Recursive formula from given explicit formula for normalized Chebyshev polynomials

Question

The Chebyshev polynomials $(T_k)_{k \in \mathbb{N}_0}$ are defined recursively by $$ T_0(x)=1 , \ \ T_1(x)=x, \ \ T_{k+1}(x)=2x\,T_k(x)-T_{k-1}(x) \ . $$ With this one can find the explicit formulas \begin{align} & T_{2l}(x) = l \sum\limits_{j=0}^{l} (-1)^{l-j} \frac{(l+j-1)!}{(l-j)!(2j)!} \, (2x)^{2j}\\ & T_{2l+1}(x) = \frac{2l+1}{2} \sum\limits_{j=0}^l (-1)^{l-j} \frac{(l+j)!}{(l-j)!(2j+1)!} \, (2x)^{2j+1} \end{align} for all $l \in \mathbb{N}$. However I needed to normalize these polynomials into \begin{align*} & \widetilde{T}_{2l}(x) = l \sum\limits_{j=0}^{l} (-1)^{l-j} \frac{(l+j-1)!}{(l-j)!(2j)!} \left( x^{2j} - \frac{1}{j+1} {2j \choose j} \right)\\ & \widetilde{T}_{2l+1}(x) = \frac{2l+1}{2} \sum\limits_{j=0}^l (-1)^{l-j} \frac{(l+j)!}{(l-j)!(2j+1)!} \, x^{2j+1} \end{align*} for $l \in \mathbb{N}$ and $\widetilde{T}_0(x)=0, \, \widetilde{T}_1(x) = \frac{x}{2}$. I'm pretty sure any link to the recursion formular of the (non-normalized) Chebyshev polynomials is gone. Can anyone help me find out if there is a new recursion formula for $\widetilde{T}_{k+1}$, depending on $\widetilde{T}_0,...,\widetilde{T}_{k}$? I think it might have some interesting applications for a certain graph theoretical setting.

Any help is much appreciated.

Answer 1

If I read your formulas correctly you define $\tilde T_{2l}(x)=T_{2l}(x/2)-C_l$ and $\tilde T_{2l+1}(x)=T_{2l+1}(x/2)$, where $$C_l=l\sum_{j=0}^l (-1)^{l-j}\frac{(l+j-1)!}{(l-j)!(2j)!}\frac 1{j+1}\binom{2j}j.$$ This can be rewritten $$C_l=(-1)^l l!\sum_{j=0}^l\frac{(-l)_j(l)_j}{j!(2)_j},$$ where $(a)_j=a(a+1)\dotsm(a+j-1)$. By the Chu-Vandermonde summation formula, $$C_l=(-1)^ll!\frac{(2-l)_l}{(2)_l}=\begin{cases}1, & l=0,\\ -1/2, & l=1,\\ 0, & l\geq 2.\end{cases}$$ So your polynomials are given by $$\tilde T_k(x)=T_k(x/2)-\delta_{0,k}-\frac 12\delta_{2,k}.$$ They satisfy the recursion $$\tilde T_{k+1}(x)=x\tilde T_k(x)-\tilde T_{k-1}(x),\qquad k\geq 4.$$