In the following proof, argument goes on based on considering $C_n$ to be nonzero then it finishes the proof for $C_n=0$ :
Also if we set $C_n=0$ in Eq. (6.10) then must $m=0,1,2,...,n- 2$ in Eq. (6.12). In either case the proof fails. Am I right? If so, how to finish the proof, that is showing $\Delta_n \ne 0$?
2026-03-25 01:30:06.1774402206
On the construction of orthogonal polynomials
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You exclude $m=n-1$ for no reason. The proof proceeds by showing that $(6.10)$ has solutions $C_0,\ldots,C_n$ with $C_n\neq 0$. This amounts to proving $\Delta_n\neq 0$. If we assume the contrary, the same system with $C_n=0$ (consisting of $n$ equations, not $n-1$) has nontrivial solutions $C_0,\ldots,C_{n-1}$. This corresponds to $(6.12)$ with the allowed values of $m$ exactly as specified there.