> How can one show for x, y in lp space with |x|,|y|=1 that |(x+y)/2|=1 => x=y. > I > am assuming here that > 1= 0 and for x >= a, define f(x) = (a^p + x^p)/2 - [(a+x)/2]^p. Then f'(x) > 0 if x > a. ==== > Here is an interesting question that I came across > given N points on a two dimensional plane, one can define a distance > matrix A = [a ] of order N x N such that a is the distance between > ij ij > the point i, and j. > The question is given this matrix can a set of points be given that > will satisfy the given distance matrix in an efficient manner? > abi A triangle given 3 sides satisfying triangle inequality in Euclidean 2D can be constructed upto one arbitrary reflection about any axis in the 2D and one arbitrary rotation, as is well known. The process can start by taking first point at origin and second point to be, say on x-axis. To determine a new point with respect to 3 given points in the triangulation process ( similar to civil engineering surveying ), we need 3 distances from them to the new point. Accordingly, if N number of points are given, p = 3 N-6 distances are needed exactly for uniquely determining their position/embedding in 2D. If more than p distances are given, then there is inconsistency/redundancy. So when these total p combination distances are given, the position of all points are intrinsically fixed in relation to each other, except for the arbitrariness involved in first three points mentioned above. HTH ==== > Here is an interesting question that I came across > given N points on a two dimensional plane, one can define a distance > matrix A = [a ] of order N x N such that a is the distance between > ij ij > the point i, and j. > > The question is given this matrix can a set of points be given that > will satisfy the given distance matrix in an efficient manner? > > abi The book by Deza and Laurent 'Geometry of cuts and metrics' is a good reference in this regard. For the specific problem you ask, I believe there is a theorem of Menger (I don't have the book handy) that states that you need to check that the generalization of the triangle inequality (and the tetrahedral inequality that Dave Rusin mentioned) is true for all subsets of d+3 points holds (in this case 5 points). The condition is necessary and sufficient. Of course being a metric is a necessary condition for any of the above to work. ==== > given N points on a two dimensional plane, one can define a distance > matrix A = [a ] of order N x N such that a is the distance between > ij ij > the point i, and j. > > The question is given this matrix can a set of points be given that > will satisfy the given distance matrix in an efficient manner? Let C = (-1/2)(I - uu'/n)B(I - uu'/n), where B be the matrix of squared distances among the n points, I is the identity matrix, u is a column vector whose elements are all 1s, and ' denotes transposition. If the points lie in an m-dimensional space then C will be positive semidefinite, with rank m. Let F be any factoring of such a C, so that FF' = C. Then F contains the coordinates of the points. This is a well-known result in multidimensional scaling. ==== Is there a simple formula for the next Farey fraction? Let a/b be any irreducible rational number. Then, next Farey fraction is defined as the next element of the Farey sequence F_b. For example, next Farey fraction of 3/5 is 2/3. Indeed: F_5 = {0/1,1/5,...,3/5,2/3,...,1/1} In other words, is there closed-form formula for elements of F_b? ==== > Is there a simple formula for the next Farey fraction? > > Let a/b be any irreducible rational number. Then, next Farey fraction is > defined as the next element of the Farey sequence F_b. For example, next > Farey fraction of 3/5 is 2/3. Indeed: > > F_5 = {0/1,1/5,...,3/5,2/3,...,1/1} > > In other words, is there closed-form formula for elements of F_b? Here's a GP-Pari program which finds Farey sequences: {term(n,s,t)= local(z); z=floor((n+denominator(s))/denominator(t)); return((z*numerator(t)-numerator(s))/(z*denominator(t)-denominator(s))); } {farey(n)= local(f); f=[]; if(n<2,return(f);); if(n==2,f=concat(f,[1/2]);return(f);); f=concat(f,[1/n]);f=concat(f,[1/(n-1)]);i=3; until(term(n,f[i-2],f[i-1])>=1,f=concat(f,[term(n,f[i-2],f[i-1])]);i=i+1); return(f); } farey(n) is the sequence with denominator n. F_5 = [1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5] It was a little bit of time to come up with the right programming for this little gem. ==== On Thu, 15 Jan 2004 14:19:41 -0800, Mikito Harakiri >Is there a simple formula for the next Farey fraction? Let a/b be any irreducible rational number. Then, next Farey fraction is >defined as the next element of the Farey sequence F_b. For example, next ^^^^^^^ Huh?!? So you're assuming that the environment in which a/b lives is the Farey sequence of order b. If that is what you want to investigate, fine! But in the general case your problem is either not well posed or badly phrased for you'd need, loosely speaking, *two* inputs: for example (but not necessarily!) a rational number a/b and an integer order, say n. Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc ==== > On Thu, 15 Jan 2004 14:19:41 -0800, Mikito Harakiri Is there a simple formula for the next Farey fraction? Let a/b be any irreducible rational number. Then, next Farey fraction is >defined as the next element of the Farey sequence F_b. For example, next > ^^^^^^^ Huh?!? So you're assuming that the environment in which a/b lives is the > Farey sequence of order b. If that is what you want to investigate, > fine! But in the general case your problem is either not well posed or > badly phrased for you'd need, loosely speaking, *two* inputs: for > example (but not necessarily!) a rational number a/b and an integer > order, say n. Michele, I don't understand your comment. Was I sloppy with the definition? Or the fact that the problem formulation is not general enough upset you? ==== > Is there a simple formula for the next Farey fraction? > > Let a/b be any irreducible rational number. Then, next Farey fraction is > defined as the next element of the Farey sequence F_b. For example, next > Farey fraction of 3/5 is 2/3. Indeed: > > F_5 = {0/1,1/5,...,3/5,2/3,...,1/1} > > In other words, is there closed-form formula for elements of F_b? > > In Knuth, _The Art of Computer Programming_, Section 1.3.2, ex. 19 shows how to generate the next element of the Farey sequence from _two_ successive elements. So you need two 'starters' to get going. It is a little harder to get the second 'starter' if you have only one fraction to begin with. But it is possible. Definition: If two fractions a/b and c/d satisfy a/b < c/d, then I will say that they are *neighbors* if bc-ad=1. Proposition: Two fractions a/b and c/d are successive elements of the Farey series of order N if and only if these conditions are met: (a) a/b and c/d are neighbors; (b) b+d > N, b<=N, d<=N. Proposition: if a/b and c/d are neighbors, then a/b and (c+ka)/(d+kb) are neighbors. So, given a/b, we want to find c/d such that bc-ad=1. If our first success at this fails to satisfy (b) in the first proposition, then we can construct a more suitable neighbor by applying the second proposition. Of course we put a/b in lowest terms, if it is not so already. Well, look at the equation to be solved in this form: ad = -1 + bc. This is as much as to say that -d is the multiplicative inverse of a in the integers modulo b. since a is relatively prime to b, there is such an inverse, an dit can be found by hacking around with euclid's algorithm. This is not a closed form, but it is a terminating algorithm. -- Chris Henrich Yes, one can rant about the program designs, but generally things keep getting more and more confused as time goes on. --Sea Wasp ==== > >>Is there a simple formula for the next Farey fraction? >Let a/b be any irreducible rational number. Then, next Farey fraction is >>defined as the next element of the Farey sequence F_b. For example, next >>Farey fraction of 3/5 is 2/3. Indeed: >F_5 = {0/1,1/5,...,3/5,2/3,...,1/1} >In other words, is there closed-form formula for elements of F_b? > > In Knuth, _The Art of Computer Programming_, Section 1.3.2, ex. 19 > shows how to generate the next element of the Farey sequence from _two_ > successive elements. So you need two 'starters' to get going. also in Graham, Knuth, Patashnik (2nd ed), ch 4, Bonus problem 61, p. 150: (where N = b?) y0 = 0 y1 = 1 yt = y(t-1) floor((t-N)/x(t-1)) - y(t-2) x0 = 1 x1 = N xt = x(t-1) floor((t-N)/x(t-1)) - x(t-2) and the t'th Farey fraction based on N is y(t)/x(t). That this works is just an application of the usual gcd and mediant properties. > It is a little harder to get the second 'starter' if you have only one > fraction to begin with. But it is possible. or the dumb slow version, just scan through F_n. -- Mitch Harris (remove q to reply) ==== Is there a simple formula for the next Farey fraction? >Let a/b be any irreducible rational number. Then, next Farey fraction is >>defined as the next element of the Farey sequence F_b. For example, next >>Farey fraction of 3/5 is 2/3. Indeed: >F_5 = {0/1,1/5,...,3/5,2/3,...,1/1} >In other words, is there closed-form formula for elements of F_b? In Knuth, _The Art of Computer Programming_, Section 1.3.2, ex. 19 > shows how to generate the next element of the Farey sequence from _two_ > successive elements. So you need two 'starters' to get going. also in Graham, Knuth, Patashnik (2nd ed), ch 4, Bonus problem 61, p. 150: > (where N = b?) y0 = 0 y1 = 1 yt = y(t-1) floor((t-N)/x(t-1)) - y(t-2) > x0 = 1 x1 = N xt = x(t-1) floor((t-N)/x(t-1)) - x(t-2) and the t'th Farey fraction based on N is y(t)/x(t). That this works is > just an application of the usual gcd and mediant properties. It is a little harder to get the second 'starter' if you have only one > fraction to begin with. But it is possible. or the dumb slow version, just scan through F_n. My first attempt was function next_farey ( l Rational ) RETURN Rational DETERMINISTIC IS exausted_fractions_search EXCEPTION; BEGIN FOR d In 1..l.den LOOP FOR c In 1..d LOOP if l.num*d+1=l.den*c then return new Rational(c,d); end if; END LOOP; END LOOP; raise exausted_fractions_search; END; but double loop is really killing: finding next_farey(17771,28345) took 300 sec. Removing inner loop (at the expence of introduced division) function next_farey ( l Rational ) RETURN Rational DETERMINISTIC IS exausted_fractions_search EXCEPTION; BEGIN FOR d In 1..l.den LOOP if mod(l.num*d+1,l.den)=0 then return new Rational((l.num*d+1)/l.den,d); end if; END LOOP; raise exausted_fractions_search; END; is more reasonable implementation. I wonder if I can find next farey fraction faster than in linear time. ==== > Is there a simple formula for the next Farey fraction? Let a/b be any irreducible rational number. Then, next Farey fraction is > defined as the next element of the Farey sequence F_b. For example, next > Farey fraction of 3/5 is 2/3. Indeed: F_5 = {0/1,1/5,...,3/5,2/3,...,1/1} In other words, is there closed-form formula for elements of F_b? Missed the obvious: min { c/d | ad+1=bc & a/b Then there is the idea that orthodox quantum theory with signal locality > is only a limited approximation like global Special Relativity and the > more general theory of the quantum information principle, including > quantum gravity, has signal nonlocality in a very essential way and > that all inner consciousness requires signal nonlocality in the sense > defined by Antony Valentini. Whoa! Did I just hear you suggest that orthodox quantum mechanics just might be bunk? :-) Bjacoby ==== james Randi owes me $1,000,000 . . . i think he is a cool guy and knows WHAT IS UP!. read my threads on the STRAIGHTDOPE-MESSAGEBOARD, i am the poster who was Question. htttp://boards.straightdope.com . . . . ignorance=education go to the SDMB and you can find some of the stupidest people on earth(not n insult, your ego creates suffering) . . the Internet is where ends will unite to form a whole. ==== Old rule, he who puts up the money makes the rules ! > james Randi owes me $1,000,000 . . . i think he is a cool guy and knows > WHAT IS UP!. read my threads on the STRAIGHTDOPE-MESSAGEBOARD, i am the > poster who was Question. htttp://boards.straightdope.com . > . > . > . > ignorance=education go to the SDMB and you can find some of the stupidest people on earth(not > n insult, your ego creates suffering) . . the Internet is where ends will > unite to form a whole. > ==== On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, reotpreeoj ishkabibbled: ^james Randi owes me $1,000,000 . . . i think he is a cool guy and knows ^WHAT IS UP!. read my threads on the STRAIGHTDOPE-MESSAGEBOARD, i am the ^poster who was Question. ^ ^htttp://boards.straightdope.com ^ ^ ^ The Straight Dope and its author, Cecil, totally ROCK! I really dig the kind of obscure info he digs up. -- The Queen of DXers, as well as Queen of the Commonwealth of Virginia, as well as The Ruler of A.D.P., as well as Saint Debbe, as well as Our Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of Seville An Ointed Minister of the Universal Life Church Reverant of the Church of the SubGenius, UnOrthodox Superior Mutha Superior of the Little Sistahs of the Politically Incorrect Worshipper of Eris, Goddess of Discord I WON'T grow up!! -- Peter Pan ==== Hey, fuck you. How about that? > james Randi owes me $1,000,000 . . . i think he is a cool guy and knows > WHAT IS UP!. read my threads on the STRAIGHTDOPE-MESSAGEBOARD, i am the > poster who was Question. htttp://boards.straightdope.com . > . > . > . > ignorance=education go to the SDMB and you can find some of the stupidest people on earth(not > n insult, your ego creates suffering) . . the Internet is where ends will > unite to form a whole. > ==== > math.GM/0312309 > I didn't know proving something about a function implies calculating > its value for every possible argument. You always learn something new. I didn't know that it was possible to show that a proof is wrong by laughing at it without even finding any holes in it. Craig ==== > 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. > > > > l8r, Mike N. Christoff > > > > I thought I'd give an extremely low detail explanation of how algebraic topology can be to model asynchronous distributed protocols. I will also note that apparently it requires very little general topology to understand as it is almost entirely combinatorial in nature. In fact, one only needs to know the AT described in the first chapter of Munkres' book on algebraic topology to understand it (but not to understand this overview which requires none). This is taken from a response I gave on the theory-edge yahoo group (also posted on comp-sci- theory). Note that this must be viewed with a fixed width font like courier or courier new. ----- Original Message ----- major applications. -------------------------------- There has already been major work done in the connection between DC and AT. Many new lower bounds have been proven for classic DC problems like k-set agreement and consensus. A quote by Fr.8ed.8eric Tronel of the French National Institute for Reseasrch In Computer Science and Control: ... the problem of consensus (distributed agreement) plays a central role in the theory of distributed computing. Since it has been proven to be unsolvable by deterministic algorithms in a purely asynchronous system, it was a real challenge to explicitly determine the border that exists between solvable and unsolvable problems in asynchronous systems. This problem remained unsolved till the publication of a seminal paper by Maurice Herlihy and Nir Shavit. In fact at the reading of this paper, one can understand that all the previous approaches where prone to fail. Indeed, all of them kept on using the traditional way of modelling an asynchronous system, namely by the mean of a graph of local states. This technic can be successful, when there is only one crash in the system. However, it does not scale well, when the number of crashes increases. On the contrary, Herlihy and Shavit have chosen to model the evolution of the system through the use of high dimensional geometrical objects. The area of distributed computing I'm interested uses asynchronous message passing. This is the type you want to look at if you're interested in technologies like peer to peer networking (which I am). The idea is very combinatorial. You start with a set of nodes, each with an initial value from some finite set. For concreteness, imagine you have three nodes arranged in a triangle (vertices = nodes, edges = network connections). Let the initial value set = {0,1}. Then this can be represented as a ----- b / / / c where a,b,c are tuples (node_id, initial_value). The number of possible initial states for the system are then 2^3 since we have 3 nodes which can each be initialized with either a 0 or a 1. This gives us 2^3 triangles. Now here is the tricky part. Connect the triangles together by matching edges whose vertices a,b are equal. For example, take the three triangles below. A) 001 (n1, 0) ------------ (n2,0) / / / (n3,1) B) 101 (n1, 1) ------------ (n2,0) / / / (n3,1) C) 000 (n1, 0) ------------ (n2,0) / / / (n3,0) This becomes: (n3,0) / / C / (n1, 0) ------------ (n2,0) / A / B / (n3,1) ----------- (n1,1) Now if we continue this process we will have generated a closed geometric figure. This one in particular looks like like two tetrahedrons connected at their bases. Or one could think of it as an approximation of a sphere using 8 triangles. This is called the input complex at step 0. As the protocol evolves in time, each face (triangle) of the object subdivides further to represent all possible states the nodes could be at step 2. So for instance, the triangle C will subdivide into a subcomplex representing all possible states of the nodes after starting with the inital inputs all 0. The system is fault tolerant, as some of the structure of the subcomplex also represents possible states of the nodes if one or more of the nodes have failed. The set of structures that develop as the protocol progresses are called protocol complexes. In theoretical computer science we have the decsion problem, so in DC we have decision tasks. A decision task is a finite process after which each node makes a decision on what its output should be. For example, in the binary consensus problem each node is initially given a number in {0,1}. After a finite number of steps, all nodes must output identical values complex of the problem. For binary consensus with three nodes, the only acceptable output complex is shown below: (n1, 1) ------------ (n2,1) / / / (n3,1) (n1, 0) ------------ (n2,0) / / / (n3,0) In other words, the final state of all nodes must either be all 1s or all 0s. A 'simplicial map' is used to map the protocol complex to the output complex. Without getting into the machinery of it (which I'm still not very clear on), the idea is to determine if the protocol complex can evolve to the required output complex. This is done by trying to find any topological obstructions (holes) to the simplicial map. If there are, then the protocol cannot work, otherwise it can. That is a bare bare bones description of the idea and I may have misexplained parts. l8r, Mike N. Christoff ==== I'm reposting this as a new thread since Outlook Express users in particular will probably never see it as a reply to the first thread since it will be buried almost a month back. Sorry to google users who see posts as they are posted in the main screen and not in threads initially. > 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. > > > > l8r, Mike N. Christoff > > > > I thought I'd give an extremely low detail explanation of how algebraic topology can be to model asynchronous distributed protocols. I will also note that apparently it requires very little general topology to understand as it is almost entirely combinatorial in nature. In fact, one only needs to know the AT described in the first chapter of Munkres' book on algebraic topology to understand it (but not to understand this overview which requires none). This is taken from a response I gave on the theory-edge yahoo group (also posted on comp-sci- theory). Note that this must be viewed with a fixed width font like courier or courier new. ----- Original Message ----- major applications. -------------------------------- There has already been major work done in the connection between DC and AT. Many new lower bounds have been proven for classic DC problems like k-set agreement and consensus. A quote by Fr[Hyphen]d[Hyphen]ric Tronel of the French National Institute for Reseasrch In Computer Science and Control: ... the problem of consensus (distributed agreement) plays a central role in the theory of distributed computing. Since it has been proven to be unsolvable by deterministic algorithms in a purely asynchronous system, it was a real challenge to explicitly determine the border that exists between solvable and unsolvable problems in asynchronous systems. This problem remained unsolved till the publication of a seminal paper by Maurice Herlihy and Nir Shavit. In fact at the reading of this paper, one can understand that all the previous approaches where prone to fail. Indeed, all of them kept on using the traditional way of modelling an asynchronous system, namely by the mean of a graph of local states. This technic can be successful, when there is only one crash in the system. However, it does not scale well, when the number of crashes increases. On the contrary, Herlihy and Shavit have chosen to model the evolution of the system through the use of high dimensional geometrical objects. The area of distributed computing I'm interested uses asynchronous message passing. This is the type you want to look at if you're interested in technologies like peer to peer networking (which I am). The idea is very combinatorial. You start with a set of nodes, each with an initial value from some finite set. For concreteness, imagine you have three nodes arranged in a triangle (vertices = nodes, edges = network connections). Let the initial value set = {0,1}. Then this can be represented as a ----- b / / / c where a,b,c are tuples (node_id, initial_value). The number of possible initial states for the system are then 2^3 since we have 3 nodes which can each be initialized with either a 0 or a 1. This gives us 2^3 triangles. Now here is the tricky part. Connect the triangles together by matching edges whose vertices a,b are equal. For example, take the three triangles below. A) 001 (n1, 0) ------------ (n2,0) / / / (n3,1) B) 101 (n1, 1) ------------ (n2,0) / / / (n3,1) C) 000 (n1, 0) ------------ (n2,0) / / / (n3,0) This becomes: (n3,0) / / C / (n1, 0) ------------ (n2,0) / A / B / (n3,1) ----------- (n1,1) Now if we continue this process we will have generated a closed geometric figure. This one in particular looks like like two tetrahedrons connected at their bases. Or one could think of it as an approximation of a sphere using 8 triangles. This is called the input complex at step 0. As the protocol evolves in time, each face (triangle) of the object subdivides further to represent all possible states the nodes could be at step 2. So for instance, the triangle C will subdivide into a subcomplex representing all possible states of the nodes after starting with the inital inputs all 0. The system is fault tolerant, as some of the structure of the subcomplex also represents possible states of the nodes if one or more of the nodes have failed. The set of structures that develop as the protocol progresses are called protocol complexes. In theoretical computer science we have the decsion problem, so in DC we have decision tasks. A decision task is a finite process after which each node makes a decision on what its output should be. For example, in the binary consensus problem each node is initially given a number in {0,1}. After a finite number of steps, all nodes must output identical values complex of the problem. For binary consensus with three nodes, the only acceptable output complex is shown below: (n1, 1) ------------ (n2,1) / / / (n3,1) (n1, 0) ------------ (n2,0) / / / (n3,0) In other words, the final state of all nodes must either be all 1s or all 0s. A 'simplicial map' is used to map the protocol complex to the output complex. Without getting into the machinery of it (which I'm still not very clear on), the idea is to determine if the protocol complex can evolve to the required output complex. This is done by trying to find any topological obstructions (holes) to the simplicial map. If there are, then the protocol cannot work, otherwise it can. That is a bare bare bones description of the idea and I may have misexplained parts. l8r, Mike N. Christoff ==== > While I know C and C++, and have at times felt hampered by certain > limitations of Java, I've concluded that it's the best programming > language for wideranging discussions, as it's accessible to people who > haven't programmed before who aren't familiar with Unix, but are > Windows people. > For a nice C/C++ compiler/IDE package, you might want to check out http://www.bloodshed.net and download dev-C++. -- Will Twentyman ==== In sci.math, James Harris on 14 Jan 2004 08:34:22 -0800 <3c65f87.0401140834.1d59810c@posting.google.com>: > While I know C and C++, and have at times felt hampered by certain > limitations of Java, I've concluded that it's the best programming > language for wideranging discussions, as it's accessible to people who > haven't programmed before who aren't familiar with Unix, but are > Windows people. > > I need as many people capable of checking things like programs that I > post as I can get, so I'm making a push for Java. > > It turns out that for most of you it's as simple as going to the Sun > website, where you can get a free download: > > http://java.sun.com/j2se/1.4.2/download.html > > I'd suggest Sun's tutorials for those who aren't at all familiar with > Java and running Java programs. > > To see some of my work, and a program that you can try running, see my > blog archives: > > > You can also do a search at Google Groups for PrimeCountH.java, > > The point for me here is to show you first that what I've discovered > works, and then I think it'll be easier for me to explain unique > features of my discovery, as I look towards finding more efficient and > effective ways of getting the word out. > > > James Harris Did you *really* want me to run Yet Another Primecounter Contest? :-P As it is, your Java implementation makes snails look fast; on my Official Test System(tm) it took 3 minutes to compute pi(10000)=1229 (which, to its credit, is the correct answer). You may want to look at your usage of Math.sqrt(). A recoding of my first LegendrePhi submission took half a second, (to run, that is; the recoding itself took the better part of a half hour) and of course also came up with the correct answer. And remember, I'm a lightweight in this department -- if Christian Bau were to really sink his teeth into this particular subproblem, he'd make both of us look ridiculous. :-) (Which is admittedly fine by me; I can't say I understand his algorithm but it's *fast*!) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >In sci.math, James Harris ><3c65f87.0401140834.1d59810c@posting.google.com>: >> While I know C and C++, and have at times felt hampered by certain >> limitations of Java, I've concluded that it's the best programming >> language for wideranging discussions, as it's accessible to people who >> haven't programmed before who aren't familiar with Unix, but are >> Windows people. >> >> I need as many people capable of checking things like programs that I >> post as I can get, so I'm making a push for Java. >> >> It turns out that for most of you it's as simple as going to the Sun >> website, where you can get a free download: >> >> http://java.sun.com/j2se/1.4.2/download.html >> >> I'd suggest Sun's tutorials for those who aren't at all familiar with >> Java and running Java programs. >> >> To see some of my work, and a program that you can try running, see my >> blog archives: >> >> >> You can also do a search at Google Groups for PrimeCountH.java, >> >> The point for me here is to show you first that what I've discovered >> works, and then I think it'll be easier for me to explain unique >> features of my discovery, as I look towards finding more efficient and >> effective ways of getting the word out. >> >> >> James Harris Did you *really* want me to run Yet Another Primecounter Contest? :-P As it is, your Java implementation makes snails look >fast; on my Official Test System(tm) it took 3 minutes >to compute pi(10000)=1229 (which, to its credit, is the >correct answer). My gosh, it's _that_ slow? I guess I wasn't paying attention during the contest... Just for giggles I whipped up a _totally_ brainless PrimePi in Python just now - it's nothing but a sieve, and not a clever sieve, a totally idiotic sieve, not particularly well written either, just top-of-the-brainless-head code: def Sieve(ran): for d in range(2, (len(ran))/2+1): for j in range(2*d, len(ran), d): ran[j]=0 def Primes(n): ran = range(n+1) Sieve(ran) return filter(lambda x: x > 1, ran) def PrimePi(n): return len(Primes(n)) This is _Python_, an extremely fabulous language, but quite slow, being interpreted instead of compiled. And (on and admittedly fast machine, 2000GH/512MB) it takes much less than a second to find Pi(10000); didn't actually time anything, but if I say print PrimePi(10000) the 1229 appears with no perceptible delay. Presumably it would be pretty slow for larger values, for reasons beyond the understanding of some of us, but it gives Pi(100000) in less than two seconds... Maybe there _is_ some significance to James' work here. Like if we had a contest for the slowest possible Pi(n), maybe. >You may want to look at your usage of Math.sqrt(). A recoding of my first LegendrePhi submission took half a second, >(to run, that is; the recoding itself took the better part of >a half hour) and of course also came up with the correct answer. And remember, I'm a lightweight in this department -- if Christian >Bau were to really sink his teeth into this particular subproblem, >he'd make both of us look ridiculous. :-) (Which is admittedly >fine by me; I can't say I understand his algorithm but it's *fast*!) ************************ David C. Ullrich ==== > This is _Python_, an extremely fabulous language, but > quite slow, being interpreted instead of compiled. I don't think you can prove programming skill by saying crazy things like Python is an extremely fabulous language. Sorry. c ==== On 15 Jan 2004 10:39:26 -0800, tb+usenet@becket.net (Thomas Bushnell, This is _Python_, an extremely fabulous language, but >> quite slow, being interpreted instead of compiled. I don't think you can prove programming skill by saying crazy things >like Python is an extremely fabulous language. Sorry. Well, I've already suggested that you should explain what your reasons for this are. It occured to me that I instead of trying to explain what's so great about Python I should just give a citation to someone else's explanation. I'm not an expert programmer, never claimed to be. On the other hand Eric Raymond is a _hugely_ respected name in open-source programming circles. Go here to see what he has to say about Python: A few quotes. First, just to show that he's not like the guy he was very skeptical at first: >I immediately tripped over the first odd feature of Python >that everyone notices: the fact that whitespace (indentation) >is actually significant in the language syntax. The language >has no analog of the C and Perl brace syntax; instead, >changes in indentation delimit statement groups. And, like >most hackers on first realizing this fact, I recoiled in >reflexive disgust. I am just barely old enough to have programmed in >batch FORTRAN for a few months back in the 1970s. Most >hackers aren't these days, but somehow our culture seems >to have retained a pretty accurate folk memory of how nasty >those old-style fixed-field languages were. Indeed, the >term ``free format'', used back then to describe the newer >style of token-oriented syntax in Pascal and C, has almost >been forgotten; all languages have been designed that way >for decades now. Or almost all, anyway. It's hard to blame >anyone, on seeing this Python feature, for initially reacting >as though they had unexpectedly stepped in a steaming pile >of dinosaur dung. He didn't start this project determined to like Python, his first reaction to one of the aspects that a lot of people find strange at first was steaming pile of dinosaur dung. A lot of people who haven't _used_ the language react that way. Now read what he has to say a little later, when he has some experience with the language (and note how much experience I'm talking about): >Of course, this brought me face to face once again with >Python's pons asinorum, the significance of whitespace. This >time, however, I charged ahead and roughed out some code >for a handful of sample GUI elements. Oddly enough, Python's >use of whitespace stopped feeling unnatural after about >twenty minutes. I just indented code, pretty much as I would >have done in a C program anyway, and it worked. That was my first surprise. My second came a couple of >hours into the project, when I noticed (allowing for pauses >needed to look up new features in Programming Python) I was >generating working code nearly as fast as I could type. When >I realized this, I was quite startled. An important measure >of effort in coding is the frequency with which you write >something that doesn't actually match your mental >representation of the problem, and have to backtrack on >realizing that what you just typed won't actually tell the >language to do what you're thinking. An important measure >of good language design is how rapidly the percentage of >missteps of this kind falls as you gain experience with >the language. When you're writing working code nearly as fast as >you can type and your misstep rate is near zero, it >generally means you've achieved mastery of the language. >But that didn't make sense, because it was still day >one and I was regularly pausing to look up new language >and library features! This was my first clue that, in Python, I was >actually dealing with an exceptionally good design. Most >languages have so much friction and awkwardness built into >their design that you learn most of their feature set >long before your misstep rate drops anywhere near zero. >Python was the first general-purpose language I'd ever >used that reversed this process. That's a few _hours_ of experience using the language, and his opinion has changed from steaming pile of dinosaur dung to an exceptionally good design. A few days later he's trying to do some metaclass hacking (there's an explanation of what that means in the original), and he determines that he can do things easily in Python that he simply can't do at all in most languages, and cannot do easily in _any_ other language that he knows (he says he knows 20 or so): >This kind of thing is called metaclass hacking and is generally >considered fearsomely esoteric--deep black magic. Most >object-oriented languages don't support it at all; in >those that do (Perl being one), it tends to be a complicated >and fragile undertaking. I had been impressed by Python's >low coefficient of friction so far, but here was a real test. >How hard would I have to wrestle with the language to get it >to do this? > I knew from previous experience that the bout was >likely to be painful, even assuming I won, but I dived into >the book and read up on Python's metaclass facilities. The >resulting function is shown in Listing 3, and the code that >calls it is in Listing 4. That doesn't look too bad for deep black magic, does >it? Thirty-two lines, counting comments. Just from knowing >what I've said about the class structure, the calling code >is even readable. But the size of this code isn't the real >shocker. Brace yourself: this code only took me about ninety >minutes to write--and it worked correctly the first time >I ran it. To say I was astonished would have been positively >wallowing in understatement. It's remarkable enough when >implementations of simple techniques work exactly as >expected the first time; but my first metaclass hack in >a new language, six days from a cold standing start? >Even if we stipulate that I am a fairly talented hacker, >this is an amazing testament to Python's clarity and >elegance of design. Could be that I'm not showing programming expertise by saying crazy things like Python is an extremely fabulous language. If so then _Eric Raymond_ is also revealing the same pathetic lack of programming expertise. [A]mazing testament to Python's clarity and elegance of design - what a moron. (Just for the record, there's some stuff towards the end that's out of date: He explains he'd still use Perl for small things that use regular expressions heavily. Evidently he was using an early version of Python - the regular expressions in current Python are essentially identical to the regular expressions in Perl.) >c ************************ David C. Ullrich ==== > Well, I've already suggested that you should explain what your > reasons for this are. It occured to me that I instead of trying to > explain what's so great about Python I should just give a citation > to someone else's explanation. Eric Raymond actually is not hugely respected by me. And heck, I'm a hugely respected name in open-source programming circles. Isn't proof by authority fun? More to the point: Python is not good for programming large systems. Most importantly here, Eric simply doesn't have experience in large programming projects. Is Python better than C? I have no idea. C sucks too, though of course for very different reasons. The things that Eric says are wonderful about Python are even more wonderful about Common Lisp or Scheme, so I would advise using those languages instead, if you think that Eric's argument is compelling. > Could be that I'm not showing programming expertise by saying > crazy things like Python is an extremely fabulous language. If > so then _Eric Raymond_ is also revealing the same pathetic > lack of programming expertise. [A]mazing testament to Python's > clarity and elegance of design - what a moron. Yes, Eric does not have much programming expertise. Sorry, but true. Thomas ==== > More to the point: Python is not good for programming large systems. Aside from the fact that this statement is complete nonsense, you have a good point. More to the point, even if your statement were true it would not mean that Python was bad. There is a lot more to programming than large systems . - William Hughes Boy, we're *WAY* off topic for sci.math now. On the other hand this is a JSH thread. Sometimes ya just gotta say, What the heck. ==== On 16 Jan 2004 09:22:14 -0800, tb+usenet@becket.net (Thomas Bushnell, Well, I've already suggested that you should explain what your >> reasons for this are. It occured to me that I instead of trying to >> explain what's so great about Python I should just give a citation >> to someone else's explanation. Eric Raymond actually is not hugely respected by me. And heck, I'm a >hugely respected name in open-source programming circles. Isn't proof >by authority fun? More to the point: Python is not good for programming large systems. _Is_ that more to the point? This started when I posted 10 lines of code for the sake of doing one simple computation. _Supposing_ that in fact Python is not good for programming large systems, how does that show that my using it to solve one tiny problem shows a lack of expertise? Really. It may well be that Python is an extremely fabulous language for programming large systems would have been a crazy thing to say. But that's not what I said - I said Python is an extremely fabulous programming language. There's a slight difference between those two statements; the vast majority of programs that people write are not part of large systems. Note that the following question is a _question_. I'm not disputing your statement that Python is not good for programming large systems (I _am_ disputing the idea that that fact, if true, shows that Python is an extremely fabulous language is a crazy thing to say.) Question, just for my information: why is Python not good for large systems? >Most importantly here, Eric simply doesn't have experience in large >programming projects. Is Python better than C? I have no idea. The statements Python is better than C and C is better than Python both strike me as ridiculous; asking whether Python is better than C is like asking whether a hammer is better than a screwdriver. For many purposes Python is much better than C. For many other purposes trying to use Python instead of C would be utterly stupid. (For example, trying to write a record-breaking PrimePi(n) calculator in Python would probably be stupid (at least doing it in straight Python would be, having no experience with numpy I couldn't say how well suited it would be for something like that. If you thought that's what I was trying to do you didn't read my post very completely, in particular you must have missed the Just for giggles I whipped up a _totally_ brainless PrimePi in Python just now - it's nothing but a sieve, and not a clever sieve, a totally idiotic sieve, not particularly well written either, just top-of-the-brainless-head code: On the other hand, if for some reason I wanted to know the value of PrimePi(10000) and didn't have a CAS handy then spending a few minutes or hours writing a C program for the purpose instead of a few seconds or minutes on a Python program seems equally stupid.)) >C sucks too, though of >course for very different reasons. The things that Eric says are >wonderful about Python are even more wonderful about Common Lisp or >Scheme, so I would advise using those languages instead, if you think >that Eric's argument is compelling. I'm somewhat skeptical of what you say about Common Lisp and Scheme - _all_ the things he says are wonderful about Python are even more wonderful about those languages? Do people talk about how they're writing working Common List or Scheme to solve actual problems they want to solve a few hours after installing the language? Do those languages allow the same sort of transparent introspection with the same sort of ease? But never mind that, I don't know, so you may well be right about those two languages. For the record, I wasn't convinced by Eric's argument - long after I started using Python I saw consistent with my own experience. (Long ago someone told me to use Perl for something. So I learned a little Perl - I hated it, the whole thing just made no sense. I heard about Python viewing a language war on a Perl group - I tried it, and it took me a few days to become as proficient at Python as I'd become in Perl in about a month. So learning enough Perl to do a tiny cgi thing took me a month, so I'm stupid, fine - but then if a moron like me could do the same thing in Python in a few days it seems to _me_ (as well as to a lot of other people) that that says something about Python. And I loved it - it all made perfect sense, once you know a few things about how it works you get it, and you can easily learn the rest because you already got it. It happens fairly often that I have some problem to solve - some are small, some look big to me but I suspect they're still very small by comparison with the large projects you mention above. Using Python, once I've figured out exactly what I want to do it's simply never been any trouble to just _do_ it. Which is why I call it an extremely fabulous language...) >> Could be that I'm not showing programming expertise by saying >> crazy things like Python is an extremely fabulous language. If >> so then _Eric Raymond_ is also revealing the same pathetic >> lack of programming expertise. [A]mazing testament to Python's >> clarity and elegance of design - what a moron. Yes, Eric does not have much programming expertise. Sorry, but true. Fine. >Thomas ************************ David C. Ullrich ==== > On 16 Jan 2004 09:22:14 -0800, tb+usenet@becket.net (Thomas Bushnell, > > Note that the following question is a _question_. I'm not disputing > your statement that Python is not good for programming large > systems On the other hand I will dispute this statement. Python is good for programming large systems. >Is Python better than C? I have no idea. > > The statements Python is better than C and C is better than Python > both strike me as ridiculous; asking whether Python is better > than C is like asking whether a hammer is better than a screwdriver. > For many purposes Python is much better than C. For many other > purposes trying to use Python instead of C would be utterly stupid. > Hey, I have an idea. Let's combine Python and C. Use Python for the complicated high level stuff where speed of development is important but execution speed isn't, and C for the simpler low level stuff where execution speed is important. Wow, what a concept, combine an interpreted language with a compiled language. I'm brilliant, I'm going to be famous, I'm going to be rich .... What do you mean someone else thought of this first? Oh well, send back the Rolls. > (For example, trying to write a record-breaking PrimePi(n) calculator > in Python would probably be stupid (at least doing it in straight > Python would be, having no experience with numpy I couldn't > say how well suited it would be for something like that. Numpy will allow you to do a lot of matrix stuff quickly without having to go to C (or equivalent). However, if you have something specialized that needs to be done quickly, Python/numpy is not the way to go. - William Hughes <878yk9m4yp.fsf@becket.becket.net> ==== This is _Python_, an extremely fabulous language, but >> quite slow, being interpreted instead of compiled. I don't think you can prove programming skill by saying crazy things > like Python is an extremely fabulous language. Sorry. You're right. Real nerds denigrate every programming tool they find. The ultimate in geek cool is hating everything involving computers (while still playing with computers 23 hours a day, of course). -- What I've learned is that [mathematicians are] the gatekeepers, and seem to have almost absolute power when it comes to mathematics. -- James Harris, on All I Really Ever Needed to Know I Learned in /Ghostbusters/. ==== > You're right. Real nerds denigrate every programming tool they find. > The ultimate in geek cool is hating everything involving computers > (while still playing with computers 23 hours a day, of course). Huh? No, I simply mean that people don't right large systems in Python, and for very good reasons. <878yk9m4yp.fsf@becket.becket.net> <87wu7s8nil.fsf@phiwumbda.org> <878yk7yfoi.fsf@becket.becket.net> ==== You're right. Real nerds denigrate every programming tool they find. >> The ultimate in geek cool is hating everything involving computers >> (while still playing with computers 23 hours a day, of course). Huh? No, I simply mean that people don't right large systems in > Python, and for very good reasons. > I didn't mean to imply that *you* had that attitude, but your comment reminded me of this tendency among geeks. Earlier today, I happened to receive this fortune, which I think is very nice. One is not superior merely because one sees the world as odious. -- Chateaubriand (1768-1848) -- Many argue that its programmers have turned out shoddy programs, but [their] objective is to make profit, not superlative programs per se. By the profit criterion, Microsoft has been one of the greatest companies in the history of this country. -- ADTI defends Microsoft ==== > > Huh? No, I simply mean that people don't right large systems in > Python, and for very good reasons. Do they left them? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > > Huh? No, I simply mean that people don't right large systems in > Python, and for very good reasons. > > Do they left them? Oh blech, my fingers slipped. :-P Perhaps they do upset them in Python, however... ==== > > This is _Python_, an extremely fabulous language, but > quite slow, being interpreted instead of compiled. > > I don't think you can prove programming skill by saying crazy things > like Python is an extremely fabulous language. Sorry. > Oh boy a language war. Can I play? Perl resembles line noise. C++ sucks moon rocks through a staw badly (but in a semi object oriented manner). Only an absolute moron would design a language (e.g. Python) where white space is significant. The existence of the Obfuscated C Contest tells you all you need to know about C. LISP is favoured by the AI comunity. That's because no one with real intelligence would ever use it. Tcl is an abomination unto the heavens. Pascal isn't even any good as a teaching language. COBOL?, ADA?, One acronym DOD! The teaching of BASIC rots the mind and as such should be considered a criminal offence. (Dykstra) FORTRAN, bringing you 1950's technology today. APL .backwards is language This Stick Java where the SUN doesn't shine. There is a persistent rumour that someone fed a SNOBOL compiler FORTRAN source and didn't get a syntax error. MATLAB. Despite decades of effort, a Lapack wrapper is still a Lapack wrapper. _______ It really doesn't matter what you put upon the list. They'd none of them be missed, They'd none of them be missed WATIAC, a decimal machine language, with op codes that look suspiciously like octal, simulated in COBOL. Now there's a language for real progammers! - William Hughes ==== >> >> This is _Python_, an extremely fabulous language, but >> quite slow, being interpreted instead of compiled. >> >> I don't think you can prove programming skill by saying crazy things >> like Python is an extremely fabulous language. Sorry. >> Oh boy a language war. Can I play? Perl resembles line noise. C++ sucks moon rocks through a staw badly >(but in a semi object oriented manner). Only an absolute moron would design a language >(e.g. Python) where white space is significant. Well of course your post was not meant to be taken seriously (or rather it seems you may have a serious point, but these comments are not meant to be taken literally). But being a zealot I nonetheless can't refrain from saying that people who actually _believe_ that the whitespace thing is bad should see . >The existence of the Obfuscated C Contest tells >you all you need to know about C. LISP is favoured by the AI comunity. That's >because no one with real intelligence would >ever use it. Tcl is an abomination unto the heavens. Pascal isn't even any good as a teaching language. COBOL?, ADA?, One acronym DOD! The teaching of BASIC rots the mind and as >such should be considered a criminal offence. (Dykstra) FORTRAN, bringing you 1950's technology today. APL .backwards is language This Stick Java where the SUN doesn't shine. There is a persistent rumour that someone >fed a SNOBOL compiler FORTRAN source and >didn't get a syntax error. MATLAB. Despite decades of effort, a Lapack >wrapper is still a Lapack wrapper. _______ It really doesn't matter what you put upon the list. > They'd none of them be missed, > They'd none of them be missed WATIAC, a decimal machine language, with op codes >that look suspiciously like octal, simulated >in COBOL. Now there's a language for >real progammers! - William Hughes ************************ David C. Ullrich ==== Only an absolute moron would design a language >(e.g. Python) where white space is significant. > > Well of course your post was not meant to be taken > seriously (or rather it seems you may have a serious > point, but these comments are not meant to be > taken literally). I didn't have a serious point, I just threw out a bunch of languages and insulted each one. In some cases I used the canonical insult for the languague. For Python the canonical insult is the use of significant white space. I am a Python fan. I don't particularly like the significant white space, but it doesn't bother me either. The biggest problem is that the significant white space issue distracts from other much more important issues. (It's a bit like what would happen if the ANSI committee decided to enforce the One True Brace Style in standard C). - William Hughes Maybe we should switch this to comp.lang.c People here just don't get upset enough about off topic posts. ==== In sci.math, William Hughes on 15 Jan 2004 14:36:10 -0800 <4d5e4663.0401151436.665cb562@posting.google.com>: >> >> This is _Python_, an extremely fabulous language, but >> quite slow, being interpreted instead of compiled. >> >> I don't think you can prove programming skill by saying crazy things >> like Python is an extremely fabulous language. Sorry. >> > > Oh boy a language war. Can I play? > > Perl resembles line noise. > > C++ sucks moon rocks through a staw badly > (but in a semi object oriented manner). > > Only an absolute moron would design a language > (e.g. Python) where white space is significant. > > The existence of the Obfuscated C Contest tells > you all you need to know about C. > > LISP is favoured by the AI comunity. That's > because no one with real intelligence would > ever use it. > > Tcl is an abomination unto the heavens. > > Pascal isn't even any good as a teaching language. > > COBOL?, ADA?, One acronym DOD! > > The teaching of BASIC rots the mind and as > such should be considered a criminal offence. (Dykstra) > > FORTRAN, bringing you 1950's technology today. > > APL .backwards is language This > > Stick Java where the SUN doesn't shine. > > There is a persistent rumour that someone > fed a SNOBOL compiler FORTRAN source and > didn't get a syntax error. > > MATLAB. Despite decades of effort, a Lapack > wrapper is still a Lapack wrapper. > > _______ It really doesn't matter what you put upon the list. > They'd none of them be missed, > They'd none of them be missed > > WATIAC, a decimal machine language, with op codes > that look suspiciously like octal, simulated > in COBOL. Now there's a language for > real progammers! > > - William Hughes Hey, you forgot C#, Visual Basic, ASP, most versions of assembly, Modula-2, Modula-3, Ruby, SQL, WATFOR and WATFIV (though they're really dialects of Fortran), Postscript (yes, it's a programming language!) and FORTH... :-) C#: Tomorrow's technology today. Maybe. Wait until the bugpatch, which will be out sometime, erm, next month. Yeah, next month. Oh, and it's sort of like Java except not really. We're not sure yet. Visual Basic: It's pretty basic, and it only can be used visually. ASP: It bites. Hard. Assembly: Sometimes it's required. Most times it should sit there in little pieces on the floor, as that's what it looks like. Modula-2: Pascal squared. Modula-3: Pascal cubed. Ruby: Is it a gem of a language? I don't know, really. SQL: Please, someone, write one. WATFOR: ... and give it what for. At least FORTRAN has an excuse. WATFIV: A SQL to WATFOR. And you know what sequels do... Postscript/FORTH: Both of these are backwards, too, although I'll admit there's a certain elegance to 1 5 + . as it doesn't need parentheses. Of course LISP took them all anyway... TeX: An interesting entry, actually, and I think it does qualify as a computer language although that's a bit like saying an automobile qualifies as a guerney. (Or vice versa.) JCL: You don't want to know. Fortunately, Unix usurped 'dd' and made it a lot clearer -- which for Unix is amazing... :-) DCL: VMS's answer to shell scripts, complete with F$PARSE(). Bourne: Unix's answer to how do I do this as cryptically as possible? REXX: IBM's answer to shell scripts, and probably someone's dog's name. There are times it looks like the end product thereof. HTML: LISP with angle brackets. XML: LISP with angle brackets and question marks. XHTML: The worst of both. MathML: Like XHTML only mathematical. XSL: Definitely in excess. ASN.1: Otherwise known as Abstruse, Strange, and Non-comprehensible. Even parsing the MIB can give one a headache. The sad thing is: this might very well be the most efficient, absent compression. ASN.2: I hope not. SOAP: Was it ever simple? How is XML an object? How does one access an object using XML? This one should be put out of its misery. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > Hey, you forgot C#, Visual Basic, ASP, most versions of assembly, > Modula-2, Modula-3, Ruby, SQL, WATFOR and WATFIV (though they're > really dialects of Fortran), Postscript (yes, it's a programming > language!) Look, if you are going to include things that weren't intended to be used as a general purpose programming language but can be so used we have to include most text editors, most word processors, most spreadsheets, etc. etc. etc. we'll never finish!. Of course we'll never finish even if we stick to general purpose programming languages but this is even worse. It's sort of like the difference between the infinity of the integers and the infinity of the reals (note the valiant attempt to get back on topic). - William Hughes P.S. Have I explained the self evident error in logic made by the evil cantorians? and FORTH... > > :-) > > C#: Tomorrow's technology today. Maybe. Wait until the bugpatch, > which will be out sometime, erm, next month. Yeah, next month. > Oh, and it's sort of like Java except not really. We're not > sure yet. > Visual Basic: It's pretty basic, and it only can be used visually. > ASP: It bites. Hard. > Assembly: Sometimes it's required. Most times it should sit there > in little pieces on the floor, as that's what it looks like. > Modula-2: Pascal squared. > Modula-3: Pascal cubed. > Ruby: Is it a gem of a language? I don't know, really. > SQL: Please, someone, write one. > WATFOR: ... and give it what for. At least FORTRAN has an excuse. > WATFIV: A SQL to WATFOR. And you know what sequels do... > Postscript/FORTH: Both of these are backwards, too, although I'll admit > there's a certain elegance to > 1 5 + . > as it doesn't need parentheses. Of course LISP took them > all anyway... > TeX: An interesting entry, actually, and I think it does qualify as > a computer language although that's a bit like saying an automobile > qualifies as a guerney. (Or vice versa.) > JCL: You don't want to know. Fortunately, Unix usurped 'dd' > and made it a lot clearer -- which for Unix is amazing... :-) > DCL: VMS's answer to shell scripts, complete with F$PARSE(). > Bourne: Unix's answer to how do I do this as cryptically as possible? > REXX: IBM's answer to shell scripts, and probably someone's dog's name. > There are times it looks like the end product thereof. > HTML: LISP with angle brackets. > XML: LISP with angle brackets and question marks. > XHTML: The worst of both. > MathML: Like XHTML only mathematical. > XSL: Definitely in excess. > ASN.1: Otherwise known as Abstruse, Strange, and Non-comprehensible. > Even parsing the MIB can give one a headache. The sad thing > is: this might very well be the most efficient, absent > compression. > ASN.2: I hope not. > SOAP: Was it ever simple? How is XML an object? How does one > access an object using XML? This one should be put out of its misery. ==== > JCL: You don't want to know. Fortunately, Unix usurped 'dd' > and made it a lot clearer -- which for Unix is amazing... :-) I was exposed to JCL once. My therapist says I am making good progress. - William Hughes ==== XML: LISP with angle brackets and question marks. > XHTML: The worst of both. these aren't even programming languages (ie. Turing complete). !Q ==== On 15 Jan 2004 10:39:26 -0800, tb+usenet@becket.net (Thomas Bushnell, This is _Python_, an extremely fabulous language, but >> quite slow, being interpreted instead of compiled. I don't think you can prove programming skill by saying crazy things >like Python is an extremely fabulous language. Sorry. Well, actually if you look back you'll see I wasn't trying to prove I was a skilled programmer, my point was almost the opposite, that it takes essentially no skill to beat the benchmark that had been mentioned. _But_ surely you're not serious. Python is in fact an extremely fabulous language. Not for every purpose, but for _many_ purposes it's the best thing out there. Before I start rambling about what's so great about Python, why don't you explain what's not so great about it? Might be more efficient if I knew what to refute. (Some complaints my reply would be no, it's not a fabulous language for that purpose, other complaints I'd insist it _was_ fabulous and try to explain why. What complaint did you have in mind?) >c ************************ David C. Ullrich ==== > >In sci.math, James Harris ><3c65f87.0401140834.1d59810c@posting.google.com>: >> While I know C and C++, and have at times felt hampered by certain >> limitations of Java, I've concluded that it's the best programming >> language for wideranging discussions, as it's accessible to people who >> haven't programmed before who aren't familiar with Unix, but are >> Windows people. >> >> I need as many people capable of checking things like programs that I >> post as I can get, so I'm making a push for Java. >> >> It turns out that for most of you it's as simple as going to the Sun >> website, where you can get a free download: >> >> http://java.sun.com/j2se/1.4.2/download.html >> >> I'd suggest Sun's tutorials for those who aren't at all familiar with >> Java and running Java programs. >> >> To see some of my work, and a program that you can try running, see my >> blog archives: >> >> >> You can also do a search at Google Groups for PrimeCountH.java, >> >> The point for me here is to show you first that what I've discovered >> works, and then I think it'll be easier for me to explain unique >> features of my discovery, as I look towards finding more efficient and >> effective ways of getting the word out. >> >> >> James Harris Did you *really* want me to run Yet Another Primecounter Contest? :-P As it is, your Java implementation makes snails look >fast; on my Official Test System(tm) it took 3 minutes >to compute pi(10000)=1229 (which, to its credit, is the >correct answer). > > My gosh, it's _that_ slow? I guess I wasn't paying attention > during the contest... > > Just for giggles I whipped up a _totally_ brainless PrimePi > in Python just now - it's nothing but a sieve, and not a > clever sieve, a totally idiotic sieve, not particularly well > written either, just top-of-the-brainless-head code: > > def Sieve(ran): > for d in range(2, (len(ran))/2+1): > for j in range(2*d, len(ran), d): > ran[j]=0 > > def Primes(n): > ran = range(n+1) > Sieve(ran) > return filter(lambda x: x > 1, ran) > > def PrimePi(n): > return len(Primes(n)) > > This is _Python_, an extremely fabulous language, but > quite slow, being interpreted instead of compiled. And > (on and admittedly fast machine, 2000GH/512MB) > it takes much less than a second to find Pi(10000); > didn't actually time anything, but if I say > > print PrimePi(10000) > > the 1229 appears with no perceptible delay. Presumably > it would be pretty slow for larger values, for reasons beyond > the understanding of some of us, but it gives Pi(100000) > in less than two seconds... > > Maybe there _is_ some significance to James' work here. Well, if I understand James' argument, _your_ program doesn't give you a partial differential equation, so _you_ won't be able to solve the Reimann Hypothesis. ;) > Like if we had a contest for the slowest possible Pi(n), > maybe. > >You may want to look at your usage of Math.sqrt(). A recoding of my first LegendrePhi submission took half a second, >(to run, that is; the recoding itself took the better part of >a half hour) and of course also came up with the correct answer. And remember, I'm a lightweight in this department -- if Christian >Bau were to really sink his teeth into this particular subproblem, >he'd make both of us look ridiculous. :-) (Which is admittedly >fine by me; I can't say I understand his algorithm but it's *fast*!) > > > ************************ > > David C. Ullrich ==== > > Well, if I understand James' argument, _your_ program doesn't give you > a partial differential equation, so _you_ won't be able to solve the Reimann > Hypothesis. ;) I'm still waiting for James to show any sign that he has the foggiest idea what the Riemann Hypothesis *is*... -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== In sci.math, David C Ullrich on Thu, 15 Jan 2004 05:25:26 -0600 : > >>In sci.math, James Harris >>><3c65f87.0401140834.1d59810c@posting.google.com>: > While I know C and C++, and have at times felt hampered by certain > limitations of Java, I've concluded that it's the best programming > language for wideranging discussions, as it's accessible to people who > haven't programmed before who aren't familiar with Unix, but are > Windows people. > > I need as many people capable of checking things like programs that I > post as I can get, so I'm making a push for Java. > > It turns out that for most of you it's as simple as going to the Sun > website, where you can get a free download: > > http://java.sun.com/j2se/1.4.2/download.html > > I'd suggest Sun's tutorials for those who aren't at all familiar with > Java and running Java programs. > > To see some of my work, and a program that you can try running, see my > blog archives: > > > You can also do a search at Google Groups for PrimeCountH.java, > > The point for me here is to show you first that what I've discovered > works, and then I think it'll be easier for me to explain unique > features of my discovery, as I look towards finding more efficient and > effective ways of getting the word out. > > > James Harris >Did you *really* want me to run Yet Another Primecounter Contest? :-P >As it is, your Java implementation makes snails look >>fast; on my Official Test System(tm) it took 3 minutes >>to compute pi(10000)=1229 (which, to its credit, is the >>correct answer). > > My gosh, it's _that_ slow? I guess I wasn't paying attention > during the contest... I was referring to his Java implementation, that he's posted http://home.earthlink.net/~ewill3/math/primecounters/ :-) It doesn't match either of his two submissions for my contest so I'm a little puzzled as to what he's thinking. To its credit it has no memoization or special cases. (Of course, that may be its problem...) > > Just for giggles I whipped up a _totally_ brainless PrimePi > in Python just now - it's nothing but a sieve, and not a > clever sieve, a totally idiotic sieve, not particularly well > written either, just top-of-the-brainless-head code: > > def Sieve(ran): > for d in range(2, (len(ran))/2+1): > for j in range(2*d, len(ran), d): > ran[j]=0 > > def Primes(n): > ran = range(n+1) > Sieve(ran) > return filter(lambda x: x > 1, ran) > > def PrimePi(n): > return len(Primes(n)) > > This is _Python_, an extremely fabulous language, but > quite slow, being interpreted instead of compiled. And > (on and admittedly fast machine, 2000GH/512MB) > it takes much less than a second to find Pi(10000); > didn't actually time anything, but if I say > > print PrimePi(10000) > > the 1229 appears with no perceptible delay. Presumably > it would be pretty slow for larger values, for reasons beyond > the understanding of some of us, but it gives Pi(100000) > in less than two seconds... My system ran it in .311 seconds. > > Maybe there _is_ some significance to James' work here. > Like if we had a contest for the slowest possible Pi(n), > maybe. Naw. Christian Bau got that, erm, honor. Of course, that's mostly because he was coding for extreme simplicity in this algorithm, and wasn't trying to be the fastest. (It's a very simple, elegant, and rather stupid algorithm; basically, it loops over the integers from 1 to N then a subloop tests for primality in the simplest way possible using the modulo ('%') operator.) But when computing pi(10000), it's faster than my sieve...and a *lot* faster than James' Java. (0.051 seconds) A recoding into Java resulted in a .61s run. don't know how well it'll work for pi(10000). http://www.cbau.freeserve.co.uk/ for the details thereon. > >>You may want to look at your usage of Math.sqrt(). >A recoding of my first LegendrePhi submission took half a second, >>(to run, that is; the recoding itself took the better part of >>a half hour) and of course also came up with the correct answer. >And remember, I'm a lightweight in this department -- if Christian >>Bau were to really sink his teeth into this particular subproblem, >>he'd make both of us look ridiculous. :-) (Which is admittedly >>fine by me; I can't say I understand his algorithm but it's *fast*!) > > > ************************ > > David C. Ullrich -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== Been too long since Ive looked at this stuff. If I have two coordinates (300, 1500) and (500, 4000), Given any x, what is the *#$@ formula to calculate the corresponding y (or x, given y) between these two points? ==== In sci.math, BCC <42GNb.11269$3q7.9922@newssvr25.news.prodigy.com>: > Been too long since Ive looked at this stuff. > > If I have two coordinates (300, 1500) and (500, 4000), > > Given any x, what is the *#$@ formula to calculate the corresponding y (or > x, given y) between these two points? > > There are a few ways of solving this; the simplest is arguably creating a parameter t and linear functions x(t) and y(t) such that: x(0) = 300 x(1) = 500 y(0) = 1500 y(1) = 4000 It turns out x(t) = 300 + (500-300)*t and y(t) = 1500+(4000-1500)*t, or x(t) = 300+200*t and y(t) = 1500+2500*t. One can then substitute: x(t) = 300+200*t implies t(x) = (x-300)/200 y = 1500 + 2500*(x-300)/200 = 1500 + 12.5*x - 3750 = 12.5*x - 2250 As a check, feeding x = 300 gives us y = 1500 and x = 500 gives us y = 4000, as it should. Another method is to equate y = ax+b and solve for a and b; one gets two equations in two unknowns by plugging in the two above coordinate points. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > Been too long since Ive looked at this stuff. > > If I have two coordinates (300, 1500) and (500, 4000), > > Given any x, what is the *#$@ formula to calculate the corresponding y (or > x, given y) between these two points? > > > This is called the two-point form of the equation of a line. If the line passes through (x1, y1) and (x2, y2), then all points on the line (x, y) satisfy (y-y1)/(x-x1) = (y2-y1)/(x2-x1). If x1=x2, then the points are (x1, y) for any y. Similarly if y1=y2. To get a feel for what is going on, plot a few cases. ==== BCC > Been too long since Ive looked at this stuff. If I have two coordinates (300, 1500) and (500, 4000), Given any x, what is the *#$@ formula to calculate the corresponding y (or > x, given y) between these two points? If the equation is y = ax + b then we need 1500 = a*300 + b 4000 = a*500 + b from which we get a = 12.5 and b = -2250. So, the equation is y = 12.5 x - 2250. LH ==== I have one number, zero, that I like better than others. 0's the solution to every algebraic equation. That's why 0's my favorite number. What's your favorite? And why? Garry Denke, Geologist Denoco Inc. of Texas ==== > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. > That's why 0's my favorite number. > What's your favorite? > And why? > > Garry Denke, Geologist > Denoco Inc. of Texas peanut butter and jelly. Charlie Volkstorf ==== > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. > That's why 0's my favorite number. > What's your favorite? > And why? > > peanut butter and jelly. Thinking about it, the closest number I get is 57, but I do not even know whether that is valid. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > > I have one number, zero, that I like better than others. > > 0's the solution to every algebraic equation. > > That's why 0's my favorite number. > > What's your favorite? > > And why? > > peanut butter and jelly. Thinking about it, the closest number I get is 57, but I do not even > know whether that is valid. 26 = number of required dimensions for string theory to work. l8r, Mike N. Christoff ==== > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. That's why 0's my favorite > number. > What's your favorite? My favorite is 17. > And why? It's the first random number. Frank ==== > > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. That's why 0's my favorite > number. > What's your favorite? > > My favorite is 17. > The third Fermat number, or the first to end in 7. > And why? > > It's the first random number. > > Frank What makes it random? -- ==== > > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. That's why 0's my favorite > number. > What's your favorite? > > My favorite is 17. > > And why? > > It's the first random number. > > Frank I was going to say 137, since it proves the existence of God and gives meaning to the universe ... but I like Frank's answer better. Rick work@ostrander.de ==== I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. That's why 0's my favorite > number. > What's your favorite? My favorite is 17. And why? It's the first random number. Frank > I was going to say 137, since it proves the existence of God and gives > meaning to the universe I thought that number was 42? Damn! My entire system of beliefs is shattered! Oh well. All praise the mighty 137!! l8r, Mike N. Christoff ==== > > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. That's why 0's my favorite > number. > What's your favorite? > > My favorite is 17. > > And why? > > It's the first random number. > > Frank Does anybody really believe Gregory Chaitin that Mathematics is random? How could computers work, then? I think he's pretty confused. 1. Non-deterministic (random) = life 2. Deterministic = science 3. Deterministic and doesn't involve your 5 senses (input) = Mathematics 4. Deterministic but not calculable = Theory of Computation Chaitin is confusing # 1 and # 4. BTW: In my Axiomatic Basis for Everything (all of Logic, Mathematics and Computer Science) the fundamental axiom is I#yes(I) that means that it's all deterministic (not random.) The # means solves or creates when evaluated and yes(I) means the set of numbers that program (Turing Machine) number I halts on, so that program I creates the set of inputs on which it halts yes. That's actually the axiom for the Theory of Computation. For English (e.g. the Liar Paradox) it's I#ts(I) which says that English sentence I evaluates to the truth of I. In general we have I#run(I) but in each case it is saying that evaluating an object is deterministic. Charlie Volkstorf ==== > 1. Non-deterministic (random) = life Not quite. Life is not random. It is merely complex. > 2. Deterministic = science Not quite. Science is not deterministic. It merely predicts with sufficient accuracy for sufficient number of cases. > 3. Deterministic and doesn't involve your 5 senses (input) = > Mathematics Not quite. Mathematics = assumptions => cognition => senses. > 4. Deterministic but not calculable = Theory of Computation Not Quite. Theory of computation is nothing more than the concept of addition. > Charlie Volkstorf Many Happy Returns of Janus. ==== > > 1. Non-deterministic (random) = life > > Not quite. Life is not random. It is merely complex. Is there anything that is random? > 2. Deterministic = science > > Not quite. Science is not deterministic. It merely predicts > with sufficient accuracy for sufficient number of cases. If something is not deterministic, then how can science relate to it? > 3. Deterministic and doesn't involve your 5 senses (input) = > Mathematics > Not quite. Mathematics = assumptions => cognition => senses. Which senses would you need to develop or perform mathematics? > 4. Deterministic but not calculable = Theory of Computation > > Not Quite. Theory of computation is nothing more than the > concept of addition. Then why don't they call it Arithmetic? > Charlie Volkstorf > > Many Happy Returns of Janus. ==== 1. Non-deterministic (random) = life Not quite. Life is not random. It is merely complex. > I'm not saying I agree/disagree but this is still an open question. > 2. Deterministic = science Not quite. Science is not deterministic. It merely predicts > with sufficient accuracy for sufficient number of cases. > Ok. > 3. Deterministic and doesn't involve your 5 senses (input) = > Mathematics Not quite. Mathematics = assumptions => cognition => senses. > I think what Charlie was getting at is that math (as it stands today) is not empirical. However the idea of 'empirical math' / 'experimental math' is starting to come together. Though largely by today's standards, the very phrase 'empirical math' is an oxymoron. > 4. Deterministic but not calculable = Theory of Computation Not Quite. Theory of computation is nothing more than the > concept of addition. > This is way off. l8r, Mike N. Christoff ==== > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. That's why 0's my favorite > number. > What's your favorite? 0/0 It's just so playful! Charlie Volkstorf So what all are the properties of 0 (and 1, for that matter) and does it have to be the same number for all of those properties? E.g. X+0=X and X*0=0 but could there be two different numbers that do that? What's it about + and * that makes it the same number? ==== In sci.logic, Garry Denke on 15 Jan 2004 16:26:13 -0800 <4e63857.0401151626.7573b27b@posting.google.com>: > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. > That's why 0's my favorite number. > What's your favorite? > And why? x^2 = 1 has a solution x = 0? Somehow, I doubt it. My favorite numbers are e, i, pi, and -1. They just interrelate so well together... :-) > > Garry Denke, Geologist > Denoco Inc. of Texas -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > In sci.logic, Garry Denke > <4e63857.0401151626.7573b27b@posting.google.com>: > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. > That's why 0's my favorite number. > What's your favorite? > And why? > > x^2 = 1 has a solution x = 0? the knowledge of nothing is everything, so nothing squared plus one equals nothing. here, i will write the algebraic equation out so i can solve it for you. x^2 + 1 = 0 solution for the equation is 0 the number (the number to the right of equals sign). challenge: post any algebraic equation in the form of it equaling 0 the number, and i will solve your algebraic equation. > Somehow, I doubt it. here are some more algebraic equations whose solution is 0 the number (the number to the right of the equals sign): radical equation... sqrt(x - 10) - 4 = 0 quadratic equation... x^2 - 5x + 3 = 0 exponential equation... e^2x - 3e^x + 2 = 0 logarithmic equation... 6log(x^2 + 1) - x = 0 trigonometric equation... 3tan^2 x - 1 = 0 > My favorite numbers are e, i, pi, and -1. > They just interrelate so well together... if you post it i will answer. garry denke, geologist denoco inc. of texas ==== > In sci.logic, Garry Denke > <4e63857.0401151626.7573b27b@posting.google.com>: > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. > That's why 0's my favorite number. > What's your favorite? > And why? x^2 = 1 has a solution x = 0? the knowledge of nothing is everything, so > nothing squared plus one equals nothing. here, i will write the algebraic equation out so i can solve it for you. x^2 + 1 = 0 solution for the equation is 0 the number > (the number to the right of equals sign). challenge: post any algebraic equation in the form of it equaling > 0 the number, and i will solve your algebraic equation. Somehow, I doubt it. here are some more algebraic equations whose solution is 0 the number > (the number to the right of the equals sign): Actually, 1 is the solution to every equation: radical equation... sqrt(x - 10) - 3 = 1 quadratic equation... x^2 - 5x + 4 = 1 exponential equation... e^2x - 3e^x + 3 = 1 logarithmic equation... 6log(x^2 + 1) - x + 1 = 1 trigonometric equation... 3tan^2 x = 1 l8r, Mike N. Christoff ==== > I have one number, zero, that I like better than others. > 0's the solution to every algebraic equation. > That's why 0's my favorite number. > What's your favorite? > And why? > > Garry Denke, Geologist > Denoco Inc. of Texas The reason garry Dense loves zero so much is that it tells you everything there is to know about him. ==== -- ZHANG Yan http://www.ntu.edu.sg/home5/pg01308021 Many thanks for the reply. The pdf of Y can be determined independently from Z but depend on the pdf of >X. For example, the sample value y of Y is uniformly distributed in [0,x] >where x is the sample value of X. The pdf of Z can be determined independently from Y but depend on the pdf of >X. For example, the sample value z of Z is uniformly distributed in [0, 10 * >x] where x is the sample value of X. That is, both of the sample values of Y and Z are reliant on the sample >value of X. Then, >1. are Y and Z dependent? >2. if pdf of Y and Z are given, can we precisely obtain the pdf of A? >3. any approximation algorithm to obtain the pdf of A from the pdf of Y and >Z by considering that Y and Z are virtually independent?? > As I said before, though Y and Z are *conditionally* independent > gilven X, they are *not* in general independent (not even > virtually). In previous posts, I have given formulae which determine > the density of A. Here is another: P{A>a} = EP(Y>a, Z>a | X) = E[P(Y>a | X) P(Z>a | X)] > = int(x=0..infty, int(w=a..infty, f(w|x)) int(w=a..infty, g(w|x)) h(x)) where f(.|x) and g(.|x) are respectively the conditional densities > of Y and Z given X=x and h is the marginal density of X. > (Again, I assume that X, Y, and Z are jointly continuous, and that > X is nonnegative.) Taking the derivative, the density of A at a is int(x=0..infty, h(x) [f(a|x) int(w=a..infty, g(w|x)) + g(a|x) > int(w=a..infty, f(w|x))]) -- > Stephen J. Herschkorn herschko@rutcor.rutgers.edu X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In , on 01/14/2004 at 09:37 AM, azilla said: > I know that the vector space R^n over R(real) is not >isomorphic to the vector space R over R. A book told me that if we >regard R^n and R as vector spaces over Q(rational), R^n is >isomorphic to R. Is that true? In what set theory? I assume that your book is using a set theory that includes the Axiom of Choice or an equivalent. >How does the book come up with that? Zorn's Lemma. You order all partial bases by inclusion, and take the union of any maximal chain. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > Dear all, > I know that the vector space R^n over R(real) is not isomorphic to > the vector space R over R. A book told me that if we regard R^n and R as > vector spaces over Q(rational), R^n is isomorphic to R. Is that true? How > > But R^n and R are isomorphic as Q-vector spaces only way out there in Axiom of Choice Land. Without AC one cannot prove it. One cannot *actually write down* an isomorphism between R^2 and R. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== >Dear all, > I know that the vector space R^n over R(real) is not isomorphic to >the vector space R over R. A book told me that if we regard R^n and R as >vector spaces over Q(rational), R^n is isomorphic to R. Is that true? How With the Axiom of Choice, there is a basis for R over Q, and similarly this gives a basis for R^n over Q. As both bases have the cardinality of the continuum, they are isomorphic. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== In , on 01/14/2004 at 09:09 AM, whopkins@csd.uwm.edu (Alfred Einstead) said: >In general, the algebra consisting of all C^{infinity} functions >(i.e. functions continuously differentiable to all orders) for a >compact Hausdorff space How do you define differentiable if you do not require that the space be a manifold? -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > Can you brilliant and gifted Mathematicians and Physicists help out > this poor soul and justify your acceptance of relativistic and quantum > mechanics by providing the axioms of a simultaneously discrete and > continuous space. Look up the paper in the following link. It posits a stationary and elastic continuous space with light as discrete wave packets in this continuous medium moving at constant c wrt to it. http://www.journaloftheoretics.com/Links/Papers/Seto.pdf Ken Seto ==== >> Can you brilliant and gifted Mathematicians and Physicists help out >> this poor soul and justify your acceptance of relativistic and quantum >> mechanics by providing the axioms of a simultaneously discrete and >> continuous space. > To go that route, you should shift focus away from point sets to > function algebras. For instance, a circle is equivalently > described by the following algebra generated from the elements: > ...,z_{-3}, z_{-2}, z_{-1}, z_0, z_1, z_2, z_3,... > with the multiplication rule: > z_a z_b = z_{a+b} > the corresponding functions being z_a <-> Z_a(x) = exp(2 pi i a x). > In general, the algebra consisting of all C^{infinity} functions > (i.e. functions continuously differentiable to all orders) for > a compact Hausdorff space comprises a commutative C^* algebra > -- and vice versa (all commutative C^* algebras are function > spaces for some compact Hausdorff space). > To generalize to something that encompasses both continuous and > discrete spaces, you generalize the space of functions from > the C^{infinity) functions to a suitable class of generalized > functions -- in particular, a class large enough to include > the delta functions. > Such a class, likewise, is to be closed under products (and > sums and multiplication by constants, as usual), closed > under differentiation, and is to include both the delta > functions and the C^{infinity} functions. Does this math properly express the physical issue ? I know that quantities, which are related to the continuum, can have discrete _values_ . I mean concretely integration constants from partial differential equations. I could demonstrate that the integration constants for most stable solutions of the Einstein- Thus, one must not provide the axioms of a simultaneously discrete and continuous space. Continuous time & space are enough to understand that all. BTW, our friend Bill Hobba points out the Kochem-Specker theorem at each occasion. If I understood it properly, it would even support my insights. May be, you have an opinion about it ? Ulrich ==== > In , on 01/12/2004 > at 07:49 PM, j.schoenfeld@programmer.net (John Schoenfeld) said: >>Can you brilliant and gifted Mathematicians and Physicists help out >>this poor soul and justify your acceptance of relativistic and >>quantum mechanics by providing the axioms of a simultaneously >>discrete and continuous space. > First, the issue is one of Physics and not Mathematics. Second, > physicists don't normally do axioms. Third, there is nothing in either > SR or QT that would require, or even allow, a simultaneously discrete > and continuous space. QFT requires a spacetime that is a real > manifold. The perceived discreteness is in certain measured > quantities, not in spacetime. Agree fully. The certain measured quantities are exactly the first integration constants of the Einstein-Maxwell equations. I could indeed demonstrate that the integration constants for http://home.t-online.de/home/Ulrich.Bruchholz/ Ulrich info at bruchholz minus acoustics dot de ==== > Dear John Schoenfeld: > > Can you brilliant and gifted Mathematicians and Physicists help out > this poor soul and justify your acceptance of relativistic and quantum > mechanics by providing the axioms of a simultaneously discrete and > continuous space. > > If the only tool you have is a hammer, then all problems look like nails. Well that's the problem. When the only tool you have is a derivative, everything sounds like an integral. > But if you have two tools, say a hammer and a saw... > one that joins problem sets into a neat whole, and > one that cuts the problem into its contituent parts, > then you can apply the tool that is required. That does works. But since it only works in Dualistic Philosophy, it only works on the weekends. > The Universe is both discrete and continuous, depending on what you > need/want to describe. > > But I am not a brilliant and gifted Mathematician and Physicist. That's easy to explain though. The only gifted Mathematician who ever lived was semi-continous, rather than continous. > David A. Smith ==== > Dear John Schoenfeld: > > Can you brilliant and gifted Mathematicians and Physicists help out > this poor soul and justify your acceptance of relativistic and quantum > mechanics by providing the axioms of a simultaneously discrete and > continuous space. > > If the only tool you have is a hammer, then all problems look like nails. > > Well that's the problem. When the only tool > you have is a derivative, everything sounds > like an integral. > > But if you have two tools, say a hammer and a saw... > one that joins problem sets into a neat whole, and > one that cuts the problem into its contituent parts, > then you can apply the tool that is required. > > That does works. But since it only works in > Dualistic Philosophy, it only works on the weekends. > > > > The Universe is both discrete and continuous, depending on what you > need/want to describe. > > But I am not a brilliant and gifted Mathematician and Physicist. > > That's easy to explain though. > The only gifted Mathematician who ever lived > was semi-continous, rather than continous. Or as modern science is usually taught. The Physicist asked the Mathematician: Quote-Unquote: Is there a doctor in the house? May Spock be with you. The Mathematician answered: Slash-Unslash: No, but if you get the brain surgeon a male nurse, he'll fake it. May Bach be with you. > > David A. Smith ==== Can you brilliant and gifted Mathematicians and Physicists help out > this poor soul and justify your acceptance of relativistic and quantum > mechanics by providing the axioms of a simultaneously discrete and > continuous space. Sure! > > Awwww... ain't that cute? The two little cranks are playing together! You appear to be undergoing a massive emotional flux. JS ==== > > > It must be pointed out that Varney can't tell the difference between > deceleration and -acceleration (and he has a PhD). > > There is no essential difference. Deceleration is just acceleration > multiplied by -1. Either one means a change in velocity wrt time. Deceleration means velocity magnitude is decreasing with time, which is satisfied by the condition v.a < 0. Putting a negative symbol infront of a does not imply decreasing velocity. > Since acceleration is a vector all that counts is magnitude and > direction. A vector is a vector is a vector. JS ==== Dear John Schoenfeld: ... > There is no essential difference. Deceleration is just acceleration > multiplied by -1. Either one means a change in velocity wrt time. Deceleration means velocity magnitude is decreasing with time, which > is satisfied by the condition v.a < 0. Putting a negative symbol > infront of a does not imply decreasing velocity. Not strictly true either, John. The actual defintion should be more along the lines of: v x a = 0 and v . a < 0 shouldn't it? David A. Smith ==== > Dear John Schoenfeld: > > ... > There is no essential difference. Deceleration is just acceleration > multiplied by -1. Either one means a change in velocity wrt time. Deceleration means velocity magnitude is decreasing with time, which > is satisfied by the condition v.a < 0. Putting a negative symbol > infront of a does not imply decreasing velocity. > > Not strictly true either, John. The actual defintion should be more along > the lines of: > v x a = 0 and v . a < 0 > shouldn't it? I don't believe so, since the cross product does not work for all dimensions n > 0 which is a requirement (as far as I know, that is) To decrease the velocity of a body, it must accelerate in a direction greater than 90degrees to the direction of velocity. Although it should be noted that a decelerating body always tends to a state of acceleration such that a body can never be infinitely decelerating. > David A. Smith JS ==== >> Dear John Schoenfeld: >> >> ... >> There is no essential difference. Deceleration is just acceleration >> multiplied by -1. Either one means a change in velocity wrt time. > Deceleration means velocity magnitude is decreasing with time, which >> is satisfied by the condition v.a < 0. Putting a negative symbol >> infront of a does not imply decreasing velocity. >> >> Not strictly true either, John. The actual defintion should be more along >> the lines of: >> v x a = 0 and v . a < 0 >> shouldn't it? I don't believe so, since the cross product does not work for all >dimensions >n > 0 which is a requirement (as far as I know, that is) How many dimensions are there on your world? Cross-products can be generalized if they need to be. But I think v.a<0 itself is enough. Actually, .a, where is a unit vector in the direction of the velocity, =v/|v|, gives the magnitude of acceleration that changes speed, as opposed to an acceleration that changes direction at constant speed. If it's less than zero you're slowing down and can call it deceleration. >To decrease the velocity of a body, it must accelerate in a direction >greater than 90degrees to the direction of velocity. Yes, that's what .a < 0 says. -- Don't try to teach a pig how to sing. You'll waste your time and annoy the pig. ==== >> Not strictly true either, John. The actual defintion should be more along >> the lines of: >> v x a = 0 and v . a < 0 >> shouldn't it? I don't believe so, since the cross product does not work for all >dimensions >n > 0 which is a requirement (as far as I know, that is) > > How many dimensions are there on your world? Cross-products can be > generalized if they need to be. > > But I think v.a<0 itself is enough. Correct. Proof: v.a < 0 |v||a|cos(x) < 0 cos(x) < 0 x > cos^-1(0) x > pi/2 x > 90 degrees > Actually, .a, where is a unit > vector in the direction of the velocity, =v/|v|, gives the magnitude of > acceleration that changes speed, Correct because you taking the component of acceleration parallel to the direction of velocity. > as opposed to an acceleration that changes direction at constant speed. Incorrect. a.v < 0 implies decreasing |v|, but does not say anything about a.v = 0 implies constant |v|, but changes a.v > 0 implies increasing |v|, but does not say anything about To decrease the velocity of a body, it must accelerate in a direction >greater than 90degrees to the direction of velocity. > > Yes, that's what .a < 0 says. You don't need , you just need v (see above proof). JS ==== Dear John Schoenfeld: >> Not strictly true either, John. The actual defintion should be more along >> the lines of: >> v x a = 0 and v . a < 0 >> shouldn't it? I don't believe so, since the cross product does not work for all >dimensions >n > 0 which is a requirement (as far as I know, that is) How many dimensions are there on your world? Cross-products can be > generalized if they need to be. But I think v.a<0 itself is enough. Correct. Proof: > v.a < 0 > |v||a|cos(x) < 0 > cos(x) < 0 > x > cos^-1(0) > x > pi/2 > x > 90 degrees Which is no proof, since any value 180¡ >= x > 90¡ meets this requirement. And no net change in the magnitude of velocity need result for other values for the angle. So v . a < 0 can be met and d|v|/dt is unconstrained. David A. Smith ==== > Dear John Schoenfeld: > >> Not strictly true either, John. The actual defintion should be more > along >> the lines of: >> v x a = 0 and v . a < 0 >> shouldn't it? I don't believe so, since the cross product does not work for all >dimensions >n > 0 which is a requirement (as far as I know, that is) How many dimensions are there on your world? Cross-products can be > generalized if they need to be. But I think v.a<0 itself is enough. Correct. Proof: > v.a < 0 > |v||a|cos(x) < 0 > cos(x) < 0 > x > cos^-1(0) > x > pi/2 > x > 90 degrees > > Which is no proof, since any value 180¡ >= x > 90¡ meets this requirement. > And no net change in the magnitude of velocity need result for other values > for the angle. So v . a < 0 can be met and d|v|/dt is unconstrained. > > David A. Smith My last part of the proof was erroneous, let me correct that. Proof: v.a < 0 |v||a|cos(x) < 0 cos(x) < 0 x > cos^-1(0) AND x < pi - cos^-1(0) pi/2 < x < 3pi/2 So to decrease the velocity of an object, an acceleration must be applied at between 90 and 270 degrees (exclusive) to the velocity vector. Trust me, I am right. The dot product of a vector function and it's derivative that is less than 0 denotes a decreasing vector function. This proof is very long now (for me), but i will finish it eventually. JS ==== Dear John Schoenfeld: > Dear John Schoenfeld: Not strictly true either, John. The actual defintion should be more > along >> the lines of: >> v x a = 0 and v . a < 0 >> shouldn't it? I don't believe so, since the cross product does not work for all >dimensions >n > 0 which is a requirement (as far as I know, that is) How many dimensions are there on your world? Cross-products can be > generalized if they need to be. But I think v.a<0 itself is enough. Correct. Proof: > v.a < 0 > |v||a|cos(x) < 0 > cos(x) < 0 > x > cos^-1(0) > x > pi/2 > x > 90 degrees Which is no proof, since any value 180¡ >= x > 90¡ meets this requirement. > And no net change in the magnitude of velocity need result for other values > for the angle. So v . a < 0 can be met and d|v|/dt is unconstrained. David A. Smith > My last part of the proof was erroneous, let me correct that. Proof: > v.a < 0 > |v||a|cos(x) < 0 > cos(x) < 0 > x > cos^-1(0) AND x < pi - cos^-1(0) > pi/2 < x < 3pi/2 So to decrease the velocity of an object, an acceleration must be > applied at between 90 and 270 degrees (exclusive) to the velocity > vector. Trust me, I am right. The dot product of a vector function and it's > derivative that is less than 0 denotes a decreasing vector function. > This proof is very long now (for me), but i will finish it eventually. You are not right. d|v|/dt < 0 is *not* the same as v . a < 0 There are too many degrees of freedom in n-dimensional space (with n > 1, and neglecting time as a dimension included in n). v x a = 0 *is* too draconian, requiring only colinear motion/action. But I cannot figure out how to express deceleration *exactly* as anything other than d|v|/dt < 0. Perhaps you can put your massive intellect to the task. David A. Smith ==== Dear John Schoenfeld: > Dear John Schoenfeld: > in ... > My last part of the proof was erroneous, let me correct that. Proof: > v.a < 0 > |v||a|cos(x) < 0 > cos(x) < 0 > x > cos^-1(0) AND x < pi - cos^-1(0) > pi/2 < x < 3pi/2 So to decrease the velocity of an object, an acceleration must be > applied at between 90 and 270 degrees (exclusive) to the velocity > vector. Trust me, I am right. The dot product of a vector function and it's > derivative that is less than 0 denotes a decreasing vector function. > This proof is very long now (for me), but i will finish it eventually. You are not right. d|v|/dt < 0 is *not* the same as v . a < 0 Correction. You *are* right only when the velocity and acceleration are colinear. > There are too many degrees of freedom in n-dimensional space (with n > 1, > and neglecting time as a dimension included in n). v x a = 0 *is* too draconian, requiring only colinear motion/action. But > I cannot figure out how to express deceleration *exactly* as anything other > than d|v|/dt < 0. Perhaps you can put your massive intellect to the task. David A. Smith ==== > There are too many degrees of freedom in n-dimensional space (with n > 1, > and neglecting time as a dimension included in n). v x a = 0 *is* too draconian, requiring only colinear motion/action. > But > I cannot figure out how to express deceleration *exactly* as anything > other > than d|v|/dt < 0. Perhaps you can put your massive intellect to the > task. > > David A. Smith Here is the proof that v.a < 0 = deceleration. let v = velocity vector let a = acceleration vector We know that decleration occurs when d|v|/dt is a negative vector. If vector a0 is the projection of a onto v, then a0 = d|v|/dt. a0 = [(a.v) / |v|] a0 = d|v|/dt d|v|/dt = (a.v / |v|) = ( |a||v|cos(x) / |v| ) = |a|cos(x) Since a0 and v are colinear, we can reduce the problem to 1 dimension, and thus we can substitute with 1, as we shall let be the positive axis of the dimension. So, d|v|/dt = |a|cos(x) We know that we are decelerating when d|v|/dt < 0, So, |a| cos(x) < 0 cos(x) < 0 pi/2 < x < 3pi/2 So we have proven that the angle between the velocity and acceleration vectors must be interior to 90 degrees and 270 degrees for deceleration to occur. Now we shall prove that (v.a) < 0 implies deceleration. |a|cos(x) = d|v|/dt |v||a|cos(x) = d|v|/dt |v| a.v = d|v|/dt |v| (a.v)/|v| = d|v|/dt (a.v)/|v| < 0 (|a||v|cos(x))/|v| < 0 cos(x) < 0 pi/2 < x < 3pi/2 Alas the same result. Interesting properties that you may not have noted: d|v|/dt = (a.v) / |v|, a 1-dimensional vector JS ==== > Trust me, I am right. The dot product of a vector function and it's its > derivative that is less than 0 denotes a decreasing vector function. > This proof is very long now (for me), but i will finish it eventually. It's very short for the rest of us. If v(t) has v(t).v'(t) < 0 then (d/dt)[|v(t)|^2] = (d/dt)[v(t).v(t)] = 2 v(t).v'(t) < 0 so |v(t)|^2 is decreasing and |v(t)| is decreasing. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== Dear Gregory L. Hansen: ... >> Not strictly true either, John. The actual defintion should be more along >> the lines of: >> v x a = 0 and v . a < 0 >> shouldn't it? I don't believe so, since the cross product does not work for all >dimensions >n > 0 which is a requirement (as far as I know, that is) How many dimensions are there on your world? Cross-products can be > generalized if they need to be. But I think v.a<0 itself is enough. Actually, .a, where is a unit > vector in the direction of the velocity, =v/|v|, gives the magnitude of > acceleration that changes speed, as opposed to an acceleration that > changes direction at constant speed. If it's less than zero you're > slowing down and can call it deceleration. If the velocity is in the x-direction, and the acceleration is in both the x and y direction (whatever the sign), the x component could decelerate, but the y would not. v . a < 1 would be satisfied, but velocity would not have to be decreasing with time. No? So v . a < 1 is not sufficient as a definition for deceleration. The cross product is *required* to assure that no component of acceleration is in any other axis (in an n-dimensional Universe). David A. Smith ==== >Dear Gregory L. Hansen: ... > Not strictly true either, John. The actual defintion should be more >along > the lines of: > v x a = 0 and v . a < 0 > shouldn't it? >I don't believe so, since the cross product does not work for all >>dimensions >>n > 0 which is a requirement (as far as I know, that is) How many dimensions are there on your world? Cross-products can be >> generalized if they need to be. But I think v.a<0 itself is enough. Actually, .a, where is a unit >> vector in the direction of the velocity, =v/|v|, gives the magnitude >of >> acceleration that changes speed, as opposed to an acceleration that >> changes direction at constant speed. If it's less than zero you're >> slowing down and can call it deceleration. If the velocity is in the x-direction, and the acceleration is in both the >x and y direction (whatever the sign), the x component could decelerate, >but the y would not. v . a < 1 would be satisfied, but velocity would not >have to be decreasing with time. No? So v . a < 1 is not sufficient as a definition for deceleration. The >cross product is *required* to assure that no component of acceleration is >in any other axis (in an n-dimensional Universe). I assume you meant v . a < 0? If the velocity in in the x-direction and acceleration is in both the x accelerating even if it's also changing direction. For instance, if you're driving in a circle and hit the brakes. You don't have to go in a straight line to change your kinetic energy. a would not, in general, be the magnitude of deceleration, unless a x v = 0. Maybe that's what you meant, that a would be the actual amount of deceleration. But for general acceleration, a.v/|v| would be the magnitude of deceleration no matter what the transverse part is. That's just picking out the component that's parallel to the velocity, throwing away the transverse part. You could also find the tranverse acceleration, a x v/|v|, if you like. -- A nice adaptation of conditions will make almost any hypothesis agree with the phenomena. This will please the imagination but does not advance our knowledge. -- J. Black, 1803. ==== Dear Gregory L. Hansen: ... >> But I think v.a<0 itself is enough. Actually, .a, where is a unit >> vector in the direction of the velocity, =v/|v|, gives the magnitude >of >> acceleration that changes speed, as opposed to an acceleration that >> changes direction at constant speed. If it's less than zero you're >> slowing down and can call it deceleration. If the velocity is in the x-direction, and the acceleration is in both the >x and y direction (whatever the sign), the x component could decelerate, >but the y would not. v . a < 1 would be satisfied, but velocity would not >have to be decreasing with time. No? So v . a < 1 is not sufficient as a definition for deceleration. The >cross product is *required* to assure that no component of acceleration is >in any other axis (in an n-dimensional Universe). I assume you meant v . a < 0? Yes. Sorry. > If the velocity in in the x-direction and acceleration is in both the x Let's put some imaginary numbers to this. v = 100i, a = -1i + 14.1j At this instant, v . a < 0, but at the end of 1 unit of time, |v| has the same magnitude as this instant. So it couldn't be said to be decelerating. > That's decelerating or > accelerating even if it's also changing direction. For instance, if > you're driving in a circle and hit the brakes. You don't have to go in a > straight line to change your kinetic energy. I agree, but John's definition does not handle multi-dimensional problems. > a would not, in general, be the magnitude of deceleration, unless > a x v = 0. Maybe that's what you meant, that a would be the actual amount > of deceleration. But for general acceleration, a.v/|v| would be the > magnitude of deceleration no matter what the transverse part is. That's > just picking out the component that's parallel to the velocity, throwing > away the transverse part. You could also find the tranverse > acceleration, a x v/|v|, if you like. I'm just trying to find an absolute definition of deceleration. In common parlance that means the magnitude of the velocity is getting smaller. Regardless of coordinate frame. David A. Smith ==== >Dear Gregory L. Hansen: ... > But I think v.a<0 itself is enough. Actually, .a, where is a >unit > vector in the direction of the velocity, =v/|v|, gives the >magnitude >>of > acceleration that changes speed, as opposed to an acceleration that > changes direction at constant speed. If it's less than zero you're > slowing down and can call it deceleration. >If the velocity is in the x-direction, and the acceleration is in both >the >>x and y direction (whatever the sign), the x component could decelerate, >>but the y would not. v . a < 1 would be satisfied, but velocity would >not >>have to be decreasing with time. No? >So v . a < 1 is not sufficient as a definition for deceleration. The >>cross product is *required* to assure that no component of acceleration >is >>in any other axis (in an n-dimensional Universe). I assume you meant v . a < 0? Yes. Sorry. If the velocity in in the x-direction and acceleration is in both the x Let's put some imaginary numbers to this. v = 100i, a = -1i + 14.1j >At this instant, v . a < 0, but at the end of 1 unit of time, |v| has the >same magnitude as this instant. So it couldn't be said to be decelerating. Well, sure, but the velocity vector will change, so a.v/|v| will change. It will be negative for some small time, and then go positive. But if a car were braking while turning a corner, the acceleration vector would also change versus time, it wouldn't be constant. That's decelerating or >> accelerating even if it's also changing direction. For instance, if >> you're driving in a circle and hit the brakes. You don't have to go in a >> straight line to change your kinetic energy. I agree, but John's definition does not handle multi-dimensional problems. Oh, well, I guess one dimension is pretty easy to handle then. Don't even need a dot product for that. a would not, in general, be the magnitude of deceleration, unless >> a x v = 0. Maybe that's what you meant, that a would be the actual >amount >> of deceleration. But for general acceleration, a.v/|v| would be the >> magnitude of deceleration no matter what the transverse part is. That's >> just picking out the component that's parallel to the velocity, throwing >> away the transverse part. You could also find the tranverse >> acceleration, a x v/|v|, if you like. I'm just trying to find an absolute definition of deceleration. In >common parlance that means the magnitude of the velocity is getting >smaller. Regardless of coordinate frame. Good luck. I'll pull a few more numbers out of my butt. Assume an initial velocity of 10i and acceleration -1i. Velocity will be (10-1*t)i, slowing down for ten seconds and then speeding up. Boost to a frame x'=x+10*t. Initial velocity v'=0, there will be no period of deceleration. -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson ==== > Dear Gregory L. Hansen: > > ... >> But I think v.a<0 itself is enough. Actually, .a, where is a > unit >> vector in the direction of the velocity, =v/|v|, gives the > magnitude > of >> acceleration that changes speed, as opposed to an acceleration that >> changes direction at constant speed. If it's less than zero you're >> slowing down and can call it deceleration. If the velocity is in the x-direction, and the acceleration is in both > the >x and y direction (whatever the sign), the x component could decelerate, >but the y would not. v . a < 1 would be satisfied, but velocity would > not >have to be decreasing with time. No? So v . a < 1 is not sufficient as a definition for deceleration. The >cross product is *required* to assure that no component of acceleration > is >in any other axis (in an n-dimensional Universe). I assume you meant v . a < 0? > > Yes. Sorry. > > If the velocity in in the x-direction and acceleration is in both the x > > Let's put some imaginary numbers to this. v = 100i, a = -1i + 14.1j > At this instant, v . a < 0, but at the end of 1 unit of time, |v| has the > same magnitude as this instant. So it couldn't be said to be decelerating. > > That's decelerating or > accelerating even if it's also changing direction. For instance, if > you're driving in a circle and hit the brakes. You don't have to go in a > straight line to change your kinetic energy. > > I agree, but John's definition does not handle multi-dimensional problems. > > a would not, in general, be the magnitude of deceleration, unless > a x v = 0. Maybe that's what you meant, that a would be the actual > amount > of deceleration. But for general acceleration, a.v/|v| would be the > magnitude of deceleration no matter what the transverse part is. That's > just picking out the component that's parallel to the velocity, throwing > away the transverse part. You could also find the tranverse > acceleration, a x v/|v|, if you like. > > I'm just trying to find an absolute definition of deceleration. In > common parlance that means the magnitude of the velocity is getting > smaller. Regardless of coordinate frame. A body is declerating when v.a < 0. Didn't they ever teach you SCHOENFELDS VECTOR MAGNITUDE THEOREM: The magnitude of the vector given by the vector function F(x) is decreasing if F(x) . F^1(x) < 0, increasing if F(x) . F^1(x) > 0, constant if F(x) . F^1(x) = 0. Proof: Is much longer than I expected, so I'll get back to you on this one. JS ==== > Awwww... ain't that cute? The two little cranks are playing together! When I was working at MIT, a professor of my acquaintance (I'll say The idea was to be kind, and yet at the same time, not waste his own time which was better spent on other things. He thought that just ignoring people or dismissing them would be unkind. And since he worked at the AI lab and was high profile in various ways, he gets look at it, see what the topic was, and pull the file of other letters about the same topic. Then he would reply with a form letter saying something like: I'm sorry, but I don't have the expertise to address your letter. The following people may be better suited to discuss your points. Then he would add the names and addresses of a few of Now this struck me as wonderfully brilliant. Simultaneously being kind, moreover, giving the nut someone to talk to who will listen and respond in the same vein, satisfying both, keeping his own time free for other things--amazing. Thomas ==== Awwww... ain't that cute? The two little cranks are playing together! When I was working at MIT, a professor of my acquaintance (I'll say The idea was to be kind, and yet at the same time, not waste his own > time which was better spent on other things. He thought that just > ignoring people or dismissing them would be unkind. And since he > worked at the AI lab and was high profile in various ways, he gets look at it, see what the topic was, and pull the file of other letters > about the same topic. Then he would reply with a form letter saying > something like: I'm sorry, but I don't have the expertise to address > your letter. The following people may be better suited to discuss > your points. Then he would add the names and addresses of a few of Now this struck me as wonderfully brilliant. Simultaneously being > kind, moreover, giving the nut someone to talk to who will listen and > respond in the same vein, satisfying both, keeping his own time free > for other things--amazing. Have you ever noticed that cranks and nutters are harder to get together than Pandas during mating season? If you watch them (the cranks) frolic on usenet, you'll notice that they seem to, for the most part, utterly ignore each other. It's as though they instinctively know to avoid their own species. ==== Awwww... ain't that cute? The two little cranks are playing together! When I was working at MIT, a professor of my acquaintance (I'll say The idea was to be kind, and yet at the same time, not waste his own > time which was better spent on other things. He thought that just > ignoring people or dismissing them would be unkind. And since he > worked at the AI lab and was high profile in various ways, he gets look at it, see what the topic was, and pull the file of other letters > about the same topic. Then he would reply with a form letter saying > something like: I'm sorry, but I don't have the expertise to address > your letter. The following people may be better suited to discuss > your points. Then he would add the names and addresses of a few of Now this struck me as wonderfully brilliant. Simultaneously being > kind, moreover, giving the nut someone to talk to who will listen and > respond in the same vein, satisfying both, keeping his own time free > for other things--amazing. > > Have you ever noticed that cranks and nutters are harder to > get together than Pandas during mating season? If you watch > them (the cranks) frolic on usenet, you'll notice that they > seem to, for the most part, utterly ignore each other. It's > as though they instinctively know to avoid their own species. That is a very astute observation. At most they will utilize the material that is posted by their fellow cranks for their flaming needs. Thus none of them have the balls to criticize anything the others post. If they did then they risk of getting into an argument. If they get into an argument then each knows that they will never be able to admit to making an error nor simply letting the other person simply disagree or have another opinion and let it rest. Instead they know that if they get into a flame war then they can't use that person's flames to flame others. For example: varney will never discuss physics to any deep extent. If he did and they disagreed then one would, without doubt, start insulting the other and claim their stupid etc. If that happens then varney will never be able to say everyone thinks your stupid because he's too scared of someone reminding him that one of the people he is referring to as everyone also thinks he is stupid. That's something that is overly obvious but rarely, if ever, stated. Pmb ==== Awwww... ain't that cute? The two little cranks are playing together! When I was working at MIT, a professor of my acquaintance (I'll say The idea was to be kind, and yet at the same time, not waste his own > time which was better spent on other things. He thought that just > ignoring people or dismissing them would be unkind. And since he > worked at the AI lab and was high profile in various ways, he gets look at it, see what the topic was, and pull the file of other letters > about the same topic. Then he would reply with a form letter saying > something like: I'm sorry, but I don't have the expertise to address > your letter. The following people may be better suited to discuss > your points. Then he would add the names and addresses of a few of Now this struck me as wonderfully brilliant. Simultaneously being > kind, moreover, giving the nut someone to talk to who will listen and > respond in the same vein, satisfying both, keeping his own time free > for other things--amazing. > > Have you ever noticed that cranks and nutters are harder to > get together than Pandas during mating season? If you watch > them (the cranks) frolic on usenet, you'll notice that they > seem to, for the most part, utterly ignore each other. It's > as though they instinctively know to avoid their own species. Fallacy of affirming the consequent. Crank, your post is self-contradictory. JS ==== Most texts include both finite and infinite (countably infinite) sets under the countable definition. A few (Rudin, e.g.) define countable sets as countably infinite only. The latter definition appears to create less confusion. For example, under the former the expression: oo U (A_j) j=1 is taken to mean the countable union of A_j's whether {A_j} is infinite or finite, with oo always present. Can someone please indicate what the benefits/drawbacks of each are, as they apply to development of the relevant theory of such fields as probability, topology, etc. Many thanks. ==== > Most texts include both finite and infinite (countably infinite) > sets under the countable definition. A few (Rudin, e.g.) define > countable sets as countably infinite only. The latter definition > appears to create less confusion. For example, under the former the > expression: > oo > U (A_j) > j=1 is taken to mean the countable union of A_j's whether {A_j} is > infinite or finite, with oo always present. Can someone please > indicate what the benefits/drawbacks of each are > Tho not much used, some make distinction 'denumerable' for 'infinitely countable'. If countable means denumerable, then uncountable would mean finite or having cardinality greater than the integers. Given a sequence a_j, the set { a_j | j in N } is countable. ==== > > Most texts include both finite and infinite (countably infinite) > sets under the countable definition. A few (Rudin, e.g.) define > countable sets as countably infinite only. The latter definition > appears to create less confusion. For example, under the former the > expression: > oo > U (A_j) > j=1 is taken to mean the countable union of A_j's whether {A_j} is > infinite or finite, with oo always present. Can someone please > indicate what the benefits/drawbacks of each are Tho not much used, some make distinction > 'denumerable' for 'infinitely countable'. > > If countable means denumerable, then uncountable would mean > finite or having cardinality greater than the integers. > > Given a sequence a_j, the set { a_j | j in N } is countable. I find that I am reluctant to label a set like {1,2,3} uncountable. ==== > Let's see. > > do). Then, by the definition of the GCD, s(m) divides > both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m). > > With that test of your imagination above, you now have that t(m) = f. > If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2. > Duh! > > Not necessarily. You see it keeps coming back to the same thing, > which is your apparent inability to comprehend a certain possibility. > > Ok, now then, can you comprehend that you're now *assuming* the very > thing that supposedly you were trying to prove? > > > James Harris Theorem: In any commutative ring with identity, (A,+,*), if (1) f divides a*b (i.e., f*g = a*b, for some g in A), and (2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A), then f*w = b for some w in A. Proof: (1) a*b = f*g given (2) (a*v)*b = f*g*v v times both sides of (1) (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2 (4) b - f*u*b = f*g*v distributing on LHS (5) b = f*u*b + f*g*v shifting terms (6) b = f*(u*b + g*v) factoring on RHS (7) b = f*w where w = u*b + g*v, in A Q.E.D. Who is assuming what, now? As the previous poster says, Duh! ==== [I changed the labels of the hypothesis from (1) and (2) to (i) and (ii), to avoid confusion with the labels of hte steps in the proof.] >Theorem: >In any commutative ring with identity, (A,+,*), if > (i) f divides a*b (i.e., f*g = a*b, for some g in A), and >(ii) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A), then f*w = b for some w in A. Proof: > (1) a*b = f*g given > (2) (a*v)*b = f*g*v v times both sides of (1) > (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2 > (4) b - f*u*b = f*g*v distributing on LHS > (5) b = f*u*b + f*g*v shifting terms > (6) b = f*(u*b + g*v) factoring on RHS > (7) b = f*w where w = u*b + g*v, in A > Q.E.D. Who is assuming what, now? The difficulty here is on the parenthetical comment on (ii): the standard use in algebraic number theory (though not necessarily elsewhere, as I have been often reminded by Bill Dubuque) is that coprime for algebraic integers means exactly what the parenthetical comment states: that there is a linear combination equal to 1; just as in the integers. However, James has repeatedly stated that he uses coprime to mean the only common factors are units. The two statements are not equivalent in general, though they are equivalent in any PID, or more generally in any Bezout Domain. Let me label them for ease: Let R be an integral domain, x and y in R; we say that x and y are coprime in R if and only if: (A) there exist r, s, in R such that rx+sy = 1. (B) if u in R divides both x and y (in R), then u is a unit in R. It is easy to verify that (A)->(B). The harder part is to show that (B)->(A) for the algebraic integers (it is not true in general; e.g., in Z[sqrt(-5)], the standard example, 2 and 1+sqrt(-5) are coprime in the sense of (B), but not in the sense of (A)). The only proof I know for (B)->(A) in the algebraic integers uses both unique factorization into prime ideals, and more importantly the finiteness of the class number, which is of course a fairly difficult result. In The Theory of Algebraic Integers, by Richard Dedekind (translated by John Stillwell, Cambridge University Library, Cambridge University Press 1996), Dedekind states in Chapter 3, General properties of algebraic integers, section 14 divisibility of integers, final paragraph (pp. 106 in my edition): A deeper investigation will enable us to see that two nonzero [algebraic] integers a and b have a ->greatest common divisor<- [emphasis in the original], which can be put in the form aa'+bb', where a' and b' are [algebraic] integers. This important theorem is NOT AT ALL EASY TO PROVE [emphasis added] with the help of the principles developed thus far, but we shall later (section 30) be able to derive it very simply from the theory of ideals. The proof does indeed appear in Section 30, the final section of the work. Here's a paraphrase, which uses finiteness of the class number. THEOREM. Any two algebraic integers a, b have a common divisor d which can be expressed in the form d = aa' + bb', where a' and b' are likewise algebraic integers. Note that the fact that it is a common divisor AND a linear combination implies that it is a greatest common divisor in the usual sense; for if e is any other common divisor of a and b, then e divides aa'+bb', hence divides d. Proof of Theorem. We may assume both a and b are nonzero. Let K be a number field containing both a and b (say, K= Q(a,b)), and let A be its ring of integers. Let h be the class number of A. Factoring the ideals (a) and (b) into prime ideals, we have (a) = A*D; (b) = B*D where A, B, D are ideals, products of prime ideals, A and B are relatively prime, and D is the gcd of (a) and (b). Since h is the class number, D^h is principal, say D^h = (e), with e in A. Since (a^h) = (a)^h and (b^h) = (b)^h are divisible by D^h = (e), it follows that both a^h and b^h are in (e), hence there exist r and s in A such that a^h = e*r, b^h = e*s; by unique factorization into primes, (r) = A^h and (s) = B^h. Since A and B are relatively prime, so are (r) and (s). Therefore, the smallest ideal containing both r and s is A itself. That is, (r,s)=A; since 1 is in A, there exist elements u and v in A such that r*u + s*v = 1. Multiplying through by e, we have e = e*1 = e*r*u + e*s*v = a^h*u + b^h*v Let d = e^{1/h}, which is an algebraic integer (being the root of x^h - e, a monic polynomial with algebraic integer coefficients). Since e is a common divisor of a^h and b^h, d is a common divisor of a and b; and since h>0, we may write u*a^{h-1} = a'*d^{h-1}, v*a^{h-1} = b'*d^{h-1} for some algebraic integers a' and b'; namely, if a = d*x, then a' = u*x^{h-1}, and if b = d*y, then b' = v*y^{h-1} aa' + bb' = a*u*x^{h-1} + b*v*y^{h-1} = d*x*u*x^{h-1} + d*y*v*y^{h-1} = d*u*x^h + d*v*y^h = d(u*x^h + v*y^h) But a=d*x implies a^h = d^h*x^h = e*x^h and b=d*y implies b^h = d^h*y^h = e*y^h. We know that e = a^h*u + b^h*v, from which we have e = a^h*u + b^h*v = e*x^h*u + e*y^h*v = e(x^h*u + y^h*v) Since a and b are nonzero, e is nonzero, so we conclude that x^h*u + y^h*v = 1. Therefore, aa' + bb' = d(u*x^h + v*y^h) = d(1) = d. We already know that d is a common divisor of a and b; and now we have shown it is an algebraic integer linear combination of a and b, as claimed. QED -- ====================================================================== 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 changed the labels of the hypothesis from (1) and (2) to (i) and > (ii), to avoid confusion with the labels of hte steps in the proof.] >Theorem: >In any commutative ring with identity, (A,+,*), if > (i) f divides a*b (i.e., f*g = a*b, for some g in A), and >(ii) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A), then f*w = b for some w in A. Proof: > (1) a*b = f*g given > (2) (a*v)*b = f*g*v v times both sides of (1) > (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2 > (4) b - f*u*b = f*g*v distributing on LHS > (5) b = f*u*b + f*g*v shifting terms > (6) b = f*(u*b + g*v) factoring on RHS > (7) b = f*w where w = u*b + g*v, in A > Q.E.D. Who is assuming what, now? The difficulty here is on the parenthetical comment on (ii): the > standard use in algebraic number theory (though not necessarily > elsewhere, as I have been often reminded by Bill Dubuque) is that > coprime for algebraic integers means exactly what the parenthetical > comment states: that there is a linear combination equal to 1; just as > in the integers. However, James has repeatedly stated that he uses coprime to mean > the only common factors are units. The two statements are not equivalent in general, though they are > equivalent in any PID, or more generally in any Bezout Domain. Let me label them for ease: Let R be an integral domain, x and y in R; we say that x and y are > coprime in R if and only if: (A) there exist r, s, in R such that rx+sy = 1. > (B) if u in R divides both x and y (in R), then u is a unit in R. It is easy to verify that (A)->(B). The harder part is to show that > (B)->(A) for the algebraic integers (it is not true in general; e.g., > in Z[sqrt(-5)], the standard example, 2 and 1+sqrt(-5) are coprime in > the sense of (B), but not in the sense of (A)). The only proof I know for (B)->(A) in the algebraic integers uses both > unique factorization into prime ideals, and more importantly the > finiteness of the class number, which is of course a fairly difficult > result. In The Theory of Algebraic Integers, by Richard Dedekind > (translated by John Stillwell, Cambridge University Library, Cambridge > University Press 1996), Dedekind states in Chapter 3, General > properties of algebraic integers, section 14 divisibility of > integers, final paragraph (pp. 106 in my edition): A deeper investigation will enable us to see that two nonzero > [algebraic] integers a and b have a ->greatest common divisor<- > [emphasis in the original], which can be put in the form aa'+bb', > where a' and b' are [algebraic] integers. This important theorem is > NOT AT ALL EASY TO PROVE [emphasis added] with the help of the > principles developed thus far, but we shall later (section 30) be > able to derive it very simply from the theory of ideals. The proof does indeed appear in Section 30, the final section of the > work. Here's a paraphrase, which uses finiteness of the class number. THEOREM. Any two algebraic integers a, b have a common divisor d > which can be expressed in the form d = aa' + bb', where a' and b' are > likewise algebraic integers. Note that the fact that it is a common divisor AND a linear > combination implies that it is a greatest common divisor in the > usual sense; for if e is any other common divisor of a and b, then e > divides aa'+bb', hence divides d. Proof of Theorem. We may assume both a and b are nonzero. Let K be a > number field containing both a and b (say, K= Q(a,b)), and let A be > its ring of integers. Let h be the class number of A. Factoring the > ideals (a) and (b) into prime ideals, we have (a) = A*D; (b) = B*D where A, B, D are ideals, products of prime ideals, A and B are > relatively prime, and D is the gcd of (a) and (b). Since h is the > class number, D^h is principal, say D^h = (e), with e in A. Since (a^h) = (a)^h and (b^h) = (b)^h are divisible by D^h = (e), it > follows that both a^h and b^h are in (e), hence there exist r and s in > A such that a^h = e*r, b^h = e*s; by unique factorization into primes, (r) = A^h and (s) = B^h. Since A > and B are relatively prime, so are (r) and (s). Therefore, the > smallest ideal containing both r and s is A itself. That is, (r,s)=A; > since 1 is in A, there exist elements u and v in A such that > r*u + s*v = 1. Multiplying through by e, we have e = e*1 = e*r*u + e*s*v > = a^h*u + b^h*v Let d = e^{1/h}, which is an algebraic integer (being the root of > x^h - e, a monic polynomial with algebraic integer > coefficients). Since e is a common divisor of a^h and b^h, d is a > common divisor of a and b; and since h>0, we may write u*a^{h-1} = a'*d^{h-1}, > v*a^{h-1} = b'*d^{h-1} for some algebraic integers a' and b'; namely, if a = d*x, then a' = > u*x^{h-1}, and if b = d*y, then b' = v*y^{h-1} aa' + bb' = a*u*x^{h-1} + b*v*y^{h-1} > = d*x*u*x^{h-1} + d*y*v*y^{h-1} > = d*u*x^h + d*v*y^h > = d(u*x^h + v*y^h) But a=d*x implies a^h = d^h*x^h = e*x^h > and b=d*y implies b^h = d^h*y^h = e*y^h. We know that e = a^h*u + b^h*v, from which we have e = a^h*u + b^h*v > = e*x^h*u + e*y^h*v > = e(x^h*u + y^h*v) Since a and b are nonzero, e is nonzero, so we conclude that > x^h*u + y^h*v = 1. Therefore, aa' + bb' = d(u*x^h + v*y^h) > = d(1) > = d. We already know that d is a common divisor of a and b; and now we have > shown it is an algebraic integer linear combination of a and b, as > claimed. QED > Rats. Now we're back to 'pages of math', an environment overflowing with opportunities for JSH to misinterpret and twist the facts. Virgil's original proof was so simple that I was sure even JSH couldn't escape it. Oh well. l8r, Mike N. Christoff ==== Let's see. do). Then, by the definition of the GCD, s(m) divides > both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m). With that test of your imagination above, you now have that t(m) = f. > If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2. > Duh! Not necessarily. You see it keeps coming back to the same thing, > which is your apparent inability to comprehend a certain possibility. Ok, now then, can you comprehend that you're now *assuming* the very > thing that supposedly you were trying to prove? > James Harris Theorem: > In any commutative ring with identity, (A,+,*), if > (1) f divides a*b (i.e., f*g = a*b, for some g in A), and > (2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A), then f*w = b for some w in A. Proof: > (1) a*b = f*g given > (2) (a*v)*b = f*g*v v times both sides of (1) > (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2 <<-- 1 - f*u > (4) b - f*u*b = f*g*v distributing on LHS > (5) b = f*u*b + f*g*v shifting terms > (6) b = f*(u*b + g*v) factoring on RHS > (7) b = f*w where w = u*b + g*v, in A > Q.E.D. Who is assuming what, now? As the previous poster says, Duh! Nice. Even I can understand this and I know nothing of algebraic integers, etc... The only thing I can think of that he'll attack is your definitions 1 and 2. ie: he'll say 'f divides a*b does not imply f*g = a*b... or something. Actually, he may simply ignore it and pretend its not there. This should be interesting :) l8r, Mike N. Christoff ==== Let G be a group of order 2^m*k where k is odd. Prove that if G contains an element of order 2^m, then the set of all elements of odd order in G is a (normal) subgroup of G. (Rotman, Introduction to the Theory of Groups, 3.30) There is a hint: Consider G as permutations via Cayley's theorem, and show that it contains an odd permutation. My thought: 1. The element of order 2^m is obviously an odd permutation; therefore half of the elements are odd permutations and of order 2^m times some odd number. 2. If it is a subgroup, it's easy to prove it's a normal subgroup because all the elements of same cycle structure is in the subgroup. However, I can't get the rest of it. Please help me:) ==== >Let G be a group of order 2^m*k where k is odd. Prove that if G contains >an element of order 2^m, then the set of all elements of odd order in G is >a (normal) subgroup of G. (Rotman, Introduction to the Theory of Groups, >3.30) There is a hint: Consider G as permutations via Cayley's theorem, and show >that it contains an odd permutation. My thought: 1. The element of order 2^m is obviously an odd permutation; >therefore half of the elements are odd permutations and of order 2^m times >some odd number. 2. If it is a subgroup, it's easy to prove it's a normal >subgroup because all the elements of same cycle structure is in the >subgroup. However, I can't get the rest of it. > It is the even permutations that form a subgroup not the odd permutations! This is because the even permutations form the intersection of G with the alternating group on the set of elements of G. So we get a normal subgroup H of index 2 in G. You can prove the original problem by induction on m. If m=1, then H is the required subgroup. If not, by induction H has a normal subgroup K of order k. You still need to prove that K is normal in G. I will leave that to you! Derek Holt. ==== > Yes, this seems obvious. > Unfortunately, I now have trouble believing proofs based on > the countability of the rationals vs the uncountability of the reals. Dare one say: here we go again? Does the term obsessive behavior raise any warning flags at all when you hear it? xanthian. -- ==== > I assumed my method was similar to known methods. > Can you give references? Pick up any book that discusses space-bounded turing machines. A standardly used text could be Papadimitriou's Complexity Theory. > I think the argument you give above may be false. > Consider a TM that prints the first ten million > digits of PI and then halts. > > 3.1415926535897932384626433832795... > > Now, assume this TM doesn't print one long list of digits. > > 314 > 951 > a=[0,1,...,9] > M=3 > Assume N=100. > N*M*(a^M) = 100*3*(10^3) = 300,000 So, you've bounded the turing machine's tape to 3 and limited it to 100 states, and you get 300,000 internal configurations... I'm afraid to say that your example is thus invalid and such a machine will never be able to print the first 10 million digits of pi, since. When an internal configuration is repeated, it means that at step t, there was some earlier step p such that: 1) the machine is in the SAME state at steps t and p 2) the tape looks exactly the same at steps t and p 3) the tape head is in the exact same position at steps t and p when an internal configuration is repeated, the TM must continue at step t in the same way it ran at step p and thus repeating itself. > Yes, this seems obvious. > Unfortunately, I now have trouble believing proofs based on > the countability of the rationals vs the uncountability of the reals. Indeed, that is unfortunate. > According to Jim Nastos' formula this SM must repeat: > > N = 3 > M = 2 > a = 2 > N * M * (a^M) = 24 > > I may have to break down and write a program > to see if this is true. I didn't look too closely at your example... ... but I hope my exaplanation to your pi thing above has convinced you that you won't need to write or do any kind of simulation... just convince yourself that when an internal configuration is repeated, the machine loops forever. > Notice that starting at step 3 this SM repeats the states: 201. > The input starting from step 3 goes: 111, 101, 110, 101. > Even if this SM repeats steps 3-27, it will never repeat > the initial segment (steps 0, 1, and 2). > As near as I can tell, there is no limit on how large > this initial non-repeating segment can be. An interesting problem related to what is mentioned above is the Busy Beaver (BB) problem. Say a machine has 2 states (using only the alphabet 0 and 1.) The BB problem asks how many (maximum) tape cells can a TM write a 1 on and halt? (If the TMs don't halt, they can write 1 forever and ever... but when you consider all halting TMs, then there is a limit as to the number of tape cells is will ever reach and write a 1 on, when limiting yourself on 2 states.) Of course, the BB problems asks about N states, and the numbers are known only up to 4 states. (Usually, a 2-way unbounded tape is used instead of the 1-way tape.) The number of 1's (marks) that can be left on an originally empty tape on: 1 state -> 1 mark 2 states -> 4 marks 3 states -> 6 marks 4 states -> 13 marks 5 states -> unknown, but known to be at least 4098 marks 6 states -> unknown, but known to be at least 866 digits long! J ==== Condition: Used but in great condition with some highlights. Fundamentals of Complex Analysis (Third Edition) by E.B. Saff & A.D. Snider ISBN #: 0-13-907874-6 Brandon ==== I have an integral that I can't solve, could anyone please help. Please click on this link to see the problem. thanks, http://www.bol.ucla.edu/~khale/spring.gif ==== In sci.math, bogiedreamer on 15 Jan 2004 20:48:45 -0800 : > I have an integral that I can't solve, could anyone please help. > > Please click on this link to see the problem. > thanks, > > > http://www.bol.ucla.edu/~khale/spring.gif That's renderable (in ASCII) as integral(sqrt(p + q*cos(t) + cos^2(t)) dt) Have you tried the substitution (cos(t) + q/2) = u? Then -sin(t)dt = du (sin(t) = sqrt(1 - cos^2(t)) = sqrt(1 - (u - q/2)^2)) and sqrt(p + q*cos(t) + cos^2(t)) = sqrt(u^2 + p - q^2/4) which therefore gives you a bit of a mess... :-) But it's worth a shot. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > > integral(sqrt(p + q*cos(t) + cos^2(t)) dt) > Maple does it. The answer is about 33K characters in size. It involves elliptic integrals. ==== integral(sqrt(p + q*cos(t) + cos^2(t)) dt) > Maple does it. The answer is about 33K characters in size. Ah. So you're saying that the size of the answer is not 0K. ;-) > It involves elliptic integrals. Certainly. I'm somewhat surprised that the current version of Mathematica doesn't give an answer. I was expecting an extremely messy answer, like the one Maple gave. But I was also expecting, based on my experience with Mathematica and elliptic integrals, that the answer would be incorrect for at least some values of the parameters. I wonder if Maple's answer (1) can be simplified significantly and (2) is correct in general. David ==== > integral(sqrt(p + q*cos(t) + cos^2(t)) dt) Maple does it. The answer is about 33K characters in size. I'm somewhat surprised that the current version of Mathematica doesn't give >an answer. course this is just be an indication that it's incorrect. >I wonder if Maple's answer >(2) is correct in general. Should be easy to check. ==== >> integral(sqrt(p + q*cos(t) + cos^2(t)) dt) > Maple does it. The answer is about 33K characters in size. I'm somewhat surprised that the current version of Mathematica doesn't >give an answer. > True. The previous version, 4.2, does give an answer. > It's also long, messy but simplifies somewhat. Of > course this is just be an indication that it's incorrect. The answer is certainly incorrect for some values of the parameters. Try, for example, substituting 1 and 2 for p and q, resp. >I wonder if Maple's answer >(2) is correct in general. Should be easy to check. I don't know why you think that. Would you care to explain? David ==== Any help with the following problems is most appreciated. I'm hoping they've already been solved, but I haven't been able to find anything about them (not sure what it'd be called). Only 2 1/2 problems since I think problems 1 and 3 are special cases of problem 2... Problem 1: ---------- Let G be a simple graph with vertex set V. Let u,v in V. Find a cycle C in G such that u,v in C and length(C) is minimised. My first thought is to find the shortest path from u to v, then remove this and find the shortest path back. However, this won't always work: v---------o /| | / | | / | | 4 5---6 | | long | | | | 1---2 3 | | / | | / | |/ | u---------o (the path marked 'long' is long enough, or has enough vertices, to ensure that any cycle containing it is suboptimal). Here the shortest cycle is (u,1,2,4,v,6,5,3,u). However, the shortest path is (u,2,5,v) which leads to the cycle (u,2,5,v,u). It is relatively straight forward to show that if the shortest path S from u to v isn't wholly contained by the shortest cycle C = (u,P1,v,P2,u), then S must have a common vertex with P1, and a common vertex with P2. (I used proof by contradiction.) Further, if S = (u,u1,...,v1,v) then C = (u,u1,...,v,v1,...,u). That is, the second vertex in S must be the second vertex in C (starting from u), and the penultimate vertex in S must be the vertex following v in C. (Again I use proof by contradiction.) Now the problem is just to find paths P1,P2 in V such that P1 and P2 don't intersect, P1 = (u1,...,v), P2 = (v1,...,u) and that minimise length(P1) + length(P2). Hence... Problem 2: ---------- Let G be a simple graph with vertex set V and let {a,b,c,d} subset V. Find paths P1 from a to b, P2 from c to d such that P1 and P2 have no common vertex and length(P1) + length(P2) is minimised. Again I started thinking about shortest paths, but in some cases the shortest path from a to b (c to d) will not be included... 2 2 a-------f-------b / / 1 1/ 1 / / / e * <------ no intersection / / 1/ 1 1/ / / c-------g-------d 2 2 Here the optimal (and only) choices for P1 and P2 are (a,f,b) and (c,g,d), but the shortest paths are (a,e,g,b) and (c,e,f,d). I really don't know how to proceed with this, any help would be most appreciated. In a similar vein... Problem 3: ---------- Let G be a simple graph with vertex set V and let {a,b,c} be a subset of V. Find paths P1 from a to b and P2 from b to c such that P1 and P2 have no common vertex and length(P1) + length(P2) is minimised. I think this is just a special case of problem 2, but I'm not sure... ==== MRK 100 Q 14,200 SBC 12,200 realestate land 3APR03 of 3 lots $19,000. science-art of pictures,porcelain etc starting JAN03 for $13,705. realestate land 30JUL03 another lot $11,500. Today I bought back my 300 shares of BCE at 22.57 that I sold last friday at 22.78 and in the process picked up 2 free shares in buying back 310. Theoretically, it is a Crossover in a switching campaign only with one company and that of cash. Last Friday my intent was to sell 3,300 BCE and buyback more of SBC but I could only fill a partial sell as both companies stock was falling. So today I bought back 310 for which 2 of those shares were freebies. And added extra cash to get the other 8 shares. The Crossover techniques is especially advantages around the ex-dividend dates because the price action is somewhat predictable and being somewhat predictable, a wise investor can generate pretty profits in the Crossover technique. For it is a pattern that SBC generally rises sharply approaching ex date and dramatically falls the days after ex. Knowledge is a gateway to profiting. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ---- Perhaps Google is having troubles. Perhaps sci.math has a censor. This is the 3rd attempt to get this post, posted. ==== could someone show me a way of obtaining the probability density function (pdf) of Z, where Z=arctan(Y/X), 0 < x < b, -b < y <0, and X and Y are independent, identically distributed. Also, if anybody knows of a good probability and random process book good for electrical engineers (not TOO much hard core math) where they have lots of examples that show more than just simple two-r.v. functions such as Z=X+Y, Z=X-Y, and Z=X/Y, I'm all ears. ==== > could someone show me a way of obtaining the probability density >function (pdf) of Z, where Z=arctan(Y/X), 0 < x < b, -b < y <0, and X and Y are independent, identically distributed. If you mean that X and Y are uniformly distributed, then we can compute the given distribution. The points (x,y) are uniformly distributed in the fourth quadrant square with corners (0,0), (b,0), (b,-b), (0,-b). z is the argument of (x,y). This means that when -pi/4 < z <= 0, the probability that (x,y) is in the wedge between z and z+dz is sec^2(z)/2 dz. When -pi/2 < z <= -pi/4, the probability that (x,y) is in the wedge between z and z+dz is csc^2(z)/2 dz. Thus, the pdf of Z is 2 pdf(z) = sec (z)/2 when -pi/4 < z <= 0 2 = csc (z)/2 when -pi/2 < z <= -pi/4 Rob Johnson take out the trash before replying <20040116.000312@whim.org> ==== correctly pointed out that X and Y cannot be identically distributed since they operate in different ranges. I was dead tired when I typed out my question last night and so didn't give the complete picture. Sorry about the confusion. And also to clarify a point Stephen brought up, I use X, Y to refer to random variables and x and y to refer to their sample values that X and Y can take respectively. That's my way of the convention. The pdf's of X and Y are as follows: f(x)=-1/(2b)^2*x+1/(2b) 0 > could someone show me a way of obtaining the probability density >function (pdf) of Z, where Z=arctan(Y/X), 0 < x < b, -b < y <0, and X and Y are independent, identically distributed. > > If you mean that X and Y are uniformly distributed, then we can compute > the given distribution. The points (x,y) are uniformly distributed in > the fourth quadrant square with corners (0,0), (b,0), (b,-b), (0,-b). > z is the argument of (x,y). > > This means that when -pi/4 < z <= 0, the probability that (x,y) is in > the wedge between z and z+dz is sec^2(z)/2 dz. When -pi/2 < z <= -pi/4, > the probability that (x,y) is in the wedge between z and z+dz is > csc^2(z)/2 dz. Thus, the pdf of Z is > > 2 > pdf(z) = sec (z)/2 when -pi/4 < z <= 0 > > 2 > = csc (z)/2 when -pi/2 < z <= -pi/4 > > Rob Johnson > ==== correctly pointed out that X and Y cannot be identically distributed since >they operate in different ranges. I was dead tired when I typed out my >question last night and so didn't give the complete picture. Sorry about >the confusion. And also to clarify a point Stephen brought up, I use X, Y >to refer to random variables and x and y to refer to their sample values >that X and Y can take respectively. That's my way of the convention. The pdf's of X and Y are as follows: f(x)=-1/(2b)^2*x+1/(2b) 0 f(y)=1/(2b)^2*y+1/(2b), -2bas: f(x,y)=-1/(2b)^4*x*y-1/(2b)^3*x+1/(2b)^3*y+1/(2b)^2, for 0Your help on this greatly appreciated. Hope to hear from you soon. I was about to post the answer below with each of the x and y density functions doubled, but I read in another subthread that this is just the density for one of four quadrants, so I have removed the doubling. Since you are dividing Y by X, we can change b without affecting Z. I will use b = 1/2: f(x) = 1-x 0 < x < 1 f(y) = 1+y -1 < y < 0 Consider the cumulative probabilities. When -pi/4 < z < 0: F(z) |1 |tan(z)x = | | (1-x)(1+y) dy dx [1] | 0 | -1 When -pi/2 < z < -pi/4 F(z) |0 |cot(z)y = | | (1-x)(1+y) dx dy [2] |-1 | 0 The density functions are the derivatives of the cumulatives: When -pi/4 < z < 0: f(z) d |1 |tan(z)x = -- | | (1-x)(1+y) dy dx dz | 0 | -1 2 |1 = sec (z) | x(1-x)(1+tan(z)x) dx | 0 2 = sec (z) (2+tan(z))/12 [3] When -pi/2 < z < -pi/4 f(z) d |0 |cot(z)y = -- | | (1-x)(1+y) dx dy dz |-1 | 0 2 |0 = -csc (z) | y(1-cot(z)y)(1+y) dy |-1 2 = csc (z) (2+cot(z))/12 [4] Notice that [4] is complementary to [3]. That is, the distribution is symmetric about z = -pi/4 (as one would expect). Rob Johnson take out the trash before replying ==== > The pdf's of X and Y are as follows: f(x)=-1/(2b)^2*x+1/(2b) 0 f(y)=1/(2b)^2*y+1/(2b), -2bas: f(x,y)=-1/(2b)^4*x*y-1/(2b)^3*x+1/(2b)^3*y+1/(2b)^2, for 0 First, note that since you are just considering the ratio of Y and X, you can assume WLOG that b = 1/2. However, we immediately see that your densities do not integrate to 1. Should your marginals be doubled and the joint density quadrupled? Since this might be homework, I give only an outline: Let T = tan(Z). By integrating over the appropriate region, determine P{T > t} for t >= -1. Take a derivative to get a density. Exploit symmetry to handle other values of t. Or you could use the two-dimensional change of variable formula to first determine the joint density of Z (or T) and a dummy variable, then integrate out the dummy. If this is not homework, and/or you need more help, post again. If it is homework, be sure to cite sources in submitted work. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu <20040116.000312@whim.org> <40088251.7040704@rutcor.rutgers.edu> ==== Stephen, no, this is not homework. I'm a Ph.D student doing some research and at this stage, working on some derivations to which my question is related. The amount of math involved is a little bit beyond me. And, to further convince you that this is no homework problem, we're still in the middle of winter recess here. :-) You're right. This is not the pdf per se. This partial pdf (for lack of a better word) describes probability density in the Quadrant IV in a Cartesian system (as evidenced by the fact that X is positive and Y is negative). So if you double-integrate f(x,y) for 0 > The pdf's of X and Y are as follows: f(x)=-1/(2b)^2*x+1/(2b) 0 f(y)=1/(2b)^2*y+1/(2b), -2bas: f(x,y)=-1/(2b)^4*x*y-1/(2b)^3*x+1/(2b)^3*y+1/(2b)^2, for 0 you can assume WLOG that b = 1/2. However, we immediately see that your > densities do not integrate to 1. Should your marginals be doubled and > the joint density quadrupled? > > Since this might be homework, I give only an outline: Let T = tan(Z). > By integrating over the appropriate region, determine P{T > t} for t > >= -1. Take a derivative to get a density. Exploit symmetry to handle > other values of t. > > Or you could use the two-dimensional change of variable formula to first > determine the joint density of Z (or T) and a dummy variable, then > integrate out the dummy. > > If this is not homework, and/or you need more help, post again. If it > is homework, be sure to cite sources in submitted work. > > > ==== > You're right. This is not the pdf per se. This partial pdf (for lack of a better word) describes probability density in the Quadrant IV in a Cartesian system (as evidenced by the fact that X is positive and Y is negative). So if you double-integrate f(x,y) for 0 So I guess you are saying that P{(X,Y) in Q. IV} = 1/4, and you have given the density of (X,Y) restricted to Q. IV. Maybe this is a good time to ask this question which I've been >pondering for a while: if f(x,y) is a partial pdf for Quadrant IV only, >can I apply it to derive a partial pdf of Z defined as Z=arctan(Y/X)? >Intuitively, I would think that whatever partial pdf of Z that may be >obtained describes only in the range of 3*pi/2 <= Z <= 2*pi. Am I correct in making the above claim? > Your partial pdf quadrupled is the conditional pdf given {(X,Y) in Q. IV}. You could work from there, I suppose. However, this concept of partiality is quite nonstanrard in probability theory. I suppose you could make better sense of it if you open your discussion to more general measures, but I am unsure that you want to do so. change of variable approach. Could you also clarify a bit on your first >suggested method? For instance, what integral region are you referring to here? > Given any continuous random variable V, the density g(v) of V is the derivative of its cdf G(v) = P{V <= v}; also, g(v) = -d/dv P{V > v}. Given X and Y independent; Z = arctan(Y/X); 0 < X < 1; -1 < Y < 0, and -pi/2 < Z < 0. For z < 0, P{Z > z} = P{Y/X > tan z} = P{Y > (tan z) X} = int(x= 0..1, y= (tan z) x..0, f(x) g(y)) when -pi/4 < z < 0. P{Z <= z} = int(y= -1..0, x = 0..(cot z) y, f(x) g(y)) when -pi/2 < z < -pi/4 Use Leibniz' rule, the fundamental theorem of calculus, and the chain rule to determine the densitiy of Z. When f(x) = 2(1-x) for 0 < x < 1 and g(y) = 2(1+y) for -1 < y < 0 (i.e., X and -Y each have Beta(0,1) distribution), I get (with the assistance of Mathematica(R)) the following density for Z: h(z) = (csc^2 z) (2 + cot z) / 3 for -pi/2 < z < -pi/4 h(z) = (sec^2 z) (2 + tan z) / 3 for -pi/4 < z < 0 > >> The pdf's of X and Y are as follows: > f(x)=-1/(2b)^2*x+1/(2b) 0 f(y)=1/(2b)^2*y+1/(2b), -2bwhere b is a positive constant. And the joint pdf of X and Y is given >as: > f(x,y)=-1/(2b)^4*x*y-1/(2b)^3*x+1/(2b)^3*y+1/(2b)^2, >for 0So again, I want to find the pdf of Z which is defined as Z=arctan(Y/X). > >>First, note that since you are just considering the ratio of Y and X, you can assume WLOG that b = 1/2. However, we immediately see that your densities do not integrate to 1. Should your marginals be doubled and the joint density quadrupled? >> -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > When f(x) = 2(1-x) for 0 < x < 1 and > g(y) = 2(1+y) for -1 < y < 0 (i.e., X and -Y each have Beta(0,1) distribution) Correction: Beta(1,2) -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > could someone show me a way of obtaining the probability density >function (pdf) of Z, where Z=arctan(Y/X), 0 < x < b, -b < y <0, and X and Y are independent, identically distributed. Also, if anybody knows of a good probability and random process >book good for electrical engineers (not TOO much hard core math) where >they have lots of examples that show more than just simple two-r.v. >functions such as Z=X+Y, Z=X-Y, and Z=X/Y, I'm all ears. > You haven't specified the distributions of X and Y, so we cannot answer your question. Also, these two r.v.'s cannot possibly be identically distributed, since they have different ranges. Or are x and y different from X and Y? If so, what does that mean? Some recommended texts for you (recongnizing your request as an engineer): Drake AW. Fundamentals of Applied Probability Theory. Papoulis A. Probability, Random Variables, and Stochastic Processes. Ross SM. A First Course in Probability. _____. Introduction to Probability and Statistics for Scientists and Engineers. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu Approved: news-answers-request@MIT.EDU Summary: All laws/equations are Galilean invariant when expressed in the generalized cartesian coordinates demanded by basic analytic geometry, vector algebra, and measurement theory. ==== Opponents of the content should first actually find out what it is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, selfless, *.answers moderators evidences ignorance and despicable netiquette. Archive-Name: physics-faq/criticism/galilean-invariance Invariant Galilean Transformations (FAQ) On All Laws (c) Eleaticus/Oren C. Webster Thnktank@concentric.net An obvious typo or two corrected. The Brittanica section revised to less 'pussy-footing' and to more directly anticipate the elementary measurement theory and basic analytic geometry that is applied to the transformation concept. ------------------------------ The purpose of this document is to provide the student of Physics, especially Relativity and Electromagnetism, the most basic princ- iples and logic with which to evaluate the historic justification of Relativity Theory as a necessary alternative to the classical physics of Newton and Galileo. We will prove that all laws are invariant under the Galilean transformation, rather than some being non-invariant, after we show you what that means. We shall also show that another primal requirement that SR exist is nonsense: Michelson-Morley and Kennedy-Thorndike do indeed fit Galilean (c+v) physics. ------------------------------ 1. Foreword and Intent 3. The Principle of Relativity 4. The Encyclopedia Brittanica Incompetency. 5. Transformations on Generalized Coordinate Laws 6. The data scale degradation absurdity. 8. What does sci.math have to say about x0'=x0-vt? 9. But Doesn't x.c'=x.c? 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations? 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent? 12. But Isn't (x'-x.c')=(x-x.c) a Tautology? 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of a Linear Transform? 14. But The Transform Won't Work On Time Dependent Equations? 15. But The Transform Won't Work On Wave Equations? 16. But Maxwell's Equations Aren't Galilean Invariant? 17. First and Second Derivative differential equations. ------------------------------ If a law is different over there than it is here, it is not one law, but at least two, and leaves us in doubt about any third location. This is the Principle of Relativity: a natural law must be the same relative to any location at which a given event may be perceived or measured, and whether or not the observer is moving. The idea of location translates to a coordinate system, largely because any object in motion could be considered as having a coordinate system origin moving with it. If you perceive me moving relative to you - who have your own coordinate system - will your measurements of my position and velocity fit the same laws my own, different measurements fit? If a law has the same form in both cases it is called covariant. If it is identical in form, var- ables, and output values, it is called invariant. What we're asking is that if the x-coordinate, x, on one coordinate axis works in an equation, does the coordinate, x', on some other, parallel axis work? Speaking in terms of the axis on which x is the coordinate, x' is the 'transformed' coordinate. The situation is complicated because we're talking about coordinates - locations - but in most mean- ingful laws/equations, it is lengths/distances (and time intervals) the equations are about, and x coord- inates that represent good, ratio scale measures of distances are only interval scale measures on the x' So, if we have an x-coordinate in one system, then we can call the x' value that corresponds to the same point/location the transform of x. In particular, the Principle of Relativity is embodied in the form of the Galilean transformation, which relates the original x, y, z, t to x', y', z', t' by the transform equations x'=x-vt, y'=y, z'=z, t'=t in the simplified case where attention is focused only on transforming the x-axis, and not y and z. In the case of Special Relativity, the x' transform is the same except that x' is then divided by sqrt(1-(v/c)^2), and t'=(t-xv/cc)/sqrt(1-(v/c)^2). In either case, v is the relative velocity of the coordinate systems; if there is already a v in the equations being trans- formed use u or some other variable name. ------------------------------ One example of the traditional fallacious idea that an equation is not invariant under the galilean transformation comes from the Encyclopedia Brittanica: Before Einstein's special theory of relativity was published in 1905, it was usually assumed that the time coordinates measured in all inertial frames were identical and equal to an 'absolute time'. Thus, t = t'. (97) The position coordinates x and x' were then assumed to be related by x' = x - vt. (98) The two formulas (97) and (98) are called a Galilean transformation. The laws of nonrelativ- istic mechanics take the same form in all frames related by Galilean transformations. This is the restricted, or Galilean, principle of relativity. The position of a light wave front speeding from the origin at time zero should satisfy x^2 - (ct)^2 = 0 (99) in the frame (t,x) and (x')^2 - (ct')^2 = 0 (100) in the frame (t',x'). Formula (100) does not transform into formula (99) using the transform- ations (97) and (98), however. ................................................. Besides the trivially correct statement of what the Galilean 'transform' equations are, there is exactly one thing they got right. I. Eq-100 is indeed the correct basis for discussing the question of invariance, given that eq-99 is the correct 'stationary' (observer S) equation. [Let observer M be the 'moving'system observer.] In particular, eq-100 is of exactly the same form [the square of argument one minus the square of argument two equals zero (argument three).] II. It is nonsense to say eq-99 should be derivable from eq-100; for one thing, the transforms are TO x' and t' from x and t, not the other way around, and the idea that either observer's equation should contain within itself the terms to simplify or rearrange to get to the other is ridiculous. As the transform equations say, the relationship of t', x' to t, x is based on the relative velocity between the two systems, but neither the original (eq-99) equation nor the M observer equation is about a relationship between coordinate systems or observers. One might as well expect the two equations to contain banana export/import data; there is no relevancy. The 'transform' equations are the relationships between x' and x, t' and t and have nothing to do with what one equation or the other ought to 'say'. The equations' content is the rate at which light emitted along the x-axes moves. III. Most remarkable, the True Believer SR crackpots who most despise the consequences of measurement theory (demonstrable fact) contained in this document are those who want to argue against our saying the Britt- anica got eq-100 right; They insist that the correct equation is derived directly from x'=x-vt and t'=t. Solve for x=x'+vt and replace t with t', then substitute the result in eq-99: (x'+vt')^2 - (ct')^2 = 0. Besides the fact that this results in an equation with arguments exactly equal to eq-99, they will insist the transform is not invariant. IV. A major justification they have for their idea of the correct M system equation on which to base the the discussion of invariance, is that the variables are M system variables, never mind the fact that the arguments are S system values. That argument of theirs is arrant nonsense. The velocity v that S sees for the M system relative to herself is the negative of what the M system sees for the S system relative to himself. In other words, x'+vt' is a mixed frame expression and it is x'+(-v)t' that would be strictly M frame notation, and that equation is far off base. [Work it out for yourself, but make sure you try out an S frame negative v so as not to mislead yourself.] V. In I. we said: given that eq-99 is the correct 'stationary' equation. Let's look at it closely: x^2 - (ct)^2 = 0 (99) This whole matter is supposed to be about coordinate transforms. Is that what t is, just a coordinate? No. It isn't, in general. Suppose you and I are both modelling the same light event and you are using EST and I'm using PST. 'Just a time coordinate' is just a clock reading amd your t clock reading says the light has been moving three hours longer than my clock reading says. Well, that's what the idea that t is a coordinate means. Eq-99 works if and only if t is a time interval, and in particular the elapsed time since the light was emitted. Thus, that equation works only if we understand just what t is, an elapsed time, with emissioon at t=0. However, we don't have to 'understand' anything if we use a more intelligent and insightful form of the equation: (x)^2 - [ c(t-t.e) ]^2 = 0, where t.e is anyone's clock reading at the time of light emission, and t is any subsequent time on the same clock. Similarly, x is not just a coordinate, but a distance since emission. (x-x.e)^2 - [ c(t-t.e) ]^2 = 0 (99a) VI. In the spirit of 'there is exactly one thing they got right', the correct M system version of eq-99a is eq-100a: (x'-x.e')^2 - [ c(t'-t.e') ]^2 = 0 (100a) Every observer in the universe can derive their eq-100a from eq-99a and vice versa, not to mention to and from every other observer's eq-99a. Now, THAT's invariance. [You do realize that every eq-100a reduces to eq-99a, when you back substitute from the transforms, right? t.e'=t.e, x.e'=x.e-vt.] ------------------------------ The traditional Gallilean transform is correct: t' = t x' = x - vt. But remember this: a transform of x doesn't effect just some values of x, but all of them, whether they are in the formula or not. This is important if you want to do things right. The crackpot position is strongly against this sci.math verified position, and the apparently standard coordinate pseudo-transformation they suggest is perhaps the result. {See Table of Let's use a simple equation: x^2 + y^2 = r^2, which is the formula for a circle with radius r, centered at a location where x=0. But what if the circle center isn't at x=0? Well, we'd want to use the form analytic geometry, vector algebra, and elementary measurement theory tells us to use, a form where we make explicit just where the circle center is, even if it is at x=x0=0: (x-x0)^2 + (y-y0)^2 = r^2. The circle center coordinate, x0, is an x-axis coordinate, just like all the x-values of points on the circle. So, in proper generalized cartesian coordinate forms of laws/equations we want to transform every occurence of x and x0 - by whatever name we call it: x.c, x_e, whatever. So, what is the transformed version of (x-x0)? Why, (x'-x0'); both x and x0 are x-coordinates, and every x-coordinate has a new value on the new axis. So, what is the value of (x'-x0') in terms of the original x data? is also true for x0'=x0-vt: (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0). In other words, when we use the generalized coordinate form specified by analytic geometry, we find that the value of (x'-x0') does not depend on either time or velocity in any way, shape, form, or fashion. Similarly for (y-y0). We can treat time the same way if necessary: (t-t0). The above is a proof that any equation in x,y,z,t is invariant under the galilean transforms. Just use the generalized coordinate form, with (x-x0)/etc, in the transformation process, not the incompetently selected privileged form, with just x/etc. [The form is privileged because it assumes the circle center, point of emission, whatever, is at the origin of the axes instead at some less convenient point. After transform the coordinate(s) of the circle center/origin are also changed but the privileged form doesn't make this explicit and screws up the calculations, which should be based on (x'-x0') but are calculated as (x'-0).] The value of (x'-x0') is the same as (x-x0). That makes sense. Draw a circle on a piece of paper, maybe to the right side of the paper. On a transparent sheet, draw x and y coordinate axes, plus x to the right, plus y at the top. Place this axis sheet so the y-axis is at the left side of the circle sheet. Now answer two questions after noting the x-coordinate of the circle center and then moving the axis sheet to the right: (a) did the circle change in any way because you moved the axis sheet (ie because you transformed the coordin- nate axis)? (b) did the coordinate of the circle center change? The circle didn't change [although SR will say it did]; that means that (x'-x0') does indeed equal (x-x0). The coordinate of the circle center did change, and it changed at the same rate (-vt) as did every point on the circle. That means that x0'<>x0, and the fact the circle center didn't change wrt the circle, means that the relationship of x0' with x0 is the same as that of any x' on the circle with the corresponding x: x'=x-vt; x0'=x0-vt. This is to prepare you for the True Believer crackpots that say 'constant' coordinates can't be transformed; some even say they aren't coordinates. These crackpots include some that brag about how they were childhood geniuses, btw. QED: The galilean transformation for any law on generalized Cartesian coordinates is invariant under the Galilean transform. The use of the privileged form explains HOW the transformed the screwed up effect of the transform is, and how use of the generalized form corrects the screwup. ------------------------------ The SR transforms and the Galilean transforms both convert good, ratio scale data to inferior interval scale data. The effect is corrected, allowed for, when the transforms are conducted on the generalized coordinate forms specified by analytic geometry and vector algebra. Both sets of transforms are 'translations' - lateral movements of an axis, increasing over time in these cases - but with the SR transform also involving a rescaling. It is the translation term, -vt in the x transform to x', and -xv/cc in the t transform to t', that degrades the ratio scale data to interval scale data. In general, rescaling does not effect scale quality in the size-of-units sense we have here. SR likes to consider its transforms just rotations, however - in spite of the fact Einstein correctly said they were 'translations' (movements) - and in the case of 'good' rotations, ratio scale data quality is indeed preserved, but SR violates the conditions of good ro- tations; they are not rigid rotations and they don't appropriately rescale all the axes that must be rescaled to preserve compatibility. The proof is in the pudding, and the pudding is the combination of simple tests of the transformations. We can tell if the transformed data are ratio scale or interval. Ratio scale data are like absolute Kelvin. A measure- ment of zero means there is zero quantity of the stuff being measured. Ratio scale data support add- ition, subtraction, multiplication, and division. The test of a ratio scale is that if one measure looks like twice as much as another, the stuff being measured is actually twice as much. With absolute Kelvin, 100 degrees really is twice the heat as 50 degrees. 200 degrees really is twice as much as 100. Interval scale data are like relative Celsius, which is why your science teacher wouldn't let you use it in gas law problems. There is only one mathematical operation interval scales support, and that has to be between two measures on the same scale: subtraction. 100 degrees relative (household) Celsius is not twice as much as 50; we have to convert the data to absolute Kelvin to tell us what the real ratio of temperatures is. However, whether we use absolute Kelvin or relative Celsius, the difference in the two temperature readings is the same: 50 degrees. Thus, if we know the real quantities of the 'stuff' being measured, we can tell if two measures are on a ratio scale by seeing if the ratio of the two measures is the same as the ratio of the known quant- ities. If a scale passes the ratio test, the interval scale test is automatically a pass. If the scale fails the ratio test, the interval scale test becomes the next in line. It isn't just the bare differences on an interval scale that provides the test, however. Differences in two interval scale measures are ratio scale, so it is ratios of two differences that tell the tale. Let's do some testing, and remember as we do that our concern is for whether or not the data are messed up, not with 'reasons', excuses, or avoidance. ------------------------------------------------------ Are we going to take a transformed length (difference) and see whether that length fits ratio or interval scale definitions? Of course, not. Interval scale data are ratio after one measure is subtracted from another. That is the major reason the SR transforms can be used in science. Let there be three rods, A, B, C, of length 10, 20, 40, respectively. These lengths are on a known ratio scale, our original x-axis, with one end of each rod at the origin, where x=0, and the other end at the coordinate that tells us the correct lengths. Note that these x-values are ratio scale only because one end of each rod is at x=0. That may remind you of the correct way to use a ruler or yard/meter-stick: put the zero end at one end of the thing you are measuring. Put the 1.00 mark there instead of the zero, and you have interval scale measures. Let A,B,C, be 10, 20, 40. Let a,b,c be x' at v=.5, t=10. x'=x-vt. A B C a b c ---------------- -------------------- 10 20 40 5 15 35 ---------------- -------------------- B/A = 2 b/a = 3 C/A = 4 c/a = 7 C/B = 2 c/b = 2.333 Obviously, the transformed values are no longer ratio scale. The effect is less on the greater values. C-A = 10 b-a = 10 C-A = 30 c-a = 30 C-B = 20 c-b = 20 Obviously, the transformed values are now interval scale. This will hold true for any value of time or velocity. (C-A)/(B-A) = 3 (c-a)/(b-a) = 3 (C-B)/(B-A) = 2 (c-b)/(b-a) = 2 Obviously, the ratios of the differences are ratio scale, being identical to the ratios of the corresponding original - ratio scale - differences. The main difference between these results and the SR results is that the differences do not correspond so neatly to the original, ratio scale, differences. This is due only to the rescaling by 1/sqrt(1-(v/c)^2). The ratios of the differences on the transformed values do correspond neatly and exactly to the ratio scale results. Using the generalized coordinate form, such as (x-x0), the transform produces an interval scale x' and an interval scale x0'. That gives us a ratio scale (x'-x0'), just like - and equal to - (x-x0). ------------------------------ It has become apparent - whether misleading or not - that the crackpot responses to the obvious derive from a common source, whether it be bandwagoning or their SR instructors. Below, in the sci.math subject, we see that all sci.math respondents agree with the basic controversial position of this faq: every coordinate is transformed, whether a supposed constant or not. Think about it, the generalized coordinate of a circle center, x0, applies to infinities upon infinities of circle locations (given y and z, too); it is a constant only for a given circle, and even then only on a given coordinate axis. And even variables are often held 'constant' during either integration or differentiation. The utility of a variable is that you can discuss all possible particular values without having to single out just one. That utility does not make particular - singled out - values on the variable's axis not values of the variable just because they have become named values. In any case, all that is preamble to the incompetent idea they have proposed for a transform of coordinates. It is based on the idea that the circle center, point of emission, whatever, has coordinates that cannot be transformed. Let there be an equation, say (x)^2 - (ict)^2 = 0. What is the transformed version of that equation? Answer: (x')^2 - (ict')^2 = 0. That's the one thing the Brittanica got right. Note that the leading crackpot just criticized this faq for presuming to correct the Britt- anica, but it then and before poses the incompetent pseudo- transform we analyze here in this section. x to x' and t to t' are obviously coordinate transforms; the x and t coordinates have been replaced by the coord- inates in the primed system. A tranform of an equation from one coordinate system to another is NOT a substitution of the/a definition of x for itself; that is not a coordinate transformation. The most that can said for such a substitution is that it is a change of variable. But the crackpots are calling this a coordinate trans- form of the original equation: (x'+vt)^2 - (ict')^2 = 0. It is not a coordinate transform, of course, except accidentally. (x'+vt) is not the primed system coordinate, it is another form/expression of x. They get that substitution by solving x'=x-vt for x; x=x'+vt. So, by incompetent misnomer, they accomplish what they have been railing against all along. It has been the generalized coordinate form in question all this time: (x-x0)^2 - (ict)^2 = 0. Here they substitute for x instead of transforming to the primed frame: (x'+vt-x0)^2 - (ict')^2. ----- ^ | ^ | It is still x ^ but see what they have accomplished by their mis/malfeasance: [x'+vt-x0]=[x'+(vt-x0)]=[x'-(x0-vt)]. =[x'-x0'] The crackpots have been bragging about how you don't have to transform the circle center's coordinate to transform the circle center's coordinate. Bragging that what they were doing was not what they said they were doing. This does give us insight as to some of the crackpot variations on their x0'<>x0-vt theme, which in all the variations will be discussed in later sections.. They are used to seeing the mixed coordinate form, (x'+vt-x0) without realizing what it respresented, so - accompanied with a lack of understanding of the term 'dependent' - they are used to seeing just the one vt term, and not the one hidden in the defi- nition of x' and are used to imagining it makes the whole expression time dependent and thus not invariant. About which, let x=10, let, x0=20, v=10, and t variously 10 and 23: (x-x0)=-10. Using their (x'+vt-x0): For t=10, we have (x'+vt-x0) = [ (10-10*10) + (10*10) - (20) ] = -90 + 100 - 20 = -10 = (x-x0) For t=23, we have (x'+vt-x0) = [ (10-10*23) + (10*23) - (20) ] = -220 + 230 - 20 = -10 = (x-x0) The result depends in no way on the value of time; we showed the obvious for a couple of instances of t just so you can see that the crackpots not only do not understand the obvious logic of the algebra { (x'-x0')=[ (-vt)-(x0-vt) ]=(x-x0) } - which shows that the transform has no possible time term effect - but they don't understand even a simple arithmetic demonstration of the facts. Oh. Their (x'+vt-x0) or (x'+vt'-x0) reduces the same way since t'=t: (x-vt+vt-x0)=(x-x0). Their process, which says (x'+vt') is the transform of x, says that (x'+vt') is the moving system location of x, but it can't be because x is moving further in the negative direction from the moving viewpoint. That formula will only work out with v<0 which is indeed the velocity the primed system sees the other moving at. However, that formula cannot be derived from x'=x-vt, the formula for transformation of the coordinates from the unprimed to the primed, ------------------------------ The crackpots' positions/arguments were put to sci.math in such a way that at least two or three who posted re- sponses thought it was your faq-er who was on the idiot's side of the questions. Their responses: ---------------------------------------------------------- I. x0' = x0. In other words: x0' <> x0-vt, or constant values on the x-axis are not subject to the transform. AA: ==================================================================== No. x0' = x0 - vt. Well, if you want, you could define constant values on the x-axis, but in the context of the question that is not relevant. The relevant fact is that if the unprimed observer holds an object at point x0, then the primed observer assigns to that object a coordinate x0' which is numerically related to x0 by x0'= x0 -vt. AA: ==================================================================== EE: ==================================================================== What does this mean? The line x=x0 will give x'=x-v*t=x0-vt', so if x0' is to give the coordinate in the (x',t',)-system, it will be given by x0'=x0-v*t': ie., it is not given by a constant. Thus, being at rest (constant x-coordinate) is a coordinate-dependent concept. EE: ==================================================================== GG: ==================================================================== Sounds very false. We can say that the representation of the point X0 is the number x0 in the unprimed system, and x0' in the primed system. Clearly x0 and x0' are different, if vt is not zero. However one may say that (though it sounds/is stupid) the point X0 itself is the same throughout the transformation. However that expression sounds meaningless, since a transform (ok, maybe we should call it a change of basis) is only a function that takes the point's representation in one system into the same point's representation in another system. It is preferrable to use three notations: X0 for the point itself and x0 and x0' for the points' representations in some coordinate systems. GG: ==================================================================== ------------------------------ That idea is one of the most idiotic to come up, and it does so frequently. And in a number of guises. The idea being that x.c' <> x.c-vt, with x.c being what we have called x0 above; the notation makes no difference. Some crackpots have managed to maintain that position even after graphs have illustrated that such an idea means that after a while a circle center represented by x.c' could be outside the circle. The leading crackpot just make that explicit, as far as one can tell from his befuddled post in response to a line about active transforms, which are actually moving body situations, not coordinate transformations: -------------------------------------------------------------------- e>An active transform is not a coordinate transform, ... Right, it is a transform of the center (in the opposite direction) done to effect the change of coordinates without a coordinate transform. ... E: Transform of the center? Center of a circle? He really is saying a circle center moves in the opposite direction of the circle! Right? -------------------------------------------------------------------- If r=10 and x.c was at x.c=0, then the points on the circle (10,0), (-10,0), (0,10) and (0,-10) could at some time become (-10,0), (-30,0), (-20,10), and (-20,-10), but with x.c'=x.c, the circle center would be at (0,0) still! The circle is here but its center is way, way over there! Indeed, although a change of coordinate systems is not movement of any object described in the coordinates, the x.c'=x.c crackpottery is tantamount to the circle staying put but the center moving away. Or vice versa. ------------------------------ One crackpot puts the (x'-x.c')=(x-vt - x.c+vt) relationship like this: (x-vt+vt - x.c). See, he says, that is transforming x (with x-vt - x.c) and then reversing the transform (x-vt+vt - x.c). That's just another crackpot form of the idiocy that x.c' <> x.c-vt. You'll have noticed the implication is that there is no transform vt term relating to x.c. ------------------------------ Time Dependent? That particular crackpottery is perhaps more corrupt than moronic, since it includes deliberately hiding a vt term from view, and pretending it isn't there. [However, we have seen above that it is a familiar incompetency, and not likely an original.] Look, the crackpots say, there is a time term in the transformed (x' - x.c+vt). The transform isn't invariant! It's time dependent! Just put x' in its original axis form, also, which reveals the other time term, the one they hide: (x'-x.c+vt) = (x-vt - x.c+vt) = (x-x.c). So, at any and all times, the transform reduces to the original expression, with no time term on which to be dependent. Then there is the fact that if you leave the equation in any of the various notation forms - with or without reducing them algebraicly - the arithmetic always comes down to the same as (x-x.c). That means nothing to crack- pots, but may mean something to you. ------------------------------ My dictionary relates 'tautology' to needless repetition. That's another form of the x.c' <> x.c-vt idiocy. The repetition involved is the vt transformation term. Apply the -vt term to the x term, and it is needless repetition to apply it anywhere again? The 'again' is to the x.c term. The x.c' = x.c crackpot idiocy. The repetition of the vt terms is required by the presence of two x values to be transformed. Be sure to note the next section. ------------------------------ a Linear Transform? Now, how on earth can we relate a tautology to a basic definition in math? we get this definition: -------------------------------------------------------------- A linear transformation, A, on the space is a method of corr- esponding to each vector of the space another vector of the space such that for any vectors U and V, and any scalars a and b, A(aU+bV) = aAU + bAV. ------------------------------------------------------------- Let points on the sphere satisfy the vector X={x,y,z,1}, and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1, and b=-1. Let A= ( 1 0 0 -ut ) ( 0 1 0 -vt ) ( 0 0 1 -wt ) ( 0 0 0 1 ) A(aX+bC) = aAX + bAC. aX+bC = (x-x.c, y-y.c, z-z.c, 0 ). The left hand side: A( x - x.c , y - y.c, z - z.c, 0 ) = ( x-x.c , y-y.c, z-z.c, 0 ). The right hand side: aAX= ( x-ut, y-vt, z-wt, 1 ). bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ). and aAX+bAC = ( x-x.c, y-y.c, z-z.c, 0 ). Need it be said? Sure: QED. On the galilean transform the definition of a linear transform, A(aU+bV)=aAU + bAV, is completely satisfied. The generalized form transforms exactly and non-redundantly - with ONE TRANSFORM, not a transform and reverse transform - and non- tautologically, just as the very definition of a linear transform says it should. And does so with absolute invariance, with this galilean transformation. ------------------------------ The main crackpot that has asserted such a thing was referring equation; for which we have shown repeatedly elsewhere that the numerical calculations are identical for any primed values as for the unprimed values. The presence - before transformation - of a velocity term seems to confuse the crackpots. It turns out there is ex- treme historical reason for this, as you will see in the subject on Maxwell's equations. ------------------------------ forms and the galilean transforms. ------------------------------ Oh? Just what is the magical term in them that prevents (x'-x.c')=(x-vt - x.c+vt)=(x-x.c) from holding true? It turns out not to be magic, but reality, that interferes with the application of the galilean transforms to the gen- eralized coordinate form(s) of Maxwell: there are no coordi- nates to transform! When True Believer crackpots are shown the simple demonstration that the galilean transform on generalized cartesian coordinates is invariant, their first defense is usually an incredibly stupid x0'=x0, because the coordinate of a circle center, or point of emission, etc, is a constant and can't be transformed. The last defense is but Maxwell's equations are not invariant under that coordinate transform. When asked just what magic occurs in Maxwell that would prevent the simple algebra (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0) from working, and when asked them for a demonstration, they will never do so, however many hundreds of times their defense is asserted. The reason may help you understand part of Einstein's 1905 paper in which he gave us his absurd Special Relativity derivation: THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED. Einstein gave the electric force vector as E=(X,Y,Z) and the magnetic force vector as B=(L,M,N), where the force components in the direction of the x axis are X and L, Y and M are in the y direction, Z and N in the z direction. Those values are not, however, coordinates, but values very much like acceleration values. BTW, the current fad is that E and B are 'fields', having been 'force fields' for a while, after being 'forces'. So, when Einstein says he is applying his coordinate transforms to the Maxwell form he presented, he is either delusive or lying. (a) there are no coordinates in the transform equations he gives us for the Maxwell transforms, where B=beta=1/sqrt(1-(v/c)^2): X'=X. L'=L. Y'=B(Y-(v/c)N). M'=B(M+(v/c)Z). Z'=B(Z+(v/c)M). N'=B(N-(v/c)Y). X is in the same direction as x, but is not a coordinate. Ditto for L. They are not locations, coordinates on the x-axis, but force magnitudes in that direction. Similarly for Y and M and y, Z and N and z. (b) the v of the coordinate transforms is in Maxwell before any transform is imposed; Einstein's transform v is the velocity of a coordinate axis, not the velocity he touched it. (c) if they were honest Einsteinian transforms, they'd be x, which means it is X and L that are supposed to be transformed, not Y and M, and Z and N. And when SR does transform more than one axis, each axis has its own velocity term; using the v along the x-axis as the v for a y-axis and z-axis transform is thus trebly absurd: the axes perpendicular to the motion are not changed according to SR, the v used is not their v, and the v is not a transform velocity anyway. (d) as everyone knows, the effect of E and B are on the direction. Both the speed and direction are changed by E and B, but v - the speed - is a constant in SR. As absurd as are the previously demonstrated Einsteinian blunders, this one transcends error and is an incredible example of True Believer delusion propagating over decades. The components of E and B do differ from point to point, and in the variations that are not coordinate free, they are subject to the usual invariant galilean trans- formation when put in the generalized coordinate form. ------------------------------------------------------------- The SR crackpots don't know what coordinates are. The various things they call coordinates include coordin- nates, but also include a variety of other quantities. ------------------------------------------------------ 1. One may express coordinates in a one-axis-at-a-time manner [like x^2+y^2=r^2] but it is the use of vector notation that shows us what is going on. In vector notation the triplet x,y,z [or x1,x2,x3, whatever] represents the three spatial coordinates, but there are so-called basis vectors that underlie them. Those may be called i,j,k. Thus, what we normally treat as x,y,z is a set of three numbers TIMES a basis vector each. 2. These e*i, f*j, g*k products can have a lot of meanings. If e, f, j are distances from the origin of i,j,k then e*i, f*j, g*k are coordinates: distances in the directions of i,j,k respectively, from their origin. That makes the triplet a coordinate vector that we describe as being an x,y,z triplet; perhaps X=(x,y,z). The e*i, f*j, g*k products could be directions; take any of the other vectors described above or below and divide the e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)]. That gives us a vector of length=1.0, the e,f,g values of which show us the direction of the original vector. That makes the triplet a direction vector that we describe as being an x,y,z triplet; perhaps D=(x,y,z). The e*i, f*j, g*k products could be velocities; take any of the unit direction vectors described above and multiply by a given speed, perhaps v. That gives a vector of length v in the direction specified. That makes the triplet a velocity vector that we describe as being an x,y,z triplet; perhaps V=(x,y,z). Each of the three values, e,f,g, is the velocity in the direction of i,j,k respectively. The e*i, f*j, g*k products could be accelerations; take any of the unit direction vectors described above and multiply by a given acceleration, perhaps a. That gives a vector of length a in the direction specified. That makes the triplet an acceleration vector that we describe as being an x,y,z triplet; perhaps A=(x,y,z). Each of the three values, e,f,g, is the acceleration in the direction of i,j,k respectively. The e*i, f*j, g*k products could be forces (much like accel- erations); take any of the unit direction vectors described above and multiply by a given force, perhaps E or B. That gives a vector of length E or B in the direction specified. That makes the triplet a force vector that we describe as being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each of the three values, e,f,g, is the force in the direction of i,j,k respectively. Einstein's - and Maxwell's - E and B are not coordinate vectors. ============================================================ There is another variety of intellectual befuddlement that misinforms the idea that Maxwell isn't invariant under the galilean transform: confusions about velocities. Velocities With Respect to Coordinate Systems. ----------------------------------------------- Aaron Bergman supplied the background in a post to a sci.physics.* newsgroup: =============================================================== Imagine two wires next to each other with a current I in each. Now, according to simple E&M, each current generates a magnetic field and this causes either a repulsion or attraction between the wires due to the interaction of the magnetic field and the current. Let's just use the case where the currents are parallel. Now, suppose you are running at the speed of the current between the wires. If you simply use a galilean transform, each wire, having an equal number of protons and electrons is neutral. So, in this frame, there is no force between the wires. But this is a contradiction. ================================================================ First of all, the invariance of the galilean transform (x'-x.c') =(x-x.c), insures that it is an error to imagine there is any difference between the data and law in one frame and in another; the usual, convenient rest frame is the best frame and only frame required for universal analysis. [Well, (x'<>x, x,c'<>x.c, but (x'-x.c')=(x-x.c).] Second, given that you decide unnecessarily to adapt a law to a moving frame, don't confuse coordinate systems with meaningful physical objects, like the velocity relative to a coordinate system instead of relative to a physical body or field. In other words, what does current velocity with respect to a coordinate system have to do with physics? Nothing. Certainly not anything in the example Bergman gave. What is relevant is not current velocity with respect to a coordinate system, but current velocity with respect to wires and/or a medium. The velocity of an imaginary coordinate sys- tem has absolutely nothing to do with meaningful physical vel- ocity. You can - if you are insightful enough and don't violate item (e) - identify a coordinate system and a relevant physical object, but where some v term in the pre-transformed law is in use, don't confuse it with the velocity of the coordinate transform. Velocities With Respect to ... What? ----------------------------------------------- Albert Einstein opened his 1905 paper on Special Relativity with this ancient incompetency: =============================================================== The equations of the day had a velocity term that was taken as meaning that moving a magnet near a conductor would create a current in the conductor, but moving a conductor near a wire would not. This was belied by fact, of course. The important velocity quantity is the velocity of the magnet and conductor with respect to each other, not to some absolute coordinate frame (as far as we know) and not to an arbitrary coordinate system. One possible cause was the idea: but the equation says the magnet must be moving wrt the coordinate system or ... the absolute rest frame. There not being anything in the equation(s) to say either of those, it is amazing that folk will still insist the velocity term has nothing to do with velocity of the two bodies wrt each other. ----------------------------------------------------------- ------------------------------ One of the intellectually corrupt ways of denying the very simple demonstration of galilean invariance of all laws expressed in the generalized coordinate form demanded by analytic geometry, vector analysis, and measurement theory [ (x'-x.c')=[ (x-vt)-(x.c-vt) ]=(x-x.c) ] is the assertion that those equations 'over there' (usually Maxwell or wave) are somehow immune to the elementary laws of algebra used to demon- strate the invariance. [Unfortunately, the assertions are never accompanied by reference to the magical math that makes elementary al- gebra invalid. Wonder why that is?] Part of the time it is based on the old lore based on the incompetent transformation of the privileged form of an equation instead of the correct form. [Evidence of this is any reference to an effect due to the velocity of the transform; it falls out algebraicly - as you see above - and cancels out arith- metically - as you can see above.] But usually it is just whistling in the dark, waving the cross (zwastika, I'd say) at the mean old vampire. The most general equation that could be conjured up is a differential with either First or Second Derivatives. Let's examine the plausibility of such magical magical, non-invariance assertions. (a) to get a Second Derivative you must have a First Derivative. (b) to get a First Derivative you must have a function to differentiate. (c) to get a Second Derivative you must have a function in the second degree. So, let us examine the question as to whether any such common Maxwell/wave equation will differ for (a) the common, privileged form, represented as ax^2, with a being an unknown constant function. (b) the generalized cartesian form, represented as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2, with a being an unknown constant function. (c) the transformed generalized cartesian form, represented as a(x-vt -x.c+vt)^2, same as for (b), = ax^2 -2ax(x.c) + ax.c^2, of course, with a being an unknown constant function. I. for (a), remembering that x.c is a constant, and that this version is only correct because x.c=0, otherwise (b) is the correct form: d/dx ax^2 = 2ax (d/dx)^2 ax^2 = 2a II. for (b), remembering that x.c is a constant. d/dx (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c (d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2a III. for (c); same as for (b). So, what we have seen so far is (1) differential equations in the second degree - the wave equations - must clearly be the same for all forms: the privileged form in x, the generalized cartesian form in x and the centroid, x.c, or the transformed generalized cartesian form. That is, anyone who imagines that correct usage gives different results for galilean transformed frames is at first showing his ignorance, and in the end showing his intellectual corruption. (2) As far as the First Derivatives are concerned, the only cases in which there really is a difference between the two forms is where x.c <> 0, and in that case, the use of the privileged form is obviously incompetent. So, how do you correctly use the differential equations? If you are using rest frame data with the centroid at x=0, etc, you can't go wrong without trying to go wrong. If you are using rest frame data with the centroid not at x=0, you must use (x-x.c) anyplace x appears in the equation. If you are using moving frame data, you must use the moving frame centroid as well as the light front (or whatever) moving frame data itself, perhaps first calculating (x'-x.c'), which equals (x-x.c) which is obviously correct, and which is obviously the plain old correct x of the privileged form. Unless, of course, there really is some magical term or expression that invalidates the obvious and elemen- tary algebra of the invariance demonstration. Or maybe you just whistle when you don't want basic algebra to hold true. Eleaticus !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ! Eleaticus Oren C. Webster ThnkTank@concentric.net ? ! Anything and everything that requires or encourages systematic ? ! examination of premises, logic, and conclusions ? !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ==== [snip] > Invariant Galilean Transformations (FAQ) On All Laws > (c) Eleaticus/Oren C. Webster [snip 1300 lines of utter crapola] Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav Psychotic ineducable boring troll Eleaticus, You see yourself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way, http://www.mazepath.com/uncleal/effete7.jpg http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg http://www.you-moron.com/ http://www.mazepath.com/uncleal/effete0.jpg http://www.mazepath.com/uncleal/effete1.png http://www.mazepath.com/uncleal/effete2.png http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/effete4.png http://www.mazepath.com/uncleal/effete5.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 Experimental constraints on General Relativity. http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf http://users.powernet.co.uk/bearsoft/Paper6.pdf http://users.powernet.co.uk/bearsoft/LPHrel.html Longitudinal and transverse mass http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) Approved: news-answers-request@MIT.EDU Summary: The Lorentz transforms themselves are proof t' and x' cannot possibly be just coordinates. Examination of their derivation verifies their identity as ==== Opponents should first actually find out what the content is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, selfless, *.answers moderators evidences ignorance and atrocious netiquette. Archive-name: physics-faq/criticism/lorentz-intervals (SR) Lorentz t', x' = Intervals (c) Eleaticus/Oren C. Webster Thnktank@concentric.net ------------------------------ of the use of 'just coordinates' in any scientific formula. Defenders of the Special Relativity faith are especially fond of telling opponents of their space-time fairy tales that they do not know the difference between coordinates and magnitudes. That may often be so, but the fault lies ultimately with SR dogma. The Lorentz-Einstein transformations cannot possibly be 'just coordinates', which is the interpre- tation required to support the many sideshow carnival acts with which the SR faithful bedazzle the public, and establish their moral and intellectual superiority. If I get in my car and drive steadily for a few hours at 50 kilometers per hour, is 50t the distance I travel? Of course not. The last time my hours-counting 'just coord- inates' clock was set to zero was when Zeno first reported one of his paradoxes to Parmenides. That was a long time ago, so my t is not useful for such purposes unless you also use my clock to established the starting time, perhaps t0, and use the formula 50(t-t0) to calculate the distance. In any case, my t is even then not 'just a coordinate' because it always represents particular elapsed times that can be used in the (t-t0) form to calculate perfectly good time intervals (elapsed times). Alternatively, I could (re)set my clock to zero at the start of some meaningful time interval, in which case my t shows a scientifically perfect current and/or end time. In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a function of an elapsed time interval (not 'just a coordinate') and a time interval (-vx/cc; the interval amount the t' clock is being screwed up at time t) and thus cannot be 'just a coordinate' since neither of the independent variables is such a 'just' thing. {Their meaning is shown below, step-by-step.] If it takes me 50 minutes to cross the Interstate highway, was x/50 my velocity crossing it? Of course not. The origin of all my axes is at the very spot where Zeno first presented his first paradox to Parmenides. That makes my x equal a couple of thousands of miles, plus, and is not useful for such purposes unless you establish the starting x value, perhaps x0, and use the formula (x-x0)/50 to calculate my velocity. In any case, even then my x is not 'just a coordinate' because it always repesents particular distance intervals that can always be used in the (x-x0) form for any and every scientific purose. Alternatively, I could move my x-axis origin to the starting (zero) point of some meaningful distance, in which case my x shows a scientifically perfect current and/or end distance. In which case, the Lorentz-Einstein x'=(x-vt)/g is a function of a current/ending distance interval (not 'just a coordinate') and a distance interval (-vt; the interval amount the x' axis is being screwed up at time t) and thus cannot be 'just a coordinate' since neither of the independent variables is such a 'just' thing. {Their meaning is shown below, step-by-step.] ------------------------------ 1. Introduction with the obvious debunking of the use of 'just coordinates' in any scientific formula. 3. The Lorentz-Einstein transforms. 4. The 'just coordinates' argument. 5. Single-system, little-purpose ambiguity. 6. Relating two coordinate measures/systems. 7. Distances and moving coordinate axes. 8. Time intervals. 9. Einstein's (1905) derivations. 10. A word about intervals. 11. Intervals versus the Twins Paradox. 12. Summary ------------------------------ Special Relativity's space-time circus is based on the 'transformation' equations by which it is believed one can relate a nominally 'stationary' system's space and time coordinates to those of an inertially (not accelerating) moving other observer. That moving observer's own physical body and coordinate system might have been identical in size to those of the stationary observer before the traveller began moving, but are 'seen' as very different by the stationary observer when the relative velocity of the two is great enough, a high percentage of the velocity of light. Concerning ourselves - as is customary - with just the spatial coordinate axis that lies parallel to the direction of motion, and with time, Einstein arrived at these formulas that relate the moving system measures or coordinates (x' and t') to the stationary system coordinates (x and t): x' = (x - vt)/sqrt(1-vv/cc) (Eq 1x) t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) The v is for the two systems' relative velocity as seen by the stationary observer, and is positive if the dir- ection is toward higher values of x. By concensus, the moving system x'-axis higher values also lie in that direction, and all axes parallel the other system's corresponding axis. We used vv to mean the square of v but might use v^2 for that purpose below. Similarly for c. Because it is believed that no physical object can reach or exceed c, the square-root term in both denominators is presumed always less than one, which means that the formulas say both x' and t' will tend to be greater than x and t, respectively. However, SRians call the x' result 'contraction' - which means shortening - and the t' result 'dilation' - which means increasing. ------------------------------ The 'just coordinates' argument is so patently ridiculous that even opponents have a hard time accepting just how simple and obvious its debunking can be, as shown in this section. However, further sections take a more arithmet- ical approach that you'll maybe find more professorial. The 'just coordinates' argument is that t is mot a duration, not a time interval; it's just an arbitrary clock reading. But what if the moving system observer comes speeding by while you make your annual 'spring forward' or 'fall back' change? The formula says that the moving system clock's 'just coordinate' reading can be calculated from yours: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Imagine the moving system oberver's confusion if his clock changes its reading while he's looking at it! If his clock doesn't change when yours does, the formula is wrong; if it is truly a 'just coordinates' formula. And then what happens if you realize you were a day early and put your clock back to what it had said previously? And suppose you are in NYC and your twin in LA and both are watching the moving observer. You'll both be using the same v because you are at rest wrt (with respect to) each other. You're on Eastern Standard Time and your twin is on Pacific Standard Time maybe. You have three hours more on your clock than does your twin. On which 'just coordinate' clock will the Lorentz transforms base the 'just coordinate' time the moving system clock says? The formula applies to both of your t-times: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Sure, the idea that you can change someone else's clock with no connection of any kind is really ridiculous, but Eqs 1x and 1t aren't MY equations. Are they yours? And we aren't the ones to say x, t, x', and t' are just coordinates. If the t' formula is actually either an elapsed time formula, or the basis of a t'/t ratio, then there is no implication that one clock's reading has anything to do with the other's. It can only be rates of clock ticking, or how one time INTERVAL compares to the other that the formula is about. ------------------------------ Since we're going to be comparing measurements on two coordinate systems in the next section, let's go to our supply cabinet and get our yard-stick (which we use to measure things in inches) and our meter-stick (which we use to measure things in centimeters). Here, I'm getting mine. Oh! Oh! There's an ant on mine, and he ... she ... sure is hanging on, right at the 3.5 inch mark of the yard- stick. Let's see if I can wave the stick around enough that she'll let go. Nope. However, before I gave up I waved the stick and the ant 'all over the place. Always, however, the ant was at the 3.5 mark on the yard-stick, and always 3.5 away from the end of the stick, however far and wide I have transported her. Neither of those 3.5 facts means very much. Of the two, the distance aspect meant almost nothing. So the distance was 3.5 from the end. So what? That length, distance, was not in use. And only maybe the ant might have been concerned with just what location, 'just coordinate', on the stick she was at. Just so with x and t. So, is the 3.5 reading just a coordinate? Or a distance/length? It's ambiguous in and of itself, and really makes no difference what you say until you try to make use of the number. Hey, my address is 5047 Newton Street. If you are looking for me and you're at 4120 Newton, it is helpful information, because it tells you which direction to go. Is that 'just coordinate'? Where it really becomes useful, perhaps, is in telling you how far away I am. That's not just a coordinate value, that's a distance, length, interval. However, it is subtracting 4120 from 5047 that tells you which direction and how far. It is only because both 5047 and 4120 are distances from the same point - ANY same point - that the result means anything. My x - my yardstick reading - is always a distance or length; it is impossible to be otherwise with an honest, competently designed yardstick. Whether or not its reading is of good use in some particular scientific formula depends on whether I put the zero end of the yardstick at some useful place. As in the introduction, we should either put it at the starting location/end, or use two readings from it: (x-x0). ------------------------------ Taking care to not damage our brave little ant, I place my yard-stick onto the table, zero end to the left, 36 end to the right. Now I place the 'just coordinate' meter-stick on the table in the same orientation, in a random location, and find that the ant's coordinate on the meter-stick is 51. The formula relating centimeters to inches is cm=i*2.54 but we want a formula similar to x'=(x-vt)/sqrt(1-vv/cc). That would be c=i/.03937 approximately, but let's use x' for the meter-stick reading, and x for the inch reading: x'=x/.3937. 3.5/.3937 = 8.89 Wait a minute. It's not just science but definition that says c=i/.3937=8.89, so something is wrong. 8.89 is not 51. We already knew that 51 cm was just an arbitrary coordinate. Arbitrary not because that point isn't 51 cm from the zero end of the meter-stick, but because the zero point was in an arbitrary position. Let's put the meter-stick in a position where it's zero point is at the yard-stick zero point. What is the centimeter coordinate now? Hey. 8.89, just like the formula says. The only way for a 'transform' like x'=x/g to work, whatever g might be, is for both coordinate systems to have their zero points aligned, in which case saying the two measures are not intervals is pure idiocy. Noe that with both zero points at the same position both x' and x are great measures for scientific purposes, in any and every case where we were smart enough to put those zero points at a useful location. There is one extension of x'=x/g that will let us use the meter-stick in arbitrary position. When the cm reading was 51, the zero point of the yard-stick read (51-8.89=) 42.11 cm. If we call that point x.z' we get x' = x.z' + x/.3937. = 42.11 + 3.5/.3937 = 42.11 + 8.89 = 51. Obviously, in this formula x/.3937 is the distance from the x' coordinate of the location where x=0. An interval. Just as obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary x' coordinate, does not mean that x' is anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (x'-x.z') in the scientific formula. (x'-x.z') equals the useful, Ratio Scale value x/.3937. So, we have discovered a basic fact: a transformation formula like x'=x/g works only if the two zero points of the coordinate systems coincide. That makes it non- sense to say the two coodinates are only coordinates and not intervals. Both must be values that represent distances from their respective zero points unless you take the proper steps to adjust for the discrepancy. Make sure you understand that although the inclusion of x.z' made it possible to correctly calculate x', the result is nonsense when it comes to use of x' for general length/distance purposes; it is x'-x.z' that is a useful number in such cases. It could be that we're measuring a sheet of paper with one end at x=0 and the other at x=3.5; x'=51 is nonsense as a centimeter measure of the paper. But, you say, the Lorentz transform contain a -vt term. ------------------------------ We discovered x'=x.z' + x/g as the correct formula for relating one coordinate to another system's. But the Lorentz transform contains another term, -vt/sqrt(1-vv/cc). What is it? Let's start with our x'=51 cm, x=3.5, x.z'=42.11 example. Every minute, let's move the meter-stick one inch to our right. At minute 0, the cm reading was 51 cm. At minute 1, the cm reading is now 50 cm. At minute 2, the cm reading is now 49 cm. In this instance, v=1 inch/minute. And t was 0, 1, 2. What has happened is that we have made our x.z' a lie, and increasingly so. -vt/.3937 is the change in x.z'. x' = (x.z - vt/.3937) + x/.3937. Obviously, vt/.3937 is not a coordinate; even most SRians wouldn't imagine it was. It is an interval, the distance over which the moving system has moved since t=0. And, of course, x/.3937 is the distance of our brave little ant from the point where x=0 and the centimeter reading is x.z'-vt/.3937. Yes, every minute the meter- stick moves to the right and the meter-stick coordinate of the spot where x=0 gets less and less - and eventually negative. Make sure you understand that every minute the x' coordinate, because of -vt/g, becomes a better measure of, say, the 3.5 paper we might be measuring with the yard-stick, given that 51 was too big a number and -vt is negative. That is, until the two origins coincide at x'=x=0, and then it gets worse and worse. With -vt positive (because v<0) the situation is different. With 51 and -vt positive, x' just gets worse and worse over time. Quite obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary x' coordinate even when the x' axis is moving, does not mean that x'is anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (x'-x.z'+vt/.3937) in the scientific formula. (x'-x.z'+vt/.3937) equals the useful, Ratio Scale value x/.3937. ------------------------------ Instead of using our sticks, let's get out two clocks. Mind you, we're not going to deal with different clock rates here, just establish the same basics as for distance. Your clock says 9:00 Eastern Standard Time (EST) and we note that t=540 minutes when we put down the clock. Blindly, let's turn the setting knob of your twin's Pacific Standard Time clock and put it down before us. According to what we see, EST's 540 minutes (9:00) corre- sponds to PST's 14:30; t'=870. We know the formula relating PST to EST is t' (pacific) = t (eastern) - 180 (minutes). Thus, it is not correct that the second clock can have an arbitrary setting, because 870 <> 540-180. We know that the two clocks are related by t' = t/1 since both are using the same second, hour, etc units. But 870 (14:30 in minutes) is not 540/1-180, so once again we know something is wrong. However, t'=t.z' + t/1 works. EST midnight equals PST 0.0 (midnite) - 180, so t.z' = -180, and t' = -180 + 540/1 = 360. Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes. We see thus that like distance measures/coordinates, time axis origins (zero points) must either be 'lined up' or adjusted for. So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must be the moving system elapsed time interval since the time axes were both at a common zero. There is no t.z' adjustment: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Make sure you understand that in the clock case, if the EST is showing a good number for elapsed time since the travelling observer passed NYC, then the PST clock is silliness. t.z' must be zero or must be taken out of time lapse calculations for the PST clock to be used intelligently, just as was true for x.z'. What is lacking as yet for Lorentz t' is the -vx/cc term that corresponds to the x' formula -vt term. Break it up into two parts: v/c and x/c. v/c is a scaling factor that changes velocity from whatever kind of unit you are using over to fractions of c. x/c is distance divided by velocity, which is time. x/c is thus the time interval since the two time axes had a common zero point - which they have to have in the Lorentz transforms which do not have the t.z' term we learned to use above. Thus, (-vx/cc)/sqrt(1-vv/cc) is the interval amount the moving system clock has been changed - since the common/ adjusted time - over and beyond the elapsed time interval represented by x/sqrt(1-vv/cc). We have discovered that the only way for t' to be t/g is for t' and t to have a common zero point, just as for x' and x. It would be otherwise if the t' formula contained an adjustment t.z' under some name or other, but the necessity to include such a term correlates 100% with t' numbers that aren't directly usable. As for x and x', our knowledge of how to setup a proper formula relating t and t' is of no use unless we use the knowledge in scientific formulas; (t'-t.z'+xv/gcc) gives us the only directly useful value: t/g. ------------------------------ When we return to Einstein's derivations of the transform formulas with a well-focused eye, we find he was a wee bit confused - or at least self-contradictory. When he set up his (at first unknown) tau=moving system time formulas, he created three particular instances of tau. Tau.0 is the time at which light is emitted at the moving origin toward a mirror to the right that is moving at rest wrt that moving origin and at a constant distance from that origin. He lets the stationary time slot have the value t, a constant, the stationary system starting time. Tau.1 is the time at which the light is reflected. He lets the stationary time be t+x'/(c-v); t is still a constant and x'/(c-v) is the time interval since t. Tau.2 is the time at which the light gets (back) to the moving origin. The stationary time value is put as t + x'/(c-v) + x'/(c+v); t is still a constant and x'/(c-v) + x'/(c+v) is the time interval since t. On the thesis that the moving observer sees the time to the mirror as the same as the time back to the origin, he sets .5[ tau.0 + tau.2 ] = tau.1. Tau.0 completely drops out of the analysis and leaves no trace, and has no effect. Further, the t you see in tau.0, tau.1, and tau.2 also completely drops out with no trace and no effect, leaving us with exactly what you'd get if you had explicilty said t' is an interval and so is t. What doesn't drop out in the stationary time values is x'/(c-v) and x'/(c+v), the time interval it takes for light to get to the fleeing mirror, and the time interval it takes for light to get back to the approaching origin. Thus, his resultant t' formula is strictly based on time intervals in the stationary system. Time intervals since some starting time, yes, but time intervals. There is absolutely nothing in the derived formulas that depends on arbitrary coordinates like the constant t in the stationary time arguments. Let's look at the x dimension; it is x'=x-vt [as x increases by vt, the effect over time is x'=(x+vt)-vt)], which Einstein explicitly sets up as a constant stationary distance. He uses that x' not just in the time interval parts of the stationary time arguments, but also in the x (distance) stationary system argument for the tau at the time light is reflected. x' can't be the stationary system coordinate of the mirror at that time. That value is x'+vt. x' is explicitly an interval, distance. Thus, the whole tau derivation of the t' formula is fully and explicitly based on x' - a spatial length/distance/interval - and the two time interals x'/(c-v) and x'/(c+v). While we're at it, if the starting t is not zero, his x'=x-vt formula is complete nonsense also. Given that there was some L that was the mirror x-location and length when the light is emitted, if t was already, say, 500, then x'=L-vt could have been a very negative length. ------------------------------ There are intervals, and there are intervals. If we put our yard stick zero point at one end of a piece of paper and read off the coordinate at the other end of the paper, we have a good measure of the paper's length, a Ratio Scale measure. [Absolute temperature scales are ratio scale.] If instead we put the one end of the paper at the one inch mark (or the zero end of the stick one inch 'into' the length of the paper) we get measures that are one inch off the true, ratio scale length. The two messed up measures are still intervals, but they are Interval Scale measures. [Household temperature scales are interval scale, which is why your physics and chemistry professors won't let you use them without first converting to the ratio scale absolute temperatures.) t'=t/g and x'=x/g represent ratio scale measures, given that t and x were ratio scalae to start with. t'=t.z'+t/g and t'=t/g-vx/gcc are both interval scale measures, even given a good ratio scale t and a good ratio scale x. x'=x.z'+x/g and x'=x/g-vt/g are both interval scale measures, even given a good ratio scale x and a good ratio scale t. Look for the (SR) Lorentz t', x' = degraded measures document soon at a newsgroup near you. ------------------------------ t'=(t-vx/cc)/g shows t' being greater than t. The reason Special Relativity will not allow the use of its basic time equation in determining what SR has to say about the twins' ages, is that t' and x' are supposedly just coordinates, and they say you have to take the coordinate pairs (t',x') and (x,t) into consideration in both the time and place the twins' separation started and the time and place the twins reunited. Since t' and x' are actually both intervals, not just coordinates, the 'excuse' is spurious, and is so even without use of the obvious (x_b-x_a) and (t_b-t_a) usages. However, SR is right to be embarrassed by their transformation formulas. Look for the (SR) Lorentz t', x' = degraded measures document at a newsgroup near you. ------------------------------ A. t'=t/g and x'=x/g can be almost 'just coordinates' in the sense that the values obtained may not be of much use except in the most primal and useless way: how long and how far since/from the time/ place they were zero. Even here, however, the zero points within each of the two scale pairs (t',t) and (x'.x) must have been lined up. If the zero points have been intelligently selected (such as at the starting point and time of a trip) they can be rationally used 'as is' in any valid sci- entific equation. B. Even the interval scale t'=t.z' - xv/gcc + t/g and x'=x.z' - vt/g + x/g are not 'just coordinates'. They can be used to good effect by establishing the relevant starting times/points and using (t'-t.z'+xv/gcc) and (x'-x.z'+vt/g), as the situation may require. C. When you see vx/gcc or vt/g in use in any guise with non-zero values, you know the resultant t' or x' is a degraded, interval scale value. E-X: Anytime you do not see what amounts to t.z' and xv/gcc in the time case, or x.z' and vt/g in the distance case, you know that the t' and/or x' in use are intervals. Period. Y: Either set your clock to zero at the start of the relevant time interval, or use (t-t0), with both being readings on the same clock. Either move your x-axis origin to the starting end or point, or use (x-x0), with both being readings on the same axis. Z: In _(SR) Lorentz t', x' = Degraded (Interval) Scales_ we see that t' and x' satisfy the mathematical tests for/of interval scales when -vt and -vx/cc are not zero; thus, they must be intervals. When -vt and -vx/cc are zero, t' and x' satisfy the much better mathematical definition of ratio scales, and are thus not just mere intervals, but (rescaled) good ones. Eleaticus !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ! Eleaticus Oren C. Webster ThnkTank@concentric.net ? ! Anything and everything that requires or encourages systematic ? ! examination of premises, logic, and conclusions ? !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ==== [snip] > (SR) Lorentz t', x' = Intervals > (c) Eleaticus/Oren C. Webster > Thnktank@concentric.net [snip 720 lines of utter crapola] Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav Psychotic ineducable boring troll Eleaticus, You see yourself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way, http://www.mazepath.com/uncleal/effete7.jpg http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg http://www.you-moron.com/ http://www.mazepath.com/uncleal/effete0.jpg http://www.mazepath.com/uncleal/effete1.png http://www.mazepath.com/uncleal/effete2.png http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/effete4.png http://www.mazepath.com/uncleal/effete5.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 Experimental constraints on General Relativity. http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf http://users.powernet.co.uk/bearsoft/Paper6.pdf http://users.powernet.co.uk/bearsoft/LPHrel.html Longitudinal and transverse mass http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics)