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Section1.2Sums and products

In this section we introduce summation notation and product notation which we will use later on. The special symbol \(\sum\) is used to denote sums. Let us consider an example

\begin{equation*} \sum_{k=1}^5 k=1+2+3+4+5\text{.} \end{equation*}

In a more general form

\begin{equation*} \sum_{k=m}^n k=m+(m+1)+\ldots+(n-1)+n\text{.} \end{equation*}

Here \(m\) is the lower bound of summation and \(n\) is the upper bound of summation. There are some other possibilities to express the above sum, e.g.

\begin{align*} \amp \sum_{m\leq k\leq n}k,\\ \amp \sum_{k\in S}k, \mbox{ where } S=\halmaz{m,m+1,\ldots,n}\text{.} \end{align*}

It is important to note that the variable used in the summation is arbitrary. That is, the values of the following summations are equal:

\begin{equation*} \sum_{k=1}^3 k^2=1^2+2^2+3^2=14 \end{equation*}

and

\begin{equation*} \sum_{m=1}^3 m^2=1^2+2^2+3^2=14\text{.} \end{equation*}

Now we write out explicitly some other summations:

(a)

\begin{equation*} \sum_{i=2}^6 (2-i)=(2-2)+(3-2)+(4-2)+(5-2)+(6-2)=10\text{,} \end{equation*}

(b)

\begin{equation*} \sum_{j=3}^5 2^{j-2}=2^{3-2}+2^{4-2}+2^{5-2}=14\text{,} \end{equation*}

(c)

\begin{equation*} \sum_{1\leq i,j\leq 2} ij= (1\cdot 1)+(1\cdot 2)+(2\cdot 1)+(2\cdot 2)=9\text{.} \end{equation*}

Now we deal with products of mathematical expressions. The symbol used in this case is \(\prod\text{.}\) Product notation is very similar to summation notation so it is straightforward to learn to use. The first example in case of summation notation was

\begin{equation*} \sum_{k=1}^5 k=1+2+3+4+5\text{.} \end{equation*}

If we change the \(\sum\) symbol to \(\prod\text{,}\) then we obtain

\begin{equation*} \prod_{k=1}^5 k=1\cdot 2\cdot 3\cdot 4\cdot 5\text{.} \end{equation*}

Let us consider the product of integers between \(m\) and \(n\text{,}\) where \(m\lt n\text{.}\) We can write it in product notation in different forms:

\begin{align*} \amp \prod_{k=m}^n k,\\ \amp \prod_{m\leq k\leq n}k,\\ \amp \prod_{k\in S}k, \mbox{ where } S=\halmaz{m,m+1,\ldots,n}\text{.} \end{align*}

It may happen that the sum or product should be evaluated on the empty set. By definition, in such situations the sum is always 0 and the product is always 1, e.g.

\begin{align*} \sum_{k \in \emptyset} k \amp = 0,\\ \prod_{k \in \emptyset} k \amp = 1\text{.} \end{align*}

If \(S\) and \(T\) be two disjoint sets, then

\begin{gather*} \sum_{k \in S} k + \sum_{k \in T} k = \sum_{k \in S \cup T} k,\\ \prod_{k \in S} k \cdot \prod_{k \in T} k = \prod_{k \in S \cup T} k\text{.} \end{gather*}

Note, that this is true even if \(S\) or \(T\) is the empty set. (This is the main reason we define the empty sum to be 0 and the empty product to be 1.)

There is a special notation for the product of positive integers up to \(n\text{,}\) that is, when we multiply the elements of

\begin{equation*} S_n = \halmazvonal{k}{k \text{ is a positive integer } , k\leq n} = \halmaz{1, 2, \dots , n}\text{.} \end{equation*}

The product of the elements of \(S_n\) is called \(n\) factorial and denoted by \(n!\text{,}\) that is,

\begin{equation*} n! = \prod_{k \in S_n} k = \prod_{k=1}^n k = 1 \cdot 2 \cdot \dots \cdot (n-1) \cdot n\text{.} \end{equation*}

We even define \(0!\text{,}\) that is, the products of elements of \(S_0\text{:}\)

\begin{equation*} 0! = \prod_{k \in S_0} k = \prod_{k \in \emptyset} k = 1\text{.} \end{equation*}

Factorials are always computed before any other operation. For example

\begin{align*} 2+3! \amp = 2 + 1\cdot 2 \cdot 3 = 2 + 6 = 8,\\ (2+3)! \amp = 5! = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120\text{.} \end{align*}

Expand the following sums.

(a) \(\sum_{i=4}^7 i\text{,}\)

(b) \(\sum_{i=1}^5 (i^2-i)\text{,}\)

(c) \(\sum_{i=1}^4 10^i\text{,}\)

(d) \(\sum_{2\leq i\leq 5} \frac{1}{2^i}\text{,}\)

(e) \(\sum_{i\in S} (-1)^i\text{,}\) where \(S=\halmaz{2,3,5,8}\text{.}\)

Write the following expressions in summation notation.

(a) \(2+4+6+8+10\text{,}\)

(b) \(1+4+7+10\text{,}\)

(c) \(\frac{1}{4}+\frac{1}{2}+1+2+4\text{,}\)

(d) \(\frac{1}{4}-\frac{1}{2}+1-2+4\text{.}\)

Expand the following products.

(a) \(\prod_{i=-4}^{-1} i\text{,}\)

(b) \(\prod_{i=1}^4 (i^2)\text{,}\)

(c) \(\prod_{i=1}^3 2^i\text{,}\)

(d) \(\prod_{-2\leq i\leq 3} \frac{1}{2^i}\text{,}\)

(e) \(\prod_{i\in S} (-1)^i\text{,}\) where \(S=\halmaz{2,4,6,7}\text{.}\)

Write the following expressions in product notation.

(a) \(1\cdot 3\cdot 5\cdot 7\text{,}\)

(b) \((-1)\cdot 2\cdot 5\cdot 8\text{,}\)

(c) \(\frac{1}{9}\cdot\frac{1}{3}\cdot 1\cdot 3\cdot 9\text{.}\)

Compute the values of \(n!\) for every \(n \in \halmaz{0, 1, 2, 3, 4, 5, 6, 7, 8}\text{.}\)

Compute the values of

\begin{align*} \amp 5+3!\\ \amp (5+3)!\\ \amp 4-2\cdot 3!\\ \amp (4-2)\cdot 3!\\ \amp 4 - (2 \cdot 3)!\\ \amp 3 \cdot 2!\\ \amp (3 \cdot 2)!\\ \amp 4 \cdot 3!\\ \amp 4! \cdot 5\text{.} \end{align*}

Prove that \(n! = n \cdot (n-1)!\) for every positive integer \(n\text{.}\) Note, that it is even true for \(n=1\text{,}\) which is one of the reasons we define \(0!\) to be equal to 1.