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Math All Day with Dr. George Sweeney
Приєднався 11 лип 2010
I love math and I love to teach math. I especially love difficult math, computer math and linear algebra. My channel is for people who are interested in mathematics, computer math (discrete structures or discrete mathematics) and all of its applications. I hope I can make the learning of this challenging and interesting subject easier and more fun.
What are reflexive, symmetric, and transitive relations?
In this video we discuss the properties of relations for binary relations:
These relations are as follows, reflexive, anti-reflexive, symmetric, anti-symmetric, and transitive relations. The relations and concepts are very valuable for doing discrete mathematics, especially graph theory, linear algebra, algorithms and functions.
If you need to review the basics of relations and graphing relations, check out this video:
ua-cam.com/video/hh5NrF7j6qw/v-deo.html
00:00 Properties of Relations: Reflexive, Symmetric and Transitive Relations
00:10 Properties of Binary Relations
00:50 Reflexive Relations
00:53 Definition of Reflexive
01:12 Reflexive Relationship Example 1
02:20 Definition of Anti-Reflexive
02:46 Anti-reflexive example 1
03:39 Reflexive Relationship 3
04:27 Review: Reflexive
04:43 Symmetric Relations
04:49 Definition of Symmetric
05:31 Symmetric Relationship Example 1
07:08 Definition of Antisymmetric
07:34 Anti-Symmetric Example 2
09:08 Symmetric Relationship 3
11:23 Both Symmetric and Anti-Symmetric
12:02 Review: Symmetry
12:38 Definition of Transitive
13:02 Transitive Example 1
14:40 Transitive Example 2
16:05 Review of the properties:
16:31 End of presentation
These relations are as follows, reflexive, anti-reflexive, symmetric, anti-symmetric, and transitive relations. The relations and concepts are very valuable for doing discrete mathematics, especially graph theory, linear algebra, algorithms and functions.
If you need to review the basics of relations and graphing relations, check out this video:
ua-cam.com/video/hh5NrF7j6qw/v-deo.html
00:00 Properties of Relations: Reflexive, Symmetric and Transitive Relations
00:10 Properties of Binary Relations
00:50 Reflexive Relations
00:53 Definition of Reflexive
01:12 Reflexive Relationship Example 1
02:20 Definition of Anti-Reflexive
02:46 Anti-reflexive example 1
03:39 Reflexive Relationship 3
04:27 Review: Reflexive
04:43 Symmetric Relations
04:49 Definition of Symmetric
05:31 Symmetric Relationship Example 1
07:08 Definition of Antisymmetric
07:34 Anti-Symmetric Example 2
09:08 Symmetric Relationship 3
11:23 Both Symmetric and Anti-Symmetric
12:02 Review: Symmetry
12:38 Definition of Transitive
13:02 Transitive Example 1
14:40 Transitive Example 2
16:05 Review of the properties:
16:31 End of presentation
Переглядів: 11
Відео
How do I transform a relation into a matrix or array?
Переглядів 2014 днів тому
In this video, I discuss a key skill for doing computer science and working with relations, writing a relation as a matrix or array. In this video I talk about the dimensions of the matrix and also how we populate the matrix given a particular representation. 00:00 Constructing a Matrix or Array from a Relation 00:17 Presentation Paused 00:18 Basic Concepts 01:16 The Dimensions of the Matrix 02...
What is a relation and how do I graph one?
Переглядів 714 днів тому
In this video, I do three things: 1. Define a relation and give examples 2. Show how to construct line graphs for relations 3. Show how to construct a matrix from a relation The concept of relations is key to understanding and doing discrete math with functions, arrays, transformations, and graphs. 00:00 Introduction to Relations 00:44 Defining a relation 02:12 Defining a Relation- Ex 2 04:40 L...
How do I solve differential equations for piecewise functions?
Переглядів 2814 місяці тому
In this video, I give an example and discuss the method for solving differential equations that have piecewise components. Piecewise differential equations are very common in engineering and circuit applications and this video gives students a basic overview of the process. 00:00 Introduction 00:56 Solving Piecewise Continous Functions 09:28 Review 10:01 Marker 1
Orthogonal Trajectories and Families of Functions
Переглядів 554 місяці тому
In this video, I show how to find orthogonal trajectories of a family of functions using differential equations. Orthogonal trajectories are a great way to introduce families of functions and the general solution to a differential equation. 00:00 Introduction 00:01 Introduction 01:01 Families of functions 04:24 Orthogonal Definition 08:39 Orthogonal Trajectories Example 1 14:16 OT Example 2 18:...
What is a parabola and how do I graph one?
Переглядів 477 місяців тому
In this video, I start with the definition of a parabola as the set of all points equidistant from the focus and directrix. I then answer the following questions: 1. What is a parabola? 2. What is the equation for a parabola? 3. How do I find the focus, directrix and graph the parabola given the equation? 4. How do I find the equation for the parabola given the focus and the directrix? 00:00 In...
What is an ellipse and how do I graph one?
Переглядів 398 місяців тому
I describe the different parts of the ellipse, including the major and minor axes, foci, and vertices. I derive the equation and then I graph ellipses using it. This is part of my conic sections lectures. 00:00 Introduction to ellipses 01:14 Defining the ellipse 03:49 Different parts of the ellipse 08:17 Deriving the equation of the ellipse 14:51 The equations for the ellipse 16:25 Ex1: Graphin...
Solving the Volterra Equation using Laplace Transforms
Переглядів 2649 місяців тому
The volterra equation is an integral equation that requires the use of the convolution theorem and Laplace transforms to solve. In this video, we show how to use the convolution theorem to solve Volterra equations. The Volterra equation is used in modelling semiconductors and in different kinds of advanced mathematical physics. I have linked to my video on the Convolution Theorem: 00:00 Definin...
Using the Convolution Theorem to Do Laplace Transforms
Переглядів 1829 місяців тому
If you want to solve initial value problems or systems with laplace transforms and you have a product of two Laplace transforms, you will want to use the Convolution theorem to solve. The Convolution theorem uses the convolution integral in order to define the inverse laplace transform. 00:00 Introduction 00:54 Defining the Convolution Product/Integra 01:23 Convolution Product Example 04:06 Pro...
Verifying Functional Completeness- Boolean Algebras
Переглядів 1359 місяців тому
To verify the completeness of a set of operations on a Boolean Algebra, we need to show that our new set of operations give equivalent functions. In this video, we show how to determine if a set of operations is functionally complete by showing how we convert the operations and then show how the new functions give us the same outputs given the same inputs. Functional completeness is valuable be...
Solving Systems of Differential Equations using Laplace Transforms
Переглядів 2649 місяців тому
Now that we have figured out how to solve initial value problems using Laplace transforms, we can use Laplace transforms to solve systems of differential equations. We will use the basic theory and Kramer's rule to solve these systems. If you need a refresher on Kramer's rule, I have linked to a video below: ua-cam.com/video/Ot87qLTODdQ/v-deo.html 00:00 Introduction 00:49 General Method for a 2...
Solving Initial Value Problems using Laplace Transforms- Basic
Переглядів 73110 місяців тому
In this video, I solve differential equations using Laplace transforms and put forward the basic theory. I only use basic fundamental Laplace transforms and inverses, but the method will hold for any functions where we will use Laplace Transforms. 00:00 Introduction 01:27 Laplace transform of the derivative 02:23 Proving the Derivative 05:12 Laplace Transform of y" 06:06 Example 1- First Order ...
The Second Shifting Theorem and the Unit Step Function
Переглядів 51010 місяців тому
This video is on how to use Laplace transforms for piecewise continuous functions without integrating. The first part of the video is on the unit step function, which is what we use to transform piecewise functions into individual functions. Then once we have used the unit step function, we can use the second shifting theorem to find the Laplace transform of piecewise functions. 00:00 Introduct...
Laplace Transfroms-The First Shifting Theorem
Переглядів 15710 місяців тому
In this video, I expand on the theory of laplace transforms to solve initial value problems by using the first shifting theorem. The First Shifting Theorem allows us to expand the number and kind of functions we can find inverse Laplace transforms for. 00:00 Introduction 00:26 Theorem and Proof 03:02 Examples for finding Laplace Transforms 06:36 Examples for finding Inverses 15:54 Techniques an...
Related Rates- Triangle Example
Переглядів 6510 місяців тому
In this video, I find the change in the area based upon the change in an angle. We use a trig formula, A=(1/2)ab sin (theta) in order to do this. This is a problem example, so if you are looking for a longer lecture on related rates, please see the lecture video:
Laplace Transforms of Piecewise Functions
Переглядів 26411 місяців тому
Laplace Transforms of Piecewise Functions
Decomposing a matrix into Elementary matrices
Переглядів 401Рік тому
Decomposing a matrix into Elementary matrices
Using Elementary Row Matrices to Do Row Reduction
Переглядів 382Рік тому
Using Elementary Row Matrices to Do Row Reduction
Basic Mixture Problems using first order linear differential equations
Переглядів 903Рік тому
Basic Mixture Problems using first order linear differential equations
Finding Big Theta with subtraction of terms
Переглядів 121Рік тому
Finding Big Theta with subtraction of terms
Thanks
You're welcome!
Bless your soul
You are welcome. You would think it would be easier.
Thank you very much sir...
You are welcome!
thank you for this beautiful video explained perfectly helped me so much
Thanks for the kind words. I am really glad it helped.
Very helpful thank you!
You are welcome, Ariana. Good luck!
Omg this was sooooo useful thanks!! First video that helped me understand this simple concept
You are very welcome. I am glad it helped.
Thank you Sir !! and Happy New Year 2025 !!
Happy new year!
this shit was ass. Didnt help atall if i fail im coming to ur house
I am sorry that didn't help Daniwal, good luck on your test and I hope you find something that was helpful. If there is a way specifically that you think I could change the video, I am open to the feedback.
thx brother,can we have video about Post's Functional Completeness Theorem
Sorry for the late reply. That is a great idea. I think that what I will need to do some studying up on it as it is not entirely obvious. Is there anything you would like me to focus on?
@@georgefsweeney to be honest, i wasnt sure if was my question was releted to this video. anyway a little background how does he came to final theorem well be great and thx for your replay.
Awesome video
Sorry for the late reply. Thank you sir, I am glad you liked it.
Thank you! This video got me right. Final exam in three days doing some review.
I am so glad! Good luck on your exams!
Very good video content but poor sound system. Please change it :)
Thanks oguza. I have changed my sound system since this video, but I will upgrade this one during my break. And I am going to pick up a new system. Thanks for the feedback!
k xloga b = c^3/2 [f x = 0-9 e f gx fgx c a3 This is the correct answer for your question ♥️
Thank you, I will look into this and make sure everything is correct.
If not, I am going to do a redo.
k xloga b = c^3/2 [f x = 0-9 e f gx fgx c a3
Really helpful video! The only thing i dont understand is how did you get the 4•1•1•-1
Hi Whatever, the values of 4, 1, 1, and -1 really come directly from the elementary matrices for the row operations. We can turn each row operation into an elementary matrix and that matrix has a determinant. If we mutliply a row by k, then the elementary matrix has a determinant of k. Addition of rows always has a determinant of 1 no matter which two we are adding or what we are multiplying the rows by. And permutations have a determinant of -1. In this case, you can see the row operations are multiplying one of the rows by 4, adding multiples of two of the rows, another addition of two rows, and switching two of the rows- so 4*1*1*-1. I have added some links in case you want a deeper dive. Here is a video about elementary matrices: ua-cam.com/video/ncU3VsrDyMo/v-deo.html Here is a video that relates the row operations: ua-cam.com/video/3gzH4Bro5Eo/v-deo.html I hope that helps.
I still don't get it. I don't understand how or why the row operations affect the matrix.
Row operations will change the values in a matrix either by adding rows or multiplying rows by scalars or permuting . This will change the values in the matrix. Essentially that is the process for Gaussian elimination. This video is focusing on how using those row operations affect the determinant and that explanation depends on elementary row matrices and their relationship to the determinant. Can you help me with a little more information about what you don't understand and maybe I can help a little more or point you to a good video?
This makes it so much easier to understand, because the worst thing about discrete math is how every concept ends with proofs and it just makes it so confusing
I totally get it. I am glad it helped.
I believe that the correct answer is 142.171 as you'll get 517.82 when you add in 14:06, when clearly its negative. Correct me if I'm wrong tho
I will give it a check. I have a list of videos for this break that need a good looking through and revision. Thanks for the heads up.
Thank you! Exactly what I needed! ☺
I am really glad it helped. Keeping plugging through multi-variable.
thanks dude
You're welcome!
incorrect. Recheck your work
You are totally right. I dropped the negative off on the second half of the first integral. It will need a revision.
Nice
I am so glad it helped!
25:11 shouldn't this intergral be (x^-p)/(-p+1)
Thank you, let me take a look at this.
Doesn’t success have 3 s’?
Yes it does. You are dead on. Looks like this one needs a revision.
Amazing video!
Thank you so much!
Super helpful god bless
Great video! In the last step when you calculated the flux out from the disc you added to be able to use divergence theorem, how did you change the double integral of S2 with dS to a double integral of D with dA? is it because 12:07 both dS and dA are the area element? Thank you
Very helpful 😊
Thank you!
how do you find the inverse of the modulo?
Minute 10:41 this is what we were talking about today in the homework example, in office hours, the reason why it's not valid is because the conclusion is false. In order to have a valid argument, we need the hypothesis to be true and conclusion to be true. If q = t hence -q = f, makes the conclusion false, if the conclusion is false its not a valid argument. If the conclusion was q then it would be valid and it would use the rule of disjunctive syllogism.
This was simple, straightforward, with clear examples. Awesome.
at 23:25 you missed the -1
Got it, I will see if I can get a revision in this weekend. Thank you!
Hey Foolish, finally got around to making the change, but if I am correct- I converted the L[-cost], but that means that I pull out the negative and take the Laplace transform of cosine. The two negatives make it positive. I wasn't explicit in moving the scalar outwards.
❤❤❤
Helpful video. Thanks man!
Glad it helped!
Let me tell you something. When I was a student and I had to take this class, I can honestly tell you that it did not make sense at all. I have been an Algebra 1 and 2 teacher for almost a ten years, and it finally clicked. Moral of the story, a person must have a strong algebra foundation to truly understand this 😅
This is true because at the end of the day, this kind of proving simply requires some algebra ability and an understanding of what the members of the domain look like.
RADIAN MODE
good explanation, thank you
You are welcome!
Hi there! Love the lecture. Thanks for the mention at 42:42. I am glad these applets I made are useful :) Regards.
Thank you, I and my students really appreciate all the great tools that people have created. Eventually I hope to pay it forward and create some great things for other teachers to use.
And specifically, thank you!
In the first example given the supply of fruits is limited if one places bars in the first two slots there will not be 5 pears to fill the rest of the slots given there are only four pears. Maybe I am missing something
Sameer, if you put the bars in the first two slots, you will have a bowl with 5 pears, no oranges, and no apples. At the bottom of the question, it says there are 5 of each type of fruit. So there should be five pears. Let me know if you have a follow up.
well I listened this from my Professor, and also checked textbook. But It hass been clear just after listening you. thanks
Shiny_apricot, I am glad that it helped.
I need clarification on why the first one is preferred to be the definition over the theorem one. Could anyone please let me know?🙏
I am not sure about the question, but if I understand you right: I use the theorem to evaluate the limits in this video and I think for most purposes this actually is the quickest and most effective way of finding the limits of vector-valued functions. So I use the theorem one for the limits problems. I think I glossed over the limit definition because I didn't want to go into evaluating distance(the two bars on either side of definition of the limit) and all of the machinery of calculating a magnitude when the theorem really was more intuitive and connected better to early ideas of single-variable calculus. So in short, the theorem is actually how we would find limits. I hoped that helped. Let me know if you have a follow up.
@@georgefsweeney Thanks!
Thanks a lot professor I follow you from Algeria.
Wonderful Bouazabachir4286. I am looking forward to hearing from a lot more people. I hope this helps.
swag
In case of Gaus elimination change matrix to a row echelon form??
That is correct. Gaussian Elimination transforms a matrix to row echelon form. Gauss-Jordan Elimination transforms matrices to their reduced row echelon form. The process is virtually identical and I think the video shows both methods. For Gaussian Elimination, just stop at the row echelon form.
@@georgefsweeney thanks for reply
thank you so much !!
You are so welcome!
My discrete math exam is in 45 minutes, and this was the last topic I needed to clean up before going in. Very concise and simple explanations with great examples for both this video and the one-to-one video. Thank you so much!
I am so glad to hear it. I hope your exam went well.
how the heck did you get your x values to plug in. you showed visuals and all but then skipped over how out got 0.1, 0.3, 0.5, 0.7, and 0.9 RIPPPP
Because we are breaking the interval up into 5 parts and the interval is from 0-1, we get each interval is a length of .2. So the first interval goes from 0-.2. Then we take the midpoint of each section, so that first midpoint is .1. Then, .3 and so on. You can see the explanation at about 1:15 in the video.
Thank you❤
You're welcome 😊
Omggg!!! I cannot thank you enough, god bless you with this video 😭💕
You're so welcome!