Trying to solve a problem I ended up working on a polynomial of degree $n-1$ in $t$, $p(t)=\sum_{k=0}^{n-1} a_k t^k$, whose coefficients are given by

$a_k=\sum_{i=1}^n\cfrac{\beta_i^k}{\prod_{j=1\\j\neq i}^n (\beta_j-\beta_i)}$

Here, $\beta_i$ are $n$ distinct positive and ordered real numbers, so that $0<\beta_1<\beta_2<\cdots<\beta_n$. What can I say about the coefficients $a_k$? I made some attempts and I think that

$a_k=\begin{cases}0 &k=0,\dots,n-2\\z\neq0 &k=n-1\end{cases}$

but I don't know if this is true, how to prove it and what $z$ would be. Do any of you know a solution or a reference for this problem? Thanks

  • $\begingroup$ Could you include the calculations in the specific case $n=3$ for the $a_k$ values? $\endgroup$ – coffeemath Dec 7 '16 at 12:18
  • $\begingroup$ How could the result come out $z$ in some case, when no $z$ appears in the definition of $a_k$? $\endgroup$ – coffeemath Dec 7 '16 at 12:23
  • $\begingroup$ I wrote $z$ to indicate an unknown. I'm sorry if it was unclear: what I meant is the only non-zero coefficient would be $a_n$. I'll try writing down the calculations for the $n=3$ case. $\endgroup$ – Muriel Dec 7 '16 at 12:27
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    $\begingroup$ Suppose $m=2$ and $n=3$. Then $a_0=\cfrac{1}{(\beta_2-\beta_1)(\beta_3-\beta_1)}+\cfrac{1}{(\beta_1-\beta_2)(\beta_3-\beta_2)}+\cfrac{1}{(\beta_1-\beta_3)(\beta_2-\beta_3)}$ $=\cfrac{\beta_3-\beta_2-\beta_3+\beta_1+\beta_2-\beta_1}{(\beta_3-\beta_2)(\beta_3-\beta_1)(\beta_2-\beta_1)}=0$ $\endgroup$ – Muriel Dec 7 '16 at 12:33
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    $\begingroup$ The degree $m$ of the polynomial and the number $n$ of points $\beta_i$ are a priori not related, so I can take $m$ to be as large as I want. $\endgroup$ – Muriel Dec 7 '16 at 15:14

Let $n=3$ and for ease of notation let the points be $a,b,c.$ Then under your conjecture, the expression using the power $k=2$ [which satisfies $k \le n-1]$, namely $$\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)},$$ should be $0.$ However this expression is identically $1,$ after a check on a symbolic algebra calculator.

Added later: Let $f(x)$ be any function, and let $S_f$ denote your expression for $a_k$ but with its numerator $\beta_i^k$ replaced by $f(\beta_i.)$ Also let $V_n$ denote the $n \times n$ Vandermonde matrix, whose $j$-th row is formed by the $(j-1)$th powers of the $\beta_i,$ for $1 \le i \le n.$ The determinant of $V_n$ is then the product of the factors $(\beta_j-\beta_i)$ for $1 \le i<j\le n.$

It can then be shown that he product $S_f \det (V_n)$ is the same as the determinant of the matrix obtained from $V_n$ when its last row is replaced by the row $[f(\beta_1),f(\beta_2),\cdots f(\beta_n.)].$

If we take $f(x)=x^{n-1}$ this shows that $a_{n-1}=1,$ since we now have the original Vandermonde matrix on both sides. And if $0<k<n-1$ we have on the right side the result of taking the Vandermonde matrix and replacing the last row with one of the rows above it, making that determinant zero.

  • $\begingroup$ You're right, I probably messed up things with the indices in my formulation above, but if I adjust them the main question still stands. Let me edit the original question, and sorry for the confusion, I'm trying to figure out the problem.. $\endgroup$ – Muriel Dec 7 '16 at 15:39
  • $\begingroup$ So, for my application I could take $m$ to be as large as I want, but my point is that, fixed the numbers $\beta_i$, I want to know what is the first non-zero coefficient $a_k$, and possibly derive its value. In the example $n=3$ we did above my updated conjecture should stand, am I right? $\endgroup$ – Muriel Dec 7 '16 at 15:43
  • $\begingroup$ What would be your updated conjecture re. the first nonzero coefficient $a_k$? Is your conjecture about what the value of the subscript $k$ should be in terms of $n$? If so my example above (with your examples from comments) shows for $n=3$ the first nonzero $a_k$ occurs when $k=2.$ [i.e. "first subscript" is not just $n-1$ in general] $\endgroup$ – coffeemath Dec 7 '16 at 15:59
  • $\begingroup$ I think that $a_k=0$ for all $k=0,\dots,n-2$ and the first non-zero coefficient is $a_{n-1}$. This agrees with our example for $n=3$ (as $a_0=0, a_1=0$ and $a_2\neq0$) but I have no idea how to prove that this is a general result. $\endgroup$ – Muriel Dec 7 '16 at 16:13
  • $\begingroup$ Agrees in $n=4$ case, first nonzero coeff is $a_{4-1}=a_3=1.$ $\endgroup$ – coffeemath Dec 7 '16 at 17:52

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