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Math Help - zeta function

  1. #1
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    zeta function

    Can someone explain how setting x = -2 in the zeta function results in the function converging to 0? The first term = 1, the second = 4, the third = 9, etc. How can the sum of the squares add up to 0? Another mystery (at least to me): setting x = -1 results in the value -1/12? The zeta function is the sum of 1/(n^s) for n = 1 to infinity, is it not?
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    Quote Originally Posted by coolmanoh View Post
    Can someone explain how setting x = -2 in the zeta function results in the function converging to 0? The first term = 1, the second = 4, the third = 9, etc. How can the sum of the squares add up to 0? Another mystery (at least to me): setting x = -1 results in the value -1/12? The zeta function is the sum of 1/(n^s) for n = 1 to infinity, is it not?
    Disclaimer: I am a novice in complex analysis. Don't take what I'm about to say as absolute truth.

    That's it's definition on a limited portion of \mathbb{C}. To consider the values you mentioned you need to analytically continue it via the usual functional equation \displaystyle \zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s). Plugging in your values gives the desired results.
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    You have gone way over my head. Everything I've read about the zeta function states that the trivial zeros are -2, -4, -6, etc. The equation you produced is never mentioned, only the one I stated. A Professor Hill of University College, London received a letter from the Indian mathematition Ramanujan where he "proved" that 1 + 2 + 3 + .... + infinity = -1/12. Ramanujan got this result by setting x = -1 in the zeta function. If they are not talking about the zeta function, then why don't they say so?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by coolmanoh View Post
    You have gone way over my head. Everything I've read about the zeta function states that the trivial zeros are -2, -4, -6, etc. The equation you produced is never mentioned, only the one I stated. A Professor Hill of University College, London received a letter from the Indian mathematition Ramanujan where he "proved" that 1 + 2 + 3 + .... + infinity = -1/12. Ramanujan got this result by setting x = -1 in the zeta function. If they are not talking about the zeta function, then why don't they say so?
    They're not. This is a common misconception. Let me lay it out for you. Consider the function 0.1]\to\mathbb{R}:x\mapsto x\sin\left(\frac{1}{x}\right)" alt="\displaystyle f0.1]\to\mathbb{R}:x\mapsto x\sin\left(\frac{1}{x}\right)" />. We can clearly continuously extend \tilde{f}:[0,1]\to\mathbb{R} by defining \displaystyle \tilde{f}(x)=\begin{cases}x\sin\left(\frac{1}{x}\r  ight) & \mbox{if}\quad x\in(0,1]\\ 0 & \mbox{if}\quad x=0\end{cases}. In which case it's true that \tilde{f}(0)=0 from where the seemingly amazing 'result' occurs that 0\sin\left(\frac{1}{0}\right)=0 even though that's clearly ludicrous. A similar thing is happening here. We have the function 1,\infty)\to\mathbb{R}:x\mapsto\sum_{n=1}^{\infty} \frac{1}{n^x}" alt="\displaystyle \zeta1,\infty)\to\mathbb{R}:x\mapsto\sum_{n=1}^{\infty} \frac{1}{n^x}" /> for which you are familiar with. Complex analysts found a particular way (in this case analytic continuation ) to take this function \zeta and 'extend' it to a function \tilde{\zeta}:\mathbb{C}\to\mathbb{C} in such a way that, in fact, \tilde{\zeta}(-2)=0. If you aren't careful, and realize that what they call \zeta when the plug in -2 is actually what I've called \tilde{\zeta} you'd get the amazing result that \displaystyle 0=\zeta\left(-2\right)=\sum_{n=1}^{\infty}\frac{1}{n^{-2}}=\sum_{n=1}^{\infty}n^2. But...that's just plain silly :P
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