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Section 6.2 More about Functions

In Section 6.1, we have seen many examples of functions. We have also seen various ways to represent functions and to convey information about them. For example, we have seen that the rule for determining outputs of a function can be given by a formula, a graph, or a table of values. We have also seen that sometimes it is more convenient to give a verbal description of the rule for a function. In cases where the domain and codomain are small, finite sets, we used an arrow diagram to convey information about how inputs and outputs are associated without explicitly stating a rule. In this section, we will study some types of functions, some of which we may not have encountered in previous mathematics courses.

Beginning Activity Beginning Activity 1: The Number of Diagonals of a Polygon
A4US

A polygon is a closed plane figure formed by the joining of three or more straight lines. For example, a triangle is a polygon that has three sides; a quadrilateral is a polygon that has four sides and includes squares, rectangles, and parallelograms; a pentagon is a polygon that has five sides; and an octagon is a polygon that has eight sides. A regular polygon is one that has equal-length sides and congruent interior angles.

A diagonal of a polygon is a line segment that connects two nonadjacent vertices of the polygon. In this activity, we will assume that all polygons are convex polygons so that, except for the vertices, each diagonal lies inside the polygon. For example, a triangle (3-sided polygon) has no diagonals and a rectangle has two diagonals.

1.

How many diagonals does any quadrilateral (4-sided polygon) have?

2.

Let \(D = \mathbb{N} - \left\{ {1, 2} \right\}\text{.}\) Define \(d\x D \to \mathbb{N} \cup \left\{ 0 \right\}\) so that \(d( n )\) is the number of diagonals of a convex polygon with \(n\) sides. Determine the values of \(d(3)\text{,}\) \(d(4)\text{,}\) \(d(5)\text{,}\) \(d(6)\text{,}\) \(d(7)\text{,}\) and \(d(8)\text{.}\) Arrange the results in the form of a table of values for the function \(d\text{.}\)

3.

Let \(f\x \mathbb{R} \to \mathbb{R}\) be defined by

\begin{equation*} f( x ) = \frac{{x\left( {x - 3} \right)}}{2}\text{.} \end{equation*}

Determine the values of \(f(0)\text{,}\) \(f(1)\text{,}\) \(f(2)\text{,}\) \(f(3)\text{,}\) \(f(4)\text{,}\) \(f(5)\text{,}\) \(f(6)\text{,}\) \(f(7)\text{,}\) \(f(8)\text{,}\) and \(f(9)\text{.}\) Arrange the results in the form of a table of values for the function \(f\text{.}\)

4.

Compare the functions in Exercise 2 and Exercise 3. What are the similarities between the two functions and what are the differences? Should these two functions be considered equal functions? Explain.

Beginning Activity Beginning Activity 2: Derivatives
A4US

In calculus, we learned how to find the derivatives of certain functions. For example, if \(f( x ) = x^2( {\sin x} )\text{,}\) then we can use the product rule to obtain

\begin{equation*} f'( x ) = 2x( {\sin x} ) + x^2( {\cos x} )\text{.} \end{equation*}

1.

If possible, find the derivative of each of the following functions:

(a)

\(f( x ) = x^4 - 5x^3 + 3x - 7\)

(b)

\(g( x ) = \cos ( {5x} )\)

(c)

\(h( x ) = \dfrac{{\sin x}}{x}\)

(d)

\(k( x ) = e^{ - x^2 }\)

(e)

\(r( x ) = \left| x \right|\)

2.

Is it possible to think of differentiation as a function? Explain. If so, what would be the domain of the function, what could be the codomain of the function, and what is the rule for computing the element of the codomain (output) that is associated with a given element of the domain (input)?

Subsection Functions Involving Congruences

Theorem 3.37 and Corollary 3.38 state that an integer is congruent (mod \(n\)) to its remainder when it is divided by \(n\text{.}\) (Recall that we always mean the remainder guaranteed by the Division Algorithm, which is the least nonnegative remainder.) Since this remainder is unique and since the only possible remainders for division by \(n\) are \(0, 1, 2, \ldots , n - 1\text{,}\) we then know that each integer is congruent, modulo \(n\text{,}\) to precisely one of the integers \(0,1,2, \ldots ,n - 1\text{.}\) So for each natural number \(n\text{,}\) we will define a new set \(R_n\) as the set of remainders upon division by \(n\text{.}\) So

\begin{equation*} R_n = \{ 0, 1, 2, \ldots , n - 1 \}\text{.} \end{equation*}

For example, \(R_4 = \{0, 1, 2, 3 \}\) and \(R_6 = \{ 0, 1, 2, 3, 4, 5 \}\text{.}\) We will now explore a method to define a function from \(R_6\) to \(R_6\text{.}\)

For each \(x \in R_6\text{,}\) we can compute \(x^2 + 3\) and then determine the value of \(r\) in \(R_6\) so that

\begin{equation*} \left( {x^2 + 3} \right) \equiv r\pmod 6\text{.} \end{equation*}

Since \(r\) must be in \(R_6\text{,}\) we must have \(0 \leq r \lt 6\text{.}\) The results are shown in the following table.

Table 6.9. Table of Values Defined by a Congruence
\(x\) \(r\) where
\(\left( x^2 + 3 \right) \equiv r \pmod 6\)
0 3
1 4
2 1
3 0
4 1
5 4

The value of \(x\) in the first column can be thought of as the input for a function with the value of \(r\) in the second column as the corresponding output. Each input produces exactly one output. So we could write \(f:R_6 \to R_6\) by

\begin{equation*} f( x ) = r \text{ where } ( {x^2 + 3} ) \equiv r \pmod 6\text{.} \end{equation*}

This description and the notation for the outputs of this function are quite cumbersome. So we will use a more concise notation. We will, instead, write Let

\begin{equation*} f\x R_6 \to R_6 \text{ by } f( x ) = \left( {x^2 + 3} \right) \pmod 6\text{.} \end{equation*}

Progress Check 6.10. Functions Defined by Congruences.

We have \(R_5 = \left\{ {0, 1, 2, 3, 4} \right\}\text{.}\) Define

\begin{align*} \amp f\x R_5 \to R_5 \text{ by } f(x) = x^4 \pmod 5, \text{ for each } x \in R_5\\ \amp g\x R_5 \to R_5 \text{ by } g(x) = x^5 \pmod 5 \text{ for each } x \in R_5 \end{align*}
(a)

Determine \(f(0)\text{,}\) \(f(1)\text{,}\) \(f(2)\text{,}\) \(f(3)\text{,}\) and \(f(4)\) and represent the function \(f\) with an arrow diagram.

Solution.

\(f(0) = 0\text{,}\) \(f(1) = 1\text{,}\) \(f(2) = 1\text{,}\) \(f(3) = 1\text{,}\) \(f(4) = 1\text{.}\)

(b)

Determine \(g(0)\text{,}\) \(g(1)\text{,}\) \(g(2)\text{,}\) \(g(3)\text{,}\) and \(g(4)\) and represent the function \(g\) with an arrow diagram.

Solution.

\(g(0) = 0\text{,}\) \(g(1) = 1\text{,}\) \(g(2) = 2\text{,}\) \(g(3) = 3\text{,}\) \(g(4) = 4\text{.}\)

Subsection Equality of Functions

The idea of equality of functions has been in the background of our discussion of functions, and it is now time to discuss it explicitly. The preliminary work for this discussion was Beginning Activity 1, in which \(D = \mathbb{N} - \left\{ {1, 2} \right\}\text{,}\) and there were two functions:

  • \(d\x D \to \mathbb{N} \cup \left\{ 0 \right\}\text{,}\) where \(d( n )\) is the number of diagonals of a convex polygon with \(n\) sides

  • \(f\x \mathbb{R} \to \mathbb{R}\text{,}\) where \(f( x ) = \dfrac{{x\left( {x - 3} \right)}}{2}\text{,}\) for each real number \(x\text{.}\)

In Beginning Activity 1, we saw that these two functions produced the same outputs for certain values of the input (independent variable). For example, we can verify that

\begin{align*} d( 3 ) \amp = f( 3 ) = 0, \qquad \amp d( 4 ) = f( 4 ) = 2,\\ d( 5 ) \amp = f( 5 ) = 5, \qquad \text{ and } \amp d( 6 ) = f( 6 ) = 9\text{.} \end{align*}

Although the functions produce the same outputs for some inputs, these are two different functions. For example, the outputs of the function \(f\) are determined by a formula, and the outputs of the function \(d\) are determined by a verbal description. This is not enough, however, to say that these are two different functions. Based on the evidence from Beginning Activity 1, we might make the following conjecture:

For \(n \geq 3\text{,}\) \(d( n ) = \dfrac{{n\left( {n - 3} \right)}}{2}\text{.}\)
Although we have not proved this statement, it is a true statement. (See Exercise 6.) However, we know the function \(d\) and the function \(f\) are not the same function. For example,

  • \(f( 0 ) = 0\text{,}\) but 0 is not in the domain of \(d\text{;}\)

  • \(f( \pi ) = \dfrac{{\pi \left( {\pi - 3} \right)}}{2}\text{,}\) but \(\pi\) is not in the domain of \(d\text{.}\)

We thus see the importance of considering the domain and codomain of each of the two functions in determining whether the two functions are equal or not. This motivates the following definition.

Definition.

Two functions \(f\) and \(g\) are equal provided that

  • The domain of \(f\) equals the domain of \(g\text{.}\) That is, \(\text{ dom} ( f ) = \text{ dom} ( g )\text{.}\)

  • The codomain of \(f\) equals the codomain of \(g\text{.}\) That is, \(\text{ codom} ( f ) = \text{ codom} ( g )\text{.}\)

  • For each \(x\) in the domain of \(f\) (which equals the domain of \(g\)), \(f( x ) = g( x )\text{.}\)

Progress Check 6.11. Equality of Functions.

Let \(A\) be a nonempty set. The identity function on the set \(\boldsymbol{A}\) denoted by \(I_A\) , is the function \(I_A\x A \to A\) defined by \(I_A ( x ) = x\) for every \(x\) in \(A\text{.}\) That is, for the identity map, the output is always equal to the input.

For this progress check, we will use the functions \(f\) and \(g\) from Progress Check 6.10. The identity function on the set \(R_5\) is

\begin{equation*} I_{R_5}:R_5 \to R_5 \text{ by } I_{R_5}(x) = x \pmod 5, \text{ for each } x \in R_5\text{.} \end{equation*}

Is the identity function on \(R_5\) equal to either of the functions \(f\) or \(g\) from Progress Check 6.10? Explain.

Solution.

\(I_{\Z_5} \ne f\) and \(I_{\Z_5} = g\text{.}\)

Subsection Mathematical Processes as Functions

Certain mathematical processes can be thought of as functions. In Beginning Activity 2, we reviewed how to find the derivatives of certain functions, and we considered whether or not we could think of this differentiation process as a function. If we use a differentiable function as the input and consider the derivative of that function to be the output, then we have the makings of a function. Computer algebra systems such as Maple and Mathematica have this derivative function as one of their predefined operators.

Different computer algebra systems will have different syntax for entering functions and for the derivative function. The first step will be to input a real function \(f\text{.}\) This is usually done by entering a formula for \(f(x)\text{,}\) which is valid for all real numbers \(x\) for which \(f(x)\) is defined. The next step is to apply the derivative function to the function \(f\text{.}\) For purposes of illustration, we will use \(D\) to represent this derivative function. So this function will give \(D(f) = f'\text{.}\)

For example, if we enter

\begin{equation*} f(x) = x^2 \sin (x) \end{equation*}

for the function \(f\text{,}\) we will get

\begin{equation*} D(f) = f', \text{ where } f'(x) = 2x\sin \left( x \right) + x^2 \cos \left( x \right)\text{.} \end{equation*}

We must be careful when determining the domain for the derivative function since there are functions that are not differentiable. To make things reasonably easy, we will let \(F\) be the set of all real functions that are differentiable and call this the domain of the derivative function \(D\text{.}\) We will use the set \(T\) of all real functions as the codomain. So our function \(D\) is

\begin{equation*} D\x F \to T \text{ by } D( f ) = f'\text{.} \end{equation*}

Progress Check 6.12. Average of a Finite Set of Numbers.

Let \(A = \left\{ a_1, a_2, \ldots, a_n \right\}\) be a finite set whose elements are the distinct real numbers \(a_1, a_2, \ldots, a_n\text{.}\) We define the average of the set \(A\) to be the real number \(\bar{A}\text{,}\) where

\begin{equation*} \bar{A} = \frac{a_1 + a_2 + \cdots + a_n}{n}\text{.} \end{equation*}
(a)

Find the average of \(A = \left\{ 3, 7, -1, 5 \right\}\text{.}\)

Solution.

3.5

(b)

Find the average of \(B = \left\{ 7, -2, 3.8, 4.2, 7.1 \right\}\text{.}\)

Solution.

4.02

(c)

Find the average of \(C = \left\{ \sqrt{2}, \sqrt{3}, \pi - \sqrt{3} \right\}\text{.}\)

Solution.

\(\dfrac{\pi + \sqrt{2}}{4}\)

(d)

Now let \(\mathscr{F}( \R )\) be the set of all nonempty finite subsets of \(\R\text{.}\) That is, a subset \(A\) of \(\R\) is in \(\mathscr{F}( \R )\) if and only if \(A\) contains only a finite number of elements. Carefully explain how the process of finding the average of a finite subset of \(\R\) can be thought of as a function. In doing this, be sure to specify the domain of the function and the codomain of the function.

Solution.

The process of finding the average of a finite set of real numbers can be thought of as a function from \(\mathcal{F} ( \R )\) to \(\R\text{.}\) So the domain is \(\mathcal{F} ( \R )\text{,}\) the codomain is \(\R\text{,}\) and we can define a function \(\text{ avg } : \mathcal{F} ( \R ) \to \R\) as follows: If \(A \in \mathscr{F} ( \R )\) and \(A = \left\{ a_1, a_2, \ldots, a_n \right\}\text{,}\) then \(\text{ avg } ( A ) = \dfrac{a_1 + a_2 + \cdots + a_n}{n}\text{.}\)

Subsection Sequences as Functions

A sequence can be considered to be an infinite list of objects that are indexed (subscripted) by the natural numbers (or some infinite subset of \(\mathbb{N} \cup \left\{ 0 \right\}\)). Using this idea, we often write a sequence in the following form:

\begin{equation*} a_1 , a_2 , \ldots , a_n , \ldots\text{.} \end{equation*}

In order to shorten our notation, we will often use the notation \(\left\langle {a_n } \right\rangle\) to represent this sequence. Sometimes a formula can be used to represent the terms of a sequence, and we might include this formula as the \(n\)th term in the list for a sequence such as in the following example:

\begin{equation*} 1, \frac{1}{2}, \frac{1}{3}, \ldots , \frac{1}{n}, \ldots\text{.} \end{equation*}

In this case, the \(n^\text{ th }\) term of the sequence is \(\dfrac{1}{n}\text{.}\) If we know a formula for the \(n\)th term, we often use this formula to represent the sequence. For example, we might say Define the sequence \(\left\langle {a_n } \right\rangle\) by \(a_n = \dfrac{1}{n}\) for each \(n \in \mathbb{N}\text{.}\) This shows that this sequence is a function with domain \(\mathbb{N}\text{.}\) If it is understood that the domain is \(\N\text{,}\) we could refer to this as the sequence \(\left\langle \dfrac{1}{n} \right \rangle\text{.}\) Given an element of the domain, we can consider \(a_n\) to be the output. In this case, we have used subscript notation to indicate the output rather than the usual function notation. We could just as easily write

\begin{equation*} a( n ) = \frac{1}{n} \text{ instead of } a_n = \frac{1}{n}\text{.} \end{equation*}

We make the following formal definition.

Definition.

An (infinite) sequence is a function whose domain is \(\mathbb{N}\) or some infinite subset of \(\mathbb{N} \cup \left\{ 0 \right\}\text{.}\)

Progress Check 6.13. Sequences.

Find the sixth and tenth terms of the following sequences, each of whose domain is \(\N\text{:}\)

(a)

\(\dfrac{1}{3}, \dfrac{1}{6}, \dfrac{1}{9}, \dfrac{1}{12}, \ldots\)

Solution.

The sixth term is \(\dfrac{1}{18}\) and the tenth term is \(\dfrac{1}{30}\text{.}\)

(b)

\(\langle a_n \rangle\text{,}\) where \(a_n = \dfrac{1}{n^2}\) for each \(n \in \N\)

Solution.

The sixth term is \(\dfrac{1}{36}\) and the tenth term is \(\dfrac{1}{100}\text{.}\)

(c)

\(\langle (-1)^n \rangle\)

Solution.

The sixth term is \(1\) and the tenth term is \(1\text{.}\)

Subsection Functions of Two Variables

In Section 5.4, we learned how to form the Cartesian product of two sets. Recall that a Cartesian product of two sets is a set of ordered pairs. For example, the set \(\Z \times \Z\) is the set of all ordered pairs, where each coordinate of an ordered pair is an integer. Since a Cartesian product is a set, it could be used as the domain or codomain of a function. For example, we could use \(\Z \times \Z\) as the domain of a function as follows: Let \(f\x \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}\) be defined by \(f( {m, n} ) = 2m + n\text{.}\)

  • Technically, an element of \(\mathbb{Z} \times \mathbb{Z}\) is an ordered pair, and so we should write \(f( {( {m, n} )} )\) for the output of the function \(f\) when the input is the ordered pair \(\left( {m, n} \right)\text{.}\) However, the double parentheses seem unnecessary in this context and there should be no confusion if we write \(f( {m, n} )\) for the output of the function \(f\) when the input is \(\left( {m, n} \right)\text{.}\) So, for example, we simply write

    \begin{align*} f( {3, 2} ) \amp = 2 \cdot 3 + 2 = 8,\text{ and }\\ f( { - 4, 5} ) \amp = 2 \cdot \left( { - 4} \right) + 5 = - 3\text{.} \end{align*}

  • Since the domain of this function is \(\Z \times \Z\) and each element of \(\Z \times \Z\) is an ordered pair of integers, we frequently call this type of function a function of two variables.

Finding the preimages of an element of the codomain for the function \(f\text{,}\) \(\mathbb{Z}\text{,}\) usually involves solving an equation with two variables. For example, to find the preimages of \(0 \in \mathbb{Z}\text{,}\) we need to find all ordered pairs \(\left( {m, n} \right) \in \mathbb{Z} \times \mathbb{Z}\) such that \(f( {m, n} ) = 0\text{.}\) This means that we must find all ordered pairs \(\left( {m, n} \right) \in \mathbb{Z} \times \mathbb{Z}\) such that

\begin{equation*} 2m + n = 0\text{.} \end{equation*}

Three such ordered pairs are \(\left( {0, 0} \right)\text{,}\) \(\left( {1, - 2} \right)\text{,}\) and \(\left( { - 1, 2} \right)\text{.}\) In fact, whenever we choose an integer value for \(m\text{,}\) we can find a corresponding integer \(n\) such that \(2m + n = 0\text{.}\) This means that 0 has infinitely many preimages, and it may be difficult to specify the set of all of the preimages of 0 using the roster method. One way that can be used to specify this set is to use set builder notation and say that the following set consists of all of the preimages of 0:

\begin{equation*} \left\{ { {\left( {m, n} \right) \in \mathbb{Z} \times \mathbb{Z} } \mid 2m + n = 0} \right\} = \left\{ { {\left( {m, n} \right) \in \mathbb{Z} \times \mathbb{Z} } \mid n = - 2m} \right\}\!\text{.} \end{equation*}

The second formulation for this set was obtained by solving the equation \(2m + n = 0\) for \(n\text{.}\)

Progress Check 6.14. Working with a Function of Two Variables.

Let \(g\x \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}\) be defined by \(g( {m, n} ) = m^2 - n\) for all \(\left(m, n \right) \in \Z \times \Z\text{.}\)

(a)

Determine \(g(0, 3)\text{,}\) \(g(3,-2)\text{,}\) \(g(-3, -2)\text{,}\) and \(g(7, -1)\text{.}\)

Solution.

\(g ( 0, 3 ) = -3\text{;}\) \(g ( 3, -2 ) = 11\text{;}\) \(g ( -3, -2 ) = 11\text{;}\) \(g ( 7, -1 ) = 50\text{.}\)

(b)

Determine the set of all preimages of the integer 0 for the function \(g\text{.}\) Write your answer using set builder notation.

Solution.

\(\left\{ (m, n) \in \Z \times \Z \mid n = m^2 \right\}\)

(c)

Determine the set of all preimages of the integer 5 for the function \(g\text{.}\) Write your answer using set builder notation.

Solution.

\(\left\{ (m, n) \in \Z \times \Z \mid n = m^2 - 5 \right\}\)

Exercises Exercises

1.

Let \(R_5 = \left\{ {0, 1, 2, 3, 4} \right\}\text{.}\) Define \(f\x R_5 \to R_5\) by \(f( x ) = x^2 + 4 \pmod 5\text{,}\) and define \(g\x R_5 \to R_5\) by \(g( x ) = (x + 1)(x + 4) \pmod 5\text{.}\)

(a)

Calculate \(f( 0 )\text{,}\) \(f( 1 ), f( 2 )\text{,}\) \(f( 3 )\text{,}\) and \(f( 4 )\text{.}\)

Answer.

\(f( 0 ) = 4\text{,}\) \(f( 1 ) = 0, f( 2 ) = 3\text{,}\) \(f( 3 ) = 3\text{,}\) \(f( 4 ) = 0\)

(b)

Calculate \(g( 0 )\text{,}\) \(g( 1 ), g( 2 )\text{,}\) \(g( 3 )\text{,}\) and \(g( 4 )\text{.}\)

Answer.

\(g( 0 ) = 4\text{,}\) \(g( 1 ) = 0, g( 2 ) = 3\text{,}\) \(g( 3 ) = 3\text{,}\) \(g( 4 ) = 0\)

(c)

Is the function \(f\) equal to the function \(g\text{?}\) Explain.

Answer.

The two functions are equal.

2.

Let \(R_6 = \left\{ {0, 1, 2, 3, 4, 5} \right\}\text{.}\) Define \(f\x R_6 \to R_6\) by \(f( x ) = x^2 + 4 \pmod 6\text{,}\) and define \(g\x R_6 \to R_6\) by \(g( x ) = (x + 1)(x + 4) \pmod 6\text{.}\)

(a)

Calculate \(f( 0 )\text{,}\) \(f( 1 ), f( 2 )\text{,}\) \(f( 3 )\text{,}\) \(f( 4 )\text{,}\) and \(f( 5 )\text{.}\)

(b)

Calculate \(g( 0 )\text{,}\) \(g( 1 ), g( 2 )\text{,}\) \(g( 3 )\text{,}\) \(g( 4 )\text{,}\) and \(g( 5 )\text{.}\)

(c)

Is the function \(f\) equal to the function \(g\text{?}\) Explain.

3.

Let \(f\x \left(\R - \{0 \} \right) \to \R\) by \(f(x) = \dfrac{x^3 + 5x}{x}\) and let \(g\x \R \to \R\) by \(g(x) = x^2 + 5\text{.}\)

(a)

Calculate \(f( 2 ), f( -2 )\text{,}\) \(f( 3 )\text{,}\) and \(f( \sqrt{2} )\text{.}\)

Answer.

\(f(2) = 9\text{,}\) \(f(-2) = 9\text{,}\) \(f(3) = 14\text{,}\) \(f(\sqrt{2} = 7\)

(b)

Calculate \(g( 0 )\text{,}\) \(g( 2 ), g( -2 )\text{,}\) \(g( 3 )\text{,}\) and \(g( \sqrt{2} )\text{.}\)

Answer.

\(g(0) = 5\text{,}\) \(g(2) = 9\text{,}\) \(g(-2) = 9\text{,}\) \(g(3) = 14\text{,}\) \(g(\sqrt{2}) = 7\)

(c)

Is the function \(f\) equal to the function \(g\text{?}\) Explain.

Answer.

The function \(f\) is not equal to the function \(g\) since they do not have the same domain.

(d)

Now let \(h\x \left(\R - \{0 \} \right) \to \R\) by \(h(x) = x^2 + 5\text{.}\) Is the function \(f\) equal to the function \(h\text{?}\) Explain.

Answer.

The function \(h\) is equal to the function \(f\) since if \(x \ne 0\text{,}\) then \(\dfrac{x^2 + 5x}{x} = x^2 + 5\text{.}\)

4.

Represent each of the following sequences as functions. In each case, state a domain, codomain, and rule for determining the outputs of the function. Also, determine if any of the sequences are equal.

(a)

\(1, \dfrac{1}{4}, \dfrac{1}{9}, \dfrac{1}{{16}}, \ldots\)

Answer.

\(\left\langle {a_n } \right\rangle\text{,}\) where \(a_n = \dfrac{1}{{n^2 }}\) for each \(n \in \mathbb{N}\text{.}\) The domain is \(\mathbb{N}\text{,}\) and \(\Q\) can be the codomain.

(b)

\(\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{{27}}, \dfrac{1}{{81}}, \ldots\)

(c)

\(1, - 1, 1, - 1, 1, - 1, \ldots\)

(d)

\(\cos( 0 ), \cos( \pi ), \cos( {2\pi } ), \cos( {3\pi } ), \cos( {4\pi } ), \ldots\)

Answer.

\(\left\langle {a_n } \right\rangle\text{,}\) where \(a_n = \cos ( {n\pi } )\) for each \(n \in \mathbb{N} \cup {0}\text{.}\) The domain is \(\mathbb{N} \cup {0}\text{,}\) and \(\left\{ -1, 1 \right\}\) can be the codomain. This sequence is equal to the sequence in Part (c).

5.

Let \(A\) and \(B\) be two nonempty sets. There are two projection functions with domain \(A \times B\text{,}\) the Cartesian product of \(A\) and \(B\text{.}\) One projection function will map an ordered pair to its first coordinate, and the other projection function will map the ordered pair to its second coordinate. So we define

\(p_1\x A \times B \to A\) by \(p_1( {a, b} ) = a\) for every \(( {a, b} ) \in A \times B\text{;}\) and
\(p_2\x A \times B \to B\) by \(p_2( {a, b} ) = b\) for every \(( {a, b} ) \in A \times B\text{.}\)
Let \(A = \left\{ {1, 2} \right\}\) and let \(B = \left\{ {x, y, z} \right\}\text{.}\)

(a)

Determine the outputs for all possible inputs for the projection function \(p_1\x A \times B \to A\text{.}\)

Answer.

\(p_1(1, x) = 1\text{,}\) \(p_1(1, y) = 1\text{,}\) \(p_1(1, z) = 1\text{,}\) \(p_1(2, x) = 2\text{,}\) \(p_1(2, y) = 2\text{,}\) \(p_1(2, z) = 2\)

(b)

Determine the outputs for all possible inputs for the projection function \(p_2\x A \times B \to B\text{.}\)

(c)

What is the range of these projection functions?

Answer.

\(\text{ range} ( p_1 ) = A\text{,}\) \(\text{ range} ( p_2 ) = B\)

(d)

Is the following statement true or false? Explain.

For all \(( {m, n} ), ( {u, v} ) \in A \times B\text{,}\) if \(( {m, n} ) \ne ( {u, v} )\text{,}\) then \(p_1( {m, n} ) \ne p_1( {u, v} )\text{.}\)

6.

Let \(D = \mathbb{N} - \left\{ {1, 2} \right\}\) and define \(d\x D \to \mathbb{N} \cup \left\{ 0 \right\}\) by \(d( n ) =\) the number of diagonals of a convex polygon with \(n\) sides. In Beginning Activity 1, we showed that for values of \(n\) from 3 through 8,

\begin{equation*} d( n ) = \frac{{n\left( {n - 3} \right)}}{2}\text{.} \end{equation*}

Use mathematical induction to prove that for all \(n \in D\text{,}\) \(d( n ) = \dfrac{{n\left( {n - 3} \right)}}{2}\text{.}\)

Hint.

To get an idea of how to handle the inductive step, use a pentagon. First, form all the diagonals that can be made from four of the vertices. Then consider how to make new diagonals when the fifth vertex is used. This may generate an idea of how to proceed from a polygon with \(k\) sides to a polygon with \(k + 1\) sides.

Answer.

Start of the inductive step: Let \(P ( n )\) be “A convex polygon with \(n\) sides has \(\dfrac{n ( n-3 )}{2}\) diagonals.” Let \(k \in D\) and assume that \(P ( k )\) is true, that is, a convex polygon with \(k\) sides has \(\dfrac{{k( {k - 3} )}}{2}\) diagonals. Now let \(Q\) be convex polygon with \(( {k + 1} )\) sides. Let \(v\) be one of the \(( {k + 1} )\) vertices of \(Q\) and let \(u\) and \(w\) be the two vertices adjacent to \(v\text{.}\) By drawing the line segment from \(u\) to \(w\) and omitting the vertex \(v\text{,}\) we form a convex polygon with \(k\) sides. Now complete the inductive step.

7.

Let \(f\x \Z \times \Z \to \mathbb{Z}\) be defined by \(f( {m, n} ) = m + 3n\text{.}\)

(a)

Calculate \(f( { - 3, 4} )\) and \(f( { - 2, - 7} )\text{.}\)

Answer.

\(f( { - 3, 4} ) = 9\text{,}\) \(f( { - 2, - 7} ) = - 23\)

(b)

Determine the set of all the preimages of 4 by using set builder notation to describe the set of all \(\left( {m, n} \right) \in \mathbb{Z} \times \mathbb{Z}\) such that \(f( {m, n} ) = 4\text{.}\)

Answer.

\(\text{ range} (f) = \left\{ { {( {m, n} ) \in \mathbb{Z} \times \mathbb{Z} } \mid m = 4 - 3n} \right\}\)

8.

Let \(g\x \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}\) be defined by \(g( {m, n} ) = \left( {2m, m - n} \right)\text{.}\)

(a)

Calculate \(g( {3, 5} )\) and \(g( { - 1, 4} )\text{.}\)

Answer.

\(g ( 3, 5 ) = ( 6, -2 )\text{,}\) \(g ( -1, 4 ) = ( -2, -5 )\text{.}\)

(b)

Determine all the preimages of \(\left( {0, 0} \right)\text{.}\) That is, find all \(( {m, n} ) \in \mathbb{Z} \times \mathbb{Z}\) such that \(g( {m, n} ) = ( {0, 0} )\text{.}\)

(c)

Determine the set of all the preimages of \(( {8, - 3} )\text{.}\)

Answer.

The set of preimages of \(( 8, -3 )\) is \(\left\{ ( 4, 7 ) \right\}\text{.}\)

(d)

Determine the set of all the preimages of \(( {1, 1} )\text{.}\)

(e)

Is the following proposition true or false? Justify your conclusion.

For each \(( {s, t} ) \in \mathbb{Z} \times \mathbb{Z}\text{,}\) there exists an \(( {m, n}) \in \mathbb{Z} \times \mathbb{Z}\) such that \(g( {m, n} ) = ( {s, t} )\text{.}\)

9.

A 2 by 2 matrix over \(\mathbb{R}\) is a rectangular array of four real numbers arranged in two rows and two columns. We usually write this array inside brackets (or parentheses) as follows:

\begin{equation*} A = \left[ {\begin{array}{*{20}c} a \amp b \\ c \amp d \end{array} } \right]\text{,} \end{equation*}

where \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are real numbers. The determinant of the 2 by 2 matrix \(A\text{,}\) denoted by \(\det( A )\text{,}\) is defined as

\begin{equation*} \det( A ) = ad - bc\text{.} \end{equation*}
(a)

Calculate the determinant of each of the following matrices:

\begin{equation*} \left[ {\begin{array}{*{20}c} 3 \amp 5 \\ 4 \amp 1 \end{array} } \right],\; \left[ {\begin{array}{*{20}c} 1 \amp 0 \\ 0 \amp 7 \end{array} } \right],\; \text{ and } \left[ {\begin{array}{*{20}c} 3 \amp { - 2} \\ 5 \amp 0 \end{array} } \right]\text{.} \end{equation*}
Answer.

\(\det \left[ {\begin{array}{*{20}c} 3 \amp 5 \\ 4 \amp 1 \end{array} } \right] = - 17, \det \left[ {\begin{array}{*{20}c} 1 \amp 0 \\ 0 \amp 7 \end{array} } \right] = 7, \text{ and det } \left[ {\begin{array}{*{20}c} 3 \amp { - 2} \\ 5 \amp 0 \end{array} } \right] = 10\)

(b)

Let \(\mathcal{M}_2( \R )\) be the set of all 2 by 2 matrices over \(\R\text{.}\) The mathematical process of finding the determinant of a 2 by 2 matrix over \(\R\) can be thought of as a function. Explain carefully how to do so, including a clear statement of the domain and codomain of this function.

10.

Using the notation from Exercise 9, let

\begin{equation*} A = \left[ {\begin{array}{*{20}c} a \amp b \\ c \amp d \end{array} } \right] \end{equation*}

be a 2 by 2 matrix over \(\R\text{.}\) The transpose of the matrix \(\boldsymbol{A}\), denoted by \(A^T\text{,}\) is the 2 by 2 matrix over \(\mathbb{R}\) defined by

\begin{equation*} A^T = \left[ {\begin{array}{*{20}c} a \amp c \\ b \amp d \end{array} } \right]\text{.} \end{equation*}
(a)

Calculate the transpose of each of the following matrices:

\begin{equation*} \left[ {\begin{array}{*{20}c} 3 \amp 5 \\ 4 \amp 1 \end{array} } \right],\; \left[ {\begin{array}{*{20}c} 1 \amp 0 \\ 0 \amp 7 \end{array} } \right],\; \text{ and } \left[ {\begin{array}{*{20}c} 3 \amp { - 2} \\ 5 \amp 0 \end{array} } \right]\text{.} \end{equation*}
(b)

Let \(\mathcal{M}_2( \R )\) be the set of all 2 by 2 matrices over \(\mathbb{R}\text{.}\) The mathematical process of finding the transpose of a 2 by 2 matrix over \(\mathbb{R}\) can be thought of as a function. Carefully explain how to do so, including a clear statement of the domain and codomain of this function.

Activity 35. Integration as a Function.

In calculus, we learned that if \(f\) is real function that is continuous on the closed interval \(\left[ {a, b} \right]\text{,}\) then the definite integral \(\int_a^b {f( x ) \, dx}\) is a real number. In fact, one form of the Fundamental Theorem of Calculus states that

\begin{equation*} \int_a^b {f( x ) \, dx} = F( b ) - F( a )\text{,} \end{equation*}

where \(F\) is any antiderivative of \(f\text{,}\) that is, where \(F \,' = f\text{.}\)

(a)

Let \(\left[ {a, b} \right]\) be a closed interval of real numbers and let \(C {\left[ {a, b} \right]}\) be the set of all real functions that are continuous on \(\left[ {a, b} \right]\text{.}\) That is,

\begin{equation*} C{\left[ {a, b} \right]} = \left\{ {f\x \left[ {a, b} \right] \to \mathbb{R} \mid f\text{ is continuous on } \left[ {a, b} \right]} \right\}\text{.} \end{equation*}
(i)

Explain how the definite integral \(\int_a^b {f( x ) \, dx}\) can be used to define a function \(I\) from \(C{\left[ {a, b} \right]}\) to \(\mathbb{R}\text{.}\)

(ii)

Let \(\left[ {a, b} \right] = \left[ {0, 2} \right]\text{.}\) Calculate \(I( f )\text{,}\) where \(f( x ) = x^2 + 1\text{.}\)

(iii)

Let \(\left[ {a, b} \right] = \left[ {0, 2} \right]\text{.}\) Calculate \(I( g )\text{,}\) where \(g( x ) = \sin ( {\pi x} )\text{.}\)

In calculus, we also learned how to determine the indefinite integral \(\int {f( x )} \;dx\) of a continuous function \(f\text{.}\)

(b)

Let \(f( x ) = x^2 + 1\) and \(g(x) = \cos( 2x )\text{.}\) Determine \(\int {f( x )} \;dx\) and \(\int {g( x )} \;dx\text{.}\)

(c)

Let \(f\) be a continuous function on the closed interval \(\left[ {0, 1} \right]\) and let \(T\) be the set of all real functions. Can the process of determining the indefinite integral of a continuous function be used to define a function from \(C{\left[ {0, 1} \right]}\) to \(T\text{?}\) Explain.

(d)

Another form of the Fundamental Theorem of Calculus states that if \(f\) is continuous on the interval \(\left[ {a, b} \right]\) and if

\begin{equation*} g( x ) = \int_a^x {f( t ) \, dt} \end{equation*}

for each \(x\) in \(\left[ {a, b} \right]\text{,}\) then \(g'\left( x \right) = f\left( x \right)\text{.}\) That is, \(g\) is an antiderivative of \(f\text{.}\) Explain how this theorem can be used to define a function from \(C{\left[ {a,\;b} \right]}\) to \(T\text{,}\) where the output of the function is an antiderivative of the input. (Recall that \(T\) is the set of all real functions.)