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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?

Simply Beautiful Art
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kodyv
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    Do you know any complex analysis? If you don't, you should learn some complex analysis first. – Qiaochu Yuan May 29 '11 at 11:44
  • 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.) – kodyv May 29 '11 at 11:48
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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 :) –  May 29 '11 at 12:01
  • @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. – Qiaochu Yuan May 29 '11 at 12:02
  • @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. :) – kodyv May 29 '11 at 12:06
  • @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. – Qiaochu Yuan May 29 '11 at 12:10
  • 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. ... – t.b. May 29 '11 at 12:27
  • ... 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 :) – t.b. May 29 '11 at 12:27
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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. – GEdgar May 29 '11 at 13:04
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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 $$\zeta (1/2)=-\left( 1+\sqrt{2}\right) \sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}.$$ – Américo Tavares May 29 '11 at 13:27
  • 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. – Niel de Beaudrap May 29 '11 at 13:57
  • Dear Koundinya, have you looked at this question and its answers: http://math.stackexchange.com/questions/39802 ? Regards, – Matt E May 29 '11 at 18:20
  • @GEdgar That’s not necessarily wrong. You’re just using a more restricted definition of summation (the Cauchy definition). – user76284 Jul 20 '19 at 04:44

3 Answers3

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I'll give you the world's simplest example. $1+x+x^2+\dots$ converges for $|x|\lt1$ only. The function $1/(1-x)$ is analytic everywhere except for a pole at $x=1$, and agrees with $1+x+x^2+\dots$ everywhere the latter is defined, so $1/(1-x)$ is the analytic continuation of $1+x+x^2+\dots$. In that sense, $1+2+4+\dots=-1$.

Gerry Myerson
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    For a suitable definition of "+", anyway. – Niel de Beaudrap May 29 '11 at 13:40
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    Is there a unique analytic function for all or most sequences? – Jus12 Jul 25 '15 at 15:15
  • I don't know what you mean by "sequences". There are no sequences in my post. – Gerry Myerson Jul 25 '15 at 23:06
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    @Jus, perhaps you are asking whether, given a sequence $a_0,a_1,\dots$, there is an analytic function $a_0+a_1z+a_2z^2+\cdots$ which has an analytic continuation to $z=1$. There are at least two obstacles. One, if the sequence grows too fast, the series won't converge at all for $z\ne0$, so you won't have an analytic function to continue. Two, even if the series does converge to an analytic function in some region, that region may have a "natural boundary" beyond which analytic continuation is impossible. – Gerry Myerson Jul 26 '15 at 00:16
  • @Jus12 Not all analytic functions have an analytic continuation. Look up the Lacunary series. – Simply Beautiful Art Dec 21 '16 at 00:58
  • @GerryMyerson what about the series $1+z^2+z^4+z^6+\dots$. Why it can't be extended anywhere beyond $|z|<1$, whereas $1+z+z^2+z^3+z^4 +\dots$ can. Is this related to the fact, that for a function to be extendable the elements $(f,D_1)$ and $(g,D_1)$ should be direct analytic continuations of each $G_1 \cap G_2 \neq \varnothing$ and $g(z)\equiv f(z),z\in G_1 \cap G_2$. Naively, I think that since since $1+z^2+z^4+z^6+\dots$ diverges for all $|z|=1$, we can't define an element that contains portions of the unit circle. Does this make sense? – Alexander Cska Mar 15 '21 at 10:36
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    @Alexander, $1+z^2+z^4+z^6+\cdots$ can be extended to the entire complex plane, excepting only $z=\pm1$; it's just $1/(1-z^2)$. Maybe you are thinking of $z+z^2+z^4+z^8+\cdots$, which doesn't converge anywhere on $|z|=1$, and can't be extended analytically beyond it. – Gerry Myerson Apr 12 '21 at 04:47
  • @GerryMyerson I have a mistake the series go as $1+z+z^2+z^4+z^8+z^{16}+z^{32}+\dots$. These series should diverge for all points on the circle $|z|=1$. Thank you for your comment. – Alexander Cska Apr 14 '21 at 10:20
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If all you're interested in is an explanation to a layman of why a thing like analytic continuation makes any sense to begin with (in particular, why it spits out a unique answer), the answer is the identity theorem for holomorphic functions, which in its stronger form says that if two holomorphic functions $f, g$ defined on a connected open subset $U$ of $\mathbb{C}$ are equal on a set of points in $U$ with an accumulation point (in particular any open subset of $U$), then in fact $f = g$. This is an extremely strong rigidity theorem generalizing the corresponding fact for polynomials, and shows that if you want to extend a holomorphic function $f$ defined on some domain $U$ to a function $\tilde{f}$ defined on some larger connected domain $V$, then there is at most one way to do it (since any two extensions agree on $U$ and hence agree on $V$).

One sense of "analytic continuation" is that it refers to any function $\tilde{f}$ with the above property, and another sense of "analytic continuation" is that it refers to methods for defining $\tilde{f}$ given $f$.

Sometimes it is the case that a function $f$ can be analytically continued to a very large domain $U$ but that a series or integral defining $f$ can't be made to converge on all of $U$. The philosophical point to take away from this is that holomorphic functions have a sort of "Platonic reality" that isn't necessarily perfectly captured by any particular series or integral definition of them, which are just imperfect "shadows" of this Platonic reality. The slogan I use in this situation and others like it is that

convergence is overrated.

In response to the edit, WolframAlpha is using "shadows" other than the standard series definition of the Riemann zeta function.

Qiaochu Yuan
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    ... and apparently under-appreciated by some :) No need to explain that as an analytically inclined person, I don't like this slogan (but as any slogan only captures part of the truth, I can live with it)... – t.b. May 29 '11 at 12:34
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    @Theo: convergence is overrated when doing combinatorics. On reflection, it's useful there, but more often there you don't want to care bout convergence. – Mitch May 29 '11 at 15:23
  • What do you mean by Wolfram uses "shadows"? What is a "shadow"? – nilo de roock Jun 27 '19 at 06:42
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Answer to the added paragraph only. I guess at least this part is simple. In the case when the argument of the zeta function is a real $x>1$ the zeta function can be expressed as $$\zeta (x)=\sum_{k=1}^{\infty }\frac{1}{k^{x}}=\frac{1}{1-2^{1-x}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{x}},\qquad(\ast)$$

where the last series is the Dirichlet eta function. Since for $x>0$ this alternating series converges (because $1/k^x\to 0$), and $1-2^{1-x}\neq 0$ for $x\in]0,1[\cup]1,\infty[$, the function $\zeta (x)$ can be continued analytically to $x\in]0,1[\cup]1,\infty[$ (note that $1$ is excluded as in the first series) as

$$\zeta (x)=\frac{1}{1-2^{1-x}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{x}}.$$

For $x=1/2$, we get $\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}\approx 0.6049$ and

$$\zeta (1/2)=-\left( 1+\sqrt{2}\right) \sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}\approx -1.4604.$$

Added: Derivation of $(\ast)$:

$$\begin{eqnarray*} \zeta (x) &=&\sum_{k=1}^{\infty }\frac{1}{k^{x}}=\sum_{k=1}^{\infty }\frac{% (-1)^{k-1}}{k^{x}}+2\sum_{k=1}^{\infty }\frac{1}{\left( 2k\right) ^{x}} \\ &=&\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{x}}+2^{1-x}\zeta (x). \end{eqnarray*}$$

The relation is as follows.

nilo de roock
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Américo Tavares
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