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Consider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $-1$, e.g. $$ M_4= \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0\\ 0 & -1 & 2 & -1\\ 0 & 0 & -1 & 2\\ \end{pmatrix} $$

I've found out that the characteristic polynomials are $$ \begin{cases} P_1(x)=2-x\\ P_2(x)=(2-x)^2-1\\ P_n(x) = (2-x)P_{n-1}(x)-P_{n-2}(x) \end{cases} $$

Or with a variable change $$ \begin{cases} Q_1(y)=y\\ Q_2(y)=y^2-1\\ Q_n(y) = y Q_{n-1}(y)-Q_{n-2}(y) \end{cases} $$

Looking at the first 8 $P_n$

graph1

I see that all eigenvalues are real (as for any symmetric matrix), they are between 0 and 4.

  1. How can I prove that all eigenvalues are between 0 and 4?
  2. Are these polynomials known (have a name)?
  3. How can I prove that the polynomial are sandwitched between $$ \frac{1}{x}+\frac{1}{4-x}\quad\text{and}\quad -\frac{1}{x}-\frac{1}{4-x} $$

graph2

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  • $\begingroup$ I’m not with my laptop right now, so i’ll just try to comment a little bit on this beautifully written question. So, your question comes down to a question of considering a real sequence with a linear recursive relation, which is not hard to control $\endgroup$ – Paresseux Nguyen Dec 4 '20 at 16:34
  • $\begingroup$ for example, for your first question, you can see that if $x_0\le 0$ you can prove easily that $P_n(x_0)>= P_{n-1}(x_0)$ etc $\endgroup$ – Paresseux Nguyen Dec 4 '20 at 16:38
  • $\begingroup$ you can even deduce a compact expression for your Pn(x) which would be $a(x)u(x)^n+b(x)v(c)^n$ something except some critical cases $\endgroup$ – Paresseux Nguyen Dec 4 '20 at 16:43
  • $\begingroup$ @ParesseuxNguyen thank you. I proved $P_n(x_0)\ge P_{n-1}(x_0)$ for $x_0\le 0$. Then using the symmetry of the polynomial is easy to show that $|P_n(x_0)|\ge |P_{n-1}(x_0)|$ for $x_0\ge 4$. Thus the first point is proven. Now I look for your last suggestion. $\endgroup$ – Agyla Dec 4 '20 at 17:04
  • $\begingroup$ Best question ever from a <1k user? $\endgroup$ – Randall Dec 5 '20 at 1:08
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The Chebyshev polynomials of the second kind satisfy the recurrence relation $$ \begin{cases} U_0(y) = 1 \\ U_1(y)=x\\ U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y) \end{cases} $$ so that $Q_n(y) = U_n(y/2)$ and $P_n(x) = U_n(1-x/2)$.

The zeros of $U_n$ are $$ y_k = \cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n $$ in the range $(-1, 1)$, so that $$ x_k = 2 - 2\cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n $$ are the zeros of $P_n$ in the range $(0, 4)$.

Also for $|x| < 1$ $$ U_n(x) = \frac{\sin((n+1)\arccos(x))}{\sqrt{1-x^2}} $$ which implies $$ |U_n(x)| \le \frac{1}{\sqrt{1-x^2}} $$ and therefore $$ | P_n(x)| \le \frac{2}{\sqrt{x(4-x)}} \le \frac 1x + \frac{1}{4-x} $$ for $0 < x < 4$, the last estimate follows from the inequality between harmonic and geometric mean.

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From a very attenuated literature search, I see that the Lucas polynomials of the second kind obey the recursion $$ L_{k+1}(x) = x \ L_k(x) - L_{k-1}(x), \quad L_1(x) = 1 \ , L_2(x)=x$$ This matches the recursion for $Q_n,$ with an index shift of 1. The one reference I've seen (I don't have access to journal papers from home) states that this polynomial set obeys the relation

$$L_k(2 \cos(\theta) ) = 2 \cos( k \ \theta) $$

This functional relationship would explain the location of the zeros, and likely the envelope property as well. I suggest that the proposer start his research with 'Lucas polynomials of the second kind.'

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  • $\begingroup$ I don't know the Lukas polynomials, but apparently $Q_n(y) = U_n(y/2)$ where $U_n$ are the Chebyshev polynomials of the second kind, and those have all zeros in the interval $(-1, 1)$. $\endgroup$ – Martin R Dec 4 '20 at 17:56
  • $\begingroup$ @MartinR Well, that will solve the problem completely. I thought with the $\cos(k \theta)$ in my second formula the Chebyshev polynomials would appear. $\endgroup$ – skbmoore Dec 4 '20 at 17:59

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