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Vincent Selhorst-Jones

Vincent Selhorst-Jones

Mathematical Induction

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Table of Contents

Section 1: Introduction
Introduction to Precalculus

10m 3s

Intro
0:00
Title of the Course
0:06
Different Names for the Course
0:07
Precalculus
0:12
Math Analysis
0:14
Trigonometry
0:16
Algebra III
0:20
Geometry II
0:24
College Algebra
0:30
Same Concepts
0:36
How do the Lessons Work?
0:54
Introducing Concepts
0:56
Apply Concepts
1:04
Go through Examples
1:25
Who is this Course For?
1:38
Those Who Need eExtra Help with Class Work
1:52
Those Working on Material but not in Formal Class at School
1:54
Those Who Want a Refresher
2:00
Try to Watch the Whole Lesson
2:20
Understanding is So Important
3:56
What to Watch First
5:26
Lesson #2: Sets, Elements, and Numbers
5:30
Lesson #7: Idea of a Function
5:33
Lesson #6: Word Problems
6:04
What to Watch First, cont.
6:46
Lesson #2: Sets, Elements and Numbers
6:56
Lesson #3: Variables, Equations, and Algebra
6:58
Lesson #4: Coordinate Systems
7:00
Lesson #5: Midpoint, Distance, the Pythagorean Theorem and Slope
7:02
Lesson #6: Word Problems
7:10
Lesson #7: Idea of a Function
7:12
Lesson #8: Graphs
7:14
Graphing Calculator Appendix
7:40
What to Watch Last
8:46
Let's get Started!
9:48
Sets, Elements, & Numbers

45m 11s

Intro
0:00
Introduction
0:05
Sets and Elements
1:19
Set
1:20
Element
1:23
Name a Set
2:20
Order The Elements Appear In Has No Effect on the Set
2:55
Describing/ Defining Sets
3:28
Directly Say All the Elements
3:36
Clearly Describing All the Members of the Set
3:55
Describing the Quality (or Qualities) Each member Of the Set Has In Common
4:32
Symbols: 'Element of' and 'Subset of'
6:01
Symbol is ∈
6:03
Subset Symbol is ⊂
6:35
Empty Set
8:07
Symbol is ∅
8:20
Since It's Empty, It is a Subset of All Sets
8:44
Union and Intersection
9:54
Union Symbol is ∪
10:08
Intersection Symbol is ∩
10:18
Sets Can Be Weird Stuff
12:26
Can Have Elements in a Set
12:50
We Can Have Infinite Sets
13:09
Example
13:22
Consider a Set Where We Take a Word and Then Repeat It An Ever Increasing Number of Times
14:08
This Set Has Infinitely Many Distinct Elements
14:40
Numbers as Sets
16:03
Natural Numbers ℕ
16:16
Including 0 and the Negatives ℤ
18:13
Rational Numbers ℚ
19:27
Can Express Rational Numbers with Decimal Expansions
22:05
Irrational Numbers
23:37
Real Numbers ℝ: Put the Rational and Irrational Numbers Together
25:15
Interval Notation and the Real Numbers
26:45
Include the End Numbers
27:06
Exclude the End Numbers
27:33
Example
28:28
Interval Notation: Infinity
29:09
Use -∞ or ∞ to Show an Interval Going on Forever in One Direction or the Other
29:14
Always Use Parentheses
29:50
Examples
30:27
Example 1
31:23
Example 2
35:26
Example 3
38:02
Example 4
42:21
Variables, Equations, & Algebra

35m 31s

Intro
0:00
What is a Variable?
0:05
A Variable is a Placeholder for a Number
0:11
Affects the Output of a Function or a Dependent Variable
0:24
Naming Variables
1:51
Useful to Use Symbols
2:21
What is a Constant?
4:14
A Constant is a Fixed, Unchanging Number
4:28
We Might Refer to a Symbol Representing a Number as a Constant
4:51
What is a Coefficient?
5:33
A Coefficient is a Multiplicative Factor on a Variable
5:37
Not All Coefficients are Constants
5:51
Expressions and Equations
6:42
An Expression is a String of Mathematical Symbols That Make Sense Used Together
7:05
An Equation is a Statement That Two Expression Have the Same Value
8:20
The Idea of Algebra
8:51
Equality
8:59
If Two Things Are the Same *Equal), Then We Can Do the Exact Same Operation to Both and the Results Will Be the Same
9:41
Always Do The Exact Same Thing to Both Sides
12:22
Solving Equations
13:23
When You Are Asked to Solve an Equation, You Are Being Asked to Solve for Something
13:33
Look For What Values Makes the Equation True
13:38
Isolate the Variable by Doing Algebra
14:37
Order of Operations
16:02
Why Certain Operations are Grouped
17:01
When You Don't Have to Worry About Order
17:39
Distributive Property
18:15
It Allows Multiplication to Act Over Addition in Parentheses
18:23
We Can Use the Distributive Property in Reverse to Combine Like Terms
19:05
Substitution
20:03
Use Information From One Equation in Another Equation
20:07
Put Your Substitution in Parentheses
20:44
Example 1
23:17
Example 2
25:49
Example 3
28:11
Example 4
30:02
Coordinate Systems

35m 2s

Intro
0:00
Inherent Order in ℝ
0:05
Real Numbers Come with an Inherent Order
0:11
Positive Numbers
0:21
Negative Numbers
0:58
'Less Than' and 'Greater Than'
2:04
Tip To Help You Remember the Signs
2:56
Inequality
4:06
Less Than or Equal and Greater Than or Equal
4:51
One Dimension: The Number Line
5:36
Graphically Represent ℝ on a Number Line
5:43
Note on Infinities
5:57
With the Number Line, We Can Directly See the Order We Put on ℝ
6:35
Ordered Pairs
7:22
Example
7:34
Allows Us to Talk About Two Numbers at the Same Time
9:41
Ordered Pairs of Real Numbers Cannot be Put Into an Order Like we Did with ℝ
10:41
Two Dimensions: The Plane
13:13
We Can Represent Ordered Pairs with the Plane
13:24
Intersection is known as the Origin
14:31
Plotting the Point
14:32
Plane = Coordinate Plane = Cartesian Plane = ℝ²
17:46
The Plane and Quadrants
18:50
Quadrant I
19:04
Quadrant II
19:21
Quadrant III
20:04
Quadrant IV
20:20
Three Dimensions: Space
21:02
Create Ordered Triplets
21:09
Visually Represent This
21:19
Three-Dimension = Space = ℝ³
21:47
Higher Dimensions
22:24
If We Have n Dimensions, We Call It n-Dimensional Space or ℝ to the nth Power
22:31
We Can Represent Places In This n-Dimensional Space As Ordered Groupings of n Numbers
22:41
Hard to Visualize Higher Dimensional Spaces
23:18
Example 1
25:07
Example 2
26:10
Example 3
28:58
Example 4
31:05
Midpoints, Distance, the Pythagorean Theorem, & Slope

48m 43s

Intro
0:00
Introduction
0:07
Midpoint: One Dimension
2:09
Example of Something More Complex
2:31
Use the Idea of a Middle
3:28
Find the Midpoint of Arbitrary Values a and b
4:17
How They're Equivalent
5:05
Official Midpoint Formula
5:46
Midpoint: Two Dimensions
6:19
The Midpoint Must Occur at the Horizontal Middle and the Vertical Middle
6:38
Arbitrary Pair of Points Example
7:25
Distance: One Dimension
9:26
Absolute Value
10:54
Idea of Forcing Positive
11:06
Distance: One Dimension, Formula
11:47
Distance Between Arbitrary a and b
11:48
Absolute Value Helps When the Distance is Negative
12:41
Distance Formula
12:58
The Pythagorean Theorem
13:24
a²+b²=c²
13:50
Distance: Two Dimensions
14:59
Break Into Horizontal and Vertical Parts and then Use the Pythagorean Theorem
15:16
Distance Between Arbitrary Points (x₁,y₁) and (x₂,y₂)
16:21
Slope
19:30
Slope is the Rate of Change
19:41
m = rise over run
21:27
Slope Between Arbitrary Points (x₁,y₁) and (x₂,y₂)
22:31
Interpreting Slope
24:12
Positive Slope and Negative Slope
25:40
m=1, m=0, m=-1
26:48
Example 1
28:25
Example 2
31:42
Example 3
36:40
Example 4
42:48
Word Problems

56m 31s

Intro
0:00
Introduction
0:05
What is a Word Problem?
0:45
Describes Any Problem That Primarily Gets Its Ideas Across With Words Instead of Math Symbols
0:48
Requires Us to Think
1:32
Why Are They So Hard?
2:11
Reason 1: No Simple Formula to Solve Them
2:16
Reason 2: Harder to Teach Word Problems
2:47
You Can Learn How to Do Them!
3:51
Grades
7:57
'But I'm Never Going to Use This In Real Life'
9:46
Solving Word Problems
12:58
First: Understand the Problem
13:37
Second: What Are You Looking For?
14:33
Third: Set Up Relationships
16:21
Fourth: Solve It!
17:48
Summary of Method
19:04
Examples on Things Other Than Math
20:21
Math-Specific Method: What You Need Now
25:30
Understand What the Problem is Talking About
25:37
Set Up and Name Any Variables You Need to Know
25:56
Set Up Equations Connecting Those Variables to the Information in the Problem Statement
26:02
Use the Equations to Solve for an Answer
26:14
Tip
26:58
Draw Pictures
27:22
Breaking Into Pieces
28:28
Try Out Hypothetical Numbers
29:52
Student Logic
31:27
Jump In!
32:40
Example 1
34:03
Example 2
39:15
Example 3
44:22
Example 4
50:24
Section 2: Functions
Idea of a Function

39m 54s

Intro
0:00
Introduction
0:04
What is a Function?
1:06
A Visual Example and Non-Example
1:30
Function Notation
3:47
f(x)
4:05
Express What Sets the Function Acts On
5:45
Metaphors for a Function
6:17
Transformation
6:28
Map
7:17
Machine
8:56
Same Input Always Gives Same Output
10:01
If We Put the Same Input Into a Function, It Will Always Produce the Same Output
10:11
Example of Something That is Not a Function
11:10
A Non-Numerical Example
12:10
The Functions We Will Use
15:05
Unless Told Otherwise, We Will Assume Every Function Takes in Real Numbers and Outputs Real Numbers
15:11
Usually Told the Rule of a Given Function
15:27
How To Use a Function
16:18
Apply the Rule to Whatever Our Input Value Is
16:28
Make Sure to Wrap Your Substitutions in Parentheses
17:09
Functions and Tables
17:36
Table of Values, Sometimes Called a T-Table
17:46
Example
17:56
Domain: What Goes In
18:55
The Domain is the Set of all Inputs That the Function Can Accept
18:56
Example
19:40
Range: What Comes Out
21:27
The Range is the Set of All Possible Outputs a Function Can Assign
21:34
Example
21:49
Another Example Would Be Our Initial Function From Earlier in This Lesson
22:29
Example 1
23:45
Example 2
25:22
Example 3
27:27
Example 4
29:23
Example 5
33:33
Graphs

58m 26s

Intro
0:00
Introduction
0:04
How to Interpret Graphs
1:17
Input / Independent Variable
1:47
Output / Dependent Variable
2:00
Graph as Input ⇒ Output
2:23
One Way to Think of a Graph: See What Happened to Various Inputs
2:25
Example
2:47
Graph as Location of Solution
4:20
A Way to See Solutions
4:36
Example
5:20
Which Way Should We Interpret?
7:13
Easiest to Think In Terms of How Inputs Are Mapped to Outputs
7:20
Sometimes It's Easier to Think In Terms of Solutions
8:39
Pay Attention to Axes
9:50
Axes Tell Where the Graph Is and What Scale It Has
10:09
Often, The Axes Will Be Square
10:14
Example
12:06
Arrows or No Arrows?
16:07
Will Not Use Arrows at the End of Our Graphs
17:13
Graph Stops Because It Hits the Edge of the Graphing Axes, Not Because the Function Stops
17:18
How to Graph
19:47
Plot Points
20:07
Connect with Curves
21:09
If You Connect with Straight Lines
21:44
Graphs of Functions are Smooth
22:21
More Points ⇒ More Accurate
23:38
Vertical Line Test
27:44
If a Vertical Line Could Intersect More Than One Point On a Graph, It Can Not Be the Graph of a Function
28:41
Every Point on a Graph Tells Us Where the x-Value Below is Mapped
30:07
Domain in Graphs
31:37
The Domain is the Set of All Inputs That a Function Can Accept
31:44
Be Aware That Our Function Probably Continues Past the Edge of Our 'Viewing Window'
33:19
Range in Graphs
33:53
Graphing Calculators: Check the Appendix!
36:55
Example 1
38:37
Example 2
45:19
Example 3
50:41
Example 4
53:28
Example 5
55:50
Properties of Functions

48m 49s

Intro
0:00
Introduction
0:05
Increasing Decreasing Constant
0:43
Looking at a Specific Graph
1:15
Increasing Interval
2:39
Constant Function
4:15
Decreasing Interval
5:10
Find Intervals by Looking at the Graph
5:32
Intervals Show x-values; Write in Parentheses
6:39
Maximum and Minimums
8:48
Relative (Local) Max/Min
10:20
Formal Definition of Relative Maximum
12:44
Formal Definition of Relative Minimum
13:05
Max/Min, More Terms
14:18
Definition of Extrema
15:01
Average Rate of Change
16:11
Drawing a Line for the Average Rate
16:48
Using the Slope of the Secant Line
17:36
Slope in Function Notation
18:45
Zeros/Roots/x-intercepts
19:45
What Zeros in a Function Mean
20:25
Even Functions
22:30
Odd Functions
24:36
Even/Odd Functions and Graphs
26:28
Example of an Even Function
27:12
Example of an Odd Function
28:03
Example 1
29:35
Example 2
33:07
Example 3
40:32
Example 4
42:34
Function Petting Zoo

29m 20s

Intro
0:00
Introduction
0:04
Don't Forget that Axes Matter!
1:44
The Constant Function
2:40
The Identity Function
3:44
The Square Function
4:40
The Cube Function
5:44
The Square Root Function
6:51
The Reciprocal Function
8:11
The Absolute Value Function
10:19
The Trigonometric Functions
11:56
f(x)=sin(x)
12:12
f(x)=cos(x)
12:24
Alternate Axes
12:40
The Exponential and Logarithmic Functions
13:35
Exponential Functions
13:44
Logarithmic Functions
14:24
Alternating Axes
15:17
Transformations and Compositions
16:08
Example 1
17:52
Example 2
18:33
Example 3
20:24
Example 4
26:07
Transformation of Functions

48m 35s

Intro
0:00
Introduction
0:04
Vertical Shift
1:12
Graphical Example
1:21
A Further Explanation
2:16
Vertical Stretch/Shrink
3:34
Graph Shrinks
3:46
Graph Stretches
3:51
A Further Explanation
5:07
Horizontal Shift
6:49
Moving the Graph to the Right
7:28
Moving the Graph to the Left
8:12
A Further Explanation
8:19
Understanding Movement on the x-axis
8:38
Horizontal Stretch/Shrink
12:59
Shrinking the Graph
13:40
Stretching the Graph
13:48
A Further Explanation
13:55
Understanding Stretches from the x-axis
14:12
Vertical Flip (aka Mirror)
16:55
Example Graph
17:07
Multiplying the Vertical Component by -1
17:18
Horizontal Flip (aka Mirror)
18:43
Example Graph
19:01
Multiplying the Horizontal Component by -1
19:54
Summary of Transformations
22:11
Stacking Transformations
24:46
Order Matters
25:20
Transformation Example
25:52
Example 1
29:21
Example 2
34:44
Example 3
38:10
Example 4
43:46
Composite Functions

33m 24s

Intro
0:00
Introduction
0:04
Arithmetic Combinations
0:40
Basic Operations
1:20
Definition of the Four Arithmetic Combinations
1:40
Composite Functions
2:53
The Function as a Machine
3:32
Function Compositions as Multiple Machines
3:59
Notation for Composite Functions
4:46
Two Formats
6:02
Another Visual Interpretation
7:17
How to Use Composite Functions
8:21
Example of on Function acting on Another
9:17
Example 1
11:03
Example 2
15:27
Example 3
21:11
Example 4
27:06
Piecewise Functions

51m 42s

Intro
0:00
Introduction
0:04
Analogies to a Piecewise Function
1:16
Different Potatoes
1:41
Factory Production
2:27
Notations for Piecewise Functions
3:39
Notation Examples from Analogies
6:11
Example of a Piecewise (with Table)
7:24
Example of a Non-Numerical Piecewise
11:35
Graphing Piecewise Functions
14:15
Graphing Piecewise Functions, Example
16:26
Continuous Functions
16:57
Statements of Continuity
19:30
Example of Continuous and Non-Continuous Graphs
20:05
Interesting Functions: the Step Function
22:00
Notation for the Step Function
22:40
How the Step Function Works
22:56
Graph of the Step Function
25:30
Example 1
26:22
Example 2
28:49
Example 3
36:50
Example 4
46:11
Inverse Functions

49m 37s

Intro
0:00
Introduction
0:04
Analogy by picture
1:10
How to Denote the inverse
1:40
What Comes out of the Inverse
1:52
Requirement for Reversing
2:02
The Basketball Factory
2:12
The Importance of Information
2:45
One-to-One
4:04
Requirement for Reversibility
4:21
When a Function has an Inverse
4:43
One-to-One
5:13
Not One-to-One
5:50
Not a Function
6:19
Horizontal Line Test
7:01
How to the test Works
7:12
One-to-One
8:12
Not One-to-One
8:45
Definition: Inverse Function
9:12
Formal Definition
9:21
Caution to Students
10:02
Domain and Range
11:12
Finding the Range of the Function Inverse
11:56
Finding the Domain of the Function Inverse
12:11
Inverse of an Inverse
13:09
Its just x!
13:26
Proof
14:03
Graphical Interpretation
17:07
Horizontal Line Test
17:20
Graph of the Inverse
18:04
Swapping Inputs and Outputs to Draw Inverses
19:02
How to Find the Inverse
21:03
What We Are Looking For
21:21
Reversing the Function
21:38
A Method to Find Inverses
22:33
Check Function is One-to-One
23:04
Swap f(x) for y
23:25
Interchange x and y
23:41
Solve for y
24:12
Replace y with the inverse
24:40
Some Comments
25:01
Keeping Step 2 and 3 Straight
25:44
Switching to Inverse
26:12
Checking Inverses
28:52
How to Check an Inverse
29:06
Quick Example of How to Check
29:56
Example 1
31:48
Example 2
34:56
Example 3
39:29
Example 4
46:19
Variation Direct and Inverse

28m 49s

Intro
0:00
Introduction
0:06
Direct Variation
1:14
Same Direction
1:21
Common Example: Groceries
1:56
Different Ways to Say that Two Things Vary Directly
2:28
Basic Equation for Direct Variation
2:55
Inverse Variation
3:40
Opposite Direction
3:50
Common Example: Gravity
4:53
Different Ways to Say that Two Things Vary Indirectly
5:48
Basic Equation for Indirect Variation
6:33
Joint Variation
7:27
Equation for Joint Variation
7:53
Explanation of the Constant
8:48
Combined Variation
9:35
Gas Law as a Combination
9:44
Single Constant
10:33
Example 1
10:49
Example 2
13:34
Example 3
15:39
Example 4
19:48
Section 3: Polynomials
Intro to Polynomials

38m 41s

Intro
0:00
Introduction
0:04
Definition of a Polynomial
1:04
Starting Integer
2:06
Structure of a Polynomial
2:49
The a Constants
3:34
Polynomial Function
5:13
Polynomial Equation
5:23
Polynomials with Different Variables
5:36
Degree
6:23
Informal Definition
6:31
Find the Largest Exponent Variable
6:44
Quick Examples
7:36
Special Names for Polynomials
8:59
Based on the Degree
9:23
Based on the Number of Terms
10:12
Distributive Property (aka 'FOIL')
11:37
Basic Distributive Property
12:21
Distributing Two Binomials
12:55
Longer Parentheses
15:12
Reverse: Factoring
17:26
Long-Term Behavior of Polynomials
17:48
Examples
18:13
Controlling Term--Term with the Largest Exponent
19:33
Positive and Negative Coefficients on the Controlling Term
20:21
Leading Coefficient Test
22:07
Even Degree, Positive Coefficient
22:13
Even Degree, Negative Coefficient
22:39
Odd Degree, Positive Coefficient
23:09
Odd Degree, Negative Coefficient
23:27
Example 1
25:11
Example 2
27:16
Example 3
31:16
Example 4
34:41
Roots (Zeros) of Polynomials

41m 7s

Intro
0:00
Introduction
0:05
Roots in Graphs
1:17
The x-intercepts
1:33
How to Remember What 'Roots' Are
1:50
Naïve Attempts
2:31
Isolating Variables
2:45
Failures of Isolating Variables
3:30
Missing Solutions
4:59
Factoring: How to Find Roots
6:28
How Factoring Works
6:36
Why Factoring Works
7:20
Steps to Finding Polynomial Roots
9:21
Factoring: How to Find Roots CAUTION
10:08
Factoring is Not Easy
11:32
Factoring Quadratics
13:08
Quadratic Trinomials
13:21
Form of Factored Binomials
13:38
Factoring Examples
14:40
Factoring Quadratics, Check Your Work
16:58
Factoring Higher Degree Polynomials
18:19
Factoring a Cubic
18:32
Factoring a Quadratic
19:04
Factoring: Roots Imply Factors
19:54
Where a Root is, A Factor Is
20:01
How to Use Known Roots to Make Factoring Easier
20:35
Not all Polynomials Can be Factored
22:30
Irreducible Polynomials
23:27
Complex Numbers Help
23:55
Max Number of Roots/Factors
24:57
Limit to Number of Roots Equal to the Degree
25:18
Why there is a Limit
25:25
Max Number of Peaks/Valleys
26:39
Shape Information from Degree
26:46
Example Graph
26:54
Max, But Not Required
28:00
Example 1
28:37
Example 2
31:21
Example 3
36:12
Example 4
38:40
Completing the Square and the Quadratic Formula

39m 43s

Intro
0:00
Introduction
0:05
Square Roots and Equations
0:51
Taking the Square Root to Find the Value of x
0:55
Getting the Positive and Negative Answers
1:05
Completing the Square: Motivation
2:04
Polynomials that are Easy to Solve
2:20
Making Complex Polynomials Easy to Solve
3:03
Steps to Completing the Square
4:30
Completing the Square: Method
7:22
Move C over
7:35
Divide by A
7:44
Find r
7:59
Add to Both Sides to Complete the Square
8:49
Solving Quadratics with Ease
9:56
The Quadratic Formula
11:38
Derivation
11:43
Final Form
12:23
Follow Format to Use Formula
13:38
How Many Roots?
14:53
The Discriminant
15:47
What the Discriminant Tells Us: How Many Roots
15:58
How the Discriminant Works
16:30
Example 1: Complete the Square
18:24
Example 2: Solve the Quadratic
22:00
Example 3: Solve for Zeroes
25:28
Example 4: Using the Quadratic Formula
30:52
Properties of Quadratic Functions

45m 34s

Intro
0:00
Introduction
0:05
Parabolas
0:35
Examples of Different Parabolas
1:06
Axis of Symmetry and Vertex
1:28
Drawing an Axis of Symmetry
1:51
Placing the Vertex
2:28
Looking at the Axis of Symmetry and Vertex for other Parabolas
3:09
Transformations
4:18
Reviewing Transformation Rules
6:28
Note the Different Horizontal Shift Form
7:45
An Alternate Form to Quadratics
8:54
The Constants: k, h, a
9:05
Transformations Formed
10:01
Analyzing Different Parabolas
10:10
Switching Forms by Completing the Square
11:43
Vertex of a Parabola
16:30
Vertex at (h, k)
16:47
Vertex in Terms of a, b, and c Coefficients
17:28
Minimum/Maximum at Vertex
18:19
When a is Positive
18:25
When a is Negative
18:52
Axis of Symmetry
19:54
Incredibly Minor Note on Grammar
20:52
Example 1
21:48
Example 2
26:35
Example 3
28:55
Example 4
31:40
Intermediate Value Theorem and Polynomial Division

46m 8s

Intro
0:00
Introduction
0:05
Reminder: Roots Imply Factors
1:32
The Intermediate Value Theorem
3:41
The Basis: U between a and b
4:11
U is on the Function
4:52
Intermediate Value Theorem, Proof Sketch
5:51
If Not True, the Graph Would Have to Jump
5:58
But Graph is Defined as Continuous
6:43
Finding Roots with the Intermediate Value Theorem
7:01
Picking a and b to be of Different Signs
7:10
Must Be at Least One Root
7:46
Dividing a Polynomial
8:16
Using Roots and Division to Factor
8:38
Long Division Refresher
9:08
The Division Algorithm
12:18
How It Works to Divide Polynomials
12:37
The Parts of the Equation
13:24
Rewriting the Equation
14:47
Polynomial Long Division
16:20
Polynomial Long Division In Action
16:29
One Step at a Time
20:51
Synthetic Division
22:46
Setup
23:11
Synthetic Division, Example
24:44
Which Method Should We Use
26:39
Advantages of Synthetic Method
26:49
Advantages of Long Division
27:13
Example 1
29:24
Example 2
31:27
Example 3
36:22
Example 4
40:55
Complex Numbers

45m 36s

Intro
0:00
Introduction
0:04
A Wacky Idea
1:02
The Definition of the Imaginary Number
1:22
How it Helps Solve Equations
2:20
Square Roots and Imaginary Numbers
3:15
Complex Numbers
5:00
Real Part and Imaginary Part
5:20
When Two Complex Numbers are Equal
6:10
Addition and Subtraction
6:40
Deal with Real and Imaginary Parts Separately
7:36
Two Quick Examples
7:54
Multiplication
9:07
FOIL Expansion
9:14
Note What Happens to the Square of the Imaginary Number
9:41
Two Quick Examples
10:22
Division
11:27
Complex Conjugates
13:37
Getting Rid of i
14:08
How to Denote the Conjugate
14:48
Division through Complex Conjugates
16:11
Multiply by the Conjugate of the Denominator
16:28
Example
17:46
Factoring So-Called 'Irreducible' Quadratics
19:24
Revisiting the Quadratic Formula
20:12
Conjugate Pairs
20:37
But Are the Complex Numbers 'Real'?
21:27
What Makes a Number Legitimate
25:38
Where Complex Numbers are Used
27:20
Still, We Won't See Much of C
29:05
Example 1
30:30
Example 2
33:15
Example 3
38:12
Example 4
42:07
Fundamental Theorem of Algebra

19m 9s

Intro
0:00
Introduction
0:05
Idea: Hidden Roots
1:16
Roots in Complex Form
1:42
All Polynomials Have Roots
2:08
Fundamental Theorem of Algebra
2:21
Where Are All the Imaginary Roots, Then?
3:17
All Roots are Complex
3:45
Real Numbers are a Subset of Complex Numbers
3:59
The n Roots Theorem
5:01
For Any Polynomial, Its Degree is Equal to the Number of Roots
5:11
Equivalent Statement
5:24
Comments: Multiplicity
6:29
Non-Distinct Roots
6:59
Denoting Multiplicity
7:20
Comments: Complex Numbers Necessary
7:41
Comments: Complex Coefficients Allowed
8:55
Comments: Existence Theorem
9:59
Proof Sketch of n Roots Theorem
10:45
First Root
11:36
Second Root
13:23
Continuation to Find all Roots
16:00
Section 4: Rational Functions
Rational Functions and Vertical Asymptotes

33m 22s

Intro
0:00
Introduction
0:05
Definition of a Rational Function
1:20
Examples of Rational Functions
2:30
Why They are Called 'Rational'
2:47
Domain of a Rational Function
3:15
Undefined at Denominator Zeros
3:25
Otherwise all Reals
4:16
Investigating a Fundamental Function
4:50
The Domain of the Function
5:04
What Occurs at the Zeroes of the Denominator
5:20
Idea of a Vertical Asymptote
6:23
What's Going On?
6:58
Approaching x=0 from the left
7:32
Approaching x=0 from the right
8:34
Dividing by Very Small Numbers Results in Very Large Numbers
9:31
Definition of a Vertical Asymptote
10:05
Vertical Asymptotes and Graphs
11:15
Drawing Asymptotes by Using a Dashed Line
11:27
The Graph Can Never Touch Its Undefined Point
12:00
Not All Zeros Give Asymptotes
13:02
Special Cases: When Numerator and Denominator Go to Zero at the Same Time
14:58
Cancel out Common Factors
15:49
How to Find Vertical Asymptotes
16:10
Figure out What Values Are Not in the Domain of x
16:24
Determine if the Numerator and Denominator Share Common Factors and Cancel
16:45
Find Denominator Roots
17:33
Note if Asymptote Approaches Negative or Positive Infinity
18:06
Example 1
18:57
Example 2
21:26
Example 3
23:04
Example 4
30:01
Horizontal Asymptotes

34m 16s

Intro
0:00
Introduction
0:05
Investigating a Fundamental Function
0:53
What Happens as x Grows Large
1:00
Different View
1:12
Idea of a Horizontal Asymptote
1:36
What's Going On?
2:24
What Happens as x Grows to a Large Negative Number
2:49
What Happens as x Grows to a Large Number
3:30
Dividing by Very Large Numbers Results in Very Small Numbers
3:52
Example Function
4:41
Definition of a Vertical Asymptote
8:09
Expanding the Idea
9:03
What's Going On?
9:48
What Happens to the Function in the Long Run?
9:51
Rewriting the Function
10:13
Definition of a Slant Asymptote
12:09
Symbolical Definition
12:30
Informal Definition
12:45
Beyond Slant Asymptotes
13:03
Not Going Beyond Slant Asymptotes
14:39
Horizontal/Slant Asymptotes and Graphs
15:43
How to Find Horizontal and Slant Asymptotes
16:52
How to Find Horizontal Asymptotes
17:12
Expand the Given Polynomials
17:18
Compare the Degrees of the Numerator and Denominator
17:40
How to Find Slant Asymptotes
20:05
Slant Asymptotes Exist When n+m=1
20:08
Use Polynomial Division
20:24
Example 1
24:32
Example 2
25:53
Example 3
26:55
Example 4
29:22
Graphing Asymptotes in a Nutshell

49m 7s

Intro
0:00
Introduction
0:05
A Process for Graphing
1:22
1. Factor Numerator and Denominator
1:50
2. Find Domain
2:53
3. Simplifying the Function
3:59
4. Find Vertical Asymptotes
4:59
5. Find Horizontal/Slant Asymptotes
5:24
6. Find Intercepts
7:35
7. Draw Graph (Find Points as Necessary)
9:21
Draw Graph Example
11:21
Vertical Asymptote
11:41
Horizontal Asymptote
11:50
Other Graphing
12:16
Test Intervals
15:08
Example 1
17:57
Example 2
23:01
Example 3
29:02
Example 4
33:37
Partial Fractions

44m 56s

Intro
0:00
Introduction: Idea
0:04
Introduction: Prerequisites and Uses
1:57
Proper vs. Improper Polynomial Fractions
3:11
Possible Things in the Denominator
4:38
Linear Factors
6:16
Example of Linear Factors
7:03
Multiple Linear Factors
7:48
Irreducible Quadratic Factors
8:25
Example of Quadratic Factors
9:26
Multiple Quadratic Factors
9:49
Mixing Factor Types
10:28
Figuring Out the Numerator
11:10
How to Solve for the Constants
11:30
Quick Example
11:40
Example 1
14:29
Example 2
18:35
Example 3
20:33
Example 4
28:51
Section 5: Exponential & Logarithmic Functions
Understanding Exponents

35m 17s

Intro
0:00
Introduction
0:05
Fundamental Idea
1:46
Expanding the Idea
2:28
Multiplication of the Same Base
2:40
Exponents acting on Exponents
3:45
Different Bases with the Same Exponent
4:31
To the Zero
5:35
To the First
5:45
Fundamental Rule with the Zero Power
6:35
To the Negative
7:45
Any Number to a Negative Power
8:14
A Fraction to a Negative Power
9:58
Division with Exponential Terms
10:41
To the Fraction
11:33
Square Root
11:58
Any Root
12:59
Summary of Rules
14:38
To the Irrational
17:21
Example 1
20:34
Example 2
23:42
Example 3
27:44
Example 4
31:44
Example 5
33:15
Exponential Functions

47m 4s

Intro
0:00
Introduction
0:05
Definition of an Exponential Function
0:48
Definition of the Base
1:02
Restrictions on the Base
1:16
Computing Exponential Functions
2:29
Harder Computations
3:10
When to Use a Calculator
3:21
Graphing Exponential Functions: a>1
6:02
Three Examples
6:13
What to Notice on the Graph
7:44
A Story
8:27
Story Diagram
9:15
Increasing Exponentials
11:29
Story Morals
14:40
Application: Compound Interest
15:15
Compounding Year after Year
16:01
Function for Compounding Interest
16:51
A Special Number: e
20:55
Expression for e
21:28
Where e stabilizes
21:55
Application: Continuously Compounded Interest
24:07
Equation for Continuous Compounding
24:22
Exponential Decay 0<a<1
25:50
Three Examples
26:11
Why they 'lose' value
26:54
Example 1
27:47
Example 2
33:11
Example 3
36:34
Example 4
41:28
Introduction to Logarithms

40m 31s

Intro
0:00
Introduction
0:04
Definition of a Logarithm, Base 2
0:51
Log 2 Defined
0:55
Examples
2:28
Definition of a Logarithm, General
3:23
Examples of Logarithms
5:15
Problems with Unusual Bases
7:38
Shorthand Notation: ln and log
9:44
base e as ln
10:01
base 10 as log
10:34
Calculating Logarithms
11:01
using a calculator
11:34
issues with other bases
11:58
Graphs of Logarithms
13:21
Three Examples
13:29
Slow Growth
15:19
Logarithms as Inverse of Exponentiation
16:02
Using Base 2
16:05
General Case
17:10
Looking More Closely at Logarithm Graphs
19:16
The Domain of Logarithms
20:41
Thinking about Logs like Inverses
21:08
The Alternate
24:00
Example 1
25:59
Example 2
30:03
Example 3
32:49
Example 4
37:34
Properties of Logarithms

42m 33s

Intro
0:00
Introduction
0:04
Basic Properties
1:12
Inverse--log(exp)
1:43
A Key Idea
2:44
What We Get through Exponentiation
3:18
B Always Exists
4:50
Inverse--exp(log)
5:53
Logarithm of a Power
7:44
Logarithm of a Product
10:07
Logarithm of a Quotient
13:48
Caution! There Is No Rule for loga(M+N)
16:12
Summary of Properties
17:42
Change of Base--Motivation
20:17
No Calculator Button
20:59
A Specific Example
21:45
Simplifying
23:45
Change of Base--Formula
24:14
Example 1
25:47
Example 2
29:08
Example 3
31:14
Example 4
34:13
Solving Exponential and Logarithmic Equations

34m 10s

Intro
0:00
Introduction
0:05
One to One Property
1:09
Exponential
1:26
Logarithmic
1:44
Specific Considerations
2:02
One-to-One Property
3:30
Solving by One-to-One
4:11
Inverse Property
6:09
Solving by Inverses
7:25
Dealing with Equations
7:50
Example of Taking an Exponent or Logarithm of an Equation
9:07
A Useful Property
11:57
Bring Down Exponents
12:01
Try to Simplify
13:20
Extraneous Solutions
13:45
Example 1
16:37
Example 2
19:39
Example 3
21:37
Example 4
26:45
Example 5
29:37
Application of Exponential and Logarithmic Functions

48m 46s

Intro
0:00
Introduction
0:06
Applications of Exponential Functions
1:07
A Secret!
2:17
Natural Exponential Growth Model
3:07
Figure out r
3:34
A Secret!--Why Does It Work?
4:44
e to the r Morphs
4:57
Example
5:06
Applications of Logarithmic Functions
8:32
Examples
8:43
What Logarithms are Useful For
9:53
Example 1
11:29
Example 2
15:30
Example 3
26:22
Example 4
32:05
Example 5
39:19
Section 6: Trigonometric Functions
Angles

39m 5s

Intro
0:00
Degrees
0:22
Circle is 360 Degrees
0:48
Splitting a Circle
1:13
Radians
2:08
Circle is 2 Pi Radians
2:31
One Radian
2:52
Half-Circle and Right Angle
4:00
Converting Between Degrees and Radians
6:24
Formulas for Degrees and Radians
6:52
Coterminal, Complementary, Supplementary Angles
7:23
Coterminal Angles
7:30
Complementary Angles
9:40
Supplementary Angles
10:08
Example 1: Dividing a Circle
10:38
Example 2: Converting Between Degrees and Radians
11:56
Example 3: Quadrants and Coterminal Angles
14:18
Extra Example 1: Common Angle Conversions
-1
Extra Example 2: Quadrants and Coterminal Angles
-2
Sine and Cosine Functions

43m 16s

Intro
0:00
Sine and Cosine
0:15
Unit Circle
0:22
Coordinates on Unit Circle
1:03
Right Triangles
1:52
Adjacent, Opposite, Hypotenuse
2:25
Master Right Triangle Formula: SOHCAHTOA
2:48
Odd Functions, Even Functions
4:40
Example: Odd Function
4:56
Example: Even Function
7:30
Example 1: Sine and Cosine
10:27
Example 2: Graphing Sine and Cosine Functions
14:39
Example 3: Right Triangle
21:40
Example 4: Odd, Even, or Neither
26:01
Extra Example 1: Right Triangle
-1
Extra Example 2: Graphing Sine and Cosine Functions
-2
Sine and Cosine Values of Special Angles

33m 5s

Intro
0:00
45-45-90 Triangle and 30-60-90 Triangle
0:08
45-45-90 Triangle
0:21
30-60-90 Triangle
2:06
Mnemonic: All Students Take Calculus (ASTC)
5:21
Using the Unit Circle
5:59
New Angles
6:21
Other Quadrants
9:43
Mnemonic: All Students Take Calculus
10:13
Example 1: Convert, Quadrant, Sine/Cosine
13:11
Example 2: Convert, Quadrant, Sine/Cosine
16:48
Example 3: All Angles and Quadrants
20:21
Extra Example 1: Convert, Quadrant, Sine/Cosine
-1
Extra Example 2: All Angles and Quadrants
-2
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D

52m 3s

Intro
0:00
Amplitude and Period of a Sine Wave
0:38
Sine Wave Graph
0:58
Amplitude: Distance from Middle to Peak
1:18
Peak: Distance from Peak to Peak
2:41
Phase Shift and Vertical Shift
4:13
Phase Shift: Distance Shifted Horizontally
4:16
Vertical Shift: Distance Shifted Vertically
6:48
Example 1: Amplitude/Period/Phase and Vertical Shift
8:04
Example 2: Amplitude/Period/Phase and Vertical Shift
17:39
Example 3: Find Sine Wave Given Attributes
25:23
Extra Example 1: Amplitude/Period/Phase and Vertical Shift
-1
Extra Example 2: Find Cosine Wave Given Attributes
-2
Tangent and Cotangent Functions

36m 4s

Intro
0:00
Tangent and Cotangent Definitions
0:21
Tangent Definition
0:25
Cotangent Definition
0:47
Master Formula: SOHCAHTOA
1:01
Mnemonic
1:16
Tangent and Cotangent Values
2:29
Remember Common Values of Sine and Cosine
2:46
90 Degrees Undefined
4:36
Slope and Menmonic: ASTC
5:47
Uses of Tangent
5:54
Example: Tangent of Angle is Slope
6:09
Sign of Tangent in Quadrants
7:49
Example 1: Graph Tangent and Cotangent Functions
10:42
Example 2: Tangent and Cotangent of Angles
16:09
Example 3: Odd, Even, or Neither
18:56
Extra Example 1: Tangent and Cotangent of Angles
-1
Extra Example 2: Tangent and Cotangent of Angles
-2
Secant and Cosecant Functions

27m 18s

Intro
0:00
Secant and Cosecant Definitions
0:17
Secant Definition
0:18
Cosecant Definition
0:33
Example 1: Graph Secant Function
0:48
Example 2: Values of Secant and Cosecant
6:49
Example 3: Odd, Even, or Neither
12:49
Extra Example 1: Graph of Cosecant Function
-1
Extra Example 2: Values of Secant and Cosecant
-2
Inverse Trigonometric Functions

32m 58s

Intro
0:00
Arcsine Function
0:24
Restrictions between -1 and 1
0:43
Arcsine Notation
1:26
Arccosine Function
3:07
Restrictions between -1 and 1
3:36
Cosine Notation
3:53
Arctangent Function
4:30
Between -Pi/2 and Pi/2
4:44
Tangent Notation
5:02
Example 1: Domain/Range/Graph of Arcsine
5:45
Example 2: Arcsin/Arccos/Arctan Values
10:46
Example 3: Domain/Range/Graph of Arctangent
17:14
Extra Example 1: Domain/Range/Graph of Arccosine
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Computations of Inverse Trigonometric Functions

31m 8s

Intro
0:00
Inverse Trigonometric Function Domains and Ranges
0:31
Arcsine
0:41
Arccosine
1:14
Arctangent
1:41
Example 1: Arcsines of Common Values
2:44
Example 2: Odd, Even, or Neither
5:57
Example 3: Arccosines of Common Values
12:24
Extra Example 1: Arctangents of Common Values
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Section 7: Trigonometric Identities
Pythagorean Identity

19m 11s

Intro
0:00
Pythagorean Identity
0:17
Pythagorean Triangle
0:27
Pythagorean Identity
0:45
Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity
1:14
Example 2: Find Angle Given Cosine and Quadrant
4:18
Example 3: Verify Trigonometric Identity
8:00
Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem
-1
Extra Example 2: Find Angle Given Cosine and Quadrant
-2
Identity Tan(squared)x+1=Sec(squared)x

23m 16s

Intro
0:00
Main Formulas
0:19
Companion to Pythagorean Identity
0:27
For Cotangents and Cosecants
0:52
How to Remember
0:58
Example 1: Prove the Identity
1:40
Example 2: Given Tan Find Sec
3:42
Example 3: Prove the Identity
7:45
Extra Example 1: Prove the Identity
-1
Extra Example 2: Given Sec Find Tan
-2
Addition and Subtraction Formulas

52m 52s

Intro
0:00
Addition and Subtraction Formulas
0:09
How to Remember
0:48
Cofunction Identities
1:31
How to Remember Graphically
1:44
Where to Use Cofunction Identities
2:52
Example 1: Derive the Formula for cos(A-B)
3:08
Example 2: Use Addition and Subtraction Formulas
16:03
Example 3: Use Addition and Subtraction Formulas to Prove Identity
25:11
Extra Example 1: Use cos(A-B) and Cofunction Identities
-1
Extra Example 2: Convert to Radians and use Formulas
-2
Double Angle Formulas

29m 5s

Intro
0:00
Main Formula
0:07
How to Remember from Addition Formula
0:18
Two Other Forms
1:35
Example 1: Find Sine and Cosine of Angle using Double Angle
3:16
Example 2: Prove Trigonometric Identity using Double Angle
9:37
Example 3: Use Addition and Subtraction Formulas
12:38
Extra Example 1: Find Sine and Cosine of Angle using Double Angle
-1
Extra Example 2: Prove Trigonometric Identity using Double Angle
-2
Half-Angle Formulas

43m 55s

Intro
0:00
Main Formulas
0:09
Confusing Part
0:34
Example 1: Find Sine and Cosine of Angle using Half-Angle
0:54
Example 2: Prove Trigonometric Identity using Half-Angle
11:51
Example 3: Prove the Half-Angle Formula for Tangents
18:39
Extra Example 1: Find Sine and Cosine of Angle using Half-Angle
-1
Extra Example 2: Prove Trigonometric Identity using Half-Angle
-2
Section 8: Applications of Trigonometry
Trigonometry in Right Angles

25m 43s

Intro
0:00
Master Formula for Right Angles
0:11
SOHCAHTOA
0:15
Only for Right Triangles
1:26
Example 1: Find All Angles in a Triangle
2:19
Example 2: Find Lengths of All Sides of Triangle
7:39
Example 3: Find All Angles in a Triangle
11:00
Extra Example 1: Find All Angles in a Triangle
-1
Extra Example 2: Find Lengths of All Sides of Triangle
-2
Law of Sines

56m 40s

Intro
0:00
Law of Sines Formula
0:18
SOHCAHTOA
0:27
Any Triangle
0:59
Graphical Representation
1:25
Solving Triangle Completely
2:37
When to Use Law of Sines
2:55
ASA, SAA, SSA, AAA
2:59
SAS, SSS for Law of Cosines
7:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
8:44
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:30
Example 3: How Many Triangles Satisfy Conditions, Solve Completely
28:32
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely
-2
Law of Cosines

49m 5s

Intro
0:00
Law of Cosines Formula
0:23
Graphical Representation
0:34
Relates Sides to Angles
1:00
Any Triangle
1:20
Generalization of Pythagorean Theorem
1:32
When to Use Law of Cosines
2:26
SAS, SSS
2:30
Heron's Formula
4:49
Semiperimeter S
5:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
5:53
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:19
Example 3: Find Area of a Triangle Given All Side Lengths
26:33
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: Length of Third Side and Area of Triangle
-2
Finding the Area of a Triangle

27m 37s

Intro
0:00
Master Right Triangle Formula and Law of Cosines
0:19
SOHCAHTOA
0:27
Law of Cosines
1:23
Heron's Formula
2:22
Semiperimeter S
2:37
Example 1: Area of Triangle with Two Sides and One Angle
3:12
Example 2: Area of Triangle with Three Sides
6:11
Example 3: Area of Triangle with Three Sides, No Heron's Formula
8:50
Extra Example 1: Area of Triangle with Two Sides and One Angle
-1
Extra Example 2: Area of Triangle with Two Sides and One Angle
-2
Word Problems and Applications of Trigonometry

34m 25s

Intro
0:00
Formulas to Remember
0:11
SOHCAHTOA
0:15
Law of Sines
0:55
Law of Cosines
1:48
Heron's Formula
2:46
Example 1: Telephone Pole Height
4:01
Example 2: Bridge Length
7:48
Example 3: Area of Triangular Field
14:20
Extra Example 1: Kite Height
-1
Extra Example 2: Roads to a Town
-2
Section 9: Systems of Equations and Inequalities
Systems of Linear Equations

55m 40s

Intro
0:00
Introduction
0:04
Graphs as Location of 'True'
1:49
All Locations that Make the Function True
2:25
Understand the Relationship Between Solutions and the Graph
3:43
Systems as Graphs
4:07
Equations as Lines
4:20
Intersection Point
5:19
Three Possibilities for Solutions
6:17
Independent
6:24
Inconsistent
6:36
Dependent
7:06
Solving by Substitution
8:37
Solve for One Variable
9:07
Substitute into the Second Equation
9:34
Solve for Both Variables
10:12
What If a System is Inconsistent or Dependent?
11:08
No Solutions
11:25
Infinite Solutions
12:30
Solving by Elimination
13:56
Example
14:22
Determining the Number of Solutions
16:30
Why Elimination Makes Sense
17:25
Solving by Graphing Calculator
19:59
Systems with More than Two Variables
23:22
Example 1
25:49
Example 2
30:22
Example 3
34:11
Example 4
38:55
Example 5
46:01
(Non-) Example 6
53:37
Systems of Linear Inequalities

1h 13s

Intro
0:00
Introduction
0:04
Inequality Refresher-Solutions
0:46
Equation Solutions vs. Inequality Solutions
1:02
Essentially a Wide Variety of Answers
1:35
Refresher--Negative Multiplication Flips
1:43
Refresher--Negative Flips: Why?
3:19
Multiplication by a Negative
3:43
The Relationship Flips
3:55
Refresher--Stick to Basic Operations
4:34
Linear Equations in Two Variables
6:50
Graphing Linear Inequalities
8:28
Why It Includes a Whole Section
8:43
How to Show The Difference Between Strict and Not Strict Inequalities
10:08
Dashed Line--Not Solutions
11:10
Solid Line--Are Solutions
11:24
Test Points for Shading
11:42
Example of Using a Point
12:41
Drawing Shading from the Point
13:14
Graphing a System
14:53
Set of Solutions is the Overlap
15:17
Example
15:22
Solutions are Best Found Through Graphing
18:05
Linear Programming-Idea
19:52
Use a Linear Objective Function
20:15
Variables in Objective Function have Constraints
21:24
Linear Programming-Method
22:09
Rearrange Equations
22:21
Graph
22:49
Critical Solution is at the Vertex of the Overlap
23:40
Try Each Vertice
24:35
Example 1
24:58
Example 2
28:57
Example 3
33:48
Example 4
43:10
Nonlinear Systems

41m 1s

Intro
0:00
Introduction
0:06
Substitution
1:12
Example
1:22
Elimination
3:46
Example
3:56
Elimination is Less Useful for Nonlinear Systems
4:56
Graphing
5:56
Using a Graphing Calculator
6:44
Number of Solutions
8:44
Systems of Nonlinear Inequalities
10:02
Graph Each Inequality
10:06
Dashed and/or Solid
10:18
Shade Appropriately
11:14
Example 1
13:24
Example 2
15:50
Example 3
22:02
Example 4
29:06
Example 4, cont.
33:40
Section 10: Vectors and Matrices
Vectors

1h 9m 31s

Intro
0:00
Introduction
0:10
Magnitude of the Force
0:22
Direction of the Force
0:48
Vector
0:52
Idea of a Vector
1:30
How Vectors are Denoted
2:00
Component Form
3:20
Angle Brackets and Parentheses
3:50
Magnitude/Length
4:26
Denoting the Magnitude of a Vector
5:16
Direction/Angle
7:52
Always Draw a Picture
8:50
Component Form from Magnitude & Angle
10:10
Scaling by Scalars
14:06
Unit Vectors
16:26
Combining Vectors - Algebraically
18:10
Combining Vectors - Geometrically
19:54
Resultant Vector
20:46
Alternate Component Form: i, j
21:16
The Zero Vector
23:18
Properties of Vectors
24:20
No Multiplication (Between Vectors)
28:30
Dot Product
29:40
Motion in a Medium
30:10
Fish in an Aquarium Example
31:38
More Than Two Dimensions
33:12
More Than Two Dimensions - Magnitude
34:18
Example 1
35:26
Example 2
38:10
Example 3
45:48
Example 4
50:40
Example 4, cont.
56:07
Example 5
1:01:32
Dot Product & Cross Product

35m 20s

Intro
0:00
Introduction
0:08
Dot Product - Definition
0:42
Dot Product Results in a Scalar, Not a Vector
2:10
Example in Two Dimensions
2:34
Angle and the Dot Product
2:58
The Dot Product of Two Vectors is Deeply Related to the Angle Between the Two Vectors
2:59
Proof of Dot Product Formula
4:14
Won't Directly Help Us Better Understand Vectors
4:18
Dot Product - Geometric Interpretation
4:58
We Can Interpret the Dot Product as a Measure of How Long and How Parallel Two Vectors Are
7:26
Dot Product - Perpendicular Vectors
8:24
If the Dot Product of Two Vectors is 0, We Know They are Perpendicular to Each Other
8:54
Cross Product - Definition
11:08
Cross Product Only Works in Three Dimensions
11:09
Cross Product - A Mnemonic
12:16
The Determinant of a 3 x 3 Matrix and Standard Unit Vectors
12:17
Cross Product - Geometric Interpretations
14:30
The Right-Hand Rule
15:17
Cross Product - Geometric Interpretations Cont.
17:00
Example 1
18:40
Example 2
22:50
Example 3
24:04
Example 4
26:20
Bonus Round
29:18
Proof: Dot Product Formula
29:24
Proof: Dot Product Formula, cont.
30:38
Matrices

54m 7s

Intro
0:00
Introduction
0:08
Definition of a Matrix
3:02
Size or Dimension
3:58
Square Matrix
4:42
Denoted by Capital Letters
4:56
When are Two Matrices Equal?
5:04
Examples of Matrices
6:44
Rows x Columns
6:46
Talking About Specific Entries
7:48
We Use Capitals to Denote a Matrix and Lower Case to Denotes Its Entries
8:32
Using Entries to Talk About Matrices
10:08
Scalar Multiplication
11:26
Scalar = Real Number
11:34
Example
12:36
Matrix Addition
13:08
Example
14:22
Matrix Multiplication
15:00
Example
18:52
Matrix Multiplication, cont.
19:58
Matrix Multiplication and Order (Size)
25:26
Make Sure Their Orders are Compatible
25:27
Matrix Multiplication is NOT Commutative
28:20
Example
30:08
Special Matrices - Zero Matrix (0)
32:48
Zero Matrix Has 0 for All of its Entries
32:49
Special Matrices - Identity Matrix (I)
34:14
Identity Matrix is a Square Matrix That Has 1 for All Its Entries on the Main Diagonal and 0 for All Other Entries
34:15
Example 1
36:16
Example 2
40:00
Example 3
44:54
Example 4
50:08
Determinants & Inverses of Matrices

47m 12s

Intro
0:00
Introduction
0:06
Not All Matrices Are Invertible
1:30
What Must a Matrix Have to Be Invertible?
2:08
Determinant
2:32
The Determinant is a Real Number Associated With a Square Matrix
2:38
If the Determinant of a Matrix is Nonzero, the Matrix is Invertible
3:40
Determinant of a 2 x 2 Matrix
4:34
Think in Terms of Diagonals
5:12
Minors and Cofactors - Minors
6:24
Example
6:46
Minors and Cofactors - Cofactors
8:00
Cofactor is Closely Based on the Minor
8:01
Alternating Sign Pattern
9:04
Determinant of Larger Matrices
10:56
Example
13:00
Alternative Method for 3x3 Matrices
16:46
Not Recommended
16:48
Inverse of a 2 x 2 Matrix
19:02
Inverse of Larger Matrices
20:00
Using Inverse Matrices
21:06
When Multiplied Together, They Create the Identity Matrix
21:24
Example 1
23:45
Example 2
27:21
Example 3
32:49
Example 4
36:27
Finding the Inverse of Larger Matrices
41:59
General Inverse Method - Step 1
43:25
General Inverse Method - Step 2
43:27
General Inverse Method - Step 2, cont.
43:27
General Inverse Method - Step 3
45:15
Using Matrices to Solve Systems of Linear Equations

58m 34s

Intro
0:00
Introduction
0:12
Augmented Matrix
1:44
We Can Represent the Entire Linear System With an Augmented Matrix
1:50
Row Operations
3:22
Interchange the Locations of Two Rows
3:50
Multiply (or Divide) a Row by a Nonzero Number
3:58
Add (or Subtract) a Multiple of One Row to Another
4:12
Row Operations - Keep Notes!
5:50
Suggested Symbols
7:08
Gauss-Jordan Elimination - Idea
8:04
Gauss-Jordan Elimination - Idea, cont.
9:16
Reduced Row-Echelon Form
9:18
Gauss-Jordan Elimination - Method
11:36
Begin by Writing the System As An Augmented Matrix
11:38
Gauss-Jordan Elimination - Method, cont.
13:48
Cramer's Rule - 2 x 2 Matrices
17:08
Cramer's Rule - n x n Matrices
19:24
Solving with Inverse Matrices
21:10
Solving Inverse Matrices, cont.
25:28
The Mighty (Graphing) Calculator
26:38
Example 1
29:56
Example 2
33:56
Example 3
37:00
Example 3, cont.
45:04
Example 4
51:28
Section 11: Alternate Ways to Graph
Parametric Equations

53m 33s

Intro
0:00
Introduction
0:06
Definition
1:10
Plane Curve
1:24
The Key Idea
2:00
Graphing with Parametric Equations
2:52
Same Graph, Different Equations
5:04
How Is That Possible?
5:36
Same Graph, Different Equations, cont.
5:42
Here's Another to Consider
7:56
Same Plane Curve, But Still Different
8:10
A Metaphor for Parametric Equations
9:36
Think of Parametric Equations As a Way to Describe the Motion of An Object
9:38
Graph Shows Where It Went, But Not Speed
10:32
Eliminating Parameters
12:14
Rectangular Equation
12:16
Caution
13:52
Creating Parametric Equations
14:30
Interesting Graphs
16:38
Graphing Calculators, Yay!
19:18
Example 1
22:36
Example 2
28:26
Example 3
37:36
Example 4
41:00
Projectile Motion
44:26
Example 5
47:00
Polar Coordinates

48m 7s

Intro
0:00
Introduction
0:04
Polar Coordinates Give Us a Way To Describe the Location of a Point
0:26
Polar Equations and Functions
0:50
Plotting Points with Polar Coordinates
1:06
The Distance of the Point from the Origin
1:09
The Angle of the Point
1:33
Give Points as the Ordered Pair (r,θ)
2:03
Visualizing Plotting in Polar Coordinates
2:32
First Way We Can Plot
2:39
Second Way We Can Plot
2:50
First, We'll Look at Visualizing r, Then θ
3:09
Rotate the Length Counter-Clockwise by θ
3:38
Alternatively, We Can Visualize θ, Then r
4:06
'Polar Graph Paper'
6:17
Horizontal and Vertical Tick Marks Are Not Useful for Polar
6:42
Use Concentric Circles to Helps Up See Distance From the Pole
7:08
Can Use Arc Sectors to See Angles
7:57
Multiple Ways to Name a Point
9:17
Examples
9:30
For Any Angle θ, We Can Make an Equivalent Angle
10:44
Negative Values for r
11:58
If r Is Negative, We Go In The Direction Opposite the One That The Angle θ Points Out
12:22
Another Way to Name the Same Point: Add π to θ and Make r Negative
13:44
Converting Between Rectangular and Polar
14:37
Rectangular Way to Name
14:43
Polar Way to Name
14:52
The Rectangular System Must Have a Right Angle Because It's Based on a Rectangle
15:08
Connect Both Systems Through Basic Trigonometry
15:38
Equation to Convert From Polar to Rectangular Coordinate Systems
16:55
Equation to Convert From Rectangular to Polar Coordinate Systems
17:13
Converting to Rectangular is Easy
17:20
Converting to Polar is a Bit Trickier
17:21
Draw Pictures
18:55
Example 1
19:50
Example 2
25:17
Example 3
31:05
Example 4
35:56
Example 5
41:49
Polar Equations & Functions

38m 16s

Intro
0:00
Introduction
0:04
Equations and Functions
1:16
Independent Variable
1:21
Dependent Variable
1:30
Examples
1:46
Always Assume That θ Is In Radians
2:44
Graphing in Polar Coordinates
3:29
Graph is the Same Way We Graph 'Normal' Stuff
3:32
Example
3:52
Graphing in Polar - Example, Cont.
6:45
Tips for Graphing
9:23
Notice Patterns
10:19
Repetition
13:39
Graphing Equations of One Variable
14:39
Converting Coordinate Types
16:16
Use the Same Conversion Formulas From the Previous Lesson
16:23
Interesting Graphs
17:48
Example 1
18:03
Example 2
18:34
Graphing Calculators, Yay!
19:07
Plot Random Things, Alter Equations You Understand, Get a Sense for How Polar Stuff Works
19:11
Check Out the Appendix
19:26
Example 1
21:36
Example 2
28:13
Example 3
34:24
Example 4
35:52
Section 12: Complex Numbers and Polar Coordinates
Polar Form of Complex Numbers

40m 43s

Intro
0:00
Polar Coordinates
0:49
Rectangular Form
0:52
Polar Form
1:25
R and Theta
1:51
Polar Form Conversion
2:27
R and Theta
2:35
Optimal Values
4:05
Euler's Formula
4:25
Multiplying Two Complex Numbers in Polar Form
6:10
Multiply r's Together and Add Exponents
6:32
Example 1: Convert Rectangular to Polar Form
7:17
Example 2: Convert Polar to Rectangular Form
13:49
Example 3: Multiply Two Complex Numbers
17:28
Extra Example 1: Convert Between Rectangular and Polar Forms
-1
Extra Example 2: Simplify Expression to Polar Form
-2
DeMoivre's Theorem

57m 37s

Intro
0:00
Introduction to DeMoivre's Theorem
0:10
n nth Roots
3:06
DeMoivre's Theorem: Finding nth Roots
3:52
Relation to Unit Circle
6:29
One nth Root for Each Value of k
7:11
Example 1: Convert to Polar Form and Use DeMoivre's Theorem
8:24
Example 2: Find Complex Eighth Roots
15:27
Example 3: Find Complex Roots
27:49
Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem
-1
Extra Example 2: Find Complex Fourth Roots
-2
Section 13: Counting & Probability
Counting

31m 36s

Intro
0:00
Introduction
0:08
Combinatorics
0:56
Definition: Event
1:24
Example
1:50
Visualizing an Event
3:02
Branching line diagram
3:06
Addition Principle
3:40
Example
4:18
Multiplication Principle
5:42
Example
6:24
Pigeonhole Principle
8:06
Example
10:26
Draw Pictures
11:06
Example 1
12:02
Example 2
14:16
Example 3
17:34
Example 4
21:26
Example 5
25:14
Permutations & Combinations

44m 3s

Intro
0:00
Introduction
0:08
Permutation
0:42
Combination
1:10
Towards a Permutation Formula
2:38
How Many Ways Can We Arrange the Letters A, B, C, D, and E?
3:02
Towards a Permutation Formula, cont.
3:34
Factorial Notation
6:56
Symbol Is '!'
6:58
Examples
7:32
Permutation of n Objects
8:44
Permutation of r Objects out of n
9:04
What If We Have More Objects Than We Have Slots to Fit Them Into?
9:46
Permutation of r Objects Out of n, cont.
10:28
Distinguishable Permutations
14:46
What If Not All Of the Objects We're Permuting Are Distinguishable From Each Other?
14:48
Distinguishable Permutations, cont.
17:04
Combinations
19:04
Combinations, cont.
20:56
Example 1
23:10
Example 2
26:16
Example 3
28:28
Example 4
31:52
Example 5
33:58
Example 6
36:34
Probability

36m 58s

Intro
0:00
Introduction
0:06
Definition: Sample Space
1:18
Event = Something Happening
1:20
Sample Space
1:36
Probability of an Event
2:12
Let E Be An Event and S Be The Corresponding Sample Space
2:14
'Equally Likely' Is Important
3:52
Fair and Random
5:26
Interpreting Probability
6:34
How Can We Interpret This Value?
7:24
We Can Represent Probability As a Fraction, a Decimal, Or a Percentage
8:04
One of Multiple Events Occurring
9:52
Mutually Exclusive Events
10:38
What If The Events Are Not Mutually Exclusive?
12:20
Taking the Possibility of Overlap Into Account
13:24
An Event Not Occurring
17:14
Complement of E
17:22
Independent Events
19:36
Independent
19:48
Conditional Events
21:28
What Is The Events Are Not Independent Though?
21:30
Conditional Probability
22:16
Conditional Events, cont.
23:51
Example 1
25:27
Example 2
27:09
Example 3
28:57
Example 4
30:51
Example 5
34:15
Section 14: Conic Sections
Parabolas

41m 27s

Intro
0:00
What is a Parabola?
0:20
Definition of a Parabola
0:29
Focus
0:59
Directrix
1:15
Axis of Symmetry
3:08
Vertex
3:33
Minimum or Maximum
3:44
Standard Form
4:59
Horizontal Parabolas
5:08
Vertex Form
5:19
Upward or Downward
5:41
Example: Standard Form
6:06
Graphing Parabolas
8:31
Shifting
8:51
Example: Completing the Square
9:22
Symmetry and Translation
12:18
Example: Graph Parabola
12:40
Latus Rectum
17:13
Length
18:15
Example: Latus Rectum
18:35
Horizontal Parabolas
18:57
Not Functions
20:08
Example: Horizontal Parabola
21:21
Focus and Directrix
24:11
Horizontal
24:48
Example 1: Parabola Standard Form
25:12
Example 2: Graph Parabola
30:00
Example 3: Graph Parabola
33:13
Example 4: Parabola Equation
37:28
Circles

21m 3s

Intro
0:00
What are Circles?
0:08
Example: Equidistant
0:17
Radius
0:32
Equation of a Circle
0:44
Example: Standard Form
1:11
Graphing Circles
1:47
Example: Circle
1:56
Center Not at Origin
3:07
Example: Completing the Square
3:51
Example 1: Equation of Circle
6:44
Example 2: Center and Radius
11:51
Example 3: Radius
15:08
Example 4: Equation of Circle
16:57
Ellipses

46m 51s

Intro
0:00
What Are Ellipses?
0:11
Foci
0:23
Properties of Ellipses
1:43
Major Axis, Minor Axis
1:47
Center
1:54
Length of Major Axis and Minor Axis
3:21
Standard Form
5:33
Example: Standard Form of Ellipse
6:09
Vertical Major Axis
9:14
Example: Vertical Major Axis
9:46
Graphing Ellipses
12:51
Complete the Square and Symmetry
13:00
Example: Graphing Ellipse
13:16
Equation with Center at (h, k)
19:57
Horizontal and Vertical
20:14
Difference
20:27
Example: Center at (h, k)
20:55
Example 1: Equation of Ellipse
24:05
Example 2: Equation of Ellipse
27:57
Example 3: Equation of Ellipse
32:32
Example 4: Graph Ellipse
38:27
Hyperbolas

38m 15s

Intro
0:00
What are Hyperbolas?
0:12
Two Branches
0:18
Foci
0:38
Properties
2:00
Transverse Axis and Conjugate Axis
2:06
Vertices
2:46
Length of Transverse Axis
3:14
Distance Between Foci
3:31
Length of Conjugate Axis
3:38
Standard Form
5:45
Vertex Location
6:36
Known Points
6:52
Vertical Transverse Axis
7:26
Vertex Location
7:50
Asymptotes
8:36
Vertex Location
8:56
Rectangle
9:28
Diagonals
10:29
Graphing Hyperbolas
12:58
Example: Hyperbola
13:16
Equation with Center at (h, k)
16:32
Example: Center at (h, k)
17:21
Example 1: Equation of Hyperbola
19:20
Example 2: Equation of Hyperbola
22:48
Example 3: Graph Hyperbola
26:05
Example 4: Equation of Hyperbola
36:29
Conic Sections

18m 43s

Intro
0:00
Conic Sections
0:16
Double Cone Sections
0:24
Standard Form
1:27
General Form
1:37
Identify Conic Sections
2:16
B = 0
2:50
X and Y
3:22
Identify Conic Sections, Cont.
4:46
Parabola
5:17
Circle
5:51
Ellipse
6:31
Hyperbola
7:10
Example 1: Identify Conic Section
8:01
Example 2: Identify Conic Section
11:03
Example 3: Identify Conic Section
11:38
Example 4: Identify Conic Section
14:50
Section 15: Sequences, Series, & Induction
Introduction to Sequences

57m 45s

Intro
0:00
Introduction
0:06
Definition: Sequence
0:28
Infinite Sequence
2:08
Finite Sequence
2:22
Length
2:58
Formula for the nth Term
3:22
Defining a Sequence Recursively
5:54
Initial Term
7:58
Sequences and Patterns
10:40
First, Identify a Pattern
12:52
How to Get From One Term to the Next
17:38
Tips for Finding Patterns
19:52
More Tips for Finding Patterns
24:14
Even More Tips
26:50
Example 1
30:32
Example 2
34:54
Fibonacci Sequence
34:55
Example 3
38:40
Example 4
45:02
Example 5
49:26
Example 6
51:54
Introduction to Series

40m 27s

Intro
0:00
Introduction
0:06
Definition: Series
1:20
Why We Need Notation
2:48
Simga Notation (AKA Summation Notation)
4:44
Thing Being Summed
5:42
Index of Summation
6:21
Lower Limit of Summation
7:09
Upper Limit of Summation
7:23
Sigma Notation, Example
7:36
Sigma Notation for Infinite Series
9:08
How to Reindex
10:58
How to Reindex, Expanding
12:56
How to Reindex, Substitution
16:46
Properties of Sums
19:42
Example 1
23:46
Example 2
25:34
Example 3
27:12
Example 4
29:54
Example 5
32:06
Example 6
37:16
Arithmetic Sequences & Series

31m 36s

Intro
0:00
Introduction
0:05
Definition: Arithmetic Sequence
0:47
Common Difference
1:13
Two Examples
1:19
Form for the nth Term
2:14
Recursive Relation
2:33
Towards an Arithmetic Series Formula
5:12
Creating a General Formula
10:09
General Formula for Arithmetic Series
14:23
Example 1
15:46
Example 2
17:37
Example 3
22:21
Example 4
24:09
Example 5
27:14
Geometric Sequences & Series

39m 27s

Intro
0:00
Introduction
0:06
Definition
0:48
Form for the nth Term
2:42
Formula for Geometric Series
5:16
Infinite Geometric Series
11:48
Diverges
13:04
Converges
14:48
Formula for Infinite Geometric Series
16:32
Example 1
20:32
Example 2
22:02
Example 3
26:00
Example 4
30:48
Example 5
34:28
Mathematical Induction

49m 53s

Intro
0:00
Introduction
0:06
Belief Vs. Proof
1:22
A Metaphor for Induction
6:14
The Principle of Mathematical Induction
11:38
Base Case
13:24
Inductive Step
13:30
Inductive Hypothesis
13:52
A Remark on Statements
14:18
Using Mathematical Induction
16:58
Working Example
19:58
Finding Patterns
28:46
Example 1
30:17
Example 2
37:50
Example 3
42:38
The Binomial Theorem

1h 13m 13s

Intro
0:00
Introduction
0:06
We've Learned That a Binomial Is An Expression That Has Two Terms
0:07
Understanding Binomial Coefficients
1:20
Things We Notice
2:24
What Goes In the Blanks?
5:52
Each Blank is Called a Binomial Coefficient
6:18
The Binomial Theorem
6:38
Example
8:10
The Binomial Theorem, cont.
10:46
We Can Also Write This Expression Compactly Using Sigma Notation
12:06
Proof of the Binomial Theorem
13:22
Proving the Binomial Theorem Is Within Our Reach
13:24
Pascal's Triangle
15:12
Pascal's Triangle, cont.
16:12
Diagonal Addition of Terms
16:24
Zeroth Row
18:04
First Row
18:12
Why Do We Care About Pascal's Triangle?
18:50
Pascal's Triangle, Example
19:26
Example 1
21:26
Example 2
24:34
Example 3
28:34
Example 4
32:28
Example 5
37:12
Time for the Fireworks!
43:38
Proof of the Binomial Theorem
43:44
We'll Prove This By Induction
44:04
Proof (By Induction)
46:36
Proof, Base Case
47:00
Proof, Inductive Step - Notation Discussion
49:22
Induction Step
49:24
Proof, Inductive Step - Setting Up
52:26
Induction Hypothesis
52:34
What We What To Show
52:44
Proof, Inductive Step - Start
54:18
Proof, Inductive Step - Middle
55:38
Expand Sigma Notations
55:48
Proof, Inductive Step - Middle, cont.
58:40
Proof, Inductive Step - Checking In
1:01:08
Let's Check In With Our Original Goal
1:01:12
Want to Show
1:01:18
Lemma - A Mini Theorem
1:02:18
Proof, Inductive Step - Lemma
1:02:52
Proof of Lemma: Let's Investigate the Left Side
1:03:08
Proof, Inductive Step - Nearly There
1:07:54
Proof, Inductive Step - End!
1:09:18
Proof, Inductive Step - End!, cont.
1:11:01
Section 16: Preview of Calculus
Idea of a Limit

40m 22s

Intro
0:00
Introduction
0:05
Motivating Example
1:26
Fuzzy Notion of a Limit
3:38
Limit is the Vertical Location a Function is Headed Towards
3:44
Limit is What the Function Output is Going to Be
4:15
Limit Notation
4:33
Exploring Limits - 'Ordinary' Function
5:26
Test Out
5:27
Graphing, We See The Answer Is What We Would Expect
5:44
Exploring Limits - Piecewise Function
6:45
If We Modify the Function a Bit
6:49
Exploring Limits - A Visual Conception
10:08
Definition of a Limit
12:07
If f(x) Becomes Arbitrarily Close to Some Number L as x Approaches Some Number c, Then the Limit of f(x) As a Approaches c is L.
12:09
We Are Not Concerned with f(x) at x=c
12:49
We Are Considering x Approaching From All Directions, Not Just One Side
13:10
Limits Do Not Always Exist
15:47
Finding Limits
19:49
Graphs
19:52
Tables
21:48
Precise Methods
24:53
Example 1
26:06
Example 2
27:39
Example 3
30:51
Example 4
33:11
Example 5
37:07
Formal Definition of a Limit

57m 11s

Intro
0:00
Introduction
0:06
New Greek Letters
2:42
Delta
3:14
Epsilon
3:46
Sometimes Called the Epsilon-Delta Definition of a Limit
3:56
Formal Definition of a Limit
4:22
What does it MEAN!?!?
5:00
The Groundwork
5:38
Set Up the Limit
5:39
The Function is Defined Over Some Portion of the Reals
5:58
The Horizontal Location is the Value the Limit Will Approach
6:28
The Vertical Location L is Where the Limit Goes To
7:00
The Epsilon-Delta Part
7:26
The Hard Part is the Second Part of the Definition
7:30
Second Half of Definition
10:04
Restrictions on the Allowed x Values
10:28
The Epsilon-Delta Part, cont.
13:34
Sherlock Holmes and Dr. Watson
15:08
The Adventure of the Delta-Epsilon Limit
15:16
Setting
15:18
We Begin By Setting Up the Game As Follows
15:52
The Adventure of the Delta-Epsilon, cont.
17:24
This Game is About Limits
17:46
What If I Try Larger?
19:39
Technically, You Haven't Proven the Limit
20:53
Here is the Method
21:18
What We Should Concern Ourselves With
22:20
Investigate the Left Sides of the Expressions
25:24
We Can Create the Following Inequalities
28:08
Finally…
28:50
Nothing Like a Good Proof to Develop the Appetite
30:42
Example 1
31:02
Example 1, cont.
36:26
Example 2
41:46
Example 2, cont.
47:50
Finding Limits

32m 40s

Intro
0:00
Introduction
0:08
Method - 'Normal' Functions
2:04
The Easiest Limits to Find
2:06
It Does Not 'Break'
2:18
It Is Not Piecewise
2:26
Method - 'Normal' Functions, Example
3:38
Method - 'Normal' Functions, cont.
4:54
The Functions We're Used to Working With Go Where We Expect Them To Go
5:22
A Limit is About Figuring Out Where a Function is 'Headed'
5:42
Method - Canceling Factors
7:18
One Weird Thing That Often Happens is Dividing By 0
7:26
Method - Canceling Factors, cont.
8:16
Notice That The Two Functions Are Identical With the Exception of x=0
8:20
Method - Canceling Factors, cont.
10:00
Example
10:52
Method - Rationalization
12:04
Rationalizing a Portion of Some Fraction
12:05
Conjugate
12:26
Method - Rationalization, cont.
13:14
Example
13:50
Method - Piecewise
16:28
The Limits of Piecewise Functions
16:30
Example 1
17:42
Example 2
18:44
Example 3
20:20
Example 4
22:24
Example 5
24:24
Example 6
27:12
Continuity & One-Sided Limits

32m 43s

Intro
0:00
Introduction
0:06
Motivating Example
0:56
Continuity - Idea
2:14
Continuous Function
2:18
All Parts of Function Are Connected
2:28
Function's Graph Can Be Drawn Without Lifting Pencil
2:36
There Are No Breaks or Holes in Graph
2:56
Continuity - Idea, cont.
3:38
We Can Interpret the Break in the Continuity of f(x) as an Issue With the Function 'Jumping'
3:52
Continuity - Definition
5:16
A Break in Continuity is Caused By the Limit Not Matching Up With What the Function Does
5:18
Discontinuous
6:02
Discontinuity
6:10
Continuity and 'Normal' Functions
6:48
Return of the Motivating Example
8:14
One-Sided Limit
8:48
One-Sided Limit - Definition
9:16
Only Considers One Side
9:20
Be Careful to Keep Track of Which Symbol Goes With Which Side
10:06
One-Sided Limit - Example
10:50
There Does Not Necessarily Need to Be a Connection Between Left or Right Side Limits
11:16
Normal Limits and One-Sided Limits
12:08
Limits of Piecewise Functions
14:12
'Breakover' Points
14:22
We Find the Limit of a Piecewise Function By Checking If the Left and Right Side Limits Agree With Each Other
15:34
Example 1
16:40
Example 2
18:54
Example 3
22:00
Example 4
26:36
Limits at Infinity & Limits of Sequences

32m 49s

Intro
0:00
Introduction
0:06
Definition: Limit of a Function at Infinity
1:44
A Limit at Infinity Works Very Similarly to How a Normal Limit Works
2:38
Evaluating Limits at Infinity
4:08
Rational Functions
4:17
Examples
4:30
For a Rational Function, the Question Boils Down to Comparing the Long Term Growth Rates of the Numerator and Denominator
5:22
There are Three Possibilities
6:36
Evaluating Limits at Infinity, cont.
8:08
Does the Function Grow Without Bound? Will It 'Settle Down' Over Time?
10:06
Two Good Ways to Think About This
10:26
Limit of a Sequence
12:20
What Value Does the Sequence Tend to Do in the Long-Run?
12:41
The Limit of a Sequence is Very Similar to the Limit of a Function at Infinity
12:52
Numerical Evaluation
14:16
Numerically: Plug in Numbers and See What Comes Out
14:24
Example 1
16:42
Example 2
21:00
Example 3
22:08
Example 4
26:14
Example 5
28:10
Example 6
31:06
Instantaneous Slope & Tangents (Derivatives)

51m 13s

Intro
0:00
Introduction
0:08
The Derivative of a Function Gives Us a Way to Talk About 'How Fast' the Function If Changing
0:16
Instantaneous Slop
0:22
Instantaneous Rate of Change
0:28
Slope
1:24
The Vertical Change Divided by the Horizontal
1:40
Idea of Instantaneous Slope
2:10
What If We Wanted to Apply the Idea of Slope to a Non-Line?
2:14
Tangent to a Circle
3:52
What is the Tangent Line for a Circle?
4:42
Tangent to a Curve
5:20
Towards a Derivative - Average Slope
6:36
Towards a Derivative - Average Slope, cont.
8:20
An Approximation
11:24
Towards a Derivative - General Form
13:18
Towards a Derivative - General Form, cont.
16:46
An h Grows Smaller, Our Slope Approximation Becomes Better
18:44
Towards a Derivative - Limits!
20:04
Towards a Derivative - Limits!, cont.
22:08
We Want to Show the Slope at x=1
22:34
Towards a Derivative - Checking Our Slope
23:12
Definition of the Derivative
23:54
Derivative: A Way to Find the Instantaneous Slope of a Function at Any Point
23:58
Differentiation
24:54
Notation for the Derivative
25:58
The Derivative is a Very Important Idea In Calculus
26:04
The Important Idea
27:34
Why Did We Learn the Formal Definition to Find a Derivative?
28:18
Example 1
30:50
Example 2
36:06
Example 3
40:24
The Power Rule
44:16
Makes It Easier to Find the Derivative of a Function
44:24
Examples
45:04
n Is Any Constant Number
45:46
Example 4
46:26
Area Under a Curve (Integrals)

45m 26s

Intro
0:00
Introduction
0:06
Integral
0:12
Idea of Area Under a Curve
1:18
Approximation by Rectangles
2:12
The Easiest Way to Find Area is With a Rectangle
2:18
Various Methods for Choosing Rectangles
4:30
Rectangle Method - Left-Most Point
5:12
The Left-Most Point
5:16
Rectangle Method - Right-Most Point
5:58
The Right-Most Point
6:00
Rectangle Method - Mid-Point
6:42
Horizontal Mid-Point
6:48
Rectangle Method - Maximum (Upper Sum)
7:34
Maximum Height
7:40
Rectangle Method - Minimum
8:54
Minimum Height
9:02
Evaluating the Area Approximation
10:08
Split the Interval Into n Sub-Intervals
10:30
More Rectangles, Better Approximation
12:14
The More We Us , the Better Our Approximation Becomes
12:16
Our Approximation Becomes More Accurate as the Number of Rectangles n Goes Off to Infinity
12:44
Finding Area with a Limit
13:08
If This Limit Exists, It Is Called the Integral From a to b
14:08
The Process of Finding Integrals is Called Integration
14:22
The Big Reveal
14:40
The Integral is Based on the Antiderivative
14:46
The Big Reveal - Wait, Why?
16:28
The Rate of Change for the Area is Based on the Height of the Function
16:50
Height is the Derivative of Area, So Area is Based on the Antiderivative of Height
17:50
Example 1
19:06
Example 2
22:48
Example 3
29:06
Example 3, cont.
35:14
Example 4
40:14
Section 17: Appendix: Graphing Calculators
Buying a Graphing Calculator

10m 41s

Intro
0:00
Should You Buy?
0:06
Should I Get a Graphing Utility?
0:20
Free Graphing Utilities - Web Based
0:38
Personal Favorite: Desmos
0:58
Free Graphing Utilities - Offline Programs
1:18
GeoGebra
1:31
Microsoft Mathematics
1:50
Grapher
2:18
Other Graphing Utilities - Tablet/Phone
2:48
Should You Buy a Graphing Calculator?
3:22
The Only Real Downside
4:10
Deciding on Buying
4:20
If You Plan on Continuing in Math and/or Science
4:26
If Money is Not Particularly Tight for You
4:32
If You Don't Plan to Continue in Math and Science
5:02
If You Do Plan to Continue and Money Is Tight
5:28
Which to Buy
5:44
Which Graphing Calculator is Best?
5:46
Too Many Factors
5:54
Do Your Research
6:12
The Old Standby
7:10
TI-83 (Plus)
7:16
TI-84 (Plus)
7:18
Tips for Purchasing
9:17
Buy Online
9:19
Buy Used
9:35
Ask Around
10:09
Graphing Calculator Basics

10m 51s

Intro
0:00
Read the Manual
0:06
Skim It
0:20
Play Around and Experiment
0:34
Syntax
0:40
Definition of Syntax in English and Math
0:46
Pay Careful Attention to Your Syntax When Working With a Calculator
2:08
Make Sure You Use Parentheses to Indicate the Proper Order of Operations
2:16
Think About the Results
3:54
Settings
4:58
You'll Almost Never Need to Change the Settings on Your Calculator
5:00
Tell Calculator In Settings Whether the Angles Are In Radians or Degrees
5:26
Graphing Mode
6:32
Error Messages
7:10
Don't Panic
7:11
Internet Search
7:32
So Many Things
8:14
More Powerful Than You Realize
8:18
Other Things Your Graphing Calculator Can Do
8:24
Playing Around
9:16
Graphing Functions, Window Settings, & Table of Values

10m 38s

Intro
0:00
Graphing Functions
0:18
Graphing Calculator Expects the Variable to Be x
0:28
Syntax
0:58
The Syntax We Choose Will Affect How the Function Graphs
1:00
Use Parentheses
1:26
The Viewing Window
2:00
One of the Most Important Ideas When Graphing Is To Think About The Viewing Window
2:01
For Example
2:30
The Viewing Window, cont.
2:36
Window Settings
3:24
Manually Choose Window Settings
4:20
x Min
4:40
x Max
4:42
y Min
4:44
y Max
4:46
Changing the x Scale or y Scale
5:08
Window Settings, cont.
5:44
Table of Values
7:38
Allows You to Quickly Churn Out Values for Various Inputs
7:42
For example
7:44
Changing the Independent Variable From 'Automatic' to 'Ask'
8:50
Finding Points of Interest

9m 45s

Intro
0:00
Points of Interest
0:06
Interesting Points on the Graph
0:11
Roots/Zeros (Zero)
0:18
Relative Minimums (Min)
0:26
Relative Maximums (Max)
0:32
Intersections (Intersection)
0:38
Finding Points of Interest - Process
1:48
Graph the Function
1:49
Adjust Viewing Window
2:12
Choose Point of Interest Type
2:54
Identify Where Search Should Occur
3:04
Give a Guess
3:36
Get Result
4:06
Advanced Technique: Arbitrary Solving
5:10
Find Out What Input Value Causes a Certain Output
5:12
For Example
5:24
Advanced Technique: Calculus
7:18
Derivative
7:22
Integral
7:30
But How Do You Show Work?
8:20
Parametric & Polar Graphs

7m 8s

Intro
0:00
Change Graph Type
0:08
Located in General 'Settings'
0:16
Graphing in Parametric
1:06
Set Up Both Horizontal Function and Vertical Function
1:08
For Example
2:04
Graphing in Polar
4:00
For Example
4:28
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Mathematical Induction

  • One of the most important ideas in mathematics is being able to prove something with total certainty. We can show not just that something works out, but that it will work every time, forever-it is mathematical fact. We show this with proof.
  • Mathematical induction is a form of proof that allows us to prove that a pattern holds true forever.
  • A good metaphor for the idea of induction is a never-ending line of dominoes. If we have two guarantees,
    1. The first domino will fall over;
    2. If a domino falls over, it will cause the next domino to fall over as well;
    then we know that all the dominoes in our infinitely long line of dominoes will fall over.
  • The principle of mathematical induction is basically the same idea as the above metaphor, except we replace "domino falling over" with "statement being true". Let P1, P2, P3, P4, ..., Pn, ... be some sequence of statements where n is a positive integer. If
    1. P1 is true, and
    2. If Pk is true for a positive integer k, then Pk+1 must also be true,
    then the statement Pn is true for all positive integers n.
  • Above, we call the first step the base case because it establishes the basis we are working from.
  • Above, the second step is called the inductive step since that is where the actual induction occurs. In the inductive step, the assumption "if Pk is true" is called the inductive hypothesis, because we must use this hypothesis to show Pk+1 will also be true.
  • It helps a lot to somehow include part of the statement for Pk when we are writing out Pk+1. This is because we will almost inevitably wind up using our inductive hypothesis (Pk is true) to show that Pk+1 is also true. But we can't use our hypothesis about Pk unless it somehow appears in Pk+1.
  • Check out the video to see a working example of the inductive hypothesis where each step is explained. Induction can be a little confusing at first, and it really helps to see it in action.

Mathematical Induction

Below is the general statement Pn.
Pn:     Buying n loaves of bread costs $(2.20 ·n).
Write out the statements P1, P7, Pk, and Pk+1.
  • A statement is just a string of words and/or symbols. In the case of the above statement, Pn is a statement that varies depending on the specific value used for n.
  • To find out what P1, P7, Pk, and Pk+1 are, we need to replace the n with the corresponding value. This is just simple substitution. Plug in the value you're working with for each place that n shows up.

    For P1, we have n=1, so substitute for every value of n in the general statement:
    P1:     Buying 1 loaves of bread costs $(2.20 ·1).
    We're close, but there are some minor issues. First off, there's a grammar problem: the plural is `loaves', the singular is `loaf'. Thus, since we're buying just one of them, we need to use the singular `loaf'. Second, because the multiplication is so simple, it doesn't really show any useful information, so we can simplify $(2.20 ·1) = $2.20. Putting these fixes in, we have
    P1:     Buying 1 loaf of bread costs $2.20.
  • To find the other statements, we just plug in the appropriate value for n:
    P7:     Buying 7 loaves of bread costs $(2.20 ·7).

    Pk:     Buying k loaves of bread costs $(2.20 ·k).

    Pk+1:     Buying (k+1) loaves of bread costs $(2.20 ·(k+1)).
    For each of these, the grammar is correct (we almost always treat variables as if they indicate the plural). For P7, it also works well to leave the cost as $(2.20 ·7), since that shows how to arrive at the price of all the bread: multiply the cost of one loaf by the number of loaves. The only minor improvement we could make is to Pk+1, where we could distribute: $(2.20 ·(k+1)) = $(2.20·k + 2.20). This makes it a little bit clearer that purchasing (k+1) loaves just adds the price of one more loaf to the cost of purchasing k loaves.
    Pk+1:     Buying (k+1) loaves of bread costs $(2.20·k + 2.20).
P1:     Buying 1 loaf of bread costs $2.20. P7:     Buying 7 loaves of bread costs $(2.20 ·7). Pk:     Buying k loaves of bread costs $(2.20 ·k). Pk+1:     Buying (k+1) loaves of bread costs $(2.20·k + 2.20).
Below is the general statement Pn.
Pn:     The number (n3−n+2) is divisible by 2.
Write out the statements P1, P7, Pk, and Pk+1.
  • A statement is just a string of words and/or symbols. In the case of the above statement, Pn is a statement that varies depending on the specific value used for n.
  • To find out what P1, P7, Pk, and Pk+1 are, we need to replace the n with the corresponding value. This is just simple substitution. Plug in the value you're working with for each place that n shows up.

    For P1, we have n=1, so substitute for every value of n in the general statement:
    P1:     The number (13−1+2) is divisible by 2.
    The above is fine: although we could simplify the expression (13−1+2), doing so will not make the structure of the later statements any clearer, so we probably want to leave it like it is. That said, we could simplify to get the equivalent statement "The number 2 is divisible by 2." Simplifying to that would be quite useful if we wanted to verify the statement, like when doing a proof by mathematical induction.
  • To find the other statements, we just plug in the appropriate value for n:
    P7:     The number (73−7+2) is divisible by 2.

    Pk:     The number (k3−k+2) is divisible by 2.

    Pk+1:     The number ((k+1)3−(k+1)+2) is divisible by 2.


    [Once again, like for the statement P1, the above statement Pk+1 could be expanded and simplified, but it doesn't help us see the pattern of the statements. Leaving it in the above, unsimplified form is fine for now. If we were proving by induction, the next step would be to expand and simplify to help us work with it, but all we want is to write out the statement, so we're fine for now.]
P1:     The number (13−1+2) is divisible by 2. P7:     The number (73−7+2) is divisible by 2. Pk:     The number (k3−k+2) is divisible by 2. Pk+1:     The number ((k+1)3−(k+1)+2) is divisible by 2.
Below is the general statement Pn.
Pn:     15  + 25  + 35  + … + n5   =   n2·(2n2+2n−1)·(n+1)2
12
Write out the statements P1, P7, Pk, and Pk+1.
  • A statement is just a string of words and/or symbols. In the case of the above statement, Pn is a statement that varies depending on the specific value used for n.
  • To find out what P1, P7, Pk, and Pk+1 are, we need to replace the n with the corresponding value. This is just simple substitution. Plug in the value you're working with for each place that n shows up.

    For P1, we have n=1, so substitute for every value of n in the general statement. The hardest part here is probably figuring out what to do with the left side (the part with the `…′ in it). To understand what to do with the sum, first make sure you understand what the notation is indicating. The expression 15  + 25  + 35  + … + n5 means that we start off with 15, then we add the next number up raised to the fifth (25), then the next number up raised to the fifth (35), and we continue to repeat this process until we reach n and add n5. However, for P1, we have n=1. Thus the summation process ends as soon as it starts. The very first value of 15 means we have already reached n5, so we only have one term on the left for P1:
    P1:     15   =   12·(2·12+2·1−1)·(1+1)2
    12
  • To find the other statements, we just plug in the appropriate value for n:
    P7:     15  + 25  + 35  + … + 75   =   72·(2·72+2·7−1)·(7+1)2
    12

    Pk:     15  + 25  + 35  + … + k5   =   k2·(2k2+2k−1)·(k+1)2
    12

    Pk+1:     15  + 25  + 35  + … + (k+1)5   =   (k+1)2·(2(k+1)2+2(k+1)−1)·((k+1)+1)2
    12


    [Remarks: The above statement Pk+1 could be expanded and simplified, but it doesn't help us see the pattern of the statements. Leave it in the above, unsimplified form for now. If we were proving by induction, the next step would be to expand and simplify to help us work with it, but all we want is to write out the statement, so we're fine. Furthermore, if we were going to prove by induction, we'd want the last statement Pk+1 to somehow be connected to Pk. We could do this by rewriting the left-side sum to show the connection:
    15  + 25  + 35  + … + (k+1)5     =     15  + 25  + 35  + … + k5  + (k+1)5
    These are exactly the same, it's just showing one more element from the part covered by the ellipsis (…). But by doing this, it makes it much easier to see that Pk is connected to Pk+1, which would allow us to plug in from the statement of Pk. However, we weren't asked to prove this by induction, so we're fine to leave it in the original form for now.]
P1:     15   =  [(12·(2·12+2·1−1)·(1+1)2)/12] P7:     15  + 25  + 35  + … + 75   =  [(72·(2·72+2·7−1)·(7+1)2)/12] Pk:     15  + 25  + 35  + … + k5   =  [(k2·(2k2+2k−1)·(k+1)2)/12] Pk+1:     15  + 25  + 35  + … + (k+1)5   =  [((k+1)2·(2(k+1)2+2(k+1)−1)·((k+1)+1)2)/12]
Using mathematical induction, prove that the below is true for every positive integer.
2 + 4  + 6  + … + 2n   =  n(n+1)
  • (Note: Make sure that you've watched the video lesson before working on these problems. The concept of mathematical induction can be rather difficult when you're first learning it, and it's important to understand the idea behind it and the steps involved. The method is carefully explained in the video, but it will be assumed in the below steps that you are already familiar with the process.) To prove something by induction, we first need a general statement Pn that makes sense for any positive integer n (notice that we haven't proven it is true yet, it just can't be total nonsense). The problem is pretty clearly set up in such a format, so we can just take the equation itself as our general statement:
    Pn:     2 + 4  + 6  + … + 2n   =  n(n+1)
  • Now that we have our general statement, we need to accomplish the two parts of mathematical induction. We begin with the base case. Base Case: Here we show that P1 is true. First, let us see what P1 is: plugging in n=1 into our general statement, we get
    P1:     2   =  1·(1+1)
    [If you're confused by why the left-side is simply `2', remember that the left-side is the sum of all the even numbers from 2 up until 2n. Since n=1, that means it's the sum of all the even numbers from 2 up until 2·1 = 2, that is, it starts where it stops, so the sum is just `2'.]
  • Now that we know the statement of the base case, we must verify that it is indeed true. Just do this with some simple arithmetic:
  • P1:     2   =  1·(1+1)
    P1:     2  =  1·(2)
    P1:     2   =  2       
  • Great, the statement P1 is clearly true, so the base case is complete.
  • Once the base case is complete, we move on to the inductive step. This is where where we will spend most of our time and effort. The inductive step begins with the inductive hypothesis, the assumption that Pk is true. Inductive Step: We assume our inductive hypothesis:
    Inductive Hypothesis-Pk:  2 + 4  + 6  + … + 2k   =  k(k+1)    is true.
    Using this inductive hypothesis, we now show that the next statement (Pk+1) must be true as well. That is, using our above assumption, we want to show the below is true:
    Pk+1:     2 + 4  + 6  + … + 2(k+1)   =  (k+1) ·⎛
    ⎝    		 (k+1)+1⎞
    ⎠
  • A key idea in induction is that we must use our inductive hypothesis of Pk to somehow prove that Pk+1 will be true as well. Thus, we need to make some portion of Pk appear in Pk+1 so that we can apply our hypothesis. Notice that we can rewrite the above formulation of Pk+1 to make it easier to see Pk in it:
    Pk+1:     2 + 4  + 6  + … + 2k  + 2(k+1)   =  (k+1) ·⎛
    ⎝    		 (k+1)+1⎞
    ⎠
    We haven't changed anything, we've just shown an extra term that was "hiding" in the ellipsis (…). From there we can plug in our inductive hypothesis of
    Pk:  2 + 4  + 6  + … + 2k   =  k(k+1).
    Let us plug in. We have
    Pk+1:    ⎡
    ⎣    		 2 + 4  + 6  + … + 2k⎤
    ⎦    		 + 2(k+1)   =  (k+1) ·⎛
    ⎝    		 (k+1)+1⎞
    ⎠    		 ,
    and we can clearly replace the bracketed [ ] part with what we know from our inductive hypothesis:
    ⎡
    ⎣    		 k(k+1)⎤
    ⎦    		 + 2(k+1)   =  (k+1) ·⎛
    ⎝    		 (k+1)+1⎞
    ⎠
  • Once we have managed to apply the inductive hypothesis, it is simply a matter of verifying that the equation is true.
  • k(k+1)  + 2(k+1)   =  (k+1) ·((k+1)+1)
    k2+k  + 2k+2   =  (k+1) ·(k+2)
    k2 + 3k + 2   =  k+ 3k + 2    
  • Excellent! Thus, from our inductive hypothesis that Pk is true, we have shown that Pk+1 must also be true.

    Therefore, by showing the base case and inductive step, the principle of mathematical induction says that the statement Pn must be true for all positive integers n. Thus we have shown that the below is true for all positive integers n:
    2 + 4  + 6  + … + 2n   =  n(n+1)       
Prove by induction. [By the nature of proof, it can't be given as just a single answer. Check out the steps to the problem if you want to see a version of the proof.]
Using mathematical induction, prove that the below is true for every positive integer.
1  + 6  + 11  + … + (5n−4)   =   n
2
 (5n−3)
  • (Note: Make sure that you've watched the video lesson before working on these problems. The concept of mathematical induction can be rather difficult when you're first learning it, and it's important to understand the idea behind it and the steps involved. The method is carefully explained in the video, but it will be assumed in the below steps that you are already familiar with the process.) To prove something by induction, we first need a general statement Pn that makes sense for any positive integer n (notice that we haven't proven it is true yet, it just can't be total nonsense). The problem is pretty clearly set up in such a format, so we can just take the equation itself as our general statement:
    Pn:     1  + 6  + 11  + … + (5n−4)   =   n
    2
    		 (5n−3)
  • Now that we have our general statement of Pn, we need to show that the base case of P1 is true. Base Case:
    P1:     1   =   1
    2
    		·(5·1 −3)
    Just simplify the equation and verify that it is true.
  • 1   =  [1/2] ·(5·1 −3)
    1   =  [1/2] ·(2)
    1   =  1     Great, the statement P1 is clearly true, so the base case is complete.
  • Once the base case is complete, we move on to the inductive step. This is where we will spend most of our time and effort. The inductive step begins with the inductive hypothesis, the assumption that Pk is true. Inductive Step: We assume our inductive hypothesis:
    Inductive Hypothesis-Pk:  1  + 6  + 11  + … + (5k−4)   =   k
    2
    		 (5k−3)    is true.
    Using this inductive hypothesis, we now show that the next statement (Pk+1) must be true as well. That is, using our above assumption, we want to show the below is true:
    Pk+1:     1  + 6  + 11  + … + (5k−4)  + ⎛
    ⎝    		 5(k+1)−4⎞
    ⎠    		  =   k+1
    2
    		⎛
    ⎝    		 5(k+1)−3⎞
    ⎠
  • Notice how we wrote Pk+1 so that it included an easily recognizable part from our inductive hypothesis of Pk. This allows us to apply our hypothesis.
    Pk+1:    ⎡
    ⎣    		 1  + 6  + 11  + … + (5k−4)⎤
    ⎦    		 + ⎛
    ⎝    		 5(k+1)−4⎞
    ⎠    		  =   k+1
    2
    		⎛
    ⎝    		 5(k+1)−3⎞
    ⎠
    We can easily substitute the bracketed portion [ ] for what we know from our inductive hypothesis Pk. Plug in:
    ⎡
    ⎣    		 k
    2
    		 (5k−3)⎤
    ⎦    		 + ⎛
    ⎝    		 5(k+1)−4⎞
    ⎠    		  =   k+1
    2
    		⎛
    ⎝    		 5(k+1)−3⎞
    ⎠
    Now we just expand and simplify both sides until it is clear that the equation is true:
  • [(5k2−3k)/2]  + (5k + 5 −4)   =  [(k+1)/2] (5k+5−3)
    [(5k2−3k)/2]  + (5k+1)   =  [(k+1)/2] (5k+2)
    [(5k2−3k)/2]  + [(10k+2)/2]   =  [((k+1))/2] ·[((5k+2))/1]
    [(5k2 + 7k + 2)/2]   =  [(5k2+7k+2)/2]     Excellent! Thus, from our inductive hypothesis that Pk is true, we have shown that Pk+1 must also be true.

    Therefore, by showing the base case and inductive step, the principle of mathematical induction says that the statement Pn must be true for all positive integers n. Thus we have shown that the below is true for all positive integers n:
    1  + 6  + 11  + … + (5n−4)   =   n
    2
    		 (5n−3)       
Prove by induction. [By the nature of proof, it can't be given as just a single answer. Check out the steps to the problem if you want to see a version of the proof.]
Using mathematical induction, prove that the below is true for every positive integer.
⎛
⎝		 1 + 1
1
⎞
⎠		⎛
⎝		 1 + 1
2
⎞
⎠		⎛
⎝		 1 + 1
3
⎞
⎠		…⎛
⎝		 1 + 1
n
⎞
⎠		  =  n+1
  • (Note: Make sure that you've watched the video lesson before working on these problems. The concept of mathematical induction can be rather difficult when you're first learning it, and it's important to understand the idea behind it and the steps involved. The method is carefully explained in the video, but it will be assumed in the below steps that you are already familiar with the process.) To prove something by induction, we first need a general statement Pn that makes sense for any positive integer n (notice that we haven't proven it is true yet, it just can't be total nonsense). The problem is pretty clearly set up in such a format, so we can just take the equation itself as our general statement:
    Pn:    ⎛
    ⎝    		 1 + 1
    1
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    2
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    3
    		⎞
    ⎠    		…⎛
    ⎝    		 1 + 1
    n
    		⎞
    ⎠    		  =  n+1
  • Now that we have our general statement of Pn, we need to show that the base case of P1 is true. Base Case:
    P1:    ⎛
    ⎝    		 1 + 1
    1
    		⎞
    ⎠    		  =  1+1
    Just simplify the equation and verify that it is true:
  • (1 + [1/1])   =  1+1
    (1 + 1)   =  2
    2   =  2     Great, the statement P1 is clearly true, so the base case is complete.
  • Once the base case is complete, we move on to the inductive step. The inductive step begins with the inductive hypothesis, the assumption that Pk is true. Inductive Step: We assume our inductive hypothesis:
    Inductive Hypothesis-Pk:  ⎛
    ⎝    		 1 + 1
    1
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    2
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    3
    		⎞
    ⎠    		…⎛
    ⎝    		 1 + 1
    k
    		⎞
    ⎠    		  =  k+1    is true.
    Using this inductive hypothesis, we now show that the next statement (Pk+1) must be true as well. That is, using our above assumption, we want to show the below is true:
    Pk+1:    ⎛
    ⎝    		 1 + 1
    1
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    2
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    3
    		⎞
    ⎠    		…⎛
    ⎝    		 1 + 1
    k
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    k+1
    		⎞
    ⎠    		  =  (k+1)+1
  • Notice how we wrote Pk+1 so that it included an easily recognizable part from our inductive hypothesis of Pk. This allows us to apply our hypothesis.
    Pk+1:    ⎡
    ⎣    		⎛
    ⎝    		 1 + 1
    1
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    2
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    3
    		⎞
    ⎠    		…⎛
    ⎝    		 1 + 1
    k
    		⎞
    ⎠    		⎤
    ⎦    		⎛
    ⎝    		 1 + 1
    k+1
    		⎞
    ⎠    		  =  (k+1)+1
    We can easily substitute the bracketed portion [ ] for what we know from our inductive hypothesis Pk. Plug in:
    ⎡
    ⎣    		 k+1⎤
    ⎦    		⎛
    ⎝    		 1 + 1
    k+1
    		⎞
    ⎠    		  =  (k+1)+1
    Now we just expand and simplify both sides until it is clear that the equation is true:
  • ( k+1) (1 + [1/(k+1)] )  =  (k+1)+1
    (k+1)  + [(k+1)/(k+1)]   =  k+2
    (k+1)  + 1   =  k+2
    k+2   =  k+2     Excellent! Thus, from our inductive hypothesis that Pk is true, we have shown that Pk+1 must also be true.

    Therefore, by showing the base case and inductive step, the principle of mathematical induction says that the statement Pn must be true for all positive integers n. Thus we have shown that the below is true for all positive integers n:
    ⎛
    ⎝    		 1 + 1
    1
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    2
    		⎞
    ⎠    		⎛
    ⎝    		 1 + 1
    3
    		⎞
    ⎠    		…⎛
    ⎝    		 1 + 1
    n
    		⎞
    ⎠    		  =  n+1       
Prove by induction. [By the nature of proof, it can't be given as just a single answer. Check out the steps to the problem if you want to see a version of the proof.]
Using mathematical induction, prove that the below is true for every positive integer.
n
∑
i=1 
 1
i(i+1)
  =   n
n+1
  • (Note1: Make sure that you've watched the video lesson before working on these problems. The concept of mathematical induction can be rather difficult when you're first learning it, and it's important to understand the idea behind it and the steps involved. The method is carefully explained in the video, but it will be assumed in the below steps that you are already familiar with the process. Note2: Furthermore, for this problem, it is necessary that you be familiar with sigma notation (Σ) for compactly writing summations. If you are not familiar with them, skim the lesson Introduction to Series where they are explained.) To prove something by induction, we first need a general statement Pn that makes sense for any positive integer n (notice that we haven't proven it is true yet, it just can't be total nonsense). The problem is pretty clearly set up in such a format, so we can just take the equation itself as our general statement:
    Pn:   n
    ∑
    i=1 
    		 1
    i(i+1)
    		  =   n
    n+1
  • Now that we have our general statement of Pn, we need to show that the base case of P1 is true. Base Case:
    P1:    1
    ∑
    i=1 
    		 1
    i(i+1)
    		  =   1
    1+1
    Just simplify the equation and verify that it is true:
  • ∑i=11 [1/(i(i+1))]   =  [1/(1+1)]
    [1/(1(1+1))]   =  [1/2]
    [1/2]   =  [1/2]     Great, the statement P1 is clearly true, so the base case is complete.
  • Once the base case is complete, we move on to the inductive step. The inductive step begins with the inductive hypothesis, the assumption that Pk is true. Inductive Step: We assume our inductive hypothesis:
    Inductive Hypothesis-Pk:  k
    ∑
    i=1 
    		 1
    i(i+1)
    		  =   k
    k+1
    		    is true.
    Using this inductive hypothesis, we now show that the next statement (Pk+1) must be true as well. That is, using our above assumption, we want to show the below is true:
    Pk+1:    k+1
    ∑
    i=1 
    		 1
    i(i+1)
    		  =   k+1
    (k+1)+1
  • At this point, since both statements Pk and Pk+1 are written in sigma notation (Σ), it's hard to see how we can apply our hypothesis of Pk to Pk+1. To deal with this, let's expand both of them into versions using an ellipsis (…) to see the pattern:
    Pk:   k
    ∑
    i=1 
    		 1
    i(i+1)
    		  =   k
    k+1
    		                                                              

           ⇒     1
    1(1+1)
    		 +  1
    2(2+1)
    		 + … +  1
    k(k+1)
    		  =   k
    k+1



    Pk+1:   k+1
    ∑
    i=1 
    		 1
    i(i+1)
    		  =   k+1
    (k+1)+1
    		                                                              

           ⇒     1
    1(1+1)
    		 +  1
    2(2+1)
    		 + … +  1
    k(k+1)
    		 +  1
    (k+1)⎛
    ⎝    		 (k+1)+1⎞
    ⎠
    		  =   k+1
    (k+1)+1
  • Notice that we wrote out the expansion of Pk+1 so that it included an easily recognizable part from our inductive hypothesis of Pk. This allows us to apply our hypothesis.
    Pk+1:  ⎡
    ⎣    		 1
    1(1+1)
    		 +  1
    2(2+1)
    		 + … +  1
    k(k+1)
    		⎤
    ⎦    		 +  1
    (k+1)⎛
    ⎝    		 (k+1)+1⎞
    ⎠
    		  =   k+1
    (k+1)+1
    We can easily substitute the bracketed portion [ ] for what we know from our inductive hypothesis Pk. Plug in:
    ⎡
    ⎣    		 k
    k+1
    		⎤
    ⎦    		 +  1
    (k+1)⎛
    ⎝    		 (k+1)+1⎞
    ⎠
    		  =   k+1
    (k+1)+1
  • Now we just expand and simplify both sides until it is clear that the equation is true:
  • [k/(k+1)]  + [1/((k+1)((k+1)+1))]   =  [(k+1)/((k+1)+1)]
    [k/(k+1)]  + [1/((k+1)(k+2))]   =  [(k+1)/(k+2)] Put the left-side fractions over a common denominator to combine them:
  • [((k+2))/((k+2))]·[k/(k+1)]  + [1/((k+1)(k+2))]   =  [(k+1)/(k+2)]
    [(k(k+2))/((k+1)(k+2))]  + [1/((k+1)(k+2))]   =  [(k+1)/(k+2)]
    [(k2+2k+1)/((k+1)(k+2))]   =  [(k+1)/(k+2)] Finally, notice that we can factor the left-side numerator to cancel out a common factor in the denominator:
  • [((k+1)(k+1))/((k+1)(k+2))]   =  [(k+1)/(k+2)]
    [(k+1)/(k+2)]   =  [(k+1)/(k+2)]     Excellent! Thus, from our inductive hypothesis that Pk is true, we have shown that Pk+1 must also be true.

    Therefore, by showing the base case and inductive step, the principle of mathematical induction says that the statement Pn must be true for all positive integers n. Thus we have shown that the below is true for all positive integers n:
    n
    ∑
    i=1 
    		 1
    i(i+1)
    		  =   n
    n+1
    		      
Prove by induction. [By the nature of proof, it can't be given as just a single answer. Check out the steps to the problem if you want to see a version of the proof.]
Prove that, for any positive integer n, the number (n2+3n) is an even number.
  • (Note: Make sure that you've watched the video lesson before working on these problems. The concept of mathematical induction can be rather difficult when you're first learning it, and it's important to understand the idea behind it and the steps involved. The method is carefully explained in the video, but it will be assumed in the below steps that you are already familiar with the process.) To prove something by induction, we first need a general statement Pn that makes sense for any positive integer n (notice that we haven't proven it is true yet, it just can't be total nonsense). The problem isn't set up as an equation, but that's okay: our general statement can use everyday language as well.
    Pn:     The number (n2+3n) is an even number.
    Notice how the above is just a re-statement of what the problem is asking us to prove. It's our job to show that the above statement is true for every positive integer n, so now we turn to mathematical induction to prove it.
  • Now that we have our general statement of Pn, we need to show that the base case of P1 is true. Base Case:
    P1:     The number (12+3·1) is an even number.
    To verify that P1 is true, let's look at the number:
    (12+3·1)     =     1 + 3     =     4
    We know the number 4 is an even number, so the statement P1 is clearly true and the base case is complete.    
  • Once the base case is complete, we move on to the inductive step. The inductive step begins with the inductive hypothesis, the assumption that Pk is true. Inductive Step: We assume that our inductive hypothesis is true:
    Inductive Hypothesis-Pk:  The number (k2+3k) is an even number.
    Using this inductive hypothesis, we now show that the next statement (Pk+1) must be true as well. That is, using our above assumption, we want to show the below is true:
    Pk+1:     The number⎛
    ⎝    		(k+1)2+3(k+1)⎞
    ⎠    		is an even number.
    In other words, we must show that ((k+1)2+3(k+1)) is indeed an even number, thus verifying the statement Pk+1.
  • Unlike previous problems, we don't have a clear way to plug in and use the hypothesis of Pk. Previously, all the statements were two-sided equations, and one of the sides contained a part that looked like part of the hypothesis Pk, which allowed us to plug in. This is not the case for this problem, though. However, we can still follow that general idea. We need to somehow make some portion of Pk appear in the statement of Pk+1. We can do this by expanding the number Pk+1 is based on, then hopefully seeing the number Pk is based on contained in that. Try expanding Pk+1's number:
    ⎛
    ⎝    		 (k+1)2+3(k+1)⎞
    ⎠

    k2+2k +1   +  3k+3

    k2 + 5k + 4
    At this point, remember that the number from Pk was (k2+3k). Notice that we can pull that part to one "side" in the number from Pk+1:
    k2 + 5k + 4     =    ⎡
    ⎣    		 k2+3k⎤
    ⎦    		 + 2k+4
  • At this point, we can finally use our inductive hypothesis. The hypothesis Pk guaranteed that (k2+3) is an even number. We have shown that the number from Pk+1 can be written as
    ⎡
    ⎣    		 k2+3k⎤
    ⎦    		 + 2k + 4.
    Thus, this number is made from adding up three numbers:
    [k2+3k],              2k,              4
    Long ago in grade school, we learned that an even number added with an even number produces an even number. Thus, if we can show that each of the three numbers above are even, it must be that Pk+1 is even as well. From the inductive hypothesis Pk, we know (k2+3k) is even. Since multiplying any number by 2 creates an even number, we also know that 2k must be even. Finally, 4 is clearly even. Therefore, since each number involved is even, we have that the whole sum must be even as well. Thus, the number in Pk+1 is even, which verifies the statement we set out to show and completes the inductive step.    

    Therefore, by showing the base case and inductive step, the principle of mathematical induction says that the statement Pn must be true for all positive integers n. Thus we have shown that the below is true for all positive integers n:
    The number (n2+3n) is an even number.       
Prove by induction. [By the nature of proof, it can't be given as just a single answer. Check out the steps to the problem if you want to see a version of the proof.]
Prove that, for any positive integer n, the number (9n−2n) is divisible by 7.
  • (Note: Make sure that you've watched the video lesson before working on these problems. The concept of mathematical induction can be rather difficult when you're first learning it, and it's important to understand the idea behind it and the steps involved. The method is carefully explained in the video, but it will be assumed in the below steps that you are already familiar with the process.) To prove something by induction, we first need a general statement Pn that makes sense for any positive integer n (notice that we haven't proven it is true yet, it just can't be total nonsense). The problem isn't set up as an equation, but that's okay: our general statement can use everyday language as well.
    Pn:     The number (9n−2n) is divisible by 7.
    Notice how the above is just a re-statement of what the problem is asking us to prove. It's our job to show that the above statement is true for every positive integer n, so now we turn to mathematical induction to prove it. [Make sure you know what the word divisible means. If you're not sure, look it up. It means that the number can be "evenly" divided by 7. In other words, we can divide out 7 and have no remainder.]
  • Now that we have our general statement of Pn, we need to show that the base case of P1 is true. Base Case:
    P1:     The number (91−21) is divisible by 7.
    To verify that P1 is true, let's look at the number:
    (91−21)     =     9−2     =     7
    The number 7 is clearly divisible by 7, so the statement P1 is true and the base case is complete.    
  • Once the base case is complete, we move on to the inductive step. The inductive step begins with the inductive hypothesis, the assumption that Pk is true. Inductive Step: We assume that our inductive hypothesis is true:
    Inductive Hypothesis-Pk:  The number (9k−2k) is divisible by 7.
    Using this inductive hypothesis, we now show that the next statement (Pk+1) must be true as well. That is, using our above assumption, we want to show the below is true:
    Pk+1:     The number ( 9k+1 − 2k+1 ) is divisible by 7.
    In other words, we must show that ( 9k+1 − 2k+1 ) is indeed divisible by 7, thus verifying the statement Pk+1.
  • Unlike past problems, we don't have a clear way that we can plug part of Pk in to the expression we have for Pk+1. However, we know that we somehow need to have the number from Pk appear in the number for Pk+1 so we can use our inductive hypothesis. With this in mind, look for a way to re-arrange the number for Pk+1 so that the number from Pk (that is, 9k−2k) will appear:
    Pk+1     ⇒     9k+1 − 2k+1
    First, we need to have 9k and 2k appear, so take the exponents down by one by "popping out" a multiplier of the appropriate base:
    9k+1 − 2k+1     =     9k·9 − 2k·2
    Next, we need to get the expression (9k−2k) to appear. We can do this by moving things around cleverly:
    9k·9 − 2k·2     =     9k ·2 + 9k ·7 − 2k ·2     =     9k·2 − 2k ·2 + 9k ·7
    Finally, we cause it to clearly appear by using the distributive property in reverse:
    9k·2 − 2k ·2 + 9k ·7     =     2·( 9k − 2k )  + 9k ·7
  • Now we have the number from Pk+1 expressed in such a way that we can use our inductive hypothesis from Pk. However, to formally prove this, we need a way to talk about what our inductive hypothesis is saying. Notice that the below
    Pk:  The number (9k−2k) is divisible by 7,
    is equivalent to the statement that
    There exists some integer number a such that   9k−2k = 7 ·a.
    Why? If there is some integer a where 9k−2k = 7 ·a, then (9k−2k) must be divisible by 7 since (7·a) is clearly divisible by 7. And if no such number a existed that would work, then (9k−2k) could not be divisible by 7, so we know the a must exist (even if we don't know its value).
  • With this re-statement of Pk in terms of (7·a) in mind, let's look again at the number from Pk+1:
    Pk+1     ⇒     ( 9k+1 − 2k+1 )     =     2·( 9k − 2k )  + 9k ·7
    Swap out the ( 9k − 2k ) for (7·a):
    2·(7 ·a)  + 9k ·7
    We can now pull out a 7 from both terms to get
    7 · (2 ·a + 9k),
    which must be divisible by 7, because we clearly have a factor of 7 contained in the number. Thus we have shown that Pk+1 is divisible by 7, which completes the inductive step.    

    Therefore, by showing the base case and inductive step, the principle of mathematical induction says that the statement Pn must be true for all positive integers n. Thus we have shown that the below is true for all positive integers n:
    The number (9n−2n) is divisible by 7.       
Prove by induction. [By the nature of proof, it can't be given as just a single answer. Check out the steps to the problem if you want to see a version of the proof.]
Below is a false theorem and its "proof" using mathematical induction. Looking at the theorem, it's clearly not true, so something must be wrong with the proof. What's the mistake in the proof?

Theorem: All horses are the same color. "Proof" (by mathematical induction): Let the general statement be
Pn:     A group of n horses are all the same color.
       Base Case: Consider the statement P1:
P1:     A group of 1 horse is all the same color.
This statement is trivially true: a single horse must be the same color as itself. Thus the base case is shown to be true.          Inductive Step: Begin by assuming the inductive hypothesis:
Inductive Hypothesis-Pk:  A group of k horses are all the same color.
Now we show that the statement Pk+1 must follow from Pk being true:
Pk+1 :     A group of (k+1) horses are all the same color.
To prove that this is true, consider the diagram below where we have lined all (k+1) horses up in some arbitrary order. Within this group of (k+1) horses, we can break it up into two sub-groups of k horses. Notice that by our hypothesis Pk, we have that each of these two sub-groups of horses must all be the same color, like in the below diagrams: Furthermore, looking at the diagrams, we see that there is some "middle horse" that appears in both sub-groups. Since this "middle horse" can not change colors, it must be that both sub-groups are the same color, and thus the entire group of (k+1) horses must all be the same color. Thus we have shown that all (k+1) horses must be the same color, so Pk+1 is true, completing the inductive step.     Therefore, by mathematical induction, we have that any group of n horses must all be the same color. Since there exists some finite number of horses in the world, they must all be the same color, so all horses are the same color.    
  • Begin by working through the proof: at first glance, everything seems to work fine. The base case is shown to be true, then the inductive step is shown to be true, so the general statement must be true. However, we know that the theorem is nonsense. If you've ever seen a group of horses, they clearly are not all necessarily the same color. Therefore there must be something wrong with the proof. Start looking through it very carefully, step-by-step.
  • To begin with, the general statement Pn is just fine. As long as the language and meaning is clear, there can be nothing inherently wrong with a statement: the job of the later steps is to prove that it is always true, but the statement we start with can be anything.
  • Next, consider the base case of P1:
    P1:     A group of 1 horse is all the same color.
    Nothing wrong here, either. A horse does have to be same color as itself, so a group with just one horse can only be one color-the color of the horse in the group.
  • Now we move on to the inductive step. The "proof" begins by putting all (k+1) horses from Pk+1 in some arbitrary order. There is nothing wrong with this: you can order any finite set, so we can put our horses in some order. Once the horses are ordered, the "proof" separates the group of (k+1) horses into two sub-groups of k horses, one where the last horse is excluded and one where the first horse is excluded. There's nothing wrong with this either: given a group, we can break it into sub-groups, and since it's ordered, we can easily choose what goes into each sub-group.
  • Finally, we make it to where the mistake occurs: the next part of the "proof" says that these two sub-groups overlap on some "middle horse". But do we have a guarantee that this "middle horse" actually exists? While the "proof" refers to the diagrams to show that the overlap is supposedly clear, we must remember that the diagrams have to represent all possible values for k. That is, k+1 ≥ 2. But what happens if we consider the case where k+1=2? That is, when there are just two horses?
  • If we consider the case of just two horses, we have a diagram like the one below: Next, we break it into two sub-groups, where each sub-group is all one color: We see that in the case of looking at only two horses, there is no overlapping "middle horse", so we cannot guarantee that the entire group must all be the same color.
  • This shows us the issue with the original "proof". While the original diagrams seemed to indicate the existence of an overlap, they failed to properly show all the cases. In the specific case of k+1=2, there cannot be an overlapping "middle horse", so we have no guarantee that both sub-groups must be the same color. Thus, we cannot show that the entire group of horses must be the same color. Because the inductive step is based on a flawed assumption, the proof does not hold.
  • The lesson we should take away from this problem is that we must be very careful in our assumptions when working on proofs. Even if something seems "obvious", we must carefully check that we can assume it before using it. When we take something for granted without carefully thinking about it, we introduce the possibility for serious error.
We cannot assume the automatic existence of an overlapping "middle horse". In the case of k+1=2 horses, there is no horse in both sub-groups, so there is no guarantee that the sub-groups have matching colors. [See the steps to the problem for a more detailed explanation along with some explanatory diagrams.]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Mathematical Induction

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:06
  • Belief Vs. Proof 1:22
  • A Metaphor for Induction 6:14
  • The Principle of Mathematical Induction 11:38
    • Base Case
    • Inductive Step
    • Inductive Hypothesis
  • A Remark on Statements 14:18
  • Using Mathematical Induction 16:58
  • Working Example 19:58
    • Finding Patterns
  • Example 1 30:17
  • Example 2 37:50
  • Example 3 42:38

Transcription: Mathematical Induction

Hi--welcome back to Educator.com.0000

Today, we are going to talk about mathematical induction.0002

The great thing about math is that we can trust it.0005

Regardless of the technology that math enables us to have in our daily lives (like the Internet), what we learn in math is entirely, beautifully true.0007

We can have certainty in this truth, because of proof.0016

In math, we start with a few simple ideas, and then we use proof to build a tower of logic.0019

Because we prove every theorem and method that we work with, we can trust the whole of mathematics.0025

Mathematics is built on this concept of truth--that we start with some things that we can just take for granted,0030

absolutely certainly--these axiomatic truths that we looked at in geometry, for example, previously--you have seen it in some courses.0037

We start with these basic axioms, and then we work up from there, using logic.0044

We can use proof to make sure that we can trust every single thing that we work with.0048

So far in this course, we have proved many things with a variety of logical arguments.0052

We have made very strong logical arguments that show, without a doubt, that something has to be true.0056

In this lesson, we are going to see a new form for proofs, mathematical induction.0061

This method can be used anywhere, from our current level of math all the way to very advanced mathematics.0066

It is this really flexible kind of proof that lets us show a lot of things; it is a very powerful way of proving things.0071

All right, let's talk about belief versus proof.0077

In math, it is important to recognize that there is a difference between believing that something is true and proving that something is true.0080

While we may strongly believe in something, that implies that there is still some level of reasonable doubt.0087

Believing in something doesn't mean that we can know it absolutely certainly; it just means that it is probably true.0092

But there is still this tiny crack of doubt that maybe it is not going to end up being true.0098

We can only be certain about something once we have proved it; we can only be certain that something is true once we have proof.0102

A proof is an irrefutable logical argument; it is something that very clearly shows (and there is no way to disagree with it)0110

that, given some something, we can show that it is always going to end up working out; it gives us total certainty.0116

This means that, when we notice a pattern in the way things work, we don't know for sure that it will always hold true.0124

We may believe that it will hold, but until we prove it, we cannot be absolutely certain.0129

So, we may believe in something, but we have to go and prove it before we can be absolutely certain.0134

Proof is this really integral part to how mathematics works.0140

As an example, consider the following: in the 1600s, a mathematician named Fermat noticed a pattern.0145

It seemed to him that all numbers of the form below were prime numbers.0150

It is pn, the nth Fermat number, 2 to the 2 to the n plus 1, where n is a natural number0154

(0, 1, 2...working our way up)...if we have p0, then that would be equal to 2 to the 2 to the 0 plus 1.0162

What is 220? Well, 2 to the 0...any number raised to the 0 is 1, so it is 2 to the 1.0168

So, we have 2 to the 2 to the 0 plus 1...2 to the 1, plus 1...that gets us 3; and indeed, 3 is a prime number.0174

Next, the next Fermat number would be 221 + 1; that gets us 5, and indeed, that is a prime number.0181

The next one would be 2 to the 2 to the 2, plus 1; that is going to come out to be 17; indeed, that is a prime number.0189

Now, we are starting to get to a place where we can't immediately see that they are prime.0194

But 223 + 1 is 257; and you can check--that does end up being prime.0197

The next one would be 2 to the 2 to the 4 plus 1, which comes out to be 65537;0203

and indeed, if you work at it, you can eventually check by hand that this does come out to be a prime number, as well.0208

So, Fermat could verify that this was true for the values of 0, 1, 2, 3, and 4--raising 2 to the 2 to the 0, to the 1, to the 2, to the 3, to the 4, plus 1.0214

Each one of these came out to be a prime number.0226

So, he could check by hand that you are able to make primes for these first five Fermat numbers (0, 1, 2, 3, 4).0229

He was able to show for sure that they do make primes.0238

However, the next Fermat number was too large for him to check, and he couldn't figure out a way to prove it prime.0242

The next one is 225, plus 1, which is equal to a huge 4 billion, 294 million, 967 thousand, 297--a big number.0246

That is going to be really hard to check by hand; and Fermat wasn't able to check by hand.0256

So, he couldn't figure out a clever way to prove for sure that it was prime without having to check by hand.0260

And it was too big for him to check by hand, so he couldn't show that it was prime.0265

Nonetheless, he had noticed a pattern; so he made a conjecture, that is, a hypothesis--an expectation,0268

of how things will work: that all numbers of the form 22n + 1 are prime.0274

But noticing a pattern does not necessarily prove it to be true.0281

This was a conjecture, but we didn't know for sure whether it was going to end up being a true conjecture0284

or a false conjecture--if it was going to end up being disproven.0288

It was not until the 1700s that we knew whether or not the conjecture was true.0292

Another mathematician, named Euler, showed that the sixth Fermat number, p5, could be factored.0296

So, 2 to the 2 to the 5, plus 1, which comes out to be 4 billion, 294 million, 967 thousand, 297, can be broken down0303

into the two numbers 641 times 6 million, 700 thousand, 417.0310

So, since that Fermat number can be broken down into two factors, that means that it is not a prime number.0316

Thus, Euler had shown that not all Fermat numbers are prime, because there is at least one of them that can be broken into factors.0323

Thus, he had proven the conjecture false; he had shown that there was a counterexample.0330

So, that conjecture that Fermat had originally thought, that 2 to the 2 to the n plus 1 was prime--that is not true.0334

We have proven it to not be true.0341

It is relatively easy to prove that something is not true; you only need a single counterexample--0344

one counterexample that cancels out and says, "No, here is a way that you can't have that work."0349

That proves that that does not work; but how can we prove that a pattern will hold forever,0353

that we have some sequence of things that is going to always end up being the case?0358

We can do this with mathematical induction.0363

There are other ways to do this; you can make logical arguments in other ways.0365

But a good way to do this is with mathematical induction; it often works well.0368

Before formally stating the principle of mathematical induction, let's look at a metaphor to help us understand how induction works.0372

Here is our metaphor: let's imagine that we have an infinitely long line of dominoes, and they are stood up on their ends, like in this picture here.0379

Domino, domino, domino, domino, domino...it is like if we were setting up a bunch of dominoes0387

to push it over and have (thump, thump, thump, thump, thump, thump, thump...) falling-over dominoes.0391

That is the image that we want to have in our mind.0395

OK, we could name the dominoes, based on their location in line.0397

The very first domino that would start on the line, on the very left side, we could call d1.0401

Here would be our d1, our first domino.0407

The next domino we could call d2; our second domino is d2, and so on.0409

Here would be d3 and d4, and it is going to just continue on in that pattern.0414

So, any domino...there exists some n that we can name any domino with.0418

We talk about some domino: there is some number that we can associate with it,0422

because we can just count up from the beginning and see what number we have to count to, to get to that domino.0425

That means that we can name any given domino.0429

All right, we have two guarantees: consider if we had two guarantees: the first guarantee0432

is that the first domino, d1, will definitely fall over.0440

We are going to be absolutely certain that d1 is going to fall over.0445

Our first domino, d1, is for sure going to fall over; we know that our very first domino in the line is going to fall over.0449

It is going to get pushed somehow; a gust of wind is going to happen...an earthquake...but something will certainly happen.0457

Who knows what it is, but we are guaranteed that it will fall over, for sure.0462

We know that our very first one is going to fall over; that is our first guarantee.0466

Our second guarantee is that, given any sequential pair, any pair of dominoes0470

(we could, for example, talk about some dk and dk + 1 for any positive integer k,0477

some dk in the line, and then the next one would be called dk + 1--0482

if we are at the k spot, the k + 1 would be the next spot), if dk falls,0486

if the domino on the left falls (dk is the domino on the left), then the next domino in line,0490

dk + 1, will also fall; this is our next guarantee.0501

So, if the domino on the left falls, then the next domino will fall, as well.0506

We are going to get falling from the next domino; if dk falls, dk + 1 must fall.0516

We have here that dk falls; and then, at the next one...we have dk falls;0522

if dk falls, dk + 1 must also fall.0527

If one domino falls, the next domino in line is always guaranteed to fall.0531

We have this guarantee that the dominoes do, indeed, always push each other over for any pair of dominoes that we end up looking at.0535

All right, we have these two things for sure: we have two guarantees and this infinite line of dominoes.0543

There is some line of dominoes, and the two guarantees: that the first domino, d1,0549

is definitely going to fall over; and that, if some domino, some dk falls over,0554

it will cause the next domino, dk + 1, to also fall over.0560

Given these two guarantees, we now know that absolutely all of the dominoes in our infinitely long line of dominoes will fall over.0564

We know for sure that all of them are going to fall over.0575

Why? Well, because, by the first guarantee, we know that d1 falls.0578

We know that the very first one in the line has to fall, because we are guaranteed that it is going to fall.0584

Then, by the second guarantee, if some domino falls, then we know that the next domino falls.0588

By our first guarantee, we know that d1 must fall.0594

But by the second guarantee, we know that d2 must fall, because d1 is the one before d2.0597

So, d2 must now fall, as well; then, by our second guarantee, once again,0602

since d2 just fell, we know that d3 must fall.0606

Then, by our second guarantee, again, we know that since d3 just fell, we know that d4 must fall.0609

And then, since d4 just fell, d5 must fall; d6 must fall; d7 must fall;0614

d8, d9, d10, forever and ever and ever.0618

Thus, all of the dominoes must fall over; we know that every single domino must fall over,0620

because we are guaranteed that the first domino is going to fall, and we know that every domino0626

after a domino that has fallen will fall as well.0630

So, since d1 falls, we know that d2 must fall; we know that d3 must fall;0633

and we can work our way out, going forever.0636

This is inductive thinking, step-by-step logic; this is the idea of mathematical induction,0638

that we have this first premise, that the definite thing is going to happen for the first thing;0644

and that, if something happens, we know that the next thing has to happen, as well.0648

We can start at d1, and from that induction idea, we can move forward forever, stepping along.0652

Mathematical induction is extremely similar; this idea of using it in math is very, very much the same thing as this line of dominoes.0658

It is just that, instead of dominoes, we will work with statements; and instead of "falls over," we will say "is true."0666

But other than that, the logic behind it is exactly the same.0673

And I know that that might seem like a crazy thing to say--that we can connect dominoes and statements,0676

and we can connect falling over with being true; but really, the idea behind it is the same.0680

And we will see why in just a second.0684

It is that we have this one thing that starts it, and then we have this other guarantee that says that it keeps going forever and ever and ever.0686

All right, now we are ready to look at the principle.0691

The principle of mathematical induction is: let p1, p2, p3, p4...0694

going on...pn...going on forever, be some infinitely long sequence of statements,0700

or just some sequence of statements, where n is a positive integer.0705

We are stepping forward, one statement at a time.0708

And we are guaranteed two things: if p1 is true, and if pk is true for a positive integer k,0711

then pk + 1 must also be true; then we know that the statement pn is true for all positive integers n.0721

Why is this the case? Well, it is very much the same thing as we just saw with the dominoes.0729

We have p1; it is definitely true--the very first one (we have p1, p2, p3,0732

p4, p5...going on forever; we effectively have this infinitely long line of sequences).0736

And we are guaranteed, by our very first thing, that p1 has to be true; p1 is true.0741

Then, our next guarantee is that, if something is true, the next thing in line must be true.0747

Our first guarantee was that p1 is true; that means that p2,0752

since it is the next thing in line, must also be true.0755

That means, by that same guarantee, that the next thing in line is true, that p3 must be true;0757

p4 must be true; p5 must be true; p6 must be true.0761

And it goes on forever and ever and ever, and that is why we now know that pn is going to be true for all positive integers n.0763

It is because, since the first thing in line was guaranteed by that first guarantee,0771

and the second thing is going to end up happening because of that second guarantee,0775

and then the third one will happen because of the second guarantee again,0778

and then the fourth thing will happen because of the second guarantee again...0782

we can keep using that step, step, step, step, step--that the second guarantee gives us.0784

We are able to step forever and ever and ever; we are able to have that infinitely many of these statements are going to end up being true.0788

All right, we often call the first step (that p1 is true) the base case,0794

because it is establishing the basis that we are working from.0801

It is the very first one in the line of true statements.0803

The second step is often called the inductive step, because it is the thing where the actual induction occurs--0806

that induction being "if 1, then 2," "if 2, then 3," so 1, then 2, then 3, then 4, then 5, then 6...0813

the fact that we can start at one place, and then step forever forward.0821

In the inductive step, the assumption, if pk is true (this part right here, the "if pk is true")0824

is called the inductive hypothesis, because it is the hypothesis that we use to show that pk + 1 will also be true.0831

We have to have this assumption that, if this thing happens, then this other thing will happen.0839

It is just like, with the dominoes, we say, "If some domino falls over, then this thing is going to happen in response."0844

But now, we are talking in terms of truth: "If this thing is true, then the next thing down the line must be true, as well."0849

All right, before we keep going, I want to have a quick remark on statements.0856

The principle of mathematical induction is based on statements.0859

We might be tempted to assume that any statement we see written out will automatically be true.0863

But that is not necessarily the case: just because you see something written out does not mean that you want to assume it will always, always be true.0868

Don't believe everything you read.0873

A statement, like in English, is just a string of words and/or symbols.0875

For example, the statement "Ice cream is made of hair" is most emphatically not true.0879

Ice cream is not made of hair; that is not true.0884

But it is still a statement, albeit a false statement.0887

Just because something is said does not mean that it is always going to end up being true.0891

Similarly, in math, the statement 3 + 5 + 7 = 10 is false; that is not true: 3 + 5 + 7 is not equal to 10.0895

But it is still a statement--just not a true statement.0903

3 + 5 + 7 is actually equal to 15, so we know that this one here is a false statement.0905

Most of the time, when we are working in math, we are always allowed to assume that the statements we are working with are true,0911

because they are given to us directly from the problem, and we are not told to question them.0916

We are saying, "Start with this thing and work with it."0919

Or we are working on the problem from the beginning (like a word problem), and we see, from the information that we are given,0922

that we can use logic to say, "Oh, this must definitely be the case; this must definitely be the case."0926

We are able to work with these things and be certain of their truth.0931

But just because something is said does not necessarily mean that it is always going to be true.0933

Everything that we have dealt with so far in math--we have been able to assume the truth of things,0937

because we know that we weren't being given lies.0941

But now, if we are just talking about statements in general, we have to evaluate their truth before we can trust them.0944

When we are working with mathematical induction, it is our job to not only set up the statements, pn,0950

but to also prove that the statements are going to be true statements.0956

We are not just going to get the statements and assume that they are true.0961

We have to get the statements, and then show for sure that they end up being true.0963

And occasionally, we might discover that the statements we have actually end up being false statements,0967

at which point we will have to go back to the beginning and figure out a way that we can state it truthfully;0971

or that will end up being the answer--the fact that it is actually not true.0974

But generally, we are going to end up showing that it is true.0977

Now, of course, if we want to show that n goes on infinitely, that all pn for any 1, 2, 3, 4, going on forever--0979

that they are all true--we can't prove each one of them by hand, because we can't prove an infinite number of things.0985

We are not going to be able to have an infinite amount of time to prove an infinite number of things.0990

So instead, we have to use some way to be able to show all of them at once.0995

We can either use some sort of clever logical argument, or we can use proof by mathematical induction,0999

where we can show that the first step will end up always being true,1003

and that, if you have some step, the next thing is going to end up being true, to work our way out infinitely,1006

like that line of dominoes, like we were just talking about a few moments ago.1011

All right, how do we use mathematical induction?1015

If we are going to use induction to prove that some statement, pn, is true for all n, we must prove two things:1017

first, the base case, that p1 is true--that our very first statement is a true statement;1022

and then, we have to do the inductive step: that if some pk is true,1029

the next one in line, pk + 1, must also be true.1033

So, that means, when we are working with proof by induction, that we have at least two separate steps.1036

We have two separate things that we have to deal with: proving the base case and proving the inductive step.1042

We have to show each one of these.1048

I am going to strongly recommend noting which one you are working on and making it so that you can keep the parts separate.1049

How you work on the base case normally isn't going to really help you deal with the inductive step.1055

How you work on the inductive step normally isn't really going to help you deal with the base case.1058

So, just write "base case" and prove your base case; then write "inductive step"; then prove your inductive step.1061

It will help you keep track of what you are doing; it will help keep it from being confusing.1066

And it will help other people who read your work in the future to understand what happened there.1069

In general, proving the base case almost always is going to be easier by far.1073

Normally, we just end up simplifying the statement--whatever this p1 statement ends up being;1078

and we verify that it is, indeed, a true statement--it is just making sure that, yes, this is true.1083

The tricky part normally ends up being dealing with the inductive step.1089

To prove the inductive step, we must first assume the inductive hypothesis, "pk is true."1093

We have to start by assuming that if pk is true...we have to say,1098

"OK, pk is true; now let's show that pk + 1 is also going to be true."1102

Once we make this assumption, we somehow have to work to prove that pk + 1 must then also be true.1108

Now, notice: we haven't actually proven pk to be true; we are just asking ourselves,1112

"If pk were true, would pk + 1 also be true?"1118

We are saying, "If we had this place that we could start from, would the next thing be true?"1122

It is like the dominoes: if this domino were to fall over, would the next one in line fall over?1125

That is all we need for mathematical induction: we don't have to prove that the one in the middle will fall over.1130

We just have to say, "Well, if it were to fall over, would the next one fall over, as well?"1133

All right, it helps a lot to somehow include part of the statement for pk when we are writing out pk + 1.1138

This is because we are almost inevitably going to end up using the inductive hypothesis.1146

We have to use this inductive hypothesis, "pk is true."1151

Otherwise, we wouldn't have needed to talk about pk being true, to show that pk + 1 is true.1154

So, we are going to end up using that inductive hypothesis, that pk is true.1159

So, we have to have some way of talking about it in our pk + 1 statement.1163

We can't use our hypothesis about pk unless it somehow appears in pk + 1.1167

That means, when we end up writing out the pk + 1 statement, that we have to figure out some way1175

to make pk, or some portion of pk, show up in some way,1178

so that we can apply that inductive hypothesis and use what we have assumed to be true.1182

It is important to have that end up showing up.1188

We will see how that occurs as we are actually working through problems and examples.1190

All right, all of this is going to make a lot more sense once we see it in action.1194

So, if it seems a little bit confusing right now, just follow it through; it is going to make much more sense as we actually work through what is going on.1197

All right, consider the following formula, which we proved in the lesson Arithmetic Sequences and Series:1203

we proved that adding the numbers 1 + 2 + 3 + ... up until we get to + n, all of the numbers consecutively,1209

from 1 up until some n that we choose, is equal to n times n + 1 divided by 2.1215

Previously, we proved this by a logical argument that involved equations and elimination.1222

Adding up two equations (if you watched that lesson), we were able to prove it through logical argument.1227

But we also could prove it through mathematical induction; let's see another way to prove this, through mathematical induction.1231

We can see that the above equation is effectively our statement, pn; we can turn this into our statement, pn, this right here.1238

How would this work? Well, our first statement, p1, would be if we went from 1 up until an n of 1 (p1, so 1 up until 1).1246

Well, that is just going to end up being 1; so 1 added to itself--is that, indeed, equal to 1 times 1 + 1 over 2?1254

That is the first statement: the first statement is that 1 is equal to 1 times 1 + 1 over 2.1261

Our next statement, at p2, would be 1 +...up until 2, so that is just going to be 1 + 2;1266

and that is going to be equal...the statement says that it is equal;1271

it will be up to us to verify that this actually ends up working out...to 2 times 2 + 1 over 2.1274

And then, the next one, p3, would be equal to 1 + 2 + 3...we get up to the third;1279

and it is saying...the statement is that that is, indeed, equal to 3 times 3 + 1 over 2.1284

And finally, the really interesting one here is pn, which is adding all the numbers from 1 up until n,1288

and that that is equal to n times n + 1 over 2.1294

So, this is our statement, pn--this way of writing it out here.1296

I didn't mean to cut through that 2.1301

To prove by induction that the statement pn = 1 added to all the numbers up until n is equal to n times n + 1 over 2--1303

to prove this as true for all positive integers, if we are going to use proof by induction, we have to prove two things.1310

We have to prove both the base case, p1, and the inductive step, that pk implies pk + 1.1314

The truth of some pk in the middle implies that pk + 1, the next one down the line, also has to be true.1322

Let's begin with the base case; the base case is normally the easier thing, by far.1328

Our p1 statement was that 1 is equal to 1 times 1 + 1 over 2.1331

Now, that is just a statement; once again, it is up to us to verify the truth of the statement.1337

So, we will write in the middle here this little equals sign with a question mark, because we don't actually know for sure that it is equal.1341

The statement said that 1 is equal to 1 times 1 + 1 over 2; but we don't know for sure that that is true.1347

The statement could be lying to us; it is up to us to figure out if this is, indeed, true;1353

are the left side and the right side actually equal to each other?1357

We work this out: we have 1; is it equal to...let's check: 1 times 1 + 1 over 2...1 compared to 1 times 2/2 (1 + 1 is 2), and then1361

that simplifies; the 2's cancel out, and we have 1 = 1; and indeed, that is true.1370

So, that means that our base case checks out; we have shown that the base case is definitely true.1374

Next, we move on to the inductive step: we write "inductive step," so that we see where we are.1380

The very first thing that we always do with the inductive step is write out our inductive hypothesis.1385

We say, "If pk is true..." so we assume that pk (which would be 1 + 2 + 3...1389

all the way up until we get to some k, is equal to k times k + 1 over 2) is true.1394

We are assuming that this thing is true: pk, which is the statement 1 + 2 + 3...up until + k = k + (k + 1)/2--1399

we can just take this as cold, hard fact; it is the assumption that if this domino were to fall,1408

now we just need to work on the next domino falling.1413

We don't have to prove for certain that this one is true; we are allowed to assume it, because it works in combination with the first guarantee.1415

All right, so we have this assumed as true.1421

With the inductive hypothesis in mind, we want to show1423

(our inductive hypothesis is right here; it is up to us to show) that pk + 1 must also be true.1426

We can write out pk + 1 as 1 + 2 + 3 +...up until...well, now we are going up to k + 1,1432

so up until + k + 1...and we can plug in k + 1 times k + 1 + 1, all over 2.1438

So, we see that this is what the statement would be.1444

However, we can't really see how pk shows up in here;1447

how does something in this portion right here show up?--because remember what we talked about before.1450

We are going to have to use our inductive hypothesis; that is what the inductive hypothesis is there for.1455

If we could prove this without an inductive hypothesis, why are we doing induction?1459

So, we have to use that inductive hypothesis somehow to be able to apply it here,1463

so that we can show that pk + 1 must be true.1467

We need to get some way to see some portion of pk appear.1472

We realize that we can rewrite what we have here: we could write pk + 1 as it is above, but we can't easily see pk in it.1476

So instead, we write it out as pk + 1 = 1 + 2 + 3 + ... + k + k + 1.1484

How can we do this? Well, we know that there had to be k in here, because it is all the numbers from 1 up until k + 1.1493

So, k has to be included in there, because it is the number before k + 1.1498

k + 1 minus 1 is k; so it has to be the number before it, so it has to be included somewhere in our ellipsis in our continuing pattern.1502

So, we can write it out; now that means that we have pk showing up right here.1510

1 + 2 + 3...up until k...look, that is this right here; so we can bring this to bear.1516

And we just simplified the right side a little bit; that just ends up being the same thing.1522

k + 1 times k + 1 + 1 is just k + 1 times k + 2, over 2; we just simplified the right side.1526

All right, we have our inductive hypothesis, that 1 + 2 + 3 + ... + k = k times k + 1 over 2.1531

We are assuming that that is true, and it is our job to show that 1 + 2 + 3 up until + k + 1 is equal to k + 1 times k + 2 over 2.1538

So, we want to verify the statement below; we want to show that this is not a question mark--1547

that it is, indeed, that the left side and the right side are equal to each other, for sure.1552

We want to verify this; we want to make that question mark be absolutely an equals sign.1556

What we notice is that we have pk here--our inductive hypothesis, pk.1561

We can substitute pk, our hypothesis; and what we have here is the same thing as what we have here.1567

So, we can plug in; we can swap out this stuff right here.1573

Now, we have gotten our hypothesis into the game.1576

We have been able to plug it in; so now, we can start using it--we can work things out.1579

We are almost absolutely, certainly, going to use that inductive hypothesis; otherwise, we wouldn't need to be doing induction.1582

We bring the hypothesis to bear somehow.1589

We can plug it in, generally; and from here, we just have to simplify the equation.1592

It is not really an equation, because we don't actually know if the two sides are equal to each other, so it is not technically an equation.1596

But we can write it as this idea of "are they equal to each other? Let's see if they are the same thing, on either side."1602

And then, we can just verify that, if it does come out to be clearly the same on both sides, then we will know that pk + 1 is true.1607

If we can get rid of that equals sign in our equation, if we can turn it into just an equals, then we will know that, indeed, pk + 1 is true.1616

And we will have shown that pk + 1 is true from pk, and we will have completed our inductive step.1627

OK, we just worked this out: what we had on the left side was k times k + 1 over 2, plus k + 1.1632

And then, on the right side, we had k + 1 times k + 2 over 2.1638

So, we expand: k times k + 1 gets us k2 + k; we can put this over a fraction, so that we can have them over common denominators.1641

So, we have 2k + 2 over 2.1648

We expand the right side, k + 1 times k + 2; that is going to be k times k, k2, plus 2k + 1k, so 3k, plus 1 times 2, 2.1650

So, we get k2 + 3k + 2, all over 2.1659

Now, we just combine these two fractions here, k2 + k over 2, plus 2k + 2 over 2.1662

It is going to be k2 + k + 2k + 2, all over 2, which simplifies to a nice k2 + 3k + 2, over 2,1667

which is equal to k2 + 3k + 2 over 2, which is, indeed, definitely always going to be true.1674

So, we have just shown that, yes, our inductive step does work; we have proved the inductive step to be true.1679

If we have pk true, then pk + 1 must be true, from our logic.1684

So, we have completed both steps: we have completed the base case (back a while), and we just completed the inductive step.1690

That means that we have completed both of the parts necessary for a proof by induction.1695

We have to prove both of those parts.1699

We now know that pn is true for all n; that is, for any positive integer, we have that 1 + 2 + 3, up until + n, is equal to n times n + 1 over 2.1701

And I like to complete my proofs with a little square, just to say, "We have completed the proof; it is done."1711

And there we are; that is an example of how proof by induction works.1718

All right, at this level in math, you will often be directly told what to prove.1721

You will be given some formula or some statement, and it is going to say, "Go prove this."1725

You will be given something, and it will be your job to prove it, by induction, since this is a thing about proof by induction.1729

But in most of the things, if you are asked to prove something else, it would also just be "here is something; go prove it," like when you were in geometry class.1734

However, sometimes you won't be given a formula.1740

Instead, it is going to be your job to discover a pattern inside of that sequence, inside of what you are looking at;1746

and then create some sort of hypothetical formula that will describe that pattern, and finally to prove that your formula will always work.1753

So, the proof itself can be challenging; that is what we are working on.1760

But sometimes the hardest part isn't the proof that the proof by induction works, but just finding the pattern and turning it into a formula.1763

Sometimes, that is the hardest part: being able to look at what you have, figure out what the pattern here is,1770

and how you can turn this into a formula, how you can turn this into a statement that you can then go on to prove.1775

If you need to do this, and you find it difficult, I recommend checking out "Tips for finding patterns,"1779

which is one of the parts in the lesson Introduction to Sequences.1784

There are a lot of connections between finding the general term of a sequence and finding a formula, because they are both based on a sequence.1787

The formula is based on some sequence of statements, effectively.1794

And a sequence is based on some sequence of numbers; so there is a lot of connection between them.1797

This "Tips for finding patterns" inside of Introduction to Sequences--you will end up seeing a lot of things there.1801

A lot of the ideas there, you can apply to looking at formulas, so that is a really good thing to check out,1807

if you are having to work on that and you find it difficult.1811

All right, we are ready for some examples.1814

Prove that the below...1816

Oh, by the way, if you just skipped directly to the examples, you definitely, definitely, definitely, in this case,1817

want to go back to the working example first; so if you just skipped to this part directly, check out the working example first,1821

because it will explain what is going on here, as opposed to being confusing and just not having any meaning.1826

The working example will explain everything that is going on.1831

If you just watched the whole lesson, I'm sorry; I didn't mean to...I'm sorry about that.1833

Anyway, let's go on.1836

Prove that the below is true for any positive integer n.1837

We have 12 + 22 + 32 + ... + n2 = n times n + 1 over 2n + 1.1840

This is effectively our statement pn...not effectively; this is our statement pn.1849

Adding all of the squares, 12 + 22...up until n2, is equal to n times (n + 1) times (2n + 1), all divided by 6.1854

Our first thing to do is to show that the base case ends up being true.1862

The base case would be what it would be if we plugged in a 1 for n, if it was just the first statement.1867

What would that end up looking like? That is going to be...proving that would be equivalent to showing1871

that these two sides are the same, that 12 is equivalent to 1 (plugging in 1 for our n),1876

1 + 1, plugging in 1 for our n...2 times...plugging in 1 for our n...plus 1...all over 6.1881

Does this end up working out to be the same?1887

12 is 1; what is on our right side? 1 times...1 + 1 is 2; 2 times 1, plus 1, would end up being 3, over 6.1889

Is 1, indeed, equal to 1 times 2 times 3...is 6...over 6?1897

And yes, indeed, that would be 1 = 1; so we just showed that our base case ends up being true.1902

The next thing that we want to work out is the inductive step.1908

What would our inductive step be? Well, always the first thing that happens in the inductive step is that we write out what our hypothesis is.1911

What is the thing we are assuming? The hypothesis: we are assuming that pk,1918

which would be 12 + 22...up until we get to some k21923

(any positive integer k) is equal to k...we plug in k for n...k + 1, times 2k + 1, all over 6.1928

So, we assume that this is true; this is our assumption.1938

We can use this later; we can plug it in.1941

All right, we have our general thing here, pn, 12 + 22 + 32...1944

up until n2, is equal to n times n + 1 times 2n + 1.1948

And our inductive hypothesis is just saying, "What if we looked at some k, and we assumed that at k, it was true?"1951

So now, we want to show--it is our job to now show--that pk + 1 is true.1957

It is our job to show that pk + 1 ends up being true.1963

Showing that pk + 1 is true is going to end up being equivalent to showing that 12 + 22 + ...1967

up until we get to (k + 1)2...now, here is the part...we don't want to put in (k + 1)2,1978

because we want to somehow have our pk show up.1984

We are going to have to apply our hypothesis in there somewhere.1987

So, we want to have it show up.1991

We realize that up in 12 + 22 ... + (k + 1)2, well, what would be right before that?1992

We go back, and instead of using (k + 1)2 as the last thing, it would be easier to see in the pattern1998

what is really there, to make it easier, to substitute this: + k2 + (k + 1)2.2003

I will break this onto a different line, because now we are starting to get a little cramped.2009

We want to show (we don't know for certain; it is up to us to verify) that that would be...2011

swap in k + 1 now for n; so k + 1 times...k + 1 in here would be k + 2, times...k + 1 in here...2 times...k + 1 + 1, all divided by 6.2016

So, at this point, we say, "We have 12 + 22 + ... up until k2,2032

so we can swap that in for what is here at pk, which is this thing right here"--we can now swap that in.2037

Now, we are going to move over here, so we have a little bit more room.2045

k times k + 1 times 2k + 1, all over 6: this is our inductive hypothesis, which we assumed is true, so we can use it as we want;2049

plus...what is the thing that we still have left over?2061

We still have (k + 1)2 left over; we didn't substitute out for that; so, plus k + 1 squared.2062

And it is now our job to show that that is, indeed, equal to what was on the other side, which is k + 1 times k + 2 times 2k + 1 + 1.2069

I will simplify that a little bit: that will be 2k + 2 + 1; so that would be 2k + 3, all over 6.2079

OK, at this point, to show that they end up being the same on either side of that...2086

"maybe-it-is," to show that each side is the same, that the sides are equivalent,2091

we can just end up simplifying each of the sides, working with each side on their own.2095

We work with expanding this part over here; k times...k + 1 times 2k + 1...k times 2k gets us 2k2;2099

k times 1 is k, plus 1 times 2k, so + 3k; 1 times 1...plus 1; all over 6; plus...2107

let's multiply this part by 6/6, because we know that we will want to put them together2115

into a single fraction, so that we can compare the two things;2120

(k + 1)2 is (k + 1) times (k + 1); it becomes k2 + 2k + 1.2122

And is that, indeed, equal to...let's expand the right side; expand the far right first.2129

k + 1 times k + 2 times 2k + 3...k times 2k gets us 2k2; k times 3 is 3k; 2 times 2k is 4k;2133

so 3k + 4k is 7k; 2 times 3 is + 6; all divided by 6.2141

So, we can keep expanding over here; k times 2k2 is 2k3, plus 3k2, plus k, all over 6.2149

Plus...6k2 + 12k + 6, all over 6; and is that equal to (k + 1) times (2k2 + 7k + 6)?2160

OK, how is this one going? I am going to do this one partially in my head.2174

If you get lost here, just write it out, and you will see how it worked.2178

k times 2k2 is going to come out to be 2k3.2181

k times 7k is 7k2; plus...1 times 2k2...7k2 + 2k2 is 9k2;2184

k times 6 is 6k; 1 times 7k is 7k; 7k + 6k is 13k; and 1 times 6 is +6, all over 6.2191

And finally, if we combine our two fractions on the left, then 2k3...2201

everything will be divided by 6 over here...2k3 plus nothing on our other fraction,2206

so that is just that...3k2 + 6k2...we get 9k2;2211

k + 12k + 13k...there are no constants on our left fraction, plus the one on our right fraction, so + 6.2216

And is that, indeed, equal to 2k3 + 9k2?2222

We see that that is, indeed, the exact same thing; thus, our inductive step right here2226

(not our inductive hypothesis, but our inductive step) just checked out.2231

So, we have now shown that the base case checks out; the inductive hypothesis checks out.2234

So, combined, they show that this here is always true.2238

We have just shown that combining these two things (the base case, p1, being true,2245

and if pk is true--our inductive hypothesis--then pk + 1 is true, which we just showed here)--2249

we combined those two things, so now we have shown that pn is true for all things:2255

that 12 + 22 + 32...up until + n2 is equal to n(n + 1)(2n + 1), all divided by 6.2258

We have completed our proof by induction.2265

The second example: Show that, for n = 1, 2, 3...forever, that n3 + 2n is divisible by 3.2269

So really, this is our statement, pn: pn is "n3 + 2n is divisible by 3."2279

The first thing that we always have to do: we have to start by showing the base case.2285

Our base case first: the base case is showing that the p1 statement would be true.2289

Verifying p1 is the same thing as verifying that 13 + 2(1) is divisible by 3.2295

13 + 2(1) is 1 + 2, or 3; that is divisible by 3--that checks out, so our base case just checked out; great.2309

Next, we want to work on the inductive step.2318

Always, the first thing we do in our inductive step is to write what our hypothesis is.2324

What is the assumption that we can work from?2328

Our hypothesis is going to be that pk ends up being true.2331

What is pk? pk would be if we plugged in k for our n here: k3 + 2k is divisible by 3.2336

This is our starting assumption: that k3 + 2k is divisible by 3.2349

At this point, we now want to show that pk + 1 is true.2353

Showing that pk + 1 is true is going to be the same thing as showing2364

that k + 1 (plugging in k + 1 for n now), cubed, plus 2(k + 1) for n now, is divisible by 3.2367

So now, we just need to show...if this is divisible by 3, then we will have proven our inductive step, as well.2378

So, we will have shown the whole thing.2384

From here on, let's just take a close look at (k + 1)3 + 2(k + 1).2385

Hopefully, we will have enough room to work this out.2397

(k + 1)3...well, we can write that as (k + 1)(k + 1)2.2399

What is (k + 1)2? That is k2 + 2k + 1.2402

Plus...2 times (k + 1)...that is 2k + 2.2407

k + 1 times k2 + 2k + 1...well, k times k2 is going to be k3;2411

k times 2k is 2k2, plus 1 times k2, so + 3k2;2416

k times 1...+ 1k; plus 1 times 2k is 2k, so + 3k total; plus 1 times 1 is 1;2421

also, in the next lesson, when we talk about the binomial theorem, we will recognize, "Oh, yes, that is definitely how it has to expand."2428

Plus 2k + 2...OK, so at this point, we can realize, "What was our hypothesis? Our hypothesis was that k3 + 2k is definitely divisible by 3."2433

Look: here is k3; here is 2k; let's bring them together.2444

Now, we have k3 + 2k; and we will put everything else on the other side.2447

So, 3k2 + 3k...and we have a 1 here and a 2 here, so that combines to + 3.2454

Now, we can look at each part of this independently.2461

k3 + 2k is a thing; plus...and over here, we have a 3 here, a 3 here, and a 3 here;2463

so we can pull a 3 out; 3 comes out of k2 + k + 1.2470

Now, at this point, we realize--look, by our inductive hypothesis, we knew that this is divisible by 3.2476

We know that that is divisible by 3; furthermore, we have 3 times k2 + k + 1;2484

well, anything that we end up multiplying by 3 on it...well, that already has a factor of 3 multiplied in.2490

So, it must be divisible by 3; so that means both parts that we are adding together are divisible by 3.2496

If both parts that we add together are divisible by 3, then it must be that the entire thing is divisible by 3.2504

If something over here is divisible by 3, and something over here is divisible by 3, well, that means that I can divide this thing by 3;2510

I can divide this thing by 3; so even when we put them together, I can still divide the thing by 3.2514

Since both parts are divisible by 3, that means that (k + 1)3 + 2(k + 1),2518

which we just showed, has to be divisible by 3.2524

So, we have just shown that, if pk is true, then pk + 1 is true.2526

So, we have now shown the inductive step; since we have shown that p1 is true,2530

and if something in the middle is true, then the next thing is true;2535

then p1 being true implies that p2 must be true, by our inductive step,2538

which implies that p3 must be true, by our inductive step; p4, p5, p6, p7...2541

So, we know that, forever and always, n3 + 2n is divisible by 3; we have completed our proof by induction.2545

The next proof, the last example: Come up with a formula that gives the value of the nth partial sum2554

of the sequence below: 1, 3, 5, 7, 9... then prove that your formula always works.2559

The first few partial sums of the sequence are below.2565

Our first partial sum would be just adding 1 together, so that is 1.2568

Our next partial sum would be adding just 3 onto what we had before, so that would be 1 + 3.2572

The next thing would be 5, so 1 + 3 + 5.2577

The next thing would be 1 + 3 + 5 + 7; the next thing would be 9; let's write that out, as well: 1 + 3 + 5 + 7 + 9.2579

If we were to keep going with this pattern, what would be next?2588

Well, that would be 11 next; we are adding by 2 each time, so 1 + 3 + 5 + 7 + 9 + 11.2590

Let's see if we can figure out what the pattern going on here is.2599

1 on its own--well, that is just 1.2601

1 + 3 is 4; 1 + 3 + 5 is 9; add on another 7: 7 + 9 is 16; add on another 9; that is 25; add on another 11; that is 36.2603

We have an idea of what the formula is: it looks like the nth partial sum is going to be equal to n2; great.2618

What we have here is 1 + 3 + 5 + ... + something...whatever we end on...is equal to n2.2629

We need to figure out what we end on; how do we call out each member of this sequence?2644

Well, we can remember what we did in arithmetic sequences.2648

We see that we have 1 as our starting place; we add by 2 each time.2651

So, we could realize that the nth term here, an, is equal to 2n - 1.2655

We verify this: n at 1 gets us 1; n at 2 would get us 3; n at 3 would get us 5; so this works out.2664

We see that what we are going up to is 2n - 1.2670

So, our statement, the pn statement, is that 1 + 3 + 5, working up until 2n - 1, is equal to n2, for any n.2674

So, that is what we have figured out that our formula is for how we are getting these partial sums.2688

As we add up more and more terms of this sequence, we have now figured out our formula.2694

We have figured out how this pattern of adding things up comes to be.2698

We see that we have n2 coming out, if we do this each time.2705

On the first term, n = 1, we have 1; at n = 2, we have 4; n = 3...2709

We see that, each time, it is just squaring that number.2714

We also have to figure out what the sequence of the things looks like, because we have to be able to figure out what we end on.2717

We have to be able to say what the last thing is, so that we can actually work with having a left side and a right side to our statement's equation.2721

At this point, we are now ready to prove this.2728

All right, so now it is just up to us to prove that this is, indeed, true: that our statement pn is true for any value of n:2731

that 1 + 3 + 5...up until + 2n - 1 is equal to n2, no matter what n we put in, as long as n is a positive integer.2739

So, the first thing to prove: always prove the base case first, because it is almost always easier.2746

It also helps you understand how the thing comes together.2750

So, the base case: showing that p1 is true is going to be the same thing as showing that 1,2752

added up until...well, we just add it up to 1, so that is just 1 on the left side--is that equal to 12?2757

Yes, 1 is equal to 1; that was pretty easy.2764

So, there is our base case, finished.2766

Next, we do the inductive step; now we want to show that the inductive step is going to end up being true.2768

The first thing that we always do: we assume our hypothesis first; what is our hypothesis going to be?2774

Our hypothesis will be that pk is true.2780

What is the statement pk? Well, that would be 1 + 3 + ... up until we get to +...plug in k for n...2k - 1 = k2.2783

So, we are assuming that 1 + 3 + ... + 2k - 1 is equal to k2.2794

OK, with that assumption that this thing is true, we need now to show that k + 1 is going to be true, as well.2800

So now, we want to show that k + 1 is true.2805

Showing that k + 1 is true is going to be equivalent to showing that 1 + 3 + ... + 2k - 1...2811

I'm sorry, it is not 2k - 1; we are going to swap out k + 1 for n here, so 2 times (k + 1) - 1, equals k + 1,2821

swapping out that n here, as well, for k + 1, squared.2830

Now, remember: we want to have our hypothesis show up somewhere.2834

We have to get our hypothesis to show up somewhere.2838

So, we realize that that is really this part up here.2840

So, we can rewrite this as 1 + 3 + ... until we get to back a step of two...so that would be 2k - 1;2843

forward a step by 2...we see that that would be 2k + 1; if we work out this value here,2855

we end up seeing that it is the same thing here; 2k + 2 - 1 is just 2k + 1.2862

And we want to show that that is equal to (k + 1)2.2866

OK, we bring this up here; and on the way, we are going to say, "Right here, from here to here, we assumed up here that this was true."2870

That means that what we have in the thing we are trying to show, pk + 1, that part of it, is going to just be equal to k2, right here.2885

So, we can swap out k2 here; so now we have k2 plus what still remains; that is 2k + 1.2894

And we want to show that this is true, so it is a question mark here.2902

We don't know for sure that it is true yet; it is up to us to show that it ends up being true.2904

Plus what remained there, 2k + 1: is that equal to (k + 1)2?2908

Well, let's just simplify it: we have k2; we expand this; well, k2 + (2k + 1)...that is just 2k + 1;2916

is that equal to...what do we get when we expand (k + 1)2?2923

Well, that is k2 + 2k + 1; we end up seeing that that is indeed true.2926

That is always going to end up being true, so we have shown that our inductive step is true.2933

With this hypothesis in mind, we know that the next thing down the line...2938

with pk being true, we know that pk + 1 must be true.2941

Because we know that our base case is true (p1 is true), and we know that if something is true,2945

then the next thing has to be true; then since p1 is true, the next thing must be true.2949

p2 is true; p3 is true; p4 is true; p5 is true; p6 is true.2953

And the fireworks keep going forever and ever; we see that everything is always true.2957

We have that this statement ends up being true for absolutely every single n that we would plug into it.2961

We have completed the proof; our proof is done--pretty cool.2968

The only really challenging part there--that wasn't that difficult of a proof by induction, compared to the ones that we did previously--2972

the challenging part there was being able to figure out what that formula was in the first place.2977

So, once again, if you have difficulty with that, I recommend checking out the Arithmetic Sequences and Series.2981

There is that section, "Tips on finding patterns," that will really help you see how to find patterns; there is a lot of good stuff in there.2985

All right, we will see you at Educator.com later--goodbye!2991

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