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It is known that the roots of Chebyshev polynomials of the second kind, denote it by $U_n(x)$, are in the interval $(-1,1)$. I have noticed that, by looking at the low values of $n$, the roots of $(1-x)U_n(x)+U_{n-1}(x)$ are in the interval $(-2,2)$. However, I don't have a clear idea how to start proving this, could anyone help me please?

*PS I have asked this question on StackExchange and set a bounty but still have not got any answer for it. see this link

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The key is to write $U_n(x)$ in terms of $x$ explicitly. Let $x = \cos t$, then $\sin t = \sqrt{1-x^2}$, where we fixed a branch of square root so that $\sqrt{-1} = i$; the value of $U_n$ is independent of the choice. Then $e^{it} = x + i \sqrt{1 - x^2}$. For $|x| \ge 1$, we can thus write $e^{it} = x - \sqrt{x^2 - 1}$.

For $|x| \ge 1$, let $y = x - \sqrt{x^2 -1}$, $A = 1-\sqrt{x^2 -1}$ and $B = (1-x)y + 1 $. Observe that $0 < y \le 1$. By analytic continuation, we can write

$$(1-x)U_n(x) + U_{n-1}(x) = \frac{ A y^{-(n+1)} -B y^n }{2 \sqrt{x^2 -1}}.$$

For $x > \sqrt{2}$, $A < 0$ and $B > 0$. So it's pretty clear the function has no zero there.

Similar argument can be made for $x \le -1$.

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    $\begingroup$ But probably the question was about complex roots? $\endgroup$
    – Fedor Petrov
    Commented Jan 1, 2016 at 9:01
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    $\begingroup$ @christian: these are not orthogonal polynomials and as the MSE link points out probably have a pair of complex roots. $\endgroup$
    – John Jiang
    Commented Jan 1, 2016 at 15:41
  • $\begingroup$ Actually my own wolfy calculation suggests all roots are real! $\endgroup$
    – John Jiang
    Commented Jan 1, 2016 at 16:45
  • $\begingroup$ One more data point: the number of real roots seems to be $\lfloor n \rfloor + 2$, for real positive $n$. $\endgroup$
    – John Jiang
    Commented Jan 1, 2016 at 21:36
  • $\begingroup$ @ChristianRemling: $U_n$'s are, but not $(1-x)U_n(x) + U_{n-1}(x)$, at least not with respect to the same weight function $\sqrt{1-x^2}$. I am looking into the possibility that they are orthogonal with respect to some other weights. Not only do they have to be orthogonal, but they need to arise from gram schmidt process based off $1, x, x^2, \ldots$. Am I missing something? $\endgroup$
    – John Jiang
    Commented Jan 1, 2016 at 22:41
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My previous answer shows that if all roots are real, then they must be contained with $[-1,\sqrt{2})$. To show all roots are real, use lemma 6.3.9 from the book Analytic Theory of Polynomials, which states that

If $P,Q$ are monic polynomials of degree $n,m$ respectively, with $n \ge 3$ and $m < n$, and $R = (z-\alpha)P - \beta Q$, with $\alpha, \beta \in \mathbb{R}$ and $\beta >0$. Then $P,Q$ have strictly interlacing zeros iff $P,R$ do. (Personally I don't see why $n = 2$ fails.)

We know that consecutive Chebyshev polynomials have strictly interlacing zeros by Proposition 6.3.10 (due to recurrence relation), so the conditions are satisfied. Note the condition on $m < n$ isn't really much more general, since interlacing implies $m = n-1$.

Real stability is a powerful tool. It also appears in the proof of Gurvits' lower bound for permanent and the proof of Kadison-Singer conjecture.

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