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Context

I’m trying to prove that the largest real root of a polynomial $p_{2}\left ( x, \ell \right )$ is greater than that of $p_{1}\left ( x, \ell \right )$ for all positive integers $\ell> 1$, where both polynomials arise from computing the capacity of $\ell$-constrained DNA composite. The polynomials have integer coefficients, and their largest real roots are close to, but less than, $5$. I’m seeking analytical solutions to compare the largest real roots for general $\ell$.

Polynomial definitions:

  • $p_{2}\left ( x, \ell \right )= \sum_{n= 0}^{3\ell}a_{n}x^{n}$: Coefficients defined in three parts:
  1. Part 1 ($n= 2\ell, \ldots, 3\ell$): $a_{n}= 3a_{n- 1}- a_{n- 2}- a_{n- 3}$, with $a_{3\ell}= 1, a_{3\ell- 1}= -2, a_{3\ell- 2}= -7$.
  2. Part 2 ($n= \ell, \ldots, 2\ell- 1$): $a_{n}= a_{n}^{\ell- 1}+ a_{n- 2}^{\ell- 1}+ a_{n- 2}^{\ell- 2}$, where $a_{n}^{\ell- 1}$ refers to coefficients of $p_{2}\left ( x, \ell- 1 \right )$.
  3. Part 3 ($n= 0, \ldots, \ell- 1$): $a_{n}= \left ( -1 \right )^{\ell}b_{n}, \quad b_{n}= -2b_{n- 1}+ b_{n- 2}- 2$, with $b_{0}= -2, b_{1}= 2$.
  • Example for $\ell= 3$: $$p_{2}\left ( x, 3 \right )= \bbox[5px, #F0FFF0]{x^{9}- 2x^{8}- 7x^{7}- 20x^{6}}\bbox[5px, #FFF0F0]{- 43x^{5}- 28x^{4}- 21x^{3}}\bbox[5px, #FFFFE0]{+ 8x^{2}- 2x+ 2}$$ Largest real root: $\lambda_{2}\left ( 3 \right )\approx 4.787559836555051$
  • Example for $\ell= 4$: $$p_{2}\left ( x, 4 \right )= \bbox[5px, #F0FFF0]{x^{12}- 2x^{11}- 7x^{10}- 20x^{9}- 51x^{8}}\bbox[5px, #FFF0F0]{- 108x^{7}- 79x^{6}- 78x^{5}- 47x^{4}}\bbox[5px, #FFFFE0]{+ 16x^{3}- 8x^{2}+ 2x- 2}$$ Largest real root: $\lambda_{2}\left ( 4 \right )\approx 4.9079904984391955$
  • Example for $\ell= 5$: $$p_{2}\left ( x, 5 \right )= \bbox[5px, #F0FFF0]{x^{15}- 2x^{14}- 7x^{13}- 20x^{12}- 51x^{11}- 126x^{10}}\bbox[5px, #FFF0F0]{- 265x^{9}- 202x^{8}- 215x^{7}- 178x^{6}- 117x^{5}}\bbox[5px, #FFFFE0]{+ 42x^{4}- 16x^{3}+ 8x^{2}- 2x+ 2}$$ Largest real root: $\lambda_{2}\left ( 5 \right )\approx 4.957983946242159$
  • $p_{1}\left ( x, \ell \right )= \sum_{n= 0}^{\ell}a_{n}x^{n}$: $a_{n- 2}= 2a_{n- 1}+ 3a_{n}$, with $a_{\ell}= 1, a_{\ell- 1}= -3$.

Question

It is easy to see that: $$p_{1}\left ( x, \ell \right )- p_{1}\left ( 5, \ell \right )= \left ( x- 5 \right )q_{1}\left ( x, \ell \right ), \quad p_{2}\left ( x, \ell \right )- p_{2}\left ( 5, \ell \right )= \left ( x- 5 \right )q_{2}\left ( x, \ell \right ),$$ with $q_{1}\left ( x, \ell \right ), q_{2}\left ( x, \ell \right )> 0$.

All largest real roots $x_{0}\approx 5- \frac{p\left ( 5, \ell \right )}{{p}'\left ( 5, \ell \right )}$ tend to $5$. So how can we compare them analytically?

I need your help. Thanks a real lot!

Dang Dang
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    Does it help that for example $p_1(x,\ell) = \frac{(x-5)x^{n+1}+c_\ell (x+1)\pm 6x }{x^2-2x-3}$? – Hagen von Eitzen Jun 18 '25 at 06:10
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    @HagenvonEitzen Ja, $p_{1}\left ( x, \ell \right )= \frac{\left ( x- 5 \right )x^{n+ 1}+ c_{\ell}\left ( x+ 1 \right )\pm 6x}{\left ( x- 3 \right )\left ( x+ 1 \right )}$ is important. For instance, we expect that $p\left ( \lambda_{1}, \ell \right )= f\left ( \lambda_{1} \right )- g_{1}\left ( \lambda_{1} \right )= f\left ( \lambda_{2} \right )- g_{2}\left ( \lambda_{2} \right )= p\left ( \lambda_{2}, \ell \right )$. If $f$ is an increasing function and $g_{1}\left ( \lambda_{1} \right )> g_{2}\left ( \lambda_{2} \right )$ even when $\ell= 1$, then $\lambda_{1}> \lambda_{2}$. Torschlusspanik. – Dang Dang Jun 18 '25 at 06:50

1 Answers1

2

Just a possible idea.

Each of the polynomials are in the form of $$P_n(x)=\sum_{k=0}^n a_n\,x^n$$ where the $a_n$ are known. Let $x=5-t$ and consider $$Q_n(t)=\sum_{k=0}^n b_n\,t^n$$ where the $b_n$ are known from the $a_n$.

Now, consider that these are $$Q_n(t)=\sum_{k=0}^n b_n\,t^n+O(t^{n+1})$$ Use power series reversion to get a new series whose coefficients are known using the explicit formula given for the $p^{\text{th}}$ term by Morse and Feshbach. You get a rational number.

Using this for the three cases (the approximation is converted to decimals) $$\left( \begin{array}{cc} \text{approximation} & \text{solution} \\ \color{red}{4.787}62036506034 & 4.78755983655505 \\ \color{red}{4.9079905}3759752 & 4.90799049843920 \\ \color{red}{4.95798394624}337 & 4.95798394624216 \\ \end{array} \right)$$

Edit

Another, probably simpler, solution could be to use the first iterate of any Newton-like method starting at $x_0=5$.

In terms of the coefficients $b_n$ the formulae for Newton, Halley and Householder methods are extremely simple (they respectively involve $(b_0,b_1), (b_0,b_1,b_2)$ and $(b_0,b_1,b_2,b_3)$).

For the three cases $$\left( \begin{array}{ccc} \text{Newton} & \text{Halley} & \text{Householder} &\text{solution} \\ 4.837773 & 4.795935 & 4.788629 & 4.787560 \\ 4.922468 & 4.909438 & 4.908096 & 4.907991 \\ 4.962138 & 4.958226 & 4.957994 & 4.957984 \\ \end{array} \right)$$

Dang Dang
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Claude Leibovici
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