Closed form for solution of degree 6 polynomial (in particular, the root of $v^6 - 6v^3 - v + 7 = 0$ near $1.7$)?

Question

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.

Answer 1

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 $3.48\times 10^{-7}$.

Answer 2

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