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The Chebyshev polynomial $U_n(x)$ of the second kind is characterized by $U_n(\cos\theta)=\frac{\sin(n+1)\theta}{\sin(\theta)}$. It seems that $$\operatorname*{Res}_x \left( U_n(x)+tU_{n-1}(x),\sum_{k=0}^{n-1}U_k(x) \right) =(-1)^{\frac{n(n-1)}{2}} t^{\left\lfloor\frac{k}{2} \right\rfloor}2^{n(n-1)},$$ where Res denotes the resultant of two polynomials.

I do not know how to prove this "equality". Is it a known result? Similar results appeared in https://www.researchgate.net/publication/228924014_Resultants_and_discriminants_of_Chebyshev_and_related_polynomials, Theorem 2.

https://www.cambridge.org/core/services/aop-cambridge-core/content/view/D3FA1C0EBDAB83534C86375A0D711BF7/S0008439500017677a.pdf/the-resultant-of-chebyshev-polynomials.pdf

The Chebyshev polynomial $U_n(x)$ of the second kind is characterized by $$ U_n(\cos\theta)=\frac{\sin(n+1)\theta}{\sin(\theta)}. $$ It seems that $$\operatorname*{Res}_x \left( U_n(x)+tU_{n-1}(x),\sum_{k=0}^{n-1}U_k(x) \right) =(-1)^{\frac{n(n-1)}{2}} t^{\left\lfloor\frac{k}{2} \right\rfloor}2^{n(n-1)},$$ where Res denotes the resultant of two polynomials.

I do not know how to prove this "equality". Is it a known result? Similar results appeared in Dilcher and Stolarsky [1] theorem 2 and in Jacobs, Rayes and Trevisan [2].

References

[1] Karl Dilcher and Kenneth Stolarsky, "Resultants and discriminants of Chebyshev and related polynomials", Transactions of the American Mathematical Society, 357, pp. 965-981 (2004), MR2110427, Zbl 1067.12001.

[2] David P. Jacobs and Mohamed O. Rayes and Vilmar Trevisan, " The resultant of Chebyshev polynomials", Canadian Mathematical Bulletin 54, No. 2, 288-296 (2011), MR2884245, Zbl 1272.12006.

The Chebyshev polynomial $U_n(x)$ of the second kind is characterized by $U_n(cos\theta)=\frac{\sin(n+1)\theta}{\sin(\theta)}$. It seems that $$Res_x(U_n(x)+tU_{n-1}(x),\sum_{k=0}^{n-1}U_k(x))=(-1)^{\frac{n(n-1)}{2}}t^{\lfloor\frac{k}{2}\rfloor}2^{n(n-1)},$$ where Res denotes the resultant of two polynomials.

I do not know how to prove this "equality". Is it a known result? Similar results appeared in https://www.researchgate.net/publication/228924014_Resultants_and_discriminants_of_Chebyshev_and_related_polynomials, Theorem 2.

https://www.cambridge.org/core/services/aop-cambridge-core/content/view/D3FA1C0EBDAB83534C86375A0D711BF7/S0008439500017677a.pdf/the-resultant-of-chebyshev-polynomials.pdf

The Chebyshev polynomial $U_n(x)$ of the second kind is characterized by $U_n(\cos\theta)=\frac{\sin(n+1)\theta}{\sin(\theta)}$. It seems that $$\operatorname*{Res}_x \left( U_n(x)+tU_{n-1}(x),\sum_{k=0}^{n-1}U_k(x) \right) =(-1)^{\frac{n(n-1)}{2}} t^{\left\lfloor\frac{k}{2} \right\rfloor}2^{n(n-1)},$$ where Res denotes the resultant of two polynomials.

I do not know how to prove this "equality". Is it a known result? Similar results appeared in https://www.researchgate.net/publication/228924014_Resultants_and_discriminants_of_Chebyshev_and_related_polynomials, Theorem 2.

https://www.cambridge.org/core/services/aop-cambridge-core/content/view/D3FA1C0EBDAB83534C86375A0D711BF7/S0008439500017677a.pdf/the-resultant-of-chebyshev-polynomials.pdf