Solution of Polynomial Equations with Nested Radicals

Home , Bring radical, Sextic equation

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d 1/d Moreover if someone deﬁnes √x := x , where d Q+ 0 , then the solution x of (4) can given in nested radicals: ∈ −{ }

q q/p x = 1 aq 1 aq q/p 1 aq q/p√1 . . .. (6) s − − − − r q Turn now into the general quintic equation

ax5 + bx4 + cx3 + dx2 + ex + f =0. (7)

This equation can reduced by means of a Tchirnhausen transform (see [1]) into

x5 + Ax + B =0. (8)

We can solve this last equation using the arguments of Lambert and Euler since it is of the form (4). This can be done deﬁning the hypergeometric function

1 2 3 4 1 3 5 3125t4 BR(t)= t 4F3 , , , ; , , ; , (9) − · 5 5 5 5 2 4 4 − 256 then according to Theorem 1:

Theorem 2. The solution of (8) is

4 55 5 4 A −A x = − BR B (10) q r 5 − 4    and for (8) also holds

5 5 5 x = B A B A B A√5 B ... (11) s− − − − − − − − r q 2 The sextic equation

It is true that in nested radicals terminology we can do much more than that. Consider the following general type of equation

ax2µ + bxµ + c = xν , (12)

2 then (12) can be written in the form 2 b ∆2 a xµ + = xν , (13) 2a − 4a where ∆ = √b2 4ac. Hence − 2 µ b ∆ 1 ν x = − + 2 + x 2a r4a a or 2 µ b ∆ 1 ν x = − + 2 + x . s 2a r4a a Hence 2 µ/ν ν b ∆ 1 ν x = − + 2 + x (14) s 2a r4a a From the above we get:

Theorem 3. The solution of (12) is

µ 2 µ/ν 2 2 v b ∆ 1 b ∆ 1 µ/ν b ∆ u v v x = u− + u 2 + − + v 2 + − + 2 + ... (15) u 2a u4a a u 2a u4a a s 2a 4a u u u u r u u u t t t t Corollary 1. The general sextic equation ax6 + bx5 + cx4 + dx3 + ex2 + fx + g = 0 (16) is always solvable with periodic nested radicals.

Proof. We know (see [1]) that all sextic equations, of the most general form (16) are equivalent by means of a Tchirnhausen transform y = kx4 + lx3 + mx2 + nx + s (17) to the form 6 2 y + e1y + f1y + g1 = 0 (18) But the last equation is of the form (12) with µ = 1 and ν = 6 and have solution

2 1 6 2 2 v / 1 6 f1 u ∆1 1 v f1 ∆1 1 / f1 ∆1 y = − + u 2 u− + v 2 v− + 2 + ... (19) 2e1 u4e1 − e1 u 2e1 u4e1 − e1 u 2e1 s4e1 u u u u u u u t t t t 3 2 where ∆1 = f1 4e1g1. Hence knowing y we ﬁnd x from (17) and get the solvability of (16) in− nested periodic radicals. More precisely it holds p kx4 + lx3 + mx2 + nx + s =

2 1 6 2 2 v / 1 6 f1 u ∆1 1 v f1 ∆1 1 / f1 ∆1 = − + u 2 u− + v 2 v− + 2 + ... (20) 2e1 u4e1 − e1 u 2e1 u4e1 − e1 u 2e1 s4e1 u u u u u u u t t t t Corollary 2. If Im(τ) > 0 and jτ denotes the j-invariant, then

− 5/3 1 2πiτ 5 2πiτ 5 5/3 R e 11 R e = jτ − −

3/5 125 12500 3/5 125 12500 = v− + v + − + + . . ., (21) u j u j2 v j j2 u τ u τ u τ s τ u u u u t t where R(q) is thet Rogers-Ramanujan continued fraction:

q1/5 q q2 q3 R(q)= ... , q < 1 (22) 1+ 1+ 1+ 1+ | |

Proof. 1/3 1/3 1/3 Equation (12) for a =1/jτ , b = 250/jτ , c = 3125/jτ and µ = 1, ν =5/3 takes the form 2 1/3 5/3 x + 250x +3125 = jr x , (23) which is a simpliﬁed form of Klein’s equation for the icosahedron (see [2] and [3]) and have solution x = Y = R(q2)−5 11 R(q2)5. From Theorem 3 we τ − − can express Yτ in nested-periodical radicals. This completes the proof.

2 2 b 5/3 Note. For more details about equation ax + bx + 20a = C1x one can see [2].

3 A solution of a tetranomial polynomial

Consider now the generalization of the Bring radical function as follows: Set p = ν and q = ν 1, x = X−1 in (4), then it becomes − − − a(ν 1)X ν + X ν+1 =1 − Multiply both sides with Xν and set a a/(ν 1), then → − Xν = X + a (24)

4 This last equation according to Theorem 1 have solution

∞ −1 1 Γ[(1 + νn)/(ν 1)] ( a)n X = − − , (25) ν 1 Γ[( 1+ νn)/(ν 1) n + 1] n! n=0 ! − X − − − which is a hypergeometric function that we call ν th order Bring radical. Also the power Xµ has expansion −

∞ −1 µ Γ[(µ + νn)/(ν 1)] ( a)n Xµ = − − . (26) ν 1 Γ[(µ + νn)/(ν 1) n + 1] n! n=0 ! − X − − A simple way to write (25),(26) is

∞ −1 νn+1 ( a)n X = BR (a)= ν−1 − (27) ν n νn +1 n=0 ! X and from Theorem 1: ∞ n µ ( 1) k1n + λ1 n X = λ1 − a , (28) k1n + λ1 n n=0 X ν −µ where k1 = ν−1 and λ1 = ν−1 . Hence we lead to the following

Deﬁnition 1. We call ”General Bring Radical” the function

∞ n µ ( 1) k1n + λ1 n BRν,µ(a) := (BRν (a)) = λ1 − a , (29) k1n + λ1 n n=0 X ν −µ where k1 = ν−1 and λ1 = ν−1 .

Assume now the following equation

kXµ = Xν X l. (30) − −

If Hν (u) is a function such that

H (u)ν H (u) l = u, (31) ν − ν − then equation (30) have solution X2 = Hν (u0) such that

ν µ H (u0) H (u0) l = u0 = kH (u0) . ν − ν − ν But H (u)= BR (l + u), u D, (32) ν ν ∀ ∈

5 where D is a suitable domain. Hence

X2 = Hν (u0)= BRν (l + u0)=

= BRν (l + kBRν,µ (l + kBRν,µ(l + ...))) (33) and we get the following

Theorem 4. The equation Xν = kXµ + X + l (34) admits solution X2 given by

µ ξ = l + kBRν (ξ) , X2 = BRν (ξ) (35)

Setting this into nested Bring radicals we have

X2 = BRν (l + kBRν,µ (l + kBRν,µ(l + ...))) (36)

Application. Assume the equation Xµ = X2 X 1. (37) − − Then µ R and ν = 2 and if ∈ 1+ √1+4x f0(x) := BR2 1(x)= (38) , 2 and µ 1+ √1+4x f(x) := BR2 (x)= (39) ,µ 2 the above equation (37) admits solution

X2 = f0 (1 + f (1 + f(1 + ...))) . (40)

Theorem 5. Assume the equation Xν = kXµ + X + l, (41) −1 ν −µ with ν>µ and k , l , kl < 1. Set k1 = , λ1 = , | | | | | | ν−1 ν−1 ∞ s k1s + λ1n n f(s; x) := xn (42) n s (s n + 1)(k1s + λ1n) n=1 X − and ∞ s −1 s g(w) := w + wλ1 ( 1) f s; kw w . (43) − s=0 X 6 Then a solution of (41) is X2 = BRν (g(l)) . (44)

Proof. It is clear that if X2 is the desired root of the equation (41), then also X2 = BRν (ξ), where ξ = l + kBRν,µ (ξ). But according to Lagrange theorem (see [6] pg 133), we have

∞ kn dn−1 ξ = l + (BR (l))n (45) n! dln−1 ν,µ n=1 X n and (BRν,µ(x)) = BRν,µn(x). Using expansion (29) we arrive to a double sum we rearrange and get the result.

Example. Assume that ν = 2, µ =1/2 and l =1/5, k =1/52, then

X1/2 = 5 25X + 25X2 (46) − − and 1+ √1+4x BR2 1(x)= (47) , 2 The polynomial f s; kl−1 is 1 sC2s−1,s 1 s s 3 1 f s; = 3F2 1 s, , 1 ; , 2 2s; + 5 50s 25 · − 2 − 2 − 2 2 − 25 −

1 2C2s− 2 ,s 1 1 s s 1 3 1 + 3F2 s, , ; , 2s; (48) 20s 5 · 2 − 2 − 2 −2 2 2 − 25 − and g will be ∞ 1 1 − 1 g(l)= ( 1)s5 sf s; , (49) 5 − 10 − 5 s=0 X where we have denoted n C := . (50) n,m m Hence a solution of (46) will be

X2 = BR2,1 (g(l)) . (51)

Corallary 3. 1) Every sextic equation can be set in the form

x6 + ax2 + bx + c = 0 (52)

7 2) Under the change of variable x = √5 bX the equation (52) takes the form − − X6 + ab 4/5X2 + X + cb 6/5 = 0 (53)

Theorem 6. Assume the equation Xν = mXλ + kXµ + X + l, (54) where ν>µ and k , l , kl−1 < 1. Assume also the equation | | | | | | Xν = kXµ + X + l. (55) Then a solution of (52) is

X3 = BRν (g(l + mBRν,λ(g(l + mBRν,λ(g(l + ...)))))), (56) with ”denested” equation

ξ = l + mBRν,λ (g(ξ)) and X3 = BRν (g(ξ)) . (57) Here the functions g and f are exactly that of Theorem 5.

Proof. Set H(u) be a function such that kH(u)µ + H(u)ν H(u) l = u. (58) − − − The function H(u) can determined using Theorem 5. Assume also that equation (52) admits solution X3 = H(u0) and for this u = u0 we have

λ µ ν mH(u0) = kH(u0) + H(u0) H(u0) l = u0. (59) − − − Then H(u)= BRν (g(l + u)) and hence λ H(u0)= BRν (g(l + u0)), with mH(u0) = u0. Hence a root of (52) is

X3 = H(u0)= BRν (g(l + mBRν,λ(g(l + mBRν,λ(g(l + ...)))))). (60)

Theorem 7. According to Lagrange inversion theorem equation (52) admits root ∞ mn dn X3 = BR g l + m BR (g(h)) (61) ν (n + 1)! dhn ν,nλ n=0 h=l!! X

8 References

[1]: J.V. Armitage and W.F. Eberlein. ’Elliptic Functions’. Cambridge University Press. (2006) [2]: N.D. Bagis. ’On a general sextic equation solved by the Rogers-Ramanujan continued fraction’. arXiv:1111.6023v2 [math.GM]. (2012) [3]: W. Duke. ’Continued fractions and Modular functions’. Bull. Amer. Math. Soc. (N.S.), 42. (2005), 137-162. [4]: B.C. Berndt. ’Ramanujan‘s Notebooks Part I’. Springer Verlag, New York. (1985). [5]: M.L. Glasser. ’The Quadratic Formula Made Hard or A Less Radical Approach to Solving Equations’. arXiv:math/94112244v1 [math.CA] 16 Nov (1994). [6]: E.T. Whittaker and G.N. Watson. ’A course on Modern Analysis’. Cambridge U.P, (1927).