>A transaction consists of adding ten balls to a bucket and removing 1. >>(Obviously a transaction is a net increase of nine balls.) Assume that >>infinitely many transactions somehow occur. Some people here think >>there will be no balls in the bucket afterward! >>There are some people who think that there will only be zero balls >>left if we label the balls in a certain way, otherwise there will be >>more than zero! >>There are other people who think that there will be none left >>because the cardinalities of the sets of balls added to the bucket is >>the same as the cardinality of the set of balls removed! >>I tend to say the heck with their theories. I know that if it were >>truly possible for infinitely many transactions to occur, the bucket >>would be VERY full! Where do theories that say othwise come >>from? Does anyone really disagree that the bucket has more balls >>after each transaction and never has less? > If the balls are labeled and a labeled ball is removed after each > time it is inserted, is that ball in the bucket at the end? If it is in the bucket at the end, how does it get there? If it is not in at the end, and being not in is true of every ball, > which (allegedly large number of) balls are in the bucket at the end? Surely if every transaction takes a finite time there is no end. Oh really, so exactly what does 1/2 sec + 1/4 sec + 1/8 sec +...+ 1/2^n +...= ? Pesonally I thought that it equals 1 sec. Gib > ==== [snip] > Surely if every transaction takes a finite time there is no end. It seems that some people have yet to resolve, in their own minds, the simple paradoxes of Zeno! Believe it or not, an infinite series of positive terms (times) can have a finite sum. For example, 1/2 sec + 1/4 sec + 1/8 sec +...+ 1/2^N sec +... = 1 sec. David ==== Poker Joker scribbled the following: > A transaction consists of adding ten balls to a bucket and removing 1. > (Obviously a transaction is a net increase of nine balls.) Assume that > infinitely many transactions somehow occur. Some people here think > there will be no balls in the bucket afterward! > There are some people who think that there will only be zero balls > left if we label the balls in a certain way, otherwise there will be > more than zero! > There are other people who think that there will be none left > because the cardinalities of the sets of balls added to the bucket is > the same as the cardinality of the set of balls removed! > I tend to say the heck with their theories. I know that if it were > truly possible for infinitely many transactions to occur, the bucket > would be VERY full! Where do theories that say othwise come > from? Does anyone really disagree that the bucket has more balls > after each transaction and never has less? I don't know if I'm responding to a troll here, but given the above, each transaction adds nine balls to the bucket, thus creating a monotonically increasing function from the index number of the transactions to the number of balls in the bucket. So it is impossible for the bucket to be empty *at any time* after the start of the experiment. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ As a boy, I often dreamed of being a baseball, but now we must go forward, not backward, upward, not forward, and always whirling, whirling towards freedom! - Kang ==== > I don't know if I'm responding to a troll here, but given the above, > each transaction adds nine balls to the bucket, thus creating a > monotonically increasing function from the index number of the > transactions to the number of balls in the bucket. So it is impossible > for the bucket to be empty *at any time* after the start of the > experiment. While it is impossible for the mathematically modeled bucket to be empty after any finite number of transactions, one of the peculiarities of infinite mathematical processes is that the bucket CAN be empty after infinitely many transactions, since each ball put into the bucket can be, then or later, removed from the bucket in an infinite mathematical process completed in finite mathematical time. It is our attempts to interpret the mathematical model of bucket and balls and transactions in our everyday non-mathematical world that troubles our understanding. ==== > Poker Joker scribbled the following: >A transaction consists of adding ten balls to a bucket and removing 1. >>(Obviously a transaction is a net increase of nine balls.) Assume that >>infinitely many transactions somehow occur. Some people here think >>there will be no balls in the bucket afterward! >>There are some people who think that there will only be zero balls >>left if we label the balls in a certain way, otherwise there will be >>more than zero! >>There are other people who think that there will be none left >>because the cardinalities of the sets of balls added to the bucket is >>the same as the cardinality of the set of balls removed! >>I tend to say the heck with their theories. I know that if it were >>truly possible for infinitely many transactions to occur, the bucket >>would be VERY full! Where do theories that say othwise come >>from? Does anyone really disagree that the bucket has more balls >>after each transaction and never has less? > I don't know if I'm responding to a troll here, but given the above, > each transaction adds nine balls to the bucket, thus creating a > monotonically increasing function from the index number of the > transactions to the number of balls in the bucket. So it is impossible > for the bucket to be empty *at any time* after the start of the > experiment. > If after adding 10 balls I remove the last one added, it is completely equivalent to adding 9 balls, and so we can say which balls are certainly going to stay in the bucket. These are the balls numbered 1-9, 11-19, 21-29 etc. In fact, we can map each natural number to a ball we definitely know is in the bucket (n -> 10n+1), and so we know there are infinitely many balls in the bucket after infinitely many transactions. However, if after adding 10 balls I remove the *first* ball (i.e., the ball which was added as early as possible and is still in the bucket), then we have a problem. At time t we remove ball number t. In this case, if you think there are infinitely many balls in the bucket, then please name one. I know that ball number t was removed in time t, and so for any natural number t, ball number t is *not* in the bucket. So the bucket must be empty. To be honest, I don't think there's any problem. All we can say is that the number of balls depends on the choice of ball to remove. If I always remove the first ball, I'll end up with no balls. If I always remove ball number t, I'll end up with t-1 balls. If I always remove ball number x_t where x_t is monotonically increasing, I'll end up with infinitely many balls. Any set theorist to resolve this? This seems to me like the paradox of f(x) = 1/x. For every natural number n, you can find a positive number for which f(n) is n+1 times larger (simply 1/(n*(n+1))). So I say that since, as n increases, we can find more and more positive numbers between 0 and f(n), how can the limit at infinity of f(n) be 0? After all, for any finite number m, we can find at least m numbers, all different, all positive, between 0 and f(n). Hence, a paradox. The problem here is that you think of the one step before infinity, but there's not such thing. It's a quantum leap. ==== > To be honest, I don't think there's any problem. All we can say is > that the number of balls depends on the choice of ball to remove. If I > always remove the first ball, I'll end up with no balls. If I always > remove ball number t, I'll end up with t-1 balls. If I always remove > ball number x_t where x_t is monotonically increasing, I'll end up > with infinitely many balls. Any set theorist to resolve this? This is right. ==== > Poker Joker scribbled the following: >A transaction consists of adding ten balls to a bucket and removing 1. >>(Obviously a transaction is a net increase of nine balls.) Assume that >>infinitely many transactions somehow occur. Some people here think >>there will be no balls in the bucket afterward! >>There are some people who think that there will only be zero balls >>left if we label the balls in a certain way, otherwise there will be >>more than zero! >>There are other people who think that there will be none left >>because the cardinalities of the sets of balls added to the bucket is >>the same as the cardinality of the set of balls removed! >>I tend to say the heck with their theories. I know that if it were >>truly possible for infinitely many transactions to occur, the bucket >>would be VERY full! Where do theories that say othwise come >>from? Does anyone really disagree that the bucket has more balls >>after each transaction and never has less? > I don't know if I'm responding to a troll here, but given the above, > each transaction adds nine balls to the bucket, thus creating a > monotonically increasing function from the index number of the > transactions to the number of balls in the bucket. So it is impossible > for the bucket to be empty *at any time* after the start of the > experiment. > If after adding 10 balls I remove the last one added, it is completely > equivalent to adding 9 balls, and so we can say which balls are > certainly going to stay in the bucket. These are the balls numbered 1-9, > 11-19, 21-29 etc. In fact, we can map each natural number to a ball we definitely know is > in the bucket (n -> 10n+1), and so we know there are infinitely many > balls in the bucket after infinitely many transactions. However, if after adding 10 balls I remove the *first* ball (i.e., the > ball which was added as early as possible and is still in the bucket), > then we have a problem. At time t we remove ball number t. In this case, > if you think there are infinitely many balls in the bucket, then please > name one. I know that ball number t was removed in time t, and so for > any natural number t, ball number t is *not* in the bucket. So the > bucket must be empty. To be honest, I don't think there's any problem. All we can say is that > the number of balls depends on the choice of ball to remove. If I always > remove the first ball, I'll end up with no balls. If I always remove > ball number t, I'll end up with t-1 balls. If I always remove ball > number x_t where x_t is monotonically increasing, I'll end up with > infinitely many balls. Any set theorist to resolve this? You will not end up with ..., because there is no such thing as after infinitely many transactions. Dirk Vdm > This seems to me like the paradox of f(x) = 1/x. For every natural > number n, you can find a positive number for which f(n) is n+1 times > larger (simply 1/(n*(n+1))). So I say that since, as n increases, we can > find more and more positive numbers between 0 and f(n), how can the > limit at infinity of f(n) be 0? After all, for any finite number m, we > can find at least m numbers, all different, all positive, between 0 and > f(n). Hence, a paradox. The problem here is that you think of the one > step before infinity, but there's not such thing. It's a quantum leap. > ==== > You will not end up with ..., because there is no such thing > as after infinitely many transactions. Sure there is; the transactions can be numbered with ordinals and there is no problem at all. ==== > Poker Joker scribbled the following: >> A transaction consists of adding ten balls to a bucket and removing 1. >> (Obviously a transaction is a net increase of nine balls.) Assume that >> infinitely many transactions somehow occur. Some people here think >> there will be no balls in the bucket afterward! >> There are some people who think that there will only be zero balls >> left if we label the balls in a certain way, otherwise there will be >> more than zero! >> There are other people who think that there will be none left >> because the cardinalities of the sets of balls added to the bucket is >> the same as the cardinality of the set of balls removed! >> I tend to say the heck with their theories. I know that if it were >> truly possible for infinitely many transactions to occur, the bucket >> would be VERY full! Where do theories that say othwise come >> from? Does anyone really disagree that the bucket has more balls >> after each transaction and never has less? > I don't know if I'm responding to a troll here, but given the above, > each transaction adds nine balls to the bucket, thus creating a > monotonically increasing function from the index number of the > transactions to the number of balls in the bucket. So it is impossible > for the bucket to be empty *at any time* after the start of the > experiment. The problem as stated is ambiguous, because it does not specify how the ball to be removed is chosen at each step. See the concurrent thread Ten Balls In - One Ball Out - Repeat - How Many Remain?. Or go to Google groups and search for the historical threads Bucket and Balls and its offshoot Transfinite Subway. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > A transaction consists of adding ten balls to a bucket and removing 1. > (Obviously a transaction is a net increase of nine balls.) Assume that > infinitely many transactions somehow occur. Some people here think > there will be no balls in the bucket afterward! There are some people who think that there will only be zero balls > left if we label the balls in a certain way, otherwise there will be > more than zero! There are other people who think that there will be none left > because the cardinalities of the sets of balls added to the bucket is > the same as the cardinality of the set of balls removed! I tend to say the heck with their theories. I know that if it were > truly possible for infinitely many transactions to occur, the bucket > would be VERY full! Where do theories that say othwise come > from? Does anyone really disagree that the bucket has more balls > after each transaction and never has less? > If you say that there's a ball left at the end, which one is it? It can't be #47, that was removed on the 47th step. And so forth. ==== A transaction consists of adding ten balls to a bucket and removing 1. > (Obviously a transaction is a net increase of nine balls.) Assume that > infinitely many transactions somehow occur. Some people here think > there will be no balls in the bucket afterward! There are some people who think that there will only be zero balls > left if we label the balls in a certain way, otherwise there will be > more than zero! There are other people who think that there will be none left > because the cardinalities of the sets of balls added to the bucket is > the same as the cardinality of the set of balls removed! I tend to say the heck with their theories. I know that if it were > truly possible for infinitely many transactions to occur, the bucket > would be VERY full! Where do theories that say othwise come > from? Does anyone really disagree that the bucket has more balls > after each transaction and never has less? > If you say that there's a ball left at the end, which one is it? It > can't be #47, that was removed on the 47th step. And so forth. You don't know that. If you remove the last ball that was put in, I'm certain that balls 1,2,3,4,5,6,7,8,10,... will always stay in the bucket. By the way, at the end of what? ==== > The length of the curve f(x) on the continuous curve [a,b] is the > integral taken from a to b of the squareroot of the sum of the > derivitive of the function, squared, plus one. This formula for arc length is probably in every elementary calculus text. Lol, not mine, but then again, mine's not all that great. Woulda saved me a good three hours though. ==== > Hey, haven't been here in a while... Anyway, Thursday and Friday I was working on a problem, basically, > given a fuciton and a continuous interval of that function, I wanted > to find the length of its curve. I eventually came up with this: > The length of the curve f(x) on the continuous curve [a,b] is the > integral taken from a to b of the squareroot of the sum of the > derivitive of the function, squared, plus one. > Bobby Simione That is a standard result in calculus. If you discovered it on your own, rather than reading it in a Calculus text, that is quite good, and shows that you have a much better than average grasp of the material. Kepp trying to discover things new to you. It does not matter if you are not the first. Martin Cohen ==== >That is a standard result in calculus. If you discovered it >on your own, rather than reading it in a Calculus text, >that is quite good, and shows that you have a much better >than average grasp of the material. Kepp trying to discover things new to you. It does not matter >if you are not the first. Martin Cohen ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ==== R is the reals. That N likely a typo, read R instead. >You claim A_n ->* A iff >A_n subset A forall n and for all p > 0, p in R exists q in N >for all a in A for all n > q exists b in A_n: d(a,b) < p For all n, let A_n = D be a dense subset of R. >Let A be any set with D subset A subset R. >Then A_n ->* A for all such A's. Your uniform convergence, and hence your convergence doesn't have unique >limits. Moreover, a sequence can even converge to different sets with >different cardinalities. -- >Remember the following part of the first >message in this series of posting? No. >You have a problem here, because a limit is not unique, if A' is a >subset of A, then A_n -> A implies A_n -> A'. All of this stuff has topology in it, contary to your title topic. For limits without topology, for A_n, A subset S, there's: limsup A_n = /{ { / Aj | j > k } | k in N } liminf A_n = /{ { / Aj | j > k } | k in N } liminf A_n subset limsup A_n; / / intersect union A_n -> A when A = limsup A_n = liminf A_n lim A_n = liminf A_n = limsup A_n descending, ascending A_n ==> lim A_n = /{ A_n | n in N }, /{ A_n | n in N } Whatever math I dream up is already old hat. -- William's Metatheorem ---- ==== > ... >am doing. in the end i am trying to show that Aut(Z16) is isomorphic >to Z2xZ8, and I dont think I want to use the fact that I know all >groups of order 8. > ... If Z16 stands for the cyclic group with 16 elements, then its rather simple to determine Aut(Z16): choose a generator, for example 1. Then an automorphism f is uniquely determined by prescribing the image f(1). This image must be another generator of Z16. So there are as many automorphisms as generators of Z16. Less than 16 by the way - so Aut (Z16) cannot be isomorphic to Z2xZ8. H actually, I made a mistake ... that should read: am doing. in the end i am trying to show that Aut(Z16) is isomorphic to Z2xZ4, and I dont think I want to use the fact that I know all groups of order 8. I already know what Aut(Z16) is, I am trying to show it by elementary means. in fact, i know that Aut(Z16) is isomorphic to the set of units in the ring (Z16,+,*) or all number relatively prime to 16 equipped with multiply. but finding a simple proof that Aut(Z16)=Z2xZ4 has eluded me. and by simple I mean not describing all groups of order 8, since |Aut(Z16)|=8. thanks ==== in message <83f64ab8.0312132314.5aaab249@posting.google.com>: [...] > I already know what Aut(Z16) is, I am trying to show it by elementary > means. in fact, i know that Aut(Z16) is isomorphic to the set of units > in the ring (Z16,+,*) or all number relatively prime to 16 equipped > with multiply. but finding a simple proof that Aut(Z16)=Z2xZ4 has > eluded me. and by simple I mean not describing all groups of order 8, > since |Aut(Z16)|=8. Well, you can get away with describing only the abelian groups of order 8 by first showing that the automorphism group of a cyclic group is abelian. -- Jim Heckman ==== > I am given a set of generators (s_1, ..., s_{n-1}) and relations for a > group G_n: > s_i^2 = 1 (i=1...n-1) > s_i s_j = s_j s_i ( |i-j|> 1) > s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1} (i=1...n-2) I have verified that these properties hold for transpositions in S_n, e.g. > for (1,2), (2,3), (n-1,n), but I don't know that there can't be other > relations for S_n. I want to show that G_n = S_n. Since phi: G_n -> S_n is > surjective, I was thinking of using Todd-Coxeter and induction to show > that > |G_n| = |S_n|. I tried this, but did not get very far. Any hints? You could show that any word in these generators is equivalent to one in a canonical form. One possible canonical form is this. Define words t(i,j) where 1 <= i <= j <= n. Then t(i,i) is empty while t(i,j) = s_i s_{i+1} ... s_{j-1} for i < j. The shw that each word in the s_i can be reduced to a word t(n-1, a_{n-1}) t(n-2, a_{n-2}) .... t(2, a_2)t(1, a_1) by means of the generators. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== Arrrghh.... this calculator is really annoying me. I've put my fx-991MS calculator into CMPLX mode, and I can get it to >calculate and show me, for example, the real and imaginary parts of 1 - 3i >(simple I know but it definitely works). However I can't get it to show me >the real and imaginary parts of, for example, 1 - sqrt3i. It says Math >ERROR. This also happens with *any* other complex numbers calculation >involving roots or powers. If you mean (sqrt(3))*i, then you can almost certainly overcome the > problem by putting in some parentheses. I still can't raise a complex expression to a power though e.g. (1 + 3i)^3. The only exception is squaring it - the calulator has a separate button specifically for squaring i.e. I don't have to put in the ^ then the 3 (or any power), I simple press the square button. Any idea on the correct parenthesis for this? I've tried (1 + 3i)^3, (1 + 3i)(^3), ((1 + 3i)(^3)) - none seem to work. > If you mean sqrt(3i) then you are up against a restriction on the > operations allowed on your calculator. (Note that sqrt is > multi-valued.) Perhaps this is the reason I can't square the expression? ==== Arrrghh.... this calculator is really annoying me. I've put my fx-991MS calculator into CMPLX mode, and I can get it to >calculate and show me, for example, the real and imaginary parts of 1 - > 3i >(simple I know but it definitely works). However I can't get it to show > me >the real and imaginary parts of, for example, 1 - sqrt3i. It says Math >ERROR. This also happens with *any* other complex numbers calculation >involving roots or powers. If you mean (sqrt(3))*i, then you can almost certainly overcome the > problem by putting in some parentheses. > I still can't raise a complex expression to a power though e.g. (1 + 3i)^3. > The only exception is squaring it - the calulator has a separate button > specifically for squaring i.e. I don't have to put in the ^ then the 3 > (or any power), I simple press the square button. Any idea on the correct > parenthesis for this? I've tried (1 + 3i)^3, (1 + 3i)(^3), ((1 + 3i)(^3)) - none seem to work. If you mean sqrt(3i) then you are up against a restriction on the > operations allowed on your calculator. (Note that sqrt is > multi-valued.) Perhaps this is the reason I can't square the expression? Does your Casio require explicit multiplication symbols? If so, then enter (1 + 3*i)^3 or (1 + (3*i))^3 with an explicit multiplication symbol, and see how it evaluates. ==== >>Arrrghh.... this calculator is really annoying me. >>I've put my fx-991MS calculator into CMPLX mode, and I can get it to >>calculate and show me, for example, the real and imaginary parts of 1 - >3i >>(simple I know but it definitely works). However I can't get it to show >me >>the real and imaginary parts of, for example, 1 - sqrt3i. It says Math >>ERROR. This also happens with *any* other complex numbers calculation >>involving roots or powers. >> If you mean (sqrt(3))*i, then you can almost certainly overcome the >> problem by putting in some parentheses. >I still can't raise a complex expression to a power though e.g. (1 + 3i)^3. >The only exception is squaring it - the calulator has a separate button >specifically for squaring i.e. I don't have to put in the ^ then the 3 >(or any power), I simple press the square button. Any idea on the correct >parenthesis for this? I've tried (1 + 3i)^3, (1 + 3i)(^3), ((1 + 3i)(^3)) - none seem to work. > If you mean sqrt(3i) then you are up against a restriction on the >> operations allowed on your calculator. (Note that sqrt is >> multi-valued.) Perhaps this is the reason I can't square the expression? > Well, probably for a similar reason. One approach to programming the ^ operation on a calculator is to take a logarithm, then do a multiplication, then do an antilogarithm. I vaguely remember one calculator I have owned (or it may have been an early implementation of BASIC) refuse to do things like (-2)^3, and I assumed that the reason is something to do with the non-existence (in reals) of log(-2). In complex numbers the log function is multi-valued. In both cases, negative reals and complex numbers, it is possible to work around the problem; my current calculator is perfectly happy with (-2)^3. I don't know the exact algorithms used on calculators. But they will have been optimised for real arithmetic. When programming the complex number operations, the designers will have had to decide when to adapt the routines to allow operations with complex numbers, and when not to bother. It seems that Casio decided that general exponentiation of complex numbers was not worth the bother. ==== >Arrrghh.... this calculator is really annoying me. >>I've put my fx-991MS calculator into CMPLX mode, and I can get it to >>calculate and show me, for example, the real and imaginary parts of 1 - >3i >>(simple I know but it definitely works). However I can't get it to show >me >>the real and imaginary parts of, for example, 1 - sqrt3i. It says Math >>ERROR. This also happens with *any* other complex numbers calculation >>involving roots or powers. >> If you mean (sqrt(3))*i, then you can almost certainly overcome the >> problem by putting in some parentheses. >I still can't raise a complex expression to a power though e.g. (1 + 3i)^3. >The only exception is squaring it - the calulator has a separate button >specifically for squaring i.e. I don't have to put in the ^ then the 3 >(or any power), I simple press the square button. Any idea on the correct >parenthesis for this? I've tried (1 + 3i)^3, (1 + 3i)(^3), ((1 + 3i)(^3)) - none seem to work. > If you mean sqrt(3i) then you are up against a restriction on the >> operations allowed on your calculator. (Note that sqrt is >> multi-valued.) Perhaps this is the reason I can't square the expression? > Well, probably for a similar reason. One approach to programming the > ^ operation on a calculator is to take a logarithm, then do a > multiplication, then do an antilogarithm. I vaguely remember one > calculator I have owned (or it may have been an early implementation > of BASIC) refuse to do things like (-2)^3, and I assumed that the > reason is something to do with the non-existence (in reals) of > log(-2). In complex numbers the log function is multi-valued. In both > cases, negative reals and complex numbers, it is possible to work > around the problem; my current calculator is perfectly happy with > (-2)^3. I don't know the exact algorithms used on calculators. But they will > have been optimised for real arithmetic. When programming the complex > number operations, the designers will have had to decide when to adapt > the routines to allow operations with complex numbers, and when not to > bother. It seems that Casio decided that general exponentiation of > complex numbers was not worth the bother. I've got a Casio CFX-9850G, and the manual specificaly states that complex operation is limited to simple addition, subtraction, multiplication and division, along with the square, square root and reciprocal functions. Complex-specific functions for the modulus, argument and conjugate are provided, so all other functions (powers, logs, trig) can be implemented using these. -- Paul V. S. Townsend Interchange the alphabetic elements to reply ==== let I=k[x_1,x_2,..,x_n] is polinomial ring over field of char=0 and > I_n - subspase gomogenius polinomial of power n. Let sl_2 - 3 - > dimesional simple lie algebra wich act at I_n in usual way. How find a > irreducible components of decomposition of this representation? Need > find something like as formulae of (Klebsh-Gordon)for tensor product. > Try finding the highest weight elements, counting dimensions (find its > character) and semi-simplicity. At present I find as follows . Let f_i - standart (i+1) -dimension representation of sl_2. The decomposition for I_2 is I_2=f_2n+f_(2n-4)+f_(2n-8)+..... I am looking for generalisation of this for arbitrary I_k ==== >> let I=k[x_1,x_2,..,x_n] is polinomial ring over field of char=0 and >> I_n - subspase gomogenius polinomial of power n. Let sl_2 - 3 - >> dimesional simple lie algebra wich act at I_n in usual way. How find a >> irreducible components of decomposition of this representation? Need >> find something like as formulae of (Klebsh-Gordon)for tensor product. >> Try finding the highest weight elements, counting dimensions (find its >> character) and semi-simplicity. At present I find as follows . Let f_i - standart (i+1) -dimension > representation of sl_2. The decomposition for I_2 is I_2=f_2n+f_(2n-4)+f_(2n-8)+..... I am looking for generalisation of this for arbitrary I_k Something struck me about this: what do you mean act on the degree n homogeneous polynomials in the usual way? I know the usual way for 2 variables. Incidentally, you're using n twice in the same statement for different things (I think). ==== > http://mathworld.wolfram.com/Determinant.html gives the distribution of |det| if the elements of the matrix are within the > unit disk. > I can't tell of this distribution depends on the matrix dimension, Of course it does, due to the mentioned upper bound... > but what > is known about it? What would be the limit when n increases? That question remains :) ==== > http://mathworld.wolfram.com/Determinant.html gives the distribution of |det| if the elements of the matrix are within the > unit disk. > I can't tell of this distribution depends on the matrix dimension, but what > is known about it? What would be the limit when n increases? My reading of this result is that is is an upper bound, not the distribution. ==== http://mathworld.wolfram.com/Determinant.html gives the distribution of |det| if the elements of the matrix are within the > unit disk. > I can't tell of this distribution depends on the matrix dimension, but what > is known about it? What would be the limit when n increases? > My reading of this result is that is is an upper bound, not the > distribution. The plots above show the distribution of determinants for random nxn complex matrices with entries satisfying |aij|<1 for n = 2, 3, and 4. Try reading more than the first line beneath a plot. How could you plot the upper bound of an 2x2 random matrix and get a non-constant function anyway.... ==== Find the values of cos(pi/6 + 3i) Am I correct in making this cos(pi/6) + cosh(3)? Probably not because the answer I've been given is 8.7189 - 5.0089i. So if not, what do I do to get the answer? ==== Use the standard formula for the cosine of a sum: cos (x+y) = cos x cos x - sin x sin y OK now suppose y is made pure imaginary as iy; then: cos (x+iy) = cos x cos iy - sin x sin iy. Now the sines and cosines of pure imaginary numbers are related to HYPERBOLIC functions of the correspondin real numbers; cos iy = cosh y and sin i = i sinh y. Plug that into the above and you should get what you are looking for. --OL ==== >Find the values of cos(pi/6 + 3i) Am I correct in making this cos(pi/6) + cosh(3)? Probably not because the >answer I've been given is 8.7189 - 5.0089i. So if not, what do I do to get >the answer? remember that cos(a+b) = cos(a)cos(b) - sin(a)sin(b) and that cos(ix) = cosh(x) and sin(ix) = i sinh(x) Rob Johnson take out the trash before replying ==== > Find the values of cos(pi/6 + 3i) Am I correct in making this cos(pi/6) + cosh(3)? Probably not because the > answer I've been given is 8.7189 - 5.0089i. So if not, what do I do to get > the answer? No, that's not correct. cos(pi/6 + 3i) = cos(pi/6)cos(3i) - sin(pi/6)sin(3i) cos(3i) = cosh(3) sin(3i) = i*sinh(3) therefor cos(pi/6 + 3i) = cos(pi/6)cosh(3) - i*sin(pi/6)sinh(3) ==== >Find the values of cos(pi/6 + 3i) Am I correct in making this cos(pi/6) + cosh(3)? No. >Probably not because the >answer I've been given is 8.7189 - 5.0089i. So if not, what do I do to get >the answer? Use the definition? (cos(z) = (exp(iz) + exp(-iz))/2.) ************************ David C. Ullrich ==== L is a langage containing only one symbol of binary relation R. T is the theory of equivalence relation with infinite many infinite classes expressed in L. M and N are two models of T and M is a sub-structure of N. It has to be proved that N is an elementary extension of M, with 3 steps: - Every equivalence class of N is represented in M. - All the equivalence classes of N have the same cardinal. - General case. Any help appreciated ! ==== You know, _crossposting_ is a better idea than just posting the same message to two different groups. With a crosspost people reading one group can see what people have said about it elsewhere... sci.math, sci.logic.) >L is a langage containing only one symbol of binary relation R. T is the theory of equivalence relation with infinite many infinite classes >expressed in L. M and N are two models of T and M is a sub-structure of N. It has to be >proved that N is an elementary extension of M, with 3 steps: You must be leaviing out part of the problem(?) >- Every equivalence class of N is represented in M. because it seems clear that this doesn't follow >- All the equivalence classes of N have the same cardinal. and while I suppose I could be overlooking something there, it's _very_ clear that _this_ doesn't follow. >- General case. Any help appreciated ! ************************ David C. Ullrich ==== Each specific point doesn't follow the other ones. They are just easier to be solve than the general case. May be my word STEP is wrong, CASE would be better. Jean-Pierre. David C. Ullrich a .8ecrit dans le message de > You know, _crossposting_ is a better idea than just > posting the same message to two different groups. > With a crosspost people reading one group can > see what people have said about it elsewhere... sci.math, sci.logic.) >L is a langage containing only one symbol of binary relation R. T is the theory of equivalence relation with infinite many infinite classes >expressed in L. M and N are two models of T and M is a sub-structure of N. It has to be >proved that N is an elementary extension of M, with 3 steps: You must be leaviing out part of the problem(?) - Every equivalence class of N is represented in M. because it seems clear that this doesn't follow - All the equivalence classes of N have the same cardinal. and while I suppose I could be overlooking something there, > it's _very_ clear that _this_ doesn't follow. - General case. Any help appreciated ! ************************ David C. Ullrich ==== >Each specific point doesn't follow the other ones. They are just easier to >be solve than the general case. May be my word STEP is wrong, CASE would be >better. Oh. Yes, the word case would have been clearer. >Jean-Pierre. David C. Ullrich a .8ecrit dans le message de >> You know, _crossposting_ is a better idea than just >> posting the same message to two different groups. >> With a crosspost people reading one group can >> see what people have said about it elsewhere... >> sci.math, sci.logic.) >>L is a langage containing only one symbol of binary relation R. >>T is the theory of equivalence relation with infinite many infinite >classes >>expressed in L. >>M and N are two models of T and M is a sub-structure of N. It has to be >>proved that N is an elementary extension of M, with 3 steps: >> You must be leaviing out part of the problem(?) >>- Every equivalence class of N is represented in M. >> because it seems clear that this doesn't follow >>- All the equivalence classes of N have the same cardinal. >> and while I suppose I could be overlooking something there, >> it's _very_ clear that _this_ doesn't follow. >>- General case. >>Any help appreciated ! >> ************************ >> David C. Ullrich > ************************ David C. Ullrich ==== KRamsay schrieb: >>Interesting how the center column progress by 12 and the two columns >>on either side progress by 3. > I'm unable really to read Perron's book on continued fractions > because I don't know any German, but Perron appears to make > a study of the numbers whose continued fraction is sort of close > to periodic in this way: [a1,...,an, b1,...,bn, b1+c1, b2+c2, > b3+c3,..., bn+cn, b1+2c1, b1+2c2,..., bn+2cn, ...], with each > element of the cycle increasing in an arithmetic progression. Gosper has explained how to compute with continued fraction > numbers, and you can calculate the expansion of multiples > of e and so on easily enough, and you'll find a bunch of > related numbers which have the same type of continued fraction. > See for example http://www.inwap.com/pdp10/hbaker/hakmem/cf.html#item101b Keith Ramsay > If you allow negative coefficients, you seem to get a pretty general scheme for 1/k e for k= ... -3,-2,-1,{0},1,2,3,... ---------------------------------------------------------------------------- -------------- - 1 3 5 7 9 11 ---------------------------------------------------------------------------- -------------- cf(e^(1/-2)): [1,-3, 1, 1, -7, 1, 1,-11, 1, 1,-15, 1, 1,-19, 1, 1,-23, 1 ] cf(e^(1/-1)): [1,-2, 1, 1, -4, 1, 1, -6, 1, 1, -8, 1, 1,-10, 1, 1,-12, 1 ] cf(e^(1/ 0)): [1,-1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1 ] divergent (oscillates on 0 and 1) cf(e^(1/ 1)): [1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1 ] cf(e^(1/ 2)): [1, 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17, 1, 1, 21, 1 ] cf(e^(1/ 3)): [1, 2, 1, 1, 8, 1, 1, 14, 1, 1, 20, 1, 1, 26, 1, 1, 32, 1 ] cf(e^(1/ 4)): [1, 3, 1, 1, 11, 1, 1, 19, 1, 1, 27, 1, 1, 35, 1, 1, 43, 1 ] cf(e^(1/ 5)): [1, 4, 1, 1, 14, 1, 1, 24, 1, 1, 34, 1, 1, 44, 1, 1, 54, 1 ] cf(e^(1/ 6)): [1, 5, 1, 1, 17, 1, 1, 29, 1, 1, 41, 1, 1, 53, 1, 1, 65, 1 ] ---------------------------------------------------------------------------- -------------- + 1 3 5 7 9 11 ---------------------------------------------------------------------------- -------------- for 2/k e for k= ... -3,-2,-1,{0},1,2,3,... ---------------------------------------------------------------------------- ------------------------------------------- delta: - 1 12 5 7 36 11 13 60 17 19 84 23 25 ---------------------------------------------------------------------------- ------------------------------------------- cf(e^(2/-1)); [1,-1, -6, -3, 1, 1, -4, -18, -6, 1, 1, -7, -30, -9, 1, 1, -10, -42, -12, 1, 1,-23 ] cf(e^(2/1)); [1, 0, 6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, 12 ] cf(e^(2/3)); [1, 1, 18, 7, 1, 1, 10, 54, 16, 1, 1, 19, 90, 25, 1, 1, 28, 126, 34, 1, 1, 37 ] cf(e^(2/5)); [1, 2, 30, 12, 1, 1, 17, 90, 27, 1, 1, 32, 150, 42, 1, 1, 47, 210, 57, 1, 1, 62 ] cf(e^(2/7)); [1, 3, 42, 17, 1, 1, 24, 126, 38, 1, 1, 45, 210, 59, 1, 1, 66, 294, 80, 1, 1, 87 ] cf(e^(2/9)); [1, 4, 54, 22, 1, 1, 31, 162, 49, 1, 1, 58, 270, 76, 1, 1, 85, 378, 103, 1, 1, 112 ] cf(e^(2/11)); [1, 5, 66, 27, 1, 1, 38, 198, 60, 1, 1, 71, 330, 93, 1, 1, 104, 462, 126, 1, 1, 95 ] ---------------------------------------------------------------------------- ------------------------------------------- delta: + 1 12 5 7 36 11 13 60 17 19 84 23 25 ---------------------------------------------------------------------------- ------------------------------------------- Also, allowing fractions for coefficients, the primary expansion of e = e^(1/1) = e^(2/2) can be inserted in the previous table: ---------------------------------------------------------------------------- ------------------------------------------- ... cf(e^(2/1)): [1, 0, 6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1 ... cf(e^(2/2)): [1, 0.5, 12, 4.5, 1, 1, 6.5, 36, 10.5, 1, 1, 12.5, 60, 16.5, 1, 1, 18.5, 84, 22.5, 1 ... cf(e^(2/3)): [1, 1, 18, 7, 1, 1, 10, 54, 16, 1, 1, 19, 90, 25, 1, 1, 28, 126, 34, 1 ... ... ---------------------------------------------------------------------------- ------------------------------------------- delta + 0.5 6 2.5 3.5 18 5.5 6.5 60 8.5 9.5 42 11.5 ---------------------------------------------------------------------------- ------------------------------------------- Perhaps this allowing of negative and/or fractional coefficients enables also to find more simple regularities for e^k with abs(k)>2 ==================================================================== With the golden-ratio (phi) you can have also some regularities: Golden-Ratio x = phi ~ 1.61803398874989484820... ------------------------------------------------------------- ?? cf(x^-2) ?? cf(x^-1); [ -1, -1, ?????? ------------------------------------------------------------ cf(x^1); [ 1, 1, 1, 1, 1, 1, 1, 1, 1 ... cf(x^3); [ 4, 4, 4, 4, 4, 4, 4, 4, 4 ... = a1*3-a_(-1) cf(x^5); [ 11, 11, 11, 11, 11, 11, 11, 11, 11 ... = a3*3-a1 cf(x^7); [ 29, 29, 29, 29, 29, 29, 29, 29, 29 ... = a5*3-a3 cf(x^9); [ 76, 76, 76, 76, 76, 76, 76, 76, 76 ... = a7*3-a5 cf(x^11); [199, 199, 199, 199, 199, 199, 199, 199 ... cf(x^0): [ 1, 0, 0, 0, 0, 0, 0, 0, 0, cf(x^2); [ 2, 1, 1, 1, 1, 1, 1, 1, 1 ... = a0 + a1 cf(x^4); [ 6, 1, 5, 1, 5, 1, 5, 1, 5 ... = a2 + a3 cf(x^6); [ 17, 1, 16, 1, 16, 1, 16, 1, 16 ... = a4 + a5 cf(x^8); [ 46, 1, 45, 1, 45, 1, 45, 1, 45 ... = a6 + a7 ----------------------------------------------------------------- ==================================================================== Maybe one can formulate a regular proof and systematize this to a greater extend. Gottfried Helms ==== I may append, that from the e^2/k -table also the e^3/-cfs can be derived, which seem patternless in their simple-c-f-representation. When negtive integers and rational *fractions* as coefficients of simple cfs allowed, seemingly all rational powers of e can be expressed completely schematically: for 2/k e for k= ... -3,-2,-1,{0},1,2,3,... ---------------------------------------------------------------------------- ------------------------------------------- delta: - 1 12 5 7 36 11 13 60 17 19 84 23 25 ---------------------------------------------------------------------------- ------------------------------------------- cf(e^(2/-1)); [1,-1, -6, -3, 1, 1, -4, -18, -6, 1, 1, -7, -30, -9, 1, 1, -10, -42, -12, 1, 1,-23 ] cf(e^(2/1)); [1, 0, 6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, 12 ] cf(e^(2/3)); [1, 1, 18, 7, 1, 1, 10, 54, 16, 1, 1, 19, 90, 25, 1, 1, 28, 126, 34, 1, 1, 37 ] cf(e^(2/5)); [1, 2, 30, 12, 1, 1, 17, 90, 27, 1, 1, 32, 150, 42, 1, 1, 47, 210, 57, 1, 1, 62 ] cf(e^(2/7)); [1, 3, 42, 17, 1, 1, 24, 126, 38, 1, 1, 45, 210, 59, 1, 1, 66, 294, 80, 1, 1, 87 ] cf(e^(2/9)); [1, 4, 54, 22, 1, 1, 31, 162, 49, 1, 1, 58, 270, 76, 1, 1, 85, 378, 103, 1, 1, 112 ] cf(e^(2/11)); [1, 5, 66, 27, 1, 1, 38, 198, 60, 1, 1, 71, 330, 93, 1, 1, 104, 462, 126, 1, 1, 95 ] ---------------------------------------------------------------------------- ------------------------------------------- delta: + 1 12 5 7 36 11 13 60 17 19 84 23 25 ---------------------------------------------------------------------------- ------------------------------------------- It seems possible to get other rational powers of e simply by interpolation. For instance, for e^3, which has a seemingly random pattern in terms of a simple continued fraction, a schematic patter can be found, if we allow negative as well as rational coefficients. First interpolate for the zero-cf, which diverges, if we try to approximate it. Let's use the symbol #+ for elementwise addition of the coefficients-lists, and analoguously #-,#*,#/ for the appropriate other operation, then cf(e^(2/0)) = [ cf(e^(2/-1)) #+ cf(e^(2/1)) ] #/ 2 Insert this into the table between the rows of cf(e^(2/-1)) and cf(e^(2/1)) ---------------------------------------------------------------------------- ------------------------------------------- ... cf(e^(2/-1)); [1,-1, -6, -3, 1, 1, -4, -18, -6, 1, 1, -7, -30, -9, 1, 1, -10, -42, -12, 1 ... cf(e^(2/0)): [1,-0.5, 0,-0.5, 1, 1,-0.5, 0,-0.5, 1, 1,-0.5, 0,-0.5, 1, 1,-0.5, 0,-0.5, 1 ... cf(e^(2/1)): [1, 0, 6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1 ... ... ---------------------------------------------------------------------------- ------------------------------------------- Also let us shorten the cf(...) notation to a c(..) notation c(-1) = cf(e^(2/-1)) c( 0) = cf(e^(2/ 0)) c( 1) = cf(e^(2/ 1)) ... and d = c(1) #- c(0) so that d = [0, 0.5, 6, 2.5, 0, 0, 3.5, 18, 5.5, 0, 0, 6.5, 30, 8.5, 0, 0, 0.5, 42, 11.5, 0 ... Then cf(e^3) can simply be interpolated as 2 cf(e^3) = c(2/3) = c(0) #+ d #* --- 3 Then cf(e^3) = cf(e^(2/(2/3)))= 1 7 11 19 23 31 35 43 =c(2/3) =[1, - -, 4, -, 1, 1, --, 12, --, 1, 1, --, 20, --, 1, 1, --, 28, --, 1] 6 6 6 6 6 6 6 6 ============================================================================ ========== And the continued-fraction for any power of e can be expressed along the schematic of that of e^2 cf(e^p) = [ a[5i+0], a[5i+1], a[5i+2], a[5i+3],a[5i+4], {for i=0..oo} ] where a[5i+0] = 1 a[5i+1] = -0.5 + ( 1 + 6*i)/p a[5i+2] = 0 + (12 +24*i)/p a[5i+3] = -0.5 + ( 5 + 6*i)/p a[5i+4] = 1 ============================================================================ ========= ----i=0------------- --------i=1----------- ------i=2------------ ... cf(e^2)=c(2/2): [1, 0, 6 , 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, ... cf(e^3)=c(2/3): [1,-1/6, 4 , 7/6, 1, 1,11/6, 12 , 19/6, 1, 1,23/6, 20,31/6, 1, ... cf(e^4)=c(2/4): [1,-2/8, 3 , 6/8, 1, 1,10/8, 9 , 18/8, 1, 1,22/8, 15,30/8, 1, ... cf(e^5)=c(2/5): [1,-3/10,24/10, 5/10, 1, 1, 9/10,72/10, 17/10, 1, 1,21/10, 12,29/10, 1, ... or even more concise: -------i=0------------- --------i=1--------- --------i=2-------- ... ... cf(e^2)=c(2/2): [ 4, 0, 24 , 8 4, 4, 12 , 72, 20, 1, 4, 24, 120, 32, 4, ... ] #/ 4 cf(e^3)=c(2/3): [ 6, -1, 24 , 7, 6, 6, 11 , 72, 19, 1, 6, 23, 120, 31, 6, ... ] #/ 6 cf(e^4)=c(2/4): [ 8, -2, 24 , 6, 8, 8, 10 , 72, 18, 1, 8, 22, 120, 30, 8, ... ] #/ 8 cf(e^5)=c(2/5): [10, -3, 24 , 5, 10, 10, 9 , 72, 17, 1, 10,21, 120, 29, 10, ... ] #/10 ... One can see the 5 interleaved arithmetic progressions a+d,a+2d, ... ; Eric Weisstein at mathworld.wolfram.com mentions some studies on such progressions in continued-fractions-coefficients. ============================================================================ =========== This ist just by trial&error; a proof would be good... Also I guess, that this interpolation-method could be applicable to the more simple representations of cf-s of e^1/p analoguously. (But I may leave to check that out to another reader's fun) Gottfried Helms (for computations the program Maxima (GNU) was used ) ==== Gottfried Helms schrieb: Also I guess, that this interpolation-method could be applicable > to the more simple representations of cf-s of e^1/p analoguously. > (But I may leave to check that out to another reader's fun) Well, it was just too simple. So here it is: with the #/-operator meaning to divide each element of the list by the divisor cf(e^1) = [1 0 1 1 2 1 1 4 1 1 6 1 1 8 1 1 10 1 ] #/1 cf(e^2) = [2 - 1 2 2 1 2 2 3 2 2 5 2 2 7 2 2 9 2 ] #/2 cf(e^3) = [3 - 2 3 3 0 3 3 2 3 3 4 3 3 6 3 3 8 3 ] #/3 cf(e^4) = [4, - 3, 4, 4, - 1, 4, 4, 1, 4, 4, 3, 4, 4, 5, 4, 4, 7, 4 ] #/4 cf(e^5) = [5, - 4, 5, 5, - 2, 5, 5, 0, 5, 5, 2, 5, 5, 4, 5, 5, 6, 5 ] #/5 ... Since these are not the common simple cf's, their convergence is not optimal; but that's ignored here for the sake of getting a simple pattern. Gottfried Helms ==== Well, a last addendum: more general relation: cf(e^x) = [x, 1-x, x, x, 3-x, x x, 5-x, x ... ] #/x cf(e^(1/x) = [1, x-1, 1 1, 3x-1, 1, 1, 5x-1, 1, ...] Putting that together, I now assume that cf(e^x) = [1, 1/x-1, 1, 1,3/x-1, 1 1,5/x-1, 1 ... ] for all rational x except x=0 and I think, this is worth to be derived algebraically and to be proven, if this is not done already. What is especially interesting, is that a multiplication over all coefficents of a continued fraction is an operation, which normally modifies the final value in a very complicated way. That with the euler-constant e such an operation comes out with such a regularity is especially astonishing to me. Anyway- now as it is compressed so much, I think, I've seen this anywhere? Does anyone know of any online-reference for that relation? Gottfried Helms ==== I'd like to get some opinions about this problem, please. The dependable capacity C available in an electric system in a month of the future, say Dec 2008, is a random variable with density function f defined on [0, Cmax]. We can admit f is continuous on this interval. On the cited month you have to meet a known and deterministic demand r<=Cmax. Define a random variable deficit, D, by D= r-C if C=r. Supposing the density function f is independent of r, the expected value of D corresponding to a demand r is given by E(r) = Integral [0,r] (r-c) f(c) dc = r*Integral [0,r] f(c) dc - Integral [0,r] c f(c) dc. Since f is continuous, E(r) exists for r<=Cmax and E is differentiable wrt r. Applying the F.T. Of Int. Calculus, we get E'(r) = r* f(r) + Integral [0,r] f(c) dc - r*f(r) = Integral [0,r] f(c) dc = Probability (C<=r) = Probability (D>=0). Though I don't have f in a closed form, I can estimate Probability (D>=0) by means of simulation models, using a process similar to Monte Carlo's. Then, for variations on r of about 5%, I can make the estimate Delta E(r) = (Delta r) * Probability (D>=0). All I need to know is that f exists and is continuous on [0, Cmax]. I don't need to know how exactly f sends c into f(c). In some situations this is reasonable, but there are cases when it's not admittable at all to suppose f is independent of r. In such cases, for each r there's a particular density function f_r. You can still suppose r is known, but it affects the distribuition of C on [0, Cmax] (Cmax is always known and independent of r]. Then I think I have something like E(r) = Integral [0,r] (r-c) g(r,c) dc where g is defined on R^2 with values on R and, for a given r, g(r,c) = f_r(c). Supposing g is continuous, is that OK if I use Leibiniz formula to compute E'(r)? Anyway, if f_r depends on r than that beautiful conclusion I came to before is no longer true, right? I'm a bit confused here. Artur ==== >The dependable capacity C available in an electric system in a month >of the future, say Dec 2008, is a random variable with density >function f defined on [0, Cmax]. We can admit f is continuous on this >interval. On the cited month you have to meet a known and >deterministic demand r<=Cmax. Define a random variable deficit, D, by >D= r-C if C=r. Supposing the density function f is >independent of r, the expected value of D corresponding to a demand r >is given by E(r) = Integral [0,r] (r-c) f(c) dc = r*Integral [0,r] >f(c) dc - Integral [0,r] c f(c) dc. Since f is continuous, E(r) >exists for r<=Cmax and E is differentiable wrt r. Applying the F.T. Of >Int. Calculus, we get E'(r) = r* f(r) + Integral [0,r] f(c) dc - >r*f(r) = Integral [0,r] f(c) dc = Probability (C<=r) = Probability >(D>=0). Though I don't have f in a closed form, I can estimate >Probability (D>=0) by means of simulation models, using a process >similar to Monte Carlo's. Then, for variations on r of about 5%, I can >make the estimate Delta E(r) = (Delta r) * Probability (D>=0). All I >need to know is that f exists and is continuous on [0, Cmax]. I don't >need to know how exactly f sends c into f(c). >In some situations this is reasonable, but there are cases when it's >not admittable at all to suppose f is independent of r. In such cases, >for each r there's a particular density function f_r. You can still >suppose r is known, but it affects the distribuition of C on [0, Cmax] >(Cmax is always known and independent of r]. Then I think I have >something like E(r) = Integral [0,r] (r-c) g(r,c) dc where g is >defined on R^2 with values on R and, for a given r, g(r,c) = f_r(c). >Supposing g is continuous, is that OK if I use Leibiniz formula to >compute E'(r)? Anyway, if f_r depends on r than that beautiful >conclusion I came to before is no longer true, right? > Your formulae seem correct. As long as the partial derivative dg/dr is continuous in some neigborhood [r0 - e, r0 + e] x [0, r0], you are allowed to use Leibniz' rule to evaluate the derivative E'(r0). You are correct that the derivative is not as pretty as in your first case. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== I've started a thread to go over some statements by Dik Winter which I say are crank statements. I'm also going to outline some crank behavior by that person. I don't mind others posting in the thread or in this one, and I may reply to people other than Dik Winter, but I want you to know where the focus is. Some of you may know that I have independent verification of the argument that he attacks, but I've been puzzled both by his persistence in making his claims against those argument, and in the acceptance of his claims by the sci.math newsgroup. So I'm doing an experiment. My guess is that despite hearing that there's independent verification of the argument Winter attacks, and despite the wackiness of his position the sci.math newsgroup will STILL either show support for Winter or fail to correct him. That's the hypothesis that I'm currently testing. James Harris ==== Some of you may know that I have independent verification of the > argument that he attacks... > If you don't want to divulge this independent source, that's fine. But at least post this person's verification. As you say, let the math speak for itself. l8r, Mike N. Christoff ==== So I'm doing an experiment. My guess is that despite hearing that > there's independent verification of the argument Winter attacks, and > despite the wackiness of his position the sci.math newsgroup will > STILL either show support for Winter or fail to correct him. Mr. Harris, great strategy! You have asked Arturo to no longer comment on your flawed math arguments. Dik Winter has also repeatedly showed you just how flawed your nonsensical dribble is and now you want him gone too? You have berated others too like Nora who has also shown the uncorrectable flaws in your so-called arguments. Do you believe that shaming these folks to not replying will make your arguments any more acceptable? You should have chosen comedy as your field of study! What a joke you are. To me, you have proven that without question! ==== >Mr. Harris, great strategy! You have asked Arturo to no longer comment on your flawed math arguments. I would suggest that asked is not the correct verb to use to describe what transpired. http://groups.google.com/groups?selm=atkubk%24bml%241%40agate.berkeley.edu http://groups.google.com/groups?selm=atrhku%242cpk%241%40agate.berkeley.edu (towards the end in both) http://groups.google.com/groups?selm=bo8tqb%241p9o%241%40agate.berkeley.edu http://groups.google.com/groups?selm=3c65f87.0311041814.5828faa8%40posting.g oogle.com -- ====================================================================== 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 ==== > Some of you may know that I have independent verification of the > argument that he attacks How are we supposed to have gotten this knowledge. This is the first mention of the existence of independent verification made on the newsgroup. I for one do not believe that independent verification exists. I suspect that what we have her is a definition problem. independent verification The rest of the world: A disinterested third party has verified this. James Harris: I thought up yet another argument. Question: How many legs does a horse have if James Harris calls a tail a leg. Answer: Four. The fact that James Harris calls a tail a leg does not make it one. - William Hughes ==== > So I'm doing an experiment. My guess is that despite hearing that > there's independent verification of the argument Winter attacks, and > despite the wackiness of his position the sci.math newsgroup will > STILL either show support for Winter or fail to correct him. But we *haven't* heard there's independent verification, have we? At least, we have no means of confirming or refuting that someone somewhere has independently verified your claim (assuming that someone here is certain what, precisely, the claim is). You've said that there is, but you have given no means to confirm or deny this claim, so that hardly counts as any particular evidence that there's been independent verification. -- Jesse F. Hughes I think the burden is on those people who think he didn't have weapons of mass destruction to tell the world where they are. -- White House spokesman Ari Fleischer ==== >I've started a thread to go over some statements by Dik Winter which I >say are crank statements. I'm also going to outline some crank >behavior by that person. I don't mind others posting in the thread or in this one, and I may >reply to people other than Dik Winter, but I want you to know where >the focus is. Some of you may know that I have independent verification of the >argument that he attacks, but I've been puzzled both by his >persistence in making his claims against those argument, and in the >acceptance of his claims by the sci.math newsgroup. So I'm doing an experiment. My guess is that despite hearing that >there's independent verification of the argument Winter attacks, and >despite the wackiness of his position the sci.math newsgroup will >STILL either show support for Winter or fail to correct him. That's the hypothesis that I'm currently testing. Um, we've only heard about this independent verification from _you_. GIven your record, nobody is going to believe anything you say just because you say it. Sorry, that's how that works. Exactly who did this independent verification? >James Harris ************************ David C. Ullrich ==== [snip] > Some of you may know that I have independent verification of the > argument that he attacks, Well, *I* haven't seen it posted in this newsgroup, and your word is no word. Where is the independent verification of the argument? Who has verified it? Your assertions are justifiably suspect, since you rarely post a single paragraph which does not contain distortions or lies. > but I've been puzzled both by his > persistence in making his claims against those argument, and in the > acceptance of his claims by the sci.math newsgroup. argument is faulty. > So I'm doing an experiment. This is 'sci.math', not 'alt.test.your.crackpot.theories'. > My guess is that despite hearing that > there's independent verification of the argument Winter attacks, and > despite the wackiness of his position the sci.math newsgroup will > STILL either show support for Winter or fail to correct him. Your assurance that independent verification exists is no assurance at all. You repeatedly misrepresent or distort everything you post. Your credibility is zero. Why? Because you have consistently defended your errors with the same vigor and passion that you exhibited when you later defended their corrections. You have no demonstrated ability to distinguish truth from error. > That's the hypothesis that I'm currently testing. The result is just in: YOU ARE A CRANK! Wacky, isn't it? But, hey, it's just basic math. Yup, yup, yup. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== [snip] Some of you may know that I have independent verification of the > argument that he attacks, Well, *I* haven't seen it posted in this newsgroup, and your word is no > word. Where is the independent verification of the argument? Who has > verified it? Your assertions are justifiably suspect, since you rarely > post a single paragraph which does not contain distortions or lies. I can tell you what he did: He printed all his arguments and put the pile of paper on one plate. Then he took a sausage and put it on the other plate. Then he told his dog: On one plate there is a printout of a proof of FLT. On the other plate there is a sausage. Go and eat the sausage. And the dog ate the sausage, which clearly demonstrates that on the plate that it didn't touch there was no sausage, but a proof of FLT. ==== > I can tell you what he did: He printed all his arguments and put the > pile of paper on one plate. Then he took a sausage and put it on the > other plate. Then he told his dog: On one plate there is a printout of > a proof of FLT. On the other plate there is a sausage. Go and eat the > sausage. And the dog ate the sausage, which clearly demonstrates that > on the plate that it didn't touch there was no sausage, but a proof of > FLT. I think the dog then crapped on the plate because what was on it was way too stinky for him to even touch! ==== > I've started a thread to go over some statements by Dik Winter which I > say are crank statements. I'm also going to outline some crank > behavior by that person. I don't mind others posting in the thread or in this one, and I may > reply to people other than Dik Winter, but I want you to know where > the focus is. Some of you may know that I have independent verification of the > argument that he attacks, but I've been puzzled both by his > persistence in making his claims against those argument, and in the > acceptance of his claims by the sci.math newsgroup. > By whom? > So I'm doing an experiment. My guess is that despite hearing that > there's independent verification of the argument Winter attacks, and > despite the wackiness of his position the sci.math newsgroup will > STILL either show support for Winter or fail to correct him. > Who has independently verified your proof? Where is this verification? > That's the hypothesis that I'm currently testing. OK, that's H_0, what's H_1? > James Harris ==== >>Are you saying we have no balls left at noon *because* pointwise >convergence holds, or are you saying that pointwise convergence holds >*because* it describes the fact (independently verifiable via ZF) that >no balls are left at noon? >>If the former, then you have not explained *why* pointwise convergence >holds. >|Definition: >>|Suppose {f_n} is a sequence of functions with domain D and that f is a >>|function also with domain D. The sequence {f_n} converges pointwise to >>|f iff for each x in D, f_n(x) converges to f(x). > Let D be the set of buckets, and f_n(x) be 1 or 0 depending on whether >> that bucket has a ball in it or not at time 12 - 2^{1-n}. > Since f_n(x) can take only two values, 1 or 0, the discrete topology is >> about the only topology we can give to the range of f_n. To converge in >> the discrete topology, all terms must be constant after some point. I do not *assume* that convergence of any type holds. I assume that >may *deduce* that pointwise convergence holds, but I do not need that >fact to solve the problem. I need only set theory. If Newton's First Law implies pointwise convergence and you assume that, then you assume even more than pointwise convergence, but that means you do assume pointwise convergence since it is implied by Newton's First Law. >>|Definition: >>|Suppose {a_n} is a sequence in a space with the discrete topology and a >>|is another point in that same space. The sequence {a_n} converges to a >>|iff there is an N so that a_n = a for all n > N. > This is why I called discrete convergence monotonous. Perhaps, due to >> the constancy of discrete convergence, it may not appear that there is >> any convergence at all, the terms are all just the same. You keep thinking that all you need to do is explain convergence more >fully and completely, and all will become clear. I know what convergence >means. I was not really sure why our views differed, but I think that your post has finally enlightened me. Hopefully I now understand where our approaches to this problem differ. I apologize if I insulted you by explaining these two types of convergence; I was merely trying to make my point clearer. >> Translating the definition of pointwise convergence to the language of >> buckets and balls and using discrete convergence at each point, we get >|A bucket is empty at noon iff it has always been empty since some time >>|before noon. A bucket has a ball in it at noon iff it has always had a >>|ball in it since some time before noon. And this follows from Newton's first law, does it not? Convergence is >merely the language you have chosen to describe what we observe; it is >not in any way a *cause* of what we observe. Newton's First Law merely describes what we observe; it is not in any way a cause of what we observe either. >> This last quoted statement is something I believe you take for granted, >> but others take it is an assumption. Whatever, it is just a restatement >> of pointwise convergence. This is why I have said that you are assuming >> pointwise convergence. Perhaps I should say you are taking pointwise >> convergence for granted. I am taking Newton's first law and ZF for granted. Pointwise convergence >is a *consequence* of those assumptions, not an assumption in its own >right. As I said, if Newton's First Law implies pointwise convergence and you assume Newton's First Law, then you are, in fact, assuming pointwise convergence. Under that assumption, as I have said before, I agree with you completely. >> Your assumption is that if a bucket is always empty after a certain time >> before noon, then that bucket is empty at noon. This is the assumption >> of pointwise convergence. >>Then you must think the assumption of pointwise convergence is equally >necessary for solving the five apples, take away two problem. > No, because there is no infinite sequence in the five apples, take away >> two problem. Convergence deals with limits of infinite sequences. You have five apples at 11:55. I take away two at 11:59. How many do >you have at noon? If f(1) = 5 and f(2) = f(1) - 2, what is f(3)? >We need to assume that Newton's first law applies, just as we do with the >buckets and balls. In math problems, especially ones that are physically unrealizable, we do not necessarily need to assume physical laws such as Newton's First Law. In order to explain the seemingly contradicatory answers that have been given for this problem, the way I have been looking at this problem has no connection with physical laws; I have been looking at it as a sequence of sets with a particular definition, which is essentially a_n = {k in Z : n+1 <= k <= 10n}. The problem then is to find the limit as n tends to infinity of a_n. Different topologies give us different limits and therefore different answers. However, if you feel we do need to assume Newton's First Law, then that is where our approaches differ. Under this added constraint, I agree with you that the limit is {}. Truce? Rob Johnson take out the trash before replying ==== >>I do not *assume* that convergence of any type holds. I assume that >>may *deduce* that pointwise convergence holds, but I do not need that >>fact to solve the problem. I need only set theory. > If Newton's First Law implies pointwise convergence and you assume that, > then you assume even more than pointwise convergence, but that means you > do assume pointwise convergence since it is implied by Newton's First > Law. If we assume a few basic axioms, are we assuming that FLT holds, merely because FLT happens to be a consequence of those axioms? > I was not really sure why our views differed, but I think that your > post has finally enlightened me. Hopefully I now understand where our > approaches to this problem differ. I apologize if I insulted you by > explaining these two types of convergence; I was merely trying to make > my point clearer. No offense taken. I was merely pointing out that the definition of convergence is not where our views differ. > Translating the definition of pointwise convergence to the language of > buckets and balls and using discrete convergence at each point, we get >|A bucket is empty at noon iff it has always been empty since some time >|before noon. A bucket has a ball in it at noon iff it has always had a >|ball in it since some time before noon. >>And this follows from Newton's first law, does it not? Convergence is >>merely the language you have chosen to describe what we observe; it is >>not in any way a *cause* of what we observe. > Newton's First Law merely describes what we observe; it is not in any > way a cause of what we observe either. True. But in solving word problems it is an unstated assumption that ordinary physical laws apply insofar as that is possible. > This last quoted statement is something I believe you take for granted, > but others take it is an assumption. Whatever, it is just a restatement > of pointwise convergence. This is why I have said that you are assuming > pointwise convergence. Perhaps I should say you are taking pointwise > convergence for granted. >>I am taking Newton's first law and ZF for granted. Pointwise convergence >>is a *consequence* of those assumptions, not an assumption in its own >>right. > As I said, if Newton's First Law implies pointwise convergence and you > assume Newton's First Law, then you are, in fact, assuming pointwise > convergence. Under that assumption, as I have said before, I agree > with you completely. I think the important point here is that assuming Newton's First Law implies pointwise convergence to the exclusion of other kinds of convergence, at least in the context of this problem. Assuming Newton's First Law allows us to deduce a unique answer. If I say outright that I am assuming pointwise convergence, then you have every right to propose other forms of convergence as equally valid approaches to the problem. >>> Your assumption is that if a bucket is always empty after a certain time >>> before noon, then that bucket is empty at noon. This is the assumption >>> of pointwise convergence. >>Then you must think the assumption of pointwise convergence is equally >>necessary for solving the five apples, take away two problem. > No, because there is no infinite sequence in the five apples, take away > two problem. Convergence deals with limits of infinite sequences. >>You have five apples at 11:55. I take away two at 11:59. How many do >>you have at noon? > If f(1) = 5 and f(2) = f(1) - 2, what is f(3)? Another common assumption regarding word problems is that there is no hidden information that would affect the answer. We are not told of any other movements of apples, and therefore none exist. > In math problems, especially ones that are physically unrealizable, we > do not necessarily need to assume physical laws such as Newton's First > Law. In order to explain the seemingly contradicatory answers that have > been given for this problem, the way I have been looking at this problem > has no connection with physical laws; I have been looking at it as a > sequence of sets with a particular definition, which is essentially > a_n = {k in Z : n+1 <= k <= 10n}. The problem then is to find the limit > as n tends to infinity of a_n. Different topologies give us different > limits and therefore different answers. I have never denied that these tools exist and that, if applied, they do indeed produce a variety of answers. I count that as an argument against assuming any particular kind of convergence without some justification. > However, if you feel we do need to assume Newton's First Law, then that > is where our approaches differ. Under this added constraint, I agree > with you that the limit is {}. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >Problem 4. Labeled balls (as in #2 and #3). At step n, remove both >balls labeled n and 10n. Switch the labels these balls, then return >the ball newly labeled 10n back in the basket. Discard the ball newly >labeled n. I think JH will insist there are no balls left in the basket >at noon. But how does this differ from Problem 3? Since this version has the same ball being added and removed an infinite > number of times, I don't see how to decide where that particular ball > might be when we finish. Rob Johnson would probably say the function > fails to converge. That's an excellent point, Dave. I lost sight of the fact that we > must look at the actual set element being reinserted, regardless of > label. Labels are helpful as a convenience for most of the other > problems given, but in this case following the labels does not > follow the balls since each ball can have a different label at > different times. Although I had first claimed that in Problem #4 the container is > empty, given Dave's argument, I would now say that it is more like > Problem #1 and indeterminant. Say we change the problem slightly. Instead of removing the two balls and then replacing one, we switch the labels while both balls are still in the basket, then remove only the ball with the smaller label. Now we don't suffer any more from the problem Dave identified. Now do you agree that the question how many balls are in the basket is identical to problem #3 and the question how many labels are in the basket is identical to problem #2? If so, the only conclusion is that there are an infinite number of balls remaining in the basket, each without a label. ==== >>>Problem 4. Labeled balls (as in #2 and #3). At step n, remove both >>>balls labeled n and 10n. Switch the labels these balls, then return >>>the ball newly labeled 10n back in the basket. Discard the ball newly >>>labeled n. I think JH will insist there are no balls left in the basket >>>at noon. But how does this differ from Problem 3? >>> >>>Since this version has the same ball being added and removed an infinite >>number of times, I don't see how to decide where that particular ball >>might be when we finish. Rob Johnson would probably say the function >>fails to converge. >> That's an excellent point, Dave. I lost sight of the fact that we >must look at the actual set element being reinserted, regardless of >label. Labels are helpful as a convenience for most of the other >problems given, but in this case following the labels does not >follow the balls since each ball can have a different label at >different times. Although I had first claimed that in Problem #4 the container is >empty, given Dave's argument, I would now say that it is more like >Problem #1 and indeterminant. > > I do not understand the basis for this claim. Your objection to Problem 1 was that it does not specify which ball we remove. In Problem 4, we specify precisely which ball we remove. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > I agree that there are no labels, but I don't agree that there are no > balls. It seems clear to me that any ball with an initial label not > divisible by 10 will never be removed. So at noon, there will still > be an infinite number of (unlabeled) balls remaining. It's not > contradictory that the balls are labeled before noon but not after > noon, just as in the original problem it's not contradictory that the > bucket is non-empty before noon but not after noon. Since by the rules of the game, no ball is reentered without a label. > I think that no labels = no balls (please, no jokes on this!) is > therefore valid. Unless you can show me a natural number n such that > 10n is not a natural number, I think the labeless ball hypothesis > cannot happen. Here are two problems: Problem 1: You have a basket and a bunch of balls. At time 12-2^-n, you put ball n in the basket and remove ball n-1. How many balls are in the basket at 12? Problem 2: You have a ball and bunch of labels. At time 12-2^n, you put label n on the ball and remove label n-1. How many labels are on the ball at 12? I don't see how the answers could possibly differ, yet you seem to claim that the answer to problem 1 is 0 and the answer to problem 2 is 1. > You have a bucket, initially with no coins in it. For n>0, at time > 1-2^(-n), you put a coin in the bucket (say coin n), flip all the > coins in the bucket, and remove all the coins that show heads. What's > the probability distribution for the number of coins in the bucket at > time 1? It feels like the bucket should be empty with probability 1, but I > haven't been able to put together a convincing argument. Hmmm...interesting problem. For any coin n, the probability that the > coin flips tails is 1/2, and thus the probability that it will never > go to heads is lim n->oo (1/2)^n = 0. As this is true for each coin, > I would think that the ending container is empty happen with > probability 1. I agree that for any given coin, it will be in the ending container with probability 0. But when you add together an infinite number of probability 0's, you don't necessarily get another probability 0 back. Here's an example: Choose a random real number x beetween 0 and 1. What's the probability that x is a real number between 0 and 1? For any specific real number y, the probability that x = y is 0. But the probability that it's ANY real number between 0 and 1 is 1. - Nate ==== >> You have a bucket, initially with no coins in it. For n>0, at time >> 1-2^(-n), you put a coin in the bucket (say coin n), flip all the >> coins in the bucket, and remove all the coins that show heads. What's >> the probability distribution for the number of coins in the bucket at >> time 1? >> It feels like the bucket should be empty with probability 1, but I >> haven't been able to put together a convincing argument. >> Hmmm...interesting problem. For any coin n, the probability that the >> coin flips tails is 1/2, and thus the probability that it will never >> go to heads is lim n->oo (1/2)^n = 0. As this is true for each coin, >> I would think that the ending container is empty happen with >> probability 1. > I agree that for any given coin, it will be in the ending container > with probability 0. But when you add together an infinite number of > probability 0's, you don't necessarily get another probability 0 back. > Here's an example: Choose a random real number x beetween 0 and 1. > What's the probability that x is a real number between 0 and 1? For > any specific real number y, the probability that x = y is 0. But the > probability that it's ANY real number between 0 and 1 is 1. Probability is countably additive. When you add together a countable number of 0's, as in the coin problem, the result is 0. The real numbers in [0,1] are uncountable, and therefore countable additivity does not apply. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >> You have a bucket, initially with no coins in it. For n>0, at time >> 1-2^(-n), you put a coin in the bucket (say coin n), flip all the >> coins in the bucket, and remove all the coins that show heads. What's >> the probability distribution for the number of coins in the bucket at >> time 1? > >> It feels like the bucket should be empty with probability 1, but I >> haven't been able to put together a convincing argument. > >> Hmmm...interesting problem. For any coin n, the probability that the >> coin flips tails is 1/2, and thus the probability that it will never >> go to heads is lim n->oo (1/2)^n = 0. As this is true for each coin, >> I would think that the ending container is empty happen with >> probability 1. > > I agree that for any given coin, it will be in the ending container > with probability 0. But when you add together an infinite number of > probability 0's, you don't necessarily get another probability 0 back. > Here's an example: Choose a random real number x beetween 0 and 1. > What's the probability that x is a real number between 0 and 1? For > any specific real number y, the probability that x = y is 0. But the > probability that it's ANY real number between 0 and 1 is 1. Probability is countably additive. When you add together a countable > number of 0's, as in the coin problem, the result is 0. The real numbers > in [0,1] are uncountable, and therefore countable additivity does not > apply. On second thought, we're not adding an infinite number of probability 0's at all -- we're multiplying an infinite number of probability 1's. Any idea if probability is countable multiplicative as well? ==== >> Probability is countably additive. When you add together a countable >> number of 0's, as in the coin problem, the result is 0. The real numbers >> in [0,1] are uncountable, and therefore countable additivity does not >> apply. > On second thought, we're not adding an infinite number of probability > 0's at all -- we're multiplying an infinite number of probability 1's. > Any idea if probability is countable multiplicative as well? An infinite product can be converted to an infinite sum by taking logarithms. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > >> Probability is countably additive. When you add together a countable >> number of 0's, as in the coin problem, the result is 0. The real numbers >> in [0,1] are uncountable, and therefore countable additivity does not >> apply. > > On second thought, we're not adding an infinite number of probability > 0's at all -- we're multiplying an infinite number of probability 1's. > Any idea if probability is countable multiplicative as well? An infinite product can be converted to an infinite sum by taking logarithms. Yes, but then you're not dealing with a countable sum of probabilities any more -- you're dealing with a countable sum of log-probabilities. Are log-probabilities countably additive? ==== > Probability is countably additive. When you add together a countable > number of 0's, as in the coin problem, the result is 0. The real numbers > in [0,1] are uncountable, and therefore countable additivity does not > apply. >> On second thought, we're not adding an infinite number of probability >> 0's at all -- we're multiplying an infinite number of probability 1's. >> Any idea if probability is countable multiplicative as well? >> An infinite product can be converted to an infinite sum by taking logarithms. > Yes, but then you're not dealing with a countable sum of probabilities > any more -- you're dealing with a countable sum of log-probabilities. > Are log-probabilities countably additive? For each n, we can compute the probability p_n that a given ball is still in the bucket after the n-th step. Since p_n -> 0 as n -> oo, we can conclude that the probability of that particular ball being in the bucket at noon is 0. After that, we use countable additivity over all the balls to show that the bucket is empty at noon with probability 1. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > > Probability is countably additive. When you add together a countable > number of 0's, as in the coin problem, the result is 0. The real numbers > in [0,1] are uncountable, and therefore countable additivity does not > apply. > >> On second thought, we're not adding an infinite number of probability >> 0's at all -- we're multiplying an infinite number of probability 1's. >> Any idea if probability is countable multiplicative as well? > >> An infinite product can be converted to an infinite sum by taking logarithms. > > Yes, but then you're not dealing with a countable sum of probabilities > any more -- you're dealing with a countable sum of log-probabilities. > Are log-probabilities countably additive? > For each n, we can compute the probability p_n that a given ball is still > in the bucket after the n-th step. Since p_n -> 0 as n -> oo, we can > conclude that the probability of that particular ball being in the bucket > at noon is 0. Right. > After that, we use countable additivity over all the balls to show that the > bucket is empty at noon with probability 1. Here I don't follow. If p_k is the probability that ball number k is in the basket at noon, the probability that the basket is empty at noon is prod(1-p_k). I don't immediately see how the fact that probabilities are countably additive helps evaluate that product. ==== >I think that 2^n = sum (nCk) (k ranges from 0 to n) (by nCk I mean n!/(k!(n-k)!)) but can't prove it. does anyone have any hints? also, is there any standard way to represent sigma notation in ascii? >the above expression is a little vague as I've written it. nCk, or as I usually write it, C(n,k), is usually called a combination. In any case, remember the recursive formula for Pascal's Triangle: C(n,k) = C(n-1,k) + C(n-1,k-1). Then try to use induction. Rob Johnson take out the trash before replying ==== Did anyone see Amir Aczel on C-SPAN2? He was talking about his book Pendulum: Leon Foucault and the Triumph of Science. I'm curious to know if anyone else watched it. Does anyone out there know him? James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ ==== > Did anyone see Amir Aczel on C-SPAN2? He was talking about his book > Pendulum: Leon Foucault and the Triumph of Science. I'm curious to know if anyone else watched it. Does anyone out there know him? I attended a talk Amir Aczel gave at the B.U. Bookstore last October. It was on his new book about Foucault. It was a good talk and it sounds like a good book. After the bookstore address, I happended to ride back on the MBTA Greenline Train with him and we chatted some more. I am waiting for the paperback edition. Bob Kolker ==== I have a monic polynomial with integer coefficients and I know that its root with greatest modulus is k and I also know that the rest of the roots lie in the interval [-a,b], where k>a>0, k>b>0. What can be said about the coefficients of the polynomial under these circumstances? Felix. as the first combinatorist. See www.nytimes.com, National News, or the url John Robertson ==== [snip] Tom Van Flandern - Washington, DC - see our web site on replacement > astronomy research at http://metaresearch.org Tom, see http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_2.0.htm and its surrounding material. Androcles. ==== The probability that a number n is divisible by a prime p is 1/p for n >> p. For n < p, the probability is zero. Is there a function that Russell ==== > The probability that a number n is divisible by a prime p is 1/p for n >> p. For n < p, the probability is zero. Is there a function that > Russell Never mind, I just realized how stupid the question was, head not screwed on right... /R ==== > Students are supposed to do homework assignments themselves. Polite question, harsh answer. Fortunately many contributors in this forum are not as rude. Students asking about their homework are welcome, IMO. ==== Does anyone know a simple proof of the fact that (at least over a > field) the rank of the symmetric matrix A^T A is the same as the rank > of A? It is easy to see that the null space of A^T A contains the null space > of A, so it is enough to see the reverse inclusion (which I haven't > been able to do). > If A^t A x = 0, then 0 = = = ||Ax||^2, therefore > Ax = 0. --Ron Bruck Can this be extended to A^T B A, where B is symmetric square and p.d., by taking the square root of B? ==== Actually, I guess that 2-by-2 example only works if p = 1 (mod 4), since to get a nontrivial solution to b^2+d^2=0 (mod p), one needs -1 to be a quadratic residue mod p. But for any prime p, there are 3-by-3 matrices A with rank(A) = 1 and rank(A^T A) = 0, since the congruence a^2+b^2+c^2=0 (mod p) has non-zero solutions for all primes p. This raises the (probably not too hard) question: If p = 3 (mod 4) and if A is a 2-by-2 matrix, is it true that A and A^T A have the same rank? Many other similar questions come to mind. For example: For which primes p (if any) do there exist 3-by-3 matrices A in characteristic p with the property that rank(A)=2 and rank(A^T A)=1? ==== >to be a quadratic residue mod p. But for any prime p, there are 3-by-3 >matrices A with rank(A) = 1 and rank(A^T A) = 0, since the congruence >a^2+b^2+c^2=0 (mod p) has non-zero solutions for all primes p. This raises the (probably not too hard) question: If p = 3 (mod 4) and if A is a 2-by-2 matrix, is it true that A and >A^T A have the same rank? Yes. It's pretty easy to see because the only counterexample would have to be rank 1, which has the general form A = [a,b]^T [c,d], and the resulting a^2+b^2 scalar factors out since p is 3 (mod 4). >Many other similar questions come to mind. For example: For which primes p (if any) do there exist 3-by-3 matrices A in >characteristic p with the property that rank(A)=2 and rank(A^T A)=1? All primes p. For p=2, we use A = [ [1 1 1] [1 1 0] [0 0 0] ]. For any odd p, let (a,b,c) be an all non-zero solution to a^2+b^2+c^2 = 0. Take A = [ [a b c] [b -a 0] [0 0 0] ]. A^T A consists of all zeroes except for one entry, a^2+b^2 = -c^2 which is necessarily non-zero. -- Erick ==== > Many people are fond of fields which contain a square root of > -1, call it i. (You might also, of course, call it 1 in case > the characteristic is 2.) Over such a field, the matrix > A = [1, i; 0, 0] may make you unhappy. That answers the question then. It's always harder to prove something > that isn't true. (Actually in this case it is A A^T that is zero, but [1, 0;i, 0] works > for the original question.) > -Chad When over C you can take A* instead of A^T to make it true. Basically, the adjoint operator w.r.t. standard inner product should do the job. Felix. ==== Can someone quote or point me to results about the inability to define reasonable non-trivial measures on big spaces? I vaugely recall a result due to Banach and Kuratowski in which reasonable was requiring that mu({x}) = 0, the sigma algebra was P(X) (where X was the space), and there were some conditions on X, I'm interested in stuff like that. Also, given a sequence of distribution functions F_n (that is, lim (t -> -oo) F_n(t) = 0, lim (t -> oo) F_n(t) = 1, F_n are monotone increasing and right continuous) am I guaranteed to have a sequence of -independent- random variables X_n on some probability space (Omega, F, P) such that F_(X_n)(x) = P(X_n <= x) = F_n(x)? Pavel ==== [...] Also, given a sequence of distribution functions F_n (that is, lim >(t -> -oo) F_n(t) = 0, lim (t -> oo) F_n(t) = 1, F_n are monotone >increasing and right continuous) am I guaranteed to have a sequence of >-independent- random variables X_n on some probability space (Omega, >F, P) such that F_(X_n)(x) = P(X_n <= x) = F_n(x)? Yes. You can take Omega to be a countable product of real lines and P to be a product measure. > Pavel ************************ David C. Ullrich ==== Happy holidays to one and all. The only 2 days of the year worth celebrating are Plutonium days where 7 November 1990 was the birth of the AtomTotality theory of 231Pu is God. For God is science and Science is God and the best bible ever written is the best physics textbook. And 14 December 1940 was the birth of the discovery of the atoms of plutonium. Unlike religions with their celebrations and holidays each year based upon some flimsy inexact date such as the birth of a Roman terrorist known as Jesus, the dates of 7 November and 14 December are precise and accurate dates. Jesus was not even born in the year 0 A.D. and the date of Christmas is pinned to the Winter solstice and not some activity of Jesus. To celebrate Plutonium day this year I am going to try to do a tradition for every year and get some photograph pictures of my years work around the home of the trees and landscaping. Analog pictures and today is a nice sunny day for doing that. The word hallelujah was a nonsense word created by the AtomTotality some 2 or 3 thousand years ago and was used in a beautiful song of the Handel's Messiah. Saved for the purpose that by 1990s this word would be replaced by the word plutonium for which god had saved since its syllables match. Another melody and song that needs to be refitted with plutonium is the Lennon song of Have a merry christmas and a happy new year,,... war is over,.... And another song that needs a refitting with plutonium is one sung by Pavorrati and some other opera singer which seems difficult to sing and have tried to locate the name of this song but unable to. I have heard it on PBS sung by some soloist a week or two ago and seems to be a crowning glory for a singer to be able to sing it-- some soloist in New Zealand of Q-2 sung it. So when I find out the title of it and the lyrics will go and replace. Sing Handel's Messiah of the Hallelujah chorus with these lyrics: A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-OMPLUTONIUM! A-T-OMPLUTONIUM! A-T-O-M-P-L-UTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-OMPLUTONIUM! A-T-OMPLUTONIUM! A-T-O-M-P-L-UTONIUM! DOT OF THE ELECTRON CLOUD, THE 5F6 A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! DOT OF THE ELECTRON CLOUD, THE 5F6 A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! DOT OF THE ELECTRON CLOUD, THE 5F6 DOT OF THE ELECTRON CLOUD, THE 5F6 A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! DOT OF THE ELECTRON CLOUD, THE 5F6 DOT OF THE ELECTRON CLOUD, THE 5F6 DOT OF THE ELECTRON CLOUD, THE 5F6 A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! ------------------------- ATOM HAS INFINITE POTENTIAL ATOM HAS INFINITE POTENTIAL ATOM HAS INFINITE POSSIBILITIES 231Pu 231Pu AND IT SHALL NUCLEOSYNTHESIZE FOREVER AND EVER AND IT SHALL NUCLEOSYNTHESIZE FOREVER AND EVER AND IT SHALL NUCLEOSYNTHESIZE FOREVER AND EVER AND IT SHALL NUCLEOSYNTHESIZE FOREVER AND EVER ATOM OF ATOMS FOREVER, AND EVER A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! And Lord of Light FOREVER, AND EVER A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! ATOM OF ATOMS FOREVER, AND EVER A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! And Lord of Life FOREVER, AND EVER A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! ATOM OF ATOMS FOREVER, AND EVER A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! And Lord of Light FOREVER, AND EVER A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! ATOM OF ATOMS FOREVER, AND EVER A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! And Lord of Life FOREVER, AND EVER A-T-O-M-PLUTONIUM! A-T-O-M-PLUTONIUM! ATOM OF ATOMS and Lord of Life FOREVER, AND EVER AND ATOMS SHALL NUCLEOSYNTHESIZE FOREVER AND EVER AND ATOMS SHALL NUCLEOSYNTHESIZE FOREVER AND EVER AND ATOMS SHALL NUCLEOSYNTHESIZE FOREVER AND EVER AND ATOMS SHALL NUCLEOSYNTHESIZE FOREVER AND EVER A-T-O-M-P-L-U-T-O-NIUM! ==== Excuse my poor english !!! Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? a and b are real or complex numbers. i have found several results: int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)) int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..inf) with E integral exponential. I don't know if theses results can help to resolve the first expression. All ideas are welcome. ==== >Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? >a and b are real or complex numbers. >i have found several results: >int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)) [ presumably for Re(a) < 0 and Re(b) < 0 ] OK, so if you have that you just need to subtract the integral from 0 to 1. int(exp(a x+b/x),x=0..1) = sum_{n=0}^infinity a^n/n! int(x^n exp(b/x),x=0..1) Note that if C_n = int(x^n exp(b/x),x=0..1) (and b < 0), C_0 = e^b + b Ei(1,-b) and (by integration by parts) C_n = e^b/(1+n) + b/(1+n) C_{n-1} so that C_n = e^b sum_{j=0}^{n-1} (n-j)! b^j/(n+1)! + b^n/(n+1)! C_0 = e^b sum_{j=0}^n (n-j)! b^j/(n+1)! + b^(n+1)/(n+1)! Ei(1,-b) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Robert Israel a .8ecrit : >>Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? >>a and b are real or complex numbers. >>i have found several results: >>int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)) > [ presumably for Re(a) < 0 and Re(b) < 0 ] yes OK, so if you have that you just need to subtract the integral from 0 to > 1. int(exp(a x+b/x),x=0..1) > = sum_{n=0}^infinity a^n/n! int(x^n exp(b/x),x=0..1) Note that if C_n = int(x^n exp(b/x),x=0..1) (and b < 0), > C_0 = e^b + b Ei(1,-b) and (by integration by parts) ok > C_n = e^b/(1+n) + b/(1+n) C_{n-1} > so that > C_n = e^b sum_{j=0}^{n-1} (n-j)! b^j/(n+1)! + b^n/(n+1)! C_0 > = e^b sum_{j=0}^n (n-j)! b^j/(n+1)! + b^(n+1)/(n+1)! Ei(1,-b) I=sum_{n=0..inf} a^n*C_n/n! b*sum_{n=0..inf} (a*b)^n/(n!*(n+1)!)=b*BesselI(1,2*sqrt(a*b))/sqrt(a*b) ok but sum_{n=0..inf} sum_{j=0..n} (n-j)! * b^j*a^n/(n!*(n+1)!) =??? i don't know. help ! > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 > ==== Excuse my poor english !!! Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? > a and b are real or complex numbers. i have found several results: > int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)) int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..inf) with E integral > exponential. I don't know if theses results can help to resolve the first expression. All ideas are welcome. > What system you are using to get the first? Maple gives -2/a*(-b)^(1/2)*(-a)^(1/2)*BesselK(1,2*(-b)^(1/2)*(-a)^(1/2)) which is ~ your answer. The second is Taylor series for b? So what do you mean by 'resolve the first'? It is a kind of 'known' function (ie: it can be computed or what ever). Hm .. what 'expression' would you like to have as answer? ==== Axel Vogt a .8ecrit : >Excuse my poor english !!! >>Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? >>a and b are real or complex numbers. >>i have found several results: >>int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)) >>int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..inf) with E integral >>exponential. >>I don't know if theses results can help to resolve the first expression. >>All ideas are welcome. > What system you are using to get the first? Maple gives > -2/a*(-b)^(1/2)*(-a)^(1/2)*BesselK(1,2*(-b)^(1/2)*(-a)^(1/2)) it's result for int(exp(a*x+b/x),x=0..inf) but i have to solve int(exp(a*x+b/x),x=0..inf). > which is ~ your answer. The second is Taylor series for b? So what do you mean by 'resolve the first'? It is a kind > of 'known' function (ie: it can be computed or what ever). > Hm .. what 'expression' would you like to have as answer? ==== > Excuse my poor english !!! Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? > a and b are real or complex numbers. i have found several results: > int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)) int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..inf) with E integral > exponential. I don't know if theses results can help to resolve the first expression. All ideas are welcome. Mathematica says that it doesn't converge. Lurch ==== >> Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? >> a and b are real or complex numbers. >Mathematica says that it doesn't converge. Even for a=-1, b=0 in which case it's simply 1/e? ==== > Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? >> a and b are real or complex numbers. >Mathematica says that it doesn't converge. ??? That sounded plausible to me. Computer algebra systems are, in my own experience, terrible with this sort of thing. However, here's what the current version of Mathematica does. In[1]:= Integrate[Exp[a*x + b/x], {x, 1, Infinity}] Out[1]= Integrate[E^(b/x + a*x), {x, 1, Infinity}] Thus, at least no unwarranted general claim of divergence is made! > Even for a=-1, b=0 in which case it's simply 1/e? Ah, at least Mathematica can handle that well. In[2]:= Assuming[a < 0 && b == 0, Integrate[Exp[a*x + b/x], {x, 1, Infinity}]] Out[2]= -(E^a/a) David ==== > [..., with reference to Mathematica] In[2]:= Assuming[a < 0 && b == 0, Integrate[Exp[a*x + b/x], {x, 1, Infinity}]] > > Assuming? That's a new one to me. It is not in my book (the second edition, 1991), nor in the version (4.1 for Sun Solaris) whereto I have access. What version do you use? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== [..., with reference to Mathematica] In[2]:= Assuming[a < 0 && b == 0, Integrate[Exp[a*x + b/x], {x, 1, >Infinity}]] Assuming? That's a new one to me. It is not in my book (the second > edition, 1991), nor in the version (4.1 for Sun Solaris) whereto I > have access. What version do you use? As I noted in the part you snipped, I'm using the _current_ version, which is version 5.0. Yes, Assuming is new to that version. But in version 4.1 I suspect that the following equivalent idea should work for you: In[1]:= Integrate[Exp[a*x + b/x], {x, 1, Infinity}, Assumptions -> a < 0 && b == 0] Out[1]= -(E^a/a) David ==== Toni Lassila a .8ecrit : >Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? >a and b are real or complex numbers. >Mathematica says that it doesn't converge. > Even for a=-1, b=0 in which case it's simply 1/e? convergence is not my problem. I choose a and b to converge. ==== Excuse my poor english !!! Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=??? > a and b are real or complex numbers. i have found several results: > int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)) int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..inf) with E integral > exponential. I don't know if theses results can help to resolve the first expression. All ideas are welcome. Mathematica says that it doesn't converge. For Re a < 0 it's easily seen to converge. But no, I don't know how to do it... the b/x looks difficult. Sorry. - Arthur Lurch ==== By computing z_1 + z_2 + z_3, or otherwise, determine the set for which the statement z_1^3 = z_2^3 = z_3^3 is false. a) z_1 = -sqrt3 + i, z_2 = sqrt3 + i, z_3 = -2i b) z_1 = 4, z_2 = -2 +sqrt12i, z_3 = -2 -sqrt12i c) z_1 = sqrt27 + 3i, z_2 = -sqrt27 - 3i, z_3 = 6i d) z_1 = -sqrt12i, z_2 = -3 +sqrt3i, z_3 = 3 +sqrt3i What is the best method of solving this problem? There must be a better method than laboriously cubing and comparing them all...? ==== Imagine that you have the polar forms: z_1 = r_1 (cos @_1 + i sin @_1) and similarly for z_2 and z_3. What do you know about r_1, r_2, and r_3 if z_1^3 = z_2^3 = z_3^3? What do you know about the @ values? --OL ==== >By computing z_1 + z_2 + z_3, or otherwise, determine the set for which the >statement z_1^3 = z_2^3 = z_3^3 is false. a) z_1 = -sqrt3 + i, z_2 = sqrt3 + i, z_3 = -2i >b) z_1 = 4, z_2 = -2 +sqrt12i, z_3 = -2 -sqrt12i >c) z_1 = sqrt27 + 3i, z_2 = -sqrt27 - 3i, z_3 = 6i >d) z_1 = -sqrt12i, z_2 = -3 +sqrt3i, z_3 = 3 +sqrt3i What is the best method of solving this problem? There must be a better >method than laboriously cubing and comparing them all...? Note that in (a): z_1+z_2 = -z_3 in (b): z_2+z_3 = -z_1 in (c): z_1 = -z_2 in (d): z_2+z_3 = 2*sqrt(3)i = sqrt(12)i = -z_1 So in (c) you know that the statement z_1^3 = z_2^3 = z_3^3 is false, since z_1 = -z_2 implies that z_1^3 = -z_2^3, and z_2^3 = -z_2^3 if and only if z_2 = 0, which it is not in this case. Are you sure you copied (c) correctly? Based on the other examples, I would expect z_1 = sqrt(27) - 3i, so that z_1 + z_2 = -z_3. So, in (a), (b), and (c), you have z_1+z_2+z_3 = 0. Cubing this gives: 0 = (z_1+z_2+z_3)^3 = (z_1+z_2)^3 + 3(z_1+z_2)^2*z_3 + 3(z_1+z_2)*z_3^2 + z_3^3 = z_1^3 + 3z_1^2*z_2 + 3z_1*z_2^2 + z_2^3 + 3z_1^2*z_3 + 6z_1z_2z_3 + 3z_1^2*z_3 + 3z_1*z_3^2 + 3z_2*z_3^2 + z_3^3 = z_1^3 + z_2^3 + z_3^3 + 3z_1^2(z_2+z_3) + 3z_2^2(z_1+z_3) + 3z_3^2(z_1+z_2) + 6z_1*z_2*z_3 = z_1^3 + z_2^3 + z_3^3 + 3z_1^2(-z_1) + 3z_2^2(-z_2) + 3z_3^2(-z_3) + 6z_1*z_2*z_3 = z_1^3 + z_2^3 + z_3^3 - 3z_1^3 - 3z_2^3 - 3z_3^3 + 6z_1*z_2*z_3 = -2(z_1^3 + z_2^3 + z_3^3) + 6z_1*z_2*z3. Which means that 3z_1*z_2*z_3 = z_1^3 + z_2^3 + z_3^3. So if z_1^3 = z_2^3 = z_3^3, then z_1*z_2*z_3 = z_1^3 = z_2^3 = z_3^3. Which means that z_1*z_2*(-z_1-z_2) = z_1^3 -z_1^2*z_2 - z_1*z_2^2 = z_1^3 -z_1*z_2 - z_2^2 = z_1^2 z_1^2 + z_1*z_2 + z_2^2 = 0 (z_1 + z_2)^2 - z_1*z_2 = 0 (z_1+z_2)^2 = z_1*z_2 (-z_3)^2 = z_1*z_2 z_3^2 = z_1*z_2 And analogous for the other three. Conversely, if z_3^2 = z_1*z_2, then z_1^3 = z_1*z_2*z_3, z_2^2 = z_1*z_3, then z_2^3 = z_1*z_2*z_3 z_1^2 = z_2*z_3, then z_3^2 = z_1*z_2*z_3. and so they are all equal. Don't know if that is simpler than cubing, 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 ==== >By computing z_1 + z_2 + z_3, or otherwise, determine the set for which the >statement z_1^3 = z_2^3 = z_3^3 is false. a) z_1 = -sqrt3 + i, z_2 = sqrt3 + i, z_3 = -2i >b) z_1 = 4, z_2 = -2 +sqrt12i, z_3 = -2 -sqrt12i >c) z_1 = sqrt27 + 3i, z_2 = -sqrt27 - 3i, z_3 = 6i >d) z_1 = -sqrt12i, z_2 = -3 +sqrt3i, z_3 = 3 +sqrt3i What is the best method of solving this problem? There must be a better >method than laboriously cubing and comparing them all...? Note that in (a): z_1+z_2 = -z_3 > in (b): z_2+z_3 = -z_1 > in (c): z_1 = -z_2 > in (d): z_2+z_3 = 2*sqrt(3)i = sqrt(12)i = -z_1 So in (c) you know that the statement z_1^3 = z_2^3 = z_3^3 is false, > since z_1 = -z_2 implies that z_1^3 = -z_2^3, and z_2^3 = -z_2^3 if > and only if z_2 = 0, which it is not in this case. Are you sure you copied (c) correctly? Based on the other examples, I > would expect z_1 = sqrt(27) - 3i, so that z_1 + z_2 = -z_3. So, in (a), (b), and (c), you have z_1+z_2+z_3 = 0. Cubing this gives: 0 = (z_1+z_2+z_3)^3 > = (z_1+z_2)^3 + 3(z_1+z_2)^2*z_3 + 3(z_1+z_2)*z_3^2 + z_3^3 > = z_1^3 + 3z_1^2*z_2 + 3z_1*z_2^2 + z_2^3 + 3z_1^2*z_3 + 6z_1z_2z_3 > + 3z_1^2*z_3 + 3z_1*z_3^2 + 3z_2*z_3^2 + z_3^3 > = z_1^3 + z_2^3 + z_3^3 + 3z_1^2(z_2+z_3) + 3z_2^2(z_1+z_3) > + 3z_3^2(z_1+z_2) + 6z_1*z_2*z_3 > = z_1^3 + z_2^3 + z_3^3 + 3z_1^2(-z_1) + 3z_2^2(-z_2) + 3z_3^2(-z_3) > + 6z_1*z_2*z_3 > = z_1^3 + z_2^3 + z_3^3 - 3z_1^3 - 3z_2^3 - 3z_3^3 + 6z_1*z_2*z_3 > = -2(z_1^3 + z_2^3 + z_3^3) + 6z_1*z_2*z3. Which means that 3z_1*z_2*z_3 = z_1^3 + z_2^3 + z_3^3. So if z_1^3 = z_2^3 = z_3^3, then z_1*z_2*z_3 = z_1^3 = z_2^3 = z_3^3. Which means that z_1*z_2*(-z_1-z_2) = z_1^3 > -z_1^2*z_2 - z_1*z_2^2 = z_1^3 > -z_1*z_2 - z_2^2 = z_1^2 > z_1^2 + z_1*z_2 + z_2^2 = 0 > (z_1 + z_2)^2 - z_1*z_2 = 0 > (z_1+z_2)^2 = z_1*z_2 > (-z_3)^2 = z_1*z_2 > z_3^2 = z_1*z_2 And analogous for the other three. Conversely, if z_3^2 = z_1*z_2, then z_1^3 = z_1*z_2*z_3, > z_2^2 = z_1*z_3, then z_2^3 = z_1*z_2*z_3 > z_1^2 = z_2*z_3, then z_3^2 = z_1*z_2*z_3. and so they are all equal. Don't know if that is simpler than cubing, > 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 > Couldn't you use De Moivre's Theorem? -- David Moran Chief Meteorologist Oklahoma Storm Team ==== [.snip.] >>By computing z_1 + z_2 + z_3, or otherwise, determine the set for which >the >>statement z_1^3 = z_2^3 = z_3^3 is false. >>a) z_1 = -sqrt3 + i, z_2 = sqrt3 + i, z_3 = -2i >>b) z_1 = 4, z_2 = -2 +sqrt12i, z_3 = -2 -sqrt12i >>c) z_1 = sqrt27 + 3i, z_2 = -sqrt27 - 3i, z_3 = 6i >>d) z_1 = -sqrt12i, z_2 = -3 +sqrt3i, z_3 = 3 +sqrt3i >Couldn't you use De Moivre's Theorem? Certainly, though I though that would imply cubing each. I'll skip (c), though I suspect it was a typo. In (a), you have that arg(z_1) = pi-arg(z_2), and arg(z_3) = -pi/2. In (b) you have that arg(z_1) = pi-arg(z_2), and arg(z_3) = -pi/2 In (d) you have arg(z_2) = pi-arg(z_3), and arg(z_1) = -pi/2. In (a), |z_1| = |z_2| = |z_3| = 2 In (b), |z_2| = |z_3| = |z_1| = 4 In (d), |z_1| = |z_2| = |z_3| = sqrt(12). In (a), you would need to verify that arg(z_2) = pi/6; in (b) that arg(z_1)=pi/6, and in (d) that arg(z_3)=pi/6. If so, then the cubes are equal; if not, then the cubes are not equal. -- ====================================================================== 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 ==== > I could be wrong, but I don't see much sign that he meant varies within > a homotopy class when he said it depends on the path. If he meant the > function element you arrive at when you continue along a path, and we take > depends on the path more literally, then that statement is correct, at > least by itself. I did in fact mean that what you get depends on the homotopy class of the path, not on the path within the homotopy class. > I would say > that it's the fact that one is taking the collection of all paths that makes > the construction independent of a choice of any one given path. It's like > making a manifold not depend on your choice of coordinate system by defining > its atlas to be all possible compatible coordinate charts. Before this sequence of posts I was not aware that there was a comstruction for Riemann surfaces which uses unrestricted analytical continuation only, bypassing cuts, sheets and boundary identifications. This may be good in geometry but my feeling is that a Riemann surface so defined is less likely be useful as a tool in the theory of functions of a complex variable. However this may be a difficult point to get across. Lee Rudolph has some interesting and relevant comments. Look in Google Groups for Re: Riemann Surface lrudolph. ==== > Before this sequence of posts I was not aware that there was a >comstruction for Riemann surfaces which uses unrestricted analytical >continuation only, bypassing cuts, sheets and boundary >identifications. This may be good in geometry but my feeling is that a >Riemann surface so defined is less likely be useful as a tool in the >theory of functions of a complex variable. However this may be a >difficult point to get across. Lee Rudolph has some interesting and >relevant comments. I do? >Look in Google Groups for Re: Riemann Surface >lrudolph. I guess you refer to http://groups.google.com/groups?selm=asvepu%243gp%241%40panix2.panix.com and, perhaps, specifically to In fact (at least for Riemann surfaces spread over a domain in C or the extended complex numbers, that is, for meromorphic functions with *numerical*, rather than another-Riemann-surface, values), there's sometimes a sort-of-canonical cut system which can be generated in a sort-of-canonical way--roughly speaking, sheets meet where real parts of branches coincide. Notice the weasel words (sometimes, sort-of-canonical, roughly) and please don't take too much comfort from my post. In fact, the two applications of such a sort-of-canonical cut system that I mentioned later in my post are (to me) themselves very much geometry and not much at all concerned with theory of functions of a complex variable (as I would understand that phrase); I really don't think they support your feeling quoted above. (Since my own interest in Riemann surfaces is primarily geometrical, indeed, topological, my ignorance of interesting and/or significant applications of such a sort-of-canonical cut system to theory of functions may only mean that I haven't run into one, not that none exists. But there are plenty of people in sci.math, several posting to this thread, whose *do* know theory of functions of a complex variable inside out, and I suspect that they'd know of such applications if there were many of them.) Lee Rudolph ==== > I've recently been looking at little and big endian memory storage but have > encountered a little confusion when it comes to transferring or copying data > between architectures of different endianness and would be grateful for some > clarification. I couldn't exactly follow what you were trying to say. First, at the conceptual level, the issue is this. Suppose I tried to tell you a number by reading the digits one at a time. We would have to agree whether I began with the most significant digit (reading the digits left-to-right) or wehther I began with the least significant digit (reading the digits from right-to-left). So would I read the number 9307 as nine-three-zero-seven or seven-zero-three-nine? Clearly, we need to agree on what we are doing or things will get very confused. Note, but the way, that there are some advantages to reading right-to-left; for example, if you were filling in a printed form where the digits went into boxes, reading right-to-left (little endian) would allow you to begin filling in the 1's place, then the 10's place, etc, while reading left-to-right (big endian), which may seem more natural, would require that you wait to hear the whole number before you know whether the first digit read goes in the box for 100's or 1000's etc. Ok, now to practical matters. Please note that there really are no guarantees in this game, but this should work--just keep an eye on things. Suppose one wanted to store the 32-bit number 0x110A0809 (in hexidecimal, base 16, right?) in memory or in a file. On a big endian machine, if you store this is a file, the first byte of the file will be 0x11 and the second byte will be 0A, the third byte will be 08, and the fourth byte will be 09. On a little endian-machine, this order will be reversed. Similarly, for memory. For example, in C unsigned int val = 0x110A0809; unsigned int *p_val = &val; unsigned char *p_char = (unsigned char*)(p_val); printf(0x%02xn, *p_char); on a big-endian machine, this should print 0x11, but on a little endian machine this should give 0x%09. If you have an unsigned int that was written to a file on an machine with one endian-ness and you have read that back into an unsigned int on a machine with a second endian-ness, then the following function would allow you to change the endian-ness: unsigned int change_endian_uint(unsigned int n) { typedef union { unsigned int value; char bytes[4]; } cheat_type; cheat_type input; cheat_type output; input.value = n; output.bytes[3] = input.bytes[0]; output.bytes[2] = input.bytes[1]; output.bytes[1] = input.bytes[2]; output.bytes[0] = input.bytes[3]; return output.value; } For other situation you may need to play other games with unions. Following up on a question in your original posts, if you did mixed 16-bit and 32 bit objects, each object nees to be handled separately (thus you might need another routine for unsigned shorts): unsigned short change_endian_ushort(unsigned short n) { typedef union { unsigned short value; char bytes[2]; } cheat_type; cheat_type input; cheat_type output; input.value = n; output.bytes[1] = input.bytes[0]; output.bytes[0] = input.bytes[1]; return output.value; } The above code fragments assume, of course, that on your machines int's are 32-bit objects and short's are 16-bit objects. Several other notes about things that can go wrong: The sizes of objects might vary between machines. For exmaple, on some older machines, int would be 16-bits, while on some high-end new machines, int might be 64-bits. It is even possible that the order of bits withing bytes can be reversed, but this is unlikely. Machines have different alignment issues, which relate to packing. For example, in C if you have struct a_struct{ int a; char b; int c; }; then on most machines, sizeof(a_struct) = 12, even though sizeof(int) + sizeof(char) + sizeof(int) = 9. This is done so that both int's can be placed in memory locations whose addresses are divisble by 4 (a requirement for reading 4-bytes at a time on some systems, and an optimization on many others). The three bytes after the char b contain undefined data. Two other useful facts. Intel is probably the only architecture you will meet that is little-endian; most other machines (SPARC, and if memory server Motorola and SGI) are big-endian. As a cultural note, the term big-endian and little-endian pay homage to Gulliver's travels, where kingdoms were at war as to whether to open eggs at the little end or the big end. Best wishes -Mike ==== -snip- ==== How to solve this equation (by elementary methods!): n^p+m^p=(2n-m)^p n>0;m>0 et p>2 ==== > M?rio Amado Alves > I've learnt how to represent a map as the set of pairs of connected > vertices. Accordingly, the object {(A, B)} represents both (1) and > (2): (1) (2) > ___________ _____ > | | | | > | A | | A | > | _____ | |_____| > | | | | | | > | | B | | | B | > | |_____| | |_____| > |___________| But clearly (1) and (2) differ with respect to the reachability of > region B from the outside [(2) reachable, (1) not]. How is this kind > of fact formally expressed usually? (I can think of a couple of ways > but I'm sure this problem has been dealt before so no need to reinvent > the wheel and/or use a 'strange' structure.) --Marius Amado Alves > (amateur mathematician playing with map coloring) > Country A is simply connected, as they say, in (2) but not in (1). If we > represent countries by vertices rather than by faces, people speak of a > bridge (resp. cutpoint) as an edge (resp. vertex) which, if omitted, causes > the graph to become disconnected. So, if there are other countries outside > A, A would be a cutpoint.... Ok. But with this I still cannot represent the fact explicitly. > Consider maps (3), (4) which are (1), (2) resp. plus a region C > connected (only) to A. The symbolic represention of *both* (3) and (4) > is {(A, B), (A, C)}. I'm playing with constructing (colored) maps by adding one region at a > time (and exchanging colors). For that I want to represent the > unreachability condition, for example expressing the fact that I can > add pair (B, C) to (4) but not to (3). Maybe I should say a litle more. I intuit that is possible to keep the > number of colors of the periphery of any map, below 4, where periphery > is the set of vertices available for connection to a newly added > vertex. I intuit this is possible by exchanging colors in the newly > created map. This is of course a quest for a different (?) proof of > the four color theorem. Maybe I'm on a beaten track. I don't know much > about four color theorem proving, other then the 'fact' that existing > proofs are not elegant enough for my taste. > The following website gives 43 proofs of the Pythagorean Theorem. http://www.cut-the-knot.org/pythagoras/index.shtml Which proof(s) do you consider to be elegant? ==== > Algebraic toplogy (AT herein) seems to be a good way of formalizing > protocols in distributed systems (DS herein) (such as decision problems like > 'consensus'). Do you know about Petri nets, process algebras such as CCS, CSP and the Pi-calculus, and more abstract characterizations such as labelled transition systems (LTSs)? If not, those would be important to know before plunging into topology. There is even a widely-accepted language (LOTOS) based on CCS and CSP that is used for formally defining communication protocols. Moreover PROMELA, the language for protocol definition which is used by the SPIN model checker, is also based partly on CSP and is also widely used. There may be some reason why these more standard formal characterizations are not adequate for your purposes, but in the absence of other information, I would say to try them first. --Jamie. (nel mezzo del cammin di nostra vita) ==== > In comp.theory Michael N. Christoff protocols in distributed systems (DS herein) (such as decision problems like > 'consensus'). Do you know about Petri nets, process algebras such as > CCS, CSP and the Pi-calculus, and more abstract > characterizations such as labelled transition systems (LTSs)? > If not, those would be important to know before plunging into > topology. There is even a widely-accepted language (LOTOS) > based on CCS and CSP that is used for formally defining > communication protocols. Moreover PROMELA, the language for > protocol definition which is used by the SPIN model checker, is > also based partly on CSP and is also widely used. > I actually have a book on CCS (that I'll be taking another look at), but if I'm not mistaken its based on an interleaving model of concurrency. Also, I am most interested in architectures that do not use shared memory in favour of distributed 'objects' with local hidden state. One of the main reason I was interested in AT is that it has given some very elegant proofs of some basic impossibility theorems (ie: impossibility of consensus with one faulty process). Are there similar proofs based on CCS, etc... ? Finally, on a more personal level, I have always been interested in topology and group theory (did a lot of research into the finite state automata / group theory connection). I have also done a small amount of work with Petri nets. Are these process algebras a prerequisite to learning AT for distributed computing? As far as AT goes, I'm going to start by learning some basic point-set topology first. The only problem is weeding out the material I need for DS from more 'pure math' aspects of it. l8r, Mike N. Christoff ==== > Algebraic toplogy (AT herein) seems to be a good way of formalizing > protocols in distributed systems (DS herein) (such as decision problems like > 'consensus'). I am interested in learning more, however AT is a huge field > and I am only interested in learning the parts directly related to > distributed computing. Can anyone suggest a book, that a) assumes no > knowledge of algebraic toplogy b) assumes no more than undergraduate level > math - ie: calculus, linear algebra, basic geometry, ability to do proofs, > etc... c) focused on showing how AT can be utilized to solve DS problems and > does not get into non-DS related aspects of AT (unless they are required > background for understanding DS related AT topics). I have found many introductions on the net, but they seem to assume at least > basic knowledge of topology, homotopy, and other topics I am not familiar > with, so I think a full book dedicated to the subject sounds more feasible > as a basis for learning AT for DS. But any links or online books you may > know of will be of great help as well. > A Tutorial on Algebraic Topology and Distributed Computing Maurice Herlihy, 1994 http://citeseer.nj.nec.com/herlihy94tutorial.html l8r, Mike N. Christoff ==== Sub: Unsolvabve Geometrical Problems > Squaring a given circle is unsolvable . There are volumes of pages on the subject which I have read - from Archimedes to SrinivasaRamanujan. > Well! I have found out a solution. Could not believe it ? My logic is very simple. > Pi is an irrational number. So is square root of 2. There is a simple geometrical construction to solve for square root of 2. Similarly I have developed a simple geometrical construction to solve for pi. > My question is Where to send it? Is there any Maths Forum where I can present my papers and answer questions of the experts? Try sending it to James Harris to verify : ) -- Paul V. S. Townsend Interchange the alphabetic elements to reply ==== > Sub: Unsolvabve Geometrical Problems > Squaring a given circle is unsolvable . There are volumes of pages on the subject which I have read - from Archimedes to SrinivasaRamanujan. > Well! I have found out a solution. Could not believe it ? My logic is very simple. > Pi is an irrational number. So is square root of 2. There is a simple geometrical construction to solve for square root of 2. Similarly I have developed a simple geometrical construction to solve for pi. > My question is Where to send it? Is there any Maths Forum where I can present my papers and answer questions of the experts? You're likely to run into the same sort of small-minded attitude among the mathematicians here, that our beloved JSH experienced. Why not follow his tack? Look into this publication of the Mega Foundation: It worked for JSH. Maybe it'll get your discoveries some clear air. BTW, the other poster suggesting a web page containing your argument, coupled by a discussion on sci.math dedicated to exposing the flaw in the argument (sorry, if your construction means to produce pi via a constriction using compass and [unmarked] straightedge only, then your are mistaken, and there is some error in the argument). Persevere, and be of good faith (not to mention good humor), and no doubt you'll learn something. Dale. ==== > Sub: Unsolvabve Geometrical Problems > Squaring a given circle is unsolvable . There are volumes of pages on the subject which I have read - from Archimedes to SrinivasaRamanujan. > Well! I have found out a solution. Could not believe it ? My logic is very simple. > Pi is an irrational number. So is square root of 2. There is a simple geometrical construction to solve for square root of 2. Similarly I have developed a simple geometrical construction to solve for pi. > My question is Where to send it? Is there any Maths Forum where I can present my papers and answer questions of the experts? > There's already a geomtrical construciton of pi, take a compass and construct a circle with a diameter of 1. ==== >Squaring a given circle is unsolvable . Note that this unsolvability is unsolvability using straightedge and compass methods. If you permit certain kinds of mechanical devices to be used, then it's possible to construct pi and square the circle. The other standard impossible constructions (trisecting an angle and duplicating a cube) are also possible if one is allowed the right kind of tool in addition to straightedge and compass. Keith Ramsay ==== Squaring a given circle is unsolvable . Note that this unsolvability is unsolvability using straightedge and > compass methods. If you permit certain kinds of mechanical > devices to be used, then it's possible to construct pi and square > the circle. The other standard impossible constructions > (trisecting an angle and duplicating a cube) are also possible > if one is allowed the right kind of tool in addition to straightedge > and compass. Keith Ramsay > Just wondering, is it possible to create these tools using a compass and a straightedge (and maybe a pair of scissors)? /R ==== > Just wondering, is it possible to create these tools using a compass and a > straightedge (and maybe a pair of scissors)? /R Well, one angle trisection requires a straightedge that you can mark; it is then slid along a circle and line. (That not being one of the operations described by Euclid, it has no impact on the proof of impossibility.) I've also seen a drawing of an angle trisection device with fairly simple moving parts. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W ==== I've actually seen a trisection device with no moving parts at all. It's basically a strip with one end having square and circular attachments at the end. By placing the object in the right position relative to a given angle, the trisectors of the angle are identified. The Archimedean trisection (using the marked straightedge) algebraiclly solves any cubic equation with three real roots, for the solution of such an equation always reduces to an angle trisection via Viete's formulation. So the method can be adapted to another old impossible (with unmarked straightedge and compass) problem, the construction of the regular heptagon. --OL ==== The problem is NOT that pi is irrational. The problem instead is that pi is not obtainable from integers by any combination of arithmetic operations plus solution of quadratic equations, which are the only operations that can be implemented via a straightedge and compass. This situation holds for any transcendental numbe, among others. You can't square a circle. But you can get close to doing so since the set of constructible numerical ratios (rational numbers and some irrational ones) is dense. --OL ==== The flaw is that pi is *not* an irrational number. It is transcendental. ==== >The flaw is that pi is *not* an irrational number. It is transcendental. That should be pi is *not* just an irrational number. It is transcendental. In other words, pi is irrational, and it is a special type of irrational number (the special type being a transcendental number). David McAnally -------------- ==== The flaw is that pi is *not* an irrational number. It is transcendental. Transcendental numbers ->are<- irrational. Not every irrational is transcendental, but every transcedental is irrational. -- ====================================================================== 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 ==== > (axiom-schema) is introduced...along with power set...pairing is >> redundant, they advocate keeping it because it is elementary and >> essential for the sequential development and even for a scheme that >> shuns replacement, pairing makes sense. I think they are suggesting >> that set theory even without replacement is strong enough for almost >> all practical purposes. What do you think? Oh, I see, I'm used to the shunning replacement version, because I'm >used to GBN set theory. So I think of replacement+comprehension as >just GBN's comprehension, but of course, that isn't quite right, and >this is an example why. I believe that in GBN set theory, pairing *is* required, is it not? Yes, because the NBG axioms do not have propositional > functions. Functions are defined only by ordered pairs. Good, that's what I thought. ==== > Mathematicians dump on cranks with extreme vitriol. No. Most mathematicians ignore them entirely. Thomas ==== > [...] >> But check this out: > http://www.bearnol.pwp.blueyonder.co.uk/Math/ Wow, James has been busy. I had no idea he'd, um, 'tackled' so many problems! >>For those who don't know Ramsden is an alias of David Rusin, but I >>don't see his point here in posting that link. >>Anyone have any idea what Rusin is up to here? Or any idea how James got the idea that Ramsden and Rusin were >the same person? Don't know, but it seems, from James's post, that James Harris may believe that James Harris = James Wanless (which is the person John Ramsden was talking about)... -- ====================================================================== 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 <3c65f87.0312131113.663a3a6c@posting.google.com> ==== >> [...] >> But check this out: >> http://www.bearnol.pwp.blueyonder.co.uk/Math/ >> Wow, James has been busy. I had no idea he'd, um, 'tackled' so many problems! >>For those who don't know Ramsden is an alias of David Rusin, but I >don't see his point here in posting that link. >>Anyone have any idea what Rusin is up to here? >>Or any idea how James got the idea that Ramsden and Rusin were >>the same person? Don't know, but it seems, from James's post, that James Harris may > believe that James Harris = James Wanless (which is the person John > Ramsden was talking about)... Yes, but also that John Ramsden = Larry Hammick, since John didn't post any links at all, but Larry did. How appropriate. The population of sci.math is plummeting because of a vengeful equivalence relation. The group sci.math / alias equivalence (the quotient of sci.math by the equivalence relation generated by the is a sock puppet of relation) won't have so many participants, I fear. I wonder which coset I'm in. -- Come on people!!! The US just blew up a lot of people in Iraq, don't you realize that a person with my exposure might just end up dead, by mysterious circumstances? --James Harris, on the dangers of proving Fermat's last theorem ==== What do you mean by properly defined? What do you mean by predicative proof step? And so on. I meant a proof technique which could be modeled or otherwise aped (the context does not require rigid precision) in a proof done within a predicative proof system. Thomas ==== Herc, please comment on the generalization of Cantor's diagonal argument. Without doing that, everything you say could be true and you would still have multiple infinity types. For all sets A, |A| < |P(A)|. Let A be a set and P(A) be the powerset of A. A can be infinite, but it doesn't need to be. There are no bijections between A and P(A) (see below). There is a bijection between A and {{a}| a in A}, subset of P(A). By our definition of |A| < |B|, it follows that |A| < |P(A)|. Let f: A -> P(A). f is not a bijection, because there is at least one element of P(A) (one subset of A) to which f does not map. (See below.) Since f is not specified beyond being a function from A to P(A), this is true of all such functions. Therefore, there are no bijections between A and P(A) and |A| < |P(A)|. Consider the set D_f, a subset of A, defined as D_f = { a in A| ~( a in f(a) ) }, which is equivalent to ( a in D_f ) iff ~( a in f(a) ). There is no d in A such that * f(d) = D_f. If there were, it would follow that * ( d in D_f = f(d) ) iff ~( d in f(d) ). But this would be a contradiction. Therefore, f is not a bijection, no such function is a bijection, and |A| < |P(A)|. See also http://mathworld.wolfram.com/CantorDiagonalMethod.html I'm going to respond to your comments below, but I think it's close to pointless, if your intention is to rule out more than one infinity-type. [...] > You say it's not more useful; I say it is more useful. How > should we decide this point? I could give you many examples > where the existence of the LUB is important to the conclusion > drawn _in real analysis_. Can you do the same with the > computability property? do any examples apply outside of number theory? what is > the contribution to technology of uncountability? anything? up a lot in quantum mechanics. I'm not sure it makes sense to talk of that if the values that the norm takes, , don't form a complete field. Completeness is the LUB guarantee. The LUB guarantee leads to uncountability. You're sitting in front of an application of quantum mechanics while you read this. Good enough? On the other hand, while I believe that thinking about computability and Turing Machines contributed to the development of today's PCs and the Internet and all, there is very little use for the distinction you are drawing, between computable-in-principle and uncomputable-in-principle. The best ciphers we have today have computable-in-principle algorithms that will crack them, but, until we have computable-in-practice algorithms, no one is going to worry about it. [...] > However, several people have pointed out that you do not need a > contradiction for this result. In fact, Cantor did not use > contradiction. In the latest version, the one you most recently > snipped, there is no contradiction, just a demonstration that > every list misses at least one real number. This is a > contradiction only to the assumption that |N| = |R|. misses an infinitely long number, at a digit that it specifies > itself must be contradictory. the number is defined in terms of > itself. The argument show that _at least_ one number is missed, yes. It does this by showing it is false that this specified number is in the list. If showing something is false is a contradiction, then, OK, that's a contradiction. I don't see anything paradoxical about that, though. Something else I've wanted to mention: your argument sounds like there is only this one diagonal number to fit into the list, and then they'll all be in there. In fact, almost all the reals do not fit in the list. A rough way to see this is to look at all the diagonal numbers, those numbers not equal to nth real at the nth decimal. Map them all to the reals in base-9, skipping over the nth digit of the nth real. 0 0.12234... (the diagonal number) 0.17842... 1 0.27843... (one off-diagonal, base-10) 0.44454... 2 0.34897... (maps to R, base-9) 0.34443... 3 0.84747... 0 1 2 3 4 5 6 7 8 9 4 0.28682... | | | | / / / / / ... 0 1 2 3 4 5 6 7 8 The number is _not_ defined in terms of itself. It is defined in terms of the list the number is not in. The number does not contradict _itself_; it contradicts _being in the list_. If you assume that all reals are _in the list_, then, yes, that would appear to be a paradox, a number that is not equal to itself at one digit. That is the point of the argument: that the number is _not_ in the list. > you treat an infinitely long sequence like a simple object you > can manipulate. If you don't like that, you'll hate how I define the reals. Anyway, what's wrong with that? And how is that different from what you do with computable numbers? > where is the contradictory bit? can you point it out? why > cannot the number be computed to any precision specified? its a > simple trick of enumeration, the number is part of the list and > poorly defined at its own referential digit, it doesn't lie > outside the list. why does logic confine itself to non self reference yet > mathematics yields it. there are simple limitations to computer > programs when they self reference, its a practical base of > limitations. natural language and set theory both get around > self referential paradoxes by specifying before hand their use in > the domains. limitations to computers are very limited in scope, > all other fields of science focus on them, they are trivial. You have created the self-referential paradox by assuming something false. It's you who says the number is part of the list, that it contradicts itself, and so on. Your reason for saying this is that _all the reals are on the list_. But they aren't. [...] > The extra number is simply this : > Given a number on a list, change it and put it back in the > list without changing the list. [...] > Would it help to tally the list of reals as a single number? DIGIT 1 2 3 4 5 6 > __________________________ > UTM(1) 4 3 6 4 2 4 > UTM(2) 7 4 3 4 3 2 > UTM(3) 0 1 0 1 1 1 > UTM(4) 1 2 2 2 2 2 > UTM(5) 7 7 7 7 7 7 Working along the diagonal in this fashion : 1 2 4 7 > 3 5 > 6 that would give the single real number : 437640.... This doesn't help at all. I still don't know what you meant before. I don't know what I'm supposed to conclude from this either. Your back-and-forth number isn't guaranteed to either be on your list UTM(n) or to not be on your list. You can only compare real numbers for equality by comparing _corresponding digits_, but your back-and-forth number mixes up the order, leading to who knows what? [...] > Original Format > alt.paranormal, alt.sci.physics I don't see anything new here. Its a diagonalisation proof of Godels theorom by Roger Penrose, > Steven Hawkings assistant. Probably accepted as a correct > proof by mainstream mathematicians to this day, yet the > definition of the diagonal function is obviously flawed and > not well defined. Read the capitalisation part of the post and tell me if its a > valid proof. I'm not sure the proof is valid, but for a different reason than you use. I think step #2 jumps too quickly from All mathematical truths are derivable to there is an algorith to derive them. The step you object to I do not see a problem with. Your own comment is just a restatement of what you said above. a does not have a paradoxical bit because it is not in the list. It is not in the list because we defined a so that we could see it's not in the list. This is only a problem if we insist, as you do, that the list is complete. : : > The pattern is not legitimately created, it is obviously : : > self referencing and a has a paradoxical bit when it evaluates : : > its own number. Just because there's two steps in seeing the : : > plausibility in a theorom, one of the steps fails so the : : > theorem fails, not the whole encapsulation of theoroms. Jim Burns ==== --------------------------------- <^> <(áÀá)> <^> ----------------------------------- Herc, please comment on the generalization of Cantor's diagonal > argument. Without doing that, everything you say could be true > and you would still have multiple infinity types. For all sets A, |A| < |P(A)|. Let A be a set and P(A) be the powerset of A. A can be infinite, > but it doesn't need to be. There are no bijections between A and P(A) (see below). There is > a bijection between A and {{a}| a in A}, subset of P(A). > By our definition of |A| < |B|, it follows that |A| < |P(A)|. Let f: A -> P(A). f is not a bijection, because there is at least > one element of P(A) (one subset of A) to which f does not map. > (See below.) Since f is not specified beyond being a function > from A to P(A), this is true of all such functions. Therefore, > there are no bijections between A and P(A) and |A| < |P(A)|. Consider the set D_f, a subset of A, defined as > D_f = { a in A| ~( a in f(a) ) }, > which is equivalent to > ( a in D_f ) iff ~( a in f(a) ). > There is no d in A such that > * f(d) = D_f. > If there were, it would follow that > * ( d in D_f = f(d) ) iff ~( d in f(d) ). > But this would be a contradiction. Therefore, f is not a bijection, > no such function is a bijection, and |A| < |P(A)|. See also > http://mathworld.wolfram.com/CantorDiagonalMethod.html I'm going to respond to your comments below, but I think it's > close to pointless, if your intention is to rule out more > than one infinity-type. [...] > You say it's not more useful; I say it is more useful. How > should we decide this point? I could give you many examples > where the existence of the LUB is important to the conclusion > drawn _in real analysis_. Can you do the same with the > computability property? do any examples apply outside of number theory? what is > the contribution to technology of uncountability? anything? up a lot in quantum mechanics. I'm not sure it makes sense to > talk of that if the values that the norm takes, , > don't form a complete field. Completeness is the LUB guarantee. > The LUB guarantee leads to uncountability. You're sitting > in front of an application of quantum mechanics while you > read this. Good enough? On the other hand, while I believe that thinking about > computability and Turing Machines contributed to the development > of today's PCs and the Internet and all, there is very little > use for the distinction you are drawing, between > computable-in-principle and uncomputable-in-principle. > The best ciphers we have today have computable-in-principle > algorithms that will crack them, but, until we have > computable-in-practice algorithms, no one is going to > worry about it. [...] > However, several people have pointed out that you do not need a > contradiction for this result. In fact, Cantor did not use > contradiction. In the latest version, the one you most recently > snipped, there is no contradiction, just a demonstration that > every list misses at least one real number. This is a > contradiction only to the assumption that |N| = |R|. misses an infinitely long number, at a digit that it specifies > itself must be contradictory. the number is defined in terms of > itself. The argument show that _at least_ one number is missed, yes. It > does this by showing it is false that this specified number is in > the list. If showing something is false is a contradiction, then, > OK, that's a contradiction. I don't see anything paradoxical > about that, though. Something else I've wanted to mention: your argument sounds like > there is only this one diagonal number to fit into the list, and > then they'll all be in there. In fact, almost all the reals do > not fit in the list. A rough way to see this is to look at all > the diagonal numbers, those numbers not equal to nth real at the > nth decimal. Map them all to the reals in base-9, skipping over > the nth digit of the nth real. 0 0.12234... (the diagonal number) 0.17842... > 1 0.27843... (one off-diagonal, base-10) 0.44454... > 2 0.34897... (maps to R, base-9) 0.34443... > 3 0.84747... 0 1 2 3 4 5 6 7 8 9 > 4 0.28682... | | | | / / / / / > ... 0 1 2 3 4 5 6 7 8 > The number is _not_ defined in terms of itself. It is defined in > terms of the list the number is not in. The number does not > contradict _itself_; it contradicts _being in the list_. If you > assume that all reals are _in the list_, then, yes, that would > appear to be a paradox, a number that is not equal to itself at > one digit. That is the point of the argument: that the number is > _not_ in the list. As the number is defined it is in the list. For a finite list you can say the number is not in the list. For an infinite list you cannot encapsulate the entire number ********* you can only provide an algorithm as such : Get the 1st digit in the grid. do a transform go down 1 and right 1 repeat for an infinite list, the only possible way to define the number is to *reference* the entire list. to reference the entire list of programs means the number will try to read a digit that it is supposed to write Cantors method works on finite lists, simple 0 order objects, constants. infinite lists require an algorithm to encapsulate the diagonal number. any algorithmic_number is automatically a part of the list by the axioms of the proof. I'm not giving you an infinite list to make a new number, I'm only giving you the algorithm, UTM(Z). you treat an infinitely long sequence like a simple object you > can manipulate. If you don't like that, you'll hate how I define the reals. Anyway, > what's wrong with that? And how is that different from what you > do with computable numbers? Every step I take can be indexed by computer. I'm not getting a paradox from simply defined 'complete models' and inferring some beyond computer knowledge technique only humans can comprehend. I am using finite methods that can all be phsically achieved. I'm calling it finitism ~ undefined commentaries on infinite lists are not accepted into proofs. where is the contradictory bit? can you point it out? why > cannot the number be computed to any precision specified? its a > simple trick of enumeration, the number is part of the list and > poorly defined at its own referential digit, it doesn't lie > outside the list. why does logic confine itself to non self reference yet > mathematics yields it. there are simple limitations to computer > programs when they self reference, its a practical base of > limitations. natural language and set theory both get around > self referential paradoxes by specifying before hand their use in > the domains. limitations to computers are very limited in scope, > all other fields of science focus on them, they are trivial. You have created the self-referential paradox by assuming > something false. It's you who says the number is part of the list, > that it contradicts itself, and so on. Your reason for saying this > is that _all the reals are on the list_. But they aren't. 'but they aren't' is assumed here. you are assuming an *algorithm* *off_the_list* to imply a new number. [...] > The extra number is simply this : > Given a number on a list, change it and put it back in the > list without changing the list. > [...] > Would it help to tally the list of reals as a single number? DIGIT 1 2 3 4 5 6 > __________________________ > UTM(1) 4 3 6 4 2 4 > UTM(2) 7 4 3 4 3 2 > UTM(3) 0 1 0 1 1 1 > UTM(4) 1 2 2 2 2 2 > UTM(5) 7 7 7 7 7 7 Working along the diagonal in this fashion : 1 2 4 7 > 3 5 > 6 that would give the single real number : 437640.... This doesn't help at all. I still don't know what you meant before. > I don't know what I'm supposed to conclude from this either. > Your back-and-forth number isn't guaranteed to either be > on your list UTM(n) or to not be on your list. You can only > compare real numbers for equality by comparing _corresponding > digits_, but your back-and-forth number mixes up the order, > leading to who knows what? > I agree, this doesn't help at all! > [...] > Original Format > alt.paranormal, alt.sci.physics I don't see anything new here. Its a diagonalisation proof of Godels theorom by Roger Penrose, > Steven Hawkings assistant. Probably accepted as a correct > proof by mainstream mathematicians to this day, yet the > definition of the diagonal function is obviously flawed and > not well defined. Read the capitalisation part of the post and tell me if its a > valid proof. I'm not sure the proof is valid, but for a different reason than > you use. I think step #2 jumps too quickly from All mathematical > truths are derivable to there is an algorith to derive them. > The step you object to I do not see a problem with. Your own comment is just a restatement of what you said above. > a does not have a paradoxical bit because it is not in the list. > It is not in the list because we defined a so that we could see > it's not in the list. This is only a problem if we insist, as you > do, that the list is complete. : : > The pattern is not legitimately created, it is obviously > : : > self referencing and a has a paradoxical bit when it evaluates > : : > its own number. Just because there's two steps in seeing the > : : > plausibility in a theorom, one of the steps fails so the > : : > theorem fails, not the whole encapsulation of theoroms. Jim Burns I'm not insisting the list is complete. I'm pointing out that if you make the axiom that the list is complete there is no contradiction. In closed knowledge the list is complete. There is nothing to define off the list. I don't follow the power set bijection proof, I'm assuming its a simliar flaw to Cantors diagonalisation. Herc ==== > ... > Maybe ... offhand ... the computable numbers are indeed countable, and > we could produce an unambiguous list giving the number of all _ finite > algorithms_ which might produce one. But the proposed algorithm of > the Cantor argument is not a finite algorithm, since it requires us to > do an infinite amount of work a priori to make the list! That just about is it. The complete list requires us to do an infinite > amount of work just to determine the n-th element of the list when n is > large enough. Suppose we define an algorithm as something that has as > input an integer and as output (when it terminates) a decimal digit. The > computable numbers are defined by all such algorithms that terminate > for each finite input value. We can enumerate the finite algorithms for > which it is obvious that they define a computable number. But that is > not sufficient, we also have to enumerate the finite algorithms for which > it is *not* obvious that they define a computable number. I'm still not sure what we are demonstrating with the Cantor argument! Accept your definitions of algorithm and computable. Assume for the sake of argument the problem algorithm defines a computable number. Then, since the problem algorithm is a valid finite algorithm it should get a number on the list, say K. Call the j-th algorithm on the list P(j) and its corresponding computable number n(j). P(K) tells us to consult the list, and augment the k-th decimal digit of n(k) by 1 mod 10 (say), in order to get the k-th decimal digit of n(K). Ok so far? So let's start computing. We realize we don't have to compute the _entire_ representation of each computable number in order to apply P(K), we only have to compute just enough ... so only a finite amount of work is required at each step after all. First we take P(1), compute exactly one digit of its corresponding number, n(1), and increment that digit by 1, modulo 10. Next we take algorithm P(2), compute the first two digits of n(2) and increment the second digit by 1 mod 10, etc. Eventually we get to K. Now P tells us to consider algorithm P itself, compute the first K digits of its number, then augment the K-th digit. We already have digits 1,2,...,K-1 , no problem; but we don't have digit K yet, because we haven't finished calculating it, so that spot is blank! So we can't carry out the instructions. The algorithm at this point simply becomes nonsense, because it also tells us that the K-th digit of n(K) is not equal to itself. So the alleged algorithm P(K) may _look_ like an algorithm, but in fact is nonsense. But maybe if we _don't_ insist on including it on the list in the first place, we _can_ compute its number, and with a finite amount of effort at each step! If the bad algoritm P _isn't_ on the list, then we can get the k-th digit of its number by running the algorithm in the k-th slot for k places, no problem. So what have we shown? If we assume a priori the valid algorithms are countable, including our problem child, the problem algorithm becomes nonsense, hence we kick it off the list. But once we don't try to count it, it becomes sense again, hence a valid algorithm which was left off the list! The family resemblence to Russell's set paradox is striking, but I'm not sure what we are supposed to conclude: I'm leaning to the conclusion that the computable numbers are not in fact countable. But how can this be? All finite symbol strings (even those drawn from a countably infinite symbol set?) form a countable class, and there can't be more computable numbers than there are finite algorithms, and there can't be more finite algorithms than there are finite symbol strings ... > For instance define the following (Collatz' hailstorm): > T(n): > c := 1; > while n != 1 do > c +:= 1; > if(n % 2 = 0) n /:= 2 > else n := n * 3 + 1 > fi > od; > return c % 10. > This is a finite algorithm. Its output is always a decimal digit. Its > input are the integers. Does it define a computable number? We do not > know, because we do not know whether it terminates for all n. Can we > determine in finite time whether it must be on the list? Perhaps, our > mathematical knowledge is not good enough to determine this. And there > are many more such finite algorithms. I don't see that this observation solves our dilemma. Non-terminating finite algorithms may pad the list, but if we can count _all_ finite algorithms, we can certainly count the subset of finite algorithms which terminate for all integer inputs. ==== ... > For instance define the following (Collatz' hailstorm): > T(n): > c := 1; > while n != 1 do > c +:= 1; > if(n % 2 = 0) n /:= 2 > else n := n * 3 + 1 > fi > od; > return c % 10. > This is a finite algorithm. Its output is always a decimal digit. Its > input are the integers. Does it define a computable number? We do not > know, because we do not know whether it terminates for all n. Can we > determine in finite time whether it must be on the list? Perhaps, our > mathematical knowledge is not good enough to determine this. And there > are many more such finite algorithms. > > I don't see that this observation solves our dilemma. Non-terminating > finite algorithms may pad the list, but if we can count _all_ finite > algorithms, we can certainly count the subset of finite algorithms > which terminate for all integer inputs. But is the algorithm above non-terminating or not? I.e. should it be on the list or not? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > > _What_ is incorrect? > > I think this has been clarified. I commented on your I think > you're thinking of. > > Well, *that's* confusing. Yup, isn't it? > Did you claim that Richard was wrong when he > said that Edward Green was thinking of, let's say, X? That is, Edward > was *not* thinking of X? Or did you allege that Richard was wrong > when he reported that he *thinks* Edward was thinking of X -- in fact, > Richard was *not* thinking that Edward was thinking of X, but was > thinking something else, and he only mistakenly reported that his > thoughts were about Edward and his thinking of X? The first, I think. But I really do not know what I am thinking of now. > I think I need a good lie-down. Yes, a good idea. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== Can a formula be found for the numbers on the right given the value on the left? 4 2 8 4 9 2 12 6 16 12 18 4 20 10 24 12 25 4 27 6 28 14 32 24 36 30 40 20 44 22 45 10 48 36 49 6 50 8 Some simple observations: 8:4, 12:6, 20:10, 24:12, 28:14, 40:20, 44:22 follow 4:2 but 16:12, 32:24, 36:30 and 48:36 do not. 18:4, 27:6, 45:10 follow 9:2 but 36:30 does not 16:12 and 32:24 and 48:36 seem to form a set. What about 64:48? 9:2, 25:4 and 49:6 have value of sqrt(n)-1 since their square roots are primes. Can anything definite be stated? ==== > Can a formula be found for the numbers on the right given the value on the left? 4 2 > 8 4 > 9 2 > 12 6 > 16 12 > 18 4 > 20 10 > 24 12 > 25 4 > 27 6 > 28 14 > 32 24 > 36 30 > 40 20 > 44 22 > 45 10 > 48 36 > 49 6 > 50 8 Yes, of course. You can find a polynomial of at least degree 18 that would work. However I do not think that is what you want. You need to state the question more clearly. Some simple observations: 8:4, 12:6, 20:10, 24:12, 28:14, 40:20, 44:22 follow 4:2 but 16:12, 32:24, 36:30 and 48:36 do not. 18:4, 27:6, 45:10 follow 9:2 but 36:30 does not 16:12 and 32:24 and 48:36 seem to form a set. What about 64:48? 9:2, 25:4 and 49:6 have value of sqrt(n)-1 since their square roots are primes. Can anything definite be stated? > ==== >Can a formula be found for the numbers on the right given the value on the left? 4 2 >8 4 >9 2 >12 6 >16 12 >18 4 >20 10 >24 12 >25 4 >27 6 >28 14 >32 24 >36 30 >40 20 >44 22 >45 10 >48 36 >49 6 >50 8 Without more knowledge, the answer is yes, but not in any useful way. For example, since you have given 19 points, one may use Lagrange interpolation to find a polynomial f(x) of degree 18 that will give you f(4)=2, f(8)=4, f(9)=2, etc. Have the pairs (x_1,y_1) = (4,2) (x_2,y_2) = (8,4) (x_3,y_3) = (9,2) . . . (x_19,y_19) = (50,8) and define f(x) as the sum of P_i(x), i=1,...,19, where P_i(x) = y_i* (product from k=1 to k=n, skipping k=i)[ (x-x_k)/(x_i-x_k)]. But while that will give you the answers you already have, it is unlikely to be what you want. (A famous example from Mathematics made difficult, for instance, points out that if you use that procedure to get a polynomial that will correspond to the points 1 2 2 4 3 8 4 16 then you will find that f(5) = 31). >Some simple observations: 8:4, 12:6, 20:10, 24:12, 28:14, 40:20, 44:22 follow 4:2 How do they follow, unless you are assuming some sort of proportionality? There is no reason to assume that if the k-th number is x_k, then then k*l-th number will be x_k*l. So in what sense do they follow? > but 16:12, 32:24, 36:30 and 48:36 do not. 18:4, 27:6, 45:10 follow 9:2 but 36:30 does not 16:12 and 32:24 and 48:36 seem to form a set. What about 64:48? 9:2, 25:4 and 49:6 have value of sqrt(n)-1 since their square roots are primes. Can anything definite be stated? There are an infinite number of functions that will take the values you specify at the points you specify; they vary in terms of their simplicity or obviousness. Without knowing more about the sequence of numbers you have or where they came from, what they are supposed to represent, etc, there is nothing definite that can be stated. Just as we cannot definitely state that the number that goes after 1, 2, 4, 8, 16 is 32. (It is, if your sequence is meant to represent subsequent powers of 2 But if, say, the n-th number gives you the number of areas into which a circle gets dividied if you place n generic points in the circumference and then draw all the lines that connect them, then the next number is 31). -- ====================================================================== 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 ==== P O S T >16 15 19 20 = 70 Oh dear. Shriner/Freemason Don Ocean started calling me a pedophile << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== >Friday April 26th 2002 116/249 16504 R I Z Z U T O >18 9 26 26 21 20 15 = 135 > << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== Friday April 26th 2002 116/249 16504 R I Z Z U T O 18 9 26 26 21 20 15 = 135 I am posting Vito Rizzuto's stats again because Shriner/Freemason Don Ocean and/or his friend James Takayama is repeatedly calling me a pedophile. I have never sexually assaulted anybody, I have never been charged with sexual assault, I have never been arrested for sexual assault, and I am not sexually attracted to children. In 1988 I criticized the phallic worship in the Protestant and Catholic churches and was repeatedly arrested and tortured at the U of S for years. Ruby would have me arrested for failing to kiss her God-damned ass and wear the clothes she was always trying to force upon me (and because Protestants and Catholics were upset with my words and lobbied her to shut me up), Ruby would have me arrested and chemically lobotomized, and then she would come to the psychiatric ward and force her choice of clothes upon me there. Eventually I found myself internet access at the U of S and posted Collecting Mail For The Coming Anti-Christ, in the essay I spoke in defense of people to wear their own choice of clothes, and this included defending the rights of women and girls to go topless if they wanted. Now Don Ocean calls me a pedophile, and as a result of Don Ocean, James Takayama in Hawaii has begun doing the same. First the fellow libels me in the WaxyOrg website using the name of Nospam, then using the name of Thomas (aka MauiCop), James Takayama quotes his own material posted as Nospam. Now another website has begun to quote the libel as being truthful, saying that I have been repeatedly arrested for sexual assault and that I was a pedophile (and it is possible that James Takayama is responsible for this website and may start others where he will claim I have been repeatedly arrested for sexual assault). Now as a result, people in Saskatoon are calling me a pedophile and are threatening to beat me up and kill me. The Saskatoon police are not in the habit of charging people for assault when they give me beatings (in fact they arrested me after James De Witt brutally assaulted me at the Seventh Day Adventist Church, I was brutally assaulted and then the police arrested me and took me to the adequate pain relievers when these people crack and break my ribs. The situation is quite unfair, and now Vito Rizzuto will have his stats posted daily until the situation is resolved. I get tortured year after year and begged people in futility for assistance to flee the country, now watch Vito spend huge piles of cash again this year turning trees into decorated idols, for his compassion is limited to traditions. 203 Nicolo Nicolo 68 Rizzuto 135 225 Libertina Libertina 90 Rizzuto 135 201 Vito 21 2 46 52/313 +4014 Vito 66 Rizzuto 135 177 Maria Maria 42 Rizzuto 135 Mom's first name adds to 90 (66th non-prime), Exodus contains 1213 verses (66+66+66th prime) and terminates at chapter 90 (66th non-prime). Vito adds to 66, mom and the little sister have first names averaging 66. The kids have first names adding together for 108 (the first 6 primes in prime positions). The kids have first names differing in value by the 24 chapters of Bible Books 6 and 10 (6th non-prime). Dad has a 6 lettered first name, there are 24 letters in all the first names, the number of chapters in Books 6 and 10 (6th non-prime). All first names add together for 266. Dad and Vito have first names adding together for 134, corresponding to Numbers 17 with 13 verses (the 6th prime). Mom and Vito have first names averaging 78 (6 times the 6th prime). Dad and Vito have full names adding together for the 404 verses of Bible Book 66, Revelation. Vito was born on the 21st, corresponding to Ecclesiastes (chapters 660 to 671). Vito has consonants adding to 132 (66+66). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 --- 108 Vito Rizzuto (135) was hot for Cammalleni (83), the names differ in value by 52, pretty as Vito was born on the 52nd day of the year. Giovanna (83) Cammalleni (83) was born with names adding to the 23rd prime and to the 23rd prime, together for 46, and she gets a husband that was born in 46. 166 Giovanna 48 Giovanna 83 Cammalleni 83 Vito married Giovanna (83) Cammalleni (83). She was born with 18 (6+6+6) letters adding to 166. Both of her names added to the 83 verses of Second Timothy (the 16th Book of the New Testament). Her names added to 83 and 83, corresponding to Exodus 33 and Exodus 33, together for 66. At birth, Vito and Giovanna had 29 letters in their names, or 6 plus the 6th prime (13) plus the 6th non-prime (13), there are 29 (6+6p+6np) chapters in Bible Book 13 (the 6th prime), there are 29 (6+6p+6np) verses in chapter 666 (Ecclesiastes 7). At birth Vito and Giovanna had 29 letters adding together for 367 (First Chronicles 29). Vito's sister's first name adds to 42 (the 29th non-prime), Bible Book 29 is Joel (42). Now Giovanna's name adds to 218 (twice the 29th prime). Nicolo's name adds to 203 (7x29). Leonardo's name adds to 219, or 3 times the 73 verses of Bible Book 29. Libertina was born in 73 (the length of Book 29 and is the Lucas numbers up to 29). Leonardo and Libertina have first names adding together for 174 (6x29). Dad and the first two kids have first names adding together for 218 (twice the 29th prime). The males were born in years adding to 182 (Deuteronomy 29 with 29 verses). 201 Vito 21 2 46 52/313 +4014 Vito 66 Rizzuto 135 218 Giovanna 48 Giovanna 83 Rizzuto 135 203 Nicolo 67 Nicolo 68 Rizzuto 135 219 Leonardo 69 Leonardo 84 Rizzuto 135 225 Libertina 73 Libertina 90 Rizzuto 135 Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime They have three kids, the first and last of the kids bear Vito's parent's names, so the daughter's first name adds to 90 (66th non-prime). The kids have first names adding together for 242 (First Samuel 6), all names in the family add together for 1066. The kids were born in 67, 69 and 73, these are the 19th prime, 50th non-prime and the 21st prime, together for 90 (66th non-prime and the value of the daughter's first name). Mom and her sons have first names adding together for 235... the 184th prime (1097) and the 184th non-prime (235) averages 666. The brothers have names averaging 211, it is approximately 66.6% of the 66th prime (317) and is the terminating chapter of Bible Book 6. Book 6 chapter 6 (193) plus the terminating chapter of Book 6 (211) adds together for the 404 verses of Bible Book 66. Daughter's first name not only adds to 90 (66th non-prime), but her first name adds to 66.666...% of her last name. This is a family of 5, the 4014 days dad is older than me is the 959 verses of Bible Book 5 short of the 666th prime (4973). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 26 61 27 67 28 71 30 73 32 79 33 83 34 89 35 97 36 101 38 103 39 107 40 109 42 113 44 127 45 131 46 137 48 139 49 149 50 151 51 157 52 163 54 167 55 173 56 179 57 181 58 191 60 193 62 197 63 199 64 211 65 223 <-48th-> 66 227 68 229 69 233 70 239 72 241 74 251 75 257 76 263 77 269 78 271 80 277 81 281 82 283 84 293 85 307 86 311 87 313 88 317 <-66th-> 90 Vito adds to 66 (48th non-prime). Vito's first name adds to 48.888...% of his last name, God gives him a wife that was born in 48. Vito and his wife have first names adding together for the 149 verses of Bible Book 48, Galatians. Vito's 201 valued name exceeds his 52nd day of birth by the 149 verses of Bible Book 48. Rizzuto (135) was hot for Cammalleni (83), the 83 value of mom's maiden name is 61.48% of the 135 value of Rizzuto. The males were born in years adding to 182, the females in years adding to 121 (66.48%). The kids are missing 15 letters from their first names, these missing letters add to 248. The 4014 days dad is older than me is 18 times the 48th prime (223), prettier as I was born on the 48th day of the year and with 317 days remaining in the year (66th prime). 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 123 <-Numbers 6, it is three times the 13th prime (41+41+41), keeping in mind that 13 is the 6th prime 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) Vito is 4014 days older than me, his name adds to 201 (Joshua 14), his mom and daughter share the same 225 valued name (Judges 14). The kids have odd valued letters in their first names adding together for 140. The letters that are neither prime nor square in the kids' given names add to 196 (14x14). Vito and his daughter have names differing in value by 14. All first names add to 391 (314th non-prime). Dad's name adds to 67+67+67. The first of the kids was born in 67 and has names differing in value by 67. These first three family members have first names adding to 66, 83 and 68, corresponding to Exodus 16, 33 and 18, together for 67. This 67 is the 19th prime, the second kid was born in 69 (Exodus 19) and has a name adding to 219. The brothers were born in years adding to 136 (Numbers 19). Vito and the last of the kids were born in years adding to 119. Perhaps mom was 19 years old when she gave birth in 67 (the 19th prime). Mom's name adds to twice 109 (Leviticus 19) while dad's name adds to 201 (3 times the 19th prime). The parents have names adding together for 419 (Nehemiah 6 with 19 verses). Vito and his kids were born in years adding to 255 (First Samuel 19). Vito and his kids have first names averaging 77 (the primes up to 19). The kids were born in years adding to 5909 (19x311). Libertina's last name adds to 150% of her first name (150 chapters in Bible Book 19). Vito was born a multiple of 19 days into the century (16853=19x887). Vito's vowels add to 69 (Exodus 19). The 2460 verses of Bible Book 19 is 19x19+19x19+19x19+19x19+19x19+19x19+19x19 minus 67 (the 19th prime). Primes Non-Primes 2 1 3 4 5 6 7 <-4th-> 8 <-Bible Book 8 contains 11 4 chapters, pretty as 13 8 is the 4th non-prime 17 19 -- 77 <-the first 8 primes plus 8 more adds to the 85 verses of Bible Book 8 Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 18 61 19 67 <- 67 <-the 8th prime in prime position R U T H <-Book 8 18 21 20 8 = 67 Vito's names differ in value by 69 and his second kid was born in 69. Vito was born on the 21st and his third kid was born in 73 (21st prime). Vito was born on the 21st and his name adds to 201, and he was likely 21 years old when the first of the kids was born. Vito was born in 46, his 201 valued name exceeds it's 155th non-prime position by 46. The kids have prime and square valued letters in their first names adding together for 46. Vito was born on day 52, it's the number of chapters in Bible Book 24, while his daughter gets a name that exceeds his by 24. Vito and I were together born 530 days closer to the beginning of our years than to the end of our years (Psalm 52). This 530 is a combination of the first three perfect numbers (6, 28 and 496), they have factors that add to form themselves: Perfects 6 - 1, 2, 3 28 - 1, 2, 4, 7, 14 496 - 1, 2, 4, 8, 16, 31, 62, 124, 248 The males have first names adding together for 218, and mom's name adds to 218. Mom was born with 18 letters and took a last name that begins with the 18th letter of the alphabet, it is a last name that adds to 135 (Numbers 18). Vito and his sister have first names adding together for 108 (Leviticus 18). Rizzuto adds to 135 (the 103rd non-prime), the 11 different letters utilized in the construction of the first names for the kids add together for 103. 3x3x3x3x3x3 3x3x3 103 <- the 3x3x3rd prime ----------- 859 <- the number of verses in Bible Book 3 The parents were born in 46 and 48, Bible Books 46 and 48 contain 437 and 149 verses. These Books contain an average of 293 verses while the 437 is 293.28% of the 149. The 437 and 149 are the 353rd non-prime and 35th prime, and so it is pretty that there would be 35 letters in the family first names, and pretty that the last name would add to 135. Mom and her sons have first names adding together for 235. The brothers have first names adding together for 152 (Numbers 35). Libertina Kabatoff adds to 152 (Numbers 35), prettier as 293 is the 62nd prime while Kabatoff adds to 62. Rizzuto adds to 135, or 57 plus the 57th non-prime (78). The main Books of end-times prophecy are Daniel with 357 verses and Revelation with 404 verses, or 57 plus the 57th prime (269) plus the 57th non-prime (78). Primes Non-Primes 2 1 3 4 5 6 7 8 11 <-5th-> 9 -- -- 28 28 Vito and his sister Maria have first names adding to 66 and 42, together these Bible Books contain 1555 verses. Vito has 11 letters and a first name adding to a multiple of 11, his kids have first names adding together for 11x11+11x11 (11 is the 5th prime). Then Vito marries Giovanna (83) Cammalleni (83), there are 83 verses in Bible Book 55, and she soon finds herself in a family of 5. The 5th of 5 family members has a name adding to 225, prettier as 25 is not only 5x5 but is 5 plus the 5th prime (11) plus the 5th non-prime (9), keeping in mind that Bible Book 5 contains 959 verses. Perhaps mom was 25 (5x5 or 5+5p+5np) years old when she gave birth to the 5th of 5 family members. Libertina's first name adds to a multiple of 5 and also a multiple of 9 (5th non-prime). Libertina's (the 5th of 5 family members) first 5 letters add together for dad's 46th year of birth. My name adds to 187 (the terminating chapter of Bible Book 5), it is 5x5x5 plus twice the 5th prime in prime position (31). If Libertina took my 62 (twice the 5th prime in prime position) valued last name, then her names would have an average value of 76 (55th non-prime). Lamentations is Bible Book 25 with 154 verses, the 154th prime is 887 while Vito was born on the 19x887th day of the century, pretty because if Libertina married me then he would be lamenting. Vito and his kids already have first names adding together for 308, or twice the 154 verses of Lamentations. Note that Bible Books 5 and 5x5 differ in length by 805 verses (the 666th non-prime). And Old Testament Book 9 (the 5th non-prime) and New Testament Book 9 (the 5th non-prime) together contain 959 verses (the number of verses of Book 5). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 Vito (66) was born on day 52, it is an average of 59 (the 17th and his sister have names averaging 189 (the first 17 primes minus the first 17 non-primes). Vito and Giovanna have names differing in value by 17 (7th prime, the primes up to 7 add to 17). Giovanna's name adds to 218 (Book 7 chapter 7). Vito and Giovanna were born with last names adding together for 218 (Book 7 chapter 7). The first of the kids gets a first name adding to a multiple of 17 and he was born in 67 (Exodus 17). The second gets a first name adding to 7 times the 7th non-prime (12). The brothers have first names adding to 68 and 84 (a span of 17). The daughter gets a name adding to 225 (Book 7 chapter 7+7). The kids were born in 67, 69 and 73, corresponding to Exodus 17, 19 and 23, together for 59 (the 17th prime). Libertina's names differ in value by 45 (45 chapters contain the length of 17 verses), she might take my name and end up with 17 letters. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 I wanted 17 French fry girls at my weddink, they would throw French fries high up into the air, the french fries would fall to the ground and get dirty, and nobody would ever eat french fries again. And I wanted 59 big nosed Greeks with their big noses throwing hamburglers at us from across the street. And I wanted American Noel Nibblett blowing his bugle and leading a marching regiment of cadets up and down the street while American Don Ocean and his Shriner friends ride circles around them on their little motorscooters. Scientists have recently discovered that tomatoes contain properties that help to prevent prostate cancer, in men, and since A&W allows one to take as much ketchup as they wanted, the Great A&W Rootbear was on the fast track to becoming an international symbol of health and fertility... so of course I wanted the Great A&W Rootbear to be the best man at my weddink. My weddink was soon approaching and it was goink to be a glorious affair, probababbly. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 207 Libertina 73 Libertina 90 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doing this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. You people spent millions of dollars having me tortured, and then annually you spend billions on your decorated trees, I begged and begged for assistance to flee the country (they tortured me for years at the U of S) and you people are so cheap that you can't even offer to buy me a cookie when I bust my ass to show you evidence that your very name is a gift from God!!! Should Libertina marry me, great, but if Marcia marries me and then Libertina marries Marcia and me, then Nicolo and Leonardo are both goink to win themselves a shiny new Cadillac!!! And if Libertina turns out to be some sort of sexual acrobat then Vito and Giovanna are both goink to win themselves a shiny new Cadillac too. Or if Melinda Jarocki sleeps with me then Vito will win himself a used Pontiac. I have Scripture to support taking seven brides (Isaiah 4:1) and I have Scripture to support sleeping with Melinda Jarocki outside of wedlock (First Kings 1:1-5), while you people have a vast multitude of Scriptures condemning your decorated trees, phallic-capped churches and your violence against me for daring to point out your pagan traditions. Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... Note that every day Freemason Don Ocean (or his friend James Takayama) libels me and calls me a pedophile (or such) on the usenet, then I will repost the stats for the following people: Daniel Bruner, Angel Cadwell, Brittanie Cecil, Carl Koopang, Jay Larson, Brian Lott, Adam Millikan, Cody Milliken, Christopher Ridsdale, Vito Rizzuto, Melissa Schultz, Erin Sorenson, Ann Wigdahl, Michael Winkler, or maybe more (more or less). ==== >Friday April 26th 2002 116/249 16504 R I Z Z U T O >18 9 26 26 21 20 15 = 135 > << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== P O S T >16 15 19 20 = 70 Oh dear. Shriner/Freemason Don Ocean started calling me a pedophile << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== I could use some helpful hints on this problem Let G be a group which operates transitively on a set S. Let H be the stabilizer of s-sub 0 in S. Let g(s-sub 1, s-sub2) = (gs1, gs2) be the action of G on S x S. Show the bijection between double cosets of H in G and G-orbits of S x S. I got this far with the problem. I want to show O: G= union from i = 1 to n of H gi H .88G-orbits. I know H = Stab(s0) = {g in S | gs0 = s0}, and also that there exist g s/t gs1 = s2 for all s1, s2 in S. But I don't understand what the orbits look like, ie I don' t understand how the elements of SxS fall into the G -orbits. If I did, I think I would be able to find representatives from each of the orbits to correspond to each HgiH, which would prove the bijection. Steven ==== > I could use some helpful hints on this problem > Let G be a group which operates transitively on a set S. Let H be the > stabilizer of s-sub 0 in S. Let g(s-sub 1, s-sub2) = (gs1, gs2) be the > action of G on S x S. Show the bijection between double cosets of H in G and > G-orbits of S x S. I got this far with the problem. I want to show O: G= union from i = 1 to n of H gi H àG-orbits. I know H = > Stab(s0) = {g in S | gs0 = s0}, and also that there exist g s/t gs1 = s2 for > all s1, s2 in S. But I don't understand what the orbits look like, ie I don' > t understand how the elements of SxS fall into the G -orbits. If I did, I > think I would be able to find representatives from each of the orbits to > correspond to each HgiH, which would prove the bijection. Steven First, let's take a particular example. Consider G= the group of integers under additions, which we will call group-Z. For S, let us take the set of integers, set-Z. For the action, for m in group-Z and k in set-Z, let m act on k to give the element m+k in set-Z under normal addition. Thus, group-Z acts on set-Z by translations (I know it is cumbersome, but I want to emphasize the fact that we are using Z is two distinct ways). Now SxS is pairs of elements (k,l), where both k and l are in set-Z. For m in group-Z, m acts on (k,l) producting (m+k, m+l). Ok, what are the orbits in SxS? There is another hint several spacings down, but I am adding spacing so that you might think about matters at this point before proceeding to the next hint. Ok, still with me. Here is another hint. First, can we find representatives for all orbits in SxS. Remember, we need to involve s 0. I had to think about this for a bit, but then I relazed that all elements form (s 0, alpha s 0) are representatives of all orbits, were alpha is an element of G. That is, any element of SxS can be written as g (s 0, alpha s 0) = (g s 0, g alpha s 0). Ok, one more push in the right direction below, but take a moment to think about this. Ok, so presumably we know that all orbits can be represented by elements of the form (s 0, alpha s 0), where s 0 is the element selected at the beginning of this discussing, and alpha ranges over all elements in G. The question is, when do (s 0, alpha s 0) and (s 0, beta s 0) represent the same orbit. Well, for them to represent the same orbit, they must be in the same orbit (right? in the discussion above, the elements 'represented' the orbits by being an element of the orbit, so that all other elements of the orbits could be obtained by applying the group action). Now, for these to be in the same orbit, this requires that there exists g in G s.t. g s 0 = s 0 and simultaneously g alpha s 0 = beta s 0. Ok, can you do some algebra to learn something about g? Best wishes, Mike ==== can someone help me with these practice problems? 1. Let f:X-->D for some connected metric space X and some discrete space D. Show f is continuous iff f is constant. 2. Let f:X-->R for some metric space X. If f is continuous, then |f| is continuous. On the first one I got one direction (namely <==) which was very easy. I might just need a slight nudge in the right direction on each, these FSB ==== grava .88 la saucisse et au marteau: > can someone help me with these practice problems? > > 1. Let f:X-->D for some connected metric space X and some discrete > space D. Show f is continuous iff f is constant. > 2. Let f:X-->R for some metric space X. If f is continuous, then |f| > is continuous. For the other direction, use the Intermediate value theorem (just hoping this is the way it is called in English) 2. I'd say that on a neighborhood of any point where the fonction is nonzero, the function is of a constant sign. If there is a such that f(a) = 0, you can divide the neighbourhood in two parts. -- Nicolas ==== [...] |> It's hard to do the kind of reality-testing that Pat referred to when |> you don't have people you let help you. | |I get help from mathematicians, much of which I don't post about, |though I have posted about some, but consistently when it comes time |to deliver, they run away. Is any of them in a position to help you test the basic issue, whether you have great mathematical discoveries, or just some amateur-level experimentation? Not as far as I can see. If they said the latter, would you actually believe them? I can't think of any way to test the quality of your work, that both you and most other people would be willing to accept the results of. Wouldn't it help a lot if there was such a way? [...] |> They don't consider you to be a world class mathematician. For a lot |> of mathematicians, the point they start trying to figure out how to |> get you to leave them alone is when they realize that you *think* of |> yourself as a world class mathematician even though you're not. | |Why would they think that? | |Basically I do a basic presentation, often I do mention that I'm an |independent researcher. | |What makes you think that they would begin thinking I'm someone who |thinks of himself as a world class mathematician? I'll get to that shortly. |Usually I mention that I'm *not* a mathematician, and am in need of |help or guidance. By mathematician I didn't mean professional mathematician in the sense of someone who gets paid to do mathematics. Obviously you don't think you're getting paid to do your stuff, yet. But you think what you're doing is worth getting paid for. |What I find troubling is how *easily* you make statements as if you |know!!! I likewise. You make generalizations about professional mathematicians. I am just describing what is common. Your idea is as far as I can see based mainly on speculations about what would cause people to treat great work as if it were nothing extraordinary. Mine is based on experience with how professionals actually deal with amateurs. For your version to be true, they would have secretly to realize you were onto something big. For mine to be true, they just have to realize that it is much like other things that they get on a routine basis. |Please tell the newsgroups if you've talked to Barry Mazur, Andrew |Granville or any of the others to get your assessment. I haven't written or talked to Mazur or Granville about you specifically. although I don't remember getting into his reasons for not feeling like opening up further dialog with you. I am aware of Granville's dealings with one fellow (of my acquaintance, although not by choice) who repeatedly and mistakenly thought he'd proven case 1 of Fermat's Last Theorem. I was told he had published papers (although not claiming to prove case 1). This seemed to be a little bit in his favor. The first time he thought he proved case 1, I think there was still some hope among people acquainted with him that he would turn out just to have gotten overexcited temporarily, but he kept slipping into the same kind of error. So I do know something about what Granville thinks of as promising and not so promising. I would tend to expect Mazur to be more tolerant than the average professional for a little eccentricity, and to be willing to take time for random strangers. On the other hand he is adept at politely wrapping things up so he can move along to the next thing he wants to do. He does a variety of interesting things. (Have a look at his popular book about visualizing numbers. I think it shows more than minimal interest in the general public.) he had one of the greatest mathematical discoveries of all time (and didn't), but I think he would know enough not to make the kinds of mistakes I've made in doing so, like arguing with you on points where you evidently have made up your mind definitely. find this to be a common occurrence, from experience with other examples of amateurs sending in their work for comment. Some professionals don't bother with amateurs in any way. A few will correspond with amateurs who they know have the idea that they've squared the circle or something like that. A lot, though, draw the line between the two, trying to maintain friendly relations with the general public in some way, but avoiding arguing with the amateurs who are going to think the professionals are prejudiced if they give their honest opinion. Obviously in individual cases they might correspond with some amateur for a bit and then stop quickly for some other reason, but finding that the amateur thinks of himself as the next Gauss (or as having discovered the next fundamental theorem of arithmetic, as it were), and just has a little trouble satisfying bureaucratic requirements like writing up his proof to publication standards, or is *only* having trouble because all the other mathematicians are prejudiced against him, often ends the conversation very soon. You make it fairly obvious that you think of yourself as having world-class mathematical abilities and accomplishments. Here's just one example: |I'm afraid of having made one of the greatest discoveries in the history of |mathematics and having to live with the consequences. One of the greatest discoveries in the history of mathematics! We've seen this kind of posting from you many times. What's more, we've seen this kind of pattern of writing over and over from others. There's a pretty clean divide between the amateurs who are amenable to feedback to a normal degree, and are aware of being at an amateur level, and the amateurs who think they are well above the people they're writing to in mathematical ability, and explain the feedback they get as the product of small, fearful, or envious minds. Dealing with the former can be a lot of fun, and dealing with the latter tends mainly to convince the amateur that one is yet another person who fails to appreciate failed attempt to prove Fermat for even exponents only, this is cute. When U.S., he is unlikely to get a reply. Do you think you succeed in passing yourself off as just a humble ordinary amateur with an interest in algebraic number theory? I don't doubt you can do this for awhile. But I doubt you ever keep it up for long. The drama of thinking of yourself as a historic figure in the history of mathematics is just a *little* hard to conceal. [...] |Also I've received replies from other notables and famous people, |which are not relevant, so I don't discuss them. Ok. |> But this is also not so unusual. Nearly any educated person who bothers |> to write to people like this, and is reasonably patient, sooner or later |> gets a response, and not just a form letter. Not all famous people ever |> answer, but enough do that just the fact you got to correspond with them |> a little doesn't mean very much. | |I never said it did. Well, okay, but why did you write the following, then? | I don't know how one sets about testing a standpoint like his | empirically. So much of it hinges on his evaluation of his own | work as being valid and valuable mathematics. As far as I can | tell, the world continuing to act just like he wasn't a world class | mathematician but just a guy who thinks he is, is consistent | with his beliefs. He just concludes his work is great but people | are choosing to pretend it isn't. You replied: |Come on Keith Ramsay, I've sent and received interesting replies from |Andrew Granville, Barry Mazur, and others. [...] So what are you claiming is the significance of it? Come on what? What's the conclusion I'm supposed to draw? You made it sound like you thought the fact they corresponded a little with you served as some kind of rebuttal to what I was saying. What they're doing is not what they do with work of world class quality. What they're doing is what they do either with ordinary interested amateurs, or with amateurs like circle squarers. There's nothing wrong with being an amateur, of course, but if you imagine that you've on the level of John Nash, mathematically, when you're closer to Russell Crowe[*], you're bound to be disappointed one way or another. [* I have no idea how much math he knows, but I assume it's somewhere in the bottom 99.9% of the population.] [...] |You're leaving out the fact that I questioned whether or not Iraq had |weapons of mass destruction back in November of last year. Well, IMO that's a point in your personal favor, no sarcasm intended, but that is of course a discussion for another newsgroup. Keith Ramsay ==== |> So you want the credibility of sources to be *more* dependent |> on their credentials and occupation than they are now? | |Erik Max Francis is a computer programmer. | |Trying to defend him as if he were a mathematician or otherwise an |expert over the fields he covers is just another example of math |society's rather sad ability to make up rules as it goes along. Not at all. I said what I meant to say. I asked a question. I haven't seen an answer to it. I don't know what Erik Max Francis has on his web site, or whether it's valid or not. I just think Tom Potter has chosen a peculiar way to criticize it. Questioning this criticism is not special treatment I reserve for mathematicians and experts. I would think you, of all people, would have a problem with the notion that in order to judge whether he, you, me, Tom Potter, or anyone else is credible, it's ever enough just to check their job titles and levels of education. On the one hand, have many times have you complained about how people will look at credentials and positions, and conclude that mathematicians who have them should be believed just because of them? Should we say, Keith Ramsay has a PhD in algebraic number theory, so whatever he says about algebraic number theory must be correct? Well, I would think it would lend *some* authority to it. But I think people should remember that the word of authorities is not the bottom line. Surely you agree with that? You wouldn't want them to start believing bullshit just because someone with a PhD said it, would you? On the other hand, should we say things like, James Harris does not have a PhD, and his BA is not in mathematics, and he does [fill in the blank] for a living, so we shouldn't listen to what he says about algebraic numbers? I don't think so. Don't you agree that this is would be a bit unfair? Maybe it's less likely that someone with less education knows what they're talking about, but one shouldn't assume that they don't. So I'll ask you too: do you really want for the credibility of sources to depend more on credentials and occupation than they do now? Keith Ramsay ==== |> [...] |> |> What third rate California college? Who rated it? What criteria? |> |> |> |Hey Wormley, |> |as you use this programmer's web site as your primary rederence, |> |it seems to me that you should know what college your resident expert |> |attended. |> |> So you want the credibility of sources to be *more* dependent |> on their credentials and occupation than they are now? | |It is interesting to see that Keith Ramsay |agrees with Sam Wormly, |that if someone puts up a web site that demeans folks |who have made a statement, or statements, |that run counter to conventional wisdom, |that such a person qualifies as an expert in the field, |even though they have not demonstrated their competence |in any way. Don't put words in my mouth. I said what I meant to say. You've been making heavy weathering of someone's level of education and occupation. Do you really think these should be used *more heavily* to judge the credibility of people's web pages? I just think this is a somewhat strange basis for criticizing. If you have some better reason to doubt whatever it is he claims, then why treat where he went to college as such a big deal? [...] |I assert that having (Or being ) an asshole |does not qualify one as an expert. | |Apparently Keith Ramsay and Sam Wormly feel otherwise. Again, don't put words in my mouth. Keith Ramsay ==== |The key to being a crank is that you can't know *nothing*...you have |to know *something*. For example, you know that Mandarin literature |comes from China, you might well know that the Mandarin's were the |bureaucratic class, and you might know that Confucius was highly |revered especially by the Mandarins. | |I think that small bit of knowledge (which just about exhausts mine, I |think) is plenty to be a crank about Mandarin literature. ;) I can see it now.... Chinese communist Mandarins are subverting freedom-loving Americans with their pagan/Maoist/Confucian doctrines of duty and respect for authority.... A Google search for communist mandarins turns up 22 hits, so there you go. Keith Ramsay ==== > As a sidenote, I've not said that my proofs are irrefutable proofs, as > I've said they are proofs. At other times I've said that I have > proofs, and that since they are proofs they are irrefutable. Ah, but they aren't proofs at all. ==== > nice post James, good general FAQ or introductory material about posting. Herc > now will anyone look at MY claim? Just look at the supporters you attract, James. Doesn't that tell you something? > Webmasters help the TRUEman by joining www.theBanner.net > Current:1 Goal:1000 What can I say? ==== WARNING ALERT At New York's Kennedy airport today, an individual later discovered to be a public school teacher was arrested trying to board a flight while in possession of a ruler, a protractor, a setsquare, a slide rule, and a calculator. At a morning press conference, Attorney general John Ashcroft said he believes the man is a member of the notorious al-gebra movement. He is being charged by the FBI with carrying weapons of math instruction. Al-gebra is a fearsome cult, Ashcroft said. They desire average solutions by means and extremes, and sometimes go off on tangents in a search of absolute value. They use secret codenames likex and y and refer to themselves as unknowns, but we have determined they belong to a common denominator of the axis of medieval with coordinates in every country. As the Greek philanderer Isosceles used to say, there are 3 sides to every triangle, Ashcroft declared. When asked to comment on the arrest, President Bush said, If God had wanted us to have better weapons of math instruction, He would have given us more fingers and toes. I am gratified that our government has given us a sine that it is intent on protracting us from these math-dogs who are willing to disintegrate us with calculus disregard. Murky statisticians love to inflict plane on every sphere of influence, the President said, adding: Under the circumferences, we must differentiate their root, make our point, and draw the line. President Bush warned, These weapons of math instruction have the potential to decimal everything in their math on a scale never before seen unless we facts of vertex. Attorney General Ashcroft said, As our Great Leader would say,read my ellipse. Here is one principle he is uncertainty of: though they continue to multiply, their days are numbered as the hypotenuse tightens around their necks. ==== Would somebody please explain the notation ___ >lim f(x) >x->c lim f(x) >~~~ >x->c These are presumably the lim sup and the lim inf. The most intuitive explanation for what the lim sup is is > probably this: You know that lim_{x->c} f(x) = L if and only if f(x_n) -> L > for every sequence x_n such that x_n <> c but x_n -> c. > The lim sup of f(x) as x -> c is the _largest_ lim f(x_n) > where x_n -> c (and x_n <> c), while the lim inf is the > smallest such lim f(x_n). (So lim_{x->c} exists if > and only if the lim sup equals the lim inf; if and only > if there is only one such lim f(x_n).) Alternately, you can define lim sup_{x->c} f(x) = lim_{d ->0+} sup{f(t) : 0 < |t - c| < d}. ==== >Would somebody please explain the notation >___ >lim f(x) >x->c lim f(x) >~~~ >x->c These are presumably the lim sup and the lim inf. The most intuitive explanation for what the lim sup is is > probably this: You know that lim_{x->c} f(x) = L if and only if f(x_n) -> L > for every sequence x_n such that x_n <> c but x_n -> c. > The lim sup of f(x) as x -> c is the _largest_ lim f(x_n) > where x_n -> c (and x_n <> c), while the lim inf is the > smallest such lim f(x_n). (So lim_{x->c} exists if > and only if the lim sup equals the lim inf; if and only > if there is only one such lim f(x_n).) Alternately, you can define > lim sup_{x->c} f(x) = lim_{d ->0+} sup{f(t) : 0 < |t - c| < d}. The reason these things are useful is that _any_ function > has a lim sup and a lim inf at _every_ point (if we allow > plus and minus infinity as a value), while not every function > has a limit at every point. > Seems same discussion can be taylored for limsup's of sequences. ==== Anyone read that book I mentioned in the subject? I am reading it and have gotten up to about page 50. It is my first time doing knot theory and it seems thus far that the exercises in that book are nasty. For example in the first chapter, called Introduction, there is Exercise 1.9 which asks to show that the knot in the figure is composite. Wow, I don't think I can even copy that scribbled knot accurately let alone show such a proof. How are people handling this exercise. In fact I have been unable to do any of the exercises so far in the book. Is this book known to have unrealistic exercises considering the books title? Even the second question in the book asks to show that there are no two-crossing nontrivial knots. Seems kind of advanced for a second question in the book? Your assistance in setting me in the right direction to solving some of these probs would be appreciated. thanks. ==== I read a bit about this group and wonder if there is a natural way to count in it. In case you dont know what it is: Let X be the set of all positive integers and let S be the group of all permutations of X; let F be the subgroup of S consisting of all permutations that move only finitely many elements of X; the Infinite Alternating group is the subgroup of F generated by all 3-cycles the main thing I'm wondering is if there's a natural and intuitive way to count in it, complete with a natural first element and a natural next for each previous........ or if one is forced to use contrived methods, like the set of rationals. Please forgive my ignorance, as I am far inferior to you in mathematical knowledge. I assure you that I am your humble servant Sniz Pilbor a) How many digits are there in 1000! (generalize, how many digits in n!) b) What is the digit in the kth position of 1000!. for any k. (generalize, same question but for n! instead of 1000!) thanks! ==== I'll try to help you with your first question. > a) How many digits are there in 1000! (generalize, how many digits in n!) In 1000!, there are 2568 decimal digits. In general the number of decimal digits in n! is just ceiling( log_10(n!) ), where log_10 denotes the decimal (i.e., base 10) logarithm. Is that OK, or do you want an expression which avoids computing n! (or a gamma function) perhaps? (If the latter, it might be tricky to get the expression exactly right.) David > b) What is the digit in the kth position of 1000!. for any k. > (generalize, same question but for n! instead of 1000!) ==== Given, where x is in the ring of algebraic integers, I've shown the factorization (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) so when x=0, a_1(0) = a_2(0) = b_3(0) = 0. I'm curious about the mental processes that allow *some* of you to claim that 49 divides off as a *variable* dependent on x, so I'm giving another opportunity for you to speak your minds. To my knowledge, in the history of mathematics, no one has ever presented such a proposition, so it is a unique one, and I must say that I'm intrigued. Speak your minds. James Harris ==== bcc bcc PZ: If Yilmaz et al. are right that static weak field n-body Einstein GR solutions don't exist, then this may turn out to be a real problem after all. JS: I think the exponential metric in the isotropic radial coordinate has serious problems independent of the precise metric model, Yilmaz or Einstein because of topological reasons. PZ: So if you are right you don't get full coordinate generality with an exponential metric? JS: I do not understand coordinate generality. I never saw that term before. This is in use by the Wolfson group at Oxford. It's used as an alternative to the more loaded general covariance which is traditionally tangled up with general relativity. The erroneous conflation of formal covariance with physical relativity is almost a founding principle of Einsteinan physics (later abandoned by Einstein). There are several legitimate issues here. Why don't you make a dictionary of these key terms and give best definitions you can to avoid confusion. That book I mentioned last time is very useful in this regard BTW. More on that anon. My debate with Hal on PV is really in relation to the the pseudo-group of passive LOCAL coordinate transformations at a fixed point P and of world crystal. This is not same as active diffeomorphisms P -> P' =/= P . The active and passive transformations must be made mutually consistent and this may solve the Kretchmann issue? Also there is the issue of whether or not the different points P are distinguishable and what is an observable in GR? There is the Einstein hole problem. BTW Joy Christian is a male. Active diffeomorphism invariants are NONLOCAL - a problem in interpretation. There is no consensus on these deep issues and others among the Pundits. It's almost as bad as the wars over the interpretation of quantum theory. Do you mean general coordinate transformations GCT? Be more precise if you use plain English. The standard terminology is loaded. When I say coordinate generality, I mean coordinate generality: the *desideratum* that laws should be formulated in such a manner that their form does not depend on the particular choice of coordinate system. Hence the use of coordinate- free devices such as tensors It is not clear to me why a modern theory of gravitation *must* be formulated in such a manner -- other than as an expression of physical general relativity of motion, which I contend does not exist. If GCT is what you mean the answer is NO. What I mean is that the manifold looks pathological and unphysical with at least a countable infinity of coordinate patches outside the turning point r* = GM/c^2 for curvature radial coordinate, which is analogous to event horizon in Einstein's GR where there are only TWO patches outside r* = 2GM/c^2 in that case (Einstein-Rosen Bridge, i.e. non-traversable wormhole in non-exotic vacuum case Ruv = 0 everywhere-when. It is clear to me that Hal is not really thinking about the topology and differential geometry in his naive engineering approach. OK, so you are saying that there is an unavoidable pathological discontinuity in the exponential PV solution for a point mass? If by point mass you mean taking the vacuum solution to the max yes. In the case of GR I mean vacuum all the way i.e. solutions of Ruv = 0 with wormhole global topology of source Mass without mass (JA Wheeler). And for PV what would correspond to that. I do not think Dicke knew the differential geometry when he introduced the exponential metric ~ 1961? I wasn't even aware that a manifold was defined in PV. A physical metric, yes; but a manifold? This is not a curved spacetime theory as far as I am aware. The model is a polarizable vacuum with physical rubber rods and clocks. The problem is that Hal is completely obscure to my mind on the fundamental world view of his model. He uses metric notation after all? PZ: Although in the alternative paradigm, general covariance looks more like a mathematical fetish, since physical general relativity is absent. JS: There is a lot about all this in Physics Meets Philosophy at the Planck Scale Callender & Huggett Cambridge Press 2001. I suggest that we temporarily cease this line of inquiry until we both digest what is in that book - some really GOOD STUFF! OK, I've ordered it. But why stop the press? Because those guys are pretty smart and have thought through a lot of the issues you are interested in. So it's time to catch up. The book was written prior to the realization of the new cosmology of dark energy/matter - there is no mention of that. However, it is good background stuff by hip philosophers and some top physicists. Mathematician (in physicist's clothing) John Baez also has a good n-categories and how they may make an interesting formal connection between GR and QM. On the other hand I see a lot of conceptual flaws in the thinking of the Pundits in both Q Gravity and M-Theory, one of which being that they all assume Lp is a constant and not a variable where perhaps Lp*/Lp ~ e^(metric engineering control parameter?) Another is that none of them seem to have read P.W. Anderson's idea More is different and how it applies to quantum measurement problem for example. Nevertheless, there are many good relevant insights in the book. What's any of that got to do with what I am talking about? I am talking about internal tensions within orthodox GR. JS: We are also talking about Hal's PV and also the book does deal also with the internal tensions. Also my focus is how to combine quantum theory and GR in order to solve the important real problems in physics today: 1. What is the Universe made of? 2. What is the physical nature of consciousness? 3. How do we achieve the kind of metric engineering we see in the UFO observations? Studying the internal consistency of this or that theory is secondary to these objectives. Such study may well be necessary however. That seems to be so. I have essentially had my eyes on this Golden Ring for 50 years and I want to get some satisfaction! :-) http://www.findmidis.com/listen.go/589 So does Hal and that is why I am not letting him rest on the issues. PZ: Also, the metric is not the field; the tensor potential phi_uv represents the physical field and the gravitational-inertial metric is derived from it. Non-linear coordinate transformations play a fundamentally different role in this alternative model. I am talking here about Yilmaz. JS: I mean there are an infinity of isotropic coordinate patches outside the turning point boundary at GM/c^2 for a single curvature coordinate. In Einstein's GR this ratio is only 2:1, i.e. 2 coordinate patches outside the event horizon at 2GM/c^2 in the Penrose-Kruskal diagram with 4 coordinate patches covering the entire vacuum manifold. It's still interesting to me that a coordinate discontinuity was originally mistaken for a physical event horizon. Even if you are right that the PV solution is pathological, this does not necessarily apply to Yilmaz's phi_uv. In Yilmaz's theory it is phi_uv that is physically fundamental, while the exponential metric is secondary and derivative. JS: Perhaps. Just what is the Yilmaz theory in your understanding? I mean what is its world view? What is the physical picture behind the obscure formalism? PZ: Yet at the same time I think you get rid of all the tricky properties of event horizons, since you get a smooth solution for a point mass with no lightcone inflection boundaries JS: There seems to be observational evidence of event horizons? I am not up on the latest on this. But I sure get the impression that competent people like Martin Rees are pretty confident on that score? JS: No because you still have the turning point where dr(isotropic)/dr(curvature) has a critical point passing through zero and changing sign. This acts spatially somewhat like an event horizon, i.e. dR ---> infinity at the turning point. dR = [1 - GM/c^2r(isotropic)]^-1dr(curvature) TURNING POINT dT = e^-GM/c^2r(isotropic) dt NO EVENT HORIZON That is, Einstein's event horizon is replaced by turning point in Hal's model. But in any case there is no fundamental reason in PV for insisting that every smooth coordinate system is good. JS: This is not the key point. I am talking about Hal's specific SSS PV model. PZ: Look, once general relativity is out of the picture, dogmatic insistence on general covariance begins to look like a mathematical fetish. I see this as an example of irrationality in contemporary physics. JS: The problem is deeper than that. You cannot throw away differential geometry. The problem is that PV's rules of the game are nebulous and shifting. PZ: Didn't Cartan produce a general covariant version of Newtonian theory? Can't you do all the metric tensor stuff within a purely Newtonian framework? The metric tensor description is a mathematical truism. It doesn't apply only to Einsteinian physics. JS: That's why I brought up the distinction between the local pseudo-group of coordinate transformations at a single P and the active P -> P' =/= P diffeomorphisms. That distinction may be important in posing the relevant question here. There is the issue of the relation of map to territory and even what is the territory? PZ: If you go to the Newtonian limit in GR (if there is such a limit), what do you get? You get a metric tensor description of the Newtonian field with non-vanishing Riemann curvature (since there are still tidal forces). JS: The Newtonian limit of GR has 2 aspects. Most important is c -> infinity so that we have Galilean relativity. Next is that mass density is not so large that radii of curvature are too small relative to scale of the measurement. Obviously the metric interpretation will still hold although it becomes much less rich. PZ: You could let c --> infinity; then you get instantaneous action at a distance. The fact that you re-write the metric tensor for non-linear coordinate transformations is, in and of itself, a mathematical truism. That you can represent an inertial field with such a transformed g_uv is in itself trivial, and I think this can also be done in Newtonian theory as well as in generalized SR. JS: The arbitrary nonlinear coordinate transformation means a dust cloud radars moving in arbitrary non-geodesic paths by firing rockets and communicating with each other on their mutual observations of the same non micro-quantum phenomena (so that Heisenberg incompatibility does not get in the way). PZ: It's just that this suggests a tensor theory of the permanent gravitational field *conceived as a physical field* -- which is exactly what Yilmaz is proposing. JS: There is something very ill-posed in what you are saying but I really can't at this moment get a handle on it. It has an ineffable vagueness to my mind. Again you do not seem to distinguish the two different ideas of the non-tensor g-force at the connection level and tidal curvature as certain tensor combinations of the partial slopes of the g-force connection field. PZ: This is also the approach taken in bimetric theories, which distinguish between the flat background geometry (Minkowski metric) and the *physical* metric field. JS: Physically what is a bimetric theory? How can you observationally tell if the world is mono or bi? There is no dT = 0, but there is dR --> infinity in Hal's model at the turning point. And he has an infinity of branches not just two like Einstein has. It's a mess! I call it the Medusa Manifold. PZ: If you're right I have to admit that this has to be taken into account in evaluating PV. JS: r(curvature) = K^1/2r(isotropic) This comes from angular part of PV metric K(r(isotropic)^2[(dtheta)^2 + sin^2(theta)(dphi)^2] in ds^2 = K^-1(cdt^2) - K(dx^2 + dy^2 + dz^2) IF dx^2 + dy^2 + dz^2 = dr(isotropic)^2 + r(isotropic)^2[(dtheta)^2 + sin^2(theta)(dphi)^2] Here I am using Ibison's argument as I recall it. This leads to K^1/2r(isotropic) = r(curvature) more than a year ago. In the PV model K = e^2GM/c^2r(isotropic) This is in effect an infinite order polynomial for the many roots r(isotropic) corresponding to one value for r(curvature). On the other hand, I think you can argue that you must really use dx^2 + dy^2 + dz^2 = dr(curvature)^2 + r(curvature)^2[(dtheta)^2 + sin^2(theta)(dphi)^2] because 4pir(curvature)^2 is area of concentric sphere by definition. Then Hal has a weird looking spatial metric e^2GM/c^2r(isotropic){dr(curvature)^2 + r(curvature)^2[(dtheta)^2 + sin^2(theta)(dphi)^2]} and the issue is, what is the functional dependence between r(curvature) and r(isotropic). The point is that Hal's metric form is ambiguous and ill-posed. It's like the tar patch in Breire Rabbit. But I think my first argument was basically Ibison's argument of over year or two ago? That is, in any SSS metric theory look at the relation between isotropic and curvature radial coordinates. You get a polynomial of degree N with the curvature radial coordinate as a control parameter. The N roots for isotropic r, at least when real, each define a coordinate patch. In the exponential case that Hal uses in his PV model, N --> infinity. In Einstein's GR N = 2. PZ: OK, this is certainly a legitimate kind of criticism and goes to the technical details, so it cannot be ignored. JS: Hal has ignored it for more than a year. I published all this in Space-Time and Beyond II a year ago. Hal? Hal Puthoff's reply to Zielinski's question was: Jack's (mis)interpretation comes from trying to force fit the PV non-curved-space results into standard GR curved-space modeling, which takes coordinate-choice pathologies (e.g., event horizons) as physical. Ibison went to a great deal of effort to educate Jack on this with a long attachment replete with spacetime graphs, etc., but Jack just blew it off because it didn't agree with his view as not even wrong. JS: Why not educate Misner, Thorne and Wheeler, or Roger Penrose, Stephen Hawking and Martin Rees on why event horizons are not physical? Good luck. ;-) The point is that I know what GR means. I have no idea what PV means from the kinds of vague statements I see above? Hal seems to have a metric without metric theory. He uses a metric formalism, but refuses to use the differential geometry that it comes from. Again, I simply cannot understand in a coherent way what Hal's 1. Physical world view is in PV? 2. What mathematical rules he is using? I mean if you do not inquire deeply into the meaning of Hal's Tables I & II and things like the meaning of r and if you ignore topology and basic metric geometry it looks like GR a little in the large r limit. But that is simply not good enough. It is not even theoretical physics as I understand it. Even if you look at Hal's action formalism, at some point one must ask what is r? I mean when GM/c^2r is no longer a small dimensionless number. HP: So what else is new?! Jack really can quote the dogma (with expertise, I might add, to cut him some slack), but doesn't seem to understand where it comes from, what it represents, how the underlying reality can be modeled from many POVs (as you are showing excellently), etc. Hal JS: I wish Hal would explain the basic assumptions and physical picture of what he thinks he is doing in PV in a way comparable to how Einstein explained what he was doing. Also I wish he would show how his math model leads to any interesting metric engineering applications, which is really his main purpose. For the record, did I correctly remember Ibison's formula r(curvature) = e^GM/c^2r(isotropic)r(isotropic) ? That is, ds^2 = e^-2GM/c^2r(isotropic)(cdt)^2 - e^2GM/c^2r(isotropic)dr(isotropic)^2 + r^2(curvature)[(dtheta)^2 + sin^2(theta)(dphi)^2] Do you accept that? If not, what is your formula in spherical coordinate form? PZ: Why an exponential metric should create such purely mathematical problems is, however, intuitively not obvious (at least to me) -- especially when it seems to solve so many others. JS: Why is not interesting really here. Simple fact is that it's algebraically trivial - 11 grade high school pre-calculus math. On Gravity Lensing The general SSS metric, for simplicity, independent of the action and/or local metrical field equation, is in the curvature radial coordinate ds^2 = gtt(r)(cdt)^2 - grr(r)dr^2 - r^2(dtheta^2 + sin^2thetadphi^2) A light ray obeys, ds^2 = 0 PZ: OK. Null geodesic. JS: Using Einstein's theory dT = gtt(r)^1/2dt dT is what a clock at rest relative to the center of symmetry in the LNIF measures between neighboring events P and P + dP. dR = grr(r)^1/2dr dR is what a radar or a measuring rod measures. where t, r are local coordinates in the rest LNIF that is a point on a timelike non-geodesic. LNIF's can only exist if there are non-gravity forces. PZ: This seems like a red herring to me. Or are you just saying that you need to be able to push a mass off a geodesic, and this push unavoidably involves the mediation of EM forces? JS: That's what I am saying. Why red herring? It seems a useful insight to me that I have not seen made explicit in textbooks? In hindsight it is obvious. PZ: It could be any force. It happens to be EM forces, as the world is presently constituted. What is important on the phenomenological level is that work has to be done to push matter off its natural trajectory. It's just that gravitational influences alter what is natural. JS: No, I mean I have a detailed model why it must be from QED (as the dominant effect). I mean you need EM both to have light cones in the first place in the off timelike geodesics. The latter creates the instability in the Dirac virtual electron-positron vacuum zero point fluctuations to create gravity and dark energy/matter exotic vacuum. PZ: The abstract observer does not need to be physically pushed, so I would think you can define an LNIF (<--> non-linear spacetime coordinate system) in the abstract without reference to such forces. JS: No you cannot. I mean Yes formally, PZ: Yes -- if your formally is my abstract. JS: No physically. PZ: Ah hah! I find it interesting that you make this distinction -- which I say is foreign to classic Einsteinian physics. How do you explain this distinction? What is your physical model? JS: Gedankenexperiments. Einstein was a Master. How do you manage to have an LNIF observer? I mean you can think of it but not make it. PZ: That is precisely the distinction between a coordinate system and a physical frame of reference in my book. This seems like a very un-Einsteinian distinction. JS: Not to me. PZ: It also parallels the difference between general covariance and general relativity. JS: Again maybe that has to do with distinction between nonlocal active diffeomorphisms and pseudo-group of local passive coordinate changes. The EEP has more to do with the latter than the former! Hal's PV Tables vaguely have to do with the former - has has P =/= P' where P' is very far away from P closer to source M. I suppose, there would be nothing that could think if there were no e and no h and only G and c. Note, if Blackett relation is true e = G*^1/2m Then e/m = G(Newton)^1/2e^(metric engineering parameter?) /2 If e^2/hc invariant, then this would control m = rest mass of the lepto-quarks. PZ: OK. JS: BTW we WANT something like at any scale L at FRW cosmic time t impose a Gaussian law of large numbers Eddington wavelet filter: 2hG*(L)/c^3 = Lp^2[1 + (Lp*/Lp)^2 e^-(Lp* - L)^2/Lp^2] where Lp*^2 = Lp^4/3(c/H(t))^2/3 i.e., world hologram idea with Mach's Principle built in. H(t) = R(t),t/R(t) in FRW metric of GR Near Big Bang c/H(t) = Lp Lp^2 = hG(Newton)/c^3 Note that as L/Lp ---> infinity G*(L) ---> G(Newton) Also as L/Lp ---> 1, G*(L) ---> G(Newton) but as L ---> Lp* G*(L) --->c^3Lp*^2/2h >> G(Newton) and at present epoch of Universe Lp* ~ 1 fermi since c/Ho ~ 10^28 cm PZ: Although I suppose when you have physical measurement devices that must physically accelerate through the vacuum, then EM forces will as a matter of fact come into play. But this points to what I regard as a fundamental conceptual problem with the Einsteinian paradigm -- the demand for total coordinate generality on the basis of the concept of spacetime as a void (leading to an arbitrariness in the choice of spacetime coordinates, and a conventionalist approach to the definition of simultaneity, etc.), combined with the evident physicality of accelerating measurement devices through the vacuum. JS: There is much discussion of these sorts of considerations in that book! JS: See the book above. Lots of good discussions on these kinds of issues! PZ: OK, but I think the arguments I have made independently are quite cogent and self-sufficient. JS: Looking at Hal's incomprehensible, to my mind, remark above, cogency is, like beauty and political truth, in the eye of the beholder. ;-) PZ: As far as I can see this contradiction can *only* be resolved by literally identifying gravitational and inertial fields -- and hence Einsteinian strict equivalence. Mach's principle is supposed to refer the accelerated motion of a mass to the average cosmic distribution of matter. But unless you are prepared to accept action at a distance, this itself implies the existence of a physical medium of propagation of the inertial influences (as later admitted by Einstein). Yet once you admit such a medium, then the original basis for Mach's approach is undermined. The snake eats its own tail. And what is the basis for Mach's principle? A Machian empiricism that refuses to attribute physical effects to unobservables (i.e. the physical vacuum) since we cannot see them, and insists that they be attributed to a relationship to astronomically distant masses -- since we *can* see them. A hopelessly naive, internally incoherent, and outdated theory of science. Like the drunk who loses his keys at the door but insists on looking for them down at the lamp post where he can *see*. Absence of evidence is not quite the same as evidence of absence. What this points to IMHO is a basic problem with the entire concept of physical relativity of motion, both special and general. That is why I believe failure of the Einstein equivalence principle (always conjectural in character) could ultimately bring down the entire house of cards (i.e. relativity taken seriously). All it would take is one reproducible experiment. And who can now imagine that inertial and gravitational fields are physically indistinguishable? The effects of accelerating a reference frame propagate through transformed spacetime at the speed of thought (Eddington) -- while time-varying gravitational effects according to Einstein's theory propagate at a *finite speed*. Apples and oranges. JS: Julian Barbour has apparently written books on this subject. Have you read them? PZ: No. JS: I suggest you do. I should too but you are more interested in this than me. I have not. Why don't you read them and report? PZ: I am more interested in developing my own line of deconstructive argument in reference to classic primary sources. What does Barbour have to add to what Eddington, Laue, Synge, etc., etc. -- and even Einstein himself -- have already written? JS: When you find out, let me know. PZ: Synge -- who was a highly regarded relativist -- thought the Einstein equivalence principle was baloney and suggested that it shouldn't even be taught. These sentiments were to some extent echoed in Ohanian and Ruffini (1995). Even in Landau and Lifshitz's Classical Theory of Fields (1951) the skepticism re: equivalence is quite evident. JS: Quotes?? PZ: What we have here is a longstanding yet muffled revolt against Einsteinian dogma (to which it appears even the born-again realist Einstein did not himself subscribe after 1920). I am simply trying to push this repressed anti-thesis to its full logical conclusions. As I said, even Ernst Mach himself later disowned classic relativity, saying that it had become too dogmatic. JS: Suppose the light ray also has a component along a tangent to a longitudinal great circle at r so that, dphi = 0 0 = gtt(r)(cdt)^2 - grr(r)dr^2 - r^2dtheta^2 c^2dT^2 = dR^2 + r^2dtheta^2 c^2 = (dR/dT)^2 + r^2(dtheta/dT)^2 Therefore, the speed of light is always c. However, we have a right triangle above; 1 = (vr/c)^2 + c^-2 r^2w^2 vr/c = cosChi w = physical angular speed of light pulse = gtt^-1/2w' w' = dtheta/dt 1 = (grr/gtt)c^-2(dr/dt)^2 + c^-2r^2gtt^-1(dtheta/dt)^2 = cos^2Chi + sin^2Chi tanChi = sinChi/cosChi = grr^1/2(dr/dt)/(dtheta/dt) = grr^1/2(dr/dtheta) Let's take Einstein's theory as a definite example. gtt = 1 - 2GM/c^2r grr = (1 - 2GM/c^2r)^-1 outside event horizon at r* where 1 - 2GM/c^2/r* = 0 only when r > 2GM/c^2 outside the throat of radius r* of this non-traversable wormhole that pinches off crushing you if you try to fly through it. as an asymptotically flat spherically symmetric static wormhole Einstein-Bridge solution of Ruv = 0 i.e. Tuv(Non- Exotic Vacuum) = 0 The gravity lens effect has one obvious measure tanChi = (1 - 2GM/c^2r)^-1/2(dr/dtheta) Note that tanChi = infinity at the event horizon, i.e. the light ray is trapped in a circular orbit confined to the surface of the event horizon. There is no radial component. Now this is pretty. Much prettier than Hal's model IMHO. Einstein is a genius in terms of pure aesthetics. PZ: Agreed. I do not question the fact of Einstein's towering creative genius. At the same time, it is well-established in the theory of science that an idea that is heuristically useful and even predictively powerful at a certain stage is not necessarily relevant to the objective evaluation of a theory, no matter how successful the program (e.g. Kekule's snake). You have to separate the context of discovery (where creative genius comes into play) from the context of evaluation (where we must rely on logical reasoning and objective criteria). JS: Beauty is Truth. PZ: That was the Florentine Platonist Galileo's error. JS: Also Dirac's? Also Ed Witten's? :-) PZ: Dirac is another good example. Of course, this kind of approach can be successful for the longest time before it eventually files for heuristic bankruptcy. The problem with this fixation on mathematical elegance and aesthetics is that it leads to a naive Platonism that does indeed confuse physical truth with perceived mathematical beauty -- leading to persistent mirages like Einstein equivalence and hypostatization of abstract chronogeometric models that I can only view as fundamentally heuristic in character. JS: Whitehead's fallacy of misplaced concreteness? Max Tegmark is a good example of maximal reification in out to be correct on this in the long run. PZ: He was totally convinced that the planets move in perfect concentric circles around the absolute center of the world, and threw Kepler's work into the trash bin. That was also Einstein's error. He really believed (at least before 1920) that the inertial and gravitational fields were one and the same, and that the monolithic gravitational-inertial field was exhaustively reducible to the Riemannian metric g_uv on a curved 4-dim spacetime manifold. Because it was all so beautiful. So much for beauty is truth. Beauty is in the mind of the beholder. JS: I still think there is equivalence between inertial and gravity fields locally. The g-force is the essentially the non-tensor connection for parallel transport. PZ: What does this mean -- the g-force is the non-tensor connection? Do you mean it can be *described* in terms of a non-tensor connection? JS: Yes this is the standard formalism in GR. http://mathworld.wolfram.com/Levi-CivitaConnection.html by non-tensor connection I mean Christoffel symbols of the second kind http://mathworld.wolfram.com/ChristoffelSymboloftheSecondKind.html What I am alluding to and what you are looking for is equation (9) in the above URL The first term on RHS of eq 9 is the non tensor or inhomogenous part of the metric connection field. I suppose you want to call that the curvilinear or inertial part? In the timelike geodesic LIF the RHS of eq (9) is zero and that is what you call cancellation? Start using this Wolfram on-line resource to add mathematical meat to your informal language. PZ: Isn't this is also true in Newtonian physics? Just set up a space-time coordinate frame and write a metric tensor expression for the invariant interval ds, i.e. ds^2 = g_uv dx^u dx^v and see what happens under accelerative coordinate transformations. Classical mechanics was fully relativistic under velocity boosts. Even the material aether was fully relativistic in this sense. In non-inertial frames, we get *fictitious forces*. Such forces can be described in terms of a metric-tensor transformation for accelerated frames (coordinate transformations that are non-linear in the time coordinate). Straight inertial trajectories are then described as geodesics that *look* curved in those frames. You even get a connection field (in flat space-time). The Newtonian fictitious force field can thus also be described in terms of this Riemann-Christoffel-Levi-Civita connection. So what? Inertial trajectories are still *really* straight. They just *look* curved in certain frames of reference -- from a certain POV. But here it is the *POV* that is curved. In other words -- formally, yes; physically, no. But the theory is still beautiful. At least in the eyes of a Newtonian. JS: d^2X^u/ds^2 = Connection^uwl(dX^w/ds)(dX^l/ds) Connection =/= 0 in LNIF PZ: OK. Connection = 0 in LIF PZ: OK. Same in covariant formulation of Newtonian theory (with no gravity). JS: Connection = 3rd rank Diff(4)Tensor part + Non-Tensor Part PZ: Yes. JS: g-force = 0 in LIF where there is mutual cancellation of the tensor with the non-tensor which defines the free float timelike geodesic extremum of the passive PZ: There is a local cancellation of the metric gradients, yes. Just as there is at a point when you add a compensating sloped line to a curve in an x-y graph. But a line is not a curve -- even at that point, where the curve has non-vanishing second derivatives, while the line doesn't. pure geometry + Matter Action + Exotic Vacuum Zero Point Stress-Energy Density Action. PZ: I won't. geometry + Matter Action. JS: This is characteristic of minimal coupling gauge theories BTW like in EM P(kinetic)u = P(canonical)u - (e/c)Au Au is also a connection in internal fiber space. Fuv(EM) is a curvature! PZ: Let me ask a naive question. What, in your theory, results in inertial effects? Do such effects arise as the result of interaction between matter and vacuum, or between matter and matter? JS: Depends on what you mean by inertial effects? Please be more specific. Do you mean origin of inertia like trying to explain m ~ 10^-27 grams for rest mass of the electron? I mean in the sense of what Bernie Haisch was trying to do? Or do you mean gravity g-force is locally equivalent to an acceleration and is like any inertial force e.g. Coriolis and centrifugal? The mass of the test out in the equation of motion of all of them! That's what they have in common. In the former I mean Wheeler's non-simply topological geometrodynamics of wormhole Mass without mass but with G* >> G(Newton) on micro-scale so that matter comes from attractive EXOTIC VACUUM, i.e. zero point stress-energy density tensors. In the latter I mean what Einstein meant as reconstructed in MTW. Bottom line, gravity lensing does give information about r the curvature radial coordinate where the concentric surface area is 4pir^2 with an entropy of 4pir^2/Lp^2 c-bits. One curvature radial coordinate has more than one isotropic radial coordinate corresponding to it. The map is not 1-1. PZ: Interesting. I'll have to give you a copy of Brian Tupper's 1974 Nuovo Cimento paper which deals with a parametrized class of geometrodynamic theories that are consistent with the four classical tests of GR. Also, a 1999 paper by Alley, Yilmaz et al. that restates the entire n-body argument in the context of more recent empirical data. ... That is what I call semantic incoherence. JS: This is to be expected from Noether's theorem! It is trivially so that the bigger theory always violates Sacred Cows of the smaller earlier theory e.g. Newton's Absolute Time as a slaughtered sacred cow. But Einstein, The Great Rabbi, did it Kosher! PZ: But then how do you explain the existence of these conserved energy-momentum integrals in Newtonian theory -- arguably the most successful physical theory of all time? JS: Trivial - action at a distance in globally flat spacetime. No problem the translation group symmetry works. PZ: : Are you saying that you don't need to recover the classical conservation principles in the Newtonian limit? JS: No I said just the opposite. I told you why it works. Curvature means violation of the translational symmetry from which conservation of momenergy comes. mainly an artifact of insisting on doing *intrinsic* geometry on higher-dimensional curved manifolds. JS: Right. PZ: OK. PZ: If you do intrinsic geometry on the surface of an ordinary 3-sphere, you have exactly the same problems. But as soon as you look at the same surface as embedded in 3-space, all the Riemannian hocus pocus about curvature connections and parallel transport becomes irrelevant, since exactly the same vectors are then directly comparable. JS: Huh? Only if the bigger space is globally flat, which it is not in M-theory and even has fermion dimensions and even may have non-commutative geometry. PZ: I am talking about an ordinary sphere in ordinary space. If you insist on doing purely intrinsic geometry on the spherical surface, you are a half-blind flatlander who sees only the shadows (connection, covariant derivatives, etc. etc), whereas viewed from 3-space, each vector has perfectly well-defined components that can be directly compared at every point on the surface. No need for parallel transport and all that. All you have to do is establish a Cartesian coordinate system in the 3-space, and all the Riemannian apparatus becomes instantly redundant -- although of course it still applies to the dimensionally challenged flatlanders on the surface. JS: In GR 4D space-time is curved and so are spacelike slices. The extrinsic curvature of the spacelike slice is a kind of time push forward observable. PZ: We can still imagine a 10-dim spacetime in which this manifold can be embedded. Then we are no longer condemned to a life of purely intrinsic surface geometry. JS: Yes but I doubt that the Calabi-Yau space of string theory is such a flat Euclidean one with simply connected global topology. 11 Dim hyperspace of M-Theory is not flat. PZ: Perhaps not. We can still *imagine* a flat higher-dimensional spacetime in which the curved 4-dim Riemannian manifold of GR is embedded, and that is sufficient for my point. It puts things in *perspective*, so to speak. The point being, let's not fetishize or fixate upon Riemannian geometry. Golden Calf. PZ: The question of how to define the energy-momentum of a physical gravitational field is not simply a question of Riemannian elegance. JS: The Question is: What is The Question? (J.A. Wheeler). PZ: Yes. Energy-momentum of gravity is a global flux integral! PZ: Of what density?! JS: I try to answer that below. But it is half-baked at this point. I seem to be getting that you need EXOTIC VACUUM to solve the problem locally. That is the missing link -- maybe? Local stress-energy density Diff(4) tensor for gravity is trivial it is tuv(Gravity) = (String Tension)Guv In non-exotic vacuum tuv(Gravity) = 0. PZ: That is exactly what is at issue here. The argument is that you do not get a satisfactory Newtonian correspondence with t_uv(vac) = 0. JS: Yeah, but maybe that is why we need exotic vacuum dark energy/matter as a consistency requirement just like when James Maxwell stuck in the displacement vacuum current to get far field EM radiation! This idea is in harmony with apparent fact that Universe is 96% EXOTIC VACUUM. I do not think those latest observations http://spaceflightnow.com/news/n0312/12darkenergy/ are being properly interpreted BTW. In exotic dark energy/matter vacuum tuv(Gravity) + tuv(Exotic Vacuum Zero Point Stress-Energy Density) = 0 PZ: OK, so you are saying that there is indeed a t_uv(vac). JS: I have been saying this all along for EXOTIC VACUUM. The effect is not there, I mean it's ZERO in the ordinary NON-EXOTIC vacuum of Einstein's Ruv = 0. Also my tuv(Exotic Vacuum) = (String Tension)/zpf guv(Curved Space-Time) IS NOT YILMAZ's! My effect is a QUANTUM EFFECT it is exactly INFINITE when quantum h -> 0 and G & c stay finite. Indeed, in large scale it gives an infinite Einstein cosmological constant when h -> 0. Vacuum Coherence -> 0 when h -> 0 and Lp -> 0 when G, c finite, therefore /zpf -> -1/Lp^2 -> -infinity (attractive dark matter limit catastrophe where Universe cannot come into Being and Becoming at all!) That is OK because there is no such thing as a consistent CLASSICAL UNIVERSE. What we live in is a MACRO-QUANTUM UNIVERSE or rather parallel brane worlds floating in hyperspace on the IT extra variable (rather than hidden variable) level. There is also BIT. PZ: How does this t_uv(vac) self-gravitate in your theory? Is it a tensor density? JS: Please Paul READ MY EQUATIONS! tuv(Geometry) + Tuv(Matter) + tuv(Exotic Vacuum) = 0 is a local tensor equation. All parts of it are tensors under at least local passive general coordinate transformations. tuv(Geometry) = (String Tension)[Ruv(Ricci) - (1/2)Rguv] Tuv(Exotic Vacuum) = (String Tension)/zpfguv /zpf = Lp^-2[Lp^3/2|Vacuum Coherence|^2 - 1] Do you mean the (-g)^1/2 Pauli tensor density formal thing? PZ: If it has a non-tensor component, then where exactly does that come from, physically speaking, in your theory? JS: You have again garbled the level distinction between g-force connection fields and tidal curvature tensor. My tuv(Exotic Vacuum) ~ guv(Curved Space-Time) i.e. a second rank GR tensor. Your it ain't my it. ;-) What about the flux integrals? Simple. tuv(Gravity) = tuv(Gravity Near Field) + tuv(Gravity Far Field) Gravity mom-energy flux integral is only for gravity waves (LIGO & LISA et-al) Pu (Far Field) = Spacelike 3D integral of tvl(Gravity Far Field)(antisymmetric 3symbol uvl)dx^w/dx^l = ?d?tvl(far field)?(u|vl)dx^v/dx^l Where in some regions contributing to the spacelike integral tuv(gravity far field) =/= 0 tuv(gravity far-field) = - tuv(gravity near field) - tuv(Exotic Vacuum Zero Point Stress-Energy Density) =/= 0 via a kind of Gauss theorem relating surface to volume integrals - there are complications doing that in curved space-time of course. PZ: OK. JS: There is an interesting prediction from this toy model I just now thought of for the first time. When tuv(gravity near field) ---> 0, one cannot see gravity waves in the far field locally unless there is an exotic-vacuum field tuv(Exotic Vacuum Zero Point Stress-Energy Density) locally present at the surface for the flux integral! This would seem to predict a lack of gravity wave signal for Kip Thorne's LISA and LIGO? Note, that is not same problem of 1913 type pulsars, which is an indirect EM measurement of gravity waves at that pulsar not a direct detection of gravity waves passing through Earth. I am not sure if any of this makes sense yet. It just popped up out of my sub-conscious mind, or maybe the Jungian Collective Unconscious Mind of P.K. Dick's VALIS? ;-) PZ: OK, this is beginning to make sense. JS: Yeah, that's AMAZING! Something like the gravity analog of Maxwell's displacement current showing why there must be dark energy/matter exotic vacuum regions - hence stargate time travel with weightless warp drive not far behind and already here in the UFO data? GR orthodoxy (many of whom are really mathematicians) are confusing an artifact of insistence on intrinsic geometry with a physical question of characterizing the real permanent gravitational field in terms of its energy-momentum content and distribution. Of course if one *defines* the physical g-field in terms of the unified gravitational- inertial Riemann-Christoffel-Levi-Civita connection, the energy carried in this field will be at least locally frame-dependent. But from my POV this is an artifice that arises from a kind of *fetishization* of Riemannian mathematics and general covariance. Think about it: take a 4-dim spacetime. Go to a 4-dim cylindrical coordinate system for which the coordinates in a 2-dim spatial submanifold are polar. You automatically get a non-vanishing connection that is purely an artifact of the choice of coordinates. Of course you have to re-write the metric components g_uv to take account of this; and inside this polar subspace coordinate system, straight lines look curved. But nowhere even in orthodox GR is it suggested that this connection field constitutes any kind of *physical* field. It is regarded as a purely mathematical animal. Yet the whole apparatus of connections, geodesics, etc. etc. applies just as it does to the physically interpreted space-time coordinate connection field. So the mere appearance of a connection field in itself need have no physical meaning at all -- even within orthodox GR. All that glisters is not gold. JS: Of course one hopes that the compensating gauge field, in this case the geometrodynamic guv field restores the broken symmetry. It does do that in the sense of locally conservation of stress-energy density currents when all dynamical degrees of freedom are included. PZ: OK. JS: Tuv(Non Exotic Vacuum) + Tuv(Exotic Vacuum) + Tuv(Matter) = 0 The total covariant 4-divergence vanishes, but not the individual terms separately in the practical metric engineering regime whose technology we see in the saucers from the brane worlds next door is my educated guess. Brian Greene thought he was joking on NOVA, the joke may be on him - let us hope. :-) More anon. JS: This is getting really interesting. Z. JS: Yup. http://stardrive.org/cartoon/MagicBean.html http://stardrive.org/cartoon/spectra.html http://stardrive.org/cartoon/USSKron.html http://stardrive.org/cartoon/bovines.html http://stardrive.org/cartoon/dan.html http://stardrive.org/cartoon/coffee.html http://stardrive.org/cartoon/Saturn.html ==== I have no idea how to even start this one. Any hints will be helpful. Let a in S_n be a fixed-point involution or inversion where a^2 = e and a ====