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Euler Zeta
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0:04
If I ask you Brady, what's 1 plus 2 plus 3 plus 4 - and if I just keep adding forever - what would you say the answer to that would be? - (Brady: Well I would say) (it would go- it would tend towards infinity.) Yeah that makes sense, doesn't it? Yeah. (Am I right?) No. It's minus 1/12, it's negative. I've added all these positive numbers together up to infinity and I've got minus 1/12. And what's even more bizarre is that this result is used in many areas of physics. In fact this result is
0:48
critical to getting the 26 dimensions of string theory to pop out. This result is critical in understanding the Casimir force in quantum electrodynamics, but I thought what I would do today is we'd show a proof of this result. - (That would be useful!) Due to Euler. It gets a little hairy, but if we just hang in there we'll- we'll do it together, we'll go slowly. So so the way it begins with is to- rather than use these 1 plus 2 plus 3 we use, let's just think of this
1:24
series: 1 plus x plus x squared plus x cubed and just keep adding higher and higher powers of x. And what we know is that this is equal to 1 over 1 minus x, that's a known result. In fact it's strictly true for x less than unity. So as it stands this is- this is it's limited, formally limited, in its value we're going to we're going to manipulate it a little bit, we're going to push the boundaries at which we were going to use it. We can either prove that now or or if
2:01
you want to accept it then- - (Let's accept it.) It's accepted. (We're accepting that.) It's not a difficult proof but we can accept it. Now we're going to do something which many of the Numberphile viewers will know about but some may not know; we're going to do something called- we're going to differentiate this. And what that means is we're work out that- what differentiation is is working out the rate of change. So for example if you know the distance as a
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function of time and I differentiate it it tells me the rate of change of the distance or the velocity. So the rule that you need to know for this is the differentiation operator is called d by dx. If it acts on something like x to the n then the only thing we need to know is that this gives me- brings down the n and it lowers the power by 1. That's all we need to know in order to be able to carry on. So we do it, well the differential of 1 is zero, the differential of x, this is x to the 1 so
2:58
n is 1 so this brings down a 1 and that becomes x to the 0, which is 1, so the first term is 1x squared; well I differentiate x squared so n equals 2 here. So it becomes 2 times x. The next term is from x cubed, so n is 3, so this becomes 3x squared ok? So you get the idea, the next term would be 4x cubed for example. And this equals - and you'll have to just accept my word for it at the moment - the right hand side when I differentiate it just becomes 1 over 1 minus x squared. Now what I'm
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going to do is I'm going to set x to be equal to minus 1 all right? This is what Euler did. - (You're just doing- what do you mean-) (you're just going to do that?) - Yeah. Well I'm just going to fit it into this, see what it tells me, this formula is going to give me. So what does it give me? It gives me 1 minus 2, because of x is minus 1, plus 3 because x squared is positive, minus 4 plus 5 etc minus- and if I set x equal to minus 1 here it becomes 1 over
4:12
2 squared, so this is 1/4. So this is already beginning to look a little strange. - (Hang on, this seems significantly-) (so what have we just shown here?) - We've just shown that 1 minus 2 plus 3 minus 4 plus 5 minus 6 plus 7 is equal all the way up to infinity is equal to 1/4. I'm now going to introduce, I think, one of the best functions ever okay? And I'm sure we're going to be seeing a lot of it in Numberphile over the next few months maybe? And it was first introduced
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I think by Euler and then it was generalised by Riemann. And it plays a huge role in understanding the distribution of prime numbers and it will play a huge role in verifying this calculation okay? - (What's it called?) It's called the Riemann zeta function, sometimes called the Euler-Riemann zeta function. And its definition is the following, it's got a funny symbol, so it's got a symbol zeta. And the argument of it is called s, and its definition is it's the
5:17
infinite sum - so I'm going to sum from of the integers from n equals 1 to infinity - of 1 over n to the s, ok? So what what is? This this is 1, well let me just 1 over 1 to the s, well that's just 1; plus 1 over 2 to the s; plus 1 over 3 to the s; plus 1 over 4 etc. Now Euler, he said that s was real, right, that it was a regular number along the real axis. What Riemann did, and I'll just quote it here for now, is that he allowed s to be made up of a real part which we'll call
6:01
sigma plus an imaginary part which we'll call t. So this is a complex ok? So this is this was the difference between Euler's approach and Riemann. Riemann generalised Euler. You can see sigma is the real bit and then t is the complex bit, this is called and it's an imaginary number because it's i times t and together they form a complex number. We're going to concentrate on the real bit here. What Euler now did, he multiplies this by 2 to the minus s times zeta of s, ok? So what is that? So
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this is just 2 to the minus s plus - another way of writing 1 over 2 to the s is 2 to the minus s. So this is equal to plus 4 to the minus s plus - this is here this is 3 to the minus s - times 2 to the minus s gives me 6 to the minus s, plus 8 to the minus s, plus all the way up ok? So he just multiplied this zeta by by this. So what Euler now did was he takes two of these away from this, ok? So let me just write it out slowly so- let me just write out the first term which
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is the zeta so it's 1, plus 2 to the minus s, plus 3 to the minus s, plus 4 to the minus s, plus etc. And now if I take away this twice this thing I get- the first thing I get is I'm taking away minus 2 lots of- and I'm going to line these up, ok? I'm lining the two up there, I'm going to line the four up there,
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and then I would line the six up, you know with the- matching the six. But now what you can see, so the 1 just lives by itself that's ok, I've got 2 to the minus s minus 2 lots of them, so this gives me minus 2 to the minus s. The 3 is by itself so that just stays as it is. 4 minus 2 lots of 4 gives me minus 1 lot of 4 - so you can see I'm beginning to alternate ok? And I need some more paper Brady. - (Alright, I'll get you some more.) So now what he does, as he did
8:42
above there when he set x equal to minus 1, he sets s equal to minus 1 ok? And now let's just ask what what does this expression look like? So on the left hand side here, when we do that we've got 1 minus 2 times 2 to the minus, minus minus 1 which is plus 1, so it's 1 minus 4 which is minus 3 okay? But what is zeta of minus 1? Well if I look up here and put s equals to minus 1 in here, can you see that this becomes 1 plus 2 plus 3 plus 4 etc?
9:28
So all the way up forever. And now on the right hand side - so I've done the left hand side, on the right hand side I do the same. I set s equal to minus 1 and you can see what I've got here is 1 minus 2 plus 3 minus 4 plus 5 etc. But he's already shown what that is, if we look here we've already seen that 1 minus 2 plus 3 minus 4 plus 5 is 1/4. And I divide out by the 3- minus 3 and I get that 1 plus 2 plus 3 plus 4 plus ... is actually equal to minus 1/12. And that is
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the Euler 'proof' in quotation marks that the sum of the integers is minus 1/12. And the- so it sounds crazy right? - (It does a bit.) It sounds crazy. And what- what you can do and what Riemann was able to show is that this actually makes sense when you look at the Riemann function, the Riemann zeta function, the complex version of it, and set s equals to minus 1. He showed that zeta of minus 1 indeed is equal to minus 1/12. And I think a way of understanding that what's going on
11:06
here is that when you look at it in terms of the Riemann zeta function there is actually- it diverges at one point, the Riemann zeta function. And that one point is when s equals 1. So zeta- so remember what zeta-
11:29
so zeta of s we said was 1, plus 1 over 2 to the s, plus 1 over 3 to the s etc. So if I set s equal to 1 then zeta of 1 is 1 plus 1/2 plus 1/3 plus 1/4 ... And this is a very slowly diverging series, this tends to infinity. And I think one way of kind of trying to understand why it is that the the actual Riemann zeta function, which we may describe in more detail when we come to think about the primes, why zeta of minus 1 is equal to minus 1/12; the reason why this is is
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consistent with the fact that we actually think when we sum over the integers they're actually infinite is that in in some way we're having to go through this point here. And what Riemann is able to demonstrate is by kind of circumventing it, which is the the physical way of saying removing the divergence, we can end up with these sensible numbers like minus 1/12. (So, Professor, ans- I mean the question I) (have is, if I sat down with a calculator) (and wrote 1 plus 2 plus 3 plus 4 plus 5,)
13:05
(and did it forever, and then press the) (equals sign, am I going to get minus 1/12?) You see I- I don't know, in reality. One of the things you have to be careful about here, that we have to be careful about, is that what you have just stated there is basically using the set of axioms that we're- that we're used to in in this in in the context of finite sums. These aren't finite. These are infinite sums. So in other words you could you could keep going to the end of the
13:38
universe typing in the numbers and you'd still have made epsilon contribution to this overall sum. - (Epsilon means) (really small?) - Really small. And I think that hidden in there is a subtlety of- you've got to deal with divergent numbers, divergent sums, very carefully. And so the logic which goes in to show that, you know, something is clearly divergent - as you were just saying there with the putting it into the calculator; it doesn't necessarily follow when you're really dealing with
14:11
these divergent sums, you have to be very careful as to understand where the divergence is emerging. And I think this is an example of a case where this Riemann zeta function is perfectly well behaved at zeta of minus 1 and it's it's it's it's called an analytic continuation, it's taken you from one regime where it clearly looks like it's divergent and moving it into a regime where it's better defined. - (I can accept) (that zero is a number of some) (significance, and that 1 is, and that)
14:47
(infinity although it's not a number is a) (significant thing. And I can accept pi, as) (you know pi, and e but minus 1/12 just) (seems so- ) - Oh, yeah, I know but but it does play a- play a role in lots of different things does the number 12 and and it's- so for example as I said the the calculations of the critical dimension in string theory, the 26 dimensions, comes from this calculation. In fact it's basically the strings live in a two-dimensional- move in a
15:20
two-dimensional plane and and then they've got orthogonal directions and there's 24 of these orthogonal directions, which is 2 times 12 and and that's- this calculation is where- and you actually get the critical number of dimensions, it gives me the 24. And there's a wonderful mathematician called John Baez or Baez, let me just write him down and people might want to go and look up some of his work. He's he's given seminars on the importance of 24 and 8. And it's used in all sorts of different
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packing formulas, it's it appears all over.
16:02
First let's let's go back and look at this and why is this answer seemingly on one level ridiculous. Well let's look at the the first n integers, that first n natural numbers. So we actually do the sum 1 to n. So you go 1 plus 2 plus 3 all the way up to n okay, right? Now we know the answer to this, it's n(n plus 1) divided by 2. What I want to emphasise here is we- this is called- this here that I've written down here is that is the partial sum, this is the nth partial sort.
16:35
Clearly as I take n to infinity this number's just getting bigger and bigger and bigger and bigger. I should be getting the answer infinity. But of course, you know, infinity doesn't really exist in nature does it? You know, you don't measure an infinity in nature you just don't do that. But you can attach a value to this sum. You can attach a number to this sum and it's unique. And there's various ways in which you can do that; one way is this is a special case of the Riemann zeta
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function. So what is the Riemann zeta function? The Riemann zeta function is the following: define like so, it's the sum from n equals 1 to infinity of n to the minus z. Ok? So if z is bigger than 1 this sum is perfectly well defined, it gives a nice finite answer, so you're basically just taking numbers that get smaller and smaller and smaller and smaller and adding them together, ok? It's intuitively feasible that you would get a finite answer from that. So from z greater than 1 you
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get this. Now what you do is you do something very important, you do this trick of analytic continuation, that's what it's called. Basically, you know that you have- you have this this formula, so you have some expression for this sum that works for z greater than 1, and then you say ok let's assume that that expression also works for different values of z; where it's not greater than 1, where it's less than 1 say. And actually assuming that the the functions that you're talking about have
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very special properties, namely that they're analytic, that means that all they're- they're continuous functions, that they're nice and smooth, that there's no kinks or anything like that. Assuming that holds then you can make this extension. When you take the analytic continuation of the Riemann zeta function and you evaluate it at minus 1 - I chose minus 1 because when I plug in minus 1 I just get the sum from n equals 1 to infinity of n, which is precisely this sum here, ok? So
18:43
at minus 1, well I do the analytic continuation, I plug it into my nice formula that I derived for z greater than 1, but I now apply it to z less than- less than 1, and I get minus 1/12. Okay? Through this process of analytic continuation. Now you're probably still skeptical, not really believing that this works, but but we can prove it without doing this ok? Just to show that we can- there are ways to get to this, there's loads of ways you can get to this answer ok? This
19:12
is the sort of the classic way using this Riemann zeta function and analytic continuation but there are a million ways to do this. Okay so shall we just prove it? (Yeah!) - Let's just- so shall I? (D'you want more paper?) - Yeah. (When in physics do you need) (to know what 1 plus 2 plus 3 plus 4 plus) (5 plus 6 plus 7 etc?) - Well, ok, so so essentially in physics what you do is- Okay so we saw one example right? In Polchinski's string theory book where he
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was trying to predict the dimensionality of the universe and actually this sum appears because you end up having to sum up all different- you can think of a string right? An actual- the strings that build string theory are built from. They have different modes of oscillation. They have the sort of single one, first harmonic, second harmonic, third harmonic, and so on and so on and you have an infinite number of harmonics. And you have to sum up all their contributions,
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and when you sum up all their contributions, naively you would get an infinite answer but actually no you don't you get minus 1/12 appearing. And that's the- actually the minus 1/12 that then feeds back into 26 dimensions. But it's not just in string theory it's- these kinds of sums which naively appear - and integrals as well, an integral is just a continuous sum - these kind of divergent sums, divergent integrals, they appear all over all over quantum field theory where you
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have seemingly infinite contributions but you can regulate them, you can make sense of that infinity. And- and the answers- how you make sense of that infinity should not affect the answer; and indeed it doesn't. And the certain properties of sums that that you can- that you can- they have to satisfy certain theorems and whatnot that- so that they do make sense, that they do give unique answer. Some sums just are too crazy. But certainly ones like this a totally totally nice
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and you can get nice finite answers out of them.
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