关于多项式$p_k(z)=z^{k+2}-z^3+z^2+z-1$正根的唯一性证明与闭式表达的技术问询

Hey Daniel, great question about the roots of your generating function's denominator—let's break down your two core concerns step by step: proving the uniqueness of the positive root $\alpha_k$, and exploring possible closed-form expressions for it.

1. Proving the Uniqueness of $\alpha_k$ #

You already made solid progress with Descartes' Rule of Signs (confirming at least one positive real root) and Bolzano's Theorem (pinpointing a root in $(0,1)$). We can formalize the uniqueness by analyzing the polynomial's behavior across all positive real numbers:

For $z \in (0,1)$: #

First, compute the derivative:
$$p_k'(z) = (k+2)z^{k+1} - 3z^2 + 2z + 1$$
Notice that the quadratic term $-3z^2 + 2z + 1$ factors to $-(3z+1)(z-1)$. Since $z < 1$, $(z-1)$ is negative, making the entire quadratic term positive. Add that to $(k+2)z^{k+1}$, which is always positive for $z>0$ and $k\geq1$, and we get $p_k'(z) > 0$ everywhere in $(0,1)$.

This means $p_k(z)$ is strictly increasing on $(0,1)$. We know $p_k(0) = -1$ and $p_k(1) = 1 - 1 + 1 + 1 - 1 = 1 > 0$, so there's exactly one root $\alpha_k$ in this interval.

For $z \geq 1$: #

Rewrite the polynomial as:
$$p_k(z) = z^{k+2} + \left(-z^3 + z^2 + z - 1\right)$$
The second term factors to $-(z-1)^2(z+1)$, which is non-positive for $z \geq 1$ (since $(z-1)^2 \geq 0$ and $z+1 > 0$). But $z^{k+2}$ grows extremely quickly for $z>1$ and $k\geq1$:

• At $z=1$, $p_k(1)=1>0$
• For $z>1$, $z^{k+2} > z^3$ (since $k+2 \geq 3$), so $z^{k+2} - z^3 > 0$. Adding $z^2 + z -1 > 1+1-1=1>0$ gives $p_k(z) > 0$ for all $z>1$.

Combining both cases, $p_k(z)$ has exactly one positive real root $\alpha_k \in (0,1)$, which matches your numerical observation that $\alpha_k \to 1$ as $k\to\infty$ (we can even derive an asymptotic approximation: $\alpha_k = 1 - \frac{\ln k}{k} + o\left(\frac{\ln k}{k}\right)$ for large $k$).

2. Closed-Form Expressions for $\alpha_k$ #

You're right that $k=1$ and $k=2$ are solvable by radicals:

• For $k=1$, $p_1(z) = z^2 + z -1$, whose positive root is the familiar golden ratio conjugate: $\frac{\sqrt{5}-1}{2}$
• For $k=2$, $p_2(z) = z^4 - z^3 + z^2 + z -1$ factors into quadratics, so its roots can also be expressed using square roots.

For $k\geq3$, the polynomial's degree is $k+2 \geq 5$. By Galois theory, general degree-5+ polynomials can't be solved with radicals unless their Galois group is solvable—and it's non-trivial to prove that for this specific family. However, we can explore alternatives:

Asymptotic Approximations #

For large $k$, we can rewrite the root equation $z^{k+2} = (z-1)^2(z+1)$. Let $z=1-t$ where $t$ is small, then $(1-t)^{k+2} \approx e^{-(k+2)t}$ and $(z-1)^2(z+1) \approx 2t^2$. Taking logs gives:
$$-(k+2)t \approx \ln 2 + 2\ln t$$
This can be rearranged to use the Lambert W function (which solves $w e^w = x$), giving an implicit form for $t$, and thus for $\alpha_k$.

Special Function Representations #

While there's no known elementary closed-form for general $k$, you could express $\alpha_k$ using iterative methods (like Newton-Raphson) or series expansions. For example, expanding around $z=1$ gives a power series in $\frac{1}{k}$ that approximates $\alpha_k$ for large $k$.

In short: for $k\geq3$, a radical-based closed-form is unlikely, but you can use special functions or asymptotic expressions to describe $\alpha_k$.

备注：内容来源于stack exchange，提问作者Daniel F. Checa