Bring your own radicals

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 a0x+a1=0 a_0 x + a_1 = 0 , the answer is trivially x=a1/a0x = -a_1 / a_0. However, the numerical answer isn't clear unless you know how to calculate a1/a0a_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 a0x2a1=0 a_0 x^2 - a_1 = 0 , we know that the answer is ±a1/a0 \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 a \sqrt{a} symbolically represent the solution of x2=ax^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 ax2+bx+c=0ax^2 + bx + c = 0 is

x=b2a±b24ac2a 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.

ax5+bx4+cx3+dx2+ex+f=0 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.

ax5+bx4+cx3+dx2+ex+f=0 ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0 \downarrow y5+gy2+hy+i=0 y^5 + gy^2 + hy + i =0 \downarrow z5+c0z+c1=0 z^5 + c_0 z + c_1 = 0 \downarrow z5z+q=0 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 ax2+bx+c=0ax^2 + bx + c =0 can be solved by "completing the square" and reducing it to the form (x+b/2a)2=b2/4a2c/a (x + b/2a)^2 = b^2/4a^2 - c/a , or y2=dy^2 = d with the transformation y=x+b/2a 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 z5z+q=0 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

z5z+q=0z=BR(q) 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=q+z5=q+q+q+555=BR(q) 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 year, 7 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

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??

Aaghaz Mahajan - 1 year, 7 months ago

Log in to reply

Of course! With the quintic x5+ax4+bx3+cx2+dx+e=0x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0 you can use the change of variables (Tschirnhaus transformation) y=x2+mx+ny = x^2 + mx + n and choose mm and nn such that the equation reduces to y5+u1y2+u2y+u3=0y^5 + u_1 y^2 + u_2 y + u_3 = 0 . Next, use the transformation z=y4+r1y3+r2y2+r3y+r4z = y^4 + r_1 y^3 + r_2 y^2 + r_3 y + r_4 which leads to z5+q1z+q2=0z^5 + q_1 z + q_2 = 0. Finally, we can bring it down with t=zq1/54t = \frac{z}{\sqrt[4]{-q_1/5}} to get t55t4st^5 - 5t - 4s and get an equation like u5u+a=0u^5 - u + a =0 . I didn't follow the full proof of the last step, so I'm hazy on the details.

Levi Walker - 1 year, 7 months ago

Log in to reply

I am confused about the last line - can you not use one without the other or you can use one of them? @Levi Walker

Log in to reply

I'm treating radicals and nested radicals as different from infinitely nested ones. Polynomials up to the 4th degree can be solved using finitely many operations and radicals, but 5th degree polynomials and higher require infinitely many to be solved.

Levi Walker - 1 year ago

Log in to reply

I meant the last line of the post, @Levi Walker

Log in to reply

@A Former Brilliant Member Infinitely nested radicals are considered "hyperradicals", the Bring radical being the simplest of them. They can definitely be numerically approximated by finitely many nested radicals (subject to the some treatment for the matter of convergence), but in a mathematical sense they have different properties. Normal radicals can break the symmetry of symmetric polynomials:

(ab)2=(ba)2 (a-b)^2 = (b-a)^2 This polynomial is symmetric since a and b can be interchanged freely, but this isn’t the case after taking a square root \text{This polynomial is symmetric since a and b can be interchanged freely, but this isn't the case after taking a square root} abba a-b \ne b - a

The ways you can interchange the variables of a symmetric polynomial forms a group structure, specifically SnS_n, the group of permutations of nn elements. Taking radicals of that polynomial results in some permutations becoming forbidden, essentially restricting SnS_n to one of its subgroups. This is the gist of Galois theory, and since the coefficients of an arbitrary polynomial are symmetric polynomials in terms of its roots, we can talk about solving polynomials in these terms. I have a post on here showing how to derive the quadratic equation with this approach in mind.

Polynomials of degree nn generally have a Galois group of SnS_n (since they have nn roots), so solving that polynomial translates to moving from SnS_n to the trivial group through a ladder of normal subgroups. "Moving" here entails taking radicals of the appropriate quantities. This can always be done for S2,S3,S4S_2, S_3, S_4, but not for S5S_5, hence quintic polynomials can't always be solved by radicals.

The Bring radical, and all other hyperradicals, "skips" the ladder of normal subgroups to go straight from S5S_5 to the trivial group, making them unlike radicals since they don't have the group \to normal subgroup property.

Levi Walker - 1 year ago

Log in to reply

@Levi Walker z=5−q+z ​=5−q+5−q+5−q+⋯ ​=BR(q)

That's the line I'm talking about - can we use one of them or do we have to use both of them?

Log in to reply

@A Former Brilliant Member You can use either one

Levi Walker - 1 year ago

Log in to reply

Log in to reply

Also, @Levi Walker, check out Nitin Kumar's discussion: A Better Way to Solve an Quadratic Equation

Log in to reply


Problem Loading...

Note Loading...

Set Loading...