3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $x^n+1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Where is the $a_n U_{-2}(x)$ coming from? The coefficient $a_n$ is unpaired. $\endgroup$ – coffeecaracal Apr 17 '18 at 1:55
  • $\begingroup$ That's right. Then that means that that sum evaluated at zero is equal to $a_n$. $\endgroup$ – user553213 Apr 17 '18 at 2:01
  • $\begingroup$ 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. $\endgroup$ – coffeecaracal Apr 17 '18 at 2:10
  • $\begingroup$ @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. $\endgroup$ – user553213 Apr 17 '18 at 2:18
  • $\begingroup$ If $n$ is even, $n-1$ is odd and certainly has real roots. $\endgroup$ – coffeecaracal Apr 17 '18 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.