Let $n\in \mathbb{N}^+$ be a positive integer. Let $L_n\colon \mathbb{R}\to \mathbb{R}$ be the $n$'th order Legendre polynomial. Let $J_n^{(\alpha,\beta)}\colon \mathbb{R} \to \mathbb{R}$ be the $n$'th order Jacobi polynomial with weights $\alpha$ and $\beta$.
How can I show that
$$2J_n^{(0,-1)}(x) = L_n(x) + L_{n-1}(x)$$ and that
$$2J_n^{(-1,0)}(x) = L_n(x) - L_{n-1}(x).$$
I have tried to prove this mainly by starting with the Rodriguez formulas and attempting to show that they are equal. Another attempt was to show that the family of polynomials $\{L_n+L_{n-1}\}_{n=1}^\infty$ is orthogonal with respect to the Jacobi differential equation's weighted inner product.
I can flesh these attempts out if anyone desires, but they seem to be dead ends.
Any help will be appreciated.
With your unusual notation (normally the Legendre polynomals are denoted with $P_n$ and $P_n^{(\alpha, \beta)}$ are the Jacobi polynomials), you have (see DLMF, equation 19.7.9 or Abramowitz/Stegun 22.5.35) $$L_n(x) = J_n^{(0,0)}(x)$$ Use the recurrence relations for Jacobi polynomials (DLMF, equation 18.9.5 or A/S 22.7.18 and 22.7.19) $$(2n+\alpha+\beta)J_n^{(\alpha-1, \beta)}(x)= (n+\alpha+\beta)J_n^{(\alpha, \beta)}(x)-(n+\beta)J_{n-1}^{(\alpha, \beta)}(x) $$ $$(2n+\alpha+\beta)J_n^{(\alpha, \beta-1)}(x)= (n+\alpha+\beta)J_n^{(\alpha, \beta)}(x)+(n+\alpha)J_{n-1}^{(\alpha, \beta)}(x) $$ For $\alpha=\beta=0,\;n\ne 0\;$ they reduce to $$2J_n^{(-1, 0)}(x)= J_n^{(0, 0)}(x)-J_{n-1}^{(0, 0)}(x) $$ $$2J_n^{(0, -1)}(x)= J_n^{(0, 0)}(x)+J_{n-1}^{(0, 0)}(x) $$ Now compute $$L_n(x)+L_{n-1}(x)=J_n^{(0, 0)}(x)+J_{n-1}^{(0, 0)}(x) = 2J_n^{(0, -1)}(x)$$ $$L_n(x)-L_{n-1}(x)=J_n^{(0, 0)}(x)-J_{n-1}^{(0, 0)}(x) = 2J_n^{(-1, 0)}(x)$$