1406.1948.tex
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\centerline{\bf\large Solution of Polynomial Equations with Nested Radicals}\vskip .2in


\centerline{Nikos Bagis}

\centerline{Stenimahou 5 Edessa}
\centerline{Pellas 58200, Greece}
\centerline{bagkis@hotmail.com}

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\centerline{\bf Abstract}

\begin{quote}
In this article we present solutions of arbitrary polynomial equations in nested periodic radicals 
\end{quote}

\textbf{Keywords}: Nested Radicals; Quintic; Sextic; Septic;  Equations; Polynomials; Solutions; Higher functions;


\section{A general type of equations}

Consider the following general type of equations
\begin{equation}
ax^{2\mu}+bx^{\mu}+c=x^{\nu}.
\end{equation}
Then (1) can be written in the form
\begin{equation}
a\left(x^{\mu}+\frac{b}{2a}\right)^2-\frac{\Delta^2}{4a}=x^{\nu},
\end{equation}
where $\Delta=\sqrt{b^2-4ac}$. Hence
\begin{equation}
x^{\mu}=\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}x^{\nu}}
\end{equation}
Set now $\sqrt[d]{x}=x^{1/d}$, $d\in\textbf{Q}_{+}-\{0\}$, then
\begin{equation}
x=\sqrt[\mu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}x^{\nu}}}
\end{equation}
hence
\begin{equation}
x^{\nu}=\sqrt[\mu/\nu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}x^{\nu}}}
\end{equation}
\textbf{Theorem 1.}\\
The solution of (1) is
\begin{equation}
x=\sqrt[\mu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}\sqrt[\mu/\nu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}\sqrt[\mu/\nu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+...}}}}}}
\end{equation}
\\
\textbf{Proof.}\\
Use relation (4) and repeat it infinite times.\\
\\
\textbf{Corollary 1.}\\
The equation 
\begin{equation}
ax^6+bx^5+cx^4+dx^3+ex^2+fx+g=0
\end{equation}
is always solvable with nested radicals.\\
\\
\textbf{Proof.}\\
We know (see [1]) that all sextic equations, of the most general form (7) are equivalent by means of a Tchirnhausen
transform 
\begin{equation}
y=kx^4+lx^3+mx^2+nx+s
\end{equation}
to the form
\begin{equation}
y^6+e_1y^2+f_1y+g_1=0
\end{equation}    
But the last equation is of the form (1) with $\mu=1$ and $\nu=6$ and have solution
\begin{equation}
y=\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}-\frac{1}{e_1}\sqrt[1/6]{\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}-\frac{1}{e_1}\sqrt[1/6]{\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}+...}}}}}
\end{equation} 
where $\Delta_1=\sqrt{f_1^2-4e_1g_1}$. Hence knowing $y$ we find $x$ from (8) and get the solvability of (7) in nested radicals. More precisely it holds
$$
kx^4+lx^3+mx^2+nx+s=
$$
\begin{equation}
=\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}-\frac{1}{e_1}\sqrt[1/6]{\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}-\frac{1}{e_1}\sqrt[1/6]{\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}+...}}}}}
\end{equation}
\\

Theorem 1 is a generalization of a theorem of Euler (see [4] pg.306-307):\\ 
\\
\textbf{Theorem 2.}\\
If the root of 
\begin{equation}
aqx^p+x^q=1
\end{equation} 
is $x$, then 
\begin{equation}
x^n=\frac{n}{q}\sum^{\infty}_{k=0}\frac{\Gamma(\{n+pk\}/q)(-q a)^k}{\Gamma(\{n+pk\}/q-k+1)k!}\textrm{, }n=1,2,3,\ldots
\end{equation}
Where $\Gamma(x)$ is Euler's the Gamma function.\\ 
Moreover the solution $x$ of (12) can given with nested radicals:   
\begin{equation}
x=\sqrt[q]{1-aq\sqrt[q/p]{1-aq\sqrt[q/p]{1-aq\sqrt[q/p]{1-\ldots}}}}
\end{equation}
\\

The general quintic
\begin{equation}
ax^5+bx^4+cx^3+dx^2+ex+f=0 
\end{equation}
can reduced also by means of a Tchirnhausen transform into
\begin{equation}
x^5+Ax+B=0
\end{equation}
Define now the hypergeometric function
\begin{equation}
\textrm{BR}(t)=-t\cdot{}_4F_3\left[\left\{\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}\right\};\left\{\frac{1}{2},\frac{3}{4},\frac{5}{4}\right\};-\frac{3125 t^4}{256}\right]
\end{equation}
\\
\textbf{Theorem 3.}\\
The solution of (16) is
\begin{equation}
x=\sqrt[4]{\frac{-A}{5}}\textrm{BR}\left(-\frac{\sqrt[4]{\frac{5^5}{-A^5}}}{4}B\right)
\end{equation}
and also holds
\begin{equation}
x=\sqrt[5]{-B-A\sqrt[5]{-B-A\sqrt[5]{-B-A\sqrt[5]{-B-\ldots}}}}
\end{equation}
\\
\textbf{Application.}\\
If $Im(\tau)>0$ and $j_{\tau}$ denotes the $j$-invariant, then
$$
\frac{1}{j_{\tau}^{5/3}}\left(R\left(e^{2\pi i \tau}\right)^{-5}-11-R\left(e^{2\pi i\tau}\right)^5\right)^{5/3}=
$$
\begin{equation}
=\sqrt[3/5]{\frac{-125}{j_{\tau}}+\sqrt{\frac{12500}{j^2_{\tau}}+\sqrt[3/5]{\frac{-125}{j_{\tau}}+\sqrt{\frac{12500}{j_{\tau}^2}+\ldots}}}},
\end{equation}
where $R(q)$ is the Rogers-Ramanujan continued fraction.\\
\\
\textbf{Proof.}\\
Equation (1) for $a=1/j_{\tau}^{1/3}$, $b=250/j_{\tau}^{1/3}$, $c=3125/j_{\tau}^{1/3}$ and $\mu=1$, $\nu=5/3$ takes the form 
\begin{equation}
x^{2}+250x+3125=j_r^{1/3}x^{5/3}
\end{equation}
Which is a simplified form of Klein's equation for the icosahedral equation [3] and have solution $x=Y_{\tau}=R(q^2)^{-5}-11-R(q^2)^5$. By Theorem 1 we can express $Y_{\tau}$ in nested-periodical radicals . This completes the proof.\\  
\\
For more details about equation $ax^2+bx+\frac{b^2}{20a}=C_1x^{5/3}$ one can see [2].

\section{The reduction of a given polynomial}

As someone can see if we assume the equation
\begin{equation}
aH_7^3+bH_7^2+cH_7+d=u,
\end{equation}
then we can rewrite (22) in the form
\begin{equation}
H_7^3=-\frac{b}{a}H_7^2-\frac{c}{a}H_7-\frac{d}{a}+\frac{u}{a}
\end{equation}
Hence
\begin{equation}
H_7=\sqrt[3]{\frac{u}{a}-\frac{\Delta_3}{4ba}-\frac{b}{a}\sqrt[1/2]{\frac{c}{2b}+H_7}}
\end{equation}
where $\Delta_3=4bd-c^2$.\\
From the above a nested radical solution of (22) is 
$$
H=H_7=H_7(a,b,c,d,u)=
$$
\begin{equation}
=\sqrt[3]{\frac{u}{a}-\frac{\Delta_3}{4ba}-\frac{b}{a}\sqrt[1/2]{\frac{c}{2b}+\sqrt[3]{\frac{u}{a}-\frac{\Delta_3}{4ba}-\frac{b}{a}\sqrt[1/2]{\frac{c}{2b}+...}}}}
\end{equation}
Let also the equation  
\begin{equation}
ax^{3\mu}+bx^{2\mu}+cx^{\mu}+d=x^{\nu}.
\end{equation}
Clearly if we drop the $x^{\nu}$ term of (23), then becomes a 3$rd$ degree equation in $x^{\mu}$ and can be written under certain conditions as 
\begin{equation}
x^{\mu}=H_7\left(a,b,c,d,x^{\nu}\right)
\end{equation}
or equivalently
\begin{equation}
x=\sqrt[\mu]{H_7(a,b,c,d,x^{\nu})}
\end{equation}
or equivalently
\begin{equation}
x^{\nu}=\sqrt[\mu/\nu]{H_7(a,b,c,d,x^{\nu})}
\end{equation}
Hence\\
\\
\textbf{Theorem 4.}\\
If $H_7$ is the function defined in (22), then equation (26) have solution in nested radicals
\begin{equation}
x=\sqrt[\mu]{H_7\left(a,b,c,d,\sqrt[\mu/\nu]{H_7\left(a,b,c,d,\sqrt[\mu/\nu]{H_7\left(a,b,c,d,\ldots\right)}\right)}\right)}.
\end{equation}
\\ 
\textbf{Note.}\\ Here we must mention that we don't interested about the conditions, since we want to establish formulas that agreed numerically. One can say that without these conditions our results are conjectures.\\
\\
\textbf{Theorem 5.}\\
Given a general septic equation 
\begin{equation}
\sum^{7}_{k=0}a_kx^k=0,
\end{equation}
after using Tchirnhausen transform can written in the form
\begin{equation}
y^7=ay^3+by^2+cy+d,
\end{equation}
which has solution
\begin{equation}
y=H_7\left(a,b,c,d,\sqrt[1/7]{H_7\left(a,b,c,d,\sqrt[1/7]{H_7\left(a,b,c,d,
\ldots\right)}\right)}\right)
\end{equation}
\\
\textbf{Proof.}\\Taking $\nu=7$ and $\mu=1$ in Theorem 4 we get the desired result.\\
\\
\textbf{Example.}\\
Consider the equation
\begin{equation}
x^7=x^3+x^2-1/2
\end{equation}
We want to find a solution of (34). For this we find $H_7$, which is solution of 
\begin{equation}
H_7^3+H_7^2-1/2=u. 
\end{equation} 
Solving the above equation with expansion (25) we get
\begin{equation}
H_7(u)=\sqrt[3]{u+\frac{1}{8}-\sqrt[3/2]{u+\frac{1}{8}-\sqrt[3/2]{u+\frac{1}{8}-...}}}
\end{equation}
Hence a solution of (34) is
\begin{equation}
x=H_7\left(\sqrt[1/7]{H_7\left(\sqrt[1/7]{H_7\left(\sqrt[1/7]{H_7\left(\ldots\right)}\right)}\right)}\right)
\end{equation}
Or if someone preferes a root of 
\begin{equation}
x=\sqrt[3]{x^7+\frac{1}{8}-\sqrt[3/2]{x^7+\frac{1}{8}-\sqrt[3/2]{x^7+\frac{1}{8}-...}}}
\end{equation}
\\
  
The next case is the octic equation which is equivalent to
\begin{equation}
x^8=ax^4+bx^3+cx^2+dx+e
\end{equation}
and with above method is again solvable in nested radicals, since the $H=H_8$ is of degree 4 and hence solvable in radicals. But, instead solving a 4-th degree equation we can use the function $H_7(u)$, since making the change of variable $x=H_7(u)$ in (39) we have
\begin{equation}
H_7^8(u)=aH_7^4(u)+u
\end{equation}
By this way
\begin{equation}
H_8(u)=H_7^8(u)-aH_7^4(u),
\end{equation}
with $H_7(u)=H_7(b,c,d,e;u)$ is the expansion (25) and 
\begin{equation}
H_8(\xi)=\xi,\textrm{ and }x=H_7(\xi)
\end{equation}
\\

The 9$th$ degree polynomial equation have $H=H_9$ and $deg(H_9)=5$. In this case we can use the hypergeometric function solution (18) or the radical solution (19).\\ 
The desired equation we want to solve is 
\begin{equation}
\sum^{9}_{k=0}a_kx^k=0
\end{equation}
This reduces to   
\begin{equation}
ay^5+by^4+cy^3+dy^2+ey+f=y^9,
\end{equation}
after a Tchirnhausen transform
\begin{equation}
y=kx^4+lx^3+mx^2+nx+s
\end{equation}
Hence we concern to construct the function $H_9=H_9(u)$ such that
\begin{equation}
aH_9(u)^5+bH_9(u)^4+cH_9(u)^3+dH_9(u)^2+eH_9(u)+f=u
\end{equation}
For to construct $H_9$ we use the Tchirnhausen transform again
\begin{equation}
G_9(u)=k'H_9(u)^4+l'H_9(u)^3+m'H_9(u)^2+n'H_9(u)+s',
\end{equation}
where $(k',l',m',n',s')$ are depending from $(a,b,c,d,e,u)$.
By this method we arrive to
\begin{equation}
G_9(u)^5+A G_9(u)+B=0
\end{equation}
Hence $G_9(u)$ is that of (18), with $A$,$B$ functions of $(a,b,c,d,e,u)$, or denoting simply $A=A(u)$ and $B=B(u)$:
\begin{equation}
G_9(u)=\sqrt[4]{\frac{-A(u)}{5}}\textrm{BR}\left(-\frac{\sqrt[4]{\frac{5^5}{-A(u)^5}}}{4}B(u)\right)
\end{equation}
Finally we substitute (49) to (47) and solve with respect to $H_9(u)$.
Hence with the above notation the next theorem holds\\
\\
\textbf{Theorem 6.}\\
i) The equation
\begin{equation}
ax^{5\mu}+bx^{4\mu}+cx^{3\mu}+dx^{2\mu}+ex^{\mu}+f=x^{\nu}
\end{equation}
have solution in periodical nested radicals
\begin{equation}
x=\sqrt[\mu]{H_9\left(\sqrt[\mu/\nu]{H_9\left(\sqrt[\mu/\nu]{H_9\left(
\ldots\right)}\right)}\right)},
\end{equation}
with $H_9$ expressible by means of (18) (or if preferred by (19)).\\ 
ii) A root of (44) is 
\begin{equation}
kx^4+lx^3+mx^2+nx+s=H_9\left(\sqrt[1/9]{H_9\left(\sqrt[1/9]{H_9\left(
\ldots\right)}\right)}\right).
\end{equation}
\\

By this method one can see that we can solve any polynomial equation with nested radicals. 



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\centerline{\bf References}\vskip .2in


\noindent

[1]: J.V. Armitage and W.F. Eberlein. 'Elliptic Functions'. Cambridge University Press. (2006)

[2]: N.D. Bagis. 'On a general sextic equation solved by the Rogers-Ramanujan continued fraction'. arXiv:1111.6023v2[math.GM]. (2012)

[3]: W. Duke. 'Continued fractions and Modular functions'. Bull. Amer. Math. Soc. (N.S.), 42. (2005), 137-162.

[4]: B.C. Berndt. 'Ramanujan`s Notebooks Part I'. Springer Verlag, New York. (1985).

[5]: M.L. Glasser. 'The Quadratic Formula Made Hard or A Less Radical Approach to Solving Equations'. arXiv:math/94112244v1 [math.CA] 16 Nov (1994).

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