from San Francisco, it's theCUBE. Covering Micron Insight 2018, brought to you by Micron. Welcome back to the Embarcadero, everybody, here in the heart of San Francisco, actually at the bay of San Francisco. Golden Gate Bridge is that way, financial district over there, Knob Hill, right up the street. You watch theCUBE, the leader in live tech coverage. I'm Dave Vellante, this is David Floyer. We're here covering the Micron Insight 2018 event. People are starting to filter in. Any minute now, we're going to start the keynotes with the executives, a lot of buzz going on. Derek Dicker is here. He's the corporate vice president and general manager of the storage business unit, emerging activity within Micron. Great to see you again. Thank you very much for having me. It's a pleasure to be here. You're very welcome. Yeah, so Micron used to be just a straight memory company. We're hearing, we heard at the investor day in May, how you guys are diversifying, finding new use cases, new applications. You run the storage business, and of course, David Floyer was one of the first, in my opinion, to predict the demise of the hard disk, the spinning disk, and it's a tailwind for you guys. But Derek, take us through your business unit, your role, and let's get into it. Sure, that sounds great. I appreciate the opportunity again to be here. The storage business unit within Micron is actually comprised across a couple of product areas, primarily NAND components, and then also SSD, solid state drives. As we like to say, and we've talked a bit more about it since Sanjay's arrival, we have a pretty material focus on accelerating what we call high value solutions. It's a big focus of ours. So not only are we developing the core technology in memory and storage, but we're attempting to build more and more products that add value to our customers in the S system space. But that's generally the storage business focus. Within the company, we have three other business units that focus on compute and networking memory, as well as the embedded business unit, and then the mobile business unit. Talk about some of the big trends that you see. I mean, we've talked about for years, the all flash data center. We clearly see that in the customers that we work with. Some of the spinning disguise don't necessarily fully buy into that, but even they have been investing in flash technologies. What are you seeing? I tell you, there is no better time in my opinion than to be in the memory and storage industry. When you look at what the trends are that are coming out in time. If you go and you stare at how memory and storage has evolved, just going back into the 80s of the PC era, a $35 billion average size of a total market, you get into the mobile space. When mobile era started with smartphones, we were looking at a $62 billion ish. And then in 17, we cleared $120 billion in size of the market. And we actually see a lot of secular trends that are going to continue to take us forward. Couple of things that are particularly noteworthy for us. The first one is the emergence of artificial intelligence and machine learning and deep learning. We're going to hear quite a bit about it here at the event. But in terms of a value driver for the consumption of both memory, DRAM, as well as storage, we see it going phenomenally up in content in every server that's purchased out in time. That's one, I think with the evolution of 5G out in time, we're also going to see that smartphone devices are going to end up having more memory to add features like facial recognition we see today, becoming mainstream, multiple cameras that drives more DRAM content. But then also on top of that, storage is increasing. We're seeing even today a terabyte being put into some of the high end phones. We know that that's going to waterfall out in time. So I think if you look at this combination of what's happening both in the devices, you look at what's happening in the infrastructure, then you couple that with the processing that needs to happen, it's just an awesome time to be affiliated with memory and storage. Yeah, well, I've been following this, the NAND marketplace for the last almost 10 years, isn't it, more than that? And it's just broken through completely in the last two or three years. What are your thoughts about pushing compute closer and closer to that memory, adding to, for example, the SSDs, the capability of doing smart work in very, very close to where the data is originally going to be placed? It's a great area of quite a bit of R&D work that's going on right now. And I actually, I think, you know, I view this as kind of two stages. One is there's the proliferation of solid state, as you suggested, it's been coming over time. I actually see it increasing dramatically as we look forward. And one of the key technologies that I think is going to enable that is QLC. The fact that we're now at a point where we're putting four bits per cell into devices, SSDs are starting to show up, I think that just creates even more opportunity. And I'll talk a little bit about that in just a minute. I want to answer your direct question as to how that's changing with AI and ML. But I think the ability once solid state is prolific to be able to architect systems where you can actually have processing take place closer to the media is a very interesting area. It's right with a ton of research going on right now. People are just starting to implement it. I think there's quite a bit of potential sitting behind it. You know, our focus, of course, is we're deploying as quickly as we can on two vectors. One is how do we proliferate more solid state into the market as an industry? And the second is how do we add value when we build a solid state drive? So I think it's definitely very viable. Let's talk about the significance of QLC. David, your forecast early on were very aggressive in terms of pricing declines for flash. We kind of maybe got caught off a little bit surprised by the demand for it. But also the supply constraints kept prices up, okay? So it didn't actually happen as fast. How does QLC change that, Derek, and what's the significance of it? Well, the thing that I think is most exciting for us as micron is we actually ended up delivering the world's first QLC device. It put a terabit of data on a single die. This is unprecedented. But then in addition to that, what we did is we actually built a solid state drive called the 5210 Ion. This is a SATA drive. It's the world's first SSD built on the technology. And by being able to develop a solution early on, it allowed us to go engage with customers and find where the right workloads were where we could add the most value. QLC technology actually is perfectly aligned for super read-intensive, very read-intensive environments. And if you look at what's happening in the data centers, we're actually seeing more and more workloads move into more read-intensive workloads. And a good chunk of that is just because there's analytics going on. The data's being collected, it's being housed in one place, but as we've talked about quite a bit here at the event, we want to be able to deliver insight out of that data, which means we're going to be reading it quite a bit and massaging it and performing analytics on it. And what we're now seeing is what in the days of the past was a four-to-one read-to-write ratio. We're seeing as high as 5,000 to one and in some cases a million to one. So we get these heavily read-intensive workloads coupled with the technology that's optimized for it. It's more power efficient than what rotating media solutions offer in certain workloads. We're starting to see these tremendous values coming out of these early engagements that we're having with customers. And does that have implications for longevity or do you just make an assumption that the read-write ratio is still going to be more right-intensive in terms of ware leveling and things like that? How does it change the reliability, if you will, of the technology? Actually, the beauty is we're able to deliver an enterprise-class SSD with these read-write capabilities that are affiliated with these read-intensive solutions and we can fit within the workloads and the needs that people are talking about. So the drive-rides per day that are required in a machine learning infrastructure, we believe we can address with QLC. Same thing with Hadoop style clusters and stuff clusters. We've actually, as we've gone out and engaged each of our earlier customers, we're able to crank out reference architecture documents that we're now posting to our websites and we're describing how we can actually leverage this technology to allow us to, in some cases, well, better optimize where an SSD was used before, but in many, many cases, we're actually in the process of displacing hard disk drives. So what are the limits of this QLC? How many more bits can we add? How many more layers can we add? So it's actually a great question, David, in terms of what does a roadmap look like? I've been asked in the recent several hours what the longevity for NAND looks like and what I'll tell you is this. QLC NAND is just getting its start. What comes after that in terms of bits per sell, I don't think anybody's made any broad claims on, but from a layer stacking perspective, which is kind of the dimension upon which the industry is growing, for the foreseeable future, we see nothing that encumbers us from going substantially higher and higher in layer count. So I think it's going to be great for our industry because it's going to allow us to deliver more bits in a given device and hopefully that'll allow us to get into markets that historically we haven't been able to approach. If you think about the demand elasticity dynamic that occurs, when we start to bring more and more costs down, the number of applications open up. Not unlike the machine learning workloads I just mentioned or Hadoop workloads, we're starting to see more and more thirst and interest for replacing with solid state just because it's more power efficient, allows for a cost structure that's better and gives better performance to it. I'm fond of saying that data's plentiful insights aren't. You guys are a $30 billion company now. You're making some interesting announcements today that we're going to hear about a little later on that I won't divulge right now, but you're putting your hands in a lot of different places. When you're that size of a company, you can't help, but as you mentioned before, adding more value, becoming more of a systems focus. How do you help the industry go from just raw data to insights? What's your role in that? Oh, it's a phenomenal question and this is a major focus of the company, not just in our business unit, but across all of the different business units in the company. We have a huge focus on sitting down with our customers and getting closer and closer to understanding what their workload needs are, where their pain points are, and then working with them to find solutions. And the beautiful part about it is, as Micron, we're the only company in the world that can combine together a 3D cross-point set of technologies, a NAN set of technologies, a DRM set of technologies. We go sit down and talk about these challenges with those in mind, plus the emerging memories that we're developing to go develop better and better solutions. But after we're able to come to a solution, we put together a reference architecture and we deploy it broadly. We've been trying to squint through 3D cross-point and understand the right fit. It seems to us that one of the big advantages of Flash was it had this behind it, right? The consumer volumes, thank you Steve Jobs. It's unclear whether or not 3D cross-point will have that and maybe have the same sort of cost advantages, but at the same time, it sounds like there's new and emerging applications. Like I said, we're trying to figure it out. Have you guys figured it out yet? You're obviously betting big on the technology. Help us understand where the fit is. Sure, I think if I look back in time, just at the storage hierarchy alone, I don't think the memory hierarchy is any different. You have these portions of the market where you build out hard disk drives and we had DRAM before and SSDs came along and people started asking, not unlike several years back when we talked about the early parts. Hey, how big is this going to get? Cost structures may be prohibitive, but as innovation unfurl the more time and investment got placed into it, we found new workloads, new use cases, we were able to drive cost out and we ended up slotting in solid state drive squarely. I think this is another tier of memory and storage. That's a beauty of the 3D XP technology. There's both memory semantics and stored semantics that are available for use. I think we're still scratching the surface on the early days, but I love what we're seeing from the customer base that we're engaging and targeting in this space. People will pay up for that performance capability relative to Flash, they'll pay down relative to DRAM. Is it, are you seeing a great answer from like the hyperscalers for example, or is it, so maybe the industrial internet, where are you seeing the... It's fair, actually I think, it's probably reasonable to say that the challenges of inserting a new memory tier into a system requires new programming algorithms, new APIs and interfaces. There's a lot of ecosystem that needs to be there as well as not to mention you got to have an ecosystem to go put memory products into a server, for instance, or any other platform. I think we're still early days of enabling all of this. And I also believe we're going to learn more and more where the value of this sits as we put it out there in a cost-effective fashion. So I would say that people who control software environments are very, very well suited for this because they can take advantage of some of those challenges without having to have a whole ecosystem in place. I think there's going to be a continued ramp and acceleration as an industry to build out that ecosystem. Well, it's been amazing to watch Micron the last several years. I mean, the last several decades, moving from just a pure memory manufacturer to this diversified, you know, gorilla in this space.  I mean, your financials have borne that out. You're really transparent to the street and have provided great guidance. And congratulations on all the success and looking forward to watching in the future. Well, thank you so much. It's a privilege to be part of the company and I really appreciate your time today. Ah, a pleasure, thanks for coming to theCUBE. All right, keep it right there, everybody will be back with our next guest right after this short break. You're watching theCUBE from Micron Insight 2018.


There are many ruins of ancient antiquity still in existence, which tell a vastly different tale for the history of human civilization than that of its contradictory counterpart, namely modern paradigm. One enforced, funded, and massively supported. A tale of events merely written by a victor. However, thanks to our ancient ancestors' incredible abilities, many of their ruins and relics so well constructed, often made from complex, elaborate, or gigantic materials, can tell us all what a well-funded modern academia simply cannot explain. Thus is staunchly rejected and ignored in favor of the illusion of an authority over our species akin to that of an oracle. One of the most popular and predictably easily argued of ancient sites for our case of events are conviction in the belief that there is a hidden history of man and, indeed, earth. One of a far greater antiquity for our species, with many now lost civilizations having once come and gone. The site is additionally one of the largest ancient anomalies of our planet. It is, of course, the great pyramids of Giza. Due to the structure's immense sizes, the evidence for several conservation efforts in the form of casing stones littering their four-sided facades. The sheer size of the stones involved in the bill, the amount of stone used, the precision of the once hermetically sealed, constellationally aligned ventilation shafts, and the many different stages of erosion present on these varying sections of the construction efforts, indeed makes the great pyramids a great avenue of study and one of strong conjecture in the face of an adverse and conspiring academia. Our hypothesis, we feel, due to our compiling of the many inexplicable factors of antiquity, has now all but been virtually proven beyond doubt. For our opinion, regarding the true history of earth is not just based upon the objection of the illogical explanation, widely given for the origin of the pyramids of Egypt. It is also based upon many other areas of ancient history, which not only support our claim, but if one sticks to mainstream historical ideology, we'll simply find the task of explaining these ancient uparts simply impossible. They simply cannot exist. And our next artifact of interest is no different. Like King Tut's dagger, the Beth Shearim glass slab. Created Switzerland by amateur archaeologists, it is known as the Hand of Preals. Not only is it the earliest, most elaborate example of bronze and goldwork ever found, but quote, we do not know either the meaning and the function attributed to it. Its gold decor suggests that it was an emblem of power, a distinct sign of the social elite, even of a deity. The hand is extended by a hollow form that suggests it was originally mounted on another object, perhaps as part of a scepter. End quote. Who made the Hand of Preals? Is it really 3,000 years old? If so, how can it be the earliest example of this form of metalworking yet be so elaborate in nature? The Hand of Preals is unquestionably an ancient upart, and could quite possibly be a surviving relic of a once highly advanced yet now lost civilization. We feel this hypothesis is far more logical, and as such, makes the Hand of Preals highly compelling.


I think we took our foot off the gas. They played harder than us, and we didn't match their energy. They found momentum, and a couple shots went in, and now we find ourselves just trying to survive. So I thought we just had that bad stretch at the end of the third start of the fourth. And frankly, I think we just took our foot off the gas. In that closing stretch, I mean, you had Bruce with the corner three to tie, and then James Joe and Kevin kind of set an amount. What do you think of the play down the stretch? Yeah, down the stretch was pretty solid. I mean, Kevin obviously went to work. We got him ball in good spots. He made a play for BG, and just overall, we executed offensively down the stretch, and our defense picked up in the last couple of minutes. But, you know, a lot of good things tonight. Just in each half, we had a period where we kind of relented our will and allowed them to get back into it. You touched on it a little bit, but I mean, when you clearly see momentum in a game like that in that fashion, just how valuable is it to have a veteran leadership like you have and to have kind of a singular talent like Kevin where you can go to the 11th couple of minutes?  I mean, I know James made a couple of plays, and Joe and BG got some buckets, but Kevin just being able to get him the ball in spots. You know, he made baskets. He also made a play for BG. He got doubled, got Joe. I mean, sorry, got James a three. So, you know, we played through him, and he delivered, of course. Steve, I know that James got to learn 15 times, but what really got James kind of going tonight? Yeah, I thought after the first time out of the first quarter, you know, he got really aggressive. He really tried to get in the paint. As we've been preaching, play early, try to attack early, try to get in the paint, and he did a great job of that. He's been kind of talking about how he's been building his confidence and building his way back.  I think it's, you know, he played similar to night as he did, you know, a couple of games on this trip for sure. I think we got to be patient with him, you know.  You know, there's the fitness, there's the rhythm, and there's the confidence, like I said. So, you know, for all those three things to come back, it does take time. He was great tonight. He's been really good the last week or so and trending in the right direction, but I don't want to get too carried away. Let's give him time to get his game back. In this day, how's it about Joe's starting in five or six threes in the first half? Yeah, I mean, just electric shooting from Joe.  So, you know, I think tonight was one of those nights where the offense was there for the most part other than a stretch at the end of the third start of the fourth. The defense can be better. That wasn't our best outing defensively.


Let's sing in honor, glory and power. Be to the one on the throne. Every triumphant nation, all of creation, sings of the glory of our God. All of heaven will join the chosen child. All of heaven we will join is the Lord. The church singing with me is real simple. Lord is the Lord.


When I agreed to talk, I thought this would be a cozy little with a couple of dozen people. So it turns out there are more people who are interested in number fish, which is a good thing. And I think it's a great thing that this seminar has been organized. It's a really nice touch that we can still keep in contact around the world. Okay, so I want to give a flavor of a BlackBall talk. So I'm using my standard Zoom teaching technique. And what I want to do is talk about a topic which began for me about 10 years ago. And I'll try to say what's going on. So first of all, listen background, I want to think about wearing's problem. So wearing's problem is to consider representations of a positive integer N, as per some of S-cake powers of natural numbers. And in particular, I'm interested in this talk in the number of representations of an integer N as per some of S-cake powers, I'll call that RSKNN. And for this classical problem, goes back to wearing in 1770. We think we know what the answer is to this representation problem. So I was conjectured and put an asterisk on this, which I'll return to in just a moment. But when S is at least as big as K plus one, the number of representations of N as per some of S-cake powers, there's a quotient of classical gamma functions times the singular series, short of finding just a moment, times N to the S of a K minus one plus a little over N to the S of a K minus one. And I call this an asymptotic polynomial. So the singular series might be zero if there are no local solutions in the problem. It's defined as actually a product of periodic densities, but it can be written in closed form in terms of these classical Gauss problems. So E is just equal to two pi I times A of a QR to the K, as usual. The only wrinkle in this, why there's an asterisk is that if S is K plus one, you have to take a little K. The singular series may not be nicely convergent. So this is what I've written down here is certainly valid. Everything makes sense if that's at least K plus two. And so with a number of variables, which is a little larger than K, you expect to understand this problem. You expect to understand this problem very well. That's the conjecture, right? So what do we know about this kind of problem? So that's where some notation comes in handy. So let's define G tilde of K for the number of variables you require to establish this asymptotic formula one, which I have at the top of the slide here. So that gives me a way of measuring what we can actually prove. We conjecture that should be just K plus one. And this is really the starting point for this talking in ways. There's a nice classical result of Poirre from 1938, which shows that with two to the K plus one variables, you get to prove this asymptotic formula. And the proof of this is if you've ever studied the circle method, your very first course in the circle method, that's what you prove. That's how you learn about the circle method most usually. It's a very classical sort of sequence of arguments really dating back to how the little word started for 20th century in the 1920s, a little before that. And Poirre introduced some very nice refinements involving counting solutions of guaranteeing equations to make the argument more efficient. Okay, so that's the ancient history. Just to complete this picture for the moment, let me tell you about what's known in the current state of play. So this is a topic which has undergone some refinements, but in the case of cubes, you want to understand where this problem in the case of cubes, then still the world record is a result of Vaughn from 1986. So you get the expected asymptotic formula with eight cubes. That's one variable less than Poirre would have gotten. When K is larger than three, then the recent progress on the integrator is the value for plays a role. So once the main conjecture and that subject have been proved by Bagan Demigloum-Gouf by myself, then some work I had in 2012 gives up a balance on these values, which supersedes those of Poirre already when K is four or more. And for large values of K, you get a result which is like K squared minus K with some stuff involving the square root of K. So this is quite a precise version of the result that was in a paper which started off life actually more or less at the Fields Institute in the 2017 program we had there and appeared last year. Bagan had a, what you might call a sketch of a less explicit bound where sketch requires some interpretation. Okay, so that sort of completes the picture there. And what I wanted to say by way of introducing the topic of this talk is that you can obtain similar results to the ones I mentioned so far for integer value polynomials. So if I replace extra K by a polynomial of degree K with rational coefficients that takes integer values and all of the results that I've explained have analogs with the same number of variables, you get an asymptotic formula. The asymptotic formula changes a little on the local solubility conditions, change a little as well. But that's all a very well understood topic. And in some sense, except for the definance of a number of variables, even the local solubility issues were very well understood back in the early 1950s thanks to work in nature and others. So you can see people are buzzing with questions maybe or complaints about what I'm saying. So I'll let Mike or Alina figure out what's going on there. Carry on, sir. Okay, I'll carry on. Right, so this brings me to a problem. So as far as I know, the first familiarity I had with this problem was a conference in Lille in 2009, I think, all of my memories can be very faulty by the same. So Ben Green posed a problem which he may well have posed earlier, which is to generalize these results from integer value polynomials to certain generalizations of polynomials which had occurred in investigations surrounding systems of linear equations and primes, so the whole nil sequence approach to this topic. So motivated by that, he posed the question of what happens if you replace as sort of a simplest example, that he replaced the squares or quadratic polynomial with rational coefficients by a bracket polynomial and by a bracket polynomial, what we mean is something like this, an integer times the integer part of that same integer times some real number and to make this interesting, it's best to keep that real number to be positive and irrational, same. So he had in mind stressed the case that way to equal to the square root of two is a special case of this problem. So for example, can you find an integer s sub zero having the property that whenever you have at least s zero variables, then all large integers n have a representation as some of s values of this bracket quadratic polynomial. That's the problem. Of course, you can imagine all kinds of variants of this problem and I'll say something about some of the variants later on, but at least that's a good problem to have in mind to begin with. And maybe I can make a couple of observations which are not new about this problem. So as Ben pointed out when he presented this problem, of course, if theta, this number theta is a natural number or rational, then basically this amounts to understanding what goes on with the polynomial case. You can take into arithmetic progressions like this. So you really should stick to the case of irrational values of theta to make this interesting. And really what makes this rather non-trivial is the observation that these polynomials, bracket quadratics, are not necessarily, don't have the same kind of equidistribution properties, but polynomials or rational coefficients have. So if you think in terms of the bio-equidistribution law, biosfilm, then you realize that polynomials or rational coefficients have nice equidistribution properties, module one, when you multiply an irrational number alpha. But this sort of thing can fail for these bracket quadratics. So basically with some examples of these bracket quadratics, as Ben observed, you can sort of relate them to values which are integer. If I multiply by two here, I'm really getting some integer here on the right-hand side. And then some term which involves the fractional parts of x squared of two squared. And x squared of two is uniformly distributed. It's equidistributed module one. So it spends a fair amount of time close to, in fact, what I mean is strictly a fair amount of time, as in not an unfair amount of time close to zero. And that means that it spends an unfair amount of time, the square of the spends an unfair amount of time close to zero. So somehow the square of something small tends to be smaller. And so this has a bias towards being close to zero. And that means that you can sort of get a failure of a distribution for these things. So that's a nice observation, which means that one would expect that some fairly routine version of the second method is not going to be very successful in tackling this problem. That's a perfectly reasonable observation or expectation. Okay, so where do we go with this? Well, there's a thesis of a student of Ben, Vicky Neal, from 2011, where she answers this question. So the answer is yes, there is such a, we'll do this, we'll get the question in the slide. Well, so there is such an S zero, you can represent large integers by values of these, sums of values of these packet quadratics. So the thesis doesn't give an explicit value for this number S zero, it involves Fourier analysis on null manifolds. And if you want to, if you're sort of familiar with the second method, which I don't expect everybody is, you don't have to be for this talk, but you might say that the method would be similar to Hardy and Littlewood's approach to Waring's problem where basically vials and quality is used on every variable. The analog of vial comes from this Fourier analysis on null manifolds, so the number of variables you need is very big. And in the thesis, if he expresses the point of view that the answer is from the methods is probably closer to a million than 10 to the 100. So probably a large number of variables, that's the upshot of that discussion. And so what I want to do in this talk, what am I gonna do in this talk? So I want to present first of all, because I think it's from a couple of results of the flavor of Hoare's lemon, which can be input into the machine that Vicky and Ben developed. And this is something that is almost 10 years old. It dates to very shortly after Vicky's thesis. It's something that at the time that I had in mind, this would probably be an appendix of some paper of Vicky and Ben. It might have become joint work at some time. Anyway, nothing's happened with it since. So I've presented it anyway. And then I'll say a bit more about some ideas I've had since then. And this gets us into the game of speculation about what methods may be used in wider generality. And so the Hoare's lemon discussion, which I'm about to enter into for these bracket quadratics is particularly elementary in a certain sense. So I think everybody should be able to get something from that. And it's kind of thought provoking. Okay, so let's get into that. So this is where my backboard comes into play. So let me say a bit about Hoare's lemon. I guess it's a good time to pause it for any questions that people have had in the meantime now. It's a very good time to ask him or carry on and wait for a little alert from the organizers. So what about the classical version of Hoare's lemon for in the simplest interesting case, which is for the squares. So that allows me to get into the ideas, which we can make use of for these bracket quadratics. So Hoare's lemon, what it does is it bounds mean values of exponential sums. And the interesting case in the case of squares is just the fourth moment. Now, of course, in fact, you can analyze this in many different ways and we can get precise certain products for the number of solutions here. That's not the direction I want to take this because I want to get a crude estimate that motivates what happens for these bracket quadratics. So biofogonality, the mean value here is counting solutions of a definite equation. And I can arrange this, yes, question. Sorry, can I interrupt you for a second? There's just a question from the audience. Question, it's going back a tiny bit. Is this uniform in theta, the bound for S naught? I think in Vicky's work it is and what I'll present, it certainly is. So, but I have a feeling that in Vicky's work, well, I mean, you can get improvements for certain values of theta, but I think there's an absolute S zero of theta. We'll see later on, but so even if I'm wrong about that, it can be made before. So, so short answer is yes, I guess. Yeah, is that enough now? Okay. Right, so I've arranged the definite equation with some suitably chosen pluses and minuses here. You can arrange it as a sum of two squares equal to a sum of two squares if you prefer. But now if I want to down the number of solutions here, you can see that for two obvious classes of solutions, there's one class where X1 is equal to X2 and if that happens, then X3 is also equal to X4 and vice versa. Of course, I can reverse the implication if I want. And here the contribution is clearly something like capital X squared solutions, like X choices for X1, X choices for X3 and then X2 and X4 are determined. And then the other case, let me use X3 and X4 as the sort of a classifying pair of variables. So maybe X3 is not equal to X4. And then what I can do is fix X3 and X4 and then X1 minus X2 and X1 plus X2 multiply together to get X1 squared minus X2 squared and that's equal to a fixed integer which I'll call in X3, X4, which is just X3 squared minus X4 squared. And that's non-zero. That's very strange, not equal to, that's not equal to zero. Okay, so that's all well and good. And so now we can see that each of X1 minus X2 and X1 plus X2, these are divisors of N and the number of divisors of N is at most N to the epsilon. Okay. So what have we done here? We've proved quasi-lemma already and that's for the case K equals two. What we have here is we've shown that this mean value which we're interested in, two classes of solutions that were being counted a diagonals contribution and then we had another type of solution. There were X squared choices for X3 and X4 in this second class of solutions and then X1 and X2 are determined by divisor function estimates. If I know X1 minus X2 and X1 plus X2 then of course I recover both X1 and X2. And so that's why it's lemma. So we've shown that the total number of solutions here is at most X to the two plus X1. And what's important here, so far as the application of the little method goes to Waring's problem is that we've saved very nearly X squared here. So the degree of this polynomial is two. We've saved very nearly X to that degree X squared over the trivial estimate which would be X to the four. And that's what allows you to prove that you get the expected number of representations of an integer as a sum of five squares of natural numbers. Of course you can save much more precise things about sums of squares with three or four squares but this is what we're doing in this setting of while's lemma. Okay, so that's the classical version all very straightforward. So what about bracket quadratics? So bracket quadratics, first of all there's no nice factorization of X1 times four function of X1 square root of two minus X2 times four function of X2 square root of two. Everything gets corrupted. So that goes away, divisor functions go away. It's all very frustrating. So it's not at all apparent how you would ever prove a lemma or a qua type. So let's deal with a bracket quadratic and following Ben's lead let's consider the case of later in square root of two. You'll see that for this argument this is not particularly special. And I'll write C of X for X times the floor function of theta times X. So for us that's X times the floor function of square root of two times X. And I'll give a proof of the following more general results in just a moment. So let's record this as a theorem so you'll see the argument that I'm explaining I'm about to explain applies if you have any theta which is the square root of a rational number and I suppose that theta is rational and positive and then we've got this analog of plus lemma. So where's my bracket quadratic X times the floor function of theta times X. And I look at the fourth moment of this and this is all bounded by X to the two plus epsilon. And what I'll do is I'll prove proof us in the case of a theta equals square root of two and you can see it's a simple exercise to convert this to a proof for more general values of theta of this type which I've listed here and the idea the strategy is to think about sums of two well sums or differences of two bracket quadratics so I suppose that n is a natural number so I've had time to sort of define this since the early days when we thought about this so you can either look at a sum or a difference of two of these things. So I claim that the number I'll call it r of n of integral solutions X1 X2 of X1 floor function of square root of two X1 minus X2 floor function of square root of two X2 equals n so this I claim satisfy this r of n less than less than n to the epsilon and if you think about what this implies for brief moments this presents the idea of applying exactly the same strategy as we had in the classical proof of Mars lemma so there we have diagonal solutions and for the non-diagonal solutions what we did was we looked at the difference of two squares and it turned out we had an argument involving the divisor function which showed that the number of representations of an integer and nonzero integers the difference of two squares grows like n to the epsilon and we're getting exactly the same result here if we can prove this result about bracket quadratics so I granted this idea the proof of this result is lemma was lemma but these bracket quadratics follow straight away and I want to actually say more about that truth other than this result some differences of bracket quadratic polynomials good so here's the strategy this is the only idea I think that really plays a role here and then you have to use the idea and the idea as with most of my ideas is very unimpressive so let's write y sub i for the proof function that's square root of two x sub i when i equals one and two and the profound implication of this is that square root of two times x i minus y i is between zero and one for i equals one and two good nobody disagrees with this so now what does this tell us this tells us that square root of two times x i minus y i sorry I should make that a one square root of two x one minus one one is squared minus square root of two x two minus y two squared well each of these objects is between zero and one so their squares are also between zero and one which means that this difference is somewhere between minus one and one good but I can expand this this expression so if I expand it and use the definition of square root of two then I get a polynomial x one x one up to y two quadratic polynomial with integer coefficients minus two times square root of two times x one y one minus x two y two and notice here that this expression is nothing other than x one times the floor function of square root of two x one minus x two times the floor function of square root of two x two so this expression here is supposed to be equal to n so so what we've done here is we've shown that this expression two x one squared plus one one squared minus two x two squared minus y two squared minus two square root of two times n is somewhere in the interval from minus one to one so far so good but that means that we've almost fixed this this integer here on the in the first bracket so it's first integer it is an integer it's in the interval from two root two n minus one to two two n plus one so the left hand side is either the ceiling function of two root two n minus one or it's the ceiling function of two root two n two choices at most okay so we almost know what this integer is so we know what some x one y one minus x two y two is its n and this other quadratic expression is also almost fixed and so what that means is that our then is bounded above by the number of choices of a system of equations let me write down that system right here so x one y one minus x two y two equals n let's say and two x one squared plus one one squared minus two x two squared minus y two squared equals m where m is in this in this interval two root two n minus one or two root two n plus one it's one of two integers okay and here's here's the punchline that means that m plus two times squared of two n I can rewrite as squared of two x one plus y one squared minus squared of two x two plus y two squared so the number of choices for this integer here on the left hand side is is at most two we know almost exactly what it is it's one of two integers in in the ring z a join square root of two and what we've got here is a product of two we've got a difference of two squares so that means that the sum and difference of these two variables so that's squared of two x one minus x two plus y one minus one plus y two and the square root of two x one plus x two plus y one plus y two these are both divisors n plus two two times n in the ring of integers and q join squared of two and what most what if I write this as big or m plus n to the epsilon choices for these divisors then that certainly gives an upper bound for or the number of choices for these two factors here and if we know about these two divisors you have a question what about units what about units well the point is that the coordinates of both of these expressions the coordinates are bounded in terms of n as well so it's a good good point so both x x one all the x i's and the y i's are bounded by a constant times square root of n thanks to thanks to the you know the setting of the original problem so we have an upper bound on the size of all of the coordinates which means that the units also feed into a sort of a factor of entry epsilon in this in this game and that's true in generality as well of course I said you could look at square root of two and think of an arbitrary quadratic irrational so things get more complicated if the plus number is big but the same conclusion holds in fact you have to work harder for it that time but it's all quite familiar because we're only looking for solutions where the coordinates are bounded in a nice way and does that answer the question absolutely okay brilliant thank you so so what we've done is we've bounded the divisors since given these divisors we can recover x one x two and y one y two this means that the big O of entry epsilon choices for for x one x two and also of course y one and y two because those defined in terms of x one and x two and that means that that this number r of n which I was interested in right at beginning is bounded by well at most two choices for m times this n to be epsilon so it's bounded by n to be epsilon which is what I what I claimed it would be bounded by and as I said that's enough to prove for this kind of example so you can see what we did in this problem and this in this argument was we started out with our mean value of course that converted by a problemality into a statement which we could say something about if we knew about numbers of representations of integers of differences of two bracket quadratics and that problem we converted by a bit of subterfuge into a bounded number of problems involving differences of two squares in differences of two squares representing fixed integers in the number field that we're working with in the ring of integers in the number field that we're working with and so secretly this problem was controlled by a problem in the number field involving actual quadratic polynomials so the argument I've just presented across generalizes from squares to more general quadratic polynomials in the number field it also generalizes to some extent to higher degree polynomials but that's a that's a different topic which depending on how much time I have I may say something about later or I will say something about it but I may not get much more details about it the other problem that you may be interested to know about is what happens if feta is not a quadratic rational what about more general values of feta and there that's sort of an interesting story so I don't know a similarly elegant proof in that case although there's a reasonably elegant proof maybe I'll give the shorter version of this the story since time is sort of running down a little so the two approaches of this but I know one involves having a machinery although neither of them are particularly easy so we have an analog for these more general rationales and maybe I can slightly short circuit the whole thing so so again this is going to involve a number of solutions of a d'affentini equation involving these bracket quadratics and on this occasion what I want to do is consider something which I'll call r star of n the number of solutions of the equation x1 times full function of x1 theta plus x2 times full function of x2 theta minus x3 times full function of x3 theta equal to n and then the claim well now we've got three variables not just two so the claim is for r star then there's less than less than n to the half plus epsilon here and to some extent we're going to we're going to use a similar strategy to the one which I used earlier so I'll write yi to be the full function of theta xi for i equals 123 so that means of course that zero is less than theta xi minus yi is less than one and then what I mean is that if I sum sorry I should be more careful about this so if I think of theta x1 minus y1 squared plus theta x2 minus y2 squared minus theta x3 minus y1 minus y3 squared then that's between minus one and two because each of the squares are between zero and one so we're just a bounded number of that times here and I can expand that whole expression so what I find is that theta squared times again a different example but so there you go theta squared times x1 squared plus x2 squared minus x3 squared minus 2 theta times x1 y1 plus x2 y2 minus x3 y3 plus y1 squared plus y2 squared minus y3 squared so that whole expression is between minus one and two so now it's not as convenient anymore right so theta has no special significance we don't know that's in the number field so it's rather hard to think of what to do in that situation so what we could do is consider the values of x1 squared plus x2 squared minus x3 squared maybe I'll call that L and x1 y1 plus x2 y2 minus x3 y3 well that's actually n that's the fixed object and y1 squared plus y2 squared minus y3 squared and I'll call that m so what do we know about L, n and m we know that L times theta squared minus 2 theta n plus m is this expression which you see at the top of the screen it's essentially fixed for most of the finite number of values of it we also know that L is kind of close to m if you think about it so well maybe I'll say that instead of actually the m is rather close to m so y1 is supposed to be theta times x1 approximately so m is roughly theta times m but there's an error and if you think about how the error works maybe it's the differences as big as square root of m times a constant m is theta n plus bigger square root of n and if I fix m then I know basically what L is so m is fixed, n is fixed L is essentially fixed up to a bigger of one choices so what that means is that the the bigger of square root of n choices for L, m and m given our fixed choice of n right at the beginning and what we have to do is to solve these three equations now for the x i's and y i's and thanks to the work of Paul and Jones and the theory of these are indefinite quadratic forms and three variables now representing binary quadratic forms the bigger of L plus m plus n to the epsilon choices for the x i and y i and again a similar observation applies concerning what happens with units in this situation as well so again because the coordinates are bounded in a sense there's an extra factor which involves the coordinate size for the x i's and y i's but they're all bounded in terms of n so you can see here that for what happens is we don't get the same simple situation it's not a problem about number fields but we can understand this in terms of what happens with these this sort of higher quadratic form problem okay and what we've shown there is that r star of m total number of choices is indeed bounded by n to the half plus epsilon and given that result remember this is all related back to a Daffin team equation in four variables for every fixed choice of that fourth variable that n to the half choices there which is like x then we're getting square root of n choices which is again like x choices for the remaining three variables so that's why we've got an extra two plus epsilon up and down here for the mean value okay so so what that means is that we have an analog of last lemma both for quadratic irrational fader which is a little simpler to describe but also for for general irrational fader as well and the fact we've saved almost everything but we should in these mean values we've saved a factor x squared means that's good input for the methods of Vicki Neal and Ben and so with an extra variable their machinery would give you a five variable asymptotic formula okay so I wanted to say in the last I think I have five minutes of rest then correctly a little bit more about where you can take this next so mean values and generalizations and I would like to see much more here than in fact I can so there is at least something that I can say which I think is some sort of interest something provable and then the rest is more speculative which I probably won't get to but so never mind so let's consider again the situation for this discussion the situation where fader squared is in q and fader is positive irrational and so if you want to generalize what we did before then one possible way that you can sort of convert a problem about racket polynomials into a genuine polynomials would be to consider special polynomials right so so I'm going to consider two polynomials cxy and cxy so these are going to be polynomials with rational coefficients and they're going to be defined this is my way of defining them so the kth power of fader x minus y is fader xy minus fader times fader xy because fader is quadratic I can always this sort of thing makes sense what I can also do is define the number s0 of k and for the moment I have a feeling I can probably improve this but let's take this to be 2 to the k, 1k is 2k and 4 or k times k plus 1, 1k is at least right and then here's a theorem to indicate what kind of thing you can prove there are some pretty clear generalizations of this but this illustrates what's possible so let me take f of x to be either fb of x, floor function of fader x or c of x, floor function of fader x with x in it x in natural number and then the claim is that if I look at the mean value of an exponential sum involving this of these bracket polynomials of order s0 of k then this saves x to the k very nearly in the exponent okay so how does this work in a nutshell so the idea is this is just a sketch proof so the mean value counts solutions of an equation that equation is closely connected with the expression you get when you look at k powers of, so let's see, so here I want some as before fader xn minus yi to be between 0 and 1 because I have a mind that yi has come to be fader times xn so when I look at this sort of difference of k powers I've got s copies of these so this expression is in absolute value smaller than s and the mean value which I have counts solutions of the sum from i equals 1 to s let's take one of the cases, let's take the case of c the number of solutions of this equation and you can see it's the same kind of game as we had before because we know about what happens when you look at these fader xn minus yi that allows me to infer something about the conjugate equation you might think of it as being a conjugate equation so we're hardly in choices for what goes on with the conjugate equation so this is absolutely smaller than fader s it's bounded just a bounded number of choices for what can happen on the right hand side here so that means that if we want to bound the number of solutions of a equation defined by the mean value what we're really doing is as we need to bound the number of solutions in the ring of integers of q join fader of the sum from i equals 1 to s and here this s is half of s0 of k by y of fader xi minus yi to the k minus fader xi minus yi to the k so back to this on the wrong side there equal to one of the fixed values which we should be choosing by trying to learn the quality of surprises to think about what happens when you represent zero now we know how to do this thanks to the recent work in vinaigretto's mean value theorem we know how to do this in number fields but that counts all of the solutions in the number fields and so what you need is for discrete restriction value into this because that nicely takes care of the feature of this problem that instead of having order x squared choices for xi and yi in a box of size x we'll just order x yi's closely linked to xi and when you make use of that discrete restriction variant which you can find in the paper on nested efficient congruencing it's part of the whole resolution of vinaigretto's mean value theorem then you get precisely the estimate that you need to save the correct amount in this problem so again one is able to obtain this variant of was lemon by reinterpreting everything in terms of the numbers of solutions of an actual polynomial equation but now thanks to the discrete restriction variant of vinaigretto's mean value theorem you get the correct bounds for this that can all be converted back into bounds for just the cave power equation and hence to the special polynomials here and it occurred to me maybe you're interested in is for are any of these polynomials in the slightest bit interesting well you know here's one example just to give a flavor of what kind of polynomials you can attack they are not arbitrary polynomials of this packet type but at least for some sort of vaguely interesting versions of these polynomials this is a degree for polynomial among ghetto resulting 17 variables for this polynomial from these sorts of ideas okay I'm out of time I'm well out of time so I can stop there taking questions or whatever I'm happy to answer questions and questions thanks for your attention everybody thanks very much Trevor so I figured out how to clap but we'll try here


Now a question here, how do I grow my capacity to have and receive a large audience and not be afraid of becoming more visible or well-known as my audience grows over time? Could you George share how your mind and heart have adapted to at each stage of your audience growth? I thought this is a wonderful question. And this is actually giving me a chance to talk about it because I've actually had some thoughts. I've actually wondered how can people don't ask about this because it is a real thing. And I feel like many people, and I appreciate the questioner because their sensitivity is one I wish more people had because as you well know, lots of influencers like on Instagram, like they get Insta famous almost overnight and many of them are quite young. And in my perspective, they probably haven't integrated how to lead well, essentially, right? Like how do you lead wisely because the bigger your platform grows? Now there's actually two issues here. I realized we need to talk about one is how do you lead wisely as your platform grows as people more and more people are looking to you and everything you do say or behave gives permission to your audience to do be say behave in that way. And so you have more and more responsibility as your audience grows. The old Spiderman quote, with great power comes great responsibility, right? There's that one aspect, the external leadership and then there's the other aspect which is the internal regulation of, oh my God, more and more people are seeing me now. How do I deal with all that attention? There's many, actually I would say many of you who are here live have more and better deeper tools than even me at dealing with that kind of emotional self-regulation, you might say mental, emotional, spiritual regulation of visibility, right? And I invite you, if you want to comment below, what is your, what's a tool that's useful for you to integrate more and more attention on the inside over time? And I guess I'll share, like I said, I'm probably less sophisticated than some of you in doing this. I kind of have a blunt instrument, you might say, and the blunt instrument is useful for me, for both my internal regulation and for my external leadership. And I guess maybe the blunt instrument is just a reflection on humility, maybe, which is, I would say generally speaking, it's kind of my spiritual practice or my spiritual growth. The practicality of it is my energy reboot. For those of you who don't know, you can Google energy reboot, you'll find my video and blog posts about that. That's the practical side of integrating my spiritual practice of humility and faith and therefore I'll just use those two words, like humility, like I don't mind being, so humility to me is like, to me humility and faith are very, very much connected for me, like humility for some people might be misconstrued or maybe misapplied so that they are, I don't wanna say too humble or they are, some people misconstrue humility as low self-esteem. I don't think that's what true humility is, but like low self-esteem is not helpful as you grow an audience, right? Because low self-esteem just means, oh, I wanna hide, because more and more people are seeing me and I'm embarrassed and I'm not worthy, right? That's the dark side of humility, you might say, so I'm not worthy to be seen by many, many people. I think the humility that I take on is connected to my faith that, to me, I'll just speak my worldview, it's gonna be different from yours, from some of yours, God used the word God, so use the universe or how yourself you wish to or consciousness.   And in other words, and that's part, one part of my faith, the other part of my faith is it's all gonna be good, like that's, I have deep, deep faith that it's all going to be good.  I genuinely believe that, I genuinely believe every single one of you and me, we're all headed towards bliss, it might not be in this lifetime. And of course, in every lifetime, we have some experiences of bliss and some experience of hope of bliss, like, oh my God, that's where I could, that's how I could, that's the place I can grow into, that's the kind of creativity I could have, that's the kind of experience I could, life and consciousness might allow me, like we're all going towards bliss. It's all gonna be amazing for all of us in my opinion. And so therefore, this life is just one blip in infinite lifetimes. And this 120, 80 years to 120 years that some of us live, it's just a tiny blink of an eye to what we're gonna experience together in our togetherness as souls. And so therefore, because of that, and because God is working through me and it's the whole of progress of becoming more perfect and becoming more wise, I don't mind making a fool of myself every day, you know, that comes back to the humility, like, I really don't mind.  You know, I don't necessarily use that language anymore, but that's still the way I think about it, is like, I don't mind if nobody likes me, not, and of course it's always a practice, we all mind if someone doesn't like us, we all mind, of course we do, right? We all mind if someone doesn't respect us or like us or, you know, want to give us attention and we all like it when people respect us and give us attention, that's normal human, you know, things we're working through. And yet the continual practices, that's true.   So anyway, that's how I see things in terms of integrating more and more of a bigger and bigger audience. Oh, good, more and more opportunity to practice being a fool in front of more people and maybe letting more of that light shine through and if any light shines through and blesses some people, my goodness, glory be to God, glory be to the divine because I'm a fool. And actually the funny thing is the more I'm a fool, the more glory there is to the divine. So that's why I don't mind being embarrassed when I need to be or telling people, you don't have to follow me, you can follow somebody else because it's all gonna be good at the end and I'm growing a lot and that's part of the process. So anyway, I hope this is helpful. It's just my perspective and I know all of us, like I said, I look forward to the comments below. How do you regulate?  And I look forward to seeing your method of doing that below or what works for you. Thanks.


Hi, everybody. Happy Thursday. I'm Joe for jaspyscasebreaks.com coming at you with 2021 Panini Falls basketball this time first off the line. We're doing both briefcases This should be pick your team number one a first off the line But it ended up being pick your team 11, but all you got to know this is first off the line if that triple the LeBron triple logo man pops Lakers heat calves, which is estimated. Yeah, maybe more than that We usually we do we definitely don't do majority rules at all. We never have but We typically just randomized just straight randomized those but that particular card will be an exception. We'll reach out to The calves owner that would be Gretchen the Lakers owner also Gretchen and and you Andrew Andrew So actually both parties are in the chat so Yeah, we'll talk to both of you and We'll see what what kind of agreement we can come up. Maybe both parties say hey, let's just let's just do let's just do a randomizer I don't know. We'll figure it out. We'll cross that route. It'd be a good problem to have First off the line should have these exclusive bronze diamond parallels number to three That should be pretty fun Andrew was also mentioning possible blockchain card in there We looked on group break checklist calm as well as Cardboard connection calm that's just an update list. We did not see blockchain on the checklist So if that pops we're gonna go by our no checklist rules, which is if it's a retired player They'll go to the team they played for the longest if they're active. They'll go to the team. They're currently on Make sure everyone knows it's first off the line 2020 2021 flawless basketball FOTL first off the line Very big thanks to everybody here. No filler Which I appreciate. Thank you. Thanks for digging deep Appreciate that Gretchen ended up with double last spot mojo. Okay, see Thunder and wizards got some extra layer security on here Right both briefcases as well looks like first off the line. He's there's the sticker right there the briefcase badge In red we'll use a couple a couple of these blanks right here to hide the encased cars Just kind of Try not to look at the names. I just want to see. Oh, is there one in the middle right here? Do we get three in first off the line? The auto is on the right side, but I think we have three here. Nice. I like that You can see on the top camera as well The birds eye view. Good luck everybody First one out of the gate is gonna be a manual quickly Momentous autograph two out of 15. It's a good start for Gretchen and the Knicks All right. Next card is draft gem four out of 25 Ruiha Chimura Scratching with the wizards last spot mojo strikes again Nice. Yeah, Andrew saying three and first off the line. That's nice. That's nice of them. All right 12 out of 25 Danny of Asia Another one for Gretchen and the wizards or wizard Gretchen. Let's break brought to you by Harry Potter on HBO Max Gryffindor, of course three out of five AI Alan Iverson It's a great patch. Nice auto AI with the braids Talk about practice Sixers Ken with Philadelphia Out of fives and under train whistle All aboard the big hit express whoop whoop horse Jersey number anything on the answer to Good eye Jonah. We got flawless finishes autograph Dominique Wilkins three out of 15 That's for Andrew and the Hawks. All right. Next is a four out of five Kendrick Nunn Would like to see more of Kendrick Nunn on my Lakers. You're playing the Clippers right now in TNT Four out of five Kendrick Nunn. This is still heat edition Andrew and All aboard the big hit express whoop whoop This was two out of five that kind of would have been sort of Jersey number Next up is theomaldon momentous ink 18 out of 25 for for Gretchen's other Last spot mojo team 70% of the time last spot mojo hits a hundred percent of the time That's what they say last one out of the slider box is a dual relic seven out of ten Dreymon green Warrior come out to pay It'll be for that scratch in as well go and say warriors All right gem. I think there was like a triple gem here and then the autograph on the far right You know Willie's thinking maybe remember the small princess. These are the gems are naturally mined not lab grown all right first First person here. See what we got Someone tell by his elbow a shirt underneath the jersey That is bam out of bio one out of three diamond So I think these these are the bronze ones that are exclusive to this set Looks really sharp the bronze on this flawless. That's classy I dig that Andrew that's for you All aboard Think this was the oh, I hope it is otherwise. I've overhyped it. Whoo three out of three there's the other bronze and It's four Wow flawless pics double diamond Pelican Timberwolves New Orleans that's for Paul Minnesota Gretchen Randomizer on that dual diamond flawless pics All right and the auto we've got some some background here on this picture we have enshrined signatures It's Charles Barkley six out of 15 Sixers that's for Ken and the Philadelphia 76ers All right, one down one more to go great start here great start. I'm sorry Rex Sir Charles Barkley can't forget the title hide the card again. I Think the auto is in the middle this time Right Gosh, I hope so I hope Zion plays at some point this season. All right, there's Blake Griffin two out of 15 That's for Brooklyn that'll be for Andrew. No, no worries. I'm glad there was no blockchain in here It's all good Emanuel quickly 11 at 15. I guess Penini kind of kind of dropped that blockchain thing kind of quickly. I thought they'd They'd lean into that a little bit more another Emanuel quickly for the Knicks. That's gonna be for Gretchen We got enshrined signatures Nate Archibald four out of five Celtics Gary with the C's all aboard the big hit express whoo-hoo on a one-of-one Sadiq Bay SA double diq Who's getting involved with the Detroit Pistons here? That's Demetrios with Motor City Nice one-of-one And all aboard the big hit express Three out of three Jamal Murray flawless finishes autograph You can see the Denver logo right there to another train whistle this time for Wayne and Denver Three and all aboard the big hit express Few more to go than the encased We got draft gem signatures Isaac a coral five out of ten Cleveland. This is for you Gretchen with the calves There is Larry bird three out of ten Team USA signatures, of course This is gonna be a Celtic Gary Gordon with Larry legend It's got some nice penmanship We got the winter Olympics happening in China right where's the next summer Olympics? Where's that happening? I know LA Los Angeles has one in 2028, but there's got to be one in between there. I want to say Anyway last one here Patrick Williams nice 15 out of 25 Another player we'd like to see a little I think he's banged up right another player We need to see a little more healthy Matt Lieber with the Chicago Bulls Maybe one to hang on to yeah, they're gonna have it there. That's a good spot. All right What do we have what if there's like no one on the front other areas? basketball maybe his hand Maybe some bronze maybe some Zion three out of three Zion Diamond Pelicans Paul Norden there you go Paul all aboard the big hit express whoop-whoop Think the on a graph is here, so Grab this over here. Yes, sir. We are pulling fire Ben Simmons. Where does Ben Simmons end up two out of three Ben Simmons for Ken? There you go Ken All aboard the big hit express whoop-whoop. All right and on this side in the center It's a 23 out of 25 Signature prime materials and it's Trey young three-color patch and autograph Trey young I think Jonah was saying that Trey young had a big game today So couple nice first off-the-line briefcases ladies and gentlemen Trey young beat the Sun's 124 115 he had 43 point six of 11 from three five assist and a steel Plus minus plus three. It's not bad. All right the case almost closed on this break We got to make one person happy and another person sad So there's a Zion one side Anthony Edwards on the other side dual diamond, so let's flip back over here to Random it out of work We got Zion so New Orleans pelicans on the left side Minnesota on the right side and a train whistle So let's roll it randomize It 11 times top team after 11 fingers crossed. Good luck one two three four five six seven eight nine Ten and 11th and final time after 11 Goes to the pelicans after five and a six eleven times Apologies to Gretchen Congrats to Paul Norton with the pelicans Dual diamond three out of three All aboard the Big Head Express There you go gang nothing else right yeah The case is officially closed on first off the line panini flawless basketball two box Pick your team number 11 is what we ended up designated as but it's the first off the line case Thanks everybody for joining for getting in and all that stuff We've done a ton of flawless over the last couple weeks or so, but um, I'm not sure this was our last case I'm not sure if we're getting any more, but keep an eye out on Jaspie's case breaks calm You never know when more will appear. I'm Joe. I'll see you next time for the next one. Bye. Bye


Hello, my name is Dave Brant. We farm in Fairfield County, 1250 acres of continuous no-till corn, soybeans and wheat. We also have 100% under cover each fall when harvest season is over, and behind is our new purchase of air seeder, and what we do with it, we go in standing corn and soybeans and sell our cover crops about 30 days before harvest. That way we have a 30-day window growth before we take the crop off, and when we take the crop off, our cover crops come right up and are going. I think it's a wise investment. We've tried to use an airplane to do aerial seedings, and have not quite been as successful as I'd like to see it. We're working real diligently with blends of 6, 8 and 10-way blends. Directly behind me is a 10-way blend, and what we're doing here with this is we're using it to loosen the soils, to build organic matter. We're presently looking at about 22,000 pounds of biomass, which will increase our organic matter here in this plot about three-quarters of 1%. The reason we're using so many is we'd like to have what I call a community working together. We have all the plants doing their thing, we have the tall species that tend to terminate themselves when the air temperature is 34 degrees, and then the smaller plants, like the winter peas and the reddish, and the crimson clover and the hairy vetch, will grow clear in and sometimes survive all winter long. This way we get lots of diversity, and we have the soil covered about 360 days a year. On this farm, we lose less than 100 pounds of soil per year because of our cover crops. That's been a benefit to us.


Are you sitting comfortably? Why not? Hello, and welcome to Doctor Who Tales. And today we will be reading...  Hey!  Hey! I'm trying! Hey! You! No!   And today we will be reading Doctor Who and the Doctor Who. Let's begin. Once upon a time Once upon a time, Lord The Doctor... No. Not that one. I don't want to go! Not that one. No. No. They found themselves in the home of the Daniels family of Southampton. And judging by appearances they are in the distant earth year of 2019. The Doctor did not steer them here. The TARDIS had taken them both by force and the Doctor had no choice but to go along with it. He didn't mind though because as we all know The Doctor is a cuck. Looking around this house he admired the bland white girl aesthetic the white and oak wood furniture the photos of loved ones on the wall none of which looked familiar at all. But most strange of all not a single person but a sight. This is strange.  Where are all the people in these photographs? Did something bad happen to them? An odd thing to be your first thought but in the Doctor's life it's usually the case. Red flag much. He looked around to the companion of this tale to realise he locked them accidentally in the TARDIS so after letting them out he asked again. Where are all these people from the photographs? The companion. No. No. Why? Dead. Dead. Fine. The companion answered. In the photos of course look I can see them right there.  But they were smart in other ways.  Doctor! Look! The door handle it's doing that thing the door handles do! It's...it's... Turning! Said the companion in a blind panic they then fainted for no clear reason. Cliffhanger. Credits. One week goes by. Titles. The Doctor gave the companion a small kick and they were back to normal. Cliffhanger solved! The focus was now back on the door handle which had been slowly turning this entire time. The Doctor got out the trusty tool the Sonic screwdriver and locked the door. The being on the other side of the door then unlocked it again with a key. With the Doctor locking it again with his Sonic screwdriver. This goes on for an excruciating amount of time before the Doctor says his classic catchphrase.  The door opens to reveal the man from the photos who looks more confused and shocked than Jackie Tyler being rejected at a swingers party.  Said the mysterious man. And the Doctor. Said the Doctor. And this is my friend slash companion slash object. Oh! Said the companion slash friend slash object It's so well that the Doctor would kick them into a supernova if corrected or challenged. What are you doing in my house? Asked the stranger and oh no! That's all the time we have for today's story. I suppose it's time to go into the forest of the night garden. But I will give you one small hint as to where this story will end. It's all based on a true story. Good night.