# Solving 5th degree or higher equations

According to this, there is a way to solve fifth degree equations by elliptic functions.

Some related questions that came to mind:

1. Besides use of elliptic functions, what other (known) methods are there for solving 5th degree or higher equations?

2. For quadratic equation, what does the solution by elliptic functions or any other non conventional method look like?

• @BalarkaSen the good old $ax^2+bx+c=0$, I think that is called the quadratic equation. Updated the question. – Arjang Dec 24 '13 at 6:48
• @BalarkaSen : Does an equation being resolved in radicals implies that it can bot be solved by other means? Or only equations that can not be resolved by radicals can be solved by elliptic functions? – Arjang Dec 24 '13 at 6:55
• @BalarkaSen : Thank you, any hints on how might one get started in these matters would be very nice – Arjang Dec 24 '13 at 7:04
• You may have a look at this article on arXiv, "On a General Sextic Equation Solved by the Rogers Ramanujan Continued Fraction" – Jean-Claude Arbaut Dec 24 '13 at 8:05

## 2 Answers

The equation $x^m + b x = 1$ has a series solution in powers of $b$: $$x = 1 - \sum_{k=1}^\infty \dfrac{\prod_{j=1}^{k-1} (jm-k-1)}{k!\; m^k} b^k$$ which can be expressed in terms of hypergeometric functions.

• One way to prove this, btw, is Lagrange inversion formula. – Grigory M Dec 24 '13 at 7:42

For general equations of deg $k>4$ one can use,

1. Fuchsian functions
2. Theta functions
3. Mellin integrals

Given positive integers $m,n$ and $m+n=k$, if the general equation of deg $k$ can be solved by a certain method, it goes to follow that the general equations of deg $m,n$ can be as well.

For example, one can multiply the general cubic by an arbitrary quadratic and solve the product as a "quintic", thus expressing the cubic roots (and quadratic ones) in terms of elliptic functions.

However, if that is considered "cheating", then one can reverse-engineer the elliptic formula and adapt it for other degrees. For the quintic, you start with a modular equation of deg six,

$$\Omega_5 = u^6 - v^6 + 5u^2v^2(u^2-v^2)+4uv(1-u^4v^4)=0\tag{1}$$

and use this to solve $x^5-x+a=0$. Details are here. If we wish to go higher, there is a modular equation of deg eight,

$$\Omega_7 = (1-u^8)(1-v^8)-(1-uv)^8 = 0\tag{2}$$

Unfortunately, we can't reduce in radicals the general septic to one-parameter form, so the general 7th deg can't be solved by elliptic functions. If we go lower, there is a modular equation of deg four,

$$\Omega_3 =u^4 - v^4 + 2uv(1 - u^2v^2) = 0\tag{3}$$

Now the cubic can be reduced to one-parameter form $x^3-x+a=0$ and one can assume analogous methods can then be applied on (3) to solve it. It seems nobody has worked out the precise details though. (Naturally enough, since solving the cubic using elliptic functions seems to be like nuking a mosquito.)

For the quadratic, it is easier. One can reduce it to the one-parameter form,

$$x^2+2ax+1=0$$

and the elliptic formula for the general quadratic is given by,

$$x_1^4 =\lambda(2\tau),\;\;\; x_2^4 =\frac{1}{\lambda(2\tau)}=\left(\frac{\eta^3(2\tau)}{\sqrt{2}\,\eta(\tau)\,\eta^2(4\tau)}\right)^8$$

$$\tau = i\,\frac{K(k')}{K(k)} = i\,\frac{\,_2F_1\big(\tfrac{1}{2},\tfrac{1}{2},1,1-\tfrac{1}{a^2}\big)}{\,_2F_1\big(\tfrac{1}{2},\tfrac{1}{2},1,\tfrac{1}{a^2}\big)}\tag{4}$$

where $\lambda(\tau)$ is the elliptic lambda function, $\eta(\tau)$ is the Dedekind eta function, $K(k)$ is the complete elliptic integral of the first kind, and $\,_2F_1$ is the hypergeometric function.

(Note: When taking the 4th root of the $x_i$, one should be careful to affix the correct power of the 4th root of unity $\zeta_4 = \exp(2\pi i/4)$, especially when the coefficients of the quadratic are complex.)

$\color{red}{Edit}$: (some time later)

Let us nuke the mosquito. The cubic can be reduced to the form,

$$x^3-x+b=0$$

The elliptic formula for the general cubic is, define,

$$w^2-(4-27b^2)w+(4-27b^2) =0$$

$$\tau = \frac{i}{\sqrt{3}}\frac{\,_2F_1\big(\tfrac{1}{3},\tfrac{2}{3},1,1-w\big)}{\,_2F_1\big(\tfrac{1}{3},\tfrac{2}{3},1,w\big)}\tag{5}$$

$$u =\left(\frac{\eta^4(3\tau)}{3\,\eta(\tau)\,\eta^3(9\tau)}\right)^3$$

then,

$$x=\pm\sqrt{\frac{4u-1}{3u-3}}$$

where the appropriate sign of the square root is chosen. (This is a simpler formula than a previous edit.)

Note the difference between (4) and (5) as the latter is Ramanujan's theory of elliptic functions of signature 3. This gives one root of the cubic, and some modifications to $\tau$ can probably give the other two roots.

• Excellently done. (+1) – Balarka Sen Feb 11 '14 at 7:32