How to solve fifth-degree equations by elliptic functions?

From F. Klein’s books, It seems that one can find the roots of a quintic equation

z5+az4+bz3+cz2+dz+e=0

(where a,b,c,d,eC) by elliptic functions. How to get that?

Update: How to transform a general higher degree five or higher equation to normal form?

Answer

To solve the general quintic using elliptic functions, one way is to reduce it to Bring-Jerrard form,

x5x+d=0

a transformation which can be done in radicals. (See this post.) To solve (1), define,

k=tan(14arcsin(16255d2))

p=iK(k)K(k)=iEllipticK[1-m]EllipticK[m]

with the complete elliptic integral of the first kind K(k) and elliptic parameter m=k2 (with p also given in Mathematica syntax above).

Method 1: For j=0,1,2,3,4, let,

Sj=uj2η(τj)η2(4τj)η3(2τj)

τj=110(p+2j)

u=ζ8=exp(2πi/8)

S5=2η(5p2)η2(10p)η3(5p)

where η(τ) is the Dedekind eta function, then,

x=±1253/4(k2)1/8k(1k2)(S0+S5)(S1+iS4)(iS2+S3)

with the sign chosen appropriately.

Remark: The five roots xncan be found by using pn=iK(k)K(k)+16n for n=0,1,2,3,4.

Method 2: For j=0,1,2,3,4, let,

Tj=(ϑ2(0,wjq1/5)ϑ3(0,wjq1/5))1/2

q=exp(iπp0)

w=ζ5=exp(2πi/5)

T5=q5/8(q5)1/8(ϑ2(0,q5)ϑ3(0,q5))1/2

with ϑn(0,q) as the Jacobi theta functions, then,

x=±ζ8253/4(k2)1/8k(1k2)(T0+T5)(T1iT4)(T2+T3)

and the sign of ζ8 chosen appropriately. Note that (Sj)8=(Tj)8.

Remark: One can also find the other roots xi, but is not as simple as in Method 1. (As one can see, you need much more than radicals to solve the general quintic.)

Example. Let,

x5x+1=0

so d=±1. Plugging it into (2) gives k0.072696, and m=k2, so p2.550572i. Since both methods use a square root, using the formulas one eventually finds,

x=1.1673039

Afterword (Added June 2015): Given the nome q=exp(iπτ), the two methods involve the function known either as the modular lambda function or elliptic lambda function, λ(τ) which has a beautiful q-continued fraction,

(λ(τ))1/8=2η(τ2)η2(2τ)η3(τ)=(ϑ2(0,q)ϑ3(0,q))1/2=2q1/81+q1+q+q21+q2+q31+q3+

studied by Ramanujan (who also had his own method to solve solvable quintics).

Attribution
Source : Link , Question Author : ziang chen , Answer Author : Community

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