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Fourier Series

Slide Duration:

Section 1: First-Order Equations
Linear Equations

1h 7m 21s

Intro
0:00
Lesson Objectives
0:19
How to Solve Linear Equations
2:54
Calculate the Integrating Factor
2:58
Changes the Left Side so We Can Integrate Both Sides
3:27
Solving Linear Equations
5:32
Further Notes
6:10
If P(x) is Negative
6:26
Leave Off the Constant
9:38
The C Is Important When Integrating Both Sides of the Equation
9:55
Example 1
10:29
Example 2
22:56
Example 3
36:12
Example 4
39:24
Example 5
44:10
Example 6
56:42
Separable Equations

35m 11s

Intro
0:00
Lesson Objectives
0:19
Some Equations Are Both Linear and Separable So You Can Use Either Technique to Solve Them
1:33
Important to Add C When You Do the Integration
2:27
Example 1
4:28
Example 2
10:45
Example 3
14:43
Example 4
19:21
Example 5
27:23
Slope & Direction Fields

1h 11m 36s

Intro
0:00
Lesson Objectives
0:20
If You Can Manipulate a Differential Equation Into a Certain Form, You Can Draw a Slope Field Also Known as a Direction Field
0:23
How You Do This
0:45
Solution Trajectories
2:49
Never Cross Each Other
3:44
General Solution to the Differential Equation
4:03
Use an Initial Condition to Find Which Solution Trajectory You Want
4:59
Example 1
6:52
Example 2
14:20
Example 3
26:36
Example 4
34:21
Example 5
46:09
Example 6
59:51
Applications, Modeling, & Word Problems of First-Order Equations

1h 5m 19s

Intro
0:00
Lesson Overview
0:38
Mixing
1:00
Population
2:49
Finance
3:22
Set Variables
4:39
Write Differential Equation
6:29
Solve It
10:54
11:47
Example 1
13:29
Example 2
24:53
Example 3
32:13
Example 4
42:46
Example 5
55:05
Autonomous Equations & Phase Plane Analysis

1h 1m 20s

Intro
0:00
Lesson Overview
0:18
Autonomous Differential Equations Have the Form y' = f(x)
0:21
Phase Plane Analysis
0:48
y' < 0
2:56
y' > 0
3:04
If we Perturb the Equilibrium Solutions
5:51
Equilibrium Solutions
7:44
8:06
Solutions Will Tend Away From Unstable Equilibria
9:32
Semistable Equilibria
10:59
Example 1
11:43
Example 2
15:50
Example 3
28:27
Example 4
31:35
Example 5
43:03
Example 6
49:01
Section 2: Second-Order Equations
Distinct Roots of Second Order Equations

28m 44s

Intro
0:00
Lesson Overview
0:36
Linear Means
0:50
Second-Order
1:15
Homogeneous
1:30
Constant Coefficient
1:55
Solve the Characteristic Equation
2:33
Roots r1 and r2
3:43
To Find c1 and c2, Use Initial Conditions
4:50
Example 1
5:46
Example 2
8:20
Example 3
16:20
Example 4
18:26
Example 5
23:52
Complex Roots of Second Order Equations

31m 49s

Intro
0:00
Lesson Overview
0:15
Sometimes The Characteristic Equation Has Complex Roots
1:12
Example 1
3:21
Example 2
7:42
Example 3
15:25
Example 4
18:59
Example 5
27:52
Repeated Roots & Reduction of Order

43m 2s

Intro
0:00
Lesson Overview
0:23
If the Characteristic Equation Has a Double Root
1:46
Reduction of Order
3:10
Example 1
7:23
Example 2
9:20
Example 3
14:12
Example 4
31:49
Example 5
33:21
Undetermined Coefficients of Inhomogeneous Equations

50m 1s

Intro
0:00
Lesson Overview
0:11
Inhomogeneous Equation Means the Right Hand Side is Not 0 Anymore
0:21
First Solve the Homogeneous Equation
1:04
Find a Particular Solution to the Inhomogeneous Equation Using Undetermined Coefficients
2:03
g(t) vs. Guess for ypar
2:42
If Any Term of Your Guess for ypar Looks Like Any Term of yhom
5:07
Example 1
7:54
Example 2
15:25
Example 3
23:45
Example 4
33:35
Example 5
42:57
Inhomogeneous Equations: Variation of Parameters

49m 22s

Intro
0:00
Lesson Overview
0:31
Inhomogeneous vs. Homogeneous
0:47
First Solve the Homogeneous Equation
1:17
Notice There is No Coefficient in Front of y''
1:27
Find a Particular Solution to the Inhomogeneous Equation Using Variation of Parameters
2:32
How to Solve
4:33
Hint on Solving the System
5:23
Example 1
7:27
Example 2
17:46
Example 3
23:14
Example 4
31:49
Example 5
36:00
Section 3: Series Solutions
Review of Power Series

57m 38s

Intro
0:00
Lesson Overview
0:36
Taylor Series Expansion
0:37
Maclaurin Series
2:36
Common Maclaurin Series to Remember From Calculus
3:35
7:58
Ratio Test
12:05
Example 1
15:18
Example 2
20:02
Example 3
27:32
Example 4
39:33
Example 5
45:42
Series Solutions Near an Ordinary Point

1h 20m 28s

Intro
0:00
Lesson Overview
0:49
Guess a Power Series Solution and Calculate Its Derivatives, Example 1
1:03
Guess a Power Series Solution and Calculate Its Derivatives, Example 2
3:14
Combine the Series
5:00
Match Exponents on x By Shifting Indices
5:11
Match Starting Indices By Pulling Out Initial Terms
5:51
Find a Recurrence Relation on the Coefficients
7:09
Example 1
7:46
Example 2
19:10
Example 3
29:57
Example 4
41:46
Example 5
57:23
Example 6
1:09:12
Euler Equations

24m 42s

Intro
0:00
Lesson Overview
0:11
Euler Equation
0:15
Real, Distinct Roots
2:22
Real, Repeated Roots
2:37
Complex Roots
2:49
Example 1
3:51
Example 2
6:20
Example 3
8:27
Example 4
13:04
Example 5
15:31
Example 6
18:31
Series Solutions

1h 26m 17s

Intro
0:00
Lesson Overview
0:13
Singular Point
1:17
Definition: Pole of Order n
1:58
Pole Of Order n
2:04
Regular Singular Point
3:25
Solving Around Regular Singular Points
7:08
Indical Equation
7:30
If the Difference Between the Roots is An Integer
8:06
If the Difference Between the Roots is Not An Integer
8:29
Example 1
8:47
Example 2
14:57
Example 3
25:40
Example 4
47:23
Example 5
1:09:01
Section 4: Laplace Transform
Laplace Transforms

41m 52s

Intro
0:00
Lesson Overview
0:09
Laplace Transform of a Function f(t)
0:18
Laplace Transform is Linear
1:04
Example 1
1:43
Example 2
18:30
Example 3
22:06
Example 4
28:27
Example 5
33:54
Inverse Laplace Transforms

47m 5s

Intro
0:00
Lesson Overview
0:09
Laplace Transform L{f}
0:13
Run Partial Fractions
0:24
Common Laplace Transforms
1:20
Example 1
3:24
Example 2
9:55
Example 3
14:49
Example 4
22:03
Example 5
33:51
Laplace Transform Initial Value Problems

45m 15s

Intro
0:00
Lesson Overview
0:12
0:14
Take the Laplace Transform of Both Sides of the Differential Equation
0:37
Plug in the Identities
1:20
Take the Inverse Laplace Transform to Find y
2:40
Example 1
4:15
Example 2
11:30
Example 3
17:59
Example 4
24:51
Example 5
36:05
Section 5: Review of Linear Algebra
Review of Linear Algebra

57m 30s

Intro
0:00
Lesson Overview
0:41
Matrix
0:54
Determinants
4:45
3x3 Determinants
5:08
Eigenvalues and Eigenvectors
7:01
Eigenvector
7:48
Eigenvalue
7:54
Lesson Overview
8:17
Characteristic Polynomial
8:47
Find Corresponding Eigenvector
9:03
Example 1
10:19
Example 2
16:49
Example 3
20:52
Example 4
25:34
Example 5
35:05
Section 6: Systems of Equations
Distinct Real Eigenvalues

59m 26s

Intro
0:00
Lesson Overview
1:11
How to Solve Systems
2:48
Find the Eigenvalues and Their Corresponding Eigenvectors
2:50
General Solution
4:30
Use Initial Conditions to Find c1 and c2
4:57
Graphing the Solutions
5:20
Solution Trajectories Tend Towards 0 or ∞ Depending on Whether r1 or r2 are Positive or Negative
6:35
Solution Trajectories Tend Towards the Axis Spanned by the Eigenvector Corresponding to the Larger Eigenvalue
7:27
Example 1
9:05
Example 2
21:06
Example 3
26:38
Example 4
36:40
Example 5
43:26
Example 6
51:33
Complex Eigenvalues

1h 3m 54s

Intro
0:00
Lesson Overview
0:47
Recall That to Solve the System of Linear Differential Equations, We find the Eigenvalues and Eigenvectors
0:52
If the Eigenvalues are Complex, Then They Will Occur in Conjugate Pairs
1:13
Expanding Complex Solutions
2:55
Euler's Formula
2:56
Multiply This Into the Eigenvector, and Separate Into Real and Imaginary Parts
1:18
Graphing Solutions From Complex Eigenvalues
5:34
Example 1
9:03
Example 2
20:48
Example 3
28:34
Example 4
41:28
Example 5
51:21
Repeated Eigenvalues

45m 17s

Intro
0:00
Lesson Overview
0:44
If the Characteristic Equation Has a Repeated Root, Then We First Find the Corresponding Eigenvector
1:14
Find the Generalized Eigenvector
1:25
Solutions from Repeated Eigenvalues
2:22
Form the Two Principal Solutions and the Two General Solution
2:23
Use Initial Conditions to Solve for c1 and c2
3:41
Graphing the Solutions
3:53
Example 1
8:10
Example 2
16:24
Example 3
23:25
Example 4
31:04
Example 5
38:17
Section 7: Inhomogeneous Systems
Undetermined Coefficients for Inhomogeneous Systems

43m 37s

Intro
0:00
Lesson Overview
0:35
First Solve the Corresponding Homogeneous System x'=Ax
0:37
Solving the Inhomogeneous System
2:32
Look for a Single Particular Solution xpar to the Inhomogeneous System
2:36
Plug the Guess Into the System and Solve for the Coefficients
3:27
Add the Homogeneous Solution and the Particular Solution to Get the General Solution
3:52
Example 1
4:49
Example 2
9:30
Example 3
15:54
Example 4
20:39
Example 5
29:43
Example 6
37:41
Variation of Parameters for Inhomogeneous Systems

1h 8m 12s

Intro
0:00
Lesson Overview
0:37
Find Two Solutions to the Homogeneous System
2:04
Look for a Single Particular Solution xpar to the inhomogeneous system as follows
2:59
Solutions by Variation of Parameters
3:35
General Solution and Matrix Inversion
6:35
General Solution
6:41
Hint for Finding Ψ-1
6:58
Example 1
8:13
Example 2
16:23
Example 3
32:23
Example 4
37:34
Example 5
49:00
Section 8: Numerical Techniques
Euler's Method

45m 30s

Intro
0:00
Lesson Overview
0:32
Euler's Method is a Way to Find Numerical Approximations for Initial Value Problems That We Cannot Solve Analytically
0:34
Based on Drawing Lines Along Slopes in a Direction Field
1:18
Formulas for Euler's Method
1:57
Example 1
4:47
Example 2
14:45
Example 3
24:03
Example 4
33:01
Example 5
37:55
Runge-Kutta & The Improved Euler Method

41m 4s

Intro
0:00
Lesson Overview
0:43
Runge-Kutta is Know as the Improved Euler Method
0:46
More Sophisticated Than Euler's Method
1:09
It is the Fundamental Algorithm Used in Most Professional Software to Solve Differential Equations
1:16
Order 2 Runge-Kutta Algorithm
1:45
Runge-Kutta Order 2 Algorithm
2:09
Example 1
4:57
Example 2
10:57
Example 3
19:45
Example 4
24:35
Example 5
31:39
Section 9: Partial Differential Equations
Review of Partial Derivatives

38m 22s

Intro
0:00
Lesson Overview
1:04
Partial Derivative of u with respect to x
1:37
Geometrically, ux Represents the Slope As You Walk in the x-direction on the Surface
2:47
Computing Partial Derivatives
3:46
Algebraically, to Find ux You Treat The Other Variable t as a Constant and Take the Derivative with Respect to x
3:49
Second Partial Derivatives
4:16
Clairaut's Theorem Says that the Two 'Mixed Partials' Are Always Equal
5:21
Example 1
5:34
Example 2
7:40
Example 3
11:17
Example 4
14:23
Example 5
31:55
The Heat Equation

44m 40s

Intro
0:00
Lesson Overview
0:28
Partial Differential Equation
0:33
Most Common Ones
1:17
Boundary Value Problem
1:41
Common Partial Differential Equations
3:41
Heat Equation
4:04
Wave Equation
5:44
Laplace's Equation
7:50
Example 1
8:35
Example 2
14:21
Example 3
21:04
Example 4
25:54
Example 5
35:12
Separation of Variables

57m 44s

Intro
0:00
Lesson Overview
0:26
Separation of Variables is a Technique for Solving Some Partial Differential Equations
0:29
Separation of Variables
2:35
Try to Separate the Variables
2:38
If You Can, Then Both Sides Must Be Constant
2:52
Reorganize These Intro Two Ordinary Differential Equations
3:05
Example 1
4:41
Example 2
11:06
Example 3
18:30
Example 4
25:49
Example 5
32:53
Fourier Series

1h 24m 33s

Intro
0:00
Lesson Overview
0:38
Fourier Series
0:42
Find the Fourier Coefficients by the Formulas
2:05
Notes on Fourier Series
3:34
Formula Simplifies
3:35
Function Must be Periodic
4:23
Even and Odd Functions
5:37
Definition
5:45
Examples
6:03
Even and Odd Functions and Fourier Series
9:47
If f is Even
9:52
If f is Odd
11:29
Extending Functions
12:46
If We Want a Cosine Series
14:13
If We Wants a Sine Series
15:20
Example 1
17:39
Example 2
43:23
Example 3
51:14
Example 4
1:01:52
Example 5
1:11:53
Solution of the Heat Equation

47m 41s

Intro
0:00
Lesson Overview
0:22
Solving the Heat Equation
1:03
Procedure for the Heat Equation
3:29
Extend So That its Fourier Series Will Have Only Sines
3:57
Find the Fourier Series for f(x)
4:19
Example 1
5:21
Example 2
8:08
Example 3
17:42
Example 4
25:13
Example 5
28:53
Example 6
42:22

• ## Related Books

### Fourier Series

Fourier Series (PDF)

### Fourier Series

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
• Lesson Overview 0:38
• Fourier Series
• Find the Fourier Coefficients by the Formulas
• Notes on Fourier Series 3:34
• Formula Simplifies
• Function Must be Periodic
• Even and Odd Functions 5:37
• Definition
• Examples
• Even and Odd Functions and Fourier Series 9:47
• If f is Even
• If f is Odd
• Extending Functions 12:46
• If We Want a Cosine Series
• If We Wants a Sine Series
• Example 1 17:39
• Example 2 43:23
• Example 3 51:14
• Example 4 1:01:52
• Example 5 1:11:53

### Transcription: Fourier Series

Hi and welcome back to the differential equations lecture here on educator.com.0000

My name is Will Murray, and we are studying a chapter on partial differential equations.0004

We will meet the differential equations behind for this lecture and were to study Fourier series.0008

Fourier series is a tool that really used to solve the heat equation in the next lecture, but Fourier is kind of a big topic by itself so, you spent all this lecture learning about Fourier series and then the next lecture were to come back and use what we have learned about Fourier series to solve the heat equation.0015

So, you will see how that works out let us go ahead and start talking about Fourier series so, before a series for function is it is kind of like a Taylor series back at calculus 2.0034

Taylor series, your expanding things into essentially big polynomials into powers of X and Fourier series your expanding a function into sins and cosine so, near the general format here is a0 / 2 + the sum from N = 1 infinity of AN x cosine e / L + BN x sin of n π X / L so, it is it is quite complicated and will break it down will study the different elements.0045

Let me mention that this L something we seen before that is a constant and when we use this to solve the heat equation, the L is going to represent the length of the rod that were measuring the heat transfer through.0078

So, that is the length of the rod in the heat equation so, that is where we are going to get the L from now, will be a constant that stays fixed throughout the study of any particular Fourier series.0100

Everything else is something that is fixed except for these coefficients a0 and 8 AN BN so, I have to give me some formulas to find those Fourier coefficients we find them is that AN BN you find them both by these integrals it is 1 of L x the integral from - L to L of FX x now, 4 and its cosine e of n π X / L the X and for BN it sin of n π X / L DX.0111

So, were to be going lots of integrals in this lecture quite a bit of integration by parts you often end up having a new integration by parts we are finding Fourier series.0146

So, if you are rusty on integration by parts you might want to go back and review that a little bit of me doing the integration by parts kind quickly here can have to or will never get through the lecture.0158

So, you want to be up on your integration by parts we do have a lecture series here on educator.com on the second semester , calculus are on that BC calculus and there is a whole lecture in there on integration by parts of the first lecture that series.0169

So, if you do not number how to do integration by parts you might want to go back watch that lecture and when you are really up to speed on integration by parts then come back and join us for the rest of this Fourier series lectures, we will be using integration by parts quite a bit.0185

So, the Fourier coefficients we do find them using integrals sometimes these formulas simplify a little bit slower show you how they get how that can happen.0200

N = 0 there is no B 0 because the B0 would be the sin of 0 X which is just 0 so, that sort of drops out before when N = 0 the formula for a 0 simplifies because remember our formula for AN was 1 / L x the integral of - L to L of FX x cosine e of N π X / L DX.0211

That was our formula now, if N = 0 then you are just taking the cosine e 0 so, that just turns into 1 and so, the formula for a 0 just turns into the 1 / L x the integral from - L to L of FX DX .0245

That simplifies nicely another note that we have to be aware of is that all of these Fourier series are periodic so, you have to start with a periodic function and it has to be it has to have period to L.0260

That that means is that the function has to repeat itself every 2 L and usually will be considering functions that are defined from - L to L so, if you have a function that is defined maybe like that from - L to L then that is a span of 2 L so, after that it has to repeat itself so, we can only look at functions that are periodic with 2 L.0275

Usually what that boils down to is will just look at functions that are defined from - L to L and then we will just assume that there periodic outside of that interval because then we can find their Fourier series and everything will work nicely .0315

So, we have to work with periodic series for with periodic functions for Fourier series .0331

There is a special case here when we talk about even and odd functions so, let me remind you what an even and odd function is.0339

This is a definition we probably had at summer back at calculus but you might not remember it so, how it goes is a function is even if F of - X is = 2 F of X for all X and its - X is = 2 - FX .0345

So, classic examples there are sin and cosine e sin of X for the graph of sin X their sin of X and it there is a value of X and there is - X you notice the - X is down below the x-axis where FX was above so, it is exactly been - of FX .0361

So, that is why sin of X is I cosine e of X is the other way around cosine e of X if you look at a particular value of X and then you look at - X - X you see that the cosine e of - X is the same as cosine e of X so, the cosine e satisfies this relation here F of - X is = 2 - F of X sin satisfies this relation here.0387

So, cosine e is even sin is odd X is odd and well that is not so, interesting to me maybe look at some more adjusting 1is X ² this assignment here sin X this as cosine e X look at X ² is even .0422

That is because if you look at - X this come out the same when you take their ²s X cubed on the other hand is odd so, there is X cubed and if you look at X and if you look at - X you get opposite values when you take X cubed or - X cube.0446

That is exactly why there named odd and even functions is because even powers of X X ² X forth and so, on those or even functions odd powers of X X cubed XX of the fifth and so, on those are those are odd functions that is why we use the words odd and even for those types of functions .0473

The 0 function as both even and odd because the 0 function is just a horizontal line now, there is a very nice geometric way you can identify odd and even functions and even function let us look at look at cosine e of X is her prototypical even function or X ² it those of both even.0494

You notice that if you drawn mirror on the y-axis then these functions are symmetric in this mirror their kind of symmetric about around the y-axis .0515

So, an even function has mirror symmetry across the y-axis and odd function on the other hand, look at her prototypical odd functions sin of X and odd function and X cubed is an odd function what those functions have in common is that if you put a pin in the graph right at the origin and then spun the graph around 180° you would end up looking the same as it did when you started.0527

Same thing here with X cubed if you put a pin and span around 180° it would look the same as when you start it so, we say that the graph has for an odd function has rotational symmetry around the origin that is how you can identify even and odd functions .0561

Even functions and this mirror symmetry and odd functions have this rotational symmetry lets us see how even and odd functions affect Fourier series.0580

If F is even, then it is fairly easy to check that F of X x sin of n π X / L since sin of X is odd will be an odd function which means it will had that rotational symmetry which means when you integrate from - L to L you are going to get 0.0593

So, the BN which remember we got the BN by integrating from - L to L is 1 / L there F of X time sin of N π X / L so, that is how we got the BN but if you had that rotational symmetry than the - parts in the + particular cancel each other out .0615

Which can come out to be 0 on the other hand the AN are going to have remember original formula was 1 / L from - L to L F of X cosine e of N π X / L that is going to be an even function and so, you can get the same area on the left is you are on the right.0643

Which means you can save a little time in by integrating from 0 to L and then just multiplied by 2 so, that is how that 1 turned into a 2 and that - L turned into a 0 just cutting off the - part and were doubling the graph reusing symmetry there .0668

The exact opposite thing happens if F is odd then X x cosine e of N π X is odd so, it be AN turns out to be 0 and the Fourier series contains only sins and then we take the BN we can change that original - L with 0 and we can double it .0687

So, again were exploiting that symmetry BN = 1 / L x integral from - L to L of F of X sin of n π X / L and so, because it turns out to be symmetric if F itself is odd then we can just cut it and integrate from 0 to L and then multiplied by 2 in order to fix that .0712

Then, all the AN comes out to be 0 so, if you have an odd function or even function it really cuts down on the work of finding a Fourier series have the coefficients come out to be 0 and the other half you have the sort of slightly simplified formula to solve for those coefficients.0739

So, even and odd functions are very nice for Fourier series. What is also nice about Fourier series is that if the function you are interested in is only to find from 0 to L and let me remind you that we did have that situation in the heat equation.0755

For the heat equation we we talked about this in the last couple lectures is where you are looking for the temperature of bar and you assume that the bar is oriented so, it is left denizen 0 it is right and is it L and that function FX was the initial temperature throughout the bar.0780

So, you really only had that as a function from 0 to L if you had that situation than what you can do is you can extend that function from 0 to - L so, let me squeeze me a graph in here there 0 there is L and maybe have a graph of F of X like that .0800

You really do not know, what FX does between 0 - L or maybe it is not even the fine so, you do not know, you do not have a definition for F of X from 0 to - L what you can do is you can create your own definition of F of X and you can define it in such a way that it ends up being either even or odd .0827

So, if you want to cosine e series then you would extend it to be even let me show what the picture will look like for that so, here 0 here is L and here is - L remember we have FX predefined from 0 to L but if we want it which we can do find ourselves on - L to L .0852

Define it to make it an even function from - L to L so, what I really did there was a look to X and I looked at - X and I said F of - X is defined to be the same as F of X this: = by the way what that means is defined to be.0876

So, what we did there was redefined F of - X to be the same as F of X and that forces it to be an even function so, the graph turns out to have symmetry across the y-axis now, let me show you what we could do if we wanted to get a sin series .0901

So, again there 0 there is L and suppose we have FX there and again we do not have a defined yet between - L in 0 and what I want to do is to find it to be an odd function so, does not agree, does not look quite symmetric, only do that again .0924

You see how the way had to fin d this function is if you spin it around the origin if you put a pin in the in a piece of paper at the origin event and then not spun around 180° this function would look the same in terms of equations what I did was when I looked at F of X and then for - X I defined F of - X to be - F of X .0926

Again that : there means I am defining F - X right now, to be - FX and so, in the end I get this function that is on function its graph has that rotational symmetry and what that means is that if I find a Fourier series for it really end up just using sins .0993

It turns out that when we want to solve the heat equation which is something order to talk about in the next lecture really need to use sin series so, we will extend our F of X to be odd.1013

Something we started to hit that in the previous lecture when we decided that our solutions to the partial differential equation all had sins in them and did not have any cosine so, when we try to solve this later on were going to extend functions to be odd so, that we find a Fourier series will be using only sins.1029

So, this is quite a lot to digest I think it is time with the practice some examples so, let us start with example 1 we have to find a Fourier series for the function below to start out by graphing this function.1051

So, the graph of this function looks like it is essentially defined between -2 and 2 between -2 and 0 FX = 0 graph this.1072

So, FX is = 0 between -2 and 0 and then between 0 and 2 FX is = 2 X so, there is FX is = 2 X and after that it looks like F of X +4 FX -4 is just defined to be F of X so, there is an open hole there .1086

What that means as that this function is repeating every 4 units is periodically 4 so, there it is that would be that would be 4 that would be 6 and and then - direction we make it periodic in the - direction as well.1117

And that would be - 4 -6 so, we have a function here that is periodically period 4 and want to find a Fourier series for its our L is remember As always 2L so, our L is going to be 2 because 2L was = 2 4.1144

So, the L were getting used Fourier series is 2 so, let us go ahead and find our Fourier coefficients.1173

Our first 1 is 0 we use the Fourier equations that we started with at the beginning a 0 is always 1 / L x integral from - L to L of F of X T X now, in this case the L is 2 so, this is 1/2 now, the integral from - 2 to 2.1181

You notice that from -20 the function is just 0 so, I am just a start this integral going from 0 to 2 that is because the function is 0 on the - part of that interval and FX on that interval is just X so, this is X the X now, that is a really easy integral that is 1/2 x the integral of X is X ² / 2 were integrating that from X = - 02 X = 2.1205

So, we get 1/4×2 is 4-0 so, just 1 so, I found my first or my 0 Fourier coefficient a 0 is = 1 so, I am going to hang onto that and then I am going to keep using that for my for when I want to set up my Fourier series later that if I know, a lot of other coefficients in the meantime.1241

So, I am going to use the next screen to do that so, let me remind you of everything we did on the screen first I just looked at this definition and I grasped my function it was 0 between -2 and 0 then it turned into FX = X and in this tells me it is periodically period 4.1271

Such a started repeating that function after that and I want to find my first Fourier coefficient so, I use my formula for a 0 here and that just sets up an integral. 2147 It started at it 0 but that is because this first part of the function is 0 so, would not contribute anything to the integral.1292

That I just flow through the integral and I worked out that a 0 was = 2 1 so, let me go ahead and find some more Fourier coefficients for this function .1312

Here is the function again so, now, are going to find our general AN we had at a 0 was always kind of a special case so, you want to suffer that we separately the general AN is will remember it is always 1 / L x integral from - L to L of F of X x cosine of n π X / L .1327

That was the formula we had at the beginning of the lecture and in this case our L is 2 s now, remember this function is 0 from -20 so, really do not need to worry about that part to start the integral and 0 and go take the integral from 0 2 of now, are between 0 and 2 our function is X so, FX = X x cosine of n π X / L .1354

Now, here is the first of our many bouts of integration by parts if you do not remember your integration by parts you really want to review that before you get too far down into the rabbit hole a Fourier series .1385

So, let me quickly do I use tabular integration works great for these kinds of functions so, quickly run through integration by parts here and that this is not looking familiar maybe go back and watch the lecture from calculus 2 the AP calculus BC part and you will find you will find a whole lecture on integration by parts.1400

So, right X and cosine of n π X / L and make little chart here and take derivatives on the left so, 1 and 0 integrals on the right and the integral of cosine is sin of n π X / L you get very good at doing this chart divided by N π / L .1429

So, that is the same as L / n π in order to plug in my L is = 2 I think it will make that a little easier so, my 2 plug-in L = 2 everywhere here so, 2/ n π here.1463

Down the integral cosine is sin the integral of sin is - cosine - cosine of n π X / 2 and again I have to multiply by 2 / n π so, put those together get 4 / N² π ² and now, I do little diagonal lines and alternating sins + and - and now, I can read off the answer .1481

The answer to the integration so, is my 1/2 from above and now, I am going to multiply along these diagonal lines that I got from integrating by parts so, I see I have 2 / n π X x sin of n π X / 2 so, + 4 / N ² π ² x cosine of n π X / 2 and I need to evaluate that from where to where.1515

From 0 to 2 so, from X = 0 to X = 2 so, I still have 1/2 here when I plug in X = 2 I noticed that the sin of n π X / 2 that which is the sin of n π so, that will be 0 and X = 2 so, + 4 / N ² π ² x cosine of N π x 2 / 2 so, cosine n π now, X = 0 I get 0 for the first term and for the second term I get - well cosine a 0 is 1 so, 4 / N² π ².1559

If I distribute that 1/2 then I see I got 2 / N ² π ² x cosine of n π -1 now, I have to think about what cosine of n π is and if think about it cosine of 0 is 1 cosine of π is -1 cosine of 2 π is 1 cosine 3 π is -1 in general and just alternating between 1 and -1 .1616

Cosine of n π is = 2 - while we say cosine of n π is = 1 if N is even and even and -1 if N is odd so, what we see is that if N is even then this term is going to come out to be 0 because we just get 1-1 so, 0 if N is even .1661

And if N is odd then we got a - 1 here for cosine of n π so, -1-1 is -2 so, - 4 / N ² π ² if N is odd so, quite a complicated formula here that is very typical for Fourier series will have a lot of formulas like that.1700

So, that is what I have for my AN coefficients so, let us hang onto that of this is AN and were to go on the next slide and were to figure out what BN is.1724

Let me recap what we did for AN I use my generic Fourier series formula AN is 1 / L x integral from - L to L but I can't simplify this formula because I do not have an even more and odd function for F of X here so, formula simplifies sometimes right is 2/ L x the integral from 0L and that does not work for this 1 because we do not have an even and odd function.1742

So, my L was 2 so, I plugged in 2 for all the L here the reason I put in 0 there is because the first half of this function is 0 so, and so, I should really only need to look at the interval from 0 to 2 of X that is from this X right there that is where that is coming from .1771

Cosine of n π X / 2 now, to solve that I integrated by parts I use my tabular integration trek and that expands into this rather unwieldy expression so, plug-in X = 2 X = 2 and these 2 terms came from X = 0 .1792

Fortunately sin of N π turned in the 0 that is what you get this term terms into 0 cosine of n π no such luck were just up with it the X here makes this term be 0 and then cosine of 0 is just 1 so, that is how we get that term right there .1811

Now, let us see what I do here 1/2×4 is where I got that 2 distributing 1/2 and I had N ² π ² in the denominator cosine e N π -1 cosine e N π worked out a pattern for the cosine here it is 1 if N is even an -1 if N is odd .1834

So, if it is even I just got 1-1 that is why get my 0 here if N is odd got -1-1 which is -2×2 is - 4 so, that is right at that - 4 / N ² π ² if N is odd.1861

That is all just to figure out AN still to figure out BN so, will go ahead on the next slide and will take a look at BN so, still working with the same function here we go to figure out BN so, my BN is to this formula this generic Fourier series formula - L to L of F of X x sin of n π X / L DX .1877

So, in this case our L is 2 that is 1 / 2 now, - L would be - 2 but were told that the function is 0 when were at that between -2 and 0 1 just can integrate from 0 to 2 of F of X is X x sin of n π X / L DX.1906

Fill in L is = 2 and again you can use integration by parts to integrate X x the sin of n π X / 2 so, I am using mild tabular integration trick X through the -X 1 through 0 the integral of sin is - cosine of n π X / 2 and we have to multiply that by 2 / n π by the chain rule.1941

And the integral of cosine is sin that - N π X / 2 multiply by 2 / n π again so, get 4 / N ² π ² and now, I am going to write diagonal lines and attach alternating sins that + and - .1977

So, if I get that integral all the way through what I end up with is -2 / N π x X x cosine of n π X / 2 - it looks like + 4 / N ² π ² x sin of n π X / 2 and this whole thing I need to evaluate it from X = 0 to X = 2 .2001

Plug-in X = 2 get 1/2 now, if X is = 2 I get I get -4 / N π x cosine of n π X / 2 so, that is just cosine e N π and then if I plug-in X = 2 to sin of n π X / 2 the sin of n π which is 0 and our X = 0 on the left since I got a factor of X there is just 0 .2041

On the right sin of 0 is 0 so, that turns into 0 and so, this whole thing simplifies down to -2 / N π now, cosine e N π had really like to find a good pattern for that and we did on the last slide remember cosine e N π was 1 if N is even and -1 if N is odd and there is a simple way we can write that we can is right that is -1 to the N.2082

So, I can simplify my cosine e N π is -1 to the N and what I get is -1 there is another - outsides so, -1×1 x 2 / N π so, that is my coefficient BN for my Fourier series and remind you on the previous size we already figured out the AN.2125

So, let me pull up those my a 0 was 1 we did that several slides ago AN we figured out was more complicated it was 0 if N is even and -4 / N ² π ² if N is odd .2151

So, let me assemble a Fourier series it is going to be quite complicated but let me remind you of the general formula first the Fourier series for a function is a 0 / 2 + the sum from N= 1 to infinity is a generic formula from the beginning of the lecture.2187

AN x cosine of n π X / L cosine up N π X / L + BN x sin of n π X / L so, I have figured out all the coefficients down just can drop them in this series and I think the pattern is too complicated to express it in a closed σ notation so, instead what we do is just right out the first few terms of the series.2210

In this case our Fourier series is the N = 0 term is just a 0 / 2 so, that is 1/2 now, let us look at the N = 1 terms if N is odd my AN is - 4 / N ² π ² so, from N = 1 I get -4 / π ² x cosine of n π while π X / 2 .2250

Now, my B term is -1 to the N +1 so, that is + since N = 1 from working out the N = 1 terms here my BN term is - 1 of this 2 set +2 / π 2 / N π some doing N = 1 x sin of n π X / L so, that is just π X / 2.2283

N = 2 looks like I get a break the even A is 0s finding nothing there within the odd 1 will be the BN will be - 1cube so, - 2 / 2pie will simplify that because they gets easier to see the pattern if we do not simplify that.2315

2/ 2 π x sin again I am in the right 2 π X / 2 even we know, that is would simplify 2 π X / 2 so, that is my N = 2 term .2345

Remember there was no cosine term because it it was 0 there is going to be N = 3 so, for N = 3 my AN where is -4/9 π ² - 4/9 π ² cosine of 3 π X / 2 and for my sin its +2 / 3 π sin of 3 π X / 2 .2360

So, that is my Fourier series for that function really quite complicated quite a bit of work to figure out each coefficient there so, let me remind you where I each of those elements came from we are finally told with that example.2395

So, we had already figured out a 0 in the generally ends of the previous slide just figuring out the BN so, BN was 1 / L x this integral formula that we had from the beginning of a lecture plug-in L = 2 everywhere and in that 0 was because the function was 0 on the - values there .2426

When that X comes from that X right there that we had to do some integration by parts so, here is me doing integration by parts over here using tabular integration so, it turns out to be this rather complicated formula .2450

So, we plug-in X = 2 and that is X = 2 the 2 terms we get for showing 1 of them 0 and even more fortunately we plug-in X = 0 both terms drop out and so, just get cosine e N π and I noticed from this pattern from the previous slide the cosine e N π is just - 1 of the end.2465

So, put - 1 of the end here for cosine e N π so, that is my BN there was another - here so, that combine with this -1 BN and that is why we got -1 + 1 here and did some finance -1 and +1×2 / N π.2490

So, is my BN and I just recalled from the previous slides the a 0 and the AN the a 0 and the AN and so, what I was doing now, is I am recalling the general formula for Fourier series is always has this formula a 0 / 2 + the sum of AN cosine n π X / LB & AN π X / L .2509

And so, then I fill N = 0 just gave me and a 0 / 2 give you 1/2 there and the N = 1 terms I filled in N = 1 everywhere here and I got my AN's in my BN's there and there that is how I got a 1 that is how I got B1 and then my A2 was 0 so, that is B2 .2540

My A2 was 0 fortunately that is same a little bit there that is my a 3 right there that is a 3 and that is my B3 and at that point I ran out of space so, I decided to say that that is my Fourier series of course keeps going for infinitely many terms there .2571

So, were going use this Fourier series in the next example so, even though it is not rather complicated you do not want to completely forget it so, let me let me show you how this comes about how this gets put to use in the next example.2594

So, example 2 reuse the Fourier series that we derived in example 1 to find the value of the series 1+1/9+1/5 what these are ²s of odd numbers so, this is 1 / 3 ² 1 / 5 ² of course 1 is 1 / 1 ² and so, on .2610

So, were adding up all the ²s of odd numbers and somehow the Fourier series is mostly useful for that let me show you how it works out let me recall the Fourier series that we figured out in the previous example this is what we went through lots of work in example 1 to find so, having just watched example 1 pull up a chair and go through in and watch that it is quite long but that we get this good answer at the end of it.2634

We get the Fourier series of our function was 1/2-4 / π ² cosine of π X / 2 +2 / π this is all coming from the previous example sin a π X / 2 π X / 2 -2 / 2 π wrote it that way so, be easier to spot a pattern that is what and cancel there.2665

Sin of 2 π X / 2 -4/9 π ² cosine of 3 π X / 2 and +2 / 3 π sin of 3 π X / 2 and it kept going but 3 π X / 2.2694

The series kept going but we did not pursue it through any more terms there and the trick we want to use that for in this example is we want to plug-in X = 02 both sides there for me just remind you what this function was back from example 1 this is a function that was just a horizontal line from -2 to 0 and then was FX = X from 0 2.2720

There is 2 there is 0 there is - 2 so, that was the function were looking at so, what were delivered a plug-in X = 0 into the original function and to the Fourier series so, F of 0 that is the original function should be = 2 the Fourier series of 0 and let us see what happens there.2757

The original function F of 0 is = to 0 so, it is good 0 on the last out for a series of 0 is much more complicated so, let us see what we get when we plugged in X = 0.2782

The nice thing here is that all asine terms are going away to 0 because sin a 00 that can go to 0 that 1s can go to 0 with the balance can go to 0%.2799

All the cosine terms the cosine terms you go to 1 so, that is 1 right there that is 1 right there and so, what were going to have on the right is 1/2 -4 / π ² -4/9 π ² what is happening is were going up by ²s of odd numbers the next 1 is 4 / 25 π ² and so, on .2817

What we can do is we can manipulate this series a bit of move all the 4 / π ²s / 2 the other side and I see that I can factor out the 1 + 1/9 +1/25 and so, on and also, have that 1/2 left over on the right-hand side.2844

So, I took all of these terms and I move them over to the other side and so, trying to solve for the value of this series so, what I see is if I multiply both sides by π ² / 4 and I will get 1 + 1 / 9 +1/25 and so, on is = to π ² / 4 x 1/2.2873

And so, if I write this in σ form the way you keep track of odd numbers is you do 21 + 1 so, the sum of the reciprocals of the odd ²s is exactly π ² / 8 really quite surprising that it should come out to be such such a nice number there π ² / 8.2900

So, we have succeeded in using that Fourier series to find the value of this curious infinite series 1 / 1 ² +1 / 3 ² +1 / 5 ² so, let me recap how we made that happen first thing I did here was I just copied the Fourier series that we derived in example 1 .2909

So, I didn't figure this out on the spot I went through lots of work back in example 1 to solve that all that Fourier series are just copied at this time and so, that what I did with that was a plug-in X = 0 I did that to the original function as a graph of the original function again that is coming from example 1.2954

Example 1 so, that is coming from example 1 so, I plug-in 0X = 0 into the original function there is right there and I also, plug it into the Fourier series while the original function we just get F 0s = 2 0 in the Fourier series we look through this see all asine terms drop out because sin of 0 is = to 0 all the cosine terms turned into 1 and so, were left with just these all these while 1/2 is kind of a special term .2977

Then all these terms with 4 / π ² and then odd-numbered ²s of odd numbers 11 925/13 ² 5 ² and so, we move those / 2 the other side we get 4 / π ² a factor that out x 1+1/9+1/5 and I still have that 1/2 over on the right and so, I am trying to solve for this odd series .3014

What I did was I multiply both sides by π ² / 4 there and so, that gave me on the right-hand side π ² / 4 ×1/2 which turned in a π ² / 8 of the left I have exactly the series that I wanted and so, I know, that my series is = 2 π ² / 8.3042

Anyone hang onto this series really use it again in the next example so, I do not forget the the value of this series so, in example 3 we have to plug-in the endpoint value X = 2 to the Fourier series above and then we want to figure out what the Fourier series converges to when the original function is discontinuous .3062

So, let me remind you what that original function looks like so, that original function it was 0 from 0 to from -2 to 2 and then we made it we made it repeat after that so, we write my scale in 2 and write my function in red .3088

So, -20 there is 2 there is 4 there is 6 there is 8 so, we made this function periodic so, there is a periodically extended function and so, the question is what happens at this end points 0 between right of the endpoint 2 what does the Fourier series do.3119

So, in order to answer this we need to go back and look at the Fourier series on the copy that down the Fourier series of X this is what we figured out in example 1 so, I am not writing this right now, this is a lot of work that we did in example 1.3158

So, 1/2 -4 / π ² cosine of π X / 2 +2 / π x sin of π X / 2 -2 / 2 / 2 π x sin of 2 π X / 2 I did not want to cancel the 2s just because I think it was easier to find a pattern that way -4/9 π ² cosine of 3 π X / 2 +2 / 3 π .3175

Cosine is sin sin cosine sin okay 2/ 3 π sin of 3 π X / 2 and then it kept going but we did not work out any farther than that in example 1 so, what I want to do is plug-in X = 2 to that Fourier series X = 2 give this X = 2.3224

Now, the Fourier series of X = 2 is = 2 1/2 now, X = 2 means that all the sins will have sin of 2 π X / 2 so, that will be sin of π sin a π is 0 sin of 2 π is still given the 0 and the sin is going to be 0 .3259

Cosine of π x / 2 so, returning to -1 here we have cosine of 3 π x 2 / 2 so, cosine 3 π is also, edition cross it off as it is really canceled but all write it as -1 here when we plug-in X = 2 remember that can be -1 .3290

So, what were going to get is 1/2 + because we have a - here - so, + 4 / π ² here we have + 4 / 9 π ² and you cannot see too much of a pattern here what is happening is getting perfect ²s of odd numbers in the denominator.3332

The next 1 is going to be 4 / 25 π ² and so, on like that so, that is what is happening there and we can simplify this event we can factor out the 4 / π ² 4 / π ² x 1+1/9 +1/25+ 1/9 next will be 49 and what she noticed here is that is exactly that odd series that we figured out what the value was back in example 2 so, you did this in example 2 .3356

Work it out on the spot here but example 2 we figured out that this was exactly = to π ² / 8 and so, what we have here is 1/2 + 4 / π ² x π ² / 8 and that simplifies really nicely 1/2 π ² cancel 4/8 is 1/2 and so, what we get for the Fourier series value it 2 is exactly 1 and let us go back and look at our original function here.3395

The original function went to let me draw a larger version of that the original function went to was FX = X so, it went up to 2 here and then because it was periodic it had to drop down to 0 again and so, what the Fourier series is doing is it says that the value is 1 which means the Fourier series is exactly picking the point exactly halfway between the left and right hand limits.3444

so, what we have maybe do that in green to really emphasize that the Fourier series so, the Fourier series order tries to mimic the function is best to can does a good job were the function is continuous but when the function is discontinuous sort of has to make a jump and what it does is it splits the difference between the left hand limits and the right hand limits .3495

So, it takes a point exactly in the middle a converges to 1 so, summarize that in words the series splits the difference and converges to appoint exactly halfway between the left and right hand limits of the original function F so, the original function F is what I got wrap there in red and the Fourier series is grasped there in green .3531

So, you can see that the green 3 much matches the red function as long as the red function is continuous for the red function makes a jump the Fourier series takes a value exactly halfway in the middle.3608

So, let us recap that example here first while I wrote out the Fourier series that we developed in example 1 I did not redo the work there and then we plug-in X = 2 which made all the sins turn out to be perfect multiples of π integer multiples of pies all the sins went to 0 all the cosine we had odd values of π cosine of π here is cosine of 3 π so, is alternately -1 so, we got 1/2 the -1 cancel out with this - .3621

So, we got + 4 / π ² + 4 / 9 π ² + 4 / 25 π ² and if you factor out 4 / π ² we got these odd on ²s building up on the bottom there we figured out in example 2 figure out what that series converges to a converges to π ² / 8 .3657

So, that is why plug-in π ² / 8 here which really nicely cancel with a 4 / π ² just end up with 1/2 +1/2 and it gave us 1 and what that is telling us is that the Fourier series were the original function the red 1 is discontinuous the Fourier series is converging to a point which is exactly halfway in between the left and the right hand limits of the original function.3681

So, example 4 were going to extend the function below in such a way that it is Fourier series will contain only cosine so, let me go ahead and draw a graph of this function.3709

FX is = 3 - X where X and between 0 and 3s with only to find between 0 and 3 sets 3 not 1 and FX = 3 - X that the line there is 3 X0 FX is 3 and is just a line straight down.3723

Try to make that line little more straight those pretty shaky lines kind of a rough angle for me as a little bit better so, there is our FX would like to extend it so, that it is Fourier series will contain only cosine .3748

Now, if you want Fourier series contain only cosine that means that you want that function to be even and so, we have to we wants F of X to be even in orders contain only cosine .3769

So, we extend and so, I am going to extend it in a way that has symmetry in the Y axis mirror symmetry and the y-axis so, that means we put my extension and read here.3792

So, that is how I am going to extend F of X to make it symmetric in the y-a Xis now, that would have to be -3 and I see that that line right there that is has slope of 1 and y-intercept of 3 so, that is why he = X +3 .3813

So, the way I am going to define my function my extended function is F of X is = 3+ X +3 4 -3 less than or = 2 X less than or = 2 0 and then I continue with the original definition 3 - X for 0 less than or = 2 X less than or = 3 and then after that I want to extend it just to make it periodic.3833

Since my L is = 3 so, I will extend it to have period.3872

What he was is 1. 2L so, in this case 2 L is 6 I will extend its at period 6 F of X -6 or X + 6 is going to be the same as F of X + or - X if X is not in between -3 and 3 so, that is my extension F of X.3883

And we go ahead and draw the periodic portion of that or that the part that extends periodically so, what that means is I have to find it from -3 to 3 and everywhere beyond that I just repeat.3913

So, that the 6 9 12 and behind that I am going to repeat the same way so, there is -6 -9 and so, on.3938

So, now, I have got a periodic function so, it will have a Fourier series and it is even it symmetric around if we just put a mirror down on the y-a Xis you can see that it symmetric on both sides of the y-axis so, it is Fourier series will contain only cosine .3971

So, we just quickly recap as soon as I see the word cosine on thinking of an even Fourier series so, I want FX to be even so, that is why I extended it to look like it has mirror symmetry in the y-Xis and that is how I got this first dotted red line here and the equation of that red line is X +3 sets wide to find my function to be X +3 from -30 .3991

And in 3 - X is the part I was already given between 0 and 3 and then I just keep on extending it whenever X is not in that interval I just add or subtract 6 from X to go back and check what he did within the interval and define it to repeat itself every 6 units after that.4021

So, hang on to this 1 because were going to use this example in the next and that the next problem and were actually going to find a Fourier series for this function see how that works out .4042

So, in example 5 were to find a Fourier series for the function that we defined back in example 4 some this comes from example 4 draw quick graph of that.4057

There is the graph of its periodic after that so, we got this dysfunction that we defined back in example 4 and we want to find a Fourier series for that contains only cosine we are the extended it to make it an even function so, we do know, that it is going to contain only cosine so, I do not even need to worry about the B ends in the Fourier formulas just getting used the AN.4070

And let me remind you of the formulas for those the Fourier coefficients the AN in general is 1 / L x the integral from - L to L of F of X x cosine of n π X / L DX but we figured out that if the function is is even then this formula simplifies a little bit you can run it from 0 to L and then due 2 / L .4101

And in the same stuff on the inside F of X x cosine of n π X / L DX and so, what I meant to do now, is figure out the coefficients A the first 1 is a 0 and we use the second form 2 / small my L is = 3 so, 2/3 x the integral from 0 to 3 of now, between 0 and 3 the function is 3 - X so, 3 - X the X.4138

I didn't worry about the cosine because cosine when N = 0 cosine e 0 is = 1 so, that is why did not worry about that part so, I get 2 thirds x now, the interval of 3 - X is 3X - X ² / 2 evaluate that from X = 02X = 3 so, 2/3 3X is when X = 3 is 9 9-9 is 9/2 so, 2/3×9/2 the 2s cancel and we just get 3 .4174

So, that is my a 0 term my other ANs are going to be a lot more complicated my other ANs are going to be 2 thirds x the integral from 0 to 3 of 3 - X x cosine of n π X / 3 DX .4224

Need a lot more space to work that out the integration by parts again so, let me take a different slide to work that 1 out let me just quickly recap what we learned on this slide.4247

My AN this is my generic Fourier series formula but we learned that if you have an even function then it simplifies a little bit you can just run it from 0 to L and by symmetry you can just double your answer so, 2 / L instead of 1 / LF of X x cosine of n π X / L when N = 0 the cosine disappears because it just turns into 1 .4262

FX is 3 - X and so, now, we get a pretty easy integral to find the 0 coefficient there just worked through the integral and it turns out to be 3 for a 0 in general AN I just took this formula I plug-in L is = 2 3 and a plug-in FX is 3 - X so, will take up that integral on the next .4287

Let us go ahead so, let me remind you what we figured out on the previous slide there we figured out that our general AN what we forgot a 0 first of all was = 3 got that on the previous side are general AN is 2/3 x the integral from 0 to 3 of 3 - X x cosine of n π X / 3 DX .4312

I want to integrate that but that is going to require integration by parts some set up my tabular integration here 3 - X and cosine of n π X / 3 so, I take derivatives of 3 - X so, that so, the derivative of 3 - X is -1 and -1 is 0 and I take integrals of cosine of n π X / 3 so, the first integral is sin of n π X / 3 .4342

And multiply that by 3 / n π by the chain rule now, the integral of sin is - cosine - cosine n π X / 3 and I multiply that by another 3 / n π so, 9 / N ² π ² like a little bit squished in their let me rewrite that.4384

9 / N ² π ² and then I do + and - some multiply down those diagonals there and so, what I get is 2/3 now, fairly complicated here 3 - X x 3 / n π x sin of n π X / 3 .4411

I see 3 - here negative so, it is the whole thing is - -9 / and ² π ² cosine of n π X / 3 and I am supposed to evaluate that whole thing from X = 0 to X = 3.4442

So, nice thing here is I see about this complicated expression right here if I plug-in X = 3 or X = 0 into this get a whole multiple of n π inside the sin .4467

So, that terms can go to 0 on both sides there for X = 3 or for X = 0 so, that is really nice on the other side it is it is not so, good so, I still got 2 thirds now, for the cosine term -9 / N² π ² x cosine of n π 3 / 3 such as cosine of n π +9 / N² π ² x cosine 0 which is just 1.4483

So, let me close that off thing to factor out the 9 / and ² π ² in 2/3 of that would give me 6 / N² π ² and then I got 1 here and - cosine of n π and let us remember the cosine of n π we worked this out earlier .4522

Cosine of n π is well when N is even then cosine e N π is 1 and when N is odd it is -1 so, if n is even then cosine of n π is just going to be 1 so, we get 1 - 1 = 0 so, this is 0 if N is even and if N is odd this then cosine e N π is - 1 we get 1 - -1 which is 3 so, 12 / N² π ² if N is odd.4553

So, that is my expression for A of N if 0s N is even and it is 12 / N ² π ² if N is odd.4604

So, let me remind you the generic form for Fourier series of X for the Fourier series of of a function it is a 0 / 2+ the sum for N= 1 new infinity of N cosine of n π X / L + BN sin of n π X / L .4620

This when we partly arrange the function so, that all the sins dropped out because we made the function be even so, I do not need to worry about the sins all those dropped out but I do still have all these cosine terms.4646

So, let me fill in what I have for the cosine terms a not that is up there and not / 2 his 3/2+ now, look to get for N = 1 that is an odd 1 so, I get 12 / π ² cosine of 1 π X / 3 so, π X / 3 N = 2 is even so, I have 0 there .4662

0 there N = 3 odd one so, 12/9 π ² and ² π ² x cosine of 3 π X / 3 go ahead and write it out that way so, it is easier to see the pattern 3 π X / 3 .4695

N = 4 is even that was 0 N = 5 is 12 / N ² π ² so, 25 π ² cosine of 5 π X / 3.4714

I see that I got a certain pattern here of odd terms so, and I try to combine those and maybe make a new series here so, I can keep track of odd terms I can start N = 0 and just used 2N + 1 to keep track of all the odd numbers that I see.4726

So, think I can factor out 12 / π ² 12 / π ² from everything there and then on the denominator all have an odd number ² so, 1 / 2 1 + 1 ² x cosine of an odd number so, 2N + 1 x π X all / 3 and that is the last word on my Fourier series using only cosine is for that function.4749

So, let me recap everything we did there we figured out that a 0 is = 3 that was on the previous slide so, not doing that again.4789

The AN we got this formula on the previous slide for my AN and of course that is in integration by parts problem so, I went to my integration by parts table here this tabular integration by parts and I worked out derivatives of 3 - X integrals of cosine e N π X / 3 and I connected them up a little diagonal lines with alternating sins .4804

Put it all together got this free horrible expression for the integral of the nice thing is that the sin term when the plug-in X = 3 or X = 0 it is can go to 0 the cosine term is not so, good if you plug-in X = 3 get cosine e N π X = 3 and if you plug-in X = 0 you get cosine e 0 which is 1 .4829

But it is being subtracted its - but it is also, the lower limits we get 2 minuses cancelled to give us a + there so, have 2/3 x looks like a factored out the 9 9×2/3 is 6 and I factored out the N ² π ² so, the 9 2/3 gave me the 6 there and the N ² π ² came out we still had a 1 here and a cosine - cosine e N π .4855

We figured this out before is it is even it is 1 if and is on its -1 and so, when it is even we just have 1 -1 so, gives a 0 if N is odd we have 1 - -1 so, without the 1 - -1 if N is odd we have 1 - -1 so, we get 2 2×6 / N² π ² gives us 12 / N² π ² so, that is our coefficient AN kind of complicated.4884

Sometimes its 0 sometimes 12 / N² π ² and I remembered my general Fourier series formula a0 / 2 + N x a cosine term BN x a sin term and we remembered that for this particular 1 we had extended the function to be even it is an even function which means that all the BN are 0 so, there is no sin terms.4925

We did that deliberately because the problem assess to find a function that had all cosine and so, my a0 was 3 that came from up here and my AN I dropped in the 12 / N² π ² there is the odd 1s right there .4950

Then we dropped in 0 for the even 1s and then I tried to rewrite this because I noticed that I had exactly the odd terms well that the 3/2 is kind of a special case.4968

But I guess you have all the odd terms here 1 3 ² 5 ² so, to find a new pattern for odd terms are used to 1 + 1 generates odd numbers so, I factored out the 12 / π ² from all of them all them have 1 / 1 1/9 1/25 so, that is 1 / 2 1 + 1 ² and enumerators there is 1 π X 3 π X 5 π X that is 1 + 1 π X / 3 .4979

So, it gives me a Fourier series for that function using only cosine which is what we are asked to do.5008

So, that wraps up this lecture on Fourier series as part of the differential equations lecture series on educator.com.5021

What we are doing here is we are trying to solve partial differential equations although we did not really get a chance to look at him in this lecture we decided a lot of background on Fourier series .5029

The whole point of that is that were going use Fourier series to solve partial differential equations so, to learn about that in the next lecture where we find a solution to the heat equation.5039

So, hope you will stick around watch the next lecture which is on using Fourier series to solve the heat equation, and that will wrap up our set of lectures on partial differential equations .5051

So again, you are watching the lectures on the differential equations here on educator.com, and my name is Will Murray, and I thank you very much for joining us today, bye bye.5062

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