arXiv:1305.1591v3 [math.GM] 27 Mar 2014 h function 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 series 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 − −