mm-426 === Subject: : Re: Naive Q: Set theory, logic - wch comes first?> I refer you to either my late wife's logic text, or the > appendix to her set theory text. Wle one does not > usually produce a formal proof, an informal proof is one> wch can be extended to a formal proof, and the notion> of a formal proof, at least in the restricted predicate> calculus, wch is what is adequate for mathematics.> There are a small number of explicit rules of inference;> one can derive others from these. Are you suggesting that the notion of a sequence of applications ofrules of inference is not exactly as problematic as the notion ofnatural number?=== === Subject: : Re: Idea for a book on popular maths> Have you read Euler: The Master of Us All by William Dunham ?Yes, and a damn fine book about someone who does not get 1/n the amount ofattention he deserves storically.But ts idea was for a book that was more mathematical per se - using theformula (and the formula's components as it's foundation, rather than Eulermself. Any biography of Euler obviously can't go into too much fine detailabout all of s mathematics, due to the sheer volume of it all.=== === Subject: : Re: Algebraic Closure w/o FTANNTP-Posting-User: [ox5vPf4X3czm/RlHAf+0tXL0YPd5Bo]> Is there a relatively simple way to prove the following two tngs about> Algebraic Closure w/o FTA:> 1) The Algebraic Closure, A, of Q is itself algebraically closed, andYes, depending on what you mean by simple. Here is a proof fromDummit & Foote wch uses some elementary field theory (but no Galoismessage.Lemma 1: given fields L and K with L containing K, if dim_K (L) isfinite, then every element of L is algebraic over K. Partialconverse: given L containing K and an element b in L, then b isalgebraic over K if and only if K(b) is finite-dimensional over K.Proof: not very hard. Left to the reader.Lemma 2: given fields L containing K containing F, if L is algebraicover K and K is algebraic over F, then L is algebraic over F.Proof: Given an element b in L, b is the root of some polynomial withcoefficients a_0, a_1, ..., a_n in K. Then you can show thatF(a_0, a_1, ..., a_n, b) is a finite extension of F, and hence b isalgebraic over F.Proof of (1): Given any polynomial f(x) with coefficients in A andany root b of f(x), A(b) is an algebraic extension of A. A is analgebraic extension of Q, so by lemma 2, A(b) is algebraic over Q.In particular, b is algebraic over Q, so b must be in A.> 2) AcC (i.e. all algebraics are complex)?I don't tnk so. There are proofs of the Fundamental Theorem ofAlgebra, hence of ts, in various algebra books (like Dummit & Foote,or Herstein's Topics in Algebra), but they tend to use Galois theoryand tngs like that. They are not what I would call relativelysimple.-- J. H. PalmieriDept of Mathematics, Box 354350 mailto:palmieri@math.wasngton.eduUniversity of Wasngton http://www.math.wasngton.edu/~palmieri/Seattle, WA 98195-4350=== === Subject: : JSH: A semantic questionRegarding the ... activity engaged in by JSH, wch name is more appropriate: Harristotelian Logic, or Masterbatics?Gib=== === Subject: : Re: Geometry terminology question> What are the proper English terms for:> - A 2D polygon with four angles and at least two right angles?Trying to exhaust all cases.. If the right angles are corrosponding, parallel sides can beconsidered as having same sign for length, it is a straighttrapezium.The diagnols intersect inside the quadrilateral.If the right angles are alternate,(anti) parallel sides can beconsidered as having oposite sign for length, it is a crossedtrapezium.The diagnols do not intersect inside the quadrilateral.[A common Structural Enginering example is: Stress distribution withand without mid-plane stretcng respectively, in a Beam.]If the right angles are on opposite sides of the diagnol, it is aregular cyclic quadrilateral.If the right angles are on same side of the diagnol, it is a crossedcyclic quadrilateral.You could easily figure them out on paper.=== === Subject: : Re: crazy cubic equation.... > 4x^3-ax-b=4(x-c)(x-d)(x-e)where a,b.c.d and e, I know to be nonzero complex conts. Now from ts I > wanted to arrive toa^3-27b^2=16(c-d)^2(c-e)^2(d-e)^2Ts is in fact the discriminant of the cubic equation....> Ts is a well-known hard problem in classical algebra. I've seen> students produce a page or two of dense Maple print-out to slog it out by> direct calculation. Here's the best short cut I know, but if anyone has a> shorter method I'd be glad to see it.> [...]Here's another method, moreorless neck and neck with yours in the lengthstakes..Starting with x^3 + p.x + q = (x - c).(x - d).(x - e), differentiatingboth sides gives: 3.x^2 + p = (x-c).(x-d) + (x-d).(x-e) + (x-e).(x-c).Plugging x = c, d, e successively into the latter gives three equations3.c^2 + p = (c - d).(c - e), ... and multiplying these and expandingthe LHS in powers of p gives: p^3 + 3.M.p^2 + 9.N.p + 27.(cde)^2 = - Dwhere: M = c^2 + d^2 + e^2 = (c + d + e)^2 - 2.(cd + de + ec) N = (cd)^2 + (de)^2 + (ec)^2 = (cd + de + ec)^2 - 2.cde.(c + d + e)Using c + d + e, cd + de + ec, cde = 0, p, -q resp, ts gives:M, N = -2p, p^2 and consequently: - D = 4.p^3 + 27.q^2-------------------------------------------------------- ------------------- R Ramsden (jr@adslate.com)---------------------------------------------- -----------------------------Eternity is a long time, especially towards the end. Woody Allen=== === Subject: : Re: Psycc Will Contact Princess Diana - WMD, Dark Energy, UFOs Conformal Group> http://www.thejameswhaleshow.co.uk> Ts is a London,U.K. radio station,10pm-1am,monday-thursday and is> recieved in Europe and by internet in the USA,last night Salla> was interviewed,http://www.exopolitics.org ,Political Implications of> Extraterrestrial Presence,tonight James Whale is interviewing a> psycc who will be directly contacting Princess Diana.As she was apparently convinced someone was planning to tamper withthe brakes of her car, perhaps the psycc will ask the stupid cowwhy she wasn't wearing a seatbelt on the night of her crash.-------------------------------------------------------- ------------------- R Ramsden (jr@adslate.com)---------------------------------------------- -----------------------------Eternity is a long time, especially towards the end. Woody Allen=== === Subject: : Tnker required- funded UK Masters degreePlease accept my apologies if you tnk ts is unsuitable traffic forts group.I'm looking for someone who wants a UK Master's degree totake part in an unusual research project- full details below. Thecandidate will have to be an EU national unless you can pay your owntuition fees (sorry!).It'll be at Cranfield University wch is 1 hour north of London. Thetopic is 'how well can a train/car driver see signs and signals in badweather or bad light', or 'how well can someone in a burning buildingsee the signs that point their escape route'. We'd like to make ourroads and buildings safer.Ts is a fully funded MSc by Research supervised jointly by theOptical Diagnostics Group and the Human Factors Group witn theSchool of Engineering of Cranfield University. A bursary of 8,000plus full tuition fees will be paid.The project will involve adapting an existing software tool tosimulate the propagation of light in visually sub-optimal conditions.For example, these include conditions of fog and rain, smoke fromfire, poor artificial lighting, or when the sun is low on the horizon.Once the code has been modified, additional work will be required tosimulate and model human perception in such conditions, consideringthe responses and limitations of the visual system. The tool will thenneed to be validated in a range of scenarios, such as the design offire escape signage and routes, and the placement of road and railwaysignals for optimal visibility.To apply, you should have experience of programming (in any language),and an interest in applying basic physics to real-world human factorsproblems. Experience in human factors is desirable but not necessary,as the successful candidate will be given the opportunity to attendpostgraduate human factors modules relevant to the research project.On successful completion of the degree, there may be the possibilityof further study or employment at Cranfield University. For furtherinformation, please contact Dr Jermy on +44 (0)1234 754680 orm.jermy@cranfield.ac.uk=== === Subject: : Re: graph theory> First qestions: How can I show that the degree of vi in G can not> exceed k?Second question: How can I show that there must be a vertex v in G> with degree less than or equal to (n-2)? Why would you want to show these? Asking these questions makes it sound J=== === Subject: : Re: Towers of Hanoi revisited ( 4 towers)>The original 3 tower problem is very simple to solve where (n) = the>number of disks then, 2^n -1 is the least number of moves for a stack>with (n) disks.>Also the smallest disk is moved starting with the 1st disk and then>every other move thereafter and repeating the same move patterns. Ts>pattern is every 3 cycles of the smallest disk wch is counting the>number of moves just for the smallest disk giving a total of 3 moves>returning it back to the original location at the end of 3 moves of>the smallest disk.>The four-tower problem is more complicated. Wle requiring many less>moves there is no set pattern to find the least number of moves for>(n).>Unlike the 3-tower problem, the 4 tower patterns are dependent on the>number of (n) disks.>The table below, I believe, gives the least possible moves for the>4-tower problem.>(n) # of >disks moves>2 3>3 5>4 9>5 13>6 17>7 25>8 33>9 41>10 ?? Not sure if (12 or 16) + 41 moves for a stack of 10>disks.>The delta between these moves is --- 2,4,4,4,8,8,8, --- (12 or 16) for>the next delta between the stack of 9 and 10 disks.>Will these deltas continue in a set pattern like --->2,4,4,4,8,8,8,16,16,16,32,32,32,64,64,64..?>Following the rules of the 3 tower problem (no larger disk can be>placed on a smaller disk and moving one disk at a time move the whole>stack to another tower) and using my table above, is it possible to>find less moves for a 4 tower for any (n) >1 and (n) <10.To move a tower of 10 from pin 1 to pin 21) Move a stack of 6 from pin 1 to pin 3 using four rods (17 moves)2) then move a stack of 3 from pin 1 to pin 4 using three rods (7 moves)3) then move the bottom disk from pin 1 to pin 2 (1 move)4) then move the stack of 3 disks from pin 4 to pin 2 (another 7 moves)5) then move the stack of 6 from pin 3 to pin 2 (another 17 moves)ts gives a total of 49 moves.In general, to move a stack of n, you need to move the top k disks using 4 rodsmove the next n-k-1 disks using 3 rodsmove the bottom diskmove n-k-1 disks to the destination using 3 rodsmove the top k to the destination.It seems ts method is optimal.Ts gives moves(n) = minimum {k=1, kThe original 3 tower problem is very simple to solve where (n) = the>number of disks then, 2^n -1 is the least number of moves for a stack>with (n) disks.>Also the smallest disk is moved starting with the 1st disk and then>every other move thereafter and repeating the same move patterns. Ts>pattern is every 3 cycles of the smallest disk wch is counting the>number of moves just for the smallest disk giving a total of 3 moves>returning it back to the original location at the end of 3 moves of>the smallest disk.>The four-tower problem is more complicated. Wle requiring many less>moves there is no set pattern to find the least number of moves for>(n).>Unlike the 3-tower problem, the 4 tower patterns are dependent on the>number of (n) disks.The table below, I believe, gives the least possible moves for the>4-tower problem.(n) # of >disks moves>2 3>3 5>4 9>5 13>6 17>7 25>8 33>9 41>10 ?? Not sure if (12 or 16) + 41 moves for a stack of 10>disks.The delta between these moves is --- 2,4,4,4,8,8,8, --- (12 or 16) for>the next delta between the stack of 9 and 10 disks.>Will these deltas continue in a set pattern like --->2,4,4,4,8,8,8,16,16,16,32,32,32,64,64,64..?Following the rules of the 3 tower problem (no larger disk can be>placed on a smaller disk and moving one disk at a time move the whole>stack to another tower) and using my table above, is it possible to>find less moves for a 4 tower for any (n) >1 and (n) <10.To move a tower of 10 from pin 1 to pin 21) Move a stack of 6 from pin 1 to pin 3 using four rods (17 moves)> 2) then move a stack of 3 from pin 1 to pin 4 using three rods (7 moves)> 3) then move the bottom disk from pin 1 to pin 2 (1 move)> 4) then move the stack of 3 disks from pin 4 to pin 2 (another 7 moves)> 5) then move the stack of 6 from pin 3 to pin 2 (another 17 moves)ts gives a total of 49 moves.In general, to move a stack of n, you need to > move the top k disks using 4 rods> move the next n-k-1 disks using 3 rods> move the bottom disk> move n-k-1 disks to the destination using 3 rods> move the top k to the destination.It seems ts method is optimal.Ts gives moves(n) = minimum {k=1, k , simply by trying all k.Ts gives1,3,5,9,13,17,25,33,41,49,65,81,97,113Ts can be found in the online encyclopedia of integer sequences:http://www.research.att.com/cgi-bin/access.cgi/as/ njas/sequences/eisA.cgi?Anum=A007664The delta's actually go 2,2,4,4,4,8,8,8,8,16,16,16,16,16,322 2's, 3 4's, 4 8's, 5 16's etc.I did not know ts existed in EIS for the 4 tower problem.I programmed the 3 tower problem where the algorithm solved up to 15disks. After answering the input prompt for the number of disks, thealgorithm gives a grapc side view of the 3 towers and tck lines ofdifferent length stacked on one another representing the disks on thefirst post. Then it proceedes showing each move wch I had to slowdown so it could be observed. It ends when all the disks are stackedon a different tower giving the precise number of moves for thatnumber of disks.Ts was simple to do because the smallest disk has a pattern,starting with the 1st move then every other move was easy to followand the moves after each smaller disk was a larger disk that had to bemoved where only one move is possible.Ts is not so easy when you add another tower though!I am still having problems finding an algorithm to solve the 4 tower.Maybe with ts new info you presented it may be possible to base thealgorithm on the number of disks and the certain patterns witn tsnumber that will give the precise number of minimum moves for thatparticular stack!tower problem.Dan=== === Subject: : Re: Uncle Al is Sadistic .> I just love being castigated in pejorative terms to trd parties by> second rate minds.>Just imagine what a thrill a first-rate mind could give you.When you run into another one on the usenet let me know.=== === Subject: : Re: Uncle Al is Sadistic .I just love being castigated in pejorative terms to trd parties by> second rate minds.>And you apparently enjoy basking in the admiration of like-minded >crackpots.As I'm sure do you. Why else would anyone even talk to you?> How do you feel about being ignored by everyone else?Oh please, please, b'rer fox! Don't tell me ts ignorance isorganized. Please don't tell me I've been blacklisted. I just couldn'td it. Whutevah will Ah do? Well frankly my dear I don't give adamn. But apparently you do.=== === Subject: : Re: Uncle Al is Sadistic .> Von Neumann, Feynman, Edward Teller, Bethe... and Enrico Fermi didn't> leave Italy just because of s wife. Put two Jews in a lunar crater> and come back in five years to find a resort. You should be scared.> Very scared.The goys have proven the following theorem...Genius isn't a monopoly. You consider individuals as individuals, oneby one. However, you mostly won't mine diamonds out of anytng butlamproite and kimberlite, plus placer deposits. If resources arelimited you invest where the odds are demonstrated best. Only a fool*continues* to commit a majority of resources in barren ground. Youcan dig down five miles deep and you won't find one natural diamond inNew Jersey.The top 20% of any university graduating class wasn't the top 20% ofSAT scores on admission. OTOH, the top 20% of SAT scores on admissionare subtially over-represented in the top 20% of the graduatingclass - especially in subjects containing objective truth. Auniversity seeking to fill its sciences, engineering, math,computer... departments would be insane to choose diversity (racialquotas) over objective performance. Look in sci.physics. What do youdo to an idiot to fix it? Education isn't relevant.If you want metal you mine ore and dispose of the gangue matter ofcourse. You sure as Hell don't dig granite if you want copper. It'sa waste of granite, too. UC/Irvine intensely, Liberally, enthusiastically courted nigger andspics - guaranteed admission no matter what and full free ride plusfat perqs. UC/Irvine intensely, Liberally, enthusiastically excludedAsians and Jews. I visit the campus. The real subjects are almostexclusively populated by Asians and Jews with astounding credentialsand demonstrated abilties. Your average Wte kid looks around anddrops those courses after the first lecture. Don't entertainhallucinations of diversity in linear algebra courses. TheUniversity of California Regents can't for the life of them figure outhow Asians and Jews can universally do so well.IT'S BECAUSE YOU ONLY ADMITTED THE MUTANT SMARTEST ONES, FOOLS, ANDTHEY PERFORMED TO SPEC. Now try applying that to the UC system as awhole and see what you get - as opposed to what you've got.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for cldren and most mammals)Quis custodiet ipsos custodes? The Net!=== === Subject: : Re: Uncle Al is Sadistic .> Von Neumann, Feynman, Edward Teller, Bethe... and Enrico Fermi didn't leave Italy just because of s wife. Put two Jews in a lunar crater and come back in five years to find a resort. You should be scared. Very scared.The goys have proven the following theorem...Genius isn't a monopoly. You consider individuals as individuals, one> by one. However, you mostly won't mine diamonds out of anytng but> lamproite and kimberlite, plus placer deposits. If resources are> limited you invest where the odds are demonstrated best. Only a fool> *continues* to commit a majority of resources in barren ground. You> can dig down five miles deep and you won't find one natural diamond in> New Jersey.The top 20% of any university graduating class wasn't the top 20% of> SAT scores on admission. OTOH, the top 20% of SAT scores on admission> are subtially over-represented in the top 20% of the graduating> class - especially in subjects containing objective truth. A> university seeking to fill its sciences, engineering, math,> computer... departments would be insane to choose diversity (racial> quotas) over objective performance. Look in sci.physics. What do you> do to an idiot to fix it? Education isn't relevant.If you want metal you mine ore and dispose of the gangue matter of> course. You sure as Hell don't dig granite if you want copper. It's> a waste of granite, too.> UC/Irvine intensely, Liberally, enthusiastically courted nigger and> spics - guaranteed admission no matter what and full free ride plus> fat perqs. UC/Irvine intensely, Liberally, enthusiastically excluded> Asians and Jews. I visit the campus. The real subjects are almost> exclusively populated by Asians and Jews with astounding credentials> and demonstrated abilties. Your average Wte kid looks around and> drops those courses after the first lecture. Don't entertain> hallucinations of diversity in linear algebra courses. The> University of California Regents can't for the life of them figure out> how Asians and Jews can universally do so well. That' just because of all California is filled with morons, not just U.C. Since Linear Algrebra was invented in ancient Eqypt, most us are still trying to figure why *any* mathematicians anywhere believe that they have the credentials necessary to grade California Home Economics Homework, nevermind anytng that appears to be politics that idiots at Harvard invented, rather than a run down compsci dump like U.C.=== === Subject: : Re: Probability question The probability is simply the ratio of combinations.> In an experiment, two cards are drawn from a deck of well-shuffled> pack C(52,2)> find the probability for the event that> (a)the two cards have a sum of 3 C(4,1) * C(4,1) / C(52,2)> (b)the two cards have a sum of 3 if one of the cards is an ace of> heart. C(4,1) / C(52,2)-- Hoa K. TranGod made the integers, all the rest is the work of man. --Leopold Kronecker=== === Subject: : Category TheoryI've been toiling over ts problem for a wle, and it seems like thereason I can't get it is either because I don't have enoughinformation in the problem (not likely), or that I'm not underdingsome fundamental concept. Let me first state the problem:Let F be a free object on a set X (i:X-->F) in a concrete category C. If C contains an object whose underlying set has at least two elementsin it, then i is an injective map of sets.To tackle ts problem, I thought it would be easiest to do thefollowing:Let p be the statement F is a free object on a set X (i:X-->F) in aconcrete category CLet q be the statement C contains an object whose underlying set hasat least two elements in itLet r be the statement i is an injective map of sets.And then attempt to prove !r and p ----> !qSo I'd need to assume i is not injective, F is free on X, and showthat every object in C contains either exactly 1 element or is theempty set.I tried to do the following:Since F is free on X, then for any A in C and any f:X-->A, there is aunique morpsm g:F-->A such that gi=f. Since i is not injective,neither is gi and hence f is not injective. Thus there does not existan injective function from X to A, and therefore the cardinality of Ais strictly less than the cardinality of X. But A is arbitrary, sothat the cardinality of X is greater than the cardinality of everyobject in C. Moreover, the definition of a function not beinginjective implies that X has at least 2 elements in it.However, ts seems to be a dead end. So I tried to throw in theextra condition q and then try to arrive at an obvious contradiction. Throwing in q you can get that the cardinality of X is at least 3(because now there's an element of C with cardinality at least 2).But now ts also seems like a dead end. Any ideas on how to proceed? Am I missing some fundamental concept here? It doesn't really seemlike you have a whole lot of information with wch to work, sothere's only so many tngs you can really do. A nt as opposed tofilling in the proof is preferred.=== === Subject: : Re: Category Theory>Let F be a free object on a set X (i:X-->F) in a concrete category C. >If C contains an object whose underlying set has at least two elements>in it, then i is an injective map of sets.>Let p be the statement F is a free object on a set X (i:X-->F) in a>concrete category C>Let q be the statement C contains an object whose underlying set has>at least two elements in it>Let r be the statement i is an injective map of sets.>And then attempt to prove !r and p ----> !qIt looks easier to me to just assume p and q and prove r usingthe defintion of injective.Take x, y in X, x <> y, and assume that i(x) = i(y). Take an objectA of C whose underlying set has at least two elements. Build amap from X to A (or rather from X to the underlying set of A)that maps x and y to distinct elements of A. [Ts is the differencewith your approach - in your approach you need/want ts map tobe injective, but that is too much]. Now you only needa tiny bit of the properties of F to derive a contradiction.-- van Rossum, | Universal law of linearity: for allDept. of Mathematics, New Mexico | f : R -> R and for all x, y in R:State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y)=== === Subject: : Re: show that> 2^(4x+1)-2^(2x)-1=0 (mod 9) for all x natural.> I tnk a slightly easier method than the others suggest is to simplyfactor ts.2^(4x+1)-2^(2x)-1= 2*[2^(2x)]^2 - 2^(2x) - 1Let y = 2^2x= 2y^2 - y - 1= (2y+1)(y-1)= (2^(2x+1) + 1)(2^(2x)-1)= (2^(2x+1) + 1)(2^x + 1)(2^x - 1)Now look at each term mod 3. The first term mod 3 is(-1)^(2x+1)+1= -1+1 (since the power is odd)= 0(hence the first term is divisible by 3)The second term is [-1]^x + 1The trd term is [-1]^x - 1Depending on whether x is odd or even, either the first or second ofthese will come out to 0, so one of them will be divisible by 3. Hence the whole tng is divisible by 9.=== === Subject: : JSH: Distributive property, math argumentIt's amazing how easily people can question very basic concepts whenthe stakes are gh, so here I am again to help you out by remindingyou just how basic the math argument I've been giving recently is.Consider f(x) = x+3, where x is an algebraic integer.Now then, if f(x) has 3 as a factor, then x *must* have 3 as a factoras well.That follows easily enough from the distributive property:a(b+c) = ab + acand for quite a few months, I've noticed people willing to debate it,as if that property can just go away at a wm.That's not mathematics, and many of you are showing a puzzling lack ofrationality by fighting mathematics itself, rather than just acceptingthe truth.After all, it's not like the math is mad at you, or that there's someconspiracy to screw you over as mathematical truth is just true.If you believed sometng else in the past that was wrong, it was justwrong.Now here's the argument, where those people questioning it havebasically been questioning the distributive property.The start is simple enough, where the x shown is in the ring ofalgebraic integers.Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078wch you'll notice has a cont term that is 1078.Well moving tngs around with P(x) gives youP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3wch is a deliberate form to allow me to factor P(x), so that I haveP(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).To use the distributive property as I wish I need to have contterms.And it *appears* that the cont terms for the three factors are all7, but that can't be right, as the cont term of P(x) is 1078.So I need to do another step, and analysis offers the simple techniqueof setting x=0 to pull out the cont terms, so setting x=0, I findthatP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's at x=0 isa^3 - 3a^2, wch has roots, 0, 0 and 3, and I've picked a_1(0) anda_2(0) to equal 0, wch leaves a_3(0) with a value of 3.So let a_3(x) = b_3(x) + 3, to keep indices matched. Then I haveP(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)and now my cont terms work out correctly.Now I can use the distributive property, as P(x) has 49 as a factor ofeach term inP(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078so P(x) has 49 as a factor, so I can divide by 49, and dividing 1078by 49 gives me 22, as the new cont term.Well that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that 49 can divide out and keep the cont termsmatcng, in what is a simple exercise dependent on the distributiveproperty.http://mathforprofit.blogspot.com=== === Subject: : Re: JSH: Distributive property, math argument[notng new]>P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3>P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)>where the a's are roots of>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).>P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)>is the only way that 49 can divide out and keep the cont terms>matcng, in what is a simple exercise dependent on the distributive>property.So, let's actually do the math.For every algebraic integer x, a_1(x), a_2(x) and a_3(x) are theroots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).So, as you observed, a_1(0), a_2(0) and a_3(0) are the roots of a^3 - 3a^2,so they are 0,0 and 3.Similarly a_1(1), a_2(1) and a_3(1) are the roots of a^3 - 144 a^2 - 110593.Therefore a_1(1)/7, a_2(1)/7 (and also a_3(1)/7, but we don't need that)are the roots of (substitute b=a/7 and multiply by 7^3) 343 b^3 - 7056 b^2 - 110593and hence of (divide by 49) 7 b^3 - 144 b^2 - 2257.Ts polynomial is irreducible, so a_1(1)/7, a_2(1)/7 (and a_3(1)/7)are not algebraic integers.-- van Rossum, | Universal law of linearity: for allDept. of Mathematics, New Mexico | f : R -> R and for all x, y in R:State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y)=== === Subject: : Re: JSH: Distributive property, math argument... > The start is simple enough, where the x shown is in the ring of > algebraic integers. > Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078Why does it not work with my polynomial? > wch you'll notice has a cont term that is 1078.Yes, also for mine. > Well moving tngs around with P(x) gives you > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3Similar for mine. > wch is a deliberate form to allow me to factor P(x), so that I have > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)Yup, also for mine. > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Yup, similar to mine. > To use the distributive property as I wish I need to have cont > terms. And it *appears* that the cont terms for the three factors are all > 7, but that can't be right, as the cont term of P(x) is 1078. > So I need to do another step, and analysis offers the simple technique > of setting x=0 to pull out the cont terms, so setting x=0, I find > that > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)Yup, also for mine. > as the cubic defining the a's at x=0 is > a^3 - 3a^2, wch has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, wch leaves a_3(0) with a value of 3.Yup, also for mine. > So let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)Exactly the same for mine. > and now my cont terms work out correctly. > Now I can use the distributive property, as P(x) has 49 as a factor of > each term in > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078Similar to mine. > so P(x) has 49 as a factor, so I can divide by 49, and dividing 1078 > by 49 gives me 22, as the new cont term. > Well that means that > P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) > is the only way that 49 can divide out and keep the cont terms > matcng, in what is a simple exercise dependent on the distributive > property.Yup the same for mine. So what exactly is wrong with my polynomialcompared to your?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/=== === Subject: : Re: Distributive property, math argument> It's amazing how easily people can question very basic concepts when> the stakes are gh, so here I am again to help you out by reminding> you just how basic the math argument I've been giving recently is.> Consider f(x) = x+3, where x is an algebraic integer.> Now then, if f(x) has 3 as a factor, then x *must* have 3 as a factor> as well.Agreed.> That follows easily enough from the distributive property:> a(b+c) = ab + acAgreed.> and for quite a few months, I've noticed people willing to debate it,> as if that property can just go away at a wm.> That's not mathematics, and many of you are showing a puzzling lack of> rationality by fighting mathematics itself, rather than just accepting> the truth.> After all, it's not like the math is mad at you, or that there's some> conspiracy to screw you over as mathematical truth is just true.> If you believed sometng else in the past that was wrong, it was just> wrong.> Now here's the argument, where those people questioning it have> basically been questioning the distributive property.> The start is simple enough, where the x shown is in the ring of> algebraic integers.> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078> wch you'll notice has a cont term that is 1078.> Well moving tngs around with P(x) gives you> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> wch is a deliberate form to allow me to factor P(x), so that I have> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)> where the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> To use the distributive property as I wish I need to have cont> terms.> And it *appears* that the cont terms for the three factors are all> 7, but that can't be right, as the cont term of P(x) is 1078.> So I need to do another step, and analysis offers the simple technique> of setting x=0 to pull out the cont terms, so setting x=0, I find> that> P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)> as the cubic defining the a's at x=0 is> a^3 - 3a^2, wch has roots, 0, 0 and 3, and I've picked a_1(0) and> a_2(0) to equal 0, wch leaves a_3(0) with a value of 3.> So let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have> P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)> P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)> and now my cont terms work out correctly.> Now I can use the distributive property, as P(x) has 49 as a factor of> each term in> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078> so P(x) has 49 as a factor, so I can divide by 49, and dividing 1078> by 49 gives me 22, as the new cont term.> Well that means that> P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)> is the only way that 49 can divide out and keep the cont terms> matcng, in what is a simple exercise dependent on the distributive> property.http://mathforprofit.blogspot.comOk, You've worked ts out for a few specific polynomials, but can you showthat ts is true for a general polynomial? It is good to see an example ofts, but it should be worthwle to see that ts works for a generalpolynomial so that one can see that it works for any polynomial. electron-dot-cloud are galaxies=== === Subject: : Re: B. Bonds test on dice of playing field Re: better to slow pitch Indeed, physics tells us that the 100mph pitch momentum is added onto the>t ball for extra long dice.> I don't know what physics text you found that one in. How can an> increase in the initial momentum lead to an increase in the final> momentum, when the two are pointed in different directions? I'mSimple. Change in direction is acceleration. Simple test is that a man on a bicycle can dupright by simply fastly changing the front wheel direction. So as the baseball of 100mph ist by the bat, it keeps its nearly 100mph addon plus thechange in direction.Second simple. You have conservation of energy. So a 100mph fastball has more energy than a5mph toss up. So, how is that extra 95mph energy conserved. Seems that your way of tnking isthat the bat carries that through the wood andinto the body of the tter, when in fact that energy is conserved by merely aredirection of that 95mph energy.Arcmedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies=== === Subject: : Re: MKC Set TheoriesMy point was the following. If we can consider both MK and MKC as first> order theories without equality and a single binary relation, but with> slightly different axioms, then we can ask if these two theories are> equivalent. By equivalent I mean that every axiom of MK is a theorem of> MKC and every axiom of MKC is a theorem of MK.Here are two bases for set theory.MKC BasisC2 AxAy[xIy -> Az(x in z <-> y in z) & Az(z in x <-> z in y)]C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)C4 AxAy[Az(z in x <-> z in y) -> {(set x & set y) <-> xIy}]MK BasisM2 AxAy[x=y -> Az(x in z <-> y in z) & Az(z in x <-> z in y)]C3 EyAx[x in y <-> Set x & A] (with y not free in A)M4 AxAy[Az(z in x <-> z in y) -> x=y] MD1: xIy means x=y & set x & set yWith the addition of CD1 to the MKC basis, the MK principlesare theorems of MKC.CD1: x=y means Az(x in z <-> y in z) & Az(z in x <-> z in y)]Likewise, with the addition of MD1 to the MK basis, the MKCprinciples are theorems of MK.MD1: xIy means x=y & set x & set yIt follows that these bases are formally equivalent.Based on the definitions CD1 and MD1, I show inthat the following MK and MKD axiom sets are formallyequivalent:MK AxiomsN1 AxAyAz[x=y -> (x in z <-> y in z)] LL 1N2 AxAyAz[x=y -> (z in x <-> z in y)] LL 2N3 EyAx[x in y <-> (Et(x in t) & P(x))] (with y not free in (x)) N4 AxAy[Az(z in x <-> z in y) -> x = y] ExtensionalityN5 AaAb[set a & set b -> Ey(set y & Ax(x in y <-> x=a v x=b))] PairsN6 Aa[set a -> Ay(Ax(x in y -> x in a) -> set y)] SubsetsN7 Aa[set a -> Ey(set y & Ax(x in y <-> Ew(w in a & x in w)))] UnionsN8 Aa[set a -> Ey(set y & Ax(x in y <-> Aw(w in x -> w in a)))] Power SetsN9 Ex(set x & Ay~(y in x)) Empty Set MKD AxiomsC1 AxAyAz[xIy -> (z in x <-> z in y)] LL1C2 AxAyAz[Az(z in x <-> z in y) <-> ((x in z <-> y in z))] LL2C3 EyAx[x in y <-> Et(x in t) & P(x)] (with y not free in P(x))C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> xIy}]C5 AaAbAy(Ax[x in y <-> (xIa v xIb)] -> yIy) PairsC6 AaAy(Ax[x in y -> x in a] -> (aIa -> yIy)) SubsetsC7 AaAy[Ax(x in y <-> Ew(w in a & x in w)) -> (aIa -> yIy)] UnionsC8 Aa(Ay[Ax(x in y <-> Aw(w in x -> w in a))) -> (aIa -> yIy]) PowerSetsFor C1-C8,CD1|-N1-N9; and N1-N9,MD1|-C1-C8Your conjecture that such frameworks are formally equivalent receivessupport from the above.In either of these theories (wch may be equivalent), we can define two > relations, usual equality (=) and Correy equality (=c), and then study > their implications. Of course, = has be well studied. The question is > what is cool about =c other that the jarring statement ~Ax(x=x).Ts is the crux of the matter. (To be continued)If one forces either of these relations (= or =c) to be primitives, > comparisons between the theories becomes somewhat trickier. I > recommmend to to consider ts approach.Dan=== === Subject: : Re: MKC Set Theories>My point was the following. If we can consider both MK and MKC as first> order theories without equality and a single binary relation, but with> slightly different axioms, then we can ask if these two theories are> equivalent. By equivalent I mean that every axiom of MK is a theorem of> MKC and every axiom of MKC is a theorem of MK.Here are two bases for set theory.MKC BasisC2 AxAy[xIy -> Az(x in z <-> y in z) & Az(z in x <-> z in y)]> C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)> C4 AxAy[Az(z in x <-> z in y) -> {(set x & set y) <-> xIy}]MK BasisM2 AxAy[x=y -> Az(x in z <-> y in z) & Az(z in x <-> z in y)]> C3 EyAx[x in y <-> Set x & A] (with y not free in A)> M4 AxAy[Az(z in x <-> z in y) -> x=y] > MD1: xIy means x=y & set x & set yWith the addition of CD1 to the MKC basis, the MK principles> are theorems of MKC.CD1: x=y means Az(x in z <-> y in z) & Az(z in x <-> z in y)]Likewise, with the addition of MD1 to the MK basis, the MKC> principles are theorems of MK.MD1: xIy means x=y & set x & set yIt follows that these bases are formally equivalent.> Based on the definitions CD1 and MD1, I show in> that the following MK and MKD axiom sets are formally> equivalent:MK AxiomsN1 AxAyAz[x=y -> (x in z <-> y in z)] LL 1> N2 AxAyAz[x=y -> (z in x <-> z in y)] LL 2> N3 EyAx[x in y <-> (Et(x in t) & P(x))] (with y not free in (x)) > N4 AxAy[Az(z in x <-> z in y) -> x = y] Extensionality> N5 AaAb[set a & set b -> Ey(set y & Ax(x in y <-> x=a v x=b))] Pairs> N6 Aa[set a -> Ay(Ax(x in y -> x in a) -> set y)] Subsets> N7 Aa[set a -> Ey(set y & Ax(x in y <-> Ew(w in a & x in w)))] Unions> N8 Aa[set a -> Ey(set y & Ax(x in y <-> Aw(w in x -> w in a)))] Power Sets> N9 Ex(set x & Ay~(y in x)) Empty Set MKD AxiomsC1 AxAyAz[xIy -> (z in x <-> z in y)] LL1> C2 AxAyAz[Az(z in x <-> z in y) <-> ((x in z <-> y in z))] LL2> C3 EyAx[x in y <-> Et(x in t) & P(x)] (with y not free in P(x))> C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> xIy}]> C5 AaAbAy(Ax[x in y <-> (xIa v xIb)] -> yIy) Pairs> C6 AaAy(Ax[x in y -> x in a] -> (aIa -> yIy)) Subsets> C7 AaAy[Ax(x in y <-> Ew(w in a & x in w)) -> (aIa -> yIy)] Unions> C8 Aa(Ay[Ax(x in y <-> Aw(w in x -> w in a))) -> (aIa -> yIy]) Power> SetsFor C1-C8,CD1|-N1-N9; and N1-N9,MD1|-C1-C8C2 should read: C2 AxAy[Az(z in x <-> z in y) -> (Az(x in z <-> y in z)] LL2It is ts C2--and not the typo-ridden C2 wch appears in the postingthat I am amending--wch figures in proofs of N1-N9 from C1-C8 (see:The typo-ridden C2 is also displayed above N2,N4 inC2 AxAyAz[Az(z in x <-> z in y) -> ((x in z <-> y in z))]N2 AxAyAz[x=y -> (z in x <-> z in y)] PremiseN4 AxAy[Az(z in x <-> z in y) -> x = y] PremiseHowever, the proof wch follows is a proof ofAxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]> 1. Show AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]> 2. Show Az(z in x <-> z in y) -> Az(x in z <-> y in z)> 3. Az(z in x <-> z in y) Assume> 4. Show Az(x in z <-> y in z)> 5. x=y 3,N4> 6. Az(z in x <-> z in y) 5,N2:Ca Sh(4)> 7. Az(z in x <-> z in y) -> Az(x in z <-> y in z) 3,4:Ca Sh(2)> 8. AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] 2:Ca Sh(1)--=== === Subject: : are Godel statements just Self Evident Truths?Out of hypothesies, postulates, theorems, axioms, primitives... did I miss one?conjectures, corrolaries, formula,the leaf level of true statements is self evident truths, they cannot be split further,the most elemental statements.Then from the S.E.T.s and other axioms that we take as granted, further theoremsare derived and considered proven.So rather than calling S.E.T.s *proven*, we could distinguish them as having a different classof truth value.So Godel's statement ts statement has no proof etc. just falls into ts class.Self evident ~ it proves itself.So the existence of Godel statements does not have an overbearing effect on completenessat all, true statements exist without proof is only true when those statements are self evident.Herc=== === Subject: : Re: are Godel statements just Self Evident Truths?> Out of hypothesies, postulates, theorems, axioms, primitives... did I missone?> conjectures, corrolaries, formula,> the leaf level of true statements is self evident truths, they cannot besplit further,> the most elemental statements.> Then from the S.E.T.s and other axioms that we take as granted, furthertheorems> are derived and considered proven.> So rather than calling S.E.T.s *proven*, we could distinguish them ashaving a different class> of truth value.> So Godel's statement ts statement has no proof etc. just falls intots class.If ts is a legal statement, then it is an inconsistent Set Theory.Therefore in order to have consistency, you must say that that phrase is nota Well-Formed Statement.> Self evident ~ it proves itself.> So the existence of Godel statements does not have an overbearing effecton completeness> at all, true statements exist without proof is only true when thosestatements are self evident.> Herc=== === Subject: : Re: The need of geodesics in physics> A metric is a rule for finding dices between points along a path, it's > sort of a generalization of Pythagorean's theorem. A metric space is an > affine space where a metric can be defined, and when the metric is > defined you can relate that to the Christoffel symbols. So I presume an > affine theory is one with a manifold that doesn't have a metric. But I've > been having trouble working up a physical picture of that.No, none of ts is outside Riemannian geometry.First of all, connection doesn't mean Christoffel symbol, in generalin differntial geometry. Connections and metrics are completelyindependent structures that a manifold can have. And connectionis also a MUCH more general concept that applies also to gauge groups.The best way to underd the indices of the usual connection Gamma_{mn}^pare: the _m is a spacetime index the _n^p is the index for a transformation matric.Ts gives you a GL(n) connection (and, when the connectionleaves the covariant derivative of the metric 0, an O(n) connection).The corresponding 1-form is: Omega^p_n = Gamma_{mn}^p dx^m[summation convention used], wch is a O(n)-valued 1-form.You can also have connections, say, for groups like U(1), T(n)the n-dimensional translation group (as here), SU(2), SU(3), etc.Likewise, curvature is a property of connections, not ofthe spacetime or metric, itself and every type of connection canhave its own curvature (or field strength). The correspondingcurvature 2-form is: Theta = d Omega + Omega ^ Omegs.So, for an ordinary connection, ts gives you the O(n)-valued2-form: Theta^a_c = d Omega^a_c + Omega^a_b ^ Omega^b_c = 1/2 (@_m Gamma_{nc}^a - @_n Gamma_{mc}^a + Gamma_{mb}^a Gamma_{nc}^b - Gamma_{nb}^a Gamma_{mc}^b) dx^m ^ dx^nwhose coefficients are just the Riemann Curvature components Theta^a_c = 1/2 R^a_{c mn} dx^m ^ dx^nThe T(4)-connection B_m^a has indices _m for spacetime indexand ^a the T(4) index (labelling the T(4) generators P_a).A SEPARATE Levi-Civita torsion-free connection (i.e. Christoffelsymbol) resides in ts affine space, wch has 0 curvature.The curvature of the T(4)-connection yields the equivalent ofa torsion, wch can be added to the Levi-Civita connection togive you a connection wch is not torsion-free.The curvature of the Levi-Civita connection is 0 in ts theory.So, it's just ordinary Minkowski space with a connection thatconsists of a Levi-Civita part, plus a non-zero torsion. Youcan equally well consider any combination of the two, includingthe one Uncle Al didn't list ... No torsion, no curvature: Locally Minkowski No torsion, curvature: Ordinary Riemannian geometry Torsion, no curvature: Locally Minkowski with an affine connection Torsion, curvature: Riemannian geometry with an affine connection=== === Subject: : Calling Jason MoxhamJason, If you read ts, can you drop me a mail please, regarding your APRCL and other tngs (your website's timing out consistently :-( )Pl-- Unpatched IE vulnerability: Extended HTML Form AttackDescription: Cross Site Scripting through non-HTTP ports, stealing cookies, etc.Published: February 6th 2002Reference: http://eyeonsecurity.org/advisories/ multple-web-browsers-vulnerable-to-extended-form-attack.htm=== === Subject: : Re: Math dependency logic REVISED > ... Notice how I'll be strongly emphasizing cont terms all the way down. P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > Works the same argument with the polynomial > Q(x) = 6125 x^3 + 6125 x^2 - 6370 x + 1078 > ? If not, what step fails? > It's a different polynomial, and you use an imperfect factorization, > wch goes into a field. Can you define the term imperfect factorisation please? I have *no* idea what you mean. What is the distinguisng feature? (And do not come with undefined terms like defaults to a unit...) I thought the factorisation was based on a1, a2 and a3 being algebraic integers, well, the same holds for my factorisation. You thought wrong. Really Dik Winter, you're trying to test my patience when I've explained the situation. > Where have you explained the difference between how you do your factorisation > and how I do my factorisation? There must be some difference, because one > is perfect, the other is not. What is the difference? Or are you not > able to answer that question? > There are an *infinity* of factorizations for any given polynomial.Yes, I know. > There's a method to how I got my factorization, wle you hacked.What *is* that method? You refuse to tell us. I ask you to tell whatthe difference is, and you only come up with there's a method. Thatis not a reply to my question. > The method I use gives a perfect factorization, wch does not go into > a field. > Wle your hack gives one that does go into a field.What *does* going into a field mean? I have a perfect factorisation thatstays for every value of x witn a ring. It is only when I say thatexactly two of the factors are divisible by 7 that I get in a *ring*(note: not a field) where 7 is a unit. You have failed to show that withyour example with the requirement that exactly two of the factors aredivisible by 7 does *not* make 7 a unit. Can you show that for *all*x there is *no* expression of a_1(x)/7 and a_2(x)/7 that results in 1/7?(Expressions here involve negation, addition and multiplication.) But eventhat is not sufficient. You have to show that there is *no* expressioninvolving all possible values of a_1(x)/7 and a_2(x)/7 such that theresult is 1/7. Until you show that that is not possible I assume thatthe ring you construct contains 1/7. (And, no, not a field, just aring that contains 1/7.) Moreover, you have to identify for eachvalue of x, wch root of the polynomial in a is a_3 (that is not divisibleby 7). You tossed together some polynomial, tried to imitate what I was doing, and wandered off the path into a subtlety. Mathematics is full of subtle points that can be difficult to handle. > Tsk. > It's not my fault that you're hacking in an area that you clearly > don't completely underd, have now gotten lost, and expect me to > work to clear up your misconceptions. > What I can do is remind you that mathematics is full of subtle points > that can be difficult to handle.Yup, but you do not seem to be aware of it. Pray give a *proof* that yourring (wch includes all possible values of a_1(x)/7 and a_2(x)/7 forinteger x) does not contain 1/7. Now I don't really care at ts point that you're hacking your way into an area that you probably find confusing, as I need people to underd the *correct* path first, and then more interesting beasties can be considered. You're clearly out of your depth and doing your best to bug me Dik Winter, and my suggestion to you is to show more maturity. > A good suggestion that could be reversed. I ask you a simple question > and you refuse to answer... > I have answered you, repeatedly.Nope, you did not. You only talk about going in a field wch has nomeaning at all in mathematics. > The mathematical argument that I give is a nice easy path into a > difficult area, and I know it's difficult because of the problems I've > been having getting people to accept my work, especially with certain > posters working to confuse. > Now I have a *very* simple path laid out, and people can *test* each > and every logical link to see that the argument is correct, wch is > what's important.Yes, and I do follow that path exactly and come to a ring that contains1/7. You follow that path, but fail to show that 1/7 is *not* in thering you get. Pray show us that 1/7 is not in your ring.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/=== === Subject: : Re: Math dependency logic REVISEDit's not a popularity contest. on the other hand,can you name a single person who accepts your work?...even Correy seems to have given up the ghost,since he should probably attend to s own ****,what ever s tng about logic is. your continual assertions to people@usenet at best,can only refer to lurkers who are into the drama of it,such as it is, since the math is strictly allover the place; only your would-be helpmeets are apparentin your endless proliferation of threads. or, like I say,the Jimmy Dean Harris Fan Club. or how about,Day Care for Jimmy? at the worst,it's just an experimnet by our handlers/wardens/realityTVwritersof the googolplex. has anyone found their policies,for divvying up the postersp on whatever basis? > The mathematical argument that I give is a nice easy path into a> difficult area, and I know it's difficult because of the problems I've> been having getting people to accept my work, especially with certain> posters working to confuse.Now I have a *very* simple path laid out, and people can *test* each> and every logical link to see that the argument is correct, wch is> what's important.--les ducs d'Enron!=== === Subject: : Re: Math dependency logic REVISED> it's not a popularity contest. on the other hand,> can you name a single person who accepts your work?...> even Correy seems to have given up the ghost,> since he should probably attend to s own ****,> what ever s tng about logic is.> your continual assertions to people@usenet at best,> can only refer to lurkers who are into the drama of it,> such as it is, since the math is strictly all> over the place; only your would-be helpmeets are apparent> in your endless proliferation of threads. or, like I say,> the Jimmy Dean Harris Fan Club. or how about,> Day Care for Jimmy?> at the worst,> it's just an experimnet by our handlers/wardens/realityTVwriters> of the googolplex. has anyone found their policies,> for divvying up the postersp on whatever basis? > The mathematical argument that I give is a nice easy path into a> difficult area, and I know it's difficult because of the problems I've> been having getting people to accept my work, especially with certain> posters working to confuse.Now I have a *very* simple path laid out, and people can *test* each> and every logical link to see that the argument is correct, wch is> what's important.--les ducs d'Enron!A sure sign of the decline of sci.math/sci.logic presidedover by David Ullrich and the Boyz is that these NGs attractloathsome mutants like you. Here one may sll for whatsoever and whosoever one pleases--including that puke, Lyndon LaRouche--as long as onefires off a few at JSH.Welcome to sci.math/sci.logic, where loathsome mutants gyreand gimble in good company.=== === Subject: : Please help explain recurrence relation solution.. Hello AllMy textbook lays out the following example but does not explain the leap atthe point I indicated. Could somebody please explain the missing link?What is the solution of the recuurence relation?a_n = 6a_(n-1) - 9a_(n-2)with initial conditions a_0 = 1 and a_1 = 6ANS:roots of r^2 - 6r + 9 = 0r = 3Solution to recuurence relation is:a_n = alpha_1 3^n + alpha_2 n3^nit follws that:(I DO NOT UNDERD FROM HERE ON...)a_0 = 1 = alpha_1 (WHY ?)a_1 = 6 = alpha_1 * 3 + alpha_2 * 3 (WHY?)Solving gives us:alpha_1 = 1 and alpha_2 = 1 (WHY?)Solution is therefore:a_n = 3^n + n3^n=== === Subject: : Re: Please help explain recurrence relation solution..> Hello All> My textbook lays out the following example but does not explain the leap at> the point I indicated. Could somebody please explain the missing link?What is the solution of the recuurence relation?> a_n = 6a_(n-1) - 9a_(n-2)> with initial conditions a_0 = 1 and a_1 = 6ANS:> roots of r^2 - 6r + 9 = 0> r = 3> Solution to recuurence relation is:> a_n = alpha_1 3^n + alpha_2 n3^nit follws that:> (I DO NOT UNDERD FROM HERE ON...)> a_0 = 1 = alpha_1 (WHY ?)a_0 = alpha_1*3^0 + alpha_2 *0*3^0> a_1 = 6 = alpha_1 * 3 + alpha_2 * 3 (WHY?)a_1 = alpha_1*3^1 + alpha_2*1*3^1> Solving gives us:> alpha_1 = 1 and alpha_2 = 1 (WHY?)Substitute alpha_0 = 1 into 6 = alpha_1*3 + alpha_2* 3> Solution is therefore:> a_n = 3^n + n3^n> -- SperryColumbia, SC (USA)=== === Subject: : Re: Brigitt Angela Blanchard - December 2nd 1971>B L A N C H A R D>2 12 1 14 3 8 1 18 4 = 63> Steve (pobular young feller) was having another party, present and>providing stats was Brigitt, she was the first of 17 people to provide>stats today.<<The following (courtesy of Waxy.org) is sort of an unofficial FAQexplaining the psychotic nonsense posted to Usenet by Shawn DarylKabatoff AKA Dar, AKA Probababbilities. And now AKA marcia andme.WARNING: Read below before even tnking about responding to tstwit.http://www.waxy.org/arcve/2002/05/21/dar_kaba.shtml# 000643Usenet has the tendency to provide a public forum for those who wouldnormally be scribbling in a closet. For example, take Daryl ShawnKabatoff. For the last few years, he's methodically gatheredstatistics from various sources, ranging from local newspaperobituary pages to the food court of the Saskatoon Midtown Plaza mall.With all the raw data he's collected, he's attempting to prove dailythat our full names are in mathematical harmony with our birthdays.s rants normally focus on a single individual he's met or readabout, starting with calculations related to their birthdate and fullnames, blending in whatever other personal information about theirfamily members, spouses, birthplace, and career he's been able tozealotry, and personal torment. I've never seen anytng like it.With all the prime numbers, Fibonacci sequences and biblicalreferences, it's like reading the notebooks of Maximillian Cohen and Nash combined. Unsurprisingly, several posts unfold to reveal astory of painful mental illness. If you have some time, take a look.I've detailed s posting story and a several sample posts below. Usenet Posting story:January 27, 1999 to July 5, 2000 as Catsco@home.comDecember 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.caOct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.caJanuary 20, 2002 to April 17, 2002 as s_kabatoff@hotmail.com (originalposts have been removed from Google Groups arcve)April 26, 2002 to Present as dar_kabatoff@hotmail.comSelected Posts:Tessa Lynne SmithDastageer Sakzai and Helen SmithBrett David MakiAndrew Meredith CottonKathryn Lee ppersonAmanda Dawn NewtonMona Marie EtcheverryTony NusplLisa Charlene McMillanGrant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my currentresidence. and every single place he's mentioned in s posts (mostnotably nervous harold's and the roastary) were either places i'vebeen (as it's a small city of 200K) or hangouts, ie. the two placesmentioned. chances are i could email some friends back home and findout if they know of m, they (my friends that is) being of thebroadway-centred slacker ilk. myself, too, until i got out of there.eh, anyways. thought it odd to see all ts. midtown mall. i ate mymeals there, wlst waiting several days in line for star wars episodeone, at the theatre across the street.posted by andy raad on May 22, 2002 06:20 PMFascinating. It's like he's trying to take chaos and bind it intowhatever rules he can find, religious, logical and otherwise. Numbersand math have a reliable pattern, sometng that can always be provento true or false. People and religion do not. It reminds me of DarrenAronofsky's movie Pi. It's the story of an paraniod genius who istrying to find a pattern in Pi. A group that takes interest in swork is convinced that the existence of Pi, a number whose existencecan be proven but no quantified, is proof of the existence of God.Kabatoff's hunt for patterns in sometng as random as name selectionis a way to reconcile s deeply logical thought process with sconflicting religious views.Exactly. I probably shouldn't have, but I e-mailed Daryl yesterday,asking m if he'd be willing to create a numerological analysis forme. I also asked m if he had seen either Pi or A Beautiful Mind, andwhat he thought of them. If he replies, I'll be sure to post it.I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbedpumpkin pie all over my breasts for m, and my breasts turned orange.I am a pumpkin for Shawn.posted by Trisha Blondie on July 24, 2002 10:41 PMUm, that's swell. So, you're in love with m?Shawn once went to a funeral for a Jehovah Witness that shot mselfand the lemon tarts were very bad, they were not only sour but wererubbery as well. Shawn said that the guy was some kind of JehovahWitness prophet, he saw in advance that the lemon tarts at s funeralwere to be very very bad, and so he shot mself. Shawn said that henever ate pumpkin pie at a funeral but would like to some day. Shawnlikes pumpkin pie and so I have been practicing to make very goodpumpkin pies.posted by Trisha Blondie on July 25, 2002 02:49 PMShawn said that the lemon tarts were sour, bitter and rubbery.I don't tnk ts guy takes notes. I tnk he has Total Recall, andit has driven m insane...Oh... I almost forgot... I didnt spend thousands of dollars a daytormenting Daryl... We got a deal on tormenting that fiscal year, itonly came to about 37cents a day....Mr. Kabatoff attempts to portray mself as a victim, but in fact heis a violent predatory pedople who is well known to s local lawenforcement. In s post to multiple newsgroups with the subjectCollecting Mail For The Coming Anti-Christ, he encourages mothers tosend m photos of their naked daughters. Mr Kabatoff explains, Ipersonally did not want photographs being mailed to (the comingAnt-Christ) that were of underage cldren unless the parent wassigning consent. He is banned from virtually all the shopping mallsin s community because he stalks young people and sexually harassesthem. He has an extensive arrest record wch includes sexualmolestation charges. He's been hospitalized in mental institutionsabout s contact with young girls in many posts. Search newsgrouparcves for posts by m containing the word nubile. As part of sharrassment, he provides personal details in a public forum, such asthe real names of real cldren, in these and other posts. About onewanted her and her sister dead.He not only curses cldren and prays for their death in s posts, healso enjoys attending the funerals of young people: And so, sincenubile sweeties are found in greatest abundance at the funerals ofgh school students, then it is the funerals of gh school studentsthat make the very very best funerals, especially if there is food...I stuff my face (and my pockets) with all the good food and look atall the pretty nubile sweeties and have the time of my life..r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039% 40twister.socal.rr.com&rnum=1 Many of s posts are sent to alt.teens.advice. However, he liberallyspams, floods and crossposts s off-topic threatening and offensivemissives to countless newsgroups. Some people HAVE problems and somefolks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. Whenhe sends these posts to any newgroup, please help by reporting m toI knew of m when I was attending the University of Saskatchewan.He'd hang out in the Arts computer lab and all you'd see is screens ofnumbers racing by on s laptop. I have an original copy of sCollecting Mail for the Coming Anti-Christ pamphlet, and have seenm be hauled away by campus security on more than one occasion. Myfriends and I refer to m as Crazy Number Man.I've been posting to (and about) Shawn for over two years with biggaps in between. He has seen Pi and didn't like it and didn't tnk itresembled m at all. (Wrong, it fits m to a tee) He doesn't havetotal recall and has stated that he travels with a lap top to notateitems. Also, he uses cut n' paste a lot if you read all the waythrough s ramblings. He is anti-social as shown by s angrystatements towards those who, by s own admission, have been kind(but not kind enough) to m. Still, he's intelligent and seems to beable to take a joke on occassion. That's where I came in. ALOHAReply to group(Unsolicited e-mail is deleted from the server unread if it comes from anyone not already in my addressbook. I'll never even see it)