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I remember from a long time ago reading a paper regarding the solution of quintic polynomial equations using hypergeometric functions. In particular, the methods are based around the solution of the so-called "Bring quintic form"

$$t^5 - t - \rho = 0$$

which has a solution

$$t = -\rho\ _4F_3\left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \frac{3125}{256} \rho^4\right).$$

However, the paper also mentioned that to solve the general quintic form

$$x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$$

you need "a piece of paper as big as a large asteroid" to write down all the formula, which is extraordinary, given that the first four degrees, while large, are not that large. Now I don't know how big a "large asteroid" is supposed to be, but if presumably that's, say, 10 km across, and treating that asteroid as a sphere, we're looking at a surface area of about 523 km^2 or somewhere around 8 billion sheets of A4 paper, using one side only, so 4 billion sheets if using both sides and, thus, if the latter were bound into 800-page (400 sheet), thus rather thick, books, 10 million such books, which is also over three times as many books as in the whole Library of Congress, most of which are not so big and thick! And not even my computer's hard disk could store that much - it sounds like quite a few TB of data, and I have only about 1-2 TB total of hard disk storage on my machine with all drives pooled. And certainly no easily affordable computer could have stored it at the time that paper was written - a small disk farm would have been needed.

And what I am wondering about is: can one do better than this? I note that the method commonly employed to reduce the quintic to the Bring form, which involves taking a "resultant" of the quintic with a quartic, can also be used to solve a cubic, and when that is done (using a quadratic, to reduce to a perfect cube), the resulting formula is considerably more "wordy" than the traditional Cardano cubic formula. And thus that suggests to me a sort of inefficiency in the method. So is there a way to improve upon it and make it maybe, if not necessarily "small", then at least small enough that it could fit in, say, one book, with everything plugged together?

It would also be acceptable, by the way, to not entirely plug it together, but to just write $\rho$ as a function of $a$, $b$, $c$, $d$, and $e$, together with $t$ (already given) and how to derive $x$ from $t$ - so long as both of those (for $\rho$ and $x$) are all, of course, plugged together themselves. Though if the Bring form must be abandoned altogether, that is fine too, so long as we don't need to introduce functions beyond a suitable complexity level, e.g. equivalent to inverting fixed polynomials with no free parameters.

The_Sympathizer
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    Note that a quintic cannot be solved in general at all, at least not by radicals. Even for degree $3$, the formula is so complicated that in practice almost always numerical methods are used. – Peter Apr 30 '22 at 08:35
  • The Bring radicals are a workaround to minimize the effort of numerical calculations. I am not sure whether such a workaround is possible for higher degrees. – Peter Apr 30 '22 at 08:43
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    @Peter You are right that for practical use a numerical algorithm is going to be much faster, but I am interested in this from a theoretical perspective. But the Bring radical or hypergeometric series does not reduce numerical complexity; if anything, a "formula as big as a large asteroid" is ludicrous to use for a numerical computation. – The_Sympathizer Apr 30 '22 at 08:48
  • For theoretical purposes , it does not matter that the formula would be almost impossible to be handled. What matters is only that we can transform the problem to find the roots exactly can be transformed to find the solutions of hypergeometric functions. Whether this concept is really worth the (apparently huge) needed effort (even for theoretical purposes) is another story. – Peter Apr 30 '22 at 08:57
  • @Peter : True, but still, I want to know if the effort to obtain a theoretical formula can be shrunken and/or what the minimum complexity of a formula under the given constraints has to be. – The_Sympathizer Apr 30 '22 at 08:58
  • I heard from Bring radicals, but I never heard that it is so complicated to use them. I wonder why they were invented at all cosndiering this (or was it not known before that it is so difficult ?). – Peter Apr 30 '22 at 09:03
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    I, for one, believe it. I remember seeing a formula once for the roots of a quartic equation and it looked like an algebraic equivalent to a zombie apocalypse. The complexity of the general formula grows extremely fast with polynomial degree. – Oscar Lanzi Apr 30 '22 at 09:05
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    By the way, the length of the formula for the discriminant also increases quickly with the degree. – Peter Apr 30 '22 at 09:07
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    @Peter : My guess is the first proof was indirect - that it was shown a certain set of substitutions would reduce the quintic, but it was never carried out. – The_Sympathizer Apr 30 '22 at 10:20
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    wiki: "in 2004, Daniel Lazard wrote out a three-page formula" for general solvable quintics. – Dmitry Ezhov Apr 30 '22 at 10:52
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    @Dmitry Ezhov : That would presumably mean ones where the solution is reducible to radicals only - whereas here we have one more additional operation. However, it's a good hint that suggests the most general case might also admit a shorter formula (though of course, far from a proof). – The_Sympathizer Apr 30 '22 at 18:13
  • A user wrote steps for simplifying a quintic in this post – Тyma Gaidash May 05 '22 at 02:22
  • It is done, a solution to the principal quintic using the Brioschi. (It is feasible to use the coefficients of the depressed quintic, get more complicated though still manageable expressions, but would just clutter the post.) – Tito Piezas III Jun 25 '23 at 15:12

1 Answers1

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Although similar equations are solvable, the following one is the most useful. Here is a closed form using the quantile function Inverse Beta Regularized $\text I^{-1}_s(a,b)$ solving:

$$x^{-4}+cx+a=0\implies x=\frac{4a}{5c}\left(\text I^{-1}_{\frac{3125c^4}{256a^5}+1}(2,4)-1\right);-1\le \frac{3125c^4}{256a^5} \le0$$

Therefore:

$$x^5+ax^4+b=0\implies x=\frac{4a}5\left(\text I^{-1}_{\frac{3125 b}{256a^5}+1}(2,4)-1\right),-1\le \frac{3125 b}{256a^5} \le0$$

Try it out here

Since many quintic equations can be put into the following form, we have another general solution:

$$\boxed{x^5+ax+b=0\implies x=\frac{5b}{4a\left(\text I^{-1}_{\frac{3125b^4}{256a^5}+1}(2,4)-1\right)};-1\le \frac{3125b^4}{256a^5} \le0 }$$

Test the formula here

Then factor to find the rest of the roots.

The solution uses a function on Wolfram Alpha, so it should be a standard one. Please correct me and give me feedback!

Тyma Gaidash
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  • This is great! You should publish all your finds with InverseBetaRegularized! – Anixx May 19 '22 at 23:28
  • Definitely! And your find of the closed form of Dottie constant should be added to Wikipedia (it currently says there is no closed form). Also, since this function can solve such different equations as quintics and $\cos(x)=x$, it looks it is something of great importance in mathematics. I wonder whether BetaRegularized functions belongs to the class of so-called "differentiated Gamma functions" (to which also belongs polygamma and Hurwitz zeta) and which are a "near elementary" and make a bridge between trigon functions and their inverses (I gave you a link to the Bernoulli umbra). – Anixx May 19 '22 at 23:46
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    You may find these posts interesting as logarithmic integral also belongs to the class of "differentiated Gamma functions": https://mathoverflow.net/questions/396216/on-the-properties-of-the-even-and-odd-components-of-the-function-f-nx-n-1 and https://mathoverflow.net/questions/396076/asymptotically-similar-functions-with-opposite-parity-were-they-considered-are and possibly the solution for the quintics relies on these "parabolas" of opposite parity to normal parabolas. – Anixx May 19 '22 at 23:49
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    Thanks. I am thinking about it now. Not sure I can propose any applications (I thought about the chain line (cosh) but its arc length is easy, so the strength of your formula is possibly in the applicability to non-e base). – Anixx May 20 '22 at 00:57
  • https://en.wikipedia.org/wiki/Catenary – Anixx May 20 '22 at 02:36
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    What happens if the condition $-1\le \frac{3125 b}{256a^5} \le0$ is not satisfied? – Anixx Jun 13 '22 at 21:15
  • @Anixx In general, $z=\text I^{-1}_p(a,b)$ solves for the value of when a certain CDF$(z)=p$. A probability is $0\le p\le1$, so the restriction follows. You seem interested in this function; can you please tell us why? – Тyma Gaidash Jun 13 '22 at 21:49
  • I am interested because of your results. So, if the condition is not satisfied, the function is not evaluated in Mathematica? But can this function be generalized to the greater range? – Anixx Jun 13 '22 at 21:56
  • @Anixx The function is just not evaluated. Yes, by using this fairly complicated sum or these identities. The last link has one formula which may not work with this answer, but solving for the inverse function in it will “generalize” the function. – Тyma Gaidash Jun 13 '22 at 22:07
  • What about other roots? Can they be expressed using a similar formula? There are known solutions for quintics in special functions but they are really, really huge! What about equations of higher order? – Anixx Jun 13 '22 at 22:21
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    @Anixx See formula $1$ in a related post here, but it is a work in progress. – Тyma Gaidash Jun 13 '22 at 22:46
  • I never knew such an animal had relevance to the quintic; however, this seems little better than inverting $x \mapsto x^5 + x$ or similar a la Bring-Jerrard, because you still have reduction to 3 degrees of freedom. Is it exponentially easier to knock out $x^3$, $x^2$, and $x$ terms (i.e. enough to get it to "sub-asteroidal size")? – The_Sympathizer Jun 27 '23 at 15:11
  • @The_Sympathizer This answer was to give another result. You should see this post where another user tries to simplify a formula for the general quintic – Тyma Gaidash Jun 27 '23 at 15:19