# About a family of orthogonal polynoms satisfying a recurrence relation

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 $$\forall n \geq 1, \forall x \in [0,1], |P_n(x)| \leq |P_n(0)|$$.

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.
• 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. – Carlo Beenakker Jun 2 at 18:20
• 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 $\forall n \geq 1$ $P_n(0)=\displaystyle \sum_{j=1}^n \frac{1}{j}$ – mamiladi Jun 3 at 17:52
• after prooving the recurrence it's easyy to proove that $\forall x$ real $P_n(x)=\displaystyle \frac{-1}{2} \displaystyle \int_{0}^1 \frac{L_n(x)-L_n(t)}{x-t}dt$ – mamiladi Jun 5 at 23:32