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I know that $$ \sum_{i=1}^n a_i = a_1 + a_2 + \dots + a_n $$ and I get how to, in general, manipulate expressions with these finite sums. This has typically been introduced to me as simply just being the sum of $a_1$ up to $a_n$. My question is

How can one rigorously define this sum?

(While not being a strict follow-up question to

Changing finite double summation $\sum_{i=1}^n\sum_{k=1}^i f(k) = \sum_{k=1}^n\sum_{i=1}^k f(n+1-k).$

this is the post that is the source of my question?)

EDIT: If there is no definition that is more rigorous, then I will accept the answer saying that. I ask because I know that things that seem fairly clear have more precise definitions. For example, I know that a function has a set theoretic definition. So I wonder if this finite sum above might have something similar.

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    $\begingroup$ Honest question: why isn't that rigorous enough? What's wrong with it? $\endgroup$ – Randall Sep 5 '18 at 17:14
  • $\begingroup$ @Randall: I was just wondering. I guess I would accept the answer saying that there is not other "more rigorous" definition. I know that some clear things (like functions) actually have set theoretic definitions. So I was wondering if there is some more rigorous definition of this finite sum. $\endgroup$ – John Doe Sep 5 '18 at 17:16
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    $\begingroup$ I think an inductive definition will do it, like $\sum\limits_{i=1}^0 a_i=0$ and $\sum\limits_{i=1}^{n+1} a_i = \left(\sum\limits_{i=1}^n a_i\right) + a_{n+1}$. $\endgroup$ – lisyarus Sep 5 '18 at 17:17
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This can be done inductively—what follows is a completely over-the-top way of doing it.

Given any additive group $(G, +, 0)$ (even a monoid would do), such as the real numbers $\mathbb{R}$ under addition, first define the set $G^n$ of $n$-tuples of elements of $G$ inductively by $$G^0 = \{ () \} \quad \text{and} \quad G^{n+1} = \{ (\vec a, a_{n+1}) \mid \vec a \in G^n,\ a_{n+1} \in G \}$$ Thus the elements of $G^n$ are lists $(a_1,a_2,\dots,a_n)$ of length $n$ of elements of $G$.$^{\dagger}$

Now we can use this inductive characterisation of $G^n$ for all $n \in \mathbb{N}$ to define, for each $n \in \mathbb{N}$, a function $$S_n : G^n \to G$$ by declaring $$S_0(()) = 0 \quad \text{and} \quad S_{n+1}((\vec a, a_{n+1})) = S_n(\vec a) + a_{n+1}$$ for each $n \in \mathbb{N}$ and each $(\vec a, a_{n+1}) \in G^{n+1}$.

Finally, define $$\sum_{i=1}^n a_i = S_n(\vec a)$$ for each $n \in \mathbb{N}$ and $\vec a \in G^n$.


${}^{\dagger}$More precisely, the lists are parenthesised to the left, so that $(a_1,a_2,a_3,a_4)$ is really $(((a_1,a_2),a_3),a_4)$, for instance.

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  • $\begingroup$ I like over-the-top! $\endgroup$ – John Doe Sep 5 '18 at 17:24
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We could also define that by recurrence/induction

  • $S_0=\sum_{i=1}^0 a_i= 0$
  • for $n\ge 0$ given $S_n=\sum_{i=1}^n a_i$ we define $S_{n+1}=\sum_{i=1}^{n+1} a_i=S_n + a_{n+1}$

but the original definition is sufficiently clear as it is.

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  • $\begingroup$ What about the empty sum? $\endgroup$ – mr_e_man Sep 5 '18 at 20:26
  • $\begingroup$ @mr_e_man Do you suggest to start from $S_0=S_0=\sum_{i=0}^0 = 0$? $\endgroup$ – user Sep 5 '18 at 20:28
  • $\begingroup$ Almost. $S_0 = \sum_{i=1}^0 a_i = 0$. $\endgroup$ – mr_e_man Sep 5 '18 at 20:28
  • $\begingroup$ @mr_e_man Could you please explain why is it necessary? Why does not suffice start from $S_1$? $\endgroup$ – user Sep 5 '18 at 20:31
  • $\begingroup$ It's not necessary, but it's more general, like defining the integral $\int_a^af(x)dx = 0$. And it makes sense of some things that otherwise are undefined. $\endgroup$ – mr_e_man Sep 5 '18 at 20:31
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You can define it recursively (actually, it's a good example of a recursive definition):

  • $\displaystyle\sum_{i=1}^1a_i=a_1$;
  • $\displaystyle\sum_{i=1}^{n+1}a_i=\left(\sum_{i=1}^na_i\right)+a_{n+1}$.
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  • $\begingroup$ What about the empty sum? $\endgroup$ – mr_e_man Sep 5 '18 at 20:26
  • $\begingroup$ @mr_e_man I assumed that, for the OP, $n\in\mathbb N$. $\endgroup$ – José Carlos Santos Sep 5 '18 at 23:21
  • $\begingroup$ It is not generally agreed whether $\mathbb N$ contains $0$. $\endgroup$ – mr_e_man Sep 5 '18 at 23:22
  • $\begingroup$ @mr_e_man Perhaps. But it is generally agreed that, in an expression of the type $\sum_{i=a}^ba_i$, $b\geqslant a$. $\endgroup$ – José Carlos Santos Sep 5 '18 at 23:24
  • $\begingroup$ I disagree that that is generally agreed! ;) Compare this with definite integrals, where $\int_a^b$ is defined regardless of the relation of $a$ and $b$. (The comparison is easier using asymmetric intervals: $\sum_{a\leq i<b} = \sum_a^{b-1}$ corresponds to $\int_a^b$, especially when $a=b$.) $\endgroup$ – mr_e_man Sep 5 '18 at 23:26
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For finite sums the only thing not rigorous is that addition is defined with two values, so you would have to write $(a_1+a_2)+a_3$ for a triple sum. Each addition then only involves two numbers. Once we prove the commutativity and associativity of addition we know that all ways of grouping the terms gives the same answer, so writing a sum with a larger number of terms is no problem. $\sum_{i=1}^n$ is just a convenient bookkeeping measure for the sum.

This does not apply to infinite sums. You can't just extend the finite approach, so we have to do something different. The something different is the limit approach you have seen, and that becomes the definition of $\sum_{i=1}^\infty$

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The concept of Sum has two basic definitions.

1) Sum over a set

$$ \sum\limits_{x\, \in \,A} {f(x)} \quad \Rightarrow \quad \sum\limits_{x\, \in \,A\, \cap \,B} {f(x)} + \sum\limits_{x\, \in \,A\,\backslash \,B} {f(x)} $$

The meaning of which is obvious, and where you can limit the sum to the $x \in A$, or consider that $f(x)$ be null for $x \notin A$ and take the sum for all the $x$ within the considered field.

If the set $A$ is partitioned into the subsets $A_k \quad | \; k \in C$ then $$ \left\{ \matrix{ A = \bigcup\limits_{k\, \in \,C} {A_{\,k} } \hfill \cr A_{\,k} \cap A_{\,j} = \emptyset \quad \left| {\;k \ne j} \right. \hfill \cr} \right.\quad \Rightarrow \quad \sum\limits_{x\, \in \,A} {f(x)} = \sum\limits_{k\, \in \,C} {\sum\limits_{x\, \in \,A_{\,k} } {f(x)} } $$

Then, in $2$D, if the summing set is partitioned into subsets where in each of them the value of $x$ is constant, then $$ \eqalign{ & \sum\limits_{\left( {x,y} \right)\, \in \,A} {f(x,y)} = \sum\limits_{k\, \in \,C} {\sum\limits_{\left( {x,y} \right)\, \in \,A_{\,k} } {f(x,y)} } = \sum\limits_{k\, \in \,C} {\sum\limits_{\left( {x,y} \right)\, \in \,A_{\,k} } {f(x_{\,k} ,y)} } = \cr & = \sum\limits_{k\, \in \,C} {\sum\limits_{y\, \in \,A_{\,k} (x_{\,k} )} {f(x_{\,k} ,y)} } = \sum\limits_{k\, \in \,C} {g(} x_{\,k} ) \cr} $$ If an analogue partition can be done for $y$, then you can take the one which is "easier", has a known closed form, ... and "manouvre" between the two.

2) Indefinite Sum

While in the definition above we make full use of the commutative and associative properties of addition, in conjunction with union, intersection and derivated concepts for sets (finite, numerable, ...), the concept of Indefinite Sum is somewhat different.

If we have that $$ \Delta _{\,x} \,F(x) = F(x + 1) - F(x) = f(x) $$ then we write $$ F(x) = \Delta _{\,x} ^{\, - \,1} \,F(x) = \sum\nolimits_{\;x\;} {f(x)} $$ in particular we have $$ F(b) - F(a) = \sum\nolimits_{\;x = \,a\,}^b {f(x)} = - \sum\nolimits_{\;x = \,b\,}^a {f(x)} $$

For example $$ \eqalign{ & F(x) = \left( \matrix{ x \cr 2 \cr} \right)\quad \Leftrightarrow \quad \left( \matrix{ x + 1 \cr 2 \cr} \right) - \left( \matrix{ x \cr 2 \cr} \right) = \left( \matrix{ x \cr 1 \cr} \right) = x\quad \Leftrightarrow \cr & \Leftrightarrow \quad \sum\nolimits_{\;x = \,a\,}^{\;b} x = \left( \matrix{ b \cr 2 \cr} \right) - \left( \matrix{ a \cr 2 \cr} \right)\quad \left| {\;a,b \in C} \right.\quad \Rightarrow \cr & \Rightarrow \quad \sum\nolimits_{\;k = \,0\,}^{\;n} k \quad \left| {\;0 \le n \in Z} \right.\quad = \sum\limits_{0\, \le \,k\, \le \,n - 1} k = \left( \matrix{n \cr 2 \cr} \right) = {{n\left( {n - 1} \right)} \over 2} \cr} $$

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