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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.

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    $\begingroup$ 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. $\endgroup$ – DMcMor May 31 '17 at 0:06
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    $\begingroup$ 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/…. $\endgroup$ – Evan Rosica May 31 '17 at 0:07
  • $\begingroup$ 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. $\endgroup$ – The Count May 31 '17 at 0:08
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    $\begingroup$ Best option is $C_2 + \sum_j(x_j + C_1)$. $\endgroup$ – Trevor Gunn May 31 '17 at 0:10
  • $\begingroup$ @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. $\endgroup$ – kon psych May 31 '17 at 0:20
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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 $\#1=\#3\neq \#2$!!!

  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.

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    $\begingroup$ Quick note: $\sum_j C_1 \neq j C_1$, since $j$ is the parameter of the sum not the number of terms. $\endgroup$ – Sebastian Schoennenbeck May 31 '17 at 6:27
  • $\begingroup$ @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 $\sum_j C_1= |dom((x_j))|\cdot C_1$ ? $\endgroup$ – Evan Rosica May 31 '17 at 7:41
  • $\begingroup$ 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). $\endgroup$ – Sebastian Schoennenbeck May 31 '17 at 8:49
  • $\begingroup$ $\sum_j C_1\ne jC_1$ $\endgroup$ – Hagen von Eitzen May 31 '17 at 8:56
  • $\begingroup$ @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. $\endgroup$ – Evan Rosica May 31 '17 at 14:40
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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

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

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  • $\begingroup$ 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. $\endgroup$ – andresgongora May 31 '17 at 13:44
  • $\begingroup$ 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... $\endgroup$ – user21820 May 31 '17 at 16:45
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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 $\displaystyle\sum_i (a_i+b+c+d_i)$.

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    $\begingroup$ 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. $\endgroup$ – Rahul May 31 '17 at 8:47
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    $\begingroup$ No, don't use the vinculum. That is dreadful advice. $\endgroup$ – TonyK May 31 '17 at 15:05
  • $\begingroup$ @Rahul - agreed. A useful guide for interpretation, but not for application. $\endgroup$ – hypergeometric May 31 '17 at 15:14
  • $\begingroup$ @TonyK - Haha what do you have against the vinculum? $\endgroup$ – hypergeometric May 31 '17 at 15:14
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    $\begingroup$ Nobody would know what it means. Well, I wouldn't, and I was pretty well trained. $\endgroup$ – TonyK May 31 '17 at 15:54
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$C_2 + \sum_{i = 1}^n C_1 + x_i$ is not ambiguous, but $C_2 + \sum_i(x_i + C_1)$ seems nicer to me.

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

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  • $\begingroup$ 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 $C$. $\endgroup$ – kon psych May 31 '17 at 15:55
  • $\begingroup$ @konpsych Then, I a line after this would contain the words where $C_2 = \sum_j \dots$ :) $\endgroup$ – Antoine May 31 '17 at 18:46
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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.

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  • $\begingroup$ 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 $N_2$. $\endgroup$ – kon psych May 31 '17 at 20:00
  • $\begingroup$ @konpsych Good point! $\endgroup$ – luchonacho May 31 '17 at 20:19

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