Talk in Göttingen

I’m currently in Göttingen, Germany at the Bernstein Sparks Workshop: Beyond mean field theory in the neurosciences, a topic near and dear to my heart.  The slides for my talk are here.  Of course no trip to Göttingen would be complete without a visit to Gauss’s grave and Max Born’s house. Photos below.


New paper on path integrals

Carson C. Chow and Michael A. Buice. Path Integral Methods for Stochastic Differential Equations. The Journal of Mathematical Neuroscience,  5:8 2015.

Abstract: Stochastic differential equations (SDEs) have multiple applications in mathematical neuroscience and are notoriously difficult. Here, we give a self-contained pedagogical review of perturbative field theoretic and path integral methods to calculate moments of the probability density function of SDEs. The methods can be extended to high dimensional systems such as networks of coupled neurons and even deterministic systems with quenched disorder.

This paper is a modified version of our arXiv paper of the same title.  We added an example of the stochastically forced FitzHugh-Nagumo equation and fixed the typos.

Talk at Jackfest

I’m currently in Banff, Alberta for a Festschrift for Jack Cowan (webpage here). Jack is one of the founders of theoretical neuroscience and has infused many important ideas into the field. The Wilson-Cowan equations that he and Hugh Wilson developed in the early seventies form a foundation for both modeling neural systems and machine learning. My talk will summarize my work on deriving “generalized Wilson-Cowan equations” that include both neural activity and correlations. The slides can be found here. References and a summary of the work can be found here. All videos of the talks can be found here.


Addendum: 17:44. Some typos in the talk were fixed.

Addendum: 18:25. I just realized I said something silly in my talk.  The Legendre transform is an involution because the transform of the transform is the inverse. I said something completely inane instead.

Analytic continuation continued

As I promised in my previous post, here is a derivation of the analytic continuation of the Riemann zeta function to negative integer values. There are several ways of doing this but a particularly simple way is given by Graham Everest, Christian Rottger, and Tom Ward at this link. It starts with the observation that you can write

\int_1^\infty x^{-s} dx = \frac{1}{s-1}

if the real part of s>0. You can then break the integral into pieces with

\frac{1}{s-1}=\int_1^\infty x^{-s} dx =\sum_{n=1}^\infty\int_n^{n+1} x^{-s} dx

=\sum_{n=1}^\infty \int_0^1(n+x)^{-s} dx=\sum_{n=1}^\infty\int_0^1 \frac{1}{n^s}\left(1+\frac{x}{n}\right)^{-s} dx      (1)

For x\in [0,1], you can expand the integrand in a binomial expansion

\left(1+\frac{x}{n}\right)^{-s} = 1 +\frac{sx}{n}+sO\left(\frac{1}{n^2}\right)   (2)

Now substitute (2) into (1) to obtain

\frac{1}{s-1}=\zeta(s) -\frac{s}{2}\zeta(s+1) - sR(s)  (3)


\zeta(s) =\frac{1}{s-1}+\frac{s}{2}\zeta(s+1) +sR(s)   (3′)

where the remainder R is an analytic function when Re s > -1 because the resulting series is absolutely convergent. Since the zeta function is analytic for Re s >1, the right hand side is a new definition of \zeta that is analytic for s >0 aside from a simple pole at s=1. Now multiply (3) by s-1 and take the limit as s\rightarrow 1 to obtain

\lim_{s\rightarrow 1} (s-1)\zeta(s)=1

which implies that

\lim_{s\rightarrow 0} s\zeta(s+1)=1     (4)

Taking the limit of s going to zero from the right of (3′) gives


Hence, the analytic continuation of the zeta function to zero is -1/2.

The analytic domain of \zeta can be pushed further into the left hand plane by extending the binomial expansion in (2) to

\left(1+\frac{x}{n}\right)^{-s} = \sum_{r=0}^{k+1} \left(\begin{array}{c} -s\\r\end{array}\right)\left(\frac{x}{n}\right)^r + (s+k)O\left(\frac{1}{n^{k+2}}\right)

 Inserting into (1) yields

\frac{1}{s-1}=\zeta(s)+\sum_{r=1}^{k+1} \left(\begin{array}{c} -s\\r\end{array}\right)\frac{1}{r+1}\zeta(r+s) + (s+k)R_{k+1}(s)

where R_{k+1}(s) is analytic for Re s>-(k+1).  Now let s\rightarrow -k^+ and extract out the last term of the sum with (4) to obtain

\frac{1}{-k-1}=\zeta(-k)+\sum_{r=1}^{k} \left(\begin{array}{c} k\\r\end{array}\right)\frac{1}{r+1}\zeta(r-k) - \frac{1}{(k+1)(k+2)}    (5)

Rearranging (5) gives

\zeta(-k)=-\sum_{r=1}^{k} \left(\begin{array}{c} k\\r\end{array}\right)\frac{1}{r+1}\zeta(r-k) -\frac{1}{k+2}     (6)

where I have used

\left( \begin{array}{c} -s\\r\end{array}\right) = (-1)^r \left(\begin{array}{c} s+r -1\\r\end{array}\right)

The righthand side of (6) is now defined for Re s > -k.  Rewrite (6) as

\zeta(-k)=-\sum_{r=1}^{k} \frac{k!}{r!(k-r)!} \frac{\zeta(r-k)(k-r+1)}{(r+1)(k-r+1)}-\frac{1}{k+2}

=-\sum_{r=1}^{k} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \frac{\zeta(r-k)(k-r+1)}{(k+1)(k+2)}-\frac{1}{k+2}

=-\sum_{r=1}^{k-1} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \frac{\zeta(r-k)(k-r+1)}{(k+1)(k+2)}-\frac{1}{k+2} - \frac{\zeta(0)}{k+1}

Collecting terms, substituting for \zeta(0) and multiplying by (k+1)(k+2)  gives

(k+1)(k+2)\zeta(-k)=-\sum_{r=1}^{k-1} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \zeta(r-k)(k-r+1) - \frac{k}{2}

Reindexing gives

(k+1)(k+2)\zeta(-k)=-\sum_{r'=2}^{k} \left(\begin{array}{c} k+2\\ r'\end{array}\right) \zeta(-r'+1)r'-\frac{k}{2}

Now, note that the Bernoulli numbers satisfy the condition \sum_{r=0}^{N-1} B_r = 0.  Hence,  let \zeta(-r'+1)=-\frac{B_r'}{r'}

and obtain

(k+1)(k+2)\zeta(-k)=\sum_{r'=0}^{k+1} \left(\begin{array}{c} k+2\\ r'\end{array}\right) B_{r'}-B_0-(k+2)B_1-(k+2)B_{k+1}-\frac{k}{2}

which using B_0=1 and B_1=-1/2 gives the self-consistent condition


which is the analytic continuation of the zeta function for integers k\ge 1.

Analytic continuation

I have received some skepticism that there are possibly other ways of assigning the sum of the natural numbers to a number other than -1/12 so I will try to be more precise. I thought it would be also useful to derive the analytic continuation of the zeta function, which I will do in a future post.  I will first give a simpler example to motivate the notion of analytic continuation. Consider the geometric series 1+s+s^2+s^3+\dots. If |s| < 1 then we know that this series is equal to

\frac{1}{1-s}                (1)

Now, while the geometric series is only convergent and thus analytic inside the unit circle, (1) is defined everywhere in the complex plane except at s=1. So even though the sum doesn’t really exist outside of the domain of convergence, we can assign a number to it based on (1). For example, if we set s=2 we can make the assignment of 1 + 2 + 4 + 8 + \dots = -1. So again, the sum of the powers of two doesn’t really equal -1, only (1) is defined at s=2. It’s just that the geometric series and (1) are the same function inside the domain of convergence. Now, it is true that the analytic continuation of a function is unique. However, although the value of -1 for s=-1 is the only value for the analytic continuation of the geometric series, that doesn’t mean that the sum of the powers of 2 needs to be uniquely assigned to negative one because the sum of the powers of 2 is not an analytic function. So if you could find some other series that is a function of some parameter z that is analytic in some domain of convergence and happens to look like the sum of the powers of two for some z value, and you can analytically continue the series to that value, then you would have another assignment.

Now consider my example from the previous post. Consider the series

\sum_{n=1}^\infty \frac{n-1}{n^{s+1}}  (2)

This series is absolutely convergent for s>1.  Also note that if I set s=-1, I get

\sum_{n=1}^\infty (n-1) = 0 +\sum_{n'=1}^\infty n' = 1 + 2 + 3 + \dots

which is the sum of then natural numbers. Now, I can write (2) as

\sum_{n=1}^\infty\left( \frac{1}{n^s}-\frac{1}{n^{s+1}}\right)

and when the real part of s is greater than 1,  I can further write this as

\sum_{n=1}^\infty\frac{1}{n^s}-\sum_{n=1}^\infty\frac{1}{n^{s+1}}=\zeta(s)-\zeta(s+1)  (3)

All of these operations are perfectly fine as long as I’m in the domain of absolute convergence.  Now, as I will show in the next post, the analytic continuation of the zeta function to the negative integers is given by

\zeta (-k) = -\frac{B_{k+1}}{k+1}

where B_k are the Bernoulli numbers, which is given by the Taylor expansion of

\frac{x}{e^x-1} = \sum B_n \frac{x^n}{n!}   (4)

The first few Bernoulli numbers are B_0=1, B_1=-1/2, B_2 = 1/6. Thus using this in (4) gives \zeta(-1)=-1/12. A similar proof will give \zeta(0)=-1/2.  Using this in (3) then gives the desired result that the sum of the natural numbers is (also) 5/12.

Now this is not to say that all assignments have the same physical value. I don’t know the details of how -1/12 is used in bosonic string theory but it is likely that the zeta function is crucial to the calculation.

Nonuniqueness of -1/12

I’ve been asked to give an example of how the sum of the natural numbers could lead to another value in the comments to my previous post so I thought it may be of general interest to more people. Consider again S=1+2+3+4\dots to be the sum of the natural numbers.  The video in the previous slide gives a simple proof by combining divergent sums. In essence, the manipulation is doing renormalization by subtracting away infinities and the left over of this renormalization is -1/12. There is another video that gives the proof through analytic continuation of the Riemann zeta function

\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}

The zeta function is only strictly convergent when the real part of s is greater than 1. However, you can use analytic continuation to extract values of the zeta function to values where the sum is divergent. What this means is that the zeta function is no longer the “same sum” per se, but a version of the sum taken to a domain where it was not originally defined but smoothly (analytically) connected to the sum. Hence, the sum of the natural numbers is given by \zeta(-1) and \zeta(0)=\sum_{n=1}^\infty 1, (infinite sum over ones). By analytic continuation, we obtain the values \zeta(-1)=-1/12 and \zeta(0)=-1/2.

Now notice that if I subtract the sum over ones from the sum over the natural numbers I still get the sum over the natural numbers, e.g.

1+2+3+4\dots - (1+1+1+1\dots)=0+1+2+3+4\dots.

Now, let me define a new function \xi(s)=\zeta(s)-\zeta(s+1) so \xi(-1) is the sum over the natural numbers and by analytic continuation \xi(-1)=-1/12+1/2=5/12 and thus the sum over the natural numbers is now 5/12. Again, if you try to do arithmetic with infinity, you can get almost anything. A fun exercise is to create some other examples.

The sum of the natural numbers is -1/12?

This wonderfully entertaining video giving a proof for why the sum of the natural numbers  is -1/12 has been viewed over 1.5 million times. It just shows that there is a hunger for interesting and well explained math and science content out there. Now, we all know that the sum of all the natural numbers is infinite but the beauty (insidiousness) of infinite numbers is that they can be assigned to virtually anything. The proof for this particular assignment considers the subtraction of the divergent oscillating sum S_1=1-2+3-4+5 \dots from the divergent sum of the natural numbers S = 1 + 2 + 3+4+5\dots to obtain 4S.  Then by similar trickery it assigns S_1=1/4. Solving for S gives you the result S = -1/12.  Hence, what you are essentially doing is dividing infinity by infinity and that as any school child should know, can be anything you want. The most astounding thing to me about the video was learning that this assignment was used in string theory, which makes me wonder if the calculations would differ if I chose a different assignment.

Addendum: Terence Tao has a nice blog post on evaluating such sums.  In a “smoothed” version of the sum, it can be thought of as the “constant” in front of an asymptotic divergent term.  This constant is equivalent to the analytic continuation of the Riemann zeta function. Anyway, the -1/12 seems to be a natural way to assign a value to the divergent sum of the natural numbers.