$1 \leq j \leq n$. Is there a formula that exists to find coefficients for this function? This is as far as i have got it using Sterling numbers of the first kind.. I was wondering if there was a more explicit formula? $$ \prod_{i=1;i \neq j}^{n} (x-i) = \frac{1}{x}\sum_{k=1;k \neq j}^{n+1} \sum_{2 \leq i_1 \leq ... \leq i_{k-1} \leq n+1} \frac{n!}{i_1 ... i_{k-1}}(-1)^{n-k+1}x^k $$

  • $\begingroup$ Which interpolation problem are you trying to solve? The coefficients of $\prod_{i=1}^{n}(x-i)$ are given by the Stirling numbers of the first kind (i.e. the elementary symmetric functions of $\{1,2,\ldots,n\}$). $\endgroup$ – Jack D'Aurizio Oct 27 '17 at 16:55
  • $\begingroup$ I believe the answer is here math.stackexchange.com/questions/500579/… $\endgroup$ – Ovi Oct 27 '17 at 16:57
  • $\begingroup$ You might want to consider deleting this question, to avoid your future question being judged a duplicate math.stackexchange.com/questions/2495709/… $\endgroup$ – Donald Splutterwit Nov 3 '17 at 22:33

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