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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 $P_n(x), 0<x<1$ but it's not very important for me, what interests me mostly is the maximum of $|P_n|$ over $[0,1]$.

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  • $\begingroup$ 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 $C^0_n$ ? $\endgroup$ – Sahil Kumar May 30 '19 at 6:42
  • $\begingroup$ yes ; binomial coefiicient and $C_n^0=1$ $\endgroup$ – mamiladi May 30 '19 at 6:45
  • $\begingroup$ $$ C_{n}^k= \displaystyle \frac{n!}{k!(n-k)!}$$ $\endgroup$ – mamiladi May 30 '19 at 6:48
  • $\begingroup$ 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 $. $\endgroup$ – user64494 May 30 '19 at 9:41
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    $\begingroup$ 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 $p_5(x)$ $\endgroup$ – mamiladi May 31 '19 at 4:45
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We can get the generating function for $P_n(x)$ and prove the property $P_n(x)=(-1)^{n+1}P_n(1-x)$ as follows.

Noticing that $\sum_p (-1)^{p+1} \frac{z^p}{p} = \log(1+z)$, we conclude that $$a_{n,k} = \sum_{p=1}^{n-k} \binom{n}{n-k-p} \frac{(-1)^{p+1}}{p} = [z^{n-k}]\, (1+z)^n\log(1+z).$$

Then, since $\binom{n+k}{n} = (-1)^k \binom{-n-1}{k}$, we get: \begin{split} P_n(x) &= [z^n]\, (1+xz)^{-n-1}(1+z)^n\log(1+z) \\ &=[z^n]\, \frac{\log(1+z)}{1+xz}\left(\frac{1+z}{1+xz}\right)^n. \end{split}

Applying Lagrange-Burmann formula, we further obtain: $$P_n(x) = [t^n]\ \frac{\log(1+\frac{t-1+\sqrt{(1-t)^2+4tx}}{2x})}{\sqrt{(1-t)^2+4tx}}.$$

It is now straightforward to verify that $P_n(1-x)=(-1)^{n+1}P_n(x)$.

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