# Upper bound over $[0,1]$ for strange family of polynomials

Let $$n$$ integer $$\geq 2,$$ $$x$$ real, and $$P_n(x)=\displaystyle \sum_{k=0}^{n-1} C_{n+k}^n (-x)^k \alpha_{n,k}$$ and where $$\forall k$$ such that $$0 \leq k \leq n-1$$ $$\alpha_{n,k}= \displaystyle \sum_{p=1}^{n-k} \displaystyle C_{n}^{n-k-p} \frac{(-1)^{p+1}}{p}$$.

Numerically I have found the following: $$\forall 2 \leq n \leq 10$$ ,

$$\max_{[0,1]} |P_n(x)|=|P_n(0)|=\sum_{j=1}^{n} \frac{1}{j}.$$

Is there an article about $$P_n$$ or similar polynomials?

I need a proof that it's true $$\forall n \geq 2$$.

Remarque $$P_n(0)=\displaystyle \sum_{p=1}^n \displaystyle \frac{(-1)^{p+1}}{p} C_{n}^{n-p}= \displaystyle \sum_{p=1}^n \displaystyle \frac{(-1)^{p+1}}{p} C_{n}^{p}=\displaystyle \int_{0}^1 \frac{1-(1-x)^n}{x}dx$$ $$= \displaystyle \int_{0}^1 \frac{1-y^n}{1-y}dy=\displaystyle \int_{0}^1 \sum_{p=0}^{n-1}y^{p}dy=\sum_{j=1}^{n} \frac{1}{j}$$

I have noticed numerically too that $$P_n(x)=(-1)^{n+1}P_{n}(1-x)$$ and $$P_n$$ have exactly $$n-1$$ zeros over $$]0,1[$$. All this properties are very similar to those of Legendre polynomial: $$L_n(x)=\displaystyle \frac{1}{n!}(x^n(1-x)^n)^{(n)}$$.

I suppose that the family of $$P_n$$ are orthogonal but I don't know to what weight, and maybe satisfy a recurrence relation...

If anyone suggests an idea for (the proof that I suppose true) thanks for his help.

I'm looking too for an integral representation of type $$\int_{0}^1 f_x^n(t).g_x(t)dt$$ for $$P_n(x), 0 but it's not very important for me, what interests me mostly is the maximum of $$|P_n|$$ over $$[0,1]$$.

• In definition of $a_{n,k}$ you have used $C^{n-k-p}_n$ which i don't understand as for example at $p=n-k$ becomes $C^0_n$ ? – Sahil Kumar May 30 at 6:42
• yes ; binomial coefiicient and $C_n^0=1$ – mamiladi May 30 at 6:45
• $$C_{n}^k= \displaystyle \frac{n!}{k!(n-k)!}$$ – mamiladi May 30 at 6:48
• Your statement "$P_n$ have exactly $n−1$ zeros over $]0,1[$" does not correspond to reality in the case $n=5$ up to Maple code n:=5:fsolve(sum(binomial(n+k, n)*(-1)^k*x^k*(sum(binomial(n, n-k-p)*(-1)^(p+1)/p, p = 1 .. n-k)), k = 0 .. n-1)) which produces $0.5066328932e-1, .2452340931$. – user64494 May 30 at 9:41
• the dominant coefficient of $p_n$ is the term on $x^{n-1}$, it's equal to $binomial(2n-1,n)*sum(binomial(n, 1-p)*\frac{(-1)^{p+1}}{p}, p = 1 .. 1)=binomial(2n-1,n),$ and in case n=5, you must have binomial(2*5-1,5)=126 as a coefficient of $x^4$and not $630$, so please check the formulea write on mapple (correct it) , and you have to find $p[5](x) := 137/60-(77/2)*x+(329/2)*x^2-252*x^3+126*x^4$ as expression for $p_5(x)$ – mamiladi May 31 at 4:45