Conjectures on the evaluation of certain functions with

Journal of Number Theory 155 (2015) 63–84

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Journal of Number Theory www.elsevier.com/locate/jnt

Conjectures on the evaluation of certain functions with algebraic properties N.D. Bagis ∗ , M.L. Glasser a r t i c l e

i n f o

Article history: Received 20 April 2014 Received in revised form 18 January 2015 Accepted 17 March 2015 Available online 6 May 2015 Communicated by David Goss Keywords: Theta functions Algebricity Special functions Periodicity

a b s t r a c t In this article we consider functions with Moebius-periodic Taylor expansion coefficients. These functions under some conditions take algebraic values and can be evaluated in terms of theta functions and the Dedekind eta function. Special cases are the elliptic singular moduli, the Rogers–Ramanujan continued fraction, Eisenstein series and functions associated with Jacobi symbol coefficients. © 2015 Elsevier Inc. All rights reserved.

1. Introduction A number of functions defined over the complex plane, which are important in Number Theory and other disciplines, have been found to have real algebraic values at certain points. The aim of this note is to explore properties common to these functions which appear to be connected to this important feature. We begin by reviewing some relevant material. * Corresponding author. E-mail address: [email protected] (N.D. Bagis). http://dx.doi.org/10.1016/j.jnt.2015.03.002 0022-314X/© 2015 Elsevier Inc. All rights reserved.

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64

An elliptic singular modulus kr is the solution x of the equation √ K( 1 − x2 ) √ = r K(x)

(1)

where  2 F1

1 1 , ; 1; x2 2 2

 =

∞  n=0

 1 2 2 n 2n x (n!)2

2 2 = K(x) = π π

π/2 0

dφ  1 − x2 sin2 (φ)

(2)

The 5th degree modular equation, which connects k25r and kr , is (see [16], Chapter 19, pp. 280–288):   kr k25r + kr k25r + 25/3 (kr k25r kr k25r )1/3 = 1

(3)

The problem of solving (3) and finding k25r reduces to that of solving, what Hermite termed the depressed equation (see [3], Section 10.11, pp. 314–316): u6 − v 6 + 5u2 v 2 (u2 − v 2 ) + 4uv(1 − u4 v 4 ) = 0 1/4

(4)

1/4

where u = kr and v = k25r . The function kr is also connected to null theta functions from the relations kr =

∞ ∞   2 θ22 (q) (n+1/2)2 , where θ (q) = q and θ (q) = qn 2 3 2 θ3 (q) n=−∞ n=−∞

(5)

√

q = e−π r . In these terms a closed form solution of the depressed equation is k25r =

θ22 (q 5 ) θ32 (q 5 )

(6)

But this is not satisfactory. For example, in the case of Ramanujan’s famous π formulas (see [13] and related references), one has to obtain from the exact value of kr the exact value of k25r in radicals. (Here we mention the fact that when r is positive rational then the value of kr is an algebraic number.) Another example is the Rogers–Ramanujan continued fraction (RRCF) R(q) =

q 1/5 q q 2 q 3 ... 1+ 1+ 1+ 1+

(7)

(see [4,5,7–9,11,18,21] and [16], Chapter 16, pp. 77–86, [17], Chapter 32, pp. 12–45), the value of which also depends on the depressed equation.

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If we know the value of (RRCF) then we can find the value of j-invariant from Klein’s equation (see [7,20] and Wolfram pages ‘Rogers Ramanujan Continued fraction’):  jr = −

3 R20 − 228R15 + 494R10 + 228R5 + 1 5

R5 (R10 + 11R5 − 1)

, where R = R(q 2 )

(8)

One can also prove that Klein’s equation (8) is equivalent to the depressed equation (4). Using the 5th degree modular equation of Ramanujan R(q 1/5 )5 = R(q)

1 − 2R(q) + 4R(q)2 − 3R(q)3 + R(q)4 1 + 3R(q) + 4R(q)2 + 2R(q)3 + R(q)4

(9)

and (8) we can find the value of jr/25 and hence from the relation jr =

256(1 − kr2 + kr4 )3 , (kr kr )4

(10)

the value of kr/25 . Knowing kr and kr/25 , we can evaluate k25r (see [6]) and specify relations of the form k25r = Φ(kr , kr/25 ) and k25n r = Φn (kr , kr/25 ), n ∈ N

(11)

Hence when we know the value of R(q) in radicals we can find kr and k25r in radicals and conversely. Also in [3], Chapter 10, pp. 276–317, and the Wikipedia entry ‘Bring Radical’ one can see how the depressed equation can used for the extraction of the solution of the general quintic equation ax5 + bx4 + cx3 + dx2 + ex + f = 0

(12)

The above equation can therefore be solved exactly in terms of theta functions and in some cases in radicals. The same holds for the sextic equation (see [7]): b2 + by + ay 2 = cy 5/3 20a

(13)

Set Y = Y (q) = Yr = R(q)−5 − 11 − R(q)5 , then (13) has the solution y=

√ b Y (q 2 ), q = e−π r , r > 0 250a

(14) 3

where r can be evaluated from the coefficients using the relation jr = 250 ac2 b .

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The Ramanujan–Dedekind eta function is defined as η(τ ) =

∞

(1 − eiπτ n ) =

n=1

∞

(1 − e−πn

√ r

), τ =

√

−r

(15)

n=1

The j-invariant can be expressed in terms of the Ramanujan–Dedekind eta function as

 jr =

q

−1/24

η(τ ) η(2τ )

16

8 3  1/24 η(2τ ) + 16 q η(τ )

(16)

Values of the Ramanujan–Dedekind eta function can found from the relation (see [12] and [22], Chapter 21, p. 488): η(τ )8 =

28/3 −1/3 q (kr )2/3 (kr )8/3 K(kr )4 π4

(17)

There are many interesting things one can say about algebricity and special functions. In this article we shall examine functions which under a transformation, their Taylor coefficients are periodic. As we will see, we can evaluate these functions at certain points by a numerical asymptotic technique and as described in the Main Theorem below the values will be algebraic. One can find them by program Mathematica, using the routines ‘Recognize’ or ‘RootApproximant’. Since this procedure is essentially a ‘black box’ the results must be viewed as conjectural. It is a challenge to find the polynomials and modular equations and build them into a general theory. Various scientists are working with special functions such as the singular moduli (j-invariants), related Hilbert polynomials, the (RRCF), theta functions, Ramanujan–Dedekind η, etc., to this end. In [6] a way is presented to evaluate the fifth singular moduli and the Rogers–Ramanujan continued √ fraction using the function wr = kr k25r . This function can replace the classical singular moduli in the case of the Rogers–Ramanujan continued fraction and Klein’s invariant. We hope that the present note will foster this work. 2. The Main Theorem We begin by giving some definitions which will help us in the statement and proof of the Main Theorem. Definition 2.1. Let a, p be positive rational numbers with a < p and q = e−π Then we define [a, p; q] :=

∞

(1 − q pn+a )(1 − q pn+p−a )

n=0

These functions have the following very interesting property.

√

r

, r > 0.

(18)

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Conjecture 1. If q = e−π

√

r

67

, with r, a, p positive rationals and a < p, then

[a, p; q]∗ := q p/12−a/2+a

2

/(2p)

[a, p; q] = Algebraic Number

(19)

The Moebius μ function is defined (see [2], Chapter 2) as μ(n) = 0 if n is not square free, and (−1)r if n has r distinct prime factors. Some values are μ(1) = 1, μ(3) = −1, μ(15) = 1, μ(12) = 0, etc. We will also use the so-called (see [2], Chapter 2) Moebius Inversion Theorem Theorem 2.1 (Moebius Inversion Theorem). If f (n) and g(n) are arbitrary arithmetic functions, then n   f (d) = g(n) ⇔ f (n) = g(d)μ (20) d d|n

d|n

Also we use the notation Γ(n) = (n − 1)!, n = 1, 2, 3, . . . for the factorial function. Definition 2.2. A rational valued arithmetic function X(n) will be called ‘Moebius Periodic’ if exist T odd positive integer such that i) X(n) is T periodic i.e. for all n, k ∈ {0, 1, 2, . . .}, we have ak := X(k + nT ) = X(k) and ii) In every period interval hold a0 = aT = 0, a1 = aT −1 , a2 = aT −2 , . . . , a(T −1)/2 = a(T +1)/2 . On this basis one has: Main Theorem. Let f be a function analytic in (−1, 1) and set X(n) =

1  f (d) (0) n μ . n Γ(d) d

(21)

d|n

Define 

A=

T −1 2



   j j2 T − + + X(j). 2 2T 12 j=1

(22)

Then if X(n) is T -Moebius periodic q A e−f (q) = Algebraic Number at the points q = e−π

√ r

where r positive rational.

(23)

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In order to prove the Main Theorem we require Lemma 2.1. If |x| < 1 and X(n) is arithmetic function such that X(n) : N → Q, then  log

∞

 n X(n)

(1 − x )

=−

n=1

∞  xn  X(d)d n n=1

(24)

d|n

provided that both sides converge. Proof. If |x| < 1, then  log

∞

 n X(n)

(1 − x )

=

n=1

=−

∞ 

∞ 

X(n) log (1 − xn ) =

n=1

X(n)

n=1

∞ 

∞ ∞   xmn xnm xn  =− X(m)m = − X(d)d. m nm n m=1 n,m=1 n=1

2

d|n

Proof of the Main Theorem. From the Taylor expansion theorem we have 

e−f (x)

∞  f (n) (0) n x = exp − n! n=1



From the Moebius inversion theorem there is an X(n) such that X(n) =

1  f (d) (0) n μ n Γ(d) d d|n

or equivalently 1 f (n) (0) = X(d)d n! n d|n

Hence from Lemma 2.1 ⎛

e−f (x)

⎞ ∞ ∞ n   x X(n) ⎝ ⎠ = exp − X(d)d = (1 − xn ) n n=1 n=1 d|n

Consequently, because of the periodic property of X(n) we have 

e−f (x) =

T −1 2



j=1



[j, T ; x]X(j)

(25)

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which is a finite product of the functions [a, p; q]. From this and Conjecture 1 there exists √ −π r a rational number A such that (23) holds, provided that x = q and r is positive rational. The number A is obtained easily from (19). 2 n is the Jacobi Example 2.1. (Conjectured from the Main Theorem.) 1) If X(n) = G m m1 m2 ms symbol and G = 2 g1 g2 . . . gs , with m, m1 , . . . , ms nonnegative integers, m = 1 and g1 < g2 < . . . < gs are primes congruent to 1 (mod 4), then exists rational A such that 

qA

∞

n

(1 − q n )

G

= qA

n=1

G−1 2





[j, G, q]X(j) = Algebraic

(26)

j=1

√

when q = e−π r , r positive rational. A special case is G = 5 which gives the Rogers–Ramanujan continued fraction. More precisely if X = {1, −1, −1, 1, 0, 1, −1, −1, 1, 0, . . .} one finds (see [5,17], Chapter 32, pp. 9–45) q 1/5

∞

n

(1 − q n )

5

= R(q) = q 1/5

n=1

1 q q2 ... 1+ 1+ 1+

(27)

and the opposite. 2) The Jacobi elliptic theta function of the 4th kind (see [3], Section 4.2, p. 79) is ϑ4 (u, q) = 1 + 2

∞ 

2

(−1)n q n cos(2nu), |q| < 1

(28)

n=1

By setting    −1  i log(x) 3/2 3 log(x) ,x f (x) = − log η ϑ4 iπ 4 and using the program Mathematica, X(n) takes the values {1, 1, 0, 1, 1, 0, √ 1, 1, · · ·}. Hence, T = 3, A = −1/12. Therefore, for r-positive rational if q = e−π r , then q −1/12 e−f (q) must be an algebraic number. As an example taking r = 1 (using the Mathematica routine ‘RootApproximant’) we obtain     √ √ −π 12 π/12 −f (e ) e e = 81 885 + 511 3 − 3 174 033 + 100 478 3 . 3) If f (x) = −

  log(x) 1 − log R(x)−5 − 11 − R(x)5 6 6

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then f (x) is Moebius periodic since the values of X(n) are {1, 1, 1, 1, 0, 1, 1, 1, 1, 0, . . .}. Hence T = 5, A = −1 6 and from our Main Theorem  1/6 x−1/6 e−f (x) = R(x)−5 − 11 − R(x)5 √

is algebraic at x = q = e−π r , r positive rational. Moreover √ i) If q = e−π 2 , then q −1/6 e−f (q) , is root of 3125 + 250v 6 − 20v 10 + v 12 = 0. Observe that this equation is of the form (13). Also from [21] we have  √

−2π 2

R(e

)=

√ √ 5(g + 1) + 2g 5 − 5g − 1 2

where √ (g 3 − g 2 )/(g + 1) = ( 5 + 1)/2. One can use the duplication formula of RRCF (see [18]) to find t = R(e−π the value of  6

Y2 = q −1/6 e−f (q) =

√

2

) and hence

 6 t−5 − 11 − t5 .

ii) If q = e−2π , then

q In general if q = e−π

√

−1/6 −f (q)

e

=

 6

 Y4 =

√ 5 5 5 + . 2 2

r

q −1/6 e−f (q) =

 6 Yr

(29)

is root of the equation 3125 + 250y 6 −

 3

jr/4 · y 10 + y 12 = 0.

(30)

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3. The representation of e−f (q) In Theorem 3.1 below we give the representation of a Moebius periodic function f in terms of certain known functions. For |q| < 1 and a > 0 the Jacobi theta functions are ∞ 

ϑ(a, b; q) :=

(−1)n q an

2

+bn

(31)

n=−∞

One can evaluate the functions [a, p; q] from the expansion found in [12] (see also [1], Section 13.2, pp. 167–174): [a, p; q] =

  ∞  2 1 1 p p − 2a ϑ , ;q . (−1)n q pn /2+(p−2a)n/2 = η(pτ ) n=−∞ η(pτ ) 2 2

(32)

Theorem 3.1. For every f analytic in (−1, 1) and X(n) =

1  f (d) (0) n μ n Γ(d) d d|n

Moebius periodic, with period T , one has  

e−f (q) = η(T τ )−

 T −1 2 j=1



X(j)

T −1 2



j=1



 ϑ

T T − 2j , ;q 2 2

X(j) (33)

for every |q| < 1. Proof. Use the expansion (32) along with relation (25). 2 Theorem 3.2. If X(n) is Moebius periodic with period T , then   ⎞ ⎛ T −1   X(j)  2 ∞ T −1  nX(n)q n  2 d T T − 2j ⎟ ⎜ , ;q = −q log ⎝η(T τ )− j=1 X(j) ϑ ⎠ (34) n 1 − q dq 2 2 n=1 j=1 Proof. If X(n) =

1  f (d) (0) μ(n/d) n Γ(d) d|n

then 

∞ 1  nX(n)q n dq = f (q). q n=1 1 − q n

From Theorem 3.1 we get the result. 2

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Remark. If R(a, b, p; q) = R(Xp , q) = q C [a,p;q] [b,p;q] denotes a Ramanujan quantity (see [5]), then we have the closed form evaluation, with theta functions ⎛ p−1  ⎞   2 X (j) p C d p p − 2j R (Xp , q) ⎜ ⎟ = + log ⎝ , ;q ϑ ⎠. R (Xp , q) q dq 2 2 j=1 



(By using (34) and the fact that is that of [5], Theorem 3.)



p−1 2

j=1

(35)



Xp (j) = 0, with X(n) = Xp (n), where Xp (n)

4. Examples and applications   The Jacobi symbol n5 is Moebius periodic, with T = 5, hence from Theorem 3.2 and relations (26), (27), we get immediately for |q| < 1   ∞  d d n nq n ϑ(5/2, 3/2; q) = −q log(q −1/5 R(q)). = −q log n 5 1 − q dq ϑ(5/2, 1/2; q) dq n=1

(36)

From Example 2.1-3 (although we have no proof that X(n) = 0 if n ≡ 0 (mod 5) and 1 otherwise) and the relation (33) above, one can prove that (see also [16], Chapter 21, Entry 4, pp. 463–467, and [10] relation (30)) ∞ ∞   d nq n nq 5n − 5 = −q log n 5n 1 − q 1 − q dq n=1 n=1



ϑ(5/2, 1/2; q)ϑ(5/2, 3/2; q) η (5τ )

2

 (37)

and ∞ ∞  q d 1  nq n nq 5n + log (Y (q)) . − 5 =− n 5n 6 n=1 1 − q 1−q 6 dq n=1

(38)

Also from [10], relation (21), we have   ∞  6 6a(r) 4K[r]2 nq n 2 L(q) := 1 − 24 = √ + 1 + kr − √ 1 − qn π2 π r r n=1

(39)

The function a(r) is the elliptic alpha function and is defined as √ π a(r) = − r 2 4K[r]



 E[r] −1 . K[r]

(40)

  (The function E[r] = E(kr ) = π2 2 F1 − 12 , 12 ; 1; kr2 is the complete elliptic integral of the second kind at a singular value, and K[r] = K(kr ) is the complete elliptic integral of the

N.D. Bagis, M.L. Glasser / Journal of Number Theory 155 (2015) 63–84

73

first kind at a singular value (see [19], Sections 5.1–5.4, pp. 152–177 and [3], Section 5.6, pp. 129–135). K(k ) We also use below the function mn (r) = K(knr2 )r which is called multiplier (see also [10,13] and the books [19], Sections 5.1–5.4, pp. 152–177, [16], Chapter 9, pp. 220–324). Both functions a(r) and mn (r) take algebraic values when r, n are positive rationals. Continuing our arguments for Y (q) of (14) and (38), for certain positive rationals r √  −π r we can find special values of Y e . Setting hr := −6m5 (r)2 a(25r) + 6a(r) +

 √  2 r 5m5 (r)2 k25r − kr2 + 5m25 (r) − 1 ,

(41)

we get from the next Theorem 4.1. If q = e−π

√ r

, r > 0 then −

1 4K[r]2 hr q Y  (q) √ . = − 6 Y (q) 6 π2 r

(42)

Note that hr is algebraic when r is positive rational. Example. With program Mathematica for r = 1, we have EE[x] := EllipticE[x2 ] EK[x] := EllipticK[x2 ] k[r]= InverseEllipticNomeQ[Exp[-Pi Sqrt[r]]]1/2 a[r] = Pi/(4 EK[k[r]]2 ) - Sqrt[r] (EE[k[r]]/EK[k[r]] - 1) m[p,r]= EK[k[p2 r]]/EK[k[r]] RootApproximant[N[-6m[5,r]2 a[25r]+6a[r]+ +Sqrt[r](5m[5,r]2 k[25r]2 -k[r]2 +5m[5, r]2 -1),1000]]//ToRadicals −6 h1 = 5



√ 1+ 5−

  √ 20 + 9 5

(43)

We do the same thing and for Y1 : 125 Y1 = 2

   √ √ 1147 + 513 5 − 2 630 810 + 1 176 534 5

(44)

Hence

 √ √  1 4 π e 5 + 5 + 1 π + 5Γ −4 Y 20 + 9 9216 − 1   Y  e−π = −  1 4 5Γ − 4

(45)

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74

From (36), (37) and (38) we also get the following evaluation for the theta function (see [9]) ρ :=

ϑ(5, 1; q)6 ϑ(5, 3; q)6 12

q 2 η (10τ )

= R(q 2 )−5 − 11 − R(q 2 )5 , q = e−π

√ r

(46)

We next have Theorem 4.2. (See [9].) Let B(x, a, b) = function, then ρ

−1 5

+∞

t1/6

√

x 0

ta−1 (1 − t)b−1 dt be the incomplete Beta

dt 1 2 = √ , 1/6, 2/3) B(k4r 2 534 125 + 22t + t

(47)

and ρ = H (k4r )

(48)

where H(x) is defined by means of an inverse integral G(x) as −1 5

G(x) 

+∞

dt √ = x and H(x) = G t1/6 125 + 22t + t2



  B x2 , 16 , 23 √ . 534

(49)

Also √ 4 π√ q 1/6 η (i r) d 3 √ B(kr2 , 1/6, 2/3) = − 4 dr 2 r and

√ 6

ρ is root of (30).

Corollary. Suppose that q = e−π −5

R(q)

√

r

with r > 0 then 

− 11 − R(q) = H(kr ) = G 5

B(kr2 , 1/6, 2/3) √ 534

√ Example. If r = 1/5, then ρ = 5 5 and

k4/5

hence

(50)

  √ 2 − 2 − 4 −2 + 5  =  √ 2 + 2 − 4 −2 + 5

 (51)

N.D. Bagis, M.L. Glasser / Journal of Number Theory 155 (2015) 63–84

⎛

75



 √ ⎞ √ 2 − 4 −2 + 5 ⎠ = 5 5.  H⎝  √ 2 + 2 − 4 −2 + 5 2−

Theorem 4.3. If g is a perfect square and g1 < g2 < . . . < gλ are the distinct primes in the factorization of g, then ∞



(1 − q n )

n g

λ

= η(τ )

n=1

η(gi τ )−1

i=1



η(gi gj τ )1

i