# Can the general septic be solved by infinitely nested radicals?

I. Quintic. The general quintic can be reduced to the form,

$$x^5=p+x\tag1$$

$$x = \sqrt[5]{p+x}$$

Hence by an iterative process,

$$x =\sqrt[5]{p+\sqrt[5]{p+\sqrt[5]{p+\sqrt[5]{p+x\dots}}}}$$

Truncating yields $x$ in terms of $p$ to any desired degree of accuracy.

II. Sextic. Likewise, the general sextic can be reduced to,

$$x^6 = x^2+2px+q\tag2$$

Or equivalently, let $D=-p^2+q$,

$$x^6 = D+(p+x)^2$$

$$x =\sqrt[6]{D+(p+x)^2}$$

By a similar iterative process,

$$x =\sqrt[6]{D+\left(p+\sqrt[6]{D+\left(p+\sqrt[6]{D+\left(p+x\dots\right)^2}\right)^2}\right)^2}$$

and so on, it can be seen that the roots of the general sextic can be expressed as an infinitely nested radical as described (albeit less concisely) in this pre-print by Nikos Bagis.

III. Septic. The general septic can be reduced to the form,

$$x^7 = x^3+px^2+qx+r\tag3$$

Bagis claims that an analogous iterative process will show that the roots of the general septic is also an infinitely nested radical. However, his eqns $22-33$ are unconvincing (especially $22-24$). An obvious way is to express $(3)$ as,

$$x^7 = (x+\alpha)^3+\beta$$

This gives us only two unknowns $\alpha,\beta$, but there are three variables $p,q,r$.

Question: Are the roots of the general septic really an infinitely nested radical analogous to the quintic and sextic cases?

P.S. This was inspired by this post.

• What is not convincing for me is the step 26 -27, ''under certain conditions''. I'm not able to figure what are those conditions. – Emilio Novati Mar 7 '15 at 17:00

From Nikos Bagis:

Yes, it seems that polynomial equations higher than 6th degree cannot be solved with method such I described in my pre-print. For example a septic equation generates multiple nested radicals, in nested radicals "struction" and is a little complicated. But then we stop thinking about nested radicals and begin thinking with nested functions. At this point, if someone wants to make analysis, has to define the lower nested radicals functions such as the BR-Bring Radicals functions then the sextic radicals (say)...etc, and build a space of such functions.

Also the generalization of the sextic equation: $x^{6n}=ax^{2m}+bx^{m}+c$: (1), which I mention in the article, maybe can give us solution of "say" $x^7=ax^{2m}+bx^m+c$ : (2), which I don't have investigate. Note that $m$ can be rational in the last equation. So, what septic equations can reduced to (2)?

Also a very interesting idea is to try solve the classical sextic equation using only the Rogers-Ramanujan continued fraction (RRCF). The equation I have solve with (RRCF) is $C_1x^{5/3}=ax^2+bx+b^2/(20a)$, which differs from the generalized sextic in the last terms $c$, $b^2/(20a)$. Then we can say in a very elegant way that: Every sextic equation is solvable by the (RRCF), which have tabulated values and can evaluated by modular equations.

The equation $$X^{\nu}=X+a$$ have solution $X_0$ such that $$X_0^{\mu}=\left(\textbf{BR}_{\nu}(a)\right)^{\mu}=\left(\frac{\mu}{\nu-1}\sum^{\infty}_{n=0}\frac{(-1)^n}{kn+\lambda}\left( \begin{array}{cc} kn+\lambda\\ n \end{array} \right)a^n\right)^{-1}$$ where $k=\frac{\nu}{\nu-1}$ and $\lambda=\frac{\mu}{\nu-1}$.

Hence if $$\textbf{BR}_{\nu,\mu}(a):=\frac{-\mu}{\nu-1}\sum^{\infty}_{n=0}\frac{(-1)^n}{kn-\lambda}\left(\begin{array}{cc} kn-\lambda\\ n \end{array}\right)a^n,$$ then equation $$kX^{\mu}=X^{\nu}-X-l$$ admits solution $$X_1=\textbf{BR}_{\nu}\left(l+k\textbf{BR}_{\nu,\mu}\left(l+k\textbf{BR}_{\nu,\mu}(l+\ldots)\right)\right)$$

Suppose we have the equation $X^{\nu}=kX^{\mu}+X+l$, $\nu>\mu$ and $|k|,|l|,|kl^{-1}|<1$ : (Eq)

Set $k_1=\frac{\nu}{\nu-1}$, $\lambda_1=\frac{-\mu}{\nu-1}$, $$\textbf{f}(s;x):=\sum^{\infty}_{n=1}\left(\begin{array}{cc} s\\ n \end{array}\right)\left(\begin{array}{cc} k_1s+\lambda_1n\\ s \end{array}\right)\frac{n}{(s-n+1)(k_1 s+\lambda_1n)}x^n$$ and $$\textbf{g}(w):=w+w\lambda_1\sum^{\infty}_{s=0}(-1)^s\textbf{f}\left(s;kw^{-1}\right)w^s.$$ Then a solution of (Eq) is $$X_2=\textbf{BR}_{\nu}\left(\textbf{g}(l)\right).$$ $\textbf{Proof.}$ It is clear that if $X_2$ is the desired root of the equation (Eq), then also $X_2=\textbf{BR}_{\nu}\left(\xi\right)$, where $\xi=l+k\textbf{BR}_{\nu,\mu}\left(\xi\right)$. But according to Lagrange theorem (see [1] pg 133), we have $$\xi=l+\sum^{\infty}_{n=1}\frac{k^n}{n!}\frac{d^{n-1}}{dl^{n-1}}\left(\textbf{BR}_{\nu,\mu}(l)\right)^n$$ and $\left(\textbf{BR}_{\nu,\mu}(x)\right)^{n}=\textbf{BR}_{\nu,\mu n}(x)$. Using expansion $$\textbf{BR}_{\nu,\mu}(a):=\left(\textbf{BR}_{\nu}(a)\right)^{\mu}=\lambda_1\sum^{\infty}_{n=0}\frac{(-1)^n}{k_1n+\lambda_1}\left(\begin{array}{cc} k_1n+\lambda_1\\ n \end{array}\right)a^n.$$ we arrive to a double sum we rearange and get the result.

Moreover

A solution of $$X^{\nu}=mX^{\lambda}+kX^{\mu}+X+l,$$ where $\nu>\mu$ and $|k|,|l|,|kl^{-1}|<1$ is $$X_3=\textbf{BR}_{\nu}(\textbf{g}(l+m\textbf{BR}_{\nu,\lambda}(\textbf{g}(l+m\textbf{BR}_{\nu,\lambda}(\textbf{g}(l+\ldots)))))),$$ and the ''denested'' equation is $$\xi=l+m\textbf{BR}_{\nu,\lambda}\left(\textbf{g}(\xi)\right)\textrm{ and }X_3=\textbf{BR}_{\nu}\left(\textbf{g}(\xi)\right).$$

References: [1]:E.T. Whittaker and G.N. Watson. 'A course on Modern Analysis'. Cambridge U.P, (1927).