# Roots of Palindromic Polynomial

I'm taking a signal processing class and we were studying windowing functions. One of the functions is the triangular function:

$$w(n) = \left\{\begin{array}{ll} \frac{2n-1}{L} & 1\le n\le L/2\\ 2-\frac{2n-1}{L} & L/2+1<n<L \end{array} \right.$$

where L is the number of points you want.

We were tasked to find L=64 points and then use the values obtained as coefficients of a polynomial and find the roots (with Matlab -- the professor isn't a sadist).

So, this polynomial will be palindromic.I found that there were no unique roots (I can post the values if needed). In fact for L=64, there are 31 roots. Investigating further, no matter what L I chose I got repeated roots.

This wasn't part of the assignment or anything but I'm just curious about it. My initial hunch would be because the polynomial is symmetric there would be repeated roots. The Binomial Theory is what came to mind. However I know that not all symmetric polynomials have repeated roots so I was curious how this can be explained.

I'm an engineer by background so I apologize if any of my nomenclature is incorrect or the question isn't worded properly.

EDIT:

So the coefficients of my polynomial are $w(n)$ where $n = 1,2,...,L$ i.e.

$w(1), w(2)....,w(L)$

$w(1)x^{L-1} + w(2)x^{L-2} +...+ w(L) = 0$

• I don't understand what polynomial we are trying to find the roots of. If you could perhaps make it more clear what the polynomial is... – Simply Beautiful Art Nov 1 '16 at 0:26
• Sorry for the confusion. w(n) where n = 1,2,3...L where L =64 in my case. So my coefficients are w(1), w(2),...,w(n). – Nabby Nov 1 '16 at 0:37
• I see. Engineer makes sense. In mathematics a symmetric polynomial is related to my research, but this is not. This is called a palindromic polynomial. – Matt Samuel Nov 1 '16 at 2:31
• The term "symmetric polynomial" is really quite entrenched, so if you don't change it you're going to get a lot of people like me who know about symmetric polynomials but not about this. Palindromic polynomials are far less popular though, so maybe unintentional traffic is what you want. In any case, it's up to you. – Matt Samuel Nov 1 '16 at 2:39
• You polynomial has following factorization $$\sum_{k=1}^L w(k) x^{L-k} = \frac1L (x+1)\left(1 + x + \cdots + x^{L-1}\right)^2$$ One way to understanding this is look at the coefficients of your polynomial. It has the shape of a sawtool. In certain sense, when you multiply two polynomials, the coefficients of the product is the convolution of the coefficients of individual polynomials. Above factorization corresponds to the fact when you convolute two box signal of width $1$, you get a sawtool with width $2$. – achille hui Nov 7 '16 at 8:05