July 2006 - mathfun@mailman.xmission.com

Re: [math-fun] Factorial n
by dasimov＠earthlink.net 04 Jul '06

04 Jul '06

I wrote: << Gene writes: < In Harvey Cohn, "Advanced Number Theory" . . . >> Thanks, Gene. If that's "Cohen" then I have the book & can look it up. >> Wrong again -- it is Cohen. Sorry, Gene. I seem to be having a bit of trouble with number theory today. --d.

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Re: [math-fun] Factorial n
by dasimov＠earthlink.net 04 Jul '06

04 Jul '06

Gene writes: << In Harvey Cohn, "Advanced Number Theory" a proof is given that the primes in arithmetic progression are equidistributed among residue classes in the sense that, for gcd(a,m) = 1, sum(1/p^s, p == a mod m) 1 ------------------------------- ==> ------ as s ==> 1. sum(1/p^s, all primes p) phi(m) >> Thanks, Gene. If that's "Cohen" then I have the book & can look it up. --Dan

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P.S. Re: [math-fun] "the lunatic asylum as my goal"
by dasimov＠earthlink.net 04 Jul '06

04 Jul '06

To blather on a bit: In fact, the function g(s) = sum {n=1,oo} of (-1)^(n+1) / n^s is easily seen to be (1 - 2/2^s)zeta(s) for s where (***) is defined: where R(s) > 1. But g(s) converges for Re(s) > 0. Thus zeta(s) may be defined as (*) zeta(s) = g(s)/(1 - 2^(1-s)) for Re(s) > 0. Let xi(s) = zeta(s) gamma(s/2) / pi^(s/2). Then the functional equation for zeta is (**) xi(s) = xi(1-s) where xi is defined, i.e., for s neither 0 nor 1. Since zeta is defined for 0 < Re(s) by (*), the functional equation (**) can be used to continue zeta from Re(s) > 0 to all s (except the pole s = 1). (If zeta were defined only for Re(s) > 1 then (**) would give another definition for Re(s) < 0, but it would not be clear whether these two functions are part of the same whole). Now from (**) we get zeta(-1) = (zeta(2)*gamma(1)/pi)/(gamma(-1/2)/pi^(-1/2)) = ((pi^2 / 6) / pi) / (-2*sqrt(pi)*sqrt(pi)) = -1/12. --Dan --------------------------------------------------- << I've seen it written in several places that since 1+2+3+ . . . = sum{n=1,oo} of 1/n^(-1), Ramanujan thought of this sum, in some sense, as zeta(-1), which turns out (using correct analytic continuation of the Dirichlet series (***) zeta(s) = sum{n=1,oo} of 1/n^s, which converges only for Re(s) > 1) to be -1/12. >>

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Re: [math-fun] "the lunatic asylum as my goal"
by dasimov＠earthlink.net 04 Jul '06

04 Jul '06

Marc writers: << This summer's AMS Notices reproduces Ramanujan's letter to Hardy where he asserts 1 + 2 + 3 + 4 + ... = - 1/12 On the hypothesis that even the errors of genius may hold value, I've always been curious what his theory was under which this result was derived. >> I've seen it written in several places that since 1+2+3+ . . . = sum{n=1,oo} of 1/n^(-1), Ramanujan thought of this sum, in some sense, as zeta(-1), which turns out (using correct analytic continuation of the Dirichlet series zeta(s) = sum{n=1,oo} of 1/n^s, which converges only for Re(s) > 1) to be -1/12. --Dan

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[math-fun] "the lunatic asylum as my goal"
by Marc LeBrun 04 Jul '06

04 Jul '06

This summer's AMS Notices reproduces Ramanujan's letter to Hardy where he asserts 1 + 2 + 3 + 4 + ... = - 1/12 On the hypothesis that even the errors of genius may hold value, I've always been curious what his theory was under which this result was derived. Is how he actually got this known? Do you have any conjectures? Or is it just an obvious common fallacy that "everybody knows"? Speculation: it reminds me how the 2-adic expansion familiar from binary computer arithmetic 1 + 2 + 4 + 8 + ... = - 1 can be interpreted by viewing it as the geometric sum y = 1/(1-x) evaluated at x=2. But, for example, dy/dx hasn't seemed fruitful. Can you construct some such way to get Ramanujan's result? Creative lunacy, anyone?

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[math-fun] Re: Factorial n
by Richard Guy 04 Jul '06

04 Jul '06

Many thanks for several replies. This one seems to contain the definitive answer. R. On Tue, 4 Jul 2006, moree(a)mpim-bonn.mpg.de wrote: > Richard Guy asked: > > Presumably 0! = 1! = 0^2 + 1^2. > 2! = 1^2 + 1^2 > 6! = 12^2 + 24^2 > > are the only integer solutions of > > n! = x^2 + y^2 > > but is there a proof? R. > > David Wilson wrote: > > No, I don't have a proof, I do have an argument following from a solid > conjecture. > > We know that for n >= 2, there exists a prime p with n appears that for n >= 4, there exists prime p == 3 (mod 4) with n I cannot prove this myself, but I am sure it is true. > > This conjecture implies that for n >= 7, there is a prime p == 3 (mod 4) > with [n/2] factorization of n, so that n! cannot be of the form x^2+y^2. > > ***** > Paul Erdos seems to have been the first to have supplied such a proof. > He namely proved that for n >= 4, there exists > a prime p == 3 (mod 4) with n arithmetic progressions'. The rest of the argument is then as David > Wilson described above. > > Reference: > P. Erdos, Ueber die Primzahlen gewisser arithmetischer Reihen, > Math. Zeitschrift 39 (1935), 473-491. > > I quote now from p. 491: > Mit Hilfe der Primzahltabellen bestimmt man nun leicht die genauen > unteren Grenzen von \xi, von welchen ab zwischen \xi (exkl.) und > 2\xi (inkl.) Primzahlen von den betrachten Formen existieren. Man > findet dass das Intervall \xi Primzahl von der Form 6k+1, fuer \xi \ge 5.5 eine Primzahl von der > Form 6k+5, fuer \zi\ge 6.5 eine Primzahl von der Form 4k+1 und endlich > fuer \zi \ge 3.5 eine Primzahl van der Form 4k+3 enthalt. Aus der > letzten Tatsache folgt auf Grund eines bekannten Satzen der elementaren > Zahlentheorie, dasz 1!, 2! und 6! die einzigen Faktoriellen sind, die > sich in die Summe zweier Quadratzahlen zerlegen lassen. > > The above article of Erdos gives an elementary method to determine > for a certain finite set of moduli d, explicitly numbers x_0(d) > and z(d) such that every interval of the form (x,z(d)x) with > x>=x_0(d) contains a prime p with p=a(mod d), for any a coprime to d. > > More precisely, for a modulus d>=2 put > sigma(d)=\sum_{p<d, p does not divide d}1/p, where p runs over the primes. > Erdos proved that if sigma(d)<1 and z>d(d-1)^{-1}(1-sigma(d))^{-1}, then > there is at least one prime from each primitive congruence class modulo > d in the interval (x,zx] for all x sufficiently large. > This result can be made effective. Erdos did this for d=4 and d=6 > (with the outcome given in the quote above). In my paper > > P. Moree, Bertrand's postulate for primes in arithmetical progressions. > Comput. Math. Appl. 26 (1993), no. 5, 35--43. > > this is done for the remaining 52 values of d satisfying \sigma(d)<1. > > Finally, I liked to note that R. Breusch earlier than Erdos published > a proof of Bertrands' Postulate for modulus 4. It is rather technical > and computational and uses results on the first zeros of L(s,\chi_4). > > R. Breusch, Zur Verallgemeinerung des Bertrandschen Postulates, dass > zwischen x und 2x stets Primzahlen liegen, Math. Zeitschrift 34 (1932), > 505-526. > > In the early 30's also G. Ricci worked on Bertrand's Postulate for > arithmetical > progressions, but I cannot comment on this paper (which I have never seen): > > G. Ricci, Sui teorem di Dirichlet relativo alla progressione arithmetica, > Bolletino della Unione Mathematica Italiana 12 (1933), 304-309. > > I hope this helps, > Pieter Moree >

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Re: Re: [math-fun] Factorial n
by dasimov＠earthlink.net 04 Jul '06

04 Jul '06

Michael writes: << If the primes congruent to a mod m have a large desert between N and (1+e)N, then the density of such primes up to (1+e)N is 1/(1+e) times what it is up to N, and so those two densities cannot both be arbitrarily close to the asymptotic value 1/phi(m) >> Aha -- what I wanted to be true would be, had I wanted the right thing to be true. -dan

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Re: [math-fun] Factorial n
by David Wilson 04 Jul '06

04 Jul '06

Michael Kleber already told us that Erdos had already proved my assertion that primes of the form 4n+1 and 4n+3 exist between n and 2n for n >= 7. I would be willing to conjecture that if there are an infinitude of primes == r (mod m), then there is a prime of this form between n and n(1+e) for sufficient n. Are there any results on the density of such primes relative to all of the primes? If so, then probably the prime number theorem could probably be used to establish the above conjecture. ----- Original Message ----- From: "Eric Bach" <bach(a)cs.wisc.edu> To: "David Wilson" <davidwwilson(a)comcast.net> Sent: Tuesday, July 04, 2006 1:53 AM Subject: Re: [math-fun] Factorial n > On Mon, 3 Jul 2006, David Wilson wrote: > >> No, I don't have a proof, I do have an argument following from a solid >> conjecture. >> >> We know that for n >= 2, there exists a prime p with n > appears that for n >= 4, there exists prime p == 3 (mod 4) with n > 2n. >> I cannot prove this myself, but I am sure it is true. > > Kevin McCurley had a paper (Math. Comp.?) giving explicit prime > number theorems for arithmetic progressions. Probably your supposition > would follow from his work plus some computer checking of small cases. > > Eric > > > > -- > No virus found in this incoming message. > Checked by AVG Free Edition. > Version: 7.1.394 / Virus Database: 268.9.8/381 - Release Date: 7/3/2006 > >

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[math-fun] Factorial n
by Richard Guy 04 Jul '06

04 Jul '06

Presumably 0! = 1! = 0^2 + 1^2. 2! = 1^2 + 1^2 6! = 12^2 + 24^2 are the only integer solutions of n! = x^2 + y^2 but is there a proof? R.

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Re: [math-fun] Sudoku levels
by Michael Reid 02 Jul '06

02 Jul '06

> I implemented the k=1 & k=8 rules, and this is "level 1". > > Interestingly, these 2 rules are somewhat dual to one another -- perhaps someone > else has better insight on this? i think this is well understood. consider the 9 x 9 "possibility matrix" whose rows are indexed by the nine cells in the row [column, box] and whose columns are indexed by the nine possible values, and whose entries are either 0 ("not possible") or 1 ("possible"). to fill in the nine cells, we need to find a "transversal" of the matrix, which is a set of nine non-zero entries, one from each row and each column. suppose the matrix has the property that the rows and columns can be permuted so that the matrix becomes "upper triangular in block form", i.e. looks like / \ | A B | | 0 D | \ / where A and D are square. then it is clear that any transversal must arise from a transversal of A and a transversal of D . the duality is between the sizes of A and D . alternatively, we can interchange the roles of "cell" and "value" which corresponds to transposing the matrix. then we seek a "lower triangular block form" of the matrix, but this is equivalent to upper triangular by reversing the order of rows and the order of columns. mike

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