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I am writing a program that computes a Lagrange polynomial using parallel processing for data points situated at every natural number. But I can have $n$ many data points, which will potentially create a polynomial of degree $n$. Doing some math and research I found out that what I need is a general form for the $ith$ coefficient of the Elementary Symmetric Polynomial specialized to the natural numbers. I just asked a question about about a double sum inside one of the solving for one of the coefficients partial sum of $\sum^h_{i=1}\sum^i_{j=1}ij$. But I thought I would ask the whole thing here.

To give some examples the coefficients of the cubic case would be $x^3-(1+2+3)x^2+(1*2+2*3+1*3)x-1*2*3$ and the quartic case would be $x^4-(1+2+3+4)x^3+(1*2+2*3+1*3+1*4+2*4+3*4)x^2-(1*2*3+1*3*4+1*2*4+2*3*4)x-1*2*3*4$

Because the polynomials would be like $(x-3)(x-2)(x-1)$ as an example.

I spent all morning coming up with only the first three partial sums of for the coefficients.

Which are $1,\frac{n(n+1)}{2},\frac{(n-1)n(n+1)(3n+2)}{24},\frac{n^2(n+1)^2(n-1)(n-2)}{48}$. Before I embark on trying to get the next coefficient then trying to generalize from there, is there a solution (I am not an expert in what seems to be combinatoric algebra) that someone can point me too for the closed form for each of the coefficients or a proof that someone can show.

I read this but this is a little over my head meaning that I don't know if my answer is in there or not. I am computer guy, so any simplified explanation is greatly appreciated.

Any help is much appreciated! Thanks in advance!

Edit: I did more research and I am came across Vieta's formula, maybe I can use this for a closed form.

Edit 2: I found this post talking about Stirling numbers Coefficients of polynomial $(x+1)(x+2)...(x+n)$ I am wondering if I can manipulate it to my answer

yosmo78
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I found the answer after a while. The coefficients are calculated using the signed Stirling numbers of the first kind.

These are the signed stirling numbers of the first kind enter image description here

In order to get the coefficients you select the row by the degree of the polynomial and read it from right to left.

To calculate it, the facts I used were

$$s(n,k)=-(n-1)\cdot s(n-1,k)+s(n-1,k-1)$$ $$s(n,n) = 1$$ $$s(n,0) = 0, n >0$$

And what is interesting is the polynomials that I found $1,\frac{n(n+1)}{2},\frac{(n-1)n(n+1)(3n+2)}{24},\frac{n^2(n+1)^2(n-1)(n-2)}{48}$ describe the diagonals.

So for instance $\frac{n(n+1)}{2}$ is the diagonal 0,1,3,6,10,15,... (at positions (2,1),(3,2),(4,3),...)

And another example would be for $\frac{(n-1)n(n+1)(3n+2)}{24}$ where the value on the diagonal are 2,11,35,...

So a future follow up question would be is there a closed form for the diagonals of the stirling numbers of the first kind?

yosmo78
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