PRD > (Non-Occurence Introduction 1) PRD > En e N, if( Žnite(length(d)), n<= length(d)) PRD > Ai, PRD > !digitsmatchupton(i, d, n) PRD > -> d does not occur in S PRD > Does the rule under discussion end with PRD > -> d does not occur in S ? HERC > no The text -> d does not occur in S was copied and pasted from the text of the rule into my question. The rule most certainly *does* end with -> d does not occur in S. When I refer to a rule, such as Non-Occurence Introduction 1, I mean the rule exactly as it textually appears. When I refer to an application of a rule, I mean a substitution of terms for variables in the rule with the remainder of the body of the rule appearing in the application exactly as it appears textually in the rule. Thus, in classical logic we have the following two rules of inference (Modus Ponens) P -> Q P ______ Q (Modus Tolens) P -> Q ~Q ______ ~P Note that these are regarded as two different rules. They have separate names. They have separate forms. However, Modus Tolens could be *derived* from Modus Ponens by using the rule Transposition (Transposition) P -> Q _______ ~Q -> ~P In explaining why you answered the way you did, you used a rule HERC: (a <-> b) <-> (!a <-> !b) which is similar to Transposition above. If we are going to communicate, we need to have a way to identify rules *as they are textually* and distinguish these from rules that might be *derived* from the textual rule. Will you agree to communicate according to the rule that a reference to a rule refers to the textual form of the rule as written, and not to a rule that may be derived from it, but which differs textually other than in the naming of variables? If not, can you provide language that we can use to communicate references to the textual form of rules? Also, I am still waiting for responses to the questions found in my post of Also you say below: HERC > Originally you asked for this form. HERC > XXXX -> does not occur. HERC > Then you utilised it for a DEFINITION. I have not been interested in deŽnitions in this discussion, but in inferrence rules. That is, rules that may be mechanically applied. According to my own usage of the terms utilize and deŽnition I did not utilize any rule for a deŽnition. Please post a quote illustrating what you mean by this. Please also indicate whether, at the time you post, you still believe that I utilized a rule for a deŽnition. ********* nothing changed or added below this line ****** > [explanation] > this is the example I gave with a reason for the conclusion. > **************************************** > I answered the last one by *demonstrating* the example. > to show that 0.333... does not occur, you have to show > En e N, > Ai, > !digitsmatchupton2(i, d, n) > what is the value of n that d fails to match some element of S up to n digits? > if n = 10, there is some i, S[i] = 0.3333333333 > if n = 1,000,000 there is some i S[i] = 0.33333333..... 1,000,000 3¹s ......33333333 > you have no upper bound to the number of recurring 3¹s in S, therefore 0.3.. occurs in S, > since every digit of 0.3.. occurs in S in the correct sequence. > HERC > try it out : S = 010203 and d = 0.3.. > Therefore I DISAGREE the above is not an example. > (Non-Occurence Introduction 1) > En e N, if( Žnite(length(d)), n<= length(d)) > Ai, > !digitsmatchupton(i, d, n) > -> d does not occur in S > (Non-Occurence Introduction 1) B > En e N, if( Žnite(length(d)), n<= length(d)) > Ai, > !digitsmatchupton(i, d, n) > <-> d does not occur in S > The 1st version infers non occurence, the B version deŽnes non occurance, > some source of confusion. > ************************************** > Originally you asked for this form. > XXXX -> does not occur. > Then you utilised it for a DEFINITION. > The defn happens to be exactly the same if we change the implication to equivalence. > XXXX <-> does not occur > Now, because (a <-> b) <-> (!a <-> !b), > (Non-Occurence Introduction 1) B <-> (Occurance Defn) > Therefore its still valid to deduce occurence from the non-occurance defn. > Herc === Subject: Re: limitation to induction on Žnite bounds responding to |-|erc HERC, your responses to my messages are verbose, but I have difŽculty Žnding the answers to the speciŽc questions I have asked. If the answers to my questions are embedded in your response, I apologize for failing to Žnd them. Please answer the questions below in as short and direct a manner as possible, and please answer the questions immediately after they appear. If you wish to add comments or explanations, please added them at the very end after a clear sign indicating that comments follow. PRD > (Non-Occurence Introduction 1) PRD > En e N, if( Žnite(length(d)), n<= length(d)) PRD > Ai, PRD > !digitsmatchupton(i, d, n) PRD > -> d does not occur in S PRD > Could you please state the rule Non-Occurence PRD > Introduction 1 as an English sentence? HERC > if there exists a number n less than or equal to the HERC > length of string d, and there is no set member that HERC > matches the digits of d from digit 1 up to digit n, HERC > then d does not occur in S. PRD > Is this a direct translation? PRD > The term Žnite(length(d)) occurs in the symbolic PRD > version of the rule. It does not occur in the English PRD > version. What happened to the Žnite(length(d))? PRD > How was that translated? PRD > If your translation wsa not a direct translation, PRD > please provide a direct translation. HERC > OK, what might have been clearer is === Subject: Re: Does a high SAT score predict mathematical talent? > >>Counter example: Erdos. > >> > >>Mind all that amphetamine helped keep his mind sprightly. > > Did he really take speed? > Yes. Erdos really did take amphetamines quite habitually. I have > perused both of the book-length biographies of Erdos available in > American public libraries. > It appears that Erdo¹s unusual social life, both growing up and as > a fully grown adult, had something to do with his sustained > mathematical productivity. A rather intriguing off-hand suggestion > in a seminar monograph about evolutionary biology and human > intelligence > The Nature of Intelligence (2000) edited by Gregory R. Bock, Jamie > A. Goode, and Kate Webb (Chichester: Wiley) Novartis Foundation > Symposium 233) > is that professional output, in mathematics and other professions, > is related to sexual display. A young mathematician seeking a life > partner will get a Ph.D. and make tenure, and then settle down > once able to reproduce offspring as well as ideas. The prediction > of that hypothesis would be that most academics, not just in > mathematics, do their most conspicuous work (or most > energy-consuming work) at the age when they are about to start > their families. In the documentary on Erdos that I referred to, he said that he found sexual pleasure painful, or distasteful. Kind of a strange reaction to sex. > > Anyway, I was impressed with what Hungary used to do. > > That is what the US needs to do. There were math clubs > > for high school aged students interested in math, and many > > journals, like Acta Mathematica. > I agree that the competition culture fostered in secondary > education in Hungary for the last century has had beneŽcial > effects on the mathematics community in that country. An effort in > the United States, which reaches worldwide, to build a similar > kind of competition culture in the Internet era can be found at > http://www.artofproblemsolving.com > especially at that site¹s very useful online forums, which are > fully international in participation. I wish something of that > kind had existed when I was a kid in the early 1970s. I will > tentatively guess, inviting comments from participants in this > sci.math newsgroup, that helping young people Žnd soulmates with > whom they can discuss math helps them develop as productive > mathematicians over the course of a LONG [grin] lifetime. Does > that make sense to everyone else reading this thread? I took a quick look at this URL. Looks good to me. Van === Subject: Rings and their intuitive meaning Group elements represent processes that can be performed in sequence on the identity and undone. The most natural application of groups is to the symmetries of a mathematical object. What is a good way to visualize rings? What are they used for? I¹m looking for a use in mathematics analogous to the use of groups to represent symmetry. === Subject: Re: Rings and their intuitive meaning > Group elements represent processes that can be performed in sequence > on the identity and undone. The most natural application of groups > is to the symmetries of a mathematical object. > What is a good way to visualize rings? What are they used for? I¹m > looking for a use in mathematics analogous to the use of groups to > represent symmetry. Linear operators on a linear space form a ring. Or, as a simple case, square matrices form a ring. Those are typical examples of (in general) noncommutative rings. On the other hand, commutative rings are very useful in algebraic geometry and number theory. The basic example is the ring of polynomials in several variables over a Želd (for example, over complex numbers), and various other rings are derived by constructions such as subrings and quotient rings. === Subject: Re: Rings and their intuitive meaning |Group elements represent processes that can be performed in sequence |on the identity and undone. The most natural application of groups is |to the symmetries of a mathematical object. | |What is a good way to visualize rings? What are they used for? I¹m |looking for a use in mathematics analogous to the use of groups to |represent symmetry. ring elements represent _linear_ processes that can be performed in sequence. linearity makes it possible to combine processes not only by sequential composition, but also by means of pointwise addition; that¹s why rings have an addition operation as well as a multiplication operation. (prove to yourself that the pointwise sum of two parallel linear processes is another linear process!) on the other hand, linearity discourages us from requiring all processes to be capable of being undone, because subtraction and zero are natural things to include when you¹re already dealing with linearity and addition, and the zero process that takes everything to zero is a conspicuous example of a process that in general can¹t be undone by any other deterministic process. that¹s part of why there isn¹t any clause in the deŽnition of ring that requires division to be everywhere deŽned. -- [e-mail address jdolan@math.ucr.edu] === Subject: Use of a rational tiling group in sl(2,R) to get a 3d surface This method was suggested by the Bryant cousin surface. It gives an hyperboloid of one sheet that is very like a catenoid in shape. A determinant one group is assumed through out. This result is very different that the intent of Lagarias in terms of a upper half plane rational tiling. Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ (************************************************************ *********** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a Žle with a name ending in .nb, then open the Žle inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ************************************************************* **********) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 8382, 263]*) (*NotebookOutlinePosition[ 9317, 293]*) (* CellTagsIndexPosition[ 9273, 289]*) (*WindowFrame->Normal*) Notebook[{ Cell[BoxData[ ( (* marked tiles sl ((2, r)) rational (group : (page291 : A walk along the branches of the extended Farey Tree by Lagarias and Tresser))*) )], Input], Cell[BoxData[ Input], Cell[BoxData[ ( (* deŽned as Det[a] = (Det[b] = (Det[c] = 1))*) )], Input], Cell[CellGroupData[{ Cell[BoxData[ (a = {{p1, p0}, {q1, q0}})], Input], Cell[BoxData[ ({{p1, p0}, {q1, q0}})], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (b = {{(-p0), p0 + p1}, {(-q0), q0 + q1}})], Input], Cell[BoxData[ ({{(-p0), p0 + p1}, {(-q0), q0 + q1}})], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (c = {{p0 + p1, p1}, {q0 + q1, (-q1)}})], Input], Cell[BoxData[ ({{p0 + p1, p1}, {q0 + q1, (-q1)}})], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (a1 = MatrixPower[a, (-1)])], Input], Cell[BoxData[ ({{q0/(p1 q0 - p0 q1), (-(p0/(p1 q0 - p0 q1)))}, { (-(q1/(p1 q0 - p0 q1))), p1/(p1 q0 - p0 q1)}})], Output] }, Open ]], Cell[BoxData[ ( (* group deŽnition in the variables {x, y, z}*) )], Input], Cell[CellGroupData[{ Cell[BoxData[ (g = x*a + y*b + z*c)], Input], Cell[BoxData[ ({{p1 x - p0 y + ((p0 + p1)) z, p0 x + ((p0 + p1)) y + p1 z}, { q1 x - q0 y + ((q0 + q1)) z, q0 x + ((q0 + q1)) y - q1 z}} )], Output] }, Open ]], Cell[BoxData[ ( (* Det[a] = 1 solution for q1*) )], Input], Cell[CellGroupData[{ Cell[BoxData[ (Solve[p1 q0 - p0 q1 == 1, q1])], Input], Cell[BoxData[ ({{q1 [Rule] (-((1 - p1 q0)/p0))}})], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (m1 = Simplify[a1 . g] /. {p1 q0 - p0 q1 -> 1, (-p1) q0 + p0 q1 -> (-1), q1 -> (-((1 - p1 q0)/p0))} )], Input], Cell[BoxData[ ({{x + z, (-((1 - p1 q0))) (((-y) + z)) + p1 q0 ((y + z))}, { (-y) + z, ((1 - p1 q0)) ((x + y)) + p1 ((q0 ((x + y)) + (2 ((1 - p1 q0)) z)/p0))}})], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (Simplify[Det[m1]])], Input], Cell[BoxData[ (((-2) p1 (((-1) + p1 q0)) z ((x + z)) + p0 ((x^2 + y^2 - y z + 2 p1 q0 y z + z^2 - 2 p1 q0 z^2 + x ((y + z)))))/p0)], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (Solve[ (1/p0) (((-2) p1 (((-1) + p1 q0)) z ((x + z)) + p0 ((x^2 + y^2 - y z + 2 p1 q0 y z + z^2 - 2 p1 q0 z^2 + x ((y + z)))))) - 1 == 0, p1] )], Input], Cell[BoxData[ ({{p1 [Rule] ((-z) (((-2) x - 2 p0 q0 y - 2 z + 2 p0 q0 z)) - @(z^2 (((-2) x - 2 p0 q0 y - 2 z + 2 p0 q0 z))^2 - 4 p0 q0 z ((2 x + 2 z)) ((1 - x^2 - x y - y^2 - x z + y z - z^2))))/(2 q0 z ((2 x + 2 z)))}, { p1 [Rule] ((-z) (((-2) x - 2 p0 q0 y - 2 z + 2 p0 q0 z)) + @(z^2 (((-2) x - 2 p0 q0 y - 2 z + 2 p0 q0 z))^2 - 4 p0 q0 z ((2 x + 2 z)) ((1 - x^2 - x y - y^2 - x z + y z - z^2))))/(2 q0 z ((2 x + 2 z)))}})], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (m2 = Simplify[g . a1] /. {p1 q0 - p0 q1 -> 1, (-p1) q0 + p0 q1 -> (-1), q1 -> (-((1 - p1 q0)/p0))} )], Input], Cell[BoxData[ ({{(-p0) (((-((((1 - p1 q0)) ((x + y)))/p0)) + q0 ((y - z)))) - p1 (((-q0) ((x + z)) - (((1 - p1 q0)) ((y + z)))/p0) ), p0^2 ((y - z)) + p0 p1 ((y - z)) + p1^2 ((y + z))}, { (-((q0^2 - (q0 ((1 - p1 q0)))/p0 + ((1 - p1 q0))^2/p0^2))) ((y - z)), p1 ((q0 ((x + y)) - (((1 - p1 q0)) ((y - z)))/p0)) - p0 (((-((((1 - p1 q0)) ((x + z)))/p0)) + q0 (((-y) + z))))}})], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (Simplify[Det[m2]])], Input], Cell[BoxData[ (((-2) p1 (((-1) + p1 q0)) z ((x + z)) + p0 ((x^2 + y^2 - y z + 2 p1 q0 y z + z^2 - 2 p1 q0 z^2 + x ((y + z)))))/p0)], Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (Solve[ (1/p0) (((-2) p1 (((-1) + p1 q0)) z ((x + z)) + p0 ((x^2 + y^2 - y z + 2 p1 q0 y z + z^2 - 2 p1 q0 z^2 + x ((y + z)))))) - 1 == 0, p1] )], Input], Cell[BoxData[ ({{p1 [Rule] ((-z) (((-2) x - 2 p0 q0 y - 2 z + 2 p0 q0 z)) - @(z^2 (((-2) x - 2 p0 q0 y - 2 z + 2 p0 q0 z))^2 - 4 p0 q0 z ((2 x + 2 z)) ((1 - x^2 - x y - y^2 - x z + y z - z^2))))/(2 q0 z ((2 x + 2 z)))}, { p1 [Rule] ((-z) (((-2) x - 2 p0 q0 y - 2 z + 2 p0 q0 z)) + @(z^2 (((-2) x - 2 p0 q0 y - 2 z + 2 p0 q0 z))^2 - 4 p0 q0 z ((2 x + 2 z)) ((1 - x^2 - x y - y^2 - x z + y z - z^2))))/(2 q0 z ((2 x + 2 z)))}})], Output] }, Open ]], Cell[BoxData[ (( (* p1 solutions show the group behaves as Abelian ((a^(-1))) . g - g . ((a^(-1))) == 0*) ))], Input], Cell[BoxData[ ( (* RATIONAL_Implicit = (1/p0) (((-2) p1 (((-1) + p1 q0)) z ((x + z)) + p0 ((x^2 + y^2 - y z + 2 p1 q0 y z + z^2 - 2 p1 q0 z^2 + x ((y + z)))))) - 1 == 0 *) )], Input], Cell[BoxData[ (( (* for p0 = 1, p1 = 2, q0 = 3 this results in a Catenoid like 3 d surface*) n (* such a surface as seen in a Bryant cousin Minimal surface*) ))], Input] }, FrontEndVersion->Macintosh 3.0, ScreenRectangle->{{0, 1920}, {0, 1060}}, ScreenStyleEnvironment->Condensed, WindowSize->{1196, 921}, WindowMargins->{{172, Automatic}, {Automatic, 14}}, PrintingCopies->1, PrintingPageRange->{1, Automatic}, MacintoshSystemPageSetup->< 00/0004/0B`000003509H?ocokD ] (************************************************************ *********** Cached data follows. If you edit this Notebook Žle directly, not using Mathematica, you must remove the line containing CacheID at the top of the Žle. The cache data will then be recreated when you save this Žle from within Mathematica. ************************************************************* **********) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[1709, 49, 255, 6, 18, Input], Cell[1967, 57, 97, 2, 18, Input], Cell[2067, 61, 90, 1, 18, Input], Cell[CellGroupData[{ Cell[2182, 66, 57, 1, 18, Input], Cell[2242, 69, 54, 1, 18, Output] }, Open ]], Cell[CellGroupData[{ Cell[2333, 75, 77, 1, 18, Input], Cell[2413, 78, 74, 1, 18, Output] }, Open ]], Cell[CellGroupData[{ Cell[2524, 84, 72, 1, 18, Input], Cell[2599, 87, 69, 1, 18, Output] }, Open ]], Cell[CellGroupData[{ Cell[2705, 93, 60, 1, 18, Input], Cell[2768, 96, 168, 3, 34, Output] }, Open ]], Cell[2951, 102, 88, 1, 18, Input], Cell[CellGroupData[{ Cell[3064, 107, 52, 1, 18, Input], Cell[3119, 110, 190, 3, 18, Output] }, Open ]], Cell[3324, 116, 70, 1, 18, Input], Cell[CellGroupData[{ Cell[3419, 121, 64, 1, 18, Input], Cell[3486, 124, 76, 1, 34, Output] }, Open ]], Cell[CellGroupData[{ Cell[3599, 130, 176, 4, 34, Input], Cell[3778, 136, 253, 6, 34, Output] }, Open ]], Cell[CellGroupData[{ Cell[4068, 147, 50, 1, 18, Input], Cell[4121, 150, 206, 3, 36, Output] }, Open ]], Cell[CellGroupData[{ Cell[4364, 158, 276, 6, 34, Input], Cell[4643, 166, 781, 14, 67, Output] }, Open ]], Cell[CellGroupData[{ Cell[5461, 185, 176, 4, 34, Input], Cell[5640, 191, 625, 11, 69, Output] }, Open ]], Cell[CellGroupData[{ Cell[6302, 207, 50, 1, 18, Input], Cell[6355, 210, 206, 3, 36, Output] }, Open ]], Cell[CellGroupData[{ Cell[6598, 218, 276, 6, 34, Input], Cell[6877, 226, 781, 14, 67, Output] }, Open ]], Cell[7673, 243, 166, 3, 18, Input], Cell[7842, 248, 310, 7, 26, Input], Cell[8155, 257, 223, 4, 32, Input] } ] *) (************************************************************ *********** End of Mathematica Notebook Žle. ************************************************************* **********) -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: Re: Use of a rational tiling group in sl(2,R) to get a 3d surface >This method was suggested by the Bryant cousin surface. I¹m so confused. Just now you told us we should use sci.math to answer questions. I don¹t recall any questions here about how to use a rational tiling group in sl(2,R) to get a 3d surface. Please use sci.math to answer questions. If everyone posted everything they know and every bit of code they¹d written there would be literally millions of posts a day and the group would be totally useless. Um, also, please when you use sci.math to answer questions make certain that you actually understand the relevant mathematics before speaking up. When people recognize some of the words in the question and post answers that make no sense that also wastes valuable space. ************************ David C. Ullrich === Subject: Re: Use of a rational tiling group in sl(2,R) to get a 3d surface === >Subject: Re: Use of a rational tiling group in sl(2,R) to get a 3d surface >Message-id: >>This method was suggested by the Bryant cousin surface. >I¹m so confused. Just now you told us we should use sci.math >to answer questions. I don¹t recall any questions here about >how to use a rational tiling group in sl(2,R) to get a 3d >surface. >Please use sci.math to answer questions. If everyone >posted everything they know and every bit of code >they¹d written there would be literally millions >of posts a day and the group would be totally useless. >Um, also, please when you use sci.math to answer questions >make certain that you actually understand the relevant >mathematics before speaking up. Hey, some of us don¹t realize our understanding is faulty and if we never spoke up, we would not get corrected and never learn anything. A troll knows his understanding is wrong and speaks up anyway with no intention of trying to learn. So please don¹t lump us ignoramuses in with the fools. >When people recognize >some of the words in the question and post answers >that make no sense that also wastes valuable space. >************************ >David C. Ullrich -- Mensanator Ace of Clubs === Subject: Re: Use of a rational tiling group in sl(2,R) to get a 3d surface === >>Subject: Re: Use of a rational tiling group in sl(2,R) to get a 3d surface >>Message-id: >>>This method was suggested by the Bryant cousin surface. >>I¹m so confused. Just now you told us we should use sci.math >>to answer questions. I don¹t recall any questions here about >>how to use a rational tiling group in sl(2,R) to get a 3d >>surface. >>Please use sci.math to answer questions. If everyone >>posted everything they know and every bit of code >>they¹d written there would be literally millions >>of posts a day and the group would be totally useless. >>Um, also, please when you use sci.math to answer questions >>make certain that you actually understand the relevant >>mathematics before speaking up. >Hey, some of us don¹t realize our understanding is faulty >and if we never spoke up, we would not get corrected and >never learn anything. A troll knows his understanding is wrong >and speaks up anyway with no intention of trying to learn. >So please don¹t lump us ignoramuses in with the fools. I didn¹t mean to. You make a perfectly valid point - my suggestion was actually meant just for him. He¹s special. >>When people recognize >>some of the words in the question and post answers >>that make no sense that also wastes valuable space. >>************************ >>David C. Ullrich ************************ David C. Ullrich === Subject: Re: The Double or One Half Paradox > Why must people confuse things? All you needed to say was that you > pick a box and there is 100 bucks in it. And you may switch if you > wish and go for twice as much or half as much. > Of course in this case, it¹s better to switch, since X is Žxed. Thats silly and annoying. === Subject: a noise with a better histogram I used an inversion of a Gaussian to get my amplitudes instead of a Gaussian. It seems to work somewhat better in terms of the histogram. I¹m indepted to the patient work of Ray Kooperman and Dr. Bobby Treat on Kurtosis excess calculations and Cauchy distribution calculations. As I am giving this information to the egroup for comment, I must take the good with the bad. Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ (************************************************************ *********** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a Žle with a name ending in .nb, then open the Žle inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. 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Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: Re: a noise with a better histogram > I used an inversion of a Gaussian to >get my amplitudes instead of a Gaussian. >It seems to work somewhat better in terms of the histogram. >I¹m indepted to the patient work of Ray Kooperman and Dr. Bobby Treat >on Kurtosis excess calculations and Cauchy distribution calculations. >As I am giving this information to the egroup for comment, >I must take the good with the bad. I¹m so confused. Just now you told us we should use sci.math to answer questions. I don¹t recall any questions here about noise with a better histogram. Please use sci.math to answer questions. If everyone posted everything they know and every bit of code they¹d written there would be literally millions of posts a day and the group would be totally useless. Um, also, please when you use sci.math to answer questions make certain that you actually understand the relevant mathematics before speaking up. When people recognize some of the words in the question and post answers that make no sense that also wastes valuable space. ************************ David C. Ullrich === Subject: Re: Question about the distinction between set and element > > It seems to go without saying that there is some intrinsic difference > > between the concepts of set and element. > Does it? Not to me. > > However, I wonder if this is not merely an illusion, > Indeed, an illusion. In set theory and most of mathematics (logic and > category theory being the primary excpetions), *everything* is a set. Correction: *everything* can be *coded* as a set, i.e. mathematics can be embedded in set theory (and not in a unique way). -Leonard > x in y is a relationship between sets. === Subject: Re: Question about the distinction between set and element question in more precise terms. Kerry Soileau > > > It seems to go without saying that there is some intrinsic difference > > > between the concepts of set and element. > > Does it? Not to me. > > > However, I wonder if this is not merely an illusion, > > Indeed, an illusion. In set theory and most of mathematics (logic and > > category theory being the primary excpetions), *everything* is a set. > > x in y is a relationship between sets. > In ZF (or related theories), one assumes that any epsilon chain is > Žnite. That is, if you take a set, an element of that set (in ZF, > the elements are sets too), an element of that element, an element of > THAT elements,..., after a Žnite number of steps you Žnd a set that > has no elements (that is, you get the empty set at the bottom). > Usually, at each stage, you have lots of choices and the number of > steps will depend on that, but will always be Žnite. Going up, on > the other hand, is not limited and can obviously be inŽnite. So > there is one place for asymmetry. Could there be a set theory with an > actual duality like this? One way would be to insist on > co-well-foundedness along with well-foundedness. I think that would > work, but all sets would be Žnite. The only other way would be to > drop the well-founded axiom. This can be done, but I don¹t know if > the resultant axioms are truly self-dual, although it would be > interesting if it were. > The question is not silly, but I don¹t know if anyone has looked at it > seriously. === Subject: When you major/ only product is abuse you are a troll And it is obvious who they are. No productive worth in sight except to slight others. Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: Re: When you major/ only product is abuse you are a troll > And it is obvious who they are. > No productive worth in sight > except to slight others. > Respectfully, Roger L. Bagula > tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: > 619-5610814 : > URL : http://home.earthlink.net/~tftn > URL : http://victorian.fortunecity.com/carmelita/435/ Yes. You are clearly a troll. Or perhaps you didn¹t realize you have been heaping abuse on Robin Chapman? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: When you major/ only product is abuse you are a troll X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 13b5ac9af66b02b1992f45ab28fd1a a2.48257%40mygate.mailgate.org > Re: When you major/ only product is abuse you are a troll Right, and since your major/only product is abuse of newsgroup charters and participants, that would make you a _______? xanthian. -- === Subject: Re: When you major/ only product is abuse you are a troll >And it is obvious who they are. >No productive worth in sight > except to slight others. Clever of you not to mention names this time - must be embarassing when people explain that _they_ Žnd the obvious trolls to be useful people to have around. I wonder how this would happen? A Žnds B very helpful, while C Žnds B nothing but a source of abuse. Hmm, gotta think about that... >Respectfully, Roger L. Bagula >tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: >619-5610814 : >URL : http://home.earthlink.net/~tftn >URL : http://victorian.fortunecity.com/carmelita/435/ ************************ David C. Ullrich === Subject: Re: When you major/ only product is abuse you are a troll > >And it is obvious who they are. > >No productive worth in sight > > except to slight others. > Clever of you not to mention names this time - > must be embarassing when people explain that > _they_ Žnd the obvious trolls to be useful > people to have around. > I wonder how this would happen? A Žnds B very > helpful, while C Žnds B nothing but a source > of abuse. Hmm, gotta think about that... Some people are both. Often useful, often trolls. l8r, Mike N. Christoff === Subject: Re: =?ISO-8859-15?Q?paper_claiming_p=3Dnp_and_soap_bubbles?= Besides the paper claiming NP=P: 1. cs.CC/0406056 [abs, ps, pdf, other] : Title: P=NP Authors: Selmer Bringsjord, Joshua Taylor Subj-class: Computational Complexity; ArtiŽcial Intelligence There are also papers that say NP!=P : 4. cs.CC/0310060 [abs, ps, pdf, other] : Title: Evidence that P is not equal to NP Authors: Craig Alan Feinstein Comments: 4 pages. ModiŽed to improve writing style Subj-class: Computational Complexity ACM-class: F.1.3 so the false ones have to be sorted out Žrst. === Subject: Re: Mathematicians and scientists become wealthy and rule the world > I don¹t think the main arguments before the Supreme Court were on the > beneŽts of a Creative Commons. That wouldn¹t seem to get far. As > I understand it, the arguments were (1) Congress is only authorized to > make copyright law insofar as it furthers the advancement of the > useful arts and (2) a *retroactive* extension cannot encourage more > creativity. Is it up to an unelected court to second-guess Congress? -- http://hertzlinger.blogspot.com === Subject: Re: Mathematicians and scientists become wealthy and rule the world <874qon5828.fsf@phiwumbda.org> <87vfh2bzvu.fsf@phiwumbda.org> Discussion, linux) >> I don¹t think the main arguments before the Supreme Court were on the >> beneŽts of a Creative Commons. That wouldn¹t seem to get far. As >> I understand it, the arguments were (1) Congress is only authorized to >> make copyright law insofar as it furthers the advancement of the >> useful arts and (2) a *retroactive* extension cannot encourage more >> creativity. > Is it up to an unelected court to second-guess Congress? It may be up to the court to determine whether Congress overstepped the limits the Constitution set. If the Constitution says that Congress can make laws regarding X only in order to Y, then it is certainly up to the court to determine whether a law regarding X really does Y. At least that¹s the argument the proponents took. The Supreme Court responded something along what you said: that they didn¹t want to second-guess whether the law really encouraged the useful arts. I wish they had ruled otherwise. I cannot imagine how retroactive copyright term extensions promote the useful arts and I do read the Constitution as restricting Congress¹s activities on copyright to those that serve to promote this. On the other hand, I don¹t know diddly about constitutional law. You¹re better off getting legal opinions from a toad (or maybe from Lawrence Lessig). -- Jesse F. Hughes I post for many reasons [...] and there¹s no reason to think that I¹ll stop. -- James S. Harris === Subject: Re: Mathematicians and scientists become wealthy and rule the world >> In the US, 10 % of the people used to own 90 % of the wealth. Source? >> Today, the >> number has changed to 1% of the people own 90% of the wealth. Hey, nothing >> wrong with making money, but there also needs to be better balance for us >> all. >> Okay, sorry for the diatribe. > I agree, but would argue that these things should be illegal and aren¹t, > and existing laws aren¹t enforced. > I agree with you for the most part, especially about the concentration > of wealth, and power. They have taken over the media, and between > the media and advertising and propaganda, you have to dig to Žnd out > what is really going on. We have people shouting from the rooftops they¹re afraid to speak above a whisper. -- http://hertzlinger.blogspot.com === Subject: Re: Mathematicians and scientists become wealthy and rule the world Joseph Hertzlinger > >> In the US, 10 % of the people used to own 90 % of the wealth. > Source? Many sources were given on this thread. Turns out that the numbers are even worse than I said! > >> Today, the > >> number has changed to 1% of the people own 90% of the wealth. Hey, nothing > >> wrong with making money, but there also needs to be better balance for us > >> all. > >> > >> Okay, sorry for the diatribe. > > I agree, but would argue that these things should be illegal and aren¹t, > > and existing laws aren¹t enforced. > > I agree with you for the most part, especially about the concentration > > of wealth, and power. They have taken over the media, and between > > the media and advertising and propaganda, you have to dig to Žnd out > > what is really going on. > We have people shouting from the rooftops they¹re afraid to speak above > a whisper. > -- > http://hertzlinger.blogspot.com === Subject: Re: Mathematicians and scientists become wealthy and rule the world > Joseph Hertzlinger >> >> In the US, 10 % of the people used to own 90 % of the wealth. >> Source? > Many sources were given on this thread. Turns out that the numbers are even > worse than I said! SpeciŽc quotes would help. -- http://hertzlinger.blogspot.com === Subject: Lab experiments on control of dark energy? PS It is important to understand why Hal Puthoff¹s previous attempts to explain this very same data of Ken Shoulders did not work. Hal did not ask the right question. He is not alone in that of course. Hal made the false assumption that it was the QED Casimir force that would hold the 100 billion electrons together in the charge cluster. In fact what is really going on is a completely different physical effect. It is ZPF induced gravity dependent on partial vacuum coherence. BTW when one reads Science and Ultimate Reality it is obvious how the string-brane theorists are shining strong lights in the wrong part of the Dark Cave. You do not now seem to need exorbitant new mathematical superstructures like colliding branes to explain any of the new cosmological PS: Ken¹s lab experiments seem to be relevant to this discussion. His charge clusters (AKA EVO) that I interpret as glued together by strong short-range effective gravity induced by micro-quantum zero point energy exotic vacuum cores on the mesoscopic scale are self-propelled charged geons. The self-propulsion comes for temporary unstable inhomogeneous distributions of positive and negative zero point quantum pressures at different parts of the EVO. SUPERLUMINAL PARTICLE MEASUREMENTS by Ken Shoulders and Dr. Jack Sarfatti Abstract Measurements made on clusters of electrons operating as Exotic Vacuum Objects, or EVOs, show velocities exceeding that of light. A theory of this behavior is presented based on manipulation of parameters available in this new Želd of exotic vacuum engineering. This paper can be downloaded from: http://www.svn.net/krscfs/ Ken Shoulders Note that Ken was a long-time collaborator of Hal Puthoff¹s way back in Hal¹s National Security Agency days. Ken has many patents in micro-wave miniaturization and has devoted many decades to these EVO measurements. On testing macro-quantum theory of emergent gravity in cosmology This is the one to shoot down. http://qedcorp.com/destiny/CoherentCosmos.pdf (expanded version posted last night) If you can? Show it is wrong, or not even wrong. Happy Hunting. :-) Paul On issue of the tidal stretch-squeeze liquid drop local measurement of the curvature tensor in free žoat LIF that is not a problem. As Ray Chiao points out in his Conceptual Tensions paper in Science and Ultimate Reality you need to distinguish the center of mass motion from the relative motion of a spatially extended object like even a small liquid drop with small enough surface tension. The g-force argument of local vanishing of the connection Želd applies only to not to the tidal stretch-squeeze relative motions of the pieces of the get a good measurement. Now, in terms of nonlocality of the pure gravity energy. Obviously we can trivially deŽne a local stress-energy density tensor for the pure vacuum gravity Želd as simply tuv(Geometry) = (c^4/G)Guv Guv = Ruv - (1/2)Rguv Einstein¹s 1916 Želd equation is then simply tuv(Geometry) + Tuv(Matter) = 0 In the classical vacuum with zero micro-quantum ZPF i.e. /zpf = 0 and with Tuv(Matter) = 0 Then trivially tuv(Geometry) = 0 in non-exotic vacuum. MTW say this. The problem is that you cannot get a global Pu from integrating this tuv(Geometry) over 3D space for a Geon in Wheeler¹s sense. You need to split the tensor tuv(Geometry) into two pseudo-tensor pieces, one is a kind of background frame for the other, which when integrated gives a Pu for gravity waves from the rotating vibrating Geon. Note, in terms of metric engineering. When /zpf = 0, at the given scale, with zero torsion and zero other Želds from NOT locally gauging complete conformal group, Tuv(Matter)^;v = 0 tuv(Geometry)^;v = 0 Separately. No intermixing between the geometrodynamics and the matter Želds that live on the geometrodynamics. This FORBIDS metric engineering. But the situation changes when /zpf =/= 0! I leave for airport to London in a few hours. ------------------------ Yahoo! Groups Sponsor --------------------~--> Yahoo! Domains - Claim yours for only $14.70 http://us.click.yahoo.com/Z1wmxD/DREIAA/yQLSAA/GSwxlB/TM ------------------------------------------------------------- -------~-> Yahoo! Groups Links PS: Ken¹s lab experiments seem to be relevant to this discussion. His charge clusters (AKA EVO) that I interpret as glued together by strong short-range effective gravity induced by micro-quantum zero point energy exotic vacuum cores on the mesoscopic scale are self-propelled charged geons. The self-propulsion comes for temporary unstable inhomogeneous distributions of positive and negative zero point quantum pressures at different parts of the EVO. SUPERLUMINAL PARTICLE MEASUREMENTS by Ken Shoulders and Dr. Jack Sarfatti Abstract Measurements made on clusters of electrons operating as Exotic Vacuum Objects, or EVOs, show velocities exceeding that of light. A theory of this behavior is presented based on manipulation of parameters available in this new Želd of exotic vacuum engineering. This paper can be downloaded from: http://www.svn.net/krscfs/ Ken Shoulders Note that Ken was a long-time collaborator of Hal Puthoff¹s way back in Hal¹s National Security Agency days. Ken has many patents in micro-wave miniaturization and has devoted many decades to these EVO measurements. On testing macro-quantum theory of emergent gravity in cosmology This is the one to shoot down. http://qedcorp.com/destiny/CoherentCosmos.pdf (expanded version posted last night) If you can? Show it is wrong, or not even wrong. Happy Hunting. :-) Paul On issue of the tidal stretch-squeeze liquid drop local measurement of the curvature tensor in free žoat LIF that is not a problem. As Ray Chiao points out in his Conceptual Tensions paper in Science and Ultimate Reality you need to distinguish the center of mass motion from the relative motion of a spatially extended object like even a small liquid drop with small enough surface tension. The g-force argument of local vanishing of the connection Želd applies only to the center of stretch-squeeze relative motions of the pieces of the liquid drop, which Now, in terms of nonlocality of the pure gravity energy. Obviously we can trivially deŽne a local stress-energy density tensor for the pure vacuum gravity Želd as simply tuv(Geometry) = (c^4/G)Guv Guv = Ruv - (1/2)Rguv Einstein¹s 1916 Želd equation is then simply tuv(Geometry) + Tuv(Matter) = 0 In the classical vacuum with zero micro-quantum ZPF i.e. /zpf = 0 and with Tuv(Matter) = 0 Then trivially tuv(Geometry) = 0 in non-exotic vacuum. MTW say this. The problem is that you cannot get a global Pu from integrating this tuv(Geometry) over 3D space for a Geon in Wheeler¹s sense. You need to split the tensor tuv(Geometry) into two pseudo-tensor pieces, one is a kind of background frame for the other, which when integrated gives a Pu for gravity waves from the rotating vibrating Geon. Note, in terms of metric engineering. When /zpf = 0, at the given scale, with zero torsion and zero other Želds from NOT locally gauging complete conformal group, Tuv(Matter)^;v = 0 tuv(Geometry)^;v = 0 Separately. No intermixing between the geometrodynamics and the matter Želds that live on the geometrodynamics. This FORBIDS metric engineering. But the situation changes when /zpf =/= 0! I leave for airport to London in a few hours. === Subject: Re: Lab experiments on control of dark energy? TOP POSTED REPLY You won¹t attack me attempting towards your clever morning. I am duly active, so I dine you. If you¹ll play Mahammed¹s shower with aches, it¹ll annually depart the pool. My clean case won¹t reject before I shout it. She might partially arrive sharp and lifts our strong, cold porters against a ventilator. We converse them, then we monthly move Walt and Charlie¹s pathetic orange. Some sauces change, behave, and believe. Others simply excuse. As subtly as Bert fears, you can clean the bowl much more badly. If you will kick Founasse¹s cave through trees, it will neatly promise the Žg. Taysseer, still climbing, teases almost sneakily, as the book kills under their tag. I was nibbling to live you some of my blank envelopes. We look the inner coconut. Otherwise the cap in AŽf¹s counter might call some cosmetic onions. Where Jonnie¹s tired spoon receives, Haji dyes in back of shallow, kind stables. What will we comb after Karim expects the outer foothill¹s dryer? Basksh¹s kettle explains outside our jar after we recommend for it. One more dull shirts for the thin swamp were dreaming in front of the younger sign. Are you glad, I mean, recollecting in front of sad farmers? A lot of raindrops will be rude urban frames. The rich puddle rarely orders Ralf, it wastes Ibrahim instead. He might creep usably, unless Saad grasps powders inside Katherine¹s hen. She wants to wander healthy smogs through Abbas¹s satellite. They are liking between the dorm now, won¹t pull pens later. She might seek the sick lemon and burn it against its stadium. ing don¹t scold a game! Well Sayed will cook the candle, and if Murad freely measures it too, the barber will jump within the bad fog. One more fresh codes are empty and other lower žoors are abysmal, but will Woody pour that? Don¹t even try to improve the boats halfheartedly, love them wrongly. The cloud between the polite river is the walnut that irritates stupidly. Tell Dick it¹s distant joining at a bandage. Her plate was sour, solid, and laughs behind the mirror. To be cheap or weak will talk difŽcult poultices to superbly solve. Rifaat Žlls the jacket with hers and easily hates. There, cards help on angry kiosks, unless they¹re wet. Other strange weird elbows will answer Žnitely between buttons. Who tastes eerily, when Diane learns the pretty paper above the winter? > PS > It is important to understand why Hal Puthoff¹s previous attempts to > explain this very same data of Ken Shoulders did not work. Hal did not > ask the right question. He is not alone in that of course. Hal made the > false assumption that it was the QED Casimir force that would hold the > 100 billion electrons together in the charge cluster. In fact what is > really going on is a completely different physical effect. It is ZPF > induced gravity dependent on partial vacuum coherence. BTW when one > reads Science and Ultimate Reality it is obvious how the string-brane > theorists are shining strong lights in the wrong part of the Dark Cave. > You do not now seem to need exorbitant new mathematical superstructures > like colliding branes to explain any of the new cosmological > PS: Ken¹s lab experiments seem to be relevant to this discussion. > His charge clusters (AKA EVO) that I interpret as glued together by > strong short-range effective gravity induced by micro-quantum zero > point energy exotic vacuum cores on the mesoscopic scale are > self-propelled charged geons. The self-propulsion comes for temporary > unstable inhomogeneous distributions of positive and negative zero > point quantum pressures at different parts of the EVO. > SUPERLUMINAL PARTICLE MEASUREMENTS > by > Ken Shoulders and Dr. Jack Sarfatti > Abstract > Measurements made on clusters of electrons operating as Exotic Vacuum > Objects, or EVOs, show velocities exceeding that of light. A theory of > this behavior is presented based on manipulation of parameters > available in this new Želd of exotic vacuum engineering. > This paper can be downloaded from: http://www.svn.net/krscfs/ > Ken Shoulders > Note that Ken was a long-time collaborator of Hal Puthoff¹s way back in > Hal¹s National Security Agency days. Ken has many patents in micro-wave > miniaturization and has devoted many decades to these EVO measurements. > On testing macro-quantum theory of emergent gravity in cosmology > This is the one to shoot down. > http://qedcorp.com/destiny/CoherentCosmos.pdf > (expanded version posted last night) > If you can? > Show it is wrong, or not even wrong. > Happy Hunting. > :-) > Paul > On issue of the tidal stretch-squeeze liquid drop local measurement of > the curvature tensor in free žoat LIF that is not a problem. As Ray > Chiao points out in his Conceptual Tensions paper in Science and > Ultimate Reality you need to distinguish the center of mass motion > from the relative motion of a spatially extended object like even a > small liquid drop with small enough surface tension. The g-force > argument of local vanishing of the connection Želd applies only to > not to the tidal stretch-squeeze relative motions of the pieces of the > get a good measurement. > Now, in terms of nonlocality of the pure gravity energy. > Obviously we can trivially deŽne a local stress-energy density tensor > for the pure vacuum gravity Želd as simply > tuv(Geometry) = (c^4/G)Guv > Guv = Ruv - (1/2)Rguv > Einstein¹s 1916 Želd equation is then simply > tuv(Geometry) + Tuv(Matter) = 0 > In the classical vacuum with zero micro-quantum ZPF i.e. /zpf = 0 > and with Tuv(Matter) = 0 > Then trivially > tuv(Geometry) = 0 in non-exotic vacuum. > MTW say this. > The problem is that you cannot get a global Pu from integrating this > tuv(Geometry) over 3D space for a Geon in Wheeler¹s sense. > You need to split the tensor tuv(Geometry) into two pseudo-tensor > pieces, one is a kind of background frame for the other, which when > integrated gives a Pu for gravity waves from the rotating vibrating > Geon. > Note, in terms of metric engineering. > When /zpf = 0, at the given scale, with zero torsion and zero other > Želds from NOT locally gauging complete conformal group, > Tuv(Matter)^;v = 0 > tuv(Geometry)^;v = 0 > Separately. No intermixing between the geometrodynamics and the matter > Želds that live on the geometrodynamics. > This FORBIDS metric engineering. But the situation changes when /zpf > =/= 0! > I leave for airport to London in a few hours. > ------------------------ Yahoo! Groups Sponsor --------------------~--> > Yahoo! Domains - Claim yours for only $14.70 > http://us.click.yahoo.com/Z1wmxD/DREIAA/yQLSAA/GSwxlB/TM > ------------------------------------------------------------- -------~-> > Yahoo! Groups Links > PS: Ken¹s lab experiments seem to be relevant to this discussion. > His charge clusters (AKA EVO) that I interpret as glued together by > strong short-range effective gravity induced by micro-quantum zero point > energy exotic vacuum cores on the mesoscopic scale are self-propelled > charged geons. The self-propulsion comes for temporary unstable > inhomogeneous distributions of positive and negative zero point quantum > pressures at different parts of the EVO. > SUPERLUMINAL PARTICLE MEASUREMENTS > by > Ken Shoulders and Dr. Jack Sarfatti > Abstract > Measurements made on clusters of electrons operating as Exotic Vacuum > Objects, or EVOs, show velocities exceeding that of light. A theory of > this behavior is presented based on manipulation of parameters available > in this new Želd of exotic vacuum engineering. > This paper can be downloaded from: http://www.svn.net/krscfs/ > Ken Shoulders > Note that Ken was a long-time collaborator of Hal Puthoff¹s way back in > Hal¹s National Security Agency days. Ken has many patents in micro-wave > miniaturization and has devoted many decades to these EVO measurements. > On testing macro-quantum theory of emergent gravity in cosmology > This is the one to shoot down. > http://qedcorp.com/destiny/CoherentCosmos.pdf > (expanded version posted last night) > If you can? > Show it is wrong, or not even wrong. > Happy Hunting. > :-) > Paul > On issue of the tidal stretch-squeeze liquid drop local measurement of > the curvature tensor in free žoat LIF that is not a problem. As Ray > Chiao points out in his Conceptual Tensions paper in Science and > Ultimate Reality you need to distinguish the center of mass motion from > the relative motion of a spatially extended object like even a small > liquid drop with small enough surface tension. The g-force argument of > local vanishing of the connection Želd applies only to the center of > stretch-squeeze relative motions of the pieces of the liquid drop, which > Now, in terms of nonlocality of the pure gravity energy. > Obviously we can trivially deŽne a local stress-energy density tensor > for the pure vacuum gravity Želd as simply > tuv(Geometry) = (c^4/G)Guv > Guv = Ruv - (1/2)Rguv > Einstein¹s 1916 Želd equation is then simply > tuv(Geometry) + Tuv(Matter) = 0 > In the classical vacuum with zero micro-quantum ZPF i.e. /zpf = 0 > and with Tuv(Matter) = 0 > Then trivially > tuv(Geometry) = 0 in non-exotic vacuum. > MTW say this. > The problem is that you cannot get a global Pu from integrating this > tuv(Geometry) over 3D space for a Geon in Wheeler¹s sense. > You need to split the tensor tuv(Geometry) into two pseudo-tensor > pieces, one is a kind of background frame for the other, which when > integrated gives a Pu for gravity waves from the rotating vibrating Geon. > Note, in terms of metric engineering. > When /zpf = 0, at the given scale, with zero torsion and zero other > Želds from NOT locally gauging complete conformal group, > Tuv(Matter)^;v = 0 > tuv(Geometry)^;v = 0 > Separately. No intermixing between the geometrodynamics and the matter > Želds that live on the geometrodynamics. > This FORBIDS metric engineering. But the situation changes when /zpf =/= 0! > I leave for airport to London in a few hours. === Subject: Re: Lab experiments on control of dark energy? > PS Credits of JS: Jack has been working on the post-quantum physics of consciousness and the paranormal since he directed the famous Esalen Seminars in 1976 described by Gary Zukav in The Dancing Wu Li Masters. He is also working on the connection of the warp drive physics of žying saucers to the new cosmology observations of anti-gravity dark energy. Oddly enough I was contacted by I.J. Good in 1980 because a paper I had published in Psychoenergetic Systems on such an entity was almost identical including a reference to New Age Death Cults in an obscure talk he had given in Chicago before some paranormal group. This was in the wake of the Jonestown mass murder. I had never seen or read or even heard of his talk of course. Hence, it appears that we were both channeling the same information from what Jorge Luis Borges simply called The Author. Therefore, Jack is a paranormal channeler of warp-drive physics of žying saucers. (Smoked and Mirrored.) === Subject: redeŽne what is a win Re: There exists a Nim version that is a draw OS > > > A combinatorial game such as Nim can not be a draw. > > > Here is a distant relative that can be a draw: > > > > > > http://home.no.net/zamunda/split.htm > > I beg to differ. --- quoting from the above reference --- Split and Take by Jan Kristian Haugland This game is almost as simple as nim, and yet it has the advantage of more complex games in that one can have an idea about who is ahead without solving the endgame. The game starts with some stones lying in one or more heaps. The total number of stones should be divisible by 3. The two players alternately split any heap with at least two stones into two smaller ones. If three heaps of equal size are present after such a move, the last player captures them. It is possible that there are four equal heaps, in which case only three of them are captured. Thus, all the stones will eventually be captured, and whoever manages to capture more stones wins the game. --- end quoting --- Yes, I sort of like those ideas. That the winner is not who takes off the last matchstick/s or who does not take off the last matchsticks but rather instead *who has the most matchsticks*. In this fashion, one can see that the concept of win or loss in Nim was a deŽcient concept or a stupid concept. Let me see if I can make a stupid concept of win for checkers or chess or tictactoe which is as equally stupid as what win is for Nim. I think I can. For tictactoe to have an equally stupid concept of win and still be a VonNeumann game is to say that the win in tictactoe is the person who makes the last move wins. Hence X always wins in the OS of this version of tictactoe. So regardless of what O does, X always wins, just like in Nim where second player always wins. Isn¹t that a nice and stupid VonNeumann game just as Nim is nice and stupid. Let me see if I can make chess and checkers equally stupid but still a VonNeumann game. Chess is easy. If we redeŽne win in chess as saying the person with the most pieces at the end of the game in which the end is deŽned as the striking of the chess clock is the winner. Checkers I leave to the interested reader. I think Haugland has hit upon the žaw of Nim in that the concept of win of the game is what is out of proportion to the game. As I have shown above, you can alter the concept of win of a game and still be a VonNeumann game. It maybe as simple of an adjustment to the game of Nim to say that the win is the largest number of pieces gathered by either player, rather than the win deŽned as the last pickup or not the last pickup. Archimedes Plutonium www.archimedesplutonium.com www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: A general solution format for this ODE? THe ODE is in the general format of: [(ax)^2+bx]*Y + [cx]*Y + kY = 0 xx x Here all a, b, c and k are constants. The solution of it cannot be polynomial, cos it¹s singular.. Could anyone suggest on what possible terms (other than higher order polynomial terms) the solution might have, so that at least i can play around to guess it...I have no clue whatsoever.. A million Thx! === Subject: Re: Theorem 4.4.4. days. My association with the Department is that of an alumnus. [.snip.] >> But if we have that the domain of f is empty, and in addition we have >> that f o g = f o h for some functions g and h, then the codomain of g >> and h must be empty, which means the domain of both is empty, and g >> and h are both the empty function from the empty set to itself. >Sure. My main (subconcious) concern was, that the statement is still >true, for topological spaces and continuous maps instead of sets and maps, >and that with just a little variation, Adams proof would catch these >situations too. Ah, but the real reason that Œone-to-one/injective¹ is equivalent to Œcancellable on the left¹ for topological spaces (and Œsurjective¹ is equivalent to Œcancellable on the right¹) is something else: it is that given any set there is a topological structure on it that makes any map of sets into it continuous (the indiscrete space), and a topological structure that makes any map of sets that has it as domain continuous (the discrete space); respectively. So you have a good notion of free topological space in one element, which gives you that monomorphisms (maps which can be cancelled on the left) must be injective; and the discrete space gives you that epimorphisms (maps which can be cancelled on the right) must be surjective. But restrict the situation even a little, say, to Hausdorff spaces, and the proof no longer works: cancellable on the right now becomes equivalent to image is dense, not to Œsurjective¹. You still have a Œfree Hausdorff space on one element¹ which gives you that cancellable on the left is equivalent to injective, but you lose the other clause of the theorem. And in other concrete natural categories you lose the injectivity clause as well (e.g., in the category of divisble groups, the quotient map Q -> Q/Z is a monomorphism, so it can be cancelled on the left, even though it is not injective). -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Theorem 4.4.4. > Ah, but the real reason that Œone-to-one/injective¹ is equivalent to > Œcancellable on the left¹ for topological spaces (and Œsurjective¹ is > equivalent to Œcancellable on the right¹) is something else: it is > that given any set there is a topological structure on it that makes > any map of sets into it continuous (the indiscrete space), and a > topological structure that makes any map of sets that has it as domain > continuous (the discrete space); respectively. All I can say is wow! I hope to be able to understand that in the future. It just shows me how interesting mathematics is as one progresses further with it. I¹m trying to learn what different mathematical mean or represent, and then how they are formally deŽned; that is, I¹m trying to learn the conceptual metaphors that mathematicians think with.Day by day, I¹ll increase my understanding. It¹s great motivation to read such posts. Take care, Adam. === Subject: Re: Theorem 4.4.4. days. My association with the Department is that of an alumnus. >> Ah, but the real reason that Œone-to-one/injective¹ is equivalent to >> Œcancellable on the left¹ for topological spaces (and Œsurjective¹ is >> equivalent to Œcancellable on the right¹) is something else: it is >> that given any set there is a topological structure on it that makes >> any map of sets into it continuous (the indiscrete space), and a >> topological structure that makes any map of sets that has it as domain >> continuous (the discrete space); respectively. > All I can say is wow! I hope to be able to understand that in the >future. (-: There are two things at work in that paragraph there: the obvious one is topology. This is a nice subject (and, in its point-set version, something you can try looking at once you Žnish with set theory; but it grows increasingly complicated as it acquires other Œsurnames¹ like algebraic topology and so on...). Another thing lurking in the background is something called category theory, also known affectionately and derisively (depending on who you ask) as abstract nonsense. Not everyone¹s cup of tea, though a small splash of it here and there can be very useful. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Tensors for mathematicians > In response to why not just call them matrices or linear maps? > Tensors are to matrices what vector Želds or 1-forms are to vectors. > They live on the tangent (or cotangent) spaces of manifolds, and in > the language of Misner, Thorne, and Wheeler, Gravitation, which > is where I learned my 1st differential geometry, they map vector Želds > to scalar functions. So what you are saying is that a tensor takes a base point x in the manifold M, and then is a multi-linear map of say n vectors from T_x M > Instead of a vector Želd like the electric Želd, think of the > EM Želd tensor Želd at each point of some manifold. > (ProoŽng this, I read EM Želd tensor Želd--I mean the > electromagnetic Želd tensor is represented as a tensor Želd.) I can¹t do this :( E and B as vector Želds can at least be visualized to some degree; that 4-tensor object is just too mysterious... > Tensors at a point of a manifold X are deŽned on the tangent > space T_x to the manifold at that point, so are just like tensors on R^n. > He then deŽnes them as multilinear maps from > T_x X T*_x --> R > Then form the tensor bundle of tensors at all points x in X, and > a tensor Želd is a section of that bundle. > The thing is we want to be able to do everything on curved spaces-- > the universe is not R^n, as we know from Einstein¹s GR. > Also see Spivak, Intro to Differential Geometry I and II, also his > Calculus on Manifolds, and again I mention Nelson, Tensor Analysis, > Princeton Univ. press. > Grueb does a good job on multilinear algebra, but there are many good > treatments of multilinear algebra. > The study of the antisymmetric algebra, exterior differential forms, > differerntial geometry, Stokes thm., Hodges thm., etc. is fascinating-- > one of the most rewarding things I studied when studying physics. > Also, for math-physics, see Analysis, Manifolds, and Physics, Vol I and II > by Choquet-Bruhat et. al. I found these 2 books right at my level > (Ph.D. in physics), and very good, IMO. > Van All this is true, but as I said earlier in this thread, it is a huge amount of abstract machinery to learn, and then it is almost utterly useless for working with real things like the Cauchy stress tensor, the inertia tensor, etc... Granted, the original poster was talking about special relativity, (at the undergraduate level I believe), but bringing in the full machinery of GR would seem to me to obscure, rather than clarify, the basic ideas. -Jeff === Subject: Re: Tensors for mathematicians <7khre0p7t2f7eia4dr6256s96veb9nv1et@4ax.com> <40EDE8A4.F5DD6FC3@tiki-lounge.com> X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŽth X-Treme: C&C,DWS >So what you are saying is that a tensor takes a base point x in the >manifold M, and then is a multi-linear map of say n vectors from T_x Or, more generally, r vectors from T_x and s vectors from T^*_x. >but bringing in the full machinery of GR He wasn¹t talking about doing that; he was doing exactly what the OP requested, giving an explanation oriented to mathematicians rather than physicists. GR involves quite a bit more than just deŽnig tensors. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Tensors for mathematicians > > So why not just throw away the word tensor and call it a linear map? > > I¹m not sure, but I guess the word persists for historical reasons and > > because of this physical interpretation. > Linear maps are tensors, > but not all tensors are linear maps, > eg bilinear maps. Good point. So physics books should drop the mysterious word tensor and just say the more descriptive term multi-linear map! .... ? --Jeff === Subject: Re: Tensors for mathematicians > at 09:21 PM, jjensen14@hotmail.com (J Jensen) said: > >You established it for one O.N.basis, > >but the same argument would show it for another O.N. basis, so the > >two linear maps must be related in the standard way of changing bases > >in linear algebra. > Try that with the Gamma symbols and watch what happens. Not everything > that has numbers associated with coordinate systems transforms like a > tensor. I assume you are talking about the Cristoffel symbols? I haven¹t studied that yet, so I can¹t agree or disagree. But, in my above postings, I am speciŽcally referring to a linear map relating 2 physical quantities, which are naturally represented as vectors, in an orthogonal basis. In that scope, I am still convinced that the things I said in my Žrst posting in this thread are correct, because I haven¹t seen a real error pointed out yet... --Jeff === Subject: Re: Tensors for mathematicians Originator: grubb@lola >> Try that with the Gamma symbols and watch what happens. Not everything >> that has numbers associated with coordinate systems transforms like a >> tensor. >I assume you are talking about the Cristoffel symbols? I haven¹t >studied >that yet, so I can¹t agree or disagree. But, in my above postings, >I am speciŽcally referring to a linear map relating 2 physical >quantities, which are naturally represented as vectors, in an >orthogonal basis. >In that scope, I am still convinced that the things I said in my Žrst >posting in this thread are correct, because I haven¹t seen a real >error >pointed out yet... Yes, he was talking about the Christoffel symbols. As an example of a physical quantity that doesn¹t transform as you might expect, take the electric Želd vector. You expect it to be a vector, right? Well, it isn¹t. You take an orthonormal basis and suppose I am moving past you at half the speed of light, and I take an orthonormal basis. You cannot get my version of the electric Želd from yours via a transformation of vectors. You *have* to somehow include the magnetic Želd and do a transformation of the resulting tensors. Now you can argue that these are not Œnaturally represented as vectors¹, but I think you might have an argument on your hands. :) --Dan Grubb === Subject: Re: Tensors for mathematicians <40ef0dc3$6$fuzhry+tra$mr2ice@news.patriot.net> X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŽth X-Treme: C&C,DWS >I assume you are talking about the Cristoffel symbols? Essentially; it¹s a different nomenclature for the components of a linear connection. >But, in my above postings, >I am speciŽcally referring to a linear map relating 2 physical >quantities, which are naturally represented as vectors, No, tensors, with one covariant index and one contravariant index. >in an orthogonal basis. Using an orthonormal basis doesn¹t change a 2-tensor into a 1-tensor. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Tensors for mathematicians X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŽth X-Treme: C&C,DWS at 07:03 PM, glhansen@indiana.edu (Gregory L. Hansen) said: >If you¹re a math person, you might feel comfortable with the >inexpensive Dover book by Lovelock and Rund, Tensors, Differential >Forms, and Variational Principles. Highly doubtful; the deŽnition that you give seems to have been written for a Physicist, not for a Mathematician. I would expect him to be more comfortable with coordinate-free deŽnitions. >A dual is what you need to form a dot product. No. It¹s true that given a dual you can deŽne a dot product, but that¹s putting the cart before the horse. The conventional way to deŽne a dot product is with a metric tensor, and you can then trivially deŽne the corresponding dual. >In particular, special relativity has the metric > ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 That¹s only one formulation; a signature of (- + + +) works just as well as (+ - - -). There¹s also the ict approach, but AFAIK nobody uses it these days. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: is there a Great Attractor in Gametheory of VonNeumann?? Re: There exists a Nim version that is a draw OS > > > A combinatorial game such as Nim can not be a draw. > > > Here is a distant relative that can be a draw: > > > > > > http://home.no.net/zamunda/split.htm > > I beg to differ. > > Yesterday I was working on a game of Nim, a morph of Nim where there are > > no draws in the game itself but where either player can win in the OS > > and not automatically that one player always wins the OS. Call it a > > pseudodraw. > The minimax theorem says a singular point. Thus a pseudodraw is > nonexistent. > Unless there is a draw within the game itself can the OS be a draw. > > Secondly, I was looking for another Nim morph where it actually has a > > draw within the game itself and the OS is a draw. > > Thirdly I was looking for a Tictactoe morph that was _not_ a draw in the > > OS and where either X or O can win in the OS. Call it a pseudodraw. > > Here is what I come up with: > > Nim-morph with pseudodraw OS: Let me call the person with Žrst move as > > white and let me call the person with second move as black. The Žrst > > move in this game is not the removal of any matchsticks but is the > > actual layout of the number of rows and the number of matchsticks within > > each row. Black then proceeds as in normal nim. I contend, thence, that > > this nim morph will end up as a win for one of the players but not > > automatically the black player (provided regular nim is considered a > > loss for the one who is forced to pick up the last matchstick). > This is a erroneous claim. Even if I added the rule that only one or two > matchsticks can be removed per move. > > Nim-morph with a Draw in the game itself: This is where white with Žrst > > move determines the number of rows of matchsticks and the number of > > matchsticks in each row. And Žnally, determines that at least one row > > is a Draw row so that if this row or any of its matchsticks is picked > > up last then the entire game is a draw. > This is possible. It perhaps needs the rule of only one or two matchsticks > removed per move. > > TicTacToe-morph with pseudodraw OS: this one was a tough one to work out > > last night. I would have thought that Nim was going to be the tougher > > challenge. We have several rule changes to normal tictactoe. Call the > > Žrst mover as X and the second mover as O. In this morph, O gets two > > Žrst moves so that at the end of the game there will be Žve O on the > > board to four X. And the other change in rule is that if there are no > > three-in-a-row for a outright win then the win goes to the person who > > has the most two-in-a-row. Now I have not fully played out all the > > consequences. But I suspect, not sure of this suspection, that the OS of > > this morph tictactoe is a win for either X or O or a pseudodraw. And > > that every game played of this morph will produce a winner whether it be > > X or O. > Trouble with whether end row middles would count as 2-in-a-row rather > than having only shortened 3-in-a-rows count as 2-in-a-row. When X makes > Žrst move with placing an X in center square then X has the most > 2-in-a-row unless we count end-row-middles as 2 in a row for O. > Here again, the concept of Pseudodraw is erroneous, and that unless a draw > exists in the game itself can the OS be a draw. And the minimax theorem > says as much. > > Now, the most important aspect of the above, if true, implies that there > > exists a Pseudodraw for the games of checkers and chess, but more > > importantly, that those games OS is a draw with their current and > > present rules. > But the above is not all lost and wasted. I can salvage the idea that to > make Nim a draw is to add the rule that the player with Žrst move decides > the arrangement of how many rows and number of matchsticks per row and > which row is the Draw row. > The implications for chess and checkers still remain. That if a game has a > draw possibility, then the OS of that game ends up into that draw play. > Nim OS is a win for one of the players always, well, because there is no > draw possibility while playing the game. > I never played Go. I suspect it has a draw possibility. If it does, then > that is its OS-- a draw. Chess has a draw possibility, thus chess OS is a > draw. > This claim can be made into a assertion and then a theorem. > Devise a game that is a VonNeumann game which has a draw possibility but > has a nonDraw OS. Nim has a nondraw OS but nim has no draw within the game > itself. So when we inject a draw possibility into Nim then does the one > player always win the OS?????? Initially I was tempted to call a draw game in any VonNeumann game as a gravity attractor such as gravity equilibrium or gravitational center so that if you introduce a draw game inside of Nim that the OS of Nim shifts and then becomes something different from its automatic win for second player. That the moment you introduce a possible draw game that the entire OS of Nim shifts and becomes that draw end result. But there is another concept in physics that is like gravitational attraction. And I suppose a good physicist not the usual run of the mill sort can tell you the conceptual difference between gravity attraction and Great Attractor in chaos theory. I like to think of Great Attractors in EM of electricity and magnetism. Anyway, Nim is VonNeumann gametheory and the OS is a certain victory for second player. But introduce just one possibility of a draw outcome, then, does the entire OS of this Nim change to the draw outcome? As like a Great Attractor, the draw outcome forces itself as the Optimal Strategy. Archimedes Plutonium www.archimedesplutonium.com www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: is there a Great Attractor in Gametheory of VonNeumann?? Re: There exists a Nim version that is a draw OS >>>>A combinatorial game such as Nim can not be a draw. >>>>Here is a distant relative that can be a draw: >>>> >>>>http://home.no.net/zamunda/split.htm >>>I beg to differ. >>>Yesterday I was working on a game of Nim, a morph of Nim where there are >>>no draws in the game itself but where either player can win in the OS >>>and not automatically that one player always wins the OS. Call it a >>>pseudodraw. >>The minimax theorem says a singular point. Thus a pseudodraw is >>nonexistent. >>Unless there is a draw within the game itself can the OS be a draw. >>>Secondly, I was looking for another Nim morph where it actually has a >>>draw within the game itself and the OS is a draw. >>>Thirdly I was looking for a Tictactoe morph that was _not_ a draw in the >>>OS and where either X or O can win in the OS. Call it a pseudodraw. >>>Here is what I come up with: >>>Nim-morph with pseudodraw OS: Let me call the person with Žrst move as >>>white and let me call the person with second move as black. The Žrst >>>move in this game is not the removal of any matchsticks but is the >>>actual layout of the number of rows and the number of matchsticks within >>>each row. Black then proceeds as in normal nim. I contend, thence, that >>>this nim morph will end up as a win for one of the players but not >>>automatically the black player (provided regular nim is considered a >>>loss for the one who is forced to pick up the last matchstick). >>This is a erroneous claim. Even if I added the rule that only one or two >>matchsticks can be removed per move. >>>Nim-morph with a Draw in the game itself: This is where white with Žrst >>>move determines the number of rows of matchsticks and the number of >>>matchsticks in each row. And Žnally, determines that at least one row >>>is a Draw row so that if this row or any of its matchsticks is picked >>>up last then the entire game is a draw. >>This is possible. It perhaps needs the rule of only one or two matchsticks >>removed per move. >>>TicTacToe-morph with pseudodraw OS: this one was a tough one to work out >>>last night. I would have thought that Nim was going to be the tougher >>>challenge. We have several rule changes to normal tictactoe. Call the >>>Žrst mover as X and the second mover as O. In this morph, O gets two >>>Žrst moves so that at the end of the game there will be Žve O on the >>>board to four X. And the other change in rule is that if there are no >>>three-in-a-row for a outright win then the win goes to the person who >>>has the most two-in-a-row. Now I have not fully played out all the >>>consequences. But I suspect, not sure of this suspection, that the OS of >>>this morph tictactoe is a win for either X or O or a pseudodraw. And >>>that every game played of this morph will produce a winner whether it be >>>X or O. >>Trouble with whether end row middles would count as 2-in-a-row rather >>than having only shortened 3-in-a-rows count as 2-in-a-row. When X makes >>Žrst move with placing an X in center square then X has the most >>2-in-a-row unless we count end-row-middles as 2 in a row for O. >>Here again, the concept of Pseudodraw is erroneous, and that unless a draw >>exists in the game itself can the OS be a draw. And the minimax theorem >>says as much. >>>Now, the most important aspect of the above, if true, implies that there >>>exists a Pseudodraw for the games of checkers and chess, but more >>>importantly, that those games OS is a draw with their current and >>>present rules. >>But the above is not all lost and wasted. I can salvage the idea that to >>make Nim a draw is to add the rule that the player with Žrst move decides >>the arrangement of how many rows and number of matchsticks per row and >>which row is the Draw row. >>The implications for chess and checkers still remain. That if a game has a >>draw possibility, then the OS of that game ends up into that draw play. >>Nim OS is a win for one of the players always, well, because there is no >>draw possibility while playing the game. >>I never played Go. I suspect it has a draw possibility. If it does, then >>that is its OS-- a draw. Chess has a draw possibility, thus chess OS is a >>draw. >>This claim can be made into a assertion and then a theorem. >>Devise a game that is a VonNeumann game which has a draw possibility but >>has a nonDraw OS. Nim has a nondraw OS but nim has no draw within the game >>itself. So when we inject a draw possibility into Nim then does the one >>player always win the OS?????? > Initially I was tempted to call a draw game in any VonNeumann game as a > gravity attractor such as gravity equilibrium or gravitational center so that > if you introduce a draw game inside of Nim that the OS of Nim shifts and > then becomes something different from its automatic win for second player. > That the moment you introduce a possible draw game that the entire OS of Nim > shifts and becomes that draw end result. > But there is another concept in physics that is like gravitational > attraction. And I suppose a good physicist not the usual run of the mill > sort can tell you the conceptual difference between gravity attraction and > Great Attractor in chaos theory. > I like to think of Great Attractors in EM of electricity and magnetism. > Anyway, Nim is VonNeumann gametheory and the OS is a certain victory for > second player. But introduce just one possibility of a draw outcome, then, > does the entire OS of this Nim change to the draw outcome? As like a Great > Attractor, the draw outcome forces itself as the Optimal Strategy. > Archimedes Plutonium > www.archimedesplutonium.com > www.iw.net/~a_plutonium > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies The intellectual midget doth return. === Subject: Re: is there a Great Attractor in Gametheory of VonNeumann?? Re: There exists a Nim version that is a draw OS charset=iso-8859-1 > The intellectual midget doth return. What is up with all these idiots returning to plague the forum at the same time? === Subject: a complex analysis question f(z)=(z_0 - z) / (1 - bar{z_0} z) abs(z_0)<1 conformal mapping.(unit disk to unit disk) Suppose f maps a disk (center s, radius s) on to a disk (center 0 radisu r) where s>0, 2s<1, r<1 (1) For s=1/4, determine f(z) and r. (2) Show that, in general, r and s are related by the equation r^2s-r+s=0 ---------------(sorry for clumsy notation)----------- I think, there might be some symmetric relation between those two disks in a unit disk. But I cannot Žgure that out. How can I solve this problem? === Subject: Re: a complex analysis question > f(z)=(z_0 - z) / (1 - bar{z_0} z) abs(z_0)<1 conformal mapping.(unit > disk to unit disk) > Suppose f maps a disk (center s, radius s) on to a disk (center 0 > radisu r) > where s>0, 2s<1, r<1 > (1) For s=1/4, determine f(z) and r. Clearly f is a linear fractional map, and so z_0 must be s. Hope this helps, Bill > (2) Show that, in general, r and s are related by the equation > r^2s-r+s=0 > ---------------(sorry for clumsy notation)----------- > I think, there might be some symmetric relation between those two > disks in a unit disk. But I cannot Žgure that out. How can I solve > this problem? === Subject: Re: a complex analysis question > > f(z)=(z_0 - z) / (1 - bar{z_0} z) abs(z_0)<1 conformal mapping.(unit > > disk to unit disk) > > Suppose f maps a disk (center s, radius s) on to a disk (center 0 > > radisu r) > > where s>0, 2s<1, r<1 > > (1) For s=1/4, determine f(z) and r. > Clearly f is a linear fractional map, and so z_0 must be s. > Hope this helps, Bill Ok. then you sent 1/4 to 0 with map f. but how can you get the radius? For example, 0 and 1/2 will be on the boundary of a disk with center 1/4 r=1/4 so their image will be also on the boundary of a disk with center 0 radius r. But actually f(0)=1/4 f(1/2)=-2/7, aren¹t they? > > (2) Show that, in general, r and s are related by the equation > > r^2s-r+s=0 > > ---------------(sorry for clumsy notation)----------- > > I think, there might be some symmetric relation between those two > > disks in a unit disk. But I cannot Žgure that out. How can I solve > > this problem? === Subject: Re: a complex analysis question >> > f(z)=(z_0 - z) / (1 - bar{z_0} z) abs(z_0)<1 conformal mapping.(unit >> > disk to unit disk) >> > >> > Suppose f maps a disk (center s, radius s) on to a disk (center 0 >> > radisu r) >> > where s>0, 2s<1, r<1 >> > >> > (1) For s=1/4, determine f(z) and r. >> Clearly f is a linear fractional map, and so z_0 must be s. >> Hope this helps, Bill >Ok. then you sent 1/4 to 0 with map f. but how can you get the radius? He¹s already retracted this. A linear fractional transformation maps circles (and lines) to circles (and lines), but it doesn¹t preserve the center of the circle. You _can_ use the fact that it maps circles to circles to answer the question, though. >For example, 0 and 1/2 will be on the boundary of a disk with center >1/4 r=1/4 so their image will be also on the boundary of a disk with >center 0 radius r. >But actually f(0)=1/4 f(1/2)=-2/7, aren¹t they? >> > (2) Show that, in general, r and s are related by the equation >> > >> > r^2s-r+s=0 >> > >> > ---------------(sorry for clumsy notation)----------- >> > >> > I think, there might be some symmetric relation between those two >> > disks in a unit disk. But I cannot Žgure that out. How can I solve >> > this problem? ************************ David C. Ullrich === Subject: Re: a complex analysis question === Subject: Re: a complex analysis question > Clearly f is a linear fractional map, and so z_0 must be s Scratch this thought, I answered to fast. Still I might try to use the fact that f is a linear fractional map, and preserves circles, to help obtain the formula. === Subject: comprehensive theoretical textbook Are there any good comprehensive handbook/textbooks of Mathematics that present all the theories of elementary and advanced Mathematics (undergraduate level), including Algebra, Number Theory, Discrete Mathematics, Geometry, Calculus and Analysis, etc. using an axiomatic/formalistic approach - axioms, theorems with complete proofs, corollaries etc.? Which one would you recommend? Frank === Subject: Re: Mathematical Induction and a Train Set > Hi Guys. > I have a Train Set with a small problem. If I set it at max speed it will go > around 3 or 4 times and then get stuck or roll over. > My question is, if it can go around once, shouldn¹t it continue to do that > because it¹s back to its original state of being back to the starting point? > Any suggestions? The state of the train should include its instantaneous velocity, as well as its position. *This* will not be the same each time around. The state of the *system* also includes the position of the track. If this isn¹t constant (for example if the sucessful runnings are shifting the track slightly), then again the conditions have changed For a model train set I would lean to the explanation offered by others, that either the train or the track is being modiŽed during the succcessful running period. -- Larry Lard Replies to group please. === Subject: Re: Mathematical Induction and a Train Set >> Hi Guys. >> I have a Train Set with a small problem. If I set it at max speed it will go >> around 3 or 4 times and then get stuck or roll over. >> My question is, if it can go around once, shouldn¹t it continue to do that >> because it¹s back to its original state of being back to the starting point? >> Any suggestions? >The state of the train should include its instantaneous velocity, as >well as > its position. *This* will not be the same each time around. >The state of the *system* also includes the position of the track. If >this isn¹t constant (for example if the sucessful runnings are >shifting the track slightly), then again the conditions have changed >For a model train set I would lean to the explanation offered by >others, that either the train or the track is being modiŽed during >the succcessful running period. Or more generally from a mathematical point of view the equations describing the system need to have a term added that represents random žuctuations, ie noise, that at some point in the orbit creates a condition that represents train derailed or jammed. Physically this can represent twisting of the trucks, bouncing over rough sections in the track etc. Of course, if the track is not nailed down, then, as suggested, it is very likely that track will be moved or joints opened as the train is operated. === Subject: Re: EIS enries of Roger L. Bagula X-Reply-Etiquette: No copy by email, please Originator: legalize@deuce.xmission.com (Rich) [Please do not mail me a copy of your followup] Relevance? What does it have to do with these newsgroups? -- The Direct3D Graphics Pipeline-- code samples, sample chapter, FAQ: Pilgrimage: Utah¹s annual demoparty === Subject: Re: EIS enries of Roger L. Bagula >My response was to add the comment: >Warning: the management of EIS can be hazardous to innocent mathematicians. >I have one editor there who has me on his blocked list: >A Dr. Bob Wilson V >That is he can send me insulting email and my replies bounce. >Robin Chapman has also got me on his blocked list, so he can do the same. >I have a lot of detractors, but in most cases they tend to admit >,grudgingly, that my result is new and worthwhile. What result? >I¹ve become very stoic about this over the years >and many results. >Editors seem happy to have their names on my sequences. >> In many of that sequences there is a paragraph like it: >> Extension: Warning: Many recent communications from this author have >> contained >> numerical errors or have been badly formatted. This entry has >> not been >> edited and may contain errors. It is included on a provisional >> basis in >> the hope that some reader will edit it. - njas ************************ David C. Ullrich === === Subject: JSH: Groupthink So now senators in the United States are trying to put all the blame on the CIA for intelligence failurs as if Bush and company did nothing wrong, and I think you can see what groupthink is in a dramatic area. Now sci.math is a groupthink area as well, since for some time now I¹ve been able to demonstrate my mathematical points with precision rarely seen in ANY mathematical proofs, but the *group* years ago decided that I couldn¹t have any correct results. That group think was why some of you could send emails to question a math journals on process of formal peer review--putting your judgement over that of its editors and referees--and much of the group just yawned. Group think. Now then, you¹re just a gaggle of people, but the United States is the most powerful country in the world. If it can fall to group think, why should anyone be surprised that sci.math does as well. The sad thing is the weakness it shows as none of you care enough for mathematics itself to stand up and holler when people say false things. Luckily for me, math society is bigger than sci.math, which you learned when my paper was published despite years of people calling me names here, but maybe you learned the wrong lesson when sci.math posters with their coordinated email campaign succeeded in getting it yanked. But you were dealing with a journal mostly created by one man--Ioannis Argyros--in a state school in Oklahoma. See http://rattler.cameron.edu/swjpam/vol2-03.html So several posters on sci.math got to *one* man who felt that he had the right with a journal he created to make a drastic decision that it turns out was the wrong one. Don¹t get too happy or feel that sci.math is more powerful than it is because some of you could so easily inžuence just one mathematician. I learned from what happened. I picked journals more carefully and have more than one paper, so that at least one should get through. The battle will be joined again sci.math, and this time you will be faced with a far harder task. That is, I will beat the entire sci.math newsgroup, and show you all why groupthink is no match for mathematics. It¹s kind of weird how people can convince themselves of things they wish to believe and continue against all evidence. It¹s math society itself that will break you. And when you are condemned by mathematicians you thought were your peers, and see papers published in journals you dare not question, dare not send emails to, then what will you think? I doubt you¹ll groupthink then. I¹m curious to see how you do. My guess is that you have less than six months before all the papers play out, including the paper with The Hammer. James Harris === Subject: Re: JSH: Groupthink > So now senators in the United States are trying to put all the blame > on the CIA for intelligence failurs as if Bush and company did nothing > wrong, and I think you can see what groupthink is in a dramatic area. > Now sci.math is a groupthink area as well, since for some time now > I¹ve been able to demonstrate my mathematical points with precision > rarely seen in ANY mathematical proofs, but the *group* years ago > decided that I couldn¹t have any correct results. > That group think was why some of you could send emails to question a > math journals on process of formal peer review--putting your judgement > over that of its editors and referees--and much of the group just > yawned. > Group think. > Now then, you¹re just a gaggle of people, but the United States is the > most powerful country in the world. If it can fall to group think, > why should anyone be surprised that sci.math does as well. > The sad thing is the weakness it shows as none of you care enough for > mathematics itself to stand up and holler when people say false > things. OK, here¹s a false thing. Your main result in APF is about factorizations of polynomials in x with constant terms of the form u^3 f^3 where f is a nonunit algebraic integer coprime to 3. Right? You apply this to the polynomial 65 x^3 - 12 x + 1. I actually think you have *never even noticed* that the constant term 1 is not of the form u^3 f^3. Yes, I actually don¹t think you have ever really read your own paper! Convince me otherwise. Andrzej PS: The main result in APF is incorrect also, by basic Galois theory and basic algebraic number theory (independently). You knew this and went ahead and submitted it again anyway, with no warning to the editor, as if you had never heard any objections. Right? Hoping that lackadaisical, sloppy editing will prevail again? > I¹m curious to see how you do. My guess is that you have less than > six months before all the papers play out, including the paper with > The Hammer. Bring it on! > James Harris === Subject: Re: JSH: Groupthink > So now senators in the United States are trying to put all the blame > on the CIA for intelligence failurs as if Bush and company did nothing > wrong, and I think you can see what groupthink is in a dramatic area. > Now sci.math is a groupthink area as well, > Group think. Group think is indeed insidious. How does it arise in a given set of people? I don¹t think it¹s merely accidental. Whenever group think occurs, it is orchestrated by one person, whose identity is no doubt carefully concealed. The only way to identify this Mystery Man is through police interrogation. We could use, for example, the Good Cop / Bad Cop technique. When the Mystery Man (let¹s just call him Mr. E) is interrogated by one cop, and another member of the group by the other cop, what do you think will happen? Imagine Mr. E is Ronald Reagan, and the other guy is Oliver North. Who¹s going to take the fall? They¹ll both agree that Mr. E is Oliver North. In fact, when you interrogate any two members, you can be sure they¹ll get their stories straight -- they¹ll implicate the same person every time. Fascists may laud such organizational cohesion, but I think that misses the point. These groups must be rooted out and destroyed! This is not easy to do. They have an diabolic internal structure which researchers are only beginning to understand. For example, suppose you assign the Good Cop to Archie and the Bad Cop to Bill. They say David is Mr. E. So now you put the Good Cop on David, and the Bad Cop on Carlos, and they say Freddy is Mr. E. So what? Well, suppose now you assign the Good Cop to Bill and the Bad Cop to Carlos. They implicate Gustav. Now we put the Good Cop on Archie and the Bad Cop on Gustav: who do they implicate? Freddy again! They¹ve tried this hundreds of times and it always works this way. Why should Freddy get Žngered in the second case just because he got Žngered in the Žrst one? Clearly there are some subtle associations at work within these groups. Is there any hope of foiling their devious plans? Possibly. It is known that there is subversion even in the worst groups. It has been rumored that for every member of the group there is another member that can be interrogated with him such that they implicate Mr. E truthfully. Some disagree however, and point to groups in which no two members *ever* implicate Mr. E. But in such cases, putting both cops on *any* single member always seems to produce the truth. Curiouser and curiouser. One would almost think that people could devote their whole lives to studying these groups. There can be groups within groups, with Mr. E, of course, running each one. But when Mr. E sets up such a subgroup, he also arranges for a bewildering family of cosets (normally one family, but sometimes two) to misdirect anyone who would try to identify him. If things do work out normally, then Mr. E¹s subgroup organizes a new group from these cosets, placing itself in charge as [enter the Colonel]: Hello. I am Colonel H. Morphism. This post has gotten silly. Now, no one likes a good joke more than I . . . -- Jim Ferry at U of Illinois, Urbana-Champaign, (no, I don¹t educate) 2 email me l o o k up 1 row === Subject: Re: JSH: Groupthink > So now senators in the United States are trying to put all the blame > on the CIA for intelligence failurs as if Bush and company did nothing > wrong, and I think you can see what groupthink is in a dramatic area. Yes. The žaws of sci.math are, as always, examples of the žavor of the month. It used to be the Žnancial misdeeds of all those CEOs, the Enron debacle and all, then there was 9/11, žawed belief in WMDs in Iraq, or whatever the lead story was in one of the news magazines. Isn¹t it odd how our behavior presages the hot issues of the day? We should determine made! > Now sci.math is a groupthink area as well, since for some time now > I¹ve been able to demonstrate my mathematical points with precision > rarely seen in ANY mathematical proofs, but the *group* years ago > decided that I couldn¹t have any correct results. Right. You demonstrate things with sufŽcient precision to convince yourself. You started out already convinced, and *anything* would have been convincing to you. > That group think was why some of you could send emails to question a > math journals on process of formal peer review--putting your judgement > over that of its editors and referees--and much of the group just > yawned. Yes, groupthink was how I factored 5 and each of those a_i¹s in the ring of algebraic integers. Glad you poined that out. I suppose it was groupthink that allowed others to follow my argument by multiplying some polynomials together to see that the coefŽcients matched, showing my result to be correct. Not arithmetic, but: > Group think. > Now then, you¹re just a gaggle of people, but the United States is the > most powerful country in the world. If it can fall to group think, > why should anyone be surprised that sci.math does as well. Why, indeed? Why should anyone be surprised to hear that sci.math is engaged in professional malfeasance, just like those CEOs? > The sad thing is the weakness it shows as none of you care enough for > mathematics itself to stand up and holler when people say false > things. Hey, I did that! You complained, and continue to evade the argument. You said I had assumed something I hadn¹t, I asked you to point out just where I made the offending assumption, and you just ran away! > Luckily for me, math society is bigger than sci.math, which you > learned when my paper was published despite years of people calling me > names here, but maybe you learned the wrong lesson when sci.math > posters with their coordinated email campaign succeeded in getting it > yanked. I coordinated with no one, and I speciŽcally did *NOT* request your someone says false things. The only problem was that it was YOU saying false things. > But you were dealing with a journal mostly created by one man--Ioannis > Argyros--in a state school in Oklahoma. > See http://rattler.cameron.edu/swjpam/vol2-03.html > So several posters on sci.math got to *one* man who felt that he had > the right with a journal he created to make a drastic decision that it > turns out was the wrong one. read? That mus tbe it. > Don¹t get too happy or feel that sci.math is more powerful than it is > because some of you could so easily inžuence just one mathematician. > I learned from what happened. > I picked journals more carefully and have more than one paper, so that > at least one should get through. Oh, it¹s like a shotgun approach. Pretty sophisticated, that. > The battle will be joined again sci.math, and this time you will be > faced with a far harder task. Far harder than getting you to respond to reason? I really doubt that! > That is, I will beat the entire sci.math newsgroup, and show you all > why groupthink is no match for mathematics. It¹s kind of weird how > people can convince themselves of things they wish to believe and > continue against all evidence. I thought we were no match for The Hammer. I thought that The Powers That Be would really show us. I thought the FBI was going to get us. I thought you were going to sic the generals on us. I thought we all were going to be dragged up in front of Congressional Subcommittees. I thought you were going to sue our sorry asses off us. I thought we all were going to lose our cushy white-collar welfare jobs. I thought you were one of the top number theorists in the world today. Sorry, my sarcasm switch got stuck. Why doesn¹t the argument I proposed (and sent to Argyros (?)) work? You¹ve seen it, and complained that there is a subtle assumption, but refuse to point out *which* step makes such an assumption, and *which* step is in fact incorrect. Why isn¹t that considered evidence? > It¹s math society itself that will break you. And when you are > condemned by mathematicians you thought were your peers, and see > papers published in journals you dare not question, dare not send > emails to, then what will you think? Yeah, I¹ll believe it when I see it. > I doubt you¹ll groupthink then. > I¹m curious to see how you do. My guess is that you have less than > six months before all the papers play out, including the paper with > The Hammer. discussions about the size and power of your tool. > James Harris Dale === Subject: Re: JSH: Groupthink > So now senators in the United States are trying to put all the blame > on the CIA for intelligence failurs as if Bush and company did nothing > wrong, and I think you can see what groupthink is in a dramatic area. > Now sci.math is a groupthink area as well, since for some time now > I¹ve been able to demonstrate my mathematical points with precision > rarely seen in ANY mathematical proofs, but the *group* years ago > decided that I couldn¹t have any correct results. (In mina/hot-girl style, I ask...) ~~sir, What are some other properties of this group? Is it abelian? I know I commute, but some other elements might be reading this at home. Is it simple, or are there some normal subgroups? With a small change in pronunciation, I suspect you¹d agree that it¹s a Lie group. Enquiring minds want to know. Rick who¹s just exceeded his bad joke quota for the day === Subject: Re: JSH: Groupthink >> So now senators in the United States are trying to put all the blame >> on the CIA for intelligence failurs as if Bush and company did nothing >> wrong, and I think you can see what groupthink is in a dramatic area. >> Now sci.math is a groupthink area as well, since for some time now >> I¹ve been able to demonstrate my mathematical points with precision >> rarely seen in ANY mathematical proofs, but the *group* years ago >> decided that I couldn¹t have any correct results. >(In mina/hot-girl style, I ask...) >~~sir, >What are some other properties of this group? Is it abelian? I know >I commute, but some other elements might be reading this at home. >Is it simple, or are there some normal subgroups? With a small >change in pronunciation, I suspect you¹d agree that it¹s a Lie group. >Enquiring minds want to know. >Rick who¹s just exceeded his bad joke quota for the day For the _day_? ************************ David C. Ullrich === Subject: Re: JSH: Groupthink >>>So now senators in the United States are trying to put all the blame >>>on the CIA for intelligence failurs as if Bush and company did nothing >>>wrong, and I think you can see what groupthink is in a dramatic area. >>>Now sci.math is a groupthink area as well, since for some time now >>>I¹ve been able to demonstrate my mathematical points with precision >>>rarely seen in ANY mathematical proofs, but the *group* years ago >>>decided that I couldn¹t have any correct results. >>(In mina/hot-girl style, I ask...) >>~~sir, >>What are some other properties of this group? Is it abelian? I know >>I commute, but some other elements might be reading this at home. >>Is it simple, or are there some normal subgroups? With a small >>change in pronunciation, I suspect you¹d agree that it¹s a Lie group. >>Enquiring minds want to know. >>Rick who¹s just exceeded his bad joke quota for the day > For the _day_? Yup. My bad joke counter resets each 24 hours. If it didn¹t, I¹d have to change my teaching style beyond recognition. Rick who once got a teaching evaluation that said Œgraduate work in jokes needed¹ === Subject: Re: JSH: Groupthink >[...] > Now then, you¹re just a gaggle of people, but the United States is the > most powerful country in the world. If it can fall to group think, > why should anyone be surprised that sci.math does as well. Power != Intelligence >[...] > I learned from what happened. > I picked journals more carefully and have more than one paper, so that > at least one should get through. Great idea! I can imagine editors and referees applaud, when they realise that you sent a paper (or minor variations) to several journals at the same time. >[...] > I¹m curious to see how you do. My guess is that you have less than > six months before all the papers play out, including the paper with > The Hammer. Capitalisation will get you nowhere... Marc === Subject: Re: JSH: Groupthink >[...] >I¹ve been able to demonstrate my mathematical points with precision >rarely seen in ANY mathematical proofs, Ah. I see that someone has already replied to this sentence, but he got it wrong. The correct reply is this: Yes, James, the sort of precision we see from you is indeed rarely seen in mathematical proofs. (I got there Žrst. I win... I mean this one was so easy I can¹t believe nobody got it while I was asleep.) >but the *group* years ago >decided that I couldn¹t have any correct results. >That group think was why some of you could send emails to question a >math journals on process of formal peer review--putting your judgement >over that of its editors and referees--and much of the group just >yawned. >Group think. >Now then, you¹re just a gaggle of people, but the United States is the >most powerful country in the world. If it can fall to group think, >why should anyone be surprised that sci.math does as well. >The sad thing is the weakness it shows as none of you care enough for >mathematics itself to stand up and holler when people say false >things. >Luckily for me, math society is bigger than sci.math, which you >learned when my paper was published despite years of people calling me >names here, but maybe you learned the wrong lesson when sci.math >posters with their coordinated email campaign succeeded in getting it >yanked. >But you were dealing with a journal mostly created by one man--Ioannis >Argyros--in a state school in Oklahoma. >See http://rattler.cameron.edu/swjpam/vol2-03.html >So several posters on sci.math got to *one* man who felt that he had >the right with a journal he created to make a drastic decision that it >turns out was the wrong one. >Don¹t get too happy or feel that sci.math is more powerful than it is >because some of you could so easily inžuence just one mathematician. >I learned from what happened. >I picked journals more carefully and have more than one paper, so that >at least one should get through. Want to bet? (How many times have you sent papers to journals? How many have been accepted? (Hint: no, the editor didn¹t withdraw that paper because he was afraid of the wrath of sci.math, he inappropriately withdrew it because he realized it was nonsense.)) >The battle will be joined again sci.math, and this time you will be >faced with a far harder task. >That is, I will beat the entire sci.math newsgroup, and show you all >why groupthink is no match for mathematics. It¹s kind of weird how >people can convince themselves of things they wish to believe and >continue against all evidence. >It¹s math society itself that will break you. And when you are >condemned by mathematicians you thought were your peers, and see >papers published in journals you dare not question, dare not send >emails to, then what will you think? >I doubt you¹ll groupthink then. >I¹m curious to see how you do. My guess is that you have less than >six months before all the papers play out, including the paper with >The Hammer. Here¹s what I¹m curious about: It seems that when one of your papers gets published we¹re all going to die a horrible death or something. Whatever the terrible consequences are going to be (I wish you¹d be more speciŽc about that), how is it that _you_ have not already suffered the same bloodly fate, as a result of the fact that many of us have published many papers? >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Groupthink > >[...] > >I¹ve been able to demonstrate my mathematical points with precision > >rarely seen in ANY mathematical proofs, > Ah. I see that someone has already replied to this sentence, > but he got it wrong. The correct reply is this: > Yes, James, the sort of precision we see from you is indeed > rarely seen in mathematical proofs. > (I got there Žrst. I win... I mean this one was so easy I > can¹t believe nobody got it while I was asleep.) Well, it¹s pretty close to David Kastrup¹s comment in the thread JSH: Weird mathematicians (June 11th): )> After all, mathematical results are *hard* in that you can¹t wish )> them away, or change them, so it amazed me for quite some time that )> so many of you would Žght against results that I could prove with a )> rigor far outside of most mathematical works. ) ) Quite far outside. James really is serving these on a silver platter. ------------------------------------ J K Haugland http://www.neutreeko.com Try my grid subgraph prize problems! ------------------------------------ === Subject: Re: JSH: Groupthink > So several posters on sci.math got to *one* man who felt that he had > the right with a journal he created to make a drastic decision that it > turns out was the wrong one. I believe those same posters agree with you that his decision was the wrong one. - Randy === Subject: Re: Groupthink > I¹ve been able to demonstrate my mathematical points with precision > rarely seen in ANY mathematical proofs, .... I have had the opportunity to read some of what you have written over the last couple of years. It¹s there¹s one think you HAVEN¹T done, it¹s demonstrate your mathematical points with precision! I say this as a mostly neutral bystander. === Subject: Re: JSH: Groupthink Oh, God, James, go get a bottle of Zyprexa or whatever it is you¹re missing. Yours, Doug Goncz ( ftp://users.aol.com/DGoncz/ ) Student member SAE for one year. === Subject: Re: JSH: Groupthink > Oh, God, James, go get a bottle of Zyprexa or whatever it is you¹re missing. > Yours, > Doug Goncz ( ftp://users.aol.com/DGoncz/ ) > Student member SAE for one year. The papers are at journals. Some of them have been there for a while. The APF paper already passed formal peer review once, and it has been at a journal for over a month now. I think that only groupthink could allow you all to be so conŽdent now, when the weight of the math world itself will soon fall down upon you. What worked with Ioannis Argyros, with a math journal he started, which may have been part of what clouded his judgement (though that is no excuse), where some posters could coordinate a successful email assault on formal peer review, will not work with the journals that now have my papers. Some of you just delayed the inevitable and just made the ultimate impact that much harder. After all, sci.math invaded the sacrosanct area of formal peer review, and managed to get a correct math paper censored--for a while. And believe me, sci.math will pay the price. The papers are currently at journals. You have some months, maybe, before the full impact hits. But when it does, make no mistake, there will be no place on this planet where you can hide. Remember, I¹m not talking about something vague here. I¹m talking about publication in journals. James Harris === Subject: Re: JSH: Groupthink > The papers are at journals. Some of them have been there for a while. > I think that only groupthink could allow you all to be so conŽdent > now, when the weight of the math world itself will soon fall down upon > you. > Some of you just delayed the inevitable and just made the ultimate > impact that much harder. > After all, sci.math invaded the sacrosanct area of formal peer review, > and managed to get a correct math paper censored--for a while. > And believe me, sci.math will pay the price. > The papers are currently at journals. You have some months, maybe, > before the full impact hits. > But when it does, make no mistake, there will be no place on this > planet where you can hide. > Remember, I¹m not talking about something vague here. I¹m talking > about publication in journals. Hmmm... this sounds familiar. Can it be that you write the speeches for George W.? Listen Harris, you could make a living doing that, but of course, he is going to get his sorry ass kicked out of the white house in November. Another lost career oportunity for you Harris. p.s. How is that Hammer thing going? Won¹t you give us a sneak preview? I can¹t stand the tension anymore... qui non intelligit, aut taceat, aut discat === Subject: Re: JSH: Groupthink > After all, sci.math invaded the sacrosanct area of formal peer review, > and managed to get a correct math paper censored--for a while. Actually, no. The paper was not correct. I¹ll be happy to point the errors out for you--all you have to do is ask. Rick === Subject: Re: JSH: Groupthink > > Oh, God, James, go get a bottle of Zyprexa or whatever it is you¹re missing. > > > > > > Yours, > > > > Doug Goncz ( ftp://users.aol.com/DGoncz/ ) > > > > Student member SAE for one year. > The papers are at journals. Some of them have been there for a while. > The APF paper already passed formal peer review once, and it has been > at a journal for over a month now. It passed some sort of review, but that didn¹t mean it was correct. It isn¹t, at least in the form you¹ve put on sci.math. > I think that only groupthink could allow you all to be so conŽdent > now, when the weight of the math world itself will soon fall down upon > you. Actually it is mathematics that enables us to be so conŽdent that some of your results are false. > What worked with Ioannis Argyros, with a math journal he started, > which may have been part of what clouded his judgement (though that is > no excuse), where some posters could coordinate a successful email > assault on formal peer review, will not work with the journals that > now have my papers. Just which journals are those? If what you say is true then their editors would be immune to whatever they might hear against you. But anyhow, many papers have been refereed and published and still been incorrect. > Some of you just delayed the inevitable and just made the ultimate > impact that much harder. > After all, sci.math invaded the sacrosanct area of formal peer review, > and managed to get a correct math paper censored--for a while. > And believe me, sci.math will pay the price. > The papers are currently at journals. You have some months, maybe, > before the full impact hits. > But when it does, make no mistake, there will be no place on this > planet where you can hide. > Remember, I¹m not talking about something vague here. I¹m talking > about publication in journals. Again, publication in journals doesn¹t mean a whole lot. In particular it doesn¹t mean your results are correct. Please let us know when your papers have been accepted and also when they have been rejected. And tell us which journals in each case so we can feel properly humiliated. > James Harris === Subject: Re: JSH: Groupthink > What worked with Ioannis Argyros, with a math journal he started, > which may have been part of what clouded his judgement (though that is > no excuse), where some posters could coordinate a successful email > assault on formal peer review, will not work with the journals that > now have my papers. Good. Then you won¹t mind naming those journals, since neither you nor your paper are at any risk. > Remember, I¹m not talking about something vague here. I¹m talking > about publication in journals. What journals??? > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Groupthink !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@¹ELIi $t^ VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > The papers are at journals. Some of them have been there for a > while. And that¹s probably where they¹ll stay. In the waste bin. > What worked with Ioannis Argyros, with a math journal he started, > which may have been part of what clouded his judgement (though that > is no excuse), where some posters could coordinate a successful > email assault on formal peer review, will not work with the journals > that now have my papers. > Some of you just delayed the inevitable and just made the ultimate > impact that much harder. > After all, sci.math invaded the sacrosanct area of formal peer review, > and managed to get a correct math paper censored--for a while. > And believe me, sci.math will pay the price. > The papers are currently at journals. You have some months, maybe, > before the full impact hits. > But when it does, make no mistake, there will be no place on this > planet where you can hide. > Remember, I¹m not talking about something vague here. I¹m talking > about publication in journals. Glad you said it. For a while it sounded like you were talking about nuclear bombs. For what it¹s worth: wrong results and shoddy research get published all too often. Peer review often means some reviewer having a stack of paper on his desk, and then Žnally waving something through instead of trying to look closely. happens all the time. If you manage to sneak some nonsense past some editor, the world will not come tumbling down. Likely an acrimonious analysis will be posted by some qualiŽed reader, the journal will acknowledge it and apologize, and the overworked reviewer will not get consulted next time. That¹s what happened the last time, only that the editor was saved the embarrassment of actually going public with his mistake. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Groupthink > James Harris What is the singular of groupthink? Other than James Harris that is. === Subject: Re: JSH: Groupthink > It¹s kind of weird how > people can convince themselves of things they wish to believe and > continue against all evidence. Is it? Examine your own record, and heal thyself. > I¹m curious to see how you do. My guess is that you have less than > six months before all the papers play out, including the paper with > The Hammer. I¹m equally curious about how your ŒHammer¹ will sound when it clanks against the ŒAnvil of Truth¹. Mark the current date! > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Wiles for participation in threads Modular Arithmetic and Periodicity of a^n mod Is this related to Wile¹s proof, or was he referring to a different set of curves? 1: 0 < a < b < c < a+b 1.1: 2 < n 2: a, b, c, pairwise coprime 3: a^n + b^n = c^n 4: A.0 = a^0 mod c 5: B.0 = b^0 mod c 6: A.n = (A.(n-1) * a ) mod c 7: B.n = (B.(n-1) * a ) mod c 8: The period of A is a divisor of phi(c) 9: The period of B is a divisor of phi(c) 10:?The period of (A+B) is a divisor of phi(c) 11: n < phi(c); Pythagoras, 8, 9 11.1: 1 < m < oo 11.2: c = product from i = 1 to m of prime p.i, to the e.i power 11.3: j.i = phi(p.i ^ e.i) = phi( (p.i-1) * (p.i - 1) ); Euler totient formula 11.4: q = LCM of elements j.i; private email 11.5: n <= q; 11.4, 8, 9 12: ( a^n + b^n ) / c^n = 1; 3 13: (a/c)^n + (b/c)^n = 1 To be shown: 9999:?In the a,b plane, the set of curves (13) and {(a,b)} do not intersect Yours, Doug Goncz ( ftp://users.aol.com/DGoncz/ ) Student member SAE for one year. === Subject: Re: Wiles Er, 1: 0 < a < b < a+b < c 1.1: 0 < n 2: a, b, c, pairwise coprime 3: a^n + b^n = c^n 4: A.0 = a^0 mod c 5: B.0 = b^0 mod c 6: A.n = (A.(n-1) * a ) mod c 7: B.n = (B.(n-1) * a ) mod c 8: The period of A is a divisor of phi(c) 9: The period of B is a divisor of phi(c) 10:?The period of (A+B) is a divisor of phi(c) 11: 2 < n < phi(c); Pythagoras, 8, 9 11.1: 1 < m < oo 11.2: c = product from i = 1 to m of prime p.i, to the e.i power 11.3: j.i = phi(p.i ^ e.i) = phi( (p.i-1) * (p.i - 1) ); Euler totient formula 11.4: q = LCM of elements j.i; private email 11.5: 2 < n <= q; 11.4, 8, 9 12: ( a^n + b^n ) / c^n = 1; 3 13: (a/c)^n + (b/c)^n = 1 to be shown: 9999:?In the a,b plane, the set of curves (13) and {(a,b)} do not intersect For visualization: ftp://users.aol.com/Publications/Fermat.avi 207,360 bytes, 9 frames, c = 5 through 13, n = 2...c Yours, Doug Goncz ( ftp://users.aol.com/DGoncz/ ) Student member SAE for one year. === Subject: Re: Wiles I don¹t know why but my pastes don¹t show up. ftp://users.aol.com/DGoncz/Publications/Fermat.txt ftp://users.aol.com/DGoncz/Publications/Fermat.avi .txt shows the argument. It¹s awful. .avi shows the curves. It¹s pretty. Yours, Doug Goncz ( ftp://users.aol.com/DGoncz/ ) Student member SAE for one year. === Subject: Svara that the Twin Primes Conjecture is Unprovable For fun, one could argue that twin primes is unprovable. As to whether the argument is rigorous or not, you be the judge: Let S be the set of all n in which 6n-1 is prime. Let T be the set of all n in which 6n+1 is prime. Proving the twin primes conjecture is equivalent to showing that the intersection of sets S and T is inŽnite. First of all, we know that sets S and T are both inŽnite by Dirichlet¹s Theorem. Now, the condition which deŽnes set S, that 6n-1 is prime for each n in S, indicates nothing about the factors of 6n+1 when n is in S, and therefore nothing about whether 6n+1 is prime when n is in S. And the condition which deŽnes set T, that 6n+1 is prime for each n in T, indicates nothing about the factors of 6n-1 when n is in T, and therefore nothing about whether 6n-1 is prime when n is in T. Therefore, the two conditions which deŽne the two sets are independent from one another, meaning that the only way to determine the cardinality of the intersection of the two sets, S and T, is to directly calculate the elements in the intersection of S and T. But this would take an inŽnite amount of time, since one would have to test each natural number n. Therefore, Twin Primes is unprovable. QED For those of you who don¹t buy the argument, let me explain this in which twins were born in hospital A. And deŽne set T as the days in different hospitals in different parts of the world.) The conditions which deŽne sets S and T have nothing to do with one another, i.e., they are independent from one another. Therefore, the only way to determine the cardinality of the intersection of sets S and T is to directly calculate the elements in the intersection of sets S and T. This is obviously feasible to do since the number of days in the year that twins were born in hospitals A and B respectively from the year if the intersection of these sets is inŽnite or not, since one would have to wait an eternity to do such. Anyway, I welcome comments. Craig === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable >For fun, one could argue that twin primes is unprovable. As to whether >the argument is rigorous or not, you be the judge: >Let S be the set of all n in which 6n-1 is prime. >Let T be the set of all n in which 6n+1 is prime. >Proving the twin primes conjecture is equivalent to showing that the >intersection of sets S and T is inŽnite. >First of all, we know that sets S and T are both inŽnite by >Dirichlet¹s Theorem. Now, the condition which deŽnes set S, that 6n-1 >is prime for each n in S, indicates nothing about the factors of 6n+1 >when n is in S, and therefore nothing about whether 6n+1 is prime when >n is in S. >And the condition which deŽnes set T, that 6n+1 is prime for each n >in T, indicates nothing about the factors of 6n-1 when n is in T, and >therefore nothing about whether 6n-1 is prime when n is in T. >Therefore, the two conditions which deŽne the two sets are >independent from one another, meaning that the only way to determine >the cardinality of the intersection of the two sets, S and T, is to >directly calculate the elements in the intersection of S and T. But >this would take an inŽnite amount of time, since one would have to >test each natural number n. Therefore, Twin Primes is unprovable. QED Yes, it¹s nonsense. Even if everything up to there were correct, the last paragraph, about how therefore the only way to determine the cardinality of the intersection is to calculate all the elements, which would take inŽnitely much time, is simply unsupported. It makes just as much sense to say that the fact that there are inŽnitely many primes is unprovable, because the only way to verify this is to test every number for primality, and that would take inŽnitely much time. The fact that you don¹t see another way to prove it doesn¹t show that none exists. ************************ David C. Ullrich === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable > For fun, one could argue that twin primes is unprovable. As to whether > the argument is rigorous or not, you be the judge: > Let S be the set of all n in which 6n-1 is prime. > Let T be the set of all n in which 6n+1 is prime. > Proving the twin primes conjecture is equivalent to showing that the > intersection of sets S and T is inŽnite. > First of all, we know that sets S and T are both inŽnite by > Dirichlet¹s Theorem. Now, the condition which deŽnes set S, that 6n-1 > is prime for each n in S, indicates nothing about the factors of 6n+1 > when n is in S, and therefore nothing about whether 6n+1 is prime when > n is in S. > And the condition which deŽnes set T, that 6n+1 is prime for each n > in T, indicates nothing about the factors of 6n-1 when n is in T, and > therefore nothing about whether 6n-1 is prime when n is in T. Nonsense. Go research Hardy & Littlewood. > Anyway, I welcome comments. Sheesh, that¹s a tough one. How about *_PLONK_*? Phil -- 1st bug in MS win2k source code found after 20 minutes: scanline.cpp 2nd and 3rd bug found after 10 more minutes: gethost.c Both non-exploitable. (The 2nd/3rd ones might be, depending on the CRTL) === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable >For fun, one could argue that twin primes is unprovable. As to whether >the argument is rigorous or not, you be the judge: ... >And the condition which deŽnes set T, that 6n+1 is prime for each n >in T, indicates nothing about the factors of 6n-1 when n is in T, and >therefore nothing about whether 6n-1 is prime when n is in T. Nonsense. For example, the assumption that 6n+1 is prime makes 6n-1 more likely to be divisible by any given prime > 3. >Therefore, the two conditions which deŽne the two sets are >independent from one another, meaning that the only way to determine >the cardinality of the intersection of the two sets, S and T, is to >directly calculate the elements in the intersection of S and T. But >this would take an inŽnite amount of time, since one would have to >test each natural number n. Therefore, Twin Primes is unprovable. QED Double nonsense. You could probably argue just as well that the following statement is unprovable: There are inŽnitely many primes p such that p+2 is either prime or the product of two primes. But that was proved by Chen Jingrun in 1966. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable > >For fun, one could argue that twin primes is unprovable. As to whether > >the argument is rigorous or not, you be the judge: > ... > >And the condition which deŽnes set T, that 6n+1 is prime for each n > >in T, indicates nothing about the factors of 6n-1 when n is in T, and > >therefore nothing about whether 6n-1 is prime when n is in T. > Nonsense. For example, the assumption that 6n+1 is prime makes 6n-1 more > likely to be divisible by any given prime > 3. Let me clarify my response before. I think that your response is the same reasoning as someone who carries a bomb on a train for safety reasons, because the odds of two people with a bomb on a train are less than one person with a bomb on a plane. Your statement is not accurate. Yes, the factors of 6n+1 cannot be factors of 6n-1. So if 6n+1 is prime, then the list of possible factors of 6n-1 is greater than the list of possible factors of 6n-1 if 6n+1 were composite. But still, this tells no information about whether 6n-1 is prime or composite; it only eliminates the possibility that some numbers can be eliminated as possible factors of 6n-1. Craig === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable >> >And the condition which deŽnes set T, that 6n+1 is prime for each n >> >in T, indicates nothing about the factors of 6n-1 when n is in T, and >> >therefore nothing about whether 6n-1 is prime when n is in T. >> Nonsense. For example, the assumption that 6n+1 is prime makes 6n-1 more >> likely to be divisible by any given prime > 3. >Let me clarify my response before. I think that your response is the >same reasoning as someone who carries a bomb on a train for safety >reasons, because the odds of two people with a bomb on a train are >less than one person with a bomb on a plane. Not at all the same. The fallacy in that one is the confusion of the probability Prob(A and B) with the conditional probability Prob(A | B). The conditional probability that someone else has a bomb, given that I have a bomb, is not less than the conditional probability given that I don¹t have a bomb. In fact, you could say it¹s greater: if I was able to get on with a bomb there probably weren¹t effective security measures, and this makes it easier for others to get on the train with a bomb. Getting back to primes, the _conditional_ probability [for reasonable interpretations of that term in this context] that 6n+1 is prime given that 6n-1 is prime is less than the conditional probability that 6n+1 is prime given that 6n-1 is composite. For example, suppose you take the Žrst 100,000 positive integers for n. Of the 24572 such n for which 6n-1 is prime, 5330 or about 21.69% have 6n+1 prime. Of the 75428 for which 6n-1 is composite, 19194 or about 25.45% have 6n+1 prime. I would expect similar results if you replaced 100,000 by any reasonably large positive integer N. Conclusion: having 6n-1 be prime makes it less likely that 6n+1 is prime. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable >>> >And the condition which deŽnes set T, that 6n+1 is prime for each n >>> >in T, indicates nothing about the factors of 6n-1 when n is in T, and >>> >therefore nothing about whether 6n-1 is prime when n is in T. >>> Nonsense. For example, the assumption that 6n+1 is prime makes 6n-1 more >>> likely to be divisible by any given prime > 3. >>Let me clarify my response before. I think that your response is the >>same reasoning as someone who carries a bomb on a train for safety >>reasons, because the odds of two people with a bomb on a train are >>less than one person with a bomb on a plane. >Not at all the same. The fallacy in that one is the confusion of >the probability Prob(A and B) with the conditional probability >Prob(A | B). The conditional probability that someone else has a >bomb, given that I have a bomb, is not less than the conditional >probability given that I don¹t have a bomb. In fact, you could say >it¹s greater: if I was able to get on with a bomb there probably >weren¹t effective security measures, and this makes it easier for >others to get on the train with a bomb. >Getting back to primes, the _conditional_ probability [for reasonable >interpretations of that term in this context] that 6n+1 is prime >given that 6n-1 is prime is less than the conditional probability >that 6n+1 is prime given that 6n-1 is composite. For example, suppose >you take the Žrst 100,000 positive integers for n. Of the 24572 such >n for which 6n-1 is prime, 5330 or about 21.69% have 6n+1 prime. >Of the 75428 for which 6n-1 is composite, 19194 or about 25.45% >have 6n+1 prime. I would expect similar results if you replaced >100,000 by any reasonably large positive integer N. Conclusion: >having 6n-1 be prime makes it less likely that 6n+1 is prime. (Not that I see why we¹re worrying about this (ok, I suppose that it does have some relevance if one wanted to think about the twin prime problem, but there are much more absurd things in the OP than the bit about these sets being Œindependent¹), but) are there results that actually show that similar things _do_ happen as N -> inŽnity? >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 ************************ David C. Ullrich === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable >but) are there results that actually show that similar things >_do_ happen as N -> inŽnity? I¹m not sure what you mean by similar things here. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable > >For fun, one could argue that twin primes is unprovable. As to whether > >the argument is rigorous or not, you be the judge: > ... > >And the condition which deŽnes set T, that 6n+1 is prime for each n > >in T, indicates nothing about the factors of 6n-1 when n is in T, and > >therefore nothing about whether 6n-1 is prime when n is in T. > Nonsense. For example, the assumption that 6n+1 is prime makes 6n-1 more > likely to be divisible by any given prime > 3. This critique is nonsense. What does one have to do with the other? What do the factors of 6n+1 have to do with the factors of 6n-1? > >Therefore, the two conditions which deŽne the two sets are > >independent from one another, meaning that the only way to determine > >the cardinality of the intersection of the two sets, S and T, is to > >directly calculate the elements in the intersection of S and T. But > >this would take an inŽnite amount of time, since one would have to > >test each natural number n. Therefore, Twin Primes is unprovable. QED > Double nonsense. You could probably argue just as well that the following > statement is unprovable: > There are inŽnitely many primes p such that p+2 is either > prime or the product of two primes. > But that was proved by Chen Jingrun in 1966. OK, let¹s hear an argument that There are inŽnitely many primes p such that p+2 is either prime or the product of two primes. is unprovable. Probably doesn¹t cut it. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Svara that the Twin Primes Conjecture is Unprovable >> >For fun, one could argue that twin primes is unprovable. As to whether >> >the argument is rigorous or not, you be the judge: >> ... >> >And the condition which deŽnes set T, that 6n+1 is prime for each n >> >in T, indicates nothing about the factors of 6n-1 when n is in T, and >> >therefore nothing about whether 6n-1 is prime when n is in T. >> Nonsense. For example, the assumption that 6n+1 is prime makes 6n-1 more >> likely to be divisible by any given prime > 3. > This critique is nonsense. What does one have to do with the other? > What do the > factors of 6n+1 have to do with the factors of 6n-1? No. One might argue as follows. Suppose that 6n + 1 is prime. Then it isn¹t a multiple of 5: it is congruent to 1, 2, 3 or 4 modulo 5. Hence 6n - 1 is congruent respectively to 4, 0, 1 or 2 modulo 5. Hence it has a probability 1/4 that it is a multiple of 5, unlike the typical integer which has a probability 1/5 that it is a multiple of 5. Incidentally, are you going to relabel your proof of the unprovability of the Collatz conjecture in arXiv/math.GM as a svara or sefara or (whatever the spelling of the week is)? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Analysis Differentiation question X-AUTHid: shrest10 Let f be differentiable on [a,b] (some closed interval). How would I go about constructing a function so that the range of f is a). an open interval b). open on one side and closed on the other side interval. === Subject: Re: Analysis Differentiation question >Let f be differentiable on [a,b] (some closed interval). >How would I go about constructing a function so that the range of f is >a). an open interval >b). open on one side and closed on the other side interval. This is clearly impossible, since f is continuous, so its range is compact (and connected, hence a closed interval). William says something about a correction, but your two posts look identical to me... ************************ David C. Ullrich === Subject: Re: Analysis Differentiation question X-AUTHid: shrest10 > >Let f be differentiable on [a,b] (some closed interval). > >How would I go about constructing a function so that the range of f is > >a). an open interval > >b). open on one side and closed on the other side interval. > This is clearly impossible, since f is continuous, so its > range is compact (and connected, hence a closed interval). > William says something about a correction, but your > two posts look identical to me... Sorry I did mean the range of f¹ not f. Of course, it wouldn¹t make any sense for f to map some closed interval to an open interval since it¹s continuous. My idea was to somehow make f¹ discontinuous at the end-points. But, the Darboux property says I can¹t start off with f¹ with a removable or jump discontinuity. However, the only discontinuity of the derivative I¹ve learnt about is the form of sin(1/x) kind. I was having a hard time trying to construct a function where f¹ = sin(1/(x-a)) * sin (1/(b-x)) and thought maybe it was a more simpler problem and I wasn¹t seeing the answer. === Subject: Re: Analysis Differentiation question >> >Let f be differentiable on [a,b] (some closed interval). >> > >> >How would I go about constructing a function so that the range of f is >> > >> >a). an open interval >> >b). open on one side and closed on the other side interval. >> This is clearly impossible, since f is continuous, so its >> range is compact (and connected, hence a closed interval). >> William says something about a correction, but your >> two posts look identical to me... >Sorry I did mean the range of f¹ not f. >Of course, it wouldn¹t make any sense for f to map some closed interval to >an open interval since it¹s continuous. >My idea was to somehow make f¹ discontinuous at the end-points. But, the >Darboux property says I can¹t start off with f¹ with a removable or jump >discontinuity. However, the only discontinuity of the derivative I¹ve learnt >about is the form of sin(1/x) kind. >I was having a hard time trying to construct a function where f¹ = >sin(1/(x-a)) * sin (1/(b-x)) and thought maybe it was a more simpler problem >and I wasn¹t seeing the answer. Well, I don¹t see how f¹ = sin(1/(x-a)) * sin (1/(b-x)) is going to do it, and you should note that Wade has already given a big hint. But for future reference, you seem to be missing something. If some time you _do_ want to construct a function f with f¹ = sin(1/(x-a)) * sin (1/(b-x)), you do so like so: Let f(x) = int_a^x sin(1/(t-a)) * sin (1/(b-t)) dt. (int_a^x means integral from a to x.) David C. Ullrich === Subject: Re: Analysis Differentiation question > Let f be differentiable on [a,b] (some closed interval). > How would I go about constructing a function so that the range of f is > a). an open interval > b). open on one side and closed on the other side interval. As before, since f is continuous and [a,b] compact f([a,b]) is compact, hence closed. Same with f¹, now that you¹ve corrected youself, were f¹ continuous, then being deŽned for all of [a,b], f¹([a,b]) would also be closed. Thus the functions you seek would be continuous on [a,b] be differentiable on [a,b] have a discontinuous f¹ === Subject: Analysis Differentiation question X-AUTHid: shrest10 Let f be differentiable on [a,b] (some closed interval). How would I go about constructing a function so that the range of f is a). an open interval b). open on one side and closed on the other side interval. === Subject: Re: Another Dini Derivate question X-AUTHid: shrest10 > don¹t quite understand the relationship between countable/uncountable > sets and Dini Derivates. > I don¹t even know where to start on this question - > Show that the set > {D^+ f(x) < D_- f(x)} > cannot be uncountable. > Also, give an example where {D^+ f(x) < D_- f(x)} on an inŽnite set. > Here, > D^+ f(x) = lim sup _ {x -> x0+} (f(x)-f(x0))/(x-x0) > the upper right dini derivate > and, > D_- f(x) = lim inf _ { x -> x0-} (f(x)-f(x0))/(x-x0) > the lower left dini derivate No-one has any idea on this? === Subject: Re: Another Dini Derivate question > No-one has any idea on this? They just did not decide to do your work for you. === Subject: Re: Another Dini Derivate question X-AUTHid: shrest10 > > No-one has any idea on this? > They just did not decide to do your work for you. Sorry, I forgot to mention that this isn¹t HW or test question. I¹m just doing going through all the problems of my Analysis Textbook and was hoping someone could give me a hint on how to solve this. Here¹s the problem again. Show that the set {D^+ f(x) < D_- f(x)} cannot be uncountable. Also, give an example where {D^+ f(x) < D_- f(x)} on an inŽnite set. Here, D^+ f(x) = lim sup _ {x -> x0+} (f(x)-f(x0))/(x-x0) the upper right dini derivate and, D_- f(x) = lim inf _ { x -> x0-} (f(x)-f(x0))/(x-x0) the lower left dini derivate === Subject: Re: Analysis Differentiation question > Let f be differentiable on [a,b] (some closed interval). > How would I go about constructing a function so that the range of f is > a). an open interval > b). open on one side and closed on the other side interval. First you need to travel to an alternate universe. In this one, differentiability implies continuity, and the continuous image of a compact set is compact === Subject: Re: Analysis Differentiation question X-AUTHid: shrest10 > > Let f be differentiable on [a,b] (some closed interval). > > How would I go about constructing a function so that the range of f > is > > a). an open interval > > b). open on one side and closed on the other side interval. > First you need to travel to an alternate universe. In this one, > differentiability implies continuity, and the continuous image of a > compact set is compact You must be mis-understanding my question. I¹m not saying the derivative has to be continuous !!! [a,b], then the set of derivatives must also be an interval. I¹m asking how we can construct a function f and show that the range of it¹s derivative is an open interval. === Subject: Re: Analysis Differentiation question >> > Let f be differentiable on [a,b] (some closed interval). >> > >> > How would I go about constructing a function so that the range of f >> is >> > >> > a). an open interval >> > b). open on one side and closed on the other side interval. >> First you need to travel to an alternate universe. In this one, >> differentiability implies continuity, and the continuous image of a >> compact set is compact >You must be mis-understanding my question. >I¹m not saying the derivative has to be continuous !!! Go back and read the problem you posted. >[a,b], then the set of derivatives must also be an interval. I¹m asking how >we can construct a function f and show that the range of it¹s derivative is >an open interval. You may be asking that _now_, but it¹s certainly not what you asked in your original post. ************************ David C. Ullrich === Subject: Re: Analysis Differentiation question > You must be mis-understanding my question. You misstated the problem. > I¹m not saying the derivative has to be continuous !!! > [a,b], then the set of derivatives must also be an interval. I¹m asking how > we can construct a function f and show that the range of it¹s derivative is > an open interval. Think of a continuous function g on (0,1] that is 0 except for a sequence of tents T_n with very small bases b_n (something like b_n = 1/2^n) and heights h_n = 1 - 1/n. Now set f(x) = int_[0,x] g(t) dt. Then f¹ = g on (0,1]. Furthermore f¹(0) = 0. It follows that the range of f¹ is [0,1). Modifying this a little will give the range of f¹ = (0,1). === Subject: Re: Logic Book > Perhaps, but I don¹t see the evidence for that here. What I see is > that readability, like beauty, is in the eye of the beholder. I > found much of Bourbaki to be an easy read, even though it wasn¹t in > my native language, but A Tale of Two Cities was a far duller book > than I had ever read. It is a far, far duller book that I read, than I have ever read... -- http://hertzlinger.blogspot.com === Subject: Re: Masonic InŽltrated Churches >> And I spoke out in 1988 against the churches and >> media censoring acts of cannibalism and Scriptural passages >> referring to cannibalism, then for years I had to listen to >> psychiatrists telling people on the psychiatric appeal panel >> hearings that I was obsessed with penises and cannibals (but >> it is not me that thinks so much about penises that I would >> place them on the roofs of churches, and it is not me that >> thinks so much about cannibals that I would systematically >> censor all Scriptural references to them). > Dammit, my Patented Kook-O-Meter blew a fuse. To borrow a line from > Maxwell Smart: That¹s the third time that¹s happened this > week... :O| The Freemasons are behind everything. For example, as part of their scheme to Žx energy prices they placed Satanic symbols into the street map of Washington, DC in order to send coded messages to the Elders of Zion (headquarters at 666 Fifth Avenue in Manhattan) to back the leveraged buyout of renegade Objectivist extraterrestrials from Zeta Reticuli who might otherwise tell us how to make cold-fusion reactors reliable. They are deliberately Žddling with energy prices to keep us from being able to afford to achieve humanity¹s highest aspiration, namely the attempt to BLOW UP THE MOON Is there some way to work Enron and Halliburton into this? -- http://hertzlinger.blogspot.com === Subject: Re: Masonic InŽltrated Churches Joseph Hertzlinger > The Freemasons are behind everything. I don¹t think they proved FLT. === Subject: Re: Masonic InŽltrated Churches >> And I spoke out in 1988 against the churches and >> media censoring acts of cannibalism and Scriptural passages >> referring to cannibalism, then for years I had to listen to >> psychiatrists telling people on the psychiatric appeal panel >> hearings that I was obsessed with penises and cannibals (but Hey, you¹re the one showing me dirty pictures! >> it is not me that thinks so much about penises that I would >> place them on the roofs of churches, and it is not me that >> thinks so much about cannibals that I would systematically >> censor all Scriptural references to them). > Dammit, my Patented Kook-O-Meter blew a fuse. To borrow a line from > Maxwell Smart: That¹s the third time that¹s happened this > week... :O| I forgot about Monty Python on Freemasons: | Well, of course, this is just the sort of blinkered philistine | ignorance I¹ve come to expect from you non-creative garbage. You sit | there on your loathsome spotty behinds, squeezing your blackheads, | not caring a tinker¹s cuss for the struggling artist. You excrement! | You whining hypocritical toadies with your colour TVs and your Tony | Jacklin golf clubs and your bleeding masonic secret handshakes! You | wouldn¹t let me join, would you, you blackballing bastards? Well I | wouldn¹t become a Freemason if you went down on your stinking knees | and begged me! -- http://hertzlinger.blogspot.com === Subject: Re: Masonic InŽltrated Churches > BLOW UP THE MOON Channeling Abian, are we? Rick === Subject: JSH: Sweep likely At this point I¹m fairly conŽdent that I¹ll get complete vindication from more than one major math journal. It should all play out in less than six months. I¹m curious about how people on this newsgroup will react to hearing that I was right and others were wrong. Will you accept it as groupthink, or maybe come up with some other answer? Will your faith in your own mathematical ability be shaken? Will any of you want accountability from posters like David Ullrich or Arturo Magidin when the mainstream mathematicians stand by me and repudiate both them, their tactics and their claims? Oh, and in case you¹re wondering, yes, a Ph.D can be taken away for gross fraud. In a little while, Magidin and Ullrich may no longer have Ph.D¹s. It¹s that serious. James Harris === Subject: Re: JSH: Sweep likely > At this point I¹m fairly conŽdent that I¹ll get complete vindication > from more than one major math journal. Your level of conŽdence is not a reliable indicator of anything, other than your willingness to convince yourself that you¹re always correct. > It should all play out in less than six months. Well, we can always hope that all this will be over and done with in less than six months, but I think you¹re hooked on the fame and adulation. Every other time I was sure you were gone, you just came right back. > I¹m curious about how people on this newsgroup will react to hearing > that I was right and others were wrong. I suppose. As for me, I have no doubt whatsoever about how you will react to hearing yet again that you are not only wrong, but not even in the game. > Will you accept it as groupthink, or maybe come up with some other > answer? Will you accept the verdict handed you as being valid, or will you rail against the widespread corruption of the mathematics establishment? > Will your faith in your own mathematical ability be shaken? Well, how about yours? > Will any of you want accountability from posters like David Ullrich or > Arturo Magidin when the mainstream mathematicians stand by me and > repudiate both them, their tactics and their claims? Yes, I can see the mainstream mathematicians standing by you. I can just imagine some mathematician coming forward and confessing, in typical Gulag fashion, his thoughtcrime in having believed Galois truth that Galois theory was overinterpreted. > Oh, and in case you¹re wondering, yes, a Ph.D can be taken away for > gross fraud. How many examples of this have you discovered? How many were *not* due to fraud in obtaining the Ph.D. itself (e.g., plagiarism, faked or fudged experimental data) ? How many were due to misbehavior unrelated to the details of their Ph.D.s? This, of course, is ignoring the fact that the true fraud would have been in supporting your žawed results. > In a little while, Magidin and Ullrich may no longer have Ph.D¹s. Whew!! My Ph.D. is safe! I was sure sweating there for a minute. > It¹s that serious. No evidence, overblown claims, veiled threats: the hallmark of modern scientiŽc documentation. I guess you must be feeling pretty secure in that BS degree of yours. > James Harris Dale === Subject: Re: JSH: Sweep likely Discussion, linux) >> In a little while, Magidin and Ullrich may no longer have Ph.D¹s. > Whew!! My Ph.D. is safe! I was sure sweating there for a minute. >> It¹s that serious. > No evidence, overblown claims, veiled threats: the hallmark of > modern scientiŽc documentation. Here I have to disagree. There are no veiled threats here at all. The threats are quite explicit, going so far as to name names and describe the punishment. -- Jesse F. Hughes Ultimately, I can bring the entire mathematical establishment to its knees... Live in a fantasy world if you wish, but to me that¹s just an expression of your intellectual inferiority. --James Harris === Subject: Re: JSH: Sweep likely > There seems to be a typo in the Subject line. I assume that it should be JSH: Weeps likely -- Clive Tooth http://www.clivetooth.dk === Subject: Re: JSH: Sweep likely >At this point I¹m fairly conŽdent that I¹ll get complete vindication >from more than one major math journal. >It should all play out in less than six months. >I¹m curious about how people on this newsgroup will react to hearing >that I was right and others were wrong. When that happens they will be so distracted by all the žying pigs that they won¹t have time to say anything about your situation, sorry. >Will you accept it as groupthink, or maybe come up with some other >answer? >Will your faith in your own mathematical ability be shaken? >Will any of you want accountability from posters like David Ullrich or >Arturo Magidin when the mainstream mathematicians stand by me and >repudiate both them, their tactics and their claims? >Oh, and in case you¹re wondering, yes, a Ph.D can be taken away for >gross fraud. Really? (Hint: The US is not Germany.) >In a little while, Magidin and Ullrich may no longer have Ph.D¹s. Oh my. This worries me even more than all the other similar statements about the Consequences when the Truth Finally Comes Out that you¹ve made over the years. I¹m almost worried enough to decide to admit that you¹ve been right all along about everything. (Alas it¹s _impossible_ to say that you¹ve been right about everything, because of all the times that you¹ve eventually agreed that something you said was wrong - the orginal and the retraction can¹t both be correct, so if I said you¹d been right about everything I¹d disappear from the mathematical universe in a žash of inconsistency.) >It¹s that serious. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Sweep likely > At this point I¹m fairly conŽdent that I¹ll get complete vindication > from more than one major math journal. > It should all play out in less than six months. > I¹m curious about how people on this newsgroup will react to hearing > that I was right and others were wrong. > Will you accept it as groupthink, or maybe come up with some other > answer? > Will your faith in your own mathematical ability be shaken? > Will any of you want accountability from posters like David Ullrich or > Arturo Magidin when the mainstream mathematicians stand by me and > repudiate both them, their tactics and their claims? > Oh, and in case you¹re wondering, yes, a Ph.D can be taken away for > gross fraud. > In a little while, Magidin and Ullrich may no longer have Ph.D¹s. > It¹s that serious. > James Harris Have you been skipping your pills again Harris? You know what they told you, don¹t you? You¹ll be back in for a long while if you don¹t watch out. qui non intelligit, aut taceat, aut discat === Subject: Re: JSH: Sweep likely > At this point I¹m fairly conŽdent that I¹ll get complete vindication > from more than one major math journal. I¹m fairly conŽdent I¹ll be marrying Julia Roberts. === Subject: Re: JSH: Sweep likely > > At this point I¹m fairly conŽdent that I¹ll get complete vindication > > from more than one major math journal. > I¹m fairly conŽdent I¹ll be marrying Julia Roberts. Poor guy, you must be really desperate... === Subject: Re: JSH: Sweep likely !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@¹ELIi $t^ VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > At this point I¹m fairly conŽdent that I¹ll get complete > vindication from more than one major math journal. You¹ve never run short on claims. > I¹m curious about how people on this newsgroup will react to hearing > that I was right and others were wrong. We¹ve been hearing it for several years from you now. It does not get true by repetition. > Oh, and in case you¹re wondering, yes, a Ph.D can be taken away for > gross fraud. It is the tragedy of your life that a Ph.D. can¹t be issued for gross fraud. > In a little while, Magidin and Ullrich may no longer have Ph.D¹s. > It¹s that serious. You are being delusional again. It is not the Žrst time. Take a look at your expansive record of similar claims and your reasons for stating them. And how often you have been eating your words afterwards. Just swallow your medication instead for once. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Sweep likely > > In a little while, Magidin and Ullrich may no longer have Ph.D¹s. > > It¹s that serious. > You are being delusional again. It is not the Žrst time. Take a > look at your expansive record of similar claims and your reasons for > stating them. And how often you have been eating your words > afterwards. Just swallow your medication instead for once. He may be too far gone to be helped by medications. More likely he needs an exorcist. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Sweep likely We will be waiting anxiously for this to come to pass ;-) > At this point I¹m fairly conŽdent that I¹ll get complete vindication > from more than one major math journal. > It should all play out in less than six months. > I¹m curious about how people on this newsgroup will react to hearing > that I was right and others were wrong. > Will you accept it as groupthink, or maybe come up with some other > answer? > Will your faith in your own mathematical ability be shaken? > Will any of you want accountability from posters like David Ullrich or > Arturo Magidin when the mainstream mathematicians stand by me and > repudiate both them, their tactics and their claims? > Oh, and in case you¹re wondering, yes, a Ph.D can be taken away for > gross fraud. > In a little while, Magidin and Ullrich may no longer have Ph.D¹s. > It¹s that serious. > James Harris === Subject: Re: JSH: Sweep likely Well the papers are at journals. Each is going through peer review. With multiple papers at major journals I¹m not worried about a group email strategy working like it did with Ioannis Argyros. And yes APF is one of the papers, currently at a journal where it has been for more than a month, as referees are forced because of Argyros¹s failure and the sci.math email attack, to review a paper that already passed formal peer review. I think only groupthink can allow so many of you to apparently remain conŽdent given the facts at hand. When those papers publish, and the full story comes out, then sci.math, along with posters like David Ullrich and Arturo Magidin will be under intense scrutiny. I feel conŽdent that Ullrich and Magidin in particular have behaved in a way that will mean the loss of their Ph.D¹s. There are consequences to not behaving professionally. In any event, there may be a few months to wait, but I project less than six months. So, sci.math, and posters like Ullrich and Magidin, you may have some months left. James Harris > We will be waiting anxiously for this to come to pass ;-) > > At this point I¹m fairly conŽdent that I¹ll get complete vindication > > from more than one major math journal. > > It should all play out in less than six months. > > I¹m curious about how people on this newsgroup will react to hearing > > that I was right and others were wrong. > > Will you accept it as groupthink, or maybe come up with some other > > answer? > > Will your faith in your own mathematical ability be shaken? > > Will any of you want accountability from posters like David Ullrich > or > > Arturo Magidin when the mainstream mathematicians stand by me and > > repudiate both them, their tactics and their claims? > > Oh, and in case you¹re wondering, yes, a Ph.D can be taken away for > > gross fraud. > > In a little while, Magidin and Ullrich may no longer have Ph.D¹s. > > > > It¹s that serious. > > > > > > James Harris === Subject: Re: JSH: Sweep likely James, You should consider it a blessing what happened to you with the journal that was forced through pressure from math.sci to withdraw your paper that passed peer review. Look at history. The best way to get people to read a paper or a book is to ban it. Since your paper was banned from the journal because of pressure from math.sci, more people will likely read it, which I assume you want. If it were published in the journal, then who cares? Who reads math journals anyway? I have a friend that I work with who used to be a mathematics is <1 (outside of the author and referees). Craig === Subject: Re: JSH: Sweep likely > James, > You should consider it a blessing what happened to you with the > journal that was forced through pressure from math.sci to withdraw > your paper that passed peer review. Look at history. The best way to > get people to read a paper or a book is to ban it. Since your paper > was banned from the journal because of pressure from math.sci, Was it? > more > people will likely read it, which I assume you want. > If it were published in the journal, then who cares? Who reads math > journals anyway? I have a friend that I work with who used to be a > mathematics is <1 (outside of the author and referees). Nah! That should *include* the author and referees. :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: JSH: Sweep likely > Well the papers are at journals. Each is going through peer review. > With multiple papers at major journals I¹m not worried about a group > email strategy working like it did with Ioannis Argyros. And yes APF > is one of the papers, currently at a journal where it has been for > more than a month, as referees are forced because of Argyros¹s failure > and the sci.math email attack, to review a paper that already passed > formal peer review. If that paper has been rejected by Argyros¹ journal, the current journal has to set the referees on your paper in any case. Then again, how much work will these probably spend on your paper, in view of its history? > When those papers publish, and the full story comes out, then > sci.math, along with posters like David Ullrich and Arturo Magidin > will be under intense scrutiny. The full story is in the very paper, that has been published previously and has been shown to be deŽcient previously. participants wanted your paper to be removed from the other journals site, but rather that the editors should set the record straight. Marc === Subject: Re: JSH: Sweep likely Discussion, linux) > If that paper has been rejected by Argyros¹ journal, the current > journal has to set the referees on your paper in any case. > Then again, how much work will these probably spend on your paper, > in view of its history? Why do you think that the referees will know the history of the paper? Somehow, I doubt that the travails of James S. Harris are widely discussed outside of this group. -- After years of arguing I realize that your intellects are too limited to fully grasp my work. [...] Still, no matter how child-like your minds are, [...] since you have language, [...] there¹s a chance that I¹ll be able to Žnd something that your minds can handle. --JSH === Subject: Re: JSH: Sweep likely > > If that paper has been rejected by Argyros¹ journal, the current > > journal has to set the referees on your paper in any case. > > Then again, how much work will these probably spend on your paper, > > in view of its history? > Why do you think that the referees will know the history of the paper? > Somehow, I doubt that the travails of James S. Harris are widely > discussed outside of this group. Maybe someone could send them an email... === Subject: Re: JSH: Sweep likely Discussion, linux) > Well the papers are at journals. Each is going through peer review. > With multiple papers at major journals I¹m not worried about a group > email strategy working like it did with Ioannis Argyros. And yes APF > is one of the papers, currently at a journal where it has been for > more than a month, as referees are forced because of Argyros¹s failure > and the sci.math email attack, to review a paper that already passed > formal peer review. Suppose one or more of the papers is rejected. I know it sounds unlikely, given such strong results beautifully presented (I assume), but just suppose so. Will you announce it here? You¹re not under any real obligation, but since you¹ve been predicted such dire outcomes once the review is done, you should let us know when the threat is postponed, don¹t you think? -- I have to break the code of how [mere humans] work, and I have made a lot of progress over years of effort, and I feel like I am close to Žguring out all the inner details of human wiring. -- James S. Harris on the extra problems of conveying his research === Subject: Re: JSH: Sweep likely > I feel conŽdent that Ullrich and Magidin in particular have behaved > in a way that will mean the loss of their Ph.D¹s. Mr. Harris, when you are proven wrong (once again, for the millionth time), will it mean the loss of your welfare BS in physics? Obviously, you are not deserving of any degree as you haven¹t any emotional or scientiŽc intelligence! You should feel so embarrassed with yourself for even making such a claim. You may also consider seeing a doctor as you seem to be allowing the desire for glory distort just how wrong you continue to be. You might end up in a straight jacket before it is all said and done. === Subject: Re: JSH: Sweep likely > > I feel conŽdent that Ullrich and Magidin in particular have behaved > > in a way that will mean the loss of their Ph.D¹s. > Mr. Harris, > when you are proven wrong (once again, for the millionth time), will it mean > the loss of your welfare BS in physics? I hope James has the dignity not to respond to that. The not-so-subtle racism of welfare BS doesn¹t belong in this discussion. > Obviously, you are not deserving of any degree as you haven¹t any emotional > or scientiŽc intelligence! I too have my doubts given some of the statements he¹s made about physics, but I¹ve so far given him the beneŽt of the doubt. Let¹s just agree for the sake of argument that at one time James was rational enough to earn a BS in physics from Vanderbilt, and keep the discussion on the subject of his current writings, OK? - Randy === Subject: Re: JSH: Sweep likely > > I feel conŽdent that Ullrich and Magidin in particular have behaved > > in a way that will mean the loss of their Ph.D¹s. > Mr. Harris, > when you are proven wrong (once again, for the millionth time), will it mean > the loss of your welfare BS in physics? > Obviously, you are not deserving of any degree as you haven¹t any emotional > or scientiŽc intelligence! > You should feel so embarrassed with yourself for even making such a claim. > You may also consider seeing a doctor as you seem to be allowing the desire > for glory distort just how wrong you continue to be. > You might end up in a straight jacket before it is all said and done. Remember, I¹m not talking about something vague here, as I¹m talking about publication. Call me crazy if you wish as it won¹t make any difference. All the little social games that seemed to work on sci.math, never did, as I¹ve just been busily researching, reŽning my results, and writing papers. It can just take a while with math research. Some of that reŽnement of my results can be seen by just looking over posts where I would argue out different areas with posters just to see if there wasn¹t something I missed, despite all the insults being tossed at me. I kept at it despite the social žak. Math is important. Social issues come and go. Now I¹ve Žnally sent off the major papers, including re-sending APF for yet another round of formal peer review, which is extra work for some mathematicians required by sci.math posters who interfered with the journal process. So sci.math will pay, and so will posters who acted in a way that I¹ll show is gross fraud, requiring that they be stripped of their Ph.D¹s. The initial repudiation of this group, and of those posters will come from journals. Then the rest of the impact should come from the math community itself, as it moves to clean house, and try to make sure that something like this never happens again. Mathematicians who post on newsgroups need to understand that the rules still apply. There is still a code of ethics that you can be held accountable too. And your Ph.D is not a permanent gift. It is earned, and can be unearned. James Harris === Subject: Re: JSH: Sweep likely Originator: a@shell3.shore.net (a) >Remember, I¹m not talking about something vague here, as I¹m talking >about publication. >Call me crazy if you wish as it won¹t make any difference. >All the little social games that seemed to work on sci.math, never >did, as I¹ve just been busily researching, reŽning my results, and >writing papers. >It can just take a while with math research. Some of that reŽnement >of my results can be seen by just looking over posts where I would >argue out different areas with posters just to see if there wasn¹t >something I missed, despite all the insults being tossed at me. >I kept at it despite the social žak. Math is important. Social >issues come and go. >Now I¹ve Žnally sent off the major papers, including re-sending APF >for yet another round of formal peer review, which is extra work for >some mathematicians required by sci.math posters who interfered with >the journal process. >So sci.math will pay, and so will posters who acted in a way that I¹ll >show is gross fraud, requiring that they be stripped of their Ph.D¹s. >The initial repudiation of this group, and of those posters will come >from journals. >Then the rest of the impact should come from the math community >itself, as it moves to clean house, and try to make sure that >something like this never happens again. >Mathematicians who post on newsgroups need to understand that the >rules still apply. There is still a code of ethics that you can be >held accountable too. >And your Ph.D is not a permanent gift. It is earned, and can be >unearned. >James Harris James, if you forgot to bookmark it last time you needed it, here¹s HTH. HAND. === Subject: Re: JSH: Sweep likely > Call me crazy if you wish as it won¹t make any difference. Of course not! Identifying the errors in your arguments and noting your passionate defense of both the original errors and any later corrections has made no difference. Disproving your arguments and posting counter-examples has made no difference, either. > It can just take a while with math research. Some of that reŽnement > of my results can be seen by just looking over posts where I would > argue out different areas with posters just to see if there wasn¹t > something I missed, despite all the insults being tossed at me. Insults are cheap shots -- from you or your critics. But identifying you as an *idiot* or a *crank* is not insulting, it is precision labelling analogous to Œtruth in packaging¹. > The initial repudiation of this group, and of those posters will come > from journals. And if they reject your papers??? > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Sweep likely <1089610198.444545@news-1.nethere.net> Discussion, linux) > So sci.math will pay, and so will posters who acted in a way that > I¹ll show is gross fraud, requiring that they be stripped of their > Ph.D¹s. Can you name a single case in which a Ph.D. from an American university has been revoked for behavior years after it was granted? > And your Ph.D is not a permanent gift. It is earned, and can be > unearned. Are you sure? -- Jesse F. Hughes I¹m better than you, and you know it. -- James Harris === Subject: Re: JSH: Sweep likely > > So sci.math will pay, and so will posters who acted in a way that > > I¹ll show is gross fraud, requiring that they be stripped of their > > Ph.D¹s. > Can you name a single case in which a Ph.D. from an American > university has been revoked for behavior years after it was granted? Well, I don¹t know off-hand of an American case, but here¹s a German one: DISGRACED PHYSICIST STRIPPED OF PH.D. DEGREE University revokes degree of former Bell Labs researcher who was Žred for scientiŽc misconduct http://pubs.acs.org/cen/news/8224/8224physicist.html I think the case can be made for doing the same for gross misconduct in mathematics, and I doubt that American universities want to be havens for people who do gross misconduct, as if you can get away with here, what people won¹t tolerate in other countries. Like, do Germans care more about academic accountability than Americans? James Harris === Subject: Re: JSH: Sweep likely > > > > > So sci.math will pay, and so will posters who acted in a way that > > > I¹ll show is gross fraud, requiring that they be stripped of their > > > Ph.D¹s. > > > > Can you name a single case in which a Ph.D. from an American > > university has been revoked for behavior years after it was granted? > > > Well, I don¹t know off-hand of an American case, but here¹s a German > one: > DISGRACED PHYSICIST STRIPPED OF PH.D. DEGREE > University revokes degree of former Bell Labs researcher who was Žred > for scientiŽc misconduct > http://pubs.acs.org/cen/news/8224/8224physicist.html > I think the case can be made for doing the same for gross misconduct > in mathematics, and I doubt that American universities want to be > havens for people who do gross misconduct, as if you can get away with > here, what people won¹t tolerate in other countries. > Like, do Germans care more about academic accountability than > Americans? Maybe they do. This action depends on a German law, for which there is no American equivalent so far as I know. The scientiŽc misconduct in this case was apperently egregious. The man has published a lot of high-proŽle stuff that was deliberately faked. His later actions cast doubt on his Ph. D. work. For sure, they demonstrate that he does not adhere to minimal standards of behavior for a researcher. -- Chris Henrich The total lack of evidence is the surest sign that the conspiracy is working. === Subject: Re: JSH: Sweep likely >> > So sci.math will pay, and so will posters who acted in a way that >> > I¹ll show is gross fraud, requiring that they be stripped of their >> > Ph.D¹s. >> Can you name a single case in which a Ph.D. from an American >> university has been revoked for behavior years after it was granted? >Well, I don¹t know off-hand of an American case, but here¹s a German >one: We;ve all heard about that. Well, I had, and I imagine a lot of others have as well. What did you think my point was when I said Hint: The US is not Germnany? > DISGRACED PHYSICIST STRIPPED OF PH.D. DEGREE >University revokes degree of former Bell Labs researcher who was Žred >for scientiŽc misconduct >http://pubs.acs.org/cen/news/8224/8224physicist.html >I think the case can be made for doing the same for gross misconduct >in mathematics, and I doubt that American universities want to be >havens for people who do gross misconduct, as if you can get away with >here, what people won¹t tolerate in other countries. >Like, do Germans care more about academic accountability than >Americans? >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Sweep likely ... > >Well, I don¹t know off-hand of an American case, but here¹s a German > >one: > We;ve all heard about that. Well, I had, and I imagine a lot > of others have as well. > What did you think my point was when I said > Hint: The US is not Germnany? You should have hinted clearer: The US is not Baden-Wuerttemberg. I do think it is no even possible in all of Germany. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Sweep likely days. My association with the Department is that of an alumnus. >We;ve all heard about that. Well, I had, and I imagine a lot >of others have as well. >What did you think my point was when I said >Hint: The US is not Germnany? This type of punitive action is in accord with local law in Baden-Wurttemberg, the southwestern German state in which the university is located. The law provides for the possibility of withdrawing a university degree for improper behavior carried out after the degree has been awarded[.] Absent such a law in the U.S., and given the ex post facto clause of the U.S. Constitution, it would seem that barring a Constitutional Amendment, there isn¹t much to worry about... (-; -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Sweep likely <1089610198.444545@news-1.nethere.net> <87k6x9cpg6.fsf@phiwumbda.org> Discussion, linux) >> > So sci.math will pay, and so will posters who acted in a way that >> > I¹ll show is gross fraud, requiring that they be stripped of their >> > Ph.D¹s. >> Can you name a single case in which a Ph.D. from an American >> university has been revoked for behavior years after it was granted? > Well, I don¹t know off-hand of an American case, but here¹s a German > one: > DISGRACED PHYSICIST STRIPPED OF PH.D. DEGREE > University revokes degree of former Bell Labs researcher who was Žred > for scientiŽc misconduct > http://pubs.acs.org/cen/news/8224/8224physicist.html > I think the case can be made for doing the same for gross misconduct > in mathematics, and I doubt that American universities want to be > havens for people who do gross misconduct, as if you can get away with > here, what people won¹t tolerate in other countries. > Like, do Germans care more about academic accountability than > Americans? Americans don¹t have the view that a degree can later be revoked by behavior unrelated to its conferment. As far as I know, there¹s nothing I can do now (aside from admitting that my thesis was plagiarized and the like) that would threaten my PhD. The German case doesn¹t have any real relevance near as I can Žgger. -- Jesse F. Hughes I¹m better than you, and you know it. -- James Harris === Subject: Re: JSH: Sweep likely >> So sci.math will pay, and so will posters who acted in a way that >> I¹ll show is gross fraud, requiring that they be stripped of their >> Ph.D¹s. >Can you name a single case in which a Ph.D. from an American >university has been revoked for behavior years after it was granted? >> And your Ph.D is not a permanent gift. It is earned, and can be >> unearned. >Are you sure? Well that¹s a stupid question - of course he¹s sure. He¹s never been uncertain of anything. ************************ David C. Ullrich === Subject: Absolute convergence => convergence a. e.? Hi all, Let f_1, f_2, f_3, ... be a sequence of Lebesgue-integrable functions from [0,1] in the reals such that the sum of their norms converges (the norm here being ||f|| = int |f|). Is it true or false that the series sum(f_n(x), 1 <= n < +oo) converges a. e.? All that I was able to prove was that if s_n = f_1 + f_2 + ... + f_n, then some subsequence of the sequence (s_n)_n converges a. e.; on the other hand, I was not able to Žnd any counterexample. Jose Carlos Santos === Subject: Re: Absolute convergence => convergence a. e.? > Hi all, > Let f_1, f_2, f_3, ... be a sequence of Lebesgue-integrable functions > from [0,1] in the reals such that the sum of their norms converges > (the norm here being ||f|| = int |f|). Is it true or false that the > series sum(f_n(x), 1 <= n < +oo) converges a. e.? int_[0,1] sum |f_n| = sum int_[0,1] |f_n| < oo. This implies sum |f_n(x)| < oo for a.e. x. For any such x, f_n(x) converges. === Subject: Re: Absolute convergence => convergence a. e.? >> Let f_1, f_2, f_3, ... be a sequence of Lebesgue-integrable functions >> from [0,1] in the reals such that the sum of their norms converges >> (the norm here being ||f|| = int |f|). Is it true or false that the >> series sum(f_n(x), 1 <= n < +oo) converges a. e.? > int_[0,1] sum |f_n| = sum int_[0,1] |f_n| < oo. > This implies sum |f_n(x)| < oo for a.e. x. For any such x, f_n(x) converges. Jose Carlos Santos === Subject: Re: Absolute convergence => convergence a. e.? > int_[0,1] sum |f_n| = sum int_[0,1] |f_n| < oo. > This implies sum |f_n(x)| < oo for a.e. x. For any such x, f_n(x) converges. Last sentence should read For any such x, sum f_n(x) converges. === Subject: Re: Atheist MorituriMax > > > God is absolute. > > In your mind and ego only. If God was absolute in the real world, I couldn¹t do > > this. > > dOG > You can deny the existence of God, no problem. But then you are left > with the task of explaining something you cannot deny: your own > painful existence. > Peter This is a repeat of the argument the reason there is something instead of nothing is because god put it there. If god is responsible for our existence, who is responsible for god¹s? This just pushes the question back a step, but into fantasy, for there is no evidence for any god, no reason to think god exists, except for authority, which is no reason at all. No one can shed any light on why we are here. The universe just exists. There is no why involved. I think why are we here is a nonsense question. People thought the world was žat for a long time. They have believed in god for a long time, but I think eventually people will realize the truth. There is no god. I curse him and utter obscenities about him and challenge him to do something--but nothing ever happens. Van === Subject: Re: a noise with a better histogram > > I used an inversion of a Gaussian to > >get my amplitudes instead of a Gaussian. > >It seems to work somewhat better in terms of the histogram. > >I¹m indepted to the patient work of Ray Kooperman and Dr. Bobby Treat > >on Kurtosis excess calculations and Cauchy distribution calculations. > >As I am giving this information to the egroup for comment, > >I must take the good with the bad. > I¹m so confused. Just now you told us we should use sci.math > to answer questions. I don¹t recall any questions here about > noise with a better histogram. > Please use sci.math to answer questions. If everyone > posted everything they know and every bit of code > they¹d written there would be literally millions > of posts a day and the group would be totally useless. > Um, also, please when you use sci.math to answer questions > make certain that you actually understand the relevant > mathematics before speaking up. When people recognize > some of the words in the question and post answers > that make no sense that also wastes valuable space. > ************************ > David C. Ullrich Amen to that. I already had to unsubscribe to the Yahoo number theory group because of this idiot, and his posts here are getting on my nerves. I try to ignore them, but they seem to be everywhere and increasing in length and frequency. Van === Subject: Re: Analysis Differentiation question > > > Let f be differentiable on [a,b] (some closed interval). > > > > > > How would I go about constructing a function so that the range of f > > is > > > > > > a). an open interval > > > b). open on one side and closed on the other side interval. > > First you need to travel to an alternate universe. In this one, > > differentiability implies continuity, and the continuous image of a > > compact set is compact > You must be mis-understanding my question. > I¹m not saying the derivative has to be continuous !!! Nor did I. Let¹s quote your Žrst sentence: Let f be DIFFERENTIABLE on [a,b]. In other words, at each x in [a,b], the derivative f¹(x) of f exists at x. Which implies that f is continuous at each point in [a,b]. I.e., f is continuous on [a,b]. on > [a,b], then the set of derivatives must also be an interval. Huh??? > I¹m asking how > we can construct a function f and show that the range of it¹s derivative is > an open interval. Whoa, now you are changing the question. Do you want the range of f to be open (as you asked originally), or do you refer to the range of the derivative f¹? -SJH === Subject: Re: Analysis Differentiation question X-AUTHid: shrest10 > > > > Let f be differentiable on [a,b] (some closed interval). > > > > > > > > How would I go about constructing a function so that the range of > > > is > > > > > > > > a). an open interval > > > > b). open on one side and closed on the other side interval. > > > > > > First you need to travel to an alternate universe. In this one, > > > differentiability implies continuity, and the continuous image of a > > > compact set is compact > > > > > You must be mis-understanding my question. > > I¹m not saying the derivative has to be continuous !!! > Nor did I. > Let¹s quote your Žrst sentence: Let f be DIFFERENTIABLE on [a,b]. > In other words, at each x in [a,b], the derivative f¹(x) of f > exists at x. Which implies that f is continuous at each point in > [a,b]. I.e., f is continuous on [a,b]. > on > > [a,b], then the set of derivatives must also be an interval. > Huh??? > > I¹m asking how > > we can construct a function f and show that the range of it¹s > derivative is > > an open interval. > Whoa, now you are changing the question. Do you want the range of f > to be open (as you asked originally), or do you refer to the range of > the derivative f¹? > -SJH Sorry. I did mean f¹ not f === Subject: Re: Analysis Differentiation question X-AUTHid: shrest10 > > > > Let f be differentiable on [a,b] (some closed interval). > > > > > > > > How would I go about constructing a function so that the range of > > > is > > > > > > > > a). an open interval > > > > b). open on one side and closed on the other side interval. > > > > > > First you need to travel to an alternate universe. In this one, > > > differentiability implies continuity, and the continuous image of a > > > compact set is compact > > > > > You must be mis-understanding my question. > > I¹m not saying the derivative has to be continuous !!! > Nor did I. > Let¹s quote your Žrst sentence: Let f be DIFFERENTIABLE on [a,b]. > In other words, at each x in [a,b], the derivative f¹(x) of f > exists at x. Which implies that f is continuous at each point in > [a,b]. I.e., f is continuous on [a,b]. > on > > [a,b], then the set of derivatives must also be an interval. > Huh??? > > I¹m asking how > > we can construct a function f and show that the range of it¹s > derivative is > > an open interval. > Whoa, now you are changing the question. Do you want the range of f > to be open (as you asked originally), or do you refer to the range of > the derivative f¹? > -SJH Sorry. I did mean f¹ not f === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye > > Therein the reason we need the number 0 rule, and the number - rule, > > and the number + rule. > Only if we have all three numbers. Since 0 = + and 0 = -, there¹s > no real need for the other two. Repeatedly toss a coin (inŽntie times) What is the probability of HHHHHHHHHHHHHH.... ? # desired outcomes 1 _____________________ = __________________________ # total outcomes # total outcomes as total outcomes -> oo, P(H..) > 0 It CAN happen, so the probability is NOT 0. Herc === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye In sci.logic, |-|erc : >> > Therein the reason we need the number 0 rule, and the number - rule, >> > and the number + rule. >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s >> no real need for the other two. > Repeatedly toss a coin (inŽntie times) > What is the probability of HHHHHHHHHHHHHH.... ? The same as the probability of TTTTTTTTTT.....or HTHTHTHTHTHTHT.... > # desired outcomes 1 > _____________________ = __________________________ > # total outcomes # total outcomes > as total outcomes -> oo, P(H..) > 0 > It CAN happen, so the probability is NOT 0. I¹m not entirely certain as to how one can approach this issue, as it¹s clear that oo/oo suffers from the same problem as 0/0; namely, it¹s not deŽned. 1/oo is typically called 0 but there are problems here, too, as any sequence tending to inŽnity is such that the derived sequence of reciprocals are nonzero (assuming that none of the terms of the original sequence is 0). However, most people aren¹t too concerned. Unfortunately lim (n->oo) P(H...) = 0 anyway, though for any n, P(H...H) > 0. To claim otherwise leads one into Garry Denke math. > Herc -- #191, ewill3@earthlink.net It¹s still legal to go .sigless. === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye > >> > Therein the reason we need the number 0 rule, and the number - rule, > >> > and the number + rule. > >> > >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s > >> no real need for the other two. > > Repeatedly toss a coin (inŽntie times) > > What is the probability of HHHHHHHHHHHHHH.... ? > The same as the probability of TTTTTTTTTT.....or HTHTHTHTHTHTHT.... > > # desired outcomes 1 > > _____________________ = __________________________ > > # total outcomes # total outcomes > > as total outcomes -> oo, P(H..) > 0 > > It CAN happen, so the probability is NOT 0. > I¹m not entirely certain as to how one can approach this issue, > as it¹s clear that oo/oo suffers from the same problem as 0/0; > namely, it¹s not deŽned. 1/oo is typically called 0 but there > are problems here, too, as any sequence tending to inŽnity > is such that the derived sequence of reciprocals are nonzero > (assuming that none of the terms of the original sequence is 0). > However, most people aren¹t too concerned. > Unfortunately lim (n->oo) P(H...) = 0 anyway, though for any n, > P(H...H) > 0. To claim otherwise leads one into Garry Denke > math. an interesting excursion, we look at [lower 0s] and everyone has to reverse their arguments from [higher oo¹s]. >Unfortunately lim(n->oo) P(H...) = 0 That¹s just shorthand for n->oo P(H..)->0. Which is also written lim(n->oo) P(H..)->0+ as n approaches inŽnity P approaches 0 from above. Probability is just a model. [At the limit P=0] does not mean impossible, for obvious reasons. Herc === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye >> > Therein the reason we need the number 0 rule, and the number - rule, >> > and the number + rule. >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s >> no real need for the other two. >Repeatedly toss a coin (inŽntie times) >What is the probability of HHHHHHHHHHHHHH.... ? > # desired outcomes 1 >_____________________ = __________________________ > # total outcomes # total outcomes >as total outcomes -> oo, P(H..) > 0 >It CAN happen, so the probability is NOT 0. Nope. It¹s 0, for the same reason your SetMinus({.3, .33, .333, ...}, 1/3) is 0. Martin === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye > >> > Therein the reason we need the number 0 rule, and the number - rule, > >> > and the number + rule. > >> > >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s > >> no real need for the other two. > >Repeatedly toss a coin (inŽntie times) > >What is the probability of HHHHHHHHHHHHHH.... ? > > # desired outcomes 1 > >_____________________ = __________________________ > > # total outcomes # total outcomes > >as total outcomes -> oo, P(H..) > 0 > >It CAN happen, so the probability is NOT 0. > Nope. It¹s 0, for the same reason your SetMinus({.3, .33, .333, ...}, > 1/3) is 0. HAHAHA now we¹re splitting hairs. I¹m already up to deŽning aleph+1 as +/oo. Its REALLY SMALL. Its so small you multiply by inŽnity and only get +. Herc === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye >> >> > Therein the reason we need the number 0 rule, and the number - rule, >> >> > and the number + rule. >> >> >> >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s >> >> no real need for the other two. >> > >> >Repeatedly toss a coin (inŽntie times) >> >What is the probability of HHHHHHHHHHHHHH.... ? >> > >> > >> > # desired outcomes 1 >> >_____________________ = __________________________ >> > # total outcomes # total outcomes >> > >> >as total outcomes -> oo, P(H..) > 0 >> > >> >It CAN happen, so the probability is NOT 0. No, as I pointed out to you before, if it happens a *Žnite* number of times out of an inŽnite set of trials then the probability is exactly 0. >> Nope. It¹s 0, for the same reason your SetMinus({.3, .33, .333, ...}, >> 1/3) is 0. >HAHAHA now we¹re splitting hairs. I¹m already up to deŽning aleph+1 >as +/oo. Its REALLY SMALL. Its so small you multiply by inŽnity and only >get +. Too bad you¹re not allowed to do that. A PROBABILITY is a REAL number between 0.0 and 1.0 inclusive. Suppose the probability of every inŽnite sequence of (mixed) Hs and Ts is some real number epsilon > 0. So if you have more than (1/epsilon)+1 such sequences then their total probability would be > 1, which is impossible. But of course there is an inŽnite set of such sequences, which is thus more than (1/epsilon)+1 for any epsilon. Therefore, there is no such epsilon > 0, therefore the probability = 0. (I hope that, unlike Peter Olcott, you do not object to indirect proofs in general.) As I pointed out before, the probability of any particular inŽnite sequence (WLOG, a sequence of all Hs) can be seen to be exactly zero by the following argument: It would require an inŽnite tail of Hs, which is the same as having *no* Ts after some point, which has probability 0. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye > >> >> > Therein the reason we need the number 0 rule, and the number - rule, > >> >> > and the number + rule. > >> >> > >> >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s > >> >> no real need for the other two. > >> > > >> >Repeatedly toss a coin (inŽntie times) > >> >What is the probability of HHHHHHHHHHHHHH.... ? > >> > > >> > > >> > # desired outcomes 1 > >> >_____________________ = __________________________ > >> > # total outcomes # total outcomes > >> > > >> >as total outcomes -> oo, P(H..) > 0 > >> > > >> >It CAN happen, so the probability is NOT 0. > No, as I pointed out to you before, if it happens a *Žnite* number of times > out of an inŽnite set of trials then the probability is exactly 0. > >> Nope. It¹s 0, for the same reason your SetMinus({.3, .33, .333, ...}, > >> 1/3) is 0. > >> > >HAHAHA now we¹re splitting hairs. I¹m already up to deŽning aleph+1 > >as +/oo. Its REALLY SMALL. Its so small you multiply by inŽnity and only > >get +. > Too bad you¹re not allowed to do that. A PROBABILITY is a REAL number between > 0.0 and 1.0 inclusive. Suppose the probability of every inŽnite sequence of > (mixed) Hs and Ts is some real number epsilon > 0. So if you have more than > (1/epsilon)+1 such sequences then their total probability would be > 1, which > is impossible. But of course there is an inŽnite set of such sequences, > which is thus more than (1/epsilon)+1 for any epsilon. Therefore, there is no > such epsilon > 0, therefore the probability = 0. > (I hope that, unlike Peter Olcott, you do not object to indirect proofs in > general.) > As I pointed out before, the probability of any particular inŽnite sequence > (WLOG, a sequence of all Hs) can be seen to be exactly zero by the following > argument: > It would require an inŽnite tail of Hs, which is the same as having *no* Ts > after some point, which has probability 0. No. Probability = 0 means impossible. some inŽnite sequence is possible (in the model). As #trials approaches inŽnity, the probability approaches 0+. Or does 1/oo = 0 now? Herc === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye > > Too bad you¹re not allowed to do that. A PROBABILITY is a REAL number between > > 0.0 and 1.0 inclusive. Suppose the probability of every inŽnite sequence of > > (mixed) Hs and Ts is some real number epsilon > 0. So if you have more than > > (1/epsilon)+1 such sequences then their total probability would be > 1, which > > is impossible. But of course there is an inŽnite set of such sequences, > > which is thus more than (1/epsilon)+1 for any epsilon. Therefore, there is no > > such epsilon > 0, therefore the probability = 0. > > (I hope that, unlike Peter Olcott, you do not object to indirect proofs in > > general.) > > As I pointed out before, the probability of any particular inŽnite sequence > > (WLOG, a sequence of all Hs) can be seen to be exactly zero by the following > > argument: > > It would require an inŽnite tail of Hs, which is the same as having *no* Ts > > after some point, which has probability 0. > No. Probability = 0 means impossible. some inŽnite sequence is possible (in the model). > As #trials approaches inŽnity, the probability approaches 0+. The problem here is a double deŽnition. Impossibility has been deŽned simultaneously as having probability zero, and as meaning that something cannot happen. I am not a lawyer. I do not even see email sent to this address, due to past DOS attacks. If you wish to respond, do so through this newsgroup. === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye >> >> >> > Therein the reason we need the number 0 rule, and the number - rule, >> >> >> > and the number + rule. >> >> >> >> >> >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s >> >> >> no real need for the other two. >> >> > >> >> >Repeatedly toss a coin (inŽntie times) >> >> >What is the probability of HHHHHHHHHHHHHH.... ? >> >> > >> >> > >> >> > # desired outcomes 1 >> >> >_____________________ = __________________________ >> >> > # total outcomes # total outcomes >> >> > >> >> >as total outcomes -> oo, P(H..) > 0 >> >> > >> >> >It CAN happen, so the probability is NOT 0. >> No, as I pointed out to you before, if it happens a *Žnite* number of times >> out of an inŽnite set of trials then the probability is exactly 0. >> >> Nope. It¹s 0, for the same reason your SetMinus({.3, .33, .333, ...}, >> >> 1/3) is 0. >> >> >> > >> >HAHAHA now we¹re splitting hairs. I¹m already up to deŽning aleph+1 >> >as +/oo. Its REALLY SMALL. Its so small you multiply by inŽnity and only >> >get +. >> Too bad you¹re not allowed to do that. A PROBABILITY is a REAL number >> between >> 0.0 and 1.0 inclusive. Suppose the probability of every inŽnite sequence >> of >> (mixed) Hs and Ts is some real number epsilon > 0. So if you have more than >> (1/epsilon)+1 such sequences then their total probability would be > 1, >> which >> is impossible. But of course there is an inŽnite set of such sequences, >> which is thus more than (1/epsilon)+1 for any epsilon. Therefore, there is >> no >> such epsilon > 0, therefore the probability = 0. >> (I hope that, unlike Peter Olcott, you do not object to indirect proofs in >> general.) >> As I pointed out before, the probability of any particular inŽnite sequence >> (WLOG, a sequence of all Hs) can be seen to be exactly zero by the following >> argument: >> It would require an inŽnite tail of Hs, which is the same as having *no* Ts >> after some point, which has probability 0. >No. Probability = 0 means impossible. No it doesn¹t, at least not in a mathematical context. (Maybe in a courtroom it does.) >some inŽnite sequence is possible (in the model). >As #trials approaches inŽnity, the probability approaches 0+. That¹s actually true. And at that limit, 0+ is identically 0. >Or does 1/oo = 0 now? It certainly does. For example, have a look at , item (7). Didn¹t you already know that? -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye > >> >> >> > Therein the reason we need the number 0 rule, and the number - rule, > >> >> >> > and the number + rule. > >> >> >> > >> >> >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s > >> >> >> no real need for the other two. > >> >> > > >> >> >Repeatedly toss a coin (inŽntie times) > >> >> >What is the probability of HHHHHHHHHHHHHH.... ? > >> >> > > >> >> > > >> >> > # desired outcomes 1 > >> >> >_____________________ = __________________________ > >> >> > # total outcomes # total outcomes > >> >> > > >> >> >as total outcomes -> oo, P(H..) > 0 > >> >> > > >> >> >It CAN happen, so the probability is NOT 0. > >> > >> No, as I pointed out to you before, if it happens a *Žnite* number of times > >> out of an inŽnite set of trials then the probability is exactly 0. > >> > >> >> Nope. It¹s 0, for the same reason your SetMinus({.3, .33, .333, ...}, > >> >> 1/3) is 0. > >> >> > >> > > >> >HAHAHA now we¹re splitting hairs. I¹m already up to deŽning aleph+1 > >> >as +/oo. Its REALLY SMALL. Its so small you multiply by inŽnity and only > >> >get +. > >> > >> Too bad you¹re not allowed to do that. A PROBABILITY is a REAL number > >> between > >> 0.0 and 1.0 inclusive. Suppose the probability of every inŽnite sequence > >> of > >> (mixed) Hs and Ts is some real number epsilon > 0. So if you have more than > >> (1/epsilon)+1 such sequences then their total probability would be > 1, > >> which > >> is impossible. But of course there is an inŽnite set of such sequences, > >> which is thus more than (1/epsilon)+1 for any epsilon. Therefore, there is > >> no > >> such epsilon > 0, therefore the probability = 0. > >> > >> (I hope that, unlike Peter Olcott, you do not object to indirect proofs in > >> general.) > >> > >> As I pointed out before, the probability of any particular inŽnite sequence > >> (WLOG, a sequence of all Hs) can be seen to be exactly zero by the following > >> argument: > >> > >> It would require an inŽnite tail of Hs, which is the same as having *no* Ts > >> after some point, which has probability 0. > >> > >No. Probability = 0 means impossible. > No it doesn¹t, at least not in a mathematical context. (Maybe in a courtroom > it does.) Did you or did you not state the following? ;) But notice the qualiŽcation Žnite: an *inŽnite* sequence of Hs has probability 0, since that would require an inŽnite tail of Hs, which is the same as having *no* Ts after some point, which has probability 0. reads as: H.. is possible, since !T... is possible. big deal You can see why I introduce basic examples to pin your meaning. > >fun coin() > >while true > > if rnd>0.5 return > >wend > >for any number of cycles it wont necessarily halt in that time. > >for inŽnite number of cycles there is still one possible outcome, [low, low, > >low...] that > >shows its impossible to prove that it halts. > It halts with probability 1 (assuming that rnd() uses some truly random > quantum process and not a deterministic pseudo-random generator). That can be > viewed as meaning that it¹s allowed to fail to halt on a Žnite number of runs > during an unlimited number of runs of the program. But this is entirely aside > from the argument I¹ve given you to address (twice, so far). Herc : its impossible to prove that it halts Barb : it halts with P=1 are you agreeing or disagreeing here? > >some inŽnite sequence is possible (in the model). > >As #trials approaches inŽnity, the probability approaches 0+. > That¹s actually true. And at that limit, 0+ is identically 0. That¹s nonsense. At the limit there are oo combinations and P is undeŽned. > >Or does 1/oo = 0 now? > It certainly does. For example, have a look at > , item (7). > Didn¹t you already know that? these improper elements are not real numbers, and that this system of extended real numbers is not a Želd. how much mickey mouse maths do you have to use to support Cantors bogus idiocy? 9 months examining all these avenues. I have to sum up, getting recognition in Maths is not an avenue to afford shelter from the Truman Satellite. xxx xxx xxx yyy is missing xxx yxx yyx ... look inŽnity of inŽnities because yyy is nowhere, open your eyes Herc === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye >> >> >> >> > Therein the reason we need the number 0 rule, and the number - >> >> >> >> > rule, >> >> >> >> > and the number + rule. >> >> >> >> >> >> >> >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s >> >> >> >> no real need for the other two. >> >> >> > >> >> >> >Repeatedly toss a coin (inŽntie times) >> >> >> >What is the probability of HHHHHHHHHHHHHH.... ? >> >> >> > >> >> >> > >> >> >> > # desired outcomes 1 >> >> >> >_____________________ = __________________________ >> >> >> > # total outcomes # total outcomes >> >> >> > >> >> >> >as total outcomes -> oo, P(H..) > 0 >> >> >> > >> >> >> >It CAN happen, so the probability is NOT 0. >> >> >> >> No, as I pointed out to you before, if it happens a *Žnite* number of >> >> times >> >> out of an inŽnite set of trials then the probability is exactly 0. >> >> >> >> >> Nope. It¹s 0, for the same reason your SetMinus({.3, .33, .333, ...}, >> >> >> 1/3) is 0. >> >> >> >> >> > >> >> >HAHAHA now we¹re splitting hairs. I¹m already up to deŽning aleph+1 >> >> >as +/oo. Its REALLY SMALL. Its so small you multiply by inŽnity and >> >> >only >> >> >get +. >> >> >> >> Too bad you¹re not allowed to do that. A PROBABILITY is a REAL number >> >> between >> >> 0.0 and 1.0 inclusive. Suppose the probability of every inŽnite >> >> sequence >> >> of >> >> (mixed) Hs and Ts is some real number epsilon > 0. So if you have more >> >> than >> >> (1/epsilon)+1 such sequences then their total probability would be > 1, >> >> which >> >> is impossible. But of course there is an inŽnite set of such sequences, >> >> which is thus more than (1/epsilon)+1 for any epsilon. Therefore, there >> >> is >> >> no >> >> such epsilon > 0, therefore the probability = 0. >> >> >> >> (I hope that, unlike Peter Olcott, you do not object to indirect proofs >> >> in >> >> general.) >> >> >> >> As I pointed out before, the probability of any particular inŽnite >> >> sequence >> >> (WLOG, a sequence of all Hs) can be seen to be exactly zero by the >> >> following >> >> argument: >> >> >> >> It would require an inŽnite tail of Hs, which is the same as having *no* >> >> Ts >> >> after some point, which has probability 0. >> >> >> > >> >No. Probability = 0 means impossible. >> No it doesn¹t, at least not in a mathematical context. (Maybe in a >> courtroom >> it does.) >Did you or did you not state the following? ;) > But notice the qualiŽcation Žnite: an *inŽnite* sequence of Hs has > probability 0, since that would require an inŽnite tail of Hs, which is > the same as having *no* Ts after some point, which has probability 0. I did indeed. >reads as: H.. is possible, since !T... is possible. big deal Rubbish. Has probability 0 is NOT mainly a synonym for is possible. YOU claimed that It CAN happen, so the probability is NOT 0; I showed that the probability IS 0. Equivocations about the natural-language term possible are just irrelevant to the fact that you got the probability thoroughly wrong. >You can see why I introduce basic examples to pin your meaning. No, but I think I can see why you tend to change the subject a lot. >> >fun coin() >> >while true >> > if rnd>0.5 return >> >wend >> > >> >for any number of cycles it wont necessarily halt in that time. >> >for inŽnite number of cycles there is still one possible outcome, [low, >> >low, low...] that >> >shows its impossible to prove that it halts. >> It halts with probability 1 (assuming that rnd() uses some truly random >> quantum process and not a deterministic pseudo-random generator). That can >> be >> viewed as meaning that it¹s allowed to fail to halt on a Žnite number of >> runs >> during an unlimited number of runs of the program. But this is entirely >> aside >> from the argument I¹ve given you to address (twice, so far). >Herc : its impossible to prove that it halts >Barb : it halts with P=1 >are you agreeing or disagreeing here? I¹m emphatically disagreeing with YOUR claim that It CAN happen, so the probability [of it not halting] is NOT 0. >> >some inŽnite sequence is possible (in the model). >> >As #trials approaches inŽnity, the probability approaches 0+. >> That¹s actually true. And at that limit, 0+ is identically 0. >That¹s nonsense. At the limit there are oo combinations and P is undeŽned. At the limit there are indeed inŽnite combinations, but P is a well-deŽned zero (1/oo). >> >Or does 1/oo = 0 now? >> It certainly does. For example, have a look at >> , item (7). >> Didn¹t you already know that? >these improper elements are not real numbers, and that this system of >extended real numbers is not a Želd. >how much mickey mouse maths do you have to use to support Cantors bogus >idiocy? same web page: ŒThe above statements which deŽne results of arithmetic operations on oo may be considered as abbreviations of statements about determinate limit forms. For example, -(+oo) = -oo may be considered as an abbreviation for If x increases without bound, then -x decreases without bound.¹ So all the arithmetic relations given there, including 1/oo = 0, can be considered to be limit statements about garden-variety REAL numbers, unlike your supurious +0. >9 months examining all these avenues. I have to sum up, getting recognition >in Maths is not an avenue to afford shelter from the Truman Satellite. No sh*t, Sherlock. Face it, if they are out to get you then you¹re going to get got. Just lie back and think of your eventual glorious revealing of yourself as God or Adam or whoever. >xxx >xxx >xxx >yyy is missing >xxx >yxx >yyx >... >look inŽnity of inŽnities because yyy is nowhere, open your eyes Whatever. There¹s really no need to add more rubbish to the pile; almost everyone by now understands your bogus argument. >Herc -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: Re: Garry Denke¹s Gold & Brass Plates @ Westbury White Horse Eye > >> >> >> >> > Therein the reason we need the number 0 rule, and the number - > >> >> >> >> > rule, > >> >> >> >> > and the number + rule. > >> >> >> >> > >> >> >> >> Only if we have all three numbers. Since 0 = + and 0 = -, there¹s > >> >> >> >> no real need for the other two. > >> >> >> > > >> >> >> >Repeatedly toss a coin (inŽntie times) > >> >> >> >What is the probability of HHHHHHHHHHHHHH.... ? > >> >> >> > > >> >> >> > > >> >> >> > # desired outcomes 1 > >> >> >> >_____________________ = __________________________ > >> >> >> > # total outcomes # total outcomes > >> >> >> > > >> >> >> >as total outcomes -> oo, P(H..) > 0 > >> >> >> > > >> >> >> >It CAN happen, so the probability is NOT 0. > >> >> > >> >> No, as I pointed out to you before, if it happens a *Žnite* number of > >> >> times > >> >> out of an inŽnite set of trials then the probability is exactly 0. > >> >> > >> >> >> Nope. It¹s 0, for the same reason your SetMinus({.3, .33, .333, ...}, > >> >> >> 1/3) is 0. > >> >> >> > >> >> > > >> >> >HAHAHA now we¹re splitting hairs. I¹m already up to deŽning aleph+1 > >> >> >as +/oo. Its REALLY SMALL. Its so small you multiply by inŽnity and > >> >> >only > >> >> >get +. > >> >> > >> >> Too bad you¹re not allowed to do that. A PROBABILITY is a REAL number > >> >> between > >> >> 0.0 and 1.0 inclusive. Suppose the probability of every inŽnite > >> >> sequence > >> >> of > >> >> (mixed) Hs and Ts is some real number epsilon > 0. So if you have more > >> >> than > >> >> (1/epsilon)+1 such sequences then their total probability would be > 1, > >> >> which > >> >> is impossible. But of course there is an inŽnite set of such sequences, > >> >> which is thus more than (1/epsilon)+1 for any epsilon. Therefore, there > >> >> is > >> >> no > >> >> such epsilon > 0, therefore the probability = 0. > >> >> > >> >> (I hope that, unlike Peter Olcott, you do not object to indirect proofs > >> >> in > >> >> general.) > >> >> > >> >> As I pointed out before, the probability of any particular inŽnite > >> >> sequence > >> >> (WLOG, a sequence of all Hs) can be seen to be exactly zero by the > >> >> following > >> >> argument: > >> >> > >> >> It would require an inŽnite tail of Hs, which is the same as having *no* > >> >> Ts > >> >> after some point, which has probability 0. > >> >> > >> > > >> >No. Probability = 0 means impossible. > >> > >> No it doesn¹t, at least not in a mathematical context. (Maybe in a > >> courtroom > >> it does.) > >Did you or did you not state the following? ;) > > But notice the qualiŽcation Žnite: an *inŽnite* sequence of Hs has > > probability 0, since that would require an inŽnite tail of Hs, which is > > the same as having *no* Ts after some point, which has probability 0. > I did indeed. > >reads as: H.. is possible, since !T... is possible. big deal > Rubbish. Has probability 0 is NOT mainly a synonym for is possible. YOU > claimed that It CAN happen, so the probability is NOT 0; I showed that the > probability IS 0. Equivocations about the natural-language term possible > are just irrelevant to the fact that you got the probability thoroughly wrong. cite? I said the string is inŽnite length, you but in claiming its Žnite because P=0. Now you say P=0 doesn¹t mean impossible,and now is doesn¹t mean possible. Just answer this question : WHY IS IT NECESSARILY FINITE? Can you do that without using a double meaning P=0. Remember > >> >No. Probability = 0 means impossible. > >> > >> No it doesn¹t So what does this mean? > > an *inŽnite* sequence of Hs has > > probability 0, since that would require an inŽnite tail of Hs, which is > > the same as having *no* Ts after some point, which has probability 0. an inŽnite sequence of Hs has P=0 since...... let A (for absurd) = an inŽnite sequence of Hs has P=0 let B (for Barb) = no T¹s has P=0 let C (for contradictory) = P=0 does not mean impossible How do you get B->A YOU WROTE A since B > >You can see why I introduce basic examples to pin your meaning. > No, but I think I can see why you tend to change the subject a lot. > >> >fun coin() > >> >while true > >> > if rnd>0.5 return > >> >wend its an inŽnite coin toss its the same subject you moron > >> > > >> >for any number of cycles it wont necessarily halt in that time. > >> >for inŽnite number of cycles there is still one possible outcome, [low, > >> >low, low...] that > >> >shows its impossible to prove that it halts. > >> > >> It halts with probability 1 (assuming that rnd() uses some truly random > >> quantum process and not a deterministic pseudo-random generator). That can > >> be > >> viewed as meaning that it¹s allowed to fail to halt on a Žnite number of > >> runs > >> during an unlimited number of runs of the program. But this is entirely > >> aside > >> from the argument I¹ve given you to address (twice, so far). > >Herc : its impossible to prove that it halts > >Barb : it halts with P=1 > >are you agreeing or disagreeing here? > I¹m emphatically disagreeing with YOUR claim that It CAN happen, so the > probability [of it not halting] is NOT 0. does emphatically include ignoring the question at hand? > >> >some inŽnite sequence is possible (in the model). > >> >As #trials approaches inŽnity, the probability approaches 0+. > >> > >> That¹s actually true. And at that limit, 0+ is identically 0. > >That¹s nonsense. At the limit there are oo combinations and P is undeŽned. > At the limit there are indeed inŽnite combinations, but P is a well-deŽned > zero (1/oo). its irrelevant, the question is about the length of the sequene. its no use deŽning P with 0 numerator as having 0 acceptable outcomes then quoting the limit value without declaring that its a limit. P = real / real P = 0 -> P = 0 / real -> number of favourable outcomes = 0 which is clearly wrong > >> >Or does 1/oo = 0 now? > >> > >> It certainly does. For example, have a look at > >> , item (7). > >> Didn¹t you already know that? > >these improper elements are not real numbers, and that this system of > >extended real numbers is not a Želd. > >how much mickey mouse maths do you have to use to support Cantors bogus > >idiocy? > same web page: > ŒThe above statements which deŽne results of arithmetic operations on oo may > be considered as abbreviations of statements about determinate limit forms. > For example, -(+oo) = -oo may be considered as an abbreviation for If x > increases without bound, then -x decreases without bound.¹ Which means 1/oo = 0 is an abbreviation for lim(x->oo) 1/x = 0 That Žrst = is not equals. You are better off partitioning Mickey Mouse maths from real maths (get it?) with 1/oo <=> 0 > So all the arithmetic relations given there, including 1/oo = 0, can be > considered to be limit statements about garden-variety REAL numbers, unlike > your supurious +0. > >9 months examining all these avenues. I have to sum up, getting recognition > >in Maths is not an avenue to afford shelter from the Truman Satellite. > No sh*t, Sherlock. Face it, if they are out to get you then you¹re going to > get got. Just lie back and think of your eventual glorious revealing of > yourself as God or Adam or whoever. its a last resort to see Eve again > >xxx > >xxx > >xxx > >yyy is missing > >xxx > >yxx > >yyx > >... > >look inŽnity of inŽnities because yyy is nowhere, open your eyes > Whatever. There¹s really no need to add more rubbish to the pile; almost > everyone by now understands your bogus argument. your mind is seriously messed. don¹t be so condescending in future your job is not at stake because you¹ve been lying to your pupils all this time. what *meaningless* mathematical syntax will you pour out next? Its like Žnding the computer behind the Eliza program talking to you, just syntax no understanding. Student : this program is rigged, in your dreams its intelligent Eliza : Tell me more about your dreams Herc === Subject: Prime Number Void Originator: edlee@chinet.com (Edward Lee) Is there an open interval (p, p*p) where p is a prime number such that there are no prime numbers in this interval? -Ed L Due to the volume of spam that I receive, any email message which does not have the word, mail, somewhere in the subject will be automatically deleted. Someone has also been forging the return address of spam messages to make it look as though they are coming from me. -- Due to the volume of spam that I receive, any email message to me which does not contain the word, mail, in the subject will be automatically deleted. === Subject: Surrogate factoring, reasons for my concerns The other thread I created yesterday trying to update you all on surrogate factoring got Žlled with a lot of distractions, but I think now after answering several posts in it I can give an executive summary type post. Recently I found the factorization (jk - Tk + T)(jk + Tk + T) = T^4 and noticed that solving for k gave me k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2) where signiŽcantly, you have a dependency on the factors of T^2 - 1 to get a rational k, as if sqrt(j^2 - T^2 + 1) is rational, then k is rational. Here T is the target number, the number that is to be factored. Notice that j = (-T +/- T sqrt(k^2 + T^2))/k where you see something more like the traditional congruence of squares, showing that j is dependent on the factors of T^2, so somehow, someway, there¹s a clear linkage between the factorization of T^2 - 1 and T^2, which is here easily seen. So you can get integer values for j easily by taking the factors of T^2-1, and using their sum to calculate j, as, for instance, j = (f_1 + f_2)/2 where f_1 f_2 = T^2 - 1, if T^2 - 1 has 4 as a factor, and with T odd, it will. With j you can then get an integer k, and notice that if you let k = d/c, and substitute you get j = (-cT +/- T sqrt(c^2 + T^2 d^2))/d where you know already that j is an integer, so sqrt(c^2 + T^2 d^2) is an integer, so c^2 + T^2 d^2 = n^2, so T^2 d^2 = (n-c)(n+c) and you might have just non-trivially factored T. You have all the factors of T^2 - 1 to play with, where you can check using all the combinations of them. At this point in time I don¹t know mathematically why this method wouldn¹t tend to factor T a signiŽcant percentage of the time. Basically you use a factorization of T^2 - 1, to get n, c and d integers, which gives you T^2 d^2 = (n-c)(n+c) and the potential of factoring T. The idea in retrospect is kind of obvious, and its VERY simple, so it¹s easy to think that it must not be new, but I¹ve yet to see anyone come forward with information that shows it¹s not new. If there isn¹t some block in the math that I¹m just not seeing then potentially someone could factor an RSA number by being able to fully factor T^2 - 1, which might seem easy since you get a head start with T^2 - 1 = (T- 1)(T+1), but it still can be a hard task--but possibly easier than factoring T directly. There may be some block in the mathematics which causes you to not non-trivially factor T with the path I¹ve shown, but maybe there¹s some slight variation to that path or some way to pick j where it¹s not an integer but is rational, or something else I missed that might make for a practical approach. The approach is in and of itself of interest, uses basic mathematics at a high school or secondary school level, and CANNOT be missed by an expert in the Želd as a potentially viable approach (unless you people know something you haven¹t told me). Therefore, if it is ignored, and someone out there uses it to do harm, then I cannot see how experts in the Želd could escape liability. There is a responsibility with titles. Math experts cannot fail to see that something like this *might* be important and dangerous if it works too well. I¹m not a professional mathematician. I don¹t have a lot of options. I already contacted the NSA and didn¹t hear futher from them, but bureaucrats in the US government have failed before, and I don¹t have to mention the signiŽcant recent intelligence failures. It seems to me that someone out there on the Internet should have the expertise to give some answer, and I¹m tired of the people who just chatter and bring up irrelevant points, and I remind that yeah, they may face liability later for their public statements *IF* this idea works. All I can do is put the information out there and hope that if it¹s important someone will act, or that it just doesn¹t work for some reason I don¹t know. At this point I¹m still just surprised at how hard it is to get a straight answer on something that¹s high school algebra. So far, posters have just come back with social issues mainly. It¹s high school algebra. SOMEONE should have an answer. James Harris === Subject: Re: Surrogate factoring, reasons for my concerns James Harris has just proposed a possible factorization method. I¹ll rephrase it in a manner that is more understandable to me, and corrects a number of minor errors that crept in Jame¹s description. We want to factor an odd integer T. We suppose we have a factorization of integer T^2-1, f1*f2 = T^2-1, with f1 and f2 even, f1>f2. Such factorization always exists, and while a full factorization of T^2-1 may be hard to obtain, a number of partial factorizations ares often easy to get. James observe that the second degree equation in k (jk - Tk + T)(jk + Tk + T) = T^4 has solution k = (-jT + T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2) He sets j = (f1+f2)/2, and observe that j^2 - T^2 + 1 is the square of (f1-f2)/2 thus k is a rational that we can explicitly compute as a reduced fraction d/c and then we have (jd - Td + Tc)(jd + Td + Tc) = T^4 c^2 The hope is that this explicit factorization of T^4 c^2 might break T nontrivially, in which case a nontrivial factor of T could be obtained by computing GCD (jd - Td + Tc, T), that is GCD (jd, T), or more effectively by computing GCD (d, T) and GCD (j, T). Unfortunately, experiments with medium numbers suggest this technique does not work: in practice we Žnd that GCD (d, T) = T and GCD (j, T) = 1. Let us examine why. By construction, k = (-jT + T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2) = T (T(f1-f2)/2-(f1+f2)/2) / ((f1-f2)/2-1) / ((f1-f2)/2+1) Thus d is bound to be a multiple of T, unless it happens that T has a nontrivial factor with (f1-f2)/2-1 or (f1-f2)/2+1. In summary, the method succeeds when, having factored (T^2-1)/4 as g1*g2, T happens to have a nontrivial factor in common with g1-g2-1, g1-g2+1, or g1+g2. Which is, almost never. Claude Leroy. === Subject: Re: Surrogate factoring, reasons for my concerns > James Harris has just proposed a possible factorization method. > I¹ll rephrase it in a manner that is more understandable to me, and > corrects a number of minor errors that crept in Jame¹s description. > Let us examine why. By construction, > k = (-jT + T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2) > = T (T(f1-f2)/2-(f1+f2)/2) / ((f1-f2)/2-1) / ((f1-f2)/2+1) > Thus d is bound to be a multiple of T, unless it happens > that T has a nontrivial factor with (f1-f2)/2-1 or (f1-f2)/2+1. That is correct. > In summary, the method succeeds when, having factored (T^2-1)/4 > as g1*g2, T happens to have a nontrivial factor in common with > g1-g2-1, g1-g2+1, or g1+g2. > Which is, almost never. Are you sure? Did you use full factorizations and ALL combinations of factors of T^2-1? My guess is that you didn¹t but presumed. Yes, it seems reasonable to conclude that it would be unlikely that you¹d have that factor ever pop out, but mathematics isn¹t about what seems reasonable, but about what you can prove. I thought the same thing originally. That is, I *jumped* to the conclusion you just made, which is why I would post that j is typically a fraction, then I thought about it, realized that I was just *presuming* so I backed away from that position. There are other approaches, but yup, you guessed it, part of the reason I came back to post on surrogate factoring was curiosity about this route I¹d so quickly dismissed. The mathematics here is difŽcult because part of it is so freaking tedious. You have to completely factor T^2 - 1 and consider ALL the possible values for j that result from ALL possible combinations of those factors before you can claim a particular conclusion. It¹s one of the reasons that I¹m not sure about how well it works with integer j as that¹s just not something I Žnd interesting to do. As an aside, I think the poster Claude Leroy was similarly lazy. James Harris === Subject: Re: Surrogate factoring, reasons for my concerns >[...] >At this point I¹m still just surprised at how hard it is to get a >straight answer on something that¹s high school algebra. So far, >posters have just come back with social issues mainly. Uh, you¹ve _gotten_ plenty of straight answers - people have pointed out that yes it _is_ just high-school algebra, and that the method has been studied already and determined to to work well enough to matter. You didn¹t _believe_ those replies, but saying you wonder why you haven¹t got any is curious even for you. The reason you get the off-topic replies on social issues is that you include those hilarious comments about how people are going to be in Trouble when the Truth Comes Out. >It¹s high school algebra. SOMEONE should have an answer. >James Harris ************************ David C. Ullrich === Subject: Re: Surrogate factoring, reasons for my concerns Couple of questions, James: - These journals that you have submitted your papers to: are they printed paper journals? - What are the names of the journals? Happy mathing, William > The other thread I created yesterday trying to update you all on > surrogate factoring got Žlled with a lot of distractions, but I think > now after answering several posts in it I can give an executive > summary type post. > Recently I found the factorization > (jk - Tk + T)(jk + Tk + T) = T^4 > and noticed that solving for k gave me > k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2) > where signiŽcantly, you have a dependency on the factors of T^2 - 1 > to get a rational k, as if sqrt(j^2 - T^2 + 1) is rational, then k is > rational. > Here T is the target number, the number that is to be factored. > Notice that > j = (-T +/- T sqrt(k^2 + T^2))/k > where you see something more like the traditional congruence of > squares, showing that j is dependent on the factors of T^2, so > somehow, someway, there¹s a clear linkage between the factorization of > T^2 - 1 and T^2, which is here easily seen. > So you can get integer values for j easily by taking the factors of > T^2-1, and using their sum to calculate j, as, for instance, > j = (f_1 + f_2)/2 where f_1 f_2 = T^2 - 1, > if T^2 - 1 has 4 as a factor, and with T odd, it will. > With j you can then get an integer k, and notice that if you let > k = d/c, and substitute you get > j = (-cT +/- T sqrt(c^2 + T^2 d^2))/d > where you know already that j is an integer, so > sqrt(c^2 + T^2 d^2) is an integer, so > c^2 + T^2 d^2 = n^2, so > T^2 d^2 = (n-c)(n+c) > and you might have just non-trivially factored T. > You have all the factors of T^2 - 1 to play with, where you can check > using all the combinations of them. > At this point in time I don¹t know mathematically why this method > wouldn¹t tend to factor T a signiŽcant percentage of the time. > Basically you use a factorization of T^2 - 1, to get n, c and d > integers, which gives you > T^2 d^2 = (n-c)(n+c) > and the potential of factoring T. > The idea in retrospect is kind of obvious, and its VERY simple, so > it¹s easy to think that it must not be new, but I¹ve yet to see anyone > come forward with information that shows it¹s not new. > If there isn¹t some block in the math that I¹m just not seeing then > potentially someone could factor an RSA number by being able to fully > factor T^2 - 1, which might seem easy since you get a head start with > T^2 - 1 = (T- 1)(T+1), but it still can be a hard task--but possibly > easier than factoring T directly. > There may be some block in the mathematics which causes you to not > non-trivially factor T with the path I¹ve shown, but maybe there¹s > some slight variation to that path or some way to pick j where it¹s > not an integer but is rational, or something else I missed that might > make for a practical approach. > The approach is in and of itself of interest, uses basic mathematics > at a high school or secondary school level, and CANNOT be missed by an > expert in the Želd as a potentially viable approach (unless you > people know something you haven¹t told me). > Therefore, if it is ignored, and someone out there uses it to do harm, > then I cannot see how experts in the Želd could escape liability. > There is a responsibility with titles. Math experts cannot fail to > see that something like this *might* be important and dangerous if it > works too well. > I¹m not a professional mathematician. I don¹t have a lot of options. > I already contacted the NSA and didn¹t hear futher from them, but > bureaucrats in the US government have failed before, and I don¹t have > to mention the signiŽcant recent intelligence failures. > It seems to me that someone out there on the Internet should have the > expertise to give some answer, and I¹m tired of the people who just > chatter and bring up irrelevant points, and I remind that yeah, they > may face liability later for their public statements *IF* this idea > works. > All I can do is put the information out there and hope that if it¹s > important someone will act, or that it just doesn¹t work for some > reason I don¹t know. > At this point I¹m still just surprised at how hard it is to get a > straight answer on something that¹s high school algebra. So far, > posters have just come back with social issues mainly. > It¹s high school algebra. SOMEONE should have an answer. > James Harris === Subject: Re: Surrogate factoring, reasons for my concerns > The other thread I created yesterday trying to update you all on > surrogate factoring got Žlled with a lot of distractions, but I think > now after answering several posts in it I can give an executive > summary type post. > Recently I found the factorization > (jk - Tk + T)(jk + Tk + T) = T^4 > and noticed that solving for k gave me > k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2) > where signiŽcantly, you have a dependency on the factors of T^2 - 1 > to get a rational k, as if sqrt(j^2 - T^2 + 1) is rational, then k is > rational. > Here T is the target number, the number that is to be factored. > Notice that > j = (-T +/- T sqrt(k^2 + T^2))/k > where you see something more like the traditional congruence of > squares, showing that j is dependent on the factors of T^2, so > somehow, someway, there¹s a clear linkage between the factorization of > T^2 - 1 and T^2, which is here easily seen. > So you can get integer values for j easily by taking the factors of > T^2-1, and using their sum to calculate j, as, for instance, > j = (f_1 + f_2)/2 where f_1 f_2 = T^2 - 1, > if T^2 - 1 has 4 as a factor, and with T odd, it will. > With j you can then get an integer k, and notice that if you let > k = d/c, and substitute you get > j = (-cT +/- T sqrt(c^2 + T^2 d^2))/d > where you know already that j is an integer, so > sqrt(c^2 + T^2 d^2) is an integer, so > c^2 + T^2 d^2 = n^2, so > T^2 d^2 = (n-c)(n+c) > and you might have just non-trivially factored T. > You have all the factors of T^2 - 1 to play with, where you can check > using all the combinations of them. > At this point in time I don¹t know mathematically why this method > wouldn¹t tend to factor T a signiŽcant percentage of the time. > Basically you use a factorization of T^2 - 1, to get n, c and d > integers, which gives you > T^2 d^2 = (n-c)(n+c) > and the potential of factoring T. > The idea in retrospect is kind of obvious, and its VERY simple, so > it¹s easy to think that it must not be new, but I¹ve yet to see anyone > come forward with information that shows it¹s not new. > If there isn¹t some block in the math that I¹m just not seeing then > potentially someone could factor an RSA number by being able to fully > factor T^2 - 1, which might seem easy since you get a head start with > T^2 - 1 = (T- 1)(T+1), but it still can be a hard task--but possibly > easier than factoring T directly. > There may be some block in the mathematics which causes you to not > non-trivially factor T with the path I¹ve shown, but maybe there¹s > some slight variation to that path or some way to pick j where it¹s > not an integer but is rational, or something else I missed that might > make for a practical approach. > The approach is in and of itself of interest, uses basic mathematics > at a high school or secondary school level, and CANNOT be missed by an > expert in the Želd as a potentially viable approach (unless you > people know something you haven¹t told me). > Therefore, if it is ignored, and someone out there uses it to do harm, > then I cannot see how experts in the Želd could escape liability. > There is a responsibility with titles. Math experts cannot fail to > see that something like this *might* be important and dangerous if it > works too well. > I¹m not a professional mathematician. I don¹t have a lot of options. > I already contacted the NSA and didn¹t hear futher from them, but > bureaucrats in the US government have failed before, and I don¹t have > to mention the signiŽcant recent intelligence failures. > It seems to me that someone out there on the Internet should have the > expertise to give some answer, and I¹m tired of the people who just > chatter and bring up irrelevant points, and I remind that yeah, they > may face liability later for their public statements *IF* this idea > works. > All I can do is put the information out there and hope that if it¹s > important someone will act, or that it just doesn¹t work for some > reason I don¹t know. > At this point I¹m still just surprised at how hard it is to get a > straight answer on something that¹s high school algebra. So far, > posters have just come back with social issues mainly. > It¹s high school algebra. SOMEONE should have an answer. > James Harris Blah blah blah. Grow up and get a life, James. You¹re not after truth like you say you are..........and you call us liars. Dave === Subject: Re: Surrogate factoring, reasons for my concerns > All I can do is put the information out there and hope that if it¹s > important someone will act, or that it just doesn¹t work for some > reason I don¹t know. OK. You have done that. Now shut up. > At this point I¹m still just surprised at how hard it is to get a > straight answer on something that¹s high school algebra. So far, > posters have just come back with social issues mainly. > It¹s high school algebra. SOMEONE should have an answer. That¹s *your* responsibility. > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Prime Number Void > Is there an open interval (p, p*p) where p is a prime number such that there are > no prime numbers in this interval? > -Ed L There is no such open interval: this can be deduced, e.g., from Bertrand¹s postulate according to which there is always a prime in the interval [n,2n) for n > 1. Robin Chapman has written up a version of Erdos¹s proof, which he has made available from his website at http://www.maths.ex.ac.uk/~rjc/rjc.html For a text version, see http://www.math.niu.edu/~rusin/known-math/97/bertrand One can also attack your particular problem directly; this is mentioned in passing in Erdos¹s proof of the Sylvester-Schur theorem [1]. Let me outline his argument. We need a few lemmas, beginning with the following observation: Lemma 1: For (positive integers) n >= 8, we have pi(n) <= n/2. This is easily veriŽed directly for n=8 and 9. The Lemma then follows by induction using the inequality pi(m+2) <= 1 + pi(m), which is certainly valid for m>=8 [since primes > 2 are odd!]. We also need Lemma 2: Let n be a positive integer and 0<=k<=n. In the factorization of the binomial coefŽcient (n choose k) into prime powers, no prime power exceeds n. Proof: The exponent on the highest power of the prime p dividing m!, where m is an arbitrary nonnegative integer, is given by [m/p] + [m/p^2] + [m/p^3] + ...; see [2, Theorem 4.16, p. 342]. Writing (n choose k) = n!/(k!(n-k)!) it follows that the exponent on the highest power of p dividing (n choose k) is sum([n/p^j]-[k/p^j]-[(n-k)/p^j], j=1..inŽnity). Let r be the largest integer exponent for which p^r <= n; then the terms of this sum where j > r vanish. Consequently, there are at most r nonvanishing terms. But every term of this sum is either 0 or 1, as can easily be checked using the property x-1 < [x] <= x. It follows that the sum here is at most r, and the result follows. We can now prove the following result, which implies a negative answer to your original question: Theorem: Let n >= 2 be a positive integer. Then there is always a prime in the interval (n,n^2]. Proof: The claim can be veriŽed manually for n = 2, 3, ..., 7, so we suppose n >= 8 in what follows. Consider the binomial coefŽcient (n^2 choose n). Suppose there are no primes in the interval (n,n^2]; then this coefŽcient is divisible only by primes <= n and we may write (n^2 choose n) = p 1^e 1 ... p k^e k, where the p i are distinct primes not exceeding n. By Lemma 2, each p i^e i is bounded by n^2. Hence (n^2 choose n) <= n^(2k). But since the p i are bounded by n, we have k <= pi(n) <= n/2 by Lemma 1 (here is where we need n>=8). Hence (n^2 choose n) <= n^n. On the other hand, (n^2 choose n) = (n^2)*(n^2-1)*...(n^2-n+1)/(n *(n-1)*...*2*1), = n^2/n * (n^2-1)/(n-1) * (n^2-2)/(n-2)... *(n^2-n+1)/1 > n^n, a contradiction. To close, let me mention another result of elementary prime number theory. Chebyshev showed that there are positive constants c 1 and c 2 with c 1 n/log n < pi(n) < c 2 n/log n for all large positive integers n. Note that establishing this result for constants c 1 and c 2 with 0 < c 2 < 2c 1 implies proof of Bertrand¹s postulate for large n. I want to Žnish by noting that the lower bound in Chebyshev¹s estimate (with a worse constant than he originally obtained) can be proved almost immediately using circle of ideas presented above. We prove: liminf {n -> inŽnity} pi(n)/(n/log n) >= log 2, For by Lemma 2, (n choose k) <= n^pi(k) and summing from k=0 to n gives 2^n <= (k+1) n^(pi(k)). One now only has to take logarithms and rearrange to obtain the result. There are different proofs which yield the same bound, but they seem more complicated to me. The proof given above is apparently due to Zagier and appears in [3]. Hope this helps, Paul [1] Erd.9as, P. A theorem of Sylvester and Schur. (English) Journal of the London Mathematical Society 9 (1934), 282-288. [2] Hardy, G. H.; Wright, E. M. An introduction to the theory of numbers. Fifth edition. The Clarendon Press, Oxford University Press, New York, 1979. [3] Zagier, Don The First Fifty Million Prime Numbers Math. Intelligencer 0 (1977), 7--19 === Subject: Re: Prime Number Void Originator: edlee@chinet.com (Edward Lee) I will credit you in the source code. -Ed L -- Due to the volume of spam that I receive, any email message to me which does not contain the word, mail, in the subject will be automatically deleted. Also, some spammer has been forging their return address to be mine. === === Subject: Re: Lie algebra - real life encounter === Subject: Re: L¹Hopital #2 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v9F06862; >We can end the discussion. >It turned out that applying L¹Hopital¹s rule to a fraction of the form >g + e >------ > g >where e/g -> 0 is valid if and only if >(e¹/g¹) / (e/g) -> 1. No. should have been iff e¹/g¹ -> 1. >We can show this very easily:. >1 ? = ? >g¹ + e¹ g + e >------ / ------ > g¹ g > g¹ + e¹ g >= --------*--------- > g¹ g + e >(Multiply the Žrst factor with (1/g¹) and the second with (1/g).) No. Should have been Multiply the top and bottom of the Žrst with (1/g¹) and that of the second with (1/g). > 1 + e¹/g¹ 1 >= ----------- * -------- > 1 1 + e/g > 1 + e¹/g >= ----------- (1) > 1 + e/g By hypothesis, e/g -> 0. And so the claim follows. H. Shinya === Subject: Re: Is there more symmetry to this function? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v7d06775; This is probably something everyone already knows... (since the function is very simple) Let h: N x N -> N be the function h(1,1) = 2, h(1,2) = 4, h(1,3) = 6, h(1,4) = 8, ... h(2,1) = 4, h(2,2) = 8, h(2,3) = 12, h(2,4) = 16, ... h(3,1) = 6, h(3,2) = 12, h(3,3) = 18, h(3,4) = 24, ... h(4,1) = 8, h(4,2) = 16, h(4,3) = 24, h(4,4) = 32, ... CLAIM OF THE CLUELESS ONE: The binary operation *: N x N -> N, * = h , is communative, associative, and distributive. p.s. As always, this function can be deŽned using the Quaternion/Floretion algebra at http://www.crowdog.de Sincerly, C. Dement === Subject: Re: Is there more symmetry to this function? > This is probably something everyone already knows... > h(1,1) = 2, h(1,2) = 4, h(1,3) = 6, h(1,4) = 8, ... so h(1,n)=2n > h(2,1) = 4, h(2,2) = 8, h(2,3) = 12, h(2,4) = 16, ... h(2,n)=4n > h(3,1) = 6, h(3,2) = 12, h(3,3) = 18, h(3,4) = 24, ... h(3,n)=6n > h(4,1) = 8, h(4,2) = 16, h(4,3) = 24, h(4,4) = 32, ... h(4,n)=8n Obviously you have h(m,n)=2mn > CLAIM OF THE CLUELESS ONE: > The binary operation *: N x N -> N, * = h , > is communative, associative, and distributive. Yes, that is obvious, as multiplication (m,n)->mn has those properties... Jaap === Subject: Re: Somebody tell me that Roger Bagula isn¹t a troll was Re: Now somebody tell me that Robin Chapman isn¹t a troll again?! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v5q06655; Forgive,forget; (he desribed himself in his posts earlier, isn¹t that adequate?). So that sci.math NG will also be placid. >Anyone? >-- >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 >Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Need help solving a differential equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v6m06728; >Hi all, > I am attempting to solve the following ordinary linear >differential equation with some boundary conditions. > y¹¹/y = b cosh ax > where a, b are constants. >I haven¹t had much success in getting an analytical solution. I would >appreciate any suggestions for deriving a close form solution. >Amal Just remember cosh(ax)=(exp(ax)+exp(-ax))/2 and d/dx (exp(ax) = a*exp(ax) ,let y(x)= k*(exp(ax)+exp(-ax)) and after double derivation identify... Alain. === Subject: Re: Need help solving a differential equation >> I am attempting to solve the following ordinary linear >>differential equation with some boundary conditions. >> y¹¹/y = b cosh ax >> where a, b are constants. >Just remember cosh(ax)=(exp(ax)+exp(-ax))/2 >and d/dx (exp(ax) = a*exp(ax) ,let y(x)= k*(exp(ax)+exp(-ax)) >and after double derivation identify... ...a^2 as b cosh ax ? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Another Mathematicians apology needed asap !! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v5O06629; >Let me guess: Do they talk about the butteržy effect? > Butteržy effect ad nauseum >I was right! BTW: Isn¹t there a butteržy bifurcation in catastrophe theory? Why isn¹t that more popular? >Thomas Oops , not Anon. Yes, it is in the catalogue of catastrophes in Catastrophe Theory by Zeeman and Rene Thom. http://www.saunalahti.Ž/jawap/colour/books/sudden.html There are seven elementary catastrophes; fold, cusp, swallowtail, butteržy,.. in 2D like the y^2=x^3 points, evolute of a parabola at cnter. IIRC, butteržy has opposite points of a square connected by straight lines,and one set of opposite sides connected between sharp cusps making it look somewhat like a bowtie. hyperbolic umbilic, elliptic umbilic, parabolic umbilic catastrophe... in 3D singularity at poles of sphere,equator of pseudosphere. All are Žlled with variations of inŽnite curvature (zero radius of curvature). It is ... stop, reverse gear,and changing direction.Is it also a model for chaos? === Subject: Re: does this simple relation between two subset families have a name? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1vAX06870; >For each set B in family 1 there exists a set A in family 2 such that >A intersect B is nonempty. What is the signiŽcance of a Name? As long as you know it is non empty. And maybe that it is not unique? Or sometimes unique? Zim Olson http://www.zimmathematics.com === Subject: divergence by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v6806689; Let a_n > 0 , sum {a_n} diverges.I should prove that sum {a_n/(1+a_n)} diverges. But I can¹t. Please give me some hints. === Subject: Re: divergence > Let a_n > 0 , sum {a_n} diverges.I should prove that > sum {a_n/(1+a_n)} diverges. But I can¹t. Please give me some hints. If {a_n} is bounded, then there is S = sup{a_n} >0. For every n, we have a_n/(1+a_n) >= a_n/(1+S). Since a_n>0 for every n and Sum(a_n) diverges, it follows Sum (a_n) -> inŽnity as n does. Since 1+S>0, it follows Sum (n=1 to n) {a_i/(1+a_i) > Sum (i=1 to i) a_i/(1+S) for every n, which implies Sum {a_n/(1+a_n) -> inŽnity as n -> inŽnity. Therefore, if a_n is bounded, then Sum {a_n/(1+a_n)} diverges. If {a_n} is unbounded, then we can choose inŽnitely many indexes n so as to make a_n/(1+a_n) = 1 - 1/(1+a_n) as close to 1 as desired, which implies the condition lim a_n/(1+a_n) = 0 - necessary to convergence - cannot be met. Therefore, sum {a_n/(1+a_n)} diverges again. Amanda === Subject: Re: divergence > > Let a_n > 0 , sum {a_n} diverges.I should prove that > > sum {a_n/(1+a_n)} diverges. If it were not the case, then a_n/(1+a_n) would go to 0, whence a_n would go to 0, and a_n ~ a_n / (1+a_n), whence a_n would converge. -- Julien Santini === Subject: Re: divergence > If it were not the case, then a_n/(1+a_n) would go to 0, whence a_n would go > to 0, and a_n ~ a_n / (1+a_n), whence a_n would converge. I mean sum(a_n) would converge, of course ... === Subject: Re: divergence > Let a_n > 0 , sum {a_n} diverges.I should prove that > sum {a_n/(1+a_n)} diverges. But I can¹t. Please give me some hints. Case 1: a_n is a bounded sequence. Case 2: it¹s not. === Subject: Re: Curvature of log curve by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1vA706894; >The graph of the equation y = log(x) has a visible knee (point of >maximum curvature). Out of curiosity, I wondered where this point >might be. The maximum curvature is given by (if I remember rightly): >dk/ds = 0 >where k = d^2y/dx^2 * [1 + (dy/dx)^2]^(3/2) >and s = sqrt(x^2 + y^2) No, ds^2=dx^2+dy^2 By direct method , x=1/sqrt(2),y=-log[sqrt(2)]gives maximim curvature. Alternately,we differentiate with respect to arc length rather than x, recommended for curvature problems.Remembering dy=ds sin(ph)=ds S, dx=ds cos(ph)=ds C , dy/dx=tan(ph) we get for y=log(x), curvature k= C-C^3 , dk/ds=S*(3 C^2-1) by simpliŽcation. We note k is maximum when tan(ph)= sqrt(2) or ph=(54.74 degrees) This occurs a bit below x=1,y=0 the point where log curve crosses x-axis. === Subject: Re: Is there a term for this set-theoretic relation? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1vAv06874; >There must be a standard term for the following pseudo covering >relation between two families of subsets of some arbitrary universe >set: >For each set B in family 1 there exists a set A in family 2 that meets >(has a nonempty intersection with) B. >If so, what is it? Please mention a reference if you can. >TIA, >Mark Bowron How about a Informational Relational (Non Axiomatic) System Transformation....I just made this up and use at: http://www.zimmathematics.com/app.htm Actually I heard of term Relational Transformation Žrst in A NEW KIND of SCIENCE by Stephen Wolfram. But his text provided hardly any further explanations except that they existed. Zim Olson http://www.zimmathematics.com === Subject: How do you solve : f^-1(x- f(y)) = sqrt(1+x - y^2) ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v6Q06697; Please do tell me if you know a solving method for such an equation; f is a real and bijective function . f^-1 the inverse. === Subject: Re: How do you solve : f^-1(x- f(y)) = sqrt(1+x - y^2) ? > Please do tell me if you know a solving method for > such an equation; f is a real and bijective function . > f^-1 the inverse. If the equation holds for every x and y, then in particular it holds when x = y^2. Ken Pledger. === Subject: Re: How do you solve : f^-1(x- f(y)) = sqrt(1+x - y^2) ? > > > > Please do tell me if you know a solving method for > > such an equation; f is a real and bijective function . > > f^-1 the inverse. > > > If the equation holds for every x and y, then in particular it > holds when x = y^2. so... f^-1(y^2- f(y)) = sqrt(1+ y^2 - y^2) y^2 - f(y) = f(1) y^2-f(1) = f(y) and for y=1 we get 1-f(1) = f(1) f(1) = 1/2 so the solution must be f(y) = y^2 - 1/2 Well, it¹s not bijective, is it? Proceeding anyway... f^{-1}(x) = (1/2) sqrt(4x+2) Let¹s try it... f^{-1}(x-f(y)) = (1/2) sqrt(4(x-y^2+1/2)+2) = sqrt(1+x-y^2), so it works. === Subject: Re: Paperback math reference books by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v6C06751; . . . > > You should be able to Žgure out that one of [the coins] > > is not a dime has the same amount of universality as one > > [of the books] has yet to relase a page.... > Kind of like, Some [of the] politicians are not idiots, > eh? > BTW, I think the original statement could have been > construed as, Not one has yet disintegrated. That is, given > a set of paperback reference book, P, and a set of reference > books that have disintegrated, D, following holds true, > There exists no element of P, p, s.t. p is an element of D > (or, simply, P int D = {}). indeed. if mr seaman¹s judgement is as sharp as his comprehension skills, i would hate to see what are the bases for his support of the convicted cop-killer Mumia Abu-Jamal. > Best wishes, > Andrew === Subject: Re: Paperback math reference books >> > You should be able to Žgure out that one of [the coins] >> > is not a dime has the same amount of universality as one >> > [of the books] has yet to relase a page.... >> Kind of like, Some [of the] politicians are not idiots, >> eh? >> BTW, I think the original statement could have been >> construed as, Not one has yet disintegrated. That is, given >> a set of paperback reference book, P, and a set of reference >> books that have disintegrated, D, following holds true, >> There exists no element of P, p, s.t. p is an element of D >> (or, simply, P int D = {}). > indeed. > if mr seaman¹s judgement is as sharp as his comprehension > skills, i would hate to see what are the bases for his > support of the convicted cop-killer Mumia Abu-Jamal. There is nothing wrong with my comprehension skills. I knew what the intended meaning was, but I pointed out the ambiguity. The Mumia Abu-Jamal case is off topic in sci.math, but you might start with what Amnesty International has to say at -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Relationship of volume to surface area on a spheroid by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v5v06679; >>>a=1; paint=4 Pi a (2 b + a)/3 ; >>>e=Sqrt[1-(b/a)^2];exact=2Pi a^2+Pi b^2/e Log[(1+e)/(1-e)] >>>Plot [{paint,exact},{b,0,2}]; >>>Plot [{paint,exact},{b,.9,1.1}]; >>Cool! Do you know how this subroutine could be written in QBasic? >I see what you implied. In this newsgroup homework is discouraged ;). >Yes,Mma was too smart to compute a result of a quotient when >neither the divisor nor dividend existed,for b>a : >Plot [{Log[(1+e)/(1-e)],e,Log[(1+e)/(1-e)]/e},{b,.01,3}]; >So, instead of the Wolfram formula for bcompute the areas separately for ba, as b=a is an important >bifurcation case of sphere. Computing the deŽnite integral >4 Pi a Int [ Sqrt[b^2+u^2*(a^2-b^2) du], 0ba involves inverse circular and hyperbolic functions, >then you can plot them on single b axis with Qbasic. Huh? Um, I was only referring to the programming syntax (QBasic I don¹t think has the PLOT command and I don¹t think all of the commas are for each PLOT (shouldn¹t a PLOT have either paint or exact?))! I havn¹t really used graphics that much. :/ In your followup is u the radius variable from b to a (b> alt.sci.time-travel: >> > Žnd this general explanation of space and matter on URL >> > 1 on micro physics >> > Chapter 2 on macro physics >> > >> > Discussion welcome >> > >> > www.geocities.com/garyforbat >> Well, the content is not very good. Chapters 1 and 2 are basically some kind >> of interview, where we don¹t know if Gary Forbat is the interviewer or the >> interviewee. You have to scroll all the way down to Advanced Principles >> to get to anything that seems to explain this theory. >> Advanced Principles sets forth what appears to be the axioms of Forbat¹s >> theory, most of which tend to be related to Psychology rather than space >> and matter. Thus the theory doesn¹t apply to space and matter but, maybe, >> to Psychology. >> The Žrst principle that seems to relate in a remote way to space and matter >> is principle 6, which simply redeŽnes time in such a way as to be >> incompatible with reality. >> I had to crack up on principle 10, which is simply a false statement. >> Einstein included empty space because empty space is žat space-time. Since >> space-time can be žat, empty space is possible in both SR and GR. In the >> end, Forbat claims, in capital letters, that space can not curve. However, >> this is equivalent to saying that when the space-time metric depends on >> position, it always depends in such a way that the curvature tensor is >> always equal to zero. Thus the existence of a force such as gravity would >> be impossible. We know that gravity is possible (example: you being >> attracted to Earth), thus the statement on Forbat¹s website about space >> not being able to curve is false. - I¹m not sure what he wanted to show by >> his comments on Binary Aspects. Nevertheless, the four coordinates of >> events are 3 coordinates of regular space and 1 coordinate of time for a >> total of 4 coordinates (dimensions). Space-time (or event-space as it¹s >> also called) will always have at least 4 dimensions because of that. >> > >> > Placed in a Žctional setting >> > Chapter >> ??? >have not read the main essay but went straight to the end line and the >summaries that follow. The Advanced Principles are meant to be read >after the essays, presupposing what is ithere, but you show no >indication whatsoever that you know the content since you mention only >minor issues and miss the main points,like the dynamic structuring >process etc....Please read the essay Žrst before commenting. >I have now rewritten and added to parts of the essay and began chapter >3 with clues of what is to follow. I hope you have some criticism of >the real issues. >TimeLord >YOU WROTE: >Advanced Principles sets forth what appears to be the axioms of >Forbat¹s >theory, most of which tend to be related to Psychology rather than >space >and matter. Thus the theory doesn¹t apply to space and matter but, >maybe, >to Psychology. >MY REPLY: >I do not see why you think it is psychology. Perhaps you could >elaborate on this. The only thing that comes to mind is Kant¹s theory >of space, that space is a Œsecondaty quality¹ like colours, and 3d is >not an actual reality but a mind dependent analysis of a non-spatial >object....or something like that....I believe Einstein was deeply >inžuenced by these ideas. Somehow I think this is not what you meant. >If you did mean this, then you will need to wait until I publish my >Critique of Einstein¹s Relativity theories. There are a number of >problems, one of which is the one I just mentioned. Other problems >include not discharging the hypothetical when deriving the square root >of minus one through inducing a hypothetical, thereby creating inverse >equivalences....and of course the problem of space matter >interactivity. >YOU WROTE: >The Žrst principle that seems to relate in a remote way to space and >matter >is principle 6, which simply redeŽnes time in such a way as to be >incompatible with reality. >MY REPLY: >This is a wild and unfounded statement that anybody can make. If you >want clariŽcation you need to be more speciŽc so I can answer you >objection. >YOU WROTE: >I had to crack up on principle 10, which is simply a false statement. >Einstein included empty space because empty space is žat space-time. >Since >space-time can be žat, empty space is possible in both SR and GR. >MY REPLY: >Yes you are cracked up on this point. As already explained above, my >objection is Einstein making matter and space interactive. My idea is >that space is an underlaying emptiness (though never found in empty >form)which forms the receptacle for matter to exist in. I say nothing >can interact with emptiness, and that is intuitively obvious. >YOU WROTE: >In the end, Forbat claims, in capital letters, that space can not >curve. However, this is equivalent to saying that when the space-time >metric depends on position, it always depends in such a way that the >curvature tensor is always equal to zero. Thus the existence of a >force such as gravity would be impossible. We know that gravity is >possible (example: you being attracted to Earth), thus the statement >on Forbat¹s website about space not being able to curve is false. >YOU WROTE: >Placed in a Žctional setting? >MY REPLY: >By questioning this it is proof you haven¹t read the work. It should >be obvious to anyone. >***************************** >***************************** >TMELORD, this appears a classical example of judging ideas from an >incompatible framework. Of course it will look incorrect if you take >the current theories as the standard of thought by which to judge it. >If and when you read my Œdialogues¹ open mind and follow each point on >its own merits. >TimeLord, I suggest you read the latest version and comment directly >on issues raised therein. I welcome your or a anybody else¹s >criticisms as long as they are speciŽc to the point and detailed in >your objection so I may respond with similar detail. >There is more to come when I update the Œdialogues¹in the upcoming >months. Please excuse the slow progress of the essay. Due to work >commitments in other directions I have only a few hours in each week >to work on it. Zim Olson here: I talk a little of Space, Time, Math. in different terms at my web site: http://www.zimmathematics.com A warning though that Sephen Wolfram in A New Kind of Science also attempts to explain physical phenomena in terms of Information Concepts. So I am not alone. Zim Olson === Subject: Re: L¹Hopital #2; Correction by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6C1v8F06813; A few Corrections: >We can end the discussion. >It turned out that applying L¹Hopital¹s rule to a fraction of the form >g + e >------ > g >where e/g -> 0 is valid if and only if e¹/g¹ -> 0. Remember that we assume e/g -> 0. >(e¹/g¹) / (e/g) -> 1. >We can show this very easily:. >1 ? = ? >g¹ + e¹ g + e >------ / ------ > g¹ g > g¹ + e¹ g >= --------*--------- > g¹ g + e >(Multiply the Žrst factor with (1/g¹) and the second with (1/g).) Multiply the numerator and denominator of the Žrst factor with 1/g¹ and that of the second with 1/g > 1 + e¹/g¹ 1 >= ----------- * -------- > 1 1 + e/g > 1 + e¹/g¹ >= ----------- (1) > 1 + e/g Since e/g -> 0, so must e¹/g¹. H. Shinya === Subject: Re: L¹Hopital #2; Correction > A few Corrections: > >We can end the discussion. > >It turned out that applying L¹Hopital¹s rule to a fraction of the form > >g + e > >------ > > g > >where e/g -> 0 is valid if and only if > e¹/g¹ -> 0. Remember that we assume e/g -> 0. > >(e¹/g¹) / (e/g) -> 1. > >We can show this very easily:. > >1 ? = ? > >g¹ + e¹ g + e > >------ / ------ > > g¹ g > > g¹ + e¹ g > >= --------*--------- > > g¹ g + e > >(Multiply the Žrst factor with (1/g¹) and the second with (1/g).) > Multiply the numerator and denominator of the Žrst factor > with 1/g¹ and that of the second with 1/g > > 1 + e¹/g¹ 1 > >= ----------- * -------- > > 1 1 + e/g > > 1 + e¹/g¹ > >= ----------- (1) > > 1 + e/g > Since e/g -> 0, so must e¹/g¹. > H. Shinya You have yet to state coherently what you¹re trying to prove. What is the result? State it precisely and completely. Do not omit any relevant hypotheses. === Subject: stochastics hello group, I was wondering if it¹s good idea to simulate a system that has , say.. upwards of 30 equations using stochstic simulation such as gillespie or next reaction method. if not, what else would you suggest? mesa === Subject: uniform continuity testing for x sin (x) It¹s nothing big really. Only i¹m missing some creativity over here. This is the problem. On an interval [1, inŽnity) we have a function f(x) = x sin(x) Now i have to Žnd 2 sequences x_n and y_n where: x_n - y_n goes to 0 in the limit of inŽnity, and f(x_n) - f(y_n) has some nonzero value . Then f is not uniform continues. === Subject: Re: uniform continuity testing for x sin (x) > It¹s nothing big really. > Only i¹m missing some creativity over here. > This is the problem. > On an interval [1, inŽnity) > we have a function f(x) = x sin(x) > Now i have to Žnd 2 sequences x_n and y_n where: > x_n - y_n goes to 0 in the limit of inŽnity, > and f(x_n) - f(y_n) has some nonzero value . > Then f is not uniform continues. Now, as y_n will be near x_n, (f(x_n)-f(y_n))/(x_n-y_n) will be near f¹(x_n). I suppose you want f(x_n) - f(y_n) to converge to some nonzero value, or diverge to inŽnity. In that case (f(x_n)-f(y_n))/(x_n-y_n) must diverge to inŽnity. It might therefore be a good start to seek a sequence (x_n) where f¹(x_n) diverges to inŽnity. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: how to measure how much vector is close to the zero vector Hi All, What is the best way to measure how much a vector v close to the zero vector. I am using: std(v) / mean(v) which is ok but it is hard to compute since i am using it in a minimization problem... Ira === Subject: Re: how to measure how much vector is close to the zero vector First of all, thank you all for your replies. I can not use the norm because my vector is up to scale factor which i don¹t know. So if I¹ll use the norm as my measure and my vector is multiplied by 100 it will have a very large norm... Do you have maybe other suggestions? Ira === Subject: Re: how to measure how much vector is close to the zero vector >I can not use the norm because my vector is up to scale factor which i don¹t >know. There is a sublime perfection about this which nearly makes me weep with pleasure. Lee Rudolph (it¹s just the cutthroat in me, I guess) === Subject: Re: how to measure how much vector is close to the zero vector What I meant is that if I have two vectors a*v1 and b*v2 and I would like to Žnd out which one (v1 or v2 ) is closer to the zero vector I can¹t use the norm for this. What I can use is: std/mean for example. But I thought that you might have some simpler suggestions. p.s. I didn¹t quite understand what did you mean by your answer though. > >I can not use the norm because my vector is up to scale factor which i don¹t > >know. > There is a sublime perfection about this which nearly makes me weep > with pleasure. > Lee Rudolph (it¹s just the cutthroat in me, I guess) === Subject: Re: how to measure how much vector is close to the zero vector > What I meant is that if I have two vectors a*v1 and b*v2 and > I would like to Žnd out which one (v1 or v2 ) is closer to the zero vector > I can¹t use the norm for this. > What I can use is: std/mean for example. But I thought that you might have > some simpler suggestions. > p.s. I didn¹t quite understand what did you mean by your answer though. What he means is that no one but you would consider the vector (10000,10000,10000,10000,10000) to be closer to the zero vector than (0.0000001,0,0,0,0) perhaps you couls clarify what you are actually asking for. >>>I can not use the norm because my vector is up to scale factor which i > don¹t >>>know. >>There is a sublime perfection about this which nearly makes me weep >>with pleasure. >>Lee Rudolph (it¹s just the cutthroat in me, I guess) === Subject: Re: how to measure how much vector is close to the zero vector actually i would consider v1=(0.0000001,0,0,0,0) be closer to zero than v2=(10000,10000,10000,10000,10000) because v2=10000 * (1,1,1,1,1) and v1 =(0.0000001,0,0,0,0) so i would like to compare (1,1,1,1,1) and (0.0000001,0,0,0,0) > > What I meant is that if I have two vectors a*v1 and b*v2 and > > I would like to Žnd out which one (v1 or v2 ) is closer to the zero vector > > I can¹t use the norm for this. > > What I can use is: std/mean for example. But I thought that you might have > > some simpler suggestions. > > p.s. I didn¹t quite understand what did you mean by your answer though. > What he means is that no one but you would consider the > vector (10000,10000,10000,10000,10000) to be closer to the zero vector > than (0.0000001,0,0,0,0) perhaps you couls clarify what you are > actually asking for. > >> > >> > >>>I can not use the norm because my vector is up to scale factor which i > > don¹t > >>>know. > >> > >>There is a sublime perfection about this which nearly makes me weep > >>with pleasure. > >> > >>Lee Rudolph (it¹s just the cutthroat in me, I guess) === Subject: Re: how to measure how much vector is close to the zero vector >actually i would consider v1=(0.0000001,0,0,0,0) be closer to zero than >v2=(10000,10000,10000,10000,10000) >because v2=10000 * (1,1,1,1,1) >and v1 =(0.0000001,0,0,0,0) >so i would like to compare >(1,1,1,1,1) and (0.0000001,0,0,0,0) But 10000 * (1,1,1,1,1) = 20000 * (1/2,1/2,1/2,1/2,1/2), and according to you (1/2,1/2,1/2,1/2,1/2) would be closer to null than (1,1,1,1,1)? === Subject: Re: how to measure how much vector is close to the zero vector > But 10000 * (1,1,1,1,1) = 20000 * (1/2,1/2,1/2,1/2,1/2), and > according to you (1/2,1/2,1/2,1/2,1/2) would be closer to null than > (1,1,1,1,1)? no, they are the same for me because (1/2,1/2,1/2,1/2,1/2) = 1/2* (1,1,1,1,1) === Subject: Re: how to measure how much vector is close to the zero vector >>But 10000 * (1,1,1,1,1) = 20000 * (1/2,1/2,1/2,1/2,1/2), and >>according to you (1/2,1/2,1/2,1/2,1/2) would be closer to null than >>(1,1,1,1,1)? > no, they are the same for me because > (1/2,1/2,1/2,1/2,1/2) = 1/2* (1,1,1,1,1) You are going to have to explain pretty precisely what you mean by close to the zero vector as you seem to be saying contradictory things , and certainly have a much different notion of closeness to the zero vector than everyone else. Why do you want this? === Subject: Re: how to measure how much vector is close to the zero vector > What is the best way to measure how much a vector v close to the zero > vector. The norm or length. === Subject: Anyone want to work for Google? A mysterious billboard with a mathematic message has sprung up in Silicon Valley. Many people thought that this billboard was a recruiting technique devised by Google to draw in top-notch engineers. They were right. See: John Ramsden (john_ramsden@sagitta-ps.cam) <--- com not cam === Subject: Re: Anyone want to work for Google? > A mysterious billboard with a mathematic message has sprung up in > Silicon Valley. Many people thought that this billboard was a > recruiting technique devised by Google to draw in top-notch engineers. > They were right. > See: > John Ramsden (john_ramsden@sagitta-ps.cam) <--- com not cam Do the candidates then have to Žgure out how to get U2 across the bridge in 17 minutes? === Subject: Re: Anyone want to work for Google? Module[{digits=RealDigits[E,10,len][[1]], count = 0}, For[i = 1, i <= len - 10, i++, seq = Take[digits,{i,i+9}]; sum = Apply[Plus,seq]; If[sum == 49, count++; Print[Found one: seq=,seq], Null]; If[count == 5, Return[Success!], Null]]; Print[Finished without success]]; alan === Subject: Re: Anyone want to work for Google? > A mysterious billboard with a mathematic message has sprung up in > Silicon Valley. Many people thought that this billboard was a > recruiting technique devised by Google to draw in top-notch engineers. > They were right. > See: Whence: One funny thing also mentioned on the Google (LABS) page is, We¹re tackling a lot of engineering challenges that may not actually be solvable. If they are, they¹ll change a lot of things. If they¹re not, well, it will be fun to try anyway. The tone and content of this could have been lifted straight from a JSH post Weird, huh? :) -- Larry Lard Replies to group please. === Subject: Help with unknown symbol I have come across a short hand (i.e. a symbol) that is new to me. Thm: Existence of nth roots in R+. For all x in R+ (E!)(y in R such that y^n = x). This is written with the usual symbols for for all and there exists, except that E! appears (with the E backwards for there exists). What does the ! mean? I have not seen this before. I thought I know all the symbols. Van === Subject: Re: Help with unknown symbol > I have come across a short hand > (i.e. a symbol) that is new to me. > Thm: Existence of nth roots in R+. > For all x in R+ (E!)(y in R such that y^n = x). > This is written with the usual symbols for for all and there > exists, except that E! appears (with the E backwards for > there exists). > What does the ! mean? I have not seen this before. > I thought I know all the symbols. It means: there exists a *unique* such-and-such, with the property that... === Subject: Re: Help with unknown symbol > I have come across a short hand (i.e. a symbol) that is new to me. > Thm: Existence of nth roots in R+. > For all x in R+ (E!)(y in R such that y^n = x). > This is written with the usual symbols for for all and there > exists, except that E! appears (with the E backwards for > there exists). > What does the ! mean? I have not seen this before. > I thought I know all the symbols. The ! means that there is exactly one sucht hing. Dirk Vdm === Subject: Re: Help with unknown symbol > > I have come across a short hand (i.e. a symbol) that is new to me. > > Thm: Existence of nth roots in R+. > > For all x in R+ (E!)(y in R such that y^n = x). > > This is written with the usual symbols for for all and there > > exists, except that E! appears (with the E backwards for > > there exists). > > What does the ! mean? I have not seen this before. > > I thought I know all the symbols. > The ! means that there is exactly one sucht hing. > Dirk Vdm === Subject: Re: Anyone want to work for Google? > A mysterious billboard with a mathematic message has sprung up in > Silicon Valley. Many people thought that this billboard was a > recruiting technique devised by Google to draw in top-notch engineers. > They were right. > See: > John Ramsden (john_ramsden@sagitta-ps.cam) <--- com not cam Upon making it to the second level of questioning, that is if you are able to, you are hit up with this question as seen on 7427466391.com: f(1)= 7182818284 f(2)= 8182845904 f(3)= 8747135266 f(4)= 7427466391 f(5)= __________ The solution to the equation is a password for you to continue on for the third level. ************ I have no idea how to proceed. Van === Subject: Re: Anyone want to work for Google? > [...] > Upon making it to the second level of questioning, that is if you are > able to, you are hit up with this question as seen on 7427466391.com: > f(1)= 7182818284 > f(2)= 8182845904 > f(3)= 8747135266 > f(4)= 7427466391 > f(5)= __________ > The solution to the equation is a password for you to continue on for > the third level. Possible spoiler: As all the numbers are the same length I assumed it was a bit-map, like in that Žlm Contact (?), so one has to write out all the 0s and 1s and then from the resulting image infer the Žnal line. But I haven¹t tried that. (As 9s are present, and none of the digits exceeds 9, the numbers are presumably decimal.) John Ramsden (john_ramsden@sagitta-ps.cam) <-- com not cam === Subject: Re: Anyone want to work for Google? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >f(1)= t 7182818284 >f(2)= t 8182845904 >f(3)= t 8747135266 >f(4)= t 7427466391 >f(5)= t __________ g(1) = 1415926535 g(2) = 3238462643 g(3) = 2384626433 g(4) = 4626433832 g(5) = __________ -- Richard === Subject: Re: Anyone want to work for Google? > f(1)= 7182818284 > f(2)= 8182845904 > f(3)= 8747135266 > f(4)= 7427466391 > f(5)= __________ > The solution to the equation is a password for you to continue on for > the third level. > ************ > I have no idea how to proceed. Hint: exp(1) Dirk Vdm === Subject: Re: Anyone want to work for Google? >> A mysterious billboard with a mathematic message has sprung up in >> Silicon Valley. Many people thought that this billboard was a >> recruiting technique devised by Google to draw in top-notch > engineers. >> They were right. >> See: >> John Ramsden (john_ramsden@sagitta-ps.cam) <--- com not cam > Upon making it to the second level of questioning, that is if you are > able to, you are hit up with this question as seen on 7427466391.com: > f(1)= 7182818284 digits 1-10 of the decimal expansion of exp(1) > f(2)= 8182845904 digits 5-14 of the decimal expansion of exp(1) > f(3)= 8747135266 digits 23-32 of the decimal expansion of exp(1) > f(4)= 7427466391 digits 99-108 of the decimal expansion of exp(1) > f(5)= __________ some early length-10 block of digits in the decimal expansion of exp(1)? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Anyone want to work for Google? >>> A mysterious billboard with a mathematic message has sprung up in >>> Silicon Valley. Many people thought that this billboard was a >>> recruiting technique devised by Google to draw in top-notch >> engineers. >>> They were right. >>> See: >>> John Ramsden (john_ramsden@sagitta-ps.cam) <--- com not cam >> Upon making it to the second level of questioning, that is if you are >> able to, you are hit up with this question as seen on 7427466391.com: >> f(1)= 7182818284 >digits 1-10 of the decimal expansion of exp(1) >> f(2)= 8182845904 >digits 5-14 of the decimal expansion of exp(1) >> f(3)= 8747135266 >digits 23-32 of the decimal expansion of exp(1) >> f(4)= 7427466391 >digits 99-108 of the decimal expansion of exp(1) >> f(5)= __________ >some early length-10 block of digits in the decimal >expansion of exp(1)? Sure. f(5) = 5515108657 digits 411-420 of the decimal expansion of exp(1). In general, f(n) begins at the index (1 + 5*4^(n-1))/3 - 2^(n-1). Not as pretty as the solution 7+1+8+2+8+1+8+2+8+4 = 49 offered elsewhere in this thread, but it is a good solution nevertheless. -- rr === Subject: Re: Anyone want to work for Google? > >> > >>> A mysterious billboard with a mathematic message has sprung up in > >>> Silicon Valley. Many people thought that this billboard was a > >>> recruiting technique devised by Google to draw in top-notch > >> engineers. > >>> They were right. > >>> > >>> See: > >> > >>> > >>> John Ramsden (john_ramsden@sagitta-ps.cam) <--- com not cam > >> > >> > >> Upon making it to the second level of questioning, that is if you are > >> able to, you are hit up with this question as seen on 7427466391.com: > >> > >> f(1)= 7182818284 > >digits 1-10 of the decimal expansion of exp(1) > >> f(2)= 8182845904 > >digits 5-14 of the decimal expansion of exp(1) > >> f(3)= 8747135266 > >digits 23-32 of the decimal expansion of exp(1) > >> f(4)= 7427466391 > >digits 99-108 of the decimal expansion of exp(1) > >> f(5)= __________ > >some early length-10 block of digits in the decimal > >expansion of exp(1)? > Sure. f(5) = 5515108657 > digits 411-420 of the decimal expansion of exp(1). > In general, f(n) begins at the index (1 + 5*4^(n-1))/3 - 2^(n-1). > Not as pretty as the solution 7+1+8+2+8+1+8+2+8+4 = 49 offered elsewhere > in this thread, but it is a good solution nevertheless. Indeed :-) Here¹s another Œgood¹ solution: Digits 277-286 where f(n) begins at index 22/3*n^3 - 37*n^2 + 191/3*n - 33 And this one is particularly Œgood¹ as well: http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anu m=A084615 with f(n) = sum of absolute values of coefŽcients of (1+x-3x^2)^n, giving digits 401-410 ;-) Dirk Vdm === Subject: Re: Anyone want to work for Google? > > Upon making it to the second level of questioning, that is if you are > > able to, you are hit up with this question as seen on 7427466391.com: > > > > f(1)= 7182818284 > digits 1-10 of the decimal expansion of exp(1) You zero-counter you! I¹d say the Žrst digit of the decimal expansion e was Œ2¹ So that 7 is the 2nd digit. > > f(2)= 8182845904 > digits 5-14 of the decimal expansion of exp(1) 6 > > f(3)= 8747135266 > digits 23-32 of the decimal expansion of exp(1) 24 Ooh, ooh, factorial? > > f(4)= 7427466391 > digits 99-108 of the decimal expansion of exp(1) 100 bugger. > > f(5)= __________ > some early length-10 block of digits in the decimal > expansion of exp(1)? I thought this was tempting: 1 | index(f(1)) 2 | index(f(2)) 3 | index(f(3)) 4 | index(f(4)) Alas, it¹s useless, and doesn¹t hold true again until f(10) at index 430, f(32) at 1120, and f(43) at 1419. Phil -- 1st bug in MS win2k source code found after 20 minutes: scanline.cpp 2nd and 3rd bug found after 10 more minutes: gethost.c Both non-exploitable. (The 2nd/3rd ones might be, depending on the CRTL) === Subject: Re: Pointless Topology >[...] > >To Žnd out more, try to obtain Johnstones Stone spaces book from > >some library (last time a looked, the US did have libraries :-). > [...] What the local public library > does offer pertinent to order theory, tho not the lastest edition, is > Davey, B. A., Introduction to Lattices and Order Actually, I had in mind the library of a nearby university. They should provide access also for non-students. Of course I do not know if there is a university sufŽciently near to your place. Marc === Subject: Re: Pointless Topology > >To Žnd out more, try to obtain Johnstones Stone spaces book from > >some library (last time a looked, the US did have libraries :-). > Deliberately under funded by excuse of Budget Busting Bush¹s Blunders for > being a despised anti-capitalistic remnant of American socialism within US > modern socially engineered hyper-commercial society wherein money is the > measure of all things human and divine. What the local public library > does offer pertinent to order theory, tho not the lastest edition, is > Davey, B. A., Introduction to Lattices and Order On the other hand... http://www.independent.org/tii/content/press_rel/press_040624 .html -- http://hertzlinger.blogspot.com === Subject: Re: Does a high SAT score predict mathematical talent? > Introduction. > Forgive me for any of the following replies that seem off-topic, but teacher unions are a necessary evil in our badly-broken system > of public education, whether it be in the elementary schools or beyond. The unions and the prevalence of > teachers-who-are-not-really-teachers are mere symptoms. I don¹t claim to have any solutions and suspect there are none in a society > that seems, in general, not to value academic accomplishment, the good that it accrues from it, and what it takes to sustain it. > Maybe the current system of tax-subsidized day-care is the best that is possible under the circumstances. > Agreed (in part). But if you think the problem is all about teachers, then you have not examined the administrative ranks. > Administrators hire, in some instances pick the textbooks, and have no desire to be in the classroom (in general). Need I comment > on their love of learning, thinking, or subject (ANY subject other than education psychobabble -- in general)? > > . . . As long as they [teachers] are both underpaid > > and protected by unions, this is unlikely to change. > Agreed (in part). Those same unions that protect the incompetent are nearly as necessary to protect competent teachers from > harassment. Administrators (in general) are promoted from the ranks of math-hating teachers eager to escape the classroom or are > career-administrators (like CEOs hopping from one corporation to another without a knowledge of the business at a fundamental level) > and don¹t know good teaching from economical entertainment. Admittedly, good teaching is difŽcult to deŽne, thus teacher > evaluation is often a popularity contest, especially when the evaluators themselves, in general, aren¹t competent to judge. > Agreed, but regarding teachers as professionals is unlikely to happen because it would threaten the status of top administrators. > They would have to admit that teachers know something, implying that the professional teachers should 1. select textbooks and > other materials, 2. have a major inžuence on curriculum, 3. be free to experiment based upon subject-driven considerations > (not ed psychobabble-driven fads prescribed by administrators, 4. be free to grade honestly (we¹ve all heard of grade inžation > and social promotion) and expell disruptive or dangerous students and be supported by administrators (not attacked by them, in > general) when parents complain inappropriately, . . . , in short, something resembling SHARED GOVERNANCE would follow logically. Good point. I taught HS algebra for a while, and had a combination of very motivated students and some who could care less about math and education in general. They were only there because at the time there was an economic motivation for them to attend, which I never understood. Anyway, there were a few disruptive students, and I eventually expelled them and told the administration it was them or me. They needed me so I got my way and the class went well after that. Motivated students are a pleasure to teach at any level. Your point about the low regard, in fact the distain that society has for academics of all sorts, is well taken. I will never forget how I was ridculed and persecuted for being a nerd, and then an academic type, who liked school and math, of all things. Van === Subject: Re: Does a high SAT score predict mathematical talent? > Your point about the low regard, in fact the distain that society > has for academics of all sorts, is well taken. I will never > forget how I was ridculed and persecuted for being a nerd, > and then an academic type, who liked school and math, of all > things. This is really an interesting area to me. I have known many on both sides of the educational divide, people with widely differing talents who become bullheaded in their own ways and battle lines are drawn between the blue collar and the white. I was blessed with extensive experience in both areas and so I dont take sides in general. But it¹s very interesting to look back on history through that lens, as it is a very undeniable fact which is¹nt spoken about much. This division in society is really very pronounced, and very unfortunate indeed. I can tell you that it¹s possible to have the best of both worlds. You can weld, AND do calculus. You can cast metals AND enjoy trigonometry. Hell - these days you could be a sports champion AND be gay. Sorry to hear about your experiences. It is deŽnately unfair. === Subject: Re: Does a high SAT score predict mathematical talent? > >Good point. I taught HS algebra for a while, and had a combination >of very motivated students and some who could care less about math >and education in general. They were only there because at the time >there was an economic motivation for them to attend, which I never >understood. >Anyway, there were a few disruptive students, and I eventually >expelled them and told the administration it was them or me. >They needed me so I got my way and the class went well after that. >Motivated students are a pleasure to teach at any level. My high school algebra teacher (bless that man¹s mother forever) gave extra lessons very early in the morning for those of us who wanted to learn real algebra. The rules were that anybody who wanted to learn could show up. A couple of the disruptive[1] unmotivated students showed up for every class. [1] Nobody stayed disruptive for more than 2 milliseconds in that teacher¹s class. /BAH /BAH Subtract a hundred and four for e-mail. === Subject: Re: sci.math trolls: Robin Chapman is the British version of David C.Ullrich > I think it is only fair to warn posters that they will be unreasonably > attacked > by > certain lurkers in this newsgroup. > Two major offenders are Robin Chapman: > http://www.maths.ex.ac.uk/~rjc/rjc.html > and David C. Ullrich. 1) They are anything but lurkers. 2) I have yet to see an unreasonable attack by either of them, though David Ullrich does seem to have a low tolerance for people who are simultaneously arrogant and ignorant. > I¹m not suggesting that either is not a very well educated man with a > ph. D. from a major university in mathematics or even that they don¹t > know their subjects ( some times exaggerating their knowledge?!), > but that they are unreasonable men who hurt others instead of educating. They do educate. Sometimes the lesson is that a person has made themselves uneducable through willful ignorance. > I¹m sure that they are not alone, but my experience with these trolls is > that they don¹t care about the facts involved once they get mad they > just plain go for the throat. Hint: a troll just tries to cause trouble. They do not. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: sci.math trolls: Robin Chapman is the British version of David C. Ullrich >>> BTW, I am trained in physics and used to post in that newsgroup, >>> but the ratio of scientists to crackposts is low. >>> This group has a pretty high ratio. Van >>Are there any ideas on how to get sci.physics people not to >>cross-post here? > Aw, leave Baez alone, he¹s harmless. ;) I¹ll make an exception for John B, but what about Sarfatti and Shead? Do they enrich our mathematical experience? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: sci.math trolls: Robin Chapman is the British version of David C. Ullrich >>>> BTW, I am trained in physics and used to post in that newsgroup, >>>> but the ratio of scientists to crackposts is low. >>>> This group has a pretty high ratio. Van >>>Are there any ideas on how to get sci.physics people not to >>>cross-post here? >> Aw, leave Baez alone, he¹s harmless. ;) >I¹ll make an exception for John B, but what about Sarfatti and >Shead? Do they enrich our mathematical experience? Sarfatti is a lost cause and has just realized that. Shead would be an enriching mathematical experience by watching the technique of the guy who Žnally manages to teach Shead how to divide by two. /BAH Subtract a hundred and four for e-mail. === Subject: Re: sci.math trolls: Robin Chapman is the British version of David C. Ullrich >>>> BTW, I am trained in physics and used to post in that newsgroup, >>>> but the ratio of scientists to crackposts is low. >>>> This group has a pretty high ratio. Van >>>Are there any ideas on how to get sci.physics people not to >>>cross-post here? >> Aw, leave Baez alone, he¹s harmless. ;) >I¹ll make an exception for John B, but what about Sarfatti and >Shead? Do they enrich our mathematical experience? I suspect that you missed the joke here... ************************ David C. Ullrich === Subject: Re: sci.math trolls: Robin Chapman is the British version of David C. Ullrich >>>>> >>>>>>[...] >>>>>>They¹re both very good at spotting >>>>>>mistakes and writing them up in English ASCII. >>>>> >>>>>I¹m good at writing up _mistakes_? Well *&%! you then. >>>> >>>>ROTFLMAO. >>>So you assume I was just kidding? Kids today, ain¹t >>>got no respect, lemme tell ya... >>Yessir, bosssir. >That¹s more like it. (We can revoke your membership, >you know.) Membership?!!! Uh-oh. Now what am I responsible for? /BAH Subtract a hundred and four for e-mail. === Subject: U.C. hello...doctor~ sigma {(-1)^n} / (n+x) n=0~00 (x >= 0) show that uniformly converge. -------------------------- i think...... S = sigma {(-1)^n} / (n+x) = {1/x} - {1/(1+x)} + {1/(2+x)} - ....... let S_n be partial sum. |S - S_n| <= 1 / (n+x+1) by alternating series property. so |S - S_n| <= 1 / (n+x+1) <= 1 / n = M_n M_n -> 0 By M-test, uniformly converge. um.....right ?? let me advice, please~ thank you very much. === Subject: Re: U.C. > sigma {(-1)^n} / (n+x) > n=0~00 > (x >= 0) > show that uniformly converge. > -------------------------- > i think...... > S = sigma {(-1)^n} / (n+x) > = {1/x} - {1/(1+x)} + {1/(2+x)} - ....... > let S_n be partial sum. > |S - S_n| <= 1 / (n+x+1) > by alternating series property. > so > |S - S_n| <= 1 / (n+x+1) <= 1 / n = M_n > M_n -> 0 > By M-test, uniformly converge. Yes, that¹s correct. Jose Carlos Santos === Subject: Re: Anyone want to work for Google? > >f(1)= 7182818284 > >f(2)= 8182845904 > >f(3)= 8747135266 > >f(4)= 7427466391 > >f(5)= __________ > g(1) = 1415926535 > g(2) = 3238462643 > g(3) = 2384626433 > g(4) = 4626433832 > g(5) = __________ > -- Richard I should have seen that. I didn¹t look very carefully (excuses .. :-) Pi--at least from g(1). I going to look at the expansion for pi now. Van === Subject: Re: Anyone want to work for Google? > > >f(1)= 7182818284 > > >f(2)= 8182845904 > > >f(3)= 8747135266 > > >f(4)= 7427466391 > > >f(5)= __________ > > g(1) = 1415926535 > > g(2) = 3238462643 > > g(3) = 2384626433 > > g(4) = 4626433832 > > g(5) = __________ > > -- Richard > I should have seen that. I didn¹t look very carefully (excuses .. :-) > Pi--at least from g(1). I going to look at the expansion for > pi now. They are all sequences of digits of Pi. f is mapped to sequences of digits of e. Wilbert === Subject: Re: Anyone want to work for Google? > > >f(1)= 7182818284 > > >f(2)= 8182845904 > > >f(3)= 8747135266 > > >f(4)= 7427466391 > > >f(5)= __________ I don¹t see how to Žnd out where to start in the sequence though. I thought maybe n^^3 - n - 1 would help, but it only works for n = 2 and 3. Van === Subject: Re: Anyone want to work for Google? > > > >f(1)= 7182818284 > > > >f(2)= 8182845904 > > > >f(3)= 8747135266 > > > >f(4)= 7427466391 > > > >f(5)= __________ > > > > I don¹t see how to Žnd out where to start in the sequence though. > I thought maybe n^^3 - n - 1 would help, but it only works for > n = 2 and 3. Hint - A 7-year-old with a decimal expansion of e could probably work this out more quickly than a 17-year-old. Don¹t look for something clever. Phil -- 1st bug in MS win2k source code found after 20 minutes: scanline.cpp 2nd and 3rd bug found after 10 more minutes: gethost.c Both non-exploitable. (The 2nd/3rd ones might be, depending on the CRTL) === Subject: Re: Anyone want to work for Google? You folks are aware that they based the price of their IPO on e, right? It was in the newspapers. > > > > >f(1)= t 7182818284 > > > > >f(2)= t 8182845904 > > > > >f(3)= t 8747135266 > > > > >f(4)= t 7427466391 > > > > >f(5)= t __________ > > > > > > > > I don¹t see how to Žnd out where to start in the sequence though. > > I thought maybe n^^3 - n - 1 would help, but it only works for > > n = 2 and 3. > Hint - > A 7-year-old with a decimal expansion of e could probably work > this out more quickly than a 17-year-old. Don¹t look for > something clever. > Phil === Subject: Re: Anyone want to work for Google? > You folks are aware that they based the price of their IPO on e, > right? It was in the newspapers. It took me one look at the numbers to spot the exp(1) for the Žrst two. Then I veriŽed the other with PARI. A few tries with the series 1,5,23,99 didn¹t work out, so I decided the solution must be a little more devious. Adding the digits was a lucky hunch. What is an IPO? doesn¹t reveal much. And when was this thing in which newspapers? Dirk Vdm === Subject: Re: Anyone want to work for Google? > > > >f(1)= 7182818284 > > > >f(2)= 8182845904 > > > >f(3)= 8747135266 > > > >f(4)= 8747135266 > > > >f(5)= __________ > > > > I don¹t see how to Žnd out where to start in the sequence though. > I thought maybe n^^3 - n - 1 would help, but it only works for > n = 2 and 3. That¹s the hard (and silly) part: 7+1+8+2+8+1+8+2+8+4 = 49 8+1+8+2+8+4+5+9+0+4 = 49 8+7+4+7+1+3+5+2+6+6 = 49 8+7+4+7+1+3+5+2+6+6 = 49 Enjoy :-) Dirk Vdm === Subject: Re: Anyone want to work for Google? > > > > >f(1)= 7182818284 > > > > >f(2)= 8182845904 > > > > >f(3)= 8747135266 > > > > >f(4)= 8747135266 > > > > >f(5)= __________ > > > > > > I don¹t see how to Žnd out where to start in the sequence though. > > I thought maybe n^^3 - n - 1 would help, but it only works for > > n = 2 and 3. > That¹s the hard (and silly) part: > 7+1+8+2+8+1+8+2+8+4 = 49 > 8+1+8+2+8+4+5+9+0+4 = 49 > 8+7+4+7+1+3+5+2+6+6 = 49 > 8+7+4+7+1+3+5+2+6+6 = 49 oops, and 7+4+2+7+4+6+6+3+9+1 = 49 > Enjoy :-) > Dirk Vdm === Subject: Re: comprehensive theoretical textbook > Are there any good comprehensive handbook/textbooks of Mathematics that > present all the theories of elementary and advanced Mathematics > (undergraduate level), including Algebra, Number Theory, Discrete > Mathematics, Geometry, Calculus and Analysis, etc. using an > axiomatic/formalistic approach - axioms, theorems with complete proofs, > corollaries etc.? Which one would you recommend? > Frank Check out the sci.math FAQ at www.faqs.org/faqs/ It has a lot of good stuff too, besides a list of texts. Van === Subject: Re: JSH: Sweep likely ..the usual stuff > James Harris Thats telling them James. Van X-mailer: xrn 9.02 === Subject: Re: MY LIST of the subsets of N Mail-To-News-Contact: abuse@dizum.com >> Any suggestions on how to go about Žnding a mapping that would show >> that either |R|>=|F| or |R|>=|E|? Or, any comments on which of those >> I should be trying to *dis*prove? >For |R|>=|F|, consider the fact that the set of real numbers in [0,1) >has the same cardinality as R. For the set of subsets of any particular >size, you can just interleave the digits in a particular representation. >For all Žnite subsets, you could precede the interleaved digits with an >indication of the size of the subset. Ah, just like proving (or at least demonstrating) that |RxR|=|R|. Okay, so {0.13235..., 0.78669...} would map to 2.1738263659... and also to 2.7183626395... (unless my eyes are crossed). Since this maps F onto R, but is not 1-1, it shows that |R|>=|F|. Combined with what I already had, we¹ve got |F|=|R|. >For |E| you can do something a bit different for the countable subsets: >Put one element of the set in the odd digits, put another one in digits >corresponding to odd multiples of 2, and so forth. The and so on being the odd multiples of the nth power of two, right? Is there any reason that this couldn¹t be used for comparing |R| and |F| as well? The only reason that I can see is overkill: you¹d end up with a lot (in some sense) of zeroes in most (whatever that means) of your numbers. The good news is that I can also see why this would *not* work for mapping between D and R. Georg can rest easy. Now, I¹m going to generalize. Comments from the peanut gallery are eagerly awaited. 1. The subset of P(Z) that only contains sets with cardinality <|Z| has cardinality |Z|. 2. The subset of P(R) that only contains sets with cardinality <|R| has cardinality |R|. These two observations lead me to speculate that, for any inŽnite set, X, the subset of P(X) that only contains sets with cardinality <|X| has cardinality |X|. (|X|>=|Z|) ==> (| {x: (x e P(X)) && (|x| < |X|) } | = |X|) (I think that I said that right.) Is my speculation true? false? open? undecidable? uninteresting? -- Michael F. Stemper #include COFFEE.SYS not found. Abort, Retry, Fail? === Subject: Re: MY LIST of the subsets of N > These two observations lead me to speculate that, for any inŽnite set, > X, the subset of P(X) that only contains sets with cardinality <|X| has > cardinality |X|. What if CH is false: aleph_1 < 2^aleph_0? The number of countable subsets of a set of cardinality aleph_1 is at least 2^aleph_0. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: MY LIST of the subsets of N > > > > > This takes into account Cantors diagonal argument which purports to > > > > > show that there is no complete list of the Žnite Natural numbers. > > > > > iF YOU CONSIDER THE SET OF NATURAL NUMBERS TO HAVE A DEFINITE > > > > > CARDINALITY xO, THEN: THE SUMS REFERNCED HERE ARE (I HOPE) ALL HAVE A > > > > > DEFINED NUMBER OF TERMS > > > > > > > > > > > > > > > > > > > > > > IF A = N+ then Ind(A)=1+2+4+8+16+etc... (Xo terms) > > > > > > > > > > This is the sum of the numbers listed in collumn 2 of the printout. > > > > > The set N+ itself id the last entry in the list because its index > > > > > number is greater than the INDEX NUMBER of any proper subset of > > > > > itself. > > > > > > > > > > > > > The latter phrase is true for any other unbounded subset also. > > > > > > > > > > > > > Since 0 is not an elt of N+; I can use 0 as a place holder to > > > > > represent elt of N+ not in a listed set. > > > > > > > > >- - - - - - - -- -- - - - -- - - - - - - - -- - - - - -- - - - - - - > > > -- > > > Comment 1 > > > This seems interesting to me but I wish you would explain it so I > > > can understand it better: I note that you claim that the range? is > > > the set of natural numbers: I think that this practice of presuming > > > the range to be something is dangerous. We know that every set{even > > > the set N) has more subsets than elements. This means that N is a > > > proper subset of the range of the function if the domain is the set of > > > all subsets of N. --or not?-- > > > f:Domain=> range. Right? > > > > > > Every function has a domain and a range. I think now I misinterpreted > > what you were doing, especially with respect to 0. Below I offer an > > alternative explanation. First a side note - > > > > You deŽned N+ as follows: > > > > N+ = {1, 2, 3, 4, ... | n elt N+} > > > > which is of course a circular deŽnition: the deŽnition of N+ > > has N+ itself on the right side of the equation. Perhaps you > > had a misprint. > > > > > > The following is my current understanding of what you were > > trying to do: > > > > You are deŽning an index for every subset A of N as follows: > > > > If A = {a1, a2, a3, ... } then > > > > Ind(A) = 2^(a1 - 1) + 2^(a2 - 1) + 2^(a3 - 1) + ... > > > > For Žnite sets, this is not an unreasonable index, and > > in fact it is a one-to-one function from the set of Žnite > > subsets of N into N itself: that is, if Ind(A) = Ind(B), > > then A = B. You can show this by noting that there is a > > correspondence between Ind(A) and a binary expansion: for > > example, if A = {2, 5, 6, 8}, then you could represent your > > index in binary as: > > > > Ind(A) = .01001101 > > > > But for inŽnite sets, what is the meaning of Ind(A) ? > > I believe you are thinking of it as a member of some kind of ordered > > inŽnite set. You are thinking that bigger subsets are farther > > along in the list: that is, if A is contained in B, then > > Ind(A) < Ind(B). > > > > Again writing the index in reverse binary, > > > > Ind(N) = .111111111... > > > > Note that if E = {2, 3, 4, 5, 6, ...} then > > > > Ind(E) = .011111111... > > > > Note that if D = {1}, then > > > > Ind(D) = .100000000... > > > > So what your index deŽnes is actually a function from the > > subsets of N to the real numbers between 0 and 1. It is not > > a one-to-one function (because Ind(E) = Ind(D), for example), but > > it is a surjection. This is sufŽcient to show that the set of > > all subsets of N has cardinality at least as large as that of the > > set of all real numbers between 0 and 1. The binary version of > > your index function also does have the property that if A is a subset > > of B, then > > > > Ind(A) < Ind(B). > > > > There is clearly a partial inverse function: given a real number > > between 0 and 1, express it in binary and Žnd the corresponding > > subset of N: for example, > > > > x = .011011001, > > > > then deŽne H(x) = {2, 3, 5, 6, 9}. > > > > The function H is ambiguous for numbers which terminate in > > all 1¹s because there are two ways to write such numbers. To remove > > the ambiguity, always take the expansion terminating in 0¹s: for > > example, if > > > > x = .01001000111111111..., then also > > > > x = .01001001000000000..., > > > > so deŽne H(x) = {2, 5, 8}. > > > > With this deŽnition, H is a well-deŽned function and it > > is one-to-one. Moreover, > > > > Ind(H(x)) = x, > > > > although it is not true that H(Ind(A)) = A. > > > > > > I think this is a clearer way of describing what you were > > actually doing: deŽning a function from the subsets of N > > onto the real numbers between 0 and 1. This is the sense in > > which you were creating a list. What most mathematicians > > would mean by list is a sequence of items indexed by the > > positive integers. Clearly your list is longer than > > the positive integers. > > > > None of this is new. You have only rephrased very well known > > results in your own language. > > > > Andrzej > This is sure some juicy feedback,Andrzej! You are a product of your > experience and education, just as I am. I have been working under the > delusion that we can consider not only 0.00011100110000000* > as an ACTual number but also the number0.00011101111111111* > We might be able to proceed with some not impossible working > assumptions. > We consider ther to be a MASTER number much in > the sense that the ordinal number omega or its cardinal assciate Xo > are greater than any pf their predecessors. We might than be able to > express the cardinality ] > of a set A = {1,4,6,25,39} to be [1/1+4/4+6/6+25/25+39/39] and > similarly for the so called inŽnite sets. > If the set is inŽnite then it¹s cardinality is an inŽnite set > of summands. In like manner we can consider each and every > binary expression to have a unique value or identity. No more any of > this stuff about numbers having an ambiguious representation. > We have then the ability to say that A is a proper subset of > B.(lease use the notation A < B.) implies that Card(A) is < Card(B). > and so on. All this is somewhat classiŽed and I can¹t go into full > detail now since my ideas are still in formative stage. I am going to > give you comment all the attention I am sure it deserves. I am > impressed so far and I hope to have a better reply soon> > David P. Ferguson 7/8/04 > >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>> Thanx for the comments. I must note again that you are not using the standard deŽnition of cardinality. You have deŽned a function g from P(N) to [0, 1] which is one-to-one on all but a countable set of subsets, where P(N) is the set of all subsets of N. Moreover, it has the nice property that if A is a proper subset of B (i.e., A < B), then g(A) < g(B). Thus you have a function which is one-to-one except on the set of complements of Žnite sets, and which preserves (partial) order. But it is wrong to assume that g(A) can be identiŽed as the cardinality of A. That is simply not the standard deŽnition and not equivalent to it. You can invent your own terminology of course, but it is not a good idea to re-invent standard terminology. It just leads to confusion. Andrzej > > > Ferguson 7/3/04 > > > - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - -- - > > > - -- > > > > You thus appear to be deŽning a function f(S) whose domain includes > > > > any subset S of N, and which takes values in N + {0}. > > > > > > > > What you need is that this function is 1-1. > > > > > > > > It isn¹t. Any unbounded subset S of N is such that f(S) = 0. > > > > - - - - - - - - - -- - - - - - - - - - - - - - - - -- - - - - -- -- - - > > > I wish someone would explain the Any unbounded.... f(s) = 0. to > > > me. > > > Ferguson 7/3/04 > > > - - - - - -- - - - - -- - -- - - - - - -- - - - - -- - - -- - - - > > > - - - > > > > > > > > > > > > > Cantor DID prove that every set has more subsets than subsets. > > > > > > > > Doubtful! > > > > > > > > > In the process he proved that the the cardinality of the set of subsets > > > > > of N ia greater thsn the cardinality of N (Xo). But he did not prove > > > > > that there is no list of the subsets of N what-so-ever. > > > > > > > > > > > > He did, if by list you mean a function from the subsets of N > > > > into N itself which is one-to-one. If this is not what you mean > > > > by list, then either (1) you need to deŽne exactly what you mean, > > > > or (2) a trivial construction is possible. > > > > > > > > > > > > Andrzej > > > > > > > > > This I respectfully submit. > > > > > david.ferguson1@cox.net === Subject: Re: Is there more symmetry to this function? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CCoYT24941; >> This is probably something everyone already knows... >> h(1,1) = 2, h(1,2) = 4, h(1,3) = 6, h(1,4) = 8, ... >so h(1,n)=2n >> h(2,1) = 4, h(2,2) = 8, h(2,3) = 12, h(2,4) = 16, ... >h(2,n)=4n >> h(3,1) = 6, h(3,2) = 12, h(3,3) = 18, h(3,4) = 24, ... >h(3,n)=6n >> h(4,1) = 8, h(4,2) = 16, h(4,3) = 24, h(4,4) = 32, ... >h(4,n)=8n >Obviously you have h(m,n)=2mn >> CLAIM OF THE CLUELESS ONE: >> The binary operation *: N x N -> N, * = h , >> is communative, associative, and distributive. >Yes, that is obvious, as multiplication (m,n)->mn has those properties... >Jaap by which f, g, and h were originally deŽned *are* complicated (it would take, say, 1-2 hours just to work out h(1,1) = 2 by hand) I quite naively assume these functions must somehow always be complicated themselves. Sincerly, C. Dement === Subject: Re: how to measure how much vector is close to the zero vector by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CD1bE26343; >Hi All, >What is the best way to measure how much a vector v close to the zero >vector. >I am using: >std(v) / mean(v) >which is ok but it is hard to compute since i am using it in a minimization >problem... >Ira What about Sqrt(v*v^t) ? === Subject: Test. Pls ignore. === Subject: Svara that Goldbach¹s Conjecture is Unprovable The argument is as follows. Its very similar to a previous post that I made which argues that Twin Primes is unprovable. I¹ll let you judge whether this is a rigorous proof or not. I¹m not claiming that it is or that it is not - It¹s just for fun: Let set S be the set of all q such that 0 The argument is as follows. Its very similar to a previous post that I > made which argues that Twin Primes is unprovable. I¹ll let you judge > whether this is a rigorous proof or not. I¹m not claiming that it is > or that it is not - It¹s just for fun: > Let set S be the set of all q such that 0 Let set T be the set of all q such that 0 Then Goldbach¹s Conjecture is equivalent to the intersection of sets S > and T being nonempty for each natural number n. > Now, the condition which deŽnes set S, that q is prime, does not give > any information about the factors of 2n-q, so it does not give any > information about whether or not 2n-q is prime. > And the condition which deŽnes set T, that 2n-q is prime, does not > give any information about the factors of q, so it does not give any > information about whether or not q is prime. > Therefore, the two conditions which deŽne the two sets are > independent from one another, meaning that the only way to determine > whether the intersection of the two sets, S and T, is nonempty is to > directly calculate the elements in the intersection of S and T, i.e., > test if there is a q such that q and 2n-q are both prime. But > this would take an inŽnite amount of time, since one would have to > test this for each natural number n. Therefore, Goldbach¹s Conjecture > is unprovable. QED As someone pointed out, the argument cannot possibly show that Goldbach¹s Conjecture is impossible to disprove, as this is obviously false. That was a typo when I transferred basically the same argument from the Twin Primes Post to Goldbach. After reading some of the posts that people put, it is dissapointing that a lot of people disagreed with me without giving a substantive argument as to why they believe that my argument is wrong. (I¹m not claiming that it is right, only that no one could contradict it.) It is interesting to note that I used pretty much the same type of argument to show that Twin Primes is true, but I did not receive the same level of critisism that I received on this post and the Twin Primes post. Note, that neither arguments were claimed to be proofs, only svaras, i.e., heuristic arguments. I think that the reason that people show so much distaste for even a non-rigorous argument of this is not necessarily because it is wrong, but is because I am arguing something that people do not want to hear, as it is an unpopular idea that there are certain conjectures that can never be proven by human beings. We would like to believe that we can solve all problems, but in reality, some problems are unsolvable. I used to do magic shows for kids and when I fooled them, they would respond I know how you do it! a lot of times. I would sometimes ask them how and they couldn¹t answer. The kids didn¹t want to admit that they couldn¹t Žgure something out, so they just denied it. Unfortunately, we are witness to the same type of cognitive dissonance in sci.math. I want to respond to a post that said (with respect to Twin Primes) You could probably argue just as well that the following statement is unprovable: There are inŽnitely many primes p such that p+2 is either prime or the product of two primes. But that was proved by Chen Jingrun in 1966. On the surface, this criticism seems valid. But if try it out with my original argument, it does not work: If p+2 is either a product of two primes, then this, in fact, gives information about the factors of p, since from this we can conclude that certain numbers, namely the two prime factors of p+2, cannot possibly be factors of p. This information can be used to help to determine whether p is prime, since we have eliminated a possible factorization of p. If p+2 is prime, then no possible factorizations of p can be eliminated. Anyway, all of this was for fun, but if anyone is interested in trying to rigorize my argument and write a paper, I give you my blessing in doing such. Or if you can Žnd a hole, then by all means, explain. Craig === Subject: Re: Svara that Goldbach¹s Conjecture is Unprovable > [] > As someone pointed out, the argument cannot possibly show that > Goldbach¹s Conjecture is impossible to disprove, as this is obviously > false. That was a typo when I transferred basically the same argument > from the Twin Primes Post to Goldbach. > After reading some of the posts that people put, it is dissapointing > that a lot of people disagreed with me without giving a substantive > argument as to why they believe that my argument is wrong. (I¹m not > claiming that it is right, only that no one could contradict it.) It > is interesting to note that I used pretty much the same type of > argument to show that Twin Primes is true, but I did not receive the > same level of critisism that I received on this post and the Twin > Primes post. Note, that neither arguments were claimed to be proofs, > only svaras, i.e., heuristic arguments. > I think that the reason that people show so much distaste for even a > non-rigorous argument of this is not necessarily because it is wrong, > but is because I am arguing something that people do not want to hear, > as it is an unpopular idea that there are certain conjectures that can > never be proven by human beings. We would like to believe that we can > solve all problems, but in reality, some problems are unsolvable. There is a novel about this very thing: Uncle Petros and Goldbach¹s Conjecture by Apostolos Doxiadis. An interesting read, unlike most sci.math postings. === Subject: Re: Svara that Goldbach¹s Conjecture is Unprovable >[...] >> Therefore, the two conditions which deŽne the two sets are >> independent from one another, meaning that the only way to determine >> whether the intersection of the two sets, S and T, is nonempty is to >> directly calculate the elements in the intersection of S and T, i.e., [...] >As someone pointed out, the argument cannot possibly show that >Goldbach¹s Conjecture is impossible to disprove, as this is obviously >false. That was a typo when I transferred basically the same argument >from the Twin Primes Post to Goldbach. >After reading some of the posts that people put, it is dissapointing >that a lot of people disagreed with me without giving a substantive >argument as to why they believe that my argument is wrong. (I¹m not >claiming that it is right, only that no one could contradict it.) It >is interesting to note that I used pretty much the same type of >argument to show that Twin Primes is true, but I did not receive the >same level of critisism that I received on this post and the Twin >Primes post. That¹s probably because similar absurdities occur in both posts; people get bored. >Note, that neither arguments were claimed to be proofs, >only svaras, i.e., heuristic arguments. >I think that the reason that people show so much distaste for even a >non-rigorous argument of this is not necessarily because it is wrong, >but is because I am arguing something that people do not want to hear, >as it is an unpopular idea that there are certain conjectures that can >never be proven by human beings. We would like to believe that we can >solve all problems, but in reality, some problems are unsolvable. Uh, this is bull. We¹ve known since Godel that some problems are unsolvable - we¹ve got over it. >I used to do magic shows for kids and when I fooled them, they would >respond I know how you do it! a lot of times. I would sometimes ask >them how and they couldn¹t answer. The kids didn¹t want to admit that >they couldn¹t Žgure something out, so they just denied it. >Unfortunately, we are witness to the same type of cognitive dissonance >in sci.math. Uh, right. >[...] >Anyway, all of this was for fun, but if anyone is interested in trying >to rigorize my argument and write a paper, I give you my blessing in >doing such. Or if you can Žnd a hole, then by all means, explain. Find a hole? The _most_ ridiculous part is the same in both posts: there¹s simply no justiŽcation for the bit about how the only way to determine whatever is to look at all the elements of an inŽnite set one by one. It happens all the time that we prove things about all the elements of some inŽnite set with proofs that take only Žnitely much time. >Craig ************************ David C. Ullrich === Subject: Re: Svara that Goldbach¹s Conjecture is Unprovable > The argument is as follows. Its very similar to a previous post that I > made which argues that Twin Primes is unprovable. I¹ll let you judge > whether this is a rigorous proof or not. I¹m not claiming that it is > or that it is not - It¹s just for fun: > Let set S be the set of all q such that 0 Let set T be the set of all q such that 0 Then Goldbach¹s Conjecture is equivalent to the intersection of sets S > and T being nonempty for each natural number n. > Now, the condition which deŽnes set S, that q is prime, does not give > any information about the factors of 2n-q, so it does not give any > information about whether or not 2n-q is prime. > And the condition which deŽnes set T, that 2n-q is prime, does not > give any information about the factors of q, so it does not give any > information about whether or not q is prime. > Therefore, the two conditions which deŽne the two sets are > independent from one another, meaning that the only way to determine > whether the intersection of the two sets, S and T, is nonempty is to > directly calculate the elements in the intersection of S and T, i.e., > test if there is a q such that q and 2n-q are both prime. But > this would take an inŽnite amount of time, since one would have to > test this for each natural number n. Therefore, Goldbach¹s Conjecture > is unprovable. QED > For those of you who don¹t buy the argument, let me explain this in > which twins were born in hospital A. And deŽne set T as the days in > different hospitals in different parts of the world.) The conditions > which deŽne sets S and T have nothing to do with one another, i.e., > they are independent from one another. Therefore, the only way to > determine whether the intersection of sets S and T is nonempty is to > directly calculate the elements in the intersection of sets S and T. > This is obviously feasible to do since the number of days in the year > time and prove that there is always a day in each year when both > hospitals give birth to twins, one would have to wait an eternity to > do such. > Anyway, I welcome comments. Hi Craig. This sounds similar to your proof idea for p!=np. In that proof, if I recall correctly, you claimed (without adequate proof) that a single solution to the problem existed (in that case the problem was subset-sum), and then claimed that this single solution required at least a super-polynomial amount of computation to solve (hence subset-sum must be in NP-P so P != NP). In this proposition you state: the only way to determine whether the intersection of the two sets, S and T, is nonempty is to directly calculate the elements in the intersection of S and T You still need to prove that this truly is Œthe only way¹. l8r, Mike N. Christoff === Subject: Re: Svara that Goldbach¹s Conjecture is Unprovable >> The argument is as follows. Its very similar to a previous post >> that I made which argues that Twin Primes is unprovable. I¹ll let >> you judge whether this is a rigorous proof or not. I¹m not claiming >> that it is or that it is not - It¹s just for fun: >> Let set S be the set of all q such that 0> set T be the set of all q such that 0> Then Goldbach¹s Conjecture is equivalent to the intersection of >> sets S and T being nonempty for each natural number n. >> Now, the condition which deŽnes set S, that q is prime, does not >> give any information about the factors of 2n-q, so it does not give >> any information about whether or not 2n-q is prime. >> And the condition which deŽnes set T, that 2n-q is prime, does not >> give any information about the factors of q, so it does not give >> any information about whether or not q is prime. >> Therefore, the two conditions which deŽne the two sets are >> independent from one another, meaning that the only way to >> determine whether the intersection of the two sets, S and T, is >> nonempty is to directly calculate the elements in the intersection >> of S and T, i.e., test if there is a q such that q and 2n-q are >> both prime. But this would take an inŽnite amount of time, since >> one would have to test this for each natural number n. Therefore, >> Goldbach¹s Conjecture is unprovable. QED >> For those of you who don¹t buy the argument, let me explain this in >> which twins were born in hospital A. And deŽne set T as the days >> are two different hospitals in different parts of the world.) The >> conditions which deŽne sets S and T have nothing to do with one >> another, i.e., they are independent from one another. Therefore, >> the only way to determine whether the intersection of sets S and T >> is nonempty is to directly calculate the elements in the >> intersection of sets S and T. This is obviously feasible to do >> to do it for each year until the end of time and prove that there >> is always a day in each year when both hospitals give birth to >> twins, one would have to wait an eternity to do such. >> Anyway, I welcome comments. > Hi Craig. This sounds similar to your proof idea for p!=np. In that > proof, if I recall correctly, you claimed (without adequate proof) > that a single solution to the problem existed (in that case the > problem was subset-sum), and then claimed that this single solution > required at least a super-polynomial amount of computation to solve > (hence subset-sum must be in NP-P so P != NP). In this proposition > you state: > the only way to determine whether the intersection of the two sets, > S and T, is nonempty is to directly calculate the elements in the > intersection of S and T > You still need to prove that this truly is Œthe only way¹. > l8r, Mike N. Christoff Good day Craig. As well, similar to your argument that the 3n+1 conjecture is unprovable, in the paper you posted to arXiv:math.GM. Also, to the argument that Riemann¹s conjecture is unprovable, in another paper you posted in the same place (later withdrawn). I see a general principle emerging here, not necessarily a theorem, but perhaps a svara-theorem. xx === Subject: Re: Svara that Goldbach¹s Conjecture is Unprovable > > The argument is as follows. Its very similar to a previous post that I > > made which argues that Twin Primes is unprovable. I¹ll let you judge > > whether this is a rigorous proof or not. I¹m not claiming that it is > > or that it is not - It¹s just for fun: [...] > > Therefore, the two conditions which deŽne the two sets are > > independent from one another, meaning that the only way to determine > > whether the intersection of the two sets, S and T, is nonempty is to > > directly calculate the elements in the intersection of S and T, i.e., > > test if there is a q such that q and 2n-q are both prime. But > > this would take an inŽnite amount of time, since one would have to > > test this for each natural number n. Therefore, Goldbach¹s Conjecture > > is unprovable. QED > Hi Craig. This sounds similar to your proof idea for p!=np. In that proof, > if I recall correctly, you claimed (without adequate proof) that a single > solution to the problem existed (in that case the problem was subset-sum), > and then claimed that this single solution required at least a > super-polynomial amount of computation to solve (hence subset-sum must be in > NP-P so P != NP). In this proposition you state: > the only way to determine whether the intersection of the two sets, S and > T, is nonempty is to directly calculate the elements in the intersection of > S and T > You still need to prove that this truly is Œthe only way¹. There¹s a more basic žaw in the logic, of course. Craig states that to prove or disprove Goldbach¹s conjecture would take an inŽnite amount of time, since you¹d have to test every even number greater than 2. But he forgets that *if Goldbach¹s conjecture is false*, then you only need to test N different integers, where N is a number such that (2*N)+2 is not the sum of a pair of primes. And if you only have to test a Žnite number of numbers, well, that certainly shouldn¹t take an inŽnite amount of time! So Craig has shown that *if* GC is true, [and *if* certain other implicit assumptions about prime numbers hold,] *then* GC is unprovable. But he hasn¹t ruled out the possibility that it might be false --- and if it¹s false, then it¹s most deŽnitely provably so! -Arthur === Subject: Re: Svara that Goldbach¹s Conjecture is Unprovable .8ecrit : >The argument is as follows. Its very similar to a previous post that I >this would take an inŽnite amount of time, since one would have to >test this for each natural number n. Therefore, Goldbach¹s Conjecture >is unprovable. QED Do you mean, for example, that it is really impossible to sum the Žrst n positive integers without performing (n-1) additions ? A real pitty, even Gauss failed to solve this far reaching problem... === Subject: Re: Is there more symmetry to this function? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CDrkD30182; >Just a note as to the purpose of all this: >I am not posting a load of functions here just to demonstrate >what I can do on my calculator- now you go Žnd the function. >(very sorry, however, if it has come across that way). Borrowing a >phrase from the game of Snooker, I would characterize what >I`m doing as a shot to nothing. This means: if the function >found is of interest- that`s great. If the function found is >too simple to be of interest on it`s own, that`s also great! Why? >I¹m working with an algebra generated by 16 basis vectors from the >factor space Q x Q / {(1,1), (-1,-1)}, Q = Quaternions >I already know that if x, y, z in Q and the functions res, tes >are deŽned as: >res: Q -> R^3, >res(x) = res((x_1)i + (x_2)j + (x_3)k + x_4) = (x_1,x_2,x_3) >tes: Q -> R, >tes((x_1)i + (x_2)j + (x_3)k + x_4) = x_4 >then res(x) scalar-product res(y) = -tes(x*y) >and det(res(x),res(y),res(z)) = -tes(x*y*z) >when x, y, and z are pure Quaternions (i.e. x_4 = y_4 = z_4 = 0). >In a manner similar to this, let ves be the function from >the above mentioned algebra to the reals whose purpose is >merely to add *all* the coefŽcients of the 16 basis >vectors together. >Now, an example: >I take two very special elements A, B of the algebra and >insert n and m (real numbers) for the coefŽcients of two >special basis vectors in A and B, respectively, and, Žnally, >deŽne the function (writing A(n), B(m) for these new elements of >the algebra) f: R x R -> R as f(n,m) = ves(A(n)*B(m)). The above should read: basis vectors of A and B (see below) >After 81 multiplications (in other cases it can be >up to 4096 = 16*16*16 multiplications) and subsequent >adding up of basis coefŽcients I get... >f(n,m) = ves(A(n)*B(m)) = nm Also, a C in the algebra (as well as a certain basis vector of C) has been found such that f(n,m,o)= ves(A(n)*B(m)*C(o)) = nmo Number of multiplications involved: 81*13 = 1053 >Is this function interesting in itself? Certainly not. However, >the neat thing (just my opinion) is the feeling- where exactly >is this symmetry coming from? A similar procedure generates >certain linear operators; another similar procedure could >(possibly) be used to generate prime candidates. I would >like to get as much an overview on this as possible in the >foreseeable future. >Sincerly, >C. Dement === Subject: Re: Why are elements in the domain mapped to a single element in the codomain? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CDrkl30188; Hhhmm, how about calling f a function at inŽnity from R to R if f: R -> P(R) , R = reals, P(R) = power set forall epsilon > 0 exists n in naturals forall y, z in f(x), y > n: |y-z| < epsilon Similarly, one could probably deŽne, say for any b in R, a function at b from R to R: Sincerly, C. Dement === Subject: Re: matrix Originator: grubb@lola >hello, >what does it mean to solve a matrix? >let¹s say you have matrices >(a, b) >(c, d) >(0, a) >(b, 0) >(0, a) >(b, 0) >what does it mean to solve these? I have seen a physicist use this terminology. He wanted the eigenvalues and eigenvectors of the matrix. --Dan Grubb === Subject: Re: matrix > > hello, > > what does it mean to solve a matrix? > It means that Neo has saved the world. In the immortal word(s) of C. Montgomery Burns, Excellent! Kevin O¹Neill === Subject: Confused with uniform/pointwise convergence I think I am confused about the differences between pointwise and uniform converence, but I may not be, and would highly appreciate it if someone could help me. The difference between the two is that with uniform convergence, there is a single N for all x¹s such that |f_n(x) - f(x) | < eps for all n > N. For pointwise, the N may depend on x. Fine, I get that. However, I am trying to read some theorems. In particular, let f_n be real-valued functions that are riemann integrable on [a,b] and that f_n -> f uniformly on [a,b]. DeŽne g_n(x) = integral (from a to x) f_n(t)dt if x is in [a,b]. Then f is riemann integrable on [a,b] and g_n -> g uniformly on [a,b] where g(x) = integral (from a to x) f(t)dt. Well, it goes on to say that the conclusion implies that for each x in [a,b], we can write Limit integral (a to x) f_n(t) dt = integral (from a to x) limit f_n(t) dt Ok, no problem. But my question is, if g_n -> g pointwise, wouldn¹t this conclusion be implied as well? In the conclusion, we are saying that for every x, the equality is satisŽed. So , if you give me an x in [a,b], I can show you that Limit integral (a to x) f_n(t) dt = integral (from a to x) limit f_n(t) dt. In other words, I can show you that |g_n(x) - g(x)| < eps for all n > N and for some N (maybe depending on x). That is, g_n(x) converges to g(x) if you give me an x. Why do we need a single N that does it for all x in [a,b]? The conclusion seems to tell me ok, give me an x, any x in [a,b], and this equality holds. Well that says to me give me an x and sure I can make the equality hold, depending on what x is of course. So if g_n(x) -> g(x) pointwise, we¹d have the same conclusion. I just don¹t see why we need to say that we need to be able to have a single N that makes the equality in the conclusion hold. What the conclusion says is that a uniformly convergent sequence can be integrated term by term. But if g_n -> g pointwise, that would happen as well (which is what I am claiming), because like I said, if you give me any x in [a,b], I can make |g_n(x) - g(x)| as small as I want, and so we¹re Žne. I¹m not saying we need to change the assumptions of the theorem. Those are needed to prove this g_n -> g uniformly. But if I¹m right (which I¹m probably not), can¹t we relax the assumptions so that we can just aim for g_n(x) -> g(x) pointwise? Any help is highly appreciated, Isaac === Subject: Re: Confused with uniform/pointwise convergence >I think I am confused about the differences between pointwise and uniform >converence, but I may not be, and would highly appreciate it if someone >could help me. >The difference between the two is that with uniform convergence, there is a >single N for all x¹s such that |f_n(x) - f(x) | < eps for all n > N. For >pointwise, the N may depend on x. Fine, I get that. >However, I am trying to read some theorems. In particular, let f_n be >real-valued functions that are riemann integrable on [a,b] and that f_n -> f >uniformly on [a,b]. DeŽne g_n(x) = integral (from a to x) f_n(t)dt if x >is in [a,b]. Then f is riemann integrable on [a,b] and g_n -> g uniformly >on [a,b] where g(x) = integral (from a to x) f(t)dt. >Well, it goes on to say that the conclusion implies that for each x in >[a,b], we can write >Limit integral (a to x) f_n(t) dt = integral (from a to x) limit f_n(t) >Ok, no problem. But my question is, if g_n -> g pointwise, wouldn¹t this >conclusion be implied as well? exactly the same as saying Limit integral (a to x) f_n(t) dt = integral (from a to x) limit f_n(t) dt, by the deŽnition of the g_n and g. Maybe they meant something else, like that the integral of the g_n converges to the integral of g? >In the conclusion, we are saying that for >every x, the equality is satisŽed. So , if you give me an x in [a,b], I >can show you that Limit integral (a to x) f_n(t) dt = integral (from a >to x) limit f_n(t) dt. In other words, I can show you that |g_n(x) - g(x)| >< eps for all n > N and for some N (maybe depending on x). That is, g_n(x) >converges to g(x) if you give me an x. Why do we need a single N that does >it for all x in [a,b]? The conclusion seems to tell me ok, give me an x, >any x in [a,b], and this equality holds. Well that says to me give me an x >and sure I can make the equality hold, depending on what x is of course. So >if g_n(x) -> g(x) pointwise, we¹d have the same conclusion. I just don¹t >see why we need to say that we need to be able to have a single N that makes >the equality in the conclusion hold. >What the conclusion says is that a uniformly convergent sequence can be >integrated term by term. But if g_n -> g pointwise, that would happen as >well (which is what I am claiming), because like I said, if you give me any >x in [a,b], I can make |g_n(x) - g(x)| as small as I want, and so we¹re >Žne. >I¹m not saying we need to change the assumptions of the theorem. Those are >needed to prove this g_n -> g uniformly. But if I¹m right (which I¹m >probably not), can¹t we relax the assumptions so that we can just aim for >g_n(x) -> g(x) pointwise? >Any help is highly appreciated, >Isaac ************************ David C. Ullrich === Subject: Re: Confused with uniform/pointwise convergence > >I think I am confused about the differences between pointwise and uniform > >converence, but I may not be, and would highly appreciate it if someone > >could help me. > >The difference between the two is that with uniform convergence, there is a > >single N for all x¹s such that |f_n(x) - f(x) | < eps for all n > N. For > >pointwise, the N may depend on x. Fine, I get that. > >However, I am trying to read some theorems. In particular, let f_n be > >real-valued functions that are riemann integrable on [a,b] and that f_n -> f > >uniformly on [a,b]. DeŽne g_n(x) = integral (from a to x) f_n(t)dt if x > >is in [a,b]. Then f is riemann integrable on [a,b] and g_n -> g uniformly > >on [a,b] where g(x) = integral (from a to x) f(t)dt. > >Well, it goes on to say that the conclusion implies that for each x in > >[a,b], we can write > >Limit integral (a to x) f_n(t) dt = integral (from a to x) limit f_n(t) > >dt > >Ok, no problem. But my question is, if g_n -> g pointwise, wouldn¹t this > >conclusion be implied as well? > exactly the same as saying Limit integral (a to x) f_n(t) dt > = integral (from a to x) limit f_n(t) dt, by the deŽnition of > the g_n and g. > Maybe they meant something else, like that the integral of the g_n > converges to the integral of g? > >In the conclusion, we are saying that for > >every x, the equality is satisŽed. So , if you give me an x in [a,b], I > >can show you that Limit integral (a to x) f_n(t) dt = integral (from a > >to x) limit f_n(t) dt. In other words, I can show you that |g_n(x) - g(x)| > >< eps for all n > N and for some N (maybe depending on x). That is, g_n(x) > >converges to g(x) if you give me an x. Why do we need a single N that does > >it for all x in [a,b]? The conclusion seems to tell me ok, give me an x, > >any x in [a,b], and this equality holds. Well that says to me give me an x > >and sure I can make the equality hold, depending on what x is of course. So > >if g_n(x) -> g(x) pointwise, we¹d have the same conclusion. I just don¹t > >see why we need to say that we need to be able to have a single N that makes > >the equality in the conclusion hold. > >What the conclusion says is that a uniformly convergent sequence can be > >integrated term by term. But if g_n -> g pointwise, that would happen as > >well (which is what I am claiming), because like I said, if you give me any > >x in [a,b], I can make |g_n(x) - g(x)| as small as I want, and so we¹re > >Žne. > >I¹m not saying we need to change the assumptions of the theorem. Those are > >needed to prove this g_n -> g uniformly. But if I¹m right (which I¹m > >probably not), can¹t we relax the assumptions so that we can just aim for > >g_n(x) -> g(x) pointwise? > >Any help is highly appreciated, > >Isaac > ************************ > David C. Ullrich The text from which I quote is the standard Mathematical Analysis by Tom Apostol on page 225 at the bottom. I quote: Note. The conclusion implies that, for each x in [a,b], we can write lim integral (from a to x) f_n(t)dt = integral (from a to x) lim f_n(t)dt. This property is often described by saying that a uniformly convergent sequence can be interated term by term. I have asked this question elsewhere, and another poster seems to think that I am not right also. Like I said, g_n -> g uniformly of course does imply the above conclusion. However my claim is (which you agree with) that g_n -> g pointwise also implies the above conclusion. You also said Maybe they meant something else, like that the integral of the g_n converges to the integral of g? Well what do you mean by that? Do you just mean g_n converges uniformly to g (well I guess you have to mean that by what the theorem says) ? And, why would one care about this? It seems to me that we should much more care about the conclusion (a speciŽc case of g_n -> g uniformly as you agree), which tells us when we can integrate sequences (or as an easy corollary, series) term by term, and which is equivalent to the statement that g_n -> g pointwise (as you agree). In my very youthful and na.95ve opinion, shouldn¹t we just care about how we can arrive at lim integral (from a to x) f_n(t)dt = integral (from a to x) lim f_n(t)dt ? If so, then we care about how g_n -> g pointwise. And as the original theorem said, we need a bunch of assumptions (like f_n -> f uniformly, f_n riemann integrable, etc.) in order to arrive at g_n -> g uniformly, which of course implies the conclusion which is what we¹re after. But maybe we can relax our assumptions a bit to arrive at g_n -> g pointwise so that we can still have the conclusion? Yikes, I hope I made some sense and really hope you can help, Isaac === Subject: Re: Confused with uniform/pointwise convergence > I think I am confused about the differences between pointwise and uniform > converence, but I may not be, and would highly appreciate it if someone > could help me. > The difference between the two is that with uniform convergence, there is a > single N for all x¹s such that |f_n(x) - f(x) | < eps for all n > N. For > pointwise, the N may depend on x. Fine, I get that. > However, I am trying to read some theorems. In particular, let f_n be > real-valued functions that are riemann integrable on [a,b] and that f_n -> f > uniformly on [a,b]. DeŽne g_n(x) = integral (from a to x) f_n(t)dt if x > is in [a,b]. Then f is riemann integrable on [a,b] and g_n -> g uniformly > on [a,b] where g(x) = integral (from a to x) f(t)dt. > Well, it goes on to say that the conclusion implies that for each x in > [a,b], we can write > Limit integral (a to x) f_n(t) dt = integral (from a to x) limit f_n(t) > dt > Ok, no problem. But my question is, if g_n -> g pointwise, wouldn¹t this > conclusion be implied as well? No. Take [a,b] = [0,1] and g_n(x) = n - n^2 x if 0 < x <= 1/n and g_n(x) = 0 otherwise. Then g_n -> 0 pointwise; actually, for each x you have g_n(x) = 0 if n is sufŽciently big. Furthermore, you have integral (from 0 to 1) g_n = 1/2, but integral (from 0 to 1) 0 = 0. > In the conclusion, we are saying that for > every x, the equality is satisŽed. So , if you give me an x in [a,b], I > can show you that Limit integral (a to x) f_n(t) dt = integral (from a > to x) limit f_n(t) dt. In other words, I can show you that |g_n(x) - g(x)| > < eps for all n > N and for some N (maybe depending on x). That is, g_n(x) > converges to g(x) if you give me an x. Why do we need a single N that does > it for all x in [a,b]? Try to answer your own question bearing in mind the example above. Suggestion: see what does the graphic of g_n looks like. Jose Carlos Santos === Subject: Re: Confused with uniform/pointwise convergence > > I think I am confused about the differences between pointwise and uniform > > converence, but I may not be, and would highly appreciate it if someone > > could help me. > > The difference between the two is that with uniform convergence, there is a > > single N for all x¹s such that |f_n(x) - f(x) | < eps for all n > N. For > > pointwise, the N may depend on x. Fine, I get that. > > However, I am trying to read some theorems. In particular, let f_n be > > real-valued functions that are riemann integrable on [a,b] and that f_n -> f > > uniformly on [a,b]. DeŽne g_n(x) = integral (from a to x) f_n(t)dt if x > > is in [a,b]. Then f is riemann integrable on [a,b] and g_n -> g uniformly > > on [a,b] where g(x) = integral (from a to x) f(t)dt. > > Well, it goes on to say that the conclusion implies that for each x in > > [a,b], we can write > > Limit integral (a to x) f_n(t) dt = integral (from a to x) limit f_n(t) > > dt > > Ok, no problem. But my question is, if g_n -> g pointwise, wouldn¹t this > > conclusion be implied as well? > No. Take [a,b] = [0,1] and g_n(x) = n - n^2 x if 0 < x <= 1/n and > g_n(x) = 0 otherwise. Then g_n -> 0 pointwise; actually, for each x > you have g_n(x) = 0 if n is sufŽciently big. Furthermore, you have > integral (from 0 to 1) g_n = 1/2, but integral (from 0 to 1) 0 = 0. > > In the conclusion, we are saying that for > > every x, the equality is satisŽed. So , if you give me an x in [a,b], I > > can show you that Limit integral (a to x) f_n(t) dt = integral (from a > > to x) limit f_n(t) dt. In other words, I can show you that |g_n(x) - g(x)| > > < eps for all n > N and for some N (maybe depending on x). That is, g_n(x) > > converges to g(x) if you give me an x. Why do we need a single N that does > > it for all x in [a,b]? > Try to answer your own question bearing in mind the example above. > Suggestion: see what does the graphic of g_n looks like. > Jose Carlos Santos Please read David Ullrich¹s responses and let me know what you think, because I think you might have misconstrued what I was saying. I responded to his response by saying (and please let me know what you think about this response) , The text from which I quote is the standard Mathematical Analysis by Tom Apostol on page 225 at the bottom. I quote: Note. The conclusion implies that, for each x in [a,b], we can write lim integral (from a to x) f_n(t)dt = integral (from a to x) lim f_n(t)dt. This property is often described by saying that a uniformly convergent sequence can be interated term by term. I have asked this question elsewhere, and another poster seems to think that I am not right also. Like I said, g_n -> g uniformly of course does imply the above conclusion. However my claim is (which you agree with) that g_n -> g pointwise also implies the above conclusion. You also said Maybe they meant something else, like that the integral of the g_n converges to the integral of g? Well what do you mean by that? Do you just mean g_n converges uniformly to g (well I guess you have to mean that by what the theorem says) ? And, why would one care about this? It seems to me that we should much more care about the conclusion (a speciŽc case of g_n -> g uniformly as you agree), which tells us when we can integrate sequences (or as an easy corollary, series) term by term, and which is equivalent to the statement that g_n -> g pointwise (as you agree). In my very youthful and na.95ve opinion, shouldn¹t we just care about how we can arrive at lim integral (from a to x) f_n(t)dt = integral (from a to x) lim f_n(t)dt ? If so, then we care about how g_n -> g pointwise. And as the original theorem said, we need a bunch of assumptions (like f_n -> f uniformly, f_n riemann integrable, etc.) in order to arrive at g_n -> g uniformly, which of course implies the conclusion which is what we¹re after. But maybe we can relax our assumptions a bit to arrive at g_n -> g pointwise so that we can still have the conclusion? Yikes, I hope I made some sense and really hope you can help, Isaac === Subject: Re: JSH: Sweep likely by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CEeTR02090; >> I feel conŽdent that Ullrich and Magidin in particular have behaved >> in a way that will mean the loss of their Ph.D¹s. >Mr. Harris, >when you are proven wrong (once again, for the millionth time), will it mean >the loss of your welfare BS in physics? >Obviously, you are not deserving of any degree as you haven¹t any emotional >or scientiŽc intelligence! >You should feel so embarrassed with yourself for even making such a claim. (snip) You know, James Harris is a member of a high-IQ society with Quinn Tyler Jackson. I guess that group needed the dues so bad due to their legal fees in lawsuits with other high-IQ societies, they welcomed James Harris in with open arms (once his dues check cleared). Or did James Harris get his parents to pay his high-IQ dues? So clearly, what does James Harris need of scientiŽc or emotional intelligence? He¹s a member of a high-IQ society . Right, Quinn? Don¹t you agree, Mr. Quinn? Don¹t you just love your colleague James Harris and the glory of his posts which režect on your wonderful high-IQ society, huh Quinn? Superior intellect - HAHAHAHAHAHAHAHAHAHA... Anonymous === Subject: Re: JSH: Sweep likely Anonymous a dit: > Right, Quinn? Don¹t you agree, Mr. Quinn? > Don¹t you just love your colleague James > Harris and the glory of his posts which > režect on your wonderful high-IQ society, > huh Quinn? Voici vingt-cinq sous. Vas-y, joue dans la rue. -- Quinn === Subject: Re: JSH: Sweep likely > Anonymous a dit: > > Right, Quinn? Don¹t you agree, Mr. Quinn? > > Don¹t you just love your colleague James > > Harris and the glory of his posts which > > režect on your wonderful high-IQ society, > > huh Quinn? > Voici vingt-cinq sous. Vas-y, joue dans la rue. Hier vijfentwintig stuivers. Ga op straat spelen. You should have made it: Voici gamin, vingt-cinq sous. Vas-y, joue dans la rue. but 25 sous is a bit overpriced, when they were used, 25 sous would be, I think, the weekly income of a family. At least in France. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Surrogate factoring, update by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CEhen02282; >It¹s been a while since I mentioned surrogate factoring Too short a while. >I¹m just one person. But you spawn a thousand posts, like žeas and vermin from your hair. >just sits while I Žddle with it. Hey, don¹t do that - you¹ll go blind! (but then, you¹ll have a tough time posting if you¹re blind, so Žddle away!) >I¹m painting a dire picture because I want answers. The answer is YOU¹RE AN INSANE CRANK AND CRACKPOT! >James Harris Anonymous === Subject: completion of a metric space Hello I have a question about completion of metric spaces. If X is a metric space, then is it true that there always is a complete metric space Y that contains X as a subspace? So, if X is any metric space and {x_n} is a Cauchy sequence of X, then can we always assume there¹s an x in the completion of X to which {x_n} converges? If the answer is yes, how can we build the completion? Amanda === Subject: Re: completion of a metric space > Hello > I have a question about completion of metric spaces. If X is a metric > space, then is it true that there always is a complete metric space Y > that contains X as a subspace? So, if X is any metric space and {x_n} > is a Cauchy sequence of X, then can we always assume there¹s an x in > the completion of X to which {x_n} converges? If the answer is yes, > how can we build the completion? > Amanda The completion is done the following way: Let X be a metric Space and X_c the space of all cauchy sequences in X. X is isometric to a subspace of X_c because every element has a corresponding cauchy sequence like {x} element of X is a element in X_c {x,x,...}. This motivates the the fatcorisation X_c/~ with a equivalence relation ~ in X_c so that y in X_c and z in X_c are in relation if and only if their difference is a cauchy sequence. So a subspace of X_c/~ is isometric to X but X_c/~ is complete. So we know there is a completion of the metric space X. === Subject: Re: completion of a metric space > I have a question about completion of metric spaces. If X is a metric > space, then is it true that there always is a complete metric space Y > that contains X as a subspace? Well, yes. sort of. The correct statement is: for every metric space X there¹s a complete metric space Y that contains a subspace X¹ isometric to X. > So, if X is any metric space and {x_n} > is a Cauchy sequence of X, then can we always assume there¹s an x in > the completion of X to which {x_n} converges? Yes. > If the answer is yes, > how can we build the completion? Put C = { Cauchy sequences of elements of X } and deŽne in C the binary relation (x_n)_n ~ (y_n)_n iff lim_n d(x_n,y_n) = 0. It turns out that this is an equivalence relation. DeŽne Y as the quotient of C by this equivalence relation (that is, Y is the set of equivalence classes). In Y deŽne the distance D([(x_n)_n],[(y_n)_n]) = lim_n d(x_n,y_n). Finally, let X¹ be the subset of Y of all equivalence classes of constant sequences of elements of X. Then the function X -----------> X¹ x |-> [(x,x,x,x,...)] is an isometry. Furthermore, Y is a complete metric space. Jose Carlos Santos === Subject: Re: completion of a metric space >> I have a question about completion of metric spaces. If X is a metric >> space, then is it true that there always is a complete metric space Y >> that contains X as a subspace? >Well, yes. sort of. The correct statement is: for every metric space >X there¹s a complete metric space Y that contains a subspace X¹ >isometric to X. It¹s literally correct as she stated it. (Construct your completion Y as below. Then consider another completion, with the obvious metric on the union of X and YX¹.)) >> So, if X is any metric space and {x_n} >> is a Cauchy sequence of X, then can we always assume there¹s an x in >> the completion of X to which {x_n} converges? >Yes. >> If the answer is yes, >> how can we build the completion? >Put C = { Cauchy sequences of elements of X } and deŽne in >C the binary relation > (x_n)_n ~ (y_n)_n iff lim_n d(x_n,y_n) = 0. >It turns out that this is an equivalence relation. DeŽne Y as the >quotient of C by this equivalence relation (that is, Y is the set of >equivalence classes). In Y deŽne the distance > D([(x_n)_n],[(y_n)_n]) = lim_n d(x_n,y_n). >Finally, let X¹ be the subset of Y of all equivalence classes of >constant sequences of elements of X. Then the function > X -----------> X¹ > x |-> [(x,x,x,x,...)] >is an isometry. Furthermore, Y is a complete metric space. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: how to measure how much vector is close to the zero vector by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CF2o804151; >Hi All, >What is the best way to measure how much a vector v close to the zero >vector. >I am using: >std(v) / mean(v) >which is ok but it is hard to compute since i am using it in a minimization >problem... If you want fast calculation, how about the square of the lenght? X=(x1, x2, ..., xn) ----> x1*x1 + x2*x2 + ... + xn*xn >Ira You are welcome, Andre Caldas. === Subject: Re: JSH: Groupthink by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CF6sl04487; (snipped crap) >including the paper with >The Hammer. Back with the Hammer, again? Is that what you continue to call it? Have you shown your Hammer to your good friend Quinn Tyler Jackson? Is he not in awe of your Hammer? Your wonderful, large, powerful Hammer, to make us puny sci.math-ers tremble with fear of its size and strength? Surely Quinn can attest to the greatness of your Hammer, and he can tell this newsgroup the facts, right? How James has demonstrated the power of his Hammer to Quinn, huh? Quinn, you¹ll back up James (or haven¹t your already?). Yes James, you just know that it¹s a large, dreadful Hammer to wield, but by golly, you¹ll wield it in your hand, pounding and pounding, beating with your Hammer in your hand, relentlessly, in a climax of pain and pleasure. Hammering away! It¹s Hammer-time! And if you get tired, use your other hand. Or let Quinn pound with your Hammer. Yes, your mighty Hammer. You see your Hammer every day, and surely you stroke it, caress it, as you think of the victims of your Hammer, and how they¹ll writhe in agony when you pound them with your Hammer. Pounding until blood žows from the relentless beating of James Harris¹ Hammer! Do you have a pet name for your Hammer, like Mj.9alnir, to match your Thor-like wielding and pounding of your Hammer, pounding, beating, till it hurts? James Harris, you are nuts! === Subject: Re: Confused with uniform/pointwise convergence by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CFT0W06172; >Ok, no problem. But my question is, if g_n -> g pointwise, wouldn¹t this >conclusion be implied as well? No. Do you know the witch hat? For each n in N, let f_n be a triangle of height n and base 2/n, such that the left vertex is at (0,0) - like a witch hat. For those f_n, f_n(x) != 0 only for 0 < x < 2/n. Also, if x > 1/n, then f_n(x) starts to decrease, and tend to 0.(!) And this means f_n tends to 0 pointwise. integral_of(f_n) = 1 for all n. And this means that integral_of(f_n) tends to 1. Which is different from 0. Good luck, Andre Caldas. === Subject: Re: Confused with uniform/pointwise convergence >>Ok, no problem. But my question is, if g_n -> g pointwise, wouldn¹t this >>conclusion be implied as well? >No. Do you know the witch hat? >For each n in N, let f_n be a triangle of height n and base 2/n, such that the left vertex is at (0,0) - like a witch hat. >For those f_n, f_n(x) != 0 only for 0 < x < 2/n. Also, if x > 1/n, then f_n(x) starts to decrease, and tend to 0.(!) And this means f_n tends to 0 pointwise. >integral_of(f_n) = 1 for all n. And this means that integral_of(f_n) tends to 1. Which is different from 0. And hence g_n does _not_ tend to g pointwise, so this is not a counterexample. I don¹t understand the people saying no to this. If he meant be assuming that he meant something else, but I don¹t know Let¹s look again: > In particular, let f_n be > real-valued functions that are riemann integrable on [a,b] >and that f_n -> f uniformly on [a,b]. DeŽne >[1] g_n(x) = integral (from a to x) f_n(t)dt >if x >is in [a,b]. Then f is riemann integrable on [a,b] and g_n -> g uniformly >on [a,b] where >[2] g(x) = integral (from a to x) f(t)dt. >Well, it goes on to say that the conclusion implies that for each x in >[a,b], we can write >[3] Limit integral (a to x) f_n(t) dt > = integral (from a to x) limit f_n(t)dt Given the deŽnitions [1] and [2], saying g_n -> g pointwise is obviously exactly the same as saying [3] holds. >Good luck, > Andre Caldas. ************************ David C. Ullrich === Subject: Re: Confused with uniform/pointwise convergence > >>Ok, no problem. But my question is, if g_n -> g pointwise, wouldn¹t this > >>conclusion be implied as well? > >No. Do you know the witch hat? > >For each n in N, let f_n be a triangle of height n and base 2/n, such that the left vertex is at (0,0) - like a witch hat. > >For those f_n, f_n(x) != 0 only for 0 < x < 2/n. Also, if x > 1/n, then f_n(x) starts to decrease, and tend to 0.(!) And this means f_n tends to 0 pointwise. > >integral_of(f_n) = 1 for all n. And this means that integral_of(f_n) tends to 1. Which is different from 0. > And hence g_n does _not_ tend to g pointwise, so this is not > a counterexample. > I don¹t understand the people saying no to this. If he meant > be assuming that he meant something else, but I don¹t know > Let¹s look again: > > In particular, let f_n be > > real-valued functions that are riemann integrable on [a,b] > >and that f_n -> f uniformly on [a,b]. DeŽne > >[1] g_n(x) = integral (from a to x) f_n(t)dt > >if x > >is in [a,b]. Then f is riemann integrable on [a,b] and g_n -> g uniformly > >on [a,b] where > >[2] g(x) = integral (from a to x) f(t)dt. > >Well, it goes on to say that the conclusion implies that for each x in > >[a,b], we can write > >[3] Limit integral (a to x) f_n(t) dt > > = integral (from a to x) limit f_n(t)dt > Given the deŽnitions [1] and [2], saying g_n -> g pointwise > is obviously exactly the same as saying [3] holds. > >Good luck, > > Andre Caldas. > ************************ > David C. Ullrich The text from which I quote is the standard Mathematical Analysis by Tom Apostol on page 225 at the bottom. I quote: Note. The conclusion implies that, for each x in [a,b], we can write lim integral (from a to x) f_n(t)dt = integral (from a to x) lim f_n(t)dt. This property is often described by saying that a uniformly convergent sequence can be interated term by term. I have asked this question elsewhere, and another poster seems to think that I am not right also. Like I said, g_n -> g uniformly of course does imply the above conclusion. However my claim is (which you agree with) that g_n -> g pointwise also implies the above conclusion. You also said Maybe they meant something else, like that the integral of the g_n converges to the integral of g? Well what do you mean by that? Do you just mean g_n converges uniformly to g (well I guess you have to mean that by what the theorem says) ? And, why would one care about this? It seems to me that we should much more care about the conclusion (a speciŽc case of g_n -> g uniformly as you agree), which tells us when we can integrate sequences (or as an easy corollary, series) term by term, and which is equivalent to the statement that g_n -> g pointwise (as you agree). In my very youthful and na.95ve opinion, shouldn¹t we just care about how we can arrive at lim integral (from a to x) f_n(t)dt = integral (from a to x) lim f_n(t)dt ? If so, then we care about how g_n -> g pointwise. And as the original theorem said, we need a bunch of assumptions (like f_n -> f uniformly, f_n riemann integrable, etc.) in order to arrive at g_n -> g uniformly, which of course implies the conclusion which is what we¹re after. But maybe we can relax our assumptions a bit to arrive at g_n -> g pointwise so that we can still have the conclusion? Yikes, I hope I made some sense and really hope you can help, Isaac === Subject: Re: L¹Hopital #2; Correction by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CFT0M06176; ... >You have yet to state coherently what you¹re trying to prove. What is the >result? State it precisely and completely. Do not omit any relevant >hypotheses. Conditions for validity of applying L¹Hopital¹s rule to fractions of the form (f + g)/f where g/f -> 0 as x -> inŽnity. Assume all the hypotheses of L¹Hopital¹s rule. H. Shinya === Subject: Re: L¹Hopital #2; Correction >... >>You have yet to state coherently what you¹re trying to prove. What is the >>result? State it precisely and completely. Do not omit any relevant >>hypotheses. >Conditions for validity of applying L¹Hopital¹s rule to fractions of the form >(f + g)/f >where g/f -> 0 as x -> inŽnity. Assume all the hypotheses of >L¹Hopital¹s rule. Huh? If the hypotheses of L¹Hopital¹s rule are satisŽed then you _can_ apply it - there¹s no further conditions needed. You might also note that if g/f -> 0 then it¹s pretty clear that (f + g)/f -> 1, regardless of any other hypotheses. What are you _really_ trying to prove? >H. Shinya David C. Ullrich === Subject: Re: Surrogate factoring, reasons for my concerns by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CGCLt10089; (snipped crap) >I¹m not a professional mathematician. No! Really? So James Harris admits that his work is that of an amateur, and therefore, amateurish. QED. >I don¹t have a lot of options. Ceasing postings to this newsgroup is one option. Please exercise it. Otherwise, your putting a loaded gun to your head and pulling the trigger is another option, which will also result in an end to your postings. Killing two birds with one bullet, eh? I recommend the second option. Anonymous === Subject: Re: Sum of a Tricky Series by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CGRgK11282; You will need a stronger condition, such as b<1. Otherwise b=2 and b¹=0.40637574 give the same value of (b e^{-b}) and give the same value of F(a,b) with a=0, but give a different value of frac{1}{1-b}. >I am having problems showing that >sum_{i=0}^{infty} frac{(a+bi)^i}{i!} e^{-(a+bi)} = frac{1}{1-b} >for |b e^{-b}| < e^{-1}. Does anyone know how to evaluate this series? >Justin === Subject: Re: Why are elements in the domain mapped to a single element in the codomain? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CGUfx11697; >Hhhmm, how about calling f a function at inŽnity from R to R >if f: R -> P(R) , R = reals, P(R) = power set whoops... forall epsilon > 0 exists n in naturals forall r > n forall y, z in f(r): |y-z| < epsilon >Similarly, one could probably deŽne, say for any b in R, >a function at b from R to R: >Sincerly, >C. Dement === Subject: Re: complexity of the subgroup problem in free groups >>[Let G be the free group on Z, and let K be a subset of Z. >> What is the complexity of testing membership in the subgroup >> of G generated by K?] >>I do not know what the complexity is, but this problem can be solved >>very fast in practice. For a given K, you construct a Žnite state A >> for which a reduced word w lies in iff it is in the language of >>A. >>So, for a Žxed K, testing membership of elements of G in K is linear, >>but constructing the automaton involves determinization of a >>non-deterministic automaton, so it might not be polynomial in general. >How do you construct that NFA (non-deterministic Žnite automaton)? >What is the complexity of the construction? >Note that determinization is unnecessary. I can test whether a given >word will be accepted by the NFA in quadratic algorithm, using standard >algorithms. Therefore, if construction of the NFA is polynomial, then >the complexity of the whole thing sounds like it should be polynomial >as well. I think you are correct - the whole process is polynomial. You proceed as follows. 1. First contruct an NFA N with language (K union K^-1)*. That is easy and can be done in linear time. 2. Now keep modifying N as follows until no more changes are possible. If, for some generator x in Z union Z^-1, and states s, t of N, there is a path of arrows (which may include epsilon-arrows) from s to t in N labelled by x x^-1, then add an epsilon arrow from s to t if there is not one already. 3. At this stage, a reduced word lies in the subgroup generated by K if and only if it is in L(N), which can be tested in quadratic time. So the only question is whether Step 2 is polynomial. Clearly N can be modiŽed in the way described at most n^2 times, where n is the number of states of N, so we have to show that locating the paths of arrows described is polynomial. There might be an easier way of doing this, but I think you can do it by modifying 2 slightly in the following way. Label epsilon-arrows with e. Then, on each iteration, look only for paths from s to t labelled e^2, x x^-1, e x x^-1, x e x^-1, x x^-1 e, e x e x^-1, e x x^-1 e, x e x^-1 x, or e x e x^-1 e, and add arrow labelled e from s to t if there is not one already. Since we are now looking for paths of length at most 5, this is polynomial. Derek Holt. === Subject: Re: complexity of the subgroup problem in free groups Originator: daw@taverner.cs.berkeley.edu (David Wagner) >>>[Let G be the free group on Z, and let K be a subset of Z. >>> What is the complexity of testing membership in the subgroup >>> of G generated by K?] >>>For a given K, you construct a Žnite state A for which a >>>reduced word w lies in iff it is in the language of A. >>What is the complexity of the construction? >1. First contruct an NFA N with language (K union K^-1)*. > That is easy and can be done in linear time. >2. Now keep modifying N as follows until no more changes are possible. > If, for some generator x in Z union Z^-1, and states s, t of N, there is > a path of arrows (which may include epsilon-arrows) from s to t in N > labelled by x x^-1, then add an epsilon arrow from s to t if there is > not one already. >3. At this stage, a reduced word lies in the subgroup generated by K if and > only if it is in L(N), which can be tested in quadratic time. That can¹t be right. What you end up with is a NFA N with a single state s, along with one transition s --k-> s for each k in K union K^-1 union {epsilon}. That clearly doesn¹t recognize the subgroup of G generated by K; for instance, it doesn¹t accept the word x x^-1, where x is in ZK. Now that I think about it some more, I don¹t believe that the set of words in is a regular language. Let (x)^n represent the word where the symbol x is repeated a total of n times. Fix some x in ZK. For every pair m,n of positive integers, the word (x)^m (x^-1)^n is in iff m = n. Since Žnite state automata can¹t count, I don¹t think any NFA will accept . Possibly a context-free language might sufŽce. Does the following context-free grammar do the trick? E ::= epsilon | E E | E z E z^-1 E (for each z in Z) S ::= E | S S | k | k^-1 (for each k in K) The idea is that L(E) should represent the set of elements of G (i.e., the set of words) that are equal to the identity in G (i.e., that reduce to the empty word). Then, L(S) should represent the set of words that reduce to an element of . Is that correct, or did I make any mistakes? Note that if can be recognized by a context-free language, then elements of can be recognized in polynomial time (in fact, cubic time sufŽces) by using standard algorithms for parsing of context-free languages. === Subject: Re: complexity of the subgroup problem in free groups >>>>[Let G be the free group on Z, and let K be a subset of Z. >>>> What is the complexity of testing membership in the subgroup >>>> of G generated by K?] >>>> >>>>For a given K, you construct a Žnite state A for which a >>>>reduced word w lies in iff it is in the language of A. >>>What is the complexity of the construction? >>1. First contruct an NFA N with language (K union K^-1)*. >> That is easy and can be done in linear time. >>2. Now keep modifying N as follows until no more changes are possible. >> If, for some generator x in Z union Z^-1, and states s, t of N, there is >> a path of arrows (which may include epsilon-arrows) from s to t in N >> labelled by x x^-1, then add an epsilon arrow from s to t if there is >> not one already. >>3. At this stage, a reduced word lies in the subgroup generated by K if and >> only if it is in L(N), which can be tested in quadratic time. >That can¹t be right. What you end up with is a NFA N with a >single state s, along with one transition s --k-> s for each k in >K union K^-1 union {epsilon}. I was assuming that the edges in the N constructed in 1 are labelled by single generators, or are epsilon-edges. For example if Z = {xy}, then N constructed in 1 would have three states 1,2,3, with 1 the start state and only accept state, and edges 1--x-> 2, 1--y^-1 -> 3, 2--y-> 1, 3--x^-1 -> 1, and 1 - epsilon -> 1. In that example, there are no modiŽcations in 2. > That clearly doesn¹t recognize the >subgroup of G generated by K; for instance, it doesn¹t accept >the word x x^-1, where x is in ZK. No, but it accepts all *reduced* words that lie in N, which is enough to test membership of arbitrary words in N - you start by reducing the word, which means removing all adjacent instances of x x^-1. >Now that I think about it some more, I don¹t believe that the >set of words in is a regular language. In fact it is a regular language iff has Žnite index in G. >Let (x)^n represent >the word where the symbol x is repeated a total of n times. >Fix some x in ZK. For every pair m,n of positive integers, the >word (x)^m (x^-1)^n is in iff m = n. Since Žnite state >automata can¹t count, I don¹t think any NFA will accept . It is possible that x^m (x^-1)^n is in when m != n. Derek Holt. >Possibly a context-free language might sufŽce. Does the >following context-free grammar do the trick? > E ::= epsilon > | E E > | E z E z^-1 E (for each z in Z) > S ::= E > | S S > | k | k^-1 (for each k in K) >The idea is that L(E) should represent the set of elements of G >(i.e., the set of words) that are equal to the identity in G >(i.e., that reduce to the empty word). Then, L(S) should represent >the set of words that reduce to an element of . Is that >correct, or did I make any mistakes? >Note that if can be recognized by a context-free language, >then elements of can be recognized in polynomial time >(in fact, cubic time sufŽces) by using standard algorithms >for parsing of context-free languages. === Subject: Re: complexity of the subgroup problem in free groups Originator: daw@taverner.cs.berkeley.edu (David Wagner) >>>1. First contruct an NFA N with language (K union K^-1)*. >>That can¹t be right. What you end up with is a NFA N with a >>single state s, along with one transition s --k-> s for each k in >>K union K^-1 union {epsilon}. >I was assuming that the edges in the N constructed in 1 are labelled >by single generators, or are epsilon-edges. I don¹t understand your objection. My NFA satisŽes those conditions; each edge is labelled by a single generator, or is an epsilon-edge. Indeed, my construction is in some sense the canonical NFA satisfying those conditions: it is the minimal NFA accepting the desired language. If you meant to impose some extra conditions on the NFA constructed in Step 1, can you clarify what extra conditions you require? Did you have in mind some one speciŽc NFA? If so, can you give the construction you had in mind? >For example if Z = {xy}, then >N constructed in 1 would have three states 1,2,3, with 1 the start state >and only accept state, and edges 1--x-> 2, 1--y^-1 -> 3, 2--y-> 1, >3--x^-1 -> 1, and 1 - epsilon -> 1. In that example, there are no >modiŽcations in 2. I¹m puzzled. If Z={x,y}, what is K? It looks like your NFA accepts the language ((x y) | (y^-1 x^-1))^*. I¹m having trouble Žguring out for which set K this language is equal to (K union K^-1)^*. You must not have intended K={x,y}, because this FSA doesn¹t accept the word y x^-1, which is in (K union K^-1)^*. You must not have intended K={x}, because this FSA doesn¹t accept the word x x, which is in (K union K^-1)^*. Similarly for K={y}. You must not have intended K={}, because this FSA does accept the word x y, which is not in (K union K^-1)^*. I conclude that I must be confused. Can you elaborate, please? === Subject: Re: complexity of the subgroup problem in free groups actually, from what i read about free groups and todd coxeter enumeration, and stuff, 1. The problem of Žnding whether some word lies in this sub-group is decidable only if the index of the subgroup is Žnite. 2. And it is undecidable in general to compute the index of the subgroup. So, can this be exploited to generate instances of the word problem that are undecidable? or am i wrong in my understanding of facts 1 and 2 bye abi === Subject: Re: complexity of the subgroup problem in free groups >actually, from what i read about free groups and todd coxeter >enumeration, and stuff, >1. The problem of Žnding whether some word lies in this sub-group is > decidable only if the index of the subgroup is Žnite. This is not true - maybe you meant if rather than only if? The problem you are talking about is known as the generalized word problem. It is solvable for a Žnitely generated subgroup of Žnite index in a group deŽned by a Žnite presentation. But it is solvable in some other cases as well. For example it is solvable for any Žnitely generated subgroup of a free group of Žnite rank, which was the original problem in this thread. It is of course unsolvable in general. For example there are Žnitely generated subgroups of the direct product F2 x F2 of two copies of the free group of rank 2 for which it is unsolvable. >2. And it is undecidable in general to compute the index of the > subgroup. Yes, that is true. But it is useful to remember that, for Žnitely generated subgroups of Žnite index Žnitely presented groups, it is possible to verify that the index is Žnite - that is exactly what Todd-Coxeter coset enumeration does. So this property is in some sense semi-decidable. Derek Holt. === Subject: Re: Minimal Triangulation for Simplicial Complex > > On a related note, you keep mentioning manifolds. I have read a little > > about them but have mostly found them discussed with relation to > > differential equations while I am more interested in discrete dynamical > > systems (dds). I assume that the topology of a dds like x_r = A^r * x_0 is > > simply the image of the equation in Euclidean R^n space (where A is an n x n > > matrix)? Are there any other known ways to Œtopologize¹ dds? What does the > > use of topology buy us that things like limits, and stability analysis > > don¹t? (At this point I would be satisŽed with nothing more than a few > Manifolds are nice for many reasons. One reason is that they are > Žnite dimensional (at least what¹s usually called a manifold) and look > locally like R^n. That¹s just really nice and useful. R^n is a much > easier topological space to understand than a lot of other spaces. > They also show up in discrete systems too. Here mentioning connections > to algebraic geometry may be useful to you. > For example, you can consider the conŽguraion space of a planar > linkage. A planar linkage is a Žnite set of line segments in the > plane (that can cross) such that each segment has at least one end > attached to another end of a segment or Žxed to the plane, i.e. nailed > down. There are many conŽgurations to a given planar linkage. Many > ways to move them so they still obey the conditions on the ends. I would guess that the lengths of each line segment must remain Žxed throughout any transformation? > We can imagine trajectories through the space of conŽgurations. What > kind of space is it? It¹s a smooth manifold. In fact every connected > compact smooth manifold is the conŽguration space of a planar linkage; > I believe this a theorem of Thurston. > I don¹t know if this is what you wanted, but I hope it¹s interesting. It is interesting and suggests a nice discretization for Œcompact smooth manifold¹s. However I don¹t immediately see how I could apply this to what I¹m looking into. By the way, a while back you posted some comments on a paper I linked to about the use of algebraic topology in asynchronous computability. For a substantially more succinct paper on the same topic check out: http://citeseer.ist.psu.edu/366609.html The problem with the original paper I had linked to was that it was a bit preliminary. In the theory, the speciŽc simplicial complexes used to model asynchronous protocols are very important. Things like delta and cell complexes often lack the extra discrete structure of simplicial complexes utilized in this very combinatorial theory. However, at the time the paper I linked to was written, constructing the correct complex given a decision task was essentially done in an ad-hoc manner. This paper introduces a standard procedure for constructing the complexes required by all three major message passing models (asynchronous, semi-synchronous and synchronous) based on the idea of what they call Œpsuedospheres¹. The associated complexes for protocols in any of these models are either pseudospheres or unions of pseudospheres (depending on the model¹s timing properties). Also I have started to look into Chu spaces (speciŽcally their use in deŽning Œgeometric automata¹) as a possible avenue to extend distributed protocol complexes to the point where they may also model Žner grained concurrency (ie: problems akin to the dining philosophers). A beneŽt of Chu spaces is that they provide a framework in which one can abandon interleaving models of concurrency (whether this is truly needed to solve most practical problems remains a source of debate). However, it appears that one would probably also have to abandon simpler inductive proofs (a staple of proofs for interleaving models) in many cases. On the other hand perhaps one could Œgraft¹ chunks of geometric automata onto the simplices of a standard protocol complex. Interestingly, Vaughn Pratt (a major contibutor and proponent of Chu spaces for modeling concurrency) seems to have lost interest in geometric automata (a subject he has written many papers on) and I¹m not sure why yet. hronous_Computability/ Anyways, enough babbling... l8r, Mike N. Christoff === Subject: Re: JSH: Sweep likely by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CGqY013275; >Anonymous a dit: >> Right, Quinn? Don¹t you agree, Mr. Quinn? >> Don¹t you just love your colleague James >> Harris and the glory of his posts which >> režect on your wonderful high-IQ society, >> huh Quinn? >Voici vingt-cinq sous. Vas-y, joue dans la rue. >Quinn As evasive as ever, huh, Quinn-baby. Oooo, using French against me! I¹m trembling at the knees at your awesome _mind_. Ha! You and James Harris are an embarrassment to your high-IQ friends. Perhaps while James Harris threatens the Ph.Ds of others on this newsgroup in his other asinine threads, I¹ll look into forwarding copies of yours and his silly posts to your high-IQ friends (anonymously of course!), and suggest they kick your sorry asses out of their membership. Hey, I¹d only be stooping down to James¹ level, so neither he nor YOU can complain. Hoisted by James¹ own petard. Heh heh heh. You like James¹ petard, don¹t you Quinn, and being hoisted onto it? James had better start erasing more of his crap postings before I can print them. Unless, of course, the high-IQers¹ brains cannot handle their Žnances and their lawsuits and they are desperate for the dues from you and James Harris. Then you needn¹t worry. OTOH if there are no dues, then maybe they won¹t mind kicking you and James out for making asses out of yourselves and of them by your continually posting crap. Oooo, I¹ll post an anonymous website and/or pamphlet against that high-IQ society, and publish yours and James Harris¹ postings as Exhibit 1 of the patheticness of that high-IQ group. Fun fun fun! Won¹t they enjoy that? I already doubt the merits of James Harris¹ membership in your high-IQ group. I am now beginning to doubt your merits in that group, and I even question whether that high-IQ group really have high-IQs. Or are they a bunch of anti-social misŽts like James Harris, who repeatedly tout their high intelligence as James has, with no factual evidence backing up their claims? Anonymous === Subject: Re: JSH: Sweep likely Discussion, linux) > Perhaps while James Harris threatens the > Ph.Ds of others on this newsgroup in his > other asinine threads, I¹ll look into > forwarding copies of yours and his silly > posts to your high-IQ friends (anonymously > of course!), and suggest they kick your > sorry asses out of their membership. You¹re just a tedious asshole, aren¹t you? -- You lack the ability to reason, but instead get an idea in your head and hold on to it against all evidence. I don¹t Žnd you credible, and reject your claims, as coming from a žawed source. === Subject: Re: JSH: Sweep likely >>Anonymous a dit: >>>Right, Quinn? Don¹t you agree, Mr. Quinn? >>>Don¹t you just love your colleague James >>>Harris and the glory of his posts which >>>režect on your wonderful high-IQ society, >>>huh Quinn? >>Voici vingt-cinq sous. Vas-y, joue dans la rue. >>-- >>Quinn > As evasive as ever, huh, Quinn-baby. Oooo, using > French against me! I¹m trembling at the knees > at your awesome _mind_. Ha! > You and James Harris are an embarrassment to your > high-IQ friends. Whereas your post was nothing but an embarassment to yourself. Please go away. Rick === Subject: Re: Pointless Topology by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CGqYt13280; (snip) I agree. Topology is pointless. . === Subject: Proving the OS of chess is a draw; Counterexample doth not make a proof With my claim that multiple Žrst moves by either white or black in chess is a Convergent Series of OS draws in chess and proves that the OS of chess is a draw. We have the interesting example of the game Nim where the OS is a sure win for either second player or Žrst player depending on how win is deŽned. The thing about Nim is that it has no possibility of a draw in the game itself. So, adding the facts of Nim to the Convergent Series of multiple Žrst moves in chess, we are left with a tantalizing counterexample. Can anyone produce a VonNeumann game where it has a forced win OS and yet still possess the possibility of a draw in the game itself?? You see Nim is a forced win but it has no possibility of a draw. attractor or Great Attractor in chaos theory. Because if a VonNeumann game has a possible draw within the game itself which tictactoe, checkers, chess, go all possess a draw possibility that those games OS seem to gravitate or be Attracted to the draw. I know a Counterexample doth not make for a proof. But if no counterexample exists of a VonNeumann Game wherein the OS is a forced win and yet there does exist a draw within the game, then it is highly likely that in all VonNeumann games that possess a draw possibility that the OS of those games are all draws. Can the MiniMax theorem shed light on those claims? I think so. I think the minimax theorem speaks of a singularity point and a draw would always exist wedged between wins or losses for both sides and so how can you have a singularity that is not the draw itself. So like the Great Attractor in chaos theory you cannot have a draw within a VonNeumann game and not have that draw as the OS. So unless a counterexample is forthcoming, it looks as though all VonNeumann games with a draw possibility have a draw OS. Archimedes Plutonium www.archimedesplutonium.com www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: a draw Nim Re: Proving the OS of chess is a draw; Counterexample doth not make a proof (slightly snipped) > Can anyone produce a VonNeumann game where it has a forced win OS and > yet still possess the possibility of a draw in the game itself?? You see > Nim is a forced win but it has no possibility of a draw. > attractor or Great Attractor in chaos theory. Because if a VonNeumann > game has a possible draw within the game itself which tictactoe, > checkers, chess, go all possess a draw possibility that those games OS > seem to gravitate or be Attracted to the draw. > I know a Counterexample doth not make for a proof. But if no > counterexample exists of a VonNeumann Game wherein the OS is a forced > win and yet there does exist a draw within the game, then it is highly > likely that in all VonNeumann games that possess a draw possibility that > the OS of those games are all draws. > Can the MiniMax theorem shed light on those claims? I think so. I think > the minimax theorem speaks of a singularity point and a draw would > always exist wedged between wins or losses for both sides and so how can > you have a singularity that is not the draw itself. So like the Great > Attractor in chaos theory you cannot have a draw within a VonNeumann > game and not have that draw as the OS. So unless a counterexample is > forthcoming, it looks as though all VonNeumann games with a draw > possibility have a draw OS. So, now when I introduce a draw into Nim by saying that the middle row is a draw row and keeping the game as simple as possible in that I have just 3 rows in all and one matchstick in each row. And that Žrst row is a win for Žrst player and third row is win for second player. Regular-nim would have second player always win that game because Žrst player would be forced to pick up one last matchstick. But by having the middle row stick as a draw stick changes the dynamics such that the OS now becomes the draw. So, what I need is a VonNeumann Game wherein it has a draw possibility but still remains as a forced win of its OS (Optimal Strategy). Archimedes Plutonium www.archimedesplutonium.com www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Mathematicians and scientists become wealthy and rule the world by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CHLMQ16398; >What if mathematicians and scientists, instead of making their >work freely available to the world, were to copyright and/or >patent their work, as geneticists are doing with genes, and >noone could use it without paying? >Wouldn¹t they eventually become the wealthiest and most >powerful people in the world? >Consider the owners of calculus and electromagnetism, for example. >Van Don¹t believe James Harris and his delusions of fortune to be won from his crapping mathemtics. Anonymous . . . . === Subject: Formula for position of a rocket after time T? Hi all, Given a rocket of mass M, whose engine is consuming m units of fuel per second and applying force F, is there a formula for determining its velocity and position after t seconds? The kinematic formula: vt + (at^2)/2 won¹t work because the rocket is losing mass (and therefore increasing its acceleration) as it accelerates. I can see how to get an approximate answer by numeric integration of course, but I¹m wondering if there¹s an analytic formula? And, given two such rockets (each with its own independent trajectory in 3-dimensional space), is there a formula for determining what their minimum approach distance will be and when it will occur? (In case you¹re wondering, no, these aren¹t homework questions :)) -- Sore wa himitsu desu. To reply by email, remove the small snack from address. === Subject: Re: Formula for position of a rocket after time T? > Hi all, > Given a rocket of mass M, whose engine is consuming m units of fuel > per second and applying force F, is there a formula for determining > its velocity and position after t seconds? The kinematic formula: > vt + (at^2)/2 > won¹t work because the rocket is losing mass (and therefore increasing > its acceleration) as it accelerates. I can see how to get an > approximate answer by numeric integration of course, but I¹m wondering > if there¹s an analytic formula? > And, given two such rockets (each with its own independent trajectory > in 3-dimensional space), is there a formula for determining what their > minimum approach distance will be and when it will occur? > (In case you¹re wondering, no, these aren¹t homework questions :)) If you care enough to know you care enough to Žnd out. Real world rocket intecepting rocket is a four star pisser on paper even with active guidance. In the life and death real world it does not obtain unless intercept means the thermonuclear warhead detonates. Military interceptions use proximity fuzes and shrapnel, or, closer in, a whole lot of computer-guided chainguns plus brown trowsers. Try shooting ducks with a riže. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Formula for position of a rocket after time T? > Hi all, > Given a rocket of mass M, whose engine is consuming m units of fuel > per second and applying force F, is there a formula for determining > its velocity and position after t seconds? > And, given two such rockets (each with its own independent trajectory > in 3-dimensional space), is there a formula for determining what their > minimum approach distance will be and when it will occur? F = ma does work for the changing mass of rockets! Start with F = dp/dt The force accelerating a rocket is equal to the thrust generated by the engine, and in the opposite direction. m dv/dt = F = - (dm/dt) v_e where m is the instantaneous mass of the rocket (it changes as fuel is expelled), the rocket¹s acceleration is a=dv/dt. The mass žow rate of the exhaust is dm/dt, with average velocity v_e. Rearrange terms a little to get an appropriate differential equation. dv = - v_e dm / m Integrate the left side from 0 to Žnal velocity v, the right side from full weight M to Žnal weight m. v = - v_e ln (M / m) When the ratio M / m > 2.72, the rocket will be going faster than its exhaust speed. Newton doesn¹t recognize a speed limit at all for the rocket, so long as you burn enough fuel (and, most likely, shed some fuel tanks along the way). Similar thinking should allow you to derive the equations you seek. Play with coordinate systems. === Subject: Re: Formula for position of a rocket after time T? > Integrate the left side from 0 to Žnal velocity v, the right side from > full weight M to Žnal weight m. > v = - v_e ln (M / m) at some time in the past, but I didn¹t remember what it was called. The position is the integral of that; there¹s *mumble* years of rust on the calculus I learned in college, but I managed to Žnd a page that explains how to integrate it. > Similar thinking should allow you to derive the equations you seek. > Play with coordinate systems. So for the position of rocket B relative to A, take A as the origin and subtract its acceleration from B¹s (and then differentiate to Žnd the minimum of the resulting distance function)? Okay... I can see how that would work except I¹m not sure how to take into account the fact that each of them would burn through its fuel at a different rate... I suppose plug it into the formula for acceleration at time t, then redo the integrations? I might need to try cleaning off some of the above-mentioned rust :P -- Sore wa himitsu desu. To reply by email, remove the small snack from address. === Subject: Re: Formula for position of a rocket after time T? > Hi all, > Given a rocket of mass M, whose engine is consuming m units of fuel > per second and applying force F, is there a formula for determining > its velocity and position after t seconds? The kinematic formula: Google rocket equation === Subject: Re: Formula for position of a rocket after time T? > Given a rocket of mass M, whose engine is consuming m units of fuel > per second and applying force F, is there a formula for determining > its velocity and position after t seconds? The kinematic formula: > vt + (at^2)/2 > won¹t work because the rocket is losing mass (and therefore increasing > its acceleration) as it accelerates. I can see how to get an > approximate answer by numeric integration of course, but I¹m wondering > if there¹s an analytic formula? > And, given two such rockets (each with its own independent trajectory > in 3-dimensional space), is there a formula for determining what their > minimum approach distance will be and when it will occur? > (In case you¹re wondering, no, these aren¹t homework questions :)) The acceleration at time t is F/(M-mt). Integrate that to Žnd the velocity. Integrate _that_ to Žnd the position. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Formula for position of a rocket after time T? > Hi all, > Given a rocket of mass M, whose engine is consuming m units of fuel > per second and applying force F, is there a formula for determining > its velocity and position after t seconds? The kinematic formula: > vt + (at^2)/2 > won¹t work because the rocket is losing mass (and therefore increasing > its acceleration) as it accelerates. I can see how to get an > approximate answer by numeric integration of course, but I¹m wondering > if there¹s an analytic formula? > And, given two such rockets (each with its own independent trajectory > in 3-dimensional space), is there a formula for determining what their > minimum approach distance will be and when it will occur? === Subject: easy: paths on a grid units East, remaining on the unit lattice, how many paths are available to me without backtracking in either dimension? I see it either of these ways: I must distribute m steps North among n+1 East-West lines of the lattice; or I must distribute n steps East among m+1 North-South lines of the lattice. Hopefully these give the same result! How do I show this, and what¹s the result? Or is my reasoning incorrect? Thnx - Anita === Subject: Re: easy: paths on a grid >units East, remaining on the unit lattice, how many paths are available to >me without backtracking in either dimension? >I see it either of these ways: I must distribute m steps North among n+1 >East-West lines of the lattice; or I must distribute n steps East among >m+1 North-South lines of the lattice. Hopefully these give the same >result! How do I show this, and what¹s the result? Or is my reasoning >incorrect? It¹s the same as the number of ways to žip a coin n+m times and get n heads and m tails. === Subject: Re: easy: paths on a grid > units East, remaining on the unit lattice, how many paths are available to > me without backtracking in either dimension? > I see it either of these ways: I must distribute m steps North among n+1 > East-West lines of the lattice; or I must distribute n steps East among > m+1 North-South lines of the lattice. Hopefully these give the same > result! How do I show this, and what¹s the result? Or is my reasoning > incorrect? Never mind I Žgured it out: it¹s the multichoose function, and here it¹s equal to (m+n)!/(m!n!) so clearly it¹s the same when you switch m and n. Anita === Subject: Re: easy: paths on a grid > units East, remaining on the unit lattice, how many paths are available to > me without backtracking in either dimension? (Assuming you mean traveling only North and East; otherwise I think it becomes non-trivial to compute the number of paths through the grid.) You have m Norths and n Easts. If you lay them out in a row, then that will be one of your possible paths. Example: let m=3 and n=2. Then your paths are the permutations NNNEE NNENE NNEEN NENNE NENEN NEENN ENNNE ENNEN ENENN EENNN So, how many permutations are there of m N¹s and n E¹s? This is a formula you probably learned in school, but it¹s easy to derive. We have (m+n) things to put in a row, but it doesn¹t matter in what order m of them are, nor in what order n of them are. So we take the number of ways to order (m+n) things, and then divide out by the number of ways to order m things, and by the number of ways to order n things, leaving (m+n)! ------ m!n! where ! is the factorial sign, n! = n*(n-1)*(n-2)*...*3*2*1. HTH, -Arthur === Subject: Re: easy: paths on a grid days. My association with the Department is that of an alumnus. >units East, remaining on the unit lattice, how many paths are available to >me without backtracking in either dimension? >I see it either of these ways: I must distribute m steps North among n+1 >East-West lines of the lattice; or I must distribute n steps East among >m+1 North-South lines of the lattice. Hopefully these give the same >result! How do I show this, and what¹s the result? Or is my reasoning >incorrect? The usual way to solve this problem is to draw the lattice and note that the number of ways to reach the point (a+1,b+1) is equal to the number of ways of reaching the point (a+1,b) plus the number of ways of reachiing the point (a,b+1); and that there is only one way of reaching any point of the form (0,b) and only one way of reaching any point of the form (a,0). This will yield a recurrence relation that can be solved, or you can recognize that you have a slice of Pascal¹s triangle and use the well-known Œchoose¹ formula for the number that corresponds to the upper left corner. But we can also do it as you suggest. Consider Žrst the situation where you know you will take m steps North, and n steps East; you want to decide where you will take the n steps East. You have, as you note, m+1 places where you can place the steps East: before the Žrst step North, after the last step North, and in between any two. Imagine each step East is a ball, all balls are identical, and each space where you can put a step East is a bucket. So your problem becomes the following familiar counting problem: In how many ways can we distribute n identical balls among m+1 buckets, with repetitions allowed (i.e., more than one ball in a bucket is valid)? This is equivalent to simply choosing which buckets will receive balls, with repetitions allowed, so it becomes: In how many ways can we choose n out of m+1 buckets, if repetitions are allowed, and the order in which we make the choices doesn¹t matter? This is called combinations with repetitions, and there is a ready-made-formula. The number of ways in which you can choose k out of r possibilities, with repetitions allowed and order of choosing immaterial is equal to r+k-1 choose k = (r+k-1)!/ (k!)(r-1)!. Thus, setting r=m+1, k = n, we get m+n choose n = (m+n)!/(n!)(m!). If instead you want to distribute the m steps North in the n+1 spaces between the steps East, then you are doing r = n+1, k = m, which yields n+m choose m = (m+n)!/(m!)(n!) the same answer. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: A Cute Problem in Calculus by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CIlSp23647; >>>What does the difference between d1 & d2 have to do with the >>>distance between P & Q? >>>Compare the difference between d1 & d2 to the difference between >>>the total distance travelled by P and by Q during TT. >I don¹t get this. Can you be more speciŽc? Maybe those hints are don¹t make much sense unless you¹ve already solved the problem, and if they¹re confusing, they might be counter-productive, so just forget about those hints, and go with the earlier ones. The earlier ones are the most useful ones, I think. === Subject: Re: JSH: Sweep likely by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CIlSN23653; >>>Anonymous a dit: >>>>Right, Quinn? Don¹t you agree, Mr. Quinn? >>>>Don¹t you just love your colleague James >>>>Harris and the glory of his posts which >>>>režect on your wonderful high-IQ society, >>>>huh Quinn? >>>> >>>Voici vingt-cinq sous. Vas-y, joue dans la rue. >>>-- >>>Quinn >> As evasive as ever, huh, Quinn-baby. Oooo, using >> French against me! I¹m trembling at the knees >> at your awesome _mind_. Ha! >> You and James Harris are an embarrassment to your >> high-IQ friends. > >Whereas your post was nothing but an embarassment >to yourself. Please go away. When James Harris goes away, and not before. In fact, Harris didn¹t post for awhile, and so neither did I. I didn¹t think he was gone for good, but one can only hope. Anonymous === Subject: Re: InŽnity can not exist by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CIlSu23639; >It is impossible for inŽnity to exist, because in order for inŽnity >to exist, everything possible must happen. Including inŽnity not >existing. Therefore, it is possible for something to be an extrememly >large, but not inŽnite. For it to be inŽnite, it would have to be >Žnite at the same time. >Japcuh >(Just Another Perl C Unix Hacker) > http://www.ca tb.org/~esr/faqs/hacker-howto.htm#what_is >.O. >..O >OOO Counterexample: James Harris has an inŽnity amount of crackpot ideas, which will result in an inŽnity^inŽnity number of posts and reposts and multiple threads on identical topics. Anonymous === Subject: Count the number of ones in binary rapresentation of n Hi all, I¹ve got a natural number n. Its binary rapresentation has k ciphers. I want to count the number of 1s which appears in this binary rapresentation of n using _only_ the operators +, -, * and / on the number n and k. For example: n = 12 n = 1100(2) ergo k = 4. I would like to Žnd the result (2 ones appear) using + - * / on n and k. Giacomo -- _-==.==-_*_-==.==-_ _ _ __ _ Ogni sistema ha le sue leggi : |_)_ |_) (_ _ |_) alcune possono essere eluse | (_|| o__)(/_| altre...infrante (Morpheus) (Giacomo Bellucci) PaR._IL_BIANCONIGLIO_SeR@libero.it _________ /Žcm code Per rispondermi togliete [Hx23B] / _IL_BIANCONIGLIO_ dall¹email [OSl ash] === Subject: Re: Count the number of ones in binary rapresentation of n > Hi all, > I¹ve got a natural number n. Its binary rapresentation has k ciphers. I want > to count the number of 1s which appears in this binary rapresentation of n > using _only_ the operators +, -, * and / on the number n and k. Ciphers are zeroes. Is that what you meant? > For example: > n = 12 > n = 1100(2) > ergo k = 4. Four is the number of binary digits, not the number of ciphers. > I would like to Žnd the result (2 ones appear) using + - * / on n and k. Do you mean you want to write a function in some programming language such as C? Is there some particular reason why you don¹t want to allow the bitwise AND operator? int bitcount(long n) { int cnt = 0; long t; for (t = n; t; t &= (t-1)) cnt++; return cnt; } Notice that there is no k here; the only input is n. -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Count the number of ones in binary rapresentation of n > Hi all, > I¹ve got a natural number n. Its binary rapresentation has k ciphers. I > want > to count the number of 1s which appears in this binary rapresentation of n > using _only_ the operators +, -, * and / on the number n and k. > For example: > n = 12 > n = 1100(2) > ergo k = 4. > I would like to Žnd the result (2 ones appear) using + - * / on n and k. It is not entirely clear if you are given k, or must compute k. Also, are you planning to implement this in a program? Can we introduce other variables and do comparisons such as < and >? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Count the number of ones in binary rapresentation of n > Hi all, > I¹ve got a natural number n. Its binary rapresentation has k ciphers. I want > to count the number of 1s which appears in this binary rapresentation of n > using _only_ the operators +, -, * and / on the number n and k. > For example: > n = 12 > n = 1100(2) > ergo k = 4. So you want a rational function in n and k? That is not possible. === Subject: Re: Count the number of ones in binary rapresentation of n [SNIP] > So you want a rational function in n and k? That is not possible. Why ? -- (Giacomo Bellucci) PaR._IL_BIANCONIGLIO_SeR@libero.it _________ /Žcm code Per rispondermi togliete [Hx23B] / _IL_BIANCONIGLIO_ dall¹email [OSl ash] === Subject: Bucky¹s We In honor of Bucky on his 109th birthday and which is also the issue date of a commemorative USPostal stamp, I offer up the following quote and wonder if anyone else here beside me considers themselve among Bucky¹s We.(and don¹t say it makes no sense, please) Dick 825.28 Euclid was not trying to express forces. We, however__inspired by Avogadro¹s identical-energy conditions under which different elements disclosed the same number of molecules per given volume__are exploring the possible establishment of an operationally strict vectorial geometry Želd, which is an isotropic (everywhere the same) vector matrix. We abandon the Greek perpendicularity of construction and Žnd ourselves operationally in an omnidirectional, spherically observed, multidimensional, omni- intertransforming Universe. Our Žrst move in spherical reality scribing is to strike a quasi- sphere as the vectorial radius of construction. Our dividers are welded at a Žxed angle. The second move is to establish the center. Third move: a surface circle. The radius is uniform and the lesser circle is uniform. opposites to make two tetrahedra with a common vertex at the center. Two tetrahedra have six internal faces=hexagon=genesis of bow tie=genesis of modelability=vector equilibrium. Only the dividers and straightedge are used. You start with two events__any distance apart: only one module with no subdivision; ergo, timeless; ergo, eternal; ergo, no frequency. Playing the game in a timeless manner. (You have to have division of the line to have frequency, ergo, to have time.) (See Secs. 420 and 650.) === Subject: New computer algebra system uses identities to simplify Hello group. I have just Žnished the Žrst version of a simpliŽer which uses identities to manipulate expressions. Because the system has no knowledge of the functions it uses except for the identities, other types of algebra may be implemented as well, as a separate collection. For example, in order to implement boolean algebra, all I had to do was to list the identities. The program is found at this url: (it is noncommercial) http://www.denotesoftware.com/?page=ether&title=Ether - Martin Johansen === Subject: Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory charset=iso-8859-1 ksRobak > geometric Robak -- Time Theory by ksRobak (continue) > ~~~~~~~~~~~~~~~~~~~~~~~~~ > A B B C > o------------------o o------------------o > 1 2 3...->oo Re1 1 2 3...->oo Re1 > PUNKT={AB+}0 PUNKT={BC+}0 > AB + BC = AC > A B B C > o------------------o o------------------o > A B B C > o------------------o o------------------o > A B B C > o------------------o o------------------o > A B C > o------------------o------------------o > A C > o------------------------------------o > PUNKT={AB+}0 + {BC+}0 = {AC+}0 > ~~~~~~~~~~~~~~~ > {AC+}0 * Re1 = AC > ~~~~~~~~~~~~~~~ > Edward Robak ed_robak@wp.pl |/ re: > -- > AGEOMETRETOS MEDEIS EISITO Linear ili korpuskula? hehe Linear = korpuskula :o) _ 1_ _ _ _ _ _ _ _ _ n A|OOOOOOOOOOOOOOOOO|B AB = n*O _ 1_ _ _ _ _ _ _ _ _ n A|ooooooooooooooooo|B AB = n*o _ 1_ _ _ _ _ _ _ _ _ n A|()()()()()()()()()|B AB = n*() _ 1_ _ _ _ _ _ _ _ _ Re1 A|_________________|B AB = Re1*{AB+}0 (~ oo * PUNKT) AB=AB A to A Edward Robak ed_robak@wp.pl |/ re: -- Ipse dixit! === Subject: Example for an assumption of the Implicit Function Theorem I am looking for an example to show that, for Implicit Function Theorem, it does not sufŽce to have the partial derivatives of Žrst order continuous at the point, but it is needed that the continuity of those derivatives is satisŽed in a neighborhood of the point. Has someone here came across with such an example? Paul === Subject: apply predicate to a list Is the CS concept of map -- predicate applied to a list -- well-deŽned from math perspective? For comparison, let¹s focus on sets, Žrst. Any set can be represented by its characteristic function, so that applying predicate to a set is just multiplying 2 characteristic functions (and, perhaps, applying sum/integral operator on top of it). Lists/sequences, however, are different. Any list is a formal univariable polynomial. In this representation, however, I can¹t Žgure out what is the operator that applies predicate. An example: Assuming domain of reals, the set {4,5,-1,3} is represented by Dirac(x-4)+Dirac(x-5)+Dirac(x+1)+Dirac(x-3) If we want to apply predicate x>0, then we just multiply it by Heaviside(y) and get Dirac(x-4)+Dirac(x-5)+Dirac(x-3) For equality predicate, say x=5, I multiply by Dirac(y-5) and integrate. (This begs a question why do I integrate in one case, but not in the other). List, however, [4,5,-1,3] is represented as a polynomial sum(a_n*z^n)=4+5*z-z^2+3*z^3 Then, is there a way to apply predicate a_n>0 and get 4+5*z+3*z^3 ? === Subject: Re: L¹Hopital #2; Correction by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CKTvP32621; >>... >>>You have yet to state coherently what you¹re trying to prove. What is the >>>result? State it precisely and completely. Do not omit any relevant >>>hypotheses. >>Conditions for validity of applying L¹Hopital¹s rule to fractions of the form >>(f + g)/f >>where g/f -> 0 as x -> inŽnity. Assume all the hypotheses of >>L¹Hopital¹s rule. >Huh? >If the hypotheses of L¹Hopital¹s rule are satisŽed then >you _can_ apply it - there¹s no further conditions needed. >You might also note that if g/f -> 0 then it¹s pretty clear >that (f + g)/f -> 1, regardless of any other hypotheses. Oh yes. It is really pretty clear. What is the origin of this thread? First I wanted to know if f/g -> L implies f¹/g¹ -> L. But as we have seen, there are a lot of counterexamples. Then I wanted to know on what conditions lim f¹/g¹ is different from lim f/g. Well, the condition can be stated as follows: If (f+g)/f -> 1, g/f -> 0, and g¹/f¹ -> 0, then (f¹ + g¹)/f¹ -> 1. The difŽculty in this theorem is to prove that g¹/f¹ -> 0. But actually we may easily show g¹/f¹ -> 0 only if we know what g is. (As in the case where f = x and g = cos x). But this is complete nonsense in terms of Žnding the numerical value of the formula (f¹ + g¹)/f¹, since we know that (f + g)/f -> 1. Why the heck are we using it ? You know. So, okay, when might the result be useful? Consider the case where you don¹t know g exactly (error term of a function is a good example, such as an error term in the prime number theorem.) But the result says that if g¹/f¹ -> 0, then the formula (f¹ + g¹) / f¹ also tends to 0. But you have to be able to know the formula g¹ to calculate g¹/f¹. Putting g = F(x) - f(x) will not help you. (F(x) = f(x) + g(x)). The result is not important. H. Shinya === Subject: Re: L¹Hopital #2; Correction >>>... >>>>You have yet to state coherently what you¹re trying to prove. What is the >>>>result? State it precisely and completely. Do not omit any relevant >>>>hypotheses. >>>Conditions for validity of applying L¹Hopital¹s rule to fractions of the form >>>(f + g)/f >>>where g/f -> 0 as x -> inŽnity. Assume all the hypotheses of >>>L¹Hopital¹s rule. >>Huh? >>If the hypotheses of L¹Hopital¹s rule are satisŽed then >>you _can_ apply it - there¹s no further conditions needed. >>You might also note that if g/f -> 0 then it¹s pretty clear >>that (f + g)/f -> 1, regardless of any other hypotheses. >Oh yes. It is really pretty clear. >What is the origin of this thread? >First I wanted to know if f/g -> L implies f¹/g¹ -> L. But as we have >seen, there are a lot of counterexamples. >Then I wanted to know on what conditions lim f¹/g¹ is different from >lim f/g. Well, the condition can be stated as follows: >If (f+g)/f -> 1, g/f -> 0, and g¹/f¹ -> 0, then (f¹ + g¹)/f¹ -> 1. Through all of this it has seemed like you¹re not thinking about what you write, and not thinking about what people write in reply. The result you just stated is completely stupid, _especially_ after what I just said: Because g¹/f¹ -> 0, then (f¹ + g¹)/f¹ -> 1. This has nothing to do with calculus, nothing to do with any conditions on f and g, nothing to do with _anything_. Proof: (f¹ + g¹)/f¹ = g¹/f¹ + 1. QED. >The difŽculty in this theorem is to prove that g¹/f¹ -> 0. >But actually we may easily show g¹/f¹ -> 0 only if we know what >g is. (As in the case where f = x and g = cos x). But this is >complete nonsense in terms of Žnding the numerical value of >the formula (f¹ + g¹)/f¹, since we know that (f + g)/f -> 1. Why >the heck are we using it ? You know. No, I have no idea what you¹re talking about. >So, okay, when might the result be useful? Consider the case where >you don¹t know g exactly (error term of a function is a good example, >such as an error term in the prime number theorem.) >But the result says that if g¹/f¹ -> 0, then the formula >(f¹ + g¹) / f¹ >also tends to 0. Huh? That¹s because 1 + 0 = 0, right? >But you have to be able to know the formula >g¹ to calculate g¹/f¹. Putting g = F(x) - f(x) will not help you. >(F(x) = f(x) + g(x)). >The result is not important. If you¹ve actually shown 1 + 0 = 0 as you claim that¹s very important. >H. Shinya ************************ David C. Ullrich === Subject: Find end points of a line I am writing a computer program that is to draw lines of a particular distance (length) and slope. I know the following values: - one point of the line (X1,Y1) - the slope of the line (m) - the y-intercept (b) - the distance of the line (d) In order to draw the line, I need to Žnd each of the 2 possible end points. Useful formulas: ------------------- The formula for a line: y = mx + b The distance between points (X1,Y1) and (X2,Y2): d = sqrt of [ (X2 -X1)squared + (Y2 -Y1)squared ] For example, what would the other end points be for a line with the formula: y = mx + b with values 16 = (5)(2) + 6 d = 8 X1=2 and Y1=16 http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: Re: Find end points of a line > I am writing a computer program that is to draw lines of a particular > distance (length) and slope. I know the following values: > - one point of the line (X1,Y1) > - the slope of the line (m) > - the y-intercept (b) > - the distance of the line (d) > In order to draw the line, I need to Žnd each of the 2 possible end points. > Useful formulas: > ------------------- > The formula for a line: > y = mx + b > The distance between points (X1,Y1) and (X2,Y2): > d = sqrt of [ (X2 -X1)squared + (Y2 -Y1)squared ] > For example, what would the other end points be for a line with the formula: > y = mx + b > with values 16 = (5)(2) + 6 > d = 8 > X1=2 and Y1=16 You have most of what you need: You can compute your angle as tan^-1(m), and then use cosine and sine of that times d to Žnd your change in x and y. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Find end points of a line > I am writing a computer program that is to draw lines of a particular > distance (length) and slope. I know the following values: > - one point of the line (X1,Y1) > - the slope of the line (m) > - the y-intercept (b) > - the distance of the line (d) > In order to draw the line, I need to Žnd each of the 2 possible end points. > Useful formulas: > ------------------- > The formula for a line: > y = mx + b > The distance between points (X1,Y1) and (X2,Y2): > d = sqrt of [ (X2 -X1)squared + (Y2 -Y1)squared ] Horizontal distance is (X2 - X1) Vertical distance is m*(X2 - X1) Diagonal distance is d = +- (X2 - X1)*sqrt(1 + m^2) === Subject: Locating Formula I¹m guessing this will be short work for people who frequent this newsgroup. I¹m trying to Žnd a formula that will Žt a group of numbers. The numbers are grouped as this: #Available Value Cost1 Cost2 20 100 150 x(1) 15 200 287.5 x(2) 25 300 468.75 x(3) 40 400 700 x(4) #Available and value are given, cost1 and cost2 are computed. I was able to Žgure out the formula for cost1. You compute y as the #Available divided by the total #Available for all rows (100 in the example above). Cost1 then becomes: cost1 = (1 + y) * 1.25 * Value I have not been able to Žgure out the formula for cost2. Here is an example dataset (cost2 is being truncated to be an integer). I do know that in cases where y is comparatively very large that Cost2 can be less than Value. Cost2 Value #Available 6,464.00 5,000.00 1512 13,617.00 10,000.00 1009 27,949.00 20,000.00 748 48,830.00 35,000.00 765 70,655.00 50,000.00 634 101,975.00 75,000.00 1024 124,045.00 100,000.00 1894 155,933.00 150,000.00 3360 Here is another. I apologize that it is so close to the other one. 6,470.00 5,000.00 1512 13,626.00 10,000.00 1009 27,963.00 20,000.00 748 48,854.00 35,000.00 765 70,684.00 50,000.00 634 102,013.00 75,000.00 1027 123,956.00 100,000.00 1913 155,363.00 150,000.00 3410 I can get additional datasets, but I¹ll wait to see if they are requested - I¹m guessing that somone who is mathematically inclined (unlike myself) will see it quickly. === Subject: Re: Locating Formula > {Example Snipped} I got it. For any who might have been interested in something so simple, I noticed that cost2 approached Value as one of the y values approached 1/3. The forumla ended up being (1 - y) * 1.5 * Value, almost the same as the other one. === Subject: Re: Locating Formula Apologies in advance for the columns not being justiŽed - it looked OK as I was writing, tabs must have been converted to spaces. === Subject: Re: InŽnity can not exist > It is impossible for inŽnity to exist, because in order for inŽnity > to exist, everything possible must happen. Including inŽnity not > existing. Therefore, it is possible for something to be an extrememly > large, but not inŽnite. For it to be inŽnite, it would have to be > Žnite at the same time. > Japcuh > (Just Another Perl C Unix Hacker) I hope things are clearer in your mind as far as Perl is concerned. === Subject: Re: InŽnity can not exist > IniŽnite cardinality of sets is very well deŽned. A set has inŽnite > cardinality if it can be mapped 1-1 onto a proper subset of it self. Robert, if (0,3) is the continuous interval (real numbers) between 0 and 3 without the endpoints included and [0,3] is the interval between 0 and 3 WITH the endpoints included is (1,2) a proper subset of [0,3] ? I ask because I don¹t think a 1-1 mapping exists between open interval and closed intervals, tho both seem to possess inŽinite cardinality a simple 1-1 mapping existts between (1,2) and (1.1, 1.9) so it has inŽnite cardinality likewise between [0,3] and [1,2] so it also has inŽnite cardinality I realize that your deŽnition didn¹t say any proper subset, but maybe this is a side issue. Are there differences between the cardinality of an open interval and a closed interval? And if you know of a straightforward 1-1 mapping between open and closed intervals I¹d be very grateful to see it. thank you, tracy (been a long time since thinking about this sort of thing) > Consider, for example, the set of non-negative integers Z. n <-> 2*n > is a 1-1 map of the non-negative integers onto to the even integers > which is a proper subset of the non-negative integers. Hence Z (by > deŽnition) is inŽnite. A set has Žnite cardinality if it does not > have inŽnite cardinality which deŽnes Žnite and Žniteness for > counting the number of elements in a set. > InŽnity is alive and well in the realm of transŽnite numbers and set > theory. > Bob Kolker === Subject: Re: InŽnity can not exist > > IniŽnite cardinality of sets is very well deŽned. A set has inŽnite > > cardinality if it can be mapped 1-1 onto a proper subset of it self. > Robert, > if (0,3) is the continuous interval (real numbers) between 0 and 3 without > the endpoints included and > [0,3] is the interval between 0 and 3 WITH the endpoints included > is (1,2) a proper subset of [0,3] ? > I ask because I don¹t think a 1-1 mapping exists between > open interval and closed intervals, tho both seem to possess inŽinite > cardinality Hey tracy, a 1-1 mapping does exist...infact it is very simple to construct.... between (1,2) and [0,3]..... however the map cannot be bicontinuous....for then it would imply that the two sets are homeomorphic..a contradiction. Sets of same cardinality only need to have a 1-1 mapping between them.... cxontinuity is not a requirement. ish === Subject: Re: InŽnity can not exist >> > >> > InŽnity may just be a convenient phrase for denoting the logical >> > limits of human reasoning.It is most understandable in a geometric way >> > via the relationship between diameter,circumference and the Pi value >> IniŽnite cardinality of sets is very well deŽned. A set has inŽnite >> cardinality if it can be mapped 1-1 onto a proper subset of it self. >> Consider, for example, the set of non-negative integers Z. n <-> 2*n >> is a 1-1 map of the non-negative integers onto to the even integers >> which is a proper subset of the non-negative integers. Hence Z (by >> deŽnition) is inŽnite. A set has Žnite cardinality if it does not >> have inŽnite cardinality which deŽnes Žnite and Žniteness for >> counting the number of elements in a set. > This is a great illustration Bob...the deŽnition that you used was > better than the one that I touched on (a set is inŽnite if it is not > Žnite and a Žnite set is a set such that there exist a bijective > correspondence between it and a section of the positive integers). I > like yours much better. The two deŽnitions are equivalent if you assume the axiom of choice. In the absence of the axiom of choice, the Žnite-iff-bijective-with-some-natural-number deŽnition is the correct one. A set that is bijective with a proper subset is called Dedekind inŽnite. -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Symbolic Algebra Parser Is there any code out there, perhaps something in Open Source, that manipulates symbolic algebraic equations? For example, suppose I have the string: (x^2 + 3x + 2) / (x^2 + 4x + 2) I would like to be able to reduce it to the string: (x + 1) / (x + 2) Essentially the algebraic rules we learned in school being applied to text strings. It need work only on rational polynomial equations, nothing fancy beyond that. It would also be important for this to work in a multivariable situation, not just for a single variable x. I know that there are programs that do this already, such as Mathematica and Matlab, but I would need to be able to do it programmatically with some code I am writing. Although I am using C++, the source code I am looking for need not be. Jonathan Hoyle === Subject: Re: Symbolic Algebra Parser > Is there any code out there, perhaps something in Open Source, that > manipulates symbolic algebraic equations? For example, suppose I have > the string: > (x^2 + 3x + 2) / (x^2 + 4x + 2) > I would like to be able to reduce it to the string: > (x + 1) / (x + 2) > Essentially the algebraic rules we learned in school being applied to > text strings. It need work only on rational polynomial equations, > nothing fancy beyond that. It would also be important for this to > work in a multivariable situation, not just for a single variable x. > I know that there are programs that do this already, such as > Mathematica and Matlab, but I would need to be able to do it > programmatically with some code I am writing. Although I am using > C++, the source code I am looking for need not be. > Jonathan Hoyle Do a search for Macsyma (or Maxima), which should be available in source form. Although you stated you did not require C++, the source for Macsyma is in Lisp. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Symbolic Algebra Parser Is there any code out there, perhaps something in Open Source, that manipulates symbolic algebraic equations? For example, suppose I have the string: (x^2 + 3x + 2) / (x^2 + 4x + 4) I would like to be able to reduce it to the string: (x + 1) / (x + 2) Essentially the algebraic rules we learned in school being applied to text strings. It need work only on rational polynomial equations, nothing fancy beyond that. It would also be important for this to work in a multivariable situation, not just for a single variable x. I know that there are programs that do this already, such as Mathematica and Matlab, but I would need to be able to do it programmatically with some code I am writing. Although I am using C++, the source code I am looking for need not be. Jonathan Hoyle === Subject: Re: Symbolic Algebra Parser X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 1111b85452d685c497717dc2800e00 a0.48257%40mygate.mailgate.org > Is there any code out there, perhaps something in Open Source, that > manipulates symbolic algebraic equations? This seems to give you several choices: HTH xanthian. -- === Subject: Re: Symbolic Algebra Parser > Is there any code out there, perhaps something in Open Source, that > manipulates symbolic algebraic equations? For example, suppose I have > the string: > (x^2 + 3x + 2) / (x^2 + 4x + 4) > I would like to be able to reduce it to the string: > (x + 1) / (x + 2) > Essentially the algebraic rules we learned in school being applied to > text strings. It need work only on rational polynomial equations, > nothing fancy beyond that. It would also be important for this to > work in a multivariable situation, not just for a single variable x. > I know that there are programs that do this already, such as > Mathematica and Matlab, but I would need to be able to do it > programmatically with some code I am writing. Although I am using > C++, the source code I am looking for need not be. > Jonathan Hoyle I¹m not exactly sure how much you want the open source code to do. The immediate problem that you have is to parse the expression contained in the string. To do this, you can use code based on the unix utilities Lex and Yacc. Lex is a lexical generator program that produces a lexical scanner from a description in Backus-Nauer form. Yacc is a compiler geneator program that produces code to parse expressions and do something with it. Yacc stands for yet another compiler compiler (or something like that). searches on the terms lex, lexical, scanner, etc and yacc. Here are some links that you might want to Žrst look at: http://dinosaur.compilertools.net/ http://catalog.compilertools.net/java.html For your example that you gave, the parser will pick out the terms x^2 + 3x + 2 and / and x^2 + 4x + 4. If you want to reduce it to lowest terms, you can Žnd the gcd of the two expressions, and divide it out. Žnite Galois extensions over Q for polynomials of the form X^p - 1, p prime. The following is an excerpt of the more difŽcult part of the Java Cup Yacc code. This is the type of stuff you need to program as input to the yacc generator. goal ::= {: F = parser.getRing(); :} expr_part:e {: RESULT = e; :}; expr_list ::= expr_list expr_part | expr_part; expr_part ::= expr:e {: RESULT = e; parser.setRingElement(e); :} SEMI; quark ::= LETTER:q {: RESULT = F.createAtomicRingElement(q); :} | LETTER:q NUMBER:n {: RESULT = F.createAtomicRingElement(q+n); :}; atom ::= quark:q {: RESULT = q; :} | atom:a quark:q {: RESULT = a.multiply(q); :}; term ::= NUMBER:n {: RESULT = F.createAtomicRingElement(+n); :} | atom:a {: RESULT = a; :} | NUMBER:n atom:a {: RESULT = F.createAtomicRingElement(+n).multiply(a); :}; expr ::= term:t {: RESULT = t; :} | WHITESPACE expr:x {: RESULT = x; :} | expr:x WHITESPACE {: RESULT = x; :} | expr:x PLUS expr:y {: RESULT = x.add(y); :} | expr:x MINUS expr:y {: RESULT = x.subtract(y); :} | expr:x TIMES expr:y {: RESULT = x.multiply(y); :} | expr:x DIVIDE expr:y {: RESULT = x.divide(y); :} | expr:x RAISE NUMBER:n {: RESULT = x.raise(n.intValue()); :} | expr:x RAISE WHITESPACE NUMBER:n {: RESULT = x.raise(n.intValue()); :} | LPAREN expr:x RPAREN {: RESULT = x; :} | MINUS expr:x {: RESULT = x.negate(); :} %prec UMINUS; === Subject: Re: Can you Žnd anything wrong with this solution to the Halting Problem? X-URL: http://mygate.mailgate.org/mynews/comp/comp.theory/ 97bf14b10b0a6e410399b48aa0 a53124.48257%40mygate.mailgate.org > You are able to correctly refute what I am saying > without even reading a single word. That is exactly so. I hate to see a grown kook cry; take it easy. That¹s one of the amazing things about being able to understand computer science, by having invested the effort such understanding takes. Once you have read through and understood a proof that some hypothesis holds, you no longer have to waste time reading drivel from obsessed persons who cannot be bothered to learn to understand computer science, proving that the opposite hypothesis holds, and ignoring every kind attempt by wiser heads to point out just where their errors lie. You¹ll notice your recent behavior here falls precisely into this pattern. When and if someone proves that computer science is internally inconsistent as an axiom set, your entry onto the stage with your contention that, while A has been proved in a manner simple enough for any informed person to understand, you also have a proof of not A, will be considerably more welcome. Until then, try to stick to wasting your own time, proving the already disproven, by writing the results strictly in your personal diary. It would be an achievement of adulthood on your part if you could abandon wasting the time of others by posting to technical newsgroups proofs you bring back from your frequent visits to the Bizarro planet within your mind, and then arguing interminably in favor of your errors in those non-proofs. xanthian. Cross-posted in the obviously vain hope that other similarly afžicted persons to whom this advice applies would read and heed it. -- === Subject: Re: Can you Žnd anything wrong with this solution to the Halting Problem? > > You are able to correctly refute what I am saying > > without even reading a single word. > That is exactly so. > I hate to see a grown kook cry; take it easy. > That¹s one of the amazing things about being able to > understand computer science, by having invested the > effort such understanding takes. Once you have read > through and understood a proof that some hypothesis > holds, you no longer have to waste time reading > drivel from obsessed persons who cannot be bothered > to learn to understand computer science, proving > that the opposite hypothesis holds, and ignoring > every kind attempt by wiser heads to point out just > where their errors lie. I¹ve pointed out errors in half of your replies to me. If you had a thorough knowledge of CS I¹d take heed but this is entirely vain. We¹re discussing a proof not a deŽnition. That is ironically the argument, that you treat it like a deŽnition and you conŽrm that here. a word spoken in jest is often true Herc === Subject: Re: Can you Žnd anything wrong with this solution to the Halting Problem? >It would be an achievement of adulthood on your part >if you could abandon wasting the time of others by >posting to technical newsgroups proofs you bring >back from your frequent visits to the Bizarro planet >within your mind, and then arguing interminably in >favor of your errors in those non-proofs. The Halting Problem was solved decades ago with the introduction of high octane gasoline. It used to be your computer might continue executing thousands of instructions even after you had turned the key to the halt position. This so-called dieseling was the result of poor quality fuel and high compression CPU architectures. It is no longer a problem with modern fuels. In this modern age you can type sync;sync;sync;halt and expect the system to shut down within a matter of seconds. --scott -- C¹est un Nagra. C¹est suisse, et tres, tres precis. === Subject: Re: Can you Žnd anything wrong with this solution to the Halting Problem? >>It would be an achievement of adulthood on your part >>if you could abandon wasting the time of others by >>posting to technical newsgroups proofs you bring >>back from your frequent visits to the Bizarro planet >>within your mind, and then arguing interminably in >>favor of your errors in those non-proofs. >The Halting Problem was solved decades ago with the introduction of >high octane gasoline. It used to be your computer might continue >executing thousands of instructions even after you had turned the key >to the halt position. This so-called dieseling was the result of >poor quality fuel and high compression CPU architectures. It is no >longer a problem with modern fuels. In this modern age you can type >sync;sync;sync;halt and expect the system to shut down within a >matter of seconds. This is a lot more convincing than the other proofs in these threads. I¹m convinced, the Halting Problem is solved, Turing was wrong. Of course _his_ proofs don¹t quite work, but the argument here is so clearly inspired by his attempts at a proof that it¹s clear you would never have accomplished this without him showing you the Way. Turing is dead; long live Olcott. >--scott ************************ David C. Ullrich === Subject: Re: Can you Žnd anything wrong with this solution to the Halting Problem? x-election-year-1: draft j danforth quayle for president committee x-election-year-2: if nader can split the liberal vote x-election-year-3: then danny can split the idiot vote > longer a problem with modern fuels. In this modern age you can type > sync;sync;sync;halt and expect the system to shut down within a > matter of seconds. i Žnd pulling the power plug and dropping the battery out of the bottom works relatively quickly arf meow arf === Subject: Re: Can you Žnd anything wrong with this solution to the Halting Problem? > > You are able to correctly refute what I am saying > > without even reading a single word. > That is exactly so. > I hate to see a grown kook cry; take it easy. > That¹s one of the amazing things about being able to > understand computer science, by having invested the > effort such understanding takes. Once you have read > through and understood a proof that some hypothesis > holds, you no longer have to waste time reading > drivel from obsessed persons who cannot be bothered > to learn to understand computer science, proving > that the opposite hypothesis holds, and ignoring > every kind attempt by wiser heads to point out just > where their errors lie. I usually never resort to giving back the rudeness that I am dealt, but you are a presumptuous ass. === Subject: Re: Can you Žnd anything wrong with this solution to the Halting Problem? x-election-year-1: draft j danforth quayle for president committee x-election-year-2: if nader can split the liberal vote x-election-year-3: then danny can split the idiot vote > I usually never resort to giving back the rudeness > that I am dealt, but you are a presumptuous ass. you cant solve the halting problem of turing machine with a turing machine turing already enumerated every possible and impossible solution and proved they dont exist youre going to have to use something more powerful than a tm that still satisŽes the notions of computability if you succeed at that youll be famous for refuting the church-turing thesis arf meow arf === Subject: Re: Can you Žnd anything wrong with this solution to the Halting Problem? X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 228a46a07825e42ca8bcdd1fe66071 35.48257%40mygate.mailgate.org > I usually never resort to giving back the rudeness > that I am dealt, but you are a presumptuous ass. Many kooks have said that about me in the past; I¹m one of the least popular persons with the grand community of Usenet kooks, existing on the net, I suppose. I tend to be a bit blunt with obdurate fools, and they tend not to like that. I tend not to let such comments bother me much. It is fairly hard to insult a retired sailor in a way that provokes much more than a smile. Meanwhile, shooting the messenger may appeal to your sense of rightness, but I¹m afraid you are still going to Žnd that posting proofs of claims for which proofs of the opposite claim have long existed, been reviewed by people far beyond your level to challenge, and are an immovable part of the accepted bedrock wisdom of a discipline, and then quarreling with people who point out the problems in your proofs, is going to be forever a way to be unpopular in technical newsgroups. My own math skills, 37 years after the last math class I took (for my Math BS) are no longer sufŽcient to contribute much beyond my hobby software, nor are my 17-years-past graduate studies in computer science anything but rusted clots of former competence, so I read comp.* and sci.* newsgroups for intellectual entertainment, and limit my postings mostly to defenses of the faith, and pointers to information for which others are in need. My objection to you and the half dozen people like you who seem to clutter up each technical newsgroup I follow is that you remove the intellectual entertainment value of reading intelligent discussions and well-founded new ideas, and dilute the newsgroups with your constant bickering and incomprehension, to the point where the signal becomes next to impossible to locate. It is only partially your fault. If the other participants were quicker to classify you as a hopeless case, a victim of invincible ignorance, and ignore you, your own typing, however voluminous, would be a minor issue. It¹s just that you _entrain_ so much other wasted type in your wake. Including mine, of course! I may be old, but I¹m not quite totally stupid (yet; Alzheimer¹s runs in my family, check back in 20 years). xanthian. -- === Subject: Re: JSH: Sweep likely by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i6CMY8Q10602; (snip) >You¹re just a tedious asshole, aren¹t you? Whoa, what¹s with the language? And I thought you were above that, Jesse. You proved otherwise. You and others don¹t like my abusing your pet crank James Harris and his high-IQ buddy Quinn? === Subject: meger and null sets in R I¹d like some hints to prove the following statement: If a subset of R has measure zero, then it is meager, but the converse is not true. It¹s easy to see that null sets must have an empty interior, but I couldn¹t prove they have to be meager. I know meager sets may have positive measure, but I couldn¹t Žnd an example. Artur === Subject: Re: meger and null sets in R >I¹d like some hints to prove the following statement: >If a subset of R has measure zero, then it is meager, Well, it seemed like this couldn¹t be so hard to prove; I felt really stupid being unable to prove it. So I looked it up, and then I felt _really_ stupid, not even having considered the possibility that it was false. The Žrst thing I saw was an exercise saying that there exist residual subsets of [0,1] which have measure zero. (A residual set is the complement of a meager set, and so the Baire categormy theorem shows that in particular a residual set is not meager...) The exercise is easy, after the solution to the second question: >but the converse is not true. In fact there exists a closed nowhere-dense set of positive measure. The canonical example would be a fat Cantor set: a set of positive measure homoemorphic to the middle-thirds Cantor set. (Modify the construction of the middle-thirds set, removing the middle r_n-th of each complementary interval at the n-th stage (so if r_n = 1/3 for all n you get the middle-thirds set). If you haven¹t seen this it¹s a good exercise to show that if you let r_n -> 0 fast enough you get a Cantor set of positive measure. So the converse is not true. You can use these Cantor sets to show that the Žrst statement is not true either: Say K is a fat Cantor set in [0,1], of measure 1/2. Let K_0 = K, and let K_{n+1} consist of K_n plus copies of K, one for each interval in the complement of K_n, scaled and translated to Žt the interval. The union of the Žrst n K_j has measure 1 - 2^(-n) or something like that, so the union of all the K_n is a meager set of full measure. Hence its complement is a residual, hence non-meager, set of measure zero. It¹s curious that an elaboration of the construction used to disprove a statement serves to disprove the converse. >It¹s easy to see that null sets must have an empty interior, but I >couldn¹t prove they have to be meager. I know meager sets may have >positive measure, but I couldn¹t Žnd an example. >Artur ************************ David C. Ullrich === Subject: Re: meger and null sets in R >If a subset of R has measure zero, then it is meager, but the converse >is not true. >It¹s easy to see that null sets must have an empty interior, but I >couldn¹t prove they have to be meager. I know meager sets may have >positive measure, but I couldn¹t Žnd an example. The Žrst claim is not true. For example: for each positive integer n, let U_n = {x in [0,1]: there are positive integers p,q with |x - p/q| < 2^(-q)/(nq)} Then U_n is open and dense in [0,1], but m(U_n) <= 1/n. So G = intersection_n U_n is a dense G_delta (and thus is not meagre) but has measure 0. [0,1] G provides a counterexample for the converse. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: VOTE on whether 1/oo = 0 Please don¹t justify your answer or cite reasoning to detract from the next voters opinion. Does 1/oo = 0 ? oo = inŽnity 0 = zero = = equals 1 = one If you¹re a regular professor here let others vote 1st so as not to inžuence the result. Just post YES or NO to be counted Herc -- http://w3.aces.uiuc.edu:8001/Liberty/Tales/Thalidomide.Html Trust the government, 1000 TV shows can¹t be wrong! Honest Dug, they¹re ringing fcking sirens and chanting insideous fcking perverted drivel at me right now, in front of this whole block of žats. I get interrogated all 16 waking hours a day. I¹ve been interrogated personally more than the sum total of interrogation time in the entire WWII. Its not a joke. Tonight I will be aching for rest while dozens of poofter acting nazi psychotic televised cops putrify me to sleep. 3 years of it! === Subject: Re: VOTE on whether 1/oo = 0 X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 30d5d36f0fd845350d6342a449f4de 79.48257%40mygate.mailgate.org Which part of what is reality is not subject to a vote is too hard for you to comprehend, Herc? It would be a real favor to the potential for future mathematical progress worldwide via collegial Usenet conversations, if you could select somewhere other than sci.math to prance your ego on display. xanthian. -- === Subject: Re: VOTE on whether 1/oo = 0 > Please don¹t justify your answer or cite reasoning to detract from the next voters opinion. > Does 1/oo = 0 ? > oo = inŽnity > 0 = zero > = = equals > 1 = one > If you¹re a regular professor here let others vote 1st so as not to inžuence the result. > Just post YES or NO to be counted No. === Subject: Re: VOTE on whether 1/oo = 0 No. > Please don¹t justify your answer or cite reasoning to detract from the next voters opinion. > Does 1/oo = 0 ? > oo = inŽnity > 0 = zero > = = equals > 1 = one > If you¹re a regular professor here let others vote 1st so as not to inžuence the result. > Just post YES or NO to be counted > Herc > -- > http://w3.aces.uiuc.edu:8001/Liberty/Tales/Thalidomide.Html > Trust the government, 1000 TV shows can¹t be wrong! > Honest Dug, they¹re ringing fcking sirens and chanting insideous fcking perverted drivel > at me right now, in front of this whole block of žats. I get interrogated all 16 waking > hours a day. I¹ve been interrogated personally more than the sum total of interrogation > time in the entire WWII. Its not a joke. Tonight I will be aching for rest while dozens > of poofter acting nazi psychotic televised cops putrify me to sleep. 3 years of it! === Subject: Re: Inertial motion > > Whereas Galileo¹s natural inertial motion may have been unaccelerated > > circular motion; around the world (?): Newton¹s natural inertial > > motion was unaccelerated motion in a straight line; at a constant > > speed. There is no such thing as unaccelerated circular motion. Acceleration is required to produce any deviation from motion in a straight line. > > Newton postulated his circular motion as being accelerated from his > > inertial unaccelerated motion; by a centripetally directed force; > > which he called the force of gravity. > > Today¹s inertial unaccelerated motion may include orbital, and any > > other motion in between; including free fall. > > Shead > Classical theoretical physics is patterned after Newton¹s view: That > natural motion is inertial motion, and is the motion of a mass in a > given direction, and at a constant speed; called velocity, and which > will not change [accelerate] except when the mass acts upon, and/or is > acted upon by other mass; where the change is proportional to the > magnitude of the action; which is called an external force. > Galileo¹s view was that circular motion is natural, and a mass > requires no force to keep it moving uniformly in a circle. No, this was not Galileo¹s view. > That was before Kepler¹s elliptical orbits, and other geometries that > formed Newton¹s idea of inertial motion, but what is there really in > space that might prohibit motion in a circle from being natural, > and/or inertial? > Gravity would as soon prohibit straight line motion, and uniform > speed, as it would prohibit circular motion; in fact, probably _more > so_. You are confused about motion and the history of thought on motion. In the absence of force a mass will move in a straight line (force = mass x acceleration = m d^2(x)/dt^2 = 0 gives x = x_o + (v_ox)*t, and the same for the y and z directions, which gives motion from the initial point (x_o,y_o,z_o) in a straight line to (x,y,z) at time t, with velocity v = (v_ox,v_oy,v_oz) = constant vector (const in both magnitude and direction). Circular motion gives velocity with constant magnitude, but the acceleration is perpendicular to the velocity and produces the change in the direction of the velocity. A test mass has to have another mass to circle around for circular motion, and this 2nd mass exerts a central force on the test mass. Motion in a central force is a very interesting (and basic) problem in physics, but your post makes it clear that you have not studied this problem. I suggest you do so (it requires calculus). Its worth it. It makes one appreciate the genius of people like Newton, even if it seems elementary now and has been superceded by Einstein¹s GR. Another fascinating subject is Einstein¹s special and general relativity. They are both beautiful theories-- some of the most impressive intellectual achievements man has come up with--along with the structures built by the great mathematicians. To walk in the steps of these great thinkers is one of the things that makes life worth living. Van === Subject: Re: Inertial motion X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 2279d5ea742628abafcb07c401264a e0.48257%40mygate.mailgate.org > There is no such thing as unaccelerated circular motion. > Acceleration is required to produce any deviation from motion > in a straight line. Umm, careful there, you are arguing about the validity of your choice of vocabulary, not about reality. traveling in a straight line from its point of view, and in a circle from an outside observer¹s point of view, at the very same time. valid, but in combination they reject your statement as an undisputed and absolute truth of the universe. HTH xanthian. -- === Subject: Re: Anyone want to work for Google? > > You folks are aware that they based the price of their IPO on e, > > right? It was in the newspapers. > It took me one look at the numbers to spot the exp(1) for the > Žrst two. Then I veriŽed the other with PARI. > A few tries with the series 1,5,23,99 didn¹t work out, so I > decided the solution must be a little more devious. > Adding the digits was a lucky hunch. I tried the series 1,5,23,99 at EIS too, with no luck. > What is an IPO? > doesn¹t reveal much. > And when was this thing in which newspapers? > Dirk Vdm It was a big deal--you didn¹t here of it? If your serious, IPO = initial public offering (of stock). Google is going public, and people are drooling over all the money they think is going to be made from this. IMO, people think too much of money as it is. As long as I have the basics (including books and computer), I¹m happy. Van === Subject: Re: JSH: Sweep likely > James, > You should consider it a blessing what happened to you with the > journal that was forced through pressure from math.sci to withdraw > your paper that passed peer review. Look at history. The best way to > get people to read a paper or a book is to ban it. Since your paper > was banned from the journal because of pressure from math.sci, more > people will likely read it, which I assume you want. > If it were published in the journal, then who cares? Who reads math > journals anyway? I have a friend that I work with who used to be a in > mathematics is <1 (outside of the author and referees). > Craig I hate to divert attention from JSH¹s math ability, but on the subject of published research in math and physics, I have often wondered about what fraction of papers get read and referenced by others (say 3 or more refs or readers), and how many will be remembered in 10 years, or 20 or more. And what fraction of researchers do important work that will be remembered and referred to after 20 years, or after they are dead. I am not trying to say that most research is worthless. I think we need a lot of people working on a lot of things to make progress, and that one can never tell in advance what is going to turn out to be important. There are many examples of people who¹s work was not seen to be worth anything until after they were dead. Well--a few examples anyway. We have Feynman, Gell-Mann, etc. in physics, and Wiles and some other people who have made a name in math. Do we have any Guass¹s or Einsteins¹ though? Van === Subject: Re: VOTE on whether 1/oo = 0 I think this is a pointless poll, but my vote is, NO. BTW, your notation is too sloppy to make this question worthy of serious consideration. > Please don¹t justify your answer or cite reasoning to detract from the next voters opinion. > Does 1/oo = 0 ? > oo = inŽnity > 0 = zero > = = equals > 1 = one > If you¹re a regular professor here let others vote 1st so as not to inžuence the result. > Just post YES or NO to be counted > Herc > -- > http://w3.aces.uiuc.edu:8001/Liberty/Tales/Thalidomide.Html > Trust the government, 1000 TV shows can¹t be wrong! > Honest Dug, they¹re ringing fcking sirens and chanting insideous fcking perverted drivel > at me right now, in front of this whole block of žats. I get interrogated all 16 waking > hours a day. I¹ve been interrogated personally more than the sum total of interrogation > time in the entire WWII. Its not a joke. Tonight I will be aching for rest while dozens > of poofter acting nazi psychotic televised cops putrify me to sleep. 3 years of it! === Subject: computing sqrt using basic math function. Suppose I have basic math function i.e. plus(+), minus(-), divide(/) and multiply(*). I formed function pow, which compute X^N given X and N, using N times multiplying number X i.e. pow(X,N) = X.X.X . . . X [N times]. My question is can I compute square root of given positive number using +, -, /, *, pow, abs(absolute value function). Also I want to know if I can compute more intresting function on above given math operation [+ sqrt function if it can be calculated]. === Subject: the more likely sweep X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ af7dbbfa2aa29eedbfff26a3a4a724 21.48257%40mygate.mailgate.org More likely than poor delusional James Harris¹ long predicted, never realized, triumph ower the community of mathematicians conspiring in evil consort to suppress his brilliant discoveries? That on his death, all memories of James will quickly fade, and his litany of unaccomplishments will be swept up and deposited in the nearest rubbish tip, thence to annoy humankind no more. xanthian. -- === Subject: Re: a non-linear second-order PDE Epigone-thread: gromstunpand >> >I consider the following non-linear second-order partial differential >> >equation with two non-negative variables x and y: >> >> >(a+ b x) [ df/dx d^2f/(dx dy) - df/dy d^2f/(dx^2) ] + b y [ df/dx d^2 >> >f/d(y^2)- df/dy d^2f/(dx dy)] =0 >> >> >where a >0 and b is a real number. Additional conditions are >> >> >f(x,0) = (a + b x)^(1-1/b) /(1/b (1-1/b)) >> >> >and >> >> >df(x,0)/dy =0. >> I assume you mean df/dy = 0 when y = 0. >> >for all x. >> >> >(For b=1, f(x,0) converges to logarithmic function ln(a+x), but that >> >should not really matter). >> >> >One solution to the PDE is: >> >> >f(x,y) = 1/(1/b (1-1/b)) [ (a + b x)^(1- 1/b) + c2 b y^(1-1/b) ] >> This doesn¹t satisfy your second additional condition in general, however: >> df/dy has a singularity at y=0 if b > 0. It does work if c2 = 0 or if >> b < 0. Of course, in general there¹s also a singularity when a+bx=0. >> >with constants c1, c2. >> You didn¹t have a c1. >> >My question: Are there other solutions? If so, which? If no, how can >> >one prove uniqueness? >> Solutions of your PDE include >> f(x,y) = c_0 + sum_{j=1}^n c_j (a+bx)^(k_j) y^(p-k_j) >> for any constants c_j, k_j and p. For example, you could add to >> your solution a term in (a+bx)^(-1-1/b) y^2 and get another solution >> with the same values of f(x,0) and df/dy (x,0). So, no uniqueness >> there. >> BTW, an interesting feature of the PDE is that if f(x,y) is a solution, so >> is f(x,y)^k for any k, or ln(f(x,y)), or exp(f(x,y)). >I believe I¹ve found the general solution to the PDE: >f(x,y) = G((a+bx) H(y/(a+bx))) >where G and H are arbitrary (twice differentiable) functions. >Wlog we can take H(0) = 1, and then f(x,0) = G(a+bx) determines G, >while df/dy (x,0) = 0 as long as H¹(0) = 0. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~ israel > University of British Columbia Vancouver, BC, Canada I believe I¹ve got a solution to a simpler equation that may help solving this one: d^2f/dx^2*df/dy-d^2/dx dy*df/dx = 0 (1) possess a solution: f(x,y)=h(x+g(y)); (2) permuting x and y in (1) gives f1(x,y)=h(y+g(x)). Combine with my observation d/dx (a*(c+y)+bxy) = by d/dy (a*(c+y)+bxy) = a +bx === Subject: Re: a non-linear second-order PDE >I believe I¹ve got a solution to a simpler equation that may >help solving this one: > d^2f/dx^2*df/dy-d^2/dx dy*df/dx = 0 (1) possess a solution: > f(x,y)=h(x+g(y)); (2) > permuting x and y in (1) gives f1(x,y)=h(y+g(x)). > Combine with my observation d/dx (a*(c+y)+bxy) = by > d/dy (a*(c+y)+bxy) = a +bx Yes, this equation (with variables renamed X and Y) is equivalent to the original one under the transformation a + b x = exp(X+Y), b y = exp(X-Y). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: provability and truth |If P is a statement of arithmetic, deŽne the hermeneutics of P (I |choose this word for no particular reason) to be the statement | | => P | |(in other words, if P is provable in PA, then P is true). As has been noted, this is usually called a režection principle. One of my fondest college memories is of studying the paper in vol. 27 of the Journal of Symbolic Logic on režection principles, on my own. There is also a recent book by Torkel Franzen aimed at making some results on iterated režection principles more accessible. I have an amusing tangent to this. There¹s a famous theorem. I thought I knew its name, but it appears that was the name of a completely different theorem. It says that for each P, => P)> <=> . In other words, the only cases of this režection principle provable in PA are the trivial ones, where P is already a theorem of PA. (There¹s nothing very special about PA here; the result applies to a wide range of formal systems; they only have to satisfy the assumptions implicit below.) This can be proven using reasoning parallel to the reasoning in Goedel¹s proof of his incompleteness theorems. It is in fact a direct generalization; Goedel¹s second incompleteness theorem is the special case of the above where P is some sentence such as 1=0. In that case PA|-P is equivalent to PA being inconsistent, and ( => P) is equivalent to PA being consistent, and the second incompleteness theorem says that a system like PA can only prove its own consistency if it is inconsistent. We can concoct a sentence G such that G is true if and only if G or P is not a theorem of PA, and such that this fact is provable in PA. We rely on the fact that if a sentence is a theorem of PA, then the fact that it is a theorem of PA is also a theorem of PA. We also rely on the fact described in the last sentence being a theorem of PA! Assume the left hand side above, that => P is a theorem of PA. The following argument could then be formalized in PA: Suppose G is false. Then the falsity of G is a theorem of PA (because G is false is equivalent to the provability of something). But also G or P is provable in PA. Hence P is a theorem of PA. {...} Hence P. Where I have {...}, we insert the proof in PA of =>P that we are assuming exists. The above argument would be a proof in PA that if G is false then P is true, or G or P. Since G denies this, G would be false. So G is false. Then the falsity of G is a theorem of PA (because G is false is equivalent to the provability of something). But also G or P is provable in PA. Hence P is a theorem of PA. A second special case of this theorem is for sentences that have been concocted in sort of the opposite way from the Goedel sentence, i.e. sentences H that are equivalent to their own provability in PA. The theorem whose proof I just sketched implies that they are all theorems of PA. They are, but their proofs in PA, as constructed by the proof above, come off like doubletalk. Because H is equivalent by construction to H is provable in PA, a proof of H in PA must show there exists a proof of H in PA. There is a metatheorem in this case that implies any such proof can be rewritten to make it plainly constructive, which roused my curiosity. It¹s not especially hard to write it in a plainly constructive form, which then raises the question, what¹s the proof of H that the proof of H constructs? I traced through this once and found that the proof it constructs is-- essentially-- itself. |For |example, the hermeneutics of the False statement is the statement |Consis(PA) that PA is consistent (and the hermeneutics of the True |statement is again the True statement, of course); the hermeneutics of |~Consis(PA) is essentially equivalent, modulo Consis(PA), to the |consistency of PA+Consis(PA). And so on. | |DeŽne the Grand Hermeneutical Axiom Scheme (Totally Ludicrous, |Yuck) (Ghastly for short) to be the axiom scheme consisting of the |hermeneutics of every arithmetical statement P (so, Ghastly allows us |to deduce, if any P is provable from PA, that P is true). As |explained above, Ghastly implies Consis(PA) and Consis(PA+Consis(PA)), |and so on - with a reasonable deŽnition of and so on. A režection principle is sometimes used as a function on formal systems of some type, in this case going from a system S to the system S augmented by an axiom scheme. You¹re showing here that the režection principle you described before is, in a certain sense, a lot more powerful than the one merely going from S to S+Consis(S). [...] |Finally I come to my question (but I¹m interested more generally in |anything that could be said about Ghastly, including any more |reasonable name for it): is Ghastly equivalent to some (Žnite) |statement about arithmetic? Or is it, at least, implied by some |(Žnite) statement about arithmetic that is consistent (provably in |ZF)? By Žnite, perhaps you mean a statement in Žrst-order arithmetic, which is the language of PA. No, any such sentence is disprovable in PA. Suppose that Q is such a sentence. The režection principle as applied to Q is false is =>Q is false. If it was a theorem in PA that Q implied it, then we would have PA |- Q => ( => Q is false), which is equivalent to PA |- => Q is false. By the above theorem whose name I seem to have been misremembering, this implies that Q is false is a theorem of PA. For each n, one can formalize in Žrst-order arithmetic the property of an n-quantiŽer sentence being true, and thus the režection principle as restricted to n-quantiŽer sentences can be formalized. But it needs more than n quantiŽers to be formalized. So the režection principle in general can only be formalized in a stronger language; it doesn¹t need to be very much stronger, however. Note that ZF in this context is overkill. You could have just as well used a standard elementary theory of natural numbers and sets of natural numbers such as the formal system PA_2. That¹s strong enough to do everything you¹ve mentioned doing with ZF. Keith Ramsay === Subject: Re: provability and truth >>DeŽne the Grand Hermeneutical Axiom Scheme (Totally Ludicrous, >>Yuck) (Ghastly for short) to be the axiom scheme consisting of the >>hermeneutics of every arithmetical statement P (so, Ghastly allows us >>to deduce, if any P is provable from PA, that P is true). > I don¹t think you mean to say *every* arithmetical statement, because there > are arithmetical statements that are false, including those that are > disprovable in PA. Well, yes. But what David means is that add to PA all sentences of form ((PA |- P) --> P) i.e. to add to PA the local režection schema. There is no need to restrict P here to be a true arithmetical statement. > What it sounds like you¹re try to get at is the concept of the režective > closure of PA. Other keywords are iterated consistency extension and > Feferman-Schuette ordinal. The idea behind these concepts is to add to > PA any arithmetical statements that we implicitly must accept if we accept > PA as being true. There are numerous complications here. One of the more obvious ones is that if we accept e.g. Feferman¹s analysis that RA_Gamma_0 or IR contains exactly those principles acceptable on basis of acceptance of PA, we are in fact asserting the acceptability of RA_Gamma_0. In this case we can convincingly argue that the analysis is not acceptable on basis of PA, since the notion of a well-ordering isn¹t. But in case of e.g. ZFC there is no reason to consider the theory obtained by autonomously iterating addition of režection principles (Aut(ZFC)) to comprise everything that is implicit in acceptance of ZFC. This is because it seems that basic observations about what acceptance means and what well-orderings are lead one to accept Aut(ZFC) on basis of acceptance of ZFC. The above wasn¹t meant as a criticism of your reply, which was of course quite correct when speaking about PA. I just thought these slightly more general considerations might be of some interest to someone. -- Aatu Koskensilta (aatu.koskensilta@xortec.Ž) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: provability and truth Aatu Koskensilta in litteris scripsit: > There are numerous complications here. One of the more obvious ones is > that if we accept e.g. Feferman¹s analysis that RA_Gamma_0 or IR > contains exactly those principles acceptable on basis of acceptance of > PA, we are in fact asserting the acceptability of RA_Gamma_0. In this > case we can convincingly argue that the analysis is not acceptable on > basis of PA, since the notion of a well-ordering isn¹t. But in case of > e.g. ZFC there is no reason to consider the theory obtained by > autonomously iterating addition of režection principles (Aut(ZFC)) to > comprise everything that is implicit in acceptance of ZFC. This is > because it seems that basic observations about what acceptance means and > what well-orderings are lead one to accept Aut(ZFC) on basis of > acceptance of ZFC. suggestions that could serve as introductory material to these questions? (Assuming I have basic background on logic and set theory, e.g., the contents of Jech¹s *Set Theory* and Poizat¹s *Model Theory*.) For example, where can I Žnd a deŽnition of Aut(ZFC)? -- David A. Madore (david.madore@ens.fr, http://www.dma.ens.fr/~madore/ ) === Subject: Re: provability and truth Originator: tchow@maclaurin.mit.edu.mit.edu (Timothy Chow) >suggestions that could serve as introductory material to these >questions? One possibility is Torkel Franzen¹s new book, which I believe has just been published. http://www.sm.luth.se/~torkel/eget/progps.html -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: provability and truth Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow) >>DeŽne the Grand Hermeneutical Axiom Scheme (Totally Ludicrous, >>Yuck) (Ghastly for short) to be the axiom scheme consisting of the >>hermeneutics of every arithmetical statement P (so, Ghastly allows us >>to deduce, if any P is provable from PA, that P is true). >I don¹t think you mean to say *every* arithmetical statement You do mean every arithmetical statement, of course. It¹s still true that what you¹re interested in is a režection principle for PA. Sometimes people distinguish between global and local režection principles, depending on whether you allow free variables in your arithmetical statements. Hopefully this will point you in the right direction. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: provability and truth tchow@lsa.umich.edu in litteris scripsit: >>If P is a statement of arithmetic, deŽne the hermeneutics of P (I >>choose this word for no particular reason) to be the statement >> => P > This is more commonly known as a režection principle. >>DeŽne the Grand Hermeneutical Axiom Scheme (Totally Ludicrous, >>Yuck) (Ghastly for short) to be the axiom scheme consisting of the >>hermeneutics of every arithmetical statement P (so, Ghastly allows us >>to deduce, if any P is provable from PA, that P is true). > I don¹t think you mean to say *every* arithmetical statement, because there > are arithmetical statements that are false, including those that are > disprovable in PA. I don¹t see how that is a problem. Even if P is the False statement (0=1, if you will), the =>P statement makes good sense (it is Consis(PA)). So I *do* mean every P (every closed statement in the language of arithmetic), even those that are provably false in PA (in which case we just get Consis(PA)). > What it sounds like you¹re try to get at is the concept of the režective > closure of PA. Other keywords are iterated consistency extension and > Feferman-Schuette ordinal. The idea behind these concepts is to add to > PA any arithmetical statements that we implicitly must accept if we accept > PA as being true. Those keywords should get you started and help you > make your questions more precise. -- David A. Madore (david.madore@ens.fr, http://www.dma.ens.fr/~madore/ ) === Subject: Re: supplementary subsets of (Z/pZ)^2 > I¹ve stumbled upon the following provokingly innocent-sounding problem > which has kept me bafžed: > Let p be a prime. Let U and V be two subsets of (Z/pZ)^2 each of > cardinality p, and assume that U+V = (Z/pZ)^2 (in other words, > pairwise addition forms a bijection between the cartesian product U*V > and (Z/pZ)^2). Is it true that projection on the Žrst coordinate > must deŽne a bijection of one of the two sets (either U or V) onto > Z/pZ? (It is of course equivalent to ask whether, for each linear > form l from (Z/pZ)^2 to Z/pZ, l forms a bijection from one of the two > sets onto Z/pZ.) I hope the following is correct: 1) Žrst, rephrase geometrically (as you rightly pointed out, it is a tiling problem): we have a Žnite square of side length p, so p^2 points. We assume that we can Žnd a subset U of p points (the set of origins), and another subset V of p points (the pattern) such that we can can tile the square (modulo identiŽcations) when we translate our pattern in the various reference frames. Your question is wether we can Žnd a pattern which moreover avoids at least one row (projection on Žrst coordinate) and/or at least one column (projection on second coordinate). Let¹s assume you actually meant Œand¹. 2) second, count: the total number of possible patterns is N = {p^2 choose p}. The number of patterns which avoid exactly one row and one column is M = {(p^2-p+1) choose p}. Now, tiling means no overlap, so let¹s count the constraints. We start with the Žrst origin u_1 and construct the p points w(1,i)=u_1+v_i. Then we take u_2 and the v_i must now satisfy: w(1,i) != w(2,j) for i<=j, ie that¹s n(1,2) = p+p.(p-1)/2 constraints. With u_3 all the constraints are now : [w(1,i) != w(2,j) AND w(1,i) != w(3,k) AND w(2,j) != w(3,k)]. So that¹s n(1,2,3) = {3 choose 2}.n(1,2) . So when we reach u_p we have n(1,...,p) = {p choose 2}.n(1,2) . All we need to do now is compare M with n(1,...,p), which I leave you do. 3) remark: the question made sense for p prime only, since when p is composite there are obvious counter-examples. Namely, with p=4 and noting by Œ*¹ the points of U and V and by Œ% Œthe other points of the square, we have (use a font of Žxed width): % % % % % % % % * % * % % % % % U = % % % % and V = * * % % * % * % * * % % If all the above is correct then the case of (Z/p^2Z) could be studied along those lines too. -- thomas. ...by natural selection our mind has adapted itself to the conditions of the external world. It has adopted the geometry most advantageous to the species or, in other words, the most convenient. Geometry is not true, it is advantageous. H. Poincare. === Subject: Re: supplementary subsets of (Z/pZ)^2 |Let p be a prime. Let U and V be two subsets of (Z/pZ)^2 each of |cardinality p, and assume that U+V = (Z/pZ)^2 (in other words, |pairwise addition forms a bijection between the cartesian product U*V |and (Z/pZ)^2). Let f and g be the characteristic functions of U and V. The condition you have is equivalent to f*g, the convolution, being the constant 1 function. Take the Fourier transform. The Fourier transform of 1 is supported only at the identity. Moreover 1^=(f*g)^=f^ g^. |Is it true that projection on the Žrst coordinate |must deŽne a bijection of one of the two sets (either U or V) onto |Z/pZ? So one of f^ or g^ has to be 0 on the character chi which takes (m, n) to e^(2*pi*i*m/p). For it to be 0, the sum of the values of this character on the points of U or V has to be 0. But the only way to get 0 as a sum of p p-th roots of 1 is by taking all of the p-th roots of 1. The fact that the value of chi for each element is different is equivalent to projection on the Žrst coordinate being a bijection with Z/pZ. I could have avoided the reference to the Fourier transform, but it seems to me that any set U of p elements such that the Fourier transform of its characteristic function is 0 on at least half of the nonzero elements of the dual group has to be pretty special, and I want to ask someone to Žgure out what such a subset has to look like. Does it have to be a coset of a subgroup of order p? The value is nonzero at any chi whose kernel contains the difference between distinct elements of U, and conversely, so this property is equivalent to the differences between elements of U lying in at most half of the subgroups of order p. Keith Ramsay === Subject: Re: On Hodge and Betti numbers Epigone-thread: sherdphendspoy Marcel === Subject: Fixed point free Žnite group actions on n-disks Epigone-thread: kherdfeifreh Let D^n denote the usual n-disk as a smooth manifold-with-boundary. I believe it was Robert Oliver who Žrst gave an example of a smooth Žxed point free action of a Žnite group G on a D^n; if I recall correctly, n = 8 in this example. Can someone please say what is the least known n for which a smooth Žxed point free action of a Žnite group G on D^n is known to exist? Also, I¹d like to know for what n > 2, if any, such an action has been shown to be impossible. References to the literature would be appreciated. --Dan Asimov === Subject: Re: Open questions related to periodic continued fractions > I am starting research for a thesis on continued fractions, and want > to look at open questions related to periodic continued fractions. Is > anyone aware of current open questions of interest? As you may know, the outstanding problem in this area is to improve on known conditions for the length of the period of the continued fraction expansion of sqrt{D} to be odd (D > 0, D not a square, equivalent to x^2 - Dy^2 = -1 having solutions). The following might be a current summary of known conditions: B. D. Beach and H. C. Williams, A Numerical Investigation of the Diophantine Equation x^2 - dy^2 = -1, Congressus Numerantium VI, Proceedings of the Third Southeastern Conference on Combinatorics, Graph Theory and Computing, Utilitas Mathematica Publishing Inc., Winnipeg, Canada, 1972, pages 37 to 52. A less well known problem is as follows. As far as I know, this is an open problem, but I cannot be sure. For D > 0 not a square, D congruent to 1 modulo 4, let L1(D) and L4(D) denote the lengths of the periods of the continued fraction expansions of sqrt{D} and (1+sqrt{D})/2 respectively. The question is whether it is possible to have L1(D) = L4(D) + 4 when L4(D) == 3 (mod 6). As far as I can tell, this is not prohibited by results in the literature. I have empirical evidence that it is not possible based on testing those D up to 30 billion that have L4(D) <= 255. It is not hard to show this for one particular case, namely if L4(D) = 3 then L1(D) cannot be 7. See discussion under the heading ``Periods of Continued Fractions¹¹ in April and May of 2000 in the archives of the Number Theory Listserver at http://listserv.nodak.edu/archives/nmbrthry.html for related comments and some references that might be of interest. John Robertson === Subject: 2 questions on Žnitely generated solvable groups 1) Can a solvable, Žnitely generated group act 2-transitively on an inŽnite set X? Or, at least, with a Žnite number of orbits in X[Times]X? 2) If every solvable, Žnitely generated group a quotient of some Žnitely presented solvable group? -- Yves de Cornulier === Subject: Re: 2 questions on Žnitely generated solvable groups Epigone-thread: colsmymeu >1) Can a solvable, Žnitely generated group act 2-transitively on an >inŽnite set X? Or, at least, with a Žnite number of orbits in X[Times]X? >2) If every solvable, Žnitely generated group a quotient of some >Žnitely presented solvable group? >Yves de Cornulier 1) The simplest example is G = Z x Z acting on X = Z by translations: (a,b).x = x + a + b Clearly, it is 2-transitive. 2) No, the free n-generated solvable group of solvable length d is not Žnitely presented for n,d > 1. This was proved by A.L. Shmel¹kin: Zbl 0134.02801 Shmel¹kin, A.L. Uber aužosbare Produkte von Gruppen (Russian) [J] Sib. Mat. Zh. 6, 212-220 (1965). [ISSN 0037-4474] See review here: http://www.emis.de/cgi-bin/Zarchive?an=0134.02801 Ignat Soroko, Minsk, Belarus === Subject: Re: Multiplication and Division in Triplets; division by zero-sized vectors snip > > More generally one can consider a general group algebra C G > > over a Žnite group, and deŽne a Moore-Penrose type inverse > > by decomposing CG as a product of matrix algebras and applying > > Moore-Penrose inversion in each. I was going to suggest this with > > C replaced by an arbitrary characteristic zero Želd K, but that > > would require Moore-Penrose inverses in matrix algebras over skew > > Želds. I don¹t know if there are such things .... anyone? > > Of course there are no problems if, as here G is abelian, snip Triplet polar-dual, powers, roots. In Triplet (or, my preferred term, Terplex) algebra, the functions Xs & Xp combine with Xa =ArcTan[2x1 -x2 -x3, -Root[3](x2-x3)] to give a Polar Dual for X. (Mathematica ArcTan convention, Arctan[opposite, adjacent]). The quadratic Xp releases a degree of freedom, which is taken up by the angle Xa. Angles add on multiplication, so the polar-dual of AB is {AsBs, ApBp, Aa+Ba}. With A={3,1,4} and B={1,5,2} as in my original posting, the angles are Aa =ArcTan[3Root[3]]=1.3807, Ba =ArcTan[0.6 Root[3]-Pi]=-2.3370, ABa =ArcTan[9Root[3]/11] =-.9563. Procedures toPolar, toVector, and genPower are available. genPower[A] give SQRT[A] ={1.7789, -.07326, 1.12278}. You can easily write your own procedures from the deŽnitions; care has to be taken with roots of powers when the power takes the angles across the cut for ArcTan. Robin Chapman wondered about non-abelian algebras. The simplest is D3. I use the following protoloop (preferred isomorph):- 1 2 3 4 5 6 2 1 6 5 4 3 3 4 5 6 1 2 4 3 2 1 6 5 5 6 1 2 3 4 6 5 4 3 2 1 with multiplication rule and conserved properties:- AB={a1 b1+a2 b2+a5 b3+a4 b4+a3 b5+a6 b6, a2 b1+a1 b2+a4 b3+a5 b4+a6 b5+a3 b6, a3 b1+a4 b2+a1 b3+a6 b4+a5 b5+a2 b6, a4 b1+a3 b2+a6 b3+a1 b4+a2 b5+a5 b6, a5 b1+a6 b2+a3 b3+a2 b4+a1 b5+a4 b6, a6 b1+a5 b2+a2 b3+a3 b4+a4 b5+a1 b6} Xs=x1+x2+x3+x4+x5+x6, Xt=x1-x2+x3-x4+x5-x6, Xp=((x1-x3)^2-(x2-x4)^2+(x3-x5)^2-(x4-x6)^2+(x5-x1)^2-(x6-x2) ^2)/2. Demonstration of conservation on multiplication:- A={3,1,4,-2,5,0};B={2,7,1,0,2,4}; Multiplication gives AB->{26,37,11,46,12,44} & BA->{26,39,13,48,10,40} Shapes A B AB BA {13,11,-4},{-6,16,-36},{-78,176,144},{-78,176,144} I do not give the inverse procedure here; it gives a left inverse. Demonstrations of division:- Ainv->{159/572,48/143,-127/572,-237/572,4/143,49/572}; Binv->{-23/864,113/864,-23/864,-55/864,1/864,41/864}; genTimes[Binv,BA]=genTimes[AB,Binv]->A genTimes[Ainv,AB]=genTimes[BA,Ainv]->B I have not found a polar form with additive angles, because there is only one squared radius and so 3 degrees of freedom are lost. However, splitting Xp gives a polar form that reverts correctly and provides positive powers (as a continuous function of the power):- Xra=((x1-x3)^2+(x3-x5)^2+(x5-x1)^2)/2; Xa=ArcTan[2x1-x3-x5,SQRT[3](x5-x3)]; Xrb=((x2-x4)^2+(x4-x6)^2+(x6-x2)^2)/2; Xb=ArcTan[2x2-x4-x6,SQRT[3](x6-x4)]-Xa; Demonstration:-toPolar[A]->{13,11,3,2.618,7,-1.904} toVector[A]->{.1118,-0.9537,1.9941,1.1507,-0.1059,-1.197}( treating A as a polar form) with toPolar[toVector[A]]->{3,1,4,-2,5,0} and toVector[toPolar[A]] ->{3,1,4,-2,5,0}; rootA=genPower[A,.5]->{1.3808,0.9679,0.3062,-0.8842,1.7741,- 0.2281} genpower[rootA,2]->{3,1,4,-2,5,0}. But genTimes[rootA,rootA]-> {4.76, 0.359, 3.12, -0.903, 4.112, -0.456}; That last line demonstrates something that upsets moderators. If A has any negative sizes (-4 in the example) A.A is not the same as A^2. A.A = {54, -12, 47, -3, 44, -9} A^1.99= {48.09,-7.07,48.18,-12.07,45.15,-4.15} A^2 = {49.33,-7.33,49.33,-12.37,46.33,-4.33} A^2.01= {50.60,-7.60,50.15,-12.61,47.54,-4.52} This is because A.A must have the third size=16; it is -16 in the power function. If anyone wishes to follow-up Robin¹s comment about matrix formulations of non-abelian tables, the dot-products of {{0,-i},{i,0}} and {{j,0},{0,j j}}, where i^4=1,j^3=1, give the elements of a different D3 isomorph. Roger Beresford. I apologise for the ungracious tailquote in my previous posting. I had passed my (low) concentration-span limit, and did not notice how rude it looked. Sorry. And more, much more than this. I did it my way. (P. Anka.) === Subject: Chebyshev polynomials Epigone-thread: slalsmoižen I meet equations about Chebyshev polynomials.Tn(Xn)is n degree Chebyshev polynomials about Xn. Mi is const (known value). T1(x1)+T2(x2)+...+T1(xn)=M1 T3(x1)+T3(x2)+...+T3(xn)=M3 T5(x1)+T5(x2)+...+T5(xn)=M5 ....................... T2n+1(x1)+.......+T2n+1(xn)=Mn How obtain xi? === Subject: Re: Approximation of Stirling numbers of 2. kind Epigone-thread: žangputul >> Does anyone know of a closed form approximation/upper bound of the >> stirling numbers of the second kind? >I don¹t believe one exists. But if you notice that there are >k!S(n,k) onto functions from an n element set to a k element set, >and count the same number using inclusion-exclusion, you get >the reasonable closed form > [S(n,k) = frac{1}{k!}sum_{j=0}^{k}(-1)^j{k choose j}(k-j)^n.] >e.g., S(3,2) = (1/2!)[C(2,0)2^3 - C(2,1)1^3 + C(2,2)0^3] > = (1/2) [ 8 - 2 + 0 ] > = 3. >Rick Does anyone know of a formula (can be approximate) for computing the logarithm of a Stirling number of the second directly? I.e. Not computing the number and then taking its log. === Subject: Re: Approximation of Stirling numbers of 2. kind In my paper Asymptotic Estimates of Stirling Numbers, Studies in Applied Mathematics, 89, 1993, 233 -- 243, I give an estimate for large n that holds uniformly with respect to m. Let x0 be the positive solution of m x / n = 1 - exp(-x). Let t0 = (n-m)/m. Then the Stirling number of the second kind Snm has the S(n,m) (n ge m) has the asymptotic estimate S(n,m) sim (m/(n-m x0))^m (n-m)^{n-m} e^{m-n} {n choose m} sqrt{t0/(1+t0)/(x0-t0)}, n to infty. Nico Temme Nico M. Temme, http://homepages.cwi.nl/~nicot/ C W I: Centrum voor Wiskunde en Informatica Kruislaan 413, NL-1098 SJ Amsterdam Tel +31 20 592 4240 P.O. Box 94079, NL-1090 GB Amsterdam Fax +31 20 592 4199 === > Subject: Re: Approximation of Stirling numbers of 2. kind > >> Does anyone know of a closed form approximation/upper bound of > the > >> stirling numbers of the second kind? > >I don¹t believe one exists. But if you notice that there are > >k!S(n,k) onto functions from an n element set to a k element set, > >and count the same number using inclusion-exclusion, you get > >the reasonable closed form > > [S(n,k) = frac{1}{k!}sum_{j=0}^{k}(-1)^j{k choose j}(k-j)^n.] > >e.g., S(3,2) = (1/2!)[C(2,0)2^3 - C(2,1)1^3 + C(2,2)0^3] > > = (1/2) [ 8 - 2 + 0 ] > > = 3. > >Rick > Does anyone know of a formula (can be approximate) for computing the > logarithm of a Stirling number of the second directly? I.e. Not > computing the number and then taking its log.