# 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

1. $$\sum_j (x_j + E_1) + E_2$$
2. $$[\sum_j x_j + E_1] + E_2$$
3. $$[\sum_j (x_j + E_1)] + E_2$$

The third one is certainly unambiguous but too verbose I think.

• I think the first is reasonably clear. The second is ambiguous because even with the brackets it's not clear whether the $C_{1}$ is included in the sum or not. Commented May 31, 2017 at 0:06
• Personally, I use #1. Its natural since we read $f(x)$ as "f of x", and thus read $\sum _j (x_j +C_1)$ as "sum of $x_j+C_1$" If you want a really good discussion of Sum notation see Knuths "Concrete Mathematics" chapter 2. google.com/…. Commented May 31, 2017 at 0:07
• I shudder at all but the first option, although the third one is also ok. There are some arguably objective problems with the second, as has been pointed out. Commented May 31, 2017 at 0:08
• Best option is $C_2 + \sum_j(x_j + C_1)$. Commented May 31, 2017 at 0:10
• @DMcMor By the same token if $C_1$ is 0 (absent) it is not clear if $C_2$ is part of the summation, unless that is the case by convention. Commented May 31, 2017 at 0:20

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

$$[\sum_{i=1}^3 x_i+C_1]+C_2 ~{= \big((x_1+x_2+x_3)+C_1\big)+C_2\\= (x_1+x_2+x_3)+C_1+C_2}$$

• This is an excelent answer. Easy to read and to keep as reference for when my students ask about it. The color highlights help a lot. Commented May 31, 2017 at 13:44
• Actually it seems the convention is that juxtaposition has higher precedence than summation or products, but that the multiplication symbol has lower precedence. Hence $\sum_{k=1}^n k \sum_{m=1}^n m^2 = \sum_{k=1}^n ( k \sum_{m=1}^n m^2 )$ while $\sum_{k=1}^n k \times \sum_{m=1}^n m^2 = ( \sum_{k=1}^n k ) \times ( \sum_{m=1}^n m^2 )$. Of course, there will always be people who adopt different precedence rules... Commented May 31, 2017 at 16:45

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 $\#1=\#3\neq \#2$!!!

1. $\sum_{j\in J} (x_j + C_1) + C_2$

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, = $|J| \cdot C_1$.

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.

1. $[\sum_{j\in J} (x_j + C_1)] + C_2$

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.

• Quick note: $\sum_j C_1 \neq j C_1$, since $j$ is the parameter of the sum not the number of terms. Commented May 31, 2017 at 6:27
• @SebastianSchoennenbeck Good catch. When the index/parameter $j$ has no bounds, are we to assume the sum ranges over all numbers such that the sequence $(x_j)$ is defined? If so, there should be $|dom((x_j))|$ terms in the sum and thus $\sum_j C_1= |dom((x_j))|\cdot C_1$ ? Commented May 31, 2017 at 7:41
• That is usually the implied meaning. Generally speaking I would always include the set over which we are taking the sum, e.g. $\sum_{j \in J}$ in which case we get $|J|\cdot C_1$ (which probably implicitly tells us that $J$ is finite). Commented May 31, 2017 at 8:49
• $\sum_j C_1\ne jC_1$ Commented May 31, 2017 at 8:56
• @SebastianSchoennenbeck and Hagen. Knuth agrees with you on the notation Sebastian. I have edited the post to explain the OPs notation for the scope of the sum is ambiguous, changed my notation to follow yours/Knuth's, fixed the algebra, and quoted Knuth's definition of the sum into the answer. Commented May 31, 2017 at 14:40

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. $$\sum_{i} \overline{a_i+b_i+c}+d$$

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 $\displaystyle\sum_i (a_i+b+c+d_i)$.

• I think this is reasonable advice for interpreting bad notation if you encounter it, but I would not recommend it at all as something you should write yourself.
– user856
Commented May 31, 2017 at 8:47
• No, don't use the vinculum. That is dreadful advice. Commented May 31, 2017 at 15:05
• @Rahul - agreed. A useful guide for interpretation, but not for application. Commented May 31, 2017 at 15:14
• @TonyK - Haha what do you have against the vinculum? Commented May 31, 2017 at 15:14
• Nobody would know what it means. Well, I wouldn't, and I was pretty well trained. Commented May 31, 2017 at 15:54

$C_2 + \sum_{i = 1}^n C_1 + x_i$ is not ambiguous, but $C_2 + \sum_i(x_i + C_1)$ seems nicer to me.

What about $C_2 + nC_1 + \sum_i x_i$?

• What if $C_2$ is sum over an index $j$. As I said don't assume it's constant. Maybe I should change the name $C$. Commented May 31, 2017 at 15:55
• @konpsych Then, I a line after this would contain the words where $C_2 = \sum_j \dots$ :) Commented May 31, 2017 at 18:46

The original problem is that you want to add $C_1$ to the sum. As you have not given a sensible reason why this is desirable/optimal, I think you should avoid it. Instead, write as

$$\sum_{j} x_j + nC_1 + C_2$$

You still preserve the $C_i$ ordering (which @Antonie's answer, albeit interesting, does not), and you are not going to confuse anyone who is familiar with summation. You hardly ever see expressions where every term in a summation is surrounded by brackets.

• An argument against adding $C_1$ out of the sum is that the set over which j is added may not be easy to enumerate. E.g j takes values in the set of all primes between $N_1$ and $N_2$. Commented May 31, 2017 at 20:00
• @konpsych Good point! Commented May 31, 2017 at 20:19