2
$\begingroup$

How to caluclate the coefficients of finite product $$\prod\limits_{i=1}^{n}(z^i-1)=\sum\limits_{k=0}^{\frac{n(n+1)}{2}}a_{n,k}z^k$$ effectively?

I have figured out a recursive rule for $n>1$:

$$a_{n,k} = a_{n-1, k} - (-1)^na_{n-1,\frac{n(n-1)}{2}-k}$$

But I feel that there should be something much efficient than this.

And also there is a sequence in OEIS: https://oeis.org/A231599

1 Answers 1

1

Let $$P= \prod_{i=1}^n (z_i-1)\ \ \ \ \ \text{ and } \ \ \ \ \ Q=\prod_{i=1}^n (t-z_i) \ . $$ The main goal here is to compute the coefficients of $p(z)=P(z, \dots, z^n) \in \mathbb{C}[z]$. Think $Q$ as a one variable polynomial in $t$ and observe that $$Q(1)=(-1)^n \prod_{i=1}^n (z_i-1)= (-1)^n P(z_1, \dots, z_n)\ .$$ On the other hand, according to Vieta's theorem, we have that $$Q(t)= \sum_{i=0}^n\sigma_i(z_1, \dots , z_n)\cdot t^n $$ where $\sigma_1, \dots , \sigma_n$ denote the elementary symmetric functions in $z_1, \dots, z_n$. Summing up $$P(z_1, \dots, z_n)=(-1)^n\sum_{i=0}^n\sigma_i(z_1, \dots , z_n) \ , $$ from where the identity $$p(z)=(-1)^n \sum_{i=0}^n \sigma_i(z, \dots, z^n) \ .$$

In order to find the coefficient $a_k$ of $p(z)=a_0+ a_1z+ \dots +a_nz^n$ you have now to count (with signs!) how many times the monomial $z^k$ appears in the sum on the right. More precisely we have that $$a_k = \sum_{i=0}^k (-1)^i \ \# \big(z^k \text{ appears in } \sigma_i(z, \dots , z^n) \big) \ .$$ By inspecting the definition of the elementary symmetric functions one can easily see that $z^k$ appears in the expression $\sigma_i(z, \dots , z^n)$ exactly $A_{k,i}$ times, where $A_{k,i}$ denotes the number of positive integer solutions of the equation $k=x_1+ \dots +x_i$ with $x_1< \dots

  • 0
    Just wondering what is the complexity in terms of n? – 2017-02-12
  • 0
    And if possible could you please provide some reference where to discover a bit more about the symmetric functions. – 2017-02-12
  • 0
    Just found it https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#Properties – 2017-02-12
  • 0
    Sorry but I was with my phone when replied. I edited to make it more clear, hope it is better now! – 2017-02-12
  • 0
    Not sure if I understand correctly, I am looking for $(z^i-1)$ products, do you mean that $z_i=z^i$ in this case the $P(z_1, \dots, z_n)=(-1)^n\sum_{i=1}^n\sigma_i(z_1, \dots , z_n) \ $ will not contain disjoint sums. – 2017-02-12
  • 0
    I think I completely misunderstood your question! You where looking for the coefficients of $p(z)$... Ok, now the combinatorial analysis at the end is a little bit more tricky. – 2017-02-13
  • 0
    This looks interesting, calculation the partition count may be less efficient than calculation of recursion: $a_{n,k} = a_{n-1, k} - (-1)^na_{n-1,\frac{n(n-1)}{2}-k}$. The other idea came from your comment: We may use the roots of unity in order for $z_1,....,z_n$. – 2017-02-13
  • 0
    And one more comment: should not it be "$A_{k,i}$ denotes the number of solutions of the equation $k=x_1+...+x_i$ with $x_1< \dots – 2017-02-13
  • 0
    I ment positive integer solutions so $x_i \leq n$ comes from the fact that $k \leq n$ – 2017-02-13
  • 0
    But the maximal order of $z$ in $\sigma_i(z, \dots , z^n) $ is greater than $n$ the highest order is $n(n+1)/2$ – 2017-02-13
  • 0
    Ok, I see your point now. Btw I think that this thing is related to Pentagonal numbers https://en.wikipedia.org/wiki/Pentagonal_number_theorem – 2017-02-13
  • 0
    Yes. I will try to investigate the case with unity roots. – 2017-02-13