Clarification on alternative expression of sum

Question

I am not familiar with alternative expression of sum shown below,

$ d_k=\sum_{i+j+l=k}a_ib_jc_l $

How it does work?

for $k = 4 $ then,

$d_{4} = \sum_{i+j+l=4}a_ib_jc_l = ...$

How do I express it in standard summation format namely $\sum_{n=0}^{\infty}$ This is from the formula to calculate the product of three summations (power series) or $\sum_{i=0}^{\infty} a_i x^i \cdot \sum_{j=0}^{\infty} b_j x^j \cdot \sum_{l=0}^{\infty} c_l x^l = \sum_{k=0}^{\infty} d_kx^k$

Answer 1

First, it is assumed that the indices are non-negative.

Second step, it means that the sum is to be taken for the triples $(i,j,l)$ for which $i+j+l=k |0 \le i,j,l$. These are the "weak" compositions of $k$ into three parts

So for e.g. $k=4$ you have to sum $a_0b_0c_4+\cdots+a_0b_1c_3+\cdots$

Answer 2

You can get explicit indices like this (assuming that the lower index of summation is $0$):

$\begin{array}\\ d_k &=\sum_{i+j+l=k}a_ib_jc_l\\ &=\sum_{i=0}^{k} a_i \sum_{j=0}^{k-i}b_jc_{k-i-j}\\ \end{array} $

If the summation is over an inequality, you get one more level of summation:

$\begin{array}\\ e_k &=\sum_{0 \le i+j+l \le k}a_ib_jc_l\\ &=\sum_{i=0}^{k} a_i \sum_{j=0}^{k-i}b_j\sum_{l=0}^{k-i-j}c_{l}\\ \end{array} $

Work out for yourself what these are if the lower index of summation is $1$ instead of $0$.

Answer 3

The representation of $d_k$ as triple sum $\sum_{i+j+l=k}a_ib_jc_l$ can be derived by applying Cauchy Series multiplication twice.

We obtain
\begin{align*} \left(\sum_{i=0}^\infty\right.&\left. a_i x^i\right)\left(\sum_{j=0}^\infty b_j x^j\right)\left(\sum_{j=0}^\infty b_j x^j\right)\\ &=\left(\sum_{n=0}^\infty\left(\sum_{{i+j=n}\atop{i,j\geq 0}}a_ib_j\right)x^n\right)\left(\sum_{j=0}^\infty b_j x^j\right)\tag{1}\\ &=\sum_{k=0}^\infty\left(\sum_{{n+l=k}\atop{n,l\geq 0}}\left(\sum_{{i+j=n}\atop{i,j\geq 0}}\right)c_l\right)x^k\\ &=\sum_{k=0}^\infty\left(\color{blue}{\sum_{{i+j+l=k}\atop{i,j,l}}a_ib_jc_l}\right)x^k\tag{2}\\ \end{align*}

From (1) we also obtain

\begin{align*} \left(\sum_{n=0}^\infty\right.&\left.\left(\sum_{{i+j=n}\atop{i,j\geq 0}}a_ib_j\right)x^n\right)\left(\sum_{j=0}^\infty b_j x^j\right)\\ &=\left(\sum_{n=0}^\infty\left(\sum_{i=0}^na_ib_{n-i}\right)x^n\right)\left(\sum_{l=0}^\infty c_lx^l\right)\\ &=\sum_{k=0}^\infty\left(\sum_{{n+l=k}\atop{n,l\geq 0}}\left(\sum_{i=0}^na_ib_{n-i}\right)c_l\right)x^l\\ &=\sum_{k=0}^\infty\left(\color{blue}{\sum_{n=0}^k\sum_{i=0}^na_ib_{n-i}c_{k-n}}\right)x^k\tag{3} \end{align*} Comparing coefficients of equal powers of $x$ in (2) and (3) shows equality of the sums.