Roots of the Chebyshev polynomials of the second kind.

Question

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?

For the second case, it seems that you can cut down the interval to (-1, $\sqrt{2}$) — ekkilop, Dec 21 '15 at 19:16
@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. — Math137, Dec 21 '15 at 19:42

marty cohen · Answer 1 · 2015-12-27T02:15:55.220

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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 $n$.

edited Dec 27 '15 at 02:15

answered Dec 20 '15 at 23:01

marty cohen

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I am working on your second case. – marty cohen Dec 20 '15 at 23:03
thank you very much for your effort – Math137 Dec 20 '15 at 23:52
The second case is not simplifying like the first. Frustrating. – marty cohen Dec 21 '15 at 04:00
do you recommend asking it in Mathoverflow? – Math137 Dec 21 '15 at 11:02
Sure can't hurt. – marty cohen Dec 22 '15 at 01:56
@Math137 I'd bet they won't accept your question at MO. It's for professional mathematicians... – draks ... Dec 23 '15 at 12:11
@draks... do you expect an easy answer for the second part? Which is basically the purpose of the question. – Math137 Dec 23 '15 at 12:14
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Note, $\cos(t/2)$ is not zero when $t=\pi/2$. Did you mean $t/2=\pi/2$? – Thomas Andrews Dec 26 '15 at 13:17
Yup. Thanks. I will change it. Upvoted. – marty cohen Dec 27 '15 at 00:07
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$t=\pi$ is not a root ($U_n(-1)+U_{n-1}(-1)=(-1)^n$). All $n$ roots are $t=\frac{k\pi}{n+\frac12}$ for $k=1,2,\ldots,n$. – Batominovski Dec 27 '15 at 01:28
Thanks for the correction. I will update my answer. Upvoted. – marty cohen Dec 27 '15 at 02:14

score 1 · Accepted Answer · edited Apr 13 '17 at 12:58

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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.

edited Apr 13 '17 at 12:58

Community

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answered Jan 01 '16 at 07:16

John Jiang

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score 0 · Answer 3 · answered Dec 27 '15 at 02:48

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.

Roots of the Chebyshev polynomials of the second kind.

3 Answers3