Then I suspect that the $\oplus$ is just an assertion that the $\sum$ is a direct sum.
Work by analogy with vector spaces. Assume $V$ is a vector space over a field $k$, and assume $W_1,...,W_n$ are sub-spaces of $V$.
Then we can define the sum $\sum W_i$, which is the smallest subspace of $V$ containing all the $W_i$. You can also define the vector space, $\oplus_i W_i$. This is an "exterior" sum, because its members are not members of $V$, but rather tuples $(w_1,...,w_n)$ with $w_i\in W_i$.
Now, there is a simple linear map, $\phi: \oplus W_i \rightarrow \sum W_i$ defined as $\phi(w_1,...,w_n) = \sum_i w_i$.
This map is not necessarily 1-1. Indeed, if $W_1\subset W_2$ are finite dimensional, then the dimension of $W_1\oplus W_2$ is the sum of the dimensions of $W_1$ and $W_2$, while the dimensions of $W_1+W_2=W_2$ is the dimension of $W_2$.
When this map is $1-1$, then the vector spaces $\oplus W_i$ and $\sum W_i$ are "isomorphic."
In that case, the statement $W=\oplus \sum_{i=1}^n W_i$ is a statement in two parts:
- $W=\sum_{i=1}^n W_i$ - that is, it is the smallest subspace of $V$ that contains all the $W_i$.
- The map $\oplus_{i=1}^n W_i \rightarrow \sum_{i=1}^n W_i$ is 1-1.
So this is one of those cases where the language is confusing, because it the $\oplus$ symbol here is an assertion about the nature of the $\sum$.
There is a way in which this analogy is precise, in that there is a notion of (left/right) $R$-modules which is analogous to the notion of a $k$-vector space, and (left/right) ideals of $R$ are sub-modules of $R$ when $R$ is considered as a (left/right) $R$-module.
Then a set of ideals has a direct sum as $R$-modules, but that sum is not a sub-module of $R$. However, as with vector spaces, there is a map from the direct sum to the sum inside $R$, and what $\oplus\sum$ is saying is that we are taking the sum inside $R$, but we are asserting that the map from the direct sum to the sum inside $R$ is 1-1.
As has been pointed out in a comment, people do this sort of
thing all the time. I have only been able to find two texts that can be used as references
for the topic. Of the two, the more readable is Terence Tao, Analysis I
(1st edition, Hindustan Book Agency 2006).
Here are some extracts:
Definition 7.1.6 (Summations over finite sets). Let $X$ be
a finite set with $n$ elements (where $n \in \mathbf{N}$), and let
$f \colon X \to \mathbf{R}$ be a function from $X$ to the real
numbers (i.e., $f$ assigns a real number $f(x)$ to each element $x$
of $X$). Then we can define the finite sum $\sum_{x \in X} f(x)$ as
follows. We first select any bijection $g$ from
$\{i \in \mathbf{N} : 1 \leq i \leq n\}$ to $X$; such a bijection
exists since $X$ is assumed to have $n$ elements. We then define
$$
\sum_{x \in X} f(x) := \sum_{i=1}^n f(g(i)).
$$
[$\ldots$]
To verify that this definition actually does give a single,
well-defined value to $\sum_{x \in X} f(x),$ one has to check that
different bijections $g$ from $\{i \in \mathbf{N} : 1 \leq i \leq n\}$
to $X$ give the same sum. In other words, we must prove
Proposition 7.1.8 (Finite summations are well-defined). Let
$X$ be a finite set with $n$ elements (where $n \in \mathbf{N}$),
let $f \colon X \to \mathbf{R}$ be a function, and let
$g \colon \{i \in \mathbf{N} : 1 \leq i \leq n\} \to X$ and
$h \colon \{i \in \mathbf{N} : 1 \leq i \leq n\} \to X$ be
bijections. Then we have
$$
\sum_{i=1}^n f(g(i)) = \sum_{i=1}^n f(h(i)).
$$
The only other useful reference I know of is Nicolas Bourbaki,
Elements of Mathematics: Algebra I, Chapters 1–3 (Hermann,
Paris 1974, PDF).
I’ll only quote one of his results, but the book might be worth exploring
further, if it is important to prove things about sums of families defined
on finite ordered sets without relying, as Tao does, on the special case
where the indices are integers.
(I haven’t tried to reproduce Bourbaki’s typography exactly, because I don’t think it’s possible in MathJax, and it would even require some hacking about in LaTeX.)
Theorem 1 (Associativity theorem). Let $E$ be an
associative magma whose law is denoted by $\odot.$ Let $A$ be a
totally ordered non-empty finite set, which is the union of an
ordered sequence of non-empty subsets $(B_i)_{i \in I}$ such that
the relations $\alpha \in B_i,$ $\beta \in B_j,$ $i < j$ imply
$\alpha < \beta$; let $(x_\alpha)_{\alpha \in A}$ be an ordered
sequence of elements of $E$ with $A$ as indexing set. Then
$$
\bigodot_{\alpha \in A} x_\alpha =
\bigodot_{i \in I}\Big(\bigodot_{\alpha \in B_i} x_\alpha\Big).
$$
Best Answer
An elegant alternative is to index over a set. For $j \in \{1,\dots,n\}$, let $I_j := \{1,\dots,n\} \backslash\{j\}$.
Now, if $$f(x) = \prod_{j=1}^n (x-a_j),$$ then $$f'(x) = \sum_{j=1}^n \prod_{k \in I_j}(x-a_k).$$