A proof that the constant-coefficient in $U_{n}$ is $0$ when $n$ is odd.

Question

Let $(U_{n}(x))_{n≥0}$ be the Chebyshev polynomials of the second kind (https://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html#:~:text=A%20modified%20set%20of%20Chebyshev,of%20the%20Gegenbauer%20polynomial%20with%20.). We know that the constant coefficient in $U_{n}$ is $(-1)^{n/2}$ when $n$ is even. It seems from the examples of the above link that the constant-coefficient in $U_{n}$ is $0$ when $n$ is odd. But I not able to find proof for this.

Answer 1

I believe this follows from the defining formula $$ U_n( \cos \theta ) \sin(\theta) = \sin ( (n + 1)\theta). $$ If $x = \cos \theta$ and $n$ is odd, then $-x = \cos (\pi + \theta)$ so that $$ U_n( -x) = \frac{ \sin( (n+1)(\theta + \pi)) }{\sin(\theta + \pi) } = \frac{\sin((n+1)\theta)}{-\sin(\theta)} = -U_n(x). $$ (when $\sin(\theta) \neq 0$; if $\sin(\theta) = 0$, then $U_n(x) = 0$). That is, $U_n$ is an odd function of $x$ when $n$ is odd. Since $U_n$ is a polynomial, this implies that the constant term is zero.

EDIT: There is a much simpler way of proving this. Since $\cos(\pi/2) = 0$, then $$ U_n(0) \sin(\pi/2) = \sin( (n+1)\pi/2), $$ i.e. $$ U_n(0) \cdot 1 = \sin\bigg( \frac{(n+1)\pi}{2} \bigg). $$ If $n$ is odd, then $n+ 1$ is even so that $$U_n(0) = \sin k\pi$$ for $n$ odd. Therefore $U_n(0) = 0$ and so the constant term of $U_n$ is $0$ when $n$ is odd.