Math 8 — Functions and Modeling

Neil Donaldson

Spring 2025

Introduction

This course aims to refresh and reinforce the conceptual foundations behind several topics commonly

encountered in grade-school mathematics. The job of a teacher is often one of selection: choosing

examples and explanations suited to the level and experience of your students. To select effectively,

and to anticipate student questions, your must understand concepts at a higher level than you’ll

likely ever teach. Not all of our topics are central to the grade-school curriculum, and it is not our

goal to teach you how to teach, though the ideas and approaches we’ll explore are often suitable for

a grade-school audience. The mathematics in this course shouldn’t present much difﬁculty for math

majors, requiring at most elementary calculus and a tiny bit of linear algebra; you should instead be

considering how to explain the material, particularly to students with less mathematical knowledge

than yourself.

We start with two motivational problems.

1. You wish to travel across the surface of a cube between two oppo-

site vertices so that your path is as short as possible.

Should you follow the path indicated?

If yes, explain why.

If not, how should you ﬁnd the shortest path?

2. Two houses are to be connected to the elec-

tricity supply using a single connection.

How should we determine where to place

the connection so as to minimize the required

length of wire?

What information do you need in order to

ﬁnd the connection point?

connection

wires

electric supply

House 1

House 2

Your goal shouldn’t only be to ﬁnd the right answer! Consider how you might discuss these problems

with grade-school students of different ability levels. Why might calculus not be a sensible approach?

Are there any similarities between the two problems? Brainstorm some strategies. . .

We are grateful to materials from UT Austin’s UTeach program for suggesting several of the examples in this course

including these motivational problems.

1 Sets & Functions

1.1 Basic Deﬁnitions

Consider how central functions are to mathematics, and how long you’ve been using them. How

would you deﬁne “function” to someone with limited mathematical knowledge? Would you use

words like rule, assign, element, domain, vertical line test, etc.? How helpful are these to your audience?

Examples 1.1. How would you explain the idea that the following do or do not represent functions?

1. y = x

2. Mon: ﬁsh, Tue: pork, Wed: fajitas, Thur: carbonara, Fri: pizza, Sat: ﬁsh, Sun: pizza

3. (3, 5), (2, 6), ( 4, 2), (3, 1).

4. x

= y

After considering the examples, perhaps you settle on a semi-formal deﬁnition:

A function f is rule which assigns to each input x exactly one output f (x)

Is this a useful deﬁnition? In what ways is it imprecise? Does the imprecision matter?

Of course the answers to these questions depend on your audience! What ideas do you want to

convey to your students and can you do so without overburdening and intimidating them? To begin

working towards a more complete picture, consider what we might allow to be inputs and outputs.

This requires a small amount of set notation.

Deﬁnition 1.2. A set A is a collection of objects, or elements.

The notation a ∈ A means that a is an

element of A, sometimes read ‘a lies in A.’ Sets are often written upper case and elements lower.

A set B is a subset of a set A, written B ⊆ A, if every element of B is also an

element of A: that is,

b ∈ B =⇒ b ∈ A

The picture illustrates sets A, B and elements a, b for which B ⊆ A, a ∈ A,

b ∈ B and a /∈ B (a does not lie in B).

Examples 1.3. 1. Suppose the elements of a set A are the numbers 1, 3, 5, 7 and 9. The simplest way

to write this is using roster notation: we list the elements (in any order) between braces

A = {1, 3, 5, 7, 9}

Subsets are commonly expressed using set-builder notation. For example, here is a subset of A:

B = {a ∈ A : 2 < a < 8}

This is read, “The set of a in A such that a lies strictly between 2 and 8.” In roster notation,

B = {3, 5, 7}. Can you express B in other ways using set-builder notation?

This is enough for our purposes, though a course in set theory will convince you that this deﬁnition has its own

problems. Selection is always at work. . .

2. We summarize several common sets of numbers using informal combinations of roster and

set-builder notation, all of which should be familiar.

Natural numbers N = {1, 2, 3, 4, . . .}. For instance, 5 ∈ N but −3 ∈ N.

Integers Z = {. . . , −2, −1, 0, 1, 2, 3, . . .}. For instance, −4 ∈ Z but

∈ Z.

Rational numbers (fractions) Q =



: p ∈ Z, q ∈ N



. For instance −

∈ Q; in this case

p = −6 is an integer, and q = 7 a natural number.

Real numbers R. For instance,

√

2 ∈ R. A formal deﬁnition is difﬁcult, though we often

informally visualize R as a ruler. Intervals are particularly important subsets, e.g.,

[−4, π) = {x ∈ R : −4 ≤ x < π}

is a half-open interval.

You should also be familiar with the Cartesian plane: R

= {(x, y) : x, y ∈ R}. The notation

(3, 4) ∈ R

here describes a point in the plane with co-ordinates x = 3, y = 4; don’t confuse

this with the interval ( 3, 4) = {x ∈ R : 3 < x < 4} which is a subset of R!

The subset relationships between these sets are in the order listed:

N ⊆ Z ⊆ Q ⊆ R

You should also have informally encountered the notion of irrationality: for instance,

√

2 and π

are real numbers but not rational numbers.

The reason we need this language when discussing functions is that the inputs and outputs of a

function are elements of sets. Here is a very formal deﬁnition of “function.”

Deﬁnition 1.4. The Cartesian product of sets A, B is the set of ordered pairs

A × B =



(a, b) : a ∈ A, b ∈ B



A function from A to B is a non-empty subset f ⊆ A ×B which satisﬁes the vertical line test

For each a ∈ A, there is a unique b ∈ B such that (a, b) ∈ f (∗)

Instead of writing f ⊆ A × B and (a, b) ∈ f , we use the more familiar notation

f : A → B and f (a) = b

To a function f : A → B are associated three useful sets:

• Domain: dom f = A is the set of inputs.

• Codomain: codom f = B is the set of possible outputs.

• Range: range f = {b ∈ B : b = f (a) for some a ∈ A} is the set of realized outputs.

This probably isn’t the deﬁnition you should give to 10

graders, or even to freshman calculus stu-

dents! But what should you do? How much of this is helpful in a a given context?

Example (1.1.2 cont.). We revisit our food-based example in this formal setting. To properly view

this as a function f : A → B, we have to carefully label the constituent sets.

A =



Mon, Tue, Wed, Thu, Fri



, B =



carbonara, fajitas, ﬁsh, pizza, pork



f =



(Mon, ﬁsh), (Tue, pork), (Wed, fajitas), (Thu, carbonara),

(Fri, pizza), (Sat, ﬁsh), (Sun, pizza)



The domain A should be clear, but we had to make a choice for the codomain B: in this case we chose

it to equal to range. Can you suggest a different choice for B? Try the other examples yourself.

Representing Functions

Functions can be represented in various ways. We illustrate a few in an example.

Example 1.5. We consider the familiar formula/rule f (x) = x

in several contexts.

Table This presentation is most helpful when the domain is very small.

The table shows the situation when dom f = {−1, 0, 1, 2, 3} and

range f = {0, 1, 4, 9}

Arrows A pictorial arrow diagram might also be helpful when the do-

main is small.

Graph This is the set of ordered pairs



x, f (x)



: x ∈ dom f



: in the

context of the formal deﬁnition (1.4), the graph is the function!

For formulæ whose inputs and outputs are real numbers, two con-

ventions are often observed:

• The domain is implied to be all real numbers for which the

formula makes sense.

• The codomain is taken to be the set of real numbers.

If no other information is provided, we’d assume that the function

deﬁned by the formula f (x) = x

has both domain and codomain

the entire set of real numbers: f : R → R.

The range of the function is the set of possible outputs, in this case

range f = {x

∈ R : x ∈ R} = [0, ∞)

is the half-open interval of non-negative real numbers.

For ‘calculus’ functions like these, the vertical line test (∗) really in-

volves vertical lines; every vertical line intersects the graph in pre-

cisely one point.

In the picture, the dots are the graph when the domain is the ﬁnite

set {−1, 0, 1, 2, 3} (as described in the table/arrow-diagram).

x −1 0 1 2 3

f (x) 1 0 1 4 9

−1

−2 0 2

Can you think of other ways to represent a function? How might you decide which to use?

Exercises 1.1. 1. Let d represent the cost in millions of dollars to produce n cars, where n is measured

in 1000s. As clearly as you can, explain what is meant by d(25) = 431.

2. A movie theater seats 200 people. For any particular show, the amount of money the theater

takes in is a function of the number of people n in attendance. If a ticket costs $25, describe the

domain and range of the function using set notation.

3. Temperature readings T were recorded every two hours from midnight to noon. Time t was

measured in hours from midnight.

t 0 2 4 6 8 10 12

T (

◦

F) 82 75 74 75 84 90 93

(a) Plot the readings and use them to sketch a rough graph of T as a function of t.

(b) Use your graph to estimate the temperature at 10:30 a.m.

4. State parts 1, 3 and 4 of Example 1.1 using the formal language of Deﬁnition 1.4. If you have a

function, state the domain and range and explain how you know you have a function. If you

don’t have a function, explain why not.

(Since insufﬁcient information is provided, there is no single correct answer!)

5. (a) Let A = {1, 3, 5, 7, 9}. Explain in words what is meant by the set

B = {x ∈ A : x

> 10}

and state B in roster notation.

(b) Find the set C = {x ∈ N : (x −1)

< 16} in roster notation.

6. Suppose that f : {−2, −1, 0, 1, 2} → R is deﬁned by the formula f (x) = x

−4x + 1.

Describe f using a table, an arrow diagram and a graph.

7. Find the implied domain and range for the functions deﬁned by each rule:

(a) f (x) =

−4

x−2

(b) g(x) =

√

−16x (c) h(x) =

√

4x −x

(What is the largest set of real numbers for which the formula makes sense?)

8. You ask your students to determine the range of the function f deﬁned by the rule f (x) = x

with domain the interval [−5, 2]. You obtain various responses, including [25, 4], [4, 25] , and

[−25, 4]. What is going wrong? What is the correct answer, and how would you explain it to

your students?

More generally, if dom f = [a, b] (where a ≤ b), what is range f ?

9. The unit circle is often represented by the implicit equation x

+ y

= 1.

(a) Draw the circle and explain why the full circle isn’t the graph of a function.

(b) Describe two functions f : [−1, 1] → R and g : [−1, 1] → R whose graphs together

comprise the circle. What are the ranges of each function?

1.2 Linear Polynomials

Perhaps the simplest functions are the linear polynomials, whose graphs are straight lines,

y = f (x) = mx + c where m, c are constants (∗)

Linear polynomials make very simple models: increase the input by ∆x and the output changes by

∆y = m∆x regardless of the starting value x. Given experimental data or a physical situation relating

two quantities x and y, a linear model is an linear polynomial (∗) relating these variables. In practice,

models are approximations to the real-world data. Later in the course we’ll consider what should be

meant by, and how to ﬁnd, a ‘good’ linear model for approximately linear data.

Some of your earliest forays into algebra likely involved ﬁnding equations of straight lines.

Example 1.6. Find the equation of the straight line through the points A = (1, 3) and B = (4, 1).

Suppose the polynomial is y = mx + c. Since both A and B sat-

isfy this equation, we start by substituting both points into the

equation to ﬁnd two relationships between m and c

(

3 = m + c

1 = 4m + c

This is a system of two linear equations in two unknowns (m, c).

By now you should know several ways to solve such, but con-

sider what might be easiest for a grade-school student. . .

0 1 2 3 4 5

Regardless of how you phrase it (solve one equation for c and substitute into the other, subtract one

question from the other, etc.), we obtain

−2 = 3m =⇒ m = −

=⇒ c = 3 − m =

whence the required polynomial is y =

(11 − 2x).

As the picture suggests, the gradient/slope m represents how far one climbs/falls on travelling one

unit to the right. The y-intercept c is the intersection of the graph with the vertical axis.

The above process works for any two points A = (x

, y

) and B = (x

, y

) provided x

= x

: is it

clear why this should be the case? The details are in Exercise 5. You might feel that such a problem

is too abstract for your students, that such a ‘proof’ might be too intimidating. Indeed it might be

counterproductive for some students, but consider several counterpoints:

• Once a student has developed comfort with concrete examples as above, Exercise 5 helps sum-

marize and unify what they’ve learned. A general/abstract discussion helps build conﬁdence

by convincing a student that any such problem can be solved the same way.

• The most helpful elementary proofs are those which essentially replicate an example abstractly.

Exercise 5 is not some abstract existence proof—it involves no trickery—it simply reinforces the

core technique by applying it in the most general situation.

• Helping and encouraging students to think abstractly is one of the overarching learning out-

comes of all mathematics. You might get push-back, but it’s part of the job. . .

Example 1.7. Often the challenge of modeling lies in converting a word problem into algebra—don’t

underestimate how hard students ﬁnd this! Here is a simple, though disguised, straight line model.

Beaker A contains a 300 ml solution of 2% acid. Beaker B contains 400 ml of acid of unknown con-

centration. The beakers are mixed together to produce an acid with concentration 6%. What was the

concentration in beaker B?

Given your mathematical experience, it should seem natural to denote the unknown concentration

(beaker B) by x. After mixing, we have a 700 ml solution containing 300 ×

100

+ 400x ml of pure

acid, whence its concentration is a linear polynomial function of x:

C(x) =

6 + 400x

700

The problem is now easily solved: C(x) =

100

⇐⇒ x =

100

= 9%.

Parametrized Lines Straight lines admit an alternative visualization. Imagine placing a ruler so that

its zero point is at the origin O = (0, 0) and the “1” lies at a point C = (c

, c

). If t (a real number) is

the measure on the ruler, then the points on the line have co-ordinates

tC = (tc

, tc

) (∗)

To describe the line through points A and B, place a ruler so that 0

corresponds to A and 1 to B. Now slide the ruler so that A moves

to the origin O: this amounts to subtracting the co-ordinates of A

from all points on the line. We obtain a parallel line through the

origin, with B transformed to the point C = B − A. Putting this

together with (∗) results in a parametrized description of the line:

(x, y) = A + tC = A + t(B − A) = (1 − t)A + tB

−

C = B − A

Contrast the parametrized description of a line with the linear polynomial approach: for instance,

one challenge is that a line may be parametrized using inﬁnitely many distinct rulers (choose any

two points on the line!), whereas the linear polynomial description is unique. Does the parametrized

approach have any advantages? Which description is easier to understand or to work with? Which

ﬁts better with your intuitive understanding of line? Which might cause a grade-school student the

greater challenge?

In the Exercises we make sure that the two descriptions of a line correspond. The discussion is little

more than the generalization of an example.

Example 1.8. The line through points A = (3, 6) and B = (−1, 4) may be parametrized by

(x, y) = (1 − t)(3, 6) + t(−1, 4) =



3 −4t, 6 −2t



To convert this to a linear polynomial, ﬁrst solve for t in terms of x,

x = 3 −4t =⇒ t =

(3 − x)

before substituting into our expression for y:

y = 6 −2t = 6 −

(3 − x) =

x −

Exercises 1.2. 1. The cost of gasoline is $4.20 per gallon on January 1

and $4.90 on March 1

. State

a linear function/model for the cost of gasoline as a function of time.

2. You have a choice of three different cell-phone plans.

(a) No monthly charge and 10¢ per minute for all calls.

(b) $10 per month and 5¢ per minute for all calls.

How should you determine which plan to purchase?

3. Revisit Exercise 1.1.3. Find an approximate linear model T(t) = mt + c for this data.

(There is no perfect answer)

4. Revisit the beakers problem (Example 1.7). This time suppose we know that the concentration

in beaker B is 9%. How much from beaker B should we pour into beaker A to obtain an acid

with concentration 5%? Would you consider this a linear polynomial problem? Why/why not?

5. Suppose points A = (x

, y

) and B = (x

, y

) are given.

(a) If x

= x

, use the method of Example 1.6 to ﬁnd the equation y = mx + c of the line

through these points.

(b) Now use the parametrized approach where A corresponds to 0 and B to 1. If, in addition,

= x

, make things match up with your answer to part (a).

What parametrization do you get if A = (0, c) and B = (1, m + c)?

polynomial description of a line is unique (‘the equation’). How might you help a student

believe this claim if the algebra is unconvincing or too intimidating?

(Think about Example 1.6)

6. A straight line is sometimes described as the set of points (x, y) ∈ R

satisfying an equation of

the form

ax + by = c

for some constants a, b, c where a, b are not both zero. How does this approach differ from our

use of linear polynomials?

7. Throughout mathematics (particularly within linear algebra), a function f : R → R is said to be

linear if it satisﬁes the condition

For all λ, x ∈ R, f (λx) = λ f (x)

Is this the same thing as a linear polynomial? Explain.

1.3 Quadratic Polynomials

Quadratic polynomials are functions of the form y = f (x) = ax

+ bx + c where a = 0. The simplest

is y = x

, the standard parabola opening upwards. Here are some commonly encountered activities:

1. Find the roots/zeros of f , the solutions x to the equation f (x) = 0.

2. Sketch the graph of the function f .

3. Use quadratic functions to model a real-world problem.

You likely know two methods for ﬁnding zeros: factorizing and the quadratic formula, each of which

has its problems. With experience it is easy to spot that

+ 2x −15 = (x −3)(x + 5) = 0 ⇐⇒ x = 3 or x = −5

though the required creativity can make this difﬁcult, particularly when coefﬁcients are large. Stu-

dents often prefer the quadratic formula since it always works, though at the cost of some intimidat-

ing algebra. We’ll think about factorization shortly. First, we see how completing the square lies behind

both the quadratic formula and the standard approach to graphing quadratic functions.

Example 1.9. Describe/graph the parabola y = −3x

+ 12x + 4.

Pay attention to the x terms; −3x

+ 12x = −3(x

−4x). Now

−3(x −2)

= −3(x

−4x + 4) = −3x

+ 12x −12

gives most of what we want: note how we divided the x-

coefﬁcient by two. To ﬁnish, just tidy everything up,

y = (−3x

+ 12x −12) + 16 = −3(x −2)

+ 16

−1 0 1 2 3 4

The parabola therefore opens downwards (−3 < 0) with its apex (maximum) at (x, y) = (2, 16) .

This is easy, if intimidating, to repeat in general:

+ bx + c = a





+ c = a



x +



−

+ c

= a



x +



−

−4ac

(∗)

The graph is that of the standard parabola which has been:

1. Vertically scaled by a;

2. Shifted horizontally by −

;

3. Shifted vertically by

4ac−b

By solving (∗) for x, we see that completing the square yields

the quadratic formula.

y = x

y = ax

+ bx + c

−

4ac−b

Theorem 1.10. If a = 0, then ax

+ bx + c = 0 ⇐⇒ x =

−b ±

√

−4ac

Example (1.9 cont). Our analysis suggests two methods for ﬁnding the roots.

1. Quadratic formula: with a = −3, b = 12, c = 4, we have

x =

−12 ±

−4(−3) ·4

2(−3)

−12 ± 4

√

+ 3

−6

= 2 ±

√

= 2 ±

√

While it is always tempting to jump for a formula, it often leads to difﬁcult surd expressions.

We simpliﬁed by noticing the common factor of 4

inside the square root. Without this, we’d

be faced with

√

144 + 48 =

√

192.

2. Use the fact that we’ve already completed the square:

−3(x −2)

+ 16 = 0 ⇐⇒ (x −2)

⇐⇒ x = 2 ±

√

In many cases it is simpler to complete the square than to use the quadratic formula—remember

that they are equivalent!

Polynomials are often employed in modelling due to their simplicity and ease of evaluation. As you

saw in calculus, the motion of a falling body, or of any projectile can be modelled using quadratic

polynomials, an observation going back to at least to Galileo in the early 1600s: the distance travelled

by a falling body is proportional to the square of the time taken y(t) − y(0) ∝ t

Example 1.11. A body is dropped from a height of 125 meters, taking exactly 5 seconds to reach the

ground. Its height at time t seconds is given by y(t) = 125 −5t

This certainly ﬁts Galileo’s observation: y(t) −y(0) = −5t

is indeed

proportional to t

Over each interval of 1 s, we may ask how far the body falls; we

summarize in a table.

t 0 1 2 3 4 5

y(t) 125 120 105 80 45 0

y(t) − y(0) 0 −5 −20 −45 −80 −125

∆y −5 −15 −25 −35 −45

100

125

0 1 2 3 4 5

Since each interval has duration 1 s, each ∆y is the average speed of the falling body over that interval.

You’ll have seen problems like this in calculus; likely you want to differentiate to ﬁnd the velocity

′

(t) = −10t m/s and acceleration y

′′

(t) = −10 m/s

. However, historically and in introductory

calculus, it is problems like these that motivate the deﬁnition of the derivative.

Armed with calculus, Galileo’s observation is that the height y(t) solves a differential equation

= −g =⇒ y

′

(t) = −gt + v

=⇒ y(t) = −

+ v

t + h

where g (approximately 32 ft/s

or 10 m/s

) is the constant acceleration due to gravity, and the con-

stants of integration h

, v

are the initial height and vertical velocity. Unless you are explicitly teach-

ing calculus or Newtonian physics, this is probably a bad place to start!

The last line of the table really does suggest that speed is a linear function!

Example 1.12. Your frisbee is stuck 15 m up a tree. Standing 10 m

from the base of the trunk, you throw a ball with the intent of knock-

ing the frisbee out of the tree.

The standard approach to modeling such problems involves consid-

ering the horizontal and vertical motions separately.

Horizontal x(t) = pt + q is a linear function of time.

Vertical y(t) = −5t

+ rt + s is a quadratic function of time.

Substituting for t yields a quadratic function for the trajectory

y(x) = ax

+ bx + c

We’ll leave the details of the solution to Exercise 6. For the present,

consider why there are multiple answers; can you explain why without

explicitly solving the problem?

start

Exercises 1.3. 1. Complete the square for each quadratic function. Use your answer to ﬁnd the

range and to graph the function.

(a) f (x) = x

−6x + 5 (b) f (x) = −x

+ x + 1

+ 8x + 5

2. For the quadratic function y = 2x

−5x + 7, produce a table for x ∈ {0, 1, 2, 3, 4, 5, 6} similarly

to that in Example 1.11. What do you observe about ∆y?

3. Find the implied domain of the function f (x) =

√

4−7x+x

4. (a) Find the equations of all quadratic polynomial functions which pass through the points

(1, 3) and (2, 4).

(b) More generally, if P = (a, b) and Q = (c, d) are given, where c = a, ﬁnd all quadratic

functions whose graphs contain P and Q.

5. Describe as best you can how the graph of the function f (x) = 3x

+ bx + 2 depends on b.

6. Consider the frisbee/tree problem (Example 1.12). Assume you’re standing at the origin and

that the frisbee is at the point ( 10, 15).

(a) Find/describe all suitable trajectories that result in the ball hitting the frisbee.

(b) (Hard) Find a formula relating the initial speed v and initial slope m of the parabola (the

initial speed/direction in which you throw the ball).

i. If you throw the ball in such a way that the initial vertical speed of the ball is twice its

horizontal speed, ﬁnd how fast you have to throw the ball in order to hit the frisbee.

ii. What is the minimum speed at which you could throw the ball if you want to dislodge

the frisbee?

(Hint: You’ll need some calculus! In the language of the original problem, the initial slope is m =

and the initial speed v =

+ r

; why?)

1.4 Polynomials, Factorization & the Rational Roots Theorem

Recall our simple example of factorization in the previous section

+ 2x −15 = (x −3)(x + 5) = 0 ⇐⇒ x = 3 or x = −5

That this approach provides all roots relies on several familiar algebraic facts:

1. Factor Theorem: f (c) = 0 ⇐⇒ x −c is a factor of f (x).

2. No zero-divisors: g(x)h(x) = 0 ⇐⇒ g(x) = 0 or h(x) = 0.

3. A quadratic has at most two distinct roots.

We’ll examine this more closely at the end of this section. For students ﬁrst learning factorization,

it isn’t the why that’s the challenge, it’s the how. Multiplying out (x − 3)(x + 5) is mechanical, but

factorizing requires some creativity; we can’t really factor without somehow knowing that 3 and −5

are roots! Beyond making a lucky guess, how might we go about this?

Example 1.13. Let’s re-examine f (x) = x

+ 2x −15 = 0 in a couple of stages.

Integer solutions The simplest type of root would be an integer n. If f (n) = 0, observe that

+ 2n −15 = 0 =⇒ n(n + 2) = 15 =⇒ 15 is divisible by n

There are only eight possible candidates for n, and it doesn’t take long to test them all:

n 1 −1 3 −3 5 −5 15 −15

n + 2 3 1 5 −1 7 −3 17 −13

Rather than computing f (n) explicitly, we listed all divisors of n in the ﬁrst, the corresponding

n + 2 in the second, and mentally checked when n(n + 2) = 15. There are precisely two integer

solutions, namely n = 3 and n = −5.

Rational Solutions If you already believe that a quadratic polynomial has at most two solutions, then

you’re done. The next simplest possibility, however, is that a solution be a rational number x =

we may assume this is in simplest terms.

Substituting into the polynomial, we see that

+ 2

−15 = 0 ⇐⇒ p

+ 2pq −15q

= 0

Remembering that p, q are integers, we rearrange this equation in two ways:

p(p + 2q) = 15q

Since the left side is a multiple of p, so also is the right. Since p, q have no

common factors, it follows that p divides into 15 (15 is a multiple of p).

= q(15q −2p) Since the right side is a multiple of q, so also is the left. Since p, q have no

common factors, we conclude that q = 1.

The upshot is that the only rational solutions to f (x) = 0 are the two integers we’ve already

found.

I.e., p ∈ Z and q ∈ N have no common factors: gcd(p, q) = 1.

Deﬁnition 1.14. A degree n polynomial is any function of the form

f (x) = a

+ a

n−1

+ ···+ a

x + a

where the coefﬁcients a

are constants with a

= 0.

A quadratic polynomial has degree 2 and a linear polynomial mx + c degree one

(if m = 0).

Our analysis in Example 1.13 generalizes to a famous result.

Theorem 1.15 (Rational Roots). Suppose f (x) = a

+ ··· + a

has integer coefﬁcients where a

and a

are non-zero. If x =

is a rational root in simplest terms, then q divides into a

and p into a

In particular, if a

= 1, then the only possible rational roots are integers.

Proof. Substitute

into f (x) and multiply by q

to obtain an equation where everything is an integer

+ a

n−1

q + ··· + a

n−1

| {z }

divisible by p

divisible by q

z }| {

+ a

= 0

By considering the braced terms we see that a

is divisible by q and a

by p. Since p, q have no

common factors, we obtain the result.

Examples 1.16. 1. If x =

is a rational root of f (x) = 2x

− x − 3 in lowest terms, then q = 1 or 2

and p = ±1 or ±3. The eight possibilities for x are easily checked:

1 −1 3 −3

−

2x −1 1 −3 5 −7 0 −2 2 −4

You may prefer to compute f (x) directly: as in the previous example, since we already know

x it is quicker to check whether x(2x − 1) = 3 rather than f (x) = 0 (consider whether this

trick would be helpful or confusing in a grade-school context). The two roots are indicated; it

is easily veriﬁed that the polynomial can be factorized f (x) = (2x −3)(x + 1).

2. If the cubic polynomial f (x) = x

−2x

+ 5 had any rational roots, the only possibilities would

be ±1 or ±5. It is quickly veriﬁed that none of these work,

f (1) = 4, f (−1) = 2, f (5) = 80, f (−5) = −170

whence f (x) = 0 has no rational roots.

Unless there are very few candidates for rational roots, checking all possibilities by hand is time-

consuming. The rational roots theorem is therefore typically used in conjunction with factorization

by providing options for how to start factorizing. This still isn’t easy, as the next example shows.

A non-zero constant polynomial has degree zero. By convention, the zero polynomial y ≡ 0 has degree −∞ so that the

theorem deg f g = deg f + deg g makes sense for all polynomials.

Example 1.17. Consider the cubic function f (x) = x

− x

− 7x + 10. The rational roots theorem

offers eight candidates for rational roots: x = ±1, ±2, ±5, ±10. It is not difﬁcult to check the ﬁrst few

of these in your head, for instance,

f (2) = 8 − 4 −14 + 10 = 0

By the factor theorem, x −2 is a factor of f (x). The factorization can be performed in various ways.

Here are three options, though all are versions of the same process.

Long/synthetic division You should have practiced this in high-school.

+ x −5

x −2



− x

−7x + 10

− x

+ 2x

−7x

− x

+ 2x

−5x + 10

5x −10

=⇒ x

− x

−7x + 10 = (x −2)(x

+ x −5)

Multiply out and solve Write f (x) = (x −2)q(x) where q(x) = ax

+ bx + c is some quadratic poly-

nomial. Now multiply out:

− x

−7x + 10 = (x −2)(ax

+ bx + c) = ax

+ (b −2a)x

+ (c −2b)x −2c

Equating coefﬁcients, we obtain the same factorization as before:

a = 1, b − 2a = −1 =⇒ b = 1, −2c = 10 =⇒ c = −5

Term-by-term factorization We construct the required quadratic factor term-by-term. Since each cal-

culation can be done in your head, with practice you’ll ﬁnd that you can factorize in one line

without showing any work. Teaching such an approach is likely a terrible idea unless your

students are already very comfortable with factorization!

(a) To create x

, the ﬁrst term of the quadratic factor must be x

−x

−7x + 10 = (x −2)(x

+ ···) = x

−2x

+ ···

(b) We have −2x

but want −x

. To correct this, add x to the quadratic (x

−2x

= −x

(x −2)(x

+ x + ···) = x

−x

−2x + ···

(x −2)(x

+ x − 5) = x

−x

−7x + 10

(d) Since the last term 10 is correct, the factorization worked!

You might have seen other approaches involving arranging the coefﬁcients in a table. Regardless, the

calculations required to complete these methods are exactly those seen above; all these methods are

versions of the same thing.

Why Does Factorization Work?

The theory of factorization relies on some algebra. Here is a brief treatment.

Theorem 1.18 (Factor Theorem). Suppose f (x) is a degree n polynomial. Then:

1. f (c) = 0 if and only if f (x) = (x −c)q(x) for some (degree n −1) polynomial q(x).

2. The polynomial has at most n distinct roots.

Proof. 1. (⇐) This is essentially trivial: f (x) = (x −c)q(x) =⇒ f (c) = (c −c)q(c) = 0.

(⇒) This relies on the division algorithm for polynomials: if f , g are polynomials, then there are

unique polynomials q, r with

f (x) = g(x)q(x) + r(x) and deg r < deg g

If g(x) = x −c is linear, r(x) must be constant. Evaluate both sides at x = c to obtain

f (x) = (x −c)q(x) + f (c) (thus f ( c) = 0 =⇒ f (x) = (x −c)q(x))

2. Suppose c

, . . . , c

are distinct real roots. By part 1, f (x) = (x −c

(x). Since

0 = f (c

) = (c

−c

) =⇒ q

) = 0

we may factor x −c

from q

(x) to obtain

f (x) = (x −c

)(x −c

(x), deg q

= n −2

Repeat this process to factor out all n linear polynomials x −c

f (x) = (x −c

) ···(x −c

, deg q

= n − n = 0

whence q

= 0 is constant. Plainly f (c) = (c − c

) ···(c −c

= 0 =⇒ c = c

for some j, so

there are no other roots.

Example (1.17 cont). We know that f (x) = x

− x

−7x + 10 = (x −2)(x

+ x −5). But then

f (x) = 0 ⇐⇒ x −2 = 0 or x

+ x −5 = 0

The former gives the root x = 2, and the latter can be attacked via the quadratic formula or complet-

ing the square; the polynomial therefore has exactly three real roots

x = 2,

−1 ±

√

For a given example, q and r may be found by synthetic division. This is similar (and may be demonstrated similarly)

to the more familiar division algorithm for integers: if m, n are integers, then there are unique integers q, r for which

m = qn + r and 0 ≤ r <

In elementary school, this is typically written m ÷n = q r r (q remainder r); e.g., 23 ÷4 = 5 r 3 corresponds to 23 = 5 ×4 + 3.

Example 1.19. We ﬁnish with a quick example of how long division (or any other factorization

method as in Example 1.17) computes the ingredients in the division algorithm.

If f (x) = x

+ 7x

−2 and g(x) = x

−2, then

x + 7

−2



+ 7x

−2

− x

+ 2x

+ 2x − 2

−7x

+ 14

2x + 12

=⇒ x

+ 7x

−2 = (x

−2)(x + 7) + (2x + 12)

Otherwise said, f (x) = g(x)q(x) + r(x), where

q(x) = x + 7, r(x) = 2x + 12 and deg r = 1 < 2 = deg g.

Exercises 1.4. 1. Apply the rational roots theorem to the polynomial x

+ 2x

− x −2 and use it to

factorize the polynomial.

2. Repeat the previous question for the polynomial 6x

+ x −2.

3. Use the rational roots theorem to prove that the polynomial 2x

−3x + 7 has no rational roots.

4. Factorize the polynomials and thereby ﬁnd their (real) roots. Explain your steps carefully.

(a) f (x) = x

+ 2x

−3x (b) f (x) = x

−13x

+ 36

−7x −6

5. Factorize the polynomial f (x) = x

−2x

− x

−4x

−4x −6 and thus demonstrate that

it has exactly two real roots.

6. Students often follow a heuristic when trying to factorize a polynomial f (x) = 0: try some

small integer values for x until you ﬁnd a root, then apply long division. For what types of

polynomial f (x) will this approach work? Explain.

7. The polynomial f (x) = 2x

−3x

+ 2x

+ 3x −9 has only one rational root. Find it and factorize

the polynomial as f (x) = g(x)q(x) where deg g = 1.

8. Find unique polynomials q(x) and r(x) for which f (x) = g(x)q(x) + r(x) and deg r < deg g.

(a) f (x) = x

+ 1 and g(x) = x + 2.

(b) f (x) = x

+ x

−2 and g(x) = x

+ 1.

9. Let f (x) = ax

+ bx

+ cx + d be a cubic polynomial. ‘Complete the cube’ by ﬁnding a constant

k such that

f (x) = a(x −k)

+ p(x −k) + q

has no (x −k)

term (here p, q are constants).

(Hint: evaluate f (x + k))

10. Suppose deg f = k and deg g = l.

(a) Show that deg( f g) = kl.

(b) Is it always the case that deg( f + g) = max(k, l)? Why/why not?

1.5 Inverse Functions & the Horizontal Line Test

The informal idea of an inverse function is that f

−1

takes the output of f and returns its input (and

vice versa).

Example 1.20. Deﬁne a simple function using a table or an arrow diagram

x 1 2 3 4

f (x) 4 2 5 7

y 4 2 5 7

−1

(y) 1 2 3 4

The inverse f

−1

is the function obtained by reversing the arrows or ﬂipping

the table upside-down.

−1

Deﬁnition 1.21. A function f : A → B is invertible if it has an inverse: a function f

−1

: B → A for

which

−1



f (x)



= x and f



−1

(y)



= y (∗)

for all possible inputs x ∈ A and y ∈ B.

Certainly Example 1.20 satisﬁes the input–output properties (∗). Our concerns are identifying when

a function is invertible, how to make it so if not, and how to compute an inverse.

Examples 1.22. 1. The function f (x) = 2x has inverse f

−1

(y) =

The input–output conditions (∗) are certainly satisﬁed.

The graph admits an interpretation of f

−1

similar to the arrow diagram.

• The function f takes an input x, moves it vertically to the graph, then

projects to the y-axis. This interpretation is precisely the vertical line

test (Deﬁnition 1.4)!

• The inverse function reverses the arrows: transport an input y horizon-

tally to the graph, then project to the x-axis.

0 1 2

−1

2. Consider f (x) = x

−1. This time, when attempting to move a real

number y horizontally to the graph, we usually encounter one of

two problems:

(a) If y > −1, there are two choices of x (two intersections).

(b) If y < −1, there is no intersection with the graph.

The na

ıve approach of reversing the arrows is insufﬁcient to deﬁne an

inverse. However, a simple remedy arises by staring at the graph:

• Problem (a) goes away if we delete the left half of the graph.

Equivalently, we restrict the domain of f to [0, ∞).

• Problem (b) disappears if we insist that y ≥ −1. Equivalently,

we restrict the codomain of f to its range [−1, ∞).

−2 2

−1

After making these restrictions so that f : [0, ∞) → [−1, ∞ ), it is easily checked that

−1

(y) =

y + 1, f

−1

: [−1, ∞) → [0, ∞)

satisﬁes the input–output conditions (∗) and is therefore the inverse of f :

x ∈ [0, ∞) =⇒ f

−1



f (x)



−1) + 1 = x

y ∈ [−1, ∞) =⇒ f



−1

(y)





y + 1



−1 = y

What makes a function invertible? The ﬁxes in the last example can be rephrased succinctly:

Horizontal line test: every horizontal line must intersect the graph exactly once

This unpacks to two conditions, each of which addresses one of the problems seen in the example.

Deﬁnition 1.23. Let f : A → B be a function. We say that f is:

(a) 1–1/one-to-one if distinct inputs x

= x

∈ A have distinct outputs f (x

) = f (x

). Equivalently,

Given x

, x

∈ A, we have f (x

) = f (x

) =⇒ x

= x

If A, B are sets of real numbers, each horizontal line intersects the graph at most once.

(b) Onto if range f = B. Equivalently,

Given y ∈ B, there is some x ∈ A for which y = f (x)

If A, B ⊆ R, the horizontal line through y ∈ B intersects the graph at least once.

Putting these ideas together, a function is both 1–1 and onto precisely when every y ∈ B corresponds

to a unique x ∈ A for which y = f (x). In summary:

Theorem 1.24. f : A → B is invertible if and only if it is both 1–1 and onto. Its inverse is the function

−1

: B → A such that f

−1

(y) = x whenever y = f (x).

Example (1.22.2, mk. II). Consider the two properties in the context of the example f (x) = x

−1:

(a) f (x

) = f (x

) =⇒ x

−1 = x

−1 =⇒ x

= x

=⇒ x

= ±x

To force f to be 1–1, it is enough to restrict the domain so that all x have the same sign: the

obvious choice is dom f = [0, ∞).

(b) range f =



−1 : x ∈ [0, ∞)



= [−1, ∞). We force f to be onto by restricting its codomain to

[−1, ∞).

The inverse function is obtained by solving y = x

−1 for x:

= y + 1 =⇒ x = f

−1

(y) =

y + 1

The non-negative square root is used since x ∈ dom f = [0, ∞).

An algorithm for inverting functions Our discussion provides an algorithmic process for making

a function f : A → B invertible and ﬁnding an inverse.

(a) Check that f is 1–1. If not, restrict the domain until it is.

(b) Check that f is onto. If not, redeﬁne B = range f .

−1

(y).

Since x is typically preferred as an input, it is common to switch x, y at the end of step 3 and write

y = f

−1

(x). If A, B ⊆ R, switching x ↔ y is equivalent to reﬂecting the graph in the line y = x.

Note also that step (a) likely involves a choice; depending on how you restrict the domain, you can

ﬁnd multiple inverse functions! To see this in action, we return once more to our example.

Example (1.22.2, mk. III). Recall that if f (x) = x

−1, then

f (x

) = f (x

) =⇒ x

= ±x

Instead of restricting the domain to [0, ∞), we can instead force f to be 1–1 by taking the other half

of the graph; by choosing dom f = (−∞ , 0]. The range/codomain remains [−1, ∞), but the inverse

function is now different:

= y + 1 =⇒ x = −

y + 1 ∈ (−∞, 0] = dom f =⇒ f

−1

(x) = −

√

x + 1

This time the new domain for f forced us to use the negative square root.

−2 2

f (x) = x

−1

dom( f ) = [0, ∞)

dom( f ) = (−∞, 0]

0 3 6 9

−1

(x) =

√

x + 1

3 6 9

−2

−1

(x) = −

√

x + 1

We could choose other domains on which f is 1–1, but these are the most natural choices.

The moral is that you cannot invert a function unless you are precise about its domain and range!

We ﬁnish with an algebraically tougher example, where you may feel that more detail is justiﬁed.

Example 1.25. Let y = f (x) =

(x−2)

. Its implied domain consists of all real numbers except 2.

The vertical line test is clearly visible on the graph: every vertical line

x = a, except x = 2, intersects the graph exactly once.

The range is the interval R

= (0, ∞) as can be seen by solving

f (x) = y ⇐⇒

x −2

= ±

√

y ⇐⇒ x = 2 ±

√

Any positive output y may be obtained via y = f



2 +

√



The ±-term shows that f fails the horizontal line test: it isn’t 1–1.

There are two natural choices for an inverse:

(a) Choose dom f = (2, ∞), then ±

√

y =

x−2

is positive. We

take the positive square root and obtain the inverse function

g : (0, ∞) → (2, ∞), g(x) = 2 +

√

(b) Choose dom f = (−∞, 2), then ±

√

y =

x−2

is negative and

we obtain a second inverse function

h : (0, ∞) → (−∞, 2), h(x) = 2 −

√

−1 0 1 2 3 4

−1

1 2 3 4 5 6

y = g(x)

y = h(x)

Exercises 1.5. 1. If dom f = R, check that f (x) = x

+ 8 passes the horizontal line test. Find f

−1

2. Consider f (x) = x

+ 2x −3. Similarly to Example 1.22, ﬁnd two inverses of f .

3. Sketch the graph of the function

f (x) =











x if 0 ≤ x < 1

x −1 if 1 ≤ x < 2

x −2 if 2 ≤ x < 3

Find three domains on which f is 1–1 and thus compute three distinct inverses.

4. Show that the following function f : R → (

, ∞) is 1–1 and onto, sketch its graph and ﬁnd f

−1

f (x) =

(

3 −

x if x ≤ 2

2 −

if x > 2

5. (Hard) Find the implied domain and range of f (x) =

x+1

. Now ﬁnd an interval on which f

is 1–1 and compute its inverse.

6. An astute student observes that Deﬁnition 1.21 only describes the properties satisﬁed by an

inverse and asks why we keep referring to the inverse. How would you respond?

2 Trigonometric Functions and Polar Co-ordinates

In this chapter we review trigonometry and periodic functions and discuss their relation to polar

co-ordinates. Some of this will be non-standard.

2.1 Deﬁnitions & Measuring Angles

Trigonometric functions date back at least 2000 years. Ancient mathemati-

cians were interested in the relationship between the chord of a circle and

the central angle, often for the purpose of astronomical measurement. It

wasn’t until 1595 that the term trigonometry (literally triangle measure) was

coined, and the functions were considered as coming from triangles.

crd θ

Here are several related deﬁnitions of sine, cosine and tangent based either on triangles or circles.

Deﬁnition 2.1. 1. (a) Given a right triangle with hypotenuse (longest

side) 1 and angle θ, deﬁne sin θ and cos θ to be the side lengths

opposite and adjacent to θ.

Deﬁne tan θ =

sin θ

cos θ

to be the slope of the hypotenuse.

(b) Given a right triangle with angle θ, hypotenuse r, adjacent x and

opposite y, deﬁne

sin θ =

cos θ =

tan θ =

2. (a) (cos θ, sin θ) are the co-ordinates of a point on the unit circle,

where θ is its polar angle measured counter-clockwise from the

positive x-axis. Provided cos θ = 0, also deﬁne tan θ =

sin θ

cos θ

(b) Repeat the deﬁnition for a circle of radius r with co-ordinates

(r cos θ, r sin θ).

cos θ

sin θ

Discuss some of the advantages and weaknesses of these deﬁnitions:

• What prerequisites are you assuming in each case?

• Is it easier to think about lengths rather than ratios?

• Where do you need basic facts from Euclidean geometry such as congruent/similar triangles?

• Convince yourself that that the triangle deﬁnitions follow from the circle deﬁnitions. What is

missing if you try to use the triangle deﬁnition to justify the circle version?

• If you were introducing trigonometry for the ﬁrst time, what would you use?

If you’ve done sufﬁcient calculus you might know of other deﬁnitions, for instance using power

(Maclaurin) series. Plainly these are not suitable for grade-school, but have the great beneﬁt of

making the calculus relationship

dθ

sin θ = cos θ very simple. Establishing this using the triangle

deﬁnition is a somewhat tricky!

Measuring Angles

There are two standard ways to measure angles (to sensibly associate a number to each angle).

Degrees A full revolution has 360° and a right-angle 90°. Degree measure dates back to ancient Baby-

lon 2–4000 years ago.

Radians The radian measure of an angle is the length of the arc subtending the angle in a circle of

radius 1. Since the circumference of a unit circle is 2π, we have the following identiﬁcations.

Degrees Radians sin θ cos θ tan θ

0° 0 0 1 0

30°

√

45°

√

60°

√

90°

1 0 n/a

180° π 0 −1 0

0°

30°

60°

90°

120°

150°

180°

210°

240°

270°

300°

330°

2π

5π

7π

4π

3π

5π

11π

In elementary mathematics, degrees are the most common way to measure angles. Do you know any

other methods?

Exercises 2.1. 1. The identity cos

θ + sin

θ = 1 is the Pythagorean Theorem in disguise. Why?

2. The word sine is the result of a long list of translations and transliterations from an ancient

Sanskrit term meaning half-chord. For the chord picture on page 21, how does the length of the

chord crd θ relate to modern trigonometric functions?

3. It is conventional not to state units when using radians since they are effectively a ratio and

therefore unitless. Think this through: if the central angle in a circle of radius r is subtended by

an arc with arc-length ℓ, what is the radian measure of the angle? What facts from Euclidean

geometry justify this observation?

4. Explain how to get the values of sine and cosine in the above table.

(Hint: Draw some triangles and use Pythagoras!)

5. Using the pictures, explain why we have the relations

sin(

−θ) = cos θ = sin(θ +

), sin(−θ) = −sin θ, sin(π −θ) = sin θ

(You cannot use multiple-angle formulæ for this!)

It is not known why they chose 360, but it ﬁts nicely with their base-60 system of counting (decimals are base-10).

The traditional subdivisions of a degree are also base-60. For instance, 34°12

′

45” is 34 degrees, 12 (arc)minutes and 45

(arc)seconds; converted to decimal notation, this becomes

34°12

′

45” = 34 +

= 34.2125°

The standard hour-minute-second measurement of time has the same origin.

2.2 Periodicity, Graphs & Inverses

One advantage of the circle deﬁnition is that it makes sketching the graphs of sine and cosine very

easy. Simply draw axes next to a unit circle and transfer the heights across.

−1

y = sin x

3π

2π

By Exercise 2.1.5, the graph of cos x = sin(x +

) is simply that of sine shifted

= 90° to the left.

Moreover, the circle deﬁnition allows us easily to extend trigonometric functions periodically since we

can measure the polar angle by looping as many times round the origin as we like: for any integer n,

sin(θ + 2nπ) = sin θ, cos(θ + 2nπ) = cos θ

Otherwise said, sine and cosine have period 2π radians (360°).

Sine and cosine are non-invertible unless we choose a domain on which they are 1–1.

−1

−

π−π

3π

5π

2π

y = sin x

f (x) = sin x is 1–1 on the domain [−

]

Inverse function f

−1

(x) = arcsin x = sin

−1

Domain dom( arcsin) = [−1, 1] = range(sin)

Range range( arcsin) = [−

] = dom(sin)

This is why your calculator always returns a value in the

interval [−

] = [−90°, 90°] when you hit the sin

−1

button.

−1

y = sin x

y = sin

−1

−

Example 2.2. If you know the graphs, then symmetry and periodicity help you solve equations. For

example, if sin θ =

then all solutions are given by

θ = sin

−1

+ 2πn or π − sin

−1

+ 2πn (n is any integer)

−1

sin

−1

−

π−π

3π

−

3π

−2π

π −sin

−1

Alternatively, we could use the circle deﬁnition directly: sin θ =

means we want angles θ corre-

sponding to the intersections of the unit circle with the horizontal line y =

Periodic Models Trig functions ﬁnd applications in modeling precisely because they are periodic.

In general, a function has period T if

f (x + T) = f (x) for all x

It is easy to ﬁnd the period of the function f (x) = sin kx just by considering what we have to add to

the input x to increase the argument kx of sine by 2π:

T =

2π

=⇒ f (x + T) = sin(kx + 2π) = sin kx = f (x)

We may therefore obtain a simple periodic model regardless of what period is required.

Example 2.3. On a given day, high tide occurs at 2:00 with a water depth of 10 ft, whereas low

tide occurs at 8:12 with a depth of 4 ft. We might model this using a periodic function with period

T = 2 ×6

hours. For instance

h(t) = 7 + 3 cos



5π

(t −4)



h( t)

0 4 8 12 16 20 24

where t is measured in hours from midnight might be suitable. In reality, tidal height is very close to

being periodic, but the magnitude of the high and low tides are somewhat variable.

In fact any periodic function may be approximated using trigonometric functions. Indeed if f (x) has

period T and we deﬁne constants

−

f (x) cos

2πnx

dx, b

−

f (x) sin

2πnx

dx (∗)

then

f (x) ≈

+ a

cos

2πx

+ b

sin

2πx

+ a

cos

4πx

+ b

sin

4πx

+ ··· (†)

This is the Fourier series of f (x). It often takes only a small number of terms to obtain a very good

approximation. Modern data-compression algorithms often employ Fourier series. Given a periodic

function f (x), one uses a computer to estimate (say) the ﬁrst 100 Fourier coefﬁcients (∗) and transmits

these values to the receiver, who recovers an approximation to the original function using (†).

Example 2.4. A square-wave function with period T = 2π is given by

f (x) =

(

1 if 0 ≤ x < π

−1 if π ≤ x < 2π

extended periodically to the real line. With a little calculus,

it is easily checked that the Fourier coefﬁcients are

−1

f (x)

π 2π

−π

f (x) cos nx dx = 0, b

−π

f (x) sin nx dx =

(

πn

if n is odd

0 if n is even

Use a graphics tool to see how the ﬁrst few terms of the series approximate the function.

Exercises 2.2. 1. f (x) = sin x is also 1–1 on the interval [

3π

]. Sketch the graph of its corresponding

inverse function.

2. Draw the graph for cosine and observe that it is invertible if we restrict the domain to the

interval [0, π]. Draw the graph of cos

−1

3. Describe all solutions to the equation cos x = −0.2.

4. Explain why the tangent function has period π; that is tan(θ + nπ) = tan θ. What facts are we

using about sine and cosine and why are they obvious from the deﬁnition?

5. Describe all solutions to the equation tan x = 5.

6. (a) Suppose θ = cos

−1

. Find the exact values for sin θ and tan θ.

(b) What changes if θ = cos

−1

−9

7. Let f (x) = csc x =

sin x

be the cosecant function. Describe a domain on which this function is

1–1 and sketch the graph of its inverse y = f

−1

(x).

8. Use a computer to sketch the curve

y = 2



sin x −

sin 2x +

sin 3x −

sin 4x +

sin 5x



What simple periodic function do you think this is approximating?

2.3 Solving Triangles

Basic trigonometry often involves ﬁnding the edges and angles of a triangle given partial data.

Example 2.5. To ﬁnd the height h of a tall tree, two angles of elevation 45° and 30° are measured a

distance 20 ft apart along a straight line from the base of the trunk.

This is easily attacked by drawing a picture and observing that we have two right-triangles. If the

(unknown) distance from the base of the tree to the nearer measurement is x, then

√

= tan 30° =

x + 20

1 = tan 45° =

Substituting the second equation into the ﬁrst returns

h =

√

3 −1

≈ 27.32 ft

45°

30°

In fact there is enough data in the problem to recover everything about the original triangle.

• The second base angle is (180° −45° = 135°).

• The third (summit) angle is 180° − 30° −135° = 15°.

• Two applications of Pythagoras compute the remaining sides of the triangle

+ h

√

2h =

√

3 −1

≈ 38.64

+ (x + 20)

+ 3h

= 2h =

√

3 −1

≈ 54.64

The example is just a disguised version of solving a triangle: computing

all six sides and angles of a triangle given three of them. The Euclidean

triangle congruence theorems tell us which combinations are sufﬁcient

to determine all the others. The example is the ASA congruence: angle-

side-angle data (30°–20–135°) is enough to compute everything else

about the triangle.

When in doubt, you can always attack basic trigonometry problems as we did in the example: create

a right-triangle, then use the deﬁnitions of sin/cos/tan and/or Pythagoras.

Example 2.6. Given the SAS (side-angle-side) combination 5–60°–9, ﬁnd the third side of the triangle.

The altitude h creates two right-triangles, from which

h = 5 sin 60°, x = 5 cos 60°

=⇒ c

= (9 − x)

+ h

= 9

+ (x

+ h

) − 18x

= 9

+ 5

−18 ·5 cos 60° = 61

=⇒ c =

√

61 ≈ 7.81

60°

Since we now know c and 9 − x =

the remaining angles could also be easily found.

In elementary situations it is typically easier to have students drop the perpendicular as we’ve done.

However, once comfortable with the method, it is helpful to have short-cuts which skip the need to

work with the perpendicular at all.

Theorem 2.7 (Sine and Cosine Rules). For any triangle,

sin A

sin B

sin C

and c

= a

+ b

−2ab cos C

The cosine rule is just the Pythagorean Theorem with a correction for non-right triangles. Both rules

follow straightforwardly by drawing an altitude as before!

Proof. Consider the picture. We have

h = a sin C = c sin A, x = a cos C, b −x = c cos A

The ﬁrst equation rearranges to

sin A

sin C

Two applications of Pythagoras give the cosine rule

= h

+ (b −x)

= h

+ x

+ b

−2bx = a

+ b

−2ab cos C

The remaining part of the sine rule and the other versions of the cosine rule are obtained by choosing

other altitudes.

Here are two examples where we use the rules instead of explicitly drawing an altitude.

Examples 2.8. 1. A triangle has sides 2 and

√

3 − 1, and the angle between them is 120°. Find the

remaining sides and angles.

We apply the cosine rule with a = 2, b =

√

3 −1 and C = 120°

= a

+ b

−2ab cos C

= 2

+ (

√

3 −1)

−2 ·2(

√

3 −1) cos 120°

= 4 + 3 + 1 −2

√

3 + 2(

√

3 −1) = 6

We have an opposite pair (c, C) = (

√

6, 120°), so the sine rule may be used

sin A =

√

sin 120° =

√

=⇒ A = 45°

We chose the acute angle since A = 180° − B − C = 60° − B < 90°.

The ﬁnal angle is then B = 180° − 45° −120° = 15°.

√

3 −1

120°

You could instead drop a perpendicular, say from the vertex A to the extension of the side of

length 2. Think about why the perpendicular has to be outside the triangle. . .

2. A triangle has one side with length 5 and its two adjacent angles are 40° and 65°. Find the

remaining data.

This time the initial data is ASA. Writing c = 5, A = 40° and

B = 65°, the remaining angle is plainly

C = 180° −40° −65° = 75°

This gives us an opposite pair (c, C), so we can apply the sine rule

a = c

sin A

sin C

= 5

sin 40°

sin 75°

≈ 3.327

A second application yields

b = c

sin B

sin c

= 5

sin 65°

sin 75°

≈ 4.691

40°

65°

3. (Discuss: courtesy of an 8 year-old contributor) Model Earth as a sphere of radius 3963 miles.

If identical vertical ladders are placed in Irvine, CA and Irvine, Scotland, 5145 miles apart by

great circle, how tall would they have to be for people at the top to ‘see’ each other?

Multiple-angle Formulae

Also useful in the context of basic trigonometry is the ability

to sum angles. The picture provides a simple justiﬁcation of

sin(α + β) = sin α cos β + cos α sin β

at least when 0 < α + β <

. If you look carefully, you

should be able to see how the same picture establishes

cos(α + β) = cos α cos β − sin α sin β

cos β

sin β

sin α cos β

cos α sin β

Exercises 2.3. 1. Find the remaining angles in the triangle in Example 2.6.

2. The other Euclidean congruence theorems are SSS and SAA. Explain how to solve triangles

using these minimal data in two ways:

(a) By drawing an altitude. (b) Using the sine/cosine rules.

3. SSA isn’t a triangle congruence theorem. For instance, there are two non-congruent triangles

with data a = 1, b =

√

3 and A = 30°. Find them.

4. Use the multiple-angle formulae to derive the familiar expressions for sin 2θ and cos 2θ.

5. Find the exact value of sin 105°.

6. (a) Find an expression for tan(α + β) purely in terms of tan α and tan β.

(b) Two wooden wedges with slope

are placed on top of each other to make a steeper slope.

What is the gradient of the new slope?

7. Carefully explain why the answer to Example 2.8.3 is approximately 1012 miles.

2.4 Polar Co-ordinates

Deﬁnition 2.1 provides an alternative way to describe points in the plane. If θ is the polar angle of

a point with Cartesian (rectangular) co-ordinates (x, y), then its polar-coordinates are precisely the

values (r, θ) seen in the deﬁnition!

Computing x = r cos θ and y = r sin θ is easy given r and θ.

Example 2.9. A point with polar co-ordinates (r, θ) = (2,

5π

) has Cartesian co-ordinates

(x, y) =



2 cos

5π

, 2 sin

5π





−

√

3, 1



Computing polar co-ordinates from Cartesian is harder, requiring some visualization.

1. Every point (x, y) has a unique radius r =

+ y

, but not polar angle. If θ is a polar angle,

so is θ + 2πn for any integer n ∈ Z. The origin (x, y) = (0, 0) is even stranger; certainly r = 0,

but any θ is a legitimate polar angle!

2. Whenever x = 0 (away from the y-axis),

(

x = r cos θ

y = r sin θ

=⇒ tan θ =

however, this doesn’t guarantee that θ = tan

−1

. Continuing the example shows us why. . .

Example (2.9, cont). If (x, y) = (−

√

3, 1), then the radius is easy

r =

(

√

+ 1

= 2

For the polar angle,

tan θ =

= −

√

= tan



−



=⇒ θ = −

Arctan has range (−

), so always returns an angle in quad-

rants 1 or 4. Our point is in the second quadrant (x < 0 < y) so we

need to adjust, using the fact that tan is π-periodic:

θ = π −

5π

= 150°

We could alternatively add any integer multiple of 2π.

−

5π

−

2π

−

5π

−

√

3−

√

−1

The example wasn’t too tricky since the polar angle was exactly computable. When you have to rely

on a calculator, it is much easier to make a mistake.

Example 2.10. The point (x, y) = (−8, −15) has polar co-ordinates (quadrant 3!)

r =

+ 15

= 17, θ = π + tan

−1

≈ 241.93°

We could summarize with formulæ describing precisely how to compute θ dependent on quadrant

(the signs of x, y), though it is better to get in to the habit of drawing a picture!

Curves in Polar Co-ordinates

Polar co-ordinates are well-suited to describing curves that encircle the origin. Indeed circles centered

at the origin with radius a > 0 have the very simple polar form r = a. Converting to rectangular

co-ordinates recovers the the natural parametrization of a circle:

x(θ) = a cos θ, y(θ) = a sin θ

This partly explains why mathematicians call sine and cosine circular functions.

General polar graphs are harder to visualize, though the major reason is lack of familiarity. Have

a little empathy: to graph polar functions, you’ll likely have to follow the same approach as new

students use to sketch Cartesian curves like y = x

! Here are a couple of examples.

Examples 2.11. 1. The curve r = θ is relatively easy to plot since r increases at exactly the same rate

as the angle; we therefore have a spiral.

To conﬁrm this, plot several points (θ, θ); we’ve done for θ in multiples of

(30°) from 0 to 2π.

It is sensible to use ‘polar graph paper’ with concentric circles separated by (say)

≈ 1.57.

−6

−3

−6 −3 3 6

7π

4π

2π

5π

11π

r = θ, 0 ≤ θ ≤ 2π

−2 −1 0 1 2

2π

5π

r = 2 sin θ, 0 ≤ θ ≤ π

2π

4π

5π

2 sin θ 1 1.73 1 1.73 1 0

2. The curve r = 2 sin θ is a little easier to work with since we know exact values for sine, assisted

√

3 ≈ 1.73.

This looks like a circle! To see this, multiply both sides by r and complete the square:

= 2r sin θ =⇒ x

+ y

= 2y =⇒ x

+ (y −1)

= 1

describes the set of points with distance 1 from the point (0, 1): a circle!

You should think about what happens in both examples if we extend the domain:

• What would r = θ look like if θ were allowed to be negative?

• What happens to r = 2 sin θ when θ > π?

Exercises 2.4. 1. Convert the following points to polar co-ordinates.

(a) (−5, 5) (b) (3, −4) (c) ( −5

√

3, −15)

(d) (−1, tan 3) (tricky—this is 3 radians!)

2. If a > 0, describe the curve with polar equation r = 2a cos θ.

(Be careful with θ >

since cosine goes negative. . . )

3. The algebraic trickery in the last example sometimes bears fruit, though you have to be lucky!

By multiplying both sides by 1 − sin θ and converting to rectangular co-ordinates, show that

the polar function

r(θ) =

1 −sin θ

is a parabola in disguise and sketch it when a = 1. How does the graph depend on a?

4. In a similar vein to the previous question, sketch the curve r =

2+sin θ

. What type of curve is

this?

5. Try to sketch the following curves.

(a) r = θ(θ − 4) (b) r = (θ − 1)

+ 1 (c) r = (θ −1)

−1

As well as plotting points directly, you should sketch the curve ﬁrst on rectangular axes (e.g.,

(a) is y = x(x −4)). What happens to (c) when θ = 1?

Once you’ve tried these, use a grapher to see if you’re right, though see how close you can get

without it!

3 Exponential and Logarithmic Functions & Models

Introducing exponential functions without calculus presents a signiﬁcant challenge. The simplest

approach is as a short-hand notation for repeated multiplication: for instance

= a · a ·a ·a ·a

analogous to how multiplication represents repeated addition

5a = a + a + a + a + a

The problem with this approach is that it doesn’t help you understand what should be meant by, say,

3/4

or a

√

: multiplying something by itself ‘

√

2 times’ sounds

insane!

To rigorously address this problem requires continuity and other ideas surrounding the foundations

of calculus which you’ll encounter in upper-division analysis; topics unsuitable for this course. In-

stead, we assume some familiarity with exponential functions via introductory calculus, where they

are unavoidable and offer two ways to introduce exponential functions and e via modelling.

3.1 The Natural Growth Model

A basic model for any variable quantity is that its rate of change be proportional to the quantity itself. This

idea necessarily needs some calculus; as a differential equation,

= ky

where k is a constant; if k > 0 this is the natural growth equation, if k < 0 the natural decay equa-

tion. This is commonly encountered when modelling population growth; an otherwise unconstrained

population seems like its growth rate should be proportional to its size (twice the people, twice the

babies. . . ). This model is hugely applicable, since population can refer to essentially any quantiﬁable

value: people, bacteria, money, reagents in a chemical/nuclear reaction, etc.

Example 3.1. The simplest natural growth equation has k = 1:

= y

If a point (x, y) lies on a solution curve, then the differential equation tells

us the direction of travel of the solution. We may visualize this by drawing

an arrow with slope

= y; the arrows are tangent to any solution.

You

should easily be able to sketch some other solution curves.

You should, of course, recognize the graph. . .

−2

−1

−2 −1 1 2

The same issue arises for multiplication: 3

√

2 =

√

2 +

√

2 +

√

2 is relatively easy to understand, but how would you

convince someone what π

√

2 means?

A similar approach is available for any ﬁrst-order differential equation

= F(x, y) : the equation deﬁnes its slope ﬁeld

(arrows), to which solution curves must be tangent.

Deﬁnition 3.2. Let a > 0 be constant. The exponential function with base a is f (x) = a

Recall the exponential laws, which are very natural when x, y, r are positive integers:

x+y

= a

x−y

)

= a

These hold for all exponents, with the same continuity caveats we saw previously. For modelling,

the crucial property of exponential functions is that they have proportional derivative.

Theorem 3.3. The rate of change of f (x) = a

is proportional to f (x). Speciﬁcally,

′

(x) = lim

h→0

x+h

− a

= a

lim

h→0

−1

so that f (x) = a

satisﬁes the natural growth/decay equation

= ky with proportionality constant

k = f

′

(0) = lim

h→0

−1

Example 3.4. We estimate the proportionality constant k = lim

h→0

−1

to 3 d.p. using a calculator for

four values of h:

a 2 2.5 2.7 2.75 3 5

0.1

−1

0.1

0.718 0.960 1.044 1.065 1.161 1.746

0.01

−1

0.01

0.696 0.921 0.998 1.017 1.105 1.622

0.001

−1

0.001

0.693 0.917 0.994 1.012 1.099 1.611

0.0001

−1

0.0001

0.693 0.916 0.993 1.012 1.099 1.610

What is happening to the proportionality constant as a increases?

It appears as if there is a special number somewhere between 2.7 and 2.75 for which the proportion-

ality constant is precisely k = 1.

Deﬁnition 3.5. The value e = 2.71828 . . . is the unique real number such that lim

h→0

−1

= 1.

The natural

exponential function exp(x) = e

has derivative

= e

Natural here means unavoidable: an old cliche suggests that if aliens were to land on Earth, they’d have to understand e

given the technology they’d require to get here. Of course they’d likely use a different symbol; ours comes from Leonhard

Euler around 1728. Like π and

√

2, the constant e is an irrational number: its decimal representation contains no repeating

pattern. There isn’t the same geeky fascination with memorizing the digits of e as there is with π, neither is there an ‘e-day’

(Feb 7

at 6:28 p.m.?).

The function f (x) =

is plotted in Example 3.1. Of course there are many other solutions to the

natural growth equation

= y: for any constants c, k,

y = ce

=⇒

= cke

= ky

In fact the converse also holds; for the details, take a differential equations course!

Theorem 3.6. The solutions to the natural growth equation

= ky are precisely the functions

y(x) = y

= y

exp(kx)

where y

= y(0) is the initial value.

Example 3.7. A Petri dish contains a population P(t) of bacteria satisfying the natural growth equa-

tion

= 0.5P where time is measured in weeks from the start of the year.

If P(0) = 100 bacteria, then P(t) = 100e

0.5t

. Speciﬁcally, at the

end of January (4

weeks) one expects there to be

) = 100 exp

= 915 bacteria

Note that the exponential doesn’t return 915 exactly; this is only

an approximation. Models like this work best for large popula-

tions where integer rounding errors are of minimal concern.

200

400

600

800

1000

0 1 2 3 4 5

Compound Interest and the Discovery of e

The ﬁrst description of e came in 1683 when Jacob Bernoulli tried to model the growth of money in a

hypothetical bank account. We give a modernized version of his approach.

Example 3.8. $1 is deposited in an account paying 100% interest per year (nice!). Bernoulli observed

that the money in the account at the end of the year depends on when the interest is paid.

• If the interest is paid once at the end of the year (this is called simple interest), you’ll have $2.

• If half the interest (50¢) is paid at six months, then the balance ($1.50) earns

· 1.50 = 75¢

interest for the rest of the year; you’ll ﬁnish the year with $2.25 in the account.

• If the interest is paid in four installments, we have the following table of transactions (data is

rounded to the nearest cent)

Date Interest Paid Balance

Jan — $1

Apr 25¢ $1.25

July

·1.25 = 31¢ $1.56

Oct

·1.56 = 39¢ $1.95

New Year

·1.95 = 49¢ $2.44

More succinctly, the year-end balance is



1 +



= $2.44.

• More generally, if the interest is paid over n equally spaced intervals, the account balance at the

end of the year would be $



1 +



. Here are a few examples rounded things to 5 d.p.

Frequency Balance after 1 year ($)

Every month



1 +



= 2.61304

Every day



1 +

365



365

= 2.71457

Every hour



1 +

8760



8760

= 2.71813

Every second



1 +

31536000



31536000

= 2.71828

As the frequency of payment increases, it appears as if the balance is increasing to $e. . .

In fact this is a theorem, though it requires signiﬁcant work (beyond this class) to prove it:

e = lim

n→∞



1 +



and more generally e

= lim

n→∞



1 +



This again shows that e arises very naturally.

Simple, Monthly & Continuous Interest In ﬁnance, interest is typically computed in one of three

ways. In each case we describe the result of investing $1 at an annual interest rate of r% =

100

Simple interest You are paid

100

dollars at the end of the year. Your invested dollar becomes 1 +

100

dollars.

Monthly interest Each month you are paid

% of your current balance. This amounts to a balance

of (1 +

1200

)

dollars at year’s end. The period need not be monthly: if interest is paid in n

installments, the balance would be ( 1 +

100n

)

Continuous interest After t years (can be any fraction of a year!) your dollar-balance is

100

= exp

100

= lim

n→∞



1 +

100n



Example 3.9. A bank account earns 6% annual interest paid monthly. To what simple annual interest

rate does this correspond? Would you perfer an account paying 6% continuously?

At the end of the year, $1 becomes

(

1 + 0.0612

)

= 1.005

≈ 1.06168 . . .

corresponding to a simple interest rate of 6.17%. By cobntrast, 6% continuous interest would result

in your dollar becoming e

6/100

≈ 1.06184, corresponding to a (marginally) higher simple interest rate

of 6.18%. You should prefer this, particularly if you have a lot of money to invest! The difference is

more noticable with an investment of $1000 over ten years:

1000 ×1.005

120

= $1819.40 versus 1000e

0.6

= $1822.12

There are several reasons for these varying approaches, not all of them consumer-friendly:

1. Simple interest is simple! It is easy to understand and compute, but hard to decide how or even

whether to compute interest for parts of a year.

2. Monthly interest ﬁts with most paychecks, so is sensible for loans, particularly mortgages.

3. Continuous interest allows the balance of an account to be found easily at any time, even be-

tween interest payment dates. It is also much easier to apply mathematical analysis (calculus).

4. A company can make an interest rate appear higher (if a savings account) or lower (if a loan) by

choosing which way to quote an interest rate.

Example 3.10. A bank quotes you a loan with a continuously compounded interest rate of 7%. If

you borrow $100,000, then at the end of the year you’ll owe

100000e

0.07

= $107, 250.82

not the $107,000 you might have expected! This corresponds to a simple interest rate (one payment

at the end of the year) of 7.25%.

Exercises 3.1. 1. Draw a slope ﬁeld for the natural decay equation

= −

y and use it to sketch

the solution curve with initial condition y(0) = 6. What is the function y(x) in this case?

2. Which of the following would you prefer for a savings account? Why?

• 5% interest paid continuously.

• 5.05% compounded monthly.

• 5.1% paid at the end of the year.

3. You invest $1000 in an account that pays 4% simple interest per year.

(a) How much money will you have after 5 years?

(b) If you close the account after 2 years and 3 months, the bank needs to decide how much

interest to credit you with. Do this in two ways (the answers will be different!):

i. Compute using the simple interest rate for 2.25 years ((1 +

100

)

2.25

ii. Suppose that interest is paid at 4% for all completed years and then at 4% paid

monthly for any completed months of an incomplete year. Find the balance of the

account at closing.

4. Explain why the proportionality constant for





is negative that for a

: that is,

lim

h→0

(

)

−1

= −lim

h→0

−1

Try to ﬁnd both an algebraic reason and a pictorial one.

5. Sketch the function f (x) = e

−x

. Where have you seen this before and what uses does this

function have?

In the US, mortgage companies typically quote an interest rate which they use to compound monthly. For example,

if the quoted rate is 7%, then the effective annual (simple) interest rate is



1 +

0.07



− 1 = 7.229%. By law, this higher

effective APR must be quoted somewhere, though it is unlikely to be as prominently posted. . .

3.2 Logarithmic Functions

Since e > 1 > 0, the natural exponential function satisﬁes several properties:

lim

x→−∞

= 0, lim

x→∞

= ∞,

= e

> 0

Thus exp : R → (0, ∞) is a differentiable (so continuous), increasing function with domain R and range

(0, ∞). It is therefore invertible.

Deﬁnition 3.11. The natural logarithm ln : (0, ∞) → R is the

inverse function to the natural exponential. That is,

• If x > 0, then e

ln x

= x;

• If y ∈ R, then ln e

= y.

−1

1 2 3 4 5 6 7 8

y = ln x

−1

ln 4

Since exp and ln are inverse functions, we can solve equations in the usual way: for instance,

3x+1

= 100 =⇒ 3x + 1 = ln 100 =⇒ x =

(ln 100 −1) ≈ 1.202

One of the great advantages of logarithms is that they allow every exponential function to be ex-

pressed in terms of the natural exponential: by the exponential laws,

= (e

ln a

)

= e

x ln a

This identity is crucial for interpreting and analyzing natural growth models.

Example 3.12. A population of rabbits doubles in size every 6 months. If there are 10 rabbits at the

start of the year, how many rabbits do we expect there to be after 9 months, and how rapidly is the

population increasing (births/month).

We are told that the population of rabbits after t months is

P(t) = 10 ·2

t/6

After 9 months the population will be approximately

P(9) = 10 · 2

3/2

= 20

√

2 ≈ 28.28 rabbits

Moreover,

0 2 4 6 8 10

P(t) =

10e

ln 2

10 ln 2

t/6

=⇒ P

′

(9) =

10 ln 2

3/2

≈ 3.27 rabbits/month

If you ask students this question, what do you expect to be the most common incorrect answers?

Why?

The Logarithm Laws and General Logarithms

The logarithm laws should be familiar; they follow immediately from the above deﬁnition and the

exponential laws (page 33)

ln x+ln y

= e

ln x

ln y

= xy = e

ln xy

=⇒ ln xy = ln x + ln y

Similarly ln

= ln x −ln y and ln x

= r ln x (∗)

These laws allow us to solve more general exponential equations.

= 5 =⇒ x ln 2 = ln 5 =⇒ x =

ln 5

ln 2

≈ 2.322

More generally, if a > 0 and a = 1, then the exponential function with base a is invertible:

y = f (x) = a

= e

x ln a

=⇒ ln y = x ln a =⇒ x =

ln y

ln a

=⇒ f

−1

(x) =

ln x

ln a

Deﬁnition 3.13. Given a > 0 and a = 1, the logarithm with base a is the function

log

x :=

ln x

ln a

As the inverse of the base a exponential function y = a

, the base a logarithm satisﬁes

• If x > 0, then a

log

= x;

• If y ∈ R, then log

= y.

The natural logarithm has base e. Unless the base is very simple (e.g. a = 2 or 10), we typically stick

to using the natural logarithm. On a calculator, the ‘log’ button means log

Exercises 3.2. 1. Find the solution to the equation 4

2−

√

= 10.

2. Find the value of x which satisﬁes the equation 4

= 8. Your answer should not contain any

logarithms. . .

3. Over one year, ﬁnd the continuous interest rate s% corresponding to a simple rate of 5%.

4. y = a

satisﬁes the natural growth equation

= ky; what is the value of k?

5. Verify the remaining logarithm laws (∗).

6. By differentiating the expression e

ln x

= x, verify that

ln x =

7. Sketch a graph of the functions f (x) = log

x and g(x) = log

0.5

x. How are they related? What

happens to the graph of log

x as a changes?

8. Logarithms were originally invented not for calculus but to simplify and multiply large num-

bers. In the pre-calculator era, it was common for students to carry a book of log tables for this

purpose. Look up a log table and investigate how to use it.

3.3 Modifying the Natural Growth Model

In this section we discuss several examples of exponential models motivated by real-world situations.

Remember that modelling always involves some guesswork and assumptions, which necessarily come

with trade-offs: simpler assumptions/models are easier to analyze, but tend to be less accurate.

Modelling is always a part of a feedback loop:

Data/theory suggest a model whose predictions are tested against real-world data, sug-

gesting changes/improvements to the model.

Applied mathematicians typically desire a ‘Goldilocks’ model: complicated enough to make accurate

predictions without being too complicated to use.

Newton’s Law of Cooling

A just-poured cup of coffee at 210°F is left outside when the air temperature is 50°F. It seems obvious

that the coffee will cool down slowly towards 50°F; but how?

To help decide how to model this, ask yourself some questions:

1. When should the rate of cooling be most rapid?

2. What happens to the rate of change in the long run (large time)?

3. Can you suggest a family of functions which behave in this manner?

Hopefully it seems reasonable to model this with a shifted exponential function, where the tempera-

ture T(t) of the coffee at time t satisﬁes

T(t) = 50 + 160e

−kt

for some positive constant k. This satisfy all our criteria:

• T(0) = 210°F.

• As t increases, e

−kt

decreases to zero, so T(t) decreases

towards 50°F.

100

150

200

0 3 6 9 12

• The rate of cooling



= 160ke

−kt

is largest at t = 0 and decreases as t increases.

To complete the model, it is enough to supply one further data point.

Suppose after 2 minutes that the temperature of the coffee is 140°F. How long does it take for the

coffee to cool to 100°F?

We know that 140 = T(2) = 50 + 160e

−2k

, whence

−2k

140 −50

160

=⇒ e

−k

=⇒ T(t) = 50 + 160





When T(t) = 100, we see that





100 −50

160

=⇒ t =

ln 16 − ln 5

ln 4 − ln 3

≈ 4.043 minutes

This is an example of a general model called Newton’s law of cooling, which asserts that the rate of

temperature change of a body is proportional to the difference between the body and its surroundings.

We have a simple modiﬁcation of Theorem 3.6

Corollary 3.14. If M and k are constant, then

= k(M −y) ⇐⇒ y(t) = M + (y

− M)e

−kt

where y

= y(0) is the initial value.

The Logistic Model

The natural growth model has one enormous drawback when applied to real-world populations: if

k > 0, then a function y(t) = y

grows unboundedly! In typical situations, environmental limitations

(availability of food, water, space) mean that populations do not explode like this. The logistic model

attempts to describe this phenomenon; it is based on two assumptions:

• When a population y is small, we want it to grow nat-

urally

∝ y.

• We want y to approach a positive value M as t → ∞.

Given constants k, M > 0, the logistic differential equation

= ky(M − y)

< 0

> 0

accomplishes both requirements. M is often referred to as the carrying capacity of the environment.

Theorem 3.15. If y

= y(0), then the solution to the logistic differential equation is

y(t) =

+ (M −y

−kMt

You can check directly that this satisﬁes the differential equation just by differentiating, though it’s a

little ugly. If you’ve studied differential equations the method of separation of variables supplies the

converse.

Example 3.16. A brewer pitches 100 billion yeast cells into a starter wort with the goal of growing it

to 200 billion cells. After one hour, the wort contains 110 billion cells.

(a) How long must the brewer wait if we use a natural growth model?

(b) How long must the brewer wait if we use a logistic model where we also assume that the wort

contains enough sugar to grow 250 billion yeast cells?

Let P(t) be the yeast population in billions at time t hours. We therefore have P(0) = 100 and

P(1) = 110, and want to ﬁnd t such that P(t) = 200.

(a) The model is

= ky, which has solution

P(t) = P

= 100e

Evaluating at t = 1 yields 1.1 = e

whence

P(t) = 100(1.1)

=⇒ t =

ln( 0.01P)

ln 1.1

≈ 7.27 hours

(b) The model is

= ky(250 −y), with solution

P(t) =

25000

100 + (250 − 100)e

−250kt

500

2 + 3e

−250kt

100

200

300

0 3 6 9 12

Evaluating at t = 1 yields

110 =

500

2 + 3e

−250k

=⇒ e

250k

whence

P(t) =

500

2 + 3





=⇒ t =

ln 6

≈ 10.91 hours

The logistic model is easily generalized.

Example 3.17. The population P(t) (in 1000s) of ﬁsh in a lake obeys the logistic equation

P(10 − P)

where t is measured in months. The ﬁrst graph shows how

the population recovers over a year if it starts at 2500 ﬁsh.

Now suppose 1000 ﬁsh are ‘harvested’ from the lake each

month. The new model is then

P(10 − P) −1 = −

−10P + 16)

= −

(P −2)(P −8)

Substituting Q = P −2, this is again logistic!

Q(6 − Q)

0 3 6 9 12

Provided the initial population P(0) = Q(0) + 2 is greater than 2000 ﬁsh, we expect the population

to eventually stabilize at 8000 ﬁsh, though it takes a long time to get close to this if we start, as in the

second graph, with only a little over 2000 ﬁsh.

Public-health Interventions

A population of 10,000 people is exposed to a novel virus. The best scientiﬁc understanding is that

1% of the susceptible population per day contracts the virus, the effects of the illness last ten days,

after which a patient recovers and is immune from reinfection.

1. Model the evolution of the sick and immune populations over the next 120 days.

Let u(t), s(t) and i(t) represent the uninfected, sick and immune populations on day t. Then

u(t + 1) = 0.99u(t), u(0) = 10000 =⇒ u(t) = 10000 · 0.99

The sick population is the sum of the previous 10 days’ decrease in the at risk population:

s(t) =

(

u(0) − u(t) = 10000(1 − 0.99

) if t ≤ 10

u(t −10) − u(t) = 10000(0.99

−10

−1)0.99

= 1057 · 0.99

if t > 10

The immune population is the difference between these and the total population

i(t) = 10000 − u(t) − s(t) =

(

0 if t ≤ 10

10000 −11057 · 0.99

if t > 10

After 120 days, we have

u(120) = 2994, s( 120) = 316, i(120) = 6690

2. Suppose 6000 vaccines are available. Discuss how these should be deployed. What should the

goal be? Discuss the following strategies; for simplicity, assume the vaccines are 100% effective

and work instantly.

(a) Use all vaccine doses immediately.

(b) Wait 30 days until some people are immune, then use all vaccines on the uninfected pop-

ulation.

(d) Wait until there are 6000 uninfected people remaining, then vaccinate them all at once.

It is much easier to analyze this problem using a spreadsheet, though we can also do things an-

alytically. Here are graphs of what happens under a non-intervention scenario and the ﬁrst three

vaccination campaigns.

Exercises 3.3. 1. A cup of coffee is left outside on a warm day when the surrounding temperature

is 90°F. Suppose the initial temperature of the coffee is 200°F and that its temperature after 2

minutes is 170°F. Find the temperature as a function of time.

2. Consider Corollary 3.14.

(a) Check that y(x) = M + (y

− M)e

−kt

satisﬁes the differential equation.

(b) A student believes that Theorem 3.6 is true. How would you convince them, in the simplest

possible way, that the only solution to y

′

= k(M −y) is as given in part (a)?

3. For both models in Example 3.16, what is the maximum growth rate of the yeast population,

and at what time does it occur?

4. In Example 3.17, suppose the initial population is 3000 ﬁsh and 1000 ﬁsh are harvested per

month, how long does it take for the population to recover to 6000 ﬁsh?

5. Plutonium-238 has a half-life of 88 years, meaning that after 88 years half of the isotope has

decayed to another element (in this case uranium-234).

(a) If you start with 100 grams of plutonium, ﬁnd a model for how much remains after t years.

(b) How long will take for the mass to decay to 10 grams, and at what rate will it be decaying?

6. (a) If public health ofﬁcials wanted to eradicate the virus on page 42 completely by using all

vaccine doses on one day, on which day should they act?

(b) Find the number of uninfected people as a function of time under the 100 per day for 60

days vaccination scenario.

4 Sequences as Functions

We’ve seen many different types of function in this course and used them to model various situations.

In practice, one is often faced with the opposite problem: given experimental data, what type of

function should you try?

4.1 Polynomial Sequences: First, Second, and Higher Differences

To begin to answer this, ﬁrst ask yourself, “What is a sequence?” Hopefully you have a decent

intuitive idea already. More formally, a sequence a function whose domain is a set like the natural

numbers, for example

f : N → R : n 7→ 3n

−2

deﬁnes the sequence



f (1), f (2), f (3), . . .





1, 10, 25, 46, 73, . . .



This is indeed the intuitive idea of a function to many grade-

school students: continuity and domains including fractions

or even irrational numbers are more advanced concepts.

Suppose instead that all you have is a data set

x 1 2 3 4 5

y 1 10 25 46 73

0 1 2 3 4 5

perhaps arising from an experiment. Could you recover the original function y = f (x) directly from

this data? You could try plotting data points as we’ve done, though it is hard to decide directly

from the plot whether we should try a quadratic model, some other power function/polynomial, or

perhaps an exponential. Of course, the physical source of real-world data might also provide clues.

A more mathematical approach involves considering how data values change:

x 1

y 1

+15

+21

+27

The ﬁrst-differences in the x-values are constant whereas those for the y-values are increasing



n+1

−y





9, 15, 21, 27, . . .



You likely already notice the pattern: the sequence of ﬁrst-differences is increasing linearly as the

arithmetic sequence

n+1

−y

= 3 + 6n

To make this even clearer, note that the sequence of second-differences in the y-values is constant

(+6). These facts are huge clues that we expect a quadratic function.

But why? Well we can certainly check the following directly:

Linear Model If f (n) = an + b, then the sequence of ﬁrst-differences is constant

f (n + 1) − f (n) = a

Quadratic Model If f (n) = an

+ bn + c, then the sequence of ﬁrst-differences is linear and the

second-differences are constant:

g(n) := f (n + 1) − f (n) = 2an + a + b, g(n + 1) − g(n) = 2a

The relationship between these results and the derivative(s) of the original function f (x) should feel

intuitive: what happens if you differentiate a quadratic twice?

Example 4.1. You are given the following data set

x 0 2 4 6 8 10

y 2 16 22 20 10 −8

The x-values have constant ﬁrst-differences while the y-values

have constant second-differences

First-differences: 14, 6, −2, −10, −18

Second-differences: −8, −8, −8, −8

2 4 6 8 10

We therefore suspect a quadratic model y = f (n) = an

+ bn + c. Rather than using the above

formulae, particularly since the x-differences are not 1, it is easier just to substitute:

2 = y(0) = c,

(

16 = f (2) = 4a + 2b + 2

22 = f (4) = 16a + 4b + 2

=⇒

(

2a + b = 7

8a + 2b = 10

=⇒ 4a = −4

whence a = −1, b = 9 and c = 2. A quadratic model is therefore

y = f (n) = −n

+ 9n + c = −n

+ 9n + 2

It is easily veriﬁed that the remaining data values satisfy this relationship.

There are at least two issues with our method:

1. The question we’re answering is, “Find a quadratic model satisfying given data.” Constant

second-differences don’t guarantee that only a quadratic model is suitable. For example,

y = −n

+ 9n + 2 + 297n(n −2)(n −4)(n −6)(n −8)(n −10)

is a very complicated model satisfying the same data set!

2. It is very unlikely that experimental data will ﬁt such precise patterns (why not?). However,

if the differences are close to satisfying such patterns, then you should feel conﬁdent that a

linear/quadratic model is a good choice.

Example 4.2. Given the data set

x 0 2 4 6 8 10

y 3 23 41 59 77 93

with sequences of ﬁrst- and second-differences

First-differences: 20, 18, 18, 18, 16

Second-differences: −2, 0, 0, −2

0 2 4 6 8 10

do you think a linear or quadratic model would be superior?

If you wanted a linear model, you’d likely be inclined to try f (x) = 9x + b for some constant b. Here

are two options:

1. f (x) = 9x + 5 ﬁts the middle four data values perfectly, but as a predictor is too large at the

endpoints: f (0) = 5 > 3 and f (10) = 95 > 93.

2. f (x) = 9x + 5 −

doesn’t pass through any of the data values but seems to reduce the net error

to zero:

x 0 2 4 6 8 10

f (x) − y −

−

=⇒

∑

f (x) − y = 0

Neither model is perfect, but then this is what you expect with real-world data!

Exercises 4.1. 1. For each data set, ﬁnd a function y = f (x) modelling the data.

(a) x 2 4 6 8

y −1 2 7 14

(b) x 2 5 8 11 14

y −6 −15 −6 21 66

y 3 15 21 33

(Be careful with (c): the x-differences aren’t constant!)

2. Suppose a table of data values containing (x

, y

) has constant ﬁrst-differences in both variables

∆x = x

n+1

− x

= a, ∆y = b

Find the equation of the linear function y = f (x) through the data.

3. What relationship do you expect to ﬁnd with the sequential differences of a cubic function

f (n) = an

+ bn

+ cn + d? What about a degree-m polynomial f (n) = an

+ bn

m−1

+ ···?

4. If f (n) = an

+ bn + c is a quadratic model for the data in Example 4.2 with constant second-

differences −1, show that a = −

. What might be reasonable values for b, c?

5. (Hard) Suppose f (x) is a twice-differentiable function and h > 0 is constant. Use the mean

value theorem from calculus to explain the following.

(a) First-differences f (x + h) − f (x) are proportional to f

′

(ξ) for some ξ ∈ (x, x + h).

(b) Second-differences satisfy



f (x + 2h) − f (x + h)



−



f (x + h) − f (x)



= f

′′

(ξ)hα for some

ξ between x and x + h and some α. Why is it unlikely that α is constant?

4.2 Exponential, Logarithmic & Power Sequences

To observe relationships between data values, you might also have to consider ratios between succes-

sive terms or skip values.

Example 4.3. From a ﬁrst glance at the given data, it is hard to decide whether an exponential or

a quadratic (or higher degree polynomial) model is more suitable. If we try to apply the constant-

difference method, we don’t seem to get anything helpful:

x 1

((

))

y 15

+120

135

+1080

1215

+9720

10935

+960

+8640

By the time we’re looking at second-differences, any conclusion

would be very weak since we only have two data values!

y (1000s)

0 2 4 6

If instead we think about ratios of y-values, then a different pattern emerges:

x 1

((

))

y 15

×9

135

×9

1215

×9

10935

The question remains: what type of function scales its output by 9 when 2 is added to its input:

f (x + 2) = 9 f (x)? This is a function that converts addition to multiplication: an exponential! If we try

y = f (x) = ba

for some constants a, b, then

f (x + 2) = ba

x+2

= ba

= a

f (x)

from which a suitable model is y = 5 ·3

We can see the pattern in the example more generally:

Exponential Model If f (x) = ba

, then adding a constant to x results in

f (x + k) = ba

x+k

= a

f (x)

If x-values have constant differences (+k), then y-values will be related by a constant ratio (×a

You might remember this as ‘addition–product’ or ‘arithmetic–geometric.’

Such a simple pattern is often disguised:

• Complete data might not be given so you might have to skip some data values to see a pattern.

For example, if our original data was

x 1 3 4 5 7

y 15 135 405 1215 10935

then the x-values are not in a strictly arithmetic sequence.

• As in Example 4.2, real-world/experimental data will only approximately exhibit such patterns.

Example 4.4. A population of rabbits is measured every two months resulting in the data set

t 0 2 4 6 8 10

P 5 7 10 14 19 28

The data seems very close to being quadratic; consider the ﬁrst and second sequences of P-differences

∆P =



2, 3, 4, 5, 9



, ∆∆P =



1, 1, 1, 4



However, the last difference doesn’t ﬁt the pattern. Instead, the fact that we expect an exponential

model is buried in the experiment: the data is measuring population growth! We therefore instead

consider the ratios of P-values:

t 0

))

P 5

×1.4

×1.43

×1.4

×1.36

×1.47

The ratios are very close to being constant, whence an exponential model is suggested! To exactly

match the ﬁrst and last data values, we could take the model

P(t) ≈ 5





t 0 2 4 6 8 10

P 5 7.057 9.960 14.057 19.839 28

Only P(8) doesn’t match when we take rounding to the nearest integer into account.

We’ve seen that addition-addition corresponds to a linear model and that addition-multiplication to

an exponential. There are two other natural combinations.

Logarithms These operate exactly as exponentials but in reverse. If f (n) = log

x + b, then multiply-

ing x by a constant results in a constant addition/subtraction to y:

f (kx) = log

(kx) + b = log

k + log

x + b = log

k + f (x)

This could be summarized as ‘product–addition.’

Power Functions If f (x) = ax

, then multiplying x by a constant will do the same to y

f (kx) = a(kx)

= ak

= k

f (x)

We have a ‘product–product’ relationship between successive terms.

Examples 4.5. Find the patterns in the following data and suggest a model y = f (x) in each case.

x 6 18 54 162

y 1 2 3 4

x 3 6 9 12

y 135 1080 3645 8640

The sequential approach in this chapter is a form of discrete calculus: using a pattern of differences to

predict the original function is similar to how we use knowledge of a derivative f

′

(x) to ﬁnd f (x).

Example 4.6. Suppose g(2) = 3 and g(4) = 9. What do you think should be the value of g(8)?

It depends on the type of model you try.

1. For a linear (addition-addition) model we know that

∆x = 2 corresponds to ∆y = 6, so

g(8) = g(4 + 2∆x) = g(4) + 2∆y = 9 + 12 = 21

2. For an exponential (addition-product) model, ∆x = 2

corresponds to a y-ratio r

= 3, so

g(8) = r

g(6) = r

g(4) = 9 · 9 = 81

3. For a power (product-product) model, r

= 2 corre-

sponds to a r

= 3, so

g(8) = g(2 ·4) = g(4r

) = r

g(4) = 3 · 9 = 27

0 2 4 6 8

1. g(x) = 3x −3

2. g(x) = 3

x/2

3. g(x) = x

log

We do not need to calculate the models explicitly(!), though they are stated below the graph for

convenience.

Exercises 4.2. 1. Find the patterns in the following data sets and use them to ﬁnd a model y = f (x).

(a) x 0 1 2 3 4

y 80 120 180 270 405

(b) x 2 4 8 10

y 1 16 256 625

y 15 5 19 57 119

(d) x 1 3 4 6

y 1 36 216 7776

(e) x 20 60 180 540

y 2 4 6 8

(f) x 2 6 54 486 4374

y 2 4 8 12 16

2. Take logarithms of the power relationship y = ax

. What is the relationship between ln y and

ln x? Use this to give another reason why the inputs and outputs of power functions satisfy a

‘product–product’ relationship.

3. How does our analysis of exponential functions change if we add a constant to the model? That

is, how might you recognize a sequence arising from a function f (x) = ba

+ c?

4. Suppose f (5) = 12 and f (10) = 18. Find the value of f (20) supposing f (x) is a:

(a) Linear function;

(b) Exponential function;

If f (20) = 39, which of the three models do you think would be more appropriate?

4.3 Newton’s Method

To ﬁnish our discussion of sequences we revisit a (hopefully) familiar technique for approximating

solutions to equations. Variations of this approach have been in use for thousands of years.

Example 4.7. We motivate the method by considering an ancient method for approximating

√

known to the Babylonians 2500 years ago!

Suppose x

√

2. Then

√

2. It seems reasonable to guess that their average

n+1





should be a more accurate approximation to

√

2. If start with an initial guess x

= 2, then we obtain

the sequence



2 +



, x

= 1.4166 . . . , x

577

408

= 1.4142 . . . , . . .

This sequence certainly appears to be converging to

√

2. . .

Since it makes use of the average, this approach is sometimes called the method of the mean. It may be

applied to any square-root

√

a where a > 0: let x

> 0 and deﬁne,

n+1





(∗)

A rigorous proof that the sequence converges requires more detail than is appropriate for us (though

see Exercise 3), but two observations should make it seem more believable:

1. If the sequence (∗) has a limit L, then the limit must satisfy

L =



L +



=⇒ 2L

= L

+ a =⇒ L

= a =⇒ L =

√

where we take the positive root since all terms x

are plainly positive.

2. The iterations have a convincing geometric interpretation.

The sequence of iterates can be found by repeatedly tak-

ing the tangent line to the curve y = f (x) = x

− a and

intersecting it with the x-axis. To see why, observe that

the tangent line at x

has equation

y = f (x

) + f

′

)(x −x

)

= x

− a + 2x

(x − x

)

= 2x

x −x

− a

which intersects the x-axis (y = 0) when

x =

+ a





= x

n+1

y = x

− a

−a

n+1

This geometric idea generalizes. . .

Deﬁnition 4.8. Given a differentiable function f (x) with non-zero derivative, the Newton–Raphson

iterates of an initial value x

are deﬁned by the recurrence formula

n+1

:= x

−

f (x

)

′

)

Our two previous observations still hold:

1. If L = lim

n→∞

exists and f

′

(L) = 0, then

L = L −

f (L)

′

(L)

=⇒ f (L) = 0

That is, the limit L is a root of the function f (x).

2. The tangent line at



, f (x

)



forms a right-triangle with

base x

− x

n+1

and height f (x

), from which its slope is

′

) =

f (x

)

− x

n+1

Rearranging this gives the formula x

n+1

= x

−

f (x

)

′

)

y = f (x)

n+1



, f (x

)



Newton’s method is particularly nice for polynomials with integer coefﬁcients, since the iterates

form a sequence of rational numbers. This approach was often used obtain rational approximations to

irrational numbers before the advent of calculators.

Examples 4.9. 1. To ﬁnd a root of f (x) = x

+ 4x −6, start with x

= 2 and iterate

n+1

= x

−

+ 4x

−6

+ 4

3(x

+ 2)

4(x

+ 1)

which yields the sequence (to 3 d.p.)



339

280

, . . .



= (2, 1.5, 1.211, 1.121, 1.114, 1.114, . . .)

You can check with a calculator that 1.114 is approximately

a root.

1 2

2. The irrational number x =

√

2 +

√

3 is a root of the polynomial

f (x) = x

−10x

+ 1

By applying Newton’s method with x

= 3, we obtain the sequence (to 3 d.p.)

n+1

= x

−

−10x

+ 1

−20x

−10x

−1

−5)

=⇒ (x

) =



= 3.167, 3.147, . . .



Newton’s method can be attempted for any differentiable function, though the sequence isn’t guar-

anteed to converge: see for instance Exercise 5. You can ﬁnd graphical interfaces online for this (for

instance with Geogebra).

Exercises 4.3. 1. Use Newton’s method to ﬁnd a root of the given function to 4 decimal places.

(Use a calculator, but explain what you are doing!)

(a) f (x) = x

−4 (b) f (x) = 2x

+ x −1 (c) f (x) = e

−

√

x −2

2. Use Newton’s method to ﬁnd a rational number approximation to

√

2 in lowest terms

where

10 < q < 100.

3. Suppose you perform Newton’s method for the function f (x) = x

− 2 starting with some

positive x

> 0.

(a) If x

> 0, show that x

n+1

−

√

2 =

−

√



−

√



−

√

2) .

(b) Explain why



−

√



−

√



. Hence conclude that the sequence of iterates (x

)

converges to

√

4. We might consider a method of the mean for approximating

√

2: given x

, deﬁne

n+1





(a) If the sequence (x

) converges, show that its limit is

√

(b) If x

√

2, show that

√

= 1. Compute x

and x

. Compare these with the values obtained using Newton’s

method for the function f (x) = x

−2 with the same initial condition x

= 1.

5. Let f (x) = x

−5x.

(a) What happens if you apply Newton’s method to this function with initial condition x

1? Draw a picture to illustrate.

(b) (Just for fun!) Investigate what happens for other values of x

. Can you make any conjec-

tures? Is is possible for x

to be positive and yet for x

→ −

√

5? Can you make any sense

of what happens if 1 < x

5 Regression Models

We’ve studied several types of function and seen how to spot whether a given data set might suit

a particular model. To get further with this analysis, we need a method for comparing how bad a

particular model is for given data.

5.1 Best-ﬁtting Lines and Linear Regression

We start with an example of some data which appears reasonably linear.

Example 5.1. At t p.m., a trail-runner’s GPS locator says that they’ve travelled y miles along a trail;

1 2 3 5

4 8 10 21

We’d like a simple model for how far the runner has travelled as

a function of t. We might use this to predict where they would be

at a given time; say at 6 p.m., or at 2 p.m. if they were to attempt

the trail on another day.

0 1 2 3 4 5

By plotting the points, the relationship looks to be approximately

linear: y ≈ mt + c. What is the

best choice of line, and how should we ﬁnd the coefﬁcients m, c?

What might be good criteria for choosing our line? What should we mean by best? Plainly, we

want the points to be close to the line, but measured how? What use do we want to make of the

approximating line?

Here are three candidate lines plotted with the data set: of the choices, which seems best and why?

0 1 2 3 4 5

y = 4t

0 1 2 3 4 5

y = 2t + 4

0 1 2 3 4 5

y = 5t −4

Since we want our model to predict the hiker’s location

y ≈

y = mt + c at a given time t, we’d like our model to

minimize vertical errors

− y

. We’ve computed these in

the table; since a positive error is as bad as a negative, we

make all the errors positive. It therefore seems reasonable

to claim that the ﬁrst line is the best choice of the three.

But can we do better?

1 2 3 5

4 8 10 21

y = 4t

−y

0 0 2 1

y = 2t + 4

−y

2 0 0 7

y = 5t −4

−y

3 2 1 0

Why should we not expect the distance traveled by the hiker to be perfectly linear?

We need a sensible deﬁnition of best-ﬁtting line for a given data set. One possibility is to minimize the

sum of the vertical errors:

∑

i=1

−y

For reasons of computational simplicity, uniqueness, statistical interpretation, and to discourage

large individual errors, we don’t do this! The standard approach is instead to minimize the sum of

the squared errors.

Deﬁnition 5.2. Let (t

, y

) be data points with at least two distinct t-values. Let

y = mt + c be a linear

predictor (model) for y given t.

• The i

error in the model is the difference e

−y

= mt

+ c −y

• The regression line or best-ﬁtting least-squares line is the function

y = mt + c which minimizes the

sum S :=

∑

(

−y

)

of the squares of the errors.

Having at least two distinct t-values (some t

= t

) is necessary for the regression line to be unique.

Example (5.1, cont). Suppose the predictor was

y = mt + c. We expand the table

1 2 3 5

4 8 10 21

m + c 2m + c 3m + c 5m + c

m + c −4 2m + c − 8 3m + c −10 5m + c −21

Our goal is to minimize the function

S(m, c) =

∑

= (m + c − 4)

+ (2m + c − 8)

+ (3m + c − 10)

+ (5m + c − 21)

This is easy to deal with if we invoke some calculus. If (m, c) minimizes S(m, c), then the ﬁrst deriva-

tive tests says that the (partial) derivatives of S must be zero.

• Keep c constant and differentiate with respect to m:

∂S

∂m

= 2( m + c −4) + 4(2m + c −8) + 6(3m + c −10) + 10(5m + c −21)

= 2

39m + 11c −155

• Keep m constant and differentiate with respect to c:

∂S

∂c

= (m + c − 4) + (2m + c −8) + (3m + c −10) + (5m + c −21)

= 11m + 4c −43

The regression line is found by solving a pair of simultaneous equations

(

39m + 11c = 155

11m + 4c = 43

=⇒ m =

, c = −

=⇒

y =

(21t −4)

By 6 p.m., we predict that the runner would have covered 24.4 miles. The sum of the squared errors

for our regression line is

∑

−y

= 4.4, compared to 5, 53 and 14 for our earlier options.

To obtain the general result for n data points, we return to our computations of the partial derivatives:

∂S

∂m

∑

∂

∂m

(mt

+ c −y

)

= 2

∑

(mt

+ c −y

) = 2



∑



m +



∑



c −

∑

∂S

∂c

∑

∂

∂c

(mt

+ c −y

)

= 2

∑

(mt

+ c −y

) = 2



∑



m + nc −

∑

These sums are often written using a short-hand notation for average:

t =

∑

i=1

, t

∑

i=1

, y =

∑

i=1

, ty =

∑

i=1

Theorem 5.3 (Linear Regression). Given n data points (t

, y

) with at least two distinct t-values, the

best-ﬁtting least-squares line has equation

y = mt + c, where m, c satisfy

(



∑



m +

(

∑

)

c =

∑

(

∑

)

m + nc =

∑

↭

(

m + tc = ty

tm + c = y

This is a pair of simultaneous equations for the coefﬁcients m, c, with solution

m =

ty − ty

−t

, c = y −mt

As the next section shows, having two distinct t-values guarantees a non-zero denominator t

− t

The expression for c shows that the regression line passes through the data’s center of mass (t, y).

Example 5.4. Five students’ scores on two quizzes are given.

If a student scores 9/10 on the ﬁrst quiz, what might we expect them

to score on the second?

Quiz 1 8 10 6 7 4

Quiz 2 10 7 5 8 6

To put the question in standard form, suppose Quiz 1 is the t-data and Quiz 2 the y-data. It is helpful

to rewrite the data and add lines to the table so that we may more easily compute everything.

Data

∑

Average

8 10 6 7 4 35 7

10 7 5 8 6 36 7.2

64 100 36 49 16 265 53

80 70 30 56 24 260 52

Q2 = y

0 2 4 6 8 10

Q1 = t

m =

52 −7 × 7.2

53 −7

1.6

= 0.4, c = 7.2 −0.4 ×7 = 4.4

=⇒

y(t) =

(t + 11)

This line which minimizes the sum of the squares of the vertical deviations. The prediction is that

the hypothetical student scores

y(9) =

·20 = 8 on Quiz 2. Note that the predictor isn’t symmetric:

if we reverse the roles of t, y we don’t get the same line!

Exercises 5.1. 1. Compute the sum of the absolute errors

∑

−y

for the regression line and

compare it to the sum of the absolute errors for

y = 4t: what do you notice?

2. Let

y = mt + c be a linear predictor for the given data.

0 1 2 3

1 2 2 3

(a) Compute the sum of squared-errors S(m, c) =

∑

−y

a function of m and c.

(b) Compute the partial derivatives

∂S

∂m

and

∂S

∂c

regression line for these data.

(d) Compare the sum of square errors S for the regression line with the errors if we use the

simple predictor y(t) = 1 +

t which passes through the ﬁrst an last data points.

3. Consider Example 5.4.

(a) Compute the sum of square-errors S =

∑

−y

for the regression line.

(b) Suppose a student was expected to score exactly the same on both quizzes; the predictor

would be

y = t. What would the sum of squared-errors be in this case?

(Warning: the answer is NOT

·8 −11 = 9. . . )

4. Ten children had their heights (inches) measured on their ﬁrst and second birthdays. The data

was as follows.

birthday 28 28 29 29 29 30 30 32 32 33

birthday 30 32 31 34 35 33 36 37 35 37

Given this data, ﬁnd a regression model and use it to predict the height at 2 years of a child

who measures 32 inches at age 1.

(It is acceptable—and encouraged!—to use a spreadsheet to ﬁnd the necessary ingredients. You can do

this by hand if you like, but the numbers are large; it is easier with some formulæ from the next section.)

5. (a) Let a, b be given. Find the value of y which minimizes the sum of squares

(y − a)

+ (y −b)

(b) For the data set



(t, y)





(1, 1), (2, 1), (2, 3)



, ﬁnd the unique least-squares linear model

for predicting y given t.

(Hint: think about part (a) if you don’t want to compute)

y = mt + c which minimize the sum of the absolute

errors

∑

i=1

−y

5.2 The Coefﬁcient of Determination

In the sense that it minimizes the sum of the squared errors S =

∑

, the linear regression model is

as good as it can be—but how good? We could use S as a quantitative measure of the model’s accuracy,

but it doesn’t do a good job at comparing the accuracy of models for different data sets. The standard

approach to this problem relies the concept of variance.

Deﬁnition 5.5. The variance of data sequence (y

, . . . , y

) is the average of the squared deviations

from their mean y =

∑

i=1

Var y :=

∑

i=1



−y



The standard deviation is σ

Var y.

Variance and standard-deviation are measures of how data deviates from being constant.

Example 5.6. Suppose (y

) = (1, 2, 5, 4). Then

y =

(1 + 2 + 5 + 4) = 3 Var y =



(−2)

+ (−1)

+ 2

+ 1



√

The square-root means that σ

has the same units as y. Loosely speaking, a typical data value is

expected to lie approximately σ

√

10 ≈ 1.58 from the mean y = 3.

To obtain a measure for how well a regression line ﬁts given data (t

, y

), we ask what fraction of the

variance in y is explained by the model.

Deﬁnition 5.7. The coefﬁcient of determination of a model

y = mt + c is the ratio

Var

Var y

Examples 5.8. We start by considering two extreme examples.

1. If the data were perfectly linear, then y

= mt

+ c for all i. The regression line is therefore

y = mt + c and the coefﬁcient of determination is precisely R

Var y

= 1. All the variance in

the output y is explained by the model’s transfer of the variance in the input t.

2. By contrast, consider the data in the table where we work

out all necessary details to ﬁnd the regression line:

m =

ty − ty

−t

= 0, c = y − mt = 2

The regression line is the constant

y ≡ 2, whence

y has no

variance and the coefﬁcient of determination is R

= 0.

data average

0 0 2 2 t = 1

1 3 1 3 y = 2

0 0 4 4 t

= 2

0 0 2 6 ty = 2

In this example, the regression model doesn’t help explain the y-data in any way: the t-values

have no obvious impact on the y-values.

In fact, the coefﬁcient of determination always lies somewhere between these extremes 0 ≤ R

≤ 1:

Exercise 6 demonstrates this and that the extreme situations are essentially those just encountered; in

practice, therefore, 0 < R

< 1. Before we revisit our examples from the previous section, observe

that the average of the model’s outputs

is the same as that of the original data:

∑

i=1

∑

i=1

(mt

+ c) = mt + c = y

This makes computing the variance of

y a breeze!

Example 5.1. Recall that

y =

(21t −4). Everything nec-

essary is in the table

Var y =

6.75

+ 2.75

+ 0.75

+ 10.25

= 39.6875

Var

y =

7.35

+ 3.15

+ 1.05

+ 9.45

= 38.5875

data average

1 2 3 5 t = 2.75

4 8 10 21 y = 10.75

3.4 7.6 11.8 20.2

y = 10.75

from which R

Var

Var y

3087

3175

≈ 97.23%. The interpretation here is that the data is very close to

being linear; the output y

is very closely approximated by the regression model with approxi-

mately 97% of its variance explained by the model.

Example 5.4. This time

y =

(t + 11).

Var y =

2.8

+ 0.2

+ 2.2

+ 0.8

+ 1.2

= 2.96

Var

y =

0.4

+ 1.2

+ 0.4

+ 0

+ 1.2

= 0.64

data average

8 10 6 7 4 t = 7

10 7 5 8 6 y = 7.2

7.6 8.4 6.8 7.2 6

y = 7.2

from which R

Var

Var y

≈ 21.62%. In this case the coefﬁcient of determination is small,

which indicates that the model does not explain much of the variation in the output.

The four examples are plotted below for easy visual comparison between the R

-values.

Perfect model R

= 1 Useless model R

= 0 Good model R

= 0.97 Poor model R

= 0.22

Efﬁcient computation of R

If you want to compute by hand, our current process is lengthy and

awkward. To obtain a more efﬁcient alternative we ﬁrst consider an alternative expression for the

variance of any collection of data:

Var x =

∑

− x)

∑

−

∑

= x

− x

Plainly Var x ≥ 0 with equality if and only if all data values x

are equal. The alternative expression

− x

justiﬁes the uniqueness of the regression line in Deﬁnition 5.2 and Theorem 5.3.

Now expand the variance of the predicted outputs:

Var

y =

∑

(

−y)

∑



+ c −(mt + c)



∑

−t)

= m

Var t

Putting these together, we obtain several equivalent expressions for the coefﬁcient of determination:

Var

Var y

= m

Var t

Var y

= m

−t

−y

( ty −ty)

( t

−t

)(y

−y

)

(∗)

Example 5.9. We do one more easy example with simple data (t

, y

) : (1, 4), (2, 1), (3, 2), (4, 0).

data average

1 2 3 4 t =

4 1 2 0 y =

1 4 9 16 t

16 1 4 0 y

4 2 6 0 ty = 3

m =

ty − ty

−t

3 −

−

100

= −

= −1.1

c = y −mt =

11 ·10

10 ·4

= 4.5

0 1 2 3 4

y = −1.1t + 4.5

121

175

= 0.69

∑

= 2.7

The regression line is

y = −

t +

= −1.1t + 4.5, and the coefﬁcient of determination is

= m

−t

−y

121

100

−

100

−

121

100

121

175

= 69.1%

The minimized square error is also easily computed:

∑

(

−y

)

= (3.4 −4)

+ (2.3 −1)

+ (1.2 −2)

+ (0.1 −0)

= 2.7

Reversion to the Mean & Correlation By (∗), the regression model may be re-written in terms of

the standard-deviation and R

y(t) = mt + c = y + m(t −t) = y +

√

(t − t) =⇒

y(t + λσ

) = y + λ

√

Deﬁnition 5.10. The correlation coefﬁcient is the value r := ±

√

(sign equal to that of m).

An input λ standard-deviations above the mean (t = t + λσ

) results in a prediction λr standard-

deviations above the mean (

y = y + λrσ

). Unless the data is perfectly linear, we have R

< 1;

relative to the ‘neutral’ measure given by the standard-deviation a prediction

y(t) is closer to the

mean than the input t

y(t) − y

= r



t − t



t − t



Example (5.9, cont). We compute the details. The correlation coefﬁcient is r = −

√

≈ −0.832;

we say that the data is negatively correlated, since the output y seems to decrease as t increases. The

standard deviations may be read off from the table:

√

Var t =

−t

√

≈ 1.118, σ

Var y =

−y

√

≈ 1.479

The predictor may therefore be written (approximately)

y(t + λσ

) =

y(2.5 + 1.12λ) = y + λrσ

= 1.75 − 1.23λ

As a sanity check,

y(2.5 + 1.12) =

y(3.62) = −1.1 × −3.98 + 4.5 = 0.52 = 1.75 −1.23

Weaknesses of Linear Regression There are two obvious issues:

• Outliers massively inﬂuence the regression line. Dealing with this problem is complicated and

there are a variety of approaches that can be used. It is important to remember that any ap-

proach to modelling, including our regression model, requires some subjective choice.

• If the data is not very linear then the regression model will produce a weak predictor. There are

several ways around this as we’ll see in the remaining sections: higher-degree polynomial re-

gression can be performed, and data sometimes becomes more linear after some manipulation,

say by an exponential or logarithmic function.

Exercises 5.2. 1. Suppose (z

) = (2, 4, 10, 8) is double the data set in Example 5.6. Find z, Var z and

. Why are you not surprised?

2. Use a spreadsheet to ﬁnd R

for the predictor in Exercise 5.1.4. How conﬁdent do you feel in

your prediction?

3. Find the standard deviations and correlation coefﬁcients for the data in Examples 5.1 and 5.4.

4. The adult heights of men and women in a given population satisfy the following:

Men: average 69.5 in, σ = 3.2 in. Women: average 63.7 in, σ = 2.5 in.

The height of a father and his adult daughter have correlation coefﬁcient 0.35. If a father’s

height is 72 in (mother’s height unknown), how tall do you expect their daughter to be?

5. Suppose R

is the coefﬁcient of determination for a linear regression model

y = mt + c. Use

one of the alternative expressions for R

(page 59) to ﬁnd the coefﬁcient of determination for

the reversed predictor

t(y)? Are you surprised?

6. Suppose that a data set {(t

, y

)}

1≤i≤n

has at least two distinct t- and y-values (some t

= t

etc.), that it has regression line

y = mt + c and coefﬁcient of determination R

(a) Show that R

= 0 ⇐⇒ m = 0.

(b) (Hard) Prove that the sum of squared errors equals S =

∑

i=1

= n(Var y −Var

y).

= 1 −

n Var y

. Hence conclude that R

≤ 1, with

equality if and only if the original data set is perfectly linear.

5.3 Matrix Multiplication & Polynomial Regression

In this section we consider how to ﬁnd a best-ﬁtting least-squares polynomial for given data. To see

how to do this, it helps to rephrase the linear approach using matrices.

We start by observing that the system of equations in Theorem 5.3 can be written in as a 2 ×2 matrix

problem. For a data set with n pairs, the coefﬁcients m, c satisfy



∑







∑



This is nice because we can decompose the square matrix on the left as the product of a simple 2 × n

matrix and its transpose (switch the rows and columns);



∑





··· t

1 1 ··· 1















=: P

We can also view the right side as the product of P

and the column vector of output values y



∑





··· t

1 1 ··· 1















=: P

A little theory tells us that if at least two of the t

are distinct, then the 2 ×2 matrix P

P is invertible;

there is a unique regression line whose coefﬁcients may be found by taking the matrix inverse





= (P

−1

y =⇒

y = mt + c = (t 1)





= (t 1)(P

−1

We can also easily compute the vector of predicted values

y(t

ˆy =



··· t

1 1 ··· 1





= P(P

−1

and the squared error

∑

−y

ˆy − y

, which leads to an alternative expression for

the coefﬁcient of determination

ˆy

−ny

where

is the length of a vector.

Matrix computations are non-examinable. The purpose of this section is to be see how the regression may easily be

automated and generalized by computer and to understand a little of how a spreadsheet calculates best-ﬁtting curves of

different types.

For those who’ve studied linear algebra, P and P

P have the same null space and thus rank, since

Px = 0 =⇒ P

Px = 0 and P

Px = 0 =⇒ x

Px = 0 =⇒

= 0 =⇒ Px = 0

For linear regression, having at least two distinct t

values means rank P = 2, whence P

P is invertible.

Examples 5.11. 1. We revisit the Example 5.9 in this language.

P =



















1 1

2 1

3 1

4 1







=⇒ P

P =



1 2 3 4

1 1 1 1









1 1

2 1

3 1

4 1









30 10

10 4



from which





= (P

−1

y =



30 10

10 4



−1



1 2 3 4

1 1 1 1















30 ·4 − 10



4 −10

−10 30







48 −70

−120 + 210





−11



The prediction vector given inputs t

is therefore

ˆy = P







1 2 3 4

1 1 1 1



−11















from which the coefﬁcient of determination is, as before

ˆy

−4y

100

(34

+ 23

+ 12

+ 1

) − 4 ·

+ 1

+ 2

+ 0

) − 4 ·

121

175

2. Given the data set {(3, 1), (3, 5), (3, 6)}, we have P =



3 1



and P

P =



27 9

9 3



which isn’t invertible: 27 ·3 −9 ·9 = 0. The

linear regression method doesn’t work!

It is easy to understand this from the picture. Since the three

data points are vertically aligned, any line minimizing the

sum of the squared errors must pass through the average

(3, 4), though it could have any slope!

This illustrates our fundamental assumption: linear regres-

sion requires at least two distinct t-values.

0 1 2 3

It is unnecessary ever to use the matrix approach for linear regression, though the method has sig-

niﬁcant advantages.

• Computers store and manipulate data in matrix format, so this method is computer-ready.

• Suppose you repeat an experiment several times, taking measurements y

at times t

. Since

P depends only on the t-data, you need only compute the matrix (P

−1

once, making

computation of the regression line for repeat experiments very efﬁcient.

• The method generalizes (easily for computers!) to polynomial regression. . .

Polynomial Regression

The pattern is almost identical when we use matrices; you just need to make the matrix P a little

larger. . . We work through the approach for a quadratic approximation.

Suppose we have a data set {(t

, y

) : 1 ≤ i ≤ n} and that we desire a quadratic polynomial predictor

y = at

+ bt + c which minimizes the sum of the squared vertical errors

S(a, b, c) =

∑

i=1

∑

i=1

(at

+ bt

+ c −y

)

This might look terrifying, but can be attacked exactly as before using differentiation: to minimize S,

we need the derivatives of S with respect to the coefﬁcients a, b, c to be zero.











∂S

∂a

= 2

∑

+ bt

+ ct

−t

= 0

∂S

∂b

= 2

∑

+ bt

+ ct

−t

= 0

∂S

∂c

= 2

∑

+ bt

+ c −y

= 0

⇐⇒











∑

+ b

∑

+ c

∑

+ b

∑

+ c

∑

+ b

∑

+ cn =

∑

As a system of equations for a, b, c this looks fairly nasty, but by rephrasing in terms of matrices, we

see that it is exactly the same problem as before!





∑

















∑





corresponds to









= P

y where P =













and y =













The only change is that P is now an n × 3 matrix so that P

P is 3 × 3. Analogous to the linear

situation, provided at least three of the t

are distinct, the matrix P

P is invertible and there is a

unique least-squares quadratic minimizer

y = at

+ bt + c =



t 1













t 1



−1

The predictions

y(t

) therefore form a vector ˆy = P





= P(P

−1

y, and the coefﬁcient of

determination may be computed as before.

ˆy

−n

−ny

The method generalizes in the obvious way: if you want a cubic minimizer, give P an extra column

of cubed t

-terms! This would be hard work by hand, but is standard fodder for computers: this isn’t

a linear algebra class, so don’t try to invert a 3 ×3 matrix!

Example 5.12. We are given data {(t

, y

)} = {(1, 2), (2, 5), (3, 7), (4, 4)}.

1. For the best-ﬁtting linear model, we use the same P (and thus P

P) from the previous example:





= (P

−1

y =



30 10

10 4



−1



1 2 3 4

1 1 1 1

















2 −5

−5 15







0.8

2.5



which yields

y(t) = 0.8t + 2.5. The predicted values and coefﬁcient of determination are then

ˆy =



1 2 3 4

1 1 1 1



0.8

2.5









3.3

4.1

4.9

5.7







84.2 −81

94 −81

≈ 0.2462

The linear model predicts only 24.6% of the variance in the output; not very accurate.

2. For a quadratic model; all that changes is the matrix P

P =







1 1 1

4 2 1

9 3 1

16 4 1







=⇒ P

P =





1 4 9 16

1 2 3 4

1 1 1 1











1 1 1

4 2 1

9 3 1

16 4 1











354 100 30

100 30 10

30 10 4





=⇒









= (P

−1

















354 100 30

100 30 10

30 10 4





−1





149









−1.5

8.3

−5





from which

y = −1.5t

+ 8.3t − 5. To quantify its accuracy, compute the vector of predicted

values

y(t

) and the coefﬁcient of determination:

ˆy = P





−1.5

8.3

−5











1.8

5.6

6.4

4.2







ˆy

−4y

93.2 −81

94 −81

≈ 0.9385

The quadratic model is far superior to the linear, ex-

plaining 94% of the observed variance.

3. We can even ﬁnd a cubic model (P is a 4 ×4 matrix!)

y =

(−4t

+ 21t

−17t + 12)

The cubic passes through all four data points, there is

no error and R

= 1.

0 1 2 3 4

For real-world data this is possibly less useful than the quadratic model—it certainly takes

longer to ﬁnd! More importantly, likely experimental error in the y-data has a strong effect

on the ‘perfect’ model—we are, in effect, modelling noise. Do you expect y(5) to be closer to −1

or −8?

Exercises 5.3. 1. Recall Example 4.2, with the following almost linear data set.

x 0 2 4 6 8 10

y 3 23 41 59 77 93

Find the best-ﬁtting straight line for the data, then use a spreadsheet to ﬁnd the best-ﬁtting

quadratic. Is the extra effort worth it?

2. You are given the following data consisting of measurements from an experiment recorded at

times t

seconds.

1 2 3 4 5 6 7 8 9 10

7 5 3 2 3 5 6 9 8 12

(a) Given the values

∑

= 55,

∑

= 385,

∑

= 60,

∑

= 385

ﬁnd the best-ﬁtting least-squares linear model for this data, and use it to predict

y(13).

(b) Find the best-ﬁtting quadratic model for the data: feel free to use a spreadsheet!

ninth-degree models and their coefﬁcients of determination.

0 2 4 6 8 10

tlinear

0 2 4 6 8 10

quadratic

0 2 4 6 8 10

tcubic

0 2 4 6 8 10

quartic

0 2 4 6 8 10

degree nine

Degree R

1 0.4264

2 0.8830

3 0.9319

4 0.9336

9 1

Which of these models would you choose for this data and why? What considerations

would you take into account?

5.4 Exponential & Power Regression Models

If you suspect that your data would be better modelled by a non-polynomial function, there are

several things you can try.

Minimizing the sum of squared-errors might be very difﬁcult for non-polynomial functions because

there is likely no simple tie-in with linear equations/algebra. Attempting this is likely to result in

a horrible non-linear system for your coefﬁcients which is difﬁcult to analyze either theoretically or

using a computer.

Log Plots The most common approach when trying to ﬁt an exponential model

y = e

mt+c

to data is

to use a log plot: taking logarithms of both sides results in

y = mt + c

If we take

Y := ln

y as a new variable, the model is now a straight-line! The idea is then to use linear

regression to ﬁnd the coefﬁcients m, ln a.

Example (4.4, cont). Recall our earlier rabbit-population P(t), repeated in the table below. We

previously considered modelling this with an exponential function for two reasons:

1. We were told it was population data!

2. The t-differences are constant (2), while the P-ratios

are approximately so (≈ 1.41).

0 2 4 6 8 10

5 7 10 14 19 28

ln P

1.61 1.95 2.30 2.64 2.94 3.33

After constructing a log-plot, the relationship is much clearer:

0 2 4 6 8 10

ln P

0 2 4 6 8 10

Since the relationship between t and ln P appears linear, we perform a linear regression calculation

to ﬁnd the best-ﬁtting least-squares line for the (t

, ln P

) data.

As an example of how horriﬁc this is, suppose you want to minimize the sum of square-errors for data (t

, y

) using an

exponential model

y(t) = ae

. The coefﬁcients of our model, a, k should minimize

S(a, k) =

∑

i=1



−y



Differentiating this with respect to a, k and setting equal to zero results in

(

∂S

∂a

= 2

∑



−y



= 0

∂S

∂k

= 2a

∑



−y



= 0

=⇒



∑



∑

2kt





∑

2kt



∑



where we substituted for a to obtain the last equation. Remember that this is an equation for k; if you think you can solve

this easily, think again!

Everything necessary comes from extending the table.

Data average

0 2 4 6 8 10 5

5 7 10 14 19 28 13.83

ln P

1.61 1.95 2.30 2.64 2.94 3.33 2.46

0 4 16 36 64 100 36.67

ln P

0 3.89 9.21 15.83 23.56 33.32 14.30

m =

t ln P − t ·ln P

−t

14.30 −5 · 2.46

36.67 −5

= 0.171

c = ln P −mt = 3.46 −0.171 ·5 = 1.609

which yields the exponential model

P(t) = e

0.171t+1.609

= 4.998(1.186)

0 2 4 6 8 10

ln P

0 2 4 6 8 10

This is very close to the model (5(1.188)

) we obtained previously by pure guesswork. The approxi-

mate doubling time T for the population satisﬁes

= 2 =⇒ T =

ln 2

= 4.06 months

When using the log plot method, interpreting errors and the goodness of ﬁt of a model is a little more

difﬁcult. Typically one computes the coefﬁcient of determination R

of the underlying linear model: in

our example,

= m

Var t

Var ln P

= 99.3%

It is important to appreciate that the log plot method does

not treat all errors equally: taking logarithms tends to re-

duce error by a greater amount when the output y is large.

This should be clear from the picture, and more formally

by the mean value theorem: if y

< y

, then there is some

ξ ∈ (y

, y

) for which

ln y

−ln y

−y

) <

−y

)

ln y

Same ∆y

Different ∆ ln y

The log plot approach therefore places a higher emphasis on accurately matching data when the

output y is small. This isn’t such a bad thing since our intuitive view of error depends on the size of

the data. For instance, misplacing a $100 bill is annoying, but a $100 mistake in escrow when buying

a house is unlikely to concern you very much! Exponential data can more easily vary over large

orders of magnitude than linear or quadratic data.

This needs more decimal places of accuracy for the log-values than what’s in our table!

Log-Log Plots If you suspect a power function model

y = at

, then taking logarithms

y = m ln t + ln a

results in a linear relationship between ln y and ln t. As before, we can apply a linear regression

approach to ﬁnd a mode; the goodness of ﬁt is again described by the coefﬁcient of determination of

the underlying model.

Exercises 5.4. 1. You suspect a logarithmic model for a data set. Describe how you would approach

ﬁnding a model in the context of this section.

2. The table shows the average weight and length of a ﬁsh species measured at different ages.

Age (years) Length(cm) Weight (g)

1 5.2 2

2 8.5 8

3 11.5 21

4 14.3 38

5 16.8 69

6 19.2 117

7 21.3 148

8 23.3 190

9 25.0 264

10 26.7 293

11 28.2 318

12 29.6 371

13 30.8 455

14 32.0 504

15 33.0 518

16 34.0 537

17 34.9 651

18 36.4 719

18 37.1 726

20 37.7 810

200

400

600

800

0 10 20 30 40

(a) Do you think an exponential model is a good

ﬁt for this data? Take logarithms of the

weight values and use a spreadsheet to ob-

tain a model

w(ℓ) = ae

mℓ

where w, ℓ are the

weight and length respectively.

(b) What happens if you try a log-log plot? Given

what we’re measuring, why do you expect a

power model to be mode accurate?

3. Population data for Long Beach CA is given.

Using a spreadsheet or otherwise, ﬁnd linear,

quadratic, exponential and logarithmic regres-

sion models for this data.

Which of these models seems to ﬁt the data best,

and which would you trust to best predict the

population in 2020?

Look up the population of Long Beach in 2020;

does it conﬁrm your suspicions? What do you

think is going on?

Year

Years since 1900 Population

1900 0 2,252

1910 10 17,809

1920 20 55,593

1930 30 142,032

1940 40 164,271

1950 50 250,767

1960 60 334,168

1970 70 358,879

1980 80 361,498

1990 90 429,433

2000 100 461,522

2010 110 462,257

4. In the early 1600s, Johannes Kepler used observational data to derive his laws of planetary motion,

the third of which relates the orbital period T of a planet (how long it takes to go round the sun)

to its (approximate) distance r from the sun.

Planet T (years) r (millions km)

Mercury 0.24 58

Venus 0.61 110

Earth 1 150

Mars 1.88 230

Jupiter 11.9 780

Saturn 29.5 1400

Uranus 84 2900

Neptune 165 4500

120

160

0 2000 4000

The table shows the data for all the planets. Use a spreadsheet to analyze this data and ﬁnd a

model relating T to r.

Kepler did not known about Uranus and Neptune and only had relative distances for the plan-

ets. Research the correct statement of Kepler’s third law and compare it with your ﬁndings.