# Abraham Bers, 1930 – 2015

I was saddened to hear that my PhD thesis advisor at MIT, Professor Abraham Bers, passed away last week at the age of 85. Abe was a fantastic physicist and mentor. He will be dearly missed by his many students. I showed up at MIT in the fall of 1986 with the intent of doing experimental particle physics. I took Abe’s plasma physics course as a breadth requirement for my degree. When I began, I didn’t know what a plasma was but by the end of the term I had joined his group. Abe was one of the best teachers I have ever had. His lectures exemplified his extremely clear and insightful mind. I still consult the notes from his classes from time to time.

Abe also had a great skill in finding the right problem for students. I struggled to get started doing research but one day Abe came to my desk with this old Russian book and showed me a figure. He said that it didn’t make sense according to the current theory and asked me to see if I could understand it. Somehow, this lit a spark in me and pursuing that little puzzle resulted in my first three papers. However, Abe also realized, even before I did I think, that I actually liked applied math better than physics. Thus, after finishing these papers and building some command in the field, he suggested that I completely switch my focus to nonlinear dynamics and chaos, which was very hot at the time. This turned out to be the perfect thing for me and it also made me realize that I could always change fields. I have never been afraid of going outside of my comfort zone since. I am always thankful for the excellent training I received at MIT.

The most eventful experience of those days was our weekly group meetings. These were famous no holds barred affairs where the job of the audience was to try to tear down everything the presenter said. I would prepare for a week to get ready when it was my turn. I couldn’t even get through the first slide my first time but by the time I graduated, nothing could faze me. Although the arguments could get quite heated at times, Abe never lost his cool. He would also come to my office after a particularly bad presentation to cheer me up. I don’t ever have any stress when giving talks or speaking in public now because I know that there could never be a sharper or tougher audience than Abe.

To me, Abe will always represent the gentleman scholar to which I’ve always aspired. He was always impeccably dressed with his tweed jacket, Burberry trench coat, and trademark bow tie. Well before good coffee became de rigueur in the US, Abe was a connoisseur and kept his coffee in a freezer in his office. He led a balanced life. He took work very seriously but also made sure to have time for his family and other pursuits. I visited him at MIT a few years ago and he was just as excited about what he was doing then as he was when I was a graduate student. Although he is gone, he will not be forgotten. The book he had been working on, Plasma Waves and Fusion, will be published this fall. I will be sure to get a copy as soon as it comes out.

2015-9-16: Here is a link to his MIT obituary.

# Hopfield on the difference between physics and biology

Here is a short essay by theoretical physicist John Hopfield of the Hopfield net and kinetic proofreading fame among many other things (hat tip to Steve Hsu). I think much of the hostility of biologists towards physicists and mathematicians that Hopfield talks about have dissipated over the past 40 years, especially amongst the younger set. In fact these days, a good share of Cell, Science, and Nature papers have some computational or mathematical component. However, the trend is towards brute force big data type analysis rather than the simple elegant conceptual advances that Hopfield was famous for. In the essay, Hopfield gives several anecdotes and summarizes them with pithy words of advice. The one that everyone should really heed and one I try to always follow is “Do your best to make falsifiable predictions. They are the distinction between physics and ‘Just So Stories.’”

# 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)

or

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

$\zeta(0^+)=-1+\frac{1}{2}=-\frac{1}{2}$

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

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

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.