arXiv:1202.6246v1 [math.GM] 22 Feb 2012
Evaluation of Fifth Degree Elliptic Singular Moduli
Nikos Bagis Stenimahou 5 Edessa, Pellas 58200 Edessa, Greece
[email protected]
Keywords: Modular equations; Ramanujan; Continued Fractions; Elliptic Functions; Modular Forms; Polynomials; Algebraic Numbers
Abstract Our main result in this article is a formula for the extraction of the solution of the fifth degree modular polynomial equation i.e. the value of k25n r0 , when we know only two consecutive values kr0 and kr0 /25 . By this way we reduce the problem of solving the depressed equation if we known two consecutive values of the Elliptic singular moduli kr .
1
Introductory Definitions
For |q| < 1, the Rogers Ramanujan continued fraction (RRCF) is defined as R(q) :=
q 1/5 q 1 q 2 q 3 ··· 1+ 1+ 1+ 1+
(1)
From the Theory of elliptic functions the complete elliptic integral of the first kind is (see [3],[4],[5]): Z π/2 1 p dt (2) K(x) = 1 − x2 sin(t)2 0 It is known that the inverse elliptic nome (singular moduli), k = kr , kr′2 = 1 − kr2 is the solution of: K (kr′ ) √ (3) = r K(k)
In what it follows we assume that r ∈ R∗+ , (when r is positive rational then kr is algebraic). The function kr can be evaluated in certain cases exactly. Continuing we define: ∞ Y f (−q) := (1 − q n ) (4) n=1
1
Also holds the following relation of Ramanujan (see [1],[2],[8]): 1 R5 (q)
− 11 − R5 (q) =
f 6 (−q) qf 6 (−q 5 )
(5)
We can write the function f using elliptic functions. It holds f (−q)8 =
28/3 −1/3 q (kr )2/3 (kr′ )8/3 K(kr )4 π4
(6)
2kr kr′ K(kr )3 π 3 q 1/2
(7)
also holds (see [3]): f (−q 2 )6 =
2
Known results on Rogers-Ramanujan continued fraction
Theorem√2.1 (see also [6]) If q = e−π r and r real positive, then we define ′ 2 r kr kr M5 (r)−3 a = ar := ′ k25r k25r Then
(8)
1/5 11 ar 1p 2 , 125 + 22ar + ar R(q) = − − + 2 2 2
(9)
where M5 (r) is root of: (5Y − 1)5 (1 − Y ) = 256(kr )2 (kr′ )2 Y . Proof. Suppose that N = n2 µ, where n is positive integer and µ is positive real then holds that K[n2 µ] = Mn (µ)K[µ], (10) where K[µ] := K(kµ ) The following formula for M5 (r) is known (5M5 (r) − 1)5 (1 − M5 (r)) = 256(kr )2 (kr′ )2 M5 (r)
(11)
Thus if we use (6),(7),(11),(12), we get: R
−5
f 6 (−q) (q) − 11 − R (q) = 6 = a = ar = qf (−q 5 ) 5
Solving with respect to R(q) we get the result.
2
kr′ ′ k25r
2 r
kr M5 (r)−3 (12) k25r
Theorem 2.2 (see [6],[7]) R−5 (q) − 11 − R5 (q) =
kr3 (kr2 − 1) w5 − kr2 w
w w′ ww′ + ′ − kr kr kr kr′
3
,
(13)
′ with w2 = kr k25r , (w′ )2 = kr′ k25r .
kr6 + kr3 (−16 + 10kr2)w + 15kr4 w2 − 20kr3 w3 + 15kr2 w4 + kr (10 − 16kr2)w5 + w6 = 0 (14) Once we know kr we can evaluate w from the above equation (15) and hence the k25r . Hence the problem reduces to √ solve the 6-th degree equation (15), which under the change of variable w = kr k25r reduces to the ’depressed equation’ (see [4]): u6 − v 6 + 5u2 v 2 (u2 − v 2 ) + 4uv(1 − u4 v 4 ) = 0 (15)
2 with u8 = kr2 , v 8 = k25r . For solving the depressed equation and the Rogers-Ramanujan Continued fraction we need a relation of the form
k25r = Φ(kr ).
(16)
But such a construction of the root of the depressed equation is not found yet. In the next section we give a relation of the form k25r = Φ(kr , kr/25 ),
(17)
which is the main result of this paper.
3
The Evaluation of the 5th Degree Modular Equation
Let q = e−π that
√ r
, r > 0 and vr = R(q), then it have been proved by Ramanujan 5 vr/25 = vr
1 − 2vr + 4vr2 − 3vr3 + vr4 , 1 + 3vr + 4vr2 + 2vr3 + vr4
also is ar =
f (−q)6 . qf (−q 5 )6
Then from Theorem 2.1 1/5 1p 11 ar + R(q) = − − . 125 + 22ar + a2r 2 2 2
Set
tr =
11 ar + and tr = i sin(θr ), 2 2 3
(18)
(19)
(20)
(21)
then ar = 2i sin(θr ) − 11 and R(q) = e−iθr /5
(22)
The function θr can take and complex values. From the above and (19) we get the following modular equation of ar : ar/25 = Q(ar ) where
11 + x y = arcsinh 2
and
(23)
5 1 2 −1 − e 5 y + e 5 y
Q(x) = 1 2 3 4 6 7 8 9 e 5 y − e 5 y + 2e 5 y − 3e 5 y + 5ey + 3e 5 y + 2e 5 y + e 5 y + e 5 y
(24)
Assume now the equation
with the solutions
√ 1 X2 5Y √ − = √ Y 3 − Y −3 2 X 5Y 5
Y = U (X) = where
s
−
5 25 x4 h(x) + 2 + + 2 3x 3x h(x) h(x) 3x2
1/3 √ p h(x) = −125 − 9x6 + 3 3 −125x6 − 22x12 − x18 s √ 1 + 18Y 6 + Y 12 1 Y4 ∗ . + U (Y ) = X = − 2 + 2Y 2 2Y 2
(25)
(26)
(27)
From [8] we have the following: Theorem 3.1 If A= then
Theorem 3.2
′ G25r f (−q 2 ) 1/6 = a4r and V = Gr q 1/3 f (−q 10 )
√ ′ ′ A2 5V 1 ′3 −3 √ ′ − √ V − V = A2 5V 5
(28)
Gr G25r = U Q1/6 U ∗6 Gr/25 Gr
(29)
4
Proof. Set A = (a4r/25 )1/6 and V ′ =
Gr , Gr/25
then from Theorem 3.1 and (24),(25),(27),(28) we have G25r Gr ∗6 1/6 1/6 1/6 U , = U [(a4r/25 ) ] = U [Q (a4r )] = U Q Gr/25 Gr which completes the proof. Continuing we have Gr = 2−1/12 (kr kr′ )−1/12 !−1/12 −1/12 !6 1/6 ′ ′ kr kr k25r k25r = U Q U ∗ ′ kr/25 kr/25 kr kr′
from
(30) (31)
k1/r = kr′ we get ′ k1/r k1/r ′ k25/r k25/r
!−1
6 1/6 12 ! −1/12 ′ k1/(25r) k1/(25r) ∗ =U Q U ′ k1/r k1/r
(32)
and setting r → 1/r: Theorem 3.3 s
12
" s " " ### ′ ′ k25r k25r k k r r ∗6 1/6 12 U =U Q ′ kr kr′ kr/25 kr/25
(33)
Once we know kr0 and kr0 /25 we can evaluate in closed form the k25r0 . Hence if we repeat the process we can find any higher or lower order of k25n r0 in closed radicals form, for n ∈ Z − {0, −1}. Example 1. R(e
−π −5
)
−11−R(e
125 ) = 2
−π 5
√ 1147 + 513 5 −
5
! r √ 2 1315405 + 588267 5
Example 2.
s
p √ √ 9 + 4 5 + 2 38 + 17 5 √ k1/5 = 18 + 8 5 s p √ √ 9 + 4 5 − 2 38 + 17 5 √ k5 = 18 + 8 5 r q √ 1 1 k125 = − 1 − (9 − 4 5)P [1]2 , 2 2
where
(34)
P (x) = P [x] = U 12 (Q1/6 (U ∗6 (x1/12 )))
(35)
Hence R(e−π
+
√
5 −5
)
− 11 − R(e−π
q
) =A
p √ √ 9 + 4 5 + 2 38 + 17 5
where
A=
5 5
105 2
p √ √ 9 + 4 5 − 2 38 + 17 5 √ 9+4 5
p √ √ 1/4 p P [1]+ [− 9 + 4 5 + 2 38 + 17 5
√ 1/4 + 9+4 5
r
√
+
√ 47 5 2
+
q
1 2
1+
1−
q
q
√ 1 + −9 + 4 5 P [1]2
√ 1 + −9 + 4 5 P [1]2
√ 1− 11111 + 4969 5 P [1]2
1/4
!3/4
1/4
+
]3 ,
q
3/4 √ 1 + −9 + 4 5 P [1]2
Example 3. It is
k25
Hence
1 k1 = √ 2 1 = r p √ √ 2 51841 + 23184 5 + 12 37325880 + 16692641 5 ′ k625 k625 =
and hence k625
h √ i 1 √ P 161 − 72 5 2 161 + 72 5
v v u u u u1 1u = t − t1 − 2 2
6
√ !2 P 161 − 72 5 √ 161 + 72 5
(36)
From Theorem 2.2 we have √ √ 1/4 R(e−5π )−5 − 11 − R(e−5π )5 = B · [p3/4 1 + 161 − 72 5 q + √ √ √ 1/4 √ p q s− + p 2p − 1 51841 + 23184 5 − 161 + 72 5 r h √ i √ 1/4 5/4 p s P 161 − 72 5 ]3 × − 51841 − 23184 5 × where we have set
−3 √ 1/4 p , 51841 + 23184 5 2p − 1
q √ √ p = 51841 + 23184 5 + 12 37325880 + 16692641 5 h √ i2 √ q = 51841 + 23184 5 − P 161 − 72 5 s = (1 − 22x + 14x2 + 2x3 + x4 )1
B = 2(1 + 2x + 14x2 − 22x3 + x4 )2 · (1 + 414728x + 414744x2 + 32x3 + 16x4 )1 × ×j −1/4 (1 − 4x + 103686x2 − 207364x3 + x4 )2
−3/2
× −1 × j · (1 − 22x + 14x2 + 2x3 + x4 )41 + (16 + 1658912x + 414744x2 + 8x3 + x4 )2 r h √ √ i2 √ 51841 + 23184 5 − P 161 − 72 5 j = 1 + −161 + 72 5 The index (H(x))h means the h-th root of H(x) = 0 with the notation of Mathematica program. Example 4. If P (n) (x) = (P ◦ |{z} . . . ◦P )(x), n
with P that of (35) then:
k25n
v s u (n−1) 2 u P (t0 ) t1 1 1− = − , 2 2 t0
(37)
√ √ in full closed form expansion, where t0 = 161 − 72 5 and t0 = 161 + 72 5. 7
References [1]: B.C.Berndt. ’Ramanujan‘s Notebooks Part III’. Springer Verlang, New York (1991). [2]: B.C.Berndt. ’Ramanujan‘s Notebooks Part V’. Springer Verlang, New York (1998). [3]: E.T.Whittaker and G.N.Watson. ’A course on Modern Analysis’. Cambridge U.P. (1927). [4]: J.V. Armitage W.F. Eberlein. ’Elliptic Functions’. Cambridge University Press. (2006). [5]: J.M. Borwein and P.B. Borwein. ’Pi and the AGM’. John Wiley and Sons, Inc. New York, Chichester, Brisbane, Toronto, Singapore. (1987). [6]: Nikos Bagis. ’The complete evaluation of Rogers-Ramanujan and other continued fractions with elliptic functions’. arXiv:1008.1304v1 [math.GM] 7 Aug 2010. [7]: Nikos Bagis. ’Parametric Evaluations of the Rogers Ramanujan Continued Fraction’. International Journal of Mathematics and Mathematical Sciences. Vol. 2011 [8]: Bruce C. Berndt, Heng Huat Chan, Sen-Shan Huang, Soon-Yi Kang, Jaebum Sohn and Seung Hwan Son. ’The Rogers-Ramanujan Continued Fraction’. (page stored in the Web). [9]: D. Broadhurst: Solutions by radicals at Singular Values kN from New Class Invariants for N ≡ 3 mod 8’. arXiv:0807.2976 (math-phy).
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