Is there a closed form solution for this polynomial? Let $1\leq j \leq n$, the polynomial is

$$ \prod_{i=1;i \neq j}^{n} (x-i) = \frac{\prod_{i=1}^{n} (x-i) }{x-j} = \frac{(x-1)(x-2) ....(x-n)}{x-j} $$

I've tried looking for it but haven't found anything on this. Wolfram only details this with the Pochammer symbol, but that isn't very useful. No, this is not homework. I would like to know if there was a paper on this that maybe derives the coefficients of this polynomial if it is even possible?

I found that Vieta's formulas for $\prod_{i=1}^{n} (x-i)$ simplify the first few terms, but afterwards it gets very complicated.

$p_n(x)=a_n x^n + a_{n-1} x^{n-1} + ... + a_0$


$a_{n-1}=\sum_{i=1}^{n} i =\frac{n(n+1)}{2}$

$a_{n-2}=\sum_{i,j;i<j}^{n} ij=\sum_{i=2}^{n} \frac{(i-1)(n-i-1)(n+i)}{2}=\frac{n(n^3-6n^2+3n+2)}{8}$



  • $\begingroup$ What is $j$ on the LHS ? Is it a double product ? $\endgroup$ – Zubzub Oct 14 '17 at 21:47
  • 2
    $\begingroup$ He defined $j$ before stating the equation - Reread more carefully $\endgroup$ – Junkyards Oct 14 '17 at 21:48

You're not going to find a closed form for these numbers. I can give you a formula for the coefficient of $x$ as proof: $$(-1)^n\frac{n!}j\left(H_n-\frac1j\right)$$ where $H_n$ is the $n$th harmonic number, for which no closed formula is known.

  • $\begingroup$ Could you explain how you obtained this? I know that for the resulting polynomial $a_1=\sum_{1 \leq r_1<r_2<...<r_{n-1} \leq n} r_1 r_2 ... r_{n-1}$ is obtained from $(x-1)...(x-n)$ even for $i \neq j$, which does not have that expression. Also, i know this sum $\sum_{i=0}^{n} i^r$ has a closed form as well $\endgroup$ – Baklava Gain Oct 15 '17 at 1:46
  • $\begingroup$ @Baklava I'm actually off by a factor of $j$. The trick is that the product is equal to $\frac{n!}j\prod{(\frac xi-1)}$. $\endgroup$ – Matt Samuel Oct 15 '17 at 1:52

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