mm-4719 === Subject: Re: cubic spline question > I guess what I'm asking here is, can I use cubic splines to produce a > formula which can be used to predict points in a trend based on > existing data? æYou can do so with any type of regression line or > curve, so can you predict points with an equation using cubic splines? > Extrapolation with splines is a terrible idea. Why is that a terrible idea? æThat's essentially what I eventually > wish to accomplish. æRather than having to store all the data points, > I can procedurally approximate them using their cubic spline > equation. æWhy is that a terrible idea? Doing purely mathematical extrapolations is a bad idea. With a spline you are only taking the last points into account. If you have some idea of what the Physics might be you are in a better position for extrapolation. If you are dealing with a dynamical system perhaps a Fourier Transform might be best. - Ian Parker === Subject: Re: cubic spline question posting-account=nKQXlgkAAABFcwCQitpPqM7oKmh-_khh Gecko/20080512 Camino/1.6.1 (like Firefox/2.0.0.14),gzip(gfe),gzip(gfe) > I guess what I'm asking here is, can I use cubic splines to produce a > formula which can be used to predict points in a trend based on > existing data? You can do so with any type of regression line or > curve, so can you predict points with an equation using cubic splines? > Extrapolation with splines is a terrible idea. > Why is that a terrible idea? That's essentially what I eventually > wish to accomplish. Rather than having to store all the data points, > I can procedurally approximate them using their cubic spline > equation. Why is that a terrible idea? Doing purely mathematical extrapolations is a bad idea. With a spline > you are only taking the last points into account. If you have some > idea of what the Physics might be you are in a better position for > extrapolation. If you are dealing with a dynamical system perhaps a Fourier Transform > might be best. - Ian Parker First of all, do you have any idea what the theoretical form of the data might be? If so, do a (possibly nonlinear) least squares fit of the data to the model. Second, if the first doesn't hold, you might try a least-squares cubic spline fit with knots chosen at locations that make some sense to you. cubic spline fit to data. It usually worked well, sometimes not so well. Martin Cohen === Subject: Re: First and Second Fundamental Forms for/of an Implicit Surface posting-account=a93YEwoAAAClHm9Euy--V39SJpP16NI8 Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > I'm doing this in order to compute the principal curvatures (k1, k2) > and principal directions (T1, T2) at a any point on the implicit > surface. I'm aware that: k1, and k2 are the eigenvalues of the 2 x 2 > matrix II/I , where I is the matrix of the first fundamental form, and > II the matrix of the second fundamental form; while T1, and T2 are > corresponding eigenvectors of (II/I). Obviously, the T1, and T2 are 2 > X 1 vectors embedded(?) in a 3D space. What then is the third > component of these vectors? Is it 1? You are thinking of the T1 and T2 as 2-tuples in the parameter > space. æYou have to lift them to the surface. æSee > æhttp://www.geometrictools.com/Documentation/PrincipalCurvature. pdf > I also have code at my web site to compute the various quantities > for implicitly defined surfaces. J = dx/du = { {1, 0, F x}', {0, 1, F y}' } , were ' denotes transpose. === Subject: Looking for two math-ish things A long time ago, I saw a joke proof of some fact wherein the key step was rotation. 1/oo was rotated to be -18, and at some point, the rotation was reversed to produce the result. If anyone knows what I'm talking about and has a copy, link or reference, I'd sure like to have it. Second, when I took sophmore DE's, G.-C. Rota told us a story about Picard. Supposedly, he had an unmotivated student who never came to class, skipped the final, and then showed up in Picard's office wanting to argue for a grade. Picard agreed to give the student an oral exam. student to solve it. The student hadn't studied a bit, so he didn't know any of the usual techniques. But in desparation, he tried something which gave Picard the idea for the method of successive approximations. Picard gave the student an A for the idea and told him to get lost. I'd would like a real reference to the story, even if it's apocryphal. Bart -- Cheerfully resisting change since 1959. === Subject: Re: Looking for two math-ish things > when I took sophmore DE's, G.-C. Rota told us > a story about Picard.... maybe the story is in it? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Looking for two math-ish things > when I took sophmore DE's, G.-C. Rota told us > a story about Picard.... maybe the story is in it? I didn't find it in Birkhoff and Rota (which doesn't seem to have a lot of stories in it). This is the sort of story you might tell a sophomore class, but is unlikely to make its way into a textbook. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Looking for two math-ish things posting-account=H1y7YgoAAADzGQwbcYaL9UvwttgsjOjp AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.1 Safari/525.18,gzip(gfe),gzip(gfe) > A long time ago, I saw a joke proof of some fact wherein > the key step was rotation. æ1/oo was rotated > to be -18, and at some point, the rotation was reversed > to produce the result. æ This probably isn't what you want, but there is the rotation joke, limit as omega goes to three of 8 is infinity. Not familiar with the Picard story. -- GM === Subject: Re: Looking for two math-ish things > On Jun 28, 4:44æam, Bart Goddard > A long time ago, I saw a joke proof of some fact > wherein the key step was rotation. æ1/oo was > rotated to be -18, and at some point, the rotation > was reversed to produce the result. æ This probably isn't what you want, but there is the > rotation joke, limit as omega goes to three of 8 is > infinity. Not familiar with the Picard story. > -- > GM Again, these two jokes are probably are not what you are looking for, but they may be close... and just as amusing ^_^ The first is the old 2 x 2 matrix rotation joke. It's lame when re-produced in ASCII, but it looks like this: http://imgs.xkcd.com/comics/matrix_transform.png . The second is a calculus (limit) joke; it looks like this: http://joi.ito.com/images2/mathjoke.gif . BTW, if you happen to find the original joke proof and/or the Picard story, please post it/them... I'd really like to see ^_^ Kyle Czarnecki === Subject: Divisibility problem Find all natural numbers a, b and c such that b|2a+1, c|2b+1 and a|2c+1. === Subject: Re: Divisibility problem > Find all natural numbers a, b and c such that b|2a+1, c|2b+1 and a|2c+1. I found the solutions 1, 1, 1 1, 1, 3 1, 3, 7 7, 15, 31 9, 19, 13 and permutations of these but no others up to 500. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Divisibility problem > Find all natural numbers a, b and c such that b|2a+1, c|2b+1 and a|2c+1. I found the solutions > 1, 1, 1 > 1, 1, 3 > 1, 3, 7 > 7, 15, 31 > 9, 19, 13 > and permutations of these > but no others up to 500. The equations 2a + 1 = m_1 b, 2b + 1 = m_2 c, 2c + 1 = m_3 a have solutions a = (4 + 2 m_1 + m_1 m_2)/(m_1 m_2 m_3 - 8) b = (4 + 2 m_2 + m_2 m_3)/(m_1 m_2 m_3 - 8) c = (4 + 2 m_3 + m_1 m_3)/(m_1 m_2 m_3 - 8) WLOG we may assume m1 <= min(m_2, m_3). Since a >= 1, we need m_1 m_2 m_3 - 8 <= 4 + 2 m_1 + m_1 m_2, i.e. m_2 (m_3 - 1) <= 12/m_1 + 2 <= 14 For m_3 > 1, there are only a few possibilities to consider. For m_3 = 1, we must also have m_1 = 1, and then c = 7/(m_2 - 8) is only a natural number for m_2 = 9 or 15. So we end up with only a few possibilities to consider, and conclude that Gerry found all the solutions. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Divisibility problem > Find all natural numbers a, b and c such that b|2a+1, c|2b+1 and a|2c+1. One can assume without loss of generality that 0 <= a <= b <= c. Then treat the two cases a=b and a Find all natural numbers a, b and c such that b|2a+1, c|2b+1 and a|2c+1. One can assume without loss of generality that 0 <= a <= b <= c. > Then treat the two cases a=b and a to the problem. > -jiw It is slightly more complicated than that. There are a few uninteresting solutions where one or more of the variables is 1. And one cannot assume without loss of generality that a <= b <= c, since if one does one misses the solution a = 9, b = 19, c = 13. === Subject: Re: Divisibility problem <8-KdnfyomOTy9_jVnZ2dnUVZ_jOdnZ2d@bresnan.com> Find all natural numbers a, b and c such that b|2a+1, c|2b+1 and > One can assume without loss of generality that 0 <= a <= b <= c. Then > treat the two cases a=b and a problem. > -jiw > It is slightly more complicated than that. There are a few > uninteresting solutions where one or more of the variables is 1. And > one cannot assume without loss of generality that a <= b <= c, since if > one does one misses the solution a = 9, b = 19, c = 13. You are right, a <= b <= c is not wolog. However, I think it would be wolog to assume { either a <= b <= c or a >= b >= c }, which allows a solution a=19, b=13, c=9 that is equivalent to a=9, b=19, c=13. -jiw === Subject: Re: Primes of the form n!+1 posting-account=nKQXlgkAAABFcwCQitpPqM7oKmh-_khh Gecko/20080512 Camino/1.6.1 (like Firefox/2.0.0.14),gzip(gfe),gzip(gfe) > For what n is n!+1 a prime? How many primes of the form n!+1 are there? > Perhaps would be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. The probability that a randomly chosen integer > near X, for large X, is prime is 1/log(X). Has it ever ocured that sum{1/log(f(x))} diverges and that are only > finitely many number x such that f(x) is a prime? f(x) = 2x+2:) === Subject: Re: Primes of the form n!+1 I think he's asking if it is possible to define f(x) such > that sum(1/log(f(x)) diverges (summming over the positive > integers x), yet only finitely many integers x exists s.t. > f(x) is prime. Of course the answer is yes, e.g. f(x) = 2x. > Almost. What I realy wanted to ask was: Has there ever been a similar (non-trivial) problem of finding how many primes are in {f(1),f(2),...} such that 1) sum{1/log(f(x))} diverges 2) it turned out that there are only finitely many primes in {f(i)|i in N} If this happend at least one time then I guess that people should not relay on probability much in this kind of problems. P.S. I appologise for eventual spelling mistakes. === Subject: Re: Primes of the form n!+1 87tzfgxf27.fsf@nonospaz.fatphil.org> posting-account=HKYBugoAAADwlzGnlrh0iplaYsMycPgv AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.1 Safari/525.18,gzip(gfe),gzip(gfe) > I think he's asking if it is possible to define f(x) such > that sum(1/log(f(x)) diverges (summming over the positive > integers x), yet only finitely many integers x exists s.t. > f(x) is prime. æOf course the answer is yes, e.g. f(x) = 2x. Almost. What I realy wanted to ask was: æHas there ever been a similar (non-trivial) problem of finding how > æmany primes are in {f(1),f(2),...} such that > æ1) sum{1/log(f(x))} diverges > æ2) it turned out that there are only finitely many primes in {f(i)|i in N} If this happend at least one time then I guess that people should not > relay on probability much in this kind of problems. P.S. > I appologise for eventual spelling mistakes. This is a subtle question. To my knowledge, all such sequences that have been proved to have only finitely many prime terms have been so proved via local arguments (typically rather more complicated than that for f(n)=2n). This is, for example, the case for f(n)=78557*2^n+1 which is always composite (as noted by Selfridge), but where naive heuristics suggest otherwise. It is instructive to consider the two cases f(n)=2^n+1 and f(n)=2^n-1. The key observation is that, in each case, local considerations place constraints upon n such that f(n) can be prime. In the first case, they are severe enough that one expects finitely many primes; in the second, one expects infinitely many. In the absence of such local constraints, I do not believe that finiteness has ever been proven when one expects otherwise. In the other direction, in many situations when a sequence grows slowly enough, sieve methods can yield the expected number of primes (such as for x^2+y^4, for instance). Extending this even to x^2+1 (let alone sequences growing exponentially) is well beyond current technology. Gaew === Subject: Re: Primes of the form n!+1 posting-account=lHNboAoAAACyasQ0uqX7OeM_tLuWGoQp CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > For what n is n!+1 a prime? How many primes of the form n!+1 are the= >re? > Perhaps would be= > of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? >Prime Number Theorem. æThe probability that a randomly chosen integer >near X, for large X, æis prime is 1/log(X). In fact, the intuitive estimate of the probability is much > greater. æSince the number is prime to 2, this gives and > extra factor of 2, prime to 3, of 3/2, and so on for any > finite number of small primes. æIf one can push this to the > limit, my calculations say the probability of primality > should be on the order of 1. æ Note: æThis is a probabilistic argument, and may or may > not indicate what is happening. Mostly correct. The argument contains an oversight. Any factor of 2^p-1 must be 1 mod p and 1 or -1 mod 8 (since 2 is a quad. residue). Thus, the density (in the integers) of the set of possible prime divisors is a lot smaller [it is 1/phi(p)] than the set of all primes less than 2^p-1. It is unclear (there are heuristics) that suggest 2^p-1 is slightly more likely than a random integer of the same size to be prime, but they are heuristics. The arguments of Wagstaff suggest that the factor is e^gamma (gamma is the Euler- Mascheroni constant). === Subject: Re: Primes of the form n!+1 > For what n is n!+1 a prime? How many primes of the form n!+1 are there? > Perhaps would be of > interest. > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. Why does that matter? > Prime Number Theorem. The probability that a randomly chosen integer > near X, for large X, is prime is 1/log(X). Has it ever ocured that sum{1/log(f(x))} diverges and that are only >finitely many number x such that f(x) is a prime? >Probabalistically, if f(n) is random and not constrained in some way >that affects the probability of f(n) being prime, then the divergence >of > oo > --- 1 > > --------- > --- log(f(n)) > n=1 >and the converse of the Borel-Cantelli Lemma > would say that there is a 100% probability of f(n) being prime for >infinitely many n. >This does not mean that it is true, just that it has a 100% chance of >being true. See > This last statement is nonsense. All one can say is that the >current intuitive probability arguments, if they are correct >in this case, would make the probability at 100%. Consider flipping a coin until a head appears. There is no way to prove a head will ever appear, but the probability that one will appear is 100%. I realize that there are either an infinite number of primes of the form n!+1 or a finite number. However, since we are only talking about probabilities here, is there a difference between this case and the nonsensical one? Rob Johnson take out the trash before replying === Subject: Re: Primes of the form n!+1 20080626.111118@whim.org> <20080627.081507@whim.org> posting-account=lHNboAoAAACyasQ0uqX7OeM_tLuWGoQp CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Although there is no proof, we expect that there are infinitely many > because sum(1/log(n!+1)) diverges. > Why does that matter? > Prime Number Theorem. æThe probability that a randomly chosen integer >This does not mean that it is true, just that it has a 100% chance of >being true. æSee > æ æThis last statement is nonsense. æAll one can say is that the >current intuitive probability æarguments, if they are correct >in this case, would make the probability at 100%. Consider flipping a coin until a head appears. æThere is no way to > prove a head will ever appear, but the probability that one will > appear is 100%. I realize that there are either an infinite number of primes of the > form n!+1 or a finite number. æHowever, since we are only talking > about probabilities here, is there a difference between this case and > the nonsensical one? Elliot's book Probabilistic Number Theory covers these issues. === Subject: Articles on Mathematics posting-account=ee2apQoAAABJNMlLTFasCJw9Nfo9FmYk Gecko/20020924 AOL/7.0,gzip(gfe),gzip(gfe) HTTP/1.1 cache-ntc-ab08.proxy.aol.com[CFC87448] (Traffic-Server/6.1.5 [uScM]) MATH (by M.Basti) and some more. You can view them on Basti Newsgroup: Sci.math. You are welcome to join: http://mathforum.org/kb/thread.jspa?threadID=1763061&tstart=0 Note: At this time the newsgroup is structured on Google site, and there is no fee attached to membership. Thus there is no problem, if issues are addressed (or advertised) about Google products. Dr.M.Basti === Subject: Probability Question posting-account=TPX07AkAAADL-Uf_yOaUNv1_FT6LrAs0 rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Suppose you have a (possibly biased) coin which has a probability p of coming up heads and 1-p coming up tails. Further suppose that Player A flips that coin n times, yielding x heads and n-x tails. How many standard deviations away from the norm would A's result be? I can compute that the probability of A's result P(A) is C(n,x)(p^x)(1- p)^(n-x) and that the population standard deviation sigma is sqrt(np(1- p)), and that these two equations will be important for the result. However, it's not obvious where I go from here. I am essentially looking for an equation of the form SD(p,n,x). In a simple case where p=0.5 and n=16, P(A,x)=C(16,x)/2^16 and sigma = 2. Since x=8 is the expected value, SD(0.5,16,8) should be 0; if x=0 or x=16, I would expect SD(0.5,16,x) to be somewhere around 3 range, since this is what I would guess how many sigma's away such a result might be. Yet I do not see any obvious (or justifiable) way to yield this result. Any suggestions? === Subject: Re: Probability Question posting-account=Kc1qAQoAAABka6KpTO102ya8Bawo-kJO AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.1 Safari/525.18,gzip(gfe),gzip(gfe) What you are asking for is the z-score (or standard score). This measures exactly the number of standard deviations away from the mean. The equation is z = (x - m) / s, where x is the measured value, m is the mean, and s is the standard deviation. For a Binomial distribution, z = (x - np)/sqrt(np(1-p)). In your example using p = 0.5 and n = 16, this reduces down to z(x) = x/2 - 4. This is about as simple a case as you can ask for, although typically these scores have more meaning when the sample size n is sufficiently large. It's all well and good to say that flipping 12 heads out of 16 tries is 2 standard deviations away from the norm, but 75% success rate overstates how badly the coin is biased. Quantization is to blame here. On the other hand if n = 1000, 2 standard deviations away from the norm is 532 heads, a mere 3% variation from from the mean. The coin may be biased, but not by the degree that small trials might suggest. Jonathan Hoyle http://www.jonhoyle.com === Subject: Re: can you have an infinitessimally small straight line? Distribution: world posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) On Jun 26, 5:16æam, mstem...@walkabout.empros.com (Michael Stemper) > can you have an infinitessimally small straight line? >or could you always cut it in half, meaning it can never be >infinitessimally small? Well, a line extends indefinitely in both directions. If you cut it > in half (which you could do at any point), you'll have a ray. You might mean to ask if it's always possible to cut a line segment in > half. A line segment is a portion of a line with finite length, that is, > it has two end-points. In that case, the answer is you can always bisect (cut in half) a line > segment. The proof is simple: One the real line, mark the endpoints of your initial segment as (a,b). > The midpoint of (a,b) is c, where c=(a+b)/2. This can be used to define > two new line segments: (a,c) and (c,b). Either of these can be cut in > half in the same manner, uzw. -- > Michael F. Stemper > The FAQ for rec.arts.sf.written is at:http://www.geocities.com/evelynleeper/sf-written > Please read it before posting. A slope of a real world curve requires two points forming a line. It is never exact because it would take infinite calculations to bring those two points infinitely close. Mitch Raemsch === Cancel-Key: sha1:2nPFkKi6rn3cUC2b+VYS5OASzJE= === Subject: On Guidelines posting-account=ee2apQoAAABJNMlLTFasCJw9Nfo9FmYk Gecko/20020924 AOL/7.0,gzip(gfe),gzip(gfe) HTTP/1.1 cache-ntc-ab08.proxy.aol.com[CFC87448] (Traffic-Server/6.1.5 [uScM]) Anyone is welcome to Basti newsgroups. me or not). You can criticize NEW EXACT MATH (or any of my/others writings) on any levels but within the guidelines of newsgroup (written on a contract). Welcome to Basti newsgroups. http://mathforum.org/kb/thread.jspa?threadID=1763061&tstart=0 Sign up on Google site. Dr.M.Basti === Subject: Re: Name this enumeration type- > What's the mathematical name of the enumeration type below? > I think this is neither combination nor permutation, > so then what is this one called? Generated by a call to a function f(v=5,k=3): > (the first column is just the index starting at 0 > and it isn't part of the generated rows). 0: 0 0 0 > 1: 0 0 1 > 2: 0 0 2 > 3: 0 0 3 > 4: 0 0 4 > 5: 0 1 1 > 6: 0 1 2 > 7: 0 1 3 > 8: 0 1 4 > 9: 0 2 2 > 10: 0 2 3 > 11: 0 2 4 > 12: 0 3 3 > 13: 0 3 4 > 14: 0 4 4 > 15: 1 1 1 > 16: 1 1 2 > 17: 1 1 3 > 18: 1 1 4 > 19: 1 2 2 > 20: 1 2 3 > 21: 1 2 4 > 22: 1 3 3 > 23: 1 3 4 > 24: 1 4 4 > 25: 2 2 2 > 26: 2 2 3 > 27: 2 2 4 > 28: 2 3 3 > 29: 2 3 4 > 30: 2 4 4 > 31: 3 3 3 > 32: 3 3 4 > 33: 3 4 4 > 34: 4 4 4 I'm looking for two computational methods: > 1.) given the index create the corrosponding row > 2.) given the row compute the corrosponding index TIA > This particular sequence lists the 35 balls packed into a regular tetrahedron with 5 balls at each edge by their coordinates (non-Cartesian) in the lattice. Top layer: 1 ball; 2nd layer: 3 balls; 3rd layer: 6 balls; 4th layer: 10 balls; 5th layer = bottom layer: 15 balls; altogether 35 balls. Terminology: figurate numbers, triangular numbers, pyramidal numbers, tetrahedal numbers, and more. http://en.wikipedia.org/wiki/Figurate_number http://en.wikipedia.org/wiki/Triangular_number http://en.wikipedia.org/wiki/Pyramidal_number http://en.wikipedia.org/wiki/Tetrahedral_number - the picture shows precisely the 35 balls of your list. Try and assign to each ball its coordinates while the whole thing still rotates! Happy counting: Johan E. Mebius === Subject: Re: Name this enumeration type > I'm looking for two computational methods: > 1.) given the index create the corrosponding row > 2.) given the row compute the corrosponding index A simple recursive formula could be used for both of these. Let I(a, b, c) be the index of a row with numbers a, b, and c, in that order. v is as you defined. It can be generalized for k, but that takes a bit more notational work. Notice the following facts: 1. I(a, b, c) = (c - b) + I(a, b, b) 2. I(a, b, b) = (v + 1 - b) + I(a, b - 1, b - 1) 3. I(a, a, a) = 1 + I(a - 1, v - 1, v - 1) 4. I(0, 0, 0) = 0 index from the row. A closed-form solution (involving summations) can be derived from these rules; doing so is left as an exercise to the reader. The reverse (computing the row from the index) is more difficult, but an algorithm should lend itself after enough study. === Subject: Re: Name this enumeration type > > I'm looking for two computational methods: > 1.) given the index create the corrosponding row > 2.) given the row compute the corrosponding index A simple recursive formula could be used for both of these. Let I(a, b, c) be the index of a row with numbers a, b, and c, in that > order. v is as you defined. It can be generalized for k, but that takes > a bit more notational work. Notice the following facts: 1. I(a, b, c) = (c - b) + I(a, b, b) > 2. I(a, b, b) = (v + 1 - b) + I(a, b - 1, b - 1) > 3. I(a, a, a) = 1 + I(a - 1, v - 1, v - 1) > 4. I(0, 0, 0) = 0 index from the row. A closed-form solution (involving summations) can be > derived from these rules; doing so is left as an exercise to the reader. The reverse (computing the row from the index) is more difficult, but an > algorithm should lend itself after enough study. I think we need one more rule because the following cases are not covered by the above 4 rules, ie. a rule for the case (a,a,b) would be needed too I think: 1: 0 0 1 2: 0 0 2 3: 0 0 3 4: 0 0 4 16: 1 1 2 17: 1 1 3 18: 1 1 4 26: 2 2 3 27: 2 2 4 32: 3 3 4 Can you verify this please. Or have I maybe made an error in my implementation? I've used these rules in that order: 4, 3, 2, 1. Is that ok? === Subject: Re: Name this enumeration type > > I'm looking for two computational methods: > 1.) given the index create the corrosponding row > 2.) given the row compute the corrosponding index > > A simple recursive formula could be used for both of these. > > Let I(a, b, c) be the index of a row with numbers a, b, and c, in that > order. v is as you defined. It can be generalized for k, but that takes > a bit more notational work. > > Notice the following facts: > > 1. I(a, b, c) = (c - b) + I(a, b, b) > 2. I(a, b, b) = (v + 1 - b) + I(a, b - 1, b - 1) > 3. I(a, a, a) = 1 + I(a - 1, v - 1, v - 1) > 4. I(0, 0, 0) = 0 > > index from the row. A closed-form solution (involving summations) can be > derived from these rules; doing so is left as an exercise to the reader. > > The reverse (computing the row from the index) is more difficult, but an > algorithm should lend itself after enough study. I think we need one more rule because the following > cases are not covered by the above 4 rules, > ie. a rule for the case (a,a,b) would be needed too I think: > 1: 0 0 1 > 2: 0 0 2 > 3: 0 0 3 > 4: 0 0 4 > 16: 1 1 2 > 17: 1 1 3 > 18: 1 1 4 > 26: 2 2 3 > 27: 2 2 4 > 32: 3 3 4 Can you verify this please. > Or have I maybe made an error in my implementation? > I've used these rules in that order: 4, 3, 2, 1. > Is that ok? Ok, problem solved. Rule 3 should read (a,a,b). But now the big question: A general solution valid for all k would be fine. You said it can be generalized for k, but that takes a bit more notational work. Would one need (k+1) such different rule sets for each k ? === Subject: Re: Name this enumeration type posting-account=Hz-TvgoAAABKrfFCvLFmTEbLVznFptlX AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.1 Safari/525.18,gzip(gfe),gzip(gfe) > What's the mathematical name of the enumeration type below? > I think this is neither combination nor permutation, > so then what is this one called? Generated by a call to a function f(v=5,k=3): > (the first column is just the index starting at 0 > and it isn't part of the generated rows). ? ?0: ? ? ?0 ?0 ?0 > ? ?1: ? ? ?0 ?0 ?1 > ? ?2: ? ? ?0 ?0 ?2 > ? ?3: ? ? ?0 ?0 ?3 > ? ?4: ? ? ?0 ?0 ?4 > ? ?5: ? ? ?0 ?1 ?1 > ? ?6: ? ? ?0 ?1 ?2 > ? ?7: ? ? ?0 ?1 ?3 > ? ?8: ? ? ?0 ?1 ?4 > ? ?9: ? ? ?0 ?2 ?2 > ? 10: ? ? ?0 ?2 ?3 > ? 11: ? ? ?0 ?2 ?4 > ? 12: ? ? ?0 ?3 ?3 > ? 13: ? ? ?0 ?3 ?4 > ? 14: ? ? ?0 ?4 ?4 > ? 15: ? ? ?1 ?1 ?1 > ? 16: ? ? ?1 ?1 ?2 > ? 17: ? ? ?1 ?1 ?3 > ? 18: ? ? ?1 ?1 ?4 > ? 19: ? ? ?1 ?2 ?2 > ? 20: ? ? ?1 ?2 ?3 > ? 21: ? ? ?1 ?2 ?4 > ? 22: ? ? ?1 ?3 ?3 > ? 23: ? ? ?1 ?3 ?4 > ? 24: ? ? ?1 ?4 ?4 > ? 25: ? ? ?2 ?2 ?2 > ? 26: ? ? ?2 ?2 ?3 > ? 27: ? ? ?2 ?2 ?4 > ? 28: ? ? ?2 ?3 ?3 > ? 29: ? ? ?2 ?3 ?4 > ? 30: ? ? ?2 ?4 ?4 > ? 31: ? ? ?3 ?3 ?3 > ? 32: ? ? ?3 ?3 ?4 > ? 33: ? ? ?3 ?4 ?4 > ? 34: ? ? ?4 ?4 ?4 I'm looking for two computational methods: > 1.) given the index create the corrosponding row > 2.) given the row compute the corrosponding index TIA It is obviously not a permutation, because a shuffling of the elements are not counted. For example you count 0 1 2 only once. If this was a permutation, you would count it 3!=6 times, or have the following 5 entries added to the list: 0 2 1 1 0 2 1 2 0 2 0 1 2 1 0 Therefore this is, in fact, a combination (so order is unimportant) where your selection of elements may contain duplicates/repetitions - you've just sorted the combination/choice in increasing order. It is not that easy, albeit possible, to calculate the index given a row and vise versa. I'm not even sure whether a closed formula exists. Rather, an iterative process can be used to establish the index/row in an element-wise fashion. Very similar to the iterative process involved in establishing the next permutation given a sequence of numbers. Some recursive example of code includes http://www.bearcave.com/random hacks/permute.html amongst others. Werner === Subject: Re: Name this enumeration type Hi Werner and the others who have replied; I just want to clarify this issue: the generation (enumeration) of the complete design is no problem at all. It is even always generated in lex order (ie. implicitly sorted), so no extra sorting is needed. But what I need are two functions for this design type (ie. combinations with repetetion / multiset; cf. http://en.wikipedia.org/wiki/Multiset ) to find the index from any given row, and vice-versa: // returns the index# of the row specified in pai // creates the corrosponding row for the given index# and puts it in pai I was hoping that maybe this is a well known problem and therefore such routines already would have been developed. But unfortunately this seems not to be the case. I've also consulted some books but unfortunately found nothing in that direction. During my research I learnt also that the book Enumerative Combinatorics by Stanley seems to be the authority in this field. Unfortunately I could not find a copy in the local library so I don't know if the two algorithms I seek are present in that book. Ok, then I'll try to develop these 2 routines myself. > What's the mathematical name of the enumeration type below? > I think this is neither combination nor permutation, > so then what is this one called? Generated by a call to a function f(v=5,k=3): > (the first column is just the index starting at 0 > and it isn't part of the generated rows). 0: 0 0 0 > 1: 0 0 1 > 2: 0 0 2 > 3: 0 0 3 > 4: 0 0 4 > 5: 0 1 1 > 6: 0 1 2 > 7: 0 1 3 > 8: 0 1 4 > 9: 0 2 2 > 10: 0 2 3 > 11: 0 2 4 > 12: 0 3 3 > 13: 0 3 4 > 14: 0 4 4 > 15: 1 1 1 > 16: 1 1 2 > 17: 1 1 3 > 18: 1 1 4 > 19: 1 2 2 > 20: 1 2 3 > 21: 1 2 4 > 22: 1 3 3 > 23: 1 3 4 > 24: 1 4 4 > 25: 2 2 2 > 26: 2 2 3 > 27: 2 2 4 > 28: 2 3 3 > 29: 2 3 4 > 30: 2 4 4 > 31: 3 3 3 > 32: 3 3 4 > 33: 3 4 4 > 34: 4 4 4 I'm looking for two computational methods: > 1.) given the index create the corrosponding row > 2.) given the row compute the corrosponding index TIA It is obviously not a permutation, because a shuffling of the elements are not counted. For example you count 0 1 2 only once. If this was a permutation, you would count it 3!=6 times, or have the following 5 entries added to the list: 0 2 1 1 0 2 1 2 0 2 0 1 2 1 0 Therefore this is, in fact, a combination (so order is unimportant) where your selection of elements may contain duplicates/repetitions - you've just sorted the combination/choice in increasing order. It is not that easy, albeit possible, to calculate the index given a row and vise versa. I'm not even sure whether a closed formula exists. Rather, an iterative process can be used to establish the index/row in an element-wise fashion. Very similar to the iterative process involved in establishing the next permutation given a sequence of numbers. Some recursive example of code includes http://www.bearcave.com/random_hacks/permute.html amongst others. Werner === Subject: Re: Name this enumeration type posting-account=McZ3aQkAAADz6LV-boDe1LcriRhf3lj3 Gecko/20080404 Firefox/2.0.0.9;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) > Hi Werner and the others who have replied; I just want to clarify this issue: > the generation (enumeration) of the complete design is no problem at all. > It is even always generated in lex order (ie. implicitly sorted), > so no extra sorting is needed. But what I need are two functions for this design type > (ie. combinations with repetetion / multiset; cf.http://en.wikipedia.org/wiki/Multiset) > to find the index from any given row, and vice-versa: // returns the index# of the row specified in pai // creates the corrosponding row for the given index# and puts it in pai I was hoping that maybe this is a well known problem and > therefore such routines already would have been developed. > But unfortunately this seems not to be the case. > I've also consulted some books but unfortunately > found nothing in that direction. During my research > I learnt also that the book Enumerative Combinatorics by Stanley > seems to be the authority in this field. Unfortunately > I could not find a copy in the local library so I don't know > if the two algorithms I seek are present in that book. Ok, then I'll try to develop these 2 routines myself. > What's the mathematical name of the enumeration type below? > I think this is neither combination nor permutation, > so then what is this one called? > Generated by a call to a function f(v=5,k=3): > (the first column is just the index starting at 0 > and it isn't part of the generated rows). > 0: 0 0 0 > 1: 0 0 1 > 2: 0 0 2 > 3: 0 0 3 > 4: 0 0 4 > 5: 0 1 1 > 6: 0 1 2 > 7: 0 1 3 > 8: 0 1 4 > 9: 0 2 2 > 10: 0 2 3 > 11: 0 2 4 > 12: 0 3 3 > 13: 0 3 4 > 14: 0 4 4 > 15: 1 1 1 > 16: 1 1 2 > 17: 1 1 3 > 18: 1 1 4 > 19: 1 2 2 > 20: 1 2 3 > 21: 1 2 4 > 22: 1 3 3 > 23: 1 3 4 > 24: 1 4 4 > 25: 2 2 2 > 26: 2 2 3 > 27: 2 2 4 > 28: 2 3 3 > 29: 2 3 4 > 30: 2 4 4 > 31: 3 3 3 > 32: 3 3 4 > 33: 3 4 4 > 34: 4 4 4 > I'm looking for two computational methods: > 1.) given the index create the corrosponding row > 2.) given the row compute the corrosponding index > TIA It is obviously not a permutation, because a shuffling of the elements > are not counted. For example you count > 0 1 2 > only once. If this was a permutation, you would count it 3!=6 times, > or have the following 5 entries added to the list: > 0 2 1 > 1 0 2 > 1 2 0 > 2 0 1 > 2 1 0 > Therefore this is, in fact, a combination (so order is unimportant) > where your selection of elements may contain duplicates/repetitions - > you've just sorted the combination/choice in increasing order. It is > not that easy, albeit possible, to calculate the index given a row and > vise versa. I'm not even sure whether a closed formula exists. Rather, > an iterative process can be used to establish the index/row in an > element-wise fashion. Very similar to the iterative process involved > in establishing the next permutation given a sequence of numbers. Some > recursive example of code includes > http://www.bearcave.com/random_hacks/permute.html > amongst others. Werner > It is rather difficult to explain, why this question seem to be > soo difficult. > Let start with the generating function n^m, here 5^3. > Instead (0,1,2,3,4), I use alphabetical symbols. This gives > 125 strings. > There are 5 strings of type aaa. > The corresponding polynomial coefficient is 5!/4!1! = 5. > Only one permutation, as 1!/1! > There are 20 strings of type aab. > The corresponding polynomial coefficient is 5!/3!1!1! 20. > Permutations inside strings as 3!/2!1! = 3. > There are 10 strings of type abc. > The corresponding polynomial coefficient is 5!/3!2!. > Permutations inside strings as 3!/1!^3 = 6. > Together 5 + 20 + 10 = 35, and 5*1 + 20*3 + 10*6 = 125, > as required. > There are two different polynomial coefficients. kunzmilan === Subject: When is a digraph... ...an activity or project digraph -- that is, one with which I could model a project and do a critical path analysis? More formally: Characterize those digraphs which are activity digraphs -- i.e. determine a necessary and sufficient condition for a digraph to be an activity digraph. Here's what I have: A digraph G is an activity digraph if and only if all of the following are true: a.) G is a weakly connected, directed acyclic graph. b.) All vertices of G are assigned a value greater than zero with the exception of exactly two vertices, call them s and e, which are zero. c.) For every vertex v of G, the longest path containing v is P: s...v...e. d.) No three vertices of G are mutually adjacent. I believe these are all true, but I'm wondering if some of these criteria don't overlap or there's not a more concise way to state it. The problem suggests that there is one condition. I should mention that the problem is out of a book, but I'm working independently. Any help is appreciated. === Subject: Re: When is a digraph... > ....an activity or project digraph -- that is, one > with which I could model a > project and do a critical path analysis? More > formally: Characterize those digraphs which are activity > digraphs -- i.e. determine a > necessary and sufficient condition for a digraph to > be an activity digraph. > The existence of a unique critical path? === Subject: Re: When is a digraph... <17654415.1214642516203.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > ....an activity or project digraph -- that is, one > with which I could model a > project and do a critical path analysis? More > formally: > Characterize those digraphs which are activity > digraphs -- i.e. determine a > necessary and sufficient condition for a digraph to > be an activity digraph. The existence of a unique critical path? No. Examples exist with more than one critical path. R.G. Vickson === Subject: Re: When is a digraph... posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) On Jun 27, 3:34 pm, contrapositive project and do a critical path analysis? More formally: Characterize those digraphs which are activity digraphs -- i.e. determine a > necessary and sufficient condition for a digraph to be an activity digraph. Here's what I have: A digraph G is an activity digraph if and only if all of the following are > true: > a.) G is a weakly connected, directed acyclic graph. > b.) All vertices of G are assigned a value greater than zero with the > exception of exactly two vertices, call them s and e, which are zero. > c.) For every vertex v of G, the longest path containing v is P: s...v...e. > d.) No three vertices of G are mutually adjacent. > I think that a directed digraph is an activity graph if, and only if it has a single start vertex, a single end vertex and has no directed cycles. Probably there are graph-theoretic criteria for no directed cycles, so you could use those. R.G. Vickson > I believe these are all true, but I'm wondering if some of these criteria > don't overlap or there's not a more concise way to state it. The problem > suggests that there is one condition. I should mention that the problem is out of a book, but I'm working > independently. Any help is appreciated. === Subject: Re: When is a digraph... > I think that a directed digraph is an activity graph if, and only if > it has a single start vertex, a single end vertex and has no directed > cycles. Probably there are graph-theoretic criteria for no directed > cycles, so you could use those. > That's part of it, but suppose I have vertices (tasks) labeled a, b and c, and a is directed to b, and b is directed to c. Also, a is directed to c. This can be part of a digraph with no directed cycles. Thus b must be completed before c and after a, but the fact that a is directed to c contradicts this. I satisfied that by stating that no three vertices of G are mutually adjacent. === Subject: Re: When is a digraph... posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) On Jun 28, 2:45 pm, contrapositive I think that a directed digraph is an activity graph if, and only if > it has a single start vertex, a single end vertex and has no directed > cycles. Probably there are graph-theoretic criteria for no directed > cycles, so you could use those. That's part of it, but suppose I have vertices (tasks) labeled a, b and c, > and a is directed to b, and b is directed to c. Also, a is directed to c. It depends in part on whether you have an activity on nodes graph, or an activity on arcs version. The former uses nodes to label the activities and the arcs to represent precedence relations. In that form, if a, b and c are activity nodes and you have arcs a-->b and b-- >c, that means that a must be completed before b can begin, and b must be completed before c can begin. It follows automatically that a must be completed before c can begin, so the arc a-->c is redundant. (This is the case even if b is a dummy node having zero duration.) On the other hand, if you intend to have an activity on arcs representation, then the nodes represent the start and end points of an activity, but again it would follow that a-->b-->c already a comes before c, so you would not want to include that redundant arc. Most project managers would try to avoid drawing an activity graph with overtly redundant arcs like your a, b, c example. Actually, you might start off drawing that arc, but it then a good idea to perform some type of redundancy test, so as to re-draw the graph in minimal terms without any obvious redundantcies. That way, the subsequent analysis is cleaner. > This can be part of a digraph with no directed cycles. Thus b must be > completed before c and after a, but the fact that a is directed to c > contradicts this. I don't see how. I guess your use of the term directed needs to be clarified. In the usual meaning of arcs there is no contradiction at all. R.G. Vickson > I satisfied that by stating that no three vertices of G > are mutually adjacent. === Subject: Re: When is a digraph... posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008061712 Fedora/3.0-1.fc9 Firefox/3.0,gzip(gfe),gzip(gfe) On Jun 27, 7:34æpm, contrapositive project and do a critical path analysis? More formally: Characterize those digraphs which are activity digraphs -- i.e. determine a > necessary and sufficient condition for a digraph to be an activity digraph. Care to elaborate on what exactly is an activity digraph? -- m === Subject: Re: When is a digraph... > Care to elaborate on what exactly is an activity digraph? Perhaps more commonly called a project digraph, one used to model a project by representing each task as a vertex with a number assigned to it representing the time it takes to complete the task. I think if you Google project digraph you'll find a more formal definition. === Subject: Re: Characterization of involutory matrices- > First of all let me thank everyone who has answered. One more > question: > suppose now that A is also orthonormal AND symmetric. A kind of Pavlov reflex of mine kicks in here: THEOREM Any two of the three properties (A) orthonormal/orthogonal, (B) symmetric and (C) involutory imply the third! (A): A-transposed = A-inverted (B): A-transposed = A (C): A-inverted = A This theorem holds in all finite-dimensional real vector spaces and also in real Hilbert spaces. Two reasons why I kick on this theorem: (1) it was a major Aha-Erlebnis during my student's years; (2) many textbooks devoting a section on calculating the Euler-Rodrigues parameters of 3D rotation matrices ignore the trouble arising in the case of half-turns. Stated differently: Half-turns in 3D space are notorious troublemakers. These troubles illustrate the mathematical fact that SO(3) and S^3 cannot be parametrised by a single chart. One needs at least two charts. Please read one of the papers mentioned at http://www.xs4all.nl/~jemebius/Eea.htm to see how to parametrise SO(3) with two charts only. > According to a Lemma in Gantmacher we have > A=3Dexp(i S) > with S symmetric and real. Since A is involutory we can write > exp(i S)*exp(i S) =3D I =3D exp(0), > where 0 is the matrix whose elements are all zeroes. > Furthermore, since (i S) and (i S) obviously commute, > exp(i 2S) =3D exp(0). > So my question is: is it possible to find a general expression > for S? Note that this is equivalent to the scalar equations > cos(2 S) =3D I > sin(2 S) =3D 0 > It seems that it will have an infinite number of solutions, > like in the scalar case exp(i theta)=3D1. > It also seems that S is independent of A and this looks a bit > surprising (at least to me!). > > The eigenvalues of S can be any integer multiples of pi. > As I mentioned before, there are two complementary subspaces U and V > of R^n, with A u = u for u in U, A v = -v for v in V. In this case > (since A is normal) U and V are orthogonal. The eigenvectors of S > for even multiples of pi are in U, and the eigenvectors for odd multiples > of pi are in V. === Subject: Re: Characterization of involutory matrices > Is there a complete characterization of all involutory matrices > (A*A=I), i.e. the square roots of > the identity? I am especially interested in the real case. > Characterizations one involution at a time were given earlier. (Blowing my horn) Years ago, I identified the set J of involutions over a (Banach) space, but think of finite dimensional ones, if you prefer. In that case, J is of course an algebraic variety described by a set of quadratic equations, if you write A*A=I entry by entry. More precisely, J is the disjoint union of differentiable manifolds (plus two isolated points I and -I) with a naturally defined affine connection; every two different components have distance 2 (in any matrix operator norm). If the fixed space (corresponding to eigenvalue 1) of an n-by-n involution A has dimension k, then the dimension of the component of J containing A is 2*k*(n-k). (And every two involutions from the same component are similar.) If we restrict ourselves to Hermitian involutions, then the component manifolds are Riemannian (induced by Hilbert-Schmidt metric of the surrounding linear space of matrices), the dimensions are half of the above, and similarity can be achieved by unitary matrices. How long ago? Manifolds of Linear Involutions, Linear Algebra and its Applications, 24:271-287 (1979). === Subject: Re: Looking for a Matrix recipe posting-account=PbSwTwgAAADCfhtjQF2Hq7jjSPSyO4QX Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) On Jun 26, 2:17 pm, Robert Israel > I would need a general (for any n) recipe to construct > a matrix P with the following properties. > 1. P is upper diagonal. > 2. P^2=unity. > 3. P*transpose(P) is a complete n*n Jordan block. I assume you mean not that it _is_ a complete block, but that its Jordan > form _has_ one complete block. > Example for n=2: > 1 2I > 0 -1 Your I, I guess, is i = sqrt(-1). I thought it was just Maple that used > I for that. > Example for n=3 > 1 0 1 > 0 1 I > 0 0 -1 > Nonexample for n=4 > 1 0 0 a > 0 1 0 b > 0 0 1 c > 0 0 0 -1 > Can't find a,b,c, it's always 3+1 Jordan blocks! I think there are no 4x4 examples. But here's a 5x5: [ 1 0 0 0 a ] > [ 0 1 0 0 aI ] > [ 0 0 1 b bI ] > [ 0 0 0 -1 0 ] > [ 0 0 0 0 -1 ] where a and b are nonzero. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada More generally, suppose P is a block matrix of the form [ I B ] [ 0 -I ] (I'm reverting to the usual practice of using I for the identity matrix, and i for sqrt(-1)). Then [ C -B ] P P' - I = [ -B' 0 ] where C = BB'. For an nxn solution where the eigenvalue of P P' is 1, we want (PP'-I)^n = 0 but (PP'-I)^(n-1) <> 0. Now it is easy to show by induction that [ f_{k}(C) -f_{k-1}(C)B ] (PP'-I)^k = [ -B'f_{k-1}(C) B'f_{k-2}(C)B ] where f_k(x) are the polynomials defined by the recurrence f_{k+1}(x) = x (f_k(x) + f_{k-1}(x)) with f_0(x) = 1, f_1(x) = x. Note that f_k(x) is divisible by x^(ceil(k/2)). My 5 x 5 example had (PP'-I)^5 = 0 because C^3 = 0 and C^2 B = 0 (implying f_5(C) = 0, f_4(C) B = 0 and f_3(C)B = 0) but (PP'-I)^4 <> 0 because C^2 <> 0 (implying f_4(C) <> 0). Similarly, for a (2m+1) x (2m+1) example, we'd want C^(m+1) = 0 and C^m B = 0 but C^m <> 0 implying f_{2m+1}(C) = 0, f_{2m}(C) B = 0 and f_{2m}(C) <> 0. This might work if B was (m+1) x m. For example, this works for a 7 x 7 example, where ab <> 0: [ 8b, 4 i + 3 b^2 i, 4 - 3 b^2 ] [ 0, -a i, a ] B = [ 0, 4 - b^2, -4 i - b^2 i ] [ 0, a, a i ] I suspect that there may be examples of this sort for all odd n, but not for even n (in your example with n=2, P P' had eigenvalue -1, and maybe this will not generalize). However, at the moment I don't see a construction that works for odd n in general. Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Looking for a Matrix recipe posting-account=PbSwTwgAAADCfhtjQF2Hq7jjSPSyO4QX Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > On Jun 26, 2:17 pm, Robert Israel > I would need a general (for any n) recipe to construct > a matrix P with the following properties. > 1. P is upper diagonal. > 2. P^2=unity. > 3. P*transpose(P) is a complete n*n Jordan block. > I assume you mean not that it _is_ a complete block, but that its Jordan > form _has_ one complete block. > Example for n=2: > 1 2I > 0 -1 > Your I, I guess, is i = sqrt(-1). I thought it was just Maple that used > I for that. > Example for n=3 > 1 0 1 > 0 1 I > 0 0 -1 > Nonexample for n=4 > 1 0 0 a > 0 1 0 b > 0 0 1 c > 0 0 0 -1 > Can't find a,b,c, it's always 3+1 Jordan blocks! > I think there are no 4x4 examples. But here's a 5x5: > [ 1 0 0 0 a ] > [ 0 1 0 0 aI ] > [ 0 0 1 b bI ] > [ 0 0 0 -1 0 ] > [ 0 0 0 0 -1 ] > where a and b are nonzero. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada More generally, suppose P is a block matrix of the form [ I B ] > [ 0 -I ] (I'm reverting to the usual practice of using I for the identity > matrix, and i for sqrt(-1)). Then > [ C -B ] > P P' - I = [ -B' 0 ] where C = BB'. For an nxn solution where the eigenvalue of P P' is 1, > we want (PP'-I)^n = 0 but (PP'-I)^(n-1) <> 0. Now it is easy to show > by induction that [ f_{k}(C) -f_{k-1}(C)B ] > (PP'-I)^k = [ -B'f_{k-1}(C) B'f_{k-2}(C)B ] where f_k(x) are the polynomials defined by the recurrence > f_{k+1}(x) = x (f_k(x) + f_{k-1}(x)) > with f_0(x) = 1, f_1(x) = x. Note that f_k(x) is divisible by > x^(ceil(k/2)). > My 5 x 5 example had > (PP'-I)^5 = 0 because C^3 = 0 and C^2 B = 0 > (implying f_5(C) = 0, f_4(C) B = 0 and f_3(C)B = 0) but > (PP'-I)^4 <> 0 because C^2 <> 0 (implying f_4(C) <> 0). > Similarly, for a (2m+1) x (2m+1) example, we'd want > C^(m+1) = 0 and C^m B = 0 but C^m <> 0 > implying f_{2m+1}(C) = 0, f_{2m}(C) B = 0 and f_{2m}(C) <> 0. > This might work if B was (m+1) x m. > For example, this works for a 7 x 7 example, where ab <> 0: [ 8b, 4 i + 3 b^2 i, 4 - 3 b^2 ] > [ 0, -a i, a ] > B = [ 0, 4 - b^2, -4 i - b^2 i ] > [ 0, a, a i ] I suspect that there may be examples of this sort for all odd n, > but not for even n (in your example with n=2, P P' had eigenvalue -1, > and maybe this will not generalize). However, at the moment I don't > see a construction that works for odd n in general. Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada To amplify a bit: Let J be the (m+1) x (m+1) Jordan block with 1's on the first diagonal above the main diagonal, 0's elsewhere. We want C=BB' to be similar to this, say C = S J S^(-1) for some invertible matrix S. Thus S J S^(-1) is symmetric, which says that J'S'S = S'SJ. Now the symmetric matrices H such that J'H = HJ are easy to characterize: they are Hankel matrices with zeros above the main antidiagonal (i.e. H_{ij} depends only on i+j, and H_{ij} = 0 for i+j < m+2). To get an example, choose such a matrix H with H_{1,m+1} <> 0 to make it invertible, then find S such that S'S = H, and finally find an (m+1)xm matrix B such that BB' = SJS^(-1). Assuming this can be done, it will give a (2m+1) x (2m+1) example for your question: note that B must have rank m, since SJS^(-1) does, so B' is surjective. Thus the fact that C^(m+1)=C^m BB'=0 implies C^m B = 0. Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Looking for a Matrix recipe > On Jun 26, 2:17 pm, Robert Israel > > I would need a general (for any n) recipe to construct > a matrix P with the following properties. > > 1. P is upper diagonal. > 2. P^2=unity. > 3. P*transpose(P) is a complete n*n Jordan block. > > I assume you mean not that it _is_ a complete block, but that its > Jordan > form _has_ one complete block. > > Example for n=2: > 1 2I > 0 -1 > > Your I, I guess, is i = sqrt(-1). I thought it was just Maple that > used > I for that. > > Example for n=3 > 1 0 1 > 0 1 I > 0 0 -1 > Nonexample for n=4 > 1 0 0 a > 0 1 0 b > 0 0 1 c > 0 0 0 -1 > Can't find a,b,c, it's always 3+1 Jordan blocks! > > I think there are no 4x4 examples. But here's a 5x5: > > [ 1 0 0 0 a ] > [ 0 1 0 0 aI ] > [ 0 0 1 b bI ] > [ 0 0 0 -1 0 ] > [ 0 0 0 0 -1 ] > > where a and b are nonzero. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada > > More generally, suppose P is a block matrix of the form > > [ I B ] > [ 0 -I ] > > (I'm reverting to the usual practice of using I for the identity > matrix, and i for sqrt(-1)). > > Then > [ C -B ] > P P' - I = [ -B' 0 ] > > where C = BB'. For an nxn solution where the eigenvalue of P P' is 1, > we want (PP'-I)^n = 0 but (PP'-I)^(n-1) <> 0. Now it is easy to show > by induction that > > [ f_{k}(C) -f_{k-1}(C)B ] > (PP'-I)^k = [ -B'f_{k-1}(C) B'f_{k-2}(C)B ] > > where f_k(x) are the polynomials defined by the recurrence > f_{k+1}(x) = x (f_k(x) + f_{k-1}(x)) > with f_0(x) = 1, f_1(x) = x. Note that f_k(x) is divisible by > x^(ceil(k/2)). > My 5 x 5 example had > (PP'-I)^5 = 0 because C^3 = 0 and C^2 B = 0 > (implying f_5(C) = 0, f_4(C) B = 0 and f_3(C)B = 0) but > (PP'-I)^4 <> 0 because C^2 <> 0 (implying f_4(C) <> 0). > Similarly, for a (2m+1) x (2m+1) example, we'd want > C^(m+1) = 0 and C^m B = 0 but C^m <> 0 > implying f_{2m+1}(C) = 0, f_{2m}(C) B = 0 and f_{2m}(C) <> 0. > This might work if B was (m+1) x m. > For example, this works for a 7 x 7 example, where ab <> 0: > > [ 8b, 4 i + 3 b^2 i, 4 - 3 b^2 ] > [ 0, -a i, a ] > B = [ 0, 4 - b^2, -4 i - b^2 i ] > [ 0, a, a i ] > > I suspect that there may be examples of this sort for all odd n, > but not for even n (in your example with n=2, P P' had eigenvalue -1, > and maybe this will not generalize). However, at the moment I don't > see a construction that works for odd n in general. > > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada To amplify a bit: Let J be the (m+1) x (m+1) Jordan block with 1's on the first diagonal > above the main diagonal, 0's elsewhere. We want > C=BB' to be similar to this, say C = S J S^(-1) for some invertible > matrix S. Thus S J S^(-1) is symmetric, which says that > J'S'S = S'SJ. Now the symmetric matrices H such that J'H = HJ > are easy to characterize: they are Hankel matrices with zeros above > the main antidiagonal (i.e. H_{ij} depends only on i+j, and H_{ij} = 0 > for i+j < m+2). To get an example, choose such a matrix H > with H_{1,m+1} <> 0 to make it invertible, then find S such that > S'S = H, and finally find an (m+1)xm matrix B such that > BB' = SJS^(-1). Assuming this can be done, it will give a > (2m+1) x (2m+1) example for your question: note that B must have > rank m, since SJS^(-1) does, so B' is surjective. Thus the fact that > C^(m+1)=C^m BB'=0 implies C^m B = 0. Robert Israel israel@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada One possibility for S is S_{i,j} = w^((i-1)(2j-1)) where w is a primitive (2m+2)'th root of unity. This makes H_{i,m+2-i} = m+1, H_{i,j}=0 otherwise. I don't have a nice solution for B, but I suspect they exist. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Looking for a Matrix recipe THX for your outline. Just for the record, here is what it's all about: - In the abstract tensor state models of knot theory, any S matrix solution allows the gauge S'=(MxM)S(M-xM-). (x tensor product, M any nonsingular,- inverse). - The creation/destruction operators then transform like P'=M+-PM- Q'=MQM+ (+ transpose). - So I must deal with congruent matrices all the time and find their simplest form. - You can immediately see that P'+P'- is similar to P+P-, so it's natural to assume it's in Jordan normal form. - I hoped that any solution for S could be gauged to a form where S=S+ (which suggest also QxP is symmetric, i.e. Q=P), but alas, the Jordan 4-block seems to refuse such a decompostion. (There is no problem to find Q such that Q+Q-={{-1,1,0,0},{0,-1... but if you ever had to check the Yang-Baxter equation with n^6 cubics in n^4 unknowns, you are grateful for any variable reduction :-) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de Er-a svo gott sem gott kve[CapitalYAcute]a .9al alda sonum, §v.92 a[CapitalYAcute] f.berra veit er fleira drekkur s.92ns til ge[CapitalYAcute]s gumi. === Subject: Re: JSH: Probabilistic quadratic residue solving posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > An off-shoot of my surrogate factoring research is a probabilistic > method to solve quadratic residues, as given k^2 = q mod p when p is an odd prime, and q is a quadratic residue modulo p, you > find k. The technique requires introduction of a few additional variables > starting with T, where T = 2q + np where n is an odd natural number so you have T = 2q mod p, but T - 2q > is also forced to have p as a factor. > Also as pointed out by an example from Chip Eastham in this thread, if T is a perfect square, then you are, of course, done, as you can simply take its square root! So you continue only if T is not a perfect square. > Next you find z, where with integer factors f 1 and f 2 where f 1*f 2 > = T: z = (f 1 + f 2)/2 and now finally you try for an answer for k, with k = 3^{-1}(2z) mod p. The method is probabilistic because if I've got the analysis right you > have a 50% probability of getting the right k, for each z that you > try. æChecking is done by looking at k^2 mod p, to see if you get q. Which is why it's a super advance on just finding T a square which is the old tech. The new tech is this kind of a trick that can greatly speed up the process. Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17. Here T=21 does not work, but T = 55 = 5(11), so z = 8 and the answer > then from 3k = 2(8) mod 17, is k = 11 mod 17. Is there any use for such a technique? I'd think there should be for anyone who likes SPEED and wants to solve a quadratic residue. But maybe none of you do, or maybe mathematicians are not into the best techniques--if one of their own doesn't find them. As I've noted many times about a field that can't be the best so it ignores the best. James Harris === Subject: Re: JSH: Probabilistic quadratic residue solving posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > An off-shoot of my surrogate factoring research is a probabilistic > method to solve quadratic residues, as given > k^2 = q mod p > when p is an odd prime, and q is a quadratic residue modulo p, you > find k. > The technique requires introduction of a few additional variables > starting with T, where > T = 2q + np > where n is an odd natural number so you have T = 2q mod p, but T - 2q > is also forced to have p as a factor. > Next you find z, where with integer factors f 1 and f 2 where f 1*f 2 > = T: > z = (f 1 + f 2)/2 > and now finally you try for an answer for k, with > k = 3^{-1}(2z) mod p. > The method is probabilistic because if I've got the analysis right you > have a 50% probability of getting the right k, for each z that you > try. æChecking is done by looking at k^2 mod p, to see if you get q. > Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17. > Here T=21 does not work, but T = 55 = 5(11), so z = 8 and the answer > then from > 3k = 2(8) mod 17, is k = 11 mod 17. > Is there any use for such a technique? > James Harris > /* notes on JSH's probabilistic square root mod p */ > Given prime p and quadratic residue q modulo p, > we seek k s.t. k^2 = q mod p. > JSH claims that the following procedure has 50% > chance of success: > Choose odd multiplier n to get T = 2q + np. > Factor T: f 1 * f 2 = 2q + np > [Here I simplify JSH's formulation, in that he > refers to the sum of factors f 1 + f 2 as 2z.] > Define z = f 1 + f 2 and k = z/3 mod p. > What has this to do with a square root of q? > Well, for k^2 = q mod p, it would require: > k^2 = (f 1^2 + 2 f 1*f 2 + f 2^2)/9 mod p > æ æ = (f 1^2 + 4 q + f 2^2)/9 mod p > 5q = f 1^2 + f 2^2 mod p > A priori I can see no reason this should be > likely. > /* Example: æFind k^2 = 3 mod 73 */ > Since 3 is a quadratic residue mod 73 if and only if 73 is a quadratic > residue mod 3 (by quadratic reciprocity), and 73 = 1 mod 3, then 3 is > a quadratic residue mod 73. æAlso, 5q = 15. > Now taking: > n = 1: T = 6 + 73 = 79 prime > æ(1 + 79^2) mod 73 = 37 > n = 3: T = 6 + 219 = 225 = 3*75 = 5*45 = 9*25 = 15 * 15 > æ(1 + 225^2) mod 73 = 37 > æ(9 + 75^2) mod 73 = 13 > æ(25 + 45^2) mod 73 = 6 > æ(81 + 25^2) mod 73 = 49 > æ 450 mod 73 = 12 > n = 5: T = 6 + 365 = 371 = 7 * 53 > æ(1 + 371^2) mod 73 = 37 > æ(49 + 53^2) mod 73 = 11 > n = 7: T = 6 + 511 = 517 = 11 * 47 > æ(1 + 517^2) mod 73 = 37 > æ(11^2 + 47^2) mod 73 = 67 > n = 9: T = 6 + 657 = 663 = 3*13*17 > æ(1 + 663^2) mod 73 = 37 > æ(9 + 221^2) mod 73 = 13 > æ(13^2 + 51^2) mod 73 = 69 > æ(17^2 + 39^2) mod 73 = 58 > n = 11: T = 6 + 803 = 809 prime > æ(1 + 809^2) mod 73 = 37 > None of the smallish multipliers n = 1 to 11 > gives a factorization T = f 1 * f 2 that > produces k such that k^2 = 3 mod 73. > k = 21, so maybe k has to be coprime to q? æBut if so, why? Correction: k = 15. > Did you try that at random or find it? æSearching for a case that > wouldn't work? I figured it out, as if T is a square then, of course, you can simply take its square root to get k. So you have to check if T is a square, which isn't a terrible thing as if it is, then, of course, you're done anyway and that's not new. What is new is a short-cut which can give you an answer much, much, much faster in general then simply looking for a perfect square. James Harris === Subject: Re: JSH: Probabilistic quadratic residue solving posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > An off-shoot of my surrogate factoring research is a probabilistic > method to solve quadratic residues, as given > k^2 = q mod p > when p is an odd prime, and q is a quadratic residue modulo p, you > find k. > The technique requires introduction of a few additional variables > starting with T, where > T = 2q + np > where n is an odd natural number so you have T = 2q mod p, but T - 2q > is also forced to have p as a factor. > Next you find z, where with integer factors f 1 and f 2 where f 1*f 2 > = T: > z = (f 1 + f 2)/2 > and now finally you try for an answer for k, with > k = 3^{-1}(2z) mod p. > The method is probabilistic because if I've got the analysis right you > have a 50% probability of getting the right k, for each z that you > try. æChecking is done by looking at k^2 mod p, to see if you get q. > Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17. > Here T=21 does not work, but T = 55 = 5(11), so z = 8 and the answer > then from > 3k = 2(8) mod 17, is k = 11 mod 17. > Is there any use for such a technique? > James Harris > /* notes on JSH's probabilistic square root mod p */ > Given prime p and quadratic residue q modulo p, > we seek k s.t. k^2 = q mod p. > JSH claims that the following procedure has 50% > chance of success: > Choose odd multiplier n to get T = 2q + np. > Factor T: f 1 * f 2 = 2q + np > [Here I simplify JSH's formulation, in that he > refers to the sum of factors f 1 + f 2 as 2z.] > Define z = f 1 + f 2 and k = z/3 mod p. > What has this to do with a square root of q? > Well, for k^2 = q mod p, it would require: > k^2 = (f 1^2 + 2 f 1*f 2 + f 2^2)/9 mod p > æ æ = (f 1^2 + 4 q + f 2^2)/9 mod p > 5q = f 1^2 + f 2^2 mod p > A priori I can see no reason this should be > likely. > /* Example: æFind k^2 = 3 mod 73 */ > Since 3 is a quadratic residue mod 73 if and only if 73 is a quadratic > residue mod 3 (by quadratic reciprocity), and 73 = 1 mod 3, then 3 is > a quadratic residue mod 73. æAlso, 5q = 15. > Now taking: > n = 1: T = 6 + 73 = 79 prime > æ(1 + 79^2) mod 73 = 37 > n = 3: T = 6 + 219 = 225 = 3*75 = 5*45 = 9*25 = 15 * 15 > æ(1 + 225^2) mod 73 = 37 > æ(9 + 75^2) mod 73 = 13 > æ(25 + 45^2) mod 73 = 6 > æ(81 + 25^2) mod 73 = 49 > æ 450 mod 73 = 12 > n = 5: T = 6 + 365 = 371 = 7 * 53 > æ(1 + 371^2) mod 73 = 37 > æ(49 + 53^2) mod 73 = 11 > n = 7: T = 6 + 511 = 517 = 11 * 47 > æ(1 + 517^2) mod 73 = 37 > æ(11^2 + 47^2) mod 73 = 67 > n = 9: T = 6 + 657 = 663 = 3*13*17 > æ(1 + 663^2) mod 73 = 37 > æ(9 + 221^2) mod 73 = 13 > æ(13^2 + 51^2) mod 73 = 69 > æ(17^2 + 39^2) mod 73 = 58 > n = 11: T = 6 + 803 = 809 prime > æ(1 + 809^2) mod 73 = 37 > None of the smallish multipliers n = 1 to 11 > gives a factorization T = f 1 * f 2 that > produces k such that k^2 = 3 mod 73. > k = 21, so maybe k has to be coprime to q? æBut if so, why? Correction: k = 15. > Did you try that at random or find it? æSearching for a case that > wouldn't work? I figured it out, as if T is a square then, of course, you can simply > take its square root to get k. So you have to check if T is a square, which isn't a terrible thing as > if it is, then, of course, you're done anyway and that's not new. What is new is a short-cut which can give you an answer much, much, > much faster in general then simply looking for a perfect square. James Harris First, I didn't do any searching. This was the first thing I tried, picking p = 73 because it's a prime congruent to 1 mod 8 (the only cases for which a fast deterministic method is unknown, so potentially where room for improvement in a probabilistic method exists). Second, I don't see the benefit of finding T is a square. Yes, this gives a square root mod p for 2q, but not for q. Here 15^2 = 225 = 6 = 2q mod p for q = 3 and p = 73.73. How would this help find a square root for 3 mod 73? It reduces the problem to finding a square root for 2 mod 73. You were right with the answer 21^2 = 441 = 3 mod 73, but I assume (since you showed no work) this answer was not obtained by the probabilistic method you outlined. === Subject: Re: JSH: Probabilistic quadratic residue solving posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > An off-shoot of my surrogate factoring research is a probabilistic > method to solve quadratic residues, as given > k^2 = q mod p > when p is an odd prime, and q is a quadratic residue modulo p, you > find k. > The technique requires introduction of a few additional variables > starting with T, where > T = 2q + np > where n is an odd natural number so you have T = 2q mod p, but T - 2q > is also forced to have p as a factor. > Next you find z, where with integer factors f 1 and f 2 where f 1*f 2 > = T: > z = (f 1 + f 2)/2 > and now finally you try for an answer for k, with > k = 3^{-1}(2z) mod p. > The method is probabilistic because if I've got the analysis right you > have a 50% probability of getting the right k, for each z that you > try. æChecking is done by looking at k^2 mod p, to see if you get q. > Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17. > Here T=21 does not work, but T = 55 = 5(11), so z = 8 and the answer > then from > 3k = 2(8) mod 17, is k = 11 mod 17. > Is there any use for such a technique? > James Harris > /* notes on JSH's probabilistic square root mod p */ > Given prime p and quadratic residue q modulo p, > we seek k s.t. k^2 = q mod p. > JSH claims that the following procedure has 50% > chance of success: > Choose odd multiplier n to get T = 2q + np. > Factor T: f 1 * f 2 = 2q + np > [Here I simplify JSH's formulation, in that he > refers to the sum of factors f 1 + f 2 as 2z.] > Define z = f 1 + f 2 and k = z/3 mod p. > What has this to do with a square root of q? > Well, for k^2 = q mod p, it would require: > k^2 = (f 1^2 + 2 f 1*f 2 + f 2^2)/9 mod p > æ æ = (f 1^2 + 4 q + f 2^2)/9 mod p > 5q = f 1^2 + f 2^2 mod p > A priori I can see no reason this should be > likely. > /* Example: æFind k^2 = 3 mod 73 */ > Since 3 is a quadratic residue mod 73 if and only if 73 is a quadratic > residue mod 3 (by quadratic reciprocity), and 73 = 1 mod 3, then 3 is > a quadratic residue mod 73. æAlso, 5q = 15. > Now taking: > n = 1: T = 6 + 73 = 79 prime > æ(1 + 79^2) mod 73 = 37 > n = 3: T = 6 + 219 = 225 = 3*75 = 5*45 = 9*25 = 15 * 15 > æ(1 + 225^2) mod 73 = 37 > æ(9 + 75^2) mod 73 = 13 > æ(25 + 45^2) mod 73 = 6 > æ(81 + 25^2) mod 73 = 49 > æ 450 mod 73 = 12 > n = 5: T = 6 + 365 = 371 = 7 * 53 > æ(1 + 371^2) mod 73 = 37 > æ(49 + 53^2) mod 73 = 11 > n = 7: T = 6 + 511 = 517 = 11 * 47 > æ(1 + 517^2) mod 73 = 37 > æ(11^2 + 47^2) mod 73 = 67 > n = 9: T = 6 + 657 = 663 = 3*13*17 > æ(1 + 663^2) mod 73 = 37 > æ(9 + 221^2) mod 73 = 13 > æ(13^2 + 51^2) mod 73 = 69 > æ(17^2 + 39^2) mod 73 = 58 > n = 11: T = 6 + 803 = 809 prime > æ(1 + 809^2) mod 73 = 37 > None of the smallish multipliers n = 1 to 11 > gives a factorization T = f 1 * f 2 that > produces k such that k^2 = 3 mod 73. > k = 21, so maybe k has to be coprime to q? æBut if so, why? > Correction: k = 15. > Did you try that at random or find it? æSearching for a case that > wouldn't work? > I figured it out, as if T is a square then, of course, you can simply > take its square root to get k. > So you have to check if T is a square, which isn't a terrible thing as > if it is, then, of course, you're done anyway and that's not new. > What is new is a short-cut which can give you an answer much, much, > much faster in general then simply looking for a perfect square. > James Harris First, I didn't do any searching. æThis was the first > thing I tried, picking p = 73 because it's a prime > congruent to 1 mod 8 (the only cases for which a fast > deterministic method is unknown, so potentially where > room for improvement in a probabilistic method exists). Second, I don't see the benefit of finding T is a square. > Yes, this gives a square root mod p for 2q, but not for q. Here 15^2 = 225 = 6 = 2q mod p for q = 3 and p = 73.73. > How would this help find a square root for 3 mod 73? æIt > reduces the problem to finding a square root for 2 mod 73. You were right with the answer 21^2 = 441 = 3 mod 73, > but I assume (since you showed no work) this answer was > not obtained by the probabilistic method you outlined. > Try these equations for your special case: If T mod 3 = 1, because p mod 3 = 1 and q is divisible by 3, then an alternate set of equations can be used as then T = 10q mod p and k = 19^{-1}(6z) mod p. JSH === Subject: Re: JSH: Probabilistic quadratic residue solving posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > An off-shoot of my surrogate factoring research is a probabilistic > method to solve quadratic residues, as given > k^2 = q mod p > when p is an odd prime, and q is a quadratic residue modulo p, you > find k. > The technique requires introduction of a few additional variables > starting with T, where > T = 2q + np > where n is an odd natural number so you have T = 2q mod p, but T - 2q > is also forced to have p as a factor. > Next you find z, where with integer factors f 1 and f 2 where f 1*f 2 > = T: > z = (f 1 + f 2)/2 > and now finally you try for an answer for k, with > k = 3^{-1}(2z) mod p. > The method is probabilistic because if I've got the analysis right you > have a 50% probability of getting the right k, for each z that you > try. æChecking is done by looking at k^2 mod p, to see if you get q. > Example: Let q=2, p=17 so T = 2(2) mod 17 = 4 mod 17. > Here T=21 does not work, but T = 55 = 5(11), so z = 8 and the answer > then from > 3k = 2(8) mod 17, is k = 11 mod 17. > Is there any use for such a technique? > James Harris > /* notes on JSH's probabilistic square root mod p */ > Given prime p and quadratic residue q modulo p, > we seek k s.t. k^2 = q mod p. > JSH claims that the following procedure has 50% > chance of success: > Choose odd multiplier n to get T = 2q + np. > Factor T: f 1 * f 2 = 2q + np > [Here I simplify JSH's formulation, in that he > refers to the sum of factors f 1 + f 2 as 2z.] > Define z = f 1 + f 2 and k = z/3 mod p. > What has this to do with a square root of q? > Well, for k^2 = q mod p, it would require: > k^2 = (f 1^2 + 2 f 1*f 2 + f 2^2)/9 mod p > æ æ = (f 1^2 + 4 q + f 2^2)/9 mod p > 5q = f 1^2 + f 2^2 mod p > A priori I can see no reason this should be > likely. > /* Example: æFind k^2 = 3 mod 73 */ > Since 3 is a quadratic residue mod 73 if and only if 73 is a quadratic > residue mod 3 (by quadratic reciprocity), and 73 = 1 mod 3, then 3 is > a quadratic residue mod 73. æAlso, 5q = 15. > Now taking: > n = 1: T = 6 + 73 = 79 prime > æ(1 + 79^2) mod 73 = 37 > n = 3: T = 6 + 219 = 225 = 3*75 = 5*45 = 9*25 = 15 * 15 > æ(1 + 225^2) mod 73 = 37 > æ(9 + 75^2) mod 73 = 13 > æ(25 + 45^2) mod 73 = 6 > æ(81 + 25^2) mod 73 = 49 > æ 450 mod 73 = 12 > n = 5: T = 6 + 365 = 371 = 7 * 53 > æ(1 + 371^2) mod 73 = 37 > æ(49 + 53^2) mod 73 = 11 > n = 7: T = 6 + 511 = 517 = 11 * 47 > æ(1 + 517^2) mod 73 = 37 > æ(11^2 + 47^2) mod 73 = 67 > n = 9: T = 6 + 657 = 663 = 3*13*17 > æ(1 + 663^2) mod 73 = 37 > æ(9 + 221^2) mod 73 = 13 > æ(13^2 + 51^2) mod 73 = 69 > æ(17^2 + 39^2) mod 73 = 58 > n = 11: T = 6 + 803 = 809 prime > æ(1 + 809^2) mod 73 = 37 > None of the smallish multipliers n = 1 to 11 > gives a factorization T = f 1 * f 2 that > produces k such that k^2 = 3 mod 73. > k = 21, so maybe k has to be coprime to q? æBut if so, why? > Correction: k = 15. > Did you try that at random or find it? æSearching for a case that > wouldn't work? > I figured it out, as if T is a square then, of course, you can simply > take its square root to get k. > So you have to check if T is a square, which isn't a terrible thing as > if it is, then, of course, you're done anyway and that's not new. > What is new is a short-cut which can give you an answer much, much, > much faster in general then simply looking for a perfect square. > James Harris First, I didn't do any searching. æThis was the first > thing I tried, picking p = 73 because it's a prime > congruent to 1 mod 8 (the only cases for which a fast > deterministic method is unknown, so potentially where > room for improvement in a probabilistic method exists). That's no the problem. q=3 was. > Second, I don't see the benefit of finding T is a square. > Yes, this gives a square root mod p for 2q, but not for q. Here 15^2 = 225 = 6 = 2q mod p for q = 3 and p = 73.73. > How would this help find a square root for 3 mod 73? æIt > reduces the problem to finding a square root for 2 mod 73. You were right with the answer 21^2 = 441 = 3 mod 73, > but I assume (since you showed no work) this answer was > not obtained by the probabilistic method you outlined. > Nope, I went and looked for it. The problem is 3. Try another q with p=73 that is a quadratic residue that is not divisible by 3 and see what happens. Oh yeah, it took two things: p mod 3 = 1 and q=3 for big problems to arise so you had a lucky set of choices, which isn't a bad thing. I find the issue curious. I am puzzling over why it's such a big deal. JSH === Subject: Re: JSH: Probabilistic quadratic residue solving What is new is a short-cut which can give you an answer much, much, much faster in general then simply looking for a perfect square. try it out, this was solved in 2000, http://www.alpertron.com.ar/ECM.HTM [ JSH, check it out! Dario already finished it. ] James Harris === Subject: Re: Number with fermat prime 2^(2^n) + 1 > Hello teacher~ Let F_n = 2^(2^n) + 1. (n = 0, 1, 2, 3, ...) Show that (F_m, F_n) = 1 for m =/= n. Hint: If an integer divides F_m and F_n, then it divides F_m - F_n ... which is ... === Subject: Re: Number with fermat prime 2^(2^n) + 1 > Hello teacher~ > Let F_n = 2^(2^n) + 1. (n = 0, 1, 2, 3, ...) > Show that (F_m, F_n) = 1 for m =/= n. > Hint: If an integer divides F_m and F_n, then it divides F_m - F_n ... > which is ... Oops [blush] it's not as easy I thought. Ignore the above, while I take a closer look. :D === Subject: Re: Number with fermat prime 2^(2^n) + 1 > message > Hello teacher~ > > Let F_n = 2^(2^n) + 1. (n = 0, 1, 2, 3, ...) > > Show that (F_m, F_n) = 1 for m =/= n. > Hint: If an integer divides F_m and F_n, then it > divides F_m - F_n ... which > is ... or much better: F_m + F_n = 2^(2m) + 2^(2n) + 2 don't we ? Ro-Bin === Subject: Lemma on continuity Sternberg, page 52. Let A be a subset of E^n. There exists a sequence of sets A_i (i>=1) and maps h_i (i>=1) such that A_0 is countable, A is included in Union of the A_r for r spanning between 0 and infinity, and for i>=1 1. h_i is a homeomorphism of the ball (B_r_i)^(m_i) into E^n with A_i included in h_i((B_r_i)^(m_i)) ((B_r)^m is the set of x in E^m such that ||x||= ||x-y|| 3. for any continuous function f vanishing on A there exist monotone functions b_i(r) that vanish for r->0 and such that |f(h_i(x))| < b_i(||x-y||) for all x, y in (B_r_i)^(m_i) with h_i(y) in A_i (and therefore f(h_i(y))=0). Now my problem is Sternberg claims that taking A_i = A / K_i (where K_i is a set of closed balls constituting a basis for E^n) and h_i to be the translation of the ball centered at zero to K_i it is clear that part (3.) is just uniform continuity. WHY IS THAT? P.S. How can a set of closed balls be a basis for E^n? Basis in what sense? I hope someone answers 'cause I'm in deep deep trouble. === Subject: Re: Lemma on continuity Nobody? === Subject: Re: Lemma on continuity > Sternberg, page 52. Let A be a subset of E^n. There exists a sequence of sets A_i (i>=1) and maps > h_i > (i>=1) such that A_0 is countable, A is included in Union of the A_r for r > spanning > between 0 and infinity, and for i>=1 1. h_i is a homeomorphism of the ball (B_r_i)^(m_i) into E^n with A_i > included in h_i((B_r_i)^(m_i)) > ((B_r)^m is the set of x in E^m such that ||x|| because r and m > obviously depend on h_i. 2. ||h_r(x)-h_r(y)||>= ||x-y|| 3. for any continuous function f vanishing on A there exist monotone > functions b_i(r) that vanish for r->0 > and such that |f(h_i(x))| < b_i(||x-y||) for all x, y in (B_r_i)^(m_i) with > h_i(y) in A_i (and therefore f(h_i(y))=0). > Now my problem is Sternberg claims that taking A_i = A / K_i (where K_i is a > set of closed balls constituting a basis for E^n) > and h_i to be the translation of the ball centered at zero to K_i it is clear > that part (3.) is just uniform continuity. > WHY IS THAT? P.S. > How can a set of closed balls be a basis for E^n? Basis in what sense? Topological basis. Any open set in E^n can be written as a union of sets in the basis. Has nothing to do with vector space basis, perhaps that's your point of confusion. See http://en.wikipedia.org/wiki/Basis_(topology) === Subject: Re: Lemma on continuity === Subject: Re: Advantages of directive newsgroups > ... member.92s > ... It's spelled with an apostrophe and not > memberRs which you spelt with an R or a ctrl_R. Why can't you use > the apostrophe key instead like you should do? It is not in my nor the > public's interest that you don't or won't. > > Hi William, > As allready mentioned, it is your newsreader that corrupts the message. > I clearly see an apostrophy looking glyph. > But your newsreader is unable to read quoted-printable and '=92' is > an apostrophy, although an unusual one. > PS : the =92 apostrophy is so unusual that I had to change the > charset to be able to *send* it. > Don't blame me. It's also the lament of others with more modern > computers. That ever since ascii was declared 'a not valued added > proddick' computers have been plagued by a hoard of formattings and > cursed like the Tower of Babel for their pride of attempting to speak > the incomprehensible all. > BTW your newsreader packs your message in some unnecessary mime > section. > That's to deal with jerks who like to send 8-bit text. > Wish I could turn if off or make to disappear everything > but printable characters. To be clear, the octets sent in any nntp message have the high bit 0. -- Michael Press === Subject: Re: [OT] Char encoding. was: Advantages of directive newsgroups ... member.92s > ... It's spelled with an apostrophe and not > memberRs which you spelt with an R or a ctrl_R. Why can't you use > the apostrophe key instead like you should do? It is not in my nor the > public's interest that you don't or won't. Hi William, > As allready mentioned, it is your newsreader that corrupts the message. > I clearly see an apostrophy looking glyph. > Don't blame me. It's also the lament of others with more modern > computers. Modern doesn't mean it can't interpret oldies like US-ASCII ! I recently had problems with old files using PC character set (remember, strange things as graphical characters to draw boxes and so) But their is a terminal font in my PC which displays that fine... Modern should mean it can do *more* that previously, not less ! > ...Tower of Babel... Yes, receiving messages in chinese on sci.math for instance ? > BTW your newsreader packs your message in some unnecessary mime > section. > That's to deal with jerks who like to send 8-bit text. > Wish I could turn if off or make to disappear everything > but printable characters. To be clear, the octets sent in any nntp message have the high bit 0. We are no more in eighties ! RFC 977 (February 1986 !!!!) 2.2. Character Codes Commands and replies are composed of characters from the ASCII character set. When the transport service provides an 8-bit byte (octet) transmission channel, each 7-bit character is transmitted right justified in an octet with the high order bit cleared to zero. But : RFC 3977 (October 2006) This document is a replacement for RFC 977, and officially updates the protocol specification. The default character set is changed from US-ASCII [ANSI1986] to UTF-8 [RFC3629] 10.1. RFC 977 [RFC977] was written at a time when internationalisation was not seen as a significant issue. As such, it was written on the assumption that all communication would be in ASCII and use only a 7-bit transport layer, although in practice just about all known implementations are 8-bit clean. [...] contain octets with the top bit set, and NNTP is only expected to operate on 8-bit clean transport paths. 10.2 [...] This specification extends NNTP from US-ASCII [ANSI1986] to UTF-8 [RFC3629]. Except in the two areas discussed below, UTF-8 (which is a superset of US-ASCII) is mandatory, and implementations MUST NOT use any other encoding. strongly recommended. [...] Accordingly, this specification the following requirements and recommendations. US-ASCII. o Header values SHOULD use US-ASCII or an encoding based on it [...] with MIME. Hence as a conclusion : And MIME meaning just the header SHOULD contain items like : for text only newsgroup ! And yes, your newsreader SHOULD allow for non US-ASCII characters, as long as the charset encoding is clearly specified in the header. And jerks who like to send 8-bit text are just those who follow the up to date nntp specification... In fact their newsreader does. They, as individual, just don't care. Do you really care how the characters you type on your keyboard are encoded inside your computer ? for ALL modern operating systems it is 16 bits ! Far far away from US-ASCII. And if the keyboard has a ' key instead of just a ' how can you type a ' ? with ALT039 ? Wouldn't it be more natural to just depress the ' key ? -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: [OT] Char encoding. was: Advantages of directive newsgroups [...] > And jerks who like to send 8-bit text are just those who follow > the up to date nntp specification... In fact their newsreader does. > They, as individual, just don't care. Do you really care how the > characters you type on your keyboard are encoded inside your > computer ? for ALL modern operating systems it is 16 bits ! > Far far away from US-ASCII. > And if the keyboard has a ' key instead of just a ' how can you > type a ' ? with ALT039 ? > Wouldn't it be more natural to just depress the ' key ? Everybody note that the quoted message is a message replying to one of mine. While Philippe 92 quotes a phrase with the epithet `jerks' in it, I did not write `jerks.' -- Michael Press === Subject: Re: [OT] Char encoding. was: Advantages of directive newsgroups > [...] > And jerks who like to send 8-bit text are just those who follow > the up to date nntp specification... In fact their newsreader does. > They, as individual, just don't care. Do you really care how the > characters you type on your keyboard are encoded inside your > computer ? for ALL modern operating systems it is 16 bits ! > Far far away from US-ASCII. > And if the keyboard has a ' key instead of just a ' how can you > type a ' ? with ALT039 ? > Wouldn't it be more natural to just depress the ' key ? Everybody note that the quoted message is a message replying > to one of mine. While Philippe 92 quotes a phrase with > the epithet `jerks' in it, I did not write `jerks.' I can see jerks in your post, and I see it in Philippe's post in exactly the same context as it was in your post; namely that of you quoting a previous poster who used it. If you don't understand how quoting works on usenet, please learn. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Advantages of directive newsgroups posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/20050915,gzip(gfe),gzip(gfe) > This is not my mistake, but within character encoding and > computer languages issues of Math Forum. It has nothing to do with Math Forum. You posted your original The problem is your Macintosh. Why do Mac Email Apostrophe's get Viewed as Question Marks by PC Users? -- Thus this will be my final note (and I will not post === Subject: Re: Advantages of directive newsgroups > On Jun 27, 5:02 am, Mehran Basti This is not my mistake, but within character > encoding and > computer languages issues of Math Forum. It has nothing to do with Math Forum. You posted > your original > using Giganews. The problem is your Macintosh. Why do Mac Email Apostrophe's get Viewed as Question > Marks by > PC Users? > -- > Thus this will be my final note (and I will not post Yes I write on Mac. You are welcome to join the newsgroup. Dr.M.Basti PS:Yes Generraly I will not post here, I will post it on my site, but sometimes I need to respond and invite you to my site. === Subject: Re: Advantages of directive newsgroups > On Jun 27, 5:02 am, Mehran Basti encoding and > computer languages issues of Math Forum. > It has nothing to do with Math Forum. You posted > your original > using Giganews. > The problem is your Macintosh. > Why do Mac Email Apostrophe's get Viewed as Question > Marks by > PC Users? > -- > Thus this will be my final note (and I will not post > Yes I write on Mac. You are welcome to join the newsgroup. > Dr.M.Basti PS:Yes Generraly I will not post here, I will post it on my site, but > sometimes I need to respond and invite you to my site. *Good move* Dr. Basti. Do you even _realize_ you have just invited to your site a person who's making fun of you with his/her signature? Sigh... Something is very very wrong here. I am beginning to suspect Dr. Basti is a bot ;o) -- I.N. Galidakis === Subject: Re: Advantages of directive newsgroups <3404079.1214599557001.JavaMail.jakarta@nitrogen.mathforum.org> <1214600623.317111@athprx03> posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > On Jun 27, 5:02 am, Mehran Basti This is not my mistake, but within character > encoding and > computer languages issues of Math Forum. > It has nothing to do with Math Forum. æYou posted > your original > using Giganews. > The problem is your Macintosh. > Why do Mac Email Apostrophe's get Viewed as Question > Marks by > PC Users? > -- > Thus this will be my final note (and I will not post > Yes I write on Mac. > You are welcome to join the newsgroup. > Dr.M.Basti > PS:Yes Generraly I will not post here, I will post it on my site, but > sometimes I need to respond and invite you to my site. *Good move* Dr. Basti. Do you even realize you have just invited to your site > a person who's making fun of you with his/her signature? Sigh... Something is very very wrong here. I am beginning to suspect Dr. Basti > is a bot ;o) If he's a bot, he passes the Turing Test! There's a reason why Basti no longer requires a fee to join his Google Group. to the personal information he needs. posted by new members. Of course, the reason is that he's been talking to a sock puppet. Programming your bot to be thick is a LOT harder than programming it to be smart. Basti balked at making the sock puppet a group manager, but I expect that's just a matter of time. > -- > I.N. Galidakis === Subject: Re: Advantages of directive newsgroups <3404079.1214599557001.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=6gwCCwoAAADEcuCVO3weXkA4SVHb4mS2 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > On Jun 27, 5:02 am, Mehran Basti This is not my mistake, but within character > encoding and > computer languages issues of Math Forum. > It has nothing to do with Math Forum. æYou posted > your original > using Giganews. > The problem is your Macintosh. > Why do Mac Email Apostrophe's get Viewed as Question > Marks by > PC Users? > -- > Thus this will be my final note (and I will not post > Yes I write on Mac. > You are welcome to join the newsgroup. > Dr.M.Basti > PS:Yes Generraly I will not post here, I will post it on my site, but > sometimes I need to respond and invite you to my site. > *Good move* Dr. Basti. Do you even realize you have just invited to your site > a person who's making fun of you with his/her signature? > Sigh... Something is very very wrong here. I am beginning to suspect Dr. Basti > is a bot ;o) If he's a bot, he passes the Turing Test! There's a reason why Basti no longer requires a fee to > join his Google Group. to the personal information he needs. posted by new members. Of course, the reason is that he's been talking to a sock puppet. Programming your bot to be thick is a LOT harder than programming > it to be smart. Basti balked at making the sock puppet a group manager, but I > expect that's just a matter of time. Who you callin' a sock puppet? > -- > I.N. Galidakis === Subject: Re: Advantages of directive newsgroups <3404079.1214599557001.JavaMail.jakarta@nitrogen.mathforum.org> <1214600623.317111@athprx03> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008061712 Fedora/3.0-1.fc9 Firefox/3.0,gzip(gfe),gzip(gfe) > On Jun 27, 5:02 am, Mehran Basti This is not my mistake, but within character > encoding and > computer languages issues of Math Forum. > It has nothing to do with Math Forum. æYou posted > your original > using Giganews. > The problem is your Macintosh. > Why do Mac Email Apostrophe's get Viewed as Question > Marks by > PC Users? > -- > Thus this will be my final note (and I will not post > Yes I write on Mac. > You are welcome to join the newsgroup. > Dr.M.Basti > PS:Yes Generraly I will not post here, I will post it on my site, but > sometimes I need to respond and invite you to my site. *Good move* Dr. Basti. Do you even realize you have just invited to your site > a person who's making fun of you with his/her signature? Sigh... Something is very very wrong here. I am beginning to suspect Dr. Basti > is a bot ;o) You are *beginning* to suspect that?! -- m === Subject: Re: Advantages of directive newsgroups > Sigh... Something is very very wrong here. I am beginning to suspect Dr. > Basti is a bot ;o) You are *beginning* to suspect that?! Yes, sorry for the delay, Mariano. My IQ approaches zero from above during hot summer days in Athens, so I am very slow at figuring out bots. In any case, isn't it about time somebody notified the MIT AI lab that sci.math figured them out? Perhaps they'll retract the 'Dr. Basti' program and try to improve it ;o) > -- m -- I.N. Galidakis === Subject: Re: Advantages of directive newsgroups <3404079.1214599557001.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=ee2apQoAAABJNMlLTFasCJw9Nfo9FmYk Gecko/20020924 AOL/7.0,gzip(gfe),gzip(gfe) HTTP/1.1 cache-ntc-ab04.proxy.aol.com[CFC87444] (Traffic-Server/6.1.5 [uScM]) Anyone is welcome to my newsgroups. me or not). You can criticize NEW EXACT MATH (or any of my/others writings) on any levels but within the guidelines of newsgroup (written on a contract). Welcome to Basti newsgroups. Dr.M.Basti === Subject: Re: Limit point, open, closed <6cjsq5F3g1l78U1@mid.individual.net>g42g56$jcl$1@localhost.localdomain Let E be a set. Prove that if E' = {x : x is a limit point of E}, E' > is closed. Maybe, T_1 is necessary. It is. Give R the topology { nulset, (-oo,x), R | x in R } Take E = {0}. If E' = (0,oo) is closed, then RE' = (-oo,0] is open, which it isn't. ---- === Subject: -- Solutions to sinh(z) = +/- sin(z) In my musings I've come across the functions f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + 1)! = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... ..and... f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + 3)! = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... I am trying to locate all the roots of f_1 and f_3 (other than the trivial zero at the origin). If we let t = e^{pi * i / 4}, then note that f_1(x) = [sinh(tx) + sin(tx)] / 2t ..and... f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. Thus, since t is a constant independent of x, locating the roots of f_1 and f_3 is equivalent to finding the solutions to sinh(z) = +/- sin(z). Any help and/or references would be greatly appreciated. Kyle Czarnecki === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) amy666 a .8ecrit : > In my musings I've come across the functions > f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + > 1)! > = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... > ...and... > f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + > 3)! > = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... > I am trying to locate all the roots of f_1 and f_3 > (other > than the trivial zero at the origin). > If we let t = e^{pi * i / 4}, then note that > f_1(x) = [sinh(tx) + sin(tx)] / 2t > ...and... > f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. > Thus, since t is a constant independent of x, > locating > the roots of f_1 and f_3 is equivalent to finding > the > solutions to sinh(z) = +/- sin(z). > Any help and/or references would be greatly > appreciated. > Kyle Czarnecki clearly about a simple inverse hypergeo function... as i have been talking about for years. Yes, you are a great one for talking. So what is the solution? === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) <25658552.1214685995575.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) > In my musings I've come across the functions > f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + > 1)! > = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... > ...and... > f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + > 3)! > = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... > I am trying to locate all the roots of f_1 and f_3 > (other > than the trivial zero at the origin). > If we let t = e^{pi * i / 4}, then note that > f_1(x) = [sinh(tx) + sin(tx)] / 2t > ...and... > f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. > Thus, since t is a constant independent of x, > locating > the roots of f_1 and f_3 is equivalent to finding > the > solutions to sinh(z) = +/- sin(z). > Any help and/or references would be greatly > appreciated. > Kyle Czarnecki clearly about a simple inverse hypergeo function... as i have been talking about for years. you are very correct -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) <25658552.1214685995575.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) > In my musings I've come across the functions > f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + > 1)! > = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... > ...and... > f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + > 3)! > = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... > I am trying to locate all the roots of f_1 and f_3 > (other > than the trivial zero at the origin). > If we let t = e^{pi * i / 4}, then note that > f_1(x) = [sinh(tx) + sin(tx)] / 2t > ...and... > f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. > Thus, since t is a constant independent of x, > locating > the roots of f_1 and f_3 is equivalent to finding > the > solutions to sinh(z) = +/- sin(z). > Any help and/or references would be greatly > appreciated. > Kyle Czarnecki > clearly about a simple inverse hypergeo function... > as i have been talking about for years. you are very correct which one can map to a generalised polynomial domain using the e^x -> x logarithmic map (respecting sheeting and all that) one can call this construction I[x] of the formal sums 3 --- --- j k i / / a x --- --- j k j=0 k e K j K c N j K finite j (polynomials with negative and imaginary exponents) this is the semigroup ring generated from the hypergeometric / - / x 4 H | ; | - | | 0 4 1/4, 1/2, 3/4, 1 4 / / and it's rotations so the problem reduces then i -1 -i x + i x - x - i x = 0 and it's splitting fields these fields are purely transcendental but easily understood -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) <25658552.1214685995575.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) > In my musings I've come across the functions > f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + > 1)! > = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... > ...and... > f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + > 3)! > = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... > I am trying to locate all the roots of f_1 and f_3 > (other > than the trivial zero at the origin). > If we let t = e^{pi * i / 4}, then note that > f_1(x) = [sinh(tx) + sin(tx)] / 2t > ...and... > f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. > Thus, since t is a constant independent of x, > locating > the roots of f_1 and f_3 is equivalent to finding > the > solutions to sinh(z) = +/- sin(z). > Any help and/or references would be greatly > appreciated. > Kyle Czarnecki > clearly about a simple inverse hypergeo function... > as i have been talking about for years. > you are very correct which one can map to a generalised polynomial domain > using the e^x -> x logarithmic map > (respecting sheeting and all that) one can call this construction I[x] of the formal sums 3 > --- --- j > k i > / / a x > --- --- j k > j=0 k e K > j > K c N > j > K finite > j (polynomials with negative and imaginary exponents) ugh this is actually just --- j / a x --- j j e J J c I J finite where I is the ring of gaussian integers (i've been dealing with these ray sums lately and i responded too quickly) > this is the semigroup ring > generated from the hypergeometric / - / x 4 > H | ; | - | | > 0 4 1/4, 1/2, 3/4, 1 4 / / and it's rotations so the problem reduces then i -1 -i > x + i x - x - i x = 0 and it's splitting fields these fields are purely transcendental > but easily understood the differential structure maintains fixpoints when the substitution D |-> xD is performed in a way that's the whole point behind it all -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) posting-account=33KaEgkAAAA9tz8WICNABjrkyMKXFbGS Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) On Jun 28, 11:31 am, Narcoleptic Insomniac In my musings I've come across the functions f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + 1)! = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... ..and... f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + 3)! = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... I am trying to locate all the roots of f_1 and f_3 (other > than the trivial zero at the origin). If we let t = e^{pi * i / 4}, then note that f_1(x) = [sinh(tx) + sin(tx)] / 2t ..and... f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. Thus, since t is a constant independent of x, locating > the roots of f_1 and f_3 is equivalent to finding the > solutions to sinh(z) = +/- sin(z). Any help and/or references would be greatly appreciated. Kyle Czarnecki We can get approximate values of roots by asymptotic expansions. Take z = al + i bt for solving sinh(z) = +/- sin(z). For first case, sin(al)cosh(bt) + i cos(al) sinh(bt) = sinh(al)cos(bt) + i cosh(al) sin(bt) Equating real and imaginary parts, and dividing tan(al) tan(bt) - tanh(al) tanh(bt) = 0 or since the limit tanh(..) -> 1 for large al and bt tan(al) tan(bt) - 1 = 0 or al + bt = (2 m -1 )pi/2. Symmetry of trig/hyp expressions gives al = bt, so al = bt = (m -1/2) pi/2; Similarly for second case anti-symmetry gives al = - bt, al = - bt = (n -1/2) pi/2, same result, all odd multiples of pi/4 are solutions; Exact roots come out by Newton-Raphson's iteration using a complex argument for integers m or n near to 1. IIRC, the equation arises in solving ODE for Plates on elastic foundation in Advanced Strength of materials. Narasimham === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) In my musings I've come across the functions f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + 1)! = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... ..and... f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + 3)! = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... I am trying to locate all the roots of f_1 and f_3 (other > than the trivial zero at the origin). If we let t = e^{pi * i / 4}, then note that f_1(x) = [sinh(tx) + sin(tx)] / 2t ..and... f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. Thus, since t is a constant independent of x, locating > the roots of f_1 and f_3 is equivalent to finding the > solutions to sinh(z) = +/- sin(z). [Apologies if part of my response duplicates part of the galathaea's material, which I haven't read.] Nontrivial solutions of sinh(z) = +/- sin(z) have the form z = (+/- 1 +/- i) x. [Those signs are not linked. Thus, for a given x, there is a solution in each quadrant.] So let's consider just the first quadrant solutions, z = (1 + i) x for positive x, knowing that the others can be obtained by symmetry. For sinh(z) = +/- sin(z), a simple approximation for x is (n +/- 1/4) pi for positive integer n. [Those signs _are_ linked, i.e., the same sign is to be used in the approximation as in the equation.] That simple approximation is merely the first term of a series, and can be much improved by using just one more term of the series: For sinh(z) = +sin(z), x ~= (n + 1/4) pi + 1/2 (1 - coth((n + 1/4) pi)); for sinh(z) = -sin(z), x ~= (n - 1/4) pi + 1/2 (1 + tanh((n - 1/4) pi)). (*) Numerical example: The solution z = (1 + i) x of sinh(z) = -sin(z) which has x close to (4 - 1/4) pi has x = 11.7809724510202275376960... For comparison, we get the approximation 11.7809724510202275376994... by using (*). As n increases, accuracy of the approximation increases quickly. The worst approximation occurs when we approximate the solution z = (1 + i) x of sinh(z) = -sin(z) which has x close to (1 - 1/4) pi; the true value of x is 2.365020..., while (*) gives 2.365097... Of course, we could get better approximations than (*) by using more terms of the series. David W. Cantrell === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) > Narcoleptic Insomniac > > In my musings I've come across the functions > > f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / > (4n + 1)! > > = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... > > ..and... > > f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / > (4n + 3)! > > = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... > > I am trying to locate all the roots of f_1 and f_3 > (other than the trivial zero at the origin). > > If we let t = e^{pi * i / 4}, then note that > > f_1(x) = [sinh(tx) + sin(tx)] / 2t > > ..and... > > f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. > > Thus, since t is a constant independent of x, > locating the roots of f_1 and f_3 is equivalent to > finding the solutions to sinh(z) = +/- sin(z). [Apologies if part of my response duplicates part of > the galathaea's material, which I haven't read.] Nontrivial solutions of sinh(z) = +/- sin(z) have the > form z = (+/- 1 +/- i) x. [Those signs are not > linked. Thus, for a given x, there is a solution in > each quadrant.] So let's consider just the first > quadrant solutions, z = (1 + i) x for positive x, > knowing that the others can be obtained by symmetry. For sinh(z) = +/- sin(z), a simple approximation for > x is (n +/- 1/4) pi for positive integer n. [Those > signs _are_ linked, i.e., the same sign is to be > used in the approximation as in the equation.] That simple approximation is merely the first term of > a series, and can be much improved by using just one > more term of the series: For sinh(z) = +sin(z), x ~= (n + 1/4) pi + 1/2 (1 - > coth((n + 1/4) pi)); > for sinh(z) = -sin(z), x ~= (n - 1/4) pi + 1/2 (1 + > tanh((n - 1/4) pi)). (*) Numerical example: > The solution z = (1 + i) x of sinh(z) = -sin(z) which > has x close to (4 - 1/4) pi has > x = 11.7809724510202275376960... > For comparison, > we get the approximation 11.7809724510202275376994... > by using (*). As n increases, accuracy of the approximation > increases quickly. The worst approximation occurs when > we approximate the solution z = (1 + i) x of > sinh(z) = -sin(z) which has x close to (1 - 1/4) pi; > the true value of x is > 2.365020..., while (*) gives 2.365097... Of course, we could get better approximations than > (*) by using more terms of the series. David W. Cantrell Kyle Czarnecki P.S. As narasimham alluded to, the functions f_1 and f_3 in question arise as solutions to the differential equation f^(4)(x) + f(x) = 0 ..which is satisfied by... f_k(x) = sum_{n = 0}^{oo} x^{4n + k} / (4n + k)!. I was looking for the roots of these functions because I was curious to see what their infinite product representations looked like. For instance, the first solution f_1(x) = 1 - x^4 / 4! + x^8 / 8! - ... = [cosh(xt) + cos(xt)] / 2 ..where t = e^{pi i / 4}, can be expressed as a product over its roots as... f_1(x) = prod_{n = 0}^{oo} {1 - 4x^4 / [(2n + 1)^4 pi^4]} = [1 - 4x^4 / pi^4] * [1 - 4x^4 / (3 pi)^4] * ... Note the similarity to the infinite product for cosine (which is obviously due to the nature of the D.E.'s these function all satisfy). === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) On Jun 29, 2:28 am, Narcoleptic Insomniac > Narcoleptic Insomniac > In my musings I've come across the functions > f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / > (4n + 1)! > = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... > ..and... > f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / > (4n + 3)! > = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... > I am trying to locate all the roots of f_1 and f_3 > (other than the trivial zero at the origin). > If we let t = e^{pi * i / 4}, then note that > f_1(x) = [sinh(tx) + sin(tx)] / 2t > ..and... > f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. > Thus, since t is a constant independent of x, > locating the roots of f_1 and f_3 is equivalent to > finding the solutions to sinh(z) = +/- sin(z). > [Apologies if part of my response duplicates part of > the galathaea's material, which I haven't read.] > Nontrivial solutions of sinh(z) = +/- sin(z) have the > form z = (+/- 1 +/- i) x. [Those signs are not > linked. Thus, for a given x, there is a solution in > each quadrant.] So let's consider just the first > quadrant solutions, z = (1 + i) x for positive x, > knowing that the others can be obtained by symmetry. > For sinh(z) = +/- sin(z), a simple approximation for > x is (n +/- 1/4) pi for positive integer n. [Those > signs _are_ linked, i.e., the same sign is to be > used in the approximation as in the equation.] > That simple approximation is merely the first term of > a series, and can be much improved by using just one > more term of the series: > For sinh(z) = +sin(z), x ~= (n + 1/4) pi + 1/2 (1 - > coth((n + 1/4) pi)); > for sinh(z) = -sin(z), x ~= (n - 1/4) pi + 1/2 (1 + > tanh((n - 1/4) pi)). (*) > Numerical example: > The solution z = (1 + i) x of sinh(z) = -sin(z) which > has x close to (4 - 1/4) pi has > x = 11.7809724510202275376960... > For comparison, > we get the approximation 11.7809724510202275376994... > by using (*). > As n increases, accuracy of the approximation > increases quickly. The worst approximation occurs when > we approximate the solution z = (1 + i) x of > sinh(z) = -sin(z) which has x close to (1 - 1/4) pi; > the true value of x is > 2.365020..., while (*) gives 2.365097... > Of course, we could get better approximations than > (*) by using more terms of the series. > David W. Cantrell > Kyle Czarnecki P.S. As narasimham alluded to, the functions f_1 and f_3 > in question arise as solutions to the differential > equation f^(4)(x) + f(x) = 0 ..which is satisfied by... f_k(x) = sum_{n = 0}^{oo} x^{4n + k} / (4n + k)!. I was looking for the roots of these functions because I > was curious to see what their infinite product > representations looked like. For instance, the first solution f_1(x) = 1 - x^4 / 4! + x^8 / 8! - ... = [cosh(xt) + cos(xt)] / 2 ..where t = e^{pi i / 4}, can be expressed as a product > over its roots as... f_1(x) = prod_{n = 0}^{oo} {1 - 4x^4 / [(2n + 1)^4 pi^4]} = [1 - 4x^4 / pi^4] * [1 - 4x^4 / (3 pi)^4] * ... Note the similarity to the infinite product for cosine > (which is obviously due to the nature of the D.E.'s these > function all satisfy). the 4 needs to be in the denominator the zeroes have periodicity 2 pi * ('/2)/2 = pi '/2 from w_8 = ('/2)/2 + ('/2)/2 i which it gets from the fourth root of the 4 all of the generalised trigonometrics can be expanded in products (by weierstrass) as they are all entire but (0,4) and (2,4) are some of the simplest due the rotated sum rules and 2 . 2 = 4 the simplest factorisation the critical values obey a discrete and differential structure much as they do in the simple case of trigonometrics but since most of the generalised forms do not have true periodicity (only asymptotic) the interaction with the product and sum rules is more involved the (1,4) and (3,4) forms you mention earlier in the thread illustrate how this affects product representation nicely -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) > On Jun 29, 2:28 am, Narcoleptic Insomniac > > Kyle Czarnecki > > P.S. As narasimham alluded to, the functions f_1 and f_3 > in question arise as solutions to the differential equation > > f^(4)(x) + f(x) = 0 > > ..which is satisfied by... > > f_k(x) = sum_{n = 0}^{oo} x^{4n + k} / (4n + k)!. > > I was looking for the roots of these functions because I > was curious to see what their infinite product > representations looked like. > > For instance, the first solution > > f_1(x) = 1 - x^4 / 4! + x^8 / 8! - ... > > = [cosh(xt) + cos(xt)] / 2 > > ..where t = e^{pi i / 4}, can be expressed as a product > over its roots as... > > f_1(x) = prod_{n = 0}^{oo} {1 - 4x^4 / [(2n + 1)^4 pi^4]} > > = [1 - 4x^4 / pi^4] * [1 - 4x^4 / (3 pi)^4] * ... > > Note the similarity to the infinite product for cosine > (which is obviously due to the nature of the D.E.'s these > function all satisfy). the 4 needs to be in the denominator No, I believe you are mistaken. The function f_1(x) has zeros of the form x = (2n + 1) pi / sqrt(+/- 2) ..for all n in Z. Thus, the infinte product is prod_{n = 0}^{oo} [1 - (2n + 1) pi / sqrt(2)] * [1 + (2n + 1) pi / sqrt(2)] * [1 - (2n + 1) pi / sqrt(-2)] * [1 + (2n + 1) pi / sqrt(-2)] ..which reduces to the product I posted earlier. > the zeroes have periodicity 2 pi * ('/2)/2 = pi '/2 > from w_8 = ('/2)/2 + ('/2)/2 i > which it gets from the fourth root of the 4 all of the generalised trigonometrics can be expanded > in products > (by weierstrass) > as they are all entire > but (0,4) and (2,4) are some of the simplest > due the rotated sum rules and 2 . 2 = 4 the simplest > factorisation the critical values obey a discrete and differential > structure > much as they do in the simple case of > of trigonometrics > but since most of the generalised forms do not have > true periodicity > (only asymptotic) > the interaction with the product and sum rules is > is more involved the (1,4) and (3,4) forms you mention earlier in the > thread > illustrate how this affects product representation > on nicely -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) <9471638.1214782052739.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) On Jun 29, 4:27 pm, Narcoleptic Insomniac > On Jun 29, 2:28 am, Narcoleptic Insomniac > Kyle Czarnecki > P.S. As narasimham alluded to, the functions f_1 and f_3 > in question arise as solutions to the differential equation > f^(4)(x) + f(x) = 0 > ..which is satisfied by... > f_k(x) = sum_{n = 0}^{oo} x^{4n + k} / (4n + k)!. > I was looking for the roots of these functions because I > was curious to see what their infinite product > representations looked like. > For instance, the first solution > f_1(x) = 1 - x^4 / 4! + x^8 / 8! - ... > = [cosh(xt) + cos(xt)] / 2 > ..where t = e^{pi i / 4}, can be expressed as a product > over its roots as... > f_1(x) = prod_{n = 0}^{oo} {1 - 4x^4 / [(2n + 1)^4 pi^4]} > = [1 - 4x^4 / pi^4] * [1 - 4x^4 / (3 pi)^4] * ... > Note the similarity to the infinite product for cosine > (which is obviously due to the nature of the D.E.'s these > function all satisfy). > the 4 needs to be in the denominator No, I believe you are mistaken. The function f_1(x) has zeros of the form x = (2n + 1) pi / sqrt(+/- 2) ..for all n in Z. Thus, the infinte product is prod_{n = 0}^{oo} [1 - (2n + 1) pi / sqrt(2)] * [1 + (2n + 1) pi / sqrt(2)] * > [1 - (2n + 1) pi / sqrt(-2)] * [1 + (2n + 1) pi / sqrt(-2)] ..which reduces to the product I posted earlier. i'm sorry you are right you still have the right period ('/2 pi) the projection you are looking at with f_1(x) is Re{1/4(e^(w_8 x) + e^(w_8^3 x) + e^(w_8^5 x) + e^(w_8^7 x)} which gives cos(x / '/2) ch(x / '/2) (which can also be calculated using the product and rotation rules) so you get zeroes at x / '/2 = pi (2n + 1)/2 or x = pi '/2 (2n + 1) / 2 = pi (2n + 1) / '/2 like you derived it's 2_t_4(x) which has the expansion i was thinking about again sorry > the zeroes have periodicity 2 pi * ('/2)/2 = pi '/2 > from w_8 = ('/2)/2 + ('/2)/2 i > which it gets from the fourth root of the 4 > all of the generalised trigonometrics can be expanded > in products > (by weierstrass) > as they are all entire > but (0,4) and (2,4) are some of the simplest > due the rotated sum rules and 2 . 2 = 4 the simplest > factorisation > the critical values obey a discrete and differential > structure > much as they do in the simple case of > of trigonometrics > but since most of the generalised forms do not have > true periodicity > (only asymptotic) > the interaction with the product and sum rules is > is more involved > the (1,4) and (3,4) forms you mention earlier in the > thread > illustrate how this affects product representation > on nicely -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) <9471638.1214782052739.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) the rotation formulae for the generalised trigonmetrics gives m t (w x) = w t (x) m n n n m n so if x is a zero of m_t_n 2 n-1 so is w x, w x, ..., w x n n n and all generalised trigonometrics have product expansions oo n --- / / x m | | | 1 - | -- | | x | | z / / j=0 j the z_j may obey some transcendental relations though for the generalised coshinusi as an example taking real parts as i showed in the first link gives n-1 2j+ 1 --- w x 1 ( 2n ) - / Re < e > = n --- ( ) j=0 n-1 2j+1 2j+1 --- Re{w } x Im{w } i x 1 ( 2n 2n ) - / Re < e e > = n --- ( ) j=0 n-1 / (2j + 1) pi / (2j + 1) pi --- cos | ----------- | x sin | ----------- | x 1 2n / ( 2n / ) - / e Re < e > = n --- ( ) j=0 n-1 / (2j + 1) pi --- cos | ----------- | x 1 2n / / / (2j + 1) pi - / e cos | sin | ----------- | x | n --- 2n / / j=0 which gives the function to invert for zeroes this product expansion has the formal series oo oo oo --- 1 --- --- 1 / -- n / ----- 2n 1 - | / n | x + | / / n n | x - ... --- z / --- --- z z / j=0 j j=0 k=0 j k so we have the formal identities oo --- 1 -- 1 / n = ---- --- z (1) j=0 j n and so on for the simple case of cosine where z_j is (2j+1) pi / 2 this gives oo --- 4 1 / -------------- = - --- 2 2 j=0 ((2j + 1) pi) or oo --- 2 1 pi / --------- = --- --- n 8 j=0 (2j + 1) a classical little sum the generalisations show the deeper structure of zeroes here -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) <9471638.1214782052739.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) > the rotation formulae for the generalised trigonmetrics gives m > t (w x) = w t (x) > m n n n m n so if x is a zero of m_t_n 2 n-1 > so is w x, w x, ..., w x > n n n and all generalised trigonometrics have product expansions oo n > --- / / x > m | | | 1 - | -- | | > x | | z / / > j=0 j the z_j may obey some transcendental relations though for the generalised coshinusi > as an example > taking real parts as i showed in the first link > gives n-1 2j+ 1 > --- w x > 1 ( 2n ) > - / Re < e > = > n --- ( ) > j=0 n-1 2j+1 2j+1 > --- Re{w } x Im{w } i x > 1 ( 2n 2n ) > - / Re < e e > = > n --- ( ) > j=0 n-1 / (2j + 1) pi / (2j + 1) pi > --- cos | ----------- | x sin | ----------- | x > 1 2n / ( 2n / ) > - / e Re < e > = > n --- ( ) > j=0 n-1 / (2j + 1) pi > --- cos | ----------- | x > 1 2n / / / (2j + 1) pi > - / e cos | sin | ----------- | x | > n --- 2n / / > j=0 which gives the function to invert for zeroes this product expansion has the formal series oo oo oo > --- 1 --- --- 1 > / -- n / ----- 2n > 1 - | / n | x + | / / n n | x - ... > --- z / --- --- z z / > j=0 j j=0 k=0 j k so we have the formal identities oo > --- 1 > -- 1 > / n = ---- > --- z (1) > j=0 j n and so on for the simple case of cosine where z_j is (2j+1) pi / 2 > this gives oo > --- > 4 1 > / -------------- = - > --- 2 2 > j=0 ((2j + 1) pi) or oo > --- 2 > 1 pi > / --------- = --- > --- n 8 > j=0 (2j + 1) a classical little sum the generalisations show the deeper structure of zeroes here and just to show an application one can start with the first f_1 of the thread taking the real parts and simplifying eventually shows that the zeroes z_j obeys th(('/2)/2 z ) = tan(-('/2)/2 z ) j j where j e Z and i will choose z_0 = 0 this is a transcendental equation then f_1 has the product form --- / / x 4 x | | | 1 - | - | | | | z / / j e Z j j =/= 0 now as a set it can be shown that --- 1 1 1 / -- = ---- = --- --- 4 (1) 120 j e Z z 5 j =/= 0 j and there are a number of ways of getting explicit representations of particular roots (i've shown the lagrange inversion method before) however the zeroes are respected by the sum rule which states (and here i use my t notation again): t (a + b) = t (a) t (b) + t (a) t (b) + 1 4 0 4 1 4 1 4 0 4 t (a) t (b) + t (a) t (b) 2 4 3 4 3 4 2 4 which shows why periodicity breaks (see a and b zeroes of 1_t_4 and leftover terms) this why despite that all m_t_4 obey the same differential equation (D^4 t = -t) some have relatively simple zero structure (periodic) and others do not it eventually comes down to the sum and product structure and the way the theory factors looking at the generalised polynomial representation (taking out the 1/4 as a distraction) -1 i -i (x + x ) (x + x ) = 1+i 1-i -1+i -1-i x + x + x + x w 8 and scaling shows the relationship to x and how the roots to the original problem can be extracted from the quadratic extension there is no such simple extension that factors i -1 -i x - i x - x + i x the discrete structure (product and sum rules) play the same role here as they do in the trigonometric sums and their relation to their period structure -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) > > In my musings I've come across the functions > > f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + 1)! > > = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... > > ..and... > > f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + 3)! > > = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... > > I am trying to locate all the roots of f_1 and f_3 (other > than the trivial zero at the origin). > > If we let t = e^{pi * i / 4}, then note that > > f_1(x) = [sinh(tx) + sin(tx)] / 2t > > ..and... > > f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. > > Thus, since t is a constant independent of x, locating > the roots of f_1 and f_3 is equivalent to finding the > solutions to sinh(z) = +/- sin(z). [Apologies if part of my response duplicates part of the galathaea's > material, which I haven't read.] Nontrivial solutions of sinh(z) = +/- sin(z) have the form > z = (+/- 1 +/- i) x. [Those signs are not linked. Thus, for a given x, > there is a solution in each quadrant.] So let's consider just the first > quadrant solutions, z = (1 + i) x for positive x, knowing that the others > can be obtained by symmetry. For sinh(z) = +/- sin(z), a simple approximation for x is (n +/- 1/4) pi > for positive integer n. [Those signs _are_ linked, i.e., the same sign is > to be used in the approximation as in the equation.] That simple approximation is merely the first term of a series, and can be > much improved by using just one more term of the series: For sinh(z) = +sin(z), x ~= (n + 1/4) pi + 1/2 (1 - coth((n + 1/4) pi)); > for sinh(z) = -sin(z), x ~= (n - 1/4) pi + 1/2 (1 + tanh((n - 1/4) pi)). (*) Numerical example: > The solution z = (1 + i) x of sinh(z) = -sin(z) which has x close > to (4 - 1/4) pi has x = 11.7809724510202275376960... For comparison, > we get the approximation 11.7809724510202275376994... by using (*). As n increases, accuracy of the approximation increases quickly. The worst > approximation occurs when we approximate the solution z = (1 + i) x of > sinh(z) = -sin(z) which has x close to (1 - 1/4) pi; the true value of x is > 2.365020..., while (*) gives 2.365097... Of course, we could get better approximations than (*) by using more terms > of the series. David W. Cantrell A notation I sometimes use when there are several sign options, some linked and others unlinked is to use constants s1, s2, ... , each representing ambiguously +1 or -1, multiplied by terms of optional sign to represent the sign choices, with linked signs having the same index. Thus a +/- b +/- c becomes a + s1*b + s1*c when the signs are to be linked, or becomes a + s1*b + s2*c for unlinked signs. === Subject: Re: -- Solutions to sinh(z) = +/- sin(z) posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) On Jun 27, 11:31 pm, Narcoleptic Insomniac In my musings I've come across the functions f_1(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 1} / (4n + 1)! = x - x^5 / 5! + x^9 / 9! - x^13 / 13! + ... ..and... f_3(x) = sum_{n = 0}^{oo} (-1)^n x^{4n + 3} / (4n + 3)! = x^3 / 3! - x^7 / 7! + x^11 / 11! - ... I am trying to locate all the roots of f_1 and f_3 (other > than the trivial zero at the origin). If we let t = e^{pi * i / 4}, then note that f_1(x) = [sinh(tx) + sin(tx)] / 2t ..and... f_3(x) = [sinh(tx) - sin(tx)] / 2t^3. Thus, since t is a constant independent of x, locating > the roots of f_1 and f_3 is equivalent to finding the > solutions to sinh(z) = +/- sin(z). Any help and/or references would be greatly appreciated. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Exciting new maths book posting-account=dSoZsgoAAAA9MM8lveB-SgYYmRSlbBhg SV1),gzip(gfe),gzip(gfe) rhwh-ip-ccache-1 (NetCache NetApp/6.1.1D2) There is an exciting new maths book that is a must have for all people doing mathematics around the world.it is a book that deals with the mind set and attitude one must have in order to excel in maths.You can find more about 'THE MATHEMATICAL GENIUS IN YOU' at www.maths-genius.co.za === Subject: Happy 2pi Day! 2pi = 6.283185307179586476925286766559... 2pi shows up in many places in math and science: * The formula C = 2pi r * Converting between fractions of a revolution and radians * Converting between frequency and angular frequency * Fourier transforms * Any physics involving waves * The formula for the complex roots of unity * The Gaussian distribution * Stirling's approximation Historical accident has given us a symbol for 3.14..., but the number twice as large has a much better claim to a place among the most important mathematical constants. -- Jim E. Black (domain in headers) How to filter out stupid arguments in 40tude Dialog: [X] Watch/Ignore works on subthreads === Subject: Re: Happy 2pi Day! Seeing were all going a little bonkers here, what about the terrorist code--- pi-(1/(pi+1)) ~ 2.9 2.9 * pi ~ 9.11 ;-) Dan === Subject: Re: Happy 2pi Day! posting-account=3ZM9rAoAAABKWkxghcKc_J4YCCsXNXXQ rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) See thats whats wrong with pi. A good pi is a 2*pi :-) Jay Bala. > Seeing were all going a little bonkers here, > what about the terrorist code--- pi-(1/(pi+1)) ~ 2.9 > 2.9 * pi ~ 9.11 ;-) Dan === Subject: Re: Happy 2pi Day! 2pi Day is over, and I missed it? Damn! I guess I'll have to send out Sorry I forgot 2pi Day cards again. -- Angus Rodgers Contains mild peril === Subject: Re: Happy 2pi Day! posting-account=3ZM9rAoAAABKWkxghcKc_J4YCCsXNXXQ rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Another day is on the rise and it has 2*pi in it. You didn't miss a thing. Jay Bala. > 2pi Day is over, and I missed it? Damn! I guess I'll have to send > out Sorry I forgot 2pi Day cards again. > -- > Angus Rodgers > Contains mild peril === Subject: Fine structure constant, a mathematical definition (was: Re: Happy 2pi Day!) posting-account=3ZM9rAoAAABKWkxghcKc_J4YCCsXNXXQ rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) In fact, a while back I showed that, alpha, the fine structure constant is a function of 2pi; which in turn part of golden ratio: Results: alpha=7.29 735 257 240 051 31... x10^-3 1/alpha=137.035 999 025 471 68... I think 2pi is more universal than pi. Jay Bala. === Subject: Re: Happy 2pi Day! posting-account=3ZM9rAoAAABKWkxghcKc_J4YCCsXNXXQ rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Everyday is a 2pi day. 2*pi*10/Phi^2= 23.9996 hours :-) Jay Bala > 2pi = 6.283185307179586476925286766559... 2pi shows up in many places in math and science: * The formula C = 2pi r > * Converting between fractions of a revolution and radians > * Converting between frequency and angular frequency > * Fourier transforms > * Any physics involving waves > * The formula for the complex roots of unity > * The Gaussian distribution > * Stirling's approximation Historical accident has given us a symbol for 3.14..., > but the number twice as large has a much better claim > to a place among the most important mathematical constants. -- > Jim E. Black (domain in headers) > How to filter out stupid arguments in 40tude Dialog: > [X] Watch/Ignore works on subthread === Subject: Re: Happy 2pi Day! posting-account=qL0m4woAAAAXbWR0iVY3aIl0MbFVyU9L Gecko/2008061015 Firefox/3.0,gzip(gfe),gzip(gfe) > Everyday is a 2pi day. 2*pi*10/Phi^2= 23.9996 hours :-) Excuse me. What's the Phi here? Jay Bala > 2pi = 6.283185307179586476925286766559... > 2pi shows up in many places in math and science: > * The formula C = 2pi r > * Converting between fractions of a revolution and radians > * Converting between frequency and angular frequency > * Fourier transforms > * Any physics involving waves > * The formula for the complex roots of unity > * The Gaussian distribution > * Stirling's approximation > Historical accident has given us a symbol for 3.14..., > but the number twice as large has a much better claim > to a place among the most important mathematical constants. > -- > Jim E. Black æ æ(domain in headers) > How to filter out stupid arguments in 40tude Dialog: > æ [X] Watch/Ignore works on subthread === Subject: Re: Happy 2pi Day! posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > 2*pi*10/Phi^2= 23.9996 hours :-) Excuse me. What's the Phi here? > ***************************************************** I aksed myself the same question, and then I answered myself: beats me! Nevertheless, after doing 20*pi/23.9996, I recognized it: Phi = the golden mean = [1 + Sqrt(5)]/2 Tonio === Subject: Re: Happy 2pi Day! posting-account=qL0m4woAAAAXbWR0iVY3aIl0MbFVyU9L Gecko/2008061015 Firefox/3.0,gzip(gfe),gzip(gfe) > 2*pi*10/Phi^2= 23.9996 hours :-) > Excuse me. What's the Phi here? ***************************************************** I aksed myself the same question, and then I answered myself: beats > me! Nevertheless, after doing 20*pi/23.9996, I recognized it: Phi = the golden mean = [1 + Sqrt(5)]/2 Tonio amazing numbers... === Subject: Re: Happy 2pi Day! posting-account=qL0m4woAAAAXbWR0iVY3aIl0MbFVyU9L Gecko/2008061015 Firefox/3.0,gzip(gfe),gzip(gfe) > 2pi = 6.283185307179586476925286766559... 2pi shows up in many places in math and science: * The formula C = 2pi r > * Converting between fractions of a revolution and radians > * Converting between frequency and angular frequency > * Fourier transforms > * Any physics involving waves > * The formula for the complex roots of unity > * The Gaussian distribution > * Stirling's approximation Historical accident has given us a symbol for 3.14..., > but the number twice as large has a much better claim > to a place among the most important mathematical constants. -- > Jim E. Black æ æ(domain in headers) > How to filter out stupid arguments in 40tude Dialog: > æ [X] Watch/Ignore works on subthreads How interesting 2pi day sounds! === Subject: Re: Happy 2pi Day! posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) > 2pi = 6.283185307179586476925286766559... > 2pi shows up in many places in math and science: > * The formula C = 2pi r > * Converting between fractions of a revolution and radians > * Converting between frequency and angular frequency > * Fourier transforms > * Any physics involving waves > * The formula for the complex roots of unity > * The Gaussian distribution > * Stirling's approximation > Historical accident has given us a symbol for 3.14..., > but the number twice as large has a much better claim > to a place among the most important mathematical constants. > -- > Jim E. Black æ æ(domain in headers) > How to filter out stupid arguments in 40tude Dialog: > æ [X] Watch/Ignore works on subthreads How interesting 2pi day sounds!- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - N.94 Hao, would you mind putting it into chinese for me? A lan === Subject: Re: Happy 2pi Day! posting-account=qL0m4woAAAAXbWR0iVY3aIl0MbFVyU9L Gecko/2008061015 Firefox/3.0,gzip(gfe),gzip(gfe) > 2pi = 6.283185307179586476925286766559... > 2pi shows up in many places in math and science: > * The formula C = 2pi r > * Converting between fractions of a revolution and radians > * Converting between frequency and angular frequency > * Fourier transforms > * Any physics involving waves > * The formula for the complex roots of unity > * The Gaussian distribution > * Stirling's approximation > Historical accident has given us a symbol for 3.14..., > but the number twice as large has a much better claim > to a place among the most important mathematical constants. > -- > Jim E. Black æ æ(domain in headers) > How to filter out stupid arguments in 40tude Dialog: > æ [X] Watch/Ignore works on subthreads > How interesting 2pi day sounds!- Masquer le texte des messages pr.8ec.8edents - > - Afficher le texte des messages pr.8ec.8edents - N.94 Hao, would you mind putting it into chinese for me? A lan Sorry, did you mean translating all Jim said into Chinese? And I'm afraid maybe Chinese is not proper for everyone here, since it is a English speaking newsgroup. === Subject: 2402 Solutions manuals posting-account=Tr_lEAoAAAADSBji3TkK41k7g4TuE_dn SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; InfoPath.2; Media Center PC 5.0),gzip(gfe),gzip(gfe) My List of Solutions Manual contact me to : mattosbw1@gmail.com mattosbw1(at)gmail.com ot to : newbergh123@yahoo.com newbergh123(at)yahoo.com .... try with both emails . If your wanted solutions manual is not in this list, also can ask me if is available . These are some only. 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Jiji) + original Ebook Fundamentals of Momentum, Heat and Mass Transfer (5th Ed., Welty) Analytical Methods for Heat Transfer and Fluid Flow Problems (Bernhard Weigand) Two-Phase Flow: Theory and Applications (Clement Kleinstreuer) Heat Tranfer (Rao) Convective Heat Transfer (kakac) An Introduction to Mass and Heat Transfer: Principles of Analysis and Design (Stanley Middleman) Fundamentals of Thermodynamics (5th Ed., Richard E. Sonntag, Claus Borgnakke & Gordon J. Van Wylen) Fundamentals of Thermodynamics (6th Ed., Richard E. Sonntag, Claus Borgnakke & Gordon J. Van Wylen) Introduction to Engineering Thermodynamics (1st Ed., Richard E. Sonntag & Claus Borgnakke) Introduction to Engineering Thermodynamics (2nd Ed., Richard E. Sonntag & Claus Borgnakke) Fundamentals of Engineering Thermodynamics, 5th Ed (Michael J. Moran, Howard N. Shapiro) + original Ebook Fundamentals of Engineering Thermodynamics, 6th Ed (Michael J. Moran, Howard N. 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Craig) Simplified Mechanics and Strength of Materials (6th Ed., James Ambrose) Engineering Applications of Dynamics (Dean C. Karnopp & Donald L. Margolis) Engineering Mechanics - Statics, 5th Ed (J. L. Meriam, L. G. Kraige) + Ebook Engineering Mechanics - Statics, 5th Ed SI Version (J. L. Meriam, L. G. Kraige) Engineering Mechanics - Statics, 6th Ed (J. L. Meriam, L. G. Kraige) Engineering Mechanics - Statics, 6th Ed SI Version (J. L. Meriam, L. G. Kraige) Engineering Mechanics - Dynamics, 5th Ed (J. L. Meriam, L. G. Kraige) Engineering Mechanics - Dynamics, 5th Ed SI Version (J. L. Meriam, L. G. Kraige) Engineering Mechanics - Dynamics, 6th Ed (J. L. Meriam, L. G. Kraige) Vector Mechanics for Engineers: Statics (7th Ed., Ferdinand P. Beer) Vector Mechanics for Engineers: Statics (8th Ed., Ferdinand P. Beer) Vector Mechanics for Engineers: Dynamics (7th Ed., Ferdinand P. Beer) Vector Mechanics for Engineers: Dynamics (8th Ed., Ferdinand P. 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Hosford) + original Ebook Mechanical Behavior of Materials (Keith Bowman) Theory and Design for Mechanical Measurements (4th Ed, Figliola & Beasley) Mechanical Measurements (6th Ed., Beckwith, Marangoni & Lienhard) Measurement and Data Analysis for Engineering and Science (Patrick F Dunn) Design and Analysis of Lean Production Systems (Askin & Goldberg) Work Systems: The Methods, Measurement & Management of Work (Mikell P. Groover) Automation, Production Systems, and Computer-Integrated Manufacturing (2nd Ed., Groover) Automation, Production Systems, and Computer-Integrated Manufacturing (3rd Ed., Groover) Fundamentals of Modern Manufacturing: Materials, Processes, and Systems (3rd Ed., Mikell P. Groover) Materials and Processes in Manufacturing (9th Ed., E. Paul DeGarmo, J. T. Black, Ronald A. Kohser) DeGarmo's Materials and Processes in Manufacturing (10th Ed., E. Paul DeGarmo, J. T. Black, Ronald A. 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Bolton) Wind Energy Explained : Theory, Design and Application (Manwell, McGowan & Rogers) Fundamentals of Renewable Energy Processes (Aldo da Rosa) + original Ebook Renewable Energy (3rd Ed., SÀrensen or Sorensen) + original Ebook Energy Technology and Directions for the Future (Fanchi) Power Generation Technologies (Paul Breeze) + original Ebook Concepts in Engineering (Holtzapple & Reece) Foundations of Engineering (2nd Ed, Holtzapple & Dan Reece) Industrial Mechanics and Maintenance (3rd Ed., Larry Chastain) Mechanical & Electrical Systems in Buildings (4th Ed., Richard Janis & William Tao) Autodesk Inventor (James M. Leake) Energy and the Environment (2nd Ed, Robert A. Ristinen & Jack P. Kraushaar) Orthopaedic Biomechanics: Mechanics and Design in Musculoskeletal Systems (Donald L. Bartel, Dwight T. Davy & Tony M. Keaveny) Science for Engineering (3rd Ed., John Bird) - Electrical, Electronics & Computer Engineering Design for Electrical and Computer Engineers (J. 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Malan) Java Made Simple (2nd Ed., McBride) Optical Networks: A Practical Perspective (2nd Ed., Ramaswami & Sivarajan) Usability Engineering: Scenario-Based Development of Human-Computer Interaction (Rosson & Carroll) The Designer's Guide to VHDL (2nd Ed., Peter Ashenden) Transactional Information Systems: Theory, Algorithms, and the Practice of Concurrency Control and Recovery (Weikum & Vossen) Parallel Computer Architecture: A Hardware/Software Approach (Culler, Singh & Gupta) Advanced Compiler Design and Implementation (Steven Muchnick) Parallel Programming with MPI (Peter Pacheco) Distributed Algorithms (Nancy Lynch) Digital Signal Processing: Fundamentals and Applications (Li Tan) Electrical and Electronic Principles and Technology (3rd Ed., John Bird) Electrical Circuit Theory and Technology (3rd Ed., John Bird) Electronic Circuits: Fundamentals & Applications (3rd Ed., Mike Tooley) Multidimensional Signal, Image, and Video Processing and Coding (John Woods) Bioelectrical Signal Processing in Cardiac and Neurological Applications (Leif S.9arnmo & Pablo Laguna) Foundations of Analog and Digital Electronic Circuits (Anant Agarwal & Jeffrey Lang) (Luis Moura & Izzat Darwazeh) Embedded Systems Architecture: A Comprehensive Guide for Engineers and Programmers (Tammy Noergaard) Bioimpedance and Bioelectricity Basics (2nd Ed., Grimnes & Martinsen) Simulation Modeling and Analysis with ARENA (Tayfur Altiok & Benjamin Melamed) The Visual Story: Creating the Visual Structure of Film, TV and Digital Media (2nd Ed., Bruce Block) The Shut Up and Shoot Documentary Guide: A Down & Dirty DV Production (Anthony Artis) Portable Video: ENG & EFP (5th Ed., Medoff & Fink) Voice and Vision: A Creative Approach to Narrative Film and DV Production (Mick Hurbis-Cherrier) Writing for Multimedia and the Web: A Practical Guide to Content Development for Interactive Media (3rd Ed., Timothy Garrand) Developing and Maintaining a Design-Tech Portfolio: A Guide for Theatre, Film & TV (Rafael Jaen) Producing for TV and Video: A Real-World Approach (Cathrine Kellison) Placing Shadows: Lighting Techniques for Video Production (3rd Ed., Gloman & Tom LeTourneau) Film Directing Fundamentals: See Your Film Before Shooting (2nd Ed., Nicholas Proferes) Introduction to Media Production: The Path to Digital Media Production (3rd Ed., Musburger & Kindem) Directing the Documentary (4th Ed., Michael Rabiger) Breslin) Prepare to Board! Creating Story and Characters for Animation Features and Shorts (Nancy Beiman) Light and Lens: Photography in the Digital Age (Robert Hirsch) The Radio Station: Broadcast, Satellite & Internet (7th Ed., Michael Keith) Developing Story Ideas (2nd Ed., Michael Rabiger) Radio Production Worktext: Studio and Equipment (5th Ed., Reese, Gross & Gross) Broadcast News Writing, Reporting, and Producing (4th Ed., Ted White) Problem Solving and Programming Concepts (7th Ed., Sprankle) Problem Solving and Programming Concepts (8th Ed., Sprankle & Hubbard) A Balanced Introduction to Computer Science (2nd Ed., David Reed) Introduction to Computing and Programming with Java: A Multimedia Approach (Guzdial & Ericson) Starting Out with Programming Logic and Design (Tony Gaddis) Tools For Structured and Object-Oriented Design (7th Ed., Bohl & Rynn) Programming with Alice and Java (Lewis & DePasquale) C How to Program (4th Ed., Harvey & Paul Deitel) C How to Program (5th Ed., Harvey & Paul Deitel) C++ How to Program (6th Ed., Harvey & Paul Deitel) Visual C++ 2008 How to Program (2nd Ed., Harvey & Paul Deitel) Internet & World Wide Web: How to Program (4th Ed., Harvey & Paul Deitel) Web Technologies: A Computer Science Perspective (Jeffrey C. Jackson) Mastering the Internet, XHTML and JavaScript (2nd Ed., Ibrahim Zeid) Weaving a Website: Programming in HTML, Java Script, Perl and Java (Susan Anderson-Freed) Simply C++: An Application-Driven Tutorial Approach (Harvey & Paul Deitel) Visual C# 2005 How to Program (2nd Ed., Harvey & Paul Deitel) Simply C#: An Application-Driven Tutorial Approach (Harvey & Paul Deitel, Hoey & Yaeger) Java: Introduction to Problem Solving and Programming (5th Ed., Savitch & Carrano) Introduction to Computing and Programming with Java: A Multimedia Approach (Guzdial & Ericson) Java How to Program (7th Ed., Harvey & Paul Deitel) Java For Students (5th Ed., Bell & Parr) Java, Java, Java, Object-Oriented Problem Solving (3rd Ed., Morelli & Walde) Java: An Eventful Approach (Bruce, Danyluk & Murtagh) Introduction to Java Programming with JBuilder (3rd Ed., Y. Daniel Liang) Starting Out with Visual Basic 2008 (4th Ed., Gaddis & Irvine) Starting Out with Python (Tony Gaddis) Object-Oriented Programming in Python (Goldwasser & Letscher) Introduction to MathCAD 11 (Ronald W. Larsen) Introduction to MathCAD 13 (2nd Ed., Ronald W. Larsen) MatLAB Programming (David Kuncicky) Introduction to Maple 8 (David Schwartz) Introduction to FORTRAN 90 (2nd Ed., Larry R. Nyhoff & Sanford Leestma) Introduction to Java (Stephen J. Chapman) Java Software Solutions for AP Computer Science A (2nd Ed., Lewis, Loftus & Cocking) Business Data Networks and Telecommunications (6th Ed., Raymond R. Panko) Business Data Networks and Telecommunications (7th Ed., Raymond R. Panko) Business Data Communications (Allen Dooley) Object-Oriented Programming in C++ (4th Ed., Robert Lafore) C++: Classes and Data Structures (Jeffrey Childs) Data Structures Outside-In with Java (Sesh Venugopal) Data Structures and Abstractions with Java (2nd Ed., Frank M. 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Hadjidakis) Managing Cultural Differences: Global Leadership Strategies for the 21st Century (7th Ed., Moran, Harris & Moran) Marketing Plans: How to prepare them, how to use them (6th Ed., Malcolm McDonald) Human Resource Management for the Hospitality and Tourism Industries (Dennis Nickson) Exploring Employee Relations: An International Approach (2nd Ed., Mike Leat) Introduction to Financial Technology (Roy Freedman) Managing Innovation: New Technology, New Products, and New Services in a Global Economy (2nd Ed., John Ettlie) Enterprise: Entrepreneurship and Innovation: Concepts, Contexts and Commercialization (Lowe & Marriott) Strategic Management: Global Cultural Perspectives for Profit and Non- Profit Organizations (Katsioloudes) Introduction to e-Business: Management and Strategy (Colin Combe) Managing People and Organizations in Changing Contexts (Graeme Martin) Sustainable Tourism (David Weaver) Knowledge Management in Theory and Practice (Kimiz Dalkir) Introduction to Human Resource Management (2nd Ed., John Stredwick) Organisations and the Business Environment (2nd Ed., David Campbell & Tom Craig) Strategic Operations Management (2nd Ed., Brown, Lamming, Bessant & Jones) Worldwide Destinations: The geography of travel and tourism (4th Ed., Boniface & Cooper) Strategic Marketing Management: Planning, Implementation and Control (3rd Ed., Wilson & Gilligan) Marketing Communications Management: Concepts and Theories, Cases and Practices (Paul Copley) Principles of Financial Engineering (Salih Neftci) Managing Finance: A Socially Responsible Approach (D. 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Bhadeshia & Honeycombe) Theory of Plasticity (3rd Ed. Jagabanduhu Chakrabarty) Engineering Materials Vol. 1 : An Introduction to Properties, Applications and Design (3rd Ed., Michael Ashby & David R H Jones) Engineering Materials Vol. 2 : An Introduction to Microstructures, Processing and Design (3rd Ed., Michael Ashby & David R H Jones) Plastics: Microstructure and Engineering Applications (3rd Ed., Nigel Mills) Laser Processing of Engineering Materials : Principles, Procedure and Industrial Application (John Ion) Sintering : Densification, Grain Growth and Microstructure (Suk-Joong Kang) Biomaterials Science : An Introduction to Materials in Medicine (2nd Ed., Buddy D. Ratner, Allan Hoffman, Frederick Schoen & Jack Lemons) Fundamentals of Solid State Engineering (2nd Ed., Manijeh Razeghi) Smart Electronic Materials: Fundamentals and Applications (Jasprit Singh) Laminar Composites (by George Staab) + original Ebook Physical Metallurgy and Advanced Materials (7th Ed., R E Smallman & A.H.W. Ngan) contact me to : newbergh123@yahoo.com or mattosbw1@gmail.com === say also you are not very sure about it. Say, a significant error-not typos. You can: 1) ask in a forum like this. The problem is that the context can be uninteresting for third parties or too complicated to explain === > say also you are not very sure about it. Say, a significant error-not typos. > You can: > 1) ask in a forum like this. The problem is that the context can be uninteresting for third parties or too complicated to explain I think (2) is a good way to go. Actually I have been on the receiving end of these kinds of emails. If you ask politely, hopefully the author will be willing to engage you in an email discussion. But excuse any disconcerted replies he might first give you. === Subject: Re: Application for membership > Ah, but now since he is advertising his group for money/commericial What does being banned from sci.math mount to? Recall also that volume, not content. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, daruber m.9fss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophics === Subject: Re: Application for membership > Welcome to Basti Newsgroup: Sci.math The topic of sci.math is not yogic colon cleansing. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, daruber m.9fss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophics === Subject: Formula sought (enumerative combinatorics) Formula sought For the below input parameters v and k, v>=k, what would be the equation that gives b ? FYI: these parameters and results are integer values from the field of enumerative combinatorics. The b below is the result of a complete enumeration for the given v,k. What is needed is to know in advance how much b will be for the given v,k without enumerating the whole set. Can one formulate an equation for this? v k b ----------------- 1 1 1 2 1 2 3 1 3 2 2 1 3 2 4 4 2 9 5 2 16 3 3 1 4 3 9 5 3 28 6 3 65 7 3 126 4 4 1 5 4 21 6 4 90 7 4 268 8 4 640 9 4 1314 5 5 1 6 5 52 7 5 298 8 5 1123 9 5 3278 10 5 7995 11 5 17104 6 6 1 7 6 136 8 6 1016 9 6 4783 10 6 16941 11 6 48895 12 6 121171 13 6 267369 7 7 1 8 7 379 9 7 3575 10 7 20720 11 7 88393 12 7 300683 13 7 861325 14 7 2160889 15 7 4884281 8 8 1 9 8 1126 10 8 13023 11 8 91412 12 8 465938 13 8 1860121 14 8 6144894 15 8 17505487 16 8 44343520 17 8 102177477 ... === Subject: Re: Formula sought (enumerative combinatorics) posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008061712 Fedora/3.0-1.fc9 Firefox/3.0,gzip(gfe),gzip(gfe) > Formula sought For the below input parameters v and k, v>=k, > what would be the equation that gives b ? FYI: these parameters and results are integer values > from the field of enumerative combinatorics. > The b below is the result of a complete enumeration > for the given v,k. > What is needed is to know in advance how much b will be > for the given v,k without enumerating the whole set. > Can one formulate an equation for this? Why don't you tell us *what* you are enumerating? The chances of coming up with something sensible will then be enormously higher... -- m === Subject: Re: Formula sought (enumerative combinatorics) > > FYI: these parameters and results are integer values > from the field of enumerative combinatorics. > The b below is the result of a complete enumeration > for the given v,k. Why don't you tell us *what* you are enumerating? > The chances of coming up with something sensible > will then be enormously higher... Sorry, there was a bug in the program, so the previously posted table is wrong. Forget the posted table please, I'll need to compile a fixed one. I'm writing program code for enumerating sets, like permutation sets and combination sets; both with and without repetetions etc. And somehow I'm getting a different number of blocks for the 'permutations without repetetion' case. For example for v=5,k=3 it is said in the literature that this would give 60 blocks, but I get only 27 blocks. What's missing or erronous in this list? Isn't the following a permutations without repetetions? Or is this a different animal? (the first column is just the block#) v=5, k=3, b=27 0: 0 1 2 1: 0 1 3 2: 0 1 4 3: 0 2 1 4: 0 2 3 5: 0 2 4 6: 0 3 1 7: 0 3 2 8: 0 3 4 9: 1 0 2 10: 1 0 3 11: 1 0 4 12: 1 2 0 13: 1 2 3 14: 1 2 4 15: 1 3 0 16: 1 3 2 17: 1 3 4 18: 2 0 1 19: 2 0 3 20: 2 0 4 21: 2 1 0 22: 2 1 3 23: 2 1 4 24: 2 3 0 25: 2 3 1 26: 2 3 4 === Subject: Re: Formula sought (enumerative combinatorics) posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 FunWebProducts; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > FYI: these parameters and results are integer values > from the field of enumerative combinatorics. > The b below is the result of a complete enumeration > for the given v,k. > Why don't you tell us *what* you are enumerating? > The chances of coming up with something sensible > will then be enormously higher... Sorry, there was a bug in the program, so the previously posted table is wrong. > Forget the posted table please, I'll need to compile a fixed one. I'm writing program code for enumerating sets, like permutation sets > and combination sets; both with and without repetetions etc. > And somehow I'm getting a different number of blocks > for the 'permutations without repetetion' case. > For example for v=5,k=3 it is said in the literature that this would > give 60 blocks, but I get only 27 blocks. > What's missing or erronous in this list? > Isn't the following a permutations without repetetions? > Or is this a different animal? > (the first column is just the block#) v=5, k=3, b=27 > 0: 0 1 2 > 1: 0 1 3 > 2: 0 1 4 > 3: 0 2 1 > 4: 0 2 3 > 5: 0 2 4 > 6: 0 3 1 > 7: 0 3 2 > 8: 0 3 4 > 9: 1 0 2 > 10: 1 0 3 > 11: 1 0 4 > 12: 1 2 0 > 13: 1 2 3 > 14: 1 2 4 > 15: 1 3 0 > 16: 1 3 2 > 17: 1 3 4 > 18: 2 0 1 > 19: 2 0 3 > 20: 2 0 4 > 21: 2 1 0 > 22: 2 1 3 > 23: 2 1 4 > 24: 2 3 0 > 25: 2 3 1 > 26: 2 3 4 Where are the permutations that begin with 3 and those that begin with 4. === Subject: Re: Formula sought (enumerative combinatorics) Where are the permutations that begin with 3 and those that begin with 4. Oopps... Another bug detected :-( Thx, fortunately I could fix it easily. Here's the output. Ok, then everything seems to be ok with these 'permutations without repetetions'. Now I have these: 1.) Permutations with repetetions 2.) Permutations without repetetions (this case) 3.) Combinations with repetetions (aka multisets) 4.) Combinations without repetetions Are there any other such enumeration cases with fixed blocksize? Name=Permutations without repetetions, v=5, k=3, b=60 0: 0 1 2 1: 0 1 3 2: 0 1 4 3: 0 2 1 4: 0 2 3 5: 0 2 4 6: 0 3 1 7: 0 3 2 8: 0 3 4 9: 0 4 1 10: 0 4 2 11: 0 4 3 12: 1 0 2 13: 1 0 3 14: 1 0 4 15: 1 2 0 16: 1 2 3 17: 1 2 4 18: 1 3 0 19: 1 3 2 20: 1 3 4 21: 1 4 0 22: 1 4 2 23: 1 4 3 24: 2 0 1 25: 2 0 3 26: 2 0 4 27: 2 1 0 28: 2 1 3 29: 2 1 4 30: 2 3 0 31: 2 3 1 32: 2 3 4 33: 2 4 0 34: 2 4 1 35: 2 4 3 36: 3 0 1 37: 3 0 2 38: 3 0 4 39: 3 1 0 40: 3 1 2 41: 3 1 4 42: 3 2 0 43: 3 2 1 44: 3 2 4 45: 3 4 0 46: 3 4 1 47: 3 4 2 48: 4 0 1 49: 4 0 2 50: 4 0 3 51: 4 1 0 52: 4 1 2 53: 4 1 3 54: 4 2 0 55: 4 2 1 56: 4 2 3 57: 4 3 0 58: 4 3 1 59: 4 3 2 === Subject: Re: Formula sought (enumerative combinatorics) posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 FunWebProducts; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Where are the permutations that begin with 3 and those that begin with 4. Oopps... Another bug detected :-( > Thx, fortunately I could fix it easily. Here's the output. > Ok, then everything seems to be ok with these > 'permutations without repetetions'. Now I have ?these: > ? 1.) Permutations with repetetions > ? 2.) Permutations without repetetions (this case) > ? 3.) Combinations with repetetions (aka multisets) > ? 4.) Combinations without repetetions Are there any other such enumeration cases with fixed blocksize? Sure. You could ask how many ways are there to but m marbles in n ordered bins such that each bin contains at least one marble? For m=5, n=4 you would have [1,1,1,2] [1,1,2,1] [1,2,1,1] [2,1,1,1] For m=6, n=3 you would have [1,1,4] [1,2,3] [1,3,2] [1,4,1] [2,1,3] [2,2,2] [2,3,1] [3,1,2] [3,2,1] [4,1,1] You could also set a constraint such that not only does each bin have a minimum, but also a maximum. So m=10, n=4 min=1 max=3 would look like [1,3,3,3] [2,3,3,2] [2,3,2,3] etc. Name=Permutations without repetetions, v=5, k=3, b=60 > ? ?0: ? ? ?0 ?1 ?2 > ? ?1: ? ? ?0 ?1 ?3 > ? ?2: ? ? ?0 ?1 ?4 > ? ?3: ? ? ?0 ?2 ?1 > ? ?4: ? ? ?0 ?2 ?3 > ? ?5: ? ? ?0 ?2 ?4 > ? ?6: ? ? ?0 ?3 ?1 > ? ?7: ? ? ?0 ?3 ?2 > ? ?8: ? ? ?0 ?3 ?4 > ? ?9: ? ? ?0 ?4 ?1 > ? 10: ? ? ?0 ?4 ?2 > ? 11: ? ? ?0 ?4 ?3 > ? 12: ? ? ?1 ?0 ?2 > ? 13: ? ? ?1 ?0 ?3 > ? 14: ? ? ?1 ?0 ?4 > ? 15: ? ? ?1 ?2 ?0 > ? 16: ? ? ?1 ?2 ?3 > ? 17: ? ? ?1 ?2 ?4 > ? 18: ? ? ?1 ?3 ?0 > ? 19: ? ? ?1 ?3 ?2 > ? 20: ? ? ?1 ?3 ?4 > ? 21: ? ? ?1 ?4 ?0 > ? 22: ? ? ?1 ?4 ?2 > ? 23: ? ? ?1 ?4 ?3 > ? 24: ? ? ?2 ?0 ?1 > ? 25: ? ? ?2 ?0 ?3 > ? 26: ? ? ?2 ?0 ?4 > ? 27: ? ? ?2 ?1 ?0 > ? 28: ? ? ?2 ?1 ?3 > ? 29: ? ? ?2 ?1 ?4 > ? 30: ? ? ?2 ?3 ?0 > ? 31: ? ? ?2 ?3 ?1 > ? 32: ? ? ?2 ?3 ?4 > ? 33: ? ? ?2 ?4 ?0 > ? 34: ? ? ?2 ?4 ?1 > ? 35: ? ? ?2 ?4 ?3 > ? 36: ? ? ?3 ?0 ?1 > ? 37: ? ? ?3 ?0 ?2 > ? 38: ? ? ?3 ?0 ?4 > ? 39: ? ? ?3 ?1 ?0 > ? 40: ? ? ?3 ?1 ?2 > ? 41: ? ? ?3 ?1 ?4 > ? 42: ? ? ?3 ?2 ?0 > ? 43: ? ? ?3 ?2 ?1 > ? 44: ? ? ?3 ?2 ?4 > ? 45: ? ? ?3 ?4 ?0 > ? 46: ? ? ?3 ?4 ?1 > ? 47: ? ? ?3 ?4 ?2 > ? 48: ? ? ?4 ?0 ?1 > ? 49: ? ? ?4 ?0 ?2 > ? 50: ? ? ?4 ?0 ?3 > ? 51: ? ? ?4 ?1 ?0 > ? 52: ? ? ?4 ?1 ?2 > ? 53: ? ? ?4 ?1 ?3 > ? 54: ? ? ?4 ?2 ?0 > ? 55: ? ? ?4 ?2 ?1 > ? 56: ? ? ?4 ?2 ?3 > ? 57: ? ? ?4 ?3 ?0 > ? 58: ? ? ?4 ?3 ?1 > ? 59: ? ? ?4 ?3 ?2 === Subject: Turing machines and context sensitive languages posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/2008061017 Firefox/3.0,gzip(gfe),gzip(gfe) does the complexity class of languages which are accepted in linear space equal the context free languages? S. === Subject: Re: Turing machines and context sensitive languages posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/2008061017 Firefox/3.0,gzip(gfe),gzip(gfe) does the complexity class of languages which are accepted in linear > space equal the context free languages? S. Sorry again, I have meant: Does the complexity class of languages which are accepted in non-deterministic linear space equal the context sensitive languages? === Subject: Re: Turing machines and context sensitive languages posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > does the complexity class of languages which are accepted in linear > space equal the context free languages? > S. Sorry again, I have meant: Does the complexity class of languages > which are accepted in non-deterministic linear space equal the context > sensitive languages? Yes, that's a standard result. The only proviso is that, according to the original definition of context sensitive languages, the empty language is not context sensitive, but many people have modified that to allow it to be. Derek Holt. === Subject: Re: Turing machines and context sensitive languages posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/2008061017 Firefox/3.0,gzip(gfe),gzip(gfe) > does the complexity class of languages which are accepted in linear > space equal the context free languages? Sorry, typo: context sensitive languages instead of context free languages. === Subject: A tiny Collatz-exercise By playing around with the collatz-problem last after- noon I found a representation, which has a nice look&feel. I tried to implement the idea of using (3x+1)/2 and (3x-1)/2 simultaneously in some way, to symmetrize the formula/the trajectory... and see, what happens. After some useless approaches I came to the following, which has at least interesting side-aspects. Short note: i usually use a compressed definition of the Collatz-transformation; so instead of 5*3+1 = 16; 16/2= 8; 8/2 = 4; 4/2=2; 2/2=1 I just write (5*3+1)/2^4 and denote C(5;4) = (5*3+1)/2^4 as one single step, and consider C(x,A) for odd x only. To intertwine the 3x+1 and 3x-1 transformations in a meaningful way I defined (for odd x) D(x) := D(x) = (3x+1)/2 if (3x+1)/2 is odd D(x) = (3x-1)/2 if (3x-1)/2 is odd If I start at x=3 I get the following sequence 3 5 7 11 17 25 37 55 83 125 187 281 421 .... a) This sequence is strictly increasing, which is obvious, since for any element D(x) > x. We find partial sequences, which obey the (compressed) Collatz-rule, for instance 3--5 7--11--17 25 37 55--83--125 187--281 421 .... indicated by the -- sign. This sequence has the nice taste, that is is somehow an upper limiting sequence for the Collatz-transformation: each partial Collatz-sequence in this sequence must be interrupted somewhere and decrease, because we cannot have infinitely many C(x,1)-steps. b) in the sequence we see, that some numbers are missing, for instance 9. So we start a new sequence beginning at 9. We get 3, 5, 7, 11, 17, 25, 37, 55, 83, 125, 187, 281, 421 ... 9, 13, 19, 29, 43, 65, 97, 145, 217, 325, 487, ... Again, this is a strictly increasing sequence for the same reason as the previous one. Well - again some numbers are missing, the first is 15. So let's append this and the first few following: 3 5 7 11 17 25 37 55 83 125 187 281 421 9 13 19 29 43 65 97 145 217 325 487 731 1097 15 23 35 53 79 119 179 269 403 605 907 1361 2041 21 31 47 71 107 161 241 361 541 811 1217 1825 2737 27 41 61 91 137 205 307 461 691 1037 1555 2333 3499 33 49 73 109 163 245 367 551 827 1241 1861 2791 4187 A bit arranged, to see how the holes are filled: 3 5 7 11 17 25 37 9 13 19 29 43 15 23 35 21 31 47 27 41 Hmm. It is obvious, that the numbers divisible by 3 must occur at the begin of a row. On the other hand: there are no other leading numbers for the rows. So the first question: Can it be shown, that there are no numbers missing, if we arrange all rows, which begin with numbers of the form 3*k? (greedy algorithm?) ---- There are a few more properties, which I found surprising/ interesting. I'll post another mail for this. Gottfried Helms === Subject: Re: A tiny Collatz-exercise Am 28.06.2008 17:00 schrieb Gottfried Helms: > Again, this is a strictly increasing sequence for the > same reason as the previous one. Well - again some numbers > are missing, the first is 15. So let's append this and the > first few following: 3 5 7 11 17 25 37 55 83 125 187 281 421 > 9 13 19 29 43 65 97 145 217 325 487 731 1097 > 15 23 35 53 79 119 179 269 403 605 907 1361 2041 > 21 31 47 71 107 161 241 361 541 811 1217 1825 2737 > 27 41 61 91 137 205 307 461 691 1037 1555 2333 3499 > 33 49 73 109 163 245 367 551 827 1241 1861 2791 4187 > A second observation is not really new, but is again nice. Increase each number of the above array by 1 - then the main property are not changed, but one more occurs: 4 6 8 12 18 26 38 56 84 126 188 282 10 14 20 30 44 66 98 146 218 326 488 732 16 24 36 54 80 120 180 270 404 606 908 1362 22 32 48 72 108 162 242 362 542 812 1218 1826 28 42 62 92 138 206 308 462 692 1038 1556 2334 34 50 74 110 164 246 368 552 828 1242 1862 2792 40 60 90 134 200 300 450 674 1010 1514 2270 3404 and then take the log to base 3/2. We get (3) 3.41902258270 (5) 4.41902258270 (7) 5.12853387405 (11) 6.12853387405 (17) 7.12853387405 (25) 8.03545477248 (9) 5.67887358727 (13) 6.50871622944 (19) 7.38838487862 (29) 8.38838487862 (43) 9.33295999642 (65) 10.3329599964 (15) 6.83804516541 (23) 7.83804516541 (35) 8.83804516541 (53) 9.83804516541 (79) 10.8074074613 (119) 11.8074074613 (21) 7.62344870507 (31) 8.54755645676 (47) 9.54755645676 (71) 10.5475564568 (107) 11.5475564568 (161) 12.5475564568 (27) 8.21822752079 (41) 9.21822752079 (61) 10.1787658235 (91) 11.1521028237 (137) 12.1521028237 (205) 13.1401594422 (33) 8.69707517145 (49) 9.64823588318 (73) 10.6151306417 (109) 11.5928110010 (163) 12.5778182286 (245) 13.5778182286 (39) 9.09789616997 (59) 10.0978961700 (89) 11.0978961700 (133) 12.0795592568 (199) 13.0672584659 (299) 14.0672584659 (45) 9.44259153236 (67) 10.4065864628 (101) 11.4065864628 (151) 12.3904139231 (227) 13.3904139231 (341) 14.3904139231 (51) 9.74496606383 (77) 10.7449660638 (115) 11.7237959470 (173) 12.7237959470 (259) 13.7143283597 (389) 14.7143283597 (57) 10.0142846557 (85) 10.9857721594 (127) 11.9665790395 (191) 12.9665790395 (287) 13.9665790395 (431) 14.9665790395 (63) 10.2570677481 (95) 11.2570677481 (143) 12.2570677481 (215) 13.2570677481 (323) 14.2570677481 (485) 15.2570677481 (69) 10.4780785254 (103) 11.4544773552 (155) 12.4544773552 (233) 13.4544773552 (349) 14.4474408213 (523) 15.4427386207 We see, that the entries, which follow the original Collatz-transformation have the same fractional digit-expansion (easy to prove). This allows us to recognize the partial Collatz-sequences in the array -in one row- by eyeballing. Since (3) 3.41902258270 (5) 4.41902258270 it is also 3->5 by the (compressed) Collatz-transformation, and since (7) 5.12853387405 (11) 6.12853387405 (17) 7.12853387405 it is also 7->17 by the compressed Collatz-transformation. However, it would be interesting to have a similar rule to see also the next steps in an analoguous fashion. But the Collatz-step from 17->13 is not obvious... (17) 7.12853387405 --> (13) 6.50871622944 --------------------------------------------------- Moreover, it seems, that (with one exception) the fractional parts are nonincreasing to the right (besides of the first difference of a row). Here are the differences of the above fractional parts. There are some negative differences in the first column (that means at transformation from 3*(2k+1) to D(3*(2k+1)) The only negative difference in the trailing columns is in the first row at the step from 25 to 37. (note: the header of the table relates only to the first row (to illustrate)!) 3 ---> 5 ---> 7 ---> 11 ---> 17 ---> 25 ---> 37 --> 55 --> ... ---------------------------------------------------------------------------- ----------------------------------------------------------------- 0 0.290488708649 0 0 0.0930791015753 -0.935936567947 0.0436525282829 0 0.170157357828 0.120331350821 0 0.0554248821974 0 0.0250388288930 0.0168349012872 0.0112874497358 0 0 0 0.0306377040839 0 0 0 0.00609716864892 0.0758922483129 0 0 0 0 0.0101703379684 0.00680360007289 0.00454618278473 0 0.0394616972619 0.0266629998193 0 0.0119433815116 0.00799450768846 0 0.00356144966202 0.0488392882647 0.0331052415207 0.0223196406767 0.0149927723861 0 0.00669282218699 0 0 0 0 0.0183369131865 0.0123007908974 0 0 0.00365649208915 0.00244067655978 0.0360050695589 0 0.0161725396712 0 0 0.00481230096559 0 0 0 0.0211701167862 0 0.00946758729765 0 0.00421951080233 0 0 -0.971487503684 0.0191931199169 0 0 0 0 0 0 0 0 0 0 0 0.00338545483822 0 0 0.0236011701742 0 0 0.00703653390971 0.00470220059003 0 0.00209274807837 0.00139615257755 I checked this with 150 columns and 24 rows without a counterexample. Proof? Disproof? Since this table can be seen as a simple reordering of the Collatz- trajectories, and the effect of equal fractional parts of logarithms is already known from the study of the Collatz-sequence, there is not much new here. Only, that the fractional parts of this table seem strictly nondecreasing (with the said exceptions) may be a remarkable property... Gottfried Helms === Subject: Re: A tiny Collatz-exercise Am 02.07.2008 19:18 schrieb Gottfried Helms: > Am 28.06.2008 17:00 schrieb Gottfried Helms: > Again, this is a strictly increasing sequence for the > same reason as the previous one. Well - again some numbers > are missing, the first is 15. So let's append this and the > first few following: > 3 5 7 11 17 25 37 55 83 125 187 281 421 > 9 13 19 29 43 65 97 145 217 325 487 731 1097 > 15 23 35 53 79 119 179 269 403 605 907 1361 2041 > 21 31 47 71 107 161 241 361 541 811 1217 1825 2737 > 27 41 61 91 137 205 307 461 691 1037 1555 2333 3499 > 33 49 73 109 163 245 367 551 827 1241 1861 2791 4187 The third observation I made is a curious one. Arrange the above tables according to the contents in idexed columns: 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 .... ------------------------------------------------------------- 3 5 7 11 17 25 37 ... 9 13 19 29 ... 15 23 35 ... 21 31 ... then this table can be created without computations by a simple search rule. Begin with the diagonal of numbers divisible by 3: 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 .... ------------------------------------------------------------- 3 9 15 21 Then begin to fill the holes, where you only consider rows above. You begin at first row. 5 is missing - fill it in 7 is missing - fill it in 9 is missing - divisible by 3: open a new row; next consider these two rows to fill the holes 11 is missing - the longest gap is in 1st row: fill it in there 13 is missing - the longest gap is in 2nd row: fill it in there we have: 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 .... ------------------------------------------------------------- 3 5 7 11 9 13 ---------------------------------------------------- 15 21 15 is missing - divisible by 3: open a new row next consider these three rows to fill the holes 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 .... ------------------------------------------------------------- 3 5 7 11 9 13 15 ---------------------------------------------------- 21 17 is missing - the longest gap is in 1st row : fill it in there 19 is missing - the longest gap is in 2nd row : fill it in there 21 is missing - divisible by 3: open a new row next consider these three rows to fill the holes 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 .... ------------------------------------------------------------- 3 5 7 11 17 9 13 19 15 21 ------------------------------------------- and so on. This looks very simple - but I'm speculating, that this is some remarkable property with further consequences. Maybe someone can find some importance in it... Gottfried Helms === Subject: Re: A tiny Collatz-exercise <6d1sjmFdsmgU1@mid.dfncis.de> posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > Am 02.07.2008 19:18 schrieb Gottfried Helms: > Am 28.06.2008 17:00 schrieb Gottfried Helms: > Again, this is a strictly increasing sequence for the > same reason as the previous one. Well - again some numbers > are missing, the first is 15. So let's append this and the > first few following: > 3 5 7 11 17 25 37 55 83 125 187 281 421 > 9 13 19 29 43 65 97 145 217 325 487 731 1097 > 15 23 35 53 79 119 179 269 403 605 907 1361 2041 > 21 31 47 71 107 161 241 361 541 811 1217 1825 2737 > 27 41 61 91 137 205 307 461 691 1037 1555 2333 3499 > 33 49 73 109 163 245 367 551 827 1241 1861 2791 4187 The third observation I made is a curious one. > Arrange the above tables according to the contents in idexed columns: > 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 .... > ------------------------------------------------------------- > 3 5 7 11 17 25 37 ... > 9 13 19 29 ... > 15 23 35 ... > 21 31 ... then this table can be created without computations by a > simple search rule. Nifty. # Python import collatz_functions as cf # the list of sequences (created by the gap_fit algorithm) seq = [] for n in xrange(3,1000,2): # start new sequence if divisible by 3 if n%3==0: seq.append([n]) else: # otherwise, find difference from last n # of all known sequences up to this point diff = [] for seq_n in seq: diff.append(n - seq_n[-1]) # find greatest difference m = max(diff) # and what sequence it came from mi = diff.index(m) # add n to that sequence seq[mi].append(n) for i in seq: if len(i)>3: # skip the short sequences print i # QC check (test above algorithm against actual 3n+-1 sequence) faults = 0 for i in seq: # start with first number in sequence n = i[0] # make sure all the numbers in sequence follow # the actual 3n+-1 sequence for s in i[1:]: n *= 3 np = n + 1 nm = n - 1 # does np have EXACTLY 1 factor of 2? f = cf.gmpy.scan1(np) if f==1: # yes, so it's next number in sequence n = np/2 else: # otherwise, nm is the next number n = nm/2 # does it match the gap_fit algorithm's next sequence number? if n!=s: faults += 1 print i,s,n print 'faults:',faults ## [3, 5, 7, 11, 17, 25, 37, 55, 83, 125, 187, 281, 421, 631, 947, 1421, 2131, 3197, 4795] ## [9, 13, 19, 29, 43, 65, 97, 145, 217, 325, 487, 731, 1097, 1645, 2467, 3701] ## [15, 23, 35, 53, 79, 119, 179, 269, 403, 605, 907, 1361, 2041, 3061, 4591] ## [21, 31, 47, 71, 107, 161, 241, 361, 541, 811, 1217, 1825, 2737, 4105] ## [27, 41, 61, 91, 137, 205, 307, 461, 691, 1037, 1555, 2333, 3499] ## [33, 49, 73, 109, 163, 245, 367, 551, 827, 1241, 1861, 2791, 4187] ## [39, 59, 89, 133, 199, 299, 449, 673, 1009, 1513, 2269, 3403] ## [45, 67, 101, 151, 227, 341, 511, 767, 1151, 1727, 2591, 3887] ## [51, 77, 115, 173, 259, 389, 583, 875, 1313, 1969, 2953, 4429] ## [57, 85, 127, 191, 287, 431, 647, 971, 1457, 2185, 3277, 4915] ## [63, 95, 143, 215, 323, 485, 727, 1091, 1637, 2455, 3683] ## [69, 103, 155, 233, 349, 523, 785, 1177, 1765, 2647, 3971] ## [75, 113, 169, 253, 379, 569, 853, 1279, 1919, 2879, 4319] ## [81, 121, 181, 271, 407, 611, 917, 1375, 2063, 3095, 4643] ## [87, 131, 197, 295, 443, 665, 997, 1495, 2243, 3365] ## [93, 139, 209, 313, 469, 703, 1055, 1583, 2375, 3563] ## [99, 149, 223, 335, 503, 755, 1133, 1699, 2549, 3823] ## [105, 157, 235, 353, 529, 793, 1189, 1783, 2675, 4013] ## [111, 167, 251, 377, 565, 847, 1271, 1907, 2861, 4291] ## [117, 175, 263, 395, 593, 889, 1333, 1999, 2999, 4499] ## [123, 185, 277, 415, 623, 935, 1403, 2105, 3157, 4735] ## [129, 193, 289, 433, 649, 973, 1459, 2189, 3283, 4925] ## [135, 203, 305, 457, 685, 1027, 1541, 2311, 3467] ## [141, 211, 317, 475, 713, 1069, 1603, 2405, 3607] ## [147, 221, 331, 497, 745, 1117, 1675, 2513, 3769] ## [153, 229, 343, 515, 773, 1159, 1739, 2609, 3913] ## [159, 239, 359, 539, 809, 1213, 1819, 2729, 4093] ## [165, 247, 371, 557, 835, 1253, 1879, 2819, 4229] ## [171, 257, 385, 577, 865, 1297, 1945, 2917, 4375] ## [177, 265, 397, 595, 893, 1339, 2009, 3013, 4519] ## [183, 275, 413, 619, 929, 1393, 2089, 3133, 4699] ## [189, 283, 425, 637, 955, 1433, 2149, 3223, 4835] ## [195, 293, 439, 659, 989, 1483, 2225, 3337] ## [201, 301, 451, 677, 1015, 1523, 2285, 3427] ## [207, 311, 467, 701, 1051, 1577, 2365, 3547] ## [213, 319, 479, 719, 1079, 1619, 2429, 3643] ## [219, 329, 493, 739, 1109, 1663, 2495, 3743] ## [225, 337, 505, 757, 1135, 1703, 2555, 3833] ## [231, 347, 521, 781, 1171, 1757, 2635, 3953] ## [237, 355, 533, 799, 1199, 1799, 2699, 4049] ## [243, 365, 547, 821, 1231, 1847, 2771, 4157] ## [249, 373, 559, 839, 1259, 1889, 2833, 4249] ## [255, 383, 575, 863, 1295, 1943, 2915, 4373] ## [261, 391, 587, 881, 1321, 1981, 2971, 4457] ## [267, 401, 601, 901, 1351, 2027, 3041, 4561] ## [273, 409, 613, 919, 1379, 2069, 3103, 4655] ## [279, 419, 629, 943, 1415, 2123, 3185, 4777] ## [285, 427, 641, 961, 1441, 2161, 3241, 4861] ## [291, 437, 655, 983, 1475, 2213, 3319, 4979] ## [297, 445, 667, 1001, 1501, 2251, 3377] ## [303, 455, 683, 1025, 1537, 2305, 3457] ## [309, 463, 695, 1043, 1565, 2347, 3521] ## [315, 473, 709, 1063, 1595, 2393, 3589] ## [321, 481, 721, 1081, 1621, 2431, 3647] ## [327, 491, 737, 1105, 1657, 2485, 3727] ## [333, 499, 749, 1123, 1685, 2527, 3791] ## [339, 509, 763, 1145, 1717, 2575, 3863] ## [345, 517, 775, 1163, 1745, 2617, 3925] ## [351, 527, 791, 1187, 1781, 2671, 4007] ## [357, 535, 803, 1205, 1807, 2711, 4067] ## [363, 545, 817, 1225, 1837, 2755, 4133] ## [369, 553, 829, 1243, 1865, 2797, 4195] ## [375, 563, 845, 1267, 1901, 2851, 4277] ## [381, 571, 857, 1285, 1927, 2891, 4337] ## [387, 581, 871, 1307, 1961, 2941, 4411] ## [393, 589, 883, 1325, 1987, 2981, 4471] ## [399, 599, 899, 1349, 2023, 3035, 4553] ## [405, 607, 911, 1367, 2051, 3077, 4615] ## [411, 617, 925, 1387, 2081, 3121, 4681] ## [417, 625, 937, 1405, 2107, 3161, 4741] ## [423, 635, 953, 1429, 2143, 3215, 4823] ## [429, 643, 965, 1447, 2171, 3257, 4885] ## [435, 653, 979, 1469, 2203, 3305, 4957] ## [441, 661, 991, 1487, 2231, 3347] ## [447, 671, 1007, 1511, 2267, 3401] ## [453, 679, 1019, 1529, 2293, 3439] ## [459, 689, 1033, 1549, 2323, 3485] ## [465, 697, 1045, 1567, 2351, 3527] ## [471, 707, 1061, 1591, 2387, 3581] ## [477, 715, 1073, 1609, 2413, 3619] ## [483, 725, 1087, 1631, 2447, 3671] ## [489, 733, 1099, 1649, 2473, 3709] ## [495, 743, 1115, 1673, 2509, 3763] ## [501, 751, 1127, 1691, 2537, 3805] ## [507, 761, 1141, 1711, 2567, 3851] ## [513, 769, 1153, 1729, 2593, 3889] ## [519, 779, 1169, 1753, 2629, 3943] ## [525, 787, 1181, 1771, 2657, 3985] ## [531, 797, 1195, 1793, 2689, 4033] ## [537, 805, 1207, 1811, 2717, 4075] ## [543, 815, 1223, 1835, 2753, 4129] ## [549, 823, 1235, 1853, 2779, 4169] ## [555, 833, 1249, 1873, 2809, 4213] ## [561, 841, 1261, 1891, 2837, 4255] ## [567, 851, 1277, 1915, 2873, 4309] ## [573, 859, 1289, 1933, 2899, 4349] ## [579, 869, 1303, 1955, 2933, 4399] ## [585, 877, 1315, 1973, 2959, 4439] ## [591, 887, 1331, 1997, 2995, 4493] ## [597, 895, 1343, 2015, 3023, 4535] ## [603, 905, 1357, 2035, 3053, 4579] ## [609, 913, 1369, 2053, 3079, 4619] ## [615, 923, 1385, 2077, 3115, 4673] ## [621, 931, 1397, 2095, 3143, 4715] ## [627, 941, 1411, 2117, 3175, 4763] ## [633, 949, 1423, 2135, 3203, 4805] ## [639, 959, 1439, 2159, 3239, 4859] ## [645, 967, 1451, 2177, 3265, 4897] ## [651, 977, 1465, 2197, 3295, 4943] ## [657, 985, 1477, 2215, 3323, 4985] ## [663, 995, 1493, 2239, 3359] ## [669, 1003, 1505, 2257, 3385] ## [675, 1013, 1519, 2279, 3419] ## [681, 1021, 1531, 2297, 3445] ## [687, 1031, 1547, 2321, 3481] ## [693, 1039, 1559, 2339, 3509] ## [699, 1049, 1573, 2359, 3539] ## [705, 1057, 1585, 2377, 3565] ## [711, 1067, 1601, 2401, 3601] ## [717, 1075, 1613, 2419, 3629] ## [723, 1085, 1627, 2441, 3661] ## [729, 1093, 1639, 2459, 3689] ## [735, 1103, 1655, 2483, 3725] ## [741, 1111, 1667, 2501, 3751] ## [747, 1121, 1681, 2521, 3781] ## [753, 1129, 1693, 2539, 3809] ## [759, 1139, 1709, 2563, 3845] ## [765, 1147, 1721, 2581, 3871] ## [771, 1157, 1735, 2603, 3905] ## [777, 1165, 1747, 2621, 3931] ## [783, 1175, 1763, 2645, 3967] ## [789, 1183, 1775, 2663, 3995] ## [795, 1193, 1789, 2683, 4025] ## [801, 1201, 1801, 2701, 4051] ## [807, 1211, 1817, 2725, 4087] ## [813, 1219, 1829, 2743, 4115] ## [819, 1229, 1843, 2765, 4147] ## [825, 1237, 1855, 2783, 4175] ## [831, 1247, 1871, 2807, 4211] ## [837, 1255, 1883, 2825, 4237] ## [843, 1265, 1897, 2845, 4267] ## [849, 1273, 1909, 2863, 4295] ## [855, 1283, 1925, 2887, 4331] ## [861, 1291, 1937, 2905, 4357] ## [867, 1301, 1951, 2927, 4391] ## [873, 1309, 1963, 2945, 4417] ## [879, 1319, 1979, 2969, 4453] ## [885, 1327, 1991, 2987, 4481] ## [891, 1337, 2005, 3007, 4511] ## [897, 1345, 2017, 3025, 4537] ## [903, 1355, 2033, 3049, 4573] ## [909, 1363, 2045, 3067, 4601] ## [915, 1373, 2059, 3089, 4633] ## [921, 1381, 2071, 3107, 4661] ## [927, 1391, 2087, 3131, 4697] ## [933, 1399, 2099, 3149, 4723] ## [939, 1409, 2113, 3169, 4753] ## [945, 1417, 2125, 3187, 4781] ## [951, 1427, 2141, 3211, 4817] ## [957, 1435, 2153, 3229, 4843] ## [963, 1445, 2167, 3251, 4877] ## [969, 1453, 2179, 3269, 4903] ## [975, 1463, 2195, 3293, 4939] ## [981, 1471, 2207, 3311, 4967] ## [987, 1481, 2221, 3331, 4997] ## [993, 1489, 2233, 3349] ## [999, 1499, 2249, 3373] ## [1005, 1507, 2261, 3391] ## [1011, 1517, 2275, 3413] ## [1017, 1525, 2287, 3431] ## [1023, 1535, 2303, 3455] ## [1029, 1543, 2315, 3473] ## [1035, 1553, 2329, 3493] ## [1041, 1561, 2341, 3511] ## [1047, 1571, 2357, 3535] ## [1053, 1579, 2369, 3553] ## [1059, 1589, 2383, 3575] ## [1065, 1597, 2395, 3593] ## [1071, 1607, 2411, 3617] ## [1077, 1615, 2423, 3635] ## [1083, 1625, 2437, 3655] ## [1089, 1633, 2449, 3673] ## [1095, 1643, 2465, 3697] ## [1101, 1651, 2477, 3715] ## [1107, 1661, 2491, 3737] ## [1113, 1669, 2503, 3755] ## [1119, 1679, 2519, 3779] ## [1125, 1687, 2531, 3797] ## [1131, 1697, 2545, 3817] ## [1137, 1705, 2557, 3835] ## [1143, 1715, 2573, 3859] ## [1149, 1723, 2585, 3877] ## [1155, 1733, 2599, 3899] ## [1161, 1741, 2611, 3917] ## [1167, 1751, 2627, 3941] ## [1173, 1759, 2639, 3959] ## [1179, 1769, 2653, 3979] ## [1185, 1777, 2665, 3997] ## [1191, 1787, 2681, 4021] ## [1197, 1795, 2693, 4039] ## [1203, 1805, 2707, 4061] ## [1209, 1813, 2719, 4079] ## [1215, 1823, 2735, 4103] ## [1221, 1831, 2747, 4121] ## [1227, 1841, 2761, 4141] ## [1233, 1849, 2773, 4159] ## [1239, 1859, 2789, 4183] ## [1245, 1867, 2801, 4201] ## [1251, 1877, 2815, 4223] ## [1257, 1885, 2827, 4241] ## [1263, 1895, 2843, 4265] ## [1269, 1903, 2855, 4283] ## [1275, 1913, 2869, 4303] ## [1281, 1921, 2881, 4321] ## [1287, 1931, 2897, 4345] ## [1293, 1939, 2909, 4363] ## [1299, 1949, 2923, 4385] ## [1305, 1957, 2935, 4403] ## [1311, 1967, 2951, 4427] ## [1317, 1975, 2963, 4445] ## [1323, 1985, 2977, 4465] ## [1329, 1993, 2989, 4483] ## [1341, 2011, 3017, 4525] ## [1347, 2021, 3031, 4547] ## [1353, 2029, 3043, 4565] ## [1359, 2039, 3059, 4589] ## [1365, 2047, 3071, 4607] ## [1371, 2057, 3085, 4627] ## [1377, 2065, 3097, 4645] ## [1383, 2075, 3113, 4669] ## [1389, 2083, 3125, 4687] ## [1395, 2093, 3139, 4709] ## [1401, 2101, 3151, 4727] ## [1407, 2111, 3167, 4751] ## [1413, 2119, 3179, 4769] ## [1419, 2129, 3193, 4789] ## [1425, 2137, 3205, 4807] ## [1431, 2147, 3221, 4831] ## [1437, 2155, 3233, 4849] ## [1443, 2165, 3247, 4871] ## [1449, 2173, 3259, 4889] ## [1455, 2183, 3275, 4913] ## [1461, 2191, 3287, 4931] ## [1467, 2201, 3301, 4951] ## [1473, 2209, 3313, 4969] ## [1479, 2219, 3329, 4993] ## faults: 0 === Subject: Re: A tiny Collatz-exercise posting-account=6gwCCwoAAADEcuCVO3weXkA4SVHb4mS2 CLR 2.0.50727),gzip(gfe),gzip(gfe) > By playing around with the collatz-problem last after- > noon I found a representation, which has a nice look&feel. I tried to implement the idea of using (3x+1)/2 and > (3x-1)/2 simultaneously in some way, to symmetrize > the formula/the trajectory... and see, what happens. > After some useless approaches I came to the following, > which has at least interesting side-aspects. Short note: i usually use a compressed definition > of the Collatz-transformation; so instead of > æ 5*3+1 = 16; 16/2= 8; 8/2 = 4; 4/2=2; 2/2=1 > I just write > æ (5*3+1)/2^4 > and denote > æ C(5;4) = (5*3+1)/2^4 > as one single step, and consider C(x,A) for odd x only. To intertwine the 3x+1 and 3x-1 transformations in a > meaningful way I defined (for odd x) > æ æD(x) := > æ æ æD(x) = (3x+1)/2 æ if (3x+1)/2 is odd > æ æ æD(x) = (3x-1)/2 æ if (3x-1)/2 is odd This implies that one or the other must have exactly one factor of two. But not both and not neither, coreect? If I start at x=3 I get the following sequence æ3 5 7 11 17 25 37 55 83 125 187 281 421 .... a) > This sequence is strictly increasing, which is obvious, > since for any element D(x) > x. No, it's not obvious from your definition, otherwise regular old 3x+1 itself would be always increasing. Because one of the two choices always has exactly one factor of two and you always pick that one, you always have (3+-1)/2**k where k is 1. Thus, the ratio is always 3/2, therefore always increasing. Looking at it another way, at the MS end of the number, the bit length always increases by 1 or possibly 2. The mean increase being 1.585 bits per iteration. At the LS end, by rigging the process to only select the one with exactly one factor of two, the number is reduced by exactly 1 bit per iteration. Thus, there is a net gain of 1.585-1 or 0.585 bits per iteration. However, in plain old 3x+1, there can be any number of factors of two. But the number of factors will be a negative binomial distribution, specifically a geometric distribution. And since the mean of a geometric distribution is the inverse of the probability, the number of bits lost at the LS end has a mean of 2. Thus, the net gain is 1.585-2 or -0.415 bits per iteration. > We find partial sequences, which obey the (compressed) > Collatz-rule, for instance æ3--5 7--11--17 25 37 55--83--125 187--281 421 .... indicated by the -- sign. > This sequence has the nice taste, that is is somehow > an upper limiting sequence for the Collatz-transformation: > each partial Collatz-sequence in this sequence must be > interrupted somewhere and decrease, because we cannot have > infinitely many C(x,1)-steps. Huh? b) in the sequence we see, that some numbers are missing, > for instance 9. Because the successor of 3x+1 is always 1 (mod 3) and the successor of 3x-1 is always 2 (mod 3). No succesor can ever be 0 (mod 3). The only way to have 0 (mod 3) in the sequence is to start with one. > So we start a new sequence beginning at 9. > We get > æ3, 5, 7, 11, 17, 25, 37, 55, 83, 125, 187, 281, 421 æ... > æ æ æ æ æ9, 13, 19, 29, 43, 65, 97, 145, 217, 325, 487, ... > Again, this is a strictly increasing sequence for the > same reason as the previous one. Well - again some numbers > are missing, Because the 0 (mod 3) numbers will always be holes in your sequence. > the first is 15. So let's append this and the > first few following: æ æ3 æ æ5 æ æ7 æ 11 æ 17 æ 25 æ 37 æ æ55 æ æ83 æ 125 æ 187 æ 281 æ 421 > æ æ9 æ 13 æ 19 æ 29 æ 43 æ 65 æ 97 æ 145 æ 217 æ 325 æ 487 æ 731 æ1097 > æ 15 æ 23 æ 35 æ 53 æ 79 æ119 æ179 æ 269 æ 403 æ 605 æ 907 æ1361 æ2041 > æ 21 æ 31 æ 47 æ 71 æ107 æ161 æ241 æ 361 æ 541 æ 811 æ1217 æ1825 æ2737 > æ 27 æ 41 æ 61 æ 91 æ137 æ205 æ307 æ 461 æ 691 æ1037 æ1555 æ2333 æ3499 > æ 33 æ 49 æ 73 æ109 æ163 æ245 æ367 æ 551 æ 827 æ1241 æ1861 æ2791 æ4187 A bit arranged, to see how the holes are filled: > æ æ3 æ æ5 æ æ7 æ 11 æ æ17 æ æ æ25 æ æ æ æ æ37 > æ æ æ æ æ æ æ æ9 æ 13 æ æ19 æ æ æ æ29 æ æ æ æ æ æ43 > æ æ æ æ æ æ æ æ æ æ æ15 æ æ æ23 æ æ æ æ æ35 > æ æ æ æ æ æ æ æ æ æ æ æ æ æ21 æ æ æ æ31 æ æ æ æ æ æ æ47 > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ27 æ æ æ æ æ æ41 Hmm. It is obvious, that the numbers divisible by 3 must occur at > the begin of a row. As I said, the only way for 0 (mod 3) to appear in a sequence is to start with 0 (mod 3) as it is not a successor of either 3x+1 nor 3x-1. > On the other hand: there are no other leading numbers for the rows. Because 1 (mod 3) and 2 (mod 3) are successors, so they will never be holes, so a new sequence will never need to be started (although nothing is preventing you from starting a sequence with such a number. So the first question: Can it be shown, that there are no numbers missing, if we arrange > all rows, which begin with numbers of the form 3*k? Well, obviously EVERY 3*k starts a new row, so they're all accounted for. As for all the rest... > (greedy algorithm?) ---- There are a few more properties, which I found surprising/ > interesting. I'll post another mail for this. Gottfried Helms === Subject: Re: A tiny Collatz-exercise Am 28.06.2008 23:43 schrieb Michael Ellis: > By playing around with the collatz-problem last after- > noon I found a representation, which has a nice look&feel. > I tried to implement the idea of using (3x+1)/2 and > (3x-1)/2 simultaneously in some way, to symmetrize > the formula/the trajectory... and see, what happens. > After some useless approaches I came to the following, > which has at least interesting side-aspects. > Short note: i usually use a compressed definition > of the Collatz-transformation; so instead of > 5*3+1 = 16; 16/2= 8; 8/2 = 4; 4/2=2; 2/2=1 > I just write > (5*3+1)/2^4 > and denote > C(5;4) = (5*3+1)/2^4 > as one single step, and consider C(x,A) for odd x only. > To intertwine the 3x+1 and 3x-1 transformations in a > meaningful way I defined (for odd x) > D(x) := > D(x) = (3x+1)/2 if (3x+1)/2 is odd > D(x) = (3x-1)/2 if (3x-1)/2 is odd This implies that one or the other must have exactly > one factor of two. But not both and not neither, coreect? Yes. take n=7, then 3x=21. Now either 22 or 20 has exactly one factor of 2, so we choose this one: 22 so D(7) = (3*7+1)/2=11 = C(7;1) This is then also the collatz-step. Next D(11) = (3*11+1)/2=17 = C(11;1) again a Collatz-step D(17) = (3*17-1)/2=25 Here the partial collatz-sequence breaks; C(17;2) = 13 > If I start at x=3 I get the following sequence > 3 5 7 11 17 25 37 55 83 125 187 281 421 .... > a) > This sequence is strictly increasing, which is obvious, > since for any element D(x) > x. No, it's not obvious from your definition, > otherwise regular old 3x+1 itself would be > always increasing. Because one of the two > choices always has exactly one factor of two > and you always pick that one, you always have > (3+-1)/2**k where k is 1. Thus, the ratio is > always 3/2, therefore always increasing. Well, the definition was, that k=1 in all cases. so it is to show that (3x+1)/2 > (3x-1)/2 > x if x=2n+1 we have (6n+2)/2 > (6n-2)/2 > 2n+1 3n+1 > 3n-1 > 2n+1 for n>1 3n+1 > 2n+1 for n=1 Looking at it another way, at the MS end of the > number, the bit length always increases by 1 or > possibly 2. The mean increase being 1.585 bits > per iteration. At the LS end, by rigging the > process to only select the one with exactly one > factor of two, the number is reduced by exactly > 1 bit per iteration. Thus, there is a net gain of 1.585-1 or 0.585 > bits per iteration. Yes. However, in plain old 3x+1, there can be any number > of factors of two. Yes, then we have the decreasing steps with k>1. But here the D()-definition includes only the step with k=1 > But the number of factors will > be a negative binomial distribution, specifically > a geometric distribution. And since the mean of > a geometric distribution is the inverse of the > probability, the number of bits lost at the LS > end has a mean of 2. I've never looked at it this way, sounds interesting... > We find partial sequences, which obey the (compressed) > Collatz-rule, for instance > 3--5 7--11--17 25 37 55--83--125 187--281 421 .... > indicated by the -- sign. > This sequence has the nice taste, that is is somehow > an upper limiting sequence for the Collatz-transformation: > each partial Collatz-sequence in this sequence must be > interrupted somewhere and decrease, because we cannot have > infinitely many C(x,1)-steps. Huh? ??? I mean: subsequent C(x;1)-steps. Assume a positive odd number x decomposed as x0 = n*2*2^k - 1 then one C(x0;1) step x1=C(x0;1) reduces this to x1 = n*2 *2^(k-1) *3 -1 , then x2 = n*2 *2^(k-2) *3^2 - 1 , then ... xk = n*2 *2^0 *3^k - 1 and then we need another step C(xk;A) where A>1. So while the D()-transformation continues to increase here (and defines the continuing sequence, the collatz-transformation steps out here (to another D() sequence) Also a sequence of x1=C(x0;1), x2=C(x1;1); ... xk=C(x_(k-1);1) cannot be infinite, since there is no number x*2*2^k - 1 with an infinite k. So we may say. Define Da to denote the sequence Da:= DaÁk = D(D(...D(a))) the k'th iterate then partial sequences may be equal to a Collatz-sequence with denominators 2, but at DaÁj with some j the Collatz-sequence requires then a higher denominator (and decreases), so I'd call this branching out. > b) in the sequence we see, that some numbers are missing, > for instance 9. Because the successor of 3x+1 is always 1 (mod 3) > and the successor of 3x-1 is always 2 (mod 3). No succesor can ever be 0 (mod 3). The only way to have 0 (mod 3) in the sequence > is to start with one. Yepp. > So we start a new sequence beginning at 9. > We get > 3, 5, 7, 11, 17, 25, 37, 55, 83, 125, 187, 281, 421 ... > 9, 13, 19, 29, 43, 65, 97, 145, 217, 325, 487, ... > Again, this is a strictly increasing sequence for the > same reason as the previous one. Well - again some numbers > are missing, Because the 0 (mod 3) numbers will always be > holes in your sequence. Yes, this is obvious. But there are also other holes; for instance 31 is still missing, which is 1 (mod 3). It is *not* obvious, that *these holes* are filled with one new D()-sequences. > On the other hand: there are no other leading numbers for the rows. Because 1 (mod 3) and 2 (mod 3) are successors, > so they will never be holes, so a new sequence > will never need to be started (although nothing > is preventing you from starting a sequence with > such a number. Well, possibly the problem is not seen: If I construct the first sequence Da:=(a,D(a),D(D(a)),...) beginning at a=3, then take the first missing odd number, call it b and start another sequence Db:=(b,D(b),D(D(b)),...) and continue the same way Dc,Dd,... Is the sequence a,b,c,d,... exactly the sequence 3,9,15,... = 6k+3 ? Or do we have some sequences beginning with other values? ------------------------------------------------- I think, this is not too difficult to show, if we consider the inverse transformation of D: for an arbitrary odd positive number x>1 define E(x) := x // if x == 0 (mod 3) E(x) := (2*x - d)/3 // if x <> 0 (mod 3) where d=+1 or -1 such that the result is integer define EÁk(x) its k'th iterate example: 17 (34-1)/3 11 11 (22-1)/3 7 7 (14+1)/3 5 5 (10-1)/3 3 then for x>1 you shall always arrive at a number of the form 6n+3 This problem sounds very similar to that of the Collatz-problem, but it seems much easier to prove. Gottfried === Subject: Re: A tiny Collatz-exercise <6comgnF3g94vnU1@mid.dfncis.de> posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > Am 28.06.2008 23:43 schrieb Michael Ellis: > By playing around with the collatz-problem last after- > noon I found a representation, which has a nice look&feel. > I tried to implement the idea of using (3x+1)/2 and > (3x-1)/2 simultaneously in some way, to symmetrize > the formula/the trajectory... and see, what happens. > After some useless approaches I came to the following, > which has at least interesting side-aspects. > Short note: i usually use a compressed definition > of the Collatz-transformation; so instead of > 5*3+1 = 16; 16/2= 8; 8/2 = 4; 4/2=2; 2/2=1 > I just write > (5*3+1)/2^4 > and denote > C(5;4) = (5*3+1)/2^4 > as one single step, and consider C(x,A) for odd x only. > To intertwine the 3x+1 and 3x-1 transformations in a > meaningful way I defined (for odd x) > D(x) := > D(x) = (3x+1)/2 if (3x+1)/2 is odd > D(x) = (3x-1)/2 if (3x-1)/2 is odd > This implies that one or the other must have exactly > one factor of two. But not both and not neither, coreect? Yes. take n=7, then 3x=21. Now either 22 or 20 has > exactly one factor of 2, I was going to ask why this is so, but then I remembered that the even numbers count of contiguous LS 0-bits is a 2-adic sequence: 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,... so every odd number has one adjacent even number with exactly 1 factor of 2. > so we choose this one: 22 > so > D(7) = (3*7+1)/2=11 = C(7;1) > This is then also the collatz-step. > Next > D(11) = (3*11+1)/2=17 = C(11;1) again a Collatz-step > D(17) = (3*17-1)/2=25 Here the partial collatz-sequence > breaks; C(17;2) = 13 > If I start at x=3 I get the following sequence > 3 5 7 11 17 25 37 55 83 125 187 281 421 .... > a) > This sequence is strictly increasing, which is obvious, > since for any element D(x) > x. > No, it's not obvious from your definition, > otherwise regular old 3x+1 itself would be > always increasing. Because one of the two > choices always has exactly one factor of two > and you always pick that one, you always have > (3+-1)/2**k where k is 1. Thus, the ratio is > always 3/2, therefore always increasing. Well, the definition was, that k=1 in all cases. > so it is to show that (3x+1)/2 > (3x-1)/2 > x if x=2n+1 we have (6n+2)/2 > (6n-2)/2 > 2n+1 > 3n+1 > 3n-1 > 2n+1 for n>1 > 3n+1 > 2n+1 for n=1 > Looking at it another way, at the MS end of the > number, the bit length always increases by 1 or > possibly 2. The mean increase being 1.585 bits > per iteration. At the LS end, by rigging the > process to only select the one with exactly one > factor of two, the number is reduced by exactly > 1 bit per iteration. > Thus, there is a net gain of 1.585-1 or 0.585 > bits per iteration. Yes. > However, in plain old 3x+1, there can be any number > of factors of two. Yes, then we have the decreasing steps with k>1. But here > the D()-definition includes only the step with k=1 > But the number of factors will > be a negative binomial distribution, specifically > a geometric distribution. And since the mean of > a geometric distribution is the inverse of the > probability, the number of bits lost at the LS > end has a mean of 2. I've never looked at it this way, sounds interesting... I should clarify: random bit patterns have a geometrice bit distribution. But, due to the way carry bits proagate, ALL numbers degenerate to a geometric distribution. For example, an n-bit Mersenne Number has a resonant pattern (all 1s) that has a net gain of +0.585 bits per iteration up to the point where the pattern is consumed, which takes n iterations at which point all that remains are the n*1.585 carry bits which are a geomtric distribution. So the iterations for a Mersenne Number is n + (1.585*n)/0.415 Keeping in mind that this is based on statistics, which doesn't take certain realities of the Collatz graph structure into consideration, which can be seen here: Those realities make the iteration counts look like quantum energy lavels. Still, they track the line of 4.819 and the percent error gets smaller as the number of bits increase. > We find partial sequences, which obey the (compressed) > Collatz-rule, for instance > 3--5 7--11--17 25 37 55--83--125 187--281 421 .... > indicated by the -- sign. > This sequence has the nice taste, that is is somehow > an upper limiting sequence for the Collatz-transformation: > each partial Collatz-sequence in this sequence must be > interrupted somewhere and decrease, because we cannot have > infinitely many C(x,1)-steps. > Huh? ??? I mean: subsequent C(x;1)-steps. Assume a positive odd number x decomposed as x0 = n*2*2^k - 1 then one C(x0;1) step x1=C(x0;1) reduces this to x1 = n*2 *2^(k-1) *3 -1 , then > x2 = n*2 *2^(k-2) *3^2 - 1 , then > ... > xk = n*2 *2^0 *3^k - 1 and then we need another step C(xk;A) where A>1. > So while the D()-transformation continues to increase > here (and defines the continuing sequence, the > collatz-transformation steps out here (to another D() > sequence) Also a sequence of x1=C(x0;1), x2=C(x1;1); ... xk=C(x (k-1);1) > cannot be infinite, since there is no number x*2*2^k - 1 with an infinite k. So we may say. > Define Da to denote the sequence Da:= but at DaÁj with some j the Collatz-sequence requires > then a higher denominator (and decreases), so I'd call > this branching out. Ok, I think Iyou're saying your D(x) sequence must eventually break with the Collatz sequence at some point. In Collatz, only contiguous 1-bits at the LS end can create continuous increase. But every number has a finite number of contiguous 1-bits, so, sooner or later, the Collatz sequence MUST start decreasing (unless the propagating carry bits can create a new, unbroken string of 1's at the MS end, but that can't happen). The increasing may resume later, but it can never increase continuously. This does not say anything about the net result of such alternating increase/decrease. > b) in the sequence we see, that some numbers are missing, > for instance 9. > Because the successor of 3x+1 is always 1 (mod 3) > and the successor of 3x-1 is always 2 (mod 3). > No succesor can ever be 0 (mod 3). > The only way to have 0 (mod 3) in the sequence > is to start with one. Yepp. > So we start a new sequence beginning at 9. > We get > 3, 5, 7, 11, 17, 25, 37, 55, 83, 125, 187, 281, 421 ... > 9, 13, 19, 29, 43, 65, 97, 145, 217, 325, 487, ... > Again, this is a strictly increasing sequence for the > same reason as the previous one. Well - again some numbers > are missing, > Because the 0 (mod 3) numbers will always be > holes in your sequence. Yes, this is obvious. But there are also other holes; > for instance 31 is still missing, which is 1 (mod 3). > It is *not* obvious, that *these holes* are filled > with one new D()-sequences. That's why I wouldn't commit to saying ALL the holes are filled when new lines are started with 0 (mod 3) numbers. > On the other hand: there are no other leading numbers for the rows. > Because 1 (mod 3) and 2 (mod 3) are successors, > so they will never be holes, so a new sequence > will never need to be started (although nothing > is preventing you from starting a sequence with > such a number. Well, possibly the problem is not seen: If I construct the first sequence Da:=(a,D(a),D(D(a)),...) beginning at a=3, then take the first missing odd number, > call it b and start another sequence Db:=(b,D(b),D(D(b)),...) and continue the same way Dc,Dd,... Is the sequence a,b,c,d,... exactly the > sequence 3,9,15,... = 6k+3 ? > Or do we have some sequences beginning with other > values? ------------------------------------------------- I think, this is not too difficult to show, if we > consider the inverse transformation of D: for an arbitrary odd positive number x>1 > define > E(x) := x // if x == 0 (mod 3) > E(x) := (2*x - d)/3 // if x <> 0 (mod 3) > where d=+1 or -1 such that the result is integer define EÁk(x) its k'th iterate example: > 17 (34-1)/3 11 > 11 (22-1)/3 7 > 7 (14+1)/3 5 > 5 (10-1)/3 3 then for x>1 you shall always arrive at a number of > the form 6n+3 This problem sounds very similar to that of the > Collatz-problem, but it seems much easier to > prove. Well, it's easy to prove in Collatz that every odd 1(mod 3) and 2(mod 3) number has not just one, but an infinite number of 0(mod 3) ancestors. Is that true of your system also? So that any non-0(mod 3) hole in any given sequence must appear an infinite number of times in the other sequences that begin with 6K+3? If not, it sounds much harder. Gottfried