Consider the double sequence $A(n,k)$ which is recursively defined by $$A(n,n)=1 \text{ for } n=0,1,2,\dots \text{ and }$$ $$A(n,k)=2\sum_{l=1}^{k+1} \binom{2n+1}{2l} A(n-l,k+1-l) \text{ for }0\leq k < n.$$ Furthermore, let $B(n,k)$ be such that $$\sum_{k=0}^m A(n,k)B(n,m-k)=\begin{cases}1 & \text{ if }m=0 \newline 0 &\text{ otherwise}.\end{cases}.$$ I'd like to prove that $B(n,k)$ is alternating for fixed $n$, i.e. $B(n,k) \cdot (-1)^k>0$ for $0 \leq k \leq n$.

Remark: The sequence $A(n,k)$ occurs when we expand the Eulerian polynomial $A_{2n+1}$ as follows $$A_{2n+1}(x)=\sum_{k=0}^n A(n,k) x^{n-k}(x-1)^{2k}.$$

This question has a long history. It has to do with approximating sphere-valued functions $f:[0,1] \rightarrow S^m$ using the projection $P(x)=x/||x||_2$ and uniform B-splines $B_n$. The problem is to find $c_{0},\dots,c_{1/h} \in S^m$ such that $|f(x)-P(\sum_{i} c_i B_n(h^{-1}x-i))|=\mathcal{O}(h^{n+1})$. If we omit the projection $P$ and the condition $c_i \in S^m$ this problem can be solved by choosing $$c_k=\sum_{i=-(n-1)/2}^{(n-1)/2} a_{n,i} f(h(k+i)),$$ where $a_{n,i}$ is a fixed double sequence of numbers, e.g. for $n=3$ it is $(-1/6,4/3,-1/6)$. However with the projection it is more difficult. There is construction which seems to work but I was not able to give a complete proof. What is missing is a proof that the Fourier transform of the coefficients $a_{n,k}$ is positive, i.e. $\sum_{k=-(n-1)/2}^{(n-1)/2}a_{n,k}e^{ikx}>0$ for all $0 \leq x <2 \pi$. The B-splines can be related to Eulerian polynomials and the coefficients $a_{n,k}$ to the polynomial defined via the $B(n,k)$ from above. Proving the above would immediately solve this problem.

Thanks for any help in advance.

  • 1
    $\begingroup$ "Prove that ..." sounds as if this is a homework. What is the context of this question? $\endgroup$ – Martin Brandenburg Apr 26 '13 at 18:29

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