Talk:Riemann zeta function

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Does anyone agree that the "Applications" section is a little silly? The connection to physics is sparse, and no connection to the use of the Riemann-Zeta function is actually motivated by it (would you actually need to use it to show that sum diverges? Surely you want to show something else). Njcnjc (talk) 03:03, 21 October 2008 (UTC)[reply]

Can anyone elaborate on the "prime numbers" section? It says "this is a consequence..." without giving even a hint or flavour of why. --Doradus 21:25, Sep 17, 2004 (UTC)

I've added what I believe to be a standard treatment of why the two expansions are equivalent. Do any real mathematicians want to do a better job? -- The Anome 23:27, 17 Sep 2004 (UTC)

I've turned the "vigorous handwaving" into an actual proof and have made a number of further changes and additions. Gene Ward Smith 08:42, 29 Jan 2005 (UTC)

There will always be a long standing question on the readability of this article; I feel the need to put my two cents in. I came to the page after reading probably 20-30 pages on abstract math, no formal training in mathematics beyond Differential Equations and Vector Calculus in college. I found it to be sufficiently readable that I was able to "get the gist" of what the Riemann Zeta function is and why it is important. No, I could not do a single calculation regarding it, but from what I gather, the masters of the topic have trouble doing more than the most basic proofs regarding the topic - so is it such a surprise that I have trouble?

Fundamentally, Riemann Zeta is not simple. Not by a long shot. But that doesn't mean its unimportant; it doesn't mean that mathematicians who can leverage the information should be denied such a tool. There's still truly trivial content on wikipedia; lets leave the profound content (like Riemann Zeta) in place. 199.46.245.230 (talk) —Preceding undated comment was added at 17:04, 26 November 2008 (UTC).[reply]

The following anonymous edit comes from an IP with a very checkered history. It needs vetting:

${\displaystyle -\zeta (s)=\prod _{p}{\frac {1}{1-p^{-s}}}+\zeta (s)=\prod _{p\in primes}{\frac {1}{1-p^{-s}}}}$

Thanks. --Wetman 13:02, 18 Apr 2005 (UTC)

This is surely wrong, as it implies that the \zeta function is zero. Oleg Alexandrov 18:15, 18 Apr 2005 (UTC)

I don't see this exact thing in the article though. Oleg Alexandrov 18:18, 18 Apr 2005 (UTC)

Wetman posted a diff; he is talking about the edit from 9:32, 8 April 2005, which Oleg reverted a few hours later anyway. No matter, the edit was fine, the product is over primes. Case closed. linas 22:09, 18 Apr 2005 (UTC)

Is there a formula out there that relates the zeta function to the perfect powers (1, 4, 8, 9, 16, 25, 27, 32, etc) in a similar way that Euler's product formula does for primes? The reason I ask is because I discovered a very simple one a while back, but I can't find information about any other such formulas. Thanks. --Vagodin 14:47, August 21, 2005 (UTC)

Is there a wikipedia policy for where pronunciations should appear? I dislike the redundancy of placing the pronunciation of Riemann both at this article and also at Bernhard Riemann. If a user wants to know the pronunciation they can simply follow the name link to find it. - Gauge 04:44, 2 September 2005 (UTC)[reply]

Also, pronunciations should probably be in International Phonetic Alphabet form, as English speakers could pronounce words very differently depending on the dialect. - Gauge 04:48, 2 September 2005 (UTC)[reply]

Remove the pronunciation from this article. linas 14:56, 2 September 2005 (UTC)[reply]

I'd like to hold a survey regarding the article Riemann zeta function, to help determine its general comprehensibility and identify areas where it may be incomplete. Please indicate your perceptions of the article below, and feel free to expand the survey or article as you see fit. ‣ᓛᖁ♀ᑐ 21:07, 9 September 2005 (UTC)[reply]

The only point I could see this survey having is to determine which users are versed in some complex analysis and those who are not. See also my rant here. Dysprosia 07:57, 11 September 2005 (UTC)[reply]

Do you currently understand this article?

Yes

No

1. ‣ᓛᖁ♀ᑐ 21:07, 9 September 2005 (UTC)[reply]
2. I got a bit lost about halfway through the second paragraph. I also have no idea why the Euler Product Formula has to be proven in the middle of the article. And the bottom of the article is very difficult to follow with only college sophomore math skills. Avocado 00:56, September 10, 2005 (UTC)
3. — Xiong熊talk* 02:20, 2005 September 12 (UTC)

Comment

1. I agree with the fact that the proof of the Euler's product formula should not be there. Oleg Alexandrov 04:10, 11 September 2005 (UTC)[reply]
2. I understood most of it up to the point it started talking about Mellin transforms (which is the point at which I lost the will to follow all the links I didn't understand). Before that point, I think the Euler's product proofs should be moved out (but linked to), and a bit more should be said (or linked to) about the "importance of the zeros", in particular what or how path integrals relate to the prime counting function. Hv 11:03, 11 September 2005 (UTC)[reply]

---

If not, do you feel you could understand it after following its internal links?

Yes

1. ‣ᓛᖁ♀ᑐ 21:07, 9 September 2005 (UTC)[reply]

No

1. Er, well, I found Analytic continuation and Meromorphic function absolutely incomprehensible. I think that if I had time and energy to wade through a huge tree of links it might begin to make a bit more sense, but the motivation is a bit lacking ;-) . On the other hand, I doubt that reading through all the other articles Wikipedia has on math would make it possible to actually follow the mathematical equations and transformations. Avocado 00:56, September 10, 2005 (UTC)

• Er, well, this article could use improvement. But what's go me stumped is, if you don't know what a meromorphic function is, why in the world are you interested in the Riemann zeta? To answer my own question: probably because of the Clay Institute prize. In which case, the answer is that we need a section, or maybe even a whole article, describing "why the Riemann zeta is important to mathematics", and said article would not make use of formulas at all.

I say this because I object to the idea of somehow "simplifying" this article; its already too simple in many ways, because it is already quite lengthly, while failing to even touch on many important properties and relations. Personally, I would like to see some of the proofs and derivations moved to something in the style of Category:Article proofs. -- linas 17:03, 11 September 2005 (UTC)[reply]

The article's lead section states the Riemann zeta function is "of paramount importance in number theory". From reading the article, do you understand why this function is important?

Yes

No

1. ‣ᓛᖁ♀ᑐ 21:07, 9 September 2005 (UTC)[reply]
2. Sort of. It has something to do with prime numbers (which I know are a knotty number theory problem) but I'm pretty sure I couldn't figure out from this article what exactly the connection is. Avocado 00:58, September 10, 2005 (UTC)

Disclosure: I am the son of a world-famous mathematician -- indeed, in her later life a number theorist. However, she mated with a whiskey-soaked advertising copywriter and I was raised on a farm by psychopathic semi-deaf-mutes shunned by their neighbors for the worldly sin of using electricity and automobiles. Mother did not rescue me from this Appalachian pastoral idyll for many years, whereupon she made my bedtime tales out of graph theory.

So, I have only half the genes and half the nurture. I have managed to stagger through a computer engineering career with nothing higher than the calculus -- which only served me once, and that indirectly. On the brighter side, I'm generally able to comprehend intelligent explanations, often in polynomial time.

As it stands, I find this article to be incomprehensible and without merit here. Wikipedia is a general reference work. For content to be included here, it has to pass the Joe test: If you had unlimited time in which to explain it (in far more detail than here) to Joe, the Everyman; if Joe was completely cooperative, intelligent, and patient; if eventually he understood all you intend, then could he imagine any possible way in which this subject might be of interest? This topic fails the Joe test.

This article bears a disturbing resemblance to Graveler. It is an isolated swatch of factoplasm lifted out of a highly specialized context, meaningless outside that context. Nothing has been done to show that it has any application to the Real World. While the subject may be a vital part of its own bubble universe, nothing in the article connects it to anything outside.

To be sure, this is longer than Graveler, and arguably factual; but I think that by the time you get past definitions of terms, it is equally valid to say that it is very like a playing card in a mathematical game. If there is a connection to anything concrete, that needs to be shown.

So, I suggest the article be downsized to a bare description of the function and merged into its parent article. (It does at least have some relationship to a larger mathematical topic, does it not?)

Another possibility is to open a new WikiBook entitled "Higher Mathematics" and expand this article into a whole chapter -- there.

I hold out one other route for improvement. I almost began to see a glimmer of light in Applications. Perhaps a diligent effort could actually find some application of this function to something tangible. If so, rewrite this section and move it to the introduction. MBAs and fools like me can read the 4 or 5 sentences that place the function in context, and then we can graze on.

— Xiong熊talk* 05:11, 2005 September 12 (UTC)

So, removing the rhetoric, your proposal for making this article more accessible seems to be to make it smaller or to remove it from Wikipedia completely. I don't see how either of these routes will improve Wikipedia. Gandalf61 11:39, September 12, 2005 (UTC)

To paraphrase, Xiong seems to be saying "I'm the product of a broken social/cultural system, and have no education. I am self-taught, having learned engineering, but otherwise, I'm ignorant. I want to stay ignorant, and everyone else should be kept ignorant too". This sounds nothing so much like the Chinese cultural revolution, where villagers melted down the strong steel of mathematical theorems into puddles of worthless self-taught folk knowledge, and starved by the millions as a result. Shall we picture the "MBA's and fools like him" holding machine guns, eating 4 or 5 rice cakes "in context", and then grazing on? This is a tragedy. linas 14:36, 12 September 2005 (UTC)[reply]

Aside from psychoanalysis of this person (is he (or she) really Chinese), he has some point. But he is questioning how wikipedia should be written, and this is probably not a place to do. There are many, too many, hopelessly technical pages already. But the consensus is we should try to revise it so that laymen can understand it, not that eliminate or downsize it from wikipedia. -- Taku 03:31, 14 September 2005 (UTC)[reply]

Agree with Taku. And I will say again: please, when you notice a page is too technical, don't just slap a template and walk away. First try to read carefully the page, and see if your criticism is justified. If yes, write at least something on the talk page explaining what you wish were better. That is, be constructive. Oleg Alexandrov 03:57, 14 September 2005 (UTC)[reply]

Guess what..the Zeta function IS technical! Maybe "Joe Test" should read a damn book about number theory.

And no matter how uninteresting "Joe" might find this, im sure he's interested by the fact that the Riemann Hypothesis one of the Millenium Problems. As for being useless, this problem is closely related to the primes. Both the distribution of primes and primality testing are two of the biggest problems in number theory, both of which are closely related to cyptanalysis, which is certainly useful. -- He Who Is^{[ Talk ]} 01:58, 29 June 2006 (UTC)[reply]

I removed the following text:

The Möbius function also relates to the zeta function and Bernoulli numbers in the coefficients in series expansion of ${\displaystyle {\zeta (n+2)} \over {\zeta (n)}}$ with the formula

${\displaystyle \sum _{d|n}\mu (d)d^{2}}$

for which A046970 gives values for the first 60 n.

I couldn't understand what it was trying to say. Its trying to describe some dirichlet series maybe??. linas 00:46, 21 December 2005 (UTC)[reply]

I spose i was trying to justify the oeis having a ref to this article. Numerao 21:33, 29 December 2005 (UTC)[reply]

I found this section very helpful for understanding the connection between the Zeta function and prime numbers. BringCocaColaBack 11:29, 13 January 2006 (UTC)[reply]

Is this function called the zeta-function or the zeta function? The article uses both. toad (t) 12:10, 10 February 2006 (UTC)[reply]

Some anon editor ran around adding hyphens recently, which no one reverted. Not just here, but in half-a-dozen articles. :-( 15:05, 10 February 2006 (UTC)

Zeta-function refers to all zeta function is general. But in this case its just Riemann zeta function. -- He Who Is^{[ Talk ]} 01:59, 29 June 2006 (UTC)[reply]

Following question from an anon contributor moved from the Globallly convergent series section on the article page. Gandalf61 08:35, 20 April 2006 (UTC)[reply]

Why is there a function of s only (zeta of s), that equals a sum which leaves behind a function of s and x?

That question applies more properly to the formula in the section above Globally convergent series, called Series expansions, which contains the formula:

${\displaystyle \zeta (s)={\frac {1}{s-1}}-\sum _{n=1}^{\infty }(\zeta (s+n)-1){\frac {x^{\overline {n}}}{(n+1)!}}}$

What's that 'x' doing there? -GTBacchus^(talk) 13:38, 20 April 2006 (UTC)[reply]

I think that's supposed to be an s not an x, then it would look almost right (?). You can check correctness by going to the article on the Hurwitz zeta function, then look at the section called "Taylor series", where it mumbles about derivatives, and with a few minor substitutions (e.g. y=-1), you should be able to derive the above (with an s where x now stands). linas 04:16, 21 April 2006 (UTC)[reply]

In the series expansion section, it's written that "Another series development valid for the entire complex plane is.." I can't figure out what the variable 'x' is supposed to be in the expansion that follows. x=s? —The preceding unsigned comment was added by 12.208.117.177 (talk • contribs) .

Ok, that's twice it's come up; I've changed it from x to s. -GTBacchus^(talk) 05:54, 11 May 2006 (UTC)[reply]

The article titled 42 (number) says:

It is believed to be the third moment of the Riemann zeta function, based partially upon evidence from quantum mechanics.

I don't know what this means. Here's a guess:

${\displaystyle \int _{(-\infty ,\infty )}s^{3}\zeta (s)\,ds=42.}$

I'm accustomed to the definition of momnets of probability measures; if ζ were a probability density function then the integral above would be the third moment of the corresponding probability distribution. But ζ is negative in some places, and from the way ζ(s) blows up at s = 1 it seems we'd have to be thinking of a Cauchy principal value or something like that.

Can someone make the article's statement clearer? Michael Hardy 17:52, 5 July 2006 (UTC)[reply]

• • • • • I have edited the article on "42" to make clearer the connection to moments of the zeta-function. It is actually the "sixth moment" that is interesting, and the moments are usually defined

${\displaystyle I_{k}(T)={1 \over T}\int _{0}^{T}\left|\zeta ({1 \over 2}+it)\right|^{2k}\,dt.}$

When k=3 you get the "sixth moment". The constant 42 comes up as a scaling factor in a conjecture by Conrey & Ghosh for the leading order term of this integral. .—Preceding unsigned comment added by 74.74.128.91 (talk • contribs) *****

I've exchanged some email with John Baez, the mathematical physicist who has edited Wikipedia articles as user:John Baez, and he reports that he cannot access Wikipedia because he is in China. He wrote:

“

So, feel free to post this comment for me: I don't know anything about the "3rd moment of the Riemann zeta function", but perhaps what's meant is the 3rd moment of the distribution of spacings between zeroes of the Riemann zeta function. There's a lot of evidence relating the distribution of these spacings to the distribution of spaces between eigenvalues of a large random self-adjoint matrix. For lots more, try this: http://www.maths.ex.ac.uk/~mwatkins/zeta/bump-gue.htm and for general connections between the Riemann zeta function and quantum mechanics, try: http://www.maths.ex.ac.uk/~mwatkins/zeta/physics1.htm Best, jb

”

My wild guess seemed so implausible that I'm both relieved to hear that it's wrong and pleased to hear that this otherwise implausible-seeming statement can be construed in such a way that it makes sense. Michael Hardy 16:20, 20 July 2006 (UTC)[reply]

No graphic of the graph in the complex plane? Surely the article should include one or perhaps three; real part, imaginary part and absolute value, as is the standard on Mathworld. Soo 14:17, 16 July 2006 (UTC)[reply]

lucky this is WP and not planetmath, or you would've gotten hooted at big time. anyway, yeah, it would be nice to have some kind of graphic. Numerao 20:18, 17 July 2006 (UTC)[reply]

The trivial zeros do not seem to yield zeros:

For example: ${\displaystyle \zeta (-2)=1+{\frac {1}{2^{-2}}}+{\frac {1}{3^{-2}}}+\cdots =1+2^{2}+3^{2}+\cdots }$;

which yields infinity.

I will appreciate if someone lets me know where I am mistaken.

As the article says, the power series only defines the Riemann zeta function for arguments x > 1. You must use the function's analytic continuation to evaluate it elsewhere. Fredrik Johansson 18:46, 26 August 2006 (UTC)[reply]

The "trivial zeros" are trivial because of the functional equation. There has to be zeros of the zeta-function to "cancel" the poles of the gamma-function, otherwise the zeta-function would not be analytic in the left half-plane.—Preceding unsigned comment added by 74.74.128.91 (talk • contribs)

Though with enough careful parsing of the article the fact that numbers with Re(s) < 1 are not in the domain of the power series is buried in the article, it is in no way clear. It took me forever to understand that that is what ya'll meant. This issue about the trivial zeros is subtle point that no one really explains well on any of the online sources. Why not start the article with the power series definition and a quantifier that explicitly says that values with Re(s) < 1 are not valid for this function, then put the analytical function next to it and use a quantifier to show that its domain covers the entire complex plane? Or at the very least add one sentence clarifying this point in no uncertain terms. I'm about 1/2 a step from adding it myself but I don't feel very qualified to talk about this topic....but if someone else doesn't do it, or they can't think of a good reason why it shouldn't be done...I'm gonna have to.

128.97.68.15 (talk) 17:25, 1 February 2008 (UTC)[reply]

By "careful parsing", do you mean reading the first and second sentences of the section labeled "Definition"? It states that the Dirichlet series converges for s > 1 and is only defined for other s through analytic continuation. "If everything else fails, read the definitions". Arcfrk (talk) 20:07, 1 February 2008 (UTC)[reply]

Although it's in the definition, someone who's not well-versed with terms like "analytic continuation" will probably skip the qualifiers, and remain serenely uninformed until they get to the assertion that s = -2, -4, -6, and so on are trivial zeros. It would be useful to repeat that particular qualifier there (s > 1), I think, and perhaps point back to that definition. I'll come back in a couple of days and do that unless someone protests. BrianTung (talk) 00:38, 15 March 2009 (UTC)[reply]

I stored in commons a graph of zeta (x) with -20 < x < 10.

see

Perhaps it can go into the article

--Brf 10:05, 31 August 2006 (UTC)[reply]

In a technical report entitled "[http://cswww.essex.ac.uk/technical-reports/2005/csm-442.pdf An elementary formulation of Riemann’s Zeta function]", myself (Riccardo Poli) and Bill Langdon provided a very simple proof that, for ${\displaystyle \Re (s)>1}$, Riemann's Zeta function can be written as ${\displaystyle \zeta (s)=1+\sum _{k=1}^{\infty }a_{k}(s)p_{k}^{-s}}$ where ${\displaystyle a_{k}(s)=\prod _{j=k}^{\infty }(1-p_{j}^{-s})^{-1}}$.

We are not experts in number theory, but we have searched widely and also asked several mathematicians: it appears that our rewrite is new. These people tell us that this is useful formulation. So, we were wondering whether it would make sense to include it in the article.

Yes, this follows from the fact that

${\displaystyle a_{k}(s)=\sum _{m\in \mathbb {P} _{k}}m^{-s}}$

where ${\displaystyle \mathbb {P} _{k}}$ is the set of integers whose smallest prime factor is greater than ${\displaystyle p_{k}}$. If this result is published in a textbook or refereed journal then it can be included it in the article. If not, however, then it falls under Wikipedia:No original research and cannot be included. Gandalf61 09:27, 27 October 2006 (UTC)[reply]

The paper mentioned above has now been published in arXiv.org in the Mathematics History and Overview section (math.HO/0701160). Perhaps the result could now be included in the article?

As Gandalf61 said, if the result is published in a textbook or refereed journal then it can be included it in the article. Papers in arXiv.org do not undergo peer-review, so they are usually not considered reliable sources. Hence, the answer is no. -- Jitse Niesen (talk) 15:04, 9 January 2007 (UTC)[reply]

Can someone add an explanation about the critical strip? This term's definition is nowhere to be found in Wikipedia. Thanks! By the way, the whole article is fine and readable; everybody with a college degree will understand at least the basics. Hugo Dufort 08:28, 14 November 2006 (UTC)[reply]

Okay, done that. Gandalf61 09:54, 14 November 2006 (UTC)[reply]

Thanks a lot! The added information greatly helps understanding some key concepts. Every specialized term needs to be defined, unless people not familiar with the subject will be lost on the way. Even before the arid math proofs! Hugo Dufort 04:16, 15 November 2006 (UTC)[reply]

I added an expanded form of the zeta function I got from Marcus de Sautoy's "Music of the Primes", because I think it helps to visualise exactly what the product is about. Comments? DavidHouse 21:21, 26 December 2006 (UTC)[reply]

I thought the Riemann zeta function referred to complex numbers - what is the justification for including the harmonic series, the Basel equality, and other zeta functions that don't involve complex numbers here? 24.61.112.3 15:12, 2 December 2007 (UTC)[reply]

The Specific Values section contains examples of series for natural numbers only. Don't these belong in an article about Zeta functions "in general"? The sudden transition from discussing natural number constants to the "Zeta zeros" - and hence complex numbers of the Riemann Zeta function - is bizarre and misleading to say the least. Michaelmross 14:53, 21 January 2007 (UTC)[reply]

I disagree. The article says right at the beginning that it's defined at all complex arguments except the number 1. Nothing in the section on specific values contradicts that. Also, it's NOT about "zeta functions in general"; it's about one function---the Riemann zeta function. Michael Hardy 03:20, 22 January 2007 (UTC)[reply]

... and now I see that that section does not give values only at natural numbers. Thus your comments above are what is misleading. Michael Hardy 03:32, 22 January 2007 (UTC)[reply]

I find it sad that a well-intended comment is misunderstood and then labeled as misleading. I didn't say the topic was about "zeta functions in general" - I clearly suggested that natural number series might belong in a *topic about zeta functions in general*. Because what I'm saying about this section of the article is that it goes from something general about natural numbers that a layperson like myself can understand to something very specific concerning complex numbers and the zeta zeros. And to me - a non-mathematician - this is very confusing. I would like to see a layperson's distinction between a generic zeta function and a Riemann zeta function. This will be my last comment on the matter, so there's no need to flame me any further. Michaelmross 20:12, 22 January 2007 (UTC)[reply]

Nobody flamed you. I find your remarks confusing. You contrast "something general about natural numbers" with "something very specific concerning complex numbers", but in fact it is the natural numbers that are specific and complex numbers that are general. As for "a layperson's distinction between a generic zeta function and a Riemann zeta function", I think you might find any definition of "zeta functions in general" to be somewhat more abstruse than an account of the Riemann zeta function. Anyway, you're being unclear; I have a hard time trying to figure out what you're saying. Michael Hardy 22:47, 22 January 2007 (UTC)[reply]

I have read, reread, and then reread this article. I am not a mathematician, but I am also not ignorant of higher mathematical concepts. This article is so technically oriented that it is virtually impossible to comprehend for a layman. In fact, I would wager that unless one already understood the Riemann zeta function that this article would be so complex that it would be useless. Please don't attack me for saying this, I only wish to improve an article that others have obviously worked very hard on. I hope someone will take up the challenge. —Preceding unsigned comment added by 70.121.7.89 (talk • contribs)

You may be right that it's impossible for non-mathematicians to understand most of it. But that may perhaps be true of nearly any article that could be written on this sort of topic.
But I think you're wrong to say that only those who already understand the zeta function can understand this article. I think most mathematicians not familiar with the zeta function would understand it. Michael Hardy 01:30, 28 January 2007 (UTC)[reply]

I would wager that unless one already understood the Riemann zeta function that this article would be so complex that it would be useless

Looking at it now, I'd say the "Definition" section, the "Relationship to primes numbers" section, and the "Specific values" subsection would be readily understood by non-mathematicians (even if not by those who simply dislike math and never study math at any level). So I think you're being a bit alarmist. Certainly there are some things here that few besides mathematicians will understand, but far from everything. Michael Hardy 23:20, 4 February 2007 (UTC)[reply]

Also there are links to other key concepts, and you might have to do some "stack-based" learning on Wikipedia to really understand it. Still there are some unclear points.67.185.99.246 08:13, 10 February 2007 (UTC)[reply]

For the given functional equation

${\displaystyle \zeta (s)=2^{s}\pi ^{s-1}\sin \left({\frac {\pi s}{2}}\right)\Gamma (1-s)\zeta (1-s)}$

it follows that

${\displaystyle \zeta (2)=0}$

but

${\displaystyle \zeta (2)={\frac {\pi ^{2}}{6}}}$

Why are these values different? 67.185.99.246 02:40, 11 February 2007 (UTC)[reply]

The gamma function has simple poles at s = −n (n = 0, 1, 2, 3, ...), and so we have to be careful while evaluating

${\displaystyle \lim _{s\rightarrow 2}\;\sin \left({\frac {\pi s}{2}}\right)\Gamma (1-s)}$.

Using Euler's reflection formula for the Gamma function I get

${\displaystyle \zeta (s)=(2\pi )^{s}{\frac {\sin(\pi s/2)}{\sin(\pi s)}}{\frac {\zeta (1-s)}{\Gamma (s)}}}$,

and from l'Hôpital's rule I get

${\displaystyle \lim _{s\rightarrow 2}\;\zeta (s)=-2\pi ^{2}\zeta (-1)}$

From this and from

${\displaystyle \zeta (2)={\frac {\pi ^{2}}{6}}}$

I get

${\displaystyle \zeta (-1)=1+2+3+\dots =-{\frac {1}{12}}}$,

a result I think already known to Euler. —Tobias Bergemann 08:54, 11 February 2007 (UTC)[reply]

That works of course, thank you for that explanation. 67.185.99.246 08:21, 12 February 2007 (UTC)[reply]

Speaking of that formula, it actually doesn't help find riemann zeta function since it uses circular definition to define it (if you don't know riemann zeta of n or 1-n you can't use it). It can be used to define factorial if you isolate (-n)! and let n=-x. Note: letting x=0 or -1 will result in zero times infinity in the formula, so here is the heads up on those: 0!=1 -1!=+-infinity srn347 —Preceding unsigned comment added by 68.7.25.121 (talk) 06:33, 19 November 2008 (UTC)[reply]

I want to add a personal support for the writers of the article. Even though the article necessarily is largely technical, the lead-in paragraph adequately establishes the backround of the function for laymen. -- Cimon Avaro; on a pogostick. 08:51, 12 February 2007 (UTC)[reply]

209.226.117.54 16:08, 17 February 2007 (UTC) Jacques Gélinas[reply]

Browsing through the talk page, it seemed to me that there had been quite a few complaints about the definition of from people not comfortable with analytic functions, or perhaps, with mathematics in general. I can definitely confirm that the so-called "introduction" to this article is too terse to be of any use. Other articles, such as Riemann hypothesis are much better in this regard. So this is certainly something that needs to be dealt with. For now, I have expanded the definition a bit, it remains a rigorous mathematical definition, so it's unclear to me how much happier would non-mathematicians be with it. Hopefully, it is somewhat gentler to those who are unsure about all the symbols and unfamiliar terminology, although to experts on Riemann zeta function it may appear to be perhaps a little too easy. I do want to point out that Enrico Bombieri, in the description of the Riemann hypothesis in the Millenium Prize book starts by mentioning that the Dirichlet series for ${\displaystyle \zeta (s)}$ is defined only for large ${\displaystyle s}$, and then explains that it is analytically continued. I definitely feel that it's not something to be taken lightly, especially since analytic continuation of general Artin L-functions is still unknown, and of course, by no means obvious! Perhaps, it would make sense to expand the definition even more, it is a judgement call (or an editorial decision), so I would wait to hear the reaction.

Incidentally, I think that in line mathematical formulas do not look very good in this case, but since it's a highly emotional issue for at least some users, I tried to preserve them. Arcfrk 06:04, 10 March 2007 (UTC)[reply]

For those who get lost in this Wikipedia article, I have found the following link to be the clearest description in layman's terms. Those who authored and are maintaining this Wiki article may want to read this to understand how a complicated math concept can be described in normal conversational English:

http://seedmagazine.com/news/2006/03/prime_numbers_get_hitched.php Overlook1977

This is definitely a worthy link! On the other hand, encyclopedias are not written in normal conversational English, and very much on purpose so. Arcfrk 17:25, 16 April 2007 (UTC)[reply]

I glanced at if for a second or two. Is there anything in this link that actually says what the zeta function is?? If not, I certainly wouldn't say it's clear to either lay persons or anyone else. Michael Hardy 20:45, 16 April 2007 (UTC)[reply]

The article (which is 3 pages long, if you did a quick glance you probably missed the meat of the article on page 2) discusses the relevance of the Riemann Zeta Function to prime numbers. Those not familiar with advanced mathmatics or number theory are probably more interested in the significance/importance of the function vs. how it is technically derived.

Im just trying to provide a non-technical explanation of the significance of the RZF, which I am still working on understanding. The article isn't 100% relevant but it does provide some insight. When I better understand the RZF myself I will contribute to the main article. The problem with this and other advanced math articles on Wikipedia is they fail to describe the overall big picture. For example, you could explain that the volume of a sphere is 4/3¶r3. But you could also say it simply the measurement of the space inside a sphere. You couldnt take the latter description and do a calculation with it, but not everyone is interested in the details. Some of us want the big picture. Overlook1977

Only the original formula of Riemann zeta function is in the article. I believe it's very important to also include its analytic continuation formula too. It is a crucial piece of information. Also it would be nice to add in the article how this analytic continuation was found. I'm searching for this information myself so if somebody could do it it would be very appreciated. SmashManiac 20:15, 19 April 2007 (UTC)[reply]

By how it was found, do you mean history, or the actual derivation? The history is very short: Riemann's revolutionary paper (link in the text) defined analytic continuation. Riemann gave two proofs of the analytic continuation, one of them using an integral representation. Is that what you mean by formula? There should definitely be a place in Wikipedia where this is explained. There are also formulas for special values (negative integer s), and there is functional equation in case you need ζ(s) for Re s < 0 in terms of ζ(s) in the convergence region. Arcfrk 00:01, 20 April 2007 (UTC)[reply]

This article needs more information on how the zeta function is connected to prime numbers. My understanding of this function and the Riemann hypothesis is that this function and prime numbers are very deeply connected, but I can not find on Wikipedia or anywhere else an explanation as to why. Sloverlord 12:36, 17 May 2007 (UTC)[reply]

Have you tried Prime number theorem? Arcfrk 19:34, 17 May 2007 (UTC)[reply]

The Mellin Tranform section has ${\displaystyle \Gamma (s)}$. Could "where ${\displaystyle \Gamma (s)}$ is the Gamma function" be added?

Sure. That was noted before, but it won't hurt to repeat it, so I added it. However, you should feel free to make such edits yourself. The "edit this page" button at the top of the article is there for a reason! Cheers, Jitse Niesen (talk) 08:23, 20 May 2007 (UTC)[reply]

The 'globally convergent series' found by Hasse appears to be essentially just the Euler transform applied to the Dirichlet eta function

Is this vandalism? At the bottom of the page there's a link to a .gz file. Even so, it should be reworded. '''T'''''o''__m__ 17:34, 7 November 2007 (UTC)[reply]

The approximation by Gergő Nemes in "The functional equation" does not work when

1. s is close to 0
2. s has a large imaginary part (in fact, according to the formula zeta has only the trivial zeros)
3. s gets very negative

In other words, it mostly makes sense as an approximation of zeta for large negative values. For these values, the derivation is simply plugging in Stirling's formula into the functional equation, which only serves to complicate the expression. —Preceding unsigned comment added by 64.3.169.42 (talk) 21:15, 3 January 2008 (UTC)[reply]

I must ask a naive question to the author.I do not understand how the continuation of the ${\displaystyle \zeta }$ function is performed when variable s is such as ${\displaystyle Re(s)<=1}$.One was speaking of an integral formula,but the formula was never shown.The only integral formula that I saw was :${\displaystyle \zeta (s)={\frac {1}{\Pi (s-1)}}{\frac {1}{e^{-i\pi s}-e^{i\pi s}}}\int _{+\infty }^{+\infty }{\frac {(-x)^{s}}{x(e^{x}-1)}}dx={\frac {1}{\Pi (s-1)}}\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}-1}}dx}$ where ${\displaystyle \Pi }$ is the well-known Pi-function. But the last integral converges for ${\displaystyle Re(s)>1}$ only,so the first integral has a meaning only on this condition.So I do not understand the fact of how the ${\displaystyle \zeta }$ function initially defined by a serie,is continued for ${\displaystyle Re(s)<=1}$.I have tried to go to the source (Riemann's work in english version),but I still not understood how the continuation was performed.So I ask to the author of the article to bring the light on this (important) point of the subject.

Excuse me not to sign my message,but I am not involved among the users. date 08.01.2008 —Preceding unsigned comment added by 90.15.195.191 (talk) 16:51, 8 January 2008 (UTC)[reply]

What is the point of this graph in the middle of the article, when the article does not mention it even once? T.Stokke (talk) 21:05, 1 March 2008 (UTC)[reply]

Yes, it's not really clear what that graph represents. Certainly not the partial sums of the series defining ζ(2). I've commented it out for the time being, it might be relevant but it's not immediately obvious from the caption. Arcfrk (talk) 03:52, 3 March 2008 (UTC)[reply]

I don't know how I missed them before, but the sections purporting to evaluate zeta(2) and zeta(4) can, perhaps, be best characterized as unencyclopaedic garbage, the word that they seem to be using quite a bit. "Zeta constant" already offers a thorough coverage of special values of zeta. The "proofs" themselves are about as tortured as one can imagine, and blatantly fail WP:NOT. I got rid of them. Arcfrk (talk) 03:45, 3 March 2008 (UTC)[reply]

Years ago I did a course in complex analysis. I got a bare pass, which doesn't seem sufficient to understand the section on trivial zeros. Could someone explain explicitly why the negative evens are zeros?RayJohnstone (talk) 15:50, 28 March 2008 (UTC) —Preceding unsigned comment added by RayJohnstone (talk • contribs) 15:47, 28 March 2008 (UTC)[reply]

I think the easiest way to see it is using the reflection formula: ${\displaystyle \zeta (s)=2^{s}\pi ^{s-1}\sin \left({\frac {\pi s}{2}}\right)\Gamma (1-s)\zeta (1-s)}$

If s is a negative even integer, then ${\displaystyle \sin \left({\frac {\pi s}{2}}\right)=0}$, so the whole thing is 0. -GTBacchus^(talk) 15:55, 28 March 2008 (UTC)[reply]

Shouldn't ζ(0)=infinity?
${\displaystyle \zeta (0)={\sum _{n=1}^{\infty }}{\frac {1}{n^{0}}}={\sum _{n\geq 1}^{\infty }}{\frac {1}{1}}=1+1+1...=\infty }$

05:47, 19 April 2008 (UTC)

The series definition of the zeta function doesn't apply unless the real part of z is greater than 1. -GTBacchus^(talk) 10:06, 19 April 2008 (UTC)[reply]

Yes this is also addressed in the article.T.Stokke (talk) 14:01, 26 April 2008 (UTC)[reply]

How does ζ(0)=-1/2? —PrecedPMajer (talk) 16:21, 8 September 2008 (UTC)ing unsigned comment added by Anon126 (talk • contribs) 05:59, 15 May 2008 (UTC)[reply]

As far as I know, the easiest way of computing ζ(0) uses the representation at the end of section 4.1 :

${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }\left({\frac {x^{s-1}}{\exp(x)-1}}\right)\ dx.\!}$

Starting from this, one computes ζ(s) for s=0, and more generally for s=-n with nonnegative integers n, in terms of the residue at x=0 of

${\displaystyle {\frac {x^{-n-1}}{\exp(x)-1}}}$,

which relates ζ(-n) with the Bernoulli numbers:

${\displaystyle \zeta (-n)=-{\frac {B_{n+1}}{n+1}}}$.

Note also that the functional equation relates ζ(n) and ζ(1-n), whence you can compute the sum of the series of ζ(n), for even n>1. Conversely, if you already computed in another way the values of the sum of the series, you can deduce, of course, the values of ζ(n) for n<0 (but not for n=0, as you observed). See also the article on the Hurwitz zeta function for a generalization. PMajer (talk) 09:24, 22 May 2008 (UTC)[reply]

In my mind, the easiest way to make sense of this is to recall the relationship between the zeta function and the Dirichlet eta function:

${\displaystyle \zeta (s)={\eta (s) \over {1-2^{1-s}}}}$:

If you plug 0 into the alternating sum for the eta function, you have 1 - 1 + 1 - 1 ... which would lend itself to being equal to 1/2, which, when divided by (1 - 2), yields -1/2. -Xylune (talk) 11:32, 12 August 2008 (UTC)[reply]

Of course, this does not make any sense... Well, sorry, I mean, one should justify the equality with the Cesaro summation. (remember that, playing breezly with non convergent series, one can easily produce many identities like "1=0" etc..) PMajer (talk) 16:21, 8 September 2008 (UTC)[reply]

As I understand it: zeta(0) = 1/1^0 + 1/2^0 + 1/3^0... = 1 + 1 + 1... = infinity != -1/2

Where do I wrong? 84.108.198.129 (talk) 23:35, 2 September 2008 (UTC)[reply]

Now I see the answer above. Sorry... 84.108.198.129 (talk) 23:43, 2 September 2008 (UTC)[reply]

zeta(-2) = 1/1^-2 + 1/2^-2 + 1/3^-2 + ... = 1 + 2^2 + 3^2 + ... = infinity !=0

Someone is wrong - me, or the rest of the world. Who? (and why?) 84.108.198.129 (talk) 23:43, 2 September 2008 (UTC)[reply]

The first sentence in the definition explains why you're wrong. The real part of -2 is not greater than 1. Ben (talk) 04:20, 3 September 2008 (UTC)[reply]

This is because of analytic continuation: a formula valid for some values may not make sense for others. As a simpler example, consider 1 + x + x² + ... = 1/(1-x) which makes sense for |x| < 1. The series definition does not make sense for |x| > 1, but the formula 1/(1-x) does. Richard Pinch (talk) 23:20, 8 September 2008 (UTC)[reply]

This article shows a kind of a common problem of mathematical articles. It seems that a lot of people are interested in the Riemann zeta function, without beeing so interested in understanding the elementary underlying facts, such as, what is a complex number, or what is the sum of a series, etc. This makes sense, of course, for the sake of a general information (in the same way, I am interested in music even though I cannot play). But in this case one should not complain if an article looks too technical, or if things are not made easier than possible, and one should not be surprised if he or she can't go beyond a row picture of the facts, because it is natural. The point is simply that a complete explanation with all details, e.g. about the zeta function, would require a good part of an elementary course of complex variables, what is beyond the scope of thes articles. On the other hand, there is a minority, made of users with a mathematical culture, that go to a math article looking for a more technical, precise and usually very short information (tipically, in areas of mathematics different from their own). For them wikipedia has become quite a useful tool, so I think it should be a mistake to keep an article to the level of the mean users. Maybe a good thing should be to keep disjoint the general and the specialistic part...PMajer (talk) 17:48, 8 September 2008 (UTC)[reply]

How can a subset of a straight line have a fractal dimension greater than 1? Or does it mean that the Riemann hypothesis is false?--Yecril (talk) 23:08, 6 October 2008 (UTC)[reply]

Surprisingly enough, it seems to be correct. See this summary. Richard Pinch (talk) 17:12, 7 October 2008 (UTC)[reply]

I don't feel there is adequate discourse of using the reciprocal form. Specifically, readers will wonder why it is used when the sum of positive integers from 1 thru N is well established as N*(N+1)/2. Is the manipulation to be able to apply specialized analysis forms or ??--Billymac00 (talk) 03:17, 27 January 2009 (UTC)[reply]

Based on a false premise, don't you think? Fergananim (talk) 20:51, 5 February 2009 (UTC)[reply]

Huh? -GTBacchus^(talk) 03:36, 15 March 2009 (UTC)[reply]

The Euler product formula is from about a hundred years before Riemann was born. Is it known who was the first to consider this function? It seems Chebyshev's work connecting prime numbers to this function was before Riemann. Why is it called the Riemann zeta function? Was Riemann the first to define it for complex numbers? A section on history would be nice. Regards, Shreevatsa (talk) 14:57, 18 March 2009 (UTC)[reply]

The article states that the function was named after Riemann because "he introduced it"... Is that the case, or is it as Shreevatsa says, that he was the first to define it for complex numbers? —Preceding unsigned comment added by 24.12.13.8 (talk) 19:07, 15 October 2009 (UTC)[reply]

See On the Number of Primes Less Than a Given Magnitude, which is already linked to in the text. Fredrik Johansson 16:51, 18 March 2009 (UTC)[reply]

The following snippet appears to be meaningless. What is s? What is the significance of the variable x? A broader question is: is this formula significant enough to be in the article? If so, could somebody please correct it and give it enough context that it becomes meaningful and manifestly notable for inclusion in the article. Sławomir Biały (talk) 03:05, 9 July 2009 (UTC)[reply]

Expansion of the logarithm of ζ(s) on the critical strip

${\displaystyle {\begin{aligned}\pi {\frac {dN}{dx}}(x)&={\frac {1}{2i}}{\frac {d}{dx}}\left(\log \left(\zeta \left({\frac {1}{2}}+ix\right)\right)-\log \left(\zeta \left({\frac {1}{2}}-ix\right)\right)\right)\\[8pt]&{}\qquad -{\frac {2}{1+4x^{2}}}-\sum _{n=0}^{\infty }{\frac {2n+{\frac {1}{2}}}{\left(2n+{\frac {1}{2}}\right)^{2}+x^{2}}}\end{aligned}}}$

here dN(x)/dx is just the derivative (as a distribution) of the number of zeros on the critical strip 0 < Re(s) < 1. A proof of this can be found on a work by Guo (see references).

- the derivatives are all respect to 'x'

- The formula is an expansion of the logarithm of ${\displaystyle \zeta (1/2+is)}$ on the critical strip, as you can see it reproduces all the poles of ${\displaystyle \zeta (1/2+is)}$

- ${\displaystyle {dN}{dx}}$ is just the expression ${\displaystyle \sum _{\gamma }\delta (x-\gamma )}$ , whith 'gamma' being a sum over the imaginary part of the zeros --Karl-H (talk) 09:45, 9 July 2009 (UTC)[reply]

I see you've gone ahead and restored the context, with fixes. Thanks for that, but I still see the second part of my question as entirely unaddressed, namely does this formula actually belong in the article at all? It was sort of vaguely referenced before, and now it is entirely unreferenced. But I get the feeling that this isn't a typical formula that one can dig out of any textbook (or mathematical encyclopedia, for that matter). And so the question remains: is inclusion of this formula consistent with the spirit of the requirements of WP:UNDUE weight? If so, then it should be possible to build a more meaningful context around the formula in question: What is it good for? Why should the reader care about it? Otherwise, the disputed content should I think be removed, as Wikipedia is WP:NOT an indiscriminate collection of information. Sławomir Biały (talk) 18:40, 10 July 2009 (UTC)[reply]

this formula is just an expansion of the logarithmic derivative of Zeta function on critical strip, it is interesting (just my opinion) since it involves a sum over the poles of Riemann zeta, i think i found in the paper refereed before or in another context in papers talking about Gutzwiller Trace formula and Riemann Hypothesis, perhaps this formula would suit better into an article about Riemann Hypothesis or Gutzwiller trace, as you wishes --Karl-H (talk) 20:19, 10 July 2009 (UTC)[reply]

If so, it wouldn't have had any zero on critical strip Re(s)=1/2, neither any trivial zero.

Clearly, '>' should be replaced by '<'.

See paragraph 1. —Preceding unsigned comment added by 151.53.136.190 (talk) 21:02, 12 August 2009 (UTC)[reply]

The zeta function itself is defined for all complex numbers except 1. The series ${\displaystyle \sum _{n=1}^{\infty }n^{-s}}$ is only convergent for Re(s) > 1, the zeta function is defined elsewhere in another way (analytic continuation). — Emil J. 09:54, 13 August 2009 (UTC)[reply]

The article says that all negative even integers are trivial zeroes, but the value of the zeta function is infinite for them, not 0.

1+4+9+16+..., 1+16+81+256+..., etc. all diverge to infinity. —Preceding unsigned comment added by 75.28.53.84 (talk) 14:51, 15 August 2009 (UTC)[reply]

Did you read the definition? The series is only for Re(s) > 1. --Zundark (talk) 15:35, 15 August 2009 (UTC)[reply]

It said that all negative even integers are trivial zeroes. --75.28.53.84 (talk) 15:43, 15 August 2009 (UTC)[reply]

Yes, and that's correct. See my reply to you at User talk:Zundark#Riemann zeta function. --Zundark (talk) 16:14, 15 August 2009 (UTC)[reply]

The "definition" section does NOT make it clear that the infinite-series expansion only applies to Re(s) > 1. The definition says, "The Riemann zeta function ζ(s) is the function of a complex variable s, initially defined by the following infinite series...". It puts no restriction on "s." The next sentence, "Leonhard Euler considered this series in 1740 for positive integer values of s, and later Chebyshev extended the definition to real s > 1," also does not make this restriction clear. It just talks about two mathematicians who "considered" and "extended" the formula. All that sentence suggests to the reader is that before Chebyshev, the formula didn't even apply to Re(s) > 1. The definition needs to be clarified. -LesPaul75_talk 05:43, 28 September 2009 (UTC)[reply]

Given the above endless confusion about the definition of the series, the definition of the function defined by the series and the definition of the zeta function itself, I've changed the wording of the definition to something that I hope is slightly clearer. I attempted to make clear what people on the talk page were saying: that the zeta function is the analytic continuation of the function defined by the series given. Please take note that I have NO UNDERSTANDING AT ALL of analytic continuations. I merely changed the definition to emphasize what has been repeatedly said on this talk page. If I got it wrong, please feel free to correct it. mkehrt (talk) 09:03, 7 October 2009 (UTC)[reply]

It's better, but in my opinion it's a little redundant now. There is "Re(s) > 1" stamped all over the definition in five different places. There's no need to beat the reader to death. Why not just make it match the definition here: Riemann_hypothesis? That definition gets right to the point in just one sentence. -LesPaul75_talk 10:05, 15 October 2009 (UTC)[reply]

The argument: (harmonic diverges) -> (euler's formula predicts infinitely many primes) is flawed since Euler's formula depends on there being infinitely many primes to sieve the infinite sum of the riemman-zeta function to '1' —Preceding unsigned comment added by 99.73.17.9 (talk) 06:17, 7 April 2010 (UTC)[reply]

No, the derivation of the formula only uses the unique factorization property, it makes no difference whether there are finitely or infinitely many primes (in fact, the derivation is easier if there are only finitely many primes, since some issues of convergence disappear).—Emil J. 10:16, 7 April 2010 (UTC)[reply]

I'm not sure I agree with this. If there are finite primes, then I don't think the sieve can reduce zeta(2) (with its infinite terms) to a finite amount of terms. but i could be wrong, since all infinities involved are countable. —Preceding unsigned comment added by 99.73.17.9 (talk) 02:47, 8 April 2010 (UTC)[reply]

First, why do you keep talking about a sieve? There is no sieving involved. Second, did you actually read the derivation of Euler's formula, or are you just making your impression up? If there are finitely many primes (lets call them p₁,...,p_k), the Euler product is still a finite product of infinite sums, which evaluates to an infinite sum by distributing the product over the sums:

I think you haven't read the method Euler used. He certainly does 'sieve' out the primes. First he sieves out multiples of 2 with 1 - 1/2^s, then he sieves out multiples of 3 with 1 - 1/3^s, etc. He does this for all primes. This is just like an Erosieve. zeta(1) diverges. You start with an infinite sum. You sieve out all the primes. You get a product for all primes. To extend the sieving method all the way ad infinum and be left with a finite result assumes that there are an infinite amount of primes to sieve out. So the fact that we have a divergence equal to a product of all primes rests on the assumption that there were an infinite amount of primes to begin with. I see now why you didn't understand me to begin with since you were unfamiliar with the content of this page and unaware that a sieving process was used to generate the formula. I now see I was right from the start.

${\displaystyle \prod _{p}{\frac {1}{1-p^{-1}}}=\prod _{p}\sum _{m=0}^{\infty }p^{-m}=\prod _{i=1}^{k}\sum _{m=0}^{\infty }p_{i}^{-m}=\sum _{m_{1},\dots ,m_{k}}(p_{1}^{m_{1}}\cdot \dots \cdot p_{k}^{m_{k}})^{-1}=\sum _{n=1}^{\infty }n^{-1}.}$

As I already wrote, the only thing used here is the unique factorization property of integers, which guarantees that the products ${\displaystyle p_{1}^{m_{1}}\cdot \dots \cdot p_{k}^{m_{k}}}$ exhaust all positive integers, each one once.—Emil J. 10:24, 8 April 2010 (UTC)[reply]