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I am well aware of the fact that a general quintic polynomial cannot be solved in radicals. However, is there a known way to obtain a formula with other functions (e.g. infinitely nested radicals)?

Betydlig
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    You might take a look at https://arxiv.org/pdf/1308.0955.pdf – Barry Cipra Oct 27 '21 at 22:03
  • Cf. this Math Overflow question – J. W. Tanner Oct 27 '21 at 22:10
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    What would be the meaning of infinitely nested radicals? Sounds like the limit of some sequence? Are you working over the reals? You need something extra from the field for the concept of a limit to make sense. – Jyrki Lahtonen Oct 28 '21 at 03:11
  • @JyrkiLahtonen "Infinitely nested radicals" is a well known term in Mathematics. There is even a section of it the Nested radical page of the Wikipedia. – jjagmath Aug 09 '24 at 22:08
  • related: https://math.stackexchange.com/questions/1828551/polynomials-with-degree-5-solvable-in-elementary-functions/4225322#4225322 https://math.stackexchange.com/questions/3538585/solvability-in-radicals-elementary-functions-and-monodromy-galois-groups/4341942#4341942 – IV_ Aug 09 '24 at 22:17
  • @jjagmath That Wikipedia page is somewhat non-rigorous, because it only gets to the convergence issue towards the end. So suffer from similar issues like in this oldie. And, as I said, it absolutely needs convergence, that is, a topology. Unless I missed something all the examples involve real numbers only as opposed to general fields. Anyway, my main point is that the asker should define what constructs they allow. Wouldn't surprise me if you can get all real numbers as some kind of infinitely nested radicals, settling the question. – Jyrki Lahtonen Aug 10 '24 at 04:08

2 Answers2

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Solving the general quintic using only radicals, but infinitely nested? Yes, it is possible. Given,

$$x^n=a+x$$

$$x = \sqrt[n]{a+x}$$

By an iterative process,

$$x =\sqrt[n]{a+\sqrt[n]{a+\sqrt[n]{a+\sqrt[n]{a+x\dots}}}}$$

For $n=2$ and $a=1$, this gives the well-known form for the golden ratio ,

$$ \phi = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1\dots}}}}$$

However, since the general quintic can be reduced to the one-parameter Bring-Jerrard form, then just substitute $n=5$ to the formula.

P.S. Surprisingly, the two-parameter form $x^n+x^2+ax+b = 0$ can be solved by infinitely nested radicals as well. See this post.

Tito Piezas III
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The general solution for $$x^n = x + a$$ is a specific case of a solution provided by a new special function I've working on. Its name is Lambert-Tsallis function. You can take a good look at solutions 1 and 2. After a few simple manipulations you will find $$x=\frac{-a}{1 + \frac{W_{r}(-r(-a)^{r}) }{r} }$$ with $r= n-1$.

Ex: $n=9,a=4$ will lead to

$W_8(z) ={ 18.7894 + 9.3288i\\ 18.7894 - 9.3288i\\ 6.6508 +23.6422i\\ 6.6508 -23.6422i\\ -11.9093 +26.9096i\\ -11.9093 -26.9096i\\ -34.6431 + 0.0000i\\ -28.2093 +17.5737i\\ -28.2093 -17.5737i}$

and $x = {-1.0653 + 0.3710i\\ -1.0653 - 0.3710i\\ -0.6060 + 0.9780i\\ -0.6060 - 0.9780i\\ 0.1692 + 1.1646i\\ 0.1692 - 1.1646i\\ 1.2011 + 0.0000i\\ 0.9016 + 0.7840i\\ 0.9016 - 0.7840i}$

ZKZ
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    For $n=5$ this is (up to some sign changes) the Bring radical. The discussion of Glasser 1994 in that article is also pertinent. – Semiclassical Aug 09 '24 at 21:55
  • I have changed from n=5 to n=9. I hope it is enough. It has been some time that every comment I do is questioned, no matter the number. I question myself if some of those who question do the calculations, specially because It is a very simple answer. – ZKZ Aug 09 '24 at 22:14