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Today, as I was flipping through my copy of Higher Algebra by Barnard and Child, I came across a theorem which said,

The series $$ 1+\frac{1}{2^p} +\frac{1}{3^p}+...$$ diverges for $p\leq 1$ and converges for $p>1$.

But later I found out that the zeta function is defined for all complex values other than 1. Now I know that Riemann analytically continued this function to fit all complex values, but how do I explain, to a layman, that $\zeta(0)=1+1+1+...=-\frac{1}{2}$?

The Wiki articles on these topics go way over my head. I'd appreciate it if someone can explain it to me what analytic continuation actually is, and which functions can be analytically continued?


Edit

If the function diverges for $p\leq1$, how is WolframAlpha able to compute $\zeta(1/5)$? Shouldn't it give out infinity as the answer?

  • 3
    Do you know any complex analysis? If you don't, you should learn some complex analysis first.2011-05-29
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    Well I don't know any complex analysis. The reason I ask this question is that I've stumbled upon a result and I want to see if I can "analytically continue" the fractional part function. (I really hope I'm not sounding terribly stupid.)2011-05-29
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    @Koundinya: Everything can't be explained to a layman. IF everything can be explained to a layman, then all layman would be Mathematicians :)2011-05-29
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    @Koudinya: I suppose you could analytically continue the fractional part function to $\mathbb{C}$ minus the horizontal lines passing through each integer on the $x$-axis, but I'm not sure why you'd want to.2011-05-29
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    @Qiaochu: So only continuous functions can be analytically continued? @Chandru: Oh well, it's just slightly unnerving when you show non-mathematicians that $1+2+4+...=\frac{-1}{12}$ and they label you immediately as "crazy". I just wanted to find out what you would say to justify that claim. :)2011-05-29
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    @Koundinya: I misspoke. Strictly speaking, analytic continuation is done relative to some domain in $\mathbb{C}$ that you want to extend the function to, and that domain doesn't have to include the entire real line. As for your other question, again, you should learn some complex analysis.2011-05-29
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    To answer your question on the fractional part and start extending it from the interval $(k,k+1)$ what you'll get is simply the function $z - k$, defined wherever it pleases you. This shows an important aspect of analytic continuation: Extending a function depends on where you start and if you come back to the initial domain of definition (e.g. the interval $(l,l+1)$ it need not coincide with the function you started with, leading to "multiply valued functions". That's why there are scary terms such as germs etc. ...2011-05-29
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    ... Note that this happens *despite* the fact that the fractional part is analytic on $\mathbb{R}\smallsetminus \mathbb{Z}$! I agree with Qiaochu's remark that learning complex analysis would be a very good idea. Since you want to understand the $\zeta$-function, you will have to do this sooner or later, anyway :)2011-05-29
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    Tell your layman these: $\zeta(0)=1+1+1+...$ is *wrong*. $1+1+1+...=-\frac{1}{2}$ is *wrong*. $\zeta(0)=-\frac{1}{2}$ is *correct*.2011-05-29
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    For $z\in\mathbb{C},\text{Re }z>1$ the zeta function can be expressed as $$\zeta (z)=\sum_{k=1}^{\infty }\frac{1}{k^{z}}=\frac{1}{1-2^{1-z}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{z}}.$$ Since the alternating series on the RHS converges for $\text{Re }z>0$, the function $\zeta (z)$ can be analitically extented to $\text{Re }z>0$ as $$\zeta (z)=\frac{1}{1-2^{1-z}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{z}}.$$ For $z=1/2$, we get $$\zeta (1/2)=-\left( 1+\sqrt{2}\right) \sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}.$$2011-05-29
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    It seems to me that people who like to identify infinite sums with analytic continuations necessarily prefer dealing with a metric in which $\lim\limits_{n \to \infty} p^n = 0$, for any positive integer $p$. If they would just explicitly state that they use such a topology in the first place, it would be much clearer to laymen how such an identification could be justified.2011-05-29
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    Dear Koundinya, have you looked at this question and its answers: http://math.stackexchange.com/questions/39802 ? Regards,2011-05-29

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