Zeroes of elementary polynomials without involving closed-form solutions

Question

Consider the following two polynomials, where $n$ is an integer: $$ p_n(x) = x^3-\frac1nx-\frac2n, \\ q_n(x) = x^2-\frac2n. $$ For any $n$, let $x_p=x_p(n)$ and $x_q=x_q(n)$ be the unique positive real numbers such that $p_n(x_p)=0$ and $q_n(x_q)=0$, respectively. These zeroes have closed-form expressions, and from these expressions, it is possible to verify that $x_p(n)>x_q(n)$ for all $n\ge1$.

My question: Is there an alternative way to prove this, without invoking either of these closed-form expressions? I confess that I am not an expert (or even a novice) on polynomials, algebra, etc.

(This pair of polynomial equations is the simplest in a sequence of [higher and higher degree] polynomials for which I conjecture a similar property holds. In trying to prove the general case, I realized I cannot even prove the simplest case without resorting to explicit solutions!)

Answer 1

Use the first iterate of a Newton-like method for $p_n(x)$ with $x_0= \sqrt{\frac{2}{n}}$ (the root of $q_n(x)$). This provides an explicit expression for $x_1$ (the first approximate solution of $p_n(x)$).

For simplicity, let $t= \sqrt{\frac{2}{n}}$ and use a Newton-like method of order $k$.

For $k>2$, the general result write

$$x_1^{(k)}=t+(2-t)\, t\,\frac{P_k(t) }{Q_k(t) }$$ where $P_k(t)$ and $Q_k(t)$ are polynomials (degree of $Q_k(t)$ being the degree of $P_k(t)$ plus $1$). In these polynomials, all coefficients are positive.

The exact solution being $$x=\sqrt{\frac{2}{3}}\, t\, \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{3 \sqrt{6}}{t}\right)\right)$$ consider the norm $$\Phi_k=\int_0^1 \Big(x-x_1^{(k)}\Big)^2\,dt$$ The results are $$\left( \begin{array}{cc} k & \log_{10}(\Phi_k) \\ 3 & -3.037 \\ 4 & -4.477 \\ 5 & -4.214 \\ 6 & -4.856 \\ 7 & -5.681 \\ 8 & -5.599 \\ 9 & -6.013 \\ 10 & -6.589 \\ 11 & -6.539 \\ 12 & -6.852 \\ \end{array} \right)$$

By increasing degrees, the coefficients of $P_{12}$ are

$$\{65536,2010624,11743680,21722240,15226400,4156972,367423\}$$ and the coefficients of $Q_{12}$ are $$\{24576,2253824,27070464,94284480,124466560,69031272,15966768,1254463\}$$

Now, the accuracy $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 1.52137971 & 1.52137971 \\ 2 & 1.16537304 & 1.16537304 \\ 3 & 1.00000000 & 1.00000000 \\ 4 & 0.89816095 & 0.89816095 \\ 5 & 0.82688697 & 0.82688697 \\ 6 & 0.77317017 & 0.77317017 \\ 7 & 0.73067452 & 0.73067453 \\ 8 & 0.69588439 & 0.69588439 \\ 9 & 0.66666667 & 0.66666667 \\ 10 & 0.64163965 & 0.64163965 \\ 20 & 0.50000019 & 0.50000000 \\ 30 & 0.43284437 & 0.43284358 \\ 40 & 0.39099845 & 0.39099661 \\ 50 & 0.36147202 & 0.36146873 \\ 60 & 0.33908159 & 0.33907653 \\ 70 & 0.32128150 & 0.32127442 \\ 80 & 0.30665013 & 0.30664084 \\ 90 & 0.29432034 & 0.29430871 \\ 100 & 0.28372793 & 0.28371387 \\ 200 & 0.22321463 & 0.22317624 \\ 300 & 0.19416582 & 0.19410899 \\ 400 & 0.17593783 & 0.17586968 \\ 500 & 0.16301227 & 0.16293889 \\ 600 & 0.15317216 & 0.15309848 \\ 700 & 0.14532363 & 0.14525359 \\ 800 & 0.13885367 & 0.13879040 \\ 900 & 0.13338732 & 0.13333333 \\ 1000 & 0.12868007 & 0.12863739 \\ \end{array} \right)$$

There is no problem going to higher orders if you wish.

You can use the same approach for the general case you mention : start from the root of $q_n(x)$ and generate the approximate root of $p_n(x)$ with the same procedure.