The Riemann Zeta Function: Extended Confusion about an Analytic ContinuationDate: 11/11/2011 at 08:33:44 From: Kesavan. Subject: Negative even numbers of Riemann zeta function How do you calculate the value of Zeta(-2)? How do you prove it equals zero? and similarly for every negative value? When I substitute -2 in place of s in Riemann Zeta function, it becomes 1 + 4 + 9 + 16 + ... This is clearly unbounded and tending to infinity. But several articles mention this equals zero. I am dying to know where I went haywire and shall not breathe my last before cracking this conundrum. Thanks for your help. R.Kesavan. Date: 11/11/2011 at 18:26:48 From: Doctor Vogler Subject: Re: Negative even numbers of Riemann zeta function Hi Kesavan, Thanks for writing to Dr. Math. The Riemann Zeta Function is defined to be the complex analytic continuation of the infinite sum you are clearly already familiar with, namely inf zeta(s) = sum n^(-s) n=1 In fact, the series for zeta(s) only converges when the real part of s is strictly bigger than 1. Indeed, the most common misunderstanding of the Riemann Zeta Function is forgetting that it is defined as the *analytic continuation* of that infinite series. The RZF is defined even where the series does not converge. Forgetting this, and saying that the RZF equals that sum, would be like saying that the function ... f(x) = 1/(1 - x) ... is equal to the sum of the geometric series ... g(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... ... and that therefore it couldn't possibly be negative when x is positive. Right? So how can you say that f(2) = -1? Well, f and g are equal when -1 < x < 1, but g does not converge when |x| >= 1, whereas f is still defined there (except at x = 1). Similarly, the Riemann Zeta Function is equal to the infinite series you know and love when the real part of s is greater than 1. At other values of s, you'll have to use a different formula. So how do you extend the domain of a function defined by an infinite sum that doesn't converge everywhere? That is where analytic continuation comes in. The technique of expanding the domain is called "analytic continuation" and is a subject from complex analysis. If you are unfamiliar with this subject, I would recommend a course or book, but I'll say a little about complex analysis here. It turns out that "analytic" is a very strong condition for a function to have. "Analytic" means that it has all of its derivatives (first derivative, second derivative, third derivative, etc.) and that it equals its Taylor Series. But you can prove that if two analytic functions are equal to one another on any open set, or even any open interval, then they are equal everywhere that they are both defined. That means that if you know an analytic function on a small interval, then there is only one way to extend it to a wider domain and still have an analytic function. And that is what we call its "analytic continuation." For example, consider the function I defined above: g(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... This is defined for all complex numbers x whenever the absolute value (or "modulus" or "norm") of x is less than 1. But this function has an analytic continuation which is defined for all complex numbers apart from x = 1. I suspect that you won't be surprised to learn that the analytic continuation is equal to f(x) = 1/(1 - x). The usual way to determine the analytic continuation of a function is to find some formula which is equal to your infinite sum in some region where both are defined, but which is itself defined in a larger region. For example, the MathWorld page on the RZF has some examples of such formulas for the RZF; look at equation (16) and below: http://mathworld.wolfram.com/RiemannZetaFunction.html The "functional equation" for the RZF (which usually refers to what the MathWorld site calls equation (13), relating zeta(1 - s) to zeta(s)) is the usual way to prove that the RZF has zeros at the negative even integers. You might also look at the Wikipedia page on the RZF, which includes a section on the functional equation: http://en.wikipedia.org/wiki/Riemann_zeta_function It's easy to show that the power series equal to the RZF is positive everywhere when the real part of s is greater than 1. This means that the RZF has no zeros or poles in that region. This fact and the functional equation are sufficient to show that there are no poles, but only the "trivial" zeros in the region where the real part of s is negative. It takes more work to show what happens where the real part of s is equal to 1 or 0 (there is a pole at s = 1, but no zeros on those lines), and it is still an unproven conjecture (The Riemann Hypothesis) that all of the other zeros are on the line where Re(s) = 1/2. If you still want more, you might also be interested in a few other related answers from our archives, such as these: http://mathforum.org/library/drmath/view/70335.html http://mathforum.org/library/drmath/view/52831.html http://mathforum.org/library/drmath/view/51929.html If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 11/12/2011 at 10:23:15 From: Kesavan. Subject: Thank you (Negative even numbers of Riemann zeta function) Dear Dr. Vogler, Hats off and encomium to the brightest explanation you have elucidated to this simpleton, as I -- in my limited knowledge of English lexicon -- could not find words enough to appreciate and acknowledge your assistance. The moment you started with a very fine example of functions 1/(1 - x) and its counterpart 1 + x + x^2 + x^3 + x^4 + ..., my doubts and uncertainties vanished into thin air, bidden adieu, or should I say were cleverly dispelled, just like the morning sun's rays clears off the mist enveloping the leaves of the tree. Kindly accept my kudos, adulations and commendations. The function, and its infinite series cannot be equal for all values. Nicely argued and well presented. The infinite series is convergent only when the real part of 's' is greater than one. Substituting -2 is itself wrong due to the wrong assumption of both being equal. Besides, the series is not convergent for -2. So substitution here is wrong. Now I understand. The beauty of mathematics could be appreciated only when it is understood properly. Once again thanking you for your time, labor and interest, R.Kesavan |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/