Zeta Explained #74: Titchmarsh's Proof of Infinitely Many Zeros on Critical Line

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[00:00:01]Hello. Welcome to Zeta Explained. Today, we're going over Titchmarsh's proof of infinitely many zeros on the critical line.

[00:00:08]Now, in episode 21, we proved that the Riemann zeta function has infinitely many non-trivial zeros, these red dots, in the critical strip, which is this shaded region between real parts zero and one.

[00:00:20]And the Riemann hypothesis is the unproven conjecture that all of these red dots, all of the non-trivial zeros, are in fact on the critical line, real part equals 1/2. So, they're not just within the strip, but in fact all on the line.

[00:00:33]And Hardy in 1914 proved that there are infinitely many zeros on the critical line, and we proved this in episode 40, or sorry, episode 60. And I'll note again that this is not a proof of the Riemann hypothesis, um for the reason that you might imagine that all uh you can prove that there are an infinite number of odd integers, for example, but that does not mean that all integers are odd.

[00:00:58]Okay, so today we're going to do a second proof, or an alternate proof, of this theorem. Um it's going to be by Titchmarsh instead of by Hardy. And this proof is going to involve the Riemann-Siegel formula, which was rediscovered by Siegel in 1932.

[00:01:12]Um and the reason I'm presenting it is that this formula, this proof, is in some ways much nicer than Hardy's original proof. At least the concept is much easier, um and more intuitive, although the math is actually, I would consider, not so much not that much easier, um as we'll see.

[00:01:30]But at least I find the intuition um easier.

[00:01:34]Okay, so here's the idea. We're going to convert the problem um into a problem about the Riemann-Siegel Z and theta functions, uh mostly the Z function. So, these are the definitions of the theta and Z functions, but I think it'll be much more enlightening to see an animation of what these two functions are, rather than just staring at the definitions.

[00:01:54]So, uh this is what I'd like to use to explain what the Z and theta functions are.

[00:01:59]Um, what we're going to do is on the top two graphs, I'm the top left is the input and the top right is the output of the zeta function.

[00:02:07]And we're going to move up the critical line and counter some zeros on these red dots. And what the zeta function is going to do is it's going to trace out a spiral looking shape like this.

[00:02:18]And on the bottom two graphs, we're going to track roughly the modulus and the angle of this point as it rotates around.

[00:02:26]Now, the bottom left is not actually a modulus.

[00:02:29]Um, it is a thing that has the same where the absolute value of Z of T is equal to the modulus of this thing, but Z of T itself can go negative. So, what you're going to see is a spinning line on this output graph. And when the spinning when the point is on the red part of the spinning line, this thing will be negative. And when it's on the green part, this Z function's going to be positive. So, here's what it looks like.

[00:02:53]So, again, we're walking up the critical line and the output of zeta traces a spiral looking shape. And the Z function is how far you are on this line and this theta function is the rotation of this line.

[00:03:09]Okay, and then the point of today's video is we're going to analyze this Z function right here.

[00:03:17]Okay, and as we've proven before, and especially in episode 58 and also in 29, that Z of T is actually a real function when T is real. Although, it might just seem obvious from this graph that this is a real function, not a complex function.

[00:03:32]And moreover, that the non-trivial zeros of zeta of 1/2 + IT correspond exactly to the zeros of Z of T.

[00:03:39]Um, and again, you can kind of see that in this graph, at least the way I've explained it here, that whenever the zeta function hits zero, then Z of T must hit zero as well and vice versa.

[00:03:52]Okay, so the goal is to show that Z of T has infinitely many real zeros and that'll prove that zeta of 1/2 + IT has infinitely many zeros when T is real. In other words, on the critical line zeta equals 1/2 + IT.

[00:04:06]Yeah, so the proof is essentially that Titchmarsh proves this theorem that's appears in his textbook, the theorem 10.6.

[00:04:14]Um he proves that as J goes to infinity, this sum of Z on a Gram point is asymptotically 2J. So let's actually go back a second. So a Gram point G sub K is defined uh by theta of G sub K equals pi times K.

[00:04:35]Where theta is the Riemann-Siegel theta function.

[00:04:38]And what Titchmarsh did is he split he evaluates the zeta function or sorry, not the zeta function, but the Z function at the Gram points. And he splits out the Gram points into into uh even and odd Gram points. And the evens are here and the odds are here.

[00:04:55]So this is G sub 2J and G sub 2J + 1.

[00:05:00]And what Titchmarsh essentially does is if you kind of look at this equation carefully or this asymptotic, what this proves is that the average value of Z of T at an even Gram point is positive two.

[00:05:14]And that the average value of Z of T on an odd Gram point is minus two. Because like each of these sums has J terms in it.

[00:05:24]And therefore Z at an even Gram point must be greater than zero infinitely often.

[00:05:31]And similarly, Z at an odd Gram point must be less than zero infinitely often.

[00:05:37]Otherwise, you can't get the average to be minus two.

[00:05:43]Okay, and then since Z of T is a real continuous function and has to alternate between positive and negative infinitely often because, you know, the even and the odd Gram points, you know, alternate.

[00:05:55]And we're saying if on average it keeps alternating between plus two and minus two, and it has to do this alternation an infinite number of times, therefore it must hit zero an infinite number of times by the intermediate value theorem.

[00:06:08]And therefore Z of T has infinitely many real zeros.

[00:06:12]And therefore that proves that zeta has infinitely many zeros on the critical line.

[00:06:16]Now, here's an actual Here's a couple of graphs of the Z function to kind of illustrate this average of plus two and average of minus two.

[00:06:24]The left graph is at low values of T where Gram's law is actually more true.

[00:06:30]So like this looks quite true, this average thing. So for example, if you just look here, the green Gram points I've drawn are the even Gram points, so we expect them to be around plus two.

[00:06:43]And you can see that at least here they're all positive.

[00:06:45]And the blue Gram points are the odd Gram points. And I've you can as you can see they're all negative in at least on the left graph.

[00:06:53]And they do appear that perhaps it's at least plausible that the average of these is minus two and the average of the green points is plus two.

[00:07:00]But this is not so obvious once you look at higher numbers. And as you get to really, really high numbers, this becomes less and less well-behaved. But you can see that even at 288, which I find would perhaps consider still a relatively low number, you have some weird things going on. For example, this green point right here is actually negative, even though we know we this theorem says that the average value of any of these green points is plus two.

[00:07:25]But here there's a negative green green point. And in fact, every single green point here, the uh all five of them, I believe. There There's five green points here. They're all lower than two.

[00:07:36]Even though we just claimed that the average is plus two on an even Gram point.

[00:07:41]And this kind of means that at some point there this graph has to like go up a bunch and then there must be another green green point that's fairly high up and can cancel out all these green points that are below two.

[00:07:54]So that's kind of what it's saying.

[00:07:57]Okay, so that's the idea of the proof.

[00:07:59]Um but obviously we still have to actually prove this thing which is quite hard to do.

[00:08:03]Um but if we know that these are true then the proof is quite quick.

[00:08:09]Um as we just explained.

[00:08:12]Okay, so the way we're going to prove this is via the Riemann-Siegel formula.

[00:08:16]Um and this is kind of why it took so long even though we proved the Hardy's original theorem in 19 uh in episode 60 but uh we kind of spent a while talking about the Riemann-Siegel formula and now we're looping back to this proof because the proof uses the Riemann-Siegel formula.

[00:08:31]Okay, so let N capital N be equal to the floor of the square root of T over 2 pi.

[00:08:36]Um then we're going to let then Z of T this is the Riemann-Siegel formula is equal to the two times this cosine sum over N uh plus a big O term of T to the minus 1/4 power.

[00:08:48]Now the weird thing about the sum is that this N here um is not just a number you're allowed to pick at will. It's actually determined by T.

[00:08:57]So just keep that in mind. Um it's not really going to come into play until like a little bit later in one of the technical slides.

[00:09:05]Um now Gram points um as we already mentioned before are when theta of G sub K is equal to pi K.

[00:09:12]And if you plug in T equals G sub K into the Riemann-Siegel formula so what we're doing is we're evaluating Z of T at a Gram point so we're plugging in T equals a Gram point.

[00:09:21]Um and we're also going to split out this cosine sum this sum from n equals 1 to n 1 to big N uh into the just the little n equals 1 term and then the n equals 2 through big N term.

[00:09:36]So this is n equals 1 and this is the rest so right here.

[00:09:41]And this is the rest of the sum.

[00:09:44]Now, when n is equal to 1 we get this g sub k times log of 1 cuz it's like g sub k minus or sorry this t becomes g sub k and the log n becomes log of 1. The log of 1 is just equal to 0 so this whole thing just becomes 2 times cosine of pi k.

[00:10:02]Um because theta of g sub k is equal to pi k.

[00:10:05]And if you think about what cosine of pi k is when k is an integer that's just negative 1 to the k-th power cuz like on a unit circle you're just like you start at plus 1 the real like the x-axis and then it goes to minus 1 and then if you go by another pi it's going to go back to plus 1 and so forth.

[00:10:23]I drew this squarely but it's going to circle around to plus 1 again.

[00:10:29]And then this thing it's going to be a little bit more complicated because the log n here doesn't cancel out but you'll get that the theta of g sub k is still equal to pi k because of this.

[00:10:42]And therefore you get negative 1 to the k cosine of g sub k log of n.

[00:10:49]And this is just due to if you kind of think about the geometry of this circle a little bit more.

[00:10:56]Now, the key step here is we're going to multiply both sides by minus 1 to the k-th power. So, therefore we have minus 1 to the k z of g sub k.

[00:11:05]But now the minus 1 to the k actually cancels out both of these minus 1 to the k's.

[00:11:10]So, we just get 2 plus 2 times this sum from n equals 2 to big n of this term here cosine of g sub k log of n over square root of n plus an error term.

[00:11:22]Um and I'll note to that Gram's law another formulation of Gram's law is just saying that negative 1 to the k-th power times z of g sub k is equal greater than 0.

[00:11:30]But we actually know that this is not always true and it's actually in fact false infinitely often. But what we want to prove is essentially that this is also true infinitely often and for both the even and odd cases.

[00:11:44]Okay, so the thing we're going to do now is we're going to split the sum um we're going to split it into k equals 2j and k equals 2j plus 1.

[00:11:57]So like think about two different scenarios.

[00:12:01]Um, so if k is an even num- even integer then we have Z of G sub 2j is equal to 2 plus like this thing just because this this thing now is negative one to an even number which is uh you know, now plus one is equal to this thing.

[00:12:21]And if k is 2j plus 1, then negative one to the to an odd number is now equal to minus one and you can kind of divide both sides by minus one here and then you get Z of G sub 2j plus 1 equal to minus 2 minus this cosine sum plus an error term.

[00:12:40]And I would claim that if you only care about an intuitive proof and you don't care about making these rigorous, then we're actually kind of already done with the proof now. And the intuition here is that this formula, these two formulas, say that if you're in even Gram point, G sub 2j then Z of G sub 2j is just 2 plus noise.

[00:13:01]And similarly for an odd term, for an odd Gram point, Z sub Z of G sub 2j plus 1 is equal to minus 2 minus some noise.

[00:13:11]And if you look at the main noise term, so of the O terms both go to zero as T or G get very large, but if you look at the cosine sum then in some sense if you sample a cosine a bunch at like if you assume these are kind of simulating random points Um And if you just like check the cosine at a bunch of random points and then kind of add them together, then you should expect that the result is on average zero.

[00:13:37]Like you might happen to sample at a very like a bunch of large values, but you there's also a chance you kind of sample at at small values, at low values, or negative values.

[00:13:49]So, what I claim is that in some intuitive sense the cosine should be on average zero in these terms.

[00:13:56]And therefore, you kind of get what you wanted to prove that Z of G sub 2J for an even term is around two um is on average two, and therefore it should be greater than zero an infinite number of times. And similarly, Z of G sub 2J + 1, so the odd gram points, is on average minus two, and therefore it's negative an infinite number of times. And therefore, there are infinitely many zeros because then Z the Z function has to keep alternating between being positive and being negative.

[00:14:28]Okay, so that's again the idea of the proof. And this kind of just seems true by staring at these two equations right here.

[00:14:35]Uh but again, this is not very rigorous.

[00:14:37]So, the next two slides are just trying to make this rigorous. And I'm not going to go through them in like too much detail, but I will present it in case anyone is curious about the rigorous proof. And this is kind of why I claim that while the idea of Titchmarsh's proof is quite simple, um as we just explained, that proving these things uh very rigorously is actually quite hard.

[00:15:01]Okay, so without loss of generality, we're going to consider only the even case, and we'll just say the odd case kind of is the same idea as the even case.

[00:15:10]And we're going to say that this is just the um formula from the last slide, and we want to prove that Z of G sub 2J is greater than zero infinitely often.

[00:15:19]And Titchmarsh's idea is that instead of trying to say something about this cosine sum by itself, we want to take a sum over both sides of this thing.

[00:15:29]So, we're going to consider the sum from J equals M plus one to J equals M plus big J. So, in other words, instead of looking at a single even Gram point, we're going to look at a bunch of even Gram points in a row.

[00:15:43]To try to say something about the average value of this Gram points.

[00:15:46]And if you just take the sum of both sides of this equation, you get this.

[00:15:50]And now, note that we now have a double sum because this part was already in a sum and we took another sum. So, we have an outer sum in terms of J and an inner sum in terms of N.

[00:16:00]And because these are both finite sums, we're going to swap the order of summation. So, now N is on the outside and J is on the inside and we're going to kind of factor this square root one over square root of N outside of the inner sum, the new inner sum, because it doesn't depend on J, it only depends on N.

[00:16:16]And we'll call this new inner sum S sub N.

[00:16:22]Okay, so remember the theorem that we're trying to prove is that as J goes to infinity, this sum right here of Z of G sub two J, so this whole sum is asymptotically two J.

[00:16:35]And it looks like, well, we already have our main term, the two J right here. So, this is what we're trying to prove.

[00:16:41]So, therefore, what remains to show is that these two terms both grow uh kind of messed up here, but that these two terms both grow slower than linear.

[00:16:54]In other words, they they have to grow slower than J. And another way of saying that is little O of J. It just means strictly slower than. So, not not like big O where it can be equal as well.

[00:17:07]Okay, so if you can first consider this error term, then this kind of logic right here shows that this thing grows like J to the 3/4 power, which So, this thing So, remember 2J is 2J, but this is roughly J to the 3/4 power, and that grows slower than J.

[00:17:25]So, as J goes to infinity, remember as J goes to infinity, then this term right here is going to dominate the J to the 3/4.

[00:17:34]Okay, so all that's left is to figure out what this middle term is, this cosine sum. And I'll claim that, as we'll see on the graph uh the math on the next page, this thing is also roughly J of to the 3/4 power, although like J to the 3/4 plus epsilon power.

[00:17:51]Okay, so now we have to deal with the cosine sum, and this is the part where the math gets very messy, um and we actually have to invoke a very, very powerful theorem um that might not seem related to what we're doing at all. So, this is a theorem about exponential sums, and at first you might not see like how this has anything to do with what we're trying to do, which is bound this cosine sum.

[00:18:14]And the trick really is that the a cosine you can also think of as just the real part of like E to the I theta. So, as you have E to the I times something, it's just going around in a circle, and the cosine is just the real part of this.

[00:18:31]And this theorem about exponential sums has this E to the 2 pi I f of J in it.

[00:18:38]Okay, so then the idea is we're going to set f of J equals g sub 2J log n over 2 pi. That's not a coincidence, this is just this thing right here in the cosine divided by 2 pi.

[00:18:50]Um and that's because this thing has a 2 pi in it, which will kind of re So, it kind of cancels out. Um and then therefore cosine of 2 pi x is equal to real part of E to the 2 pi I x.

[00:19:04]So, we're just invoking we're just kind of rewriting the sum into an exponential sum like this.

[00:19:12]Uh where the f of J is equal to g of g sub 2 j log n over 2 pi. And then using some like you know, this is due to uh like definition of Gram points, then do some calculus and find some first and second derivatives cuz this theorem requires the second derivative.

[00:19:30]Uh then we do a bunch of stuff and we find that to use this theorem, we need to find a of suitable lambda and we figure out this lambda is uh figure out a suitable lambda of log of n over t log cubed of t.

[00:19:46]Okay, and then once you apply the theorem, then you get s sub n is equal to like this these two bounds right here.

[00:19:55]And when you evaluate them, you get two things that are roughly similar in growth rate. Um but you like this one on the right is actually growing faster than the one on the left. So then you kind of absorb this growth rate into this growth rate and you're just left with this.

[00:20:12]And then recall in the Riemann-Siegel formula that the big n is is approximately the square root of t over 2 pi.

[00:20:20]And then if you just kind of follow again, I I mentioned that I'm not going to go through this like super carefully, but I hopefully written enough steps to follow if you do care about uh this the technical details of this proof. Um we're going to kind of drop this square root of log of n on the denominator uh cuz it's a bit too hard to sum, but this thing is just O of 1.

[00:20:43]And then using this thing, then we can have this chain of big O's that'll give us eventually the O of j to the 3/4 * log to the 3/4 of j, which is roughly j to the 3/4 cuz log grows slower than any power of j. And then therefore this bound is little O of j as desired because j to the 3/4 definitely grows slower than j.

[00:21:08]Okay, so obviously I skipped a lot of steps here, um but uh hopefully if the text as written is enough to like fill in most of the gaps.

[00:21:18]Okay, so therefore the full sum for the even terms is that this sum of Z of G sub 2 J is equal to 2 J plus little O of G.

[00:21:27]So in other words, as J goes to infinity, then this like this term here is going to dominate this term over here.

[00:21:37]Okay, and then I'm going to claim that a similar argument can be made for the odd terms since we only went through this math for the even terms, uh but it's the same idea and therefore this proves Titchmarsh's theorem that as J goes to infinity, then the average value of the even terms is two. In other words, the sum of the even terms is asymptotically 2 J.

[00:21:57]And the sum over the odd terms is asymptotically negative 2 J.

[00:22:04]And therefore, by this theorem, it must be the case that Z of G sub 2 J is greater than zero infinitely often and Z of 2 G sub G 2 J plus 1, the odd Gram points, must be less than zero infinitely often.

[00:22:19]And therefore, because Z of T is a real continuous function, it must alternate between being positive and negative infinitely often due to this.

[00:22:27]Then it follows that Z of T has infinitely many real zeros.

[00:22:31]Cuz it has to keep going back and forth between being positive and negative.

[00:22:35]Okay, so uh I think we're done now with the proof. So we now have an alternate proof of Hardy's theorem from 1914 that there are infinitely many zeros on the critical line.

[00:22:45]And I guess I'll pose a question. So if you have seen episode 60 before where we where we went over Hardy's original proof, um like which one do you like better? Um remember here is a summary of that proof. Uh Hardy used this interval with the xi function, um and the xi function is kind of an of the factorial of not sorry I put I was about to say factorial but I meant the zeta function um that Riemann used.

[00:23:11]And Hardy proved that this following integral has to converge as alpha goes to pi over 4.

[00:23:17]Uh but he also showed that if there only finitely many zeros then this integral has to diverge which forms a contradiction so therefore there must be infinitely many zeros.

[00:23:28]And his argument looked something like this when you try to visualize it and we again this diagram is just straight from episode 60.

[00:23:35]Um that on the right is the actual like integrand that you're integrating over.

[00:23:40]And the integral of this thing you can think of it just as the normal calculus sense it's the area of all the shaded regions here.

[00:23:47]And the blue is like positive shaded region and the red is negative shaded region so you just do the blue minus the red and Hardy showed that this thing has to converge.

[00:23:58]Um and he also showed that if there are only finitely many zeros then the graph would look like something on the left.

[00:24:03]And that on average in some sense of average that this function even though it can oscillate some somewhat it has to be roughly this value here and if you take the area under this curve it does not converge.

[00:24:18]And therefore that forms a contradiction and therefore the graph must look like something on the left so the only way for this thing to converge is if it keeps alternating between being positive and negative.

[00:24:30]Okay so that's Hardy's proof.

[00:24:32]And then Titchmarsh's proof is this thing that we already showed you from before that on average the even Gram points when you evaluate the Z function are at plus two.

[00:24:42]And on average the odd Gram points which are the blue dots are at minus two.

[00:24:48]And therefore the Z function must keep crossing between being positive and negative an infinite number of times.

[00:24:55]Okay so um yeah let me know which one you like better. I personally find Titchmarsh's proof to be a little bit more intuitive, uh, but the math, as you can see, as you saw from before the those slides, is not very easy.

[00:25:08]Um, although the math for Hardy's proof is not particularly easy, too. So, in some sense, this is a fairly deep theorem.

[00:25:17]Okay, and then I finally want to end on a note about this video series. So, we're done with our saga on the Riemann-Siegel formula.

[00:25:25]Um, and I'll mention that there are actually more advanced things you can do. So, we're not done with um, the critical line or anything yet, um, or Hardy's or Littlewood's proofs. Um, for example, what we proved today is just an alternate proof of this thing, that there are infinitely many zeros on the critical line.

[00:25:42]Um, so, let me just draw a like zeta function with a critical line on it.

[00:25:46]Uh, let's say this blue is the critical line.

[00:25:49]Now, just proving that there are infinitely many zeros doesn't really tell you how well the zeros are spaced.

[00:25:55]So, for example, we infinitely many zeros could mean that you found a zero here, and then there's a zero here, and then there's a zero up here, and then they get exponentially spaced apart. And that's not very good.

[00:26:07]Now, Hardy and Littlewood together proved in 1921, we haven't gone to either of these two yet, that there are at least KT zeros on the critical line below given height of T. So, in other words, they are roughly they're at least like linearly like spaced.

[00:26:22]And by at least, I mean like they're at least as many zeros as in as linearly spaced like this.

[00:26:27]And then Selberg in 1942 proved that there are at least KT log T zeros on the critical line.

[00:26:33]Um, so, in fact, the distribution the zeros actually get more clustered as you go up the line.

[00:26:40]Now, the crucial thing about this KT log T is that we already know from the Riemann-von Mangoldt formula about the zeros in the critical strip, so not just on the not on the critical line, but just anywhere in the critical strip, is exactly T over 2 pi log of T, or at least roughly T over 2 pi log of T zeros in the critical strip. And because these two things have the same order of growth, the KT log T and T over 2 pi log of T, just differing by a constant, then we then this proof by Selberg is the first proof that there are at least some finite some positive proportion of zeros on the critical line.

[00:27:21]Okay, but again, so I'm not really going to talk much about this because this is actually quite advanced and it would like to kind of prove either of these things we would need multiple episodes to introduce exponential sums and various other topics.

[00:27:35]So instead of doing that in as the next few videos, we're going to take a breather from all this heavy calculus and we're going to do some more fun arithmetic episodes.

[00:27:45]So hopefully that'll be cool.

[00:27:48]So thanks for watching. We'll see you next time.