It is known that there is no algebraic formula for the solution to a general polynomial, in terms of its coefficients. Of course, there are formulas for low degrees (1,2,3,4).

Are there `closed form' expressions for any particular classes of polynomials? By closed form, I mean using the usual arithmetic operations and extracting roots, not extending the toolkit to modular functions or whatever fancy things. By class of polynomials, I guess I mean some arithmetic or algebraic conditions on the coefficients.

Concretely, is there an algebraic expression for the solution to $$v^6 - 6v^3 - v + 7 = 0$$ near $v\approx 1.7$, and is there a way I might tell why/why not by looking at the equation?

Pointers to relevant literature would also be welcome.

  • 1
    $\begingroup$ Not really an answer, but perhaps Galois theory can help here. All polynomials whose Galois groups are solvable can be solved by radicals. I recommend reading Abstract Algebra by Dummit and Foote for this $\endgroup$
    – IAAW
    Apr 28, 2023 at 3:19
  • 2
    $\begingroup$ Maple says the Galois group of polynomial $v^6 - 6v^3 - v + 7 = 0$ is $S6$, so it cannot be solved in terms of radicals. $\endgroup$
    – GEdgar
    Apr 28, 2023 at 8:24

2 Answers 2


Just for the fun, a good approximation of the largest real root of $$v^6 - 6v^3 - v + 7 = 0$$ is $$v\sim\Gamma \left(\frac{7}{24}\right)^{\tau \log (\pi )}$$ where $$\tau=\frac{1}{2}-\frac{1}{4} \prod _{k=0}^{\infty } \left(1-2^{-2^k}\right)$$ is the Prouhet-Thue-Morse constant (aka the parity constant) which, in binary, is $$.01101001100101101001011001101001100101100110100101101001100101\cdots_2$$ giving an absolute error of $9.17\times 10^{-9}$

For the smallest root, which is very close to $1$, use a Newton-like method of order $n$ to generate the sequence $$\left\{\frac{14}{13},\frac{185}{172},\frac{2433}{2261},\frac{31973}{29712 },\frac{210071}{195215},\frac{5520843}{5130413},\frac{72546144}{67415731},\frac{953286035}{885870304}\right\}$$

The last given in the table is in an absolute error of $5.30\times 10^{-10}$.

For the fun, I generated the longest one fitting in the page $$\frac{33350558429714365013837861800107299668298414993455168656}{309920299 21166592119409492197999281452819325161116276645}$$ which in an absolute error of $1.79\times 10^{-51}$.

For the largest one, starting with $\sqrt 3$, the sequence would be (for Newton, then Halley, then Householder methods) $$\left\{\frac{4 \left(302+1233 \sqrt{3}\right)}{5723},\frac{13 \left(506479+841473 \sqrt{3}\right)}{15006142},\frac{17439624851+12169797708 \sqrt{3}}{22641070127}\right\} $$ The last given in the table is in an absolute error of $7.26\times 10^{-6}$.

The next one (no name for the method) is $$\frac{26642967117244-3287933442786 \sqrt{3}}{12313339596484}$$ is in an absolute error of $3.48\times 10^{-7}$.

  • $\begingroup$ Why is that a good approximation (for ~1.7)? How did you find it? $\endgroup$
    – Ryan
    Apr 29, 2023 at 3:33
  • $\begingroup$ @Ryan. Just as for the first. This is the result of Householder method starting at $\sqrt 3$. I shall add the next one (no name for the method). Cheers :-) $\endgroup$ Apr 29, 2023 at 4:21
  • $\begingroup$ I'm confused as to how this 'parity constant' is appearing $\endgroup$
    – Ryan
    Apr 29, 2023 at 20:45
  • $\begingroup$ @Ryan. I wrote "Just for the fun ..." $\endgroup$ Apr 30, 2023 at 2:50

PadeApproximant[v^6 - 6 v^3 - v + 7,{v,170125/100000,{2,2}}] // InputForm

(-17984351127839/262144000000000000 + (537467662544822153367156895381570813952101*(-1361/800 + v))/ 16584085431328650429992902737920000000000 + (2296073069834313956049534347544383686129*(-1361/800 + v)^2)/ 41460213578321626074982256844800000000)/(1 - (247662144285247281973665600*(-1361/800 + v))/ 202442449112898564819249301 + (148527346121768016406400000*(-1361/800 + v)^2)/202442449112898564819249301)

So the root near 1.7 can be approximated

(5174975570999358281632518703252670765738936 + Sqrt[1155502672193280238447965011461004738056701178969225525537075081952466238134479523866]) / 3673716911734902329679254956071013897806400

1.70125211686161088289767122781 Original root near 1.7 1.70125211686161088289767122784 Pade root

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Apr 28, 2023 at 8:23
  • $\begingroup$ It is short clear answer to the question and gives a good estimate! $\endgroup$
    – Ytrewq
    Apr 29, 2023 at 6:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .