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It is known that the roots of the Chebyshev polynomials of the second kind, denote it by $U_n(x)$, are in the interval $(-1,1)$ and they are simple (of multiplicity one). I have noticed that the roots of $U_n{(x)}+U_{n-1}(x)$ (by looking at the law ranks of $n$) also lies in $(-1,1)$, I also noticed that for $(1-x)U_n{(x)}+U_{n-1}(x)$ the roots lie in $(-2,2)$. But I don't have any idea how to prove that in general, I wonder, first, if these claims are true? and how can I start proving them?

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  • $\begingroup$ For the second case, it seems that you can cut down the interval to (-1, $\sqrt{2}$) $\endgroup$
    – ekkilop
    Dec 21, 2015 at 19:16
  • $\begingroup$ @ekkilop I just wanted to be safe that is why I rounded to these numbers, otherwise you are right low rank calculations confirm what you are saying. $\endgroup$
    – Math137
    Dec 21, 2015 at 19:42

3 Answers 3

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For your first case:

Since $U_n(x) =\frac{\sin((n+1)t)}{\sin(t)} $ where $x = \cos(t) $,

$\begin{array}\\ U_n(x)+U_{n-1}(x) &=\frac{\sin((n+1)t)}{\sin(t)}+\frac{\sin(nt)}{\sin(t)}\\ &=\frac{\sin((n+1)t)+\sin(nt)}{\sin(t)}\\ &=\frac{2\sin((n+1/2)t)\cos(t/2)}{\sin(t)}\\ \end{array} $

and this is zero when $t(n+1/2) =k\pi $ for some integer $k$, or $t =\frac{k\pi}{n+1/2} $ for $1 \le k \le n$.

This gives $n$ real roots, and that is all since $U_n(x)+U_{n-1}(x)$ is of degree $n$.

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    $\begingroup$ I am working on your second case. $\endgroup$ Dec 20, 2015 at 23:03
  • $\begingroup$ thank you very much for your effort $\endgroup$
    – Math137
    Dec 20, 2015 at 23:52
  • $\begingroup$ The second case is not simplifying like the first. Frustrating. $\endgroup$ Dec 21, 2015 at 4:00
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    $\begingroup$ Note, $\cos(t/2)$ is not zero when $t=\pi/2$. Did you mean $t/2=\pi/2$? $\endgroup$ Dec 26, 2015 at 13:17
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    $\begingroup$ $t=\pi$ is not a root ($U_n(-1)+U_{n-1}(-1)=(-1)^n$). All $n$ roots are $t=\frac{k\pi}{n+\frac12}$ for $k=1,2,\ldots,n$. $\endgroup$ Dec 27, 2015 at 1:28
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See my answer here by writing chebyshev polynomial explicitly in terms of $x = \cos t$. Update: I also show there that the roots are all real.

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For your second case:

Since $U_n(x) =\frac{\sin((n+1)t)}{\sin(t)} $ where $x = \cos(t) $,

$\begin{array}\\ (x-1)U_n(x)+U_{n-1}(x) &=(\cos(t)-1)\frac{\sin((n+1)t)}{\sin(t)}+\frac{\sin(nt)}{\sin(t)}\\ &=\frac{(\cos(t)-1)\sin((n+1)t)+\sin(nt)}{\sin(t)}\\ \end{array} $

Putting this into Wolfy, this has this many roots from $0$ to $\pi$ for these values of $n$:

$2: 1; 3: 2; 4: 3; 5:4 $

This seems to show that this has $n-1$ real roots. Since this is a polynomial of degree $n+1$, there should be one complex pair of roots. This is confirmed by Wolfy directly by entering "$(x-1)ChebyshevU[n, x]+ChebyshevU[n-1, x]=0 $" for various values of $n$.

In all the cases I have tries, all $n-1$ real roots are between $-1$ and $1$.

Other than this, I haven't been able to make any real progress proving the result.

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