Say that you're given a polynomial of any arbitrary degree and asked to find its roots. This is no problem computationally, since you can use modifications of Newton's method or the Durand-Kerner method to an arbitrary precision. But what if you're asked to find those roots in closed form, as exact answer instead of an approximation?

Let's start slow and easy. If you're given a first-degree polynomial $a_0 x + a_1 = 0$, the answer is trivially $x = -a_1 / a_0$. However, the numerical answer isn't clear unless you know how to calculate $a_1 / a_0$. We can do this, of course, but if it were your first time looking at a fraction (or a rational number) then you would have some more work to do.

Let's move up to quadratics. Given $a_0 x^2 - a_1 = 0$, we know that the answer is $\pm \sqrt{a_1 / a_0}$. This is where things start to get nontrivial - the square roots of rational numbers aren't usually rational numbers. Thus we need to find a way to calculate square roots numerically. Maybe all of this is cheating. We're looking for a closed-form solution, after all, not a numerical one. But we can argue past this - as long as we have a well-defined and unambiguous way to find roots then we can say that roots are adequate for closed-form solutions. In other words, square roots like $\sqrt{a}$ symbolically represent the solution of $x^2 = a$.

So we can find the solution of a general quadratic equation by using a square root, or equivalently a "radical". Concretely, the general solution of $ax^2 + bx + c = 0$ is

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

Likewise, you can find the solutions of general cubic and quartic polynomials in terms of radical expressions (though not necessarily with square roots, but cube roots and above). This means that you can find a formula like the one above which solves 3rd and 4th degree polynomials. But this is as far as we can get with radicals: 5th-degree polynomials cannot be solved with radicals alone.

$ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0$

Proving that this equation can't be solved with radicals requires Galois theory, and I won't cover the details here. Does this mean that there's no general formula to solve this? Not at all. Just as we defined radicals to "solve" quadratic equations, we can define a "hyperradical" to solve the quintic. First, we can introduce a few coordinate transformations known as Tschirnhaus transformations to reduce the equation in complexity.

$ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0$ $\downarrow$ $y^5 + gy^2 + hy + i =0$ $\downarrow$ $z^5 + c_0 z + c_1 = 0$ $\downarrow$ $z^5 - z + q = 0$

This gives us a simplified quintic to solve, which we can use to find the roots of the original quintic by using the inverse Tschirnhaus transformations on the simplified roots. To give an idea of what is actually going on, the quadratic equation $ax^2 + bx + c =0$ can be solved by "completing the square" and reducing it to the form $(x + b/2a)^2 = b^2/4a^2 - c/a$, or $y^2 = d$ with the transformation $y = x + b/2a$. This is just a Tschirnhaus transformation in disguise, since they are the generalization of this method to high-degree polynomials.

Anyway, back to the topic. The quintic equation $z^5 - z + q = 0$ cannot be solved in radicals because it is just the general quintic under a series of Tschirnhaus transformations (again, Galois theory gets involved here). So, we can define a new radical, known as the "Bring radical" after the person who discovered it. It is defined as

$z^5 - z + q = 0 \to z = BR(q)$

Is there a way to relate the Bring radical to the radicals we all know and love? It turns out we can. We can rearrange the quintic into

$z = \sqrt[5]{-q + z} = \sqrt[5]{-q + \sqrt[5]{-q + \sqrt[5]{-q + \cdots}}} = BR(q)$

While the general quintic cannot be solved in terms of radicals, it can be solved in terms of infinitely nested radicals.

Note by Levi Walker
1 month, 3 weeks ago

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Woahh!! So the Bring radical is simply a manipulation of the quintic which turns it into an infinitely nested radical??!! Also, a small doubt......could you please elaborate on how you reduced the quintic into the equivalent forms??

- 1 month, 3 weeks ago

Of course! With the quintic $x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$ you can use the change of variables (Tschirnhaus transformation) $y = x^2 + mx + n$ and choose $m$ and $n$ such that the equation reduces to $y^5 + u_1 y^2 + u_2 y + u_3 = 0$. Next, use the transformation $z = y^4 + r_1 y^3 + r_2 y^2 + r_3 y + r_4$ which leads to $z^5 + q_1 z + q_2 = 0$. Finally, we can bring it down with $t = \frac{z}{\sqrt[4]{-q_1/5}}$ to get $t^5 - 5t - 4s$ and get an equation like $u^5 - u + a =0$. I didn't follow the full proof of the last step, so I'm hazy on the details.

- 1 month, 3 weeks ago