多项式n阶导相等

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Rodrigues’ Formula取$g(x)=x(1-x),w(x)=x$得\[\frac{\frac{\rmd^n}{\rmd x^n}\left(x^{n+1} (1-x)^n\right)}{x}=-\frac{\frac{\rmd^{n+1}}{\rmd x^{n+1}}\left(x^n (1-x)^{n+1}\right)}{n+1}\]

1. Table[D[x (x (1 - x))^n, {x, n}]/x + D[x^n (1 - x)^(n + 1), {x, n + 1}]/(n + 1), {n, 4}] // Factor

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战巡

硬来呗，怕啥啊
\[x^{n+1}(1-x)^n=\sum_{k=0}^nC_n^k(-1)^{n-k}x^{2n-k+1}\]
\[\frac{d^n}{dx^n}(x^{n+1}(1-x)^n)=\sum_{k=0}^nC_n^k(-1)^{n-k}\frac{(2n-k+1)!}{(n-k+1)!}x^{n-k+1}\]
\[\frac{\frac{d^n}{dx^n}(x^{n+1}(1-x)^n)}{x}=\sum_{k=0}^nC_n^k(-1)^{n-k}\frac{(2n-k+1)!}{(n-k+1)!}x^{n-k}\]
\[=\sum_{k=0}^n(-1)^{n-k}\frac{n!(2n-k+1)!}{k!(n-k)!(n-k+1)!}x^{n-k}\]

另一边
\[x^n(1-x)^{n+1}=\sum_{k=0}^{n+1}C_{n+1}^k(-1)^{n+1-k}x^{2n-k+1}\]
\[\frac{d^{n+1}}{dx^{n+1}}(x^n(1-x)^{n+1})=\sum_{k=0}^{n}C_{n+1}^k(-1)^{n+1-k}\frac{(2n-k+1)!}{(n-k)!}x^{n-k}\]
\[=\sum_{k=0}^{n}(-1)^{n+1-k}\frac{(n+1)!(2n-k+1)!}{k!(n+1-k)!(n-k)!}x^{n-k}\]
\[-\frac{\frac{d^{n+1}}{dx^{n+1}}(x^n(1-x)^{n+1})}{n+1}=\sum_{k=0}^{n}(-1)^{n-k}\frac{(n)!(2n-k+1)!}{k!(n+1-k)!(n-k)!}x^{n-k}\]
于是明显相等