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This is a continuation of How to solve fifth-degree equations by elliptic functions?

How to transform a general higher degree five or higher equation to normal form? For example, a quintic equation to Bring-Jerrard form?

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To reduce the general quintic,

$$x^5+ax^4+bx^3+cx^2+dx+e=0\tag{1}$$

to Bring-Jerrard form,

$$x^5+x+k = 0\tag{2}$$

is done in two steps.

Step 1: Transform $(1)$ to principal quintic form (which is missing the $x^4,x^3$ terms) using a quadratic Tschirnhausen transformation,

$$y=x^2+mx+n\tag{3}$$

and eliminate $x$ between $(1)$ and $(3)$ using resultants. Nowadays, this is easily done by Mathematica or Maple. In wolframalpha.com, the command is,

  Collect[Resultant[x^5+ax^4+bx^3+cx^2+dx+e, y-(x^2+mx+n), x],y] 

which eliminates $x$ and collects the new variable $y$ yielding,

$$y^5+c_1y^4+c_2y^3+c_3y^2+c_4y+c_5=0\tag{4}$$

where,

$$c_1 = -a^2 + 2 b + a m - 5 n$$

$$c_2 = b^2 - 2 a c + 2 d - a b m + 3 c m + b m^2 + 4 a^2 n - 8 b n - 4 a m n + 10 n^2$$

and so on. The two unknowns $m,n$ allow you to eliminate two $c_i$. One can see that solving $c_1 = c_2 = 0$ will need only a quadratic. Thus, $(1)$ becomes the principal quintic form,

$$y^5+uy^2+vy+w=0\tag{5}$$

Step 2: To transform this to Bring-Jerrard, the impulse is to use a cubic Tschirnhausen. But this involves a composition of 1st, 2nd, 3rd-deg equations which will result in a sextic. Bring and Jerrard cleverly found a way around that by using a quartic Tschirnhausen,

$$z = y^4+py^3+qy^2+ry+s\tag{6}$$

and the extra parameter prevents elevation of degree. Eliminating $y$ between $(5)$ and $(6)$, we get,

$$z^5+d_1z^4+d_2z^3+d_3z^2+d_4z+d_5=0\tag{7}$$

where,

$$d_1 = -5 s + 3 p u + 4 v$$

$$d_2 = 10 s^2 - 12 p s u + 3 p^2 u^2 - 3 q u^2 + 2 q^2 v - 16 s v + 5 p u v + 6 v^2 + 5 p q w - 4 u w + r \color{brown}{(3 q u + 4 p v + 5 w)}$$

and so on. Similar to the first step, solving $d_1 = d_2 = 0$ will need only a quadratic. One then uses 3 variables $p,q,s$ to solve the 3 equations,

$$\color{brown}{3 q u + 4 p v + 5 w} = 0\tag{8}$$

$$d_1 = d_2 = 0\tag{9}$$

But notice that by solving $(8)$, it causes $r$ to disappear from $d_2$ and it remains a free parameter. Since the third term of $(7)$ has form,

$$d_3 = e_3r^3+e_2r^2+e_1r+e_0$$

where the $e_i$ are polynomials in the other variables, one can then use $r$ to solve $d_3 =0$ merely as a cubic. (If the general quintic wasn't reduced to principal form first, it would be harder to make $r$ disappear from $d_2$. Bring and Jerrard were clever, weren't they?)

What remains is,

$$z^5+d_4z+d_5 = 0$$

We can make a further simplification $d_4 = \pm1$ by scaling variables $z = t/f$,

$$t^5+d_4f^4t+d_5f^5 = 0$$

and solving for $f$ in $d_4f^4 =\pm1$. Thus, we end up with the Bring-Jerrard quintic,

$$t^5\pm t+k = 0\tag{10}$$

P.S. The same approach can be used to eliminate the $x^{n-1},x^{n-2},x^{n-3}$ terms simultaneously from the general equation of degree $n>3$.

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  • $\begingroup$ If I may be so impulsive, what prevents performing a quartic Tschirnhausen transformation at the beginning? $\endgroup$ – Simply Beautiful Art Jun 25 '18 at 0:55
  • $\begingroup$ Never mind, you may ignore my impulsive question lol. $\endgroup$ – Simply Beautiful Art Jun 25 '18 at 14:33
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I believe Piezas' answer above covers pretty much every bit of the question, although it might not be entirely worthless to present a plausibility argument for such a transformation.

Start with the principle quintic $x^5 + ax^2 + bx + c$. It is evident that the first two elementary symmetric polynomials vanishes if the variables are all roots of the above quintic, i.e.,

$$x_1 + x_2 + x_3 + x_4 + x_5 = 0\tag{1}$$

$$x_1x_2 + x_1x_3 + x_1x_4 + x_1x_5 + x_2x_3 + x_2x_4 + x_2x_5 + x_3x_4 + x_3x_5 + x_4x_5 = 0 \tag{2}$$

For roots of the principle quintic $x_i$, $i = 1, \ldots, 5$

From these equalities, we see that the second-order newton polynomial of the roots vanishes too, i.e.,

$$x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 0\tag{3}$$

Let $f(x) = x^3 + px^2 - s_3/5$ and $g(x) = x^4 + qx^2 - s_4/5$ for $s_3$ and $s_4$ being the third and the fourth of the Newton-Girard symmetric polynomials. Applying $(1)$ and $(3)$, we have $\sum f = \sum g = 0$ where the argument is running through all the roots of the principle quintic.

Furthermore,

$$\begin{align}\sum x f &= \sum_{i} x_i f(x_i) \\ &= \sum_{i} x_{i}^4 + p \sum_{i} x_i^3 - s_3\sum_{i} x_i \\ &= s_4 + ps_3 \end{align} \tag{4}$$

$$\begin{align}\sum x g &= \sum_{i} x_i g(x_i) \\ &= \sum_{i} x_{i}^5 + q \sum_{i} x_i^3 - s_4\sum_{i} x_i \\ &= s_5 + qs_3 \end{align} \tag{5}$$

Hence, setting $p = -s_4/s_3$ and $q = -s_5/s_3$ forces $\sum f = \sum g = 0$.

Now consider the shifted quintic derived by substituting $y = c_1 x + c_2 f + c_3 g$ into the principle quintic. We first calculate the first few Newton polynomials to get an idea of it's form :

$$ \sum_i y_i = c_1 \sum_i x_i + c_2 \sum_i f(x_i) + c_3 \sum_i g(x_i) = 0 \tag{6}$$

$$ \sum_i y_i^2 = c_1^2 \sum_i x_i^2 + c_2^2 \sum_i f(x_i)^2 + c_3^2 \sum_i g(x_i)^2 + 2c_1c_2 \sum_i x_i f(x_i) + 2c_2c_3 \sum_i f(x_i)g(x_i) + 2c_1c_3 \sum_i x_i g(x_i) = c_2^2 \sum_i f(x_i)^2 + c_3^2 \sum_i g(x_i)^2 + 2c_2c_3 \sum_i f(x_i)g(x_i) \tag{7}$$

Suppose we want to make $(7)$ vanish so that the coefficient of the term $x^3$ in transformed quintic becomes $0$. This implies we have to find a solution to the homogeneous quadratic equation in $(7)$ for some non-zero $c_2$ and $c_3$. It can be done by first dividing out through $c_3^2$ to transform it into a usual quadratic equation in $c_2/c_3$ and noting that

$$\sum f^2 = -\frac{2s_4s_5}{s_3} + \frac{s_4^3}{s_3^2} - \frac{2s_3^2}{5} + \frac{2s_4s_2}{5} + \frac{s_3^2}{25} + s_6 \tag{8}$$ $$\sum fg = -\frac{s_4s_6}{s_3} - \frac{s_5^2}{s_3} - \frac{s_3s_4}{5} + \frac{s_4^2s_5}{s_3^2} - \frac{s_4s_3}{5} + \frac{s_4^2s_2}{5s_3} + \frac{s_5s_2}{5} + \frac{s_3s_4}{25} + s_7 \tag{9}$$ $$\sum g^2 = -\frac{2s_5s_6}{s_3} + \frac{s_5^2s_4}{s_3^2} - \frac{2s_4^2}{5} + \frac{2s_4s_5s_2}{5s_3} + \frac{s_4^2}{25} + s_8 \tag{10}$$

Where $s_1$ and similar notations are Newton-Girard polynomials, and can readily be expressed in terms of coefficients of the principal quintic. We ensure that the shifted quintic is the Bring-Jerrard form by noting that $\sum y^3 = 0$ is just a cubic equation in $c_1$ which can be satisfied for non-zero choice of $c_1$.

As a remark, I think it would have been more descriptive if I had given the coefficients in terms of Newton-Girard polynomials as well as a complete expression of $c_1, c_2$ and $c_3$ in terms the coefficients of the principal quintic but computing them in general form is very hard, even using some CAS at hand. This is one of the beauties of quintic. As you can see, even though a general quintic is not solvable by radicals, it takes a lot more radical than a quartic to even get reduced into a normal form. As for the Bring-Jerrard form, we see a solution to two quadratics (one from the principle form) and a solution to a cubic is required for this transformation, giving a total of three square roots and one cube root.

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  • $\begingroup$ I found a special cubic Tschirnhausen transformation where the coefficients are in radicals will actually work. Can you look at it and see if you can make it more rigorous? I'll post the question in about 20 minutes or so. $\endgroup$ – Tito Piezas III Dec 22 '13 at 17:49
  • $\begingroup$ No, one avoids a sextic altogether. Typing this takes some time. :) $\endgroup$ – Tito Piezas III Dec 22 '13 at 18:12
  • $\begingroup$ It's finally done. Kindly see math.stackexchange.com/questions/615801/… $\endgroup$ – Tito Piezas III Dec 22 '13 at 18:49

protected by Community Dec 3 '18 at 4:50

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