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Here’s what’s really going on: You can define a function f from a subset of the complex numbers to the complex numbers where f(z) = \sum_{n=1}^\infty 1/n^z. Be

Here’s what’s really going on: You can define a function f from a subset of the complex numbers to the complex numbers where f(z) = \sum_{n=1}^\infty 1/n^z. Be careful with the domain of this function: the series does not converge for all z. You can plug in -1 and see that symbolically, f(-1) = 1 + 2 + 3 + .... But the series does not converge for z = -1, and it is simply not true that \sum_{n=1}^\infty n = -1/12; the series does not converge; equating it to something is a nonsensical thing to do. What is going then? Even though f is not defined for all complex numbers, there exist functions from the complex numbers to the complex numbers that -- restricted to the domain of f -- are equal to f. They "continue" f to all of the complex numbers. And if one imposes a restriction on these continuations (namely that they are analytic), then it turns out that there is a unique analytic continuation of f: the Riemann zeta function. And zeta(-1) = -1/12. Don’t confuse the definitions here. The Riemann zeta function can be defined as the analytic continuation of the series, the series is not defined in terms of the Riemann zeta function!

nwjtkjn 10y ago
What bothers me is that it seems there could be many ways of writing 1 + 2 + 3 + ... as the "specialization" of a formal series depending on a variable z, and for which the formal series converges to an analytic function on some domain away from the specialization. I can imagine there's a way of doing this in such a way that the function's analytic continuation to the specialization evaluates to any number you want, not just -1/12. However, I'm having trouble cooking up such an example.
- noobermin 10y ago
  It indeed makes you wonder whether it is true or not and whether it justifies the usage. This is one of the many reasons I left particle physics, so I have a biased perspective.
- Chinjut 10y ago
  This is a good concern. FWIW, though, there ARE natural summability methods which turn mere indexed sequences of terms into sums, without any particular choice as to how to interpret those sequences as specializations of formal series, which happen to assign the sequence whose n-th term is n a sum of -1/12. For an example of one such method, see the end of my post elsewhere in this thread.
- DavidSJ 10y ago
  A trivial and artificial example is g(z), defined as the sum of G(n, z) for n from 1 to infinity, where G(n, z) = 0 except if z is -1, in which case it is n. Thus g(-1) is 1 + 2 + 3 + ... while g(z) is otherwise 0 + 0 + 0 + ... = 0. And then the zero function is the unique analytic continuation of g. The reason people care about the Riemann zeta function is because of its deep connections to analysis, number theory, and physics.
  - nwjtkjn 10y ago
    Hm yeah, I was thinking of something more natural, like requiring the G(n,z)'s themselves be analytic on some domain. Anyway, sure the Riemann zeta is important, but I'm not sure how it's canonical.
- fmap 10y ago
  I can't speak for applications in physics, but in combinatorics and analytic number theory there is no magic involved. The idea in combinatorics is that you start by looking at the ring of infinite sequences over C with componentwise addition and Dirichlet multiplication as the product. That is, if a, b : nat -> C we define (a <*> b)(n) = \sum_{k | n} a(k) * b(n/k) It is easy to check that this is a ring, and it has wonderful properties which make it very easy to solve many equations of interest in this ring. This is usually called the ring of "arithmetic functions". However, it is difficult to get asymptotic estimates for the coefficients of a series by purely algebraic means. This is the first and last time that complex valued functions enter the picture, but it's a very neat trick. Let's consider two series a, b and define the functions A(s) = \sum_{n >= 1} a(n) * n^-s B(s) = \sum_{n >= 1} b(n) * n^-s then we have A(s) * B(s) = \sum_{n >= 1} (\sum_{k | n} a(k) * b(n/k)) * n^-s = \sum_{n >= 1} (a <*> b)(n) n^-s So this mapping, from the ring of arithmetic functions to the ring of (partial) complex functions with pointwise addition and multiplication. Glossing over some details for now, this allows you to analyze the function belonging to a sequence to gain information about the sequence itself. In particular, you can use the theory of complex integration and Cauchy's residue theorem to gain information about (partial sums of) coefficients. Unfortunately, the world is not quite this simple. The functions we are mapping into typically aren't very well behaved and usually aren't defined on large parts of the complex plane (consider a(n) = n^n). This means that all of our nice tools from complex analysis actually won't work very well! The whole idea behind "analytic continuations" is that we aren't actually using this mapping! We are constructing a different (partial, injective) ring homomorphism from arithmetic functions to meromorphic complex functions. The idea behind this is that meromorphic functions are rather restricted in what they can do. In particular, there is at most one meromorphic function A with A(s) = \sum_{n >= 1} a(n) n^-s for s with Re(s) > k, for some k. We define our mapping from sequences to functions by mapping the sequence a(n) to the meromorphic function A(s) with A(s) = \sum_{n >= 1} a(n) n^-s for Re(s) > k for some k, if this function exists. By the same argument as above, this is a ring homomorphism, and since it is injective we can still use information about the functions to gain information about the underlying sequences. For example, the Riemann zeta function is not really defined by the equation Zeta(s) = \sum_{n >= 1} n^-s. It is defined to be the unique meromorphic function with Zeta(s) = \sum_{n >= 1} n^-s for all s with R(s) > 1. In particular, Zeta(-1) has nothing to do with \sum_{n >= 1} n. The latter expression doesn't define a complex number at all, but for the former it is not too difficult to show that Zeta(-1) = -1/12. The main advantage, though, is that meromorphic functions are very well behaved. This allows us to use Cauchy's residue theorem and Mellin transforms to get very deep results about the underlying sequences. If you play this game with the "von Mangoldt" sequence you can, for instance, derive an asymptotic bound on the density of the prime numbers. This is a surprisingly simple derivation, given that this problem had the worlds greatest mathematicians stumped for a hundred years! Summing up, the mapping or "continuation" you use is choosen (!) so that multiplication of functions corresponds to your chosen multiplication in the ring of sequences and so that you get functions which are as well-behaved as possible. There is a large design space here, and you can find different "analytic continuations" for a given sequence.
noobermin 10y ago
This equating the analytic continuation to the divergent series is done throughout physics, specifically for a process known as integral regularization. For example, it's been a while since I really learned string theory, but IIRC it's this very summation that is relevant to the derived result that space time needs to have more than 4 dimensions in order for the photon predicted from string theories to be massless (which experimentally, they are). It makes one wonder if there can be a better theory then that doesn't require magic. There is at least one book I know of (before I became a plasma physicist) which apparently avoids the regularization altogether, but people still do regularization like this in particle physics.
- iheartmemcache 10y ago
  This is a very approachable explanation by [mathematician] John Baez https://youtu.be/vzjbRhYjELo?t=426 https://youtu.be/vzjbRhYjELo?t=426 and going into this with a bit of depth mentioning the 'rigorous' (i.e., analytically 'acceptable' form of the summation of Riemann Zeta function and the radius of convergence and all) along with the naive Euler summation. He goes on to mention a few times as to why it's fundamental in string theory (not to be confused with super string nor m-theory, which are of course different). A bunch of people had issues with the Numberphile episode where these restrictions where elided (due to time, I'd imagine), but they had another episode featuring the zeta function and a math professor (rather than physics professors, who were featured on the one that went viral) explaining the summation with more context: https://www.youtube.com/watch?v=0Oazb7IWzbA https://www.youtube.com/watch?v=0Oazb7IWzbA.
cohomologo 10y ago
There is a much simpler mathematical treatment of this series that appeared in my class on quantum field theory in the computation of the vacuum energy. In that case, the series was treated as lim \epsilon -> 0+ ( \sum_{n=1}^\infty n e^{- \epsilon n} + ...), that is the series was multiplied by a decaying exponential function with a rate of decay that goes to zero. This sum can easily be evaluated for small epsilon takes the form sum = 1/epsilon - 1/12 + ... Crucially, there was another term in the calculation that naturally appeared that canceled the 1/epsilon. Without that other term, the sum would of course be infinite when epsilon -> 0. This is much simpler than analytic continuation through the complex plane, and again, this is how the physics calculation appears in QFT courses. There is no need to appeal to complex analysis here, which leads to all of this mysticism and confusion.
- archgoon 10y ago
  That's completely unreadable. So how exactly is it that by multiplying a regular sum of positive numbers by a decaying exponent do you get a negative number when you take the limit?
  - tamana 10y ago
    You don't. You get -1/12 plus a positive infinite expression, which is then reduced by a positive infinite expression later. Its theoretical physics, not math.
    - archgoon 10y ago
      > which leads to all of this mysticism and confusion. Funny, I thought the entire point of this 'simplification' was not to be mystical and hand wavey.
  - bzbarsky 10y ago
    I think the argument goes like this: 1) You introduce a family of series, parametrized by epsilon, whose terms, for small epsilon, closely approximate your original series. Sort of; obviously for large n they don't. But the idea is that if you pick any N and delta you can pick epsilon such that for n < N the approximation is within delta of the actual terms. 2) You show that the series in this family can all be summed and the sum of each one is 1/epsilon - 1/12 + O(epsilon). Now of course what this means is that the sums blow up as you approximate your original series better and better, since 1/epsilon gets large. That's good, because your original series totally diverges off to infinity. ;) The part after this point I'm less clear on, but it sounds like in the computation involved what you actually have is your (divergent) series 1+2+3+... plus some _other_ (also divergent, going off to negative infinity) stuff. And that you might be able to arrange things such that the other divergent stuff looks like -1/epsilon, cancels out the 1/epsilon from your approximation, and you come out with the sum of the two things being -1/12. The obvious issue here is that once you start adding up divergent things by rearranging terms and telescoping you can come up with whatever answer you want: see <https://en.wikipedia.org/wiki/Riemann_series_theorem> https://en.wikipedia.org/wiki/Riemann_series_theorem>. So this procedure all only makes sense if there are some sort of fundamental reasons to think that this particular rearrangement is the "right" one in some sense.
fryguy 10y ago
I would say it's much like f(x) = sqrt(x) over real numbers. sqrt(-16) doesn't exist, but it is still convenient to think about the square root of negative number (for say factoring polynomials). In this case, our function happens to have a representation at f(-1) being equivilent to 1 + 2 + 3 + ..., and being able to calculate f(-1) = -1/12 much like sqrt(-16) = 4i when we introduce i = sqrt(-1).

nwjtkjn 10y ago

What bothers me is that it seems there could be many ways of writing 1 + 2 + 3 + ... as the "specialization" of a formal series depending on a variable z, and for which the formal series converges to an analytic function on some domain away from the specialization. I can imagine there's a way of doing this in such a way that the function's analytic continuation to the specialization evaluates to any number you want, not just -1/12. However, I'm having trouble cooking up such an example.

noobermin 10y ago
It indeed makes you wonder whether it is true or not and whether it justifies the usage. This is one of the many reasons I left particle physics, so I have a biased perspective.
Chinjut 10y ago
This is a good concern. FWIW, though, there ARE natural summability methods which turn mere indexed sequences of terms into sums, without any particular choice as to how to interpret those sequences as specializations of formal series, which happen to assign the sequence whose n-th term is n a sum of -1/12. For an example of one such method, see the end of my post elsewhere in this thread.
DavidSJ 10y ago
A trivial and artificial example is g(z), defined as the sum of G(n, z) for n from 1 to infinity, where G(n, z) = 0 except if z is -1, in which case it is n. Thus g(-1) is 1 + 2 + 3 + ... while g(z) is otherwise 0 + 0 + 0 + ... = 0. And then the zero function is the unique analytic continuation of g. The reason people care about the Riemann zeta function is because of its deep connections to analysis, number theory, and physics.
- nwjtkjn 10y ago
  Hm yeah, I was thinking of something more natural, like requiring the G(n,z)'s themselves be analytic on some domain. Anyway, sure the Riemann zeta is important, but I'm not sure how it's canonical.
fmap 10y ago
I can't speak for applications in physics, but in combinatorics and analytic number theory there is no magic involved. The idea in combinatorics is that you start by looking at the ring of infinite sequences over C with componentwise addition and Dirichlet multiplication as the product. That is, if a, b : nat -> C we define (a <*> b)(n) = \sum_{k | n} a(k) * b(n/k) It is easy to check that this is a ring, and it has wonderful properties which make it very easy to solve many equations of interest in this ring. This is usually called the ring of "arithmetic functions". However, it is difficult to get asymptotic estimates for the coefficients of a series by purely algebraic means. This is the first and last time that complex valued functions enter the picture, but it's a very neat trick. Let's consider two series a, b and define the functions A(s) = \sum_{n >= 1} a(n) * n^-s B(s) = \sum_{n >= 1} b(n) * n^-s then we have A(s) * B(s) = \sum_{n >= 1} (\sum_{k | n} a(k) * b(n/k)) * n^-s = \sum_{n >= 1} (a <*> b)(n) n^-s So this mapping, from the ring of arithmetic functions to the ring of (partial) complex functions with pointwise addition and multiplication. Glossing over some details for now, this allows you to analyze the function belonging to a sequence to gain information about the sequence itself. In particular, you can use the theory of complex integration and Cauchy's residue theorem to gain information about (partial sums of) coefficients. Unfortunately, the world is not quite this simple. The functions we are mapping into typically aren't very well behaved and usually aren't defined on large parts of the complex plane (consider a(n) = n^n). This means that all of our nice tools from complex analysis actually won't work very well! The whole idea behind "analytic continuations" is that we aren't actually using this mapping! We are constructing a different (partial, injective) ring homomorphism from arithmetic functions to meromorphic complex functions. The idea behind this is that meromorphic functions are rather restricted in what they can do. In particular, there is at most one meromorphic function A with A(s) = \sum_{n >= 1} a(n) n^-s for s with Re(s) > k, for some k. We define our mapping from sequences to functions by mapping the sequence a(n) to the meromorphic function A(s) with A(s) = \sum_{n >= 1} a(n) n^-s for Re(s) > k for some k, if this function exists. By the same argument as above, this is a ring homomorphism, and since it is injective we can still use information about the functions to gain information about the underlying sequences. For example, the Riemann zeta function is not really defined by the equation Zeta(s) = \sum_{n >= 1} n^-s. It is defined to be the unique meromorphic function with Zeta(s) = \sum_{n >= 1} n^-s for all s with R(s) > 1. In particular, Zeta(-1) has nothing to do with \sum_{n >= 1} n. The latter expression doesn't define a complex number at all, but for the former it is not too difficult to show that Zeta(-1) = -1/12. The main advantage, though, is that meromorphic functions are very well behaved. This allows us to use Cauchy's residue theorem and Mellin transforms to get very deep results about the underlying sequences. If you play this game with the "von Mangoldt" sequence you can, for instance, derive an asymptotic bound on the density of the prime numbers. This is a surprisingly simple derivation, given that this problem had the worlds greatest mathematicians stumped for a hundred years! Summing up, the mapping or "continuation" you use is choosen (!) so that multiplication of functions corresponds to your chosen multiplication in the ring of sequences and so that you get functions which are as well-behaved as possible. There is a large design space here, and you can find different "analytic continuations" for a given sequence.

noobermin 10y ago

This equating the analytic continuation to the divergent series is done throughout physics, specifically for a process known as integral regularization. For example, it's been a while since I really learned string theory, but IIRC it's this very summation that is relevant to the derived result that space time needs to have more than 4 dimensions in order for the photon predicted from string theories to be massless (which experimentally, they are). It makes one wonder if there can be a better theory then that doesn't require magic. There is at least one book I know of (before I became a plasma physicist) which apparently avoids the regularization altogether, but people still do regularization like this in particle physics.

iheartmemcache 10y ago
This is a very approachable explanation by [mathematician] John Baez https://youtu.be/vzjbRhYjELo?t=426 https://youtu.be/vzjbRhYjELo?t=426 and going into this with a bit of depth mentioning the 'rigorous' (i.e., analytically 'acceptable' form of the summation of Riemann Zeta function and the radius of convergence and all) along with the naive Euler summation. He goes on to mention a few times as to why it's fundamental in string theory (not to be confused with super string nor m-theory, which are of course different). A bunch of people had issues with the Numberphile episode where these restrictions where elided (due to time, I'd imagine), but they had another episode featuring the zeta function and a math professor (rather than physics professors, who were featured on the one that went viral) explaining the summation with more context: https://www.youtube.com/watch?v=0Oazb7IWzbA https://www.youtube.com/watch?v=0Oazb7IWzbA.

cohomologo 10y ago

There is a much simpler mathematical treatment of this series that appeared in my class on quantum field theory in the computation of the vacuum energy. In that case, the series was treated as lim \epsilon -> 0+ ( \sum_{n=1}^\infty n e^{- \epsilon n} + ...), that is the series was multiplied by a decaying exponential function with a rate of decay that goes to zero. This sum can easily be evaluated for small epsilon takes the form sum = 1/epsilon - 1/12 + ... Crucially, there was another term in the calculation that naturally appeared that canceled the 1/epsilon. Without that other term, the sum would of course be infinite when epsilon -> 0. This is much simpler than analytic continuation through the complex plane, and again, this is how the physics calculation appears in QFT courses. There is no need to appeal to complex analysis here, which leads to all of this mysticism and confusion.

archgoon 10y ago
That's completely unreadable. So how exactly is it that by multiplying a regular sum of positive numbers by a decaying exponent do you get a negative number when you take the limit?
- tamana 10y ago
  You don't. You get -1/12 plus a positive infinite expression, which is then reduced by a positive infinite expression later. Its theoretical physics, not math.
  - archgoon 10y ago
    > which leads to all of this mysticism and confusion. Funny, I thought the entire point of this 'simplification' was not to be mystical and hand wavey.
- bzbarsky 10y ago
  I think the argument goes like this: 1) You introduce a family of series, parametrized by epsilon, whose terms, for small epsilon, closely approximate your original series. Sort of; obviously for large n they don't. But the idea is that if you pick any N and delta you can pick epsilon such that for n < N the approximation is within delta of the actual terms. 2) You show that the series in this family can all be summed and the sum of each one is 1/epsilon - 1/12 + O(epsilon). Now of course what this means is that the sums blow up as you approximate your original series better and better, since 1/epsilon gets large. That's good, because your original series totally diverges off to infinity. ;) The part after this point I'm less clear on, but it sounds like in the computation involved what you actually have is your (divergent) series 1+2+3+... plus some _other_ (also divergent, going off to negative infinity) stuff. And that you might be able to arrange things such that the other divergent stuff looks like -1/epsilon, cancels out the 1/epsilon from your approximation, and you come out with the sum of the two things being -1/12. The obvious issue here is that once you start adding up divergent things by rearranging terms and telescoping you can come up with whatever answer you want: see <https://en.wikipedia.org/wiki/Riemann_series_theorem> https://en.wikipedia.org/wiki/Riemann_series_theorem>. So this procedure all only makes sense if there are some sort of fundamental reasons to think that this particular rearrangement is the "right" one in some sense.

fryguy 10y ago

I would say it's much like f(x) = sqrt(x) over real numbers. sqrt(-16) doesn't exist, but it is still convenient to think about the square root of negative number (for say factoring polynomials). In this case, our function happens to have a representation at f(-1) being equivilent to 1 + 2 + 3 + ..., and being able to calculate f(-1) = -1/12 much like sqrt(-16) = 4i when we introduce i = sqrt(-1).