# Palindromic combinations of Chebyshev Polynomials share common roots?

Suppose that the real polynomial below $$p(x)=\sum_{k=0}^{2n}\alpha_{k}x^k$$ is a palindromic polynomial of even degree; that is, $p_{2n-k}=p_k$ for $0\leq k\leq 2n$ and $\alpha_0\neq 0$.

Is it true that the combination of Chebyshev polynomials (of the second kind) below $$\sum_{k=0}^{2n} \alpha_k U_{k-1}(x)\quad\text{and}\quad \sum_{k=0}^{2n}\alpha_k U_{k-2}(x)$$ share a common real root?

I have been able to show that these two polynomials share a common factor of degree $n$. Thus, I have been able to show that this is true when $n$ is odd. Investigations with graphing software seems to indicate that this is true in general. Somebody may be able to manipulate the common factor I've discovered to find a root.

If we extend the definition of $U_n(x)$ to integral indices (which we implicitly do above) and still demand that $$U_{n+1}=2x U_n(x)-U_{n-1}(x)$$ we discover that $$U_{-1}(x)\equiv 0\quad\text{and}\quad U_{-n}(x)=-U_{n-2}(x)$$ for nonnegative $n$. This gives a certain symmetry about $U_{-1}(x)$ that we will exploit by pairing with the palindromic combinations. We wish to pair $\alpha_n$ with $U_{-1}(x)$. Going through the Euclidean Algorithm and exploiting the recursive property, we find that

\begin{align} \gcd\left(\sum_{k=0}^{2n} \alpha_k U_{k-1}, \sum_{k=0}^{2n} \alpha_k U_{k-2}\right) & =\gcd\left(\sum_{k=0}^{2n} \alpha_k (2xU_{k-2}-U_{k-3}), \sum_{k=0}^{2n} \alpha_k U_{k-2}\right)\\ & =\gcd\left(\sum_{k=0}^{2n} \alpha_k U_{k-3}, \sum_{k=0}^{2n} \alpha_k U_{k-2}\right)\\ & =\vdots\\ & =\gcd\left(\sum_{k=0}^{2n} \alpha_k U_{k-(n+1)}, \sum_{k=0}^{2n} \alpha_k U_{k-n}\right) \end{align}

(ignoring some unit multiples along the way). Now by the palindromic condition and the symmetry about $U_{-1}$, we have that $$\sum_{k=0}^{2n} \alpha_k U_{k-(n+1)}(x)=0\,,$$ which in turn shows that the two polynomials above have a common factor of $$\sum_{k=0}^{2n} \alpha_k U_{k-n}(x)\,.$$ This polynomial has degree $n$ as one can verify. However, I've had difficulty in showing that this polynomial must have real roots. Evaluating at $0$ gives an alternating sum of the $\alpha_k$ which is negative only in some conditions.

## 1 Answer

From your formulas I get that \begin{align}\sum_{k=0}^{2n}a_kU_{k-n}&=a_0(U_n-U_{n-2})+a_1(U_{n-1}-U_{n-3})+...+a_{n-1}(U_1-U_{-1})+a_{n}U_0\end{align}

The first equality is from the symmetry $a_k=a_{2n-k}$ and $U_{-n}=-U_{n-2}$.

Since $U_n-U_{n-2}$ has degree $n$, $U_{n-1}-U_{n-2}$ has degree $n-1$, ..., and $U_1-U_{-1}$ has degree $1$, and $U_0$ degree zero (a non-zero constant), then these polynomials span all the polynomials of degree up to $n$ (degree exactly $n$ because $a_0\neq0$).

If $n$ is odd there is always a real root.

If $n$ is even, then the coefficients $a_0,...,a_n$ can be chosen such that the polynomial is one that doesn't have real roots, like $x^n+1$.

• Where is the $a_n U_{-2}(x)$ coming from? The coefficient $a_n$ is unpaired. – coffeecaracal Apr 17 '18 at 1:55
• That's right. Then that means that that sum evaluated at zero is equal to $a_n$. – user553213 Apr 17 '18 at 2:01
• The recursive formula is $U_{n+1}(x)=2xU_n - U_{n-1}$. You have applied the formula $U_{n+1}-U_{n-1}=2xU_n$ which is not equivalent. And if that were note a problem, this still does not procure a root. Yes the polynomial inside has degree $n-1$ but I want a root for the entire polynomial. I can evaluate the inside to zero. But $a_n$ will still be tagging along. – coffeecaracal Apr 17 '18 at 2:10
• @Robert Eliminated the use of the recursive formula. The argument is that the expression spam all polynomials of degree $n-1$. In particular, you can choose the $a_0,...,a_n$ such that you get a polynomial of degree $n-1$ without real roots. – user553213 Apr 17 '18 at 2:18
• If $n$ is even, $n-1$ is odd and certainly has real roots. – coffeecaracal Apr 17 '18 at 2:33