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Products

Sums and products · Chapter 2 · Lecture notes (PDF) · Exercises: Products · Slides (PDF) · App: Sums and products

1. Definition

Definition 1: product

Given \(n\) values \(a_j \in \R\) with \(j \in \{1,2,\dots,n\}\), the product

\[ a_1 \; a_2 \; {\rm \dots} \; a_n \]

can be indicated in compact form with the product symbol:

\[ \prod_{j=1}^n a_j \]

which reads: “product for \(j\) from \(1\) to \(n\) of \(a_j\)”. The symbol \(j\) is called product index.

  • The product symbol is therefore very useful when the terms \(a_j\) are explicitly defined as a function of the product index \(j\).

Example 1: products

\[\begin{align*} \prod_{j=1}^{10} \frac{1}{j} &~~=~~ 1 \; \frac{1}{2} \; \frac{1}{3} \; \frac{1}{4} \; \frac{1}{5} \; \frac{1}{6} \; \frac{1}{7} \; \frac{1}{8} \; \frac{1}{9} \; \frac{1}{10} \\[2ex] \prod_{j=3}^{n} j^2 &~~=~~ 3^2 \; 4^2 \; 5^2 \; {\rm \dots} \; n^2 \end{align*}\]

2. Properties

Observation 1: product with constant term

For every natural number \(n\ge 1\) and real number \(c\), we have:

\[\begin{equation} \label{P2} \prod_{j=1}^n c = c^n \end{equation}\]
Proof

We have:

\[ \underbrace{c\: \; c \; {\rm \dots} \; c\:}_{=\prod_{j=1}^n c, {\rm ~~~} c {\rm ~multiplied~} n {\rm ~times}} = c^n \]

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Observation 2: product by a constant

For every product \(\prod_{j=1}^n a_j\) and real number \(c\), we have:

\[\begin{equation} \label{P1} \prod_{j=1}^n (c \; a_j) = c^n \: \prod_{j=1}^n a_j \end{equation}\]
Proof

From the commutative and associative properties of the product, we have:

\[ \underbrace{c\: a_1 \; c\: a_2 \; {\rm \dots} \; c\: a_n}_{=\prod_{j=1}^n (c \; a_j)} = \underbrace{\underbrace{(c\: \; c \; {\rm \dots} \; c)\:}_{=\prod_{j=1}^n c, {\rm ~~~} c {\rm ~multiplied~} n {\rm ~times}} \; (a_1 \; a_2 \; {\rm \dots} \; a_n)}_{= c^n \: \prod_{j=1}^n a_j} \]

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Observation 3: product of products

For every pair of products \(\prod_{j=1}^n a_j\) and \(\prod_{j=1}^n b_j\), we have:

\[\begin{equation} \label{P3} \prod_{j=1}^n a_j \; \prod_{j=1}^n b_j = \prod_{j=1}^n (a_j \; b_j) \end{equation}\]
Proof

We have:

\[ \underbrace{a_1 \; a_2 \; {\rm \dots} \; a_n \; b_1 \; b_2 \; {\rm \dots} \; b_n}_{=\prod_{j=1}^n a_j \; \prod_{j=1}^n b_j} = \underbrace{a_1 \; b_1 \; a_2 \; b_2 \; {\rm \dots} \; a_n \; b_n}_{=\prod_{j=1}^n (a_j \; b_j) } \]

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Observation 4: decomposition

For every product \(\prod_{j=1}^{n+m} a_j\), we have:

\[\begin{align} \label{P4} \prod_{j=1}^{n+m} a_j &= \prod_{j=1}^n a_j \; \prod_{j=n+1}^{n+m} a_j \end{align}\]

Observation 5: index translation

For every product \(\prod_{j=1}^{n} a_j\) and natural number \(m \ge 1\), we have:

\[\begin{align} \label{P5} \prod_{j=1}^n a_j &= \prod_{j=1+m}^{n+m} a_{j-m} = \prod_{j=1-m}^{n-m} a_{j+m} \end{align}\]

Observation 6: index reflexion

For every product \(\prod_{j=1}^{n} a_j\), we have:

\[\begin{align} \label{P6} \prod_{j=1}^n a_j &= \prod_{j=1}^n a_{n-j+1} = \prod_{j=0}^{n-1} a_{n-j} \end{align}\]
Proof

The three properties are simply different notations and/or arrangements of the terms in the products. □

3. Factorial

Definition 2: Factorial

The factorial of a natural number \( n \), denoted as \( n! \), is the product of the first \( n \) positive natural numbers. By convention, the factorial of 0 is defined as \( 1 \).

  • Formally, the factorial is defined as follows:

    \[\begin{equation} n! = \begin{cases} 1 & \text{if~~} n = 0, \\[2ex] \prod_{j=1}^n j & \text{if~~} n \geq 1. \end{cases} \end{equation}\]

  • A closed formula for the factorial is not known but Stirling's approximation provides an asymptotic approximation.

Observation 7

For every pair of natural numbers \(n \ge 0\) and \(k \in \{0,1,\dots, n\}\), we have:

\[\begin{equation} \label{MM} \frac{n!}{(n-k)!} = \prod_{j=1}^{k} (n-j+1) = \prod_{j=n-k+1}^{n} j \end{equation}\]

  • Formula \(\eqref{MM}\) is the product of \(k\) factors, from \(n\) and decreasing by one unit at a time.

  • By convention, a product with no factors (i.e., whose upper index is smaller than its lower index, as \(\prod_{j=1}^{0} a_j\)) is equal to \(1\). This convention is consistent with \(0!=1\) and makes formula \(\eqref{MM}\) valid also for \(k=0\) and \(k=n\).

Proof

We have:

\[ \frac{n!}{(n-k)!} = \frac{\prod_{j=1}^n j}{\prod_{j=1}^{n-k} j} = \frac{\prod_{j=1}^{n-k} j \; \prod_{j=n-k+1}^{n} j }{\prod_{j=1}^{n-k} j} = \prod_{j=n-k+1}^{n} j = \prod_{j=1}^{k} (n-j+1) \]

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Example 2: use of the observation

\[ \frac{100!}{98!}= \frac{100!}{(100-2)!}=\prod_{j=1}^{2} (100-j+1)=100 \cdot 99 = 9.900 \]

Try it in the lab

the same computation step by step: change the matrix and watch the steps change.

Exercises and lab