arXiv:1305.1591v3 [math.GM] 27 Mar 2014 h The where qainatrnmdb emt se[3]): (see Hermite by find named and after (3) equation solving of problem The h t eremdlreuto hc connects which equation modular degree 5th The where nw eut nAgbacFunctions moduli Algebraic singular on elliptic The Results Known 1 odicity; k 2 r Keywords F = 1 itdwt aoismo coefficients. symbol Rogers- as Jacobi functions the with and moduli, ciated singular Eisenstein elliptic fraction, e the continued Dedekind are Ramanujan the cases and algebraic functions Special take theta by function. conditions recovered be some can rational under and values Moebius-periodic functions with These functions consider coefficients. we article this In  u θ θ 2 3 2 2 2 1 = ( ( , q q k 2 1 ) ) r 1 1; ; , / k 4 where ht ucin;Agbacfntos pca ucin;Peri- functions; Special functions; Algebraic functions; Theta : r x and u sas once oteafntosfo h relations the from functions theta to connected also is 2 k 6  r − k v = θ 25 v = 2 6 n r X ( nAgbacFunctions Algebraic On ∞ =0 q 5 + + k = ) 25 1 / 2 k ( r 4 F u n r ′ 1 2 k . 2 tnmhu5Edessa 5 Stenimahou k θ 2 1 [email protected] !)  el 80,Greece 58200, Pella F r v 25 2 ′ n 2 2 1 2 stesolution the is = 2 1 r x ( , u 2 2 + 2 1 ..Bagis N.D. n n Abstract 2 1 2 , = X ;1 1; ; − = ∞ 2 1 −∞ k 5 1; ; / 25 v 3 π 1 2 2 ( r q − K 4 + ) k x ( eue ota ovn h depressed the solving that to reduces r 2 n ( x k  +1 x 2 25 = )  uv / r x 2) k = (1 r ′ 2 fteequation the of π k 2 √ and k 25 ′ − Z 25 r r 0 u r ) π/ 1 4 θ and / v 2 3 3 4 ( q (4) 0 = ) q (3) 1 = = ) k 1 r − s(e [13]): (see is θ x 3 dφ 2 = sin n = 2 X ∞ ( −∞ so- φ ta ) q n (2) (1) (5) 2 π√r q = e− . Hence a closed form solution of the depressed equation is

2 5 θ2(q ) k25r = 2 5 (6) θ3(q ) But this is not satisfactory. For example in the case of π formulas of Ramanujan (see [12] and related ref- erences), one has to know from the exact value of kr the exact value of k25r in radicals. (Here we mention the concept that when r is positive rational then the value of kr is algebraic number). Another example is the Rogers-Ramanujan continued fraction (RRCF) which is

q1/5 q q2 q3 R(q)= ... (7) 1+ 1+ 1+ 1+ (see [4],[5],[6],[8],[9],[10],[13],[14],[16],[21]), the value of which depends form the depressed equation. If we know the value of (RRCF) then we can find the value of j-invariant from Klein’s equation (see [19],[8] and Wolfram pages ’Rogers Ramanujan Continued fraction’):

20 15 10 5 3 R 228R + 494R + 228R +1 2 jr = − , where R = R(q ) (8) − R5 (R10 + 11R5 1)5 −  One can also prove that Klein’s equation (8) is equivalent to depressed equation (4). Using the 5th degree modular equation of Ramanujan 1 2R(q)+4R(q)2 3R(q)3 + R(q)4 R(q1/5)5 = R(q) − − (9) 1+3R(q)+4R(q)2 +2R(q)3 + R(q)4 and (8) we can find the value of jr/25 and hence from the relation

2 4 3 256(kr + kr′ ) jr = 4 . (10) (krkr′ ) kr/25. Knowing kr and kr/25, we have evaluated k25r (see [7]) and give relations of the form

k = Φ(k , k ) and k n =Φ (k , k ),n N (11) 25r r r/25 25 r n r r/25 ∈

Hence when we know the value of R(q) in radicals we can find kr and k25r in radicals and the opposite.

Also in [3] and Wikipedia ’Bring Radical’ one can see how the depressed equa- tion can used for the extraction of the solution of the general quintic equation

ax5 + bx4 + cx3 + dx2 + ex + f = 0 (12)

2 The above equation can solved exactly with theta functions and in some cases in radicals. The same holds and with the sextic equation (see [8])

b2 + bY + aY 2 = cY 5/3 (13) 20a which have solution

b 2 5 2 5 π√r Y = Y = R(q )− 11 R(q ) , q = e− , r> 0 (14) r 250a − −

 3 c and r can evaluted from the constants using the relation jr = 250 a2b , in order to generate the solution. The Ramanujan-Dedekind eta function is defined as

∞ η(τ)= (1 qn), q = eiπτ , τ = √ r (15) − − n=1 Y The j-invariant can be expresed in terms of Ramanujan-Dedekind eta function as 16 8 3 1/24 η(τ) 1/24 η(2τ) j = q− + 16 q (16) r η(2τ) η(τ) "    # The Ramanujan-Dedekind eta function is (see [11],[22])

8/3 8 2 1/3 2/3 8/3 4 η(τ) = q− (k ) (k′ ) K(k ) (17) π4 r r r There are many interesting things one can say about algebricity and special functions. In this article we examine Moebius-periodic functions. If the Taylor coeffi- cients of a function are Moebius periodic, then we can evaluate prefectly these functions, by taking numerical values. This can be done using the Program Mathematica and the routine ’Recognize’. By this method we can find values coming from the middle of nowhere. However they still remaining conjectures. Also it is great challenge to find the and modular equations of these Moebius-periodic functions and united them with a general theory. Many various scientists are working for special functions (mentioned above) such as the singular moduli (j-invariant) and the related to them Hilbert polynomials, (RRCF), theta functions, Dedekind-Ramanujan η and other similar to them. In [7] is presented a way to evaluate the fifth singular moduli and the Rogers- Ramanujan continued fraction with the function wr = √krk25r. This function can replace the classical singular moduli in the case of Rogers-Ramanujan con- tinued fraction and Klein’s invariant. We are concern to construct a theory of such functions and characterize them.

3 2 The Main Theorem

We begin by giving a definition and a conjecture which will help us for the proof of the Main Theorem.

Definition 1. π√r Let a,p be positive rational numbers with a

0. We call ”agiles” the quantities

∞ pn+a pn+p a [a,p; q] := (1 q )(1 q − ) (18) − − n=0 Y The ”agiles” have the following very interesting conjecture-property.

Conjecture. π√r If q = e− , r is positive rational and a,b are positive rationals with a

Assuming the above unproved property we will show the following

Main Theorem. Let f be a function analytic in ( 1, 1). Set by Moebius theorem bellow X(n) to be − 1 f (d)(0) n X(n)= µ (20) n Γ(d) d d n X|   If X(n) is T periodic sequence, rational valued and catoptric in the every period- interval i.e. if for every n N is a = X(k + nT )= X(k) with a = 0 we have ∈ k T a1 = aT 1,a2 = aT 2,...,a(T 1)/2 = a(T +1)/2, then exist rational a number A, such that− − − A f(q) q e− = Algebraic Number, (21) π√r in the points q = e− , r positive rational. The number A is given from

T −1 [ 2 ] j j2 T A = + + X(j) (22) −2 2T 12 j=1 X  

For to prove the Main Theorem we will use the next known (see [2]) Moebius inversion Theorem

4 Theorem. (Moebius inversion Theorem) If f(n) and g(n) are arbitrary arithmetic functions, then n f(d)= g(n) f(n)= g(d)µ (23) ⇔ d d n d n X| X|   The Moebius µ function is defined as µ(n) = 0 if n is not square free, and ( 1)r if n have r distinct primes. −

Hence some values are µ(1) = 1, µ(3) = 1, µ(15) = 1, µ(12) = 0, etc. For to prove the Main Theorem we will use− the next

Lemma 1. If x < 1 | | ∞ ∞ xn log (1 xn)X(n) = X(d)d (24) − − n n=1 ! n=1 d n Y X X| Proof. It is x < 1, hence | | ∞ ∞ log (1 xn)X(n) = X(n) log (1 xn)= − − n=1 ! n=1 Y X ∞ ∞ xmn ∞ xnm ∞ xn = X(n) = X(m)m = X(d)d. − m − nm − n n=1 m=1 n,m=1 n=1 d n X X X X X| Proof of Main Theorem. From Taylor expansion theorem we have

∞ (n) f(x) f (0) n e− = exp x − n! n=1 ! X From the Moebius inversion theorem exists X(n) such that

1 f (d)(0) n X(n)= µ n Γ(d) d d n X|   or equivalent f (n)(0) 1 = X(d)d n! n d n X| Hence from Lemma 1

∞ n ∞ f(x) x n X(n) e− = exp X(d)d = (1 x ) − n  − n=1 d n n=1 X X| Y   5 and consequently because of the periodicity and the catoptric property of X(n), we get T −1 [ 2 ] f(x) X(j) e− = [j, T ; x] (25) j=1 Y which is a finite product of ”agiles” and from the Conjecture exist A rational π√r such that (21) hold, provided that x = q− and r positive rational.

Examples.

n m m1 m2 ms 1) For X(n)= G , where G = 2 g1 g2 ...gs , m,m1,...,ms not neg- ative integers, m = 1 and g < g <...

G−1 [ 2 ] ∞ n qA (1 qn)( G ) = qA [j, G, q]X(j) = Algebraic − n=1 j=1 Y Y π√r when q = e− , r positive rational. A special case is G = 5 which gives the Rogers-Ramanujan continued fraction. More precisely is X = 1, 1, 1, 1, 0, ... and evaluations can given. { − − } 2 ∞ n 1 q q q1/5 (1 qn)( 5 ) = R(q)= q1/5 ... − 1+ 1+ 1+ n=1 Y 1 2) If X = 1, 1, 0, 1, 1, 0,... , then T = 3 and A = 12 . Hence we get that π { } − if q = e− , then

1/12 f(q) 12 q− e− = 81 885 + 511√3 3 174033 + 100478√3 − s  q  1 3) If X = 1, 1, 1, 1, 0, 1, 1, 1, 1, 0,... , then T = 5 and A = −6 we get that π√2{ 1/6 f(q) } i) If q = e− , then q− e− , is root of 3125 + 250v6 20v10 + v12 =0 − We can solve the above equation observing that is of the form (13):

6 12 1/3 10 3125 + 250Yr + Yr = jr Yr , (eq) where jr is the j-invariant. Hence

1/6 f(q) 6 q− e− = Y1/2 q then see [21]:

5(g +1)+2g√5 √5g 1 2π√2 R(e− )= − − q 2

6 where (g3 g2)/(g +1)=(√5+1)/2 − π√2 One can use the duplication formula of RRCF (see [16]) to find R(e− ) in radicals and hence the value of Y1/2 in radicals. 2π ii) If q = e− , then

1/6 f(q) 6 5 5√5 q− e− = Y = + 1 s 2 2 p ...etc π√r If q = e− 1/6 f(q) 6 q− e− = Yr/4 q f(q) 3 The Representation of e−

We give in Theorem 1 bellow the representation of a Moebius periodic function f in terms of known functions. For q < 1 the Jacobi theta functions are | |

∞ 2 ϑ(a,b; q) := ( 1)nqan +bn (26) − n= X−∞ One can evaluate the agiles by the theta function

∞ 1 cq c(q q2) cq3 c(q4 q2) M(c, q) := cnqn(n+1)/2 = − − − − − − ... (27) 1+ 1+ 1+ 1+ 1+ n=0 X then from [11] a way to express the agiles is

a p a a p M( q− , q ) q M( q , q ) [a,p; q]= − − − (28) η(τp) however we shall use the general theta functions evaluation see relation (32) bellow for more concentrated forms. The reader can change from one form to the other.

A first result in the agiles also given in [11] was the evaluation of the dupli- cation formula 2 [a,p; q ]∗ = τ ∗(a,p; q), (29) [a,p; q]∗ for which if a,b are positive reals and n integer, then

τ ∗(a,p; q)= τ ∗(np a,p; q) (30) ±

7 Theorem 1. For every f analytic in ( 1, 1) with − 1 f (d)(0) n X(n)= µ n Γ(d) d d n X|   periodic-symmetric with period T (Moebius periodic) and real-valued, then hold

T −1 T −1 [ 2 ] X(j) [ 2 ] f(q) X(j) T T 2j e− = η(T τ)− j=1 ϑ , − ; q (31) 2 2 j=1 P Y   A f(q) for every q < 1. In case that X(j) are rational, then q e− is algebraic | | π√r when q = e− , r positive rational.

Proof. Use the expansion found in [11]:

2 1 ∞ n pn /2+(p 2a)n/2 1 p p 2a [a,p; q]= ( 1) q − = ϑ , − ; q (32) η(pτ) − η(pτ) 2 2 n=   X−∞ along with relation (25).

Theorem 2. If X(n) is real T -periodic and catoptric, then

T −1 T −1 [ 2 ] X(j) n [ 2 ] ∞ nX(n)q d X(j) T T 2j = q log η(T τ)− j=1 ϑ , − ; q 1 qn − dq  2 2  n=1 − P j=1   X  Y   (33) Proof. If 1 f (d)(0) X(n)= µ(n/d) n Γ(d) d n X| then it holds 1 ∞ nX(n)qn dq = f(q) q 1 qn n=1 Z X − From Theorem 1 we get the result.

Remark. C [a,p;q] If R(a,b,p; q)= q [b,p;q] denotes a Ramanujan quantity (see[5]), then we have the next closed form evaluation, with theta functions

p−1 [ 2 ] Xp(j) R′ (X , q) C d p p 2j p = + log ϑ , − , q (34) R (X , q) q dq  2 2  p j=1    Y    8 by using (33).

4 Examples and Applications

n 4) The Jacobi symbol 5 is 5 periodic and symmetric, hence  ∞ n n nq d ϑ(5/2, 3/2; q) d 1/5 = q log = q log(q− R(q)). (35) 5 1 qn − dq ϑ(5/2, 1/2; q) − dq n=1 X   −   From Example 3 and the above relation (35), one can prove that

∞ nqn ∞ 5nq5n d ϑ(5/2, 1/2; q)ϑ(5/2, 3/2; q) = q log (36) 1 qn − 1 q5n − dq 2 n=1 n=1 η (5τ) ! X − X − and 1 ∞ nqn ∞ nq5n q d + 5 = log (Y (√q)) (37) 6 1 qn − 1 q5n −6 dq n=1 n=1 X − X − In view of [5] relation (92) and the expansion of L1(q) in the same paper, we get

1/2 1/2 2 2 2 q d q Y ′ q 1 K[r] a(r)K[r] 5K[25r] log Y q1/2 = − = + + −6 dq 12 Y q1/2 −6− 6π2 π2√r 6π2 −     a(25r)K[25r]2 K[r]2k2 5K[25r]2k2 r + 25r (38) − π2√r − 6π2 6π2 where a(r) is the elliptic alpha function (see [17]) and K[r]= K(kr) is the com- plete of the first kind at singular values. Hence for certain r we π√r can find special values of Y ′ e− .   From (36) and (37) we get the following evaluation for the theta function

6 6 ϑ(5, 1; q) ϑ(5, 3; q) 2 5 2 5 θ = = R(q )− 11 R(q ) (39) q2η (10τ)12 − − and in view of [9] we get the following, similar to inverse elliptic nome theorem

Theorem 3.

1 θ dt 1 − = B(k4r, 1/6, 2/3) (40) 1/6 2 3 5 + t √125+ 22t + t 5√4 Z ∞ and θ = H (k4r) (41)

9 where

1 G(x) dt B x, 1 , 2 − = x and H(x)= G 6 3 (42) 1/6 2 3 5 + t √125+ 22t + t 5√4 Z ∞ ! Also 1/6 4 d π 3 q η (τ) B(k2, 1/6, 2/3) = √4 (43) dr r − 2 √r and θ1/6 is root of (eq).

For example with r =1/5, then θ =5√5 and

2 2 4 2+ √5 k = − − − 4/5 q 2+ 2 4p 2+ √5 − − q p 2 2 4 2+ √5 5√5= H − − − .  q  2+ 2 4p 2+ √5  − −  q p

n m0 m1 m2 ms Continuing we have, if X(n)= G0 have period p and G0 =2 p1 p2 ...ps , with pj -primes of the form 1(mod4), ms,s,j =0, 1, 2,... and m0 = 1, then 6 G0−1 [ 2 ] j ∞ n nqn d G G 2j G0 = q log ϑ 0 , 0 ; q (44) n  −  G0 1 q − dq 2 2  n=1   − j=1   X  Y    Also

Conjecture 2. If g is perfect square and p1 < p2 <...

λ ∞ n n ( ) 1 1 1 (1 q ) g = η(τ) η(p τ)− η(p p τ) η(p p p τ)− ... (45) − i i j i j k n=1 i=1 i

∞ nqn 6 K2[r] 6α(r)+ √r 1+ k2 1 24 = +4 − r , (46) − 1 qn π√r π2√r n=1  X −

α(r) is the elliptic alpha function, kr the elliptic singular moduli.

We also have

Theorem 4. If p is prime then

p−1 2 [ 2 ] π √r d p−1 p p 2j 1+ p 24q log η(pτ)− 2 ϑ , − ; q = 4K[r]2 − − dq  2 2  j=1     Y     2 2 2 2 =6α(r) √r 1+ k + m 2 6α(p r)+ p√r 1+ k 2 . − r p r − p r K(k 2 ) where m 2 = n r is the multiplier and is algebraic valued function when n r K(kr ) r, n are in Q+∗ , (see [13],[17]).

5 The corespondence of theta functions and sin- gular modulus

π√r In [11] we have shown that if m is integer and q = e− , r> 0 then

2 2 ∞ n +2mn m 2K[r] q = q− (47) π n= r X−∞ which is a classical result (see [3]). Also we have shown that

2 2 1/6 ∞ n +(2m+1)n 5/6 (2m+1) /4 (k11k12k21) K[r] q =2 q− (48) 1/3 π n= k22 r X−∞ 2 2 2 k11 2k12 2 where k11 = kr, k12 = 1 k11, k21 = − 2− , k22 = 1 k12. − k11 − The above relations (47)p and (48) give us evaluations of allp

∞ 2 qn +mn (49) n= X−∞

11 with m integer. We define ki(x) the inverse function of the singular modulus kx. Then it must holds from relation (1)

2 K √1 x2 ki(x)= − (50) K(x) ! From all the above algebraic properties of the theta functions of the previous m paragraphs one might think what will happen if we replace r with ki n , where m,n positive integers? Is there a nome that says that theta functions are alge- braic functions of the singular modulus? The answer is ”yes” for a given theta function 2 p a + a 12 2 2p ∞ 2 q − n pn /2+(p 2a)n/2 [a,p; q]∗ = ( 1) q − = η(pτ) − n= −∞ 2 X p a + a q 12 − 2 2p p p 2a = ϑ , − ; q , (51) η(pτ) 2 2   there exists a unique Q a,p (x) such that { }

π√ki(x) [a,p; e− ]= Q a,p (x) (52) { } and Q a,p (x) has the property that is root of a of a fixed degree depending{ } only on a and p. For example the theta function [1, 4, q] has

2 12 4(1 x ) Q 1,4 (x)= − (53) { } r x Also the function [1/2, 4, q] has

4 12 48 4(1 x) (2 + x 2√1+ x) Q 1/2,4 (x)= − − (54) { } s x13(1 + x)2 Hence one may lead to the evaluations

2 12 4(1 kr ) [1, 1/2, q]∗ = − (55) s kr and 4 12 48 4 (1 kr) 2+ kr 2√1+ kr [1/2, 4, q]∗ = − − (56) v 13 2 u kr (1 + kr) u  for every r > 0. Other numericalt results can also show us that this is conjec- turaly true. The function Q a,p (x) is always algebraic and its degree is the { } π√ki(x) same for all the values of qx = e− , x positive rational and π√r [a,p; q]∗ = Q a,p (kr), q = e− , r> 0. (57) { } ∀

12 Hence finding the expansion of a theta function only requires the knowlege of Q a,p (x). Instant cases of Q can be found for rational values of x with the routines{ } ’Recognize’ or ’RootApproximant’ with program Mathematica. An example of evaluation is

6 1 3 π√ki(1/5) 2/3 [1, 3; e− ]∗ = [ 182 689224 148230 3 √10+ 90 − − − ·   q

2 3 + 2 92571934 + 74115 32/3 √10 + 689224 ] v 2/3 3 u s344612 74115 3 √10 · ! u − · t (58) which is root of the equation 45x4 + 364x3 21870x2 885735=0 (59) − −

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