Summing over set of sets

Question

I am having trouble trying to understand an expression based on the multivariate Taylor series which is used to generate a system of residual equations. The expression is from Chauhan (2017) Benchmarking Approaches for the Multidisciplinary Analysis of Complex Systems Using a Taylor Series-Based Scalable Problem and the expression is

\begin{align} R_i(v_1,\dots,v_n)=\sum\limits_{r=1}^{d_i}\frac{1}{r!}\sum\limits_{(j_1,\cdots,j_r)\\1\leq j_k\leq n\\j_1,\cdots,j_r \in A(i)} \frac{\partial^rR_i}{\partial v_{j_i}\cdots\partial v_{j_r}} \prod\limits_{k=1}^{r}v_{j_k} \end{align}

where $n$ is the number of variables, $d_i$ is the degree of the polynomial, and $A(i)$ are "arguments of the $i^{th}$ equation; $A:\{1,\cdots,n\}\rightarrow P(\{1,\cdots,n\})$ where $P()$ is the power set". From this description I assume $A(i)$ is a function that returns the power set for a given list of elements, e.g. $A([1,2])=\{\{\},\{1\},\{2\},\{1,2\}\}$.

My point of confusion is the summation term and its three conditions. Is it a single sum over all elements $(j_1,\cdots,j_r)$ in the power set, giving four terms in the sum? Or is it a nested sum $\sum\limits_{j_1}\sum\limits_{j_2}\cdots\sum\limits_{j_r}$ where each sum is performed over the sets in the power set? How would the empty set or a set like $j_k=\{1,2\}$ be substituted as a single index into $v_{j_k}$?

Answer 1

Here are some hints which might be useful.

The following representations are equivalent \begin{align*} \sum_{1\leq j_1,j_2\leq n}a_{j_{1},j_{2}}&=\sum_{j_1=1}^n\left(\sum_{j_2=1}^n a_{j_1,j_2}\right)\\ &=\sum_{j_2=1}^n\left(\sum_{j_1=1}^na_{j_{1},j_{2}}\right)\\ &=\sum_{{(j_1,j_2)}\atop{1\leq j_k\leq n}}a_{j_1,j_2}\tag{1}\\ &=\sum_{{1\leq j_k\leq n}\atop {1\leq k\leq 2}}a_{j_1,j_2}\tag{2} \end{align*}

• Note that (1) tells us to sum up over all tupels $(j_1,j_2)$ which gives us just the information how many different summation indices are in use.

• Precisely the same is stated in (2) when we list the valid indices $j_k$ via $1\leq k\leq 2$.

Let's have a look at the inequality chain $1\leq j_k\leq n$.

• The inequality chain implies that $j_k$ are positive integer values fulfilling an order relation. So, the $j_k$ cannot be sets.

• The relation $j_1,\ldots,j_r\in A(i)$ tells us due to the element-operator $\in$ that $A(i)$ is a set containing integer values. It is not a set of sets.

Example: We take $n=5,r=3, i=1, A(1)=\{1,3\}$.

We obtain \begin{align*} \sum\limits_{{{(j_1,\cdots,j_r)}\atop{1\leq j_k\leq n}}\atop{j_1,\cdots,j_r \in A(i) }}a_{j_1,j_2,\ldots,j_r} &=\sum\limits_{{{(j_1,j_2,j_3)}\atop{1\leq j_k\leq 5}}\atop{j_1,j_2,j_3\in \{1,3\} }}a_{j_1,j_2,j_3}\\ &=a_{1,1,1}+a_{1,1,3}+a_{1,3,1}+a_{1,3,3}\\ &\qquad+a_{3,1,1}+a_{3,1,3}+a_{3,3,1}+a_{3,3,3} \end{align*}

Note that although $1\leq j_k\leq 5$ with $1\leq k\leq 3$ we have only a small number of terms due to $j_k\in\{1,3\}$.