mm-3009 === Subject: Re: (n to the 5th power) mod 10 First, to prove a congruence modulo 10, it's enough to prove that it's true both mod 2 and mod 5. Now it's clearly true mod 2, since a^k = a (mod 2) for any a and any power k. For the mod 5 part, use Fermat's little theorem: if p is prime and p does not divide a, then a^(p-1) = 1 (mod p) So in your example, there are two cases. If 5 does not divide a, then a^(4b+1) = ((a^4)^b) * a = (1^b) * a = a (mod p). And if 5 does divide a, then both a^(4b+1) and a are congruent to 0 mod 5. So that covers all cases. As Virgil says, for a problem mod 10, you can simply check all the possible values for a. But using Fermat's little theorem gives a nice generalization: If p is prime, then for all integers a and b, a^((p-1)b+1) = a (mod b). More generally, if m is any squarefree integer, i.e., m is a product of distinct primes, then a^(i(m)b+1) = a (mod m). Here i(m) is Euler's i function. If m=p1*p2*...*pr, then i(m) = (p1-1)*(p2-1)*...*(pr-1). In particular, i(10) = 4, which gives your example. > > Is there an easy way to prove that, when a is an integer, > the equation `(a ^ 5) mod 10 = a mod 10' holds? > > By the way, it also seems to work for powers of 9, 13, 17, > etc. That is: `(a ^ (4b + 1)) mod 10 = a mod 10'. > > > Mauro > For a modulus as small as 10, one can simply calculate a^5 for each > possible residue, 0,1,2,3,4,5,6,7,8 and 9, > or even for 0, 1, 2, 3, 4 and 5 and note that 6 = 10-4, 7 = 10-5, 8 > = 10-2 and 9 = 10-1. === Subject: Aggregate of Two Geometric Series I have a real-life math problem. I have a product that cusers subscribe to. Every year, on average, a certain percentage of subscribers cancel. For example, if I have 10 subscribers, and the cancellation rate is 50% per year, the number of subscribers with time is a geometric series: 10, 5, 2.5, etc. To calculate the average length of a subscription, I can apply the formula: Average Length of Subscription = 1/Cancellation Rate, which is a modification of the formula for the sum of a geometric series. So in the example above, the average length of subscription is 2 years To calculate the average lifetime value of the subscription, I simply multiply this value by the price of subscription. So, if the subscription costs $100, the average lifetime value of the subscription in the example above will be $200. Now for my problem. I want to factor-in price inflation. Every year, I increase my prices by 25%. # subscribers 10 5 2.5 1.25 Price $100 $125 $156 $195 Revenue $1000 $625 $391 $244 I am assuming that since the series of revenue with time is a function with only three factors: - Initial number of subscribers - Cancellation Rate - Price Inflation and that as long as the series revenue with time tends towards 0, it seems to me to be an infinite geometric series, and a combination of two other geometric series. So, my problem is, how can I work out the formula to calculate the average life time value, including price inflation? Any suggestions? === Subject: Re: Aggregate of Two Geometric Series je v sporo.8filu novic > I have a real-life math problem. > I have a product that cusers subscribe to. Every year, on average, > a certain percentage of subscribers cancel. > For example, if I have 10 subscribers, and the cancellation rate is > 50% per year, the number of subscribers with time is a geometric > series: > 10, 5, 2.5, etc. > To calculate the average length of a subscription, I can apply the > formula: > Average Length of Subscription = 1/Cancellation Rate, which is a > modification of the formula for the sum of a geometric series. So in > the example above, the average length of subscription is 2 years In formula for the sum of geometric series is s = a[1]/(1 - q); q is qoutient (a[n+1] / a[n]) > To calculate the average lifetime value of the subscription, I simply > multiply this value by the price of subscription. So, if the > subscription costs $100, the average lifetime value of the > subscription in the example above will be $200. > Now for my problem. I want to factor-in price inflation. Every year, I > increase my prices by 25%. > # subscribers 10 5 2.5 1.25 Here is q[a] = 0.5 and a[1] = 10 > Price $100 $125 $156 $195 Here is q[b] = 1.25 and b[1] = $100 > Revenue $1000 $625 $391 $244 r[n] = a[n] * b[n] Theorem; q[r] = q[a]*q[b] Prove: q[r] = r[n+1]/r[n]=(a[n+1] * b[n+1])/(a[n] * b[n]) = (a[n+1] / a[n]) *(b[n+1] / b[n]) =q[a]*q[b] Q.E.D. > I am assuming that since the series of revenue with time is a function > with only three factors: > - Initial number of subscribers > - Cancellation Rate > - Price Inflation > and that as long as the series revenue with time tends towards 0, it > seems to me to be an infinite geometric series, and a combination of > two other geometric series. > So, my problem is, how can I work out the formula to calculate the > average life time value, including price inflation? > Any suggestions? LOOK UP I ope it helps, Peter > === Subject: Re: algrithm to list all possible partitioning of a network > Pardon me for ignorance. > Lets say, I want to generate all possible partition on a set networked > nodes. Is there an algrithm to do that? I understand that for a > general set, this would be related to Bell Number. But in my case, a > network relation is defined on the set. Theofore, using a set > partition algrithm is very ineffecient because it would generate a lot > of invalid partitioning. I assume a fairly large network of 50 nodes > but each node is connected to just a few (usually less than 3) other > nodes. > Does anyone know of an algrithm to list all partitioning of a network? What do you mean by a partition on a networked set (gra)? Do you mean all partitions of the node set into subsets such that all points in the subset are part of the same connected component of the gra? Or do you mean that all points in a subset must be directly connected (shared edge) with every other point in the subset? -- Paul ************************************************************************* Paul A. Rubin one: (517) 432-3509 Department of Management Fax: (517) 432-1111 The Eli Broad Graduate School of Management E-mail: rubin@msu.edu Michigan State University http://www.msu.edu/~rubin/ East Lansing, MI 48824-1122 (USA) ************************************************************************* Mathematicians are like Frenchmen: whenever you say something to them, they translate it into their own language, and at once it is something entirely different. J. W. v. GOETHE === Subject: a matrix problem related to eigevalues I have a matrix A(n), where A(2)= [0 1/2 1 1/2] A(3)= [0 0 1/3 0 1/2 1/3 1 1/2 1/3] A(4)= [0 0 0 1/4 0 0 1/3 1/4 0 1/2 1/3 1/4 1 1/2 1/3 1/4] And so on. My purpose is to find the eigevalues. Through matlab simulation I found they should be: 1, -1/2, 1/3, -1/4,..., (-1)^n/n But I cannot prove it. Climber === Subject: Re: binomials over negative reals > >>The binomial coefficients, I'll call it n choose m, have all sorts of >>definitions, all agreeing when both params are positive integers. It is >>usually defined by fiat to be 0 when m is negative. >>However, if you define it (er.. compute it) using factorials, that is: >> n! >>--------- >>(n-m)! m! >>and then translate to Gamma (the analytic continuation of factorials), >>you get something a bit different when both n and m are negative >>(integral or not). Since Gamma(n) has poles only at negative integers, >>1/Gamma is entire, so the only problem in computing is when n is a >>negative integer. >>Does there exist a formula for the binomials whose computation does not >>involve such singularities? The Beta function B(n,m) (which happens to >>equal Gamma(n)Gamma(m)/Gamma(n+m) ) seems close with the corresponding >>formula Integral[u^n (1-u)^m, {u,0,1}], but again that is undefined for >>both n and m negative. > but is just fine as 1/Beta(...) > First, Gamma(n)Gamma(m)/Gamma(n+m) = > integral{0 to 1} u^(n-1) (1-u)^(m-1) du > where defined. > > (m and n replaced with (m-1) and (n-1) in beta-integral.) > oops, yes. > If the m in binomial(n,m) (ie. n choose m) is a nonnegative integer, > then we can write binomial(n,m) as > > n*(n-1)*(n-2)*...(n-m+1)/m!, > > where n can be any complex number. > ... > Using this definition, then: > > binomial(-n,m) = > > (-n)*(-n-1)*(-n-2)*...(-n-m+1)/m! = > > (-1)^m n*(n+1)*(n+2)*...(n+m-1)/m! = > > (-1)^m binomial(n+m-1,m). > Yes, that's the most accepted version of binomial(n, m). it implies the > accepted definition, which, if binomial(n, m) = 0 for n pos and m > negative, then so is binomial(-n,m). Unfortunately, the above also > depends on m being an integer. > If you compute using > n! Gamma(n+1) 1 > --------- = ---------------------- = ------------------------ > (n-m)! m! Gamma(n-m+1)Gamma(m+1) Beta(n-m+1, m+1) (n + 1) > Then you get singularities in the middle of computation despite it being > total. > I'm looking for a nice way to compute this nonstandard binomial coeffs > (meaning not defined to be zero for negative m) when negative reals in > both params. I suspect there is a way to transform 1/Beta(..) into an > integral directly (that is transform 1/Integral(f(n,m)) to > Integral(g(n,m)), but my analysis skills are not up to it. Any hints? > I've tried modifying the derivation in: > http://mathworld.wolfram.com/BetaFunction.html > but I was unsuccessful. > There are many ways to calculate binomial coefficients, I am sure, but > I am in too much of a hurry today to list any more, if any others > would even came to mind now. > Of all the ones I've seen, they all seem to have difficulty (in > computation) with negative integers in both n and m > Improvising... Since binomial(n-1,m-1) = binomial(n,m)*m/n, for m and n not = 0, we can perhaps, if we want binomial(-n,-m), do something like this: (unrigorously...) We can take the limit, as x -> 0, of binomial(x-n,x-m), where binomial(x-n,x-m) = binomial(x,x-m+n) ((x-m+n)/x)((x-m+n-1)/(x-1))...(x-m+1)/(x-n+1), the k_th term in the product of n terms (following the binomial-coefficient) being (x-m+n+1-k)/(x+1-k). Now, the (1/x) in the first term cancels out with the numerator of the (n-m+1)th term if n > m >= 0. In this case, since the rest of the product is defined when we let x = 0, and so (if we assume binomial(x,x-m+n), as x ->0, approaches binomial(0,n-m) = 0 for n > m) binomial(-n,-m) = 0. (This does not seem right.) If m > n >= 0, we have only a problem with the limit as far as the binomial(x,x-m+n)/x is concerned. But this is just binomial(x-1,x-m+n-1)/(x-m+n), which is (by binomial(-n,m) equality in my 1st reply), as x -> 0, (-1)^(m+n+1) binomial(n-m-1,n-m-1)/(n-m) = (-1)^(m+n)/(m-n) So, multiply all of these together, with x = 0, to get: For m > n >= 1, binomial(-n,-m) = (-1)^(m+n+1) ((n-m-1)/(-1))((n-m-2)/(-2))...(-m+1)/(-n+1) = (-1)^(m+n+1) binomial(m-1,n-1). (I think). And, of course, binomial(-n,-n) = 1. Hopefully, these are right. Leroy Quet === Subject: Re: Calculus is irrational? Hamish Reid > Well, I'm in favour of using the common definition(s) of rational and > rationality, none of which touch on real world concerns. This is what our disagreement all comes down to. So God is a rational notion, eh? The bible-thumpers would love to hear this. > How about experiential? Experimental is observation without the subsequent reason. It doesn't adequately describe what I'm going for. > Make that does not lie in our experiences of the 'real world' and I > probably wouldn't disagree. But because I say observe instead of lies in our experiences of the 'real world' you do disagree? I honestly don't see the difference in the terms. > But since we can reason about unicorns in > detailed and coherent ways, it is -- *by definition* -- within our > rationality. If this is true God must be somewhere thanking you. Mike Helland === Subject: Re: Calculus is irrational? > Hamish Reid > Well, I'm in favour of using the common definition(s) of rational and > rationality, none of which touch on real world concerns. > This is what our disagreement all comes down to. Not at all, but since you snipped the meat of my discussion, and don;t appear to have understood the majority of it, I'll just pass on in silence. > So God is a rational notion, eh? Most notions of God are amenable to rational analysis, yes. Do you really think otherwise, or is it just blind faith that tells you someone's god doesn't exist in the real world? Is three a rational notion? You said it was, but it doesn't seem any more real world than a Christian God. I've certainly seen a three exactly as many times as I've seen a god (as long as you ignore the fact that my neighbours used to think their cat was a god...). >The bible-thumpers would love to hear > this. And...? Does your measure of what's true and what isn't depend on whether people you dislike agree with you or not? Would the fundemantalist Christian God suddenly become a rational notion for you if Fundamentalist bible-thumpers didn't want to hear this? > How about experiential? > Experimental is observation without the subsequent reason. Did I say experimental? >It doesn't > adequately describe what I'm going for. Then, rather than reusing an existing word (rational) to mean something quite different to what it means formally and informally for the rest of the world, why not (as I've suggested) coin a new word or rase, or use one of the many existing words or rases that already describe it? > Make that does not lie in our experiences of the 'real world' and I > probably wouldn't disagree. > But because I say observe instead of lies in our experiences of the > 'real world' you do disagree? You really don't get any of this, do you?... > I honestly don't see the difference in the terms. ... no, you don't. Is it rational to keep insisting that we should all adopt your wildly non-standard definition of rational just because you can't be bothered using existing (other) words for what you want to say? Should I proclaim that inifinity is now to be defined as 1/0 because Mike Helland says so? Do *you* (still) think that's the definition of infinity? Is it rational (or even reasonable) to insist that calculus is irrational because it depends on infinity? (I notice you keep snipping these reminders of what you originally claimed to believe when you barged into this group -- a wise move, I suspect). > But since we can reason about unicorns in > detailed and coherent ways, it is -- *by definition* -- within our > rationality. > If this is true God must be somewhere thanking you. What an odd non-sequitur. Why on earth would any God or god thank me for using a long-standing definition of rational? I should think any decent god would be above all that, no? Hamish === Subject: Re: Calculus is irrational? Someone asked me the question why does your rationality include > unicorns? > It doesn't. Because a unicorn has never been observed, it does not lie > in our rationality. We can theorize about one, suggest that it might > look like, but ONLY IF our defintion of rationality demands that what > we reason about must be observed is the unicorn excluded from > rationality. You're reasoning about imaginary unicorns right there. Since according to your definition of rationality, you cannot do this, and you just did, your definition is obviously wrong. Here's a pretty good definition of rationality: imagination bounded by logic. If you can imagine something, you can reason about it. why it isn't useless. But I'm not about to admit that its 100% > rational. > You agree with me that there are elements of calc that lie outside the > real world. If not irrational, what is the adjective you give to > something that lies outside the real world? > Mike Helland Don't tell me... english major? hoop dica === Subject: chaos f: R -> R. orbit{x, f(x), (f^2)(x), ...} Let {(f^i)(x)} be the rbit of x. {(f^i)(x)} is chaotic if (1) {(f^i)(x)} is not asymptotic periodic. (2) Lipunov exponent > 0. What does (1) mean? What is the definition of asymptotic periodic? ex. Let n be a positive integer and f(x) = nx (mod 1) on [0,1]. Which orbits of f are chaotic? Please give me a hint. -- === Subject: Re: Collatz Conjecture > Here is a DC project attacking the Collatz Conjecture. I am sure they will > get into more interesting projects but its not bad. They basically take > your Program/Executable, make some small changes to it and run it as a dc > project (I think you also need to supply the work), very cool... > Anyways: > http://distributed.redirectme.net/ I can't get this stupid webpage to load. Is this the same project as http://personal.computrain.nl/eric/wondrous/search.html or is it someone else trying the same thing? Eric is currently at 2^53. How far do you think he'll get before the sun burns out? === Subject: Re: Collatz Conjecture I got it going, you have to sign up here: http://distributed.redirectme.net/adduser.htm and you can get the client here: http://distributed.redirectme.net/grid.zip > Here is a DC project attacking the Collatz Conjecture. I am sure they will > get into more interesting projects but its not bad. They basically take > your Program/Executable, make some small changes to it and run it as a dc > project (I think you also need to supply the work), very cool... > Anyways: > http://distributed.redirectme.net/ > I can't get this stupid webpage to load. Is this the same project as > http://personal.computrain.nl/eric/wondrous/search.html > or is it someone else trying the same thing? > Eric is currently at 2^53. How far do you think he'll get before the sun burns out? === Subject: Re: Collatz Conjecture > or is it someone else trying the same thing? > Eric is currently at 2^53. How far do you think he'll get before the sun burns out? They are doing the same thing, the interesting thing is they are running 3x+1 because someone made an executable, added a few lines of code, provided these guys with the work, and they threw it on a grid... I guess they will run programs for people on a grid... interesting... it would be cool to get some other projects going on their grid. === Subject: Re: Collatz Conjecture Has anyone read this book How to Prove The Collatz Conjecture by Danny Fleming? anything, but was just wondering if someone had read it and what their take on it was? The review section has no entries, so that is not a good sign! Dan === Subject: Re: Collatz Conjecture === >Subject: Re: Collatz Conjecture >Has anyone read this book How to Prove The Collatz Conjecture by >Danny Fleming? >anything, but was just wondering if someone had read it and what their >take on it was? >The review section has no entries, so that is not a good sign! I didn't see it listed on Amazon, but I found it on Lulu (?). Note the description. Possibly it's good for a laugh. Title: How to Prove The Collatz Conjecture Author: Danny Fleming Category: Books > Math & Science Version: 2 Description: The first proof of the Collatz Conjecture. (210 pages) >Dan -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm electron-dot-cloud are galaxies === Subject: Re: Coulomb barrier becomes Fusion Barrier Principle; compounding of Maxwell Equations In past years when trying to reconcile the Fusion Barrier Principle with its limit of 2/3 breakeven due to the math-theorem of sere enlosed by cylinder with maximum 2/3 volume and surface area. I placed the mechanism on the idea that of the geometry shape of protons and that nuclei as they fused were sere and cylinder enclosing exercises. Something akin to the s, p electron orbitals where s is serical and the p orbitals elongated into cylinders, but these are nuclei and not electron orbitals. And that the Fusion Barrier Principle was on the microscale of 2 protons trying to fuse and so this 2/3 limit appears. But now I tend to see the FBP not as a reflection of nuclei geometry as to sere and cylinder, but rather on Maxwell Equations as being either sere or cylinder. Before I was thinking that the microscale of nuclei was creating the sere enclosed cylinder relationship, but now I am lenient on the idea that the Maxwell Equations create the sere, cylinder relationship. And it may come about that both are the cause of FBP in that the nuclei have shape as to either sere or cylinder and that the Maxwell Equations also have a sere, cylinder cause. So far, the 53 years of attempts at fusion success have all been failures. Strong evidence in support of FBP. Question: in the Nagamine muon catalyzed fusion, the point of weakness is the stickyness of muons. Which of the Maxwell Equations is equivalent to the stickyness of muons in catalyzed fusion? Is the stickyness of muons a result of both Gauss's law and also Faraday's law? So in the past 6 years working on FBP, I have moved from the geometry of nuclei as seres and cylinders in the math enclosing theorems, and moved to the Maxwell Equations as the rise of seres and cylinders. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big a where dots of the electron-dot-cloud are galaxies === Subject: Re: Criss-crossed Dodecagon: (better?) puzzle (First, do not confuse this puzzle with another I posted a few days ago, Dodecagon algorithm maze.) I am replying so as to give those who have not yet seen this post a chance to attempt to solve the version where the sequence, found by Waldby, gives a unique solution. (The original sequence I gave had many solutions.) I have copied selected parts from this thread and pasted them below: >First, start with a dodecagon (12-gon), the vertexes numbered as a >clock's face. >... we make a path that moves from vertex to >vertex, one straight line-segment between each pair of connected >vertexes, visiting each vertex exactly one time. >And the path returns to the starting-point. >But in this puzzle, consecutive vertexes MAY be connected by a >segment. >So, I give a list of nonnegative integers below. As the path is drawn >(as opposed to after the path is completed), the n_th segment crosses >a(n) previously drawn segments, where a(n) is the n_th term of the >integer-list. >The path starts at 12. And the first segment goes from 12 to 8. >... > For the 12-gon there are 893597 different sequences that begin 12,8, > and only 315027 of them belong to unique paths; I don't know if any > of those paths are interesting. :) For example, the > sequential-crossing-counts list 0 0 0 1 1 1 1 0 6 7 8 2 belongs to > a unique path. >... > Interesting. > So a puzzle (which *I* could NEVER solve, most probably) for > rec.puzzles and sci.math readers is to find the solution for the path > found by James' computer, the *unique* path which follow the rules of > the original puzzle and has the crossing-count sequence: > 0 0 0 1 1 1 1 0 6 7 8 2 > ;) > (I have no idea personally if this path is at all interesting from a > puzzle view-point..) I still have not yet solved this, but I wonder if *you* can solve the puzzle. (And if there is interest at all in this thread, perhaps can post some of the other sequences with unique solutions, if he still has them handy.) ;) === Subject: Re: Decidability in mathematics > However, I'm correct in noting that there is no field of rationals, as > the term is a social fossil. OK, mate, which of these axioms does NOT apply to the set of rational numbers? 1) For all a, b in the set, a+b is in the set 2) For all a, b, a+b=b+a 3) For all a, b, c, a+(b+c)=(a+b)+c 4) For all a, 0+a=a 5) For all a, there exists -a s.t. a+(-a)=0 6) For all a, b in the set, ab is in the set 7) For all a, b, ab=ba 8) For all a, b, c, a(bc)=(ab)c 9) For all a, 1a=a 10) For all a!=0, there exists 1/a s.t. a(1/a)=1 11) For all a, b, c, (a+b)c=ac+bc If the answer is they all apply, then the set of rationals is a field. Jack Rudd === Subject: Re: Decidability in mathematics > However, I'm correct in noting that there is no field of rationals, as > the term is a social fossil. himself has demonstrated that he is so socially inept that his judgement on the social quality of anything should convince everyone of the opposite. === Subject: Re: derivations of the simple continued fraction for e [re-posted] > (3) http://mathworld.wolfram.com/e.html claims the desired result > follows immediately from re-writing the usual power series as > e = 2 + (1/2)(1 + (1/3)(1 + (1/4)(1 + (1/5)(1 + ...)))). > I would appreciate a hint on how that works, as I must be missing > something obvious. Is that a false claim by MathWorld? === Subject: Re: Direct sum = free product for non-abelian groups??? Visiting Assistant Professor at the University of Montana. >I'm not much up on the current rage of category theory. >I always thought that the direct sum in group theory occurs when we >have a group G, with subgroups H and K; and >(i) H intersect K = {1} >(ii) HK = G >(iii) H and K are both normal subgroups of G >in which case we write G = H + K, and say that G is the direct sum of >subgroups H and K. >Someone has written me that in current mathematical usage, the above >terminology is incorrect; and that direct sums with this definition >are _only_ used in the category of _abelian_ groups. Well, when there are only finitely many terms, the direct sum and the direct product are isomoric. So most people reserve the direct sum symbol to groups which are written additively, that is, abelian groups. >Furthermore, the direct sum of non-abelian groups H and K is >apparently now considered to be what was formerly termed the free >product of H and K. >Is this correct? Almost. What happens is that the direct sum of abelian groups is a coproduct in the categorical sense for the category of abelian groups; the analogous construction for not-necessarily-abelian groups is the free product. To understand this, consider the following definition of the direct product of groups: DEF. Let {G_i}_{i in I} be a family of groups. A group P is called the direct product of the G_i if and only if: (i) For each i there exists a homomorism pi_i: P -> G_i (called the projections); and (ii) UNIVERSAL PROPERTY: Given any group K and any family of group homomorisms f_i:K->G_i, there exists a unique homomorism f:K->P such that (pi_i)f = f_i for each i. If you want to think of P as the cartesian product of the G_i with coordinate-wise multiplication, define pi_i to be the projection onto the i-th coordinate, and define f:K->P to be the map given by: f(k) is the tuple whose i-th coordinate is f_i(k). The advantage of this definition is that it can be used in many other contexts, such as semigroups, rings, abelian groups, etc., and so many of the proofs go over. Now, consider the dual definition, obtained by reversing the arrows: DEF. Let {G_i}_{i in I} be a family of groups. A group C is called the coproduct of the G_i if and only if: (i) For each i there exists a homomorism inc_i:G_i -> C (called the inclusions); and (ii) UNIVERSAL PROPERTY: Given any group K and any family of group homomorisms f_i:G_i->K, there exists a unique homomorism f:C->K such that f(inc_i) = f_i for each i. If you restrict all groups to be abelian groups, then it is not hard to verify that the direct sum of the G_i satisfies the properties. The maps inc_i are given by letting inc_i(g) be the element which has g in the i-th coordinate and 0 in all others; the map f:C->K is given by taking an element (g_i) in the direct sum, which will therefore contain only finitely many nonzero entries, and map it to the sum of f_i(g_i) (only finitely many terms are nonzero, so this makes sense). However, if you do not restrict to abelian groups, the direct sum no longer satisfies the universal property. The group that does is the free product of the G_i. If you place the category of abelian groups as part of a larger family of categories with similar properties (the abelian categories), then they all have an operation similar to the direct sum with the same universal property; so the direct sum symbol and terminology is sometimes reserved for this kind of category; the analogous construction for arbitrary groups (the subgroup of the direct product consisting of elements with finite support) is then often called the restricted direct prouct, and the usual direct product the cartesian direct product or unrestricted direct product. And the construction that has the same universal property is the free product. Note that the restricted and unrestricted direct products coincide when the family is finite. >How is the situation I define above as direct sum >currently named? Surely not that G equals the direct product of H and >K (although it is obviously isomoric to such a product)? I have seen it called G is the INTERNAL direct product of H and K, to distinguish it from the situation where H and K are arbitrary groups and their direct product H x K = {(h,k) : h in H, k in K} is called the external direct product. Never found much use for the distinction myself, though. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Easy Subtraction Hello Everyone, My name is Daniel and I'm a Junior who his undergrad in Computer Engineering and Premed. Maybe this is the most silly math concept that you have read ever. Maybe it is fascinating. Each man for themselves. If you do like it, do share you comments with me. I take constructive criticism quite well. This works only if you are subtracting a 2 digit number (like 23) from a three digit number (like 234) (or vice-versa), or subtracting a three digit number (234) from a four digit number (2345). The basic idea is that one number should be one digit greater than the other. ok heres an example, say I'm computing 329-78. you subtract 329 by 100 i.e. 329-100=229, also 100-78=22. Then you add the two up, 229+22=251. which is exactlyequal to 329-78. Now you can do subtracting mentally. You should already be comfortable with a little mental adding and subtracting for this to help. It also works the other way. 78-329. Do same as above, and put a negative sign in front. It becomes more tougher if you are subtracting a three digit number from a four digit number. But it still works. === Subject: Re: Easy Subtraction > Hello Everyone, > My name is Daniel and I'm a Junior who his undergrad in Computer > Engineering and Premed. > Maybe this is the most silly math concept that you have read ever. > Maybe it is fascinating. Each man for themselves. If you do like it, > do share you comments with me. I take constructive criticism quite > well. > This works only if you are subtracting a 2 digit number (like 23) from > a three digit number (like 234) (or vice-versa), or > subtracting a three digit number (234) from a four digit number > (2345). The basic idea is that one number should be one digit greater > than the other. > ok heres an example, > say I'm computing 329-78. > you subtract 329 by 100 i.e. 329-100=229, > also 100-78=22. > Then you add the two up, 229+22=251. which is exactlyequal to 329-78. > Now you can do subtracting mentally. You should already be comfortable > with a little mental adding and subtracting for this to help. > It also works the other way. 78-329. Do same as above, and put a > negative sign in front. > It becomes more tougher if you are subtracting a three digit number > from a four digit number. But it still works. 329 - 78 = 329 - any way you feel like writing 78. Since 78 = 100 - 22 we can calculate 329 - ( 100 - 22 ) as 329 - 100 + 22 = (329 - 100) + 22. You can also do 329 -78 as 329 - (70 + 8) ie 329 - 70 - 8 or even (remember - 70 is the same as -100 + 30) 329 - 100 + 30 -8. Personally I hardly ever add or subtract the numbers that are given to me. === Subject: Re: Easy Subtraction > Hello Everyone, > My name is Daniel and I'm a Junior who his undergrad in Computer > Engineering and Premed. > Maybe this is the most silly math concept that you have read ever. I've always found subtraction rather silly. > ok heres an example, > say I'm computing 329-78. > you subtract 329 by 100 i.e. 329-100=229, > also 100-78=22. > Then you add the two up, 229+22=251. which is exactlyequal to 329-78. Indeed (329 - 100) + (100 - 78) = 329 + (100 - 100) - 78 = 329 - 78 If this helps you to subtract numbers then, by all means, use it. I use something kind of similar when squaring numbers near 50: 52^2 = (50 +2)*(50 + 2) = 50^2 + 2*100 + 2*2 = 2500 + 200 + 4 for some reason I find it easier to think about this way (probably because I already know what 50 squared is). Whatever works for ya. > Now you can do subtracting mentally. You should already be comfortable > with a little mental adding and subtracting for this to help. > It also works the other way. 78-329. Do same as above, and put a > negative sign in front. > It becomes more tougher if you are subtracting a three digit number More tougher? you mean Tougherer. > from a four digit number. But it still works. Good luck with the subtraction. adam === Subject: Eigenvector problem Suppose I have a stochastic matrix A. Define B = A + A^2 + ... (note that lim(k->oo) A^k is the zero matrix). I want to solve for the eigenvectors (& values) of B. Any ideas? Slightly easier version: A is an n x n lower-triangular stochastic matrix; define C = A + A^2 + ... + A^n (since A^(n+1) is the zero matrix). Now can I solve for the eigenvectors (& values) efficiently? Any and all help (including, or especially, pointers to relevant literature) would be greatly appreciated! -Shlomo- === Subject: Re: Eigenvector problem > Suppose I have a stochastic matrix A. Define B = A + A^2 + ... (note > that lim(k->oo) A^k is the zero matrix). I want to solve for the > eigenvectors (& values) of B. Any ideas? Is a stochastic matrix the transition matrix for a Markov chain? So all the rows (or columns, depending on how you write it), add up to one. Then, by a Perron-Frobenius theorem, the maximum eigenvalue of A is unity. I believe that this implies I + A + A^2 + ... = (I - A)^(-1) So B = A (I - A)^(-1) I think you might find something else helpful in a reference that describes the Perron-Frobenius theorems, both for indecomposable and decomposable matrices. Can you reformulate your problem such that only the maximum eigenvalue and corresponding eigenvector are of interest? -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Eigenvector problem >Suppose I have a stochastic matrix A. Define B = A + A^2 + ... (note >that lim(k->oo) A^k is the zero matrix). Huh? If by a stochastic matrix you mean a nonnegative matrix whose rows sum to unity, then A^k is stochastic for all k. -- Steen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Eigenvector problem >Suppose I have a stochastic matrix A. Define B = A + A^2 + ... (note >that lim(k->oo) A^k is the zero matrix). > Huh? If by a stochastic matrix you mean a nonnegative matrix whose rows > sum to unity, then A^k is stochastic for all k. Right. Oops! Here's a corrected statement of my actual problem: Given non-negative matrix A with zero diagonal and rows sum to < 1, define B = A + A^2 + ... Problem: efficiently find all eigenvalues/eigenvectors of B (an algorithm is fine). -Shlomo- === Subject: Re: Eigenvector problem >Suppose I have a stochastic matrix A. Define B = A + A^2 + ... (note >that lim(k->oo) A^k is the zero matrix). Huh? If by a stochastic matrix you mean a nonnegative matrix whose rows > sum to unity, then A^k is stochastic for all k. > Right. Oops! Here's a corrected statement of my actual problem: > Given non-negative matrix A with zero diagonal and rows sum to < 1, > define B = A + A^2 + ... Problem: efficiently find all > eigenvalues/eigenvectors of B (an algorithm is fine). > -Shlomo- Since I + B = (I - A)^-1, the problem can be solved in general position (ie. assuming A has a complete set of eigenvectors) by noting: x in Spec(B) <==> 1/(x + 1) in Spec(I - A) <==> 1 - 1/(x+1) in Spec(A) and this correspondance in eigenvalues extends to a common basis of eigenvectors for A and B. Regarding the eigenvalues of A, note that since the diagonal of A contains only zero entries, the trace(A) = sum of eigenvalues of A will necessarily be zero. === Subject: Re: Eigenvector problem >> Given non-negative matrix A with zero diagonal and rows sum to < 1, >> define B = A + A^2 + ... Problem: efficiently find all >> eigenvalues/eigenvectors of B (an algorithm is fine). >Since I + B = (I - A)^-1, the problem can be solved in general position >(ie. assuming A has a complete set of eigenvectors) by noting: This doesn't require a complete set of eigenvectors. >x in Spec(B) > <==> 1/(x + 1) in Spec(I - A) > <==> 1 - 1/(x+1) in Spec(A) >and this correspondance in eigenvalues extends to a common basis >of eigenvectors for A and B. Even without a complete set of eigenvectors, v is an eigenvector of B for eigenvalue x iff v is an eigenvector of A for eigenvalue 1-1/(x+1). In general, for any analytic function f in a neighbourhood of Spec(B), Spec(f(B)) = f(Spec(B)), and if v is an eigenvector of B for eigenvalue x then v is an eigenvector of f(B) for eigenvalue f(x). Here f(B) is defined via the holomoric functional calculus. For rational functions, it's easy to prove directly. In this case, f(z) = 1 - 1/(x+1) is one-to-one from C {-1} onto C {1}, its inverse being also a rational function, so we have if and only if. Numerically finding all the eigenvectors and eigenvalues of A (or B) is not, I think, much different from finding the eigenvectors and eigenvalues of a matrix in general. There are a number of standard techniques. In practice, use Maple, Matlab or other software packages which have good algorithms. === Subject: Exponential/Dirichlet-combined Series Let, for any sequence {a(k)} where the sum below converges for some range of x's, A(x) = sum{m=1 to oo} a(m) *x^m /(m! *m^x). (So, with this definition, A(0) = 0 if the sum converges uniformly{necessary?} in the neighborhood of x = 0.) Now, we can get the sequence {a(k)} recursively from A(x) if A is analytic at x = 0. (More precisely, we can get ONE sequence {a(k)} which gets A(x) when used in the sum above.) a(1) = A'(0) (ie. the derivative of A at x = 0); And a(m) = A{m}(0) - sum{j=1 to m-1} binomial(m,j) (-ln(j))^(m-j) a(j), where A{m}(0) is the m_th-order derivative of A(x) at x = 0. (I believe..maybe) Is there a direct way to calculate the a's based upon the derivatives of A? And, is the sequence {a(k)} the unique sequence where A(x) = sum{m=1 to oo} a(m) *x^m /(m! *m^x) ? In other words, is this sum a generating-function? === Subject: Re: finding a function >> given that: >> h=7c; 2h=18c; and 3h=35c; >> 20c=a; >> 10a=l; and >> 3l=s >> is it possible to express 's' as a function of 'h' and if so what is >> the function and how did you arrive at that please ? > Well, if h=7c then 2h CAN NOT also equal 18c nor can 3h be 35c, > unless > both h and c are 0. Yes, stupid Ivan :( What I think I meant (hope I get it right this time) is that there is a function of h such that: h| 1 | 2 | 3 | -------------- c| 7 | 18| 35| As you explained so politely and clearly, s is no more complicated than 600c ? So the problem now is determining f(h) for those values above. Presumably, if c=f(h) and s=600c, I can calculate h for any value of s as s=600[f(h)] ? Ivan. === Subject: Re: Focus on point of dispute, more math >> There is a much more general question. Why does anyone read these >> Harris threads at all? >> >> I read the replies to JSH, knowing that most of the math is totally >> beyond me, in the hope and belief that as I study more mathemathics >> (which I am doing even now) that even such a casual and meaningless >> (to me) exposure to the terminology and presentation will benefit me. >> Already I have learned things like the existence of rings and >> fields and Galois Theory and I even know that these are to do >> with algebra :) > Notice the community aspect, as I've noted repeatedly. > The math community is rife with people who know that pleasing the > group IS considered substance. > It's part of the math fashion show, where people believe belief is > substance. > I must admit I still find it fascinating. Let's see about tweaking > things a bit more. >> I read the actual JSH posts because I have recently found some >> definite pleasure in watching a grasshopper wriggle after being >> skewered by a pin :) (and even without being able to follow the >> maths, I can tell by the nature of the JSH replies (or lack of) when >> he has been well and truly skewered) >> ivan. > What's telling to me is that animosity, which is rather primitive. > There's an anger response from people who go from admitting, like this > poster did, a basic ignorance of mathematics, to making the claim they > apparently think the group wants, which is their perception that I'm > harmed in some way. > So this poster who's opinion, by his own admissions in terms of > mathematics is worthless, still feels that his opinion is of value, > clearly because he's at least learned the true nature of the math > community. > My assessment is that for mathematicians the *claim* of logic and > mathematical preciseness is part of the show, and not even people > outside of the math community--who want in, like this fellow--actually > believe it. > Harris James, I don't know why you replied to this - I was answering a question posed by another person. But to respond to your unwarranted and baseless attack on my motivation in posting my reply, I say only that I don't particularly want to be part of any group other than my immediate family and that even when I have been a part of a group I am noted for following the party line only when the party line agrees entirely with my own feelings. Furthermore, I am, as you noted, one of those non-mathematicians that you regularly appeal to, that they may step forward and give their opinion of your work. I repeat, based on your reply(ies) and, more often your -lack- of reply(ies), I can only judge that you have been thoroughly discredited. The logic isn't difficult to follow: 1. The burden of proof is on the proponent of an argument 2. You claim you have a proof of something 3. Therefore the burden of proof lays with you 4. People who would normally be regarded as qualified to present counter- aruguments to you have done so 5. You have not refuted the counter arguments(*) 6. Therefore your claim is invalid (*) Amongst other things, there was a post by Arturo Magidin dated 12 August that he claimed showed your proof to be invalid that you have not responded to. I actually went to the trouble of printing this and showing it to a mathematician (the one who is actually attempting to force some basic knowledge of maths through my thick skull :)) who told me that she didn't have the knowledge of the issue and so couldn't say whether your argument has been disproved but that she couldn't see any problem with the work of Arturo Magidin. That is, a mathematician who (presumably) is not a part of the cabel that you claim exists has said that the questions asked and the definitions given by Arturo Magidin are completely valid. Ivan. === Subject: Re: Fraud in Computer Science Publishing > > programs. How do you define programming? > > Writing: > > FAC(x,I)^LT(x,I)^~(exists A)FAC(A,I)^LT(A,I)^LT(x,A) > > Is programming, no matter what system translates that input into a program. > > Sam > 1. Do you consider Predicate Calculus wffs to be programs? It may or may not be a program, depending on the context. (If it is written in a logic textbook, it is not a program; if it is passed to a Prolog interpreter, it is.) > 2. How do you define a program? A program is a system of controls which cause a computer to behave in a desired fashion. > 3. Do you know of a better way to sepcify the largest proper factor > of a given number? Yes: the largest proper factor of a given number. > 4. Do you know of a way to specify it that isn't programming? Yes: the largest proper factor of a given number. This may still be used as the input to a program, of course. > 5. Do you know of a simpler way to specify it? No. > 6. How do you think that a Mathematician would specify it? The largest proper factor of a given number. > 7. Did you know that the state-of-the-art in Program Synthesis is to > specify the program requirement as a Predicate Calculus wff? Not surprising. I doubt it's really state-of-the-art though; I expect people having doing that for decades. > 8. Do you see a qualitative difference between a Predicate Calculus > wff and a computer program? Yes; a wff is an assertion, while a program is a command. > 9. How about the fact that a Predicate Calculus wff has no assignment, > conditional execution, loops or the possibility of not terminating? Many programming languages don't have any of these things, except the last. And what would it mean for a wff to not terminate anyway? > 10. Do you think that Predicate Calculus wffs and computer programs > are at the same level of abstraction? No, wffs are at a higher level of abstraction, because they need not be associated with any sort of computer. > 11. Do you think that Predicate Calculus wffs and computer programs > are in a one-to-many relationship in that one wff can be implemented > by multiple programs based on different algorithms, but for one > program there is essentially only one wff that represents the > functionality that it provides (ignoring permutations of the > conjuncts/disjuncts and other logical redundancy)? No. There could be many different wffs which represent the functionality of a given computer program. > 12. Do you think that computer programs have to be analyzed to > determine what function they compute, and in general you cannot do > that? No. The (partial) function determined by program x is already determinate, independently of any analysis. > 13. Do you think that there is a corresponding process of analysis to > detemine what a Predicate Calculus wff is doing, or is it the final > word as to the definition that it is conveying? No. The statement expressed by wff x is already determinate, independently of any analysis. > 14. Do you think that a simple wff may require complex programs to > implement it (defining simple and complex informally or > intuitively)? Possibly; if the wff involves a complicated system of objects it may be difficult or impossible to evaluate it. > 15. Do you see value in being able to determine the wff that a > particular program computes? Maybe; it depends on the program. > 16. Do you see value in being able to determine computer programs that > implement a given predicate calculus wff? Maybe, it depends on the wff. > 17. Does my system determmine programs that compute a given wff? I wouldn't know. > 18. Do you know of any system other than mine that determines programs > that compute a given wff? Yes; Prolog springs to mind immediately. === Subject: Re: Fraud in Computer Science Publishing > Some Basic Facts > 1. The research community has failed miserably in its attempts to > solve the problems of computer programming. > a. After years of attempting to develop a program synthesis system, > they have either given up or are fraudulently presenting new > programming languages as program synthesis (that don't even output > programs in the first place!) What exactly is program synthesis? The way you use it, it doesn't seem to refer to the synthesis of programs. I would guess that the standard definition is different than both. > b. Relational DataBases [...are the model referred to in (c)...] > c. Incredibly, they declared that the myriad of database structures > that their model did not model were off limits and undesirable, > despite their use for good benefit in existing systems. That's standard scientific practice: If you can't answer a question, ignore it until you can. (Actually, I bet there's a lot of research into non-relational database structures.) > d. Even more incredibly, the American computer industry bought into > this farce and invested inordinate amounts of money in this pitifully > limited model. > e. For practical techniques they present foolish procedures such as > Extreme Programming (XP) that advocates irresponsible procedures such > as making changes as soon as they are requested by the user, rather > than batching multiple requests (in which the user changing his mind > is cancelled out and we avoid implementing it.) You can hardly blame the research communnity for these last two things. They would be the fault of the industry, which consists of overly gullible programmers and managers, most of whom are not computer scientists. > f. Theoretical branches of Computer Science, e.g. Theory of > Computation (Computability) and Recursion Theory, lack a formalization > of even the most primitive concepts, such as a recursive or > recursively enumerable set or function. Not this twaddle again! They've been formalized since the 1920s, at least. If you can't understand the formalism, that's your problem, not theirs. [...] > h. When faced with solutions to these problems, there is strong > rejection by the publishing community, I don't doubt that.... >with fake reasons presented: > claiming that they contain the material that is being presented. > Examples are never given of the actual material as it appears in > published literature. Of course not! That would violate the copyright. > ii.) Endless delays with vague demands such as formalize everything > even when the presentation is rigorous Sounds like you... === Subject: Re: Golbach, Godel & Murray Gell-Mann [about Goldbach's conjecture] : I'm assuming it can't be proven to be unprovable (since that would prove : it to be true) but is it possible that it could be something like : Goodstein's sequence (i.e. we can prove that it's undecidable, but not : within Peano arithmetic)? Or am I babbling? Undecidability is always with respect to some system of axioms, such as (for example) Peano Arithmetic. If GC is false, then it is decidable within PA, since any purported counterexample can be verified by a terminating program. But it's possible that GC is true yet undecidable in PA (since there's no known way to establish the truth of GC by a terminating program). Ted === Subject: Golbach, G.9adel & Murray Gell-Mann There is no reason to believe that Goldbach's conjecture is undecidable, but suppose it is. It would then be true, even though unprovable, because there could be no exception to it. The existence of any even number greater than 2 that is not the sum of two odd primes would disprove the conjecture and therefore contradict its undecidability. Is this correct? Gell-Man suggests there could be no exception. Why not? I believe there could be, but with no hope of finding it, should it exist. Am I correct in this? L === Subject: Re: Golbach, =?ISO-8859-1?Q?G=F6del_=26_Murray_Gell-Mann?= > There is no reason to believe that Goldbach's conjecture is undecidable, > but suppose it is. It would then be true, even though unprovable, because > there could be no exception to it. The existence of any even number greater > than 2 that is not the sum of two odd primes would disprove the conjecture > and therefore contradict its undecidability. > Is this correct? Gell-Man suggests there could be no exception. Why not? > I believe there could be, but with no hope of finding it, should it exist. > Am I correct in this? If Goldbach's conjecture is undecidable, it means that it cannot be proved or disproved in finite time on a Turing machine (or equivalent). If Goldbach's conjecture is false, there is an even number greater than 2 that is not the sum of two odd primes. This is something that could be discovered by an exhaustive search, since both primes are bounded above by the even number. It can be programmed and a Turing machine would find it in finite time. This contradicts the undecidability assumed above. Therefore: Goldbach's conjecture being undecidable -> Goldbach's conjecture is true by contradiction. Gell-Man is correct because the nature of the conjecture is decidable if it is false. It is easy to write an algorithm such as the following: x=4 Not found Repeat until found: { for i=1 to x/2, see if i and x-i are both prime. If they are, note that this x doesn't work. If this x doesn't work, set x=x+2 } The above could be formalized as a computer program easily. The only issue would be waiting for it to finish and hoping it doesn't run out of memory first (something a Turing machine doesn't need to worry about). -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Golbach, G.9adel & Murray Gell-Mann > There is no reason to believe that Goldbach's conjecture is undecidable, > but suppose it is. It would then be true, even though unprovable, because > there could be no exception to it. The existence of any even number greater > than 2 that is not the sum of two odd primes would disprove the conjecture > and therefore contradict its undecidability. > Is this correct? Gell-Man suggests there could be no exception. Why not? > I believe there could be, but with no hope of finding it, should it exist. Why? If an exception exists, you can in principal find it, by testing all even numbers until you reach the exception. BTW, I reckon Goldbach is an excellent candidate for true but unprovable. There are very strong probabalistic arguments for believeing it to be true because larger numbers have lots of smaller primes to select from in forming the sum, and it has been tested right out to some very large numbers. The probability of a given number being an exception drops fast enough (as the size of the number increases) to make it very likely that there are no exceptions. But despite 200 years of effort no actual proof has been found. Goldbach may just happen to be true for probabilistic reasons, but there is no actual proof for all even numbers. > Am I correct in this? === Subject: Re: Golbach, G.9adel & Murray Gell-Mann >> There is no reason to believe that Goldbach's conjecture is undecidable, >> but suppose it is. It would then be true, even though unprovable, because >> there could be no exception to it. The existence of any even number >greater >> than 2 that is not the sum of two odd primes would disprove the conjecture >> and therefore contradict its undecidability. >> Is this correct? Gell-Man suggests there could be no exception. Why >not? >> I believe there could be, but with no hope of finding it, should it >exist. >Why? If an exception exists, you can in principal find it, by testing all >even numbers until you reach the exception. >BTW, I reckon Goldbach is an excellent candidate for true but unprovable. >There are very strong probabalistic arguments for believeing it to be true >because larger numbers have lots of smaller primes to select from in forming >the sum, and it has been tested right out to some very large numbers. The >probability of a given number being an exception drops fast enough (as the >size of the number increases) to make it very likely that there are no >exceptions. But despite 200 years of effort no actual proof has been found. >Goldbach may just happen to be true for probabilistic reasons, but there is >no actual proof for all even numbers. I'm assuming it can't be proven to be unprovable (since that would prove it to be true) but is it possible that it could be something like Goodstein's sequence (i.e. we can prove that it's undecidable, but not within Peano arithmetic)? Or am I babbling? Alan -- Defendit numerus === Subject: Re: Golbach, G.9adel & Murray Gell-Mann >> There is no reason to believe that Goldbach's conjecture is undecidable, >> but suppose it is. It would then be true, even though unprovable, because >> there could be no exception to it. The existence of any even number >greater >> than 2 that is not the sum of two odd primes would disprove the conjecture >> and therefore contradict its undecidability. >> Is this correct? Gell-Man suggests there could be no exception. Why >not? >> I believe there could be, but with no hope of finding it, should it >exist. >Why? If an exception exists, you can in principal find it, by testing all >even numbers until you reach the exception. >BTW, I reckon Goldbach is an excellent candidate for true but unprovable. >There are very strong probabalistic arguments for believeing it to be true >because larger numbers have lots of smaller primes to select from in forming >the sum, and it has been tested right out to some very large numbers. The >probability of a given number being an exception drops fast enough (as the >size of the number increases) to make it very likely that there are no >exceptions. But despite 200 years of effort no actual proof has been found. >Goldbach may just happen to be true for probabilistic reasons, but there is >no actual proof for all even numbers. > I'm assuming it can't be proven to be unprovable (since that would prove it to > be true) No. It could be false; that would be a good reason why we can't prove it. The problem lies more with the converse statement there exists an even number which is NOT the sum of two primes. This cannot be true and unprovable, as if it is true it is easy to prove by finding the even number. > but is it possible that it could be something like Goodstein's sequence > (i.e. we can prove that it's undecidable, but not within Peano arithmetic)? Yes, good point. It is provable (or unprovable) only in some axiom system. Its certainly provable in the axiom system PA + GC, where GC is the Goldbach conjecture. Trouble is, I don't know if PA + GC is a consistent system .... > Or > am I babbling? Not at all. Or rather, I wouldn't know ;) > Alan > -- > Defendit numerus === Subject: =?ISO-8859-1?B?UmU6IEdvbGJhY2gsIEf2ZGVsICYgTXVycmF5IEdlbGwtTWFubg==?= [...] |>BTW, I reckon Goldbach is an excellent candidate for true but unprovable. |>There are very strong probabalistic arguments for believeing it to be true |>because larger numbers have lots of smaller primes to select from in forming |>the sum, and it has been tested right out to some very large numbers. The |>probability of a given number being an exception drops fast enough (as the |>size of the number increases) to make it very likely that there are no |>exceptions. But despite 200 years of effort no actual proof has been found. |>Goldbach may just happen to be true for probabilistic reasons, but there is |>no actual proof for all even numbers. It's possible. The heuristic evidence is good enough to convince people that it's almost certainly true. I think one can be more optimistic about it's being provable, however. There's a famous proof that every sufficiently large even integer is either a sum of two primes or a sum of a prime and the product of two primes. It seems as though something of a similar kind but cleverer might well do the job. It was also proven that all large enough odd numbers are sums of three primes. It certainly would be a pain if it turned out those things could be proven but Goldbach could not. |I'm assuming it can't be proven to be unprovable (since that would prove |it to be true) but is it possible that it could be something like |Goodstein's sequence (i.e. we can prove that it's undecidable, but not |within Peano arithmetic)? Or am I babbling? I re-word-wrapped that. I think you mean we can prove it's true, but not in PA, and we can prove that it's undecidable in PA (also not in PA). Yes, in principle it could be like that; it has the same logical type as the Goedel sentence for a theory like PA which is like that, namely, it has the form of P(n) holds for every integer n where P is primitive recursive computable. The Goodstein sequence result is more complex in the sense that it says for each n there exists an N large enough to make some primitive-recursive property of the pair (n,N) true. As I recall it's independent of PA because the N grows faster as a function of n than any of the functions which PA is able to recognize as computable functions. That's like the Paris-Harrington result, a theorem unprovable in PA for the same reason. Goldbach could not be unprovable for that reason, because it only involves the one unbounded quantifier. Keith Ramsay === Subject: horseshoe map How to construct the periodic table for the horseshoe map for periods up to 6? -- === Subject: Re: How do you discuss mathematics with a person who thinks mathematics is a part of ysics? > ... He calls the mathematics that is done without any connection to the > ysical world imaginary. > Like it or not, every major development in mathematics ( or in any > science for that matter} is either inspired/provoked by reference, > directly or indirectly, to a real ysical situation. You've quantified yourself into a trivial statement. A *major* development by definition implies a real ysical situation. > This applies perhaps as well to already realized or systemematized > work. The form of equations in elecagnetic theory for example,( > that once made Boltzmann to remark if it was not God's work afterall ) > when ysically/geometrically arranged on paper - are visualizations > of beautiful relationships and concepts with symmetry. If information > is ordered into a set, it is also a real world exercise. I would argue a G.U.T. would use 1% of the known maths for a complete ysical model. What ysical construct does 'there is no largest prime' relate to? The closest would be that our encryption is unlimited, as opposed to unbreakable, a theoretic difference. Herc === Subject: Re: How do you solve this? You haven't solved for x in terms of a. -Justin : 2^x = a*x : => ln_2 (2^x) = ln_2 (a*x) : => x = ln_2 (a*x) === Subject: Re: How do you solve this? Ok, fair enough. Determine an a and you can see if it solvable or not. But for the general case, it is not solvable. It is however solvable for say a = 10. Lurch > You haven't solved for x in terms of a. > -Justin > : 2^x = a*x > : => ln_2 (2^x) = ln_2 (a*x) > : => x = ln_2 (a*x) === Subject: Re: How do you solve this? Obvously by inspection, when a =2, x =1 and x=2 are solutions. > Ok, fair enough. Determine an a and you can see if it solvable or not. But > for the general case, it is not solvable. It is however solvable for say a > = 10. > Lurch > You haven't solved for x in terms of a. > -Justin > : 2^x = a*x > : => ln_2 (2^x) = ln_2 (a*x) > : => x = ln_2 (a*x) === Subject: Re: how to calculate a paralell line > Lines, being infinite in extent, don't have a beginning nor an end. > Let m = slope of lines, d = distance between lines > First line y = mx + a > Second line y = mx + a + d/cos t Yes, that's better. m = tan t; 1 + m^2 = csc^2 t; cos t = 1/sqr(1+m^2) y = mx + a + d.sqr(1+m^2) ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: how to make a (fired) bullet stay put Hello to all, Just discovered Google Groups and think this is probably the best place to settle a dispute that my friends and I have. I appologize in advance if I don't explain this very well, so here goes... The question is: How fast would an aircraft have to fly to make a bullet fall straight down if shot from the rear of the aircraft? Now just for the sake of argument lets say that the bullet travels at 400m. I think the aircraft would have to travel at 400m or slightly above that. My reasoning is this, if an aircraft is flying 200m (feel free to interchange aircraft with car, truck, or bicycle) and it fires a (400m) bullet forward, the initial speed is 600m because the bullet is already traveling at 200m. I realize that drag will slow the bullet very quickly but initially it is traveling at 600m Is that correct or not? If that is correct, would it not be true that shooting this 400m bullet out the rear of an aircraft traveling at 400m cause the bullet fall straight down? My friends disagree but don't even try to come up with an answer. I take that back, one says that for the fraction of a second, when the bullet is fired, it just acts as though it is standing still and is therfore shot out in the opposite direction at 400 m. That is wrong to me because the bullet is constantly traveling at 400m when it is in the aircraft, even for that split second. Any help is greatly appreciated!!! === Subject: Re: how to make a (fired) bullet stay put > Hello to all, > Just discovered Google Groups and think this is probably the best > place to settle a dispute that my friends and I have. I appologize in > advance if I don't explain this very well, so here goes... > The question is: How fast would an aircraft have to fly to make a > bullet fall straight down if shot from the rear of the aircraft? Now > just for the sake of argument lets say that the bullet travels at > 400m. I think the aircraft would have to travel at 400m or > slightly above that. My reasoning is this, if an aircraft is flying > 200m (feel free to interchange aircraft with car, truck, or bicycle) > and it fires a (400m) bullet forward, the initial speed is 600m > because the bullet is already traveling at 200m. I realize that > drag will slow the bullet very quickly but initially it is traveling > at 600m Is that correct or not? If that is correct, would it not > be true that shooting this 400m bullet out the rear of an aircraft > traveling at 400m cause the bullet fall straight down? My friends > disagree but don't even try to come up with an answer. I take that > back, one says that for the fraction of a second, when the bullet is > fired, it just acts as though it is standing still and is therfore > shot out in the opposite direction at 400 m. That is wrong to me > because the bullet is constantly traveling at 400m when it is in the > aircraft, even for that split second. Any help is greatly > appreciated!!! This solution will assume ideal conditions--the aircraft will not affect the bullet other than the aircraft causing the bullet to be moving at the rate of X m.For the bullet to fall straight down it must be true that the horizontal component of the velocity of the bullet is 0. The force of gravity is only in the vertical direction so we can ignore this force for now. The only horizontal forces acting on the bullet are the initial velocity of the bullet and the velocity of the aircraft as the bullet leaves the barrel of the gun. We want these forces to equal but in opposite direction. So the aircraft should be travelling at the rate of 400m. === Subject: Re: how to make a (fired) bullet stay put > slightly above that. My reasoning is this, if an aircraft is flying > 200m (feel free to interchange aircraft with car, truck, or bicycle) > and it fires a (400m) bullet forward, the initial speed is 600m > because the bullet is already traveling at 200m. I Handgun bullets are usually in the range of 800 fps to 1300 fps while rifle bullets are usually in the range of 1700 fps to 3200 fps...The 400 m bullet would be about 587 fps and I think that even a 45 acp would beat that. === Subject: Infinite Sum = Another Sum (A Power Series) Let {a(k)} and {b(k)} be any sequences defined for all positive integer indexes k, and where the sums below converge absolutely. Then: oo --- m m x b(m) a(m) exp(x b(m)) / ----------------------------- = --- m! m=1 oo m --- m --- x / m m / --- / | | a(k) b(k) , --- m! --- k / m=1 k=1 which is, in linear-mode: sum{m=1 to oo} x^m b(m)^m a(m) exp(x b(m)) /m! = sum{m=1 to oo} (x^m/m!) sum{k=1 to m} binomial(m,k) a(k) b(k)^m. Now, this identity is actually obvious, and easily found by simply rearranging the sums a little. But the general result leads to interesting specific examples. (Maybe you can find some on your own...) For instance, if I did not err: sum{m=1 to oo} x^m m^m exp(-x m) /m! = x/(1-x). sum{m=1 to oo} x^m h(m,m) /m! = sum{m=1 to oo} x^m (-1)^(m+1) exp(x/m) /(m^m m!), where h(0,m) = 1, h(n,m) = sum{k=1 to m} h(n-1,k)/k. (I am a little uncertain that I did not make a mistake here.) And: The n_th derivative of A_n(x) at x= 0, where A_n(x) = sum{m=2 to oo} x^m (ln(m))^(m-n) /(m^(x+1) m!), is (-1)^(n+1) (H(n) -n), where H(n) = 1 +1/2 +1/3 +..+1/n, the n_th harmonic number. (By the way, H(n) = h(1,n).) Leroy Quet === Subject: Re: Harris > Or perhaps everything is a gigantic theatrical spam. I've skimmed a few of the posts and recently noticed he is not getting any benefit of doubt. He makes several mathematical statements, people rebuke the maths, then corrects them and that's the end of the thread. Noone admits when their counters to his proof were incorrect, and has to repeatedly address the same attacks. It won't go anywhere until the sequences of posts are as systematic as the posts should be themselves. Try a new tactic sci.math isolate where his proof if ~right~ first. http://tinyurl.com/ko46 heard of the hammer?, this is KO Herc === Subject: Re: Harris >>Or perhaps everything is a gigantic theatrical spam. > I've skimmed a few of the posts and recently noticed he is not getting any benefit of doubt. > He makes several mathematical statements, people rebuke the maths, then > corrects them and that's the end of the thread. Noone admits when > their counters to his proof were incorrect, and has to repeatedly address > the same attacks. It won't go anywhere until the sequences of posts are as > systematic as the posts should be themselves. Try a new tactic sci.math isolate > where his proof if ~right~ first. > http://tinyurl.com/ko46 heard of the hammer?, this is KO > Herc We have. He tends to post the exact same *incorrect* material repeatedly. The post you reference above has almost the exact same content as many other times. The correct portions were already agreed to. The incorrect portions were illustrated. He dismissed the objections and did a copy/paste of the entire argument again. There comes a point when replying is futile because there's nothing new to say. He just racks up unanswered objections with only a few addressed or clarified. The critical ones are never truly dealt with. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Harris >>Or perhaps everything is a gigantic theatrical spam. > I've skimmed a few of the posts and recently noticed he is not getting any benefit of doubt. > He makes several mathematical statements, people rebuke the maths, then > corrects them and that's the end of the thread. Noone admits when > their counters to his proof were incorrect, and has to repeatedly address > the same attacks. It won't go anywhere until the sequences of posts are as > systematic as the posts should be themselves. Try a new tactic sci.math isolate > where his proof if ~right~ first. > http://tinyurl.com/ko46 heard of the hammer?, this is KO > Herc > We have. He tends to post the exact same *incorrect* material > repeatedly. The post you reference above has almost the exact same > content as many other times. The correct portions were already agreed > to. The incorrect portions were illustrated. He dismissed the > objections and did a copy/paste of the entire argument again. There > comes a point when replying is futile because there's nothing new to > say. He just racks up unanswered objections with only a few addressed > or clarified. The critical ones are never truly dealt with. Fair enough, even if he is refining his work it sometimes comes across as very reasonable argument. And he is 1 against the will of 20 men! Perhaps fair systems of clarifying assertion aren't scaling, I've had similar problems to having to repeat arguments to new members and when you do finally get somewhere its all of a sudden out of scope for the group you can't just win. To go on through this much detail there must be something to his work, in comp sci it would have been decided faster, the proof would have to be labelled 1.1.1 Item 1 1.1.2 1.2.1 1.3.1 1.3.2 2.1.1 Atleast a status of the proof would be available to stop redundant argument by identifying sections. Herc === Subject: leech lattice generator matrix Hi there I am reading the paper Amrani, O. et al, The Leech Lattice and the Golay Code: Bounded-Distance Decoding and Multilevel Constructions, IEEE Transactions on Information Theory, vol. 40, No. 4, pp. 1030-1043, Jul. 1994. It is about leech lattice decoding I wonder what the generation matrix for this kind of leech lattice definition. And in SPLAG chap 4, there is a leech lattice generator matrix based on the MOG, I wonder whether it is the generator matrix based on Golay code. And I am also wondering how to get the generator matrix by the construction === Subject: Re: leech lattice generator matrix > And in SPLAG chap 4, there is a leech lattice generator matrix > based on the MOG, I wonder whether it is the generator matrix > based on Golay code. And I am also wondering how to get the > generator matrix by the construction It's there on p.133 (1st ed.). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: LOGIC ^ ATHEISM This is a sci.logic post that may hold some interest here, as it starts on the basic assumptions of this group about logical reasoning in theology. Re: Logic : Form v Substance English speakers employ exists as a predicate when asking such > questions as Does God exist?, which is (intuitively and > pre-theoretically) understood by all *normal* folk as a question > *about* God. Unfortunately, first-order logic with identity does > not--because it cannot--treat Does God exist > as asking about either God or existence, it being a presupposition > of this logic that although no property is designated by exists, > existence is something that every logical subject of any meaningful > proposition must have! > Right. On the other hand FOPL is simply not the only true logic (TM). Alternative systems -free logics- have been devised. Yeah, that word 'exists' is ambiguous between the normal-language > sense (of existence in reality) and the logical sense (of existence in > a domain of discourse). That can be a problem when - as in another > thread going on here - one is debating a question like does god > exist?. > Right. In a system of _free logic_ we could formulate the question: (Ex)(x = God)? Now (Ex)(x = a) can be abbreviated to E!a. Hence, we may ask: E!God ? Does God exist? IME, the best way to handle it is to translate normal-language > statements like X exists or X does not exist into ones containing > predicates [...] > By now most will agree that Iraqui WMDs do not--and did not--exist. > Such weapons have no more purchase on existence than the golden > mountain, the largest prime or Ullrich's sense of fair play > on sci.logic. If their non-existence does not prevent us > from discoursing about these things, why should non-existents > be excluded (as such) from logic? We don't exclude them, normally. There are fundamentally three ways to develop free logic. In negative free logic one assumes that all aic sentences involving non-referential terms are false. In positive free logics aic sentences involving non-referential terms are sometimes true. This seems to be a natural requirement, at least to the extent that it is possible to make a case that sentences such as Pegasus is a horse, Vulcan is a planet, or the present king of France = the present King of France, are not false. [...] (G. ALDO ANTONELLI) ------ On the other hand... > ...to talk about hypothetical Iraqi WMDs using FOL, we'd > translate There are Iraqi WMDs as > (Ex)Wx ( W = _is an Iraqi WMD ) > It would be perfectly sensible to say: > ~(Ex)Wx Right. > using this vocabulary, because that would be saying the same thing as: > (x)~Wx > or, IOW, Everything that exists is not an Iraqi WMD > which would be a statement about existing objects. Yes. > If division by zero is allowed, Ax~(x/0 = x/0) is no > less well-formed than Ax(x/0 = x/0). Sure... Though there are several _different_ possibilities to accomplish this task. Mathematicians usually tend to render x/y as not defined for y = 0. (Probably the most straightforward solution how to deal with these expressions.) Herc -- >>add this link to your sig for $5 a month>> I chat at www.YeOldeCoffeeShoppe.com electron-dot-cloud are galaxies === Subject: Re: math answers on a daily rise percentage Re: 100 free SBC; VonNeumann Gametheory how to play StockMarket I do not know if there is anything in the natural world of biology to say something that a 5% rise or more per one day in food or in something biological is a rise that is worthy of acting upon and securing. Or whether in ysics or chemistry of a 5% rise or more in one day is such as to capitalize upon because in the days and weeks to follow that the tendency will be to move in the direction of the price before the 5% one day rise. You see, I do not know if there is something in Nature and biology in specific that a 5% rise or more in a day is above normal and so one should take advantage of it. Is there a ysical process where a 5% rise or more in a day is something to act upon. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big a where dots of the electron-dot-cloud are galaxies === Subject: Math Trick Here is a good Math Trick that I invented. Check this web site with a very unique Math Game http://w4u.sytes.net/MagicMath You can send me your comments at gonzalep@.84scientist.com don't forget to remove the .84 from the e-mail. === Subject: Re: Math Trick > Here is a good Math Trick that I invented. > Check this web site with a very unique Math Game > http://w4u.sytes.net/MagicMath > You can send me your comments at gonzalep@.84scientist.com don't forget to > remove the .84 from the e-mail. Regardless of how many and what operations you ever do at first, at the end once you multiply by 9 the sum of the digits will always equal a multiple of 9, so removing any one digit and telling me what that digit was is very easy. J === Subject: Re: Math Trick >> Here is a good Math Trick that I invented. >> Check this web site with a very unique Math Game >> http://w4u.sytes.net/MagicMath >> You can send me your comments at gonzalep@.84scientist.com don't forget to >> remove the .84 from the e-mail. > Regardless of how many and what operations you ever do at first, at the > end once you multiply by 9 the sum of the digits will always equal a > multiple of 9, so removing any one digit and telling me what that digit > was is very easy. Unless of course the number was 0 or 9. This makes the program give false results, e.g. for the starting number 1231, choosing 66 and final result 332766X. Note, that the starting number is always guessed correctly, only the number choosen and the number under the X may be wrong. Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de === Subject: Re: Most consistent prime function > don't have to be primes) it generates 48 primes (in linear time (I > think that's the term) (i.e. the time it takes to execute the function > does not increase/decrease as the size of the variable n changes.)). > I'm not trying to sound cocky and don't want to be flamed but I don't > know much about math and what functions are already out there. Does > anyone know which ones are better than this one? I would like to take > a look at them. > D. Oops. After doing some more checking it already exists. I was playing with circles and the function n^2 -n +1 emerged. I thought since it seemed like a bit of an odd connection it might be new. Oh well. I figured it was too good to be true. D. === Subject: Re: Most consistent prime function My favorite is the alternating sum of factorials: n! - (n-1)! + (n-2)! - (n-3)! ... Example: 6! - 5! + 4! - 3! + 2! - 1! = 619 The first non-prime is: 9! - 8! + 7! - 6! + 5! - 4! + 3! - 2! + 1! = 326981 (79 * 4139) Russell - 2 many 2 count === Subject: Re: My (Un)Originality (was Re: Reminder: Wages, Employment Not Determined By Supply, Demand) > Robert Vienneau thinks he's shown factor demand curves can slope > up. Consider Appendix B in (I am no longer committed to calling any curve there a demand function.) > ...that isn't what the literature [stupidity deleted] claims. the pattern of activities adopted in the face of long-run factor-price changes can be complicated and counterintuitive. Consequently, the long-run demand for factors can be badly behaved functions of factor prices -- Michael Mandler, 1999. However, as was argued in Section 3 with regard to 'perversely' shaped, that is, upward sloping, factor-demand functions, this possibility would question the validity of the entire economic analysis in terms of demand and supply. -- H. D. Kurz and N. Salvadori, _Theory of Production: A Long Period Analysis_, Cambridge University Press, 1995. The essential point of the criticism concerns the factor demand curves. The discovery that factor demand curves may be positively sloped in the relevant range, not negatively... -- Bertram Schefold, Joint Production, Intertemporal Preferences, and Long-Period Equilibirum, _Political Economy: Studies in the Surplus Approach_, V. 6, 1990, pp. 162-163. there is not necessarily an inverse monotonic relation between the cost-minimizing quantity of an input and its price... Figures 6.17a-6.17c can be interpreted as demand curves for labour... in Figure 6.17b, ...the sectoral demand curve is upward-sloping... I have shown in Figure 6.17c that the aggregate demand curve is not downward-sloping in the presence of reswitching: indeed, like the sectoral demand curve, it is not even monotonic. Reswitching is sufficient, not necessary, for the aggregate demand curve for labour not to be downward-sloping: to see this, consider Figure 6.18... -- J. E. Woods, _The Production of Commodities: An Introduction to Sraffa_, Humanities Press, 1990. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: New Tensor Calculus Package for Mathematica Renan Cabrera, Jean-Francois Gouyet and I have completed a new Version 3.0 of the Tensorial package. It is available for download at http://home.earthlink.net/~djmp/Mathematica.html The package should be useful both as an introduction to tensor calculations and for advanced calculations. Some of its features are: 1) There is complete freedom in the choice of symbols for tensor labels and indices. 2) Base indices may be any set of integers or symbols. Thus you could use {0,1,2,3} for relativity problems, or {t,x,y,z}, or {[Rho],[Theta],[i]} for serical coordinates. 3) Flavored indices (colored or annotated symbols) to represent different coordinate systems. 4) Tensor shortcuts for easy entry of tensors. 5) Easy methods to store and substitute tensor values. 6) Partial, covariant, total, absolute and Lie derivative routines. 7) Complete documentation, with a Help page and numerous examples for each command. In addition there are tutorial and extended example notebooks. Park djmp@earthlink.net http://home.earthlink.net/~djmp/ === Subject: Re: ordered n-tuples as sets > But it's still a problem, if he's worried about unbounded > nesting levels, because the elements of S (where S = N) are > themselves nested sets. Or do I miss your point? > I guess the upshot must be that he shouldn't worry about unbounded > nesting. At least, not if he wants N as his index set, since it > happens already in N and there's nothing he can do about that. This relies on canonical definition of N with unrestricted set nesting. The same question, however, applies to N too: Can elements of N be defined as sets with bounded nesting? (Clearly, we have to give up the property n is element of n+1 in favor of something else. n is contained in n+1, for example). === Subject: Re: ordered n-tuples as sets > An ordered n-tuple can be defined as (a1,a2,a3,...,a_n-1,an) = {(a1,a2),(a2,a3),..., (a_n-1,an)} > But that doesn't work. Consider: > (1,2,1) = {(1,2),(2,1)} ?? > (2,1,2) = {(2,1),(1,2)} ?? > Any counter example for transitevely closed definition as well? > (1,2,1) = {(1,2),(2,1),(1,1)} != > != (2,1,2) = {(2,1),(1,2),(2,2)} Sure. Try (1,1,2) and (2,1,1) for example. Any similar definition is bound to fail in many, many cases where numbers are repeated in the ordered n-tuple. === Subject: Re: ordered n-tuples as sets > But it's still a problem, if he's worried about unbounded > nesting levels, because the elements of S (where S = N) are > themselves nested sets. Or do I miss your point? > I guess the upshot must be that he shouldn't worry about unbounded > nesting. At least, not if he wants N as his index set, since it > happens already in N and there's nothing he can do about that. > This relies on canonical definition of N with unrestricted set nesting. The > same question, however, applies to N too: > Can elements of N be defined as sets with bounded nesting? (Clearly, we have > to give up the property n is element of n+1 in favor of something else. n > is contained in n+1, for example). I just don't understand your concern over this. If you want to be able to write an expression with ... in it, then *something* has to be unbounded. Might as well be curly braces as something else, it seems to me. === Subject: Re: ordered n-tuples as sets > Any counter example for transitevely closed definition as well? > Sure. Try (1,1,2) and (2,1,1) for example. Any similar > definition is bound to fail in many, many cases where > numbers are repeated in the ordered n-tuple. I dont understand. All the sets below are distinct: (1,1,1) = {(1,1)} (1,1,2) = {(1,1),(1,2)} (1,2,1) = {(1,1),(1,2),(2,1)} (2,1,1) = {(1,1),(2,1)} === Subject: Re: ordered n-tuples as sets > An ordered n-tuple can be defined as (a1,a2,a3,...,a_n-1,an) = {(a1,a2),(a2,a3),..., (a_n-1,an)} But that doesn't work. Consider: (1,2,1) = {(1,2),(2,1)} ?? > (2,1,2) = {(2,1),(1,2)} ?? > > Any counter example for transitevely closed definition as well? > > (1,2,1) = {(1,2),(2,1),(1,1)} != > != (2,1,2) = {(2,1),(1,2),(2,2)} > Sure. Try (1,1,2) and (2,1,1) for example. Any similar > definition is bound to fail in many, many cases where > numbers are repeated in the ordered n-tuple. The correct way to define such ordered n-tuples with sets is by (a1,a2,a3,...,an) = { {a1}, {a1,a2}, {a2,a3}, ..., {an-1,an} } , as - for example - Russell uses in his Introduction to Mathematical ilosoy. J === Subject: Re: ordered n-tuples as sets >An ordered n-tuple can be defined as >>(a1,a2,a3,...,a_n-1,an) = {(a1,a2),(a2,a3),..., (a_n-1,an)} >>But that doesn't work. Consider: >>(1,2,1) = {(1,2),(2,1)} ?? >>(2,1,2) = {(2,1),(1,2)} ?? >Any counter example for transitevely closed definition as well? >(1,2,1) = {(1,2),(2,1),(1,1)} != >!= (2,1,2) = {(2,1),(1,2),(2,2)} >>Sure. Try (1,1,2) and (2,1,1) for example. Any similar >>definition is bound to fail in many, many cases where >>numbers are repeated in the ordered n-tuple. > The correct way to define such ordered n-tuples with sets is by > (a1,a2,a3,...,an) = { {a1}, {a1,a2}, {a2,a3}, ..., {an-1,an} } , as > - for example - Russell uses in his Introduction to Mathematical > ilosoy. (1,2,1,3,1) = {{1},{1,2},{2,1},{1,3},{3,1}} = (1,3,1,2,1) -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: ordered n-tuples as sets > Any counter example for transitevely closed definition as well? > Sure. Try (1,1,2) and (2,1,1) for example. Any similar > definition is bound to fail in many, many cases where > numbers are repeated in the ordered n-tuple. > I dont understand. All the sets below are distinct: > (1,1,1) = {(1,1)} > (1,1,2) = {(1,1),(1,2)} > (1,2,1) = {(1,1),(1,2),(2,1)} > (2,1,1) = {(1,1),(2,1)} > In (1,2,1) = {(1,1),(1,2),(2,1)} natural question arises why transitive closure doesn't include (2,2). Formal definition: Transitively closed ordered n-tuple is (a_1, a_2, ..., a_n) = set of all (a_i, a_j) such that i < j. In the above example: a_1=1, a_2=2, a_3=1 and (a_2,a_2) pair doesn't meet the criteria. === Subject: Re: ordered n-tuples as sets > Formal definition: > Transitively closed ordered n-tuple is > (a_1, a_2, ..., a_n) = set of all (a_i, a_j) such that i < j. > In the above example: > a_1=1, > a_2=2, > a_3=1 > and (a_2,a_2) pair doesn't meet the criteria. (1,1,2,1) = {(1,1),(1,2),(2,1)} = (1,2,1,1) === Subject: Re: ordered n-tuples as sets > Any counter example for transitevely closed definition as well? > Sure. Try (1,1,2) and (2,1,1) for example. Any similar > definition is bound to fail in many, many cases where > numbers are repeated in the ordered n-tuple. > I dont understand. All the sets below are distinct: > (1,1,1) = {(1,1)} > (1,1,2) = {(1,1),(1,2)} > (1,2,1) = {(1,1),(1,2),(2,1)} > (2,1,1) = {(1,1),(2,1)} > Sorry, my mistake; I was using a different (and inferior) definition that happened to fit your expansion for the two special cases of the previous post. Yes, this does work for 3-tuples. But I think that's as far as it can go; you can't do the same thing with 4-tuples defined in terms of 2-tuples. That is, I think this method may require n-tuples to be defined in terms of (n-1)-tuples. For example, suppose you define a 4-tuple as (a,b,c,d) = {(a,b),(a,c),(a,d),(b,c),(b,d),(c,d)} Then (1,1,2,1) and (1,2,1,1) both yield {(1,1),(1,2),(2,1)}. And the same goes for at least one other possible generalization of your 3-tuple rule; I'm not quite sure what you mean to use for general n since your definition of a few posts back was garbled. If I'm getting it wrong, please clarify. === Subject: Re: ordered n-tuples as sets <3F43D911.7BAE7BA6@mdli.com> <3F43E6F5.F4C2B14A@mdli.com> <3f42a6ad_2@newsfeed The correct way to define such ordered n-tuples with sets is by > (a1,a2,a3,...,an) = { {a1}, {a1,a2}, {a2,a3}, ..., {an-1,an} } , as > - for example - Russell uses in his Introduction to Mathematical > ilosoy. > > (1,2,1,3,1) = {{1},{1,2},{2,1},{1,3},{3,1}} = (1,3,1,2,1) Haha, I suppose I extrapolated a little too quickly there... Russell presented it, I believe, only for ordered pairs ( (a,b) = {{a}, {a,b}} ) and perhaps he was allowing his sets to be multisets if he went further... so nevermind. J === Subject: Re: ordered n-tuples as sets > > Since Kuratowski's definition works for ordered pairs, let's make an > ordered triple > (a,b,c) = (a,(b,c)) > = {{a}, {a,(b,c)}} = {{a},{a,{{b},{b,c}}}} and similarly for an ordered n-tuple (a1,a2,...,an) = (a1, (a2,...,an)) Recursively nesting ordering pairs works, but, unfortunately, set nesting > level is unbounded. > But it's still a problem, if he's worried about unbounded > nesting levels, because the elements of S (where S = N) are > themselves nested sets. Or do I miss your point? Thus, my comment: > I suppose that for some, all of mathematics is founded on {}, {{}}, > {{{}}}, ..., {{{...}}}, ..., {{{...w...}}}, etc. === Subject: Re: ordered n-tuples as sets > Recursively nesting ordering pairs works, but, unfortunately, set > nesting > level is unbounded. On the other hand, the alternative definition (a,b,c,...) = {{a},{a,b},{a,b,c},...} has nesting level 2, but clearly doesn't work. Is there an alternative > ordered n-tuple definition which require sets nesting no more than 3 > levels? > If 3 is not giong to cut it, some bigger number, maybe? > <... the S-tuple (a, b, c, ...) can be considered as a set of ordered > pairs (s,d) > <... That was the key to the answer, thank you! > An ordered n-tuple can be defined as > (a1,a2,a3,...,a_n-1,an) = {(a1,a2),(a2,a3),..., (a_n-1,an)} > which clearly has a limit on curly brakets nesting. I was actually thinking of an n-tuple as a function from [1..n] -> D; so I would write the above as: (a1, a2, a3, ..., an) = {(1, a1), (2, a2), (3, a3), ... (n, an)} or, using the definition of (x,y) = {x, {{x}, y}}, (1, 1, 2, 1) = { {1, {{1},1}}, {2, {{2},1}}, {3, {{3},2}}, {4, {{4},1}} } === Subject: Primes and the Collatz conjecture! For any Collatz sequence path find only the prime start numbers that creates an odd prime for each of its odd integers except 1. Here are the first few prime start numbers that meet these criteria --- 3,5,7,11,13,17,19,29,37,53,59,67,89,101,131,149,157,179,181,197,241,269,277, 349,397,739,853,1109,1237,1429,1621,1861,1877,2161,2389,2531,2957,3413,3797, 4549... I believe I found them all up too prime 4549. What is interesting also, only primes from the (above list) are possible in any sequence where all the odd integers are prime except 1 and where this prime list above ------>oo. If you do not believe this to be true, please show a counter example. Also, the all odd integer = prime paths (except start # 3 and 5) all seem to merge at this point 40,20,10,5,16,8,4,2,1 Dan === Subject: Re: Primes and the Collatz conjecture! > For any Collatz sequence path find only the prime start numbers that > creates an odd prime for each of its odd integers except 1. > Here are the first few prime start numbers that meet these criteria > --- > 3,5,7,11,13,17,19,29,37,53,59,67,89,101,131,149,157,179,181,197,241,269,277, > 349,397,739,853,1109,1237,1429,1621,1861,1877,2161,2389,2531,2957,3413,3797, 4 5 > 49... > I believe I found them all up too prime 4549. > What is interesting also, only primes from the (above list) are > possible in any sequence where all the odd integers are prime except 1 > and where this prime list above ------>oo. > If you do not believe this to be true, please show a counter example. > Also, the all odd integer = prime paths (except start # 3 and 5) all > seem to merge at this point 40,20,10,5,16,8,4,2,1 I think k = 22,369,621 is a prime, and 3k+1 = 2^26. The odd numbers on the path are 22,369,621 -> 1 === Subject: Re: Primes and the Collatz conjecture! > For any Collatz sequence path find only the prime start numbers that > creates an odd prime for each of its odd integers except 1. > Here are the first few prime start numbers that meet these criteria > --- > 3,5,7,11,13,17,19,29,37,53,59,67,89,101,131,149,157,179,181,197,241,269,277, > 349,397,739,853,1109,1237,1429,1621,1861,1877,2161,2389,2531,2957,3413,3797, 4 5 > 49... > I believe I found them all up too prime 4549. > What is interesting also, only primes from the (above list) are > possible in any sequence where all the odd integers are prime except 1 > and where this prime list above ------>oo. > If you do not believe this to be true, please show a counter example. > Also, the all odd integer = prime paths (except start # 3 and 5) all > seem to merge at this point 40,20,10,5,16,8,4,2,1 Just wanted to say: A novel and interesting idea. === Subject: Re: Product space > > Let X be a metric space. Let U, V be two subsets in X. > Is the following statement true? > > The product space (U,V) is open in (X cross X) if and only if U, V are > both > open sets in X. > > If it is true, would kindly provide a concise roadmap of showing it ? > > K. Yam > > > U,V open implies UxV is open: actually for topological spaces, this follows > by > definition. > UxV open implies U and V are open: show that the projections p,q:XxX->X, > p(x,y)=x, q(x,y)=y are conitinuous. Are you sure the second part is right? The constant map that sends (u,v) to 0 is continuous, but {0} is not an open set. A continuous map need not send open sets to open sets. === Subject: Re: Product space Let X be a metric space. Let U, V be two subsets in X. > Is the following statement true? The product space (U,V) is open in (X cross X) if and only if U, V > are both open sets in X. > U,V open implies UxV is open: actually for topological spaces, this > follows by definition. > UxV open implies U and V are open: show that the projections > p,q:XxX->X, p(x,y)=x, q(x,y)=y are conitinuous. > Are you sure the second part is right? The constant map that sends > (u,v) to 0 is continuous, but {0} is not an open set. A continuous map > need not send open sets to open sets. Amost. Projections in addition to being continuous are also open. Thus if UxV is open in XxY, then p1(UxV) = U is open in X and p2(UxV) = V is open in Y By continuity, if U open in X, V open in Y UxY = p1^-1(U), XxV = p2^-1(V) are open in XxY. Thus UxV = UxY / XvV is open in XxY. ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Publishing > Serre, as editor, does not referee papers he does not want to see > published, but instead quickly returns a letter stating that he will > pretend that the already formally submitted paper has not been submitted > to him, and if the submitter insists on having the paper refereed, it will > not become published anyhow. I think there must be quite a few such Serre > letters around. (In the cases I know of, the papers where published > elsewhere and eventually judged to be of utmost quality, but Serre's > unusual editorial practises caused a long delay.) Why would you appoint someone as editor in the first place if you didn't want him to exercise his judgment about what should and should not be published? And once the editor has determined that something is not going to be published, why should he waste the time of a referee by sending it out anyway? As a referee, I'd be very annoyed to learn that an editor had sent me a paper without first exercising some preliminary judgment as to whether it was publishable. Steven E. Landsburg www.landsburg.com/about2.html === Subject: Re: Push or Pull problem >Question:In a rectangle made up of 11*12 square tile,I can choose to >either push or pull a square. >But when pushing or pulling,ALL adjacent tile(one to the left,to the >right,above,and below) will also be affected. >For all affected square,if it is now up,it would be pushed to become >down.If it is down,it would be pulled to become up >I have to make ALL square tile change from ALL down to ALL up Is there a difference between pushing and pulling? What happens if you pull a square that is already up, or push one that is down? In what you write below, you don't talk about push or pull, just press. I think what you're trying to say is that choosing a square causes it and each adjacent square to switch between up and down. You want to switch each square an odd number of times. It's convenient to use arithmetic in Z_2: down is 0 and up is 1, and adding 1 interchanges these. A state of the system is a vector in (Z_2)^(11*12), i.e. 11*12 components, each 0 or 1. Addition of vectors is done in Z_2, i.e. 0+0=1+1=0, 0+1=1+0=1. If A_i is the vector you get by starting at all 0's and choosing square i, then after choosing i_1, i_2, ..., i_k you get A_{i_1} + A_{i_2} + ... + A_{i_k}. So the problem becomes one of linear algebra: you want to write the vector [1,...,1] as a linear combination of the A_i. This can be done by solving a 132 x 132 system of linear equations over Z_2, but that's a bit much to do by hand. Here's the solution as found by Maple 9's modular linear algebra package. A 1 indicates a square that is pressed, a 0 a square that is not pressed. 1 0 1 1 0 0 0 0 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0 0 1 1 0 0 1 0 0 1 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 0 0 1 1 0 0 1 1 1 1 0 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 0 1 1 0 0 0 0 1 1 0 1 === Subject: Re: Push or Pull problem http://www.albartus.com/forum/viewtopic.p?t=313 There,many people (include me) has got frustrated with this puzzle.May I post it at there the solution(and explaintion) ? > Is there a difference between pushing and pulling? What happens if you > pull a square that is already up, or push one that is down? > In what you write below, you don't talk about push or pull, just > press. I think what you're trying to say is that choosing a square > causes it and each adjacent square to switch between up and down. > You want to switch each square an odd number of times. > It's convenient to use arithmetic in Z_2: down is 0 and up is 1, > and adding 1 interchanges these. A state of the system is a vector in > (Z_2)^(11*12), i.e. 11*12 components, each 0 or 1. Addition of vectors > is done in Z_2, i.e. 0+0=1+1=0, 0+1=1+0=1. > If A_i is the vector you get by starting at all 0's and choosing square i, > then after choosing i_1, i_2, ..., i_k you get > A_{i_1} + A_{i_2} + ... + A_{i_k}. So the problem becomes one of linear > algebra: you want to write the vector [1,...,1] as a linear combination > of the A_i. This can be done by solving a 132 x 132 system of linear > equations over Z_2, but that's a bit much to do by hand. Here's the > solution as found by Maple 9's modular linear algebra package. A 1 > indicates a square that is pressed, a 0 a square that is not pressed. > 1 0 1 1 0 0 0 0 1 1 0 1 > 0 1 1 1 0 1 1 0 1 1 1 0 > 1 1 1 0 1 1 1 1 0 1 1 1 > 1 1 0 0 1 1 1 1 0 0 1 1 > 0 0 1 0 0 1 1 0 0 1 0 0 > 0 1 0 0 1 0 0 1 0 0 1 0 > 0 0 1 0 0 1 1 0 0 1 0 0 > 1 1 0 0 1 1 1 1 0 0 1 1 > 1 1 1 0 1 1 1 1 0 1 1 1 > 0 1 1 1 0 1 1 0 1 1 1 0 > 1 0 1 1 0 0 0 0 1 1 0 1 > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Push or Pull problem >http://www.albartus.com/forum/viewtopic.p?t=313 >There,many people (include me) has got frustrated with this puzzle.May >I post it at there the solution(and explaintion) ? Sure. === Subject: Rest mass of the electron from zero point fluctuations typo GRW should be FRW in earlier version My non-analytic (as z -> 0) non-perturbative BCS Nambu-Jona Lasino inspired mass formula m = (e/c)^2|/zpf|^1/2 = (h/cLp)e^-1/z Therefore, using Susskind's world hologram formula (e^2/hc) (Lo/c)^1/3 = e^-1/z So that we can numerically compute z z ~ -1/loge[(e^2/hc) (Lo/c)^1/3] ~ - 1/loge[10^-63] using data from both quantum gravity units and cosmology. The final theory should have some kind of polynomial in z whose roots determine the lepto-quark rest masses that go into QCD. One uses QCD to get the hadronic rest masses that are mostly kinetic in origin as in the bag model of QCD Lite (Frank Wilzcek, APS March & What Haisch, Rueda and Puthoff talk about should only be a small correction to this. reformatted & typo corrected This is essentially the core of the talk I will give in Paris at the International Vigier Conference at Pierre et Marie Universite to be published by Kluwer Academic Publishers. Re: Timothy Ferris on C-Span LA TIMES Book Fair ñSeduction of Scienceî Ferris is a good writer, though not a ysicist, he has a good understanding of the conventional wisdom in mainstream ysics today. The Question is: What is The Question? Ferris describes the wrong question asked by all of the pundits in modern precision cosmology, which is the dark matter? This is an important issue because, as Ferris was quick to point out, contrary to John Maddox's (NATURE) comment, inflationary cosmology is an excellent theory since 3D space is now measured to be flat on the large scale > 10 megaparsecs and the black body cosmic microwave background temperature variations ~ 10^-5 over the whole celestial sere with scale invariant Gaussian fluctuations are now, since WMAP, measured with good accuracy and a precision ~ 2%. Heisenberg uncertainty principle that, for example, the product of the momentum and position uncertainties is greater than a quantity proportional to Planck's quantum of action. Similarly, for energy and time, for action momentum and angle, for number and ase, electric and magnetic fields, generalized to any pair of incompatible non-commuting matrix observables in quantum bit space etc. interior Heisenberg uncertainty principle that the products of uncertainties in conflicting measurements is smaller than the There are two kinds of ysical vacuum spacetime regions, exotic and non-exotic. The non-exotic vacuum does not gravitate and it is close to (including real otons forming radiation, and even giant quantum states of virtual otons forming near fields like magnetic fields in MHD plasmas, in pulsars etc.). The nonexotic vacuum on the large scale has zero cosmological constant / = 0. Gravity lensing, type 1a supernovae, WMAP et-al all show that the ysical vacuum on a large scale is only about 4% in terms of the Omega of standard FRW cosmology from Einstein's theory of gravity. Our expanding 3D space is accelerating rather than slowing down. This means, from type 1s supernovae data, a / > 0 that is an anti-gravity space warp field with an Omega that is ~ 73%. Flat 3D space then requires a dark matter component whose Omega is ~ 23%. The three main components of the stuff of our local universe then adds up to 100% with ordinary as and radiation in non-gravitating vacuum only ~ 4% of the universe. Our local universe is what we can see in our telescopes in all regions of the electromagnetic spectrum. That will include future gravity wave and neutrino detectors as well. That is, our local universe the past light cone in curved spacetime on the large scale. There are an infinity of parallel universes outside our past light cone. Indeed, one needs to use Cantor's theory of transfinite numbers with an infinity of infinities as explained by Max Tegmark in several recent papers like May geometrodynamical in the sense of Wheeler's IT not the same as Deutsch's informational BIT multiverse (Fabric of Reality). Wheeler says IT FROM BIT which requires delving into the alternative meanings of quantum theory. I suggest Wheeler's idea is incomplete needing also BIT FROM IT to get his Oroborus of the universe as a self-excited circuit. http://www.metanexus.net/ultimate_reality/agenda.htm Einstein's large-scale cosmological constant is really a limiting case of the local complex scalar field responsible for inflation. The Dirac electron vacuum dominates to a good approximation but it is unstable from the exchange of a virtual oton between a virtual electron-positron pair at the Fermi surface of zero energy with momentum 10^19Gev/c. The micro-quantum Dirac vacuum collapses to a macro-quantum coherent vacuum of lower entropy. The local complex scalar field is the vacuum polarization coherence. The coherence length is c/H(t) where the Hubble parameter is H(t) = R(t),t/R(t) in the large scale limit of the GRW metric including / =/= 0. This local complex numbered scalar field PSI(x) has a Goldstone ase and a Higgs amplitude. Einstein's geometrodynamic c-number field guv(x) of superfluid curved spacetime is simply emergent from the modulation of the Goldstone ase according to the IT FROM BIT Bohm pilot wave guidance equation guv(x) = nuv + (1/2)Lp^2 (Goldstone ase(x)){,u,v} , means ordinary partial derivative, { }means symmetrized anti-commutator and [ ] means commutator, nuv is the globally flat Minkowski metric tensor Lp^2 = hG/c^3 consistent with black hole thermodynamics. Torsion fields correspond to the non-integrable anholonomy (Goldstone ase(x))[,u,v] =/= 0 Susskind's world hologram corresponds to using Lp* = Lp^2/3(c/Ho)^1/3 ~ 1 fermi ~ 10^-13 cm rather than Lp ~ 10^-33 cm The local exotic vacuum zero point energy fluctuation field is then /(x)zpf = Lp*^-2[1 - Lp*^3|Higgs Amplitude(x)|^2] where /(x) > 0 is anti-gravitating dark energy exotic vacuum and /(x) < 0 is gravitating dark matter exotic vacuum. The zero point energy induced electron rest mass is m~ (e^2/c^2)|/zpf|^1/2 where |/zpf |~ (1 fermi)^-2 ~ Lp*^-2 on the scale of 1 fermi with /zpf < 0 dark matter exotic vacuum core to prevent the explosion of the self charge ~ e. The string theory ala' corresponds to this same value for /zpf on the microscale. Dark matter ~ 23% of the stuff of our local universe is simply exotic vacuum with positive zero point energy pressure. In contrast, dark energy ~ 73% of our local universe is simply exotic vacuum with negative zero point energy pressure that is equivalent to Kip Thorne's exotic matter for traversable wormholes and Alcubierre timelike geodesic warp drive vacuum propulsion when we learn how to locally metric engineer the ysical vacuum. Potential applications of this new ysics are described in Chapter Nine of Sir Martin Rees's Our Final Hour. Jack Sarfatti American for background. Imagine a single Level 1 post inflationary bubble which (suppress 1 space dimension temporarily) is like an infinite flat rubber sheet Euclidean plane that is not only stretching isotropically, but is accelerating in its expansion rate. Our local universe is the interior of an expanding circle on the plane. The radius of the circle is c/H(t). The circumference of the circle is singular in the sense that the vacuum coherence PSI = 0 on the circumference where the micro incoherent --> macro quantum coherent vacuum inflationary ase transition happened. That is, the circumference is analogous to a vortex line in Type II superconductor with a lattice of vortex lines. Each vortex line is analogous to a local Level I parallel universe. The strong force hadronic resonances lie on a set of parallel lines of identical slope when spin J is plotted against square of mass m of the resonance peak. Each line corresponds to a fixed assignment of strong, weak and electric charge. The universality of the slope suggests strong short-range gravity induced by the zero point fluctuations of all the basic quantum fields. In my theory, the basic data of string theory for the strong sub-nuclear color force of SU(3) becomes simply J ~ |/zpf|^-1(mc/h)^2 = Lp*^2(mc/h)^2 = Lp^4/3(c/Ho)^2/3 = G*m^2/hc (+ intercept depending on the charges) where for the real electron (lepto-quark) on mass shell, for example, m ~ m(Planck)e^-1/z as in BCS theory. Ho is the large-scale cosmological Hubble parameter for our particular local universe in the sense of <0|e+e-|0> is the inflation complex scalar field. z is a dimensionless measure of the binding of virtual electron-positron pairs at the zero energy Fermi volume edge of the Dirac Sea vacuum with momentum 10^19 Gev/c in the micro -> macro quantum vacuum ase transition that lowers the nonclassical entropy kick-starting the arrow of time for emergent Einstein gravity with the unified dark energy/matter field. /zpf = Lp*^-2[1 - Lp*^3|<0|e+e-|0>|^2] = -1/Lp*^2 in the dark matter exotic vacuum core of the electron That is, 1 - Lp*^3 |<0|e+e-|0>|^2 = -1 in that case. The Hubble horizon of radius c/H(t) of our local universe section on the large scale spatially flat expanding accelerating post-inflationary bubble is analogous to the core of the superconducting vortex singularity where the local macro-quantum coherent order parameter PSI vanishes. That is where the micro -> macro quantum vacuum ase transition happened after all in which the non-classical entropy (log of volume of ase space of the vacuum) is lowered to set the cosmic clock for the irreversible arrow of time in which the initial singularity big bang starts out with low entropy compared to the later universe. This Hubble horizon where PSI = 0 is a 2D surface out of which our 3D space of Einstein's curved 4D spacetime emerges like the projected holograic image in Lenny Susskind's sense. That's why Lp* = Lp^2/3(c/Ho)^1/3 ~ 1 fermi (10^-13 cm) where Lp ~ 10^-33 cm and c/Ho ~ 10^28 cm You can think of the infinity of Tegmark's Level I parallel universes on a single post-inflationary bubble as a kind of type II superconductor lattice of vortices. Each vortex line maps to a Hubble horizon. Then there are, in the chaotic inflationary model of Tegmark's Level II, an infinity of such bubbles floating in hyperspace (in Kip Thorne's sense) like a glass of champagne (brewed by Rashi de Troyes 1000 years ago ;-)) These are not the QUBIT parallel worlds of Deutsch's multiverse. The QUBIT multiverse is a landscape for the ensemble of IT system points of geometrodynamic post-inflationary bubbles to roll on like in vs = (h/m)Grad(arg psi) in deBroglie-Bohm-Vigier theory in 3D, which for 4D geometrodynamics becomes du = Lp*^2 argPSI,u ,u is the ordinary partial derivative of the Goldstone ase argPSI, u = 0,1,2,3 PSI = <0|e+e-|o> is the inflation complex scalar field |PSI| = Higgs field. du = Hagen Kleinert's world crystal lattice distortion field in which curvature and torsion are disclination and dislocation defect densities as shown for ordinary crystals by Bullough, Bilby et-al at Harwell AERE Theoretical ysics Division in studies of breakdown of nuclear reactor control rods. I was at Harwell in that division in the summer of 1966 after spending time at Heisenberg's Max Planck Institute in Munich at a Wheeler seminar. http://qedcorp.com/APS/Vigier4.pdf === Subject: Re: Set nomenclature question > F is a field and its elements are denoted by B_n_m. Does this mean > there is a one-to-one and onto function from the pairs of positive > integers to F? Or, for example, could B_n_m = A (some element of F)for > all pairs(n,m)? > This is an effort to understand the concept of all possible countable > unions of intersections (and intersections of unions) of B_n's. A > simple illustration involving a 2-element F would be useful indeed. I should have written B_(nm)and not B_n_m. So second paragra should read .....all possible countable unions of intersections (and intersections of unions) of B_(nm)'s. My apologies. === Subject: set theory with as In the first exercise on p.51 of Jech's book, The Axiom of Choice, he asks the reader to prove the consistency of ZFC+there are infinitely many as. He also gives a generous hint on how to construct a model. His chapter notes don't attribute this construction to anyone. What I would like to know is a reference to the publication in which this construction first appeared. Ignorantly, Allan Adler ara@zurich.ai.mit.edu **************************************************************************** * * * Disclaimer: I am a guest and *not* a member of the MIT Artificial * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * **************************************************************************** === Subject: Re: set theory with as >In the first exercise on p.51 of Jech's book, The Axiom of Choice, he asks >the reader to prove the consistency of ZFC+there are infinitely many as. >He also gives a generous hint on how to construct a model. His chapter notes >don't attribute this construction to anyone. What I would like to know is a >reference to the publication in which this construction first appeared. I don't have an answer to your question, but do really need all of ZFC, or even choice at all? Given as a1, a2, ..., a_n, a different a would be {{a1, a2,..., a_n}}. Does Jech include Foundation in ZF? Or perhaps I am misinterpreting the term a. -- Steen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: solving algebraic equation > That is, what you are doing is a series of steps, some > of which involve if and only if transformations > (and are therefore reversible), and some of which involve > implies, i.e. only if transformations which are > not one way. > [maths elided] > Is that an improvement? track of the sign of x (in the present case) as a remedy, but it is merely another way of checking both solutions at the end. In the meantime I dug out a math textbook which states that for this kind of algebraic equations one always needs to resubstitute the solutions to tell the 'correct' one... Martin === Subject: Re: Synergetics coordinates >Do You mean that I have to explain my reply more detailed to your earlier >post before You can consider, or do You accept my arguments as such? >Of course - You are welcome to express counterarguments. There is a model for a correspondence of xyz Cartesian and the xyz axes of the equilateral triangle. The Cartesian axes could go through the centers of the six square faces of the cuboctahedron. I can see the ninety degree angles between the Cartesian axes and the one hundred and twenty degree angles between the axes of the triangle at the same time. I can't see ninety degree angles between the four dimensional Cartesian xyzw axes or how they would correspond to the four axes of the regular tetrahedron, so, Cartesian coordinates are an afterimage that just gets in my way. >I consider synergetics coordinates is nothing but a new name for old facts >as a final goal to support somebody's effort to lift his profile as a great >inventor. >(Please notice this message is no way against You and the honest information >You wanted to share with sci.math. community). My criticism was dedicated >against the original reference and against the inventor's ambitions.) >Tapio I've been looking for web sites or books about something like Synergetics coordinates for years. Cliff Nelson >>Keywords: fourth dimension, simplex, tetrahedron, coordinate geometry, >>sixty degree coordinate system, sere packing, linear algebra, >>operations research, linear programming, Newton's method, roots of >>unity, derivative, field, cycloic fields, B numbers, Bucky Numbers, >>Synergetics, R. Buckminster Fuller. >>I've read too much about most of the keywords from too many books. An >>afterimage, persistence of vision, time lag, stops me from seeing simple >>things from a new point of view for awhile. >> Cliff Nelson >> > Fyi: The nice property that was mentioned in the given link, is based > on the Viviani's theorem. > http://mathworld.wolfram.com/VivianisTheorem.html > Ternary ase diagrams have been used during the last century - > perhaps longer. See for example: > >http://www.sv.vt.edu/classes/MSE2094_NoteBook/96ClassProj/experimental/tern a > ry2.html > Btw: MS-Excel has radar plot property and, on the other hand, > xyz-cartesian plot, which can be viewed from any angle so that you > can result in synergetics view. (45-45 degrees rotation of the > point of view) Imho: I didn't know a certain angle of view in the xyz > cartesian coordination system is called synergetics coordinates. > ;-) > Tapio > >>What Eric W. Weisstein calls the x,y,z axes of the triangle are >>rotated one position to the right from mine in his description, >>but, it looks like at least some version of Synergetics coordinates >>might become respectable in academia now that they are mentioned >>at: >>http://mathworld.wolfram.com/SynergeticsCoordinates.html >>Cliff Nelson >> > === Subject: trace of matrix Let A, B be nxn complex matrix. infinity Let f(t) = exp( sigma ((tr(A^k) - tr(B^k))t^k)/k ) k=1 Show that f(t) = det(I - tB)/det(I - tA). Pleaese give me a hint. -- === Subject: Re: trace of matrix > Let A, B be nxn complex matrix. > infinity > Let f(t) = exp( sigma ((tr(A^k) - tr(B^k))t^k)/k ) > k=1 > Show that f(t) = det(I - tB)/det(I - tA). > Pleaese give me a hint. First of all, these sums probably don't converge for all t. But it will work if |t| is small enough. Second hint: do you know what sum_{k=1}^infinity z^k/k is? Third hint: exp(tr(M)) = det(exp(M)) === Subject: Re: Triples search >Note that if any two of x, y, t are even they can't >be coprime, Sorry, I forgot that is x, y, t even and OTHERWISE coprime. And to be clear, there needs to be only one common factor of two between them, not multiple common factors of two. Two must appear exactly once in each number's list of factors. sci.crypt. In a raster sequencing of pixels, quantization and blurring apply horizontally. In discrete Lissajous patterns built on lattices that have singly-even and otherwise coprime sides, an offset provides a continuous loop serialization without retrace. I use this currently for feature extraction and image compression. It could be the basis of a new family of CRT monitors, but I don't have the hardware expertise. If a picture of x by y pixels is stacked with a total of t pictures to form a video chunk, and x, y, and t are all single even and relatively coprime, then a repeating serialization of the information can be made, leading to simultaneous manipulation of spatial and temporal information, and a balanced compression extracting the major spatial and temporal features of the video clip. I wish to find sets of x, y, and t, and start processing actual video. I have the ability to import *.avi files into Mathcad with a hack. For me, ideally, z = ( x + y ) / 2 but this is not a condition for processing, it's just appealing and reduces the total number of chunk sizes I need to try to balance features, compression, and processing time. I use the Fourier transform of the repeated serialization for feature extraction, and linear algebra to go from RBG to YIQ. I don't have an HSV yet. I don't know and can't find AOL's Save text with line breaks and have had complaints. How do they look? Current screen resolution is 1024 by 768, AOL is maximized, and the Reply window is maximized. The default window is smaller. Have the line breaks tended to be too short lately? I had complaint of very long lines as well. I do not know is save text with line breaks is off or on, and aol does not say what the control does, only that to turn it on, you check it! The transform of a raster serilization has artifacts that inhibit feature extraction. Also a raster isn't looped so a retrace or a step is inevitable. I used to call this nonretrace video. Sure it could up the pixel count a little, but it's really a manufacturing dodge. Linearity is inherent with such a system. Triangle waves have much less harmonic content than the sawtooth raster drive wave. Musically, they are highly dissonant ratios, but for some aspect ratios, they are nearly consonant. Does that help? I should have mentioned that earlier. Yours, Doug Goncz, Replikon Research, Seven Corners, VA The hormones work at different speeds. In a fight-or-flight scenario, glucocorticoids are the ones drawing up blueprints for new aircraft carriers; epinerine is the one handing out guns. === Subject: Re: Trying to understand Languages and Structures > Let L = Lab. Show that for every L-term t(x1, ..., xn) there are > integers k1, ..., kn such that for every abelian group A = (A, 0, -, > +) and every a1, ..., an belonging to An we have tA(a1, ..., an) = > k1a1 + ... + knan. You made no attempt to describe the context. What's Lab, tA ? > Solution: > Let x = (x1, ..., xn), and a = a1, ..., an. > Case 1: t(x) is the variable xi (has arity 0) > Then tA(a) = ai by the definition of tA. > Define kj = 0 if j != i and ki = 1. > Then tA(a) = k1a1 + ... + knan. > Case 2: t(x) = - x (has arity 1) > Then tA(a) = - ai by the definition of tA. > Define kj = 0 if j != i and ki = -1. > Then tA(a) = k1a1 + ... + knan. > Case 3: t(x) = + xixj (has arity 2) > Then tA(a) = + aiaj by the definition of tA. > Define km = 0 if m != i and m != j and km = 1 if m = i or m = j. > Then tA(a) = k1a1 + ... + knan. > Case 4: t(x) = Ft1(x)t2(x)...tn(x) (has arity > 2) > Then tA(a) = FAt1A(a)t2A(a)...tnA(a) by the definition of tA. > Since each function of L is either nullary, unary, or binary > (definition of Lab), each ti will fall under case 1, case 2, or case > 3. Hence each ki will be either 0, 1, or -1. > Then tA(a) = k1a1 + ... + knan. > Since every term falls into either case 1, 2, 3, or 4, it follows that > the proposition is true. > QED. > I tried to put the proof in html, but I cant display html in these > groups. The actual proof with superscripts, subscripts and such is > listed at www.geocities.com/blakman211/lgprf1.htm x sub n = x_n x1, x2, ... alright if it's clear it's x_1, x_2, ... x to the nth = x^n F sub a super b = F_a^b Tho it's best to present problems in body of post, to be useful, present web addresses as http://www.geocities.com/blakman211/lgprf1.htm ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Typing Monkeys Don't Write Shakespeare BricksInTheWall8 > What the hell is the point of this? Is it just supposed to be really funny or > something? I thought it was funny, if only for this line: illips said the results showed that monkeys are not random generators. They're more complex than that. More complex than randomness. Hmmmm.... that sounds so simple... thats its complex... so its wrong.... which means illips doesn't know what he's talking about. Therefore, illips is just a monkey putting words together at random and making a paradoxially worthwhile statement. Which is what the story was about... Oh well, I think it's funny. Mike Helland === Subject: what are the essential features for a image compression transform basis such as 2D DCT? I am facing with a question in transform based image compression. Let me ask in this way, if I want to design my own image compression transform basis, consider the transform as 8x8 matrix multiplication. In 2D case, the transform can be regarded as Y=T*X*T'. For 1D, the transform is Y=T*X. Now if I want to design my own transform basis matrix, other than DCT, DWT, DFT, DHT, etc... what enabling features does my basis matrix must have? Orthogonality is one, but I don't think orthogonality is essential... -Walala === Subject: Re: What cardinality may a ring have? > > > Given a non-empty set S of cardinality kappa (finite or infinite) can > one always define +, * and 0 so that is a ring? > > It can certainly be done for any S having any natural number > cardinality greater than one. >>Why _greater_ than one? Surely S = {0}, with +, *, 0 defined in the >>only way possible, is a ring? Rather perversely such an S is even a >>ring with a multiplicative identity! >In my old-fashioned world, all rings had at least two members, an >additive identity and a distinct multipicative identity. It seems to me that this point came up a few weeks ago. While it is purely a matter of definition, what do you call something which fullfils all the axioms, but does not have a multiplicative identity? It seems much easier to me to define 'ring' without the necessity of a multiplicative identity, and then say 'ring with unit' when you need to specifiy that there is an identity. There are 'interesting' rings without unit. (One that comes immediately to mind is the ring of function with compact support (on some non-compact space. >> Indeed, it is the terminal object of the category of rings. >> But some people demand that their rings have a 1, and moreover that 1 >> be different from 0, so it is possible for some not to consider the >> zero ring as a ring. >> It's not denial. I'm just very selective about >> what I accept as reality. >> --- Calvin (Calvin and Hobbes) >> Arturo Magidin >> magidin@math.berkeley.edu (this space unintentially left blank ..... make obvious deletion for email === Subject: Re: Why not anonymous publication? >Well, why not type-A space? Sure, but we'd run out of letters pretty soon, and it's usually easier to remember names than arbitrary codes. I really don't understand your problem with using names. > What to do when cultures meet up (what happens when >aliens come?). If you name things after people, you'll argue that you >want your culture's name, and it's a pointless battle. We may well have pointless battles with aliens, but it won't be over nomenclature. Obviously (except on Star Trek) the aliens don't speak English. If they even use anything that sounds remotely like words, the onemes would be completely different and probably don't map well into our alabet. Nobody will be surprised that they don't use the same words we do for anything non-mathematical, so why should they use the same words for mathematical objects? As for other cultures on Earth, although mathematical words that sound similar to the corresponding terms in mathematical English are pretty common, they are not universal. If the mathematicians of Lower Slobovia want to use the term Hcanab ecaps (in honour of some Slobovian mathematician) instead of Banach space, they are free to do so, and nobody else will care. The English-Slobovian mathematical dictionary will contain an entry: Banach space: Hcanab ecaps So what? This won't cause any problems. What _does_ cause problems is when the same word is used in different cultures, but with slightly different meanings. For example, the French positif means non-negative (i.e. >= 0) rather than positive (> 0). === Subject: Re: Why not anonymous publication? > Hello again, Herman. Ok, so I assume Hilbert is some person, then. > Well, why not type-A space? I think the example of a commutative > group being called Abelian is at least one example of something that > is far more useful with a descriptive label. Certainly that only goes > so far (as systematic chemical naming has found). But the problem > remains of honoring people when nobody creates math, they just > discover it. What to do when cultures meet up (what happens when > aliens come?). If you name things after people, you'll argue that you > want your culture's name, and it's a pointless battle. You may not realize this, but discovering math is hard work which requires a lot of insight. If you don't believe me, try constructing the real numbers, developing calculus (and maybe some linear algebra on R^n), and proving the generalized Stokes' theorem, in that order, building only on what you have already done. Every math major has or will do this. This is actually fairly easy to do, once you know how. But developing such machinery from scratch is far from easy. People like Riemann, Gauss, Stokes, Cauchy, et al did this, which is why substantial portions of analysis are named after them. By the way, talk of space aliens doesn't help your argument. hoop dica === Subject: Re: Why not anonymous publication? > However, if we carried this idea out for all definitions, > think of how clumsy the notation would be. A Hilbert space > would be a complete normed topological vector space with > the norm given by an inner product, and these terms could > be expanded. In this particular example, couldn't we just use inproduct space? Would you like to use topological space for > which all open coverings have countable refinements > instead of Lindelof spaces. countably-compact spaces? Or how about Stone-Cech > compactification? What is that? It could be that i actually know the construction, but under another name... > Or the Stone space of a Boolean algebra? Canonic space, Prime filter space? Herman Jurjus === Subject: Re: Why not anonymous publication? > Hello again, Herman. Ok, so I assume Hilbert is some person, then. > Well, why not type-A space? I think the example of a commutative > group being called Abelian is at least one example of something that > is far more useful with a descriptive label. Certainly that only goes > so far (as systematic chemical naming has found). But the problem > remains of honoring people when nobody creates math, they just > discover it. What to do when cultures meet up (what happens when > aliens come?). If you name things after people, you'll argue that you > want your culture's name, and it's a pointless battle. > You may not realize this, but discovering math is hard work which > requires a lot of insight. And one more crucial point: There is very little emotional satisfation in doing math. (Not in the same way as in, say, being a musician, teacher, doctor in a hospital, preacher, ...) And the money and you can make is also not comparable to that of other businesses. Same for social status or power. So most mathematicians get their satisfaction from two things: being part of an age old culture and being able to do things that not everyone can. The hope that your own name will one day be part of that culture is analogous to the hope of, say, a young tennis player hoping to win Wimbledon one day. (And equally idle, in most cases.) Without a culture that honours it's heros, why would anyone bother? Herman Jurjus === Subject: Re: Why not anonymous publication? > Would you like to use topological space for >> which all open coverings have countable refinements >> instead of Lindelof spaces. > countably-compact spaces? Bad example. Countably compact already has an accepted meaning. A Lindelof space is one where each open covering has a countable subcovering. A countably compact space is one where each countable open covering has a finite subcovering. Not the same thing at all! -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Why not anonymous publication? > Without a culture that honours it's heros, why would anyone bother? its heroes. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Why not anonymous publication? > And one more crucial point: > There is very little emotional satisfation in doing math. > (Not in the same way as in, say, being a musician, teacher, > doctor in a hospital, preacher, ...) Sorry, but the above is medieval mentality nonsense. Math is, predominantly, beauty. Anybody who hasn't realized that, _of course_ doesn't feel emotional satisfaction. To those of us who see it for what it is, a beautiful game about structures, complete with playing rules and all the excitement similar to engaging into any other game in which a kid derives the emotional pleasure from delving into it and gradually exploring the areas that interest them, it can be most exciting, emotionally, to the point of excilaration. I am not interested in any of the things you quote below. My only satisfaction is discovering the elegance that lurks behind Math's structures. Understanding Math, even to a small degree, offers a unique insight into the world, and honestly, in my not so humble opinion, it tells you a lot about Nature's taste, which is, The Taste. > And the money and you can make is also not comparable to that > of other businesses. Same for social status or power. > So most mathematicians get their satisfaction from two things: > being part of an age old culture and being able to do > things that not everyone can. Not all of us do. Some of us derive pleasure from having at our disposal an almost infinite resource, which keeps our minds busy with beautiful thoughts. These thoughts elevate the mind, prune it and keep it at a state of constant excitement for what is yet to come. The concept of discovery in Math, particularly when it occures in self-study or research (under favourable cirmustances and not under the supervision of some indifferent supervisor who could care less about your thesis), is emotionally excilarating. I'd say it almost touches the divine, but in a different way than, say, music does. Music can literally cause ecstasis, where most of one's emotional resources are given into the experience. Math causes a different kind of ecstasis, which is not as involved emotionally as that for music, but is, nevertheless, a very intense experience, in and of itself. Don't confuse quality with quantity: Music offers quantitativelly, more emotional highs. But when you discover something in Math, the high is almost unbearable. There have been times when I have been unable to sleep for two days in a row because of sheer excitement, after my intuition leads me into an unexplored area. Besides Music, Math is perhaps the absolute forbidden high. And I say forbidden, because the sense of awe that one gets from quality Math, is perhaps not for everyone. Or should I say, not everyone can appreciate it. If all Mathematicans viewed Math as a beautiful game, (actually if everybody saw _everything_ as a game) then Math's social status would be much higher and there would be many more students studying it. Unfortunatelly, so many get stuck on the rules and fuss over unimportant details, which often obfuscate the grander picture behind Math. And the grander picture is the sense of awe one gets when one first discovers his first new theorem. > The hope that your own name will one day be part of that > culture is analogous to the hope of, say, a young tennis player > hoping to win Wimbledon one day. (And equally idle, in most cases.) > Without a culture that honours it's heros, why would anyone bother? Peer review and acceptance is probably one of the most important aspects of modern day professional Math. But it doesn't even compare to the personal satisfaction one gets, after discovering a beautiful theorem or lemma. ___________________________________________ .