Roots of polynomial $z^{k+2}-z^3+z^2+z-1$

186 Views Asked by At

Hello :) I'm working in a combinatorics problem where I need to describe the roots of the denominator of a generating function, specifically, $p_k(z)=z^{k+2}-z^3+z^2+z-1$, with $(k\geq1)$. The cases $k=1,2$ are solvable by radicals, and I would like to be able to express the roots of $p_k(z)$ in therms of $k$. I don't know too much about Galois Theory, but I found some results:

  • By Descartes' rule of signs, $p_k(z)$ has at least one positive real root.
  • By Bolzano's Theorem, $p_k(z)$ has at least one positive root on the interval $(0,1)$. Numerically, it seems to be the only positive root, and if $\alpha_k$ is this root then $\alpha_k\to1$ as $k\to\infty$.

I'm interested on the positive root. I'd like to know if there is any way to prove the unicity of $\alpha_k$ and giving a closed expression. Of course, this formula may be not expressable by radicals, but I'm open to use another trascendental functions like we do in the Viète's formula for the depressed cubic.

2

There are 2 best solutions below

0
On BEST ANSWER

Let $f_k(x)=x^{k+2}$. Let $g(x)=x^3-x^2-x+1=(x-1)^2(x+1)$. Then $p_k=f_k-g$.

We first show that there is a unique positive root $\alpha_k$ of $p_k$ and $\alpha_k\in (0,1)$ for all $k\ge 1$. The existence can be proved by intermediate value theorem, as noted by OP. For uniqueness, we have $f'_k(x)=(k+2)x^{k+1}$ and $g'(x)=3x^2-2x-1<3x^2$ for all $x>0$. Note that $f_k$ and $-g$ are strictly increasing on $(0,1)$, and so is $p_k$. Then there is only one root of $p_k$ in $(0,1)$. Also, $p_k(1)=1$, so $1$ is not a root. For $x>1$, we have $f_k(1)>g_k(1)$ and $f'_k\ge g'$ on $(1,\infty)$. By racetrack principle, we have $p_k>0$ on $(1,\infty)$. This shows that $\alpha_k$ is the unique root of $p_k$ on $(0,\infty)$.

Now we prove that $\displaystyle\lim_{k\to\infty} \alpha_k=1$. Let $\varepsilon>0$. Note that $g(x)>0$ for all $x\in (0,1)$. Then there exists $N>0$ such that $f_N(1-\varepsilon)<g(1-\varepsilon)$. Let $k>N$. We have $f_k(x)<f_N(x)$ for all $x\in (0,1)$, so $f_k(1-\varepsilon)<g(1-\varepsilon)$. Then $p_k(1-\varepsilon)<0$ and $p_k(1)=1$. By intermediate value theorem and uniqueness of $\alpha_k$, we have $\alpha_k\in (1-\varepsilon, 1)$. This shows that $|1-\alpha_k|<\varepsilon$ for all $k>N$. Thus, $\displaystyle\lim_{k\to \infty} \alpha_k=1$.


As for closed form of $\alpha_k$, it depends on what kind of expressions you are looking for. It most probably does not have a finite radical form. For $1\le k\le 30$, by a Magma computation, the Galois group of $p_k$ is solvable only if $k=1,2,4$. That means $\alpha_1,\alpha_2,\alpha_4$ are the only roots of $p_k$ for $k\le 30$ that can be expressed in finite radical form.

0
On

Beside the solution proposed by @TymaGaidash, if we were able to use a Newton-like method of infinite order, we could have the exact solution for the positive root.

Limiting to some order $n$, the approximate solution is $$z_{(n)}=1-(n+1)\frac{\sum_{m=0}^n a_m\,k^m}{\sum_{m=0}^{n+1} b_m\,k^m}$$ where all coefficients are positive.

For $n=8$, the coefficients $a_m$ are $$\{25643520,41466192,26678476,8695316,1557689,137648,7994,44,1\}$$ and the corresponding $b_m$ are $$\{518555520,930785184,680029128,261385172,56543886,7128345,436212,1 9698,54,1\}$$

Limited to this small order, for sure, the results are not extremely good $$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 5 & 0.737888 & 0.738036 \\ 10 & 0.801907 & 0.801905 \\ 15 & 0.838271 & 0.837155 \\ 20 & 0.862444 & 0.860164 \\ 25 & 0.879670 & 0.876618 \\ 30 & 0.892479 & 0.889087 \\ 35 & 0.902309 & 0.898926 \\ 40 & 0.910051 & 0.906925 \\ 45 & 0.916283 & 0.913580 \\ 50 & 0.921398 & 0.919218 \\ \end{array} \right)$$

For sure, the results are better and better increasing $n$. For example, for $k=50$, the estimates are $$z_{(7)}=0.916725 \qquad \text{while} \qquad z_{(8)}=0.921398$$

Edit

If you use the above as $z_0$ and perform $\color{red}{\text{one single iteration}}$ of Newton, Halley or Householder method (the derivatives are so simple), the results would be quite good . $$\left( \begin{array}{ccccc} k & \text{Newton} & \text{Halley} & \text{Householder} & \text{solution} \\ 5 & 0.738036 & 0.738036 & 0.738036 & 0.738036 \\ 10 & 0.801905 & 0.801905 & 0.801905 & 0.801905 \\ 15 & 0.837161 & 0.837155 & 0.837155 & 0.837155 \\ 20 & 0.860200 & 0.860164 & 0.860164 & 0.860164 \\ 25 & 0.876699 & 0.876618 & 0.876618 & 0.876618 \\ 30 & 0.889208 & 0.889086 & 0.889087 & 0.889087 \\ 35 & 0.899067 & 0.898925 & 0.898926 & 0.898926 \\ 40 & 0.907063 & 0.906924 & 0.906925 & 0.906925 \\ 45 & 0.913696 & 0.913579 & 0.913579 & 0.913580 \\ 50 & 0.919302 & 0.919217 & 0.919218 & 0.919218 \\ \end{array} \right)$$