Kevin Knudson: Welcome to My Favorite Theorem, the math podcast with no quiz at the end. I'm Kevin Knutson, professor of mathematics at the University of Florida. And today I am flying solo while I am at Texas Christian University in Fort Worth, where I'm serving as the Green Honors Chair for the week. And I've been given some talks and meeting the fine folks here at TCU. And today, I have the pleasure to talk with some of their students. And they're going to tell us about their favorite theorems and what they pair well with. And we're just going to jump right in. So my first guest is Aaryan. Can you introduce yourself?
Aaryan Dehade: My name is Aaryan. I'm a sophomore computer science major at TCU, and I'm from India. And I've chosen to go with the fundamental theorem of calculus.
KK: Okay. The fundamental calculus.
AD: Yeah.
KK: Okay, so now there are two parts of the fundamental theorem of calculus. Do you have a favorite part?
AD: So that's what I like about it. Honestly, I can't choose a favorite part because one of the parts is very important, and the other is interesting.
KK: Yes.
AD: So the first part, basically, it tells us the relationship between the integral and derivative. The second part tells us — like, basically, you can use that second part supply and to get solutions for questions in calculus. Does that makes sense?
KK: Sure.
AD: So what I like about this theorem is that it's not like other theorems where it has two parts. So it's kind of interesting how everything in calculus is based off of these two theorems. And if you didn't know what the relationship between the derivative and the integral was, probably you wouldn't be able to do anything with mathematics with it.
KK: Sure, maybe. Yeah. So you said one part was useful, and one part was interesting. Which part do you think is interesting?
AD: I feel like the first part is interesting, because it's almost intuitive. Like, you know that should happen, like the relationship between integrals and derivatives should be, like, one is an inverse of the other. And that is intuitive. So it's almost given. And it's interesting that we have to say that.
KK: Really, you think that’s intuitive? I mean, I’m not — I don’t know, when I first learned the fundamental theorem, I thought it was kind of shocking that somehow this this thing, this integral, which is sort of defined as in terms of these Riemann sums, somehow that went, you know, if you let x be the upper limit there, and you differentiate that function, you get your original function back. That’s intuitive? That’s amazing.
AD: Yeah, basically what it is, it's just an area of a rectangle. So I just thought of it as just decreasing the width of the rectangles, and then you get smaller and smaller rectangles, so you get the area. And then if you take the function at that point…that’s what I think.
KK: You’re cleverer than I am. I was just kind of dim, I guess, and I didn't think it was so intuitive. I mean, I saw the proof and believed it. But then, yeah, then the second part is how you actually evaluate integrals.
AD: Yeah, so that's what you use to calculate the area between two points.
KK: Yeah. Although I guess the problem is, right. So the theorem says that, you know, if you want to find the, the integral, the value of this definite integral, all you have to do — and our listeners can't see me doing the scare quotes — “all” you have to do is find an antiderivative of the function. Right?
AD: Right.
KK: Yeah. And then you spend all of Calc II learning how to find antiderivatives.
AD: Yeah.
KK: And even then, if I hand you an arbitrary function, you can't even do it, right? Like that's the sort of disappointing part of that theorem, is that most functions, you can't find a closed form antiderivative for. And so what do you do? But you’re a computer science major. You know what you do, right? You do it numerically, right?
AD: Yeah.
KK: Okay. Very cool. So you've known this theorem for quite some time, I guess.
AD: Yeah, I've done it since high school.
KK: So you love the theorem.
AD: I really, yeah, I do. Because when I was in high school, I used to sit at my dining table and study because I wanted to have some snacks at the same time.
KK: Sure. As we all do.
AD: I would just spend hours just doing sums on integrals, or basically just integrals. That was difficult at that point.
KK: Sure.
AD: And yeah, it was interesting, because I got used to that at some point. And then it got easier. And I just started liking the satisfaction of being able to do this. That was fun.
KK: Cool. All right. So on this podcast, we also like to ask our guests to pair their theorem or something. So what pairs well with the fundamental theorem?
AD: So as I said, I used to sit at the dining table and have snacks. And there's this really, really popular biscuit in India called Parle-G. And I used to have that with tea while doing my sums. So that was the highlight of it, that's why I used to look forward to studying, just for those biscuits.
KK: Okay, so I assume there's an Indian market in town somewhere, right? Can you get these?
AD: Yeah, I do have them in my dorm right now. Yeah. I have them every day.
KK: All right. So what are these called again?
AD: Parle-G.
KK: Parle-G. Okay. So I'll have to go to the Indian market when I get back home and see if I can find these because I am always on the lookout for a good new biscuit.
AD: Yeah, they’re amazing. Okay, so you should you should know, our very first episode of this podcast, aur guest, who was Amie Wilkinson, who is on the faculty at University of Chicago, chose the fundamental theorem as her favorite theorem. So you're in very good company, because she's a phenomenal mathematician. And okay, thanks.
AD: Awesome. Thanks so much.
KK: All right, up next, we have Toan. So why don't you tell us about yourself and what your favorite theorem is?
Duc Toan Nguyen: Okay. My name is Duc Toan Nguyen. People usually call me Toan. I’m an international student from Vietnam, and I'm a sophomore majoring in math and computer science
KK: Okay, great. So, favorite theorem. What’ve you got?
DTN: Yeah. So as my peer Aaryan, he chose the fundamental theorem of calculus, right?
KK: Yes.
DTN: But I want to bring another fundamental theorem in analysis, which is the mean value theorem.
KK: Oh, okay. So I have a theory, okay. I call the mean value theorem, the real fundamental theorem of calculus.
DTN: Yeah, me too!
KK: So why do you like it so much?
DTN: Yeah, I think I have the same idea with you of why it is called the real fundamental theorem. I think, because to prove the fundamental theorem of calculus, you need the mean value theorem.
KK: You absolutely do.
DTN: Also for analysis, the most popular and common tool in calculus, which is a derivative test, also has the mean value theorem behind it.
KK: That's right. Yeah, that's right.
DTN: So when I first so I first approached the mean value theorem when I was in high school. I took the Math Olympiad in Vietnam. So I had to prepare for that, and there is a section about that.
KK: Okay.
DTN: So it's called the Lagrange Theorem, it was kind of very fancy. Yeah. And it usually applies to — so you know, in the exam, we had some of the problems related to the continuous version, and f(a) minus f(b), something like that. Most of time, we used the mean value theorem. So yeah, it was kind of cool at the time, but I really enjoyed that until last semester, I took real analysis. So I could see the whole process was using the mean value theorem. That's why it can be taught in one lecture or one unit. Even today, this semester, I’m taking multivariate analysis. And it's also a very fundamental thing in proof, everything from differentiability on. It also even has a mean value theorem in it higher-dimensional space.
KK: So maybe we should remind our listeners what the mean value theorem actually says,
DTN: Oh, okay. So, let f be a function defined on an interval [a,b], so that f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). So the theorem say that there exists a point c between a and b that is not is not inclusive so that f(b)−f(a) is equal to f’(c)(b−a). So I think the mean value theorem, the name comes from the quantity f(b)−f(a) divided by (b−a).
KK: Yeah. Right. So the average rate of change over the interval is equal somewhere to the instantaneous rate.
DTN: Yeah.
KK: Yeah, that's right. That's how you prove the fundamental theorem, too, because it's just a telescoping sum when you write it out correctly. And the mean value theorem sort of pushes everything away, and then you're done.
DTN: Yeah, that's also my favorite part. Because it can tell you the relationship between the integral of functions and their derivatives.
KK: Right, right. So what do you want to pair with your theorem, what pairs with the mean value theorem?
DTN: Yeah, I want to pair with something really weird. Which is a phone with FaceTime. Okay, so I'm here. I study. I'm far from my home. My home is in Vietnam, which is on the other side of the Earth.
KK: Almost exactly opposite, right?
DTN: Actually it takes 20 hours from this time to my country time. So it's actually like, opposite, it's more. So yeah, and the FaceTime, why? Because through FaceTime, I can see what people in my home are doing and they also [can see me]. So it's kind of a bridge or relationship that connects what I'm doing here and what my family is doing there. And, you know, my family is always wants the best for me and hopes everything is good for me here. And me too. So that's a very meaningful thing for me.
KK: Yeah. That's great. And I'm glad that technology exists. When I was in college, the internet didn't exist. So you know, I had to call people on the phone and phone calls to Vietnam, I imagine, would be — I can't imagine what that would cost. It was expensive enough to call my girlfriend who lived four hours away.
DTN: It’s kind of more romantic. And you can give them a love letter.
KK: We wrote letters too. All right. Well, Toan, thanks so much. That was great.
DTN: Yeah. Thank you.
KK: Up next, we have Maiyu Diaz. You can introduce yourself.
Maiyu Diaz: My name is Maiyu Diaz. I'm a second-year graduate student here at the department of mathematics at TCU.
KK: Cool, and you have the best shirt. You win the shirt contest today.
MD: Oh, thank you.
KK: I’m actually kind of wishing you would like give me that shirt. [Editor’s note: How dare you say that and not send us a picture?!] All right, so you’re a second-year grad student here in the math department. Okay, great. And so yeah, so what your favorite theorem?
MD: So my favorite theorem is — I don't see it really formally presented, but Stirling's formula where n factorial can be approximated by n^n e^−n times square root of 2πn.
KK: Yes.
MD: That is my favorite theorem.
KK: Yeah. So that’s a really interesting approximation for factorials. So okay, where did you come across this?
MD: I first came across this, I want to say, when I was first studying the factorial back in grade school, and it was just more of like, looking up on a Wikipedia page, which was something that I really relied on when it came to writing essays for my English classes. And when I found out that it could also be used as a resource for mathematics as well, I thought, oh, factorial, let's learn interesting things about this. And that's where I found an example about Stirling's formula, under that. It wasn’t until much later that I was able to understand how that was derived, and there were a variety of proofs for proving Stirling's formula. One of them relies on probability distributions, which are looking at how the factorial works. And then there's another way of finagling a little bit with the integral formula for n factorial that comes from the gamma function, and that's another way of deriving Stirling's formula.
KK: Do you have a favorite proof of your favorite theorem?
MD: I do, but it's a very uncommon proof.
KK: Okay.
MD: The proof relies on a contour integral involving the derivative of the Riemann zeta function.
KK: Oh! I don't know if I know this proof, but what contour do you use?
MD: So you're going to go ahead and do a contour that is on the half strip, so you're just going to fix a number like σ and since the derivative of the Riemann zeta function, you just want to pick a σ just a little bit to the right of where it converges. So real part greater than one, and that's going to be a line integral from σ − i infinity to σ + i infinity, okay? And what you want to do is that you kind of want to start pushing that back a little bit so you can start picking up the residues of the derivative of the zeta function. You’re going to go ahead and possibly the non trivial zeros you could possibly hit, though. You’re definitely going to hit the pole at s=1. That's where the n log n minus n term comes from, when you look at the log of n factorial.
KK: Right, okay, sure.
MD: You keep pushing that more, and then you're going to be picking up the rest of the terms from the Stirling’s formula as well. It's pretty interesting.
KK: Yeah. All right. I haven't seen this proof. That's very cool. Yeah, okay. Cool. All right. So what do you think pairs well with Stirling's formula?
MD: I am going to say chicken tikka masala.
KK: Chicken Tikka Masala. Okay. I do like chicken tikka masala. What in particular makes you want to link those two things together?
MD: So Stirling's formula, I would say is a little bit spicy. And I underestimated it at first.
KK: Right.
MD: Because that was just something I would not have been — I just, if you’d asked me if the formula was intuitive, it’s just like, absolutely not. Where's where does this e term come from? Where does the square root of two pi come from?
KK: Sure.
MD: It’s not until you start familiarizing with the proof more. It's like, okay, at this point, it's not that I'm used to that, I’ve just seen this too many times.
kk: Sure. Sure. Yeah.
MD: But I want to say it’s spicy because of a paper, a PhD thesis, that was published in 2014 by Matthew Lamoureux, whose advisor is Keith Conrad, who wrote a paper on Stirling's formula, just devoting a series of notes on it, you can see all the things he has compiled and I like reading through his notes. And this PhD thesis, he was looking at a generalization of the factorial, which is the factorial for number fields. So instead of looking at the derivative of the zeta function, what it was looking at was a modification of the derivative of the Dedekind zeta function because they had a function defined over number fields. And that's the same technique as well. You want to keep pushing it along the line, be able to pick up non-zeros, possible poles, and it just more or less the same outline as well, and spicy because number one, there is a ton of information coming about the factorial just from the location of zeros and poles of these zeta functions. I was like, Whoa, this is some pretty advanced stuff. I don’t want to be messing with this. But then it's also really delicious because one, I think chicken tikka masala is very delicious.
KK: It is. Agreed.
MD: But also delicious in the context of these formulas, because the approximation formula for factorials is reliant on the poles and zeros of these zeta functions. So it's like, whatever I want to know about on the left side over here, the approximation for the factorials I'm looking at, all I’ve got to know is the information about the zeros and poles of these zeta functions.
KK: Right.
MD: That’s the spicy and delicious part.
KK: Very cool. All right. That's a good pairing. I like that. Cool. Thanks, Maiyu.
MD: Thank you.
KK: Up next, we have Hope Sage. Why don’t you introduce yourself?
Hope Sage: Hey, I'm Hope. I'm a junior physics major at TCU.
KK: Cool.
HS: My favorite theorem is Bell's theorem.
KK: Okay, I don't think I know this theorem.
HS: Okay. It’s kind of Physics-y. It's from quantum mechanics. And John Bell wrote it as a response to the EPR [Einstein-Poldosky-Rosen] paper, which is kind of a famous paper in quantum physics, where it talks about how quantum physics is probably incomplete, and there's probably some hidden variable that's underlying it. And Bell uses this inequality to calculate the probabilities based on what you would expect classically. So if the quantum particles are not entangled, then you would get this expected probability. And then he shows that experimental results kind of conflict with that. And so the underlying assumption is that from that, local realism isn't a thing. So then the universe is like, super wacky.
KK: I think we knew that, right?
HS: Yeah.
KK: So okay, so Alright, so my quantum mechanics is — well, calling it rusty would be, like, an insult to rusty things. So the probability of what? The state that a particle is in?
HS: Yeah, so whenever you have two entangled particles, one might be spin up, one might be spin down, okay? And you can run an experiment. They have a beam splitter and two detectors, and you measure the different states. It's kind of a traditional example. And the probabilities depend on the angle that everything is situated at. So you have nine, and then you were to take all those probabilities, he basically proves this inequality that just shows that mathematically, it can't be possible for there to be hidden variables, which means that things are paired and they would have to be, like, communicating at faster than the speed of light, which doesn't happen. So then there are all these theories about what could theoretically be the underlying nature of the universe from that okay, in different interpretations.
KK: Okay. So which is your favorite interpretation of what might be going wrong here? Or right, whatever the right word is.
HS: There’s this interpretation kind of extended from this called the many worlds interpretation of quantum mechanics. I don't know if I’d necessarily say it is the most likely to be correct. But I think it's the most fun one. It's also fun because you can read cool science fiction books about it. Like, every action you take, there's a different universe, different paths. I think it's kind of fun.
KK: Sure, right. So so right, so like, right now, what we're doing, we could, like, split into any number of paths. And there are all these weird different outcomes that could happen, depending on whether or not some quantum state is what it is or not.
HS: Yeah, basically. One of my favorite books is Dark Matter. It's by Blake Crouch. And it's about, like, every action that you take, there’s a different universe. And then there's infinitely many possibilities based on every single decision you make, which is kind of interesting, because every decision that you make does create a different next possible decision.
KK: Sure.
HS: But yeah.
KK: Okay. Well, all right, so our minds are getting blown. All right, what pairs well with this theorem?
Well, sometimes it's called “spooky action at a distance” and so I was going to go with Halloween candy. Because spooky.
KK: Yeah, yeah. All right. Okay, so what what's your favorite Halloween candy though?
HS: Probably Reese’s.
KK: What? Okay, I have strong Reese’s opinions. So which, like the full size or the miniatures or what?
HS: Okay, a Reese's Peanut Butter Cup, but the dark chocolate version.
KK: Okay. All right. I can respect that. I am team miniature. I think that's the correct ratio of chocolate to peanut butter. But the dark, I get it, I understand. Okay. All right. That was great. Thanks, Hope. Jonah, why don’t you introduce yourself?
Jonah Morgan: All right. Well, I'm Jonah Morgan. I'm a freshman engineering major here at TCU, and my favorite theorem is Gödel’s incompleteness theorem.
KK: Gödel’s incompleteness theorems. So that's more than one theorem. All right, so let's remind our listeners what what at least one of them is.
JM: Sure. So the first one: The first of Gödel’s incompleteness theorems is effectively any — I'll call it interesting, okay — any sufficiently interesting or complex set of axioms, it fundamentally has theorems or statements that cannot be proven, but which are true.
KK: Cannot be proven inside the system, right?
JM: Yeah, cannot be proven inside the system. Right.
KK: And okay, you hedged around, but I think “sufficiently complicated” just means, like, you can do arithmetic.
JM: Yeah, you can add numbers. Because if you just can't do anything, then well, you can’t say anything.
KK: Okay. Yeah. So that's the first one. What's the second one? I think I don't even remember the second one.
JM: So the second one was, I think — I'll make some background. I just think background’s kind of fun to know.
KK: Sure.
JM: His first theorem kind of says, Okay, well, if you have a set of axioms, you can have effectively a statement that says this statement cannot be proven by the axioms. So if you have that statement, then well, if that statement is true, then it's a true statement within the system that cannot be proven by the axioms. But if it's false, then it is a statement which cannot be proven. Or it is a false statement, which cannot be proven. I'm a little rusty on that aspect of it. But effectively, the idea is, so there's this weird statement that you can have in any system, which makes it he says incomplete, where incomplete is the word for it. But then, so mathematicians are like, “Okay, well, we want to prove things. And you gave one example, it's a bit of a weird example. I don't want to prove that a statement is unprovable.” So then he has a second theorem that says, well, also there are true statements which are unprovable which we cannot prove or unprovable.
KK: Yes.
JM: And so that's the second theorem. And that's like, okay, so you can spend your life working on a theorem or working on a problem, and then you can't even know whether or not you can know the answer to this problem within within the set of axioms that you're working with.
KK: Right. So it's hopeless, in some sense. Yeah. You can't fix this issue.
JM: No. It’s — some people say math is broken. But really, it just incomplete. There are certain things —
KK: Yeah, I mean, I think it caused a crisis amongst certain elements of the mathematical community, but I'm with you, I just sort of view it as, well, okay, so there are unprovable statements. It doesn't mean the bridges that we build are going to fall down,
JM: Right. Everything we have proven still stands. You can still prove a lot of things.
KK: All right, still lots more to prove. Where did you come across this?
JM: I think some time in grade school, middle school, high school, I got really into just watching videos about mathematics. And at first they were just little conjectures and little fun things and then Gödel’s Incompleteness Theorem stuck out because it's like, I was diving into the world of math for the first time. And there are so many things that you can prove, and even these, like, crazy things that I never thought were provable, or like things that are so complex. Fermat’s last theorem took 300 years to be proven.
KK: Right.
JM: And I don't understand any part of that proof, to be honest with you.
KK: Same.
JM: But we did it. And it took a long time, but sort of my idea after seeing all of this was that, well, anything can be proven if you have a sufficiently — or maybe we're not smart enough to find the proof, but everything should be provable. Mathematics, it’s a language, you should be able to explain things in that language. And then Gödel’s incompleteness theorem says no. And also, there are things that you just — it kind of changes your perspective on that.
KK: Right.
JM: And so then you get questions like, well, you know, the Riemann zeta hypothesis, one of the most famous unsolved hypotheses. And a lot of people see this and they're like, Oh, well, does this mean that this this million dollar problem, one of the millennium problems, could just be unsolvable? And we can't even know that it's unsolvable? And I think that was the first thing I looked up when I heard about this theorem. And something that was even more interesting to me was that if the Riemann zeta hypothesis is false, it’s provably false. Because it is equivalent to saying that there exists some number on the real part one half line, yeah. So you can write an algorithm, and given infinite time, you will find sure if there is one, you will find it. So it is provably false, which also means that proving that the Riemann zeta hypothesis cannot be proven means that it must be true.
KK: Okay. Yeah.
JM: So I can't give you the formal explanation on that. I don't know what it is. But the general idea is that you can prove that something is true by proving that you cannot prove it.
KK: Yeah, that's a little mind-twisting. But I can see why this would appeal to you, as you’re coming into your own intellectual being sort of state and moving out of being a kid. Yeah, that's really great. Okay, so. So what do you think pairs well with the incompleteness theorems?
JM: So, this isn't sponsored, but GrubHub, or Uber Eats or whatever.
KK: Okay.
JM: Because, occasionally, I order food there. It's easy, it's convenient. Most of the time, you get what you ordered. And sometimes your driver takes a nugget and you don’t — and, you know, that's gone. And sometimes he just doesn't show up at the door, and you're left wondering, where's the stuff that I paid my money for? And I think the feeling is similar there that you can you can pay for something, like you can spend your time working on this theorem, and it just unprovable, and you'll never know where it went or where it goes. And also you can pay for your food and just have no idea where.
KK: Yeah, well, that's a good pairing right there.
JM: Thank you.
KK: Thanks a lot, Jonah.
JM: Thank you.
KK: All right, up next, we have Anna Long. Anna?
Anna Long: Yeah, so my name is Anna. I'm a senior math and French double major. Actually.
KK: Nice!
AL: I’ll stick with English for you.
KK: Je ne parle pas bien le français.
AL: Tres bien!
KK: So I can say I don't speak the language really well in several languages.
AL: Well, that’s all you really need.
KK: That’s right. It was only that and “toilette,” you know, yeah.
AL: Yeah, so my favorite theorem I picked is the invertible matrix theorem from linear algebra.
KK: Okay.
AL: And it's really a pretty big theorem. It's 24 equivalent statements for a square matrix, that'll call A. So just a few of my favorite little statements in there. So obviously, we have that the matrix is invertible, that the columns then form a linearly independent set. So there's always a solution. And then that, for it to equal zero, it's only the trivial solution, that your vector is zero. That it has n pivot positions, or that it has full rank. And then that the linear map is both one-to-one and onto, and that the determinant is nonzero and that zero is not an eigenvalue.
KK: Right. And that's only, like, six of the equivalent conditions. So yeah, if you open up a linear algebra book, there will usually be at some point, some page where they list all these, I had forgotten there were 24. I can probably — so I'm teaching our senior-level analysis course this year. And part of it, we do some stuff with operator theory, and they remind, in our text that we wrote ourselves, they use of 11 of the equivalent conditions, but not all of them. So yeah, I can't imagine. I don't even think I know what some of the other ones are. I'm sure I would if you told me.
AL: A lot of them are, like, jumbled together. So, like, n pivot positions and full rank I've seen defined as two different things, but they’re really the same. But essentially, because of the theorem, they're all the same.
KK: Yeah. So yeah. So why do you love this theorem so much?
AL: I just love it because it's so useful. I’ve had two linear algebra classes now, and it just makes life so much easier trying to prove that any various things in class. So yeah, it's just great being able to find the easiest statement and prove that one, and then you just know all of them are true.
KK: Right. Yeah, I do like those things when, like, 50 things are equivalent. that's really nice. Yeah. Okay. So I guess you learned this in your linear algebra course. You had a second linear algebra course? What’s the second one?
AL: I’m in applied linear algebra right now.
KK: Okay. All right. So you're doing, like, singular values and things like that?
AL: Yes, we are.
KK: Okay. All right. So that's super useful stuff. Linear algebra, of course, I think is one of those things that we don't teach enough of. And basically, any problem in math comes down to either making some estimate, like analysis, or some linear algebra problem, it seems to me. so yeah, the more you learn, the more you know, then the better off you'll be. So you're a senior. What’s next for you?
AL: Graduate school.
KK: In what? In French?
AL: No. In math. Yeah. So I haven't decided on a school yet, but I'm looking at PhD programs in applied or computational math.
KK: Excellent. That’s great. Well, good luck to you. So what pairs well with this theorem?
AL: Yeah, so I picked chicken tortilla soup, mostly because it's my favorite soup. But you kind of have all sorts of different things piled in there. You've got your spices, you've got your chicken, you got your, you know, pieces of tortilla, you've got maybe chives or your different little vegetables in there. So you have a whole bunch of things that may or may not look very similar or different to each other. But you get one scoop of it, and you have the whole thing. So it's kind of like a 24-for-one deal.
KK: It’s like 24 equivalent soups in one.
AL: Right!
KK: Okay. Very cool. Yeah. So are you from Texas?
AL: I’m from Oklahoma.
KK: Okay. So this region, yeah, that sort of makes sense. That’s a popular sort of soup. Yeah. Okay. All right. Excellent. Well, thanks so much, Anna, that was great.
AL: Thank you.
KK: Up next, we have Matthew Bolding. Matthew, welcome!
Matthew Bolding: Hey there. Thank you. Yeah. So my name is Matthew. I am a senior dual degree student for mathematics and computer science, and I'm from around here as well.
KK: Okay. Cool. All right. So what's your favorite theorem?
MB: My favorite theorem today, or really, I guess all time, is the four color theorem.
KK: The four color theorem. Oh boy. Okay.
MB: Are you familiar?
KK: Oh, yeah, I am. So there's a lot to be said about this theorem. So yeah, tell us what it is, and then we'll unpack it.
MB: Sure. So I suppose in the most dry mathematical language, the four color theorem states that the chromatic number for a graph, for a simple planar graph specifically, is no more than four. And I guess there are some things to unpack there.
KK: Yes. Right.
MB: So a planar graph is a graph that can be constructed, drawn, if you will, such that no two edges cross one another. And a simple graph is one that does not have any self loops. And I guess the other part of it is, what's a chromatic number, right? And so a chromatic number is essentially the smallest k for which the graph is k-colorable. And then that also takes us down the rabbit hole with what is k-colorable? So a k-coloring of a graph is an assignment of at most k colors to the vertices of the graph in such a way so that no two adjacent vertices are the same color.
KK: Right? Okay. So most people might know this in terms of maps.
MB: Yes, that is correct. That actually what got me interested in it. It seems so deceivingly simple, I guess, you know, four colors, all you need are really at most four colors. You could do it in three or two, depending on on the map or graph.
KK: Sure.
MB: But, you know, you can ask people, what do you think? How many colors you think you might need to color this graph under these conditions and constraints? Oh, six, seven? No, you only need four.
KK: Right.
MB: And although I'm no cartographer, you know, I've never really colored a map maybe since pre-K, I think it is just so interesting, and sort of out of the blue, that you only need four colors. And with that, I also think it is really interesting that the proof that the chromatic number is no more than four hasn't been proved by humans, by hand.
KK: Right.
MB: And we've had to rely on computers to facilitate that proof. And being a computer science, or within the computer science field to study, I think that's really, really interesting.
KK: I thought that was part of part of your motivation here. I mean, so although, yeah, recently there, some people announced a proof by hand of the four color theorem. And it's wrong.
MB: Really?
KK: So I mean, anybody who's tried to prove this thing just by hand has come up short. And, you know, so the question of the maps, right? So, for a map, it's like, you know, you don't want two states that share a border to be colored the same, you know, and you can draw examples where you need four, but it's interesting that you can always do it with four, but then it does turn into a graph or a question, because how do you how do you create a graph out of this? Well, you stick a vertex for each state, and you join them if they share a border, and now you've converted it to a graph theory question. So that's how they come along.
MB: Exactly. And actually, I found this, or was presented this theorem in graph theory last spring. And I mean, I really — of course, with a computer science background, I mean, I just, you know, jumped in headfirst. I thought it was the coolest class. You know, it maybe doesn't have the same rigor as, like, really analysis or something.
KK: Oh no, it’s hard.
MB: Well, don't get me wrong. There are some difficult concepts. You know, I guess, with a computer science mindset, you know, I mean, of course, not every topic, you know, had roots in computer science, but maybe with Huffman encoding, or shortest path algorithms, you know, it was just so interesting and fascinating.
KK: Yeah, graph theory stuff is vital in computer science. I mean, it's everywhere, and having good algorithms for that is really important. You know, decision trees and all these kinds of things that you need to know. Cool. All right. So, yeah, but it's true that the first proof was given in, what, 1976?
MB: Around there.
KK: And yeah, and basically, it reduces to some couple hundred special cases that you just check. And then you get a computer to check it. And yeah, so for mathematicians, that's unsatisfying, right? We would just like a nice clean, wordy proof that works instead of relying on computer code. But I mean, I'm okay with it personally.
MB: I am too, but according to the Wikipedia page for this theorem, there are still many doubters. I guess we just have to prove it by hands.
KK: I guess. So that's how it goes. All right. So what pairs well with the four-color theorem?
MB: I really struggled trying to think about something that went along with this, but I landed on something that you could actually, you know, show the four color theorem with, and that would be Skittles.
KK: Okay.
MB: You know, you could lay them out all flat, you know, make it a planar collection of Skittles, basically. you could arrange the Skittles in such a way that no adjacent Skittles share the same color. Of course, there are more than four colors in a Skittle pack.
KK: Right.
MB: I would think it's been a while since I've had Skittles.
KK: Yeah, too sweet. Although, so the five color theorem is really not so hard to prove. Apparently, I've been told. I think I may have even read the five color theorem proof. It's not so bad. But four is tricky.
MB: I looked back at my graph theory notes, and we worked from a chromatic number no greater than six to five. And then we sort of just had a blank, you know, statement. Well, you can prove that the chromatic number is no more than four. But right, the ones for five and six aren't, too, too — I mean, compared to having to do it on a computer.
KK: Right. Right. Compared to people, you know, not necessarily believing the computer proof, right? Yeah. Yeah. We’re convinced about five, so we’re good. All right. Well, Matthew, that was great. Thank you.
MB: Thank you.
KK: All right, up next we have Brandon Isensee. Brandon?
Brandon Isensee: I'm Brandon Isensee. I'm a math major at TCU. I'm a senior. I'll be graduating this semester.
KK: What’s next?
BI: Grad school at Rice University.
KK: In math?
BI: Yeah. Computational and applied mathematics.
KK: Okay, that’s That’s great. It’s a terrific university. You're going to have a great time there. Well, I mean, grad school is what it is. It's fun and work and all those things. But it will really a great experience. All right. So, what’s your favorite theorem?
BI: So my favorite theorem is called Sharkovskii's theorem.
KK: Sharkovskii's theorem?
BI: Yes.
KK: Okay.
BI: Have you heard of it?
KK: I have. But let's tell our listeners.
BI: So kind of the theoretical way of saying it is that the theorem relates to discrete equations that are in, it’s one-dimensional discrete equation, so you only have one variable, and if a certain period exists — if a certain periodic orbit exists — that implies the existence of other periodic orbits. And if there's a three-cycle in particular, that implies the existence of all the other cycles
KK: Yes.
BI: And so that's more of the theoretical way of saying it. But if we put this in more concrete terms, if you have an equation that models a population over time, and it's discrete time, so it's years 0, 1, 2, 3, etc. So this equation tells you the population values for each year, right? And say there's a population growth parameter within this equation that you can vary, so we'll call it K. And this parameter tells you how fast the population is growing. And so say your growth parameter is two. And for this growth parameter, no matter which population value you choose, your population ends up oscillating between three different values across time.
KK: Right.
BI: So that's the long term behavior. So as time goes on, your population oscillates between, say, four individuals, five individuals, six individuals, and it keeps repeating. So it's 4, 5, 6, 4, 5, 6, 4, 5, 6. So if that's the case, then that implies that there is a four-cycle. So maybe it's 8, 9, 10, 11, there's a five-cycle, there's a six-cycle, there’s a 1-million-cycle, all the other cycles exist.
KK: Right.
BI: And so it's quite interesting, because when you look at particular examples of equations, like the discrete logistic map, and you look at where the three-cycle exists, you only see that three-cycle. So you may be wondering why? Why am I only seeing that three-cycle and not the four-cycle or the five-cycle? That's because Sharkovskii's theorem tells you that these cycles exist, but doesn't tell you whether or not they're stable. And so when you see the three cycle, it's stable, because that's where your populations are oscillating between. But when it's unstable, well those unstable cycles repel the population values away from them, and it ends up settling at that three cycle. So it's almost like there's an infinite number of fixed points, if I'm understanding this correctly. It’s like there's an infinite number of fixed points where you see the three-cycle, but all of them are unstable, except for the three-cycle, because that's what you're seeing on the graph.
KK: Right. Yeah. So sometimes this theorem is stated as “period three implies chaos.” Right?
BI: Right. So there, so there was actually like a difference between those two. So Sharkovskii's theorem is the one that's stronger, because that tells you exactly which periods imply the existence of other periods.
KK: Yes, right.
BI: And so period three, that relates to the three-period, right, the three-period implies everything, but Sharkovskii's theorem tells you exactly which periods imply the existence of others.
KK: Yeah. And if I remember right, he puts some weird order on the natural numbers.
BI: Yes. So the order the order is, if you're doing this in rows, the first row is your odd numbers, so 3,5,7…. Your second row is your odd numbers times two. And then the next row is your odd numbers times two to the second power.
KK: Sure. Right.
BI: So it's odd numbers times powers of twos. It’s very interesting. It took me some time to understand the order, but now I get it.
KK: And then, like, one is at the end or something, right?
BI: Yeah. Your two-cycles and such, you know, your periods of twos are all the way at the bottom.
KK: And then this is very important for discrete dynamics, right? It's just kind of the whole story. That's very cool. It's very, very cool. All right, so what do you think pairs well, with Sharkovskii's theorem?
BI: So maybe it's very a superficial connection.
KK: That doesn’t matter.
BI: There's a story with Alan Turing. He's the mathematician — for those that don't know, he solved the Enigma code during World War Two, the Germany Enigma code, and because of that, I forget the exact estimates, but it's like at least a million lives were saved because of that in, like, two years.
KK: It was vitally important.
BI: That shortened the war by two years. And so there's a story with Alan Turing where he rode his bicycle, and after a certain number of revolutions of the bike wheels, the bike chain would fall off. But instead of him just fixing it, he would just count the number of revolutions as he's riding the bike, and right before the bike chain would fall, he’d get off the bike and he would just put the bike chain back on.
KK: I wonder if that’s true.
BI: I suppose so.
KK: It’s a good story either way. Yeah. Okay. That's that's a good pairing. Excellent. Thanks.
BI: Thank you.
KK: All right, and our last willing volunteer today — I think they were all willing — is Julia Goldman.
Julia Goldman: Hi. Yeah, I'm Julia. I'm in my first year of grad school here at TCU.
KK: In math?
JG: Yes.
KK: Okay. How do you like it so far?
JG: I like it so far.
KK: Math’s pretty cool.
JG: I think so.
KK: All right. So what's your favorite theorem?
JG: So my favorite theorem today is Brouwer’s fixed point theorem.
KK: Brouwer! All right, good. Finally a topology theorem. Good.
JG: So last week, I was trying to come up with a theorem talk about, and one of my professors suggested this one. And when I was going online, and reading it and learning about it, I was looking at all these proofs of it that seem fairly technical, and I’d probably want to take a topology class to really get into it.
KK: Sure, right.
JG: But the theorem itself is, I think, very understandable. And I came across so many of these cute little fun real world examples that make the theorem pretty explainable to anyone of any math background, and I really appreciated that aspect of it is approachable.
KK: Okay, so what's the theorem? Let’s remind everyone.
JG: Oh yeah. The theorem is for any continuous function of a convex compact set onto itself, there's going to be at least one fixed point.
KK: Right. One point that doesn't move.
JG: Exactly. I think that’s really fun. There’s a couple examples. Like if I had a map of Fort Worth right now, and I laid it on the floor, there would be at least one point on that map lying directly on top of the point it's supposed to represent.
KK: That’s right.
JG: That’s kind of fun.
KK: Yeah. That's a good example.
JG: Then my other favorite example is a cup of tea. You stir the cup of tea. When you're done stirring, there’s going to be one little bit of your tea that's in the same spot as when you started stirring.
KK: That’s right. Okay. Are you a tea drinker or a coffee drinker?
JG: A little bit of both. Maybe it's because I'm not a topologist, but I read that example, and I just thought, I feel like I could stir my tea enough.
KK: Sure.
JG: But Brouwer says I'm wrong.
KK: That’s right. That's right. Okay. So maybe you don't know enough topology to prove this yet? Did you find a favorite proof that sort of made sense to you?
JG: Not a favorite. I just glanced over it.
KK: Right. So the algebraic topology proof involves, like, something with the homotopy groups, or homology groups and things like that. So that's kind of weird. There are sort of analysis-type proofs. So we just talked about dynamical systems a little bit. So if you just pick any point, and you start iterating the function, right? Just keep pushing it around, then eventually it will converge to a fixed point. Well, some subsequence of it will, because you're in a compact set. So that sort of analysis thing. So that's another way to think about it. But yeah, this is a popular theorem among topologists. You know, we all really dig this theorem a lot. It's probably one of our favorite examples. It is a fan favorite. Absolutely. Yeah. Okay, good. I like the map example, too, because that's really illustrative. Okay. So what do you think pairs well with Brouwer’s fixed point theorem?
JG: Well, I wanted to pair my favorite theorem with my current favorite TV show, which I'm only a little bit embarrassed to say is a reality show called Love Island.
KK: Okay. I've heard of it. I have not watched it.
JG: I think they must have some topologists on set there because I think the show itself is kind of an example of a theorem if you stretch some definitions a little bit.
KK: Okay. Let’s hear it.
JG: So if you're unfamiliar with the show, it’s a dating show. The very first episode, all the participants are put into couples. And then there’s, like, 65 episodes of just, like, fighting and breaking up and getting into other couples, whatever. From the seasons I've seen, at the end of the show, there's always at least one couple that ends up back in their original pairing.
KK: Right.
JG: So thinking of the participants as our convex set, and all the show drama as a function, then there’s your example of the theorem.
KK: Yeah, it's funny how there are always two people who were like, you know, we were we were right all along.
JG: Exactly.
KK: Do they do it randomly? I’m sure the producers don’t actually do it randomly.
JG: I mean, they kind of mix it up. They let the girls choose the first time, or the boys.
KK: All right. Okay. Well, I’ll have to check this out.
JG: Great show. Highly recommended.
KK: Okay. All right. Excellent. Okay, well, thanks, Julia.
JG: Thank you so much.
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