Good morning TJV fans. This is the A9 towards Inverness in Scotland, gloomy day. But it's gonna be a good day. Don't forget to like and subscribe. Good morning, everybody. This is our load for today. I'm just tying it down. I've got one more strap to throw. I'll take you along with me for it.  I've got the one strap here. I'm gonna throw it over from the other side. Reason being is you'll see it from the back here. This side has a straight edge down, right? Obviously this side doesn't. Now I don't want to throw my strap from there and pull the load that way. I want the strap to go over from here, hold that bottom piece in, go over here and pin this whole piece straight down. Not pulling it over, but pinning it straight down. Does that make sense? We've got old blue sitting at the front of this train. Ready to work. So I'm gonna throw this one right over here. The thing is they put these right up to the edge of the rub rail here, and I can't fit my hook through here, right? So I'm gonna unroll the whole strap and feed it through here where I want it. Right about here, yeah, because the strap has to be inside the rub rail. That's very important. If you don't want to attract the attention, unnecessary attention at the scale or DOT, MOT, wherever you are, never put your straps over the rub rail, always inside the rub rail. It's called a rub rail for a reason.  Rub rail's there to protect your straps. If at all possible also, you don't want to be attached in these hooks to these rub rails. On these trailers especially, because this is a very, this is just welded on. It's not as strong of a rub rail. You can see it over here. You never want to secure your load to this. You always want to go under to the frame on here, or in this case, this trailer has these where I hook them into. You can see this one is hooked in right here, and that's where it's supposed to be hooked into. That's another way to get bad attention if you strap your load to that, because that's not rated for very much weight. That'll just rip right off. You know, you hit one bump, the trailer goes like this.  That's what they're trying to prevent anyways, I think. This one last strap to throw over, and I'm not claiming to be perfect or all-knowing either, just so you know. I'm learning every day too. Little tips like that are just that.  I've been driving trucks since 2006. So I've learned the thing or two in there, but you know, there's always more to learn. You always got to be open to learn. Sometimes I'm wrong, but if you go to DOT and they're trying to give you a ticket, or if they come to you and they're trying to give you a ticket, and you try to get out of it by saying, well, Trucker Josh told me to do it this way. Well, guess what? They're not gonna buy it. You're getting a ticket anyway. So I try not to give you any false information. I do my absolute best. Coming down here. Got this one coming down, right over this here. So we're here, and these straps here will pin this piece here. Straight down to here, and the other side will hold these bottom pieces together like that. And then through the magical powers of friction, this load isn't going anywhere. Park Rapids, where this freight is going to, is about a five-hour drive from here. So I'm gonna quickly finish tying this all down. And we'll start making our way south. Hopefully I can get unloaded today yet. And most times there's more than just one way of doing things too. It's not like I'm telling you this is the way you do it. It's my way or the highway. I just show you how I do it, and what works for me. There are other ways of doing things too. I always say there's no right or wrong way of tying down a load. I should phrase that carefully though. There is a wrong way of tying down a load. As long as the load is secure, there's not always just one way to do it. I mean, there's many different variations of techniques you could use. You sort of just gotta find out what works for you. And as long as the load is legal and good to go down the highway, don't overthink it sometimes. Don't second-guess yourself. But just make sure you're confident in your load, because if anything falls off that load, it's your fault. I want to see something again. I think I might have missed something actually. I'm going to be there 440. Just to their closing time, I like to give them a heads up and ask them if it's all right, if I show up then. Because sometimes they use that time a day to clean up and get ready to close. Leaving new freight at that time, even though they are open, it's understandable. They close five o'clock, and that's a decent time. Right there for me, and they're going to unload me when I get there, which is awesome. After that, I have to get unloaded today, because I have to get back to work tonight yet. I have to hook on to a van trailer. We're going to be a barn door swinger for a couple of days. We're picking up a load in Portage La Prairie. That load is taking us up north of Thompson Manitoba, to Northern Manitoba. Never been that far north of my own province. I've got to be there first thing Monday morning, so I'll be headed out there over the weekend. A little bit of extra cash on my week. We're trying to catch up with a whole bunch of stuff and get ready for the baby. And save up to build a house. And it just seems there's surprise expenses keep popping up. And then there's the price of food and inflation. And now our government is tripling our tax on our home heat. So this winter, our heating bills are going to be so much higher. They're tripling it. So overwhelming sometimes. So trying to get an extra day in here wherever I can, just to make ends meet and still try to advance forward with all of our plans. And once again, empty.  It's nice of them to stay just long enough to unload me. Got here at 4.40pm. I think it's about five o'clock exactly now, so they're all headed home. And I got to get out of their yard so they can close their gate. But I'm going to leave this piece of wood here with them. Because this was part of their load. Looks like it's broken. Some of the dunnage, I guess. It's theirs. Okay. Looking good. Looking good. That was a rush. That was not the good kind of rush either.  I really hate rushing. I like enjoying the ride, but this is trucking. And let's be honest, almost everything is a rush. Everybody wants everything yesterday, right? So we're headed back to the yard now. Gonna grab that van trailer and head over to Porter's little prayer. Get ready for tomorrow morning. Busy busy. Back to the north we go. Just barely made it. Just barely. And it worked out perfect that I was there for exactly 30 minutes. So I got my 30 minute break. It was a good day. Ready to be harvested now. Next time I come down, that corn will be gone. But it looks like a Christmas tree farm or something. That looks fun.  Next time I'm in this area, there might be snow on the ground. It's the best thing about this job. Something different every day, regardless of where you are or where you're going. Helping majority. I'm talking like 99% of people you meet and talk to and deliver to and pick up from. One kilometer. We're good kind people. Township road nine and then. Turn left in 16 meters. Even these guys here, I mean they didn't have to receive me today. But I called, I was polite. I tried to be as polite as I could be. Asked them, told them the situation, what my ETA was. I was honest about my ETA. Don't go and ruin it for the rest of us.  Turn right on. Township road nine and then. Turn left in 16 meters. I've had a couple of receivers actually surprised when I show up. When I say, when I say I'll be there. I say, oh most truckers lie and say they'll be here like a 15 minutes earlier or something. No, I try to be as precise as possible. That next time when I call them, hopefully they recognize me. They'll trust me when I say I'll be there. If you wait an extra two minutes for me, I'll be there. You know me. This is Bagley, Minnesota. Diesel price here is $5.20 US per US gallon. That's probably about $1.88 to $1.90 Canadian per liter. Just for comparison, back at home, diesel is going for about $2.20 a liter right now in Manitoba. So still considerably cheaper down here in the US. Just like I said before, I'm going to pull it up to Portage La Prairie and grab a load with it in the morning. Bring it back here. And next week I'm going to pull it up into Northern Manitoba. It looks a little strange to me with a box behind it. In order for it to work for me to pull a box behind me or a van trailer, I have to slide my fifth wheel all the way to the back as far as I can. Because the pin on the trailer is further in than on the step decks, there's more overhang. So it comes closer to my truck. And if I leave my fifth wheel, where I usually have it when I'm pulling quad beds, when I make a corner, that comes around and hits my headache rack and could cause some pretty expensive damage. So now there's just enough room in there. Not much. I got no extra space, but just enough room to make a corner and not wreck my truck. Still, let's get going.  Let's make sure it's not going to fall off first. Okay, one more time. All forward and spike. And triple triple quadruple check. That thing is attached. Nice. Peace of mind. I still always watch it like a hawk for the first like half hour. I don't know why I'm paranoid. You guys want to see how close the trailer comes to my truck when we turn? I know I do. I want to double check on that. I'm going to make a sharp corner right here and then we'll go check it out. Okay, turn. Let's go see how close it is. Say we still got plenty of space in there. I might have even been able to leave it where it was. I usually have the fifth wheel centered between my axles here. And now let's just set back just a little bit. But I could have left it where it was. But just in case if I hit a driveway connection and the truck goes up like this, I want to make sure that the top doesn't touch that. So yeah, it'll be good. And yes, I'm still waiting for my decals that I ordered four weeks ago. Haven't heard back from them. I'm kind of upset about it. I might go to a different company. Four weeks. Haven't heard one peep. Not nothing. Usually they would send me a template of the design for my approval. Nothing. And I called them last week too. After three weeks, I called them and said, Hey, what's going on? I said, yeah, yeah, we'll get right back to you. It's been another week. Haven't heard back from them. So creative print all on Steinback. I'm waiting to hear from you. I'm anxious to get my new decals. Don't forget about me, please. What is this thing? That's a big tank for something. It feels so weird. I had to put my tarps inside the trailer because there was no room for them on the back of my truck. Open says me. Excuse me. Excuse me. Do you not see this big glorious hood here? Got to get right up close and personal with it, eh? Come on. Today, Junior.  We're good to go. I'll talk to you when we get there. Manitoba found one of the very last possible swallows we could park. Their sign looks like you could use some help there, doesn't it? This is where we spent the night. It's the next morning for me already, so I got to start tomorrow's video. I hope you tune in for it. Don't forget to subscribe if you haven't already. Make new videos of my life on the road as a truck driver across North America.  We're leaving Sunday. Getting an early start to the week and we're headed way up into Northern Manitoba, further north than I've ever been in my own home province. Pass Thompson. So it's going to be interesting. I hope we'll be alright. I know we'll be alright. I don't know what kind of cell service they got up there. There's not much up there, but tune in then. It'll be a lot of fun. Until then, we're going to see what we have up this weekend. I got to go pick up this load first in tomorrow's video and then go home Friday, Saturday. And see what adventure awaits. See you later.


Alright, thanks everybody for making it out today and thanks Grant for agreeing to give a talk. Today we have our own graduate student Grant Fix who will be talking about recurrence ranks of sequences. Go ahead and take us away Grant. Thanks for the invitation drew and thanks everybody for tuning in and giving me an hour of your time to listen to what Josh and I had to work on so. I'd just like to take a quick minute to thank Josh for the opportunity to work with him and for his endless patience and encouragement through our entire process so I look forward to what we can continue to do. And in the years to come so this title the presentation recurrence ranks of sequences so what we're going to do is we're going to develop a couple definitions for different kinds of ranks. As it pertains to sequences so the context for. The context for this work is so Josh just does a lot of works in a lot of different areas but one of them being hypergraphs and and spectral hypergraph theory and that was the motivation for these ideas so characteristic polynomials are something that we can look at in terms of graphs we can look at the adjacency matrix of a graph and like any matrix we can look at its, its characteristic polynomial so we call that the characteristic polynomial of a graph. And there's a lot of information known about these characteristic polynomials for graphs because there are a lot of linear algebra tools that you can apply to gain information about a graph based on its characteristic polynomial. So for hypergraphs, not nearly as much as is known about these characteristic polynomials so a couple examples of some things that are known for example, a three uniform hyper edge now what that looks like is for a simple 10 like me it looks like a triangle. So three vertices, three uniform just means that all of the edges in the graph are three subsets of the vertices so in just standard graphs we have a vertex set an edge set, and every edge is a two subset of the vertices. In hypergraphs we relax that condition and any subset of the vertex set can define an edge, but the most convenient hypergraphs have all of the edges are the same size or connect the same number of vertices. So we call that that cardinality of an edge set or the number of vertices that every edge connects we call that uniform. So three uniform hyper edge is three vertices, and there's one edge that connects them it may be drawn like a triangle, you might think of it as like a face, for example. But that's that's what it looks like it has characteristic polynomial, given by this so if our variable is lambda than the characteristic polynomial that graph is is lambda cubed times lambda cubed minus one cube. So right away you notice that the degree of this polynomial is 12 there's there's a lambda cubed and then the term lambda cubed minus one cube that that contributes nine to the degree. And in graphs just by themselves, the degree of the characteristic polynomial is given by just the number of vertices, or the, or the order of the graph that we're working in razz. Here right away we see with hypergraphs that the characteristic polynomial degree is much, much larger, typically then the actual the number of vertices, or the order of the graph now a K uniform hyper edge in general. So again, just trying to picture this if you're not familiar with hypergraphs. It looks to me like a case cycle if we're just thinking about it in terms of graphs but really what it is, is it's K vertices all connected with one edge. And sometimes it's drawn like a case cycle sometimes it's, it's kind of shaded in so it looks like the face of some kind of surface. This is the characteristic polynomial of the K uniform hyper edge and what I really like you to notice is, there's this this large power of lambda. So, zero is a root of this characteristic polynomial and then it has the K through to be as well to some relatively high degrees. So there are other results known about characteristic polynomials of hypergraphs. Greg Clark, who worked under Josh as well he graduated two years ago when is off at Oxford at a postdoc there. He did a couple results which are looked at a couple different hypergraphs. Some of them I have included here just for some reference for us. This is called the rolling hypergraph, because JK Rowling wrote the Harry Potter books and this is a symbol that comes up in there. But he found he and Josh work together, found the characteristic polynomial for this hypergraph. And again we can see that the degree is quite large, and the, the, the, the sheer size of this degree is is one of the reasons that it's a huge characteristic polynomials of these hypergraphs. The computation is NP hard in general. And just thinking about it the way we compute characteristic polynomials for grabs we're taking the determinant of some matrix. The adjacency matrix for hypergraphs is tensor or a hyper matrix however you want to think about it so it's this for three uniform, it's a three dimensional block or box of numbers that we have to try to take determinant and still notions of determinant for those tensors. I won't get into exactly what those are but it's much more complicated to compute compared to the nice methods that we have to compute determinants of just general matrices. Now, the, the reason I include this in in here is if I draw in an edge here and edge there. What I'm drawing now is the phanoplane. So it just has two additional edges so this graph has seven vertices it's a three uniform hypergraph. So all of the edges connect three different vertices in the graph if I just add these two additional edges. Then we turn it into the phanoplane. It's actually unknown what the characteristic polynomial of the phanoplane is so just, just adding those two extra edges takes us from a hypergraph whose characteristic polynomial we know to a hypergraph whose characteristic polynomial is still unknown. So that kind of provides us with a little bit of context on just how hard these characteristic polynomials are to compute. And this particular example that I want to consider is this, this thing called the hummingbird hypergraph that looks something like this. This is this hypergraph falls into a very nice or very simple class of hypergraphs namely linear trees. So linear just means that any two edges that you choose to take the intersection of their vertex sets, and the size of that intersection is at most one. Any two edges have at most one vertex in common. They play nicely in terms of different things and trees just in the graph contexts are our graphs connected with no cycles. So you can kind of see here that this graph is connected. And then there are there are a couple different notions of what it means to be a tree in in the hypergraph context because there are a couple different notions of what it means to be a cycle but no definition you choose. There's no cycle in this particular graph hypergraph because no matter what two vertices you choose there's really only one way to get between them again noting that each of these triangle looking things are really just one edge. But anyway this this hypergraph has characteristic polynomial given by this quite large. So let's look at a lot of different routes but there's the one route that we do see here is zero. So here, the factor lambda has degree or multiplicity almost 21,000. Quite large. But anyway, the thing that all three of these examples have in common is that zero was a root of all of the characteristic columns we saw. The uniform hyper edge had this factor lambda the to the K times K minus one to the K minus one minus K to the K minus one. A decent factor there of just lambda so there's the zero root in that characteristic polynomial. The rolling hypergraph characteristic polynomial had this landed to the 133rd power. And then as we just saw in the hummingbird hypergraph, the multiplicity of zero there was almost 20,000. And it seems like zero is pretty common as a root of many of these hypergraphs so what we wanted to do was kind of study and see if we could figure out some bounds or maybe exact ways to compute the multiplicity of the zero eigenvalue for a particular graph. So one technique that comes up fairly often in trying to prove results for hypergraphs and hypergraph characteristic polynomials is to take some notion from graphs and see if you can extend it to something interesting in regards to hypergraphs. The fortunate thing in this setting is there isn't a nice result for graphs that allows us to compute or gives us interesting bounds on the multiplicity of zero for a given graph so we couldn't really pursue that that avenue so we had to come up with something new we had to try something else. And that's where these, these sequences come into play and we'll see how this, this comes up here so let's just suppose we have some hypergraph it's got some characteristic polynomial I'm going to call it f of x. Then what we're interested in at the moment is the multiplicity of the eigenvalue zero as a root of this polynomial so there's a couple ways to think of that if you just have some polynomial for example maybe you have like x to the fourth plus x cubed. So after an x cube data there and you're left with x plus one. So one way to think about the multiplicity of the zero root of a given polynomial is just the largest non negative integer. So that when you take your original polynomial f of x, divide out x to whatever this this power is we'll call it M. So that's one way to think about the multiplicity of this zero root so I'm going to let G bar be the reciprocal of G so G is defined to be essentially f of x, what's left of that polynomial after you divide out all of the copies of just x. So the constant term of G is non zero. And but G is, she is still a polynomial so the reciprocal of a polynomial. Here's a here's a formula for it essentially the way you can think about it is take G as a polynomial just rearrange all the coefficients. Flip them just reverse their order so the leading coefficient and the constant term flip the linear term coefficient and the code degree one or the degree minus one coefficient they flip you just do that the whole way through. But more rigorously you can think about just taking the polynomial G, plugging in one over x and then multiplying everything by x to the degree of the original polynomial we have that's the algebraic way to reverse the order of all of the coefficients. But no matter how you'd like to think about it. I'm going to let G bar of x be the reciprocal of this polynomial G. And I'm going to look at this this difference here so because G bar is the reciprocal polynomial of G, then the constant term or G of zero is actually the leading coefficient of G bar so remember that subtraction outside of two logarithms turns into division inside of the log so really what I'm doing here is this expression takes G bar and divides by the leading coefficient. And just then if I if I combine these in that way, I have a monoc polynomial, or the, I guess rather the log of a monoc polynomial but overseas as polynomial factors. And I guess I want to look at the factorization of just the polynomial G over the complex numbers that way I can write G bar as a product of one minus these roots times x, because then again multiplication inside of a log turns into addition. So outside the log so I break that up. Look at the sum and that's that's one way to represent those. The nice thing about reciprocal polynomials is if I have a root of the original polynomial G, as long as that root is non zero, then one over that root is a zero of G bar. And that allows me to factor it in this way, noting that zero is not a root of G, because we took f and divided out all of the, the powers of x that all of the terms had in common so zero is not a root of G so we don't have to worry about anything here. This is one minus some non zero constant times x. And so there's there's no like extra terms or extra information here. We really do have degree of G terms in this, in this sum. In the log I can I can look at the, the Taylor series expansion of log. And you might say well, what about convergence throughout the entire, the entire presentation we're going to look at some some power series concepts. I'm not terribly concerned with convergence. In, I think all of the cases you could take a small enough disk around zero and we get convergence but we're really just concerned about these formal power series. So if we look at the power series expansion of log, we get something like this. So then what we're interested in is notice if we would flip the order of summation here. The coefficient on x to the J for some particular J is then the sum of the degree of G terms, but these these coefficients are all some complex number be I raised to the J power and then we're dividing by J but essentially, aside from this little divisor. All of the coefficients of the powers of X and this Taylor series are the sum of some finite number of J powers and complex numbers. So sometimes we can be given information about a sum that has this kind of form. And what we would like to do is express these coefficients in this form in this, this is a sum of some finite number of J powers, and then we can take all this information and go backwards and try to gain some information about the original characteristic polynomial of the graph because we know, we know the degree of the characteristic polynomial. It's actually, if we have an n vertex hypergraph uniformity K, I believe the degree is n times k minus one to the n minus one. So we know the degree of F. If we could gather some information like this about either G or G bar essentially we're the same kind of thing just the context makes them look a little bit different. But if we have some information like this, and we can find whatever this this positive number needs to be the smallest, the smallest positive integer, so that we can express all of these coefficients as J powers. As the sum of that many J powers, and we can determine the degree of G and G bar. And then, comparing that with the degree of F, I can tell us what this, this non negative integer m needs to be, or maybe we can provide balance or something like that. But that's the game that we're going to play. We're going to find the smallest integer are so that those coefficients times J can be written as a sum of these are different J powers and then that tells us some about the degree of G bar. Or I guess really G itself. Okay, so why is this useful so there are some situations like I was mentioning where we can get information about the polynomial log of G or maybe G itself, and that provides us with some extra information So, we'll go through we'll develop the tools that we that we found that we can use to help us get this information, and we'll apply it towards the end. Okay, so let's take some complex complex sequence. Okay, I'm going to let's see be the sequence where this will be preserved the whole way through I'll keep defining it where we need it. But we'll let's see be the sequence that we're looking at. And what does it mean for it to satisfy an art order linear recruits so I'm sure all the faculty have seen all this before but for the grad students I know from personal experience that it's nice to be able to understand a little bit of a talk. So, I'll go through, kind of slowly, maybe, and go through these definitions and talk about them a little bit. But if I have some sequence, we know what recurrences are. We define sequences using a recursive definition all the time. This is an art order linear recurrence me well think about the Fibonacci sequence okay if I have two consecutive terms of the sequence add them together you get the one after that. So, if I call that fn for my for the Fibonacci numbers fn plus fn plus one is fn plus two. So that would be a second order recurrence, because it takes two of the previous terms to define the next one. So if we think about moving that, all of the terms over to one side, then it's a second order recurrence but we really have three terms over there. So that's what this expression is. We really have r plus one terms here. But because we've said it equal to zero. So what we're saying is the terms with index zero through r minus one are provide us with enough information to define the arth term of the sequence but again moving that all over to one side we get really r plus one terms on that side so all of these are complex coefficients some complex constants. And again our sequences is complex that's why the constants are complex but essentially what we've done here is, we have a non trivial linear combination of r plus one terms of our sequence, I get a zero. So that's what we mean by an art order linear recurrence linear just because all of these coefficients are just constants. That's what we're going to do with with anything else. We can look at the characteristic polynomial associated with a particular sequence. And essentially all we do is replace C to the I, or C sub I with X to the I. So if we just think about this for one second. We're going from from zero to our. So the first term is a zero C zero and then a one C one. Essentially what we're doing is AI is going to be the coefficient of X to the I in this polynomial we call this the characteristic polynomial associated with the particular recurrence that we have. And we call the original recurrence simple. If this characteristic polynomial has distinct complex roots. It's a complex polynomial so we can factor it into completely factor it over C, get all these linear factors determine all of the roots. And if it has all distinct roots, then we call the original recurrence simple. The reason that we're concerned with these simple linear occurrences is because there's a really nice age old result in in this theory of linear recurrence is that if we have a simple linear recurrence like this or find an order linear recurrence where the characteristic polynomial has all distinct roots, then we can express an arbitrary term of our sequence. As the sum of the our roots to these these n plus first powers times some appropriate complex. So, the nice thing again about these simple linear recurrences is we get this form for any term of the sequence. So we define the moment rank of some complex sequence and we denote it in this way. It's the smallest positive integer. So there exists non zero alphas distinct betas that have this form so essentially what I've done is, I've taken the form from the last slide and use that as the definition for moment rank. So it might be possible to write or express the terms of our sequence in this kind of form for different values of R maybe an R one and an R two or three or something like that. But we're going to take the smallest or the minimum possible value for this for the top end of this series or finite song. So some of the game that we're going to play here is try to find some equivalent conditions to a sequence having moment rank R. And then we'll look at an example then of how this helps us in the work or the case that we're considering. So what we already talked about highlights for us that if a particular complex complex sequence has moment rank R. It's equivalent to R being the smallest positive integer so that our sequence satisfies and Arthur or simply recurrence. So, almost from the definition, we see that those two things are equivalent. Definitely moment rank R implies the recurrence business. And it's not too difficult to see that the, the implication goes the other way as well proving that these two things are equivalent so before we continue any farther, it would be really really nice to know that the moment rank of a particular sequences is well defined. When I started working with Josh on this. I know that it was possible for a particular sequence to satisfy multiple linear recurrences simultaneously so this question by itself I, I didn't even consider because I didn't realize that was even a thing like example if I go back to the, the fibbing the fibonacci numbers. We all know the recurrence that they satisfy, but they satisfy infinitely more linear recurrences as well if you look at the characteristic polynomial for a particular recurrence. If you multiply in some other distinct linear term. So take another root for another value that's outside the roots of the characteristic polynomial multiply in the factor that's associated with that distribute everything out you get another polynomial degree is one higher. And if you go back and take that that polynomial and go back to a recurrence, we are original sequence is going to satisfy that as well. So actually you can take one linear recurrence and generate infinitely many other recurrences that the sequence that you have satisfied. That was a foreign concept to me but once I was introduced to that, that makes me realize that this question is actually valid and important. Let's take a sequence suppose that we have some kind of form like this so what I've done is, I've assumed that our sequence values can be described in the moment rank definition they can be described in two different ways. So I'm going to take the sum from equals one to R of the alphas times betas to the m plus one but let's also suppose that there's another positive integer s. So that the original sequence terms can be described as the sum from Jake was one to s of alpha prime beta prime again to the power. Let's take the ordinary generating function of the sequence that we're looking at so we're multiplying in by z to the n and taking the sum from n equals zero to infinity on both sides, and by doing this what we can, we can use all of our complex analysis tools. So now we have all of that at our disposal to kind of give us some information on what's really going on here so anytime we have multiple sums again we're working over formal power series but I think if you take a small enough disk. It all works in in regular power series. Anyway, but we can switch the order of summation, and now the this interior sum is the sum from n equals zero to infinity. You notice we have a z to the n. We have a beta the n plus one. So if we pull the alpha and one copy of beta out of this internal sum, because this this interior some only depends on can pull alpha and one copy of beta out and now we have z times beta to the end power. So that's one over, that's the form x to the n one over one minus x. So we, we look at those we make that that simplification and now we have two finite sums of these rational expressions these fractions. So if you stare at it for 10 seconds or so and think about a little bit of complex analysis. So all of the betas have to be the same because the poles of both of these functions have to be the same set. And then once we we make that conclusion. The alphas are going to be the same thing as well so R and s have to be the same value and the multi set of betas. I guess they're just sets because they're just so the set of betas and the set of beta primes are also the same. So we assume that there were two different representations using the moment rank definition for a particular sequence and derive that well really they are the same thing. So what this tells us is moment rank is is well defined for a particular sequence and it makes sense to continue this endeavor of trying to find conditions or settings where we can find equivalent conditions to the moment rank of the sequence being some number. So, there's a variety of different contexts where we where we turn to and could find equivalent conditions. One of my favorites was this one here so we looked at an algebra setting because we said okay we have these, these recurrences, and they immediately correspond to So it's, it can be interesting to study these polynomials. I already mentioned briefly that if I have some recurrence, I can multiply extra factors in to get recurrent recurrences of other orders. So that that immediately makes us realize that this if we kind of take the set of all of the characteristic polynomials associated to linear recurrences that a sequence satisfies. It's going to be closed under multiplication but also closed under multiplication by just complex polynomials in general. So that kind of leads us towards some kind of algebraic structure. So I'm going to let this the script R sub C be a set for now in the complex is a joint X. So all complex polynomials in one variable. And again, these are these are all of the polynomials characteristic polynomials for all of the linear recurrences that are original sequence C satisfies it turns out that this set isn't ideal. So that's kind of what I was leaning towards. It's closed under multiplication by by any complex polynomial. And it's closed under addition that's that's not too bad to see if I have two linear recurrences, and I add them together. Well, that's going to give me another linear recurrence, because each one by themselves. So in our definition where everything all of the terms are on one side and it's equal to zero, add two things that are both equal to zero you still get a sum of zero. And then that creates a new linear recurrence for us and the polynomials reflect that in the same exact way. So it's really not too bad to see that this set is closed under addition and closed under multiplication by any complex polynomial so we get that this set of these recurrence characteristic polynomials is an ideal and the nice thing to see is that the field so see join X is a PID. So there's some unique, but not unique up to scalar multiplication, but there's some polynomial that generates this ideal. So this was this was a really interesting thing for us. And how do we then use that to adapt it and apply it to the situation that we're trying to look at. So if we have a sequence with moment rank R. Then, essentially what that's going to correspond to is this polynomial F has to have degree R. It has to have distinct roots. And that's essentially what we say here. The other interesting thing about that is, if F has those properties so just if F. Regardless of its degree if it has distinct roots, then this ideal is radical radical just means that if you have some power of a polynomial within your ideal then you have the original polynomial self. If for example P is some polynomial, if P to the 12th power is in this ideal or just a radical ideal general. But if P to the 12th power is in here then what that tells us is P by itself needs to be in there as well. So this was one context where we were able to come up with an equivalent condition. So the place that we turned to were Hankle matrices. So I'm going to develop conditions on some finite Hankle matrices, but then we're also going to look at some infinite Hankle matrices as well. So Hankle matrices has a special have a special form. Essentially what these are going to look at is some finite portion of whatever infinite sequence we're looking at. We use this notation to define the matrices. So m, this first parameter is going to give us the size of our matrix. We're actually going to define it to be of size m plus one by m plus one. So it's a square matrix. So things will work out nicely. But yeah, we're going to, we're going to let n be the parameter, but the size be m plus one by m plus one we'll see here in a second why that makes the most sense. So the second parameter just kind of gives us a shift in the sequence. So we might not always want to look at the initial segment of the sequence we might want to look at some part farther down the line. So we're going to let T denote the first index of the sequence that we're going to consider and then I'll be some finite part of the sequence starting from there. So what do you handle matrices look like the formally the IJ entry is given by that value there, but if we're just looking at an example, maybe I have a sequence, just essentially the identity sequence so CN is the value and, and we're going to define this for non negative integers. And here's what a couple of our ankle matrices look like so our first parameter the size is going to be one bigger than that. So the parameters one so we're looking at a two by two matrix. And we're going to start with the, the first term of the sequence the one with index zero. So, essentially, these ankle matrices are identical on anti diagonals. So typically our main diagonal is given this we go down into the right anti diagonals go the opposite way. But any anti diagonal, the ankle matrices are constant. But essentially what we do is we take the first m plus one or the first parameter plus one terms of wherever we're starting our sequence, write those down in the row and then shift everything over by one to get subsequent rows. So in this way, this is the, the C zero terms C one, but then C one C two so we get zero one one two. So the first parameter is to so we're working with three by three matrices and we're starting with the fifth term in the sequence. So we get this matrix five six seven six seven eight and seven eight nine just take the first row. Start with this term in the sequence. And then once you have the entire row shift everything. So these are what our final ankle matrices look like. What are we going to do with this well if we think about matrices and they're determined we know that the determinant of a matrix is zero. If we go all the way back to our introductory linear algebra class we learned a laundry list of different equivalent conditions to a matrix having determined zero. One of them being that there's some non trivial linear combination of the columns which give us the zero vector with zero. And we're going to use that and exploit that in this particular setting so let's suppose that we have some sequence. Normal definition let's let's say it has my rank are, I'm going to look at the, we know that because it has my rank are we've already said that there's some are the order simple linear occurrence that the sequence satisfies. I'm going to let the vector a or the column a contain all of the coefficients of that recurrence. And we know that this product is going to be zero. Because if we think about a row of our ankle matrix, it's just because the parameter is are, it's going to be r plus one consecutive values of our sequence. And we know just based on this moment rank are that gives us our linear recurrence. So the dot product of that row of the ankle matrix and this vector here, or, yeah, that vector is going to give us zero. What it tells us is that we get some non trivial linear combination of the columns, which gives us zero so the determinant, the determinant of this matrix is zero and some interesting other consequences I won't go through in and prove them, but other things that happen here. The null space, which is the vector space, which contains all vectors that have this property so all null vectors so I know that you're just being when you multiply it by the original matrix that you have you get zero that's a null vector, take the set of all those they're actually a vector space. The null space of the particular matrix the ankle matrix that we're looking at here has dimension one. So this column here which we generated from the coefficients of the linear recurrence satisfied by the sequence we have. That's actually a generator where the span of that vector gives us the null space of this final ankle matrix. And the, we can kind of see then so no matter what t value we choose the null space always has to mention one. So because of that, we already know a vector that's in the null space of all of these matrices so it turns out that the null spaces of all of these matrices no matter what shift that we use in the sequence, all of those null spaces are the same. And if I take some size ankle matrix for anything smaller than this moment rank of the sequence, the determinant is non zero for all of those as well so this is also kind of a consequence of the first two. Because if the determinant was zero for some smaller size or parameter. And we would see the null space would have higher dimension. So the null space having dimension one.  So if I take all three of these properties, put them together and then also specify that. So formally I've said here that a is in the complement of the affine hard discriminative variety. All that means is that if I look at the characteristic polynomial of the recurrence that this that these coefficients define that it has distinct roots. Because we need that to correspond to the, the simple mess of the linear recurrence that were that we're looking at so put all these together. And a defines a polynomial which, which has distinct roots, and that's equivalent to the original sequence having memory bar. Okay, so I also said that we can consider infinite ankle matrices. So let's go ahead and do that. This infinite ankle matrix is constructed in the same way. Just taking the first row that's that's the sequence itself, and then shifting things over as we go down through subsequent rows we just now instead of having some finite matrix. It's an infinite one. So we can also explore some ideas here. So, because our sequence has moment rank are, we know that the terms of the sequence, take this form where the alphas are non zero and the betas are all staked. So, Van der Maan matrices, hopefully most of us have seen those before their first, their first column is all ones, then their their second column is filled with with some value and subsequent columns are higher powers of the values in the second column can take a Van der Maan matrix you can generate it by some particular values I hope most of us have seen those before. But you can generate a Van der Maan matrix from some collection of complex numbers or real numbers in general. But if I take the Van der Maan matrix generated by the betas that I have. I take a diagonal matrix where the diagonal entry is alpha i times beta i. Then when I look at this product here I actually get the sequence terms back. So, the ankle matrix that we have that's populated by the sequence terms is going to be equal to this, we call this a Van der Maan decomposition. And actually we've we've decomposed this matrix this ankle matrix into this product here where D is again a diagonal matrix and V is a Van der Maan matrix generated by the distinct roots of the characteristic polynomial associated with the linear recurrence. So there's we can translate between this setting and the roots of the characteristic polynomial seamlessly. So D is diagonal, all of the diagonal entries are non zero. So it clearly has rank R. The same is true for the Van der Maan matrix. The Van der Maan matrix actually has R rows and infinitely many columns, but it also has rank R because all of the betas are distinct. So both of these matrices have rank R and a little bit a little bit of work is required but we can see, and the ideas are kind of there that the infinite ankle matrix H also has rank R. And this this ankle matrix having rank R and giving a Van der Maan decomposition where the size of V is R by infinity or just having R rows. That's also equivalent to our original sequence. Another place that we can turn to are the the notion of generating functions. So again I have my my sequence to see it satisfies some rth order linear recurrence. I'm going to let fee denote the ordinary generating function of this sequence. What I'm going to do is I'm going to multiply this ordinary generating function by this finite sum here might say well that's kind of odd. Why would you do that. Well, when you actually do the multiplication. If we look at something in the product determine the product of these two are this this polynomial and the generating function, if the degree of the monomial that we're looking at is less than R. Then we only get some of the terms out of this polynomial providing some kind of contribution and that that chunk of this power series is given by this some here this finite some notice that even though we have double sums they're both finite. Whereas if the degree of the monomial that we're looking at is at least are then we're going to get a contribution from all of these terms, and that's going to look something like this. Now it takes a little bit of thought and some index chasing to realize that this is just a restatement of the recurrence that we satisfy things are shifted a little bit just depending on the actual value of J. But this is just the linear recurrence that we satisfy. So this double sum here is really just zero. So we see that this product the product of this polynomial and our ordinary generating function is equal to this polynomial notice that the power of X is is J and J ranges up to R minus one. So this is an R minus one degree polynomial. This is an R degree polynomial. So if we take this polynomial and divide it from both sides we see that the the ordinary generating function of the sequence we have is a rational function and the degree of the denominator is exactly the moment rank of the sequence that I started with. And that's actually so not just that denominator having degree R but it factoring into distinct roots meaning that the generating function itself has our simple poles, that's equivalent to our sequence having moment rank R. So to just summarize some of these equivalent conditions for us. If we have a sequence with moment rank R, that's the same as R being the smallest positive manager. So that C satisfies an R for a simple linear recurrence. We also had that finite ankle matrices condition. And that's summarized here. So we looked at this ordinary generating function of associated with the particular sequence that we started with. We looked at Van der Monde compositions of infinite ankle matrices, we looked at our recurrence ideal, and that the set of all the polynomials associated to the sequences of our sequence actually form an ideal in C adjoin X and that that ideal is radical. And lastly I didn't talk about this at all. But it shouldn't, it might not be too hard to see that our sequence having moment rank R is the same thing as satisfying a complex or atomic measure or B, sorry, the complex are the moments, it's the moment sequence for a complex or atomic measure. So the finite sum of some direct delta functions, where the betas are alphas are just the constants that go out for the right. So the title wasn't something about recurrence ranks the plural we've only talked about memory. The other one that we define we specialized it to this context of unitary rank. So unitary rank was actually the original idea that we were going to employ to apply to these, these hypergraph characteristic polynomials. But in working with it we saw that there was this more general thing happening that's where the moment rank ideas coming from. So unitary rank is really just a specialization of moment rank. Now we take our sequence to start with index one. We require that all of the alphas from the moment rank definition be one. And with that, we no longer require that the betas are distinct. So this in general can be something different. For example, if we have the sequence defined as two times three to the end, just as an example, this is going to have moment rank one, but unitary rank to because of this this extra coefficient here. This breaks into three to the end plus three to the end, which is the form required for unitary rank. But for a moment rank, we can just let that alpha be to the only beta is going to be three. So we see that this can be different depending on whatever sequence. We're looking at so a lot of the statements that were equivalent for moment rank, we can translate them in some careful way, and we get an equivalent statement for unitary rank but there's one new one. So again, let's take our sequence again we're going to start with index one, and we'll see why actually right here, we're going to consider this, this object here. So, e to this integral where fee is the ordinary generating function of particular sequence that we're looking at. So, because the sequence starts with index one, it's constant term is zero. So we can look at this use power rule to integrate everything and we get something that looks like this. Okay, now notice right away, this kind of looks like log. Right, so the, the, the exponential in the log are going to seemingly cancel. So when we exponentiate this, this is what happens because our sequence we suppose that it has unitary rank are we can replace our sequence terms with this finite sum of these are and powers. So because the sequence now starts with, with index one, we've shifted the power of betas down to match the index of the sequence so we get that there are sequence values are the same as the sum for Michael's one to R of these betas to the end power. And like any good proof with two sums in it we're going to flip the order of summation. And by doing this now we've turned things into logs. So a finite sum in the exponent of the exponential turns into a product of those exponentials. Now this is log. This is log of one minus beta times x, either the log well we know what happens there. We got a product of some linear factors. So this is a polynomial of degree R. And those are actually equivalent as well. So if our sequence has unitary rank are that's equivalent to this object here. Being a polynomial degree R with some non zero moves. So in the last couple minutes that I have I would like to go all the way back to the beginning, and help us go through a very almost trivial example, but to illustrate the concepts and the ideas of how we can use these ideas to help us with graph and hyper graph characteristic So, we're just going to, I'm just going to consider a graph we're going to look at just the complete graph on inverteces. But the idea is then we can try to to generalize to the hyper graph case because I mentioned that one technique is to take graph results and generalize them to hypergraphs, but there was no graph result for this so this is an attempt. Maybe to look at a graph result and then help us extend it to a hyper graph result. So if I have some graph with adjacency matrix a, I have all of this linear algebra that we can apply to that adjacency matrix. So one of them being just the just the rank only theorem so the rank plus not only the graph is equal to its order, assuming that we have a square adjacency matrix. So, the multiplicity of zero as an eigenvalue for our JC adjacency matrix is the same thing as the nullity. So that's the order or the size of a minus its rank. The size of a is very easy to obtain just looking at the graph. That's the number of vertices that we have again in the graph setting. But what about the rank is it is it possible to look at, or, or take properties of the graph and determine the rank of this adjacency matrix. The rank is actually the same thing as the number of non zero. So you can take all of the eigenvalues, you can look at the non zero ones, the zero ones zeros correspond to the nullity non zero corresponds to the rank. And the other interesting thing is that if I take the trace of some power of the adjacency matrix trace again is just the sum of the diagonal entries. That is actually the sum of the jth powers of all of the non zero eigenvalues. And if you think about the graph context. The jth power of the adjacency matrix, the ij entry gives us the number of distinct closed walks from vertex. So again, we're looking at the trace. So this is the diagonal entries. So this gives us the number of closed walks, meaning walks where the starting and ending vertex of the same thing. But this gives us the number of closed walks of length j particular graph. So we want to try to take this, this sequence see if we can find a form for the sequence, given by the trace of a to the j or these, the number of closed walks of particular lengths, and see if we can apply some of our conditions to it to determine either moment rank or unitary rank for it so in the last couple minutes. Let's look at just the complete graph. We know that some recurrence, some finite recurrence of some finite order, less than the number of vertices that we have were less than or equal to I should say we can use that recurrence to define this sequence so just looking at the first couple values. As with any graph on inverteces the number of closed walks of length zeros is the number of vertices so that's n, but the closed walks of length one are going to there aren't any, because if you traverse an edge. You're going to be different than where you started. There's this many choices for a to, but let's try to determine a recurrence from here so if I have some closed walk of, of order and then there's two different, there's two cases. Either the last vertex and two from the end so WN minus two either the same vertex, where they're different. They're the same vertex. Then I have a closed walk of length and minus two, and essentially I can traverse one edge forward and backwards. However, I'd like and that gets me all of the. The closed walks of that form of length and so because there are the complete graph is regular of degree and minus one. I have this contribution to this recurrence that I'm trying to build up. Now if that wouldn't be the case so if WN and WN minus two aren't the same value, then essentially what I'm going to do is I'm going to take a walk of my then minus one. I'm going to delete the last edge, and then use two other edges to get to my destination. And there are n minus two ways to do this. So this contributes an extra, some end of n minus two times a n minus one or the number of closed walks of length and minus one my graph, putting these two together I've covered every case. So that gives us a linear recurrence for the number of closed walks in the complete graph on it. You can look at the characteristic polynomial associated to this recurrence. We removed these two terms over to the other side, set it equal to zero, replace the an with x squared and minus one with x and then minus two with just x to the zero. I factors in this way and so we see that the two different roots are n minus one and negative one. And based on the initial conditions that we have. It's not too challenging to see that the coefficient on n minus one has to be one and the coefficient on negative one has to be n minus one. So what this tells us is the moment rank of the trace of a to the J sequences to but the unitary rank is actually n, because we have to break up these this n minus one coefficient into n minus one ones. So we get n different terms for the unitary right definition. So what that tells us is the multiplicity of zero as an eigenvalue zero. Just because this unitary rank is actually the same as the rank of the adjacency matrix. So the sizes and now we know the rank of the adjacency matrix is in. So what this tells us that the, the nullity or the multiplicity of zero as an eigenvalue zero. So again this is nothing. This is nothing new or revolutionary for this particular example but the idea is that we can do something like this for other kinds of graphs. And then hopefully at some point, extended type of graphs. So that's all I have. Thank you very much for sticking around and being attentive what questions do you all have. Thanks Grant, we could all thank our speaker and then we'll go ahead and open it up for some questions. Thank you so much speaker. Grant, what is the largest. Have a graph you can compute. What are the science. Well, I, I don't, I don't have a good answer to that. I, well, the, the, so by sizing the number of edges Lincoln or. Either way. So just the trivially the K unit, the, well, the K uniform hyper edge. We know that one so as the uniformity grows so does the number of vertices so you can take that one infinitely that's kind of trivial probably not a very satisfactory or satisfied answer. But it depends on the structure of the edges. For example, the Hummingbird graph that Greg and Josh had worked on. I think that was like 13 vertices if I go way back here. But that was one of those linear trees which behave really nicely. But I think it was 13 vertices and so there's 246, 810 13 vertices and six edges. And that one's known. But the phantom plane is seven vertices and one, two, three, four, five, and then two more so seven edges, but that one we don't know. So it kind of depends on the structure of the graph itself. If not, thanks again, Grant. Thanks everybody for making it out and have a good weekend everybody. Bye bye.


Let's get together and sleep it all.  So 2024, I cannot believe that we are here. I am really excited. I am running late as always. New year, same me, but we are going to get started on our day. I woke up super early this morning. Like at 6.30, I washed my hair, I styled my hair, I did these little twists just because I have yoga tonight and yoga sculpt tomorrow and I just didn't want to straighten my hair. Like I thought about it and I was like I'm gonna straighten my hair just to sweat it out tonight and tomorrow. Plus, my hair is getting long and wearing these protective styles just helps your hair stay long. So I feel really good about that. I am heading up to the Upper West side because I have a little team lunch with my diversity equity inclusion team at Columbia University. I'm really excited because we haven't seen each other in person. Probably, well, definitely not since last year, but even last semester, I think we only all met up once. After that, we're going to go to Strand because I have a bunch of books that I want to donate. If you follow me on Instagram, you'll see I had like two really huge piles of books that I was like, okay, like these have to go. I just don't have the space for them and I'm trying to clean out and decolor my apartment for 2024. I got this really cute sweater yesterday. I went shopping with my friend, Tumaya. She's like TikTok famous. You guys might know her. Oh my god, I feel like I always talk about Tumaya on the vlog and I feel like she's very, very late on the vlog as of late. But anyway, the title of this vlog is doing nothing and gaining everything in 2024. And we're going to get into that a little bit later. But right now, like I said, I'm running late. We have to go to the Upper West Side. And if you're new here, my name is Nika. I am a 29 year old graduate student at Columbia University Teachers College and the clinical psychology program. Make sure you subscribe and I will check in with you guys later. But let's get this party moving because we have a lot to do today. Okay, Hume loves my little bun bun. I just finished our little lunch brunch with my team at Columbia. I'm here, but I'm going to head back downtown because I don't feel like hanging out on campus. I also did not bring my laptop, so I wouldn't have anything to do anyways. But I think I need to start reading newspapers. Like it was kind of crazy the conversations we're having. And like, obviously all the people that I work with are much older because they're like, senior and seasoned professors. But actually, um, first of all, I didn't even know Columbia was like in the news. And that like the president had been subpoenaed or summoned by Congress and just like didn't go. Is that not like insane? So there's this thing called the Columbia like some spectator, which essentially I guess is just gossip girl for Columbia. So I'm going to start reading that scene and just see what's going on. Because I was very shocked. I love my DEI AA team because I feel like it's honestly, I just get so much news and insight about like what it truly means to be a professor at Columbia and like what it truly means to do diversity, equity and inclusion, and like large institutions like this. So I'm talking too much, but let's go downtown to Strand and donate some books. I am so tired. We did so much, you know, we had our little team meeting or team lunch and we made, we sold all of our to Strand. It was like 40 books. That within itself was a workout then because they offered me $16 or like $24 store credit. And I just took the 24 store credit because I was like $16 is just offensive. So I'm really, really happy. If you guys follow me on Instagram, you may have seen, I like offered my friends the forced choice of the books and two of my friends reached out to me. So I gave them like probably 10 books each. So that just makes me feel really good because Strand did not offer me like anything for all the books that I donated to them. I was going to make a little salad, this kale like ultimate winter salad that I absolutely love, but I opened up my bag of kale and all of it was bolted. It was absolutely disgusting, but luckily I actually still have, I'll go ahead and make those instead. I'm also making a little cup of tea because I have a workout class at seven. So I'm trying to eat something like flight and healthy. So I was thinking like a little salad, but yeah, I need to kind of wake up cause I'm really tired. I really want to go to sleep and we made the little shoe rack. So like we did a lot of stuff today and I'm really happy because like I want to put a larger piece of furniture there, like a proper piece of furniture shoe rack there. But before I did that, I was like, let me see how I feel just like having something there in general. Cause obviously before that it was empty and it does feel weird. Like I walked in from like throwing something out and I was like, oh, like what is this? It's kind of big. And plus because I bought that on Amazon and I can see the dimensions of it and Amazon now it also makes it easier to kind of picture what dimensions like a shoe cabinet that goes there would need to be because I don't think I would want anything bigger than that. Okay. So I'm a bit exhausted. It's been a while since I actually vlogged a little bit rusty, but let's get this tea going and get the salad going. It's also an absolute mess. Love you. So I moved everything around again. I feel like I post everything on Instagram because I haven't been posting on YouTube lately. So this is how dark my room is at night. Luckily on camera it doesn't look that dark, but in person like this portion of my room is like absolutely like pitch black. And the only light I have is this lamp right here. So I do want to get some light fixture of some sort somewhere on this wall. And I moved my desk over here and I think I want to get a bigger desk here as well and get rid of this laundry basket and somehow fit it in one of my closets. So I have a lot of little home projects to do. Oh my god. I'm supposed to be making a salad. I literally get so distracted. All right. Let me have a cup of tea and make a salad and we can catch up a little bit later. Brussels sprouts actually came as like a little salad kit, but obviously I'm having more than just a Brussels sprout salad. So I'm actually going to use the dressing that came with the kit, the cheese and then the nuts is like a little natural crouton. I'm currently warming up some like impossible nuggets to add on for a little bit of extra protein, but I'm really excited about this salad. It has the Brussels sprouts, the orange things really look like sweet potatoes, but they're actually was that called butternut squash. And then you saw I added the bell peppers and cucumbers. And I think it's just gonna be nice healthy early dinner. So I feel properly fulfilled and fueled fueled for my yoga class tonight at seven. So just waiting for the chicken to get done. And while I'm doing that, I'm actually going to pack my yoga bag. I typically go to a Y seven and I was using gym pass before. So like a yoga mat, yoga towel was included. So I only had to bring my water bottle, but I just started with class pass and I'm not sure if a mat and a towel are included. And I'd rather be safe than sorry because I'm trying to be really tight my budget. I have not been good. I went shopping to my yesterday and I got all this really cute stuff. Even though I didn't need to go shopping, like I was not going to go shopping. I was just assisting her and I ended up buying a lot of stuff for my trip coming up in February. And yeah, so that happened. And then also today I actually invested in my YouTube channel. So you guys have probably noticed new branding and stuff. And I just paid for that today. And that was not cheap. I'll tell you that. But I really love it. The person that helped me out with everything. Her name is Tape Haley and I'll leave all of her information in the comment section below in the description box below. And the most chaotic turn of events, I just feel like it's six o'clock now. And if I eat, I'm just gonna be too full. And I'm also not like starving. Like I'm not super hungry. Like I feel like the chicken's done now. And the salad is basically made. But I think I'm just gonna put it in the fridge. And then that way when I come back from class, I can shower really fast and eat. And like my dinner's done. So not too bad. I also realize you know how people are like, what's your favorite color? Like I'm always like, oh, I love like all the colors. I love red. I love blue. I love green. I love purple, bro. I'm literally wearing a purple sweater. My yoga mat is purple. It's also purple. And it's like, I don't do these things like intentionally.   Like it's a surprise to me. So I think I need to stop telling people that my favorite color is like the rainbow because clearly it's purple. I have you and we have like a little hour or so before we have to leave. We'll night in an hour. We have like 20 minutes before we need to leave for yoga. I do want to talk to you guys a little bit just about like my plans for this year. So the title of this video like doing nothing and getting everything is kind of my vibes right now. I'm at a point in my life and in my journey. If you guys watch my kind of my PhD journal, maybe that's why I should actually call the series. I was really stressed last semester applying to all my PhD programs and I did it. And now it's done. And honestly, all I can do at this point is just wait and see what happens. Hi, cutie. Hello. Yeah, so all I can really do is just wait and see what happens. And in the meantime, worst case scenario, if I don't get in anywhere, I want to take some time to really strategically pivot into user experience research so I can still conduct research and work in tech companies and get paid more than a typical research job, as in one that's kind of in my field that I want to specialize in. I think psychology, like clinical psychology in general, is notorious for being very underpaying essentially until you have your license or your doctorate or like whatever your end goal is. So I just, I live in New York City. My rent is expensive. Like I cannot afford to work a $25 an hour research job because literally at $25 an hour, even if it was $30 an hour full time, it just wouldn't be enough to cover my lifestyle. Before I went to Columbia, I was making $100,000 a year and I can't leave grad school and make less than that. Like I absolutely refuse. So I'm not taking any classes this semester. That's my huge like do nothing. I'm really taking my last semester of graduate school just to transition back into adulthood because these last two years, I have just been chilling. Like I started my YouTube channel and I'm really proud about how much it's grown and how I've gotten so much better at video editing. And I really want to focus on that and see where it goes. I want to start my podcast.  But yeah, so 2024 you're actually not going to see a lot of Day in the Life of Columbia student vlogs because at most I'll be going up to Columbia once a week on Tuesday evenings, maybe twice a week because I do kind of want to audit a coaching class that takes place on the weekends. But other than that, I am not taking any classes. I'm fully focusing on relaxing, manifesting, attracting and hopefully pivoting into user experience and user research. So this is going to be a very interesting year. I'm so nervous. And I got dinner with my friend last night who graduated last spring and she was telling me like in the grand scheme of things, it's going to work out.  And this last semester is just going to be a blip in the grand scheme of life. And it's so true. But when you're in it, like we're in it right now, I'm just so filled with anxiety and fear. And it's just like doing nothing and attracting everything. That is just my theme of 2024 because I've planted all the seeds. And when you plant the seeds, you water the seeds, the sun is on the seeds, you have to just sit back and let the seeds grow because if you try to keep watering them, trying to keep flowering them, keep like hoeing them or raking them, the seeds mess up. Like, you know, you got to let the seeds just kind of do what they do. And so I feel like that's where I'm at right now in life. Like I've planted all the seeds, I've done all the watering, all the sunlight, all the nurturing and I just have to sit back and watch the seeds grow and just see where I end up. So this is kind of a huge transition, especially a weird year because I also turned 30 in November at the end of the year. So I'm really excited to see where life goes. But I am also really nervous at this time because I could either get into a PhD program and start really my journey to becoming a doctor, a PhD, like a clinical psychologist. It's so crazy to say. Or I could be going back to corporate America.  There's so many possibilities. So I don't even know which one I want to come through the most. I just know that I feel very anxious right now. And I just know that this is the perfect opportunity to practice surrendering to the universe. So with all that being said, wow, I've really been preaching at you guys, I am going to drink my tea, which I think I left over there and head to yoga. Anybody else's face? Just give me eyes too in particular. I think it's like the heat heating system and stuff. Anyways, this Javi coffee and it's actually really, really good. I've been using it instead of Starbucks because I am now boycotting Starbucks. One of you guys left a comment on one of my videos and I had no idea that Starbucks was on like the support Israel list. Like you should ban if you're protesting Ghana. So now I'm protesting Starbucks. I'm kind of annoyed because I still have like $5 on my Starbucks card, but maybe it's like not enough to really get a coffee. But I guess maybe I can get a little snack with it. But I'm also trying not to spend money, but the money's already spent. I don't know. We'll see. But anyway, this Javi coffee is a really good replacement for Starbucks. It tastes delicious and that little container that you guys saw contains like a million cups of coffee. I'll probably look it up and give you the exact number on the screen. Anyway, today is Friday and a lot of people get like the Sunday scaries, but I get the Thursday scaries because I really don't be doing anything. And I feel like on Friday I'm like always scrambling to be like, okay, let me at least be like a little productive in the week. Reflect, like set my motion, do like make a to-do list. Do the to-do list and then we'll get our day started. We do have a workout later today around two o'clock. And then my friend asked me to sub her 5 p.m. yoga class, but like 5 p.m. on a Friday. I don't know if I really want to do that. I also realized I have my like little music playing in the background. I hope that's not terrible for you guys. All right. I also have a thing of laundry going and that just ended. So I'm going to go pick up my laundry, hang it up to dry, and then do a little bit of work. And yeah, I am really excited for this year on YouTube honestly and like when I say doing nothing, I also mean like not doing stuff for other people. Like I've spent a lot of my life working diligently to make other people's dreams come true, to work in other professors labs, just really using all of my skills, energy, and effort to help elevate either a corporation or an academic institution. And so I really want to dedicate a lot more of my energy and to doing stuff for me. And that means really showing up for you guys on YouTube. I want to start like a meditation challenge on my TikTok and just like posting more mindfully on Instagram. If you guys follow me on Instagram, I post a lot of like what I'm eating on my stories and like random things on my feed, but I want to get back into consistently posting on my feed. And if you're a really OG subscriber, then you know I used to do a lot of budgeting and financial content, but when I started graduate school, I just felt so insecure about my finances because I went from making $100,000 to making like negative $80,000 because I was taking out a lot of student loans. So this is my year to transition back into making money, hopefully. And I wanted to just get a better handle on my finances and maybe we'll have some like mindful money content soon. Anyway, once again, I am rambling. I'm going to go get my laundry and then do some work and I'll check them with you guys in a bit. Transition. My camera has to Jenny. She literally how pretty these are. My kale was like wilted and I used all the greens in that salad. So I got another one of those. The camera's going to die again. So I might have to say good. It's really sad because I actually was charging that battery for the entire time I was at YogaSculpt and Trader Joe's. So probably like two, three hours and I put in my camera. It was only at 6%, which is weird. I probably just need to get a new like camera charging station, which is annoying. But anyway, I got this cold press organic green juice from Trader Joe's. I'm really interested to see this because I just felt like I should be having more veggies and hopefully this can help me get there. Also, it's better to eat whole veggies and like juice veggies, but this is better than nothing. So not going to get ourselves up about it. And then last thing I got was just some hash browns because some of my girlfriends are coming over tomorrow and we're going to have like a little brunch at home, which is really sweet because my friend was like, do you want to hang out? And I was like, oh, I just paid my graphic designer for my YouTube channel and like, I'm tapped out. Let's do something affordable. And she was like, let's just do brunch at your place. So that's why I got the eggs and the hash browns just so we have something. I feel like those are breakfast staples and I literally have one left egg left in my refrigerator. So that's that. And then I am currently making a little bubble bath. I'm going to take a bath. And then after my bath, probably edit this YouTube video, see where we're at because honestly, if I have to do this all day tomorrow, I don't know if we're going to film. So I'm going to let my camera battery charge, edit these videos, take a bath, and then I'll probably check in with you guys tomorrow morning. Guys, I was literally freaking out for like no reason. My charger is fine. Everything that's fine. We have a new battery now. I'm taking out these like little ear beads. I got like these little ear bead acupuncture things. They're real cool. I highly recommend if you are able to, oh my God, I literally have like craters in my ear. Okay, that is kind of crazy. Honestly, this is why I really just need to take a bath. You know, like I was tripping. The bath calmed me down and now we feel good. I'm going to do this aloe mask by Innisfree. I haven't done this one before, but I've done a lot of their masks and I really like them. It's Friday and I really have no plans. As you can see, I've already put my pajamas on. So we're just gonna have a nice like little self-care evening because why not, you know? I think this is what being in 29 is like. Like if I was 22, I'd be like hitting people up, like trying to find something to do tonight, but now I'm 29 and I'm like, ooh, thank God, I don't have to leave my apartment. Love that for me. Oh my God, this feels so good. Again, it's just to decrease my caffeine intake. We had such a productive morning today, like we journal, productivity train going. So I didn't have time to do my financial planner or my content planning for February last night. So I want to take some time to do that this morning. Then my friends are actually at SoulCycle right now. I was invited, but I wanted my hair to be cute for you guys. So I didn't want to go. But yeah, after SoulCycle, they should be coming over here. We're going to do a little brunch. I feel bad because I accidentally double booked myself. Don't you guys hate when that happens? So I need to send a fine text and ask if we can like reschedule our plans. So I need to do that as well. And the weather has taken a turn. Like it was very rainy and gloomy and cloudy this morning. But now the sun is out, the clouds have dispersed and it looks like a really good day. So hopefully we can go on a walk or something later. I'm trying to be a little bit financially conscious because I've spent a lot of money so far and I need to pay some bills. So yeah, let's get this day started. Oh, also subscribe if you haven't already. Guys, because I'm starting to get hungry, but they said they're on their way. I'm quite sure knowing Tina, she's probably like grocery shopping for soft breast to cook and eat. But like at this point, eggs and toast will be fine for me because I'm starving. I'm going to try this green juice from Trader Joe's. I've never tried it before. I usually get like their better one, but I don't like that one that much because it's super grainy. So this one looked a little bit more fresh and definitely not grainy. Let's do a little taste test. Vegetable green juice. So the other juice, it is more grainy, but it definitely tastes like sugar. So I actually kind of like this one. I think I want to be a little bit more healthier because I'm trying to slim down honestly for my trip coming up soon. This might be my new green juice. Super healthy. Love this for me. All right. In the meantime, I might make myself a little something to eat. Actually, I'm just, I'm going to have a banana. And if I'm still hungry after the banana slash, if they're still not here after I get hungry again, then I'm just going to make my little brunch and help them. So I will check in with you guys once Tina and Ellie and I get here. Culture. The girls are going to bed you scramble. We got fruit. We have bobca.


Welcome to a screencast on a new concept, our last concept we're going to see in our study of set theory on a thing called a Cartesian product. Now a Cartesian product sounds fancy, but you're actually very familiar with this idea. Just think back to where you first learned about R2, the X, Y plane here. R2, we call it R2 because it's really two copies of the real numbers and sometimes you'll see it as R cross R. You've got one real number line going left to right and one real number line going up and down. And if we put those two real number lines together, we can create a grid and use that Cartesian plane as it's called because it was first conceptualized by Rene Descartes back in the day. That Cartesian plane can be used to plot ordered pairs of numbers. Okay, so every time you plot a point here, let's assume for example that every grid mark here is one unit. If I put a point right there on the grid, then that's at 1, 2, 3, 4 units on this real number line and then 2 units up on that real number line. So I have 4 out this way, 2 up this way and we call that point 4, 2. So points that are in this plane or in this set here are ordered pairs of points and the first number is a real number and so is the second one. And they do matter. If I looked at the ordered pair 2, 4 instead and that would be a completely different point in the plane. I would go to 2 here and then 1, 2, 3, 4 up here. So that's what 2, 4 is. So there are ordered pairs and each coordinate, we had a first coordinate and a second coordinate was a real number and that's why we called it R cross R. So if you wanted to write this in set notation, the Cartesian plane, the XY plane here, is the set of all such ordered pairs where the first coordinate is a real number and the second coordinate is a real number. So this idea of the Cartesian plane here is a way of grouping 2 numbers together in order. And if you think about it, grouping 2 objects in order comes up a lot in life. I think in thinking about teaching, if you're teaching you might have a class here for example that's got say maybe a lot of students but here are 4 students and here are their grades on the last test. Now if you're entering these grades into a database, say like a blackboard type system, then you're going to want to put these things in order.  So really you can think of the data here in our little database as an ordered pair. I could represent each of these data points here as an ordered pair. Let's say Alice, comma 90. And so the first coordinate of this ordered pair is a name of the student in the class and the second coordinate here of this ordered pair is a number between 0 and 100 that would represent that student's grade. Bob would be represented as the ordered pair Bob comma 92. In fact databases really work this way. They pair things off or actually there could be several different data fields, not an ordered pair but an ordered triple or quadruple where you have a bunch of different data collected together in order. So Clarence is ordered pair, Clarence comma 73 and so on. So what we're doing here is we're extending the idea of the Cartesian plane, the xy plane, to include things that aren't necessarily numbers. This is right, Dora's ordered pair out here, 88. So you could go through the entire class roster, maybe there are 20 some odd students beyond just these four and you know every student and their test grade could be represented as an ordered pair say s comma g where s belongs to this set, the set of all students in the class and g belongs to this set right here, the set 0 through 100 that represents a grade. So this notion of ordered pairs actually is pretty applicable to things beyond just numbers. We can pair off text with numbers or any kind of object with any other kind of object. And that gives us to the idea of the Cartesian product. So let's let a and b be any two sets whatsoever. Maybe there are sets of numbers, maybe one of them is a set of names and the other is a set of numbers, maybe they're both names, maybe they're one is a set of happy faces and frowny faces and the other is a set of states in the United States. We don't care just any two sets whatsoever. Then we're going to define the Cartesian product and we're going to read this thing here as a cross b. The Cartesian product is the set of all ordered pairs whose first coordinate belongs to a and whose second coordinate belongs to b. That is if we're going to write this out as a set, a cross b, the Cartesian product is the set of all ordered pairs where x belongs to a and that should say y, y belongs to b. And so this is just a generalization of what you already have learned about the xy plane and it also comes in very handy to describe things like the student grade database that we just constructed here, the Cartesian product, set of all ordered pairs with the first coordinate and the first set and the second coordinate and the second set. So let's look at an example here and I'm thinking about kids. I have three kids of my own and they often fight over things at home. So let's suppose I have three kids Jack, Bill, and Annie and put those in the set k. And let's let t be the set of these three toys, a ball, a car, and some blocks. Now I would be wondering if I were taking care of these kids, what are all the possible ways that I could pair off a kid with a toy? Well that would, the set of all possible pairings there would be the Cartesian product k cross t. This is going to be the set of all ordered pairs, let's say little k, little t, such that k belongs to the set of kids and t belongs to the set of toys. So what is that Cartesian product? What does it consist of? Well let's start listing out the ordered pairs. This is going to be a large set consisting of ordered pairs. Well one ordered pair would be Jack and ball. Any ordered pair where that has two coordinates with the first coordinate being a kid's name from the set k, and the second coordinate being a toy from the set t is an element of k cross t. So there are probably, you can think ahead there are going to be nine of these things. I could pair Jack off of the ball, I could give Jack the car, I could give Jack the blocks. I'm running out of room here so let me just tuck that in there. I could also, I have a pairing where Bill has the ball, Bill has the car, or Bill has the blocks. And then finally all the pairings that involve Annie, and that would just be Annie could have the ball, Annie could have the car, or Annie could have the blocks. And that set would close off this last ordered pair and then put a giant set brace up here to pair it off with this set brace that we started with. That set of nine things is the Cartesian product. I could even plot them if I wanted to, I guess, kind of like an R2. We don't often do that. We just sort of think of these set of pairs. And those pairings mean something, okay? The pairings are important. They do come in order. I need to have a kid's name first and a toy second. And the kid's name has to actually come from k. The toy's name has to come from t. The set of all such pairs would be my Cartesian product. So let's do a couple of concept checks here to see how well we're understanding these basic ideas. So take a look at the ordered pair that consists of two numbers, the square root of two comma the number one. What set or sets could that ordered pair be considered an element of? Here are five choices and I want you to pause the video and select all that apply. Well, let's mark off some things that are definitely incorrect to begin with. A can't be right. A is just the set of real numbers. And an ordered pair is not a real number. It's a couple of real numbers. It's a pair of real numbers. So the only thing that's in R are individual real numbers like radical two by itself would be an element of R. But not an ordered pair with radical two in it. I need to be in a Cartesian product somewhere. Now B could work because two one, radical two one is an ordered pair and the first coordinate is a real number. So it belongs to this and one is a real number and it belongs to this. So that would certainly work. R cross Z also works because it's an ordered pair. This is an ordered pair where the first coordinate belongs to the real numbers and the second coordinate belongs to the integers. So that would work as well. D does not work and why is that? D does not work because the first coordinate here is not an element of the rational numbers. As we've proven and seen before, square root of two is an irrational number. So although one does belong to the integers, square root of two does not belong to Q and so this ordered pair does not belong to Q cross Z. Here are on the other hand some things that do belong to Q cross Z. Anything that's an ordered pair where I have a rational number in the first position and an integer in the second position. So something like 3 halves comma 17 would be an element of Q cross Z. But radical two comma one is not. And finally this last set here doesn't work either. Even though this belongs to R cross C, it does not belong to Z cross R because the order matters. This is a square root of two comma one. Square root of two does not belong to the integers. And so this ordered pair does not belong to that Cartesian product. So the only correct answers here are B and C. Now speaking of this last issue here that we just discussed, here's another concept check. What's the relationship between A cross B and B cross A? Similar Cartesian product but I'm reversing the order here. Is A cross B a subset of B cross A? Is it equal to B cross A? Or is there just really no relationship between that set and B cross A? Well the answer here is essentially C. There's not really any relationship in terms of subsets or equality between A cross B and B cross A. And I think we saw that in the last concept check. We have this ordered pair of square root of two comma one that was an element of R cross the integers. Okay? Because two is an element of R and one is an element of the integers. But if I keep the same ordered pair, that is not an element of Z cross R. Because or square root of two is not an element of the integers. And so that fails. So certainly there's no subset in relationship and therefore there's no equality relationship. Now there is a sort of symmetry. There is a sort of relationship between those two sets that we're going to explore a little further in chapter six when we talk about functions. They are very similar to each other. The only difference is that the order is reversed. For example, the ordered pair of one comma radical two would belong to the set Z cross R. So maybe there's some sort of a operation or a transformation I can do on an element of one Cartesian product that gets me into the Cartesian product I have by reversing the order. But for now there's no relationship that's really evident there in terms of subsets of equality. So that's the idea of the Cartesian product is generalizing the notion of R2 in your old fashioned algebra XY plane. And it's very useful and very handy. And in the next video we're going to look at how do we prove things about Cartesian products. So thanks for watching.