A Direct Proof of the Riemann Hypothesis - Part 3: Prospecting for a Proof of the Ri emann Hypothesis

Good evening, mathematicians and lay math enthusiasts, that's a term. I am so excited about this video. I have been waiting to put this video out for about a year and a half. At least the information in it had to go through a long process in writing the paper, which is entitled The Direct Proof of the Riemann Hypothesis, with co-authors Greg Bulk and Dennis P. Allen. But this is the information I've been dying to get out, because none of it really is in the paper, only a little bit at the very end. But this is how I made the discovery that I presented to my co-authors, which led to this paper. And this, I think, is the most interesting, and for various reasons we took all of it out. And I do think that's the right decision, but I'd love to discuss it here. So let's get into it. If you watched the other two videos, you found where to find the preprint of this paper at researchgate.net. I did put a link in the description below to this video, and I went back and put it in the other two videos as well. Or you can look it up on Google. As you know from the last two videos, I am a big fan of dark humor, especially in times like these today. If you need a good laugh, and you want to just kind of walk away, this paper was written back in the early 90s, so it'll bring back maybe some memories for those of you who were older at that time, or younger at that time. Everyone was younger at that time, unless you were older at that time. So those of you my age and older who remember the early 90s and the world of just coming out with pagers and cell phones, you might get a kick out of this. Eh, give it a try. It does have a couple good reviews. Grady Hart is a top 50 reviewer at Amazon. He wrote that, Jeff Cook pounces onto the literary scene with a fascinating first novel that is one of the more interesting concepts in storytelling to grace bookshelves in some time. This is my only not fiction novel, but I'm a big fan of literary fiction. Also got a really good review from Daniel Jolly. Gave me five stars. He wrote, Jeff Cook's first novel really stands out from the crowd in terms of both originality and excellent, insightful writing. I am working on another book. I've got about one chapter left. This is nonfiction. It's a book on physics, maybe a little bit, maybe natural philosophy would be the better way to describe it. It's a study on all of the findings of light throughout history. So Einstein, string theory, Faraday, Maxwell, and it's not saying that any of those had it nailed down. And you'll see when it comes out. Well, one more chapter to write. OK, let's get into what I think is going to be the most exciting of these videos, at least for me. Now, it's going to be overwhelming if you get through this. If you can get through this, you're going to the paper is a breeze. The paper is a breeze anyway. And so this one, there's some areas. I mean, it goes deep. Obviously, I've been working on the Riemann hypothesis for about 17 years. There's obviously a lot I turned up, and a lot of that doesn't necessarily apply directly to the proof. But let's just start getting into it and show you where we're going. I did have an analogy running in the last two videos comparing prospecting for golds and the zeros, which you saw in the last video. Riemann related the roots of the Riemann zeta function to the prime numbers. And we're going to continue this further. So if Euler was the prospector, and he went and he kind of, you know, he poured out his sieve, all right? He got us some gold nuggets. He found the mine. I would say that Riemann is the one that took dynamite to the mountain and threw all the gold ore and nuggets all over the place. This is the refinement of the ore, okay? And you'll see why I call that analogy, I take that analogy as we go forward. All right. How to prove the Riemann hypothesis is what I brought up in the last video. All right. So there's probably many ways to do it. But this is the way I got to it and succeeded, I believe, very much with the help of my co-authors. But this is what I discovered. All right. We're going to step back. We're going to go way back to the, well, to Riemann's xi function. And he defined it over T and demanded that the real part of S equals 1 half. And we did kind of roll up with this. And let me just see something real quick. All right. And he said T can be real, complex, anything, quaternion, anything you want. He didn't say that, but it can be any value, any type of number. But his hypothesis, Riemann hypothesis, is that T is real for all of the zeros of xi of T. Okay? So we're going to kind of take this and use some property of it, and we're going to apply it a different way. All right. But first, let's kind of follow the path back. We're going way back before he ever got to his final result of the prime counting function. Okay. We're going to define it over all S. We're going to ignore the fact that he demanded that we set the real part to 1 half. All right. We're going to solve back for zeta. And there you see in the numerator pi to the power of S over 2, which I mentioned in the first video. And this is exactly why, and you're going to see further, exactly why you get all of those positive real values of pi to the power of S in the numerator. So the values of zeta function. So let's keep moving because we have a lot to cover. All right. We're going to set this equal to the abel plana form of zeta function. Now, as I mentioned before, mathematicians have exited the xi function state left. Well, we're going to do that to the zeta function instead and see where it takes us, if anywhere. Okay. So they're equal, and we are going to, I like the integral, but we're going to give that third term a name because it's a little clunky. I just want to get out of the way, treat it, let it be a function in its own right. All right. We're going to call it gamma of S for its very striking value at S equals 1, which is gamma minus one half. Gamma is the Euler-Mascaroni constant, which you can get by integrating e to the power of minus t times the natural logarithm of t. All right. Good. Now, let's just do some prospecting really quick. We're going to kind of got to go look at the different or here. In the left column, I have some integer arguments for S. I'm going to start minus three, just kind of looking at the basic flow of things here. The second column, we have zeta of S, and it looks like, well, it's got this nice rational value out for the negative and zero integer arguments. And we see that the negative even arguments are zero. We only have minus two on here, but as we keep going, we would see. And we get a big old infinity at one, and we get all the even positive integer arguments. We get this pi to the power of S in the numerator. So it's times some rational number. And for the odds, we don't know a whole lot. Now, let's take a look over at xi. All right. What do we see? You're not going to see it necessarily here on this page, but you can trust me or look into it yourself. It does share all of the trivial and the non-trivial zeros. And more importantly, and more interesting to me, is that it has all these pi's to some power in the numerator for the negatives and drops out to pi to the power of zero at the S equals zero. And then it starts putting them in the denominator for all positive arguments. So that the relationship between the zeta function and xi function is simply pi to some power times some rational number. Okay. And so, good. All right. It's just an observation. Gamma of S. That was the integral from the Abel-Plana form of zeta function. And we see that it's just a difference. Of course, we expect it to be. It's just a difference of the zeta function. And with the exception of S equals one, which is that nice little gamma. It's our single exception where it's not a difference of the zeta function. And that gamma of S is the only zero that I found. You can go and hammer it. It's not of too much importance for me. But it's the only one I could find. It definitely doesn't share any of the trivial or non-trivial zeros. And so I couldn't find any zeros in there except for S equals zero. All right. So does this tell us anything? Not really. Does it lead us to the Riemann hypothesis? Not really. But it does kind of illustrate the problem of why it may be difficult. All right. So we have this. For the xi column, we have a purely multiplicative factor of the zeta function. And for the gamma, we have a difference. We want, instead of either of those two, something between the two of them. We want a new function that's going to be a rational factor of the zeta function. And we have a couple other. See, we don't have the resources here. We don't have the resources to move forward. We need a new resource that's going to be a new function. And it has to fulfill a lot of restrictions. All right. So we're not going to create one. So we don't really create mathematics. But we do begin with the end in mind. So we have to know what we're looking for to have a desired result. And then we can go looking for it. And when it arises, we say that's it. That's how you go prospecting for gold. You've got to know what it looks like before you go looking for it. And so that's what this approach is. We're going to begin with the end in mind. We're going to look at what is the solution we want. Well, if we had the perfect ideal function that might be some algebraic factor of the Riemann zeta function, and it had a couple other neat little requirements, then it makes proving the Riemann hypothesis fairly easy, as we'll see. All right. So the alternate form of this zeta function is going to involve this function, this new function in one of the terms or some factor, as we'll see. So what do we want from this alternate form of the zeta function? Well, we want it to consist of just three terms. Why? Well, because there are only two types of zeros for a plus b plus c equals zero. There's only two types. The first one is where one of the terms are equal to zero, and the other two terms negate each other. So in this case, we have b equals zero, and then a equals minus c. This has a geometric representation of a line. Now, if you take that line, and I'm looking at it and going, that's a really simple solution, and all the solutions would lie somewhere on that line. Remind you anything? What if we were to rotate it and translate it over to the critical line, and this type one solution contained all of the critical zeros? What if we could isolate the zeros into two different groups? All right? We want two different groups. Why two different groups? Well, the Riemann hypothesis suggests that there are only two types of zeros for the Riemann zeta function. There's the critical zeros and the trivial zeros. Critical going straight up and down on the critical line, and the trivial going horizontally to the left, right? So we need a function. Well, ideally, it'd be great to have a function that had only two types of zeros. We could say there are this type of zeros and this type of zeros, and all of the zeros fit into this, a line, and the other fit into the next type, which is where none of the terms are equal to zero, but two terms negate the third. This has a geometric representation of a plane. Now count the number of points with me, please. One, two, three, four, five, six. Six points, six sides. It is a hexagon, right? Not too unlike the zeros of the Riemann zeta function. Count them with me. One, two, three, four, five, six. Okay, they're grouped in this form. Yes, well, we discussed that all of the non-trivial zeros off the critical line are going to get the vertices of rectangles that are symmetric across the critical line and symmetric across the real line, and they're going to be internally tangent to the circle, which is right there centered on the real point one-half. And we also mentioned in the other videos that these zeros are reflections of each other through that point. So we draw a line from this one over to here. They're going to go right through the point one-half, and this one here is going to go right through this point here. But we're also going to have critical zeros above and below these four. So it's six points. Whether or not they're all related, you might only get a solution, a planar solution, where you get this one and this one, or this one and this one. But that's okay because if we get any one of these, we get four for the price of one. And now, whether or not, at this point, we don't know if the critical zeros are going to be— if we can get the critical zeros from these four vertices, or if we can get these four vertices from the critical zero. We don't know that yet, except that in the ideal perfect expression of the zeta function, we're going to have all the critical zeros on a line. So we might double up and have some in the planar solution as well. All right, let's go before we get into anything else. Now, wouldn't it be nice to be able to rotate this planar solution, a plus b plus c equals zero, to the complex plane lined up on the critical line. That would be nice. We can do it. All right. Now, hold on one second. Okay. All right, so how are we going to do it? All right. You have to have a little background in tangent half-angle substitution. All right? Now, this presentation— I made a mistake. I should have put them before I did this. All right, but we're going to do this, and then I'm going to have the description of the tangent half-angle substitution below it. Okay, so I'm going to go through a lot of this, because it's very quick, and this is in the paper. It's the very first part of the proof. This is what we do. In the paper, I'm not describing why we're doing this. I will show some of it, though, as we go here. Okay, first, begin with Riemann's Functional Equation. Now, remember, these are five different functions, all multiplicative factors of each other, okay? If you're not familiar with it, go back to the first and second videos and check it out. All right, and if, as I discussed by analogy, this is ore, gold ore, and we don't see any gold in it, but we have to refine it in order to bring out the gold. All right, the gold are the primes and zeros, or one or the other, because they both connect to each other, all right, by Riemann's final result in his paper on the number of primes less than a given magnitude. Let's begin breaking away, or I should say, we're going to perform a chemical reaction now. It's a better analogy. Okay, let's do it. Multiply both sides by S minus one to the power of three. Watch the left-hand side as we're going. I could reduce the right-hand side, but we're not going to do that this way, because we're going to do something else with it. All right, subtract both sides by one, multiply by I, add minus two times the imaginary part of S to both sides. Multiply both sides by sine of argument of S. This is where we're going to kind of bring up some of the tangent half-angle substitution here. All right, when we're doing this, we get the absolute value of S, the denominator and imaginary part of the numerator. Not a big deal. As we keep refining, you're going to get rid of all of that and leave behind just the gold. Now we got a big old rock here. We're going to add two times the imaginary part of S times, in parentheses, two times the imaginary part of S plus I times cosine of argument of S. Both sides, okay? And we're going to divide by sine of argument of S times the complex conjugate of S, and that reduces the absolute value and the denominator and the imaginary part. All right, this is going somewhere. You're going to see in about two seconds. Last step, divide both sides by two times S minus one and give that right-hand side a name. This is our new function with all of the properties. You don't see it yet, but you will soon. All right? Solve for the zeta function. Expand, and there it is. This representation or form of zeta function is going to tell us everything about the zeros of zeta function. Now, we do have a little bit of a problem here with the S equals zero. Okay, so when S equals zero, it is not going to equal the zeta function. Not a big deal. We have to work around to get a value for every single argument that's going to be equal to the zeta function. At this point, there's only one argument that doesn't equal the zeta function. But we're going to clean up this here a little bit. I like the I there in complex conjugates, so because I times complex conjugate of S squared is equal to minus complex conjugate of S squared, we can just kind of call them the same thing, but we have the squared in the term here. We're going to give that a name because everyone's going to ask, what is is? And I don't want anyone to ask that, so we're simply going to call it omega of S. And a little bit, we just got to remember that it's simply I times complex conjugate of S. Very simple. Just to get it out of the way, clean up a little bit. All right. Now, as you notice in each of these terms, we have something really interesting going on. You may not see it at first, but we talked about the analytic continuation of geometric series in the first video, and here we have some other geometric series as analytic continuation of geometric series as well. They each can be described as a geometric series when the absolute value of S is less than one. Okay, so how are they connected? Well, first let's give them some names just to show what we're doing, and we're going to come back and we're going to see how they're connected. All right, because it is kind of interesting. All right, we're giving names. What do you think? A, B, and C, because we're looking for the form of zeta function where A plus B plus C plus zero rotated the bit, which these coefficients will be doing, and this function, upsilon, will do the rest. Now, line it right up to the critical line. All right. Now, let's talk about upsilon. I know a lot about upsilon now, and this was, I went into every corner of this function as best I could. Okay, there's, I still don't have an Euler product for it, but I can do, I can express it as functional equation, analytic continuation, or the Abel-Plana form, infinite series for when the real part of S is greater than one. I can do a lot with upsilon. We're going to, I'm going to simplify some of this to get just the Abel-Plana continuation of it, so we could put any argument S in, and we're going to get value for upsilon, okay? Rather than that big, large, remember that quality back, let me go show you again, how big it was getting. And so, you don't need to know a lot about upsilon for the proof. You don't need to know anything about it, actually, except its relationship to the zeta function. But you remember that everything to the right of that equation, I mean, that reduces, it's just, we see where the functional equation's going to come out, but, you know, that's ridiculous. We're not going to work with that. So we're going to do something else. All right, here we go. All right, let's get back to my slide. All right, so, first we're going to start by getting rid of this little problem of S equaling zero, and then it's no longer equal to the zeta function. So, like we absorbed all of the irrational properties into upsilon, irrational properties of the functional equation, Riemann's functional equation, we're going to do the same thing with this little guy, one last little bugger to get in there, and we're just going to simply call him upsilon 2 of S, just for the time being. Okay, so we don't need to worry about him going to zero. All right, interestingly enough, when we do that, there is a cancellation in there. So let's go ahead now. Let's show where we're going. We're going to now think of the infinite series. Now, I'm not going to put it on the screen because it'll take up the entire screen with infinite series, okay? So you mathematicians, you are going to be able to envision this, and lay mathematicians, you will thank me for not bringing this all up here. All right, so this is what we do. Since the zeta function can be thought of as an infinite series when the real part, you know, it converges for the real part, that's greater than one, and the geometric series, which, you know, not hard to look up, or you can just go to Wolfram Alpha and type this in, and it'll tell you what the infinite series is for it in that region. But the region where it converges is where the absolute value of S is less than one. So all we need to do to make them equal in terms of infinite series, we're going to make this give epsilon of S an infinite series so that it can apply to the Abel-Plana formula. All right, so, like I said, I'm not going to show all the infinite series, but to do that, we're going to use the alternating zeta function, which also has its infinite series. And this converges whenever the real part of S is greater than zero. So that gets us now, has some equality, and dividing that by one minus two to the power of one minus S, which you might recognize as a slight variation of the Euler product, the function governing it, anyway, we then get the equality in the critical step. All right? And so we can have, we can then express, rearrange the infinite series. So we're going to take all, imagine this is an infinite series now. Everything in here is an infinite series with the exception of the multiplicative factor, you know, of two, and I times the complex factor of S squared, but that's just multiples. We're going to solve for epsilon two of S, and we're going to apply it to the Abel-Plana formula. Okay, so we take all the series, we apply it to the Abel-Plana formula, and this is what we get. So now we have, for any argument S, we get a value output for epsilon two of S. And since it is simply a multiplicative factor, we can actually go ahead and do the same thing with epsilon, which is the function used in the paper. Epsilon two of S is not used in the paper. It's not at all necessary for proof. Just show a little background. Like I said, this is the area I've been wanting to share because this is where I think all of the beauty leads up to proof. Okay, so you see, we just kind of then build this little relationship here, and we can get the analytic continuation of epsilon of S. So, good. And again, we cannot have zero. S cannot be zero. You see, still. Epsilon of S cannot be zero. Epsilon two of S can. All right, S. There's, yeah, okay. I'm sorry, it's unscripted. All right, good. And it's getting late, but I'm having fun. Okay, now, observe. All right, back to this. What do we got? So, how are they related? I'll show you how these geometric series are related to each other. And this is going to bring out our gold refiner, I guess you could say. All right, they're all connected by this, what you could call a regular expression. All right, so if we could think of this function as a finite state machine, and finite meaning there's only three terms, and we can apply this to each one of these. And this is how we apply it. Not terribly important to see, but this is, we're going to expand it out so you can see. You have, for every N, you're going to get, you know, this little sequence here to the right. And if you take the first one and integrate it, you get the first term, right? And we're going to integrate the second one, and we get the second term, and we integrate the next one, and we get to the third. Now, why is that important? Because it says that they are all derivatives of C. You see, which is this term over here. All right, neat. Electrical engineers, physicists, step in when you see it. This is in the form of a driven harmonic oscillator, okay? At an infinitesimal T. So we can say that the zeta function can be thought of as a driven harmonic oscillator at an infinitesimal T. It is a thermodynamic system. Now, this is really neat. Check this out. Well, it gets really kind of, eh, that's pretty intense. You're going to do fine. All right, here we go. Let's just kind of take a little side note into electrical engineering physics, okay? All right, all this was taken out of the paper. We went, I and my co-authors went back and forth. Greg and Dennis both are very big fans of physics and electrical engineering, but Danny, he's more, he understands number theory. He taught it at university, and he studied it, and Greg, you know, studied electrical engineering and these aspects. So we're a bit torn, like, do we include this? I mean, this is where the real, you know, interesting, the luster of the paper is, is in the physics behind this. We decided, and I do believe Danny was right, that the answer is no. It all needs to be taken out. We don't want to have people focusing on a number, a simple algebraic proof and get distracted by the physics, okay? That's beneath it. But if you have interest in that, which I do, so let's take a look at it. So actually, because this is what made the proof so easy to come to, all right? You have the Riemann zeta function in the form of driven harmonic oscillator, all right? Then you could do all sorts of, you can know everything about it, all right? One, driven harmonic oscillators only have zeros, two types of zeros. They have, if the driven harmonic oscillator, let's go back to the, if F is a force, then T is representative of time and M is representative of mass and this would be the omega naught, which is the fundamental frequency of the system. I mean, you got all this is in there, all right? Now, what you can do with oscillators, oh, by the way, the zeros, yeah. So when F equals zero, the force is zero, there's only two conditions. It is when omega or not omega, epsilon is equal to zero, which in form of driven harmonic oscillator would be the damping ratio and when the damping ratio is zero, that's one time you're gonna get F equals zero and the other is when the two terms negate each other or the two terms negate the third. So now we can do something really neat and this is how it was reduced to a geometric, the last step of the proof is a geometric, a simple geometric result. All right, none of this is in the paper, but this is how it was done. So you can, harmonic oscillators are often graphed in phase space using trigonometry as harmonic motion is identical to continuous circular motion. There are some really neat videos on YouTube if you're interested in this, okay? And you can see a much better description of all these areas. This is new, it's very interesting. So we can have, we can do a number of things here. My thought is we like to plot it in a complex plane, all right? Instead of velocity, we're gonna make other things instead of orbit, we're gonna have time or T, not time because we're not gonna deal with force, just gonna call it T, are gonna be points on this, on a circle and they're gonna rotate around and we're gonna get points on the zeta function, it's really neat, okay? And so position, we're not gonna treat it as position, not gonna treat it as velocity, we're gonna treat it as the real line and the imaginary line and everything else can fall into place, okay? We're gonna begin and I'm not gonna go through all the steps because I wanna save this for another video, but we're gonna use R.J. Walker, begin with R.J. Walker's geometric proof of the tangent half angle formula on the unit circle. So our way of refining this goal is by using the tangent half angle substitution and I forget the mathematician, oh, it's terrible because he's still alive and he may watch this someday. The mathematician name who came up with it, this expression I'm gonna tell you, I'll find it, I'll put the link. And he said that the tangent half angle substitution is the sneakiest trick in the literature and it really is. It's very, very beautiful, sneaky tool and all mathematicians should learn it right now. All right, so with this, you could say that here's the proof and this was done by R.J. Walker. He was editor of a journal and he just released this as an image. So it's a proof without words. So let's see, cosine of this angle right here is equal to one minus u squared over one plus u squared. Okay, so what I did, and I had these steps in the paper originally, is we're going to generalize this tangent half angle formula and so it's not on the unit circle, but we could do it for AdS, all right? So for unit circle, meaning the radius is equal to one. Okay, good, all right. So and then we're going to plot Riemann's data function onto it, all right? So tangent half angle formula. Now take a look, if you remember, we had a t squared plus one or minus squared in Riemann's final result and then I said this has some trigonometric aspects to it. So here's how we got it and also if you remember in the first, no, it was the last video, I was discussing how to apply a function f of n to the Abel-Plana formula, all right? Here's what it is. That is the zeta function and this is the result that I got from the tangent half angle substitution applied to Riemann's zeta function. Now we can go ahead and make it two other, that's how all the, you saw the multiplication when I was doing those trigonometric and algebraic manipulations to Riemann's zeta function. That's what all of that was doing, okay? So this is a very cumbersome diagram, okay? But let's kind of, I'll just kind of show you what happened. Basically now for any complex S argument, all right, where are we at? All right, any complex argument S, we plot its point right here. We can get the Riemann oscillator, which is, remember the alternating zeta function, right, you get its points kind of taking the spiral, okay? So how this works, or how you find the zeta function from this. So you remember how the alternating zeta function is simply a multiplicative factor of the Riemann zeta function, right? And at infinitesimal t, you have the, you end up getting the Riemann zeta function. But in this, because I converted in this right here, I converted all of the real u's to n's, and of course there's some changes signs and so on and so forth, and it's inverted. But now you can get the sum of these integer points and they end up all summing up to the alternating zeta function of which you can get the Riemann zeta function out of. All right, so what do I mean? All right, so if we're going to do just the integer arguments of t, okay? And we have a given s, so this is going to be fixed, s is going to be fixed, the argument of s is going to be fixed. The only thing that is moving and changing is this alternating zeta function. It's actually the minus, the negative of it. So you could be looking at this whole image, and I'm not even going to explain it because it's just too complicated. I wish I had a 3D view of it, I can show you. All right, anyway, so it's the minus. All right, take one at t equals one. Let's just say t, switch it out, t for n. t equals one, t equals two, t equals three, t equals four, t equals five, and then you're going to go all the way down to t equals infinity, you take the sum of those, and you get the values for the alternating zeta function. Isn't that neat? For me, I think it is neat, but it's obviously a little bit of a click in your head around. Now, time, I had mentioned we're going to have time and space. All right, so time would be points for every t. We're going to get a different point, and it's going to go around this circle. So time is going to have some values, and that circle has radius one divided by the imaginary part of s. This geometric is related to it by that. And this is tricky, but it's very interesting. If you know anything about spirals, there's a, what is it called, generalization of all spirals. They can all be determined by a simple power. And in this case, the distance to the points, which are the values at the alternating zeta function, all have the radius of t, which are these points on the circle. t to the power of minus three, a part of s. So that governs the radius. And of course, then you can find the angle tangent to that radius with this function. You should kind of have to subtract it by pi. Anyway, whoa, okay. So you didn't know this is all related geometrically. You remember the u, all right? So u for R.J. Walker's, all right? So all of that is the same. There's a couple of different things. The difference is, is the radius of this is now determined by the inverse of the imaginary part of s. So this, well, actually this was one, but this, everything kind of is, engineers say it's constrained to that argument. So the only thing that changed, you have two points, two freedoms of, degrees of freedom. You have u and the imaginary part, or the one divided by the imaginary part of s. Too much to go through. I'm gonna do another video and I'm gonna break it all down one at a time, it'll be much easier. But that is how you get the oscillator and see that actually is a functioning oscillator. Good, now we're going to find the zeros with it. All right, check for proper rotation and translation. We are so close to the proof. You're gonna see it right here. Check for proper rotation and translation. Set our new form of the zeta function equal to zero. So this could be for any zero of the zeta function, trivial, non-trivial, it doesn't matter. And we're gonna expand it out so we can see all the s's. And we see that this is a true statement right here. Whenever epsilon of s equals zero, real part of s equals one half. Now this is a similar result as Riemann had with his xi function, but I didn't break his down. I am gonna break this down and show you how it comes about. Remember that the imaginary part of t only equals zero when the real part equals one half. All right, that's the whole Riemann hypothesis. This is one part of it. This is just going to show us something. How do we know that this is the case here? Well, we're gonna set the second term, epsilon, in the second term to equal to zero. Riemann's zeta function is already equal. And epsilon now is zero. And you see that second term just goes away. And this very manageable sum here equals the right-hand side of this. And we said this right-hand side equals the right-hand side of this. So in the numerator, i times one, in the left parenthesis, i times one minus two times i times the imaginary part of s cannot equal zero for any s, right? Doesn't make any sense by the rules of complex arithmetic. So it must be in the second parenthesis. Set that equal to zero, add one to both sides, divide by two, and we see that the real part equals one half. So therefore we have our important first result, that. Type I solution contains only critical zeros. We have effectively isolated all the critical zeros into the type I. So we were able to rotate the entire, remember that A plus B plus C equals zero. We were able to rotate that to the critical line. And that is our first result. Now, there's only one other type of zero. We are very, very close to the proof. And the other type of zero is not when B goes to zero, but when B does not equal zero. And if you remember, we had epsilon in the second term. So epsilon cannot go equal to zero. And two terms negate the third. And this is, has the geometric representation of a plane. You remember this from earlier. And we're going to check for rotation and translation for this, all right? So here we go. Epsilon of s cannot be equal to zero. Solve for the real part of s. Now you can apply this, you can apply this equation in Wolfram Alpha, wolframalpha.com. And it handles it very well. Derive five, we don't work with that. You can apply it to most calculators. And since they're free, you'll get them. All right, solve for the real part of s. And you will get, now you see why epsilon of s cannot be zero in the planar solution, because it is in the denominator and they'll be undefined, right? So this is for whenever epsilon of s does not equal zero. Now, you remember we had the able planar continuation, all right? So let's go and grab some trivial zeros just to see. We don't know of any non-trivial zeros off the critical lines. We can't test that. So let's go ahead and simply just see if it's aligned for the trivial zeros. And of course, you know, pretty much assume it will be. We don't want to assume anything. We assume it will be because the critical zeros all lined up. All right, but let's just check the trivial and make sure that we have this all perfectly aligned. And of course, it should be a Zeta function, but let's just do it. Just broadening our minds with it. All right, apply the first non-trivial zero to the able planar continuation, and it reduces to minus i times five over 12, and apply that to type two solution. Now, we do not need to assume that the imaginary part of s equals zero for both, because maybe we'll find another non-trivial zero off the critical line with this, right? So we don't want to assume that the imaginary part equals s. We just want to apply Upsilon of s, that value in there, and let's do the positive solution first, and we get minus two. Oh, well, we already knew that because we got Upsilon from minus two. What's the other one? The negative solution. All right, let's do it. And no, real part equals three-fifths. Cool. Does that mean we have a non-trivial zero off the critical line with real part three-fifths? No, because the imaginary parts cancel out. You see, we didn't use them, and it just gave us minus two. They cancel out. So it's saying that we have a real three-fifths, which there are no positive real zeros of zeta function. Therefore, the negative solution is extraneous in this case. All right. Now, everything's lined up. We have the resources to prove the Riemann hypothesis. All we need to do at this point is solve this little geometric problem. If the zeta function equals zero, and we can prove this line here from here to the critical line and this line from here to the critical zero cannot be equal by the rules of mathematics, cannot be equal, it's impossible, then we have proved the Riemann hypothesis. Or we could do it a couple different ways. Or we could say, if the zeta function equals zero, and we can prove that the line from here to the real point one-half and the line from here to the real point one-half, the distance of that line, the length of it, cannot be equal, it's mathematically impossible for them to be equal, we have reproven the Riemann hypothesis. Or we could do the third way, which is the way, a third way, there's probably more. A third way, which was brought to the table by Denny, one of his many contributions, and that is that we're going to, it has the fewer equations you have to go to get it proved. Each one of these three proofs is very simple at this point. This one has the fewest number of equations, so we like that. And that is where, if you look at it, these are symmetric across the critical line, and they're within the critical strip, then the distance from here, zero to here, and zero to here, must add up to one, right? Because the distance from here to here, and the distance from here to here, must sum to one. If we can prove that it's mathematically impossible when the zeta function equals zero, we have proven the Riemann hypothesis. Why? Because we have proven already that there are only two types of zeros, the first type only are critical zeros, they only can have real part one-half, the rules are complex arithmetic, and all of the other zeros must occur when epsilon doesn't equal zero, and we solve for the quadratic equation, and we got this, right? There's only two solutions. All we need to use is this equation right here to prove that geometry is possible. And we'll do that in the next video. Thank you so much for watching this, this is long. I hope I'll see you in the next one, probably gonna be uploaded in about a week or so. Like I said, you got through this one, going through over the paper is going to be the next one. I'm just gonna go over the proof line by line, so much easier. You have all the information, if you watch all three of these videos, you have all the information to get you through the paper and understand everything. And we will do that and summarize in the next one. Thank you so much. I do have a Patreon account, if you do like my research and you wanna support me, even if you don't support this proof, if you'd like to see research coming out and more videos, you can do that. And I really appreciate that. All right, thank you, have a good night.