Solve $x^5+3x^3+x+\frac{1}{x^2}=0$

Question

I'm trying to solve this equation $$x^5+3x^3+x+\frac{1}{x^2}=0.$$ It's possible to notice that if we write $3x^3$ as $2x^3+x^3$, then $$ x^3\left(x^2+2+\frac{1}{x^2}\right)+x^3+\frac{1}{x^2}=0, $$ which equals $$ x^3\left(x+\frac{1}{x}\right)^2+x^3+\frac{1}{x^2}=0. $$ But I have no idea how to continue further.

Answer 1

The rational equation:

$$ x^5 + 3\,x^3 + x + \frac{1}{x^2} = 0 $$

shares the same solutions of the polynomial equation:

$$ \underbrace{x^7 + 3\,x^5 + x^3 + 1}_{p(x)} = 0\,. $$

So, once the seven first-attempt solutions on the unit circle have been calculated:

$$ x_k = \cos(2\,k\,\pi/7) + \text{i}\,\sin(2\,k\,\pi/7) \quad \quad \text{with} \; k = 1,2,\dots,7 $$

it's sufficient to update them about ten times by applying the Aberth-Ehrlich method:

$$ x_k' = x_k - \left(\frac{p'(x_k)}{p(x_k)} - \begin{aligned}\sum_{j=1;\,j\ne k}^7\end{aligned}\frac{1}{x_k-x_j}\right)^{-1} $$

i.e

p[u_] := u^7 + 3 u^5 + u^3 + 1
d = Exponent[# /. u -> v RandomReal[], v] &[p[u]];
x = N@Table[Cos[2 k π/d] + I Sin[2 k π/d], {k, d}];
Do[x[[k]] = x[[k]] - (p'[x[[k]]]/p[x[[k]]] - 
   Sum[If[j == k, 0, 1/(x[[k]] - x[[j]])], {j, d}])^(-1), {n, 8}, {k, d}]

from which:

$$ \begin{aligned} & x_1 = 0.598251 + 0.491251\,\text{i}\,; \\ & x_2 = -0.274401 + 0.90337\,\text{i}\,; \\ & x_3 = -0.71158 - 7.35225\cdot 10^{-19}\,\text{i}\,; \\ & x_4 = 0.0319404 - 1.62172\,\text{i}\,; \\ & x_5 = -0.274401 - 0.90337\,\text{i}\,; \\ & x_6 = 0.598251 - 0.491251\,\text{i}\,; \\ & x_7 = 0.0319404 + 1.62172\,\text{i}\,. \\ \end{aligned} $$

Of course this doesn't prove whether or not it's possible to express them by radicals.

Answer 2

Totally useless but funny (at least to me)

For the only real root, write $$f(x)=x^7+3 x^5+x^3+1$$ as $$f(x)=-4+25 (x+1)-54 (x+1)^2+66 (x+1)^3-50 (x+1)^4+$$ $$24 (x+1)^5-7(x+1)^6+(x+1)^7$$

Now, consider the above as a series expansion to $O\left((x+1)^{p+1}\right)$ with $p\geq 7$ and use power series reversion to get $$x=-1+\sum_{n=0}^p a_n\,(f(x)+4)^n+O((f(x)+4)^{p+1}$$ Since we want $f(x)=0$ then $$x_{(p)}=-1+\sum_{n=0}^p a_n\,4^n$$

Using $p=8$ which is exactly the polynomial, you should get, as an estimate $$x_{(8)}=-\frac{1075280545512027477}{1490116119384765625}=-0.721609$$

Now, repeat and you will get (exact fraction reprted as decimal numbers) $$\left( \begin{array}{cc} p & x_{(p)} \\ 7 & -0.721609 \\ 8 & -0.718758 \\ 9 & -0.716772 \\ 20 & -0.711791 \\ 30 & -0.711595 \\ 40 & -0.711581 \\ 50 & -0.711580 \\ \end{array} \right)$$

Edit (for more fun !)

Transform the series to their equivalent $[1,n]$ Padé approximants $P_n$, that is to say $$P_n=\frac {-4+ a_{(n)}\,(x+1) } {1+\sum_{p=1}^n b_p\,(x+1)^p }\quad \implies x_{(n)}=-1+\frac 4{a_{(n)}}$$ which is the same as using the first itterate of a Newton-like method of order $(n+2)$.

This will generate the converging sequence $$\left( \begin{array}{c} -\frac{21}{25} , -\frac{309}{409}, -\frac{4245}{5881}, -\frac{58357}{81881} , -\frac{807253}{1134777} , -\frac{3730663}{5243699}, -\frac{155226133}{218150521}, -\frac{2152852853}{3025454937} \end{array} \right)$$

The last one in the list is $$x_{(7)}= -\frac{2152852853}{3025454937}={\large{\color{red}{-0.711579}}}88$$