I am currently reading a book called The Music of the Primes which is about prime numbers and in particular the Riemann Hypothesis. It discusses the work of the Indian mathematician Ramanujan and, almost in passing, mentions that his equation:
appeared at first to be the ramblings of a madman. It doesn’t really elaborate, so I can only assume that the equation is in fact cromulent.
I looked on Wikipedia and found a page on Ramanujan summation, but that doesn’t really help much. I gather that it’s some kind of “cheat” to sum a series that doesn’t really have a sum as such.
I can see, for instance, why 1-2+3-4+… = ¼ – the partial sums kind of “average out” to this value as they diverge either side – but with a divergent series like 1+2+3+4+… I can’t see how this happens.
Can anyone explain in as near to layman’s terms as possible?
The key is the Riemann zeta function. If you take any s > 1, then the series Sum(1/n[sup]2[/sup],n=1…infinity) converges, and the value of that sum is defined as zeta(s). So zeta is a perfectly well-defined function for s > 1. But it turns out that there’s a very powerful and somewhat surprising theorem that if you have a function which is defined on some region, there is at most one function which is smooth on the complex plane and which matches up with that function. This complex function is called the analytic continuation of the original function, and since folks like analytic functions, there’s a tendancy to think of the analytic continuation as being the “same” as the original function. So, if we take the analytic continuation of the Riemann zeta function, we can evaluate it anywhere we like, not just for s real and greater than 1. Suppose, for instance, that we evaluate zeta(-1): It happens that that gives us 1/12. But if that’s really the same zeta, then from the original definition, zeta(-1) is (or should be) also equal to 1 + 2 + 3 + 4 + …
To clarify my last post a little, here is what I am now wondering:
Why is there no letter ‘s’ in the expresion “Sum(1/n[sup]2[/sup],n=1…infinity)”? If that is supposed to give me Zeta(s), why does it not mention s?
When you mention a function “defined on some region,” I imagine that by region, you mean something like a segment of the x axis. But I am not confident this is anything but imagination on my part. Can you clarify what “region” means here?
What does it mean for one function (smooth on the complex plane) to “match up” with another function (defined on some region)? Specifically, I mean to ask what’s “match up with” mean?
What is an analytic function and why do folks like them?
Why is not the result that 1+2+3+4+… = 1/12 taken to be a reductio against something, instead of just a suprising result of something? Those two quantities can not be equal, can they?
This is related to question 1. How do you get from “Sum(1/n[sup]2[/sup],n=1…infinity)” to 1 + 2 + 3 + 4 +… when s = -1? Where does s figure in “Sum(1/n[sup]2[/sup],n=1…infinity)”?
Analytic continuation is easiest to understand in the context of power series. Suppose that you have a power series sum( f(n)x[sup]n[/sup], 0 < n ) that converges when |x| < r. If we denote the sum as g(x), we have a function defined on a disk in the complex plane. By the theorem that Chronos mentioned, there is exactly one analytic function h(x) which is defined almost everywhere in the complex plane and has the same values as g(x) where they’re both defined. So we can treat h(x) as the sum of the series outside the disk we had initially.
For instance, consider a geometric series sum( [symbol]r[/symbol]x[sup]n[/sup], 0 < n ), where [symbol]r[/symbol] is not equal to 1. This sums to 1/(1 - [symbol]r[/symbol]x) for |x| < [symbol]r[/symbol]. The analytic continuation of that function is 1/(1 - [symbol]r[/symbol]x) everywhere on C, so we take that as the sum. At x = 1, the sum is equal to 1/(1 - [symbol]r[/symbol]), which is the formula you learned back in highschool without the restrictions on the absolute value of [symbol]r[/symbol]. As a point of trivia, 1 + 1 + 1 + … = -1/12.
There are other methods for summing divergent series. Analytic continuation is one of the most powerful in the sense that it can be used to sum more series than many others, but it doesn’t have some of the nice properties that, say, the Cesàro mean does.
Why are we licensed to treat h(x) as giving a “sum?” How is this use of the word “sum” continuous with the normal use of the term? Until all of what you’ve described came to light, the word “sum” meant (AFAIK) “the result of several additions.” But the result of an infinite number of additions of the number one is not -1/12. The number -1/12 is arrived at by a means other than repeated addition. So why call it a sum? Why not give it a new name?
Remind me in which course and in what context I learned the formula “1/(1 - [symbol]r[/symbol]),” in high school.
How do I read “sum( [symbol]r[/symbol]x[sup]n[/sup], 0 < n )”? I would have thought it means the sum of the numbers 2, 2, 2,… when rho is 2 and x is 1, the sum of the numbers 4, 8, 16… when rho is 2 and x is 2, and so on. But it appears from your subsequent comments that this is not the correct way to read the notation.
As I asked Chronos, though I now see that this question is related to my question 4 to you, why do we not say that since our theory gives us -1/12 as the sum of 1 + 1 + 1 +…, our theory is apparently wrong, since the claim that the sum of all those ones is -1/12 is absurd?
No, it’s just -1/12. The book in question is sitting on my “to be read” shelf, so I went and found the formula in question (it’s on p. 135 and 137, if anyone wants to look it up).
And no, it doesn’t look like the book does a very good job of explaining it as far as I can tell, though I haven’t actually read the book yet. But I’m pretty sure analytic continuation, as explained by Chronos and ultrafilter, is what the book had in mind.
Incidentally, the best attempt at explaining the mathematics behind the zeta function and the Riemann hypothesis at a popular level (to the extent that such a thing is possible) is Prime Obsession by John Derbyshire, which I highly recommend if you’re really interested.
I’ll answer this one only, and just briefly. “Analytic” is a basic concept from the theory of functions of complex variables which means something analogous to “differentiable” when talking about functions of real variables.
Frylock, maybe ultrafilter can answer your questions, but I’ll try to simplify it down even more than he did.
Consider the inifinite series 1 + x + x[sup]2[/sup] + x[sup]3[/sup] + x[sup]4[/sup] + …
If you replace x with a number strictly between -1 and 1 (like 1/2 for example), the series converges; it adds up to a sum of 1/(1-x).
(For example, 1 + 1/2 + 1/4 + 1/8 + 1/16 + … = 1/(1 - 1/2) = 1/(1/2) = 2.
If x is not between -1 and 1, the series 1 + x + x[sup]2[/sup] + x[sup]3[/sup] + x[sup]4[/sup] + … diverges (rather than getting closer and closer to some particular number as you add more and more of the terms, the running totals keep getting bigger and bigger without bound, or else bouncing around. But 1/(1-x) still works fine (as long as x is anything other than 1).
So, since 1 + x + x[sup]2[/sup] + x[sup]3[/sup] + x[sup]4[/sup] + … and 1/(1-x) are the “same function” for some x’s, it sorta makes sense to treat them as “the same” for other x’s where the series doesn’t really, actually add up. If, based on the fact that 1 + x + x[sup]2[/sup] + x[sup]3[/sup] + x[sup]4[/sup] + … = 1/(1-x) when -1 < x < 1, you say they’re equal for other x’s too, that would mean, for example, that 1 + 2 + 2[sup]2[/sup] + 2[sup]3[/sup] + 2[sup]4[/sup] + … = -1.
It’s an infinite sum; look at it as “The sum of all the values of f(n)x[sup]n[/sup] where 0 < n [and n is an integer]”. So it’s the sum of the infinite series f(0) * x^0 + f(1)*x^1 + f(2)*x^2 + …
You remember Taylor series from calculus? Where you approximate a function’s values near some point with a conceptual “infinite degree” polynomial “centered about” that point, whose nth coefficient is determined from the nth derivative of the function at that point? Well, Taylor series don’t always actually exist (if the function isn’t differentiable enough) or converge or give values agreeing with the function they approximate. Still, if your function is nice enough, this works well enough, which gives us the concept of the analytic function: An analytic function is one such that, for every point in its domain, there’s some neighborhood of that point such that the function agrees with its Taylor series everywhere in that neighborhood. (A neighborhood of a point, in case you don’t know what I mean by that, is just some set containing the point and some “wiggle room”; i.e., S is a neighborhood of x iff there’s some positive distance epsilon such that everything within epsilon of x is in S.)
If it’s hard to wrap your head around that definition, you’re in luck. We’re working in the context of complex numbers here, where, amazingly enough, the analytic functions are precisely those which are complex differentiable throughout their domain. [Complex differentiability is defined in basically the same way as real differentiability, but, I should note, it comes out to a much stronger condition than you might naively think or be used to from mere real differentiability. Specifically, if F is complex differentiable at the point p in the complex plane, then the “rate of change” of F (as a ratio of change in output to change in input) has to be the same no matter which direction you move away from p in. Obviously, this directionality aspect wasn’t such a concern when dealing with the mere one-dimensional real numbers and real differentiability.]
This one’s easy. A property holds “almost everywhere” in the complex plane if the set of points where it doesn’t hold has a total area of 0. (Thus, for example, the property of being a complex number but not a real number holds almost everywhere in the complex plane, since it fails on only one measly little line, which has area 0).
Because the method by which we arrived at it is through taking a very natural notion of generalization and applying it to the conventional “sum”, because the concept of the new notion of sum has some family resemblance with the conventional concept of sums, and because various useful properties of the old notion of sum carry on to the new notion. Though I don’t think most mathematicians are generally all that loose and liberal to freely assert things like “The sum of the infinite series 1+2+3+4+… is -1/12” without the appropriate context being set, the way they would freely say “The sum of the infinite series 1+1/2+1/4+… is 2”. I think most would take some care to make it clear that they are referring to a generalization based on analytic continuation.
For the summation of a geometric series. Consider a sum whose terms are in a constant ratio, so as to be A + Ar + Ar^2 + … + A*r^n. For simplicity, let’s pull the A out and get 1 + r + r^2 + r^3 + … + r^n. You probably learnt in precalculus or some such course that the sum of this series was [1 - r^(n+1)]/(1 - r). Furthermore, if we take the infinite analogue (the series 1 + r + r^2 + r^3 + …), and just apply the formula, we get something like [1 - [limit of r^n as n goes to infinity]]/(1 - r). If |r| happens to be less than 1, we know that repeated multiplications of r bring it closer and closer to 0, so that 0 is the limit of r^n as n goes to infinity. This gives us that the infinite geometric series 1 + r + r^2 + r^3 + … = 1/(1-r), whenever, as noted, the condition |r| < 1 is satisfied. You probably learnt that in a precalculus course alongside the formula for the finite sums. If not… well, you know it now.
As noted above, read “sum(f(n), some constraint on n)” as the sum of all the values f(n) for which (the integer) n satisfies the constraint. (With some implicit understood notion of how you are to order those values). Thus, “sum( [symbol]r[/symbol]x[sup]n[/sup], 0 < n )” denotes the infinite sum [symbol]r[/symbol]x[sup]0[/sup] + [symbol]r[/symbol]x[sup]1[/sup] + [symbol]r[/symbol]x[sup]2[/sup] + …
It’s less a theory and more an extension of the old definition. It’s absurd relative to some notions of what a sum should be, but, as I noted above in response to your question 4, I don’t think many mathematicians use the word “sum” in this special way without first making it clear that they are doing so. And this extension of the old definition has its uses towards some purposes and notions of what a sum should be, for which it is less manifestly “absurd”, giving us some justification for adopting it in those contexts (again, noted above in response to question 4).
(Oh, and, also, I reiterate my nitpick that 1 + 1 + 1 … should equal -1/2 and not -1/12, using the methods of this thread. It’s 1 + 2 + 3 + 4 + … that equals -1/12.)
This definition is correct; let me help a bit with its interpretation.
In taking the derivative of a real function f(x), we’re all familiar with the limit [f(x+h) - f(x)]/h; evaluate this under the assumption that h is a real number tending toward zero, and you get the derivative f’(x).
Now let’s look at the complex function f(z), where z is a complex number. How do you take the derivative of this? It’s not a trivial question because there are an infinte number of paths in the complex plane for a now-complex h to approach zero.
For example, suppose we wish to evaluate the derivative of f(z) = z^2. Using h = x, where x is a real number, x->0 means h -> 0 along the path of the real axis, and you get the usual derivative f’(z) = 2z. But what if h=ix? x->0 then means h tends toward zero along the imaginary axis. It turns out (if you are careful with the math and apply the chain rule judiciously) that you will get the same answer f’(z)=2z. This can be shown to be true for any path you take for h->0, and this fact is what makes f(z) analytic.
The fact that any polynomial in a complex variable is analytic in this way–it’s derivative does not depend on the path taken to get h->0–is why you can define an analytic function as one that agrees with its Taylor series in a neighborhood around any point in the complex plane.
A simple example of a non-analytic fnction (for variable z=x+iy) is f(z) = Real(z) = x. Obviously, taking h->0 along the path of the real axis yields f’(z)=1, but taking it along the path of the imaginary axis yields f’(z)=0.