From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Riemann zeta function Date: 11 May 2000 19:12:22 -0400 Newsgroups: sci.math Summary: [missing] In article <02558b7c.3fed7ea2@usw-ex0104-087.remarq.com>, coolman wrote: :What did I start?? All I wanted to know was how a simple function :becomes zero if s = -2. (I'm not a mathematician.) : And I made an attempt to answer - it looks like my reply dropped out. A quick summary: If you take the sums a_n(s) = 1/1^s + 1/2^s + ... + 1/n^s and consider their limit as n goes to infinity, you will get zeta(s) for Re(s) > 1 only. Now take the expressions b_n(s) = a_n(s) + (1/(s-1)) * 1/n^(s-1) - (1/2)/n^s + (s/12)/s^(n+1) The sequence {b_n(s)} has the same limit as {a_n(s)} for Re(s)>1 , namely zeta(s). But - and this is the "analytic continuation" trick: The sequence {b_n(s)} will also converge for Re(s) > -3, except when s=1. So, one has extended zeta to a larger domain, Re(s)>-3, z different from 1. (And the extension will be analytic, and the domain remains connected, if that is a concern). Check out that for s = -2, the sequence b_n(s) is actually constant 0. Cheers, ZVK(Slavek). ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Riemann zeta function Date: 11 May 2000 18:01:27 GMT Newsgroups: sci.math In article <391ab43c.12353216@news.ecn.ab.ca>, John Savard wrote: >On Mon, 08 May 2000 17:33:21 -0700, Mike Oliver >wrote, in part: > >>it has a unique analytic continuation to the >>entire complex plane, > >That's probably not quite true, since with functions like e^-(x^2) one >can construct functions that are analytic, but are identically zero >over large regions, to make any continuation non-unique. Um, no. The example you mentioned is the usual one used to construct functions which are C^\infty and vanish on a nonempty open set. But it is a fundamental feature of analytic functions that the only (entire) one which vanishes on a nonempty open set is the zero function. Think of it this way: one way to define C^\infty functions is that they _have_ a Taylor series at each point; one way to define analytic functions is that on a neighborhood of each point this Taylor series actually _converges_ to the original function. So you look where the function is already defined, compute its Taylor series near the edge, use that as the definition of the function on a slightly larger region, and, well, (analytically) continue! Actually the case of the Riemann zeta function is a bit easier: the series Sum(1/n^s) converges in a half-plane, and there is a functional equation which then instantly defines zeta(s) in most of the other half-plane, too. You do need some analytic continuation to get across the critical strip, although for real values of s between 0 and 1 you may use the convergent series Sum( (-1)^n/n^s ) to define a function which is easily related to zeta. (Maybe this trick even works for complex s in the critical strip but it wasn't obvious to me that this series converges for non-real s.) >But, as with >the gamma function, a particular analytic continuation can be simply >defined in terms of some integral, and that particular continuation, >in addition to being the simplest, happens to have the most >interesting mathematical properties. The situation with Gamma is that the "factorial" definition only extends to the non-negative integers, a set without a limit point, so that there is not a unique analytic continuation. (For example, you could replace Gamma(z) with Gamma(z)+sin(2 pi z).) The Bohr-Mollerup characterization defines Gamma by insisting not only on analyticity but on other properties as well (log-convexity). dave ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Riemann zeta function Date: 10 May 2000 03:43:41 -0400 Newsgroups: sci.math In article <39175CD1.7B702DB1@math.ucla.edu>, Mike Oliver wrote: :coolman wrote: :> :> Can somebody clarify something regarding the Riemann zeta :> function: Supposedly, the value of the function is zero for all :> negative even integers, -2, -4, -6, etc. Does this mean that if :> s = -2, then the function 1 + 1/(2^s) + 1/(3^s) + .... equals :> zero? I can't see how that can be true since each succeeding term :> will be larger than the previous. If that isn't what is meant, :> then what is the correct interpretation? : :The definition of zeta that you're using is valid only when the :real part of s is greater than 1 (otherwise it doesn't converge, :at least not absolutely). : :However if you look at the function defined by the above series :for (Re s > 1), it has a unique analytic continuation to the :entire complex plane, except maybe for a pole or two (must :have a pole at s=1, I guess). And it's this continuation :which equals zero at the negative integers. : :Not really my field, so if I've made a boo-boo somewhere, :someone please correct. No need to correct; there is actually a neat procedure how to extend zeta to sets where Re(s) < 1, except for s=1 itself. I will do the part where Re(s) > -3, with s different from 1. The infinite sum, for Re(s)>1, is of course the limit of the sequence of partial sums: a_1(s) = 1/1^s a_2(s) = 1/1^s + 1/2^s a_3(s) = 1/1^s + 1/2^s + 1/3^s ... a_n(s) = 1/1^s + 1/2^s + ... + 1/n^s ... Now add some terms which go to 0 as n goes to infinity: b_n(s) = a_n(s) + (1/(s-1)) * 1/n^(s-1) and not only does b_n(s) converge to zeta(s) faster than a_n(s), but it will also converge for Re(s) > -1, except for s=1. [Secret: The extra term is inspired by the proof of the Integral test.] Still, for s=-1, the sequence will diverge. So, we try harder, adding another "accelerating" term (in the limit, we are adding zero, thus not changing the sum wherever it converged before). c_n(s) = b_n(s) - (1/2) * 1/n^s + (s/12) / n^(s+1) [This correction was inspired by the Trapezoidal Rule for integration, with error estimate.] Careful estimates, using integration tricks, reveals that the sequence c_n(s) converges for Re(s) > -3, other than s=1. In particular, it will converge for s = 0, giving a limit -1/2 , for s = -1, with limit -1/12 , for s = -2, with limit 0. Further extensions can be made by adding more correction terms; there is a systematic way of doing it, called Euler-Maclaurin Formula. Hope it helps, ZVK(Slavek).