# How can I write this series in $\sum$ & $\prod$ notation

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 $$x_k^l$$

• This series of terms and the last look a bit odd to me, if they are correct then never mind: $c_{123} x_1^2 x_2 + ... + c_{n2n} x_1^2 x_n$ Commented Sep 3, 2019 at 3:08
• @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 Commented Sep 3, 2019 at 3:14
• What are the restrictions to the number of distinct variables and their respective exponents? Commented Sep 3, 2019 at 14:54
• Are $n,m$ known apriori? It seems like we might be able to boil to a problem where you have $\sum_i c_i \prod_k x_k ^{(l_{i,k})}$ Commented Sep 3, 2019 at 14:56

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:

$$n=2, m=2$$

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

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