0
$\begingroup$

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 $[0,1]$.

$\endgroup$
9
  • $\begingroup$ 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 $C^0_n$ ? $\endgroup$ May 30, 2019 at 6:42
  • $\begingroup$ yes ; binomial coefiicient and $C_n^0=1$ $\endgroup$
    – mamiladi
    May 30, 2019 at 6:45
  • $\begingroup$ $$ C_{n}^k= \displaystyle \frac{n!}{k!(n-k)!}$$ $\endgroup$
    – mamiladi
    May 30, 2019 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, 2019 at 9:41
  • 1
    $\begingroup$ 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, 2019 at 4:45

1 Answer 1

1
$\begingroup$

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 $P_n(1-x)=(-1)^{n+1}P_n(x)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.