Calculus Illustrated. Volume 1 Precalculus

1.6. The algebra of exponents 50

proportionIn general, we de ne a sequence to represent the of the largest possible population as

follows:

an+1 = ran(1 − an) ,

r > 0where is a parameter. So, the rst factor tries to, say, double the population, while the second

‚‘EI“gholds it back. For the spreadsheet, the formula refers to the previous state, :

a‚PgPB‚‘EI“gB@IE‚‘EI“gA

‚PgPwhile ris the cell that contains the value of the parameter . We present a computation with a

speci c parameter r = 3.9 (and a1 = .5):

logistic sequenceSuch a sequence is called a. Its dynamics dramatically depends on the choice of the

parameter r:

1/2Above, the sequence starts with and then shows several di erent types of dynamics: decay, equi-

nlibrium, oscillating convergence to equilibrium, and chaos. There is no th-term formula. The chaotic

one looks similar to a computer-generated random sequence below:

Exercise 1.5.22

What if new bacteria are introduced to the jar at a constant rate?

1.6. The algebra of exponents

Let's review our analysis from earlier where the exponents by analogy with products come from:

1.6. The algebra of exponents 51

Analogy: repeated addition vs. repeated multiplication

Multiplication: Exponentiation:

n = 1, 2, 3, ... a + a + a + ... + a = a · n a · a · a · ... · a = an

n=0 n times n times

a·0 =0 a0 = 1

Rules: 1. a(n + m) = a · n + a · m an+m = an · am

2. (a + b)n = a · n + b · n (ab)n = anbn

3. a · (nm) = (a · n) · m anm = (an)m

Exercise 1.6.1

Represent as a power of 3: (a) 33 · 9 , (b) (9 · 27)5 .

factoringThis new algebra helps us to understand of integers. For example, the last factorization below is

best:

20 = 4 · 5 = 2 · 2 · 5 = 2251 .

But we can improve it even further:

20 = 223051 .

Let's review what we know about factoring:

Theorem 1.6.2: Rules of Factoring Integers

• > 1A prime number cannot be factored into smaller integers :

2, 3, 5, 7, 11, 13, ...

• Fundamental Theorem of Arithmetic: Every integer can be repre-

sented as the product of primes.

• Fundamental Theorem of Arithmetic: This representation is unique,

up to the order of the factors:

q = 2m · 3n · 5k · ...

• The multiplicity of a prime factor is how many times it appears in the

factorization.

Then, in the prime factorization

20 = 223051 ,

the multiplicity of 2 is 2, of 5 is 1, and the rest are 0s. And there are no other factorizations!

Example 1.6.3: prime factorization

theLet's nd prime factorization of 1176.

Our strategy is to divide by larger and larger primes, switching to the next one when we can't. We

2 2start with and then go up. We try to divide by and discover that it is possible:

1176/2 = 588 .

2 2We start the list of factors with this . We will continue to try to divide by . It is possible:

588/2 = 294 ,

1.6. The algebra of exponents 52

and we add this 2 to our list of factors: 2, 2. We continue to divide by 2:
294/2 = 147 .

It works and we have: 2, 2, 2. But the result is odd! We, therefore, move on to the next prime, 3:
147/3 = 49 .

It works and our list becomes: 2, 2, 2, 3. The result is not divisible by 3 and we, therefore, move on to
5 5the next prime, . We fail to divide by . The next prime works:

49/7 = 7 .

This last number is also prime and we are done with our list:

2, 2, 2, 3, 7, 7 .

The prime factorization is as follows:

1176 = 2 · 2 · 2 · 3 · 7 · 7 = 23315072 .
Here, 3, 1, 0, 2 are the respective multiplicities.

Exercise 1.6.4

Factorize each into the product of powers of primes: (a) 500 , (b) 660 , (c) 1024 .

We are familiar with geometric progressions. They start with index n = 0:

nHowever, there are other values of that might also interest us!

Example 1.6.5: bacteria multiplying, the past

Suppose our bacteria double in number every day and the current population is 1000. Then, tomorrow
it will be 1, 000 · 2, and so on. After n days, it's

1, 000 · 2n .

yesterdayBut how many were there ? Half of it, 500! Does this number t into our geometric progres-

sion an? Can we choose n = −1? Let's try:

a−1 = 500 = 1, 000/2 ==? 1, 000 · 2−1 .

divideIt seems that if we go into the future, we multiply, and if we go to the past, we'll need to.

Just as earlier in this chapter we face new circumstances and, once again, we ask ourselves: Can we proceed

without changing the rules?

Can the exponents be negative?

We start with n = −1. Can −1 be the exponent, such as 2−1? And what would be the outcome and the

minus timesmeaning of repeating an algebraic operation
1 ?!

1.6. The algebra of exponents 53

We will need another convention! We choose, for a = 0, the following notation:

a−1 = 1 .
a

The choice is dictated by our desire for the three properties to be satis ed!

Theorem 1.6.6: Exponent Equal to −1
The rules of exponents hold for a−1 = 1/a but fail for any other choice of a−1.

Proof.

Let's check. We plug in n = ±1 and m = ±1, use our convention, and then apply one of the properties

presented above:

1: an+m = an · am n = −1, m = 1 =⇒ a−1+1 = a−1 · a1 ⇐⇒ a0 = 1 · a „‚…i
a

2: anbn = (ab)n n = −1 =⇒ a−1b−1 = (ab)−1 ⇐⇒ 1 · 1 = 1 „‚…i
a b ab

3: anm = (an)m n = −1, m = 1 =⇒ a(−1)1 = (a−1)1 ⇐⇒ a−1 = 1 „‚…i
n = 1, m = −1 =⇒ a1(−1) = (a1)−1 a „‚…i

⇐⇒ a−1 = a−1

Exercise 1.6.7

Provide the rest of the proof.

So, this is our convention: Multiplying by a−1 means dividing by a.

Exercise 1.6.8

Represent as exponents: (a) .2 , (b) .25 , (c) .0001 .

Now, the rest of the negative numbers.

First, let's take another look at our analogy. While multiplication by a positive integer means repeated

negative subtractionaddition, multiplication by a
integer means repeated (the inverse of addition):

a(−n) = −an = 0 − a − a − a − ... − a .

Similarly, while exponentiation by a positive integer means repeated multiplication, exponentiation by a

negative divisioninteger means repeated (the inverse of multiplication):

a−n = 1 ÷ a ÷ a ÷ a ÷ ... ÷ a .

Our convention for any n = 1, 2, 3, ... will be, once and for all, the following:

De nition 1.6.9: negative exponent

The power of the negative of a number is the division by the power of that
number:

a−n = 1
an

1.6. The algebra of exponents 54

To con rm, we observe this: a−1 n

a−1 · a−1 · a−1 · ... · a−1 = ||
1n
n times a

||

1 ÷a ÷ a ÷ a ÷ ... ÷ a =

n times

Theorem 1.6.10: Negative Exponent Rule

The rules of exponents hold for a−n = 1/an, where n > 0, but fail for any other
choice of a−n.

Exercise 1.6.11

Provide the proof of the theorem.

Example 1.6.12: formulas are shortcuts

shortcutsOnce again, we see these properties as :

25 = 25 · 2−2 = 25−2 = 23 .
22

Exercise 1.6.13

Represent as a power of 4: (a) 43 · .25 , (b) .25/42 . Represent as a power of 2.

Here is the summary of what we have established:

Analogy: repeated addition vs. repeated multiplication

Multiplication: Exponentiation:

Conventions: a + a + a + ... + a =a·n a · a · a · ... · a = an
=0
n = 1, 2, ... n times = a · (−n) n times =1
n=0 = a−n
a·0 = (−a) · n a0
n = −1, −2, ... 0 −a − a − a − ... − a 1 ÷a ÷ a ÷ a ÷ ... ÷ a 1n
= −(a · n) =
Rules: n times n times
=a·n+a·m a
1. = (−a) + (−a) + ... + (−a) =a·n+b·n = 1 · 1 · ... · 1 1
2. = (a · n) · m aa a = an
3. n times
n times = an · am
= −(a + a + a + ... + a) = anbn
= 1 ÷ (a · a · a · ... · a) = (an)m
n times
n times
a · (n + m)
(a + b) · n an+m
a · (n · m) (a · b)n

an·m

So, the three properties are still satis ed, and we will continue to use them with no regard for a possible
negativity of the exponent. There is no need to count anymore.

nThese are the possible values of from now on:

..., −3, −2, −1, 0, 1, 2, 3, ...

1.6. The algebra of exponents 55

multiply divideIn the graph, we notice that we
aby as we move right (increase) and by a as we move left

(decrease):

However, if we start at any point and proceed to the right, we only multiply. This is our conclusion.

Theorem 1.6.14: Monotonicity of Geometric Progression

A geometric progression rn is
• increasing if r > 1, and
• decreasing if 0 < r < 1.

This is what the graphs look like:

Exercise 1.6.15

Prove the theorem.

Exercise 1.6.16

If our bacteria double in number every day and the current population is 1024, how many were there

two days ago?
The integers as the possible moments of time still miss some of the numbers that might interest us in these

1models! For example, suppose our bacteria double in number every day starting with . What happens
after 10.5 days? We'll need to gure out the meaning of this:

n = 10.5, 210.5 = ?

We address the fractional exponents in Chapter 4.

1.7. The Binomial Formula 56

1.7. The Binomial Formula

powers over additionWe know that we aren't supposed to distribute (an unforgivable mistake!):

(a + b)2 = a2 + b2 .

Instead, we nd what it is equal to by distributing multiplication over addition:

(a + b)2 = (a + b) · (a + b)
= (a + b) · a + (a + b) · b
= (a · a + b · a) + (a · b + b · b)
= a2 + 2ab + b2 .

Warning!

We repeat the equal sign = every time because it

stands for the verb is and we want our sentences
to be grammatically correct.

area of a squareThe meaning of the formula is revealed if we interpret (a + b)2 as the with side a + b:

The end terms correspond to the two smaller squares, a × a and b × b, and the intermediate terms
correspond to the rectangles, a × b and b × a.

The result is a handy formula:

(a + b)2 = a2 + 2ab + b2

factorIt is very useful, especially when we need to an expression; we simply read the formula from right to

left:

a2 + 2ab + b2 = (a + b)2 .

Example 1.7.1: a complete square

To factor 4x2 + 4x + 1, we match it with the left-hand side of the formula above (we assume that the

numbers are positive) as follows:

a2 + 2ab + b2 = (a + b)2 We match vertically.

4x2 + 4x + 1 = ?

=⇒ a2 = 4x2 ... b2 = 1 The equations that come from the matching.

=⇒ a = 2x ... b = 1 The equations are solved.

=⇒ 4x2 + 4x + 1 = (2x + 1)2 The middle term also checks out.

1.7. The Binomial Formula 57

We have a complete square:

4x2 + 4x + 1 = (2x + 1)2 .

It is more compact than the original and we can notice certain facts about it that used to be invisible.
For example, the expression can't be negative!

Warning!

We do not put the equal sign = between equa-

tions because each is a separate sentence, and the
quantities aren't equal.

Exercise 1.7.2

Factor 9x2 + 12xy + 4y2.

higher powersWhat about the ? To get the cubic power and above, we simply continue multiplying by (a + b)

and applying the Distributive Property:

(a + b)3 = (a + b)2 · (a + b)
= (a2 + 2ab + b2) · (a + b)
= (a2 + 2ab + b2) · a + (a2 + 2ab + b2) · b
= (a2 · a + 2ab · a + b2 · a) + (a2 · b + 2ab · b + b2 · b)
= (a3 + 2a2b + ab2) + (a2b + 2ab2 + b3)
= a3 + 3a2b + 3ab2 + b3 .

The result is another handy formula:

(a + b)3 = a3 + 3a2b + 3ab2 + b3

Example 1.7.3: a complete cube

To factor x3 + 3x2 + 3x + 1, we match it with the left-hand side of the formula above:

a3 + 3a2b + 3ab2 + b3 = (a + b)3

x3 + 3x2 + 3x + 1 = ?

=⇒ a3 = x3 .. .. b3 = 1

=⇒ a = x .. .. b=1

=⇒ x3 + 3x2 + 3x + 1 = (x + 1)3

The intermediate terms also checks out.

Exercise 1.7.4

Interpret (a+b)3 as the volume of a cube with side a+b and illustrate the meaning of the intermediate

terms in the above formula.

binomialThe two-term expression a + b is called a . In the formula (m = 2 and m = 3 above),

(a + b)m = am + ... + bm ,

we call the left-hand side a binomial power and the right-hand side the binomial expansion.

mWhere these terms come from? We pick one term from each binomial and there are a total of them:

(a + b) · (a + b) · ... · (a + b) = sum of terms of the type an · bk .

1.7. The Binomial Formula 58

aAs we arrange the terms according to the decreasing powers of , a pattern starts to emerge! As the power

combined power of andof a is decreasing, that of b is increasing. Moreover, the
a b, i.e., the sum of the two

powers, is m = 3, just as it was m = 2 in the last formula. Let's make a table for the binomials that starts

with m = 1:

m = 1 (a + b)1 = a1b0 + a0b1 a0b2 a0b3
powers: 1 = 1 + 0 = 0+1 0+2 0+3
m = 2 (a + b)2 = a2b0 + 2a1b1 + 3a1b2 + ? a1b3 + a0b4
powers: 2 = 2 + 0 = 1+1 = 1+2 = 1+3 = 0+4
m = 3 (a + b)3 = a3b0 + 3a2b1 + ? a2b2 +
powers: 3 = 3 + 0 = 2+1 = 2+2 =
m = 4 (a + b)4 = a4b0 + ? a3b1 +
powers: 4 = 4 + 0 = 3+1 =

We also included as a guess the case m = 4. We see that m + 1 is the number of ways we can represent
m as a sum of two non-negative numbers. Below we put the results in a spreadsheet with the formula:

a‚gPC‚Pg

The expansions above follow the diagonal as marked:

a b 1In each, the combined power of and remains the same and it grows by as we move to the next diagonal.

coe cientsThe behavior of the powers is clear; it's the that present a challenge. Let's describe the former

rst.

Theorem 1.7.5: Number of Terms in Binomial Expansion

The expansion of the mth power (a + b)m of the binomial (a + b) consists of m + 1
terms of the form anbk, each of which has the sum of powers of a and b equal to

m.

Proof.

If all of the terms of the expansion of (a + b)m have the sum of powers equal to m, then what happens

in the next:

(a + b)m+1 = (a + b)m · (a + b) ?

According to the Distributive Property, each of the terms of the expansion of (a + b)m is multiplied by

a or by b. Therefore, the total power goes up by 1!

Exercise 1.7.6

Provide the missing parts of the proof.

Once again, we have:

(a + b)m = sum of terms of the type an · bk .

1.7. The Binomial Formula 59

how many timesWe want to know each term appear in the expansion.

coe cientsIn the summary below, we show the powers but the of the terms remain unknown:

What are the coe cients?

(a + b)m = amb0 + ? am−1b1 + ? am−2b2 +...+ ? a1bm−1 + a0bm
m = m + 0 = (m − 1) + 1 = (m − 2) + 2 = ... = 1 + (m − 1) = 0 + m

mAs you can see, the last row just shows all possible ways to represent as the sum of two non-negative

integers. Let's take a look at the coe cients:

m=1 (a + b)1 = a1 + b1
1
coe cients: 1 2a1b1 +
2
m=2 (a + b)2 = a2 + 3a2b1 + b2
3 1
coe cients: 1 ? a3b1 + 3a1b2 +
4 3
m=3 (a + b)3 = a3 + ? a2b2 + b3
? 1
coe cients: 1 ? a1b3 + b4
41
m=4 (a + b)4 = a4 +

coe cients: 1

We've made a guess for the last row, but the middle term remains unknown.

6An actual computation reveals that it's . The table of coe cients becomes:

m=1 1 1
m=2 1 2 1
m=3 1 3 3 1
m=4 1 4 6 4 1

An examination of two consecutive rows allows us to observe how things develop in this table:

m=3 1 + 3 3+3 3+1
|| || ||

m=4 1 4 46 64

We can guess that every two adjacent coe cients in each row are added and placed in the next row.

sequence of sequencesThe table is a, and the terms in the next row are determined by the ones in the last.

It is more convenient, however, to arrange the terms in an isosceles instead of a right triangle as above. This

is how we build the 4th row from the 3rd:

known: m=3 1 3 3 1

1+3 3+3 3+1

unknown: m = 4 1 4 6 41

Pascal TriangleEach entry is the sum of the two entries above it. The result is called the :

m=0 1

m=1 11

m=2 121

m=3 13 31

m=4 14 6 41

m = 5 1 5 10 10 5 1

... . . . . . . .

1.7. The Binomial Formula 60

Exercise 1.7.7

Explain the entry at the top of the triangle.

The computation may continue inde nitely following this procedure:

XY

X +Y

We summarize the result below.

Theorem 1.7.8: Binomial Theorem

If two consecutive terms of the expansion of (a + b)m are the following:
X · anbm−n and Y · an−1bm−n+1 ,

then the expansion of (a + b)m+1 contains the following term:
(X + Y ) · anbm−n+1 .

Proof.

(Xanbm−n + Y an−1bm−n+1)(a + b)
= Xanbm−na + Y an−1bm−n+1a + Xanbm−nb + Y an−1bm−n+1b
= Xan+1bm−n + Y anbm−n+1 + Xanbm−n+1 + Y an−1bm−n+2
= Xan+1bm−n + (X + Y )anbm−n+1 + Y an−1bm−n+2 .

Exercise 1.7.9

Provide the missing parts of the proof.

Example 1.7.10: Pascal Triangle, computed

5This is how the Pascal Triangle is used. We take its th row and produce a new binomial expansion:

m=5 1 5 10 10 51

(a + b)5 = 1 · a5 + 5 · a4b + 10 · a3b2 + 10 · a2b3 + 5 · ab4 + 1 · b5

One can easily compute, recursively, a large part of the Pascal triangle with a spreadsheet:

We use the formula:

a‚‘EI“g‘EI“C‚‘EI“g‘I“

So, for each m, the coe cients of an expansion form a sequence of m + 1 integers. They are simple in the

very beginning and the very end:

n 0 1 ... m − 1 m

1 m ... m 1

1.7. The Binomial Formula 61

Let's examine the rest.

De nition 1.7.11: binomial coe cient

The nth term of this sequence is called the binomial coe cient and is denoted

by

m
n

It reads m choose n .

The rst number gives the vertical location and the second horizontal location within the Pascal triangle.

A certain symmetry of the Pascal triangle is seen in the following formula.

Exercise 1.7.12

Show that mm
n = m−n .

Exercise 1.7.13 m
= m.
Show that
2

In the new notation, the Pascal Triangle takes the form:

0
0
11
01
222
012
3333
0123
.. . ..

Therefore, the rst number indicates the row, and the second number the position within the row.

In the new notation, the Binomial Theorem is restated as follows.

Corollary 1.7.14: Sum of Binomial Coe cients

For every positive integers m and n ≤ m, we have:

m m m+1
+=

n n+1 n+1

That's a recursive formula! What is the nth-term formula for this sequence, or rather sequences? It can be

factorialexpression in terms of the . The following is an important result:

1.7. The Binomial Formula 62

Theorem 1.7.15: Binomial Coe cient

For every positive integers m and n ≤ m, we have:

m m!
n = n!(m − n)!

Proof.

We con rm the recursive formula:

mm m! m!
+ = n!(m − n)! + (n + 1)!(m − n − 1)!

n n+1 m! m!
= n!(m − n − 1)!(m − n) + n!(m − n − 1)!(n + 1)

m!(n + 1) + m!(m − n)
= n!(m − n − 1)!(m − n)(n + 1)

m! (n + 1) + (m − n)
= (n + 1)!(m − (n + 1))!

m! m + 1
= (n + 1)!(m − (n + 1))!

m+1
=.

n+1

Exercise 1.7.16

Why does this computation prove the formula?

Warning!

Even though a fraction, a binomial coe cient is an
integer.

Example 1.7.17: binomial coe cients, computed

To test the formula, we substitute m = 5 and n = 3:

5 = 5! 3)! = 5! = 2·3·4·5 = 2 · 5 = 10 .
3 3!(5 − 3!2! 2·3·2

The result matches the one in the Pascal Triangle.

The binomial coe cients have another interpretation. Since the power is just a repeated multiplication,

(a + b)m = (a + b) · (a + b) · ... · (a + b) ,

m times

a bthere will be one term in the binomial expansion for each choice of either or from each of the factors:

anbm−n = a · a · ... · a · b · b · ... · b .

n times m−n times

We draw the following conclusion:

1.8. The sequence of di erences: velocity 63

Theorem 1.7.18: Choose n From m

m
n nThe binomial coe cient
is the number of ways we can choose objects

from m objects.

Example 1.7.19: number of ways to choose a team

In how many ways can one form a team of 5 from 20 players? We compute:

20 20!
=

5 5!15!
15! · 16 · 17 · 18 · 19 · 20

=
5!15!

16 · 17 · 18 · 19 · 20
=

2·3·4·5
16 · 17 · 18 · 19
=

2·3
= 16 · 17 · 3 · 19
= 15, 504 .

Exercise 1.7.20

How many di erent hands of 5 are there in a deck of 52 cards?

1.8. The sequence of di erences: velocity

Numbers are subject to algebraic operations: addition, subtraction, multiplication, and division. Since the

terms of sequences are numbers, a pair of sequences can be added (subtracted, etc.) to produce a new one.

newIn addition, there are two operations that apply to a single sequence and produce a sequence that

originaltells us a lot about the sequence. These operations are: subtracting the consecutive terms of the

sequence and adding its terms repeatedly. We saw them in action in the rst section:

This is the summary:

velocities• If each term of a sequence represents a location, the pair-wise di erences will give you the ,

and

location• If each term of a sequence represents the velocity, their sum up to that point will give you the

1.8. The sequence of di erences: velocity 64

(or displacement).

In this section we start on the path of development of an idea that culminates with the rst fundamental

concept of calculus, the derivative (Chapter 2DC-3).

changeThe pairwise di erences represent the within the sequence, from each of its terms to the next.

Example 1.8.1: sequence given by list

When a sequence is given by a list, we subtract the last term from the current one and put the result
in the bottom row as follows:

a sequence: 2 4 7 1 ...

...

its di erences: 4−2 7−4 1−7 ...

|| || || ...

a new sequence: 2 3 −6 ...

We have a new list.

Example 1.8.2: sequence given by graph

In the simplest case, a sequence takes only integer values, then on the graph of the sequence, we just
count the number of steps we make, up and down:

These increments then make a new sequence plotted on the right.

De nition 1.8.3: sequence of di erences

For a sequence an, its sequence of di erences, or simply the di erence, is a new

sequence, say dn, de ned for each n by the following:

dn = an+1 − an .

It is denoted by

∆an = an+1 − an

Warning!

wholeThe symbol ∆ applies to the sequence an,

and ∆a should be seen as the name of the new

sequence; the notation for the di erence is an ab-

breviation for (∆a)n.

This is what the de nition says:

1.8. The sequence of di erences: velocity 65

Sequence of di erences

a sequence: a1 a2 a3 a4 ...
...
its di erences: a2 − a1 a3 − a2 a4 − a3 ...
a new sequence: || || || ...
...
the notation: d1 d2 d3 ...
|| || || ...

∆a1 ∆a2 ∆a3

Warning!

If the original sequence starts with n = q, then the
new sequence starts with n = q + 1 but it could
also be arranged to start with n = q or any other

index.

formulasNow, what about sequences given by ? Let's consider a couple of speci c sequences.

The rst one is the arithmetic progression and it is very simple.

Theorem 1.8.4: Di erence of Arithmetic Progression

mThe sequence of di erences of an arithmetic progression with increment is a
mconstant sequence with the value equal to .

Proof.

We simply compute from the de nition:

dn = ∆(a0 + mn) = an+1 − an = (a0 + b(m + 1)) − (a0 + mn) = m .

The theorem can be recast as an implication, an if-then statement:

sp an is an arithmetic progression with increment m, „rix its sequence of di erences is a
mconstant sequence with the value equal to .

We can also use our convenient abbreviation:

an is an arithmetic progression with increment m =⇒ aits sequence of di erences of n is a
mconstant sequence with the value equal to .

This is what the graphs of this pair of sequences may look like, zoomed out, for the following three choices

of the increment m:

Let's take a look at a geometric progression an = arn with a > 0 and r > 0.

1.8. The sequence of di erences: velocity 66

Example 1.8.5: geometric progression an = 3n
This is a geometric progression with ratio r = 3. Let's compute the di erence:

∆an = an+1 − an
= 3n+1 − 3n
= 3n · 3 − 3n
= 3n(3 − 1)
= 3n · 2 .

It's a geometric progression with r = 3, again!

Is there a pattern? Let's plot the graph of a geometric progression an = arn with a > 0 and r > 0. There
rare two cases, depending on the choice of ratio (growth or decay):

What do the sequence of di erences (second row) look like? We notice the following:

• rIt is positive and increasing, with speeding up when the ratio is larger than 1.
• It is negative and increasing, with slowing down when 0 < r < 1.

It also resembles the original sequence!

Theorem 1.8.6: Di erence of Geometric Progression

The sequence of di erences of a geometric progression is a geometric progression
with the same ratio.

Proof.

If we have a geometric progression with ratio r and initial term a, its formula is an = arn. Therefore,
dn = ∆(arn) = an+1 − an = arn+1 − arn = a(r − 1) · rn .

But that's the formula of a geometric progression with ratio r and initial term a(r − 1).

The theorem can be restated as an implication:

sp an is a geometric progression with ratio r, „rix its sequence of di erences is a geometric
progression with ratio r.

Also:

an is a geometric progression with ratio r =⇒ its sequence of di erences is a geometric
progression with ratio r.

1.8. The sequence of di erences: velocity 67

Example 1.8.7: alternating sequence

The sequence of di erences of the alternating sequence an = (−1)n is computed below:

∆ ((−1)n) = (−1)n+1 − (−1)n = (−1) − 1, n is even = −2, n is even = 2(−1)n+1 .
1 − (−1), n is odd 2, n is odd

Exercise 1.8.8

What is the relation between the sequence above and its di erence?

Example 1.8.9: di erences are velocities

We can use computers to speed up these computations. For example, one may have been recording

one's locations and now needs to nd the velocities. Here is a spreadsheet formula for the sequence of

di erences (velocities):

a‚g‘EI“E‚‘EI“g‘EI“

Whether the sequence comes from a formula or it's just a list of numbers, the formula applies equally:

As a result, a curve has produced a new curve:

While the rst graph tells us that we are moving forward and then backward, it is easier to derive
better description from the second: speed up forward, then slow down, then speed up backward.

Exercise 1.8.10

Describe what has happened referring to, separately, the rst graph and the second graph.

Exercise 1.8.11

Imagine, instead, that the rst column of the spreadsheet above is where you have been recording the
monthly balance of your bank account. What does the second column represent? Describe what has
been happening with your nances referring to, separately, the rst graph and the second graph.

theoryThis is the time for some .

Consider this obvious statement about motion:

sp I am standing still, „rix my velocity is zero.

1.8. The sequence of di erences: velocity 68

We can also say:

sp my velocity is zero, „rix I am standing still.

converseWe see the implications going both ways; the latter is the of the original statement (and vice

versa!). Let's use symbols to restate these statements more compactly:

I am standing still =⇒ my velocity is zero.
I am standing still ⇐= my velocity is zero.

The abbreviation of the combination of the two is an equivalence, an if-and-only-if statement:

I am standing still ⇐⇒ my velocity is zero.

It's just another way of saying the same thing.

If we set the motion point of view aside, here is the general statement.

Theorem 1.8.12: Di erence of Constant Sequence

A sequence is constant IF AND ONLY IF its sequence of di erences is zero.

In other words, we have:

an is constant ⇐⇒ ∆an = 0 .

Proof. an = c for all n =⇒ an+1 − an = c − c = 0 =⇒ ∆an = 0 .
an+1 = an = c for all n ⇐= an+1 − an = 0 ⇐= ∆an = 0 .
Direct:

Converse:

Exercise 1.8.13

Prove that the di erence of an arithmetic progression is constant and, conversely, that if the di erence
of a sequence is a constant sequence, then the sequence is an arithmetic progression.

Consider another obvious statement about motion:

sp I am moving forward, „rix my velocity is positive.

And, conversely:

sp my velocity is positive, „rix I am moving forward.

In other words, we have this pair of statements:

Original: I am moving forward =⇒ my velocity is positive.
Converse: I am moving forward ⇐= my velocity is positive.

Exercise 1.8.14

What is the converse of the converse?

We combine these two in the following far reaching result.

Theorem 1.8.15: Monotonicity Theorem for Sequences

A sequence is increasing/decreasing IF AND ONLY IF the sequence of di er-

ences is positive/negative or zero, respectively.

1.8. The sequence of di erences: velocity 69

In other words, we have: ⇐⇒ ∆an ≥ 0 .
⇐⇒ ∆an ≤ 0 .
an is increasing ⇐⇒ ∆an = 0 .
an is decreasing
an is constant

Proof.

an+1 ≥ an for all n ⇐⇒ an+1 − an ≥ 0 ⇐⇒ ∆an ≥ 0 .

It's just another way of saying the same thing.

twoSuppose now that there are runners; then we have a less obvious fact about motion:

sp „rixthe distance between two runners isn't changing, they are running with the same

velocity.

And vice versa:

sp „rixtwo runners are running with the same velocity, the distance between them isn't

changing.

It's as if they are holding the two ends of a pole:

The conclusion holds even if they speed up and slow down all the time. In other words, we have:

The distance between two runners isn't changing sp exh yxv‰ sp they are running with

the same velocity.

a b nOnce again, for sequences n and n representing their respective positions at time , we can restate this

idea mathematically in order to con rm that our theory makes sense.

Corollary 1.8.16: Di erence under Subtraction

Two sequences di er by a constant IF AND ONLY IF if their sequences of

di erences are equal.

In other words, we have:

an − bn is constant ⇐⇒ ∆an = ∆bn .

Proof.

The corollary follows from the Di erence of Constant Sequence Theorem above.

Example 1.8.17: shift of sequence

We shift the sequence an below by 1 unit up to produce a new sequence bn (top):

1.8. The sequence of di erences: velocity 70

Because the ups and downs remain the same, the sequences of di erences of these two sequences are
identical (bottom).

Exercise 1.8.18

What if the two runners holding the pole also start to move their hands back and forth?

We can use the latter theorem to watch after the distance between the two runners. A matching statement
about motion is the following:

sp „rixthe distance from one of the two runners to the other is increasing, the former's

velocity is higher.

Conversely:

sp „rixthe velocity of one runner is higher than the other, the distance between them is

increasing.

Exercise 1.8.19

Combine the two statements into one.

We can restate this mathematically.

Corollary 1.8.20: Monotonicity and Subtraction

The di erence of two sequences is increasing IF AND ONLY IF the former's

di erence is bigger than the latter's.

In other words, we have:

an − bn is increasing ⇐⇒ ∆an ≥ ∆bn ,
an − bn is decreasing ⇐⇒ ∆an ≤ ∆bn .

Proof.

The corollary follows from the Monotonicity Theorem for Sequences above.

Example 1.8.21: three runners

nThe graph below shows the positions of three runners in terms of time, . Describe what has happened:

1.8. The sequence of di erences: velocity 71

They are all at the starting line together, and at the end, they are all at the nish line. Furthermore,

howA reaches the nish line rst, followed by B, and then C (who also starts late). This is each did

it: It becomes slower

• A starts fast, then slows down, and almost stops close to the nish line.
• B maintains the same speed.
• C starts late and then runs fast at the same speed.
A BWe can see that is running faster because the distance from is increasing.

later, which is visible from the decreasing distance. We can discover this and the rest of the facts by

di erencesexamining the graphs of the of the sequences:

Exercise 1.8.22

Suppose a sequence is given by the graph for velocity above. Sketch the graph of the di erence of this
sequence. What is its meaning?

Exercise 1.8.23

Plot the location and the velocity for the following trip: I drove fast, then gradually slowed down,
stopped for a very short moment, gradually accelerated, maintained speed, hit a wall. Make up your
own story and repeat the task.

Exercise 1.8.24

Draw a curve on a piece of paper, imagine that it represents your locations, and then sketch what your
velocity would look like. Repeat.

Exercise 1.8.25

Imagine that the rst graph represents, instead of locations, the balances of three bank accounts.
Describe what has been happening.

Example 1.8.26: di erence quotient of sequence

velocity1How do we treat motion when the time increment isn't ? What is the then?

x yFirst, we de ne the time and the location by two separate sequences, say, n and n. Then the

1.9. The sequence of the sums: displacement 72

velocity is the increment of the latter over the increment of the former. We notice that those two are

di erence quotientthe di erences of the two sequences. The of two sequences of xn and yn is de ned

y xto be the sequence that is the di erence of n divided by the di erence of n:

∆yn = yn+1 − yn ,
∆xn xn+1 − xn

rateprovided the denominator is not zero. It is the relative change the of change of the two

sequences (for each consecutive pair of points, it is the slope):

The di erence quotient of the sequence of the location with respect to the sequence of time is the
velocity. (We will continue this study in Chapter 2.)

Example 1.8.27: di erence of random sequence

−1 1If we take a sequence of random numbers, with its values spread between
and , its di erence is

also random but the values are spread between −2 and 2:

Exercise 1.8.28

Point out and explain the di erence between the two graphs.

1.9. The sequence of the sums: displacement

locations and velocitiesIn the rst section, we saw how the sequences of interact. We took a closer look at

the transition from the former to the latter and now in reverse:

1.9. The sequence of the sums: displacement 73

In this section, we start on the path of development of an idea that culminates with the second fundamental

the integralconcept of calculus, (Chapter 3IC-1).

The sum represents the totality of the beginning of a sequence, found by adding each of its terms to the
next, up to that point.

Example 1.9.1: sequences given by lists

We just add the current term to what we have accumulated so far:

sequence: 2 4 7 1 −1 ...

↓↓ ↓ ↓ ↓ ...

sums: 2

2+4=6

6 + 7 = 13

13 + 1 = 14

14 + (−1) = 13

↓↓ ↓ ↓ ↓ ...

new sequence: 2 6 13 14 13 ...

We have a new list!

Example 1.9.2: sequences given by graphs

We treat the graph of a sequence as if made of bars and then just stack up these bars on top of each
other one by one:

These stacked bars or rather the process of stacking make a new sequence.

recursiveUnlike the di erence, the sum must be de ned (and computed) in a manner.

De nition 1.9.3: sequence of sums

sequence of sumsFor a sequence an, itss, or simply the sum, is a new sequence n

de ned and denoted for each n ≥ m within the domain of an by the following

(recursive) formula:

sm = 0, sn+1 = sn + an+1

1.9. The sequence of the sums: displacement 74

In other words, we have:

sn = am + am+1 + ... + an

Example 1.9.4: alternating sequence

a sLet's do some algebra. Here n is the original sequence and n is the new one:

n an sn = sn

111 =1

2 −1 1 − 1 =0

3 1 1−1+1 =1

... ... ... ...

n (−1)n 1 − 1 + 1 − ... + (−1)n = 1 or 0

The resulting sequence is also alternating !

A commonly used notation is the following:

Sigma notation for summation

n

sn = am + am+1 + ... + an = ak

k=m

Let's take a closer look at the new notation. The rst choice of how to represent the sum of a segment

from m to n of a sequence an is this:

am +am+1 +... +ak +... +an .

step 1 step 2 step k step n−m

This notation re ects the recursive nature of the process but it can also be repetitive and cumbersome. The

second choice is more compact:

n

ak .

k=m

ΣHere the Greek letter stands for the letter S meaning sum .

Sigma notation

3 −→ beginning and end values for k = 20
↑
k2 + k = 20 ↓ k2 + k
3 ↑
k=0
k=0

a speci c sequence a speci c number

Warning!

It would make sense to have k = 0 above the

sigma:

k=3

k2 + k .

k=0

1.9. The sequence of the sums: displacement 75

Example 1.9.5: expanding from sigma notation

The computation above is expanded here:

k k2 + k

3 0 02 + 0 =0 +
12 + 1 =2 +
k2 + k 1 22 + 2 =6 +
= 32 + 3 = 12
k=0
2

3

= 20

Exercise 1.9.6: contracting to sigma notation

How will the sum change if we replace k = 0 with k = 1, or k = −1? What if we replace 3 at the top
with 4?

Example 1.9.7: contracting summation

contractThis is how we the summation:

n

12 + 22 + 32 + ... + 172 = k2 .

k=1

This is only possible if we nd the nth-term formula for the sequence; in this case, ak = k2. And this

expandis how we back from this compact notation, by plugging the values of k = 1, 2, ..., 17 into the

formula: 17

k2 = 12 + 22 + 32 +... + 172 .

k=1 k=1 k=2 k=3 k=17

Similarly, we have:

11 1 1 10 1 .
1 + + + + ... + =
2 22 23 210 2k
k=0

Exercise 1.9.8

Con rm that we can start at any other initial index if we just modify the formula:

11 1 1 ?1 ?1
1 + + + + ... + = 2k−1 = 2k−2 = ...
2 22 23 210
k=? k=?

Exercise 1.9.9 11 1
1+ + + = ?
Contract this summation:
3 9 27

Exercise 1.9.10

Expand this summation:

4

(k/2) = ?

k=0

1.9. The sequence of the sums: displacement 76

Exercise 1.9.11

Rewrite using the sigma notation:

1. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
2. .9 + .99 + .999 + .9999
3. 1/2 − 1/4 + 1/8 − 1/16
4. 1 + 1/2 + 1/3 + 1/4 + ... + 1/n
5. 1 + 1/2 + 1/4 + 1/8
6. 2 + 3 + 5 + 7 + 11 + 13 + 17
7. 1 − 4 + 9 − 16 + 25

Example 1.9.12: binomials

In this notation, the Binomial Theorem reads:

m m am−nbn .
n
(a + b)m =

n=0

For example: 4 4 a4−nbn .
n
Then we have: (a + b)4 =

= 4 a4−0b0 + n=0
0
4 a4−1b1 + 4 a4−2b2 + 4 a4−3b3 + 4 a4−4b4
= 1 · a4b0 + 1 2 3 4
= a4 + + + +
4 · a3b1 + 6 · a2b2 + 4 · a1b3 + 1 · a0b4

4a3b 6a2b2 4ab3 b4 .

The notation applies to all sequences, both nite and in nite. For in nite sequences, recognized by ... at
the end, the sum sequence is also called partial sums as well as series (to be discussed in Chapter 3IC-5).

This is the de nition of the sequence of sums written with the sigma notation:

Sequence of sums

a sequence: a1 a2 a3 a4 ...
...
↓↓ ↓ ↓
...
its sums: a1 ...
a1 + a2 = s2 ...
...
s2 + a3 = s3 ...
s3 + a4 = s4
s4 ...

↓↓↓↓

the sequence of sums: s1 s2 s3 s4

|| || || ||

1234

the sigma notation: ak ak ak ak

k=1 k=1 k=1 k=1

Below is the simplest result about the sums. It is still considerably more challenging than most results about
the di erences that we saw in the last section.

1.9. The sequence of the sums: displacement 77

Theorem 1.9.13: Sum of Arithmetic Progression

m 0The sum of an arithmetic progression with increment and a initial term is

a (quadratic) sequence given by the following:

n n(n + 1) m.
(mk) =
2
k=1

Exercise 1.9.14

Prove the theorem. Hint: Use the example about round robins presented earlier in this chapter.

Example 1.9.15: from sequences to series

What does the sum 111
+ + + ...

248

compute? We can see these terms as the areas of the squares below:

1On the one hand, adding the areas of these squares will never go over (the area of the big square), and,

in niteon the other, these squares seem to exhaust this square entirely. So, even the sum sometimes

makes sense (to be considered in Chapter 3IC-5). Of course, this is a geometric progression.

Let's take a look at a geometric progression with an = arn with a > 0 and r > 0. There are two cases,

rdepending on the choice of ratio (growth or decay):

1.9. The sequence of the sums: displacement 78

What about the sequence of sums (second row)? We notice the following:

• rIt is increasing with speeding up when its ratio is larger than 1.
• It is increasing with slowing down when 0 < r < 1.

It also resembles the original sequence!

Theorem 1.9.16: Sum of Geometric Progression

The sequence of sums of a geometric progression with ratio r = 1 is a geometric

progression with the same ratio and a constant sequence.

In other words, we have:

n

ark = Arn + C ,

k=1

for some real numbers A and C .

Proof.

clever trickBelow, we use a to get rid of ... . We write the nth sum sn,

sn = ar0 +ar1 +ar2 +... +arn−1 +arn

and then multiply it by r:

rsn = r ar0 +ar1 +ar2 +... +arn−1 +arn
= ar1 + ar2 + ar3 +... + arn + arn+1

Now we subtract these two:

sn = ar0 +ar1 +ar2 +... +arn−1 +arn

rsn = ar1 + ar2 + ar3 +... + arn + arn+1

sn − rsn = ar0 − ar1 +ar1 − ar2 +ar2 − ar3 +... +arn−1 − arn +arn − arn+1

= ar0 −arn+1

We cancel the terms that appear twice in the last row and ... is gone! Therefore,

sn(1 − r) = a − arn+1 .

nThus, we have an explicit formula for the th term of the sum:

sn = 1 a (1 − rn+1) = − a r · rn+1 + 1 a r .
− r 1 − −

1.9. The sequence of the sums: displacement 79

The former term is the geometric part, and the latter is the constant:

A = − 1 ar r , C = a
− 1−r .

Exercise 1.9.17

Find the explicit formula for the sequence of sums of the alternating sequence an = (−1)n.

Exercise 1.9.18

Use the trick to prove the theorem about arithmetic progressions.

Warning!

Our ability to produce an explicit formula for the

nth terms of the sequence of the sum is an excep-

tion, not a rule.

Example 1.9.19: sums are displacements

We can use computers to speed up these computations. For example, one may have been recording
one's velocities and now looking for the location. This is a formula for a spreadsheet (the locations):

a‚‘EI“gC‚g‘EI“

Whether the sequence comes from a formula or it's just a list of numbers, the formula applies:

As a result, a curve has produced a new curve:

Exercise 1.9.20

Describe what has happened referring to, separately, the rst graph and the second graph.

Exercise 1.9.21

Imagine that the rst column of the spreadsheet is where you have been recording your monthly
deposit/withdrawals at your bank account. What does the second column represent? Describe what

1.9. The sequence of the sums: displacement 80

has been happening referring to, separately, the rst graph and the second graph.

theoryThis is the time for some .

Recall from the last section this pair of obvious statements about motion:

I am standing still sp exh yxv‰ sp my velocity is zero.

If the velocity is represented by a sequence, its sum is the location. We can then restate the above mathe-
matically.

Theorem 1.9.22: Constant Sequence as Sum

The sequence of sums of a sequence is constant IF AND ONLY IF the sequence
Nhas only zero values starting from some index .

In other words, we have:

n ⇐⇒ an = 0 for all n ≥ N .

ak is constant

k=m

Proof.

n n+1 n

ak = c for all n ⇐⇒ an+1 = ak − ak = c − c = 0 ⇐⇒ an+1 = 0 .

k=m k=m k=m

Here is another equivalence statements about motion:

I am moving forward sp exh yxv‰ sp my velocity is positive.

We can restate this mathematically using the sums.

Theorem 1.9.23: Monotonicity of Sum

The sequence of sums of a sequence is increasing IF AND ONLY IF the terms

of the sequence are non-negative.

In other words, we have:

n is increasing ⇐⇒ an ≥ 0 .
is decreasing ⇐⇒ an ≤ 0 .
ak

k=m
n

ak

k=m

Proof.

n+1 n n+1 n

ak ≥ ak for all n ⇐⇒ an+1 = ak − ak ≥ 0 .

k=m k=m k=m k=m

twoNow suppose just like in the last section, that there are runners:

1.9. The sequence of the sums: displacement 81

Then, we have:

the distance between two runners isn't changing sp exh yxv‰ sp they are running with

the same velocity.

We restate this mathematically.

Corollary 1.9.24: Subtracting Sums of Sequences

The sequences of sums of two sequences di er by a constant IF AND ONLY IF
Nthe sequences are equal starting with some term .

In other words, we have:

nn

ak − bk is constant ⇐⇒ an = bn for all n ≥ N .

k=m k=m

Proof.

The corollary follows from the Constant Sequence as Sum above.

Example 1.9.25: shift of sequence

We have below two di erent sequences an and bn that become identical after 3 terms:

The result is that the sum of the latter sequence is just a vertical shift of the sum of the former:

nTo state this algebraically, we have for each :

nn

ak = bk + C ,

k=m k=m

C a bwhere is some number. The outcome is the same when the two sequences n and n are identical

but the computation of their sequences of sums starts at di erent points:

1.9. The sequence of the sums: displacement 82

Exercise 1.9.26

CWhat is the meaning of the number ?

We can use the theorems to watch for the distance between the two runners:

the distance from one of the two runners to the other is increasing sp exh yxv‰ sp the

former's velocity is higher.

We can restate this mathematically using the sums.

Corollary 1.9.27: Subtracting Sums: Monotonicity

IF ANDThe di erence of the sequences of sums of two sequences is increasing
ONLY IF the corresponding terms of the former are larger than or equal to

those of the latter.

In other words, we have:

nn

ak − bk is increasing ⇐⇒ an ≥ bn .
is decreasing ⇐⇒ an ≤ bn .
k=m k=m
n n

ak − bk

k=m k=m

Proof.

The corollary follows from Monotonicity of Sum above.

Here is another way to look theat the statement faster covers the longer distance. It is about comparing

the values of two sums. Consider this simple algebra:

a ≤b
A ≤B
a+A ≤ b+B

The rule applies even if we have more than just two terms:

am ≤ bm
am+1 ≤ bm+1

... ... ...

aq ≤ bq
am + ... + aq ≤ bm + ... + bq

The summation is illustrated below:

1.9. The sequence of the sums: displacement 83

Example 1.9.28: three runners, continued

nThe graph shows the velocities of three runners in terms of time, :

howIt's easy to describe they are moving:

• A starts fast and the slows down.

• B maintains the same speed.

• C starts late and then runs fast.

whereBut are they, at every moment? There are several possible answers:

Which one is the right one depends on the starting point. Of course, a simple examination of the rst
graph doesn't prove that the three runners will arrive at the nish line at the same time.

Furthermore, if the requirement that they all start at the same location is lifted, the result will be
di erent, for example:

Exercise 1.9.29

Suggest other graphs that match the description above.

1.9. The sequence of the sums: displacement 84

Exercise 1.9.30

Plot the location and the velocity for the following trip: I drove slowly, gradually speed up, stopped
for a very short moment, and started but in the opposite direction, quickly accelerated, and from that
point maintained the speed. Make up your own story and repeat the task.

Exercise 1.9.31

Draw a curve on a piece of paper, imagine that it represents your velocity, and then sketch what your
locations would look like. Repeat.

motionHere is another trivial statement about :

distance covered during the 1st hour

+ distance covered during the 2nd hour
= distance during the two hours

The statement is about the fact that when adding, we can change the order of terms freely; this is called

Associativity Propertythe of addition. At its simplest, it allows us to remove the parentheses:

(am + am+1 + ... + aq−1 + aq) + (aq+1 + aq+2 + ... + ar−1 + ar)

= am + am+1 + ... + aq−1 + aq + aq+1 + aq+2 + ... + ar−1 + ar

= am + am+1+ ... +ar−1 + ar .

The three sums are shown below:

An abbreviated version of this identity is as follows.

Theorem 1.9.32: Additivity for Sums

The sum of the sums of two consecutive segments of a sequence is the sum of
the combined segment.

In other words, for any sequence an and for any m, q, r with m ≤ q ≤ r, we have:

qr r

ak + ak = ak

k=m k=q+1 k=m

Example 1.9.33: Riemann sums of sequences

displacementHow do we deal with motion when the time moments aren't integers? What is the then?

x vSuppose n is the sequence of locations and n the sequence of velocities:

1.10. Sums of di erences and di erences of sums: motion 85

Riemann sumThen their vis de ned to be the sequence of sums of the sequence of the product of n

and the di erence of xn: n

vn ∆xn .

k=1

yWe know from the last section that if n is the position, then the velocity is vn = ∆yn/∆xn. There-

fore, the Riemann sum of the sequence of the velocity with respect to the sequence of time is the

displacement. Of course, this is also the total area of the rectangles.

Example 1.9.34: sum of random sequence

We take a computer-generated random sequence (the values spread between −1 and 1):

Then its sum is also random but might exhibit apparent periods of growth and decline ( streaks ).
The numbers in the original sequence can represent the outcome of playing a hand of cards with the

$1possibility of winning or losing an amount within . Then the sequence of sums represents the amount

the person has at every moment of time.

1.10. Sums of di erences and di erences of sums: motion

In this section, we start on the path of development of an idea that culminates with the cornerstone result

of calculus, the Fundamental Theorem of Calculus (Chapter 3IC-1).
We know that addition and subtraction undo each other ; it makes sense then that the operations of making

the sequence of di erences and making the sequence of sums will cancel each other too!

Example 1.10.1: broken odometer broken speedometer

We know how to get the velocity from the location, and the location from the velocity. We expect
that executing these two operations consecutively should bring us back where we started.

twoLet's take another look at the example of computations about motion a broken odometer and

1.10. Sums of di erences and di erences of sums: motion 86

a broken speedometer presented in the beginning of this chapter. The terminology has now been
developed: Every time we speak of a sequence, we also speak of the sequence of its di erences and
the sequence of its sums.

In the rst diagram, one rst takes the velocity data and acquires the displacements via the sums,
then someone else takes this displacement data and acquires the velocities by using the di erences:

We are back to the original sequence.

In the second diagram, one rst takes the location data and acquires the velocities via the di erences,
then someone else takes this velocity data and acquires the locations by using the sums:

We are back to the original sequence (provided we start at the same initial value).

Example 1.10.2: sequences given by lists

Below we have a sequence given by a list. We compute its sequence sums and then compute the

1.10. Sums of di erences and di erences of sums: motion 87

sequence of di erences of the result:

a sequence: 3 1 0 −2 . . .

↓ ↓ ↓ ↓ ...

its sums: 3

3+ 1 =4

4 + 0 = 4 ...

4 + (−2) = 2 . . .

2 ...

↓ ↓ ↓ ↓ ...

the sequence of sums: 3 4 4 2 ...

...

the di erences: 4−3 4−4 2−4 ...

|| || || . . .

a new sequence: 1 0 −2 . . .

We are back to the original sequence!

Exercise 1.10.3

What happened to the very rst term?

Exercise 1.10.4

Start with the sequence in the last example and use the diagrams to show that the sums of the
di erences give us the original sequence.

Example 1.10.5: sequences given by graphs

Just comparing the illustrations above demonstrates that the two operations the di erence and the
sum undo the e ect of each other. The two operations are shown together below:

As you can see in the picture, the sum (left to right) stacks up the terms of the sequence on top of
each other, while the di erence (right to left) takes these apart.

Let's take care of the algebra.

These are the two facts we will be using:

di erencea1. Suppose we have a sequence, n. We compute its , a new sequence:

bn+1 = an+1 − an .

sumc2. Suppose we have a sequence, k. We compute its , a new sequence:

n

dn = ck ,

k=1

or, recursively:

dn+1 = dn + cn+1 .

1.10. Sums of di erences and di erences of sums: motion 88

We use this setup to answer the following two questions.
The rst question we would like to answer is:

What is the di erence of the sum?

cWe start with n. Then, we have from (2) and (1), respectively:
dn+1 = dn + cn+1 and bn+1 = dn+1 − dn .

We substitute the rst formula into the second (and then cancel):

bn+1 = dn+1 − dn = (dn + cn+1) − dn = cn+1 .

As we can see, the answer is:

The original sequence.

The second question we would like to answer is:

What is the sum of the di erence?

aWe start with n. Then, we have from (1) and (2), respectively:

n

bk = ak − ak−1 and dn = bk .

k=1

We substitute the rst formula into the second (and then cancel):
nn

dn = bk = (ak − ak−1) = (a2 − a1) + (a3 − a2) + (a4 − a3) + ... + (an − an−1) = −a1 + an .

k=1 k=1

As we can see, the answer is:

The original sequence plus a number.

We summarize these results in the form of the following two far-reaching theorems.

Theorem 1.10.6: Fundamental Theorem of Calculus of Sequences I

nThe di erence of the sum of a sequence is that sequence; i.e., for all , we have:

n

∆ ak = an

k=1

cancelThe two operations each other!

Theorem 1.10.7: Fundamental Theorem of Calculus of Sequences II

The sum of the di erence of a sequence is that sequence plus a constant number;

i.e., for all n, we have:

n

∆bk = bn + C

k=1

The two operations almost cancel each other, again!

1.10. Sums of di erences and di erences of sums: motion 89

Example 1.10.8: fundamental theorems, computed

For larger sets of data, we use a spreadsheet. Recall the formulas:

• From a sequence to its sum:
a‚‘EI“gC‚g‘EI“

• From a sequence to its di erence:

a‚g‘EI“E‚‘EI“g‘EI“

What if we combine the two consecutively? From a sequence to its di erence to the sum of the latter:

It's the same curve! Now in the opposite order, from a sequence to its sum to the di erence of the
latter:

It's the same curve!

Exercise 1.10.9

What would the resulting curve look like if we started at another point?

Example 1.10.10: falling ball, acceleration

In an example from earlier in this chapter, we viewed the experimental data of the heights of a ping-
pong ball falling down:

1.10. Sums of di erences and di erences of sums: motion 90

locationJust as before, we use a spreadsheet to plot the psequence, n (green). We then compute the

velocitydi erence of pn, i.e., the , vn (purple):

It looks like a straight line. But this time, we take one more step: We compute the di erence of the

accelerationvelocity sequence. It is the a, n (blue). It appears constant! There might be a law of

nature here.

Example 1.10.11: shooting a cannon

Let's accept the premise put forward in the last example, that the acceleration of free fall is constant.

Then we can try to predict the behavior of an object shot in the air from any initial height and
with any initial velocity.

The direction of our computation is opposite to that of the last example: We assume that we know

the acceleration, then derive the velocity, and then derive the location (altitude) of the object in time.

While we used di erences in the last example, we use sums now:

100Above we show a projectile launched from a -meter tall building vertically up in the air with a

speed of 100 meters per second (the gravity causes acceleration of −9.8 meters per second squared).

1.11. The algebra of sums and di erences 91

20We can see that it will reach its highest point in about seconds and will hit the ground in about

40 seconds.

Exercise 1.10.12

How high does the projectile go in the above example?

Exercise 1.10.13

downUsing the above example, how long will it take for the projectile to reach the ground if red ?

Exercise 1.10.14

Use the above model to determine how long it will take for an object to reach the ground if it is
dropped. Make up your own questions about the situation and answer them. Repeat.

Exercise 1.10.15

Suppose the time moments are given by another sequence (an arithmetic progression). Compute the
velocity and the acceleration from the table below:

time height

n tn an
1 .00 36
2 .05 35
3 .10 32
4 .15 25
5 .20 20
6 .25 11
7 .30 0

This study of motion continues throughout the book.

1.11. The algebra of sums and di erences

di erences sumsWhat happens to their after an algebraic operation is carried out with a pair of
and their

sequences? There are a few shortcut properties.

motionHere is an elementary statement about :

sp „rixtwo runners are running away from a post, their relative velocity is the sum of

their respective velocities.

It's as if the one runner is standing still while the other is running with the combined speed:

addThe idea why we their di erences when we add sequences is illustrated below:

1.11. The algebra of sums and di erences 92

Here, the bars that represent the change of the values of the sequence are stacked on top of each other. The

heights are then added to each other, and so are the height di erences. The algebra behind this geometry

is very simple:

(A + B) − (a + b) = (A − a) + (B − b) .

It's the Associative Rule of addition.

The idea above is equally applicable to runners who change how fast they run; we speak of sequences:

Theorem 1.11.1: Sum Rule for Di erences

The di erence of the sum of two sequences is the sum of their di erences.

In other words, for any two sequences an, bn, their sequences of di erences satisfy:

∆(an + bn) = ∆an + ∆bn

Proof.

∆(an + bn) = (an+1 + bn+1) − (an + bn)
= (an+1 − an) + (bn+1 − bn)
= ∆an + ∆bn .

Example 1.11.2: di erence of sum

Consider the sum of an arithmetic progression and a geometric progression:

∆(a + mn + arn) = ∆(a + mn) + ∆(arn) = m + arn(r − 1) .

Now, di erences and sums are matched up according to the Fundamental Theorems of Calculus of Sequences

presented in the last section! It should be no surprise then that there is a matching theorem about sums.

When two sequences are added to each other, what happens to their sums? This simple algebra, the

Associative Property combined with the Commutative Property, tells the whole story:

a + b = (a + b),

re-arrange the sum of four numbers: ++ +

A + B = (A + B)

= (a + A) + (b + B) = a + A + b + B

The rule applies even if we have more than just two terms; it's about re-arranging terms of sequences:

ap + bp = (ap + bp)+

re-arrange the sum ap+1 + bp+1 = (ap+1 + bp+1)+
of two sequences:
... ... ... ...

aq + bq = (aq + bq)

= (ap + ... + aq) + (bp + ... + bq) = (ap + bp)+ ... +(aq + bq)

1.11. The algebra of sums and di erences 93

The summation is illustrated below:

An abbreviated version of this formula is as follows.

Theorem 1.11.3: Sum Rule for Sums

The sum of the sums of two sequences is the sum of the sequence of the sums.

In other words, if an and bn are sequences, then, for any p, q with p ≤ q, their

sequences of sums satisfy:

qqq

an + bn = (an + bn)

n=p n=p n=p

Exercise 1.11.4

Derive this theorem from the last one, reverse.

motionHere is another simple statement about :

sp „rixthe distance is re-scaled, such as from miles to kilometers, so is the velocity at

the same proportion.

proportionalThe idea why a change causes the same proportional change in the di erences is illustrated

below (tripling):

1.11. The algebra of sums and di erences 94

Here, if the heights triple, then so do the height di erences. The algebra behind this geometry is very simple:

kA − ka = k(A − a) .

It's the Distributive Rule. This is how it applies to sequences.

Theorem 1.11.5: Constant Multiple Rule for Di erences

The di erence of a multiple of a sequence is the multiple of the sequence's
di erence.

aIn other words, for any sequence n, the sequence of di erences satis es:

∆(kan) = k∆an

Proof.

∆(kan) = kan+1k − kan
= kan+1k − kan
= k∆an .

The theorem can also be interpreted as follows: If the distances are proportionally increased, then so are
the velocities needed to cover them, in the same period of time.

Is there a matching statement about sums? Yes, but let's rst look at motion again: If your velocity is

tripled, then so is the distance you have covered.

DistributiveWhen a sequence is multiplied by a constant, what happens to its sums? This simple algebra, the
Property, tells the whole story:

take out a common factor: k · ( a + b)
= ka + kb

The rule applies even if we have more than just two terms; it's about factoring:

k · ap = k · ap+

k · ap+1 = k · ap+1+

take out a common factor: ... ... ... ...

k · aq = k · aq

= k · ap+ ... +k · aq = k · (ap + ... + aq)

1.11. The algebra of sums and di erences 95

This summation is illustrated below:

An abbreviated version of this formula is as follows.

Theorem 1.11.6: Constant Multiple Rule for Sums

The sum of a multiple of a sequence is the multiple of its sum.

In other words, if an is a sequence, then for any p, q with p ≤ q and any real k,

its sequence of sums satis es:

qq

(kan) = k an

n=p n=p

Exercise 1.11.7

Derive this theorem from the last one, reverse.

Now we go beyond addition and multiplication by a constant.
Let's imagine this:

If two groups of runners are unfolding a tarp (or unfurling a ag) while running east and
north, respectively, what is happening to the area of this rectangle?
They may be running at di erent speeds:

areasThen, the product of two sequences is interpreted as the of the rectangles formed by the sequences,

illustrated below:

1.11. The algebra of sums and di erences 96

As the width and the depth are increasing, so is the area of the rectangle. We can see that the increase of
the area cannot be expressed entirely in terms of the increases of the width and depth! This increase is split
into two parts corresponding to the two terms in the right-hand side of the formula below.

Theorem 1.11.8: Product Rule for Di erences

The di erence of the product of two sequences is found as a combination of
these sequences and either of the two di erences.

Speci cally, for any two sequences an, bn, the sequence of di erences of the prod-

uct satis es:

∆(an · bn) = an+1 · ∆bn + ∆an · bn

Proof.

∆(an · bn) = an+1 · bn+1 − an · bn Insert terms.
= an+1 · bn+1 − an+1 · bn + an+1 · bn − an · bn Factor.
= an+1 · (bn+1) − bn) + (an+1 − an) · bn
= an+1 · ∆b + ∆an · bn .

Example 1.11.9: di erence of product

Consider the product of an arithmetic progression and a geometric progressions:

∆(mn · arn) = ∆(mn) · arn + mn∆(arn) = marn + mnarn(r − 1) .

Example 1.11.10: di erence of square

Product RuleThe di erence of the square sequence an = n2 is computed with the :

∆(n2) = ∆(n · n) = (n + 1) · ∆(n) + ∆(n) · (n) = (n + 1) + (n) = 2n + 1 .

It's an arithmetic progression, con rmed by the second graph below:

Exercise 1.11.11

a = nFind a formula for the di erence of the power sequence, nm, using the Binomial Theorem.

Example 1.11.12: di erence of sequence of reciprocals

Let's nd the formula for the di erence of the reciprocals:

1 = 1 − 1 = n − (n + 1) = − 1 .
∆
n n + 1 n n(n + 1) n(n + 1)

1.11. The algebra of sums and di erences 97

The sequence decreases but slower and slower:

Theorem 1.11.13: Quotient Rule for Di erences

The di erence of the quotient of two sequences is found as a combination of
these sequences and either of the two di erences.

Speci cally, for any two sequences an, bn, the sequence of di erences of the quo-

tient satis es:

∆ an = an+1 · ∆bn + ∆an · bn
bn bnbn+1

Exercise 1.11.14

Prove the theorem.

sumsThere are no matching statements for as simple as these two.

Here precalculus as a preview of calculus ends, and precalculus as a prerequisite for calculus starts.

Chapter 2: Sets and functions

Contents

2.1 Sets and relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
2.2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
2.3 Sequences are numerical functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
2.4 How numerical functions emerge: optimization . . . . . . . . . . . . . . . . . . . . . . . . . 117
2.5 Set building . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

2.6 The xy-plane: where graphs live... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

2.7 Linear relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
2.8 Relations vs. functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
2.9 A function as a black box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
2.10 Give the function a domain... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
2.11 The graph of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
2.12 Linear functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
2.13 Algebra creates functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
2.14 The image: the range of values of a function . . . . . . . . . . . . . . . . . . . . . . . . . . 192

2.1. Sets and relations

setIn mathematics, we refer to any loose collection of objects or entities of any nature as a .

For example, is this a circle of marbles that we see in a bag? No, the marbles it is made of aren't connected
to each other or to any location. One shake and the circle is gone:

Example 2.1.1: sets as lists

Sets given explicitly as lists are the simplest ones:

• A roster of students: Adams, Adkins, Arrows, ...
• A list of numbers: 1, 2, 3, 4, ...
• A list of planets: Mercury, Venus, Earth, Mars, ...

The order at which they appear on the list is not a part of the information we care about when we
speak of sets. Here is a bag of numbers:

2.1. Sets and relations 99

Example 2.1.2: sets

The idea of set contrasts with such expressions as a set of silverware when the word set suggests
a certain structure: speci c types of knives and forks with a speci c place in the box. It is the same
set, mathematically, whether the items are arranged in a box or piled up on the counter. A set of
encyclopedia consists of books that can be arranged alphabetically or chronologically or randomly.

Warning!

Even though we try to provide a precise de nition
of every new concept, the idea of set is so general
that we will have to rely on examples.

belongs or does not belongWhat creates a set is our knowledge or ability to determine whether an object to

it. A list is one such possibility. Another is a condition to be veri ed.

Example 2.1.3: sets via conditions

A roster of a class produces a set of the students in this class. It's a list! On the other hand, the

female students in the class also form a set even if there is no such list; we can just go down the roster

and determine if a student belongs to this new set. Similarly, the students with an A on the last test

also implicitly form a set.

Example 2.1.4: sets in math

numbers evenA lot of sets examined early in this book will be sets of
. For example, take the set of

numbers 2 3; then we know that belongs to it but does not. We simply check the condition: Is the

points2number divisible by ? Another example from familiar parts of mathematics is sets of on the

plane: straight lines, triangles, circles and other curves, etc.:

We can always tell whether a point belongs to the set!

Example 2.1.5: non-sets

If the condition is vague, we don't have a set: interesting novels , bad paintings , etc. When the
condition is nonsensical, we don't have a set either: fast trees , blue numbers , etc.