The sigma Greek letter is often used to denote summation. For example, $\sum_{i=2}^{5} i^2=2^2+3^2+4^2+5^2=4+9+16+25=54$. In general, the sigma (summation) notation is $\sum_{i=b}^{e} f(i)$ where $i$ is the index variable, $b$ is the start integer value of $i$, $e$ is the inclusive end value of $i$, and $f(i)$ is a function of $i$ where the value of the function is being added.
In our previous example, we can think of $b=2$, $e=5$, and $f(i)=i^2$.
In general, summation is difficult to define because of the way it expands. Sure, most people get the idea of what it means, but can summation be defined in a really clear way?
Let’s give it a try.
First, the start value of $i$ should be less than or equal to the $e$ value, we can express this as follows:
$b > e \Rightarrow \sum_{i=b}^{e}f(i)=0$
This takes care of the weird case where the start value of $i$ is greater than the end value. Then, we can think of the special case when $b=e$:
$b = e \Rightarrow \sum_{i=b}^{e}f(i)=f(b)=f(e)$
Ah, that kind of makes sense! Now we have the tricky case when $b < e$. We know that in this case, $f(b)$ has to be part of the sum, but what about the rest? Let’s check out the following:
$b < e \Rightarrow \sum_{i=b}^{e}f(i) = f(b)+\sum_{i=b+1}^{e}f(i)$
That’s right, we just defined summation in a recursively way! In fact, using the ternary operator, we can clean up the definition into something very neat:
$\sum_{i=b}^{e}f(i) = (b > e) ? 0 : \left(f(b)+ \sum_{i=b+1}^{e} f(i) \right)$
The following illustrates how this recursive definition expands in our example:
$\begin{eqnarray} \sum_{i=2}^{5} i^2 & = & 2^2 + \sum_{i=3}^{5} i^2 \ & = & 2^2 + 3^2 + \sum_{i=4}^{5} i^2 \ & = & 2^2+3^2+4^2+\sum_{i=5}^{5} i^2 \ & = & 2^2+3^2+4^2+5^2+\sum_{i=6}^{5} i^2 \ & = & 2^2+3^2+4^2+5^2+0 \end{eqnarray}$
Summation (addition) is not the only operator that is used in computer science:
Note that for multiplication, the base-case default value is 1 because 1 is the identity of multiplication.
$\Pi_{i=b}^{e}f(i) = (b > e) ? 1 : \left(f(b) \cdot \Pi_{i=b+1}^{e} f(i) \right)$
For disjunction, the base-case default value is false (0).
$\bigvee_{i=b}^{e}f(i) = (b > e) ? 0 : \left(f(b) \vee \bigvee_{i=b+1}^{e} f(i) \right)$
For conjunction, the base-case default value is true (1).
$\bigwedge_{i=b}^{e}f(i) = (b > e) ? 1 : \left(f(b) \wedge \bigwedge_{i=b+1}^{e} f(i) \right)$
In an interesting twist, the existential quantifier can also be thought of as a “big operator”, iterating through elements:
$\left( \exists e \in X\left(f(e)\right) \right) = \bigvee_{e \in X} f(e)$
From here, the issue is that the iterations through members of $X$ are not well stated. Because elements in a set have no inherent ordering, it is not possible to directly address an element in a set. To solve this problem, we need to create a new notation $d(X)$ to “deterministically draw an element from a non-empty set $X$”. This means that given $X$ does not change, $d(X)$ draws exactly the same element every time the function is called.
With this notation, we can define the following:
$\left(\exists e \in X\left(f(e)\right) \right) = \left( \bigvee_{e \in X} f(e) \right) = (X = \{\}) ? 0 : \left( f \left( d(X) \right) \vee \bigvee_{e \in (X-\{d(X)\})} f(e) \right)$
The universal quantifier can be seen as a “big operator”:
$\left( \forall e \in X\left(f(e)\right) \right) = \bigwedge_{e \in X}f(e)$
Similar to the existential quantifier, we can use the notation $d(X)$ as follows:
$\left(\forall e \in X \left(f(e) \right) \right) = \left( \bigwedge_{e \in X} f(e) \right) = (X = \{\}) ? 1 : \left( f \left( d(X) \right) \wedge \bigwedge_{e \in (X-\{d(X)\})} f(e) \right)$
Since we have introduced $d(X)$ at this point, we can also define other big operators based on iterating through elements in a set. In general, if $\square$ is an operator, and $i(\square)$ is the identity of the operator such that $\forall x(x \square i(\square) = x)$, then we can define the following:
$\square_{e \in X} f(e) = (X=\{\}) ? i(\square) : f(d(x)) \square \left( \square_{e \in (X-\{d(x)\}} f(e) \right)$
Let us apply this abstract big operator format to the big union. Union ($\cup$) has an identity of an empty set (${}$). This means the following:
$\bigcup_{e \in X}f(e) = (X = \{\}) ? \{\} : \left(f\left(d(X)\right) \cup \left(\bigcup_{e \in X-\{d(X)\}}f(e) \right) \right)$
Yes, for two reasons.
The first reason is that there is no better way to explain the summation (sigma) notation. The recursive definition only relies on existing operators (addition, etc.) and the very same definition being defined. The key is the base-case where the summation returns the identity of addition, zero.
In other words, this recursive definition of summation is concise (short, does not take many words), complete (does not depend on something else), and sound (correct).
The second reason is that this is a good example of recursive definitions. The non-recursive definition itself is fairly intuitive, and most people do not use a recursive definition. But this is why it is a good example for teaching the use of recursive definitions.