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For $n\geq 1$, given a polynomial \begin{equation*} \begin{aligned} f(x)=&\frac{2+(x+3)\sqrt{-x}}{2(x+4)}(\sqrt{-x})^n+\frac{2-(x+3)\sqrt{-x}}{2(x+4)}(-\sqrt{-x})^n \\ &+\frac{x+2+\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x+\sqrt{x(x+4)}}{2} \right )^n+\frac{x+2-\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x-\sqrt{x(x+4)}}{2} \right )^n. \end{aligned} \end{equation*} Using Mathematic $12.3$, when $n$ is large enough, we give the distribution of the roots of $f(x)$ in the complex plane as follows

enter image description here

In this figure, we can see that the closure of the real roots of $f(x)$ may be $\left [ -4,0 \right ]$.

So we have the following question

Question: all roots of $f(x)$ are real? It seems yes! But we have no way of proving it.

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    $\begingroup$ Is it also true that $f_n$ for small $n$ have roots only in the real segment $[0,4]$? Also the explicit (polynomial - type) form of $f_n$ for small $n$ could be helpful $\endgroup$ Commented Feb 17, 2023 at 6:23
  • $\begingroup$ It seems to me that the change of variable $x=2\cos(\theta)-2$ transforms your polynomial into a simple trigonometric expression, but I am too lazy to work it out. $\endgroup$ Commented Feb 18, 2023 at 9:53
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    $\begingroup$ Are the roots of $f_n$ and $f_{n-1}$ intertwined? By the way, no need to compute roots numerically, using Sturm's Theorem you can check that all roots are in $[0,4]$ using only algebra. $\endgroup$ Commented Feb 18, 2023 at 10:15
  • $\begingroup$ Why is there a denominator of $2(x+4)$ in every term of $f$? It would seem polite to remove that common factor. $\endgroup$
    – user44143
    Commented Feb 18, 2023 at 21:20
  • $\begingroup$ @HenriCohen, On the first glance this substitution gives $f_{4k} = 2^{2k}(\cos\theta - 1)^{2k}(1 + 2\sum_{m = 1}^k (-1)^m\cos(m\theta))^2$. Some similar relation also seem to hold for $ 8k + 2$ and $8k + 6$. However I struggle to guess the pattern for odd numbers $\endgroup$ Commented Feb 18, 2023 at 21:46

6 Answers 6

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Assume that we have $x = - (t + 1/t)^2 = -t^2 - 1/t^2 - 2$ for some $t \in \mathbb{C}$. Then $$ \sqrt{-x} = t + \frac{1}{t} = \frac{t^2 + 1}{t},\quad x + 4 = -\left(t - \frac{1}{t}\right)^2, \quad \sqrt{x(x + 4)} = t^2 - \frac{1}{t^2}, \\ x + \sqrt{x(x + 4)} = -2\left(1 + \frac{1}{t^2}\right) = -2\frac{1 + t^2}{t^2}, \quad x - \sqrt{x(x + 4)} = -2(1 + t^2) $$ Consequently, we get $$ f_n(x)= \tilde f_n(t) = \frac{(1 + t^2)^n(t^{n + 2} - 1)^2}{t^{2n}(t^2 - 1)^2},\quad\text{if } n\text{ is even}, \\ f_n(x) = \tilde f_n(t) = -\frac{(1 + t^2)^n(1 - t^{n -1})(1 - t^{n+5})}{t^{2n}(t^2 - 1)^2},\quad\text{if } n\text{ is odd}. $$ Hence the only roots of $\tilde f_n$ are the suitable roots of unity and the roots of $f_n$ are all in the real segment $[-4, 0]$ and their closure is the whole segment for large $n$.

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  • $\begingroup$ Thanks, but the interval of the root of $f_n$ seems to be $\left [ -4,0 \right ]$? $\endgroup$
    – Connor
    Commented Feb 20, 2023 at 2:20
  • $\begingroup$ Oh, you right! That is a misprint, fixed now $\endgroup$ Commented Feb 20, 2023 at 6:30
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    $\begingroup$ And I have a question for this answer, how to guarantee $\sqrt{x(x+4)}=t^2-\frac{1}{t^2}$ rather than $\sqrt{x(x+4)}=-t^2+\frac{1}{t^2}$, and so on. $\endgroup$
    – Connor
    Commented Feb 20, 2023 at 6:58
  • $\begingroup$ Formula is symmetric with respect to $\sqrt{x(x + 4)}$ and $-\sqrt{x(x + 4)}$, so it does not matter how to extract the square root. As for $\sqrt{-x}$, for even $n$ there is $\sqrt{-x}^{n} = (-x)^{n/2}$ and for odd $n$ there is $\sqrt{-x}^{n + 1} = (-x)^{(n + 1)/2}$, hence one can actually make the substitution without computing $\sqrt{-x}$ $\endgroup$ Commented Feb 20, 2023 at 7:25
  • $\begingroup$ That's nice! Thanks for your careful answer. :) $\endgroup$
    – Connor
    Commented Feb 20, 2023 at 7:32
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EDITED. First let us define $$g_n(x) := \sum_{k=0}^{\lfloor n/2\rfloor} {n-k\choose k} x^{n-k}.$$ Then empirically, $$\eqalign{f_{2n}(x) &= g_n(x)^2\cr f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_{n+2}(x)\cr}$$ where $h_n(x)$ is a Lucas polynomial. I think it should be possible to prove these formulas using the recurrence $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in Ira Gessel's comment below, the desired properties of the roots of $f_{2n}(x)$ and $f_{4n+1}(x)$ follow from the properties of the Chebyshev polynomials. Presumably the roots of Lucas polynomials are also well understood, but I'm not so familiar with them.

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    $\begingroup$ So $f_{2n}(x) = (-x)^n U_n(\sqrt{-x}/2)^2$, where $U_n$ is the Chebyshev polynomial of the second kind. Most likely there is some sort of expression for $f_{2n+1}(x)$ in terms of Chebyshev polynomials but it will not be as simple. $\endgroup$
    – Ira Gessel
    Commented Feb 19, 2023 at 0:22
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    $\begingroup$ I also just discovered that the connection between the Lucas polynomials and $T_n$ is given in Section 6.4 of On the characteristic polynomial of Cartan matrices and Chebyshev polynomials by Pantelis A. Damianou. $\endgroup$ Commented Feb 19, 2023 at 4:18
  • $\begingroup$ Thanks for Timothy Chow and Ira Gessel's answers, amazing construction! At first, we tried to use the Sturm sequence to prove the real zeros, but it seem to be impossible :( $\endgroup$
    – Connor
    Commented Feb 19, 2023 at 8:26
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    $\begingroup$ The OEIS page for the Lucas polynomials uses a different normalization. Here is a corrected version of my comment on them. The polynomials $h_n(x)$ in Tim's answer have the generating function $\sum_{n=0}^\infty h_n(x) y^n = (2-xy)/(1-xy-xy^2)$. They are given by $h_0(x)=2$ and $$h_n(x) = \sum_{k=0}^{\left\lfloor n/2\right\rfloor}\frac{n}{n-k}\binom{n-k}{k}x^{n-k}$$ for $n>0$. They are related to the Chebyshev polynomials of the first kind $T_n(x)$ by $$h_n(x) = 2(-\sqrt{-x})^{n} T_n(\sqrt{-x}/2)$$ so their roots can be determined from the roots of the Chebyshev polynomials. $\endgroup$
    – Ira Gessel
    Commented Feb 19, 2023 at 15:47
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    $\begingroup$ In the process of deleting his incorrect comment, Ira Gessel also deleted the math.SE link, Algorithm for computing Hadamard product of two rational generating functions. This is another way to prove the correctness of the empirically derived formulas: take Hadamard products of (suitably rescaled) generating functions for the Chebyshev polynomials, and confirm that they give you the generating function of the $f$ polynomials. $\endgroup$ Commented Feb 19, 2023 at 18:20
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(This is a comment, not an answer.)

If $f_n(x)$ is your polynomial, starting with $f_0(x)=1$, then $$ \sum_{n=0}^\infty f_n(x) y^n = \frac{1-xy+x^2y^2+x^2y^3}{(1+xy^2)(1-xy-xy^2)} = 1 + \frac{x^2y^2(1+y)^2}{(1+xy^2)(1-xy-xy^2)}. $$ Also, I noticed that $f_n(x)-x f_{n-1}(x)-x f_{n-2}(x)$ only has one or two terms, so a recurrence is possible. That there are no positive real zeros follows from the fact that there are no negative coefficients. The rest of your question is another matter.

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    $\begingroup$ Thanks for your comment! I'm sorry that I forgot to indicate the $n \geq 1$. Based on your comment, then by the generating function of $f_n(x)$, we get the recurrence relation of $f_n(x)$, that is $f_n(x)=xf_{n-1}(x)+x^2f_{n-3}+x^2f_{n-4}$ (for $n \geq 5$). $\endgroup$
    – Connor
    Commented Feb 16, 2023 at 12:55
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    $\begingroup$ But in this recurrence relation, we can't seem to be sure that $f_n(x)$ is all real roots. $\endgroup$
    – Connor
    Commented Feb 16, 2023 at 12:59
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Here is a proof, using Maple calculations, of Tim Chow's empirical observations. We use Hadamard products of power series. The Hadamard product (with respect to the variable $y$) is defined by $$ \sum_{n=0}^\infty a_n y^n *\sum_{n=0}^\infty b_n y^n= \sum_{n=0}^\infty a_n b_n y^n. $$ The Hadamard product of two rational power series is rational, and I did the following computations with a Maple program I wrote using the method described here .

For any power series $A(y) = \sum_{n=0}^\infty a_n y^n$ and integers $m$ and $i$, let \begin{equation*} A_{m,i}(y) = \sum_{n=0}^\infty a_{mn+i}y^i, \end{equation*} where we take $a_n=0$ if $n<0$. Following Brendan McKay, we define the generating function. $$F=F(y) = \sum_{n=0}^\infty f_n(x) y^n = 1+\frac{x^{2} y^{2} \left(1+y \right)^{2}}{\left(1+x y^{2}\right) \left(1-xy-x y^{2}\right)} $$ We also define generating functions for Timothy Chow's polynomials $g_n(x)$ and $h_n(x)$: \begin{gather*} G=\sum_{n=0}^\infty g_n(x) y^n = \frac{1}{1-xy-xy^2}\\ H=\sum_{n=0}^\infty h_n(x) y^n = \frac{2-xy}{1-xy-xy^2}. \end{gather*} Then we want to prove \begin{gather} F_{2,0}=G*G\tag{1}\\ F_{4,1}=G_{2,2}*G_{2,-1}\tag{2}\\ F_{4,3}=G_{2,0}*G_{1,1}*H_{1,2}\tag{3} \end{gather} Multisections, can be computed by Hadamard products (or in other ways). For example, $F(y)*y/(1-y^4) = yF_{4,1}(y^4)$. We find that \begin{gather*} F_{2,0} =\frac{1-x y }{\left(1+x y \right) \left(1-2xy -x^{2} y +x^{2} y^{2}\right)}\\ F_{4,1}=\frac{x^3y(1+3x+x^2 -x^2y)}{(1-x^2y)(1-(2x^2+4x^3+x^4) y +x^{4} y^{2})}\\ F_{4,3}=\frac{x^2(2+x -\left(3x^2+4x^3+x^4\right) y +x^{4} y^{2})}{(1-x^2y)(1-(2x^2+4x^3+x^4) y +x^{4} y^{2})}\\ G_{1,1}=\frac{x(1+y)}{1-xy-xy^2}\\ G_{2,-1}=\frac{xy}{1-2xy-x^2y+x^2y^2}\\ G_{2,0}=\frac{1-xy}{1-2xy-x^2y+x^2y^2}\\ G_{2,2}=\frac{x(1+x-xy)}{1-2xy-x^2y+x^2y^2}\\ H_{1,2}=\frac{x(2+x+xy)}{1-xy-xy^2} \end{gather*} We can then verify $(1)$$(3)$ directly.

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(another comment, not an answer.)

Experimentally, with the following code

from sympy import *
var('x')
var('n', integer = True)

f = ( (2+(x+3) * sqrt(-x))/(2*(x+4)) *
      sqrt(-x)**n + ((2-(x+3) * sqrt(-x)) / (2*(x+4))) * (-sqrt(-x))**n +
      + (x+2+sqrt(x*(x+4)))/(2*(x+4)) *
        ( (x+sqrt(x*(x+4)))/2 )**n +
        (x+2-sqrt(x*(x+4)))/(2*(x+4)) *
        ( (x-sqrt(x*(x+4)))/2 )**n
        )
pprint(f)
for i in range(10):
   eq = simplify(f.subs(n,i))
   print ('========== n = ', i)
   pprint(eq)
   print ('have solutions')
   sols = solve(eq)
   pprint (sols)
   pprint ('approx. = ')
   pprint ([s.evalf() for s in sols])

we see the following: $$ \begin{array}{l} f_{0} = 1\\ f_{1} = 0\\ f_{2} = x^{2}\\ f_{3} = x^{2} \left(x + 2\right)\\ f_{4} = x^{2} \left(x^{2} + 2 x + 1\right)\\ f_{5} = x^{3} \left(x^{2} + 3 x + 1\right)\\ f_{6} = x^{4} \left(x^{2} + 4 x + 4\right)\\ f_{7} = x^{4} \left(x^{3} + 5 x^{2} + 7 x + 3\right)\\ f_{8} = x^{4} \left(x^{4} + 6 x^{3} + 11 x^{2} + 6 x + 1\right)\\ f_{9} = x^{5} \left(x^{4} + 7 x^{3} + 16 x^{2} + 13 x + 2\right)\\ \end{array} $$

enter image description here

And it seems that already $f_9$ have some imaginary roots, albeit very small. Do these imaginary roots really converge to 0 when $n\to \infty$ or they stay on the same magnitude ?

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    $\begingroup$ This looks like a calculation error of some type. $P(x) = x^4 + 7x^3 + 16x^2 + 13x + 2$ has four real zeroes $\endgroup$ Commented Feb 18, 2023 at 9:16
  • $\begingroup$ hum... You are right. Softwares are not intelligent enough to simplify stuff. The roots of $f_9$ are -$ \frac{5}{3} + \frac{2 \sqrt{7} \sin{\left(- \frac{\operatorname{atan}{\left(3 \sqrt{3} \right)}}{3} + \frac{\pi}{6} \right)}}{3}; - \frac{5}{3} + \frac{2 \sqrt{7} \sin{\left(\frac{\operatorname{atan}{\left(3 \sqrt{3} \right)}}{3} + \frac{\pi}{6} \right)}}{3}; - \frac{5}{3} - \frac{2 \sqrt{7} \cos{\left(\frac{\operatorname{atan}{\left(3 \sqrt{3} \right)}}{3} \right)}}{3}$ $\endgroup$
    – kerzol
    Commented Feb 18, 2023 at 11:47
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    $\begingroup$ If you compute more terms using the recurrence then the polynomials $f_n(x)$ seem to factor nicely; see A054142 and A145324 for example. $\endgroup$ Commented Feb 18, 2023 at 13:55
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Take the first two terms in your sum, and multiply by $x+4$, then you get a sequence of polynomials which satisfy $P_n = -x P_{n-2}$, and if you do the same for the last two terms, you get a sequence determined by $Q_n = x Q_{n-1}+x Q_{n-2}$. Now, $Q_n$ interlaces the roots of $Q_{n+1}$ so these are easy to show are real-rooted. Same goes for $P_n$ and $P_{n+2}$.

So, perhaps one can combine these somehow in order to show that the sum is real-rooted.

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    $\begingroup$ Thank you for your answer. We have also tried this method, but the sign pattern of $P_n$ is $--++\cdots $, and the existing results cannot seem to get $P_n+Q_n \in RZ$ $\endgroup$
    – Connor
    Commented Feb 19, 2023 at 10:51

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