Unambiguous summation notation
Is there a way to unambiguously express which term is inside a summation and which one not? Are there any universally used conventions? For example let's say a want to sum terms $x_j$ and $E_1$ over $j$ and add to them $E_2$. Consider that the terms $E_1$, $E_2$ can be expressions that are independent of $j$ and not necessarily constants (they can be summations themselves over another index). Some alternatives are
- $\sum_j (x_j + E_1) + E_2$
- $[\sum_j x_j + E_1] + E_2$
- $[\sum_j (x_j + E_1)] + E_2$
The third one is certainly unambiguous but too verbose I think.
Solution 1:
The scope of the series operator is the following term; it encompassed any multiplication (or division).
$$\sum_i \color{blue}{x_iy^i} = \sum_i \color{blue}{y^ix_i}$$
The scope of the series operator is just one term; it ends at a '$+$'.
$$\sum_i \color{blue}{x_iy^i} + z = z+ \sum_i \color{blue}{x_iy^i}$$
Parenthesis may be added just emphasis if you wish; around the term or the whole series. This is just used to add clarrity.
$$\sum_i \color{blue}{(x_iy^i)}+z ~=~ \left(\sum_i \color{blue}{x_iy^i}\right)+z ~=~ \sum_i \color{blue}{x_iy^i}+z$$
Note enclosing an entires expression in parenthesis does not change the operator precedence within.
$$\left(\sum_i \color{blue}{x_iy^i} + z\right) = \left(z+ \sum_i \color{blue}{x_iy^i}\right)$$
However, if such is required, parenthesis are employed to enclose a term which is itself a sum, thusly.
$$\sum_i \color{blue}{\Big(x_iy^i + z_i\Big)} = \sum_i \color{blue}{\Big(z_i+x_iy^i\Big)}$$
tl:dr$$\color{silver}[\sum_{i=1}^3 (x_i+C_1)\color{silver}]+C_2 ~{= \color{silver}[(x_1+C_1)+(x_2+C_1)+(x_3+C_1)\color{silver}]+C_2 \\ = (x_1+x_2+x_3)+3 C_1+ C_2}$$
While
$$[\sum_{i=1}^3 x_i+C_1]+C_2 ~{= \big((x_1+x_2+x_3)+C_1\big)+C_2\\= (x_1+x_2+x_3)+C_1+C_2}$$
Solution 2:
Short Answer:
The sum $\sum_{j\in J} a + b +c+\cdots $ is taken over the first term following the $\sum$ operator. That is, everything before the first plus or minus sign. In my example, the first term is $a$. If you want to include more things in the first term, use parentheses. For instance, if you want to sum over the sum of $a$ and $b$ (but not $c$, etc), you write $\sum_{j\in J} (a + b) +c+\cdots$
Also, suppose $(x_j)_{j\in J}=x_1,x_2,\ldots,x_j$ is a sequence with domain $J$. It has been pointed out in the comments by SebastianSchoennenbeck, if one wants to take the sum over all numbers $j$ such that the sequence $(x_j)_{j\in J}$ is defined, it is clearer to write $\sum_{j\in J} (x_j + C_1)$ than $\sum_{j} (x_j + C_1)$. Knuth's definition of the $\sum$ operator agrees, stating
Formally, we write $\sum _{P(j)} x_j$ as an abbreviation for the sum of all terms $x_j$ such that $j$ is an integer satisfying a given property $P(k)$. A 'property $P(k)$' is any statement about $k$ that can be either true or false (Knuth, Concrete Mathematics, 2e, p.23).
Observe that $j\in J$ is a statement, while $j$ is not, thus only $\sum_{j\in J}$ is correct notation.
Now, i'll discuss each of your suggestions. Note $\#1=\#3\neq \#2$!!!
- $\sum_{j\in J} (x_j + C_1) + C_2$
This is the correct notation for taking the sum over the sum of $x_j + C_1$. The parenthesis indicates that (the sum of) both terms $x_j,C_1$ are included in the summand (ie the thing to be summed). Since the $\sum $ operator is linear, we have that: \begin{align} \sum_{j\in J} (x_j + C_1) + C_2 &= \sum _{j\in J} x_j +\sum _{j\in J} C_1 +C_2 \\ &= \sum _{j\in J} x_j +|J|\cdot C_1 +C_2 \\ \end{align}
with the last equality because $C_1$ is a constant, and any constant (say $C_1$) plus itself $|J|$ times, = $|J| \cdot C_1$.
- $[\sum_{j\in J} x_j + C_1] + C_2$
This notation only takes the sum of $x_j$. This expression is not equal to the other two because here $C_1$ is not in first term ($x_j$) and thus not in the summand. Thus your middle term is only a $C_1$ term instead of a $|J|\cdot C_1$.
This gets to the heart of your question: Only the first term after the $\sum$ is considered to be in the summand (the first term being the stuff before any $+,-$). If we want to take the sum over $x$ terms we enclose them in parentheses.
Writing out some simple sums may help you understand the notation. Suppose we have a sequence $(a_j)=a_1,a_2,a_3$ (whose domain is $J$={1,2,3}), and a constant C.
$$\sum_{j\in J} a_j + C= a_1 + a_2 +a_3 + C$$
Observe that since $C$ is not in the first term, and there are no parentheses, it is not in the summand, and we do not sum over it. In contrast:
\begin{align} \sum_{j\in J} (a_j + C) &= (a_1+C) + (a_2+C) +(a_3 + C)\\ &= \sum_{j\in J} (a_j) + 3C \\ &= \sum_{j\in J} (a_j)+ \sum_{j\in J}(C) \end{align}
Finally, note that that $\sum _{j\in J} (x_j)$ is just a real number (real numbers are closed under addition), call it $R_1$. Then you have $[\sum_{j\in J} x_j + C_1] + C_2=(R_1+C_1)+C_2$. Of course $R_1+C_2$ is also a real number say $R_2$, thus making these substitutions, you wrote $(R_2)+C_2$. Thus the parentheses are clearly redundant.
- $[\sum_{j\in J} (x_j + C_1)] + C_2$
This is just #1 with a redundant pair of outer parentheses. That is $\sum_{j\in J} (x_j + C_1)$ is just a real number say $R_3$. Thus you wrote $ (R_3)+C_1$.
References:
Stewart, Calculus, 7e, p, A34-A37. Knuth, Concrete Mathematics, 2e, 21-33.
Solution 3:
Good question.
If the terms carry the index, then it can be implicitly assumed that they are part of the summand, e.g.
$$\sum_{i} a_i+b_i+c=\sum_{i}(a_i+b_i)+c$$ even though it is not clear if $c$ is part of the summand. Here it is assumed not.
Perhaps one could use the vinculum if one doesn't want the summand to be cluttered with brackets, e.g. $$\sum_{i} \overline{a_i+b_i+c}+d$$
Addendum
One way to think about it is to consider the summand to be the first "object" immediately after the summation sign. "Object" here can be a single term (e.g. $a_i$), a product of two or more terms (e.g. $a_i b_i$), or terms included in brackets (e.g. $(a_i+c)$ ). Other terms are not to be summed.
However, one may choose to intepret a stray indexed summand as implying parentheses e.g. taking $\displaystyle\sum_i a_i+b+c+d_i$ to mean $\displaystyle\sum_i (a_i+b+c+d_i)$.