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

$$z^5+az^4+bz^3+cz^2+dz+e=0$$

(where $a,b,c,d,e\in\Bbb C$) by elliptic functions. How to get that?

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

ziang chen
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    First you have to reduce it to Bring-Jerrard form. This is the one solvable in elliptic functions. Give me time to get my notes. (The one in Mathworld is missing some details.) – Tito Piezas III Oct 27 '13 at 03:08
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    If you don't know how to transform it to Bring-Jerrard form, ask it in the forum and I'll answer. – Tito Piezas III Oct 27 '13 at 15:49
  • @TitoPiezasIII Thanks! Can you add another answer here, or edit your solution? I don't think it is necessary to ask one more question – ziang chen Oct 27 '13 at 20:58
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    The combined answer will be too long. Besides, how to transform the general quintic to Bring-Jerrad form will be a good question, and will turn up in a google search for those looking. – Tito Piezas III Oct 27 '13 at 21:15
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    @TitoPiezasIII Thanks! see http://math.stackexchange.com/questions/542108/how-to-transform-a-general-higher-degree-five-or-higher-equation-to-normal-form – ziang chen Oct 27 '13 at 22:39
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    You may be interested in this post on how to solve the Brioschi quintic. – Tito Piezas III Sep 07 '15 at 15:08
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    The method is explained there p.36, it is based on inverting the doubly periodic function $f_C(z) = \wp_j'(z)\wp_j(z)+i \sqrt{c}\wp_j(z)+2i\sqrt{c}$ (with $j=e^{2i\pi /3}$) to find its zeros $z_k$ and solve $x^5+cx+x=0$ as $x=\wp_j(z_k)$ – reuns Oct 29 '19 at 08:59

1 Answers1

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

$$x^5-x+ d = 0\tag{1}$$

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

$$k = \tan\left(\tfrac{1}{4}\arcsin\Big(\frac{16}{25\sqrt{5}\,d^2}\Big)\right)\tag{2}$$

$$p = i\frac{K(k')}{K(k)} = i\,\frac{\text{EllipticK[1-m]}}{\text{EllipticK[m]}}\tag3$$

with the complete elliptic integral of the first kind $K(k)$ and elliptic parameter $m=k^2$ (with $p$ also given in Mathematica syntax above).

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

$$S_j={u}^j \frac{\sqrt{2}\,\eta(\tau_j)\,\eta^2(4\,\tau_j)}{\eta^3(2\,\tau_j)}$$

$$\tau_j = \tfrac{1}{10}(p+2j)$$

$$u=\zeta_8 = \exp(2 \pi i/8)$$

$$S_5=\frac{\sqrt{2}\,\eta(\frac{5p}{2})\,\eta^2(10p)}{\eta^3(5p)}$$

where $\eta(\tau)$ is the Dedekind eta function, then,

$$x = \frac{\pm 1}{2\cdot 5^{3/4}}\frac{(k^2)^{1/8}}{\sqrt{k(1-k^2)}}(S_0+S_5)(S_1+i\,S_4)(i\,S_2+S_3) \tag{4}$$

with the sign chosen appropriately.

$\color{blue}{\text{Remark}}$: The five roots $x_n$can be found by using $p_n = i\frac{K(k')}{K(k)}+16n$ for $n = 0,1,2,3,4$.

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

$$T_j =\left(\frac{\vartheta_2(0,w^j q^{1/5})}{\vartheta_3(0,w^j q^{1/5})}\right)^{1/2}$$

$$q = \exp(i \pi p_0)$$

$$w = \zeta_5 = \exp(2 \pi i/5)$$

$$T_5 =\frac{q^{5/8}}{(q^5)^{1/8}}\left(\frac{\vartheta_2(0,q^{5})}{\vartheta_3(0, q^{5})}\right)^{1/2}$$

with $\vartheta_n(0,q)$ as the Jacobi theta functions, then,

$$x = \frac{\pm{\zeta_8}}{2\cdot 5^{3/4}}\frac{(k^2)^{1/8}}{\sqrt{k(1-k^2)}}(T_0+T_5)(T_1-i\,T_4)(T_2+T_3) \tag{5}$$

and the sign of $\zeta_8$ chosen appropriately. Note that the two methods are connected via $(S_j)^8 = (T_j)^8$ as can be seen more in the Afterword.

$\color{blue}{\text{Remark}}$: One can also find the other roots $x_i$, 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,

$$x^5-x+1 = 0\tag6$$

so $d = \pm1$. Plugging it into $(2)$ gives $k \approx 0.072696$, and $m=k^2$, so $p \approx 2.550572\,i$. Since both methods use a square root, using the formulas one eventually finds,

$$x = \mp\,1.1673039\dots$$

$\color{blue}{\text{Afterword}}$ (Added June 2015): Given the nome $q = \exp(i \pi \tau)$, the two methods involve the function known either as the modular lambda function or elliptic lambda function, $\lambda(\tau)$ which has a beautiful q-continued fraction,

$$\big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)} = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^{1/2} = \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$

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

Tito Piezas III
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  • This gives just one of the five complex roots, right? It's clear that at most three are real, in which case, is this the largest, smallest, or middle one? – Ryan Reich Dec 23 '13 at 07:16
  • @Ryan: As you implied, the Bring-Jerrard quintic has $x_1^2+x_2^2+\dots+x_5^2=0$, hence not all its roots are real. I do not know if Hermite's method always gives the largest root (in absolute value), but you can tweak it to produce ALL five. I revised my answer to include an alternative method using the Dedekind eta function which easily yields all the roots. – Tito Piezas III Dec 23 '13 at 22:51
  • @Tito Wow this is beautiful. Besides the fact that this brings together (apparently) distant fields of math, is there a reason why such solution is interesting besides what one can obtain with, I don't know, Newton's method? – lcv Nov 26 '15 at 21:42
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    @lcv: Thanks. I believe the primary utility of such results is to show connections (and in my opinion, beautiful connections) between different mathematical objects. They are certainly not useful for numerical evaluations better than Newton's method. :) – Tito Piezas III Nov 27 '15 at 07:42
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    @TitoPiezasIII thanks for confirming my impression :) PS I agree: beautiful connections – lcv Nov 27 '15 at 08:13
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    Some notes for anyone trying this method, based on my own mistakes. (1) Mathematica and Wolfram Alpha expect the argument of the elliptic integral function to be $k^2$, not $k$. (2) In the formula for $S_j$, the power of $u$ is not doubled but the term in $\tau$ is. (3) Not accounting for their signs, the set of real and imaginary parts of $S_j$, for a given $n$, are simply permuted when $n$ is changed so it is only necessary to calculate them accurately once. – Jam Dec 19 '19 at 18:10
  • Can you link ramanujans own methods? – Alexander Conrad Apr 29 '21 at 14:51
  • @TitoPiezasIII do you have any references or explanations as to how to derive the methods 1 and 2 for solving the fifth degree polynomial? I've been trying to read Felix Klein's Lectures on the ikosahedron (which I'm guessing these methods may come from), but I still don't have a clear picture of how the icosahedron shows up in these methods. – fsuna064 Oct 20 '21 at 02:15
  • @TitoPiezasIII If Method 1’s solution were written out only in terms of $x,d,j$, would anything simplify and how would the formula look? – Тyma Gaidash Jun 15 '22 at 15:35
  • @TymaGaidash Yes, you can write the root $x$ of the Bring quintic $x^5-x+d=0$ explicitly in terms of its sole parameter $d$, but you would have to go through the elliptic function $K(k)$ and the Dedekind eta function $\eta(\tau)$ so the solution WON'T be in radicals. And it would look horribly messy. – Tito Piezas III Nov 14 '22 at 12:45
  • Could you link a pdf to how Ramanujan solved quintics? – Alexander Conrad Feb 18 '23 at 15:22