Analytic Number Theory: Factorials, Primes, and the Riemann Hypothesis

[00:00:02]Hello, and welcome to zetamath! What I'd like to  do today is talk about some unexpected connections   between what I'm going to call arithmetic  functions; that is, functions which come   from number theory, and what I'll call analytic  functions; that is, functions from calculus.

[00:00:24]First, I want to point out that these are  two very different kinds of functions.

[00:00:31]Here, for example, is a graph of the arithmetic  function "is prime" which gives the value   1 if the number you put in is prime, and the  value 0 if the number you put in isn't prime.   Notice its graph is a bunch of disconnected  dots, and the reason for that is because   what I mean by "the number" when I say the  number is prime or the number isn't prime,   is really an integer—it doesn't make any  sense to ask whether a number like π,   that is a real number that's not an integer,  is prime or not. This is very typical   of an arithmetic function. On the other hand,  here is a graph of the analytic function log(x),   which is the sort of smooth curve you're probably  used to from a calculus course. Indeed, probably   every graph you looked at in such a course looked  roughly something like this. The study of the   connections between arithmetic and analytic  functions is called analytic number theory,   and it is at the heart of many of the most  important unsolved problems in mathematics,   including both the Riemann Hypothesis and  the Birch and Swinnterton-Dyer conjecture,   both of which carry a million dollar bounty  courtesy of the Clay Mathematics institute. Both   of these problems are very difficult to formally  state, but by the end of today's video I hope   to give you at least an intuitive understanding  of what it is that the Riemann Hypothesis says.

[00:02:07]So, whenever I show you an arithmetic  function—that is, a function which is really only   meaningfully taking values on the integers—I'm  always going to in this video show it to you   displayed in these red dots, and that's to remind  you that these functions only take values at   discrete points. Even when I push all the points  really really close, because by the time we're   done this video we're going to be talking about  graphs with scales like a hundred million, still   when I show you graphs of arithmetic functions I'm  going to thin out a lot of the dots just to remind   you again in every graph that you see that you're  looking at a graph of an arithmetic function,   as opposed to if I'm showing you the graph of an  analytic function I'll show it to you in a thick   black line, again like this graph of natural log.  The particular arithmetic function I'm going to be   most interested in today is what I'm going to call  𝛿(x) which gives the proportion of integers near   x that are prime. For example, if I say  𝛿(100) = 0.2, that means that approximately   20% of the integers near 100 are prime. You might  rightly critique this as not a real definition,   and you might even question if the notion  makes sense at all. For now, though,   I ask you to suspend your disbelief, and  reserve judgment until later in the video.   Here, I have given you a graph of 𝛿(x)  from 0 to 50, which I have estimated   by finding the proportion of integers from 0.75  times x to 1.25 times x that are prime. The seeming   randomness of this graph highlights the difference  between arithmetic and analytic functions.   Zooming out to a maximum x of 500 we  start to see some structure forming,   and if I zoom out to a million,  everything calms down substantially.

[00:04:23]This curve representing 𝛿(x), the density of  the primes near x, is what I want to focus on   today. Now just as a little teaser for what we're  going to do later, and to convince you that maybe   there is some meaningful connection here between  the ideas and functions of calculus and the ideas   and functions of number theory, I'm going to show  you a graph of a particular arithmetic function;   namely at each prime what I'm going to take  is the sum of the natural logarithms of that   prime and all the primes less than it. So  that is to say, for example if I evaluate   this function at 7 it's going to give  me log(7) + log(5) + log(3) + log(2),   and I want you to think in your head,  "what do I think the graph of this   function is going to look like?" Well,  I have no idea, it sounds frankly awful   but I at least know it's going to be increasing.  But that's really all I can sort of figure out.   But I'm going to go ahead and show you a graph of  this function and you'll notice this graph is   way nicer than I'd expect! Namely, it looks  almost identical to the graph of a line.   And so in order to convince you that that's really  the case, I'm going to draw in on top of the graph   the graph of the line y = x, and you'll see  beautifully the line y = x almost exactly   matches up with this weird function where we  sum the value of the logarithm of the primes for   primes less than this particular prime. It's not  at all obvious why these two things should have   anything to do with each other, but my hope  is that by the end of today's video you'll   be able to understand in some sense why there's  this connection—that is why it's true that x is   approximately equal to the sum of the logarithms  of the primes that are less than or equal to x.

[00:06:32]So how are we going to get a handle on the density  of the primes? The method that we're going to   use is that we're going to look at, as you might  have guessed from the video title, the factorial.   And namely what I'm going to be interested  in is sort of understanding how big n! itself   is in two different ways: namely on the one  hand what I'm going to do is I'm going to   factor n! and when I factor n! that's going to  give me a relationship between n! and the primes.   On the other hand I'm going to  estimate the size of n! using calculus,   and that's going to give me a version of the size  of n! that has nothing to do with the primes.   But because both of these are  approximations of the same thing,   they're going to have to be close to equal, and  we're going to use that to sort of wrangle out   a sense of how dense the primes have to be  at different points along the number line.   So, to begin the first piece—that is in  order to figure out how to factor n!—what   I really need to do is answer this question: how  many times does a particular prime go into n!?   So I want to answer this question because  ultimately what I want to do is I want to   write down the prime factorization  of n! as explicitly as I can.   So let's start with n = 100 and p = 3. So we want  to know how many times the prime 3 divides 100!.   If we write out all the digits of 100!  we get this. At the very least we can   factor this, and it looks like  this which is a little better.   So what I want to think about is where the  threes that go into 100! are going to come from.   Well, all the threes that are ever going to  appear in the factorization of 100!, they're   in one of these numbers. Now most of these  numbers, they aren't even divisible by three,   and so if they're not divisible by 3—if  they're not a multiple of 3—they're not   going to contribute any threes to 100!, so we're  just going to fade out those numbers that aren't   divisible by 3. And so these numbers, all the  numbers circled in red will contribute threes.   Which might lead your first guess for  how many times the prime three divides n!   being, well probably, how many  of these numbers are there?   Well there's one of these numbers every third  number along the line, so if I go up to 100 there   should be 100/3 of them. And then maybe if you  think about it you realize 100/3 isn't an integer,   and so you need to round down because this  would be about 33.3 something like that,   and this would give you just 33. This  guess turns out to not quite be correct,   and you can actually see why if you look  back closely at our list of numbers.   Namely, I counted the red circles, and it's true  all the threes come from the red circle numbers,   but some of these numbers actually include  bonus threes! So for example, this 9,   it actually gives us 2 threes. And then  if I look over this 18 gives us 2 threes.   And then this 27, it actually gives  us 3 threes. I can't just count the   numbers that have at least one three in  them, I need to count all of the threes So what I'm going to do is I'm going to put a dot  below each of these numbers for each factor of 3   that appears in it, and you'll notice it has this  very nice structure: it goes dot, dot, dot dot;   dot, dot, dot dot and so on down this  line. And so the number of dots here   is exactly the number I'm looking for—that is  the number of threes that divide 100!. And so   if you look at this page  and you look at all the dots   the dots exactly correspond to all of the threes,  and so you could just count them. Of course,   if you want to answer this question in general  just counting them here is not super useful,   but it turns out there's a rather easy way to  count them, which is to sort of think about what   we did before: namely, this number here that's  the number of numbers listed here, there are 33   numbers from 3 to 99 in multiples of 3, that's the  33. This is the number of dots in the first row.

[00:11:55]I could also now count the number of dots in the  second row. When is there a dot in the second row?   Well there's a dot in the second row if our number  is divisible by 9, which you can see this kind   of fractal pattern here actually that sort of  every third of our original numbers had a dot,   now every third of these numbers has two dots,  every third of the two dot numbers has three dots,   and so on down the line. But really you  don't even have to think about it that hard,   you can just think about when is there a second  dot—at least a second dot—and that's every time   the number's visible by nine, so 9, 18, 27, 36  and so on. Well how many of those are there?   Well that will be the floor of 100/9 for much  the same reason that this was the floor of100/3.   So the floor of a hundred divided by nine, what  is that? That's 11. So the first row has 33 dots;   the second row has 11 dots.

[00:13:05]Well what about the third row? Well in  order to have a dot in the third row,   the number has to be divisible by 27.   So how many of those are there? Well that  would be the floor of 100/27, and the floor of 100/27 is 3.

[00:13:26]Now in fact, here if you want to check  that we have not done anything crazy   you might notice, "hey I can actually count them,   one two three, those are my dots in  the third row; there are no others."

[00:13:42]Now I can ask about dots in the fourth row,  because I actually have one of those. That'll   be the floor of 100/81; there's only one of those,  and I could keep going and think about okay   well what about dots in the fifth row, but there  won't actually be a dot in the fifth row at all   until I get to 243, which is much larger than 100  and so really I can stop there. So actually the   answer to this question, "how many times does 3 go  into 100!?", it is 33 + 11 + 3 + 1, which is 48.   Okay, now I want to go ahead and write down  the general form, because it's not so hard   to generalize this: namely, here's a theorem  that we have effectively just proved, which is   that if p is prime, then p goes into n! this  many times, so I'm going to call this number   v_p(n). What is it? It's the floor of n/p plus  the floor of n/p^2 plus the floor of n/p^3, and so on.

[00:15:03]Now I'm going to write this as an infinite sum,  which it kind of is but just like in the case   that we just did, it's an infinite sum almost all  of whose terms are 0, because once I get to the   point where p to the whatever power is bigger  than n then all the terms after that are 0.   So I want to note that we figured  out how many threes there are in 100!   but there's nothing to stop us from doing  that for every prime that divides 100!.   In fact, what are the primes that  divide 100!? Well if you think about it,   it's exactly the primes that are less than 100,  because every prime less than 100 is, well,   in the product, but on the other hand no primes  bigger than 100 could possibly be in the product.   And so with a little work, we can write  down the prime factorization of 100!,   and here it is. Now, you'll go ahead and notice  this power on the 3 here, this 48, this is exactly   the number we calculated just now, and so in  general it's always going to be the case that   n! is going to be the product across the primes  ≤ n of exactly that prime to this v_p(n) power,   just like we saw for 100 a second ago. So I want  to make an observation about how big v_p(n) is:   well, on the one hand how small can v_p(n) be?  Well, v_p(n), it's at least as big as the floor of n/p,   which is certainly at least as big as (n/p)  - 1, because if this thing is an integer,   well then it just is n/p, otherwise if this  thing is really close to the next integer,   like if this is something that ends in.99,  then we end up knocking off almost a whole   1 (so in fact actually it's strictly bigger but  we're not going to worry about this too much).

[00:17:25]So this thing is at least as big as (n/p) - 1.   On the other hand I can also put an  upper bound on it without too much work:   namely, v_p(n), this is less than or equal to,   well, floors only make things smaller so this is  less than or equal to just n/p + n/p^2 + …   This thing is a geometric series with  first term n/p and common ratio   1/p, so by the sum of an infinite geometric  series formula, it's this, which works out to be   n/(p - 1) at the end of the day once you  do a little bit of fraction flippiness,   the ps end up canceling and you get exactly this.

[00:18:22]So to summarize, this v_p(n) number,   it is on the one hand bigger than (n / p) - 1 and  on the other hand it is smaller than n / (p - 1),   and I want to note that in the grand scheme of  things these numbers are not that far apart, so   the thing I'm going to note here is that roughly  v_p(n), it's not so far off from just being n / p.   Can't be too much bigger, can't be too much  smaller, it's not that far off from n / p.   So what does that mean? Well, that means n!  which we said is just the product of p ^ v_p(n),   so n! is not so far off from being the product  across primes < n of p ^ (n / p).

[00:19:35]Again, this isn't exactly  right but it's not so far off,   and you might notice here we have this  product and we could actually make this   product a sum if we take logarithms, and  that's going to make my life much nicer.   And logarithms are going to turn products into  sums, so this is going to work out to be the sum   over p ≤ n of log( p^(n/p) ), and remembering  my properties of logarithms I'm allowed to pull   this n/p down, so this is the sum across p  ≤ n of (n/p) · log(p) and if you want, the n   is actually not changing term to term, so this  is n times the sum across p ≤ n of log(p) / p.   (I lied before, that was approximately  equal.) And so here's my conclusion: log(n!),   it's approximately equal to n times the  sum across the primes ≤ n of log(p) / p So as promised, what we're going to  do now is we're going to estimate   log(n!) in a completely different way, one that  doesn't in fact involve any real number theory.   The reason why I said we're using log before is  that it makes life a little nicer in that it turns   products into sums, but I want to point out  as we do this, another reason why log(n!)   is a lot nicer to work with than n! is that  log is an incredibly forgiving function,   and what I mean by that is that log is a  function which changes very slowly, so when   I make approximations, even if the approximations  introduce a fairly large change in the thing I'm   taking the logarithm of, they're going to produce  a very very small change in the logarithm.

[00:21:52]As I'm doing this, I'm going to be making  all of these approximations just like we   did in the previous iteration, but the hope is  that because they're buried inside a logarithm   that ultimately those things  aren't going to matter very much.

[00:22:07]How are we going to do this? Well, the first  thing I'm going to do is I'm going to notice   that since n! is itself a product, that this  whole idea that logarithms turn products into sums   means that in fact log(n!), that's the same thing  as log(n) + log(n - 1) + … so on down to the   log(1), which already I can think of as rewriting  as the sum as j goes from 1 to n of log(j).   Now so far I haven't done any approximating, but  I do want to do a little bit of approximating:   namely, I want to approximate this sum by this  integral. If you haven't seen this before it   might seem a little weird to approximate a sum by  an integral, but if you recall actually the very   definition of an integral involves some sort of a  limit of sums that approximate it, so usually when   you start defining integrals you start by defining  them as some limit of sums to approximate them,   but that actually means the opposite's also true,  that integrals are an approximation of sums.

[00:23:41]Now on screen is a visualization of the connection  between the sum, log(1) + log(2) + … and I went up   here to log(6), and that's represented as  the combined area of the blue rectangles.   Now I'm showing you the integral from 2 to  7 of log(x) and that's now represented as   the area of this pink region. Thus we can see  the difference between the value of the sum   and the value of the integral is just the area  of all these little red triangulish regions.   If I show you the same graph for even a modestly  large n, so here n = 30, you can see that the   red error is minuscule by comparison to  the blue area that we're approximating,   so we'll call this approximation  good enough for our purposes,   although if you wanted you could certainly be  much more precise about how large this error is.   So, summary, where are we now? log(n!) it's  approximately the value of this integral.   But, if you remember your integral formulas,  this integral, the integral of log(t), it is t   · log(t) - t, so we're going to evaluate that as  t goes from 2 to n + 1. This is going to be equal   to (n + 1) · log(n + 1) - n - 1 - 2 · log(2) + 2.

[00:25:28]Okay, so let's simplify this a little bit.

[00:25:32]First of all, I'm going to distribute this,  and also because of that fact that log is a   very slow growing function that I mentioned as to  why it's nice that we're taking logs of things,   I'm going to just go ahead and note that  log(n) and log(n + 1) are very close,   so I'm just going to replace this log(n + 1)  with a log(n), so this will become n · log(n)   and so that means I'm now approximating rather  than saying exactly equal because I've replaced   the log(n + 1) with a log(n). Plus log(n) - n  + 1 - 2 · log(2). Now, logs are really small,   so I'm going to throw away the logs. Also these  are constants so they're really small, and so   I'm going to say at the end of the day that what  we've just proven is that log(n!) ≈ n · log(n) - n As promised, we have now produced two different  approximations of log(n!). This first one,   which involves the primes, and this second one  which is just what we got by integrating the   function natural log. So because these two things  are approximately equal to the same thing they   must be approximately equal to each other, and so  we get that n times the sum across the primes ≤ n   of log(p) / p, that's approximately  equal to n · log(n) - n Now I'm just going to divide everything  in sight through by n, so I'm going to   get that the sum across p ≤ n of log(p) /  p, that should be approximately log(n) - 1.

[00:27:40]The nice thing about this claim is that this  claim is readily verifiable empirically:   I can just go ahead and graph the sum of  log(p) / p across primes ≤ n for a bunch of n,   and log(n) - 1. Indeed, the dots really do  track the logarithm very closely. What you might   notice is that it looks like there's a little  bit of consistent error in the same direction,   and you might worry that if I went on a much  larger scale that that would get worse. In fact,   it doesn't get worse; we're just missing a  little bit of this constant here because of   how we've done this, and of course it's not  at all surprising that we're missing some   piece of what's happening because we've done an  awful lot of rounding throughout this process.   So summarizing what we have so far, we've shown  that log(x) - 1 ≈ the sum of the primes ≤ x   of log(p) / p. I only made two changes  here, one is I changed n to x, and   the other is that I swapped which side of this was  which. The reason I changed n to x is because the   next thing I want to do here is I want to take  the derivative of both sides of this equation.

[00:29:03]Now on this side, taking the derivative is easy:  so on this side the derivative of log(x) is 1 / x;   the derivative of 1 is 0. And if you believe  me that there was some constant we were missing   before here, well that also ends up getting  obliterated when we take the derivative,   so this also disappears. Now things  get more difficult when it comes to   taking the derivative of this  thing on the right hand side   now I want to remind you that this thing is one of  those arithmetic functions that we talked about,   so really its graph it's just a bunch of dots  but I'm going to try and convince you that if   you were forced to think about what it meant to  take the derivative of this, that you would come   up with something and you would really go back to  what is it a derivative means: a derivative really   means if I increase the input by some amount,  how much do I expect to change the output by?   And I want to remind you when I say "the input"  here, the thing that's the input is not p—this p   is a dummy variable in the sum—the thing that's  the input is this x here so if I thought about   only changing this x by 1, either the new x that I  put in is prime, in which case I've added a whole   log(x) / x, or the new x I've added isn't  prime in which case I've added nothing.

[00:30:32]And so what I want to think about is kind  of the average of those cases: that is,   sometimes I add a whole log(x)  / x and other times I add 0.   Now, those two cases don't happen equally  often, so I can't just literally average   the two of them; I have to think about  what proportion of the time they happen.

[00:30:56]Well, what proportion of the time do I get a new  log(x) / x? Well, that's exactly the proportion of   the numbers near x that are prime. And so here  what I'm really saying is that if you take the   derivative of the right hand side and think  about how that increases, well that should   increase by about this thing that we called 𝛿(x):  the density of the primes near x times log(x) / x.

[00:31:25]So just to recap what's happening here, when  my input x goes up by 1, either I add a term   of log(x) / x or I add nothing. If  I add nothing, well then of course   the value changes by zero, but some proportion  of the time it goes up by a full log(x) / x,   and so in order to compensate for that sometimes  it's zero and sometimes it's this relatively large   compared to zero thing log(x) / x, I am going  to include basically a fraction of a log(x) /  x for each 1 that we go up. And what fraction am  I going to include? Well the fraction I'm going   to include is going to be the proportion of the  numbers near x that are prime, and so I should get   this identity. Recall from the introduction that  I stated our goal was to say something meaningful   about the density of primes near x. So,  that is, something meaningful about 𝛿(x),   and so in fact I can solve  this equation here for 𝛿(x).

[00:32:39]𝛿(x) ≈ 1 / log(x). That is to say, the density  of the primes near a number x is approximately   1 divided by the natural log of that x. Again,  I showed you some graphs of the density earlier   and now I can show you those graphs with the  graph of the analytic function 1 / log(x)   going next to them, and you'll see actually  there's an almost uncannily good fit here,   given how messy this whole function seems to be.   And so the moral of the story here is that  even though we've made lots of approximations,   because again so many things have been buried  in logarithms we've ended up not that far   off, and we actually have a really realistic  answer for the density of the primes near x   and so if there's one thing I want you to take  away from this video, this is the key point:   the density of the primes near x is  uncannily accurately given by 1 / log(x).

[00:33:52]So I want to use this to talk about the  function π(x), which is in some sense the   central object of analytic number theory:  this is just the number of primes ≤ x.   In Riemann's famous paper where he originally laid  out the Riemann Hypothesis, which is sort of a   motivation for talking about the zeta function  and this big unsolved problem in mathematics,   what Riemann did (and this is a preview  for what I'm going to talk about   in later videos), Riemann showed that you could  give an explicit formula for this number of   primes ≤ x exactly in terms of the zeros of  the zeta function (but we'll talk about that   later). What I want to talk about now is that  we can approximate this function π(x) by using   this idea of density, and just  before we even approximate π(x)   let's make sure we understand what exactly  it is that π is doing. So π(10) for example,   this is going to be the number of primes  ≤ 10. So, remember 1 is not prime, but 2,   3, 5, and 7 are, so this is 4. Likewise you can  calculate π(100) is 25, and I'll go ahead and put   up a table here for π for various other powers of  ten so that you can see how this thing grows.

[00:35:31]But what I'd like to point out is that if you have  this idea of what the density of the primes is,   you should have a pretty good idea of how many  primes less than or equal to x you should expect.   What is that relationship? Well broadly, the  relationship between density and mass (which   is really how you should think of this number of  primes ≤ x, this is really just the mass of all of   the primes if the primes have weight 1) well then  you should integrate the density to get the mass,   and so what we should expect is that  π(x) should be approximately given   by the integral from 2 to x of 1 / log(t) dt, and the only reason I used t instead of  x is because I can't use x both for the limit   on the integral and also the dummy variable  for the integral. And so you can look   at this and this should give you a pretty good  idea of how many primes there are less than x.   Now you're perfectly welcome to say "well,  we could look at this," and so I'll show you   again some graphs that give you this integral's  value versus the value of π(x) at various points   so that you can see in fact this is  actually a very good approximation.   Now if you've thought about  this or seen this stuff before   you've probably seen something  called the Prime Number Theorem,   which is usually stated as saying that  the number of primes ≤ x ≈ x / log(x),   and one thing I want to point out in this  video and hopefully you'll take away,   is that this x / log(x), it's actually a bad  approximation of this integral, and in fact   this integral (if you're willing to grant  this integral as being a fundamental object   in mathematics or number theory), this  integral is a much better approximation   of the number of primes less than or  equal to x than this function x / log(x).

[00:38:00]Again you might be inclined to ask "okay,  how is this a theorem?" because this T   in PNT stands for "theorem", to say  that this thing is approximately equal to   this or approximately equal to this, and so the  theorem here is a much stronger statement than   this "approximately equal." In fact, what the  prime number theorem actually says is that if   you take the limit as x goes to ∞ of π(x) divided  by this integral from 2 to x of 1 / log(t) dt,   then this limit is in fact exactly equal to 1.

[00:38:46]And I'll note, even if this limit had  worked out to be say 1.01, that would still   in my mind be saying that the number of primes  less than or equal to x is approximately equal   to the value of this integral, because that would  say the error is asymptotically about one percent.   Even if it had been 1.07 I would say that still  would justify this approximately equal because   that's saying there's about seven percent  error, and counting the number of primes   less than or equal to x even with seven  percent error would be pretty impressive.

[00:39:20]Or even maybe what you should really expect is  you should expect this limit doesn't exist at all; that's probably the real expectation,  and so maybe this value of this if this is an   approximation maybe bounces back and forth between  say 0.91 and 1.1 or something like that. Even if   that were true—even if this expression bounced  back and forth like that between numbers around  1— I'd still say this "approximately equals"  is justified. But no, this limit is exactly 1,   and what that says is that the error in  this approximation is asymptotically 0%.

[00:40:03]But I want to be clear, just so that you  don't take away the wrong message from this,   saying that the error is asymptotically 0%  is a much different statement   from saying that the error is asymptotically 0.   Really what you should take this away as saying is  that the error is asymptotically of a lower order   of magnitude than the thing you're approximating.  That is, the error is fundamentally smaller in   size than the number of primes less than or equal  to x, so the limit of the error over π(x) is 0.

[00:40:46]The error is still very much a real positive  number that gets bigger as x gets bigger,   and so I'll show you a chart here  of x through some powers of 10, π(x)   through some powers of 10, and then I'll  show you the error in this integral here   which is called the "logarithmic  integral" and abbreviated by the way,   this thing is often abbreviated as Li(x), and the  error versus this x / log(x), and so the things   I want you to take away from this chart are, one,  wow Li(x) is a way better approximation than   x / log(x), so if you've internally thought of  x / log(x) you should probably start thinking of   this integral instead as being what the  Prime Number Theorem is really saying.

[00:41:38]Two, wow, this is actually a pretty good  approximation; you can see that here;   but even though it's a pretty good approximation  there's still some mystery left here in terms of   how big this error is. So the question "how big is  this error" is actually a very big open question.   So when I say this is a big open question  what exactly do I mean? Well in fact it   turns out that this big open question is exactly  equivalent to answering the Riemann Hypothesis.   So, if you prove the Riemann Hypothesis is true,   it gives you a very specific answer to how  big is this error; if you prove the Riemann   Hypothesis is false, it also gives you a very  specific answer to how big this error is.

[00:42:32]So I promised to give you some insight  into what the Riemann Hypothesis says,   and I think the best way of thinking about the  Riemann Hypothesis is that it's a strengthening of   the statement of the Prime Number Theorem. Here's  a graph of the absolute value of π(x) - Li(x) all   divided by x. Note the numerator here is the error  in the estimate given by the Prime Number Theorem,   and then the denominator here is the input, so  you can think of this fraction here as telling you   the size of the error as a percentage of x.  What the prime number theorem says is precisely   that the limit of this ratio is exactly 0, and  that should be pretty believable from this graph.

[00:43:31]So in order to understand the sense  in which the Riemann Hypothesis   is a strengthening of the Prime Number Theorem, I  want to play around with this idea a little bit,   and so what I'm going to do is I'm going to  add to the graph a similar ratio. I'm going to   leave the numerator the same, but I'm going to  make the denominator x^0.75 instead of just x,   so since I've made the denominator  smaller because it's a smaller power of x,   that's going to make the whole value of the  ratio larger, and so you can see the green   dots that I've added, they all lie above  the red dots that I showed you before,   but I want you to note from the graph that  this ratio is still less than 0.01, even on   this interval that I'm showing it to you now, and  thus it's reasonable for us to conjecture that the   limit of this ratio—the ratio of the error divided  by x^0.75—that that limit is also 0.   Now that's what I was talking about before: that  would be a strengthening of the Prime Number   Theorem. That would be something that implies  the Prime Number Theorem but says something   even more. So for comparison I want to  show you a graph of another similar ratio   but now I'm replacing the x^1 or the  x^0.75 in the denominator with an x^0.25,   and you'll notice that now the y-axis has  completely shifted and now this ratio in no way   seems to be approaching 0. In fact, if anything,  it seems to be approaching infinity. Now   just in case you don't believe me, and you  think "okay well he's just showing us a really   small axis and there's going to be a surprise  reveal that this really does go to 0", I'll   go ahead and bring the axis here up to a hundred  million just to show you it does not quiet down.

[00:45:51]To shorthand sort of the discussion that  we've been having we're going to talk about   the order of the error and so this graph I'm  showing you now purports to be showing you   that the error has order larger than x^0.25—that  is, the error grows faster than x^0.25.   On the other hand you can think of  the prime number theorem as saying   that the order of the error is less than x^1,  again because the limit of that ratio is 0.

[00:46:32]And if you go back to this graph where  I showed you the ratio with x^0.75,   then what we conjectured from that  was that the order of the error   was less than x^0.75. In other words,  the error should grow slower than x^0.75.   So then it's a natural question (and indeed it's  a question that people have thought about a lot),   "what is the order of the error?" or  put a different way, what numbers p   have the property that if I take the limit of the  error divided by x^p then I'll get 0? And so far   from what we've seen it seems like 0.25  doesn't have that property, and 0.75 seems to,   and I've told you that what the Prime Number  Theorem says is that 1 does have that property.

[00:47:32]Let me show you the graphs for a  few values of p. So here's p = 0.4;   this is pretty clearly still in the diverge camp,  so if we look at this graph and we are forced to   guess I think we're going to guess that 0.4  doesn't have this property. Here's p = 0.5.   This is a little less clear but this  still doesn't look like it's converging;   if it's converging it's converging very slowly.  And now here is p is 0.6 and here from the shape   of the graph and the values on the y axis it sure  looks like this limit really is headed towards 0.

[00:48:19]So what does the Riemann Hypothesis  say? What the Riemann Hypothesis says   is precisely that the magic cutoff  for this happens at p = 0.5.   If p is greater than 0.5, strictly greater, then  this limit of the error divided by x^p is 0.

[00:48:44]So this might not be the version of the  Riemann Hypothesis you're used to seeing   if you've thought about the Riemann Hypothesis  before—but if you have thought about it before   and you think it says something about  the zeros of some complex function   having real part 1/2, I want to  point out this 1/2 is the same 1/2.   Otherwise you don't have to think about it, that  part is a little more complicated to get into.

[00:49:12]So as noted earlier in the video, the Riemann  Hypothesis is unproven: we certainly don't know   this is true, although we very strongly suspect  that it is true. Now, there's a million dollar   prize, so if you can prove that statement  I just put on the screen, congratulations   you can win a million dollars, but probably  there's easier ways to win a million dollars.

[00:49:36]So, you might ask though, "what do we know now,"  that is, what do we know about the ps which work   or don't work? So first of all we know the Prime  Number Theorem, that is, p = 1 works; this limit   is 0. That's what the Prime Number Theorem says;  we know that's true, that was proven around the   turn of the 20th century. Also we know that if you  take p to be 0.5, that this limit is undefined.  So, in other words we know that the order of the  error, we know it's at least as big as x^0.5,   or if you want to think about it  x^0.5 is the square root of x,   so we know the order of the error is at  least as large as the square root of x.

[00:50:28]Okay, but what don't we know? Well, we don't  know anything else! That is, we don't know any p   strictly less than 1 for which we can prove that  this limit is 0. So, I showed you a suggestive   graph earlier that definitely seems to imply that  if p is 0.75 that this limit is 0 but no one can   prove that; that is still open, that would still  be huge progress if we were able to prove that.

[00:51:03]Okay, so for the final flourish of this video I  want to go back to what I teased in the intro:   namely, I want to ask the question "how big is the  sum across the primes less than or equal to some   number x of the logarithm of that prime? I'll note  this question is actually pretty similar to the   question earlier when I said okay π(x) is really  counting the mass of the primes. π(x) is counting   the mass of the primes ≤ x if you say each prime  is mass 1; in some sense this is a little more natural because this is now saying what's the  mass of the primes less than or equal to x if   the mass actually goes up as the prime goes up;  in particular here the mass we're counting of   a prime p is log(p) so if p is a big prime it  "weighs more", that's really what this is doing Now if we take our notion of the density  of the primes, then how big should this be?   Well the sum across p ≤ x of log(p),  well this thing should be really close   to the sum across let's say all integers less than  or equal to x of the density of primes near there   times the log of that number. Really here I'm  interpreting this density as being almost like   a probability. The probability that n is  prime tells me what percentage of a log(n)   I end up getting here. Now probability is of  course nonsense because every number either is   prime or isn't prime, so we can't really think of  the probability that a particular number is prime,   but over a large interval we could make  sense of the density over that interval   being close to the number of primes and so hence  the contribution of the number of logs that we get   of about that size. Repeating a theme from  earlier, if we look over a long interval this   change where we replaced a bunch of discrete  logarithms that were large with a bunch of   𝛿 fractions of logarithms introduces very little  change because logarithm itself changes so slowly,   and so if I have a good understanding  of the density (which I claim we do)   then I have a good understanding of the  size of this sum, which means I have a   very good understanding of the size of this sum.

[00:53:37]But if you keep in mind what we said  before about what the density was,   the density was just 1 /  log(n), so that times log(n),   which says that this is the sum across n ≤ x  of 1, which gives you the statement that if   you sum the logarithms of the primes less than  or equal to x, you should get very close to x.   Which is a pretty surreal statement to be  making, but it's one I've already shown you   this fairly striking visualization  of at the beginning of the video.

[00:54:26]If you'd like to see this with numbers  I'll show you on screen the sum across the   primes less than 10 million of the log of that  prime, and you can see just by looking at it   that the difference of this from 10 million  itself is less than a tenth of a percent.

[00:54:47]I hope you've enjoyed this little  romp through analytic number theory.   If you have any questions please  let me know in the comments.   I intend to post more videos like this, so if you  liked what you saw please consider subscribing   so you'll get a notification when I post more, and  doing all those other YouTube things; press the   thumbs up button, ring the bell, you know the  drill. But for now, until next time, be well.