Part I. Given any constant $a,b$, the equation in $x$,


is solvable in radicals for any degree $k$. The general solution is,

$$x = \frac{-a}{\beta^{1/k}}+\beta^{1/k},\quad\text{where}\;\beta = \frac{b+\sqrt{b^2+4a^k}}{2}$$

For example, expanding at $k=5$, we get the DeMoivre quintic,


Part II. Given any constant $a,b$, the equation in $x$,


is also solvable in radicals for any $k$. For example, for $k=5$, we get,


Question: What is the general solution of $(2)$?

  • $\begingroup$ I have a doubt now since my answer is rather trivial... Did you want the polynomial or did you hope an explicit solution $x=\cdots$ ? (I should probably erase my answer!) $\endgroup$ Dec 23 '15 at 14:10
  • $\begingroup$ @RaymondManzoni: The polynomials are easy to get by expansion. I was hoping for the explicit solution $x$. But as I look at your answer, it seems you are on the right track. $\endgroup$ Dec 23 '15 at 14:14
  • $\begingroup$ In the mean time I had reverted $(2)$ in my solution. Now are there others? $\endgroup$ Dec 23 '15 at 14:18
  • $\begingroup$ @RaymondManzoni: I think all $k$ roots can be found by using something similar to this. (P.S. Maybe you can delete everything after $(3)$? The solution $x$ is what I'm after.) $\endgroup$ Dec 23 '15 at 14:22
  • 1
    $\begingroup$ @RaymondManzoni: I figured out the expression for all $k$ roots. Do you want me to do the edit, or did you figure it out also? :) $\endgroup$ Dec 23 '15 at 14:29

Let's rewrite your equation $\,(2)\,$ for $\;r:=\dfrac {\sqrt{b}}x\,$ as : $$\tag{1}\dfrac{(1+r)^k-(1-r)^k}{(1+r)^k+(1-r)^k}=-\frac {\sqrt{b}}a$$ then from Ron Gordon's recent answer : $$\tag{2}\tanh(k\;\operatorname{tanh^{-1}}r)=\frac{e^{(2 k \tanh^{-1} r)}-1}{e^{(2 k\tanh^{-1} r)}+1}=-\frac {\sqrt{b}}a$$

We may revert this to get one real solution : $$\tag{3}\dfrac {\sqrt{b}}x=-\tanh\left(\frac 1k\tanh^{-1}\frac {\sqrt{b}}a\right)$$

Other solutions may be obtained by noticing that $\,e^{\large{(2 k \tanh^{-1} r+2\pi in)}}\,$ will give the same expansion in $(2)$ so that we have too : $$\tag{4}\tanh(k\;\operatorname{tanh^{-1}}r+\pi i n)=-\frac {\sqrt{b}}a$$ and the $k$ roots $x_n$ (for $n=0..k-1$) given by : $$\tag{5}\boxed{\displaystyle x_n=-\frac 1{\sqrt{b}}\;\operatorname{cotanh}\left(\frac 1k\tanh^{-1}\frac {\sqrt{b}}a+\frac{\pi in}k\right)}$$ We may too revert the passage from $(1)$ to $(2)$ (with $k$ replaced by $\dfrac 1k$ and for $\alpha:=\dfrac {\sqrt{b}}a$ )
and rewrite $(5)$ as :
$$\tag{6}\frac {\sqrt{b}}x=-\dfrac{\left(\frac{1+\alpha}{1-\alpha}\right)^{1/k}-1}{\left(\frac{1+\alpha}{1-\alpha}\right)^{1/k}+1}$$ that is : $$\tag{7}x=\sqrt{b}\;\dfrac{1+R^{1/k}}{1-R^{1/k}}$$

where $\;R^{1/k}\;$ is one of the $k\;$ $k$-th roots of $\;R:=\dfrac{1+\alpha}{1-\alpha}=\dfrac{a+\sqrt{b}}{a-\sqrt{b}}$.

  • $\begingroup$ I'm glad you noticed that post with R. Gordon's answer. $\endgroup$ Dec 23 '15 at 14:53
  • $\begingroup$ @TitoPiezasIII: so am I. Excellent continuation! $\endgroup$ Dec 23 '15 at 14:55
  • $\begingroup$ One thing I would like though. This family has a solvable Galois group. Do you know a way to express your $x_n$ in terms of plain trigonometric functions? (Hence in radicals.) $\endgroup$ Dec 23 '15 at 14:56
  • $\begingroup$ I'll try to edit (and convert the log in powers if possible...) $\endgroup$ Dec 23 '15 at 15:03
  • 1
    $\begingroup$ Oh, I already tested it with Mathematica. Besides, the form is verification enough for me. If it had been ugly and complicated, I would be doubtful. But since it is beautiful and simple, then it has to be true. (Admittedly not the most rigorous test.) :) $\endgroup$ Dec 23 '15 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.