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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 $v_{j_k}$?

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  • $\begingroup$ When faced with dauntingly complicated or seemingly ambiguous summation (and/or product) notation, I always pick the simplest case I can think of, plug it in, and evaluate things to an explicit sum (and/or product). If I succeed in doing so, then I try what looks like the next simplest case. Sometimes in doing so, you can figure out what the notation must mean. At the very least, it can clarify where the confusion or ambiguity is. $\endgroup$ – Barry Cipra Mar 19 '19 at 16:39
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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 $j_k\in\{1,3\}$.

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