A Direct Proof of the Riemann Hypothesis - Part 2: What is the Riemann Hypothesis?

Good evening, mathematicians and lay mathematical enthusiasts. Good to have you joining me in this second video. If you haven't watched the first, you might want to go find Direct Proof of Riemann Hypothesis Part 1. Okay? If you've already watched that, then you know what this one is going to be about. All right. Again, my name is Jeff Cook, and about a year and a half ago, I approached Greg Bulk and Dennis Allen with some discovery I made, and this paper that is in preprint now is the result of a year and a half of work, of hammering on it, banging away at it. Have another mathematician look at it. It is entitled A Direct Proof of the Riemann Hypothesis. Okay. You can find this preprint by going to Google, do a search for ResearchGate Cook Riemann Hypothesis, and it should be the first link that comes up. You should be able to get the PDF from there. You do not need to be a member of ResearchGate to download it. It is public. If you can't find it on Google, just kind of go to ResearchGate.net and do a search by my name, Jeffrey Cook, or Dennis Allen. Greg Allen is not there. If you want a little bit more background, go watch the first video, Background on Us, or you can check out some of my and Dennis' research there if it is public. I am not sure if all of this is public, but I do believe his profile is. Okay. Good. All right. Before we get in, as you know from my other videos, I am a big fan of dark humor, like Joseph Heller, Catch-22, all of this. I really like looking at the world this way because, you know, when something is not allowed to be laughed at, it just makes it all the funnier. Okay. So, if you're interested in that as well, I am not only into math. I think the world doesn't need any more stuffy mathematicians. We have computers for that. If you want a good laugh, please do check out Woe to the Hunted, the second edition. It is available at Amazon and pretty much everywhere else. Okay. It does have some good reviews from top reviewers at Amazon. So, Grady Harp is a Hall of Fame top 50 reviewer. He writes, Jeff Cook pounces on the literary scene with a fascinating first novel that is one of the more interesting concepts in storytelling to grace bookshelves in some time. It is not only my first novel, it is my only novel. So, I am mainly into nonfiction, but I do love literary fiction, particularly dark humor. Daniel Jolly, super nice guy, and Hall of Fame, Vine Voice gave five stars, and he writes Jeff Cook's first novel. Really stands out from the crowd in terms of both originality and excellent, insightful writing. Thank you. Alright. I do have another book. It is nonfiction. It covers, it's called In Light of What We Are. It covers about everything kind of from ancient concept of light to today and even covers a little bit on Einstein and his concept of matter being the condensation of electromagnetic fields. Also, a lot on Faraday, Maxwell, and Euler. Also, who I spoke about in the last video. A lot about Euler as well. As you know, I love Euler. Anyway, cool. Alright, part two. The Riemann Hypothesis. In the last video, we were talking about prospecting for gold, particularly in mathematics. Okay, so, the sieve of rap, and if you remember from the video, I'm glad this is the last time I have to mention this. Eratothenes. I don't even know how to say his name. I don't care. He didn't know how to say my name either. Alright, it's Jeff Cook. In this part, we're going to get into the Riemann Hypothesis itself. I'm going to cover, there are many videos out there on the zeta function. There are many videos out there on the Riemann Hypothesis, but none that I could reference to give the background of where we go with the proof. So, like, Euler was prospecting for his findings in one area, and everyone else is prospecting for the Riemann Hypothesis, proof of the Riemann Hypothesis in other areas, I went a different direction. So, I kind of have to give a little background on the Riemann Hypothesis that touches on the areas where we go. So, there is a lot of ground to cover. I'm not going to cover the areas that do not lead up to the proof. But, what is... Oh, before we get into the Riemann Hypothesis... So, in the last video, also, we kind of got up to the Riemann zeta function, which is the analytic continuation of the generalization of the harmonic series. So, the generalization of the harmonic series, if you didn't watch the last video, is this. And this is the Euler product of it, where he used something like the Ziv of, and I'm not going to say his name, and in his paper on the number of primes less than a given magnitude, Riemann came up with what is often called the functional equation of the Riemann zeta function. And it is a product of five different functions. 2 to the power of s times pi to the power of s minus 1 times psi of pi times s over 2 times the gamma function of 1 minus s, which I'm going to touch on a little bit in this video, and times the zeta of 1 minus s. And this relationship between s and 1 minus s is going to be very important, as you see. So, what is the Riemann Hypothesis? If you go to any other video, or pretty much any paper, Wikipedia, you're going to see the statement the Riemann Hypothesis is that all the non-trivial zeros of the Riemann zeta function have a real part 1 half. I might call that into question a little bit and say that this statement is an equivalent consequence to the Riemann Hypothesis, and you will see why. Alright, let's talk a little bit about analytic continuation. Now, I have this belief that given enough time and the proper resources, any problem can be solved. So, how much time did it take for me to make this discovery and then to be able to etch out a proof with these other co-authors? 17 years for me. 17 years. And many of those were 24-7, 365. Okay, waking up at 4 in the morning, going to bed at 3, you know, just getting 2 hours. A little bit obsessive for many of the years. And then sometimes 10 months off. But I did look at it for about that long. And what I found fairly recently, like I said about 2 years ago, a year and a half to 2 years ago, I started becoming clear of what I needed. I needed a new resource. Not a new, not some new mathematics, but something you will see as we go. Now, let's talk a little bit about the resources that Riemann had. He did have the Abel-Plana formula, where you can take certain functions, f of n, and when you need to know the sum of them from n equals 0 to infinity, many times, many functions will work. We can apply them to this formula, which I'm going to show you here. And at the top, up here, is the Abel-Plana continuation of the Riemann zeta function. So, any argument we want to apply at s, s here, s here, it's going to give us some output, right? So, how does this formula work? It's very big. It produces very large, I mean, it's, I mean, look, it's pretty clunky, you know, equation. But it's also very manageable. And the fact that it gives any output for any input is very nice. Do I? Alright. In other words, it converges for all s. It's an entire function. Except for s equals 1. So, how does it work? Alright, so, f of n would be, in this case, n to the power of minus s. And they're asking us to put n equals 0 there. Well, we could turn this n equals 0 and change this to n plus 1. And this was discovered by Plana in 1826 and about 10-15 years later by Abel, and Riemann wrote his paper on the number of primes up to a given magnitude in 1859. So, you see, he did have this at his disposal. But this is how it works when we use this, and we're going to compare it with Riemann's analytic continuation. Alright. So, I'm not going to use n to the power of minus s, okay? I'm not going to. I'm going to use a different f of n so we can see that it works for just about any. Alright, and this is an alternate form of the generalization of the harmonic series, okay? Now, my f of n that I came up with, and I didn't prove it or it's not in the paper, but I'm using it here just to show how this works. Alright, so my f of n is not equal to n to the power of minus s, but as we sum it to infinity, it does result in the same sum. Alright, so we're going to take this, and we're going to replace n with x, and we're going to integrate it, alright? And we're going to end up with 1 over s minus 1, okay? See how that works? Now we go to the second term, and we're going to take that. We're going to swap out n for 0, and we're just going to just consider this to the right in the red, and everything reduces to 1 at f of 0, and so we just end up with 1 half in the second term. And now we're going to do the last one. It's a little trickier, but nothing you can't handle. We're just going to replace n with the i t and minus i t, take the difference of those two functions, integrate it, and we get the final result above, okay? So that's how that works. Neat. Now, in doing this, I'm particularly interested in this first term. Now, if you have ever taken a college course, first-year college course on analytic continuation, and you probably have seen this, and you probably use this, or 1 over 1 minus s, and it's a pretty easy function to work out with students. Alright, it is the analytic continuation of the geometric series minus s to the power of n. Geometric series converge for the absolute value of s less than 1, as does this one, and basically anything geometric series is anything times s to the power of n. So here we have minus 1 times s to the power of n. We could do 58 times s to the power of n, whatever it is, and that's going to come out, that's going to be a geometric series. Alright, well, why is that interesting? Well, note here also we can get this really quick, before I go, why it's interesting. It also, you can get this by integrating fn in the traditional sense of n to the power of minus s. But why is this interesting to me? Alright, so if Euler was kind of like you saw last video, and if not, go listen to it for this analogy, because I'm going to carry it through a little bit. So if Euler is a kind of, how would you say, kind of approaches mathematics like a gold prospector would, and other mathematicians have as well. Riemann, he does things a little differently. He's going for a different result, and right now we have a very different problem. Alright, so Riemann hypothesis is a very different problem, and we're going to show it. So I'm going to give an introduction before I show how this is done. I'm going to give you the insight that I had into this. If this is a big, chunky rock, and it has gold in it, I don't see any primes, right? So if it has gold in it, then I want to break this apart a little. I want to chip away at this to reveal the gold beneath. Alright, now the zeros of the zeta function, as you're going to learn, relate to the prime numbers. And so I think you can consider both of them as significant and valuable, something we're looking for. Alright, so we're going to chip away out of here, break off a piece, and this is showing what's underneath. We see that geometric series. Now this is the way I see it. Alright, so I'm flipping back between these two, and I see this as chipping away the irrational functions and irrational properties of this functional equation. And what I want to do is chip it away to leave just the skeleton of this function underneath. And when I look at the Apoplano formula in comparison, I see geometric series underneath. I understand there are rational functions, 1 over s minus 1 is analytic continuation of geometric series, but I see that in there. And yes, of course, 2 is not a function. Maybe there's something else that cancels out that leads to here. And what my thought is, after going down every other road, is what if we just keep chipping this away and take all of these rational functions and put them into one function, and leave that function an algebraic factor of the zeta function. And maybe if we can do that, it can tell us something about the zeros. Maybe we will see the gold underneath. And indeed, that is what we get. Alright, so by returning to Riemann's functional equation, we have to understand what we're dealing with here. Alright, so when he gets his functional equation, what do you think the first number he'd apply to s would be? Okay, so it would probably be the first number ever applied to the generalization of the harmonic series, which is 2. Alright, so we know that zeta of 2 equals pi squared over 6, but applying that to the functional equation, we end up with minus infinity. It's a little bit of a problem since how do we say that pi squared over 6 is equal to minus infinity? Okay, so this is a problem. We know how it's a problem the way Riemann wrote about it. He didn't just gloss over it and say, well, we have all of these simple poles and all of the positive arguments. No, he wanted to eliminate it. And I'll show you why I think that's neat and important because by doing so, he was led to the Riemann hypothesis and his final result. So let's see what he did. Alright, so in order to do this, he had to not eliminate the gamma function, but do something which you might think is more ingenious, but I'm going to say not really. Anyone could do it, and you're going to do it right about now. Alright, so take a look. We're finding out where the problem is. The problem is the gamma function of 1 minus s, which happens to equal the factorial of minus s, which ends up being minus infinity for all natural numbers greater than 0. Alright, so what does he do? Well, everything you're going to do right now is pick a number between 0 and 1. The number that comes to mind is 1 half, doesn't it? I think so. Maybe you're thinking of maybe some other number between 0 and 1 like the Euler-Mascaroni constant. I don't know. Probably not. Probably you're going to say, well, 1 and 2 and 3 and all of these don't work, but maybe right down the middle, let's go try those. Of course, we know that gamma function converges for these, and so what is the gamma function of 1 minus 1 half? It is the square root of pi. Now, I have a hunch that he was going to try to reapply this square root of pi in there to cancel out the pi's, get them out of the way, also maybe reduce this to purely rational functions, but it's not that easy, and what he got stuck at is something far more important and far more beautiful, and that is this. Alright, so he took S, he divided it by 2 for all, and he's just redefining the function here by dividing S's by 2 in this way, and what he found is that this function cannot see the difference between S and 1 minus S. They end up being the same. So this function doesn't care. You can plug in S or 1 minus S, and it's going to give you the same result, and why is that so extremely important? Because it says that the zeros of the zeta functions are reflections of each other through the real point 1 half right here. So, if this is over here, rho h is a hypothetical non-trivial zero. We draw a line between it and this other hypothetical non-trivial zero, 1 minus rho h. We're going to go right through the point 1 half. See that? And of course, with any complex number, you have this complex conjugate, and it's going to go to his little partner over here, right through the real point 1 half. Now, let's give a little background on complex arithmetic really quick for anyone who's coming here and going, I have no idea what I'm looking at here. Alright, you know all of the numbers, right? The real numbers. 1, 0, 1, 2, 3, minus 1, 1 half. Pi is a real number, so pi would be somewhere on the real line just past 3 somewhere. Okay? So those are all real numbers, and you've seen these number lines. Alright, so a complex number has two parts. Think of it as a combo. Alright, it's a combination. Two parts. A real part and an imaginary part. And this imaginary, calling it an imaginary part is a very bad term, but you will get used to it. So a complex number has a real part plus an imaginary part. And what makes it imaginary? It's multiplied by the letter i. And the letter i is equal to the square root of minus 1. Does it have a value? No, there's no value for the square root of minus 1. We just call it i. So, anything, any real number plus an imaginary number is a complex number. And with those complex numbers, we can do anything that we can do with real numbers. Add, subtract, raise to powers, all of that. We can do so much more. So instead of plotting on a real line a complex number, we count over its real part and then up or down its imaginary part. So, let's suppose we have a number, a complex number, 1 plus i1, or 1 plus i. Okay? So we would just simply count over 1 and up 1. And we put a little point there and say that number is plotted on the complex plane. So, real 1 is also on the complex plane right here. And the real number 1 half is right here. And 0 is here and so on and so forth. So, this area between 0 and 1 is a very important area in terms of the Riemann Hypothesis. It's called the critical strip. And so you imagine this area going all the way up to positive imaginary infinity and all the way down to negative imaginary infinity. And it's all within that area of, between 0 and 1. Now, the dead center of that critical strip is called the critical line, obviously, because it's kind of centered. And it's, all of these would have a real part 1 half. Okay? So they have a real part 1 half. So let's say we go over a real part 1 half plus i. Then we count up 1. So we plot it right around here somewhere. Okay? So every complex number has what's called the complex conjugate. So, if this is 1 third plus 1 fifth, well then 1 third minus 1 third plus i 1 fifth and 1 third minus i fifth would be its complex conjugate. I'm just kind of guessing based on where we're looking at here. And we denote its complex conjugate by putting this little line over the top of it. Alright? So, we see the importance of that function right here. This equality. Because we can actually, and we're going to show how, see the relationship between the zeros of the zeta function. Now, they're not just the zeros. Alright? So, there's apparently in Mathematics there's many types of zeros. They're not just zeros. Alright? So, there are certain types of zeros that are referred to as roots. And these tell a lot about the function itself. And in the case of the zeta function, the roots tell us a lot about the prime numbers. Alright? Now, there are other zeros. They look like zeros, but they're not non-trivial zeros. Alright? So, they look just like a zero. They arrive at the negative even integers. Minus 2, minus 4, minus 6, all the way out there. Those also send the zeta function to 0, but they tell us nothing about the prime numbers. Okay? So, you see the little difference. Those are called the trivial zeros. The non-trivial zeros are all in the critical strip. Okay? And so far today, all of them are known to have a real part one-half. So, that's where people get the equivalent consequence of that statement, of the real part all in one-half. We're going to show you what the Riemann hypothesis actually states, which, again, is interesting if you want to take the search for prospecting in a different area. Alright. The Riemann hypothesis is that t is real whenever xi of t equals 0. Well, there's no t in there, so what am I talking about? Well, so he liked this function so much, you know, he kind of, a little modification multiplied by s minus 1. But he defined this as xi function. He demanded and mandated that all of the real parts of s equal one-half, and then he defines xi over a function of t. So, t is the input. And t could be real. t could be a complex number. It could be, I don't know, quaternion. It could be any number. Alright? But it's that number that is multiplied by i in this thing. Now, why is this a very important equation? Particularly to Riemann, and it should be to you as well. Solve for the imaginary part of t, and come back at about pause this video, and come back at about a thousand years when you've realized that there are no solutions. Actually, it didn't take a thousand years. By the rules of complex arithmetic, you can determine that there are no solutions exist. And so, I'm not going to go ahead and rearrange to show you because I'm going to do it in a different way with something new in the next video. Okay? So we're going to remember this equation, how very important it is, and we're going to do something similar which will lead to the proof. Alright? So the hypothesis then is that, you know, he demanded that the real part equals one-half. Alright. He defined, and his final result will end up having the real part in there. But, the hypothesis is that, I'm being distracted by coffee, I'm sorry. The hypothesis is that t is real for all of the zeros of psi. Okay? So there's zeros of the zeta function, and there's zeros of the psi function. And there's also trivial and non-trivial. In fact, psi of t shares all of the trivial and non-trivial zeros. It is just the multiplicative factor of the zeta function. But this is the function that is of most interest to Riemann, and it is because of this property how the imaginary part ends up being equal to zero when the real part of S equals one-half. Alright? Now. Okay. So. Makes sense? Are you lost? Well, just remember that you can't be found until you're first lost. So, you'll be found. Don't worry. Let's first ask the question, why might t be real when psi of t equals zero? Why does he suspect it? At this point in the paper, he didn't have his final result. He went on to explain. You can't have a hypothesis without a reasoning. So, you can't just say, I believe that, you know, big newtons will grow wings and fly to the moon. Okay? Why? Well, you have to have a reason why. So, his reason why was this. A value of S whose real part is greater than one and log of zeta function of S equals minus sum of log times one minus pi to the power of minus S p to the power of minus S, sorry, not pi remains finite, and since the same holds for the logarithms of the other factors of psi of t, psi of t only vanishes with this region minus i over two and i over two. If you weren't lost before, come on and admit it. Alright, I'm going to just kind of summarize a little bit of this. Alright, so the Riemann hypothesis is a hypothesis on absolutes. Okay, what is the dead center region between here? Alright, forget all this. I didn't mean to do all this. What is the dead center region? Alright? It is zero. Alright? So, if the imaginary part goes to zero, that complex number becomes real. Alright? So, we can plot real numbers on the complex plane, but the dead center is zero. Now, there's some So, he's saying that the building blocks of everything that is going to lead to all of, connect the zeros to the primes, all of that is going to be an absolute. It's not going to balance. It's symmetric, but it's only going to be symmetric on the dead center of zero. Alright? Yes, it's in this region. You know, you can prove that it's within this region, and that region connects to the critical strip of the zeta function, but how do you prove it? That's his question. He worked on it a little bit, and like I said before, I do believe he had all of the resources at hand to prove the Riemann hypothesis, but I don't believe he applied enough time to it. So, remember I said I believe that having enough resources and enough time, you can solve any problem. And like Euler, it's just a matter of kind of sometimes getting lucky or having the right resources. I think it has to do with having the right resources. So, this is Riemann's final result. Pretty big equation. Very important. Alright, this is his, I guess you could say, equivalent to prime counting function. We can give x any value, and it's going to output the number of primes up to that value. Isn't that amazing? I think it is. And let's kind of break it down one term by term. Okay, so the first term here is the logarithmic integral, and you see it's nothing you can't handle. It's just an integral. You can plug this into any online calculator, and it's going to give you a value if you apply x to here. So, you integrate up to x, and it's going to give you some value. Alright, we also have the logarithmic integral in the second term, and this is the one to look at. Alright, so these halves, you saw them in the other, they're cemented in. You can't change them. Alright, they're locked in. Alright, we can't manipulate them, call them, give them another real part, nothing. Alright, so what is alpha here? Those are the roots of xi function. Alright, and he says that he thinks they're all real. That's the Riemann hypothesis. Alright, he believes alpha is always real. Got it? But here, he's not saying that. He's just saying, I just know that this works. So he's going to start applying, and he knows because he found the first few of them, so he was able to see that his function worked. Also, because this first term is so very close already to the prime counting function. Immediately you add this other sum, and you're going to, this is the second term, and you're going to get closer, and before I get to these next two terms, let's talk a little bit more about this. Alright, so you take all of the zeros and you add some of these two logarithmic integrals. You got the complex number, and you got the complex conjugate. See, 1 half plus i alpha, and then 1 minus s, you know, is equivalent in this sense. So we have to do the complex conjugate 1 over 2, 1 half minus i alpha. And we're going to add those together, and for every alpha, we're going to sum that to infinity, or at least he doesn't know if there's an infinite number of non-trivial zeros, but for all the zeros that they are, are going to add up and give them some value, and we're going to add this to that. Now, let's continue. The next term, I'm not going to get into it. There's some elements of trigonometry in this, actually. I'm not going to get into it here, but I may do another video on it. And this is very closely related to the logarithmic integral as well. But we're not going to get into it too much, because we don't have too much time. But the value that is output from here is very small, and it works very rapidly. This term, I do want to talk a little bit. Now, you remember when we did the Abel-Plana continuation, and it asked for, we let go of n, and we made it equal to zero. A lot similar way is resulting here in his analytic continuation, where he has xi of zero. So we drop out its value of t, which we remember, and we're going to just replace t with zero, and its value is one-half. Okay, so the log of one-half also is equal to minus log of two, which isn't terribly interesting, except that that is how mathematicians today have replaced the xi function, sent it offstage left, and to replace it with the zeta function. And this is the result of doing that. Pretty neat. We can, instead of taking the zeros of the xi function, we now can take the zeros of the zeta function, which we're going to give a name, call them rho. Alright, so now we're not making the assumption the real part in there is one-half, because we don't know for sure at this point. So we're just calling them rho. We do know that all of the roots do have real part one-half up until the trillions. I think they're now coming up with so many of them. So we do know a lot of them have real part one-half far enough towards infinity that we can at least calculate our number of primes less than a given magnitude. Okay, so does that make sense? Got it? Alright, so again, it's kind of like the analytic continuation, we're stuck with all these terms that give a result for any argument x. But what he has done here, this is the big thing, the gold is in here, the prime numbers, now he knows the number of primes up to a given magnitude. Very important. And what did he connect it to? The zeros of the zeta function. Alright? The zeros of the zeta function relate to the prime numbers, namely the number of primes. But if you know the number of primes, you can actually manipulate to find if a number is prime or not. So it's just a matter of basic arithmetic at that point. But knowing them up to a very large number is complicated solved with this wonderful equation. Alright? Make sense? Alright, how do you prove it? How do you prove the Riemann Hopf balance? That I'll have to show in a different video. Thank you so much. Please check the next video. That is where it's going to have all of the information that is new. This right here is a review of the hypothesis and the history of the zeta function. The next video, we're going to get into the result that leads to the proof. And I thank you so much for watching.