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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?

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    $\begingroup$ 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.) $\endgroup$ – Tito Piezas III Oct 27 '13 at 3:08
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    $\begingroup$ If you don't know how to transform it to Bring-Jerrard form, ask it in the forum and I'll answer. $\endgroup$ – Tito Piezas III Oct 27 '13 at 15:49
  • $\begingroup$ @TitoPiezasIII Thanks! Can you add another answer here, or edit your solution? I don't think it is necessary to ask one more question $\endgroup$ – ziang chen Oct 27 '13 at 20:58
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    $\begingroup$ 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. $\endgroup$ – Tito Piezas III Oct 27 '13 at 21:15
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    $\begingroup$ @TitoPiezasIII Thanks! see math.stackexchange.com/questions/542108/… $\endgroup$ – ziang chen Oct 27 '13 at 22:39
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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 $(S_j)^8 = (T_j)^8$.

$\color{blue}{\text{Remark}}$: 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).

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  • $\begingroup$ 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? $\endgroup$ – Ryan Reich Dec 23 '13 at 7:16
  • $\begingroup$ @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. $\endgroup$ – Tito Piezas III Dec 23 '13 at 22:51
  • $\begingroup$ @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? $\endgroup$ – lcv Nov 26 '15 at 21:42
  • $\begingroup$ @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. :) $\endgroup$ – Tito Piezas III Nov 27 '15 at 7:42
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    $\begingroup$ @TitoPiezasIII thanks for confirming my impression :) PS I agree: beautiful connections $\endgroup$ – lcv Nov 27 '15 at 8:13

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