  where we intend to concentrate on developing the algorithm for determining the roots of polynomials belonging to the  −2 − a cos 13 arctan ∆∗ ,√ p  selected families of polynomials of higher orders, for which = p a 3 − 3 cos 13 arctan ∆∗ ± 3 sin 13 arctan ∆∗ ,  such description of the roots is known (the examples are (4) quartic polynomials, the modified Chebyshev polynomials [35] and the selected quintics [8], [26]). or in the equivalent form ∗  p a 1   −2 p − 3 cos 3 arctan ∆ ,  z =u+v = 2p− a3 cos 13 (π + arctan ∆∗ ) , (5) II. Z EROS OF CUBIC POLYNOMIALS – FINAL FORMULAE  2 − a3 sin 16 (π + 2 arctan ∆∗ ) , − 3 cos 31 arctan ∆∗ ,   p a  −2 From the Cardano procedure for finding the (complex) zeros = 2 − a3 sin π6 ± 31 arctan ∆∗ . p of real cubic polynomial (see for example [11], [16], [20], [28], [35] – two last papers consider also because of their historical The other form of zeros of p(z) can be deduced if we introduce connotations): a new parameter ϕ setting p(z) := z 3 + az + b ∆∗ = tan ϕ ≥ 0. 3  2 Then − a3 b = 2 cos ϕ . Hence, if we set a = −3α2 , it follows that p(z) = q 0 if and only if z = u + v where q √ √ 2 3 b = 2β 3 cos ϕ, then α = ±β. Finally we obtain the following u ∈ − 2 + ∆, v ∈ − 2b − ∆ and ∆ = 2b + a3 , 3 b 3 implication: if tan ϕ ≥ 0 and 2α cos ϕ > 0, then and where the respective complex roots (of the second and third order) are kept in mind. Let us discuss three cases with z 3 − 3α2 z + (2 cos ϕ)α3 = 0 ⇔ −2α cos ϕ3 , √  respect to the sign of discriminant ∆. (6) ⇔z= α cos ϕ3 ± 3 sin ϕ3 .  First we assume that ∆ < 0 (which implies a < 0). Then r r For example, we get b √ b √ −2α cos π9 , √  3 3 u∈ − + i −∆, v∈ − − i −∆ 3 2 3 z − 3α z + α = 0 ⇔ z = 2 2 α cos π9 ± 3 sin π9 ,   k  2 π and p(z) possesses three real roots. = 2α cos , k = 1, 2, 3, 9 Now let us discuss two cases with respect to the sign of coefficient b. √ z 3 − 3α2 z + 2α3 = 0 First q case, when b > 0: Additionally let us set ∆∗ :=  √  2 π 3+1 − 2b ∆. We suppose that ∆∗ ∈ [0, ∞), since −∆ ≥ 0 ⇔   −2α cos 12 = − √ 2 α,  ( √ (∆∗ )2 = −1 − a3 2a 2 ≥ 0. ⇔z= √ 2α, π π √   α cos 12 ± 3 sin 12 = 3b  3−1 We have √ α,  2 ( √ √ r 2α √ b a 3   = − ± i −∆ = − exp ± i (π − arctan ∆∗ ) ± √3−1 α, 2 3 2 √ π  and (each sign from the circle corresponds with two signs ± −2α cos 18 ,√ z 3 − 3α2 z + 3α3 = 0 ⇔ z = π π  not included in the circle): α cos 18 ± 3 sin 18 ,  k  2 π r √ = 2α(−1)k−1 sin , k = 1, 2, 3. b 3 9 − ± i −∆ = r 2 Second case, when b < 0: We get     a 0 1 ∗ r = − exp i ± (π − arctan ∆ ) , b √ a 3   3 ± 23 π 3 − ± i −∆ = − exp ± i arctan ∆∗ 2 3 72 q √ q √ and next where u := 3 − 2b + ∆, v := 3 2b + ∆. Hence, on the basis of the same formulae, if ∆ = 0, then r b √ 3 − ± i −∆ = we get 2 r r 3 b 3 b r     a 0 1 ∗ z1 = −2 , z2 = z3 = . = − exp i ± arctan ∆ , 2 2 3 ± 32 π 3 III. U NEXPECTED IDENTITIES which implies r    By using the cubic polynomials (and their decompositions) a 0 1 ∗ z = u + v = 2 − cos + arctan ∆ (7) from papers [17], [18], [28], [30], [33], [34], [35] and by 3 ± 23 π 3 applying the discussed here Cardano’s formulae for the roots 2 − 3 cos 13 arctan ∆∗ , √  p a  of cubic polynomials, we obtained the completely unexpected = − − 3 cos 13 arctan ∆∗ ∓ 3 sin 13 arctan ∆∗ , p a  trigonometric identities (of course we present here just few (8) selected examples denoted by A – H). Let us explain only that  p a  2 p 1 − 3 cos 3 arctan ∆ ,∗  all the presented equalities are the consequence of the fact that = −2p− a3 cos 13 (π + arctan ∆∗ ) ,  (9) the decompositions of cubic polynomials, given in the cited  −2 − a3 sin 16 (π + 2 arctan ∆∗ ) , above papers, result, in the first place, from the appropriate trigonometric transformations (also by geometric discussion) − 3 cos 13 arctan ∆∗ ,  p a  2 p and not from the application of the Cardano’s formulae. = −2 − a3 sin π6 ± 31 arctan ∆∗ .  We obtain Similarly as in the previous case, if we set ∆∗ = tan ϕ ≥ 0, A) a = −3α2 , b = 2α3 cos ϕ < 0, then Y3  2k π  3 2 z + z − 2z − 1 = z − 2 cos , −2α cos ϕ3 , √  3 2 3 7 z −3α z+(2 cos ϕ)α = 0 ⇔ z = k=1 α cos ϕ3 ± 3 sin ϕ3 .  which implies For example, we obtain (below we assume that α > 0, √ √   moreover, the respective values of sine and cosine functions 2π 1 6 cos = −1 + 2 7 cos arctan(3 3) , can be found in [20], [21], [28]): 7 3 √ z 3 − 3α2 z − 2α3 = 0 √ √     4π 1 √ 6 cos = −1 − 2 7 cos π + arctan(3 3) ,  π 3+1 7 3  2α cos 12 = √2 α,  ( √ ⇔z= √ √ √     π π  −√ 2α, 8π 1  −α cos 12 ± 3 sin 12 =  − √3+1 6 cos = −1 − 2 7 sin π + 2 arctan(3 3) , 2 α, 7 6 ( √ − √2α, B) = k 3 ! 1± √ 3 α, 3 2 Y cos 2 7π 2 z − 3z − 4z − 1 = z− k+2 , k=1 cos 2 7 π √ 3 2 5+1 3 which follows z − 3α z − α =0 2 cos 8π √ r √ p √     π α 7 7 1 1  2α cos 15 = 4 5 − 1 + 3 10 + 2 5 , = 1 + 2 cos arctan √ , √ cos 4π 3 3    π π  7 3 3 −α( cos15 ± 3 sin 15 =  ⇔z= √ √ p √  cos 4π r    α 7 1 1  5 − 1 − 3 10 + 2 5 , 7  = 4 √ , =1−2 cos π + arctan √ ,   cos 2π 3 3  α 3 3 2 (1 − 5), 7 ( α √ 2 (1 − 5), As an example we propose to examine the following polynomial = α  √ √ p √  4 5 − 1 ± 3 10 + 2 5 . 2    Y   sin(9α/2) x − 2 cos 2k α = x3 + 1 + 2 cos 3α − x2 We only need to discuss two more (rather trouble-free) k=0 sin(α/2) cases.  sin(13α/2)  sin 8α + 2 cos 3α − 2 cos 4α − 1 + x− , If ∆ > 0, then we have precisely one real zero and two sin(α/2) sin α conjugate complex roots which gives for α = 2π/7, 2π/9 the polynomial of type A and polynomial z1 = u − v, 3  √ x3 − 3x + 1 = Y  x − 2 cos 2k π/9  i 23 π −i 23 π 1 3 z2 = e u−e v = − (u − v) + i (u + v), k=1 2 2 which can be simply generated on the basis of discussion presented item E, z3 = z2 , respectively. 73 cos 2π r    7 7 1 1 To the contrast, we note that = 1 − 2 sin π + 2 arctan √ . cos 8π 3 6 3 3 √ 7 3 √ 2 3 We note that (see [34]): z + 3z − 3z −   3   v 2π 4π 8π 3 uu cos 2k π = z − cot z + cot z − cot , 9 9 9 X 3 7 t 2k+2 π = 0. k=1 cos 7 which implies the relation C) √ 3  2k π 2k π √ 2k π (−1)k−1 3 cot = 4 cos − 1, Y  z 3 − 7z 2 + 7z + 7 = z − 7 cot , 9 9 7 k=1 for k = 1, 2, 3, or the equivalent equalities which implies 2π 4 cos 2π −1 4π π 1 − 4 sin 18 √ 8π √  1 √  cot = √9 , cot = √ , 3 7 cot = 7 + 4 7 cos arctan 3 3 , 9 3 9 3 7 3 8π π 4 cos 9 + 1 − cot = √ . √ 2π √   1 √   9 3 3 7 cot = 7 − 4 7 cos π + arctan 3 3 , 7 3 F) √ √   √   √ √ 4π 1 1− 13 2 13 + 3 3 7 cot = 7 − 4 7 sin π + 2 arctan 3 3 ; z +3 z −z+ 7 6 2 2     D) 2π 6π 8π = z−2 cos z−2 cos z−2 cos , 3  13 13 13 2k π (10)  1 Y 4 z3 − z2 − z − = z − 1 + √ sin , which implies 7 7 7 k=1 √ √ q which follows 2π 12 cos = −1 + 13 + 2 26 − 2 13× √ ! 13 6 8π 1 3 3 √ ! √ sin = 1 − 2 cos arctan , 1 5 + 2 13 7 7 3 13 × cos arctan √ , 3 3 3 √ !! 6 2π 1 3 3 √ sin = 1 + 2 cos π + arctan , 6π √ q √ 7 7 3 13 12 cos = −1 + 13 − 2 26 − 2 13× 13 √ !! √ !! 6 4π 1 3 3 1 5 + 2 13 √ sin = 1 + 2 sin π + 2 arctan . × cos π + arctan √ , 7 7 6 13 3 3 3 To the contrast, we note that √ √ q 3 √ ! 8π 1 Y 7 k 2 π 12 cos = −1 + 13 − 2 26 − 2 13× z3 + z2 − z + = z− tan . 13 √ !! 7 7 7 1 5 + 2 13 k=1 × sin π + 2 arctan √ . E) 6 3 3 √ √ z 3 + 3 3z 2 − 3z − 3 G)  2π  4π  8π  √ √ = z − tan z + tan z − tan , 3 1+13 2 13 + 3 9 9 9 z + z −z−  2  2   which implies the relation 4π 10π 12π = z−2 cos z−2 cos z−2 cos , √ 13 13 13 2k π 2k π (11) tan = 4 sin + (−1)k 3, 9 9 from which we get for k = 1, 2, 3, or the equivalent equalities √ q √ 2π 2π 2π √ 4π π √ 12 cos = −1 − 13 + 2 26 + 2 13× tan = 4 sin − 3, tan = 4 cos + 3, 13 9 9 9 18 √ ! 1 5 − 2 13 8π π √ × cos arctan √ , tan = 4 sin − 3. 3 3 3 9 9 74 √ √ q 10π Furthermore, from the equality z 3 = z+1 for z = τ0−1 we 12 cos = −1 − 13 − 2 26 + 2 13× 13 √ !! obtain the following decomposition of τ0−1 in the nested 1 5 − 2 13 third roots × cos π + arctan √ , 3 3 3 r 1√ 1√ r r √ q 3 1 3 1 3 3 − 69+ + 69 = 1 + 1 + 3 1 + . . . 12π √ q √ 2 18 2 18 12 cos = −1 − 13 − 2 26 + 2 13× 13 We note that the convergence of the above nested radical √ !! 1 5 − 2 13 from Theorem 3.2 in [15] holds. Moreover, we observe × sin π + 2 arctan √ . 6 3 3 that the similar equality is fulfilled also for the only positive root zk of theqpolynomial z k − z − 1, k ≥ 3, Moreover, surprisingly the following relation holds k p √ √ having the form zk = 1 + k 1 + k 1 + . . .. 13 − 4 8π In order to emphasize the importance of this section let us − 2 cos ≈ 0.577727 3 13 introduce two more decompositions of the cubic polynomials, ≈ C = 0.57721, which we have not found in literature till now and which where C is the Eulerian constant called also as the Euler- essentially complete the decompositions presented in items F) Mascheroni constant (see [13]). and G). H) So from (10) and (11) we can obtain the decompositions √ √ z 3 − z − 1 = (z − τ0−1 )(z − i τ0 eiΨ )(z + i τ0 e−iΨ ).     4π 10π 12π z − 2 sin z − 2 sin z − 2 sin 13 13 13 The above polynomial is called the Perrin polynomial, s √ s √ also called as the Siegel’s polynomial (see [7], [34]). 3 13 + 3 13 2 √ 13 − 3 13 Constant τ0−1 plays an important role in estimating the =z − z + 13z − 2 2 Mahler measure M (f ) of polynomials f over C, which are not reciprocal. This estimation is of the form M (f ) ≥ and τ0−1 and is optimal (see [23], [25]). To the contrast let     us note that τ0 is the only positive root of polynomial 2π 6π 8π z − 2 sin z − 2 sin z + 2 sin τ 3 + τ 2 − 1 (the same fact holds for the polynomial 13 13 13 τ 5 + τ − 1 = (τ 3 + τ 2 − 1)(τ 2 − τ + 1)) and s √ s √ 13 − 3 13 2 √ 13 + 3 13 = z3 − z − 13z + . 1 2 2 Ψ := arcsin p 3 . 2 τ0 Remark III.1. In this section we have dealt with expressions The number −τ0 is the only real root of polynomial τ 3 − of the form cos 31 arctan ∆∗ and sin 13 arctan ∆∗ . Let us  τ 2 + 1. Furthermore, we get notice that one can find in literature some very interesting √ 3 q 3 √ q 3 √  decompositions of such expressions on the nested square roots. 6τ0 = −2 + 4 25 − 3 69 + 25 + 3 69 , For example, in [3], [4] we can find the following Ramanujan formula √ √ √ q q 3 3 18τ0−1 = 3 9 + 69 + 9 − 69, A − 1 2√  1 2A + 1  lim an = + 4a + A sin arctan √ , √ 1 √ 1 q n→∞ 6 3 3 3 3 τ0 sin Ψ = τ0−1 , τ0 cos Ψ = τ0−1 4τ03 − 1 2 2 √ where A := 4a − 7, a ≥ 2, and and   r √ iΨ 1 −1 √ √ √ q q q i τ0 e = τ0 3 −1 + i 4τ0 − 1 , 2 a1 = a, a2 = a − a, a3 = a − a + a, s r √ q √ √ √ q  √ q 3 iΨ 3 3 a4 = a − a + a + a, . . . i2 18 τ0 e =− 9 + 69 + 9 − 69 √ √ √ q q  3 3 and where the sequence of signs −, +, +, . . ., appearing in +i 3 9 + 69 − 9 − 69 . this nested radicals, has period 3. In case a = 44 we obtain (see equalities in examples A, C, F for possible connections): Hence we deduce the equality p √ p √ √  1 √  q √ 3 9 + 69 − 3 9 − 69 lim an = 2 + 2 21 sin arctan 3 3 . 3 4τ0 − 1 = 3 p 3 √ p 3 √ . n→∞ 3 9 + 69 + 9 − 69 75 √ √ √ q p 3 IV. R AMANUJAN ’ S CUBIC POLYNOMIALS 3 x1 x2 + x1 x3 + x2 x3 = q + 6r2/3 − 3 3 9r2 − pqr 3 3 V. Shevelev and R. Wituła in papers [1], [24], [29], [31]   = sgn q + 6r2/3 − 3 sgn(9r2 − pqr)|9r2 − pqr|1/3 have distinguished and discussed the so called Ramanujan’s cubic polynomials and Ramanujan’s cubic polynomials of 1/3 × q + 6r2/3 − 3 sgn(9r2 − pqr)|9r2 − pqr|1/3 , the second kind, denoted for shortness by RCP and RCP2, respectively. And all this could happen thanks to the great as well as the so called Shevelev’s formula [24], [31]: Indian mathematician Srinivasa Ramanujan who proposed the r x1 r x2 r x1 r x3 r x2 r x3 proof of the following equalities (see [22]): 3 + 3 + 3 + 3 + 3 + 3  1/3  1/3  1/3 x2 x1 x3 x1 x3 x2 1 2 4 √ 1 q q q q = ( 2 − 1)1/3 , 3 − + = √ 3 x 2 x + 3 x x2 + 3 x2 x + 3 x x2 1 3 1 3 1 2 1 2 9 9 9 3 x x x 1 2 3   pq r pq q q  1/3  1/3  1/3 + 3 x22 x3 + 3 x2 x23 = sgn −9 3 −9 .  2π 4π 8π cos + cos + cos r r 7 7 7 √ !1/3 To the contrast, if instead of conditions (12) the following 5−337 conditions are fulfilled = , 2   r 6= 0, p3 r + 27r2 + q 3 = 0, (13)  1/3  1/3  1/3  b 2 a 3  2π 4π 8π a + 3 < 0, cos + cos + cos 9 9 9 then Q(x) is a RCP2 and the following relations hold √ !1/3 339−6 p3 p p  √ = . ξ1 + 3 ξ2 + 3 ξ3 = − p − 6 3 r 2 1/3 q √ √ q 3 √ √ 3 It is easy to connect the above equations with the following −3 3 3 3 r(q + p 3 r) − 3 (p + 3 3 r)(q + 3 r2 ) , problem: for which cubic polynomials Q(x) (with all real roots) of the form as well as s s s s s s 3 2 ξ ξ ξ ξ ξ ξ3 Q(x) = (x − ξ1 )(x − ξ2 )(x − ξ3 ) = x + px + qx + r 3 1 2 1 3 2 + 3 + 3 + 3 + 3 + 3 ξ2 ξ1 ξ3 ξ1 ξ3 ξ2 there exists a function f (p, q, r) that possesses possibly ”sim- √ r r ple” algebraic form and for which the following equality holds 3 3 √ 3 pq 3 √ 3 = (q + p 3 r) + + (q + p 3 r + 3 r2 ). r2/3 r r2/3 p3 p p p ξ1 + 3 ξ2 + 3 ξ3 = 3 f (p, q, r). Let us notice that a cubic polynomial p(x) = x3 + px2 + Although some attempts to solve this interesting problem have qx + r, which is either RCP or RCP2, is simultaneously RCP been undertaken (see for example the respective Ramanujan and RCP2 if and only if pq = 0 which implies that either Theorem in [4] or in the second Notebook of Ramanujan [22]), only the mentioned above V. Shevelev and R. Wituła √3  √ 2π  3 2 p(x) = x − 3 r x + r = x − 2 r cos 3 × succeeded in distinguishing the appropriate families of cubic 9 √ √    polynomials and in describing properties of these polynomials 4π 8π (see also the paper [1]). Let us only recall that if the following × x − 2 3 r cos x − 2 3 r cos 9 9 conditions are satisifed or 6= 0, √   r√ √ 1√   2π 3 p 3 r + 3 r2 + q = 0, (12) p(x) = x3 − 3 3 rx2 + r = x − 3 r sec ×  b 2 2 9 a 3  2 + 3 < 0,  1√ 4π  1√ 8π  × x − 3 r sec x − 3 r sec , where 2 9 2 9 p2 2 3 1 a := q − , b := p − pq + r, where r ∈ R \ {0}. On the other hand we know that the 3 27 3 polynomials belonging to families RCP and RCP2 share many then Q(x) is a RCP and the following identities hold (the last common analytical-algebraic properties (see [14], [24]). expressions in all three formulae below are prepared for the Let us present now the examples of cubic polynomials with algorithmic applications): indicating the Ramanujan √ classes of polynomials they belong: √ √ √  1/3 1◦ p(x) = x3 +3x2 −3 3 2x+1 is the RCP2 polynomial which 3 x1 + 3 x2 + 3 x3 = −p − 6r1/3 + 3(9r − pq)1/3   is not the RCP one. It has the following zeros (we apply here = sgn −p − 6r1/3 + 3 sgn(9r − pq)|9r − pq|1/3 our formulae (5)): ! √ 1/3 q 1 3 × −p − 6r1/3 + 3 sgn(9r − pq)|9r − pq|1/3 3 , x1 = −1 − 2 1 + 2 cos arccot p √ , 3 432−5 76 !! √ q 3 1 3 (this identity results directly from formula (15)). It means that x2 = −1 + 2 1+ 2 cos π + arccot p √ , 3 3 4 2−5 the following ”unexpected” polynomial identity holds √ √ !! q  √ q 3 1 3 p 3 3 3 3( 2 + 1)x − 1 = 3( 2 + 1)P (x). (16) x3 = −1 + 2 1 + 2 sin π + 2arccot p √ . 6 3 4 2−5 Additionally we observe that Moreover, by ”numerical experiment” we find that √ √ √ √ √ √ √ √ 3 √ q 3 3( 3 2 + 1) q 3 3 p √ = 3(25 + 20 3 2 + 16 3 4) = √ 3 , 3 x1 x2 + x1 x3 + x2 x3 + 2 + 3 3 2−1 432−5 2−1 √ q √ q √ 3 3 3 3 3 = 2+ 2 − 1 − 3( 2 + 1) ≈ 0.00545. so by identity arctan x = arccot x1 for x > 0 we get √ √ 3 3( 3 2 + 1) 2◦ q(x) = x3 +x2 −2x−1 = (x−2 cos 2π 4π 7 )(x−2 cos 7 )(x− arccot p √ = arctan √ 3 . 8π 2 cos 7 ) – an example of the RCP polynomial which is not 432−5 2−1 the RCP2 one. Using formulae (3)–(5) from [29] (or the respective ones from [31]) we obtain the following Ramanujan type equalities (for 3◦ The polynomials from examples B) – H) presented real roots of third order): in Section 3 are neither RCP polynomials nor RCP2 ones. √ √ √ 3 x1 + 3 x2 + 3 x3 = 0, r r r r r r Announcement x1 x2 x1 x3 x2 x3 3 + 3 + 3 + 3 + 3 + 3 = −3. (14) While preparing this section we have solved unexpectedly x2 x1 x3 x1 x3 x2 one more problem. We have proven that the polynomials We also have x1 + x2 + x3 = −3 and  3 q √ 6 √ √ √ 3 3 R(x; p) := x3 + 9px2 + 23 √ p2 x 3 x1 x2 + x1 x3 + x2 x3 = − 3( 2 + 1). 3 3 1 ± 93 3 (17) Hence we deduce the relation  6 √ √ √ − √ p3 = (x − x1 )(x − x2 )(x − x3 ), P (x) = (x − 3 x1 )(x − 3 x2 )(x − 3 x3 ) 1 ± 93 q √ 3 (15) = x3 − 3( 2 + 1)x + 1 3 where p ∈ R \ {0} and (for the upper signs): q √ and next, by applying the Cardano’s formulae (5) we get √ x1 6 6( 93 − 1) arccot 31 √ √   √ q q 2 6 3 1 1 3 = −3 − cos , 3 x1 = − √ 1 + 2 cos arctan 4 2−5 , p 23 3 3 3 3 3 √ √ √ x2 6 6( 93 − 1) π + arccot 31 √ 2 6 q √   1 1 q √  = −3 + cos , (18) 3 x2 = √ 1 + 3 2 cos π + arctan 4 3 2 − 5 , p 23 3 3 3 3 3 √ √ √ x3 6 6( 93 − 1) π + 2arccot 31 √ 2 q √   1 1 q √  = −3 + sin 3 x3 = √ 6 3 1 + 2 sin π + 2 arctan 3 4 2−5 . p 23 6 3 3 6 3 are the only RCP such that polynomial By comparing these relations with the formulae for values of √ √ √ x1 , x2 , x3 we obtain the identity (x − 3 x1 )(x − 3 x2 )(x − 3 x3 ) q √ xk + 1 3 3 √ = 3( 2 + 1), k = 1, 2, 3, belongs to the RCP2 family (see [2]). By applying ”our” 3 x k Cardano’s formulae and a bit of Mathematica ”magic sim- It is well known that each polynomial q(x) from RCP2 class has the from plifications” we obtain the following relations (see [31]): √ p √ 1 + 93 + 4 106 − 9 93 cos 13 ϕ r  √ q x 1 3 q(x) = x3 + 3 krx2 − 3 3 (k + 1)r2 x + r, 3 =− p √ , p 3 2 182 93 − 1497 where k, r ∈ R, r 6= 0. Then for the roots x1 , x2 , x3 of q(x) the following Shevelev’s identity holds √ p √ 93 + 4 106 − 9 93 cos 13 (π+ϕ)  −1 − r x2 3 = √ , r r r r r r x1 x2 x1 x3 x2 x3 p 3 + 3 + 3 + 3 + 3 + 3 p 3 2 182 93 − 1497 x2 x1 x3 x1 x3 x2 3 √ √ q √ q √ √  = 9 3 3 3 3 k − k + 1 + ( k + 1)(1 − 3 k + 1) . 3 √ p √ 1  −1 − 93 + 4 106 − 9 93 sin (π+2ϕ) r x 3 6 3 = p √ , Hence and from (14) we deduce the equality p 3 2 182 93 − 1497 √ 3 √ 3 q 3 √ 3 √ √  3= 2+1 2 − 1. where ϕ = arctan 21 3 9 + 93 x1 , x2 , x3 are defined by formulae (18). 77 V. C ONCLUSION [15] C.D. Lynd: Using difference equations to generalize results for periodic nested radicals, Amer. Math. Monthly 121 No 1 (2014), 45–59. In this paper we have presented the algorithms for [16] G.P. Matvievskaya: René Descartes 1596–1650, Nauka, Moscow 1976 determining the roots of cubic complex polynomials. The (in Russian). [17] P.S. Modenov: Problems in Geometry, Mir Publishers, Moscow 1981. proposed algorithms generate the descriptions of these roots [18] S.G. Moreno, E.M. Garcia-Caballero: A Geometric proof of Morrie’s with the aid of radicals of trigonometric functions. The Law, Amer. Math. Monthly 122 (2015), 168. examples of testing polynomials, presented in this paper, have [19] T.J. Osler: Cardan polynomials and the reduction of radicals, Math. Mag. 74 (2001), 26–32. been originally generated by using the direct methods different [20] F.W.I. Olver, D.W. Lozier, R.F. Boisvert, C.W. Clark: NIST Handbook than the given here algorithms. In consequence, by applying of Mathematical Functions, Cambridge Univ. Press, 2010. the algorithms presented here we received, the most often, the [21] A.P. Prudnikov, A.J. Bryczkov, O.I. Mariczev: Integrals and series, Elementary Functions, Nauka, Moscov 1981 (in Russian). new and different symbolic descriptions of the sought roots [22] S. Ramanujan: Notebooks (2 volumes), Tata Institute of Fundamental of the given cubic polynomials. It gave us the possibility, Research, Bombay 1957. by comparing the obtained and the testing descriptions, to [23] A. Schinzel: Solved and unsolved problems on polynomials, 149–159 in Panoramas of Mathematics Banach Center Publications, vol. 34, reveal many new identities and relations. Additionally some Warszawa 1995. conjecture arose about the possible existence of description of [24] V. Shevelev: On Ramanujan cubic polynomials, South East Asian J. the values of functions cos 31 arctan α , sin 13 arctan α in  Math. & Math. Sci. 8 (2009), 113122. [25] C.J. Smyth: On the product of the conjugates outside the unit circle of the form of radicals of variable α, where α is also a radical an algebraic integer, Bull. London Math. Soc. 3 (1971), 169175. defined on the set of rational numbers. We intend to discuss [26] B.K. Spearman, K.S. Williams: Characterization of solvable quintics this problem in a separate paper. Moreover, let us notice that x5 + ax + b, Amer. Math. Monthly 101 No. 10 (1994), 986992. [27] J.-P. Tignol: Galois’ Theory of Algebraic Equations, World Scientific, our algorithm verifies whether the given cubic polynomial New Jersey 2001. belongs to the RCP or RCP2 class. [28] R. Wituła: Complex Numbers, Polynomials and Partial Fractions De- compositions, Parts 1, 2 and 3, Silesian University of Technology Press, Gliwice 2010 (in Polish). [29] R. Wituła: Full description of Ramanujan cubic polynomials, J. Integer Also in a separate paper (see [36]) we intend, as we Seq. 13 (2010), article 10.5.7. declared at the end of Introduction, to extend the discussion [30] R. Wituła: On Some Applications of Formulae for Sums of Unimodular Complex Numbers, PKJS Publishers, Gliwice 2011. of the symbolic description of the roots of cubic polynomials, [31] R. Wituła: Ramanujan cubic polynomials of the second kind, J. Integer undertaken in this paper, for the roots of quartic polynomials Seq. 13 (2010), article 10.7.5. and the selected polynomials of higher order. Next, we plan to [32] R. Wituła: Ramanujan type trigonometric formulae, Demonstratio Math. 45 No. 4 (2012), 779–796. adapt them numerically since we count on some better results [33] R. Wituła, E. Hetmaniok, D. Słota, N. Gawrońska: Sums of the rational then the ones obtained with the aid of Mathematica software. powers of roots of the polynomials, International Journal of Pure and Applied Mathematics 85 No 1 (2013), 179–191. R EFERENCES [34] R. Wituła, P. Lorenc, M. Różański, M. Szweda: Sums of the rational powers of roots of cubic polynomials, Zesz. Naukowe [1] S. Barbero, U. Cerruti, N. Murru, M. Abrate: Identities involving zeros Politechniki Śla̧skiej, Seria: Matematyka Stosowana z.4 (2010), 17–34, of Ramanujan and Shanks cubic polynomials, J. Integer Seq. 16 (2013), http://mat.polsl.pl/zn/zeszyty/z4/znps mat stos 04 2014 str 017- article 13.8.1. 034.pdf. [2] B. Bajorska-Harapińska, M. Pleszczyński, R. Wituła, When RCP gener- [35] R. Wituła, D. Słota: Cardano’s formula, square roots, Chebyshev poly- ates RCP2, that is one more property of the Ramanujan polynomials, in nomials and radicals, J. Math. Anal. Appl. 363 (2010), 639–647. preparation. [36] A. Wróbel, E. Hetmaniok, M. Pleszczyński, R. Wituła: Symbolic de- [3] B.C. Berndt: Ramanujan’s Notebooks, Part IV, Springer, New York 1994. scription of roots of polynomials and their numerical implementation – [4] B.C. Berndt, S. Bhargava: Ramanujan – for Lowbrows, Amer. Math. better than in Mathematica software?, in preparation. Monthly 100 No 7 (1993), 644–656. [5] W.A. Beyer, J.D. Louck, D. Zeilberger: Math Bite: A generalization of a curiosity that Feynmann remembered all his life, Math. Magazine 69 No 1 (1993), 43–44. [6] B.N. Delone: Algebra (Theory of algebraic equations), in Mathematics, its subject, methods and significance, Part I, Academy of Science Press, Moscow 1956. [7] A. Dubickas, K.G. Hare, J. Jankauskas: There are no two non-real conju- gates of a Pisot number with the same imaginary part, arXiv:1410.1600v1 [math.NT]. [8] D.S. Dummit: Solving solvable quintics, Math. Comp. 57 (1991), 387– 401. [9] H.B. Dwight: Tables of Integrals and Other Mathematical Data, The MacMillan Company, New York 1961. [10] S.K. Gupta, W. Szymański: Cubic polynomials with rational roots and critical points, College Math. J. 41 No 5 (2010), 365–369. [11] A. Heefer, T. Rothman: On remembering Cardano anew, Math. Intelli- gencer 36 No 4 (2014), 53–66. [12] E. Hetmaniok, P. Lorenc, M. Pleszczyński, R. Wituła: Iterated integrals of polynomials, Appl. Math. Comput. 249 (2014), 389–398. [13] J.C. Lagarias: Eulers constant: Eulers work and modern developments, Bull. Amer. Math. Soc. 50 (2013), 527-628; arXiv:1303.1856. [14] P. Lorenc, M. Pleszczyński, R. Wituła: Representation of the sum of powers of the first n positive integers with multiple integral, [in review] 78