Who knows this formula for polynomial interpolation?

Question

For my high school math project I studied polynomial interpolation: given a set of points $(x_0,y_0),...,(x_n,y_n)$, find the polynomial of degree $n$ that passes through all points. The solutions by Newton and Lagrange are well known, but I wanted to know the coefficients of the polynomial in standard form (powers of $x$). Using elementary algebra I solved the cases up to $n=3$ and generalized this to

$$ V(x) = \sum^n_{i=0}(-1)^{n-i}\sum^n_{j=0}y_j\frac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{\prod^n_{k=0,k\neq j}(x_j-x_k)}x^i $$

where $e_k(S)$ is the elementary symmetric polynomial (ESP). I also proved this. Later, I found out that V(x) is equivalent to the solution using Vandermonde's matrix, see the answer to exercise 40 in Knuth's The Art of Computer Programming, vol. 1, section 1.2.3. Sums and products. The only difference is that Vandermonde doesn't use the ESP. Lagrange's solution is expressed in elementary algebra, Newton's solution uses the divided differences (DD) $[y_0,...,y_n]$, and Vandermonde's solution can be expressed using ESP. So I wondered whether there is a solution using both DD and ESP. And there is. My solution is

$$H(x) = \sum^n_{i=0}\sum^n_{j=i}[y_0,...,y_j](-1)^{j-i}e_{j-i}(x_0,...,x_{j-1})x^i$$

It can be derived from Newton's solution and from Vandermonde's solution. After finding the solution tried to find it in (online) literature but I couldn't. I asked my math teacher and a university math professor and both did not know where it can be found. (Question) Who knows where to find the solution $H(x)$ in math literature?

Answer 1

Hint:

By Cramer, the coefficient of the $n^{th}$ power is the ratio of the Vandermonde determinant dividing a modified determinant where the $n^{th}$ column is replaced by the RHS. The latter can be expanded using the modified column and corresponding minors, which are of a quasi-Vandermonde type (there is a discontinuity in the exponents). These minors are still factored with the differences between two $x$'s, but there is an extra factor.

Answer 2

Note that you can pass from Newton or Lagrange , expressed in terms of products in $(x-x_k)$ to $x^k$ through the Vieta's formulas.
That is, if given the vector $\bf x_h$ indexed $0 \cdots h$ (the abscissas of the $h+1$ interpolating points) we construct the square matrix $ {\bf A}_{\,h} ({\bf x})$ according to Vieta's formulas as below $$ \begin{array}{l} {\bf A}_{\,h} ({\bf x}) = \left\| {\;a_{\,n,\,m} ({\bf x})\;} \right\|_{\,h} = \\ = \left\| {\;\left[ {m \le n} \right]\left( { - 1} \right)^{\,n - m} \sum\limits_{\begin{array}{*{20}c} {} \\ {0\, \le \,k_{\,0} \, < \,k_{\,1} \, < \, \cdots \, < \,k_{\,n - m - 1} \, \le \,n - 1} \\ \end{array}} {\prod\limits_{0\, \le \,j\, \le \,n - m - 1} {x_{\,k_{\,j} } } } \;} \right\|_{\,h} \\ \end{array} $$ and if you accept the following notation (nothing standard, it's just my personal elaboration) $$ \begin{array}{l} z^{\,{\bf x}_{\,h} } = \left\| {\;\prod\limits_{0\, \le \,k\, \le \,n - 1} {\left( {z - x_{\,k} } \right)} \;} \right\|_{\,h,0} \quad \Rightarrow \\ \Rightarrow \quad z^{\,{\bf o}_{\,h} } = \left\| {\;\prod\limits_{0\, \le \,k\, \le \,n - 1} {\left( {z - 0} \right)} \;} \right\|_{\,h,0} = \left\| {\;z^{\,n} \;} \right\|_{\,h,0} \\ \Rightarrow \quad z^{\,{\bf n}_{\,h} } = \left\| {\;\prod\limits_{0\, \le \,k\, \le \,n - 1} {\left( {z - k} \right)} \;} \right\|_{\,h,0} = \left\| {\;z^{\,\underline {\,n\,} } \;} \right\|_{\,h,0} \\ \end{array} $$ where $ z^{\,{\bf x}_{\,h} } $ is thus the Newton basis polynomial and where $x^{\,\underline {\,k\,} }$ represents the Falling Factorial

then $$ \bbox[lightyellow] { z^{\,{\bf x}_{\,h} } = {\bf A}_{\,h} ({\bf x})\;z^{\,{\bf o}_{\,h} } }$$

Since $$ z^{\,{\bf n}_{\,h} } = \left\| {\;z^{\,\underline {\,n\,} } \;} \right\|_{\,h,0} = {\bf A}_{\,h} ({\bf n})\;z^{\,{\bf o}_{\,h} } = {\bf A}_{\,h} ({\bf n})\left\| {\;z^{\,n} \;} \right\|_{\,h,0} $$ and, representing in square brackets the (unsigned) Stirling N. of 1st kind and in curly brackets those of 2nd kind, it is $$ x^{\,\underline {\,n\,} } = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,n - k} \left[ \matrix{ n \cr k \cr} \right]x^{\,k} } $$ then $$ \begin{array}{l} {\bf A}_{\,h} ({\bf n}) = \left\| {\;\left[ {m \le n} \right]\left( { - 1} \right)^{\,n - m} \sum\limits_{\begin{array}{*{20}c} {} \\ {0\, \le \,k_{\,0} \, < \,k_{\,1} \, < \, \cdots \, < \,k_{\,n - m - 1} \, \le \,n - 1} \\ \end{array}} {\prod\limits_{0\, \le \,j\, \le \,n - m - 1} {\,k_{\,j} } } \;} \right\|_{\,h} = \\ = \left\| {\;\left( { - 1} \right)^{\,n - m} \left[ \begin{array}{c} n \\ m \\ \end{array} \right]\;} \right\|_{\,h} = {\left\| {\;\left\{ \begin{array}{c} n \\ m \\ \end{array} \right\}\;} \right\|_{\,h}} ^{\; - \,{\bf 1}} = {\bf S}_{{{\bf t2}} _h} ^{ - \,{\bf 1}} \\ \end{array} $$

so that the Vieta' formulas provide a generalization of the Stirling Numbers.

Some references you may find interesting to start are this related post and this paper.