Tchebyshev polinomial T_k(x) = cos(k arccos(x)), k>=1.
Lanczos polinomial L_{k-1}(x) = \frac{1 - T_k(x)}{k^2 (1 - x)}, k>=2.
How to find the first derivative d/dx of L_{k - 1}(x) at the x = 1 ?
My result is equal to zero, but this is not correct.
The proper result is smth. like (k-1)(k+1)/6 .
I have tried a lot of approaches like Taylor series expansion, but all of them failed.
Any thoughts are welcome!
Lanczos polinomial L_{k-1}(x) = \frac{1 - T_k(x)}{k^2 (1 - x)}, k>=2.
How to find the first derivative d/dx of L_{k - 1}(x) at the x = 1 ?
My result is equal to zero, but this is not correct.
The proper result is smth. like (k-1)(k+1)/6 .
I have tried a lot of approaches like Taylor series expansion, but all of them failed.
Any thoughts are welcome!