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let $P_0(x)=0$;$P_1(x)=1$ Let $\forall n $ integer $ \geq 2$, $\forall x$ real, $$P_n(x)=\displaystyle \sum_{k=0}^{n-1} C_{n+k}^n (-x)^k \alpha_{n,k}$$ and where $\forall k$ such that $0 \leq k \leq n-1 $ $$ \alpha_{n,k}= \displaystyle \sum_{p=1}^{n-k} \displaystyle C_{n}^{n-k-p} \frac{(-1)^{p+1}}{p}$$ So it's easy to check that $ \forall n $ integer $ \geq 1$, degree of $P_n$ is $n-1$.

I have found the following using Mapple: $ \forall 0 \leq n \leq 20,$

$(n+2).P_{n+2}(x)-(2.n+3).(1-2.x).P_{n+1}(x)+(n+1).P_{n}(x)=0$

I think i can proove that this recurrence relation is true $ \forall n $ integer with working on the coefficient of the polynom $P_n$.

What interess me is to know the weight $w$ such that $\int_{0}^1 P_n(x)w(x)x^idx=0$, ( PS:i'm not sure for the existence of $w$)

and i need to proove that $ \forall n \geq 1, \forall x \in [0,1], |P_n(x)| \leq |P_n(0)|$.

Thanks for your help

Ps: the familiy of legendre polynomial $L_n(x)=\displaystyle \frac{1}{n!}(x^n (1-x)^n)^{(n)}$ satisfy exactly the same recurrence relation but obviously with not with others initials conditions.

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    $\begingroup$ I don't understand the "PS": the recursion relation for the Legendre polynomials has a factor $x$ instead of the factor $1-2x$ in your recursion relation. $\endgroup$
    – Carlo Beenakker
    Jun 2, 2019 at 18:20
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    $\begingroup$ @CarloBeenakker probably OP means shifted Legendre polynomials. $\endgroup$
    – Martin Nicholson
    Jun 2, 2019 at 18:58
  • $\begingroup$ thanks @Nemo -- that clears up my confusion ; I wonder if generalized Legendre polynomials defined by the same three-term recursion but different starting values have ever been considered. $\endgroup$
    – Carlo Beenakker
    Jun 2, 2019 at 19:47
  • $\begingroup$ yes $ \forall n $ intger $ \forall x $ real,$ (n+2).L_{n+2}(x)−(2.n+3).(1−2.x).L_{n}+1(x)+(n+1).L_{n}(x)=0, $ $L_0=1,L_1=1−2x$. We have for the family polynom $(L_n) $ the following: $ ∀x \in [0,1] ,$ $∀n $ integer; $ |L_n(x)| \leq L_n(0)=1$. Numerically we have $P_n(x)=(-1)^{n+1}P_{n}(1-x)$ and $P_n$ have exactly $n-1$ zeors over $]0,1[$ and i can proove that $\forall n \geq 1$ $P_n(0)=\displaystyle \sum_{j=1}^n \frac{1}{j}$ $\endgroup$
    – mamiladi
    Jun 3, 2019 at 17:52
  • $\begingroup$ after prooving the recurrence it's easyy to proove that $\forall x $ real $P_n(x)=\displaystyle \frac{-1}{2} \displaystyle \int_{0}^1 \frac{L_n(x)-L_n(t)}{x-t}dt$ $\endgroup$
    – mamiladi
    Jun 5, 2019 at 23:32

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