# Talk:Möbius function

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## Move from "Moebius arithmetical function"

The following was moved here from the "Moebius arithmetical function" article:

In number theory is very important another sum, defined by:

M(n) = ∑ μ(n) .

This function is closely linked with the positions of zeroes of Euler - Riemann ζ- function. The connection between M(n) and Riemann conjecture was known to Thomas Joannes Stieltjes.

How exactly is M(n) defined. What is the connection with the Riemann conjecture? AxelBoldt

Whuh Axel. Let me first say you didn't just put M(n) from the Möbius arithmetical function article - but you changed the whole article I had started, (I guess so). Who's gonna say that now a new article is more comprehensible? I think now μ(n) is not 'well defined'. It is hard to calculate its values by hand, isn't it? And perhaps it is just because someone does not fully understand the meaning of such great and innovative function. The best definition is I suppose that one I gave - but you just put it away. In TeX notation it would be even better, but no-one can say which way is the best for writting formulas in Wikipedia. I am new herein but I have at least about 15 years experience in math related stuff and futhermore I am not educated mathematician (but in the end it doesn't matter for the Feynman's curious character, right...)

Yes, μ(n) is multiplicative function, but basically it is arithmetic function - the most important one, as it is written in appropriate article. Let us not split hairs for everything here - because such free encyclopedia won't be very good after all. I know it mainly from number theory, combinatorics and in connection with the famous Riemann hypothesis. For L-systems see bellow...

Do you know the answer about definition range of M(n), or are you just asking so? I can't answer this without any preparations. And in the other side I don't quite understand your claim. I guess it is defined for every natural n if this answers the question? The first values for M(n) are:

M(1) = 1
M(2) = 0
M(3) = -1
M(4) = -1
M(5) = -2
M(6) = -1
M(7) = -2
M(8) = -2
M(9) = -2
M(10) = -1
M(11) = -2
M(12) = -2
M(13) = -3
M(14) = -2
M(15) = -1
M(16) = -1
M(17) = -2
M(18 = -2
M(19) = -3
M(20) = -3

Is M(n) for every further number negative? No! M(39) = 0 is the first counterexample to this foolish assumption. And here's a few more values to consider:

M(92) = -1
M(93) = 0
M(94) = 1
M(95) = 2
M(96) = 2
M(97) = 1
M(98) = 1
M(99) = 1
M(100) = 1

Stieltjes wrote to his collaborator Hermite 1885 that he succeeded to proove for every "big" n:

| M(n) | < √ n.

Stieltjes never published his 'proof' because he probably found an error. It is valid onward, if for every constant number a with every natural n:

| M(n) | < √ n,

then Riemann conjecture is correct. Stieltjes announcement had pulled the trigger of researching wave about function M(n). Franz Mertens 1897 published 50 pages long table with values for M(n) for number up to 10000. With this table he thought that Stieltjes inequality is very likely. We say today this is Mertens conjecture. Another one was R. D. Sterneck who from 1897 to 1913 in several articles published tables for function M(n) for selected values n up to 5,000,000. On this basis he conjectured that for every numbers n larger than 200 is:

| M(n) | < (1/2) √ n

W. B. Jurkat found this conjecture is wrong not until 1960. The least n for above inequality is wrong is n = 7725038629, because | M(n) | = 43947 > 43946.0994544. Futher researches was made by H. Cohen and Dress in 1979. They showed that for every first 7,800,000,000 numbers is valid:

| M(n) | < (3/5) √ n

Many computer calculations indicated the truthfulness of Mertens conjecture and some theoretical results showed just the opposite. The final judgement was said 1983 by H. te Riele and A. Odlyzko. Their 8 years joint research work via electronic mail finally canceled Mertens conjecture. Te Riele mainly worked in Mathematical Centre in Amsterdam and Odlyzko in Bell Laboratories in New Jersey.

How te Riele and Odlyzko disproved Mertens conjecture? Not in the way that they found certain number n, for which is | M(n) | > √ n. We don't know today the number and it is very likely that it must be larger than 10 30. They used a method, which was discovered by A. E. Ingham 1942. Ingham defined a function h(x), which has this property: Let x be arbitrary real number and let A < h(x) - at that time there is a natural number n, that | M(n) | > A √ n. Let us say if we know a number x for which is | M(n) | > 1.06 √ n. This would disprove Mertens conjecture. But Ingham method does not tell us for which number n this inequality is valid.

Te Riele and Odlyzko decided to find a number x for which is h(x) > 1. It soon showed up this is very hard exploit. Calculating of function h(x) is very difficult and it needs supercomputers. Among other we need for calculating of h(x) many exact values of zeroes of Euler - Riemann ζ- function. And this is not the only trouble. Even if we have a computer programme for h(x), we should find a certain real number x, for which is h(x) > 1. And this is much harder because function h(x) is almost much smaller than 1. The search for suitable number x is very similar to searching of the needle in a hay. They have realised this fact and till 1979 they have found the largest value for x 0.86. They had abandoned their own research.

The further research on this field came with an algorithm from A. K. Lenstra, H. W. Lenstra junior and L. Lovasz. First application of this algorithm on Mertens conjecture enabled successful search of number x.

The first task was to calculate (non-trivial ??) zeroes of Euler - Riemann ζ-function. It was made by computer CDC CYBER 750 at Amsterdam's Mathematical Centre. Within 40 hours the computer defined with accuracy of 100 decimals 2000 zeroes. The other task - calculation of h(x) was running on CRAY-1 at Bell's Laboratories 10 hours with newest Lenstra's algorithms. They had found a number x for which is h(x) = 1.061 545 and thus disproved Mertens conjecture. This so wanted number x is:

- 14 045 289 680 592 998 046 790 361 630 399 781 127 400 591
999 789 738 039 965 960 762.521 505 .


Te Riele and Odlyzko publicated their work in an article Disproff of the Mertens conjecture in Journal für die reine und angewandte Mathematik, Vol. 357 (1985), p. 138-160. We know today that

| M(n) | < 1.06 √ n

is not correct for every natural n. We can ask if there is a certain constant A that it would be:

| M(n) | < A √ n .

Te Riele and Odlyzko think this constant does not exist. We can not disprove above inequality as an example for A = 2 with today's algorithms and computers.

What does it mean for Riemann conjecture a fact that Mertens conjecture is wrong? Strictly speaking - nothing. From regularity of Mertens conjecture follows the regularity of Riemann conjecture. But in the other side irregularity of Mertens conjecture does not say anything about Riemann conjecture. Mertens and Riemann conjectures are not equivalent.

Riemann conjecture is equivalent to weak form of Mertens conjecture, that for every real r, which is larger than 1/2 exists a real constant A that for every natural n is valid:

| M(n) | < A n r .

And this is something very different than Mertens conjecture. Okay, this is this all about connection of M(n) with Riemann conjecture in breaf. Mainly this comes from Keith Devlin's book from 1988 Mathematics: The New Golden Age in Slovene translation from 1993. Since then had passed 13 years and I believe that mathematicians and computer scientists didn'l sleep a lot... We should see herewith another similar result with Bieberbach conjecture and some more.

I deeply hope I gave you some clue on what you meant. And futhermore more I would like to writte that I would be very glad if my contributions would be accepted as they are mentioned to be, otherwise I won't participate in a run for a "giant question" and a "final quest of everything". And I won't put this lines to very nice article on Riemann hypothesis until I am not shure that someone won't totaly change them. It is very relative what one article should contain. I say it is better if contains more rather than less... I believe Axel, you won't go to my article on leading Slovene mathematician Josip Plemelj and throw everything out just because. You should consider other opinions and knowledge too.

I have another questions for everyone who can help on. Let us say that Möbius (multiplicative arithmetical) function μ(n) behave like an L-system. What is then its Besicovitch - Hausdorff dimension? Let someone say too, if this question is senseless? I should say I had written years ago a computer graphical programme that draws Möbius function μ(n) as a L-system and I still have no clue what is its BH dimension. Remember that every modern computer shows that Hilberts curve fills a square already with n ~ 10. And in the other side I haven't found what is BH dimension for Mandelbrodt set either? Does anyone know how should such questions be solved? I also had made a programme which extends Ulam sheet with Möbius function μ(n).

Axel, how we should name such curve: a Möbius curve or Möbius function in plane? I look forward for every little suggestion.

In and out over the hills...
XJamRastafire

Sorry, but I didn't mean to remove your definition and replace it with a "better" one. I simply noticed that there was already a Moebius function article there and I redirected your new one over there and got rid of what I thought was duplication.

Feel free to add your definition back in. You can still get it if you go to http://wikipedia.com/wiki/M%F6bius_arithmetical_function&action=history

M(n) = ∑ μ(n)

I think you probably mean

M(n) = ∑1≤kn μ(k)?

The above story of the Mertens conjecture is really interesting. It deserves its own article, and links from Moebius function and Riemann hypothesis.

Cheers, AxelBoldt

Ahah, Axel I get it. I first thought you just canseled my effort to explain μ(n). I haven't found any articles before I started, so that's why I am apologizeing. I will get used to keep track of all of these meanings. Okay, let that definition of μ(n) in an article stay. Yes, as it is obvious from the first values of k the sum is really

M(n) = ∑1≤kn μ(k),

defined also for all non negative natural numbers.

I can rewrite the story of Mertens conjecture and put it out. But this takes time and I shall endeavour. Thank you also for correcting my "false" redirections. I think in this way this FE (free encyclopedia) has a great future. You gave me futher stiff doze of mettle.

Best regards.
XJamRastafire

I have found Axel that you are changing articles quite a bit. But the questions is - if all your changes are strictly necessary. What is the main difference between words inverse and converse in common English language. Yes, I think you meant the usage of it in the context as in inverse of matrix but, anyway you can use it, as I have, and so and so on. I admit my English is not fully 100 %, so you should too as German Wikipedian, 'richtig'. Please correct sensible or otherwise I won't write a word anymore. Are you some kind of master's tutor here or are your intentions better as it seems. Fist you put away almost all my 'free' stuff, after that you commend my talkings about Mertens conjecture and after that you again put away 1/3 of my 'free' work on Möbius arithmetical function μ(n) and on twin prime conjecture. Perhaps I don't understand quite conceptions of GNU license and fair plays.

Let me say, for example, that if I have fully understandable article on Cantor's great work and I would like to put it in Wikipedia. But first I have to translate it from my native language, which is by chance to be Slovene one. Second this will take me a lot of time. Okay. And then would come one "funny" Axel from Paderborn or so, ha, ha, who shall fix this 'boted' stuff. After all I shall get tired of it in the end. But as Nikolaj Pečenko likes to say: "I could be wrong, too..." A solution of this tricky situation can be in sending e-mails about particular subject. Please, Axes if you shall 'superfix' some more articles, I think it is better for us to hear via e-mail and make things clearer.

To the infinity and back...
XJamRastafire [2002.02.25] 1 Monday

I don't think that negotiating an article over email would be easier than negotiating it on Wiki. Of course articles get changed all the time, including the ones I started. If you take a more relaxed attitude, it's a lot easier. If somebody makes an edit that you don't agree with, for instance because you think it was more correct or clearer before, just change it back and discuss on the talk page. Generally, it's not a matter of "which changes are strictly necessary" but "which changes improve the article".

If you intend to write about Cantor's work, you may want to check first the material we already have, to see where your additions fit in: Georg Cantor, Cantors Diagonal argument, Continuum hypothesis, countable, cardinal number. AxelBoldt

I know articles are changing all the time. But in the end with these changes we shall be in a strange situation, let me say, to change Euler arithmetical function and all of his work. What will rest in the far end. I'll be very wanton if I would change one's whole article. My job is just to put dot points on i's and nothing more. If I shall find in some near future, for example, that π is over-super-real number I will all let you know. I'll keep in mind your words on improveing of articles. I think I shall see the whole picture later when I will get used of Wikipedia. There are a lot of GNU based computer programmes all over and that's for shure they must have firm frames to work properly. That's how I imagine good articles in FE (free encyclopedia). Axel check the Paderborn sails if they all work properly :-)
XJamRastafire

XJamRastafire, could you please clarify this paragraph:

Very simple numbers to above ones are numbers by exactly 5 different primes, but they differ in values of μ(n) for it is μ(4620) = 0, as 4620 = 2 2. 3. 5. 7. 11 (SIDN A051270): 2310,2730,3570,3990,4290,4620,4830,5460,5610,6006,6090,6270,6510,6630,6930,7140,

Exactly what class of numbers are you listing here?

Also, could you add a description of how you generated the pictures? I don't think "colored L-systems" is a very well-known term. AxelBoldt

Yes, wellcome to the "strange" and beautiful world of numbers. But let me first say (as I already had somewhere above) that I am not "well educated" mathematician, whatever that really could mean and after all I am not Hilbert to chop Hydra's heads here. I have some knowledge in number theory mainly from the readings of leading Slovene mathematician in this field Jože Grasselli which once lectured at high school in my birth place. And I have done some own 'laic' researches. Gauss was probably up to his 16 years one laic youth from Brunswick and finally he became famous all over with his extraordinary work. It is known too for him that he started to teach himself Russian language when he was already over 60. These 'guys' can teach us all a lot. I am not ashame that I love mathematics very much and such. I put Cantor just for an example. Yes, for shure, I have to check all the material which is available here, than all my material and after that do the rest. We all know how many problems Cantor had outside with his understandings of infinity and such. His case is not alone (let me think - I know some more from head: Johhny (a.k.a Russell Crowe) Forbes Nash, Paul Ehrenfest and perhaps many more with their visionary mind...) I work in a tool factory and I know what a tool (or die) means. If you want to solve one problem you need some tools. In math is the same. And these tools is mathematical language itself, which changes with time too. Beautiful example for this is Hawking which is by Thorne's words the only one in the world who can speculate about topology and thermodynamics of black holes. Deep inside me there are a lot of proper thoughts about math, but I am not able to solve a great deal of problems in an ordinary fashion. It is quite well known that Feynman's (path) integral can not be defined in a strictly mathematical way even today, but physicists use it happily and with great amount. I dream of a day when some more unsolved problems will be revealed to all... We owe a lot to hard work of an army of mathematicians, physicists and astronomers and in recent times to computational scientiest.
I dont know what you don't understand with above numbers. Is it their mathematical nature or general one? I don't know either for their class if there is one.
μ(4620) = 0, as 4620 = 2 2. 3. 5. 7. 11 (only five different primes but 2 comes twice...)
μ(4830) = -1 because 4830 = 2. 3. 5. 7. 23 (just five distict primes and noone comes more than once...)
I don't know for the class. I just put it here to distinguish the properties of μ(n). I even hardly believe that there is such tricky classification for this weird numbers. The significance of μ(n) is, as I think, that it tells us so much more about natural numbers (or maybe I am wrong - the time will tell). Perhaps this is another somekind of higher dimension view on the numbers. And strange fact is that Möbius is known to be (theoretical) astronomer. He must have observed very well. He also discovered his famous strip which today may be a child's toy. We can say this was just a lucky chanse but, we should know more about his relations to numbers.
Yes, the pictures are all 'mine'. I made all the 'tools' to produce them except for graphical grasping. I haven't found themlike on the net. I just found Ulam's prime spirals or Ulam's square (prime) spirals which in my native language are called Ulam's cloth or sheet (Slovene = Ulamov prt). (I haven't found this term in English either). Check for Ulam's prime spiral at URL I gave in the article:
We just draw numbers in a spiral as for example (number 1 lies at the center and than we go in the circle left or right (or clockwise and vice versa):
      17  16  15  14  13    ^
18   5   4   3  12    .
19   6   1   2  11    .
20   7   8   9  10    .
21  22  23  24  25 ....

Then we check the numbers if they are primes (or they have some different arithmetical or other properties) and sieve or pan them out. What remains is Ulam's cloth:
      17              13    ^
5       3        .
19       1   2  11    .
7                .
23         ....

I had made a C programme to do the same with μ(n) and colours the definition range {-1, 0, 1} with three different colours. Now (as I call it Ulam - Möbius cloth) with values (colours) for μ(n) looks like:
      -1   0   1   1  -1    ^
0  -1   0  -1   0    .
-1   1   1  -1  -1    .
0  -1   0   0   1    .
1   1  -1   0   0 ....

I think a pattern which comes from a programme's graphical output is really most beautiful but we won't discuss about its meaning here. Let us just say that primes come mainly on diagonals of such Ulam's spiraled square. This is all just OK mathematically I think.
But about Möbius coloured L-systems I am not shure. Typical L-systems (I don't know their definition origin right now, but it is highly implemented in very good computer programme fractint) are a Hilbert's curve or a dragon's curve. My (semi) definition for Möbius L-system (or something else - representation on a plane) is perhaps not quite the same as in L-system, because it is not recursive or selfgenerated over the limit as real L-systems are. We start with particular point, let us say 1 at the center and we do almost the same as in Ulam's prime spiral. We go upwards for a unit (but units does not count in L-systems I guess). We calculate μ(n+1) and we get μ(2)= -1. If it changes the sign, we go left, otherwise we go right -just like a turtle with a supermind would go to catch Zenon on some strange sphere and Universe, ha, ha. Then we calculate and get μ(3)= -1. If μ(n)= 1 we dont change a course of our clever turtle and we turn in the same temporal direction. And if μ(n)= 0, we even don't turn left or right but we let our L-turtie go straight on for a unit. I have to say that my programme does not use turtle graphic or strictly L-system "macro" language but ordinary and "more" understandable Cartesian coordinates. I think it even does not need it, because its main thing is after all just an arithmetical function which is in this case to be our famous and loving μ(n).
Now you must tell me if this worths something, mathematically speaking. Hardy spent all of his life doing just the same kind of things as they say, so this is my argument. Please let me know about your opinion. And finally if we "knew" all the values of μ(n) how this strange curve would look like? In my programme monitor x,y coordinates for certain n run out at a certain window and a scaling should be done - this is not implemented in it. And what would be its BH dimension. Perhaps someone knows the answer - I don't. I even don't know if this is the right mathematical question. Help me out. Would this function stretch all over (the flat 2D Universe, R2) or does it have a certain orbit or somekind of atractor to form itself? If we look to another starting point, we get different shape of function - but perhaps there are some selfsimilarities with its edge as in ordinary fractal functions as Mandelbrodt's or others are. But I don't yell for fractal nature of such Möbius function.
I hope you understand better now. I must say that an idea to define Möbius function and Ulam prime square spiral are properties of a visionary mind for me. Let me say here too that the similar thing happend when Jožef Stefan defined his Stefan - Boltzmann law of a radiation of a black body where there temperature T strangely came in fourth degree and many physicists thought and perhaps still think he just guessed that. But he was among the first in Europe who fully understand Maxwell's electromagnetic field equations and the whole theory at that time so he must have been a real physicist. --XJamRastafire [2002.02.26] 2 Thuesday (2nd ed.)

Ok, I understand the number spiral and the turtle graphics approach. Maybe we should add an explanation to the main article.

These numbers

(SIDN A051270): 2310,2730,3570,3990,4290,4620,4830,5460,5610,6006,6090,6270,6510,6630,6930,7140

are still mysterious to me: where do they come from, how are they defined, what do they have to do with μ? AxelBoldt

Edit conflict - hope everything is placed as it should be XJamRastafire [2002.02.26] 2 Thuesday (3rd ed.)

At first glance above Sloane's sequence A0512270 for me does not seem very special. They are well defined with a simple request.

How are they defined and why did you add them to the article about the μ function? AxelBoldt

I think it really doesn't matter where they "come from". We can form another sequence which would represent μ(k) of above sequence. First such sequence would be then:

(-1,-1,-1,-1,-1),0,(-1),0,(-1,-1,-1,-1,-1,-1),0,0,(-1,-1,-1,-1),0,0,0,0,(-1,-1,-1,-1),0,(-1,-1),0,(-1,...


or if we write shorter:

5,1,1,1,6,2,4,4,4,1,2,1,... defined for k>0


we get another "strange" sequence which is not in the base o Sloane's On-Line Encyclopedia of Integer Sequences. But we just speculating and we never know where we would come. But what bothers me is for example according to Peter Meyer's home page what are the similarities and connections with such properties found on his page as:

• Cluster colour of primes, (Similar property would have above sequence 5,1,1,6,... because numbers are somehow grouped with μ(k)),
• Sequence colour of primes,
• Palmen's colour of natural numbers and so on

with μ(n)? This is nice challenge for some more computations and thinkings with pure μ(n)... Axel if you think that those pictures don't fit in an article we can throw them away. And in the other side if you think they are just fine, we shall find some time to put futher explanations. Perhaps this free encyclopedia some day becomes a research center for some serious and alternative approaches to natural sciences. After all mathematicians are not the only ones who love mathematics and who knows what really are they thinking. :-)
XJamRastafire [2002.02.26] 2 Thuesday (4rd ed.)

Dear friends check this animated *.gif which merges the representations of μ(n)) if we observe it around and along sequenced prime numbers.

Representation of μ(n) on a plane as a coloured L-system around prime points n=1,2,3,5,7,11 cycled.

This picture clearly shows the "strange" behaviour of μ(n). It looks different for every single point but in the other side shows some similarities on the edges. Axel, sorry but I still don't understand your question about Sloane's sequence A0512270! The definition of it states existence of exactly 5 different primes but every one of them can be arbitrary numbered. So everything in these sets is correct for such sequences (I guess so):

{ 21761. 1417, 3, 11, 21, 1093. 18}
{ 21761. 1418, 3, 1111, 21, 1133. 18}
{ 29, 3, 11, 21, 127}
{ 5, 3, 11, 21, 109}
{ 5, 7, 11, 21, 131}
{ 7, 11, 13, 21, 137}

as long there are just and exactly 5 primes. I think this should be enough. Let me know if this is still not good for ya.
XJam [2002.02.27] 3 Wednesday (0)

Axel can we agree on one another thing. You use asterix (*) for multiplication of natural numbers. I was tought already at elementary school to use period (.). Generally we use (*) for other kind of multiplications excpecially to multiplicate matrices where there multiplication has some different properties of natural multiplication. We also use multiplication sign (x) in HTML (&times). I use as in:

1. 11. 111. 11111. 1111111. 111111111. ...


you use:

1 * 11 * 111 * 11111 * 1111111 * 111111111 * ...


The best it would be I guess (but it is a little bit longer):

1 · 11 · 111 · 11111 · 1111111 · 111111111 · ...


and in sequences:

11,
3 · 11 · 37,
3 · 11 · 37 · 41 · 271,
3 · 11 · 37 · 41 · 239 · 271 · 4649,
3 · 11 · 37 · 41 · 239 · 271 · 4649 · 21649  · 513239,
3 · 11 · 37 · 41 ·  53 ·  79 ·  239 ·   271 ·    4649 · 21649 · 513239 · 265371653



XJam [2002.02.27] 3 Wednesday (1st ed. )

I agree that we shouldn't use * for multiplication, since it's just computerese. × and · should display correctly for almost everyone, so there's no reason to avoid them. (And even x and . are better than *.) --Zundark, 2002 Feb 27
Ok, I vow to drop *; I'm happy with × and ·, but I don't like x and .: x is often a variable, and . looks too much like the decimal point used in many countries. AxelBoldt

In the computational world μ(n) is sometimes designated as M(n), which must not be mistaken with a common mathematical notation of another sum (see below).

thrown out from the article again? This statement clearify the mess with math notation. -- XJam [2002.06.14] 5 Friday (0)

The letter used to denote a function is not fixed; you can use any letter you like as long as you say what function you are talking about. μ is used for lots of functions which aren't the Mobius function, and lots of letters which aren't μ are used for the Mobius function. There are also many functions called M around, not just the one that the article mentions. The name of a function is fixed, not the letter. There are way too few letters around. Especially if you are on a system without greek letters. AxelBoldt, Friday, June 14, 2002

I've moved the relevant info from here to Ulam spiral, and removed the info that doesn't seem related to this page. If anyone wants to resurrect the info here, please add text explaining how it relates to this page.

## M(infinity)=-2 ?

I believe I have found $M({\infty })$ . It should equal -2. Here's why:

I use this well known formula:
${\frac {1}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}$ ,
and insert s=0. This reduces to
${\frac {1}{\zeta (0)}}=\sum _{n=1}^{\infty }\mu (n)$ and so because
${\zeta (0)}={\frac {-1}{2}}$ and $\sum _{i=1}^{n}\mu (i)=M(n)$ $M({\infty })={\frac {1}{\frac {-1}{2}}}=-2$ .
can anybody verify this, can this be used to prove that

$M(n)=O({\sqrt {n}})$ and so too the Riemann hypothesis?

Ozone 20:12, 24 January 2006 (UTC)

You must have meant
$\sum _{k=1}^{n}\mu (k)=M(n).$ The meaning of the identity
${\frac {1}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}$ seems clear when s > 1. To say that it continues true when s = 0, I suspect you'd have to talk about "summation methods" or the like. I'd be very surprised if anything as simple as this argument could be used to prove the Riemann hypothesis. Whether, and in what sense, it makes sense to say that M(∞) = −2 is something I'll leave to those who know more that what little I know about this. Michael Hardy 21:48, 24 January 2006 (UTC)

I was assuming it would continue to work. I wonder... Yes I thought I fixed the error about Merten's function, maybe I didn't click save. The reason I said that it might help to prove the RH is that this shows that M doesn't go off to infinity as it increases. I don't think it will but there's a chance, allbeit a very very small one. It is still a strange result though, there are more -1 values than 1 values, weird...Ozone 02:07, 25 January 2006 (UTC)

The correct way to compute $\sum _{n}^{\infty }f(n)$ of any arithmetic function f is to compute one of the following limits (which are all identical):

$\lim _{t\to 0}\sum _{n}f(n)e^{-tn}$ or

$\lim _{s\to 0}\sum _{n}f(n)n^{-s}$ Here, you need to substitute M(n) for f(n), and not μ(n) for f(n). See zero-point energy for a real-life example, or also zeta function regularization. linas 14:42, 25 January 2006 (UTC)

But wouldn't this way still work, Your way may be the "official way" but how would you calculate M(infinty) your way? On the other hand I think I have found reason to believe that I am wrong. Riemann Zeta Function says that this only works for s>1. Also it gives very strange results as you use lower values of S. Ozone(sorry wrong computer for sig.)
Perhaps I was too breif. M(infty) should really be understood to be given by either one or the other of the two sums above. The exponential sum will have divergences at t=0, these are to be "ignored", if you just want the finite part. The dirichlet series which is given in zeta function regularization will have simple poles at places other than s=0, and so taking the s=0 value is a valid thing to do. These two are related by a Mellin transform, again, as explained on the zeta function regularization page.
I am not sure what you thought the Riemann Zeta Function page may have said. The zeta can be analytically continued to s values less than 1. The zeta function has a pole at s=1, but nothing else. I suspect that what you really need to do is to study analytic continuation in greater detail: it is a technique for defining functions on the complex plane even if the series that defines them is divergent or infinite in some way. The very very simplest example is
$\sum _{n=0}^{\infty }x^{n}$ which fails to converge for any value of x greater than 1. Indeeded, for x=1, we have 1+1+1+1+... and for x=2, we have 2+4+8+16+.... yet, famously, we know that this sum is just plain old 1/(1-x). and so there we have it: 2+4+8+16+...= -1. linas 01:54, 26 January 2006 (UTC)
Oops, I had made several misleading statements last time, which are now corrected, I beleive. linas 02:09, 26 January 2006 (UTC)
Also, I did not mean to imply you were incorrect or wrong, or anything like that. Your maniplation was entirely valid, and you got the right answer. I guess I'm pointing out the technique has a whole bundle of terminology and theorems wrapped around it, and that one must be careful to use this termiology correctly: otherwise you get "silliness" like 2+4+8+16+...= -1 or perhaps more interestingly, 1+2+3+...=-1/12, etc. linas 02:09, 26 January 2006 (UTC)

So Linas, you agree that M(infinty)=-2? I wonder if there are any results from this. (Yes I knew about the analytic continuation, I was just using the value of zeta(0)=-1/2 from a website.)

## Exponential generating function?

Silly me, the exponential generating function

$f(x)=\sum _{n=1}^{\infty }\mu (n)e^{-nx}$ seems to be non-trivial? linas 03:28, 2 July 2006 (UTC)

This form is especially inteeresting,ssince its Mellin transform is just the Gamma function dvivided by the Riemann zeta function (note that I have changed the independent variable in the above formula from 'y' to 'x') 81.102.15.200 (talk) 13:27, 1 May 2009 (UTC)

## Moebius of 0

I understand, of course, that the sentence claiming that μ(0) is generally left undefined is an undisputable fact, so instead I wish to argue the reasoning behind that belief. It would seems to me it would be obvious that the Maple system's definition of -1 is correct. 0 is clearly not square-free, since its divisible by all spmplex numbers. That leaves only 1 and -1. Since 0 is divisible by all complex numbers, and inherently all primes, it has an infinte number of primes divisors. Since infinity is not an integer, and it is also half of itself, it can be shown not to be even. Thus, 0 is non-square-free integer with and odd number of distinct prime divisors, which gives us μ(0) = -1. I may have overlooked something, but if not I simply wonder why it is generally left undefined. -- He Who Is[ Talk ] 22:58, 5 July 2006 (UTC)

As you say, 0 is not square-free, so the value of Moebius of 0 should be 0.

## Physics connection needs citation

the connection w/physics and pauli exclusion principle sounds neat but it's possibly a clever physicist messing with our heads.It was put in by 70.16.52.150 on 10/26/05. At about same time that editor stated similar stuff on Mobius inversion formula, later removed by Linas, I think.Rich 10:34, 25 October 2006 (UTC)

I think the current version is correct, and that the primon gas article provides the citations, as well as enough details to verify it yourself. linas (talk) 13:13, 6 May 2009 (UTC)

## Strictly positive natural numbers

It seems like this phrase near the beginning of the article overspecifies the numbers we're talking about. If we want to avoid the 0-v-no-0 business mentioned at the natural number page, wouldn't it make more sense to say "positive integers," and just avoid the poorly-defined label "natural" in the first place? I didn't want to change it myself, because there may be a reason I'm not aware of for keeping it. 129.2.56.220 (talk) 14:09, 15 February 2008 (UTC)

Thanks. I agree, and I changed it. You may want to read WP:BOLD. -- Dominus (talk) 14:36, 15 February 2008 (UTC)

## Distribution at $\infty$ What is the overall distribution of the values of the mobius function? Is it -1 1/3 of the time 0 1/3 of the time and 1 1/3 of the time? Woscafrench (talk) 22:28, 29 April 2010 (UTC)

I'm (quite possibly mistakenly) under the impression that the Mertens function is chaotic. Therefore, I think that that implies that as n goes to infinity, the chances of the mobius function being any one of the three values is equal, namely 1/3.My 2 Cents' Worth (talk) 12:13, 25 May 2010 (UTC)

The article Square-free integer has a short proof that the density of square-free integers is asymptotically ${\frac {6}{\pi ^{2}}}.$ These numbers have non-zero values of μ. I do not know if they are equally divided between plus and minus one. Thus $1-{\frac {6}{\pi ^{2}}}$ of the integers have μ equal to zero. Virginia-American (talk) 00:43, 4 January 2011 (UTC)

## 0 is even?

What makes 0 even? Is it because 0/2 = 0? 0 is the absence of a number, so can you give to 0 properties of a real number? My 2 Cents' Worth (talk) 12:16, 25 May 2010 (UTC)

0 is commonly considered as an integer, and hence a real number. 0 is not the absence of a number, it is the number that represents the absence of something. If you have 1 apple, and you eat it, you can still ask what number of apples you have, the answer being that you have zero apples. As for the evenness of zero, I'll refer you to the wiki article Parity of zero for a long discussion (though I'll also add that zero is divisible by two (since 2×0=0), so it satisfies one of the defining properties of even numbers). RobHar (talk) 18:00, 25 May 2010 (UTC)

## Hello

"There is a formula for calculating the Möbius function without knowing the factorization of its argument"

That formula requires knowing the k's such that gcd(k,n) = 1, which requires knowing the factorization of the argument.

" μ(n) is the sum of the primitive nth roots of unity."

This is misleading because the sum of the primitive nth roots of unity is zero which is not always μ(n) [edit: for n>1]

77.1.191.252 (talk) 18:14, 3 January 2011 (UTC)

To address your first point: finding a gcd is easier than factoring, see Euclidean algorithm. To address your second point: the sum of all nth roots of unity is zero, but the sum of the primitive nth roots of unity need not be. Take for example n = 3, then the three third roots of unity are 1, ${\frac {-1+i{\sqrt {3}}}{2}},{\frac {-1-i{\sqrt {3}}}{2}}$ , and only the last two are primitive. Their sum is clearly −1, which is μ(3). Anyway, there is a reference for this statement in the article, namely (16.6.4) of Hardy and Wright's book, so you should take a look there before claiming the formula is wrong. Cheers. RobHar (talk) 19:56, 3 January 2011 (UTC)
There is a proof of this formuula in the article Root of unity
Virginia-American (talk) 00:52, 4 January 2011 (UTC)

## Incoherent sentence

Toward the end of the article this passage occurs:

"The Möbius function also arises in the primon gas or free Riemann gas model of supersymmetry. In this theory, the fundamental particles or "primons" have energies log p. Under second-quantization, multiparticle excitations are considered; these are given by log n for any natural number n. This follows from the fact that the factorization of the natural numbers into primes is unique. In the free Riemann gas, any natural number can occur, if the primons are taken as bosons."

Aha! So any natural number can occur. As what????? Readers trying to understand the last sentence of this passage will not be able to get even a clue as to its meaning if an essential piece of information is omitted. It makes as much sense as if someone had written just: "In mathematics, any natural number can occur."Daqu (talk) 04:46, 10 February 2013 (UTC)

## Product formula relation

"The Dirichlet series that generates the Möbius function is the (multiplicative) inverse of the Riemann zeta function

\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}. This is easy to see from its Euler product"

Why is it clear? Am i being stupid here? — Preceding unsigned comment added by 109.153.163.53 (talk) 23:23, 29 September 2013 (UTC)

## Unsourced material

I have removed material sourced to a blog post on math.stackexchange.com. Material added must be verified by independent reliable sources. This does not qualify. Deltahedron (talk) 15:46, 12 April 2014 (UTC)

## Möbius function of 1

‹See TfM›

If you are still confused, please forward your question to Wikipedia:Reference desk/Mathematics; this is off-topic here per WP:NOTAFORUM
The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

I'm a little confused as to why the function returns positive 1 for the input 1. Wouldn't it be considered to have repeating prime factors (1x1x1x...) and output to 0? Or simply an odd number of factors, a single factor, which is "1". I'm not understanding how 1 has an even number of prime factors. 67.186.128.135 (talk) 06:07, 28 January 2020 (UTC)

1 is not prime...--Jasper Deng (talk) 06:14, 28 January 2020 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

## Typographical consistency

Regarding this revert...

To answer the question in your edit summary, "more consistent with what?", the consistency is between the inline and standalone formulas. For example, the μ in these formulas:

μ(ab) = μ(a) μ(b)
$\sum _{d\mid n}\mu (d)={\begin{cases}1&{\text{if }}n=1,\\0&{\text{if }}n>1.\end{cases}}$ They look slightly different. It's possible some question is left in the mind of the reader as to whether or not they represent the same thing. Sometimes typography does make a difference in math formulas. For example, consider the potential differences and non-differences among R, R, ℝ, r, and r. It would be better if math symbols that mean the same thing retain the same appearance, so the reader has fewer questions to puzzle through. -- Beland (talk) 14:19, 21 September 2020 (UTC)