Month: February 2014

Analytic continuation

Carson Chow Mathematics, Physics

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

Carson Chow Mathematics, Physics

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?

Carson Chow Mathematics, Physics

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.