Full Description of Ramanujan Cubic Polynomials

Roman Wituła Institute of Mathematics Silesian University of Technology Kaszubska 23 Gliwice 44-100 Poland

                     roman.witula@polsl.pl

Dedicated to Vladimir Shevelev – for his inspiration

Abstract We give a full description of the Ramanujan cubic polynomials, introduced and first

investigated by V. Shevelev. We also present some applications of this result.

1 Introduction

Shevelev [2] called the cubic polynomial

x^{3} + p x^{2} + q x + r  (1)

a Ramanujan cubic polynomial (RCP), if it has real roots x_{1}, x_{2}, x_{3} and the condition

p r^{1/3} + 3 r^{2/3} + q = 0  (2)

is satisfied. It should be noticed, that if x_{1}, x_{2}, x_{3} are roots of RCP of the form (1), then the following formulas hold (see [2, 5]):

x_{1}^{1/3} + x_{2}^{1/3} + x_{3}^{1/3} = (−p − 6 r^{1/3} + 3 (9 r − p q)^{1/3})^{1/3},  (3)
(x_{1} x_{2})^{1/3} + (x_{1} x_{3})^{1/3} + (x_{2} x_{3})^{1/3} = (q + 6 r^{2/3} − 3 (9 r^{2} − p q r)^{1/3})^{1/3},  (4)

and Shevelev’s formula [2]:

(\frac{x_{1}}{x_{2}} )^{1/3} + (\frac{x_{2}}{x_{1}} )^{1/3} + (\frac{x_{1}}{x_{3}} )^{1/3} + (\frac{x_{3}}{x_{1}} )^{1/3} + (\frac{x_{2}}{x_{3}} )^{1/3} + (\frac{x_{3}}{x_{2}} )^{1/3} = (\frac{p q}{r} − 9)^{1/3}. (5)

We note that (3) easily implies all three Ramanujan equalities

(\frac{1}{9})^{1/3} − (\frac{2}{9})^{1/3} + (\frac{4}{9})^{1/3} = (\sqrt[3]{2} − 1)^{1/3},  (6)
( cos \frac{2 π}{7} )^{1/3} + ( cos \frac{4 π}{7} )^{1/3} + ( cos \frac{8 π}{7} )^{1/3} = (\frac{5 − 3 \sqrt[3]{7}}{2})^{1/3},  (7)
( cos \frac{2 π}{9} )^{1/3} + ( cos \frac{4 π}{9} )^{1/3} + ( cos \frac{8 π}{9} )^{1/3} = (\frac{3 \sqrt[3]{9} − 6}{2})^{1/3},  (8)

since the following decompositions of polynomials hold: (19), which implies (6) after some algebraic transformations for every r ∈ ℝ \ {0} (the equality (6) we obtain by setting r = 8/729), (28), which implies (7) and at last (10), which implies (8).

In [2] many interesting and fundamental properties of RCP’s are presented.

The object of this paper is to prove the following fact

Theorem 1. All RCP’s have the following form

x^{3} − \frac{P (γ − 1)}{(γ − 1) (γ − 2)} r^{1/3} x^{2} − \frac{P (2 − γ)}{(1 − γ) (2 − γ)} r^{2/3} x + r =

= (x − \frac{r^{1/3}}{2 − γ} ) (x − (γ − 1) r^{1/3}) (x − \frac{2 − γ}{1 − γ} r^{1/3}), (9)

where r ∈ ℝ \ {0}, γ ∈ ℝ \ {1, 2}, and

P (γ) := γ^{3} − 3 γ + 1 = (γ − 2 cos \frac{2 π}{9} ) (γ − 2 cos \frac{4 π}{9} ) (γ − 2 cos \frac{8 π}{9} ).  (10)

Corollary 2. From formulas (3), (4) and (5) for the sums of the real cube root of the roots of polynomial (9), the following equalities can be generated

γ^{3} − 9 (γ − 1)^{2} + 3 (γ^{2} − 3 γ + 3) \sqrt[3]{(γ − 1) (γ − 2)} =

= (\sqrt[3]{1 − γ} − \sqrt[3]{(2 − γ) (1 − γ)^{2}} + \sqrt[3]{(2 − γ)^{2}})^{3}, (11)

γ^{3} − 9 γ + 9 − 3 (γ^{2} − 3 γ + 3) \sqrt[3]{(γ − 1) (γ − 2)} =

= (\sqrt[3]{2 − γ} − \sqrt[3]{(1 − γ) (2 − γ)^{2}} − \sqrt[3]{(1 − γ)^{2}})^{3}, (12)

which, after replacing γ := 3 − γ, is equivalent to (11);

(\frac{1}{\sqrt[3]{(2 − γ) (1 − γ)}} + \sqrt[3]{(2 − γ) (1 − γ)} − \sqrt[3]{\frac{1 − γ}{(2 − γ)^{2}}} + \sqrt[3]{\frac{2 − γ}{(1 − γ)^{2}}} +

+ \sqrt[3]{\frac{(1 − γ)^{2}}{2 − γ}} − \sqrt[3]{\frac{(2 − γ)^{2}}{1 − γ}} )^{3} = 9 − \frac{P (γ − 1) P (2 − γ)}{(γ − 1)^{2} (2 − γ)^{2}}, (13)

i.e.,

(γ^{2} − 3 γ + 3)^{3} = 9 (γ − 1)^{2} (2 − γ)^{2} − P (γ − 1) P (2 − γ).  (14)

The above relations essentially supplement the set of identities presented in [1]. Furthermore, (11)–(14) entail Ramanujan’s equalities (6)–(8), as well as all the other expressions of this type discussed in [2, 4, 5].

In the second part of this paper we will discuss an important Shevelev parameter \frac{p q}{r} of RCP’s having the form (1). We note, that from (17) the following Shevelev inequality follows:

\frac{p q}{r} ≤ \frac{9}{4}.  (15)

We remark that for every a ∈ ℝ, a ≤ \frac{9}{4}, there exist at most six different sets of RCP’s, depending only on values r and having the same value of \frac{p q}{r}, equal to a. In the sequel, there exist only two sets of RCP’s, depending on r ∈ ℝ, having the value \frac{p q}{r} = 2 (see the descriptions (40) and (41)). However, there is only one family of RCP’s, depending on r ∈ ℝ with \frac{p q}{r} = \frac{9}{4} (see the descriptions (19)). This fact is proven in Section 2, but it independently results from (31), (9), (14) and from the following identity

\frac{p q}{r} = 9 − \frac{((γ − 1) (γ − 2) + 1)^{3}}{((γ − 1) (γ − 2))^{2}} .  (16)

From (16) we get

\frac{p q}{r} = \frac{9}{4} ⇔ t := (γ − 1) (γ − 2) ∈ \{−\frac{1}{4}, 2\} ⇔ γ ∈ \{0, \frac{3}{2}, 3\},

since we have

\frac{d}{dt}(9 − \frac{(t + 1)^{3}}{t^{2}} ) = t (2 − t) \frac{(t + 1)^{2}}{t^{4}} .

All three values γ ∈ \{0, \frac{3}{2}, 3\} generate the same set of RCP’s of the form (19).

2 Proof of Theorem 1

Let us indicate that from condition (2) the following equality follows (see [2]):

\frac{9}{4} − \frac{p q}{r} = (\frac{3}{2} + \frac{p}{r^{1/3}} )^{2}.  (17)

Let

p := (α − \frac{3}{2}) r^{1/3}.

By (17) we have

\frac{p q}{r} = \frac{9}{4} − α^{2},
q = −(α + \frac{3}{2}) r^{2/3}.

In other words, an RCP has the form

x^{3} + (α − \frac{3}{2}) r^{1/3} x^{2} − (α + \frac{3}{2}) r^{2/3} x + r  (18)

for some α, r ∈ ℝ. If α = 0, the following decomposition holds

x^{3} − \frac{3}{2} r^{1/3} x^{2} − \frac{3}{2} r^{2/3} x + r = (x − \frac{1}{2} r^{1/3}) (x + r^{1/3}) (x − 2 r^{1/3}).  (19)

Accordingly, the roots x_{1}, x_{2}, x_{3} of the polynomial (18) have the form (r ≠ 0):

x_{1} = (\frac{1}{2} + β) r^{1/3}, x_{2} = (−1 + γ) r^{1/3}, x_{3} = (2 + δ) r^{1/3}  (20)

for certain β, γ, δ ∈ ℝ. Then from Vieta’s formulae the following equations can be obtained

α = −(β + γ + δ),  (21)
(\frac{1}{2} + β) (−1 + γ) + (\frac{1}{2} + β) (2 + δ) + (−1 + γ) (2 + δ) = −α − \frac{3}{2},  (22)
(\frac{1}{2} + β) (−1 + γ) (2 + δ) = 1.  (23)

From (21) and (22) we receive

β = \frac{\frac{3}{2} (δ − γ) − δ γ}{δ + γ} ,  (24)

which, by (23), implies

δ^{2} (γ^{2} − 3 γ + 2) + δ (3 γ^{2} − 7 γ + 3) + 2 γ^{2} − 3 γ = 0.

Hence, after some manipulations, we get

Δ_{δ} = (γ^{2} − 3 γ + 3)^{2},

and next

δ = \frac{γ}{1 − γ} or δ = \frac{3 − 2 γ}{γ − 2} . (25)

If we choose δ = \frac{γ}{1−γ}, then by (24) we have β = \frac{γ}{2 (2−γ)}, and by (20) we obtain

\begin{matrix}x_{1} = \frac{r^{1/3}}{2 − γ} , x_{2} = (γ − 1) r^{1/3}, x_{3} = \frac{2 − γ}{1 − γ} r^{1/3}, \\ α = −(\frac{γ}{2 (2 − γ)} + γ + \frac{γ}{1 − γ} ) = \frac{−2 γ^{3} + 9 γ^{2} − 9 γ}{2 (γ − 1) (γ − 2)} . \end{matrix}  (26)

Finally

x^{3} + \frac{−γ^{3} + 3 γ^{2} − 3}{(γ − 1) (γ − 2)} r^{1/3} x^{2} + \frac{γ^{3} − 6 γ^{2} + 9 γ − 3}{(γ − 1) (γ − 2)} r^{2/3} x + r =

= (x − \frac{r^{1/3}}{2 − γ} ) (x − (γ − 1) r^{1/3}) (x − \frac{2 − γ}{1 − γ} r^{1/3}), (27)

which is compatible with (9).

On the other hand, if we choose δ = \frac{3−2 γ}{γ−2} , then β = \frac{γ−3}{2 (γ−1)}, and we obtain the same values of x_{1}, x_{2}, x_{3} and α as in (26) above. □

Example 3. Since

(x − 2 cos \frac{2 π}{7} ) (x − 2 cos \frac{4 π}{7} ) (x − 2 cos \frac{8 π}{7} ) = x^{3} + x^{2} − 2 x − 1  (28)

is the RCP [4], then, from (9) the following relations can be deduced

\begin{matrix}γ = 1 − 2 cos \frac{2 π}{7} , r = −1, \\ \frac{P (γ − 1)}{(1 − γ) (2 − γ)} = 1 and \frac{P (2 − γ)}{(1 − γ) (2 − γ)} = 2, \end{matrix}

which implies the equalities

\frac{1}{γ − 2} = 2 cos \frac{4 π}{7} , \frac{γ − 2}{1 − γ} = 2 cos \frac{8 π}{7} ,

\frac{( cos \frac{2 π}{7} + cos \frac{2 π}{9}) ( cos \frac{2 π}{7} + cos \frac{4 π}{9}) ( cos \frac{2 π}{7} + cos \frac{8 π}{9})}{cos \frac{2 π}{7} (1 + 2 cos \frac{2 π}{7})} = −\frac{1}{4},  (29)

and the equivalent one

\frac{(\frac{1}{2} + cos \frac{2 π}{7} − cos \frac{2 π}{9}) (\frac{1}{2} + cos \frac{2 π}{7} − cos \frac{4 π}{9}) (\frac{1}{2} + cos \frac{2 π}{7} − cos \frac{8 π}{9})}{cos \frac{2 π}{7} (1 + 2 cos \frac{2 π}{7})} = \frac{1}{2}.  (30)

3 Values of \frac{p q}{r} for RCP’s

By (9) we obtain

\frac{p q}{r} = \frac{P (γ − 1) P (2 − γ)}{(γ − 1)^{2} (2 − γ)^{2}}.  (31)

The examples of RCP’s, which are given in [4, 5] (see also [2]), are produced by \frac{p q}{r} equal only to 2, −40, −180.

The following theorem holds.

Theorem 4. For every a ≤ \frac{9}{4} there exist at most six different sets of RCP’s, depending on r ∈ ℝ, having the same value of \frac{p q}{r}, equal to a.

Proof. The proof of this theorem results easily from inequality (15) and from relation (31).

We will present now a series of remarks, connected with the parameter a = \frac{p q}{r}.

Remark 5. Let us consider the following equation

\frac{P (γ − 1) P (2 − γ)}{(γ − 1)^{2} (2 − γ)^{2}} = a (a ∈ ℝ).  (32)

This equation, by (16), after substitution t := (γ − 1) (γ − 2), is equivalent to the following one

R(t) := t^{3} + (a − 6) t^{2} + 3 t + 1 = 0.  (33)

If we replace t in (33) by τ − \frac{a−6}{3}, then the canonical form of R(t) can be generated

τ^{3} + (3 − \frac{1}{3} (a − 6)^{2} ) τ + \frac{2}{27} (a − 6)^{3} − (a − 6) + 1.  (34)

But the polynomial (34) has only one real root, if and only if

\frac{1}{4} (\frac{2}{27} (a − 6)^{3} − (a − 6) + 1)^{2} + \frac{1}{27} (3 − \frac{1}{3} (a − 6)^{2} )^{3} > 0 ⟺

⟺ \frac{4}{27} (a − 6)^{3} − \frac{1}{3} (a − 6)^{2} − 2 (a − 6) + 5 > 0 ⟺ (a − 9)^{2}(a − \frac{9}{4}) > 0.

Since the case a = \frac{9}{4} was discussed in (19), the polynomial R(t) has three real roots for every a ≤ \frac{9}{4}.

Remark 6. If γ_{0} ∈ ℂ is a root of equation (32) (for fixed a ∈ ℂ) then also γ = 3 − γ_{0} and γ = \frac{γ_{0}}{1−γ_{0}} are roots of this one. We note, that the last fact derives from the following identities

(1 − γ)^{3} P (\frac{1}{γ − 1}) = P (2 − γ)

and

(1 − γ)^{3} P (\frac{γ − 2}{γ − 1}) = P (γ − 1).

Consequently, the roots of (32) are also

γ = \frac{3 − γ_{0}}{1 − (3 − γ_{0})} = \frac{3 − γ_{0}}{γ_{0} − 2}, γ = 3 − \frac{γ_{0}}{1 − γ_{0}} = \frac{3 − 4 γ_{0}}{1 − γ_{0}} , γ = 3 − \frac{3 − γ_{0}}{γ_{0} − 2} = \frac{4 γ_{0} − 9}{γ_{0} − 2}.

Remark 7. Let us separately discuss equation (32) for a = 2. After substitution t = 1−τ in (33), the following equation is derived

τ^{3} + τ^{2} − 2 τ − 1 = 0,  (35)

i.e. (see [4]):

(τ − 2 cos \frac{2 π}{7} ) (τ − 2 cos \frac{4 π}{7} ) (τ − 2 cos \frac{8 π}{7} ) = 0.  (36)

Hence, equation (32) for a = 2 is equivalent to each of the following three equations

(γ − 1) (γ − 2) = 1 − 2 cos \frac{2 π}{7} ⟺ γ − 1 = −2 cos \frac{4 π}{7} ∨ γ − 2 = 2 cos \frac{4 π}{7} , (37)

or

(γ − 1) (γ − 2) = 1 − 2 cos \frac{4 π}{7} ⟺ γ − 1 = −2 cos \frac{8 π}{7} ∨ γ − 2 = 2 cos \frac{8 π}{7} , (38)

or

(γ − 1) (γ − 2) = 1 − 2 cos \frac{8 π}{7} ⟺ γ − 1 = −2 cos \frac{2 π}{7} ∨ γ − 2 = 2 cos \frac{2 π}{7} . (39)

For the values

γ ∈ \{1 − 2 cos \frac{2^{k} π}{7} : k = 1, 2, 3\},

we obtain the same set of RCP’s

x^{3} + r^{1/3} x^{2} − 2 r^{2/3} x − r,  r ∈ ℝ.  (40)

On the other hand, for values

γ ∈ \{2 + 2 cos \frac{2^{k} π}{7} : k = 1, 2, 3\},

we obtain the following set of RCP’s

x^{3} − 2 r^{1/3} x^{2} − r^{2/3} x + r,  r ∈ ℝ.  (41)

We note, that RCP of the form (see [2, 4]):

x^{3} + 7 x^{2} − 98 x − 343 =

= (x − 128 cos \frac{2 π}{7} ( sin \frac{2 π}{7} sin \frac{8 π}{7} )^{3}) (x − 128 cos \frac{4 π}{7} ( sin \frac{2 π}{7} sin \frac{4 π}{7} )^{3}) ·

· (x − 128 cos \frac{8 π}{7} ( sin \frac{4 π}{7} sin \frac{8 π}{7} )^{3})

belongs to the set (40) of RCP’s with \frac{p q}{r} = 2 for r = 7^{3}, because of the following remark.

Remark 8. Suppose, that α ∈ \{\frac{2π}{7}, \frac{4π}{7}, \frac{8π}{7}\}. Then, we have sin α = sin 8α, which implies

14 cos α = 7 \frac{sin 2α}{sin α} = (8 sin α sin 2α sin 4α)^{2} \frac{sin 2α}{sin α} = 64 \frac{sin α}{sin 4α}( sin 2α)^{3} ( sin 4α)^{3} =

= 64 \frac{sin 8α}{sin 4α}( sin 2α)^{3} ( sin 4α)^{3} = 128 cos 4α ( sin 2α sin 4α)^{3}.

References

[1] B. C. Berndt, H. H. Chan and L.-Ch. Zhang, Radicals and units in Ramanujan’s work, Acta Arith. 87 (1998), 145–158.

[2] V. Shevelev, On Ramanujan Cubic Polynomials, preprint,

    http://arxiv.org/abs/0711.3420, 2007.

[3] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, available electronically

    at http://www.research.att.com/~njas/sequences/, 2010.

[4] R. Wituła and D. Słota, New Ramanujan-type formulas and quasi-Fibonacci numbers of order 7, J. Integer Seq. 10 (2007), Article 07.5.6.

[5] R. Wituła, Ramanujan type trigonometric formulas: the general form for the argument \frac{2 π}{7}, J. Integer Seq. 12 (2009), Article 09.8.5.

2000 Mathematics Subject Classification: Primary 11C08; Secondary 11B83, 33B10. Keywords: Ramanujan cubic polynomial.

Received December 1 2009; revised versions received March 2 2010; May 4 2010. Published in Journal of Integer Sequences, May 5 2010.

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