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I know that

$x_1 + x_2 + ... + x_n$ can be re-written as $\sum_{i=1}^{n} x_i$

and

$x_1 x_2,\dots, x_n$ can be re-written as $\prod_{i=1}^{n} x_i$

I am trying to write this in form of $\sum$ & $\prod$

$c_0 + c_1 x_1 + c_2 x_2 + ... + c_n x_n + c_{12} x_1 x_2 + c_{13} x_1 x_3+ ... + c_{1n} x_1 x_n + ...+ c_{23} x_2 x_3 + ...+ c_{n-1 n} x_{n-1}x_n + c_{123} x_1^2 x_2 + ... + c_{n2n} x_1^2 x_n + .. + c_{542} x_1^5 x_3^2 x_8^2 x_n +... + c_{nmm}x_1^m x_2^m...x_n^m $ where c_i is constant

in other words all combinations of $x_i^j$ multiplied by $x_k^l$

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  • $\begingroup$ This series of terms and the last look a bit odd to me, if they are correct then never mind: $c_{123} x_1^2 x_2 + ... + c_{n2n} x_1^2 x_n $ $\endgroup$
    – NoChance
    Commented Sep 3, 2019 at 3:08
  • $\begingroup$ @NoChance I just wanted to say that C value is different for each term, the numbers attached to the C123, Cm2m are just random to say that each one is different than the other $\endgroup$
    – asmgx
    Commented Sep 3, 2019 at 3:14
  • $\begingroup$ What are the restrictions to the number of distinct variables and their respective exponents? $\endgroup$ Commented Sep 3, 2019 at 14:54
  • $\begingroup$ Are $n,m$ known apriori? It seems like we might be able to boil to a problem where you have $\sum_i c_i \prod_k x_k ^{(l_{i,k})}$ $\endgroup$ Commented Sep 3, 2019 at 14:56

1 Answer 1

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There are $(m+1)^{n}$ possible ways of having $x_i^j$ over all $i\in\{1,2,3,4,...,n\}$ and $j\in\{0,1,2,3,4...,m\}$ thus we need the indices of $c_{k}$ to follow $k\in\{0,1,2,3,...,(m+1)^n-1\}$.

We also need a way to convert $k$ into powers for the $x_i$. This can be done by turning $k$ into a base $m$ number with each digit representing the power of an $x_i$. We can define such a power as $l_{k,i}= \lfloor\frac{k}{(m+1)^i}\rfloor\mod{m+1}$

This gives us a final sum/product as: $$\sum_{k=0}^{(m+1)^n-1} c_k \prod_{i=0}^n x_i^{l_{k,i}}$$


As a sanity check:

$n=2, m=2$

$$\sum_{k=0}^{3^{2}-1} c_k \prod_{i=0}^n x_i^{l_{k,i}}$$ $$ =c_0 x_1^0 x_2^0 + c_1x_1^1 x_2^0+c_2x_1^2x_2^0 +c_3 x_1^0 x_2^1 + c_4 x_1^1 x_2^1+c_5 x_1^2 x_2^1 + c_6x_1^0x_2^2+c_7x_1^1x_2^2+c_8 x_1^2x_2^2$$

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  • $\begingroup$ @asmgx Is this what you are lookin for? $\endgroup$ Commented Sep 3, 2019 at 18:00
  • $\begingroup$ Thanks @Kitter Catter i think this is it. $\endgroup$
    – asmgx
    Commented Sep 3, 2019 at 21:32

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