mm-385 === Subject: : Re: Riemann Hypothesis and P vs NPOriginator: If the Riemann Hypothesis is true, then there is a fast way to>discover the primes used to build the RSA codes on which the security>of e-business currently relies.That's a pretty confused description. Here's what he probably meant.There are known algorithms for picking a random prime whose running time has been proven to be polynomial under the assumption that RH is true.(Their running time might or might not be polynomial if RH is false; wedon't know, and we don't have any proofs. There are also other algorithmsfor choosing a random prime whose running time is provably polynomial underno assumptions, i.e., whether or not RH is true; however, these alternativesmay be less attractive in practice.)One of the steps in generating a RSA public/private keypair is choosinga random prime, so this is a helpful fact to know.As far as we know, being able to choose a random prime doesn't help youfactor and doesn't help you crack RSA. As far as we know, the RH haslittle or nothing to do with the security of RSA.>My question is: If the RH is proved to be true does it automatically>follow that P=NP?No.=== === Subject: : Re: Riemann Hypothesis and P vs NP>If the Riemann Hypothesis is true, then there is a fast way to>discover the primes used to build the RSA codes on which the security>of e-business currently relies.That's a pretty confused description. Here's what he probably meant.> There are known algorithms for picking a random prime whose running > time has been proven to be polynomial under the assumption that RH is true.(Their running time might or might not be polynomial if RH is false; we> don't know, and we don't have any proofs. There are also other algorithms> for choosing a random prime whose running time is provably polynomial under> no assumptions, i.e., whether or not RH is true; however, these alternatives> may be less attractive in practice.)One of the steps in generating a RSA public/private keypair is choosing> a random prime, so this is a helpful fact to know.As far as we know, being able to choose a random prime doesn't help you> factor and doesn't help you crack RSA. As far as we know, the RH has> little or nothing to do with the security of RSA.Right. I think others in this thread may have misunderstood the original, confused description. RH gives you a fast way to discover some primes that you can use to build an RSA code; it doesn't give you a fast way to discover the particular primes that someone else has used to build her RSA code. All you really need to build an RSA code is two humongous primes; pretty much any two will do, and while some pairs of primes are more secure than others, there are lots and lots and lots of good pairs of primes out there.-- === === Subject: : Re: New jokes on math> New jokes on math added on www.bymath.com!!!Pretty poor jokes imho. However, there is one that reminds me of thatSimpson's episode where Lisa steals the teachers' answer books: At the parents' meeting a teacher of mathematics complains of his pupils totheir parents:- Your children are very dull. Today I explained them a new theorem. Iexplained once, they didn't understand. I explained once again, they didn't understand. I explained the third time, Iunderstood myself, but they still didn't understand. N. Christoff=== === Subject: : Re: New jokes on math Pretty poor jokes imho. However, there is one that reminds me of that> Simpson's episode where Lisa steals the teachers' answer books:> At the parents' meeting a teacher of mathematics complains of his pupils to> their parents:> - Your children are very dull. Today I explained them a new theorem. I> explained once, they didn't understand. I explained> once again, they didn't understand. I explained the third time, I> understood myself, but they still didn't understand.> N. ChristoffMust be one of the universal jokes: I heard it in Kosice, Slovakia,from my colleague sometimes in 1965.A military counterpart:Colonel: An officer who cannot explain to his subordinates what he means is an idiot. Do you understand?Major: No, sir.Cheers, ZVK(Slavek).=== === Subject: : chernoff bound or sth elseSorry to bother you guys. Here I have a question about a randomizedalgorithm.Backgroud: Random variable X is independently drawn from {1, -1} with equalprobability, i.e., Pr(x=1)=0.5, Pr(x=-1)=0.5, then we have E(X)=0.Or we can write frac{sum_{i=1}^{n} x_i} {n} = 0 when n is largeenough.Here the question is: Can we define a bound for n such that | frac{sum_{i=1}^{n} x_i} {n} - 0 | < epsilon ? That is, we need to find the number of iterations nsuch that the average of x_i is less than some parameter epsilon.I was wondering if chernoof bound can do this job or some othertechniques.=== === Subject: : Re: chernoff bound or sth else>Sorry to bother you guys. Here I have a question about a randomized>algorithm.>Backgroud: >Random variable X is independently drawn from {1, -1} with equal>probability, i.e., Pr(x=1)=0.5, Pr(x=-1)=0.5, then we have E(X)=0.>Or we can write frac{sum_{i=1}^{n} x_i} {n} = 0 when n is large>enough.No we can't, but the limit is 0 almost surely.>Here the question is: >Can we define a bound for n such that | frac{sum_{i=1}^{n} x_i} {n} >- 0 | < epsilon ? That is, we need to find the number of iterations n>such that the average of x_i is less than some parameter epsilon.You can find bounds for n such that the probability of this event isat least 1 - delta (for some given delta > 0). For example,Chebyshev's Inequality will do that. But there's always a nonzeroprobability for the complement.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Order/Chaos Question> I would like to know of a number distribution> generator that can be adjusted to like a rheostat> to produce numbers between order or chaos.The problem is similar to mutating a perfect> sine wave into less and less order, eventually> turning it into random noise.I would like a randomness index built into it> that goes from 0 to 1, where zero is chaos and 1 is> order. All fractions in between can also be input.> Luis Rodriguez answer:The simplest form is to utilize the iteration : X = k X^2 - 1with 1=< k <= 2 . Initial X between 0.51 and .99Index of order = 2-k (If k = 1 then total order, until k = 2that means total chaos)If you want integers of n digits, multiply X by 10^n and then take its integer part.Who that are interested in this generator write me and I will send my program KAOS.EXE that have other generators of Chaos, inclusivea curve that can be turned from order to chaos by changing only the position of a variable. So refuting the Chaitin's theory of the lengthof program as mesure of complexity.=== === Subject: : Re: Proving Hall's Theorem using Tutte's> Can someone tell me how to use tutte's theorem to derive hall's> theorem(marriage theorem).I know Hall's Marriage Theorem. I don't know Tutte's Theorem. What is it? Whatever it is, what you're looking for is very likely to be in this lovely little book: MR0781348 (86j:05001)Reichmeider, Philip F.The equivalence of some combinatorial matching theorems.Polygonal Publ. House, Washington, NJ, 1984. i+123 pp. ISBN 0-936428-09-005-01 (05C70)It is well known that various theorems concerning bipartite matchings and network flows are all equivalent, in the sense that any one may be deduced easily from any other. This expository work painstakingly organises and presents various proofs of these results and of the relationships. It focusses on theorems of K.9anig and Egerv.87ry, Hall, Dilworth, Menger, Ford and Fulkerson, and Hoffman. The book deals only with the finite case, and does not consider matroids or algorithms.Reviewed by Colin J. H. McDiarmid-- === === Subject: : Re: I need a site about number theory .> I want to learn about number theory , If you know a good website about> that please let me know.www.google.comNot a bad suggestion. A liitle more focussed is http://www.numbertheory.org/ntw/web.h-- === === Subject: : Re: I need a site about number theory .> I want to learn about number theory , If you know a good website about> that please let me know.www.google.comNot a bad suggestion. A liitle more focussed is http://www.numbertheory.org/ntw/web.h=-=-=-=-Have you evere gone to www.themathpages.com ??It is a site where you learn precalculus,trigonometry,arithmetic .Thetexts there are so simple and learning there is so good.I need a site such as mathpages.com but about number theory and Idon't have enough time to search by google.=== === Subject: : Re: I need a site about number theory .> I want to learn about number theory , If you know a good websiteabout> that please let me know.> www.google.comNot a bad suggestion. A liitle more focussed ishttp://www.numbertheory.org/ntw/web.h> =-=-=-=-> Have you evere gone to> www.themathpages.com ??> It is a site where you learn precalculus,trigonometry,arithmetic .The> texts there are so simple and learning there is so good.> I need a site such as mathpages.com but about number theory and I> don't have enough time to search by google.You want to learn number theory, but you haven't got time to search withGoogle?=== === Subject: : Re: I need a site about number theory .> I want to learn about number theory , If you know a good website> about> that please let me know.> www.google.com> Not a bad suggestion. A liitle more focussed is> http://www.numbertheory.org/ntw/web.h=-=-=-=-> Have you evere gone to> www.themathpages.com ??> It is a site where you learn precalculus,trigonometry,arithmetic .The> texts there are so simple and learning there is so good.> I need a site such as mathpages.com but about number theory and I> don't have enough time to search by google.You want to learn number theory, but you haven't got time to search with> Google?Why when somebody else has done that before I have to waste my time ?When you search with Google you will finde thousands of site that manyof them are so boring that ....=== === Subject: : Request for comments on antiquated algebraic topology online-bookOriginator: israel@math.ubc.ca (Robert Israel)Algebraic Topologyby Solomon LefschetzNumber of Pages: 389 pp.ISBN: 0-8218-3397-9COLL/27.Ehttp://www.ams.org/online_bks/coll27/--- I was wondering if anyone familiar with algebraic topology (assuming youhave the time and/or inclination) could skim through this text and let meknow what they think of it. I'm particularly interested in knowing of anydifferences in terminology or definitions that may be present in this bookwhen compared to more modern treatments. Of course, I would be grateful forany comments at all that people may have.Note: one need only download the first chapter. In the pdf for ch1, youonly need to click on the subsequent chapters in the table of contents, andacrobat reader (at least v6+) will automatically download that chapter intothe pdf viewer.At first glance it seems quite good (starts from the very bottom), and iseasily an order of magnitude easier to understand than Hatcher's 'AlgebraicTopology' - also a free ebook, but much more modern. N. Christoff=== === Subject: : Re: Request for comments on antiquated algebraic topology online-bookOriginator: israel@math.ubc.ca (Robert Israel)> Algebraic Topologyby Solomon LefschetzNumber of Pages: 389 pp.> ISBN: 0-8218-3397-9> COLL/27.Ehttp://www.ams.org/online_bks/coll27/---I was wondering if anyone familiar with algebraic topology (assuming you> have the time and/or inclination) could skim through this text and let me> know what they think of it. I'm particularly interested in knowing of any> differences in terminology or definitions that may be present in this book> when compared to more modern treatments. Of course, I would be grateful for> any comments at all that people may have.Note: one need only download the first chapter. In the pdf for ch1, you> only need to click on the subsequent chapters in the table of contents, and> acrobat reader (at least v6+) will automatically download that chapter into> the pdf viewer.At first glance it seems quite good (starts from the very bottom), and is> easily an order of magnitude easier to understand than Hatcher's 'Algebraic> Topology' - also a free ebook, but much more modern.> N. ChristoffLefschetz was a great pioneer in this subject and this book is aclassic. Your title question uses the word antiquated and that truelyapplies to this book. As a text to learn the basic ideas it has longbeen superceded. On the other hand you can learn how to think about alot of stuff from this book. A major caveat: Lefschetz was (is)notorious for incorrect proofs. A humorous remark about him which isonly a slight exaggeration states that his papers never contained afalse theorem and never contained a correct, rigorous proof. All ofthis suggests that this would not be a good first book to learn thesubject.=== === Subject: : Re: Request for comments on antiquated algebraic topology online-bookOriginator: israel@math.ubc.ca (Robert Israel)> Algebraic Topologyby Solomon Lefschetzhttp://www.ams.org/online_bks/coll27/---I was wondering if anyone familiar with algebraic topology (assuming you> have the time and/or inclination) could skim through this text and let me> know what they think of it. I'm particularly interested in knowing of any> differences in terminology or definitions that may be present in this book> when compared to more modern treatments. Of course, I would be grateful for> any comments at all that people may have.At first glance it seems quite good (starts from the very bottom), and is> easily an order of magnitude easier to understand than Hatcher's 'Algebraic> Topology' - also a free ebook, but much more modern.I only had time to take a quick glance at it, but it looks pretty outof date. I think that at the time this book was written, not all ofthe material that constitutes a standard first-year algebraic topologycourse today had been worked out, and much of that which had beenworked out was still being polished. The book emphasizes some topicswhich are not emphasized so much anymore, and uses some obsoleteterminology, or modern terminology in slightly different ways. Forexample, if you look at the definition of complex at the beginningof Chapter 3, it is (almost) equivalent to the modern definition of achain complex (over the integers), but looks completely different. Ididn't see singular homology theory in there (but I didn't have timeto download and scan the whole book) while this is an essential partof most modern treatments. Exact sequences, which simplify and unifyso many formulas in algebraic topology, are apparently not in the book(and maybe hadn't been invented yet). One could probably say muchmore about how this book differs from modern approaches but I don'twant to do a historical study. Suffice it to say that when flippingthrough it, not too much of what I saw was immediately recognizable.Anyway, while reading old-fashioned texts won't do you any harm, Ithink that for learning well-established material, in general it isbetter to get a modern introduction first, because things getsimplified and improved so much over time (and of course you need toknow the modern terminology and approaches if you want to understandcurrent developments and if you don't want people to look at you funnywhen you talk to them). After you have done that, then it can be funand enlightening to look at older literature to get some idea of howpeople originally thought of things (because some intuition can getlost in the polishing process).For an introduction to algebraic topology, I would look for somethingwritten after 1960 or so. Hatcher's book is great, but quitesophisticated. The book by Greenberg and Harper gives a prettyconcise introduction. I also like the book by Bredon a lot, if youcan find a copy.=== === Subject: : Re: Request for comments on antiquated algebraic topology online-book Originator: israel@math.ubc.ca (Robert Israel)> For an introduction to algebraic topology, I would look for something> written after 1960 or so. Hatcher's book is great, but quite> sophisticated. The book by Greenberg and Harper gives a pretty> concise introduction. I also like the book by Bredon a lot, if you> can find a copy.I also found J. Rotman's An Introduction to Algebraic Topology,(Springer Verlag, 1988) pretty good; as readable as it gets, IMHO. === === Subject: : Re: Request for comments on antiquated algebraic topology online-bookOriginator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)Originator: israel@math.ubc.ca (Robert Israel)>For an introduction to algebraic topology, I would look for something>written after 1960 or so. Hatcher's book is great, but quite>sophisticated. The book by Greenberg and Harper gives a pretty>concise introduction. I also like the book by Bredon a lot, if you>can find a copy.One advantage of Greenberg and Harper is that they introduce homotopygroups early on, which many elementary texts don't. Apart from that,though, I think it's too terse to serve as the primary text for self-study.I rather like Munkres's Elements of Algebraic Topology, partly becauseit spends quite a lot of time on simplicial homology. Some people mightdislike it for the same reason, feeling that singular homology is morerelevant for the budding topologist and that simplicial homology is justa distraction. But I think that simplicial homology is easier for thebeginner to grasp, and it is still an important tool today in (forexample) combinatorics.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences=== === Subject: : Re: JSH: End of argument? Decker example revisited> It turns out that I can prove my case using the pattern argument quite> simply.First assume(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2with y_1(x) and y_2(x) algebraic integer functions, that is, when x is> an algebraic integer they result in algebraic integers.That's an unjustified assumption, but let's continue... > Now consider z_2(x), where z_2(x) = y_2(x) - 1, so it's also an> algebraic integer function.Then y_2(x) = z_2(x) + 1, and substituting gives(5y_1(x) + 1)(5z_2(x) + 5 + 2) = 25x^2 + 30x + 2, which is(5y_1(x) + 1)(5z_2(x) + 7) = 25x^2 + 30x + 2.Now multiply both sides by 7, multiplying through the first factor on> the left by 7 to get(5(7)y_1(x) + 7)(5z_2(x) + 7) = 7(25x^2 + 30x + 2).Now let a_1(x) = 7y_1(x), and a_2(x) = z_2(x), substituting gives(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)which is the Decker example (see below for reference information).Here solving for the a's requires noticing that you can re-group on> the right side to get(5a_1(x) + 7)(5a_2(x) + 7) = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2(from [1] in Decker's post, reference at bottom).Now finishing up is easy as the a's are roots ofa^2 - (x - 1)a + 7(x^2 + x).This is a correct derivation. Now let's work backwards to your y's:y_1(x) = a_1(x) / 7, z_2(x) = a_2(x)y_1(x) = a_1(x) / 7, y_2(x) = 1 + a_2(x)Now it is certainly true that y_2(x) is an algebraic integerfor every rational integer x, but what reason do you havefor asserting that y_1(x) is an algebraic integer? In fact,y_1(x) will in most cases *not* be an algebraic integer.All one has to do is consider the case x = 1, where thea's are +/- sqrt(-14). Your y_1(1) is then +/- sqrt(-14)/7which is certainly not an algebraic integer. The fact isthat except in the two cases x = 0 and x = -1, neitherof a_1 and a_2 will be divisible by 7. It proves that mathematicians need a bigger category as the ring of> algebraic integers is too small to include ALL the values of y_1(x),> even when x is only integers.Err, no. What it does mean is that your choice of factorizationis not a factorization in the algebraic integers. Simply put,your proposed factorization at the very top of this post is wrong. === === Subject: : permutation involving even and oddI'm having a problem understanding even and odd with permutations. The proofsmake no sense because the concept itself makes little sense to me.One Lemma says that if g=S1 S2 ******Sr where the S's are 2 cycles then r iseven. Another theorem is Always even or always odd - If a permutation g can beexpressed as a product of an even number of 2 cycles then every decompositionof g into a product of 2 cycles must hae an even number of 2 cycles. Insymbols, if g=S1 S2********Sr and g=T1 T2********Ts, where the S's and the T'sare 2-cycles then r and s are both even or both odd.=== === Subject: : Re: permutation involving even and odd Adjunct Assistant Professor at the University of Montana.>I'm having a problem understanding even and odd with permutations. The proofs>make no sense because the concept itself makes little sense to me.>One Lemma says that if g=S1 S2 ******Sr where the S's are 2 cycles then r is>even. I'll bet it says that IF g is an even permutation, THEN anydecomposition into 2-cylces will have an even number of 2-cycles; notthat any permutation written as a product of 2-cycles will necessarilybe written as a product of an even number of cycles, which is what you>Another theorem is Always even or always odd - If a permutation g can be>expressed as a product of an even number of 2 cycles then every decomposition>of g into a product of 2 cycles must hae an even number of 2 cycles. In>symbols, if g=S1 S2********Sr and g=T1 T2********Ts, where the S's and the T's>are 2-cycles then r and s are both even or both odd.Take any permutation, say (1,3,4,2,6). There are many ways of writingit as a product of 2-cycles. For example, one such way (I am composingright to left) is(1,6)(1,2)(1,4)(1,3)Another, different way, is to write it as(3,1)(3,6)(3,2)(3,4); or (4,3)(4,1)(4,6)(4,2). or (2,3)(1,6)(1,3)(1,4)(1,2)(2,3); or many other ways. But no matter which way you do, the number of 2-cycles will always beeven: because if ONE way of writing it uses an even number of2-cycles, then ALL ways of writing it will use an even number of2-cycles. If ONE way uses an odd number of cycles, then ALL ways willuse an odd number of cycles.-- ============================================================== ============ === Subject: : Re: Sequence A001511>>If anyone had looked up the sequence 1, 2, 1, 3, ... , 5 in the>>On-line Encyclopedia of Integer Sequences, they would have found it>>listed as sequence A001511.>> So what?!?>I believe there was originally some context to his comment ('What is the>next term in this sequence' or some such) but this has been snipped and===>the === Subject: : header changed.>That context, alas, has now been lost, as apparently neither you nor I>have the time or inclination to work out which particular thread the post>belongs to.To be fair, I *strongly* suspected that the message was in referenceto some other post/thread. Maybe I've been both too rude and crypticto the OP who in turn should better read some posting guideline orIt would have been enough to give some context either by explicitquoting or by means of a brief abstract *and* to include somethinglike was(:...) in the === Subject: line.Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc=== === Subject: : Re: Concentric-Circles Game (to help teach basic number theory)A few notes below within copied original message: > I made this simple game up today, in part so as to help teach kids> basic number-theory concepts (such a congruences and divisibility),> but it might be far more useful as useless entertainment...if even> that...Start with m concentric circles drawn on paper,> where n is from 5 to 12 (or higher perhaps).Should be:where *m* is from 5 to 12 (or higher perhaps). > The rings between the circles are subdivided so that the inner circle> is subdivided into 2 equal-area sections, the next ring out is> subdivided into 3 equal sections, the next ring into 4, etc..until the> outer ring is subdivided into (m+1) equal sections.> (And the sections are aligned so that one boundry of each is along a> single line.)By ring, I of course mean a GEOMETRIC ring, ie. an annulus. > Players start by each placing a game-piece (small enough to fit in the> smallest section) into any unoccupied section on the board.A spinner or dice are used to pick a random integer n, where each n> should be between 1 and somewhere near m.> On each move, a player can place a new game-piece (leaving earlier> pieces) n rings inward (if possible) from their last game-piece's> position, or n rings outwards (if possible).> (Outward and inward are along a chain of sections which touch. Do> not pass center circle or outer circle.)> Okay, there is a possible problem with the above situation.For example:If you have placed your last piece on the 3rd ring,which is subdivided into 4 sections,and (on your next move) the random number n is 4, and m is 6,then you have *automatically* lost...I have a rule-change which keeps in the spirit of the game,and should make it more fun/educational as well:If you choose to move outward/inward,you may move a number (k) of rings outward/inward,where k is any *divisor* of n.(So, if n = 4, for example, you can move 1, 2, or 4{if possible} rings outward or inward.)So, as for a move, a player can choose, after n is picked, to either:Move counterclockwise or clockwise any *multiple* of n segments;ORMove outwardly or inwardly any *divisor* of n rings. > OR the player can place a game-piece clockwise or counterclockwise (in> the same ring)> a *multiple* of n spaces from their last-placed piece.But in any case, a player can't change directions during his/her move.> (And the player's {counter}clockwise moves may indeed circle around a> single ring as many times as needed.)A player can jump over occupied spaces when measuring the number of> spaces from their last piece, but they must ONLY place a piece on an> unoccupied section.And the winner is the last player able to place a piece on the board.Leroy QuetFinally, I will give an example of where a player can move,using this rectified gameboard:(Use fixed-width font.Imagine board curved, and left and right sides joined so game-board is circle.)-----------------------------------! * ! * ! o ! @ ! @ ! * ! o ! !------------------------------------! ! * ! @ ! @ ! ! ! !------------------------------------! @ ! ! x ! * ! @ ! o !------------------------------------! * ! @ ! @ ! o ! !------------------------------------! @ ! @ ! * ! !------------------------------------! * ! ! o !------------------------------------! @ ! o !....................................Game is player-* vs player-o (or sometimes x, sometimes @).It is o's move. His last piece is the 'x'.He has just rolled a 4 .So, I have denoted the sections he can put his next piece at by '@'s.Leroy Quet=== === Subject: : Re: Point-set Topology Hints Wanted . Adjunct Assistant Professor at the University of Montana.>Well , great now you have jumped ahead of me in the book.Just be sure to note that my initial answer used the wrong definitionof A', so the answer was false. That A' U A'' = A' is true under someassumptions on the topology, but false in general.The stuff about T0 and T1 are ways to codify those extraassumptions. They are called separation axioms, and intuitively youcan think of them as telling you how much the topology distinguishesbetween points. A space is (at least) T0 if and only if given any twodistinct points x and y, there is EITHER an open set U which containsx and does not contain y; OR ELSE there is an open set V whichcontains y and does not contain x. A space is (at least) T1 if andonly if given two distinct points x and y, there is BOTH an open set Uwhich contains x and does not contain y; AND an open set V whichcontains y and does not contain x. If you have topologywhere every open set that contains x also contains y and vice-versa,then topologically, you will not be able to tell the differencebetween x and y... The 'separation axioms' have a number of standardconsequences; for example, if your is T1, then every singleton {x} hasto be a close set (for each point y different from x, consider an openset U_y such that y is in U_y but x is not; then let C_y = X-U_y,which is closed, and take the intersection of all C_y as y ranges overall points in X other than x...)-- ============================================================== ============ === Subject: : Puzzle/Problem Based Upon Game(this post: a math problem/puzzle deserving of its own thread,hopefully..)Regarding my game at:http://mathforum.org/discuss/sci.math/m/578094/578398(Above link is post:Re: Concentric-Circles Game (to help teach basic number theory))What if this was a one-person game, the goal simply being to fill inall of the sections;and n simply went, in order, 1, 2, 3, 4,...,(m+1)(m+2)/2 -1.(And the first piece could be placed in any section we desire.)For which m would it then be possible so we could somehow, with properplay, place a game-piece on each of the(m+1)(m+2)-1 sections, following the game's rules (except we have oneplayer only, and n is not random) ??(I have not thought about this much, so I do not know if this questionis trivial.)Perhaps I should ask too, what if the center circle had 1 section, andnot 2 sections, and the outer circle had m, not (m+1) sections?(ie. the number of sections was instead 1, 2, 3, 4, ...,m.)I repost my reply (with the game's corrected rules) below:>A few notes below within copied original message: >> I made this simple game up today, in part so as to help teach kids>> basic number-theory concepts (such a congruences and divisibility),>> but it might be far more useful as useless entertainment...if even>> that...>>> Start with m concentric circles drawn on paper,>> where n is from 5 to 12 (or higher perhaps).>Should be:>where *m* is from 5 to 12 (or higher perhaps).>> The rings between the circles are subdivided so that the innercircle>> is subdivided into 2 equal-area sections, the next ring out is>> subdivided into 3 equal sections, the next ring into 4, etc..untilthe>> outer ring is subdivided into (m+1) equal sections.>> (And the sections are aligned so that one boundry of each is alonga>> single line.)>By ring, I of course mean a GEOMETRIC ring, ie. an annulus.>> Players start by each placing a game-piece (small enough to fit inthe>> smallest section) into any unoccupied section on the board.>>> A spinner or dice are used to pick a random integer n, where each n>> should be between 1 and somewhere near m.>>> On each move, a player can place a new game-piece (leaving earlier>> pieces) n rings inward (if possible) from their last game-piece's>> position, or n rings outwards (if possible).>> (Outward and inward are along a chain of sections which touch.Do>> not pass center circle or outer circle.)>>Okay, there is a possible problem with the above situation.>For example:>If you have placed your last piece on the 3rd ring,>which is subdivided into 4 sections,>and (on your next move) the random number n is 4, and m is 6,>then you have *automatically* lost...>I have a rule-change which keeps in the spirit of the game,>and should make it more fun/educational as well:>If you choose to move outward/inward,>you may move a number (k) of rings outward/inward,>where k is any *divisor* of n.>(So, if n = 4, for example, you can move 1, 2, or 4>{if possible} rings outward or inward.)>So, as for a move, a player can choose, after n is picked, to either:>Move counterclockwise or clockwise any *multiple* of n segments;>OR>Move outwardly or inwardly any *divisor* of n rings.>> OR the player can place a game-piece clockwise or counterclockwise(in>> the same ring)>> a *multiple* of n spaces from their last-placed piece.>>> But in any case, a player can't change directions during his/hermove.>> (And the player's {counter}clockwise moves may indeed circle arounda>> single ring as many times as needed.)>>> A player can jump over occupied spaces when measuring the numberof>> spaces from their last piece, but they must ONLY place a piece onan>> unoccupied section.>>> And the winner is the last player able to place a piece on theboard.>>> Leroy Quet>Finally, I will give an example of where a player can move,>using this rectified gameboard:>(Use fixed-width font.>Imagine board curved, and left and right sides joined so game-boardis>circle.)>----------------------------------->! * ! * ! o ! @ ! @ ! * ! o ! !>------------------------------------>! ! * ! @ ! @ ! ! ! !>------------------------------------>! @ ! ! x ! * ! @ ! o !>------------------------------------>! * ! @ ! @ ! o ! !>------------------------------------>! @ ! @ ! * ! !>------------------------------------>! * ! ! o !>------------------------------------>! @ ! o !>....................................>Game is player-* vs player-o (or sometimes x, sometimes @).>It is o's move. His last piece is the 'x'.>He has just rolled a 4 .>So, I have denoted the sections he can put his next piece at by '@'s.Leroy Quet=== === Subject: : Re: existence of bounded linear functional>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars,>x_1,...,x_n some elements of X. How do you show that>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>for some M for each n and for scalars b_1,...,b_n implies that there exists>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M.>i don't know how to show this without explicitly constructing the bounded>linear functional on X, and of course i'm stuck on constructing it (hence>the problem in the first place)..>if X is finite dimensional, then each x in X can be written x = b_1 x_1 +>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + ... + b_n>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given constraint.Careful: the x_n's were given, and might not be a basis. >But what to do if X is infinite-dimensional???Use the Hahn-Banach Theorem.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: existence of bounded linear functionalAh, of course. I still have one lingering question, though. Given {x_k} asequence in X and {a_k} a sequence of scalars, define a bounded linearfunctional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... +b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k.How does the requirement that| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do youassert equality?>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars,>x_1,...,x_n some elements of X. How do you show that>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>for some M for each n and for scalars b_1,...,b_n implies that thereexists>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M.>i don't know how to show this without explicitly constructing the bounded>linear functional on X, and of course i'm stuck on constructing it (hence>the problem in the first place)..>if X is finite dimensional, then each x in X can be written x = b_1 x_1 +>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + ... +b_n>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the givenconstraint.> Careful: the x_n's were given, and might not be a basis.>But what to do if X is infinite-dimensional???> Use the Hahn-Banach Theorem.> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: existence of bounded linear functional>Ah, of course. I still have one lingering question, though. Given {x_k} a>sequence in X and {a_k} a sequence of scalars, define a bounded linear>functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... +>b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k.>How does the requirement that>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you>assert equality?Oh. It doesn't, of course, and I don't. If x_1,...,x_n alreadyspan X, you're clearly out of luck, and your statement is false.Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don'tspan X (e.g. if X is infinite-dimensional), it also gives you y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t, x' + t y' will be what you want.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 >>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars,>>x_1,...,x_n some elements of X. How do you show that>>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>>for some M for each n and for scalars b_1,...,b_n implies that there>exists>>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M.>>i don't know how to show this without explicitly constructing the bounded>>linear functional on X, and of course i'm stuck on constructing it (hence>>the problem in the first place)..>>if X is finite dimensional, then each x in X can be written x = b_1 x_1 +>>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + ... +>b_n>>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given>constraint.>> Careful: the x_n's were given, and might not be a basis.>>But what to do if X is infinite-dimensional???>> Use the Hahn-Banach Theorem.=== === Subject: : Re: existence of bounded linear functionalHmmm...another sticking point (apart from this norm stuff). Suppose| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||for some M for each n and for scalars b_1,...,b_n. We want to construct afunctional x' in X' such that x'(x_j)=a_j for ALL j = 1,2,....Consider the subspace spanned by ALL the x_k, and call it S. Then any x in Scan be writtenx = b_1x_1 + b_2x_2 + ....Define f(x) = b_1a_1 + b_2a_2 + ....Then f(x_j)=a_j for all j = 1,2...But is f bounded? The given inequality only works for finite n, and|f(x)| = |b_1a_1 + ... | = |lim b_1a_1 + ... + b_na_n| = lim |b_1a_1 + ...+ b_na_n| < = lim M ||b_1x_1 + ... b_nx_n||, butlim ||b_1x_1 + ... + b_nx_n|| = ||lim b_1x_1 + ... b_nx_n||, so how do youshow f is bounded???>Ah, of course. I still have one lingering question, though. Given {x_k} a>sequence in X and {a_k} a sequence of scalars, define a bounded linear>functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... +>b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k.>How does the requirement that>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you>assert equality?> Oh. It doesn't, of course, and I don't. If x_1,...,x_n already> span X, you're clearly out of luck, and your statement is false.> Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don't> span X (e.g. if X is infinite-dimensional), it also gives you> y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t,> x' + t y' will be what you want.> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2>>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars,>>x_1,...,x_n some elements of X. How do you show that>>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>>for some M for each n and for scalars b_1,...,b_n implies that there>exists>>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M.>>i don't know how to show this without explicitly constructing thebounded>>linear functional on X, and of course i'm stuck on constructing it(hence>>the problem in the first place)..>>if X is finite dimensional, then each x in X can be written x = b_1x_1 +>>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + ...+>b_n>>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given>constraint.>> Careful: the x_n's were given, and might not be a basis.>>But what to do if X is infinite-dimensional???>> Use the Hahn-Banach Theorem.=== === Subject: : Re: existence of bounded linear functionalDuh, the norm is continuous. I'm officially an idiot.> Hmmm...another sticking point (apart from this norm stuff). Suppose> | b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||> for some M for each n and for scalars b_1,...,b_n. We want to construct a> functional x' in X' such that x'(x_j)=a_j for ALL j = 1,2,....> Consider the subspace spanned by ALL the x_k, and call it S. Then any x inS> can be written> x = b_1x_1 + b_2x_2 + ....> Define f(x) = b_1a_1 + b_2a_2 + ....> Then f(x_j)=a_j for all j = 1,2...> But is f bounded? The given inequality only works for finite n, and> |f(x)| = |b_1a_1 + ... | = |lim b_1a_1 + ... + b_na_n| = lim |b_1a_1 +...> + b_na_n| < = lim M ||b_1x_1 + ... b_nx_n||, but> lim ||b_1x_1 + ... + b_nx_n|| = ||lim b_1x_1 + ... b_nx_n||, so how doyou> show f is bounded???>Ah, of course. I still have one lingering question, though. Given {x_k}a>sequence in X and {a_k} a sequence of scalars, define a bounded linear>functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... +>b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k.>How does the requirement that>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you>assert equality?Oh. It doesn't, of course, and I don't. If x_1,...,x_n already> span X, you're clearly out of luck, and your statement is false.> Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don't> span X (e.g. if X is infinite-dimensional), it also gives you> y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t,> x' + t y' will be what you want.> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars,>>x_1,...,x_n some elements of X. How do you show that>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>for some M for each n and for scalars b_1,...,b_n implies that there>exists>>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M.>i don't know how to show this without explicitly constructing the> bounded>>linear functional on X, and of course i'm stuck on constructing it> (hence>>the problem in the first place)..>if X is finite dimensional, then each x in X can be written x = b_1> x_1 +>>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) +...> +>b_n>>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given>constraint.> Careful: the x_n's were given, and might not be a basis.>But what to do if X is infinite-dimensional???> Use the Hahn-Banach Theorem.=== === Subject: : Re: existence of bounded linear functionalO.K., so you've got f(x)=b_1*a_1 + ... for x = b_1*x_1 + ... in thesubspace generated by all of the x_i.Then |f(x)| <= M ||x|| for all x in the subspace.Before we even bother extending this functional to the rest of X, howdo we know ||f|| = M? The OP must have misstated his problem... (maybehe means ||x|| <= M )> Duh, the norm is continuous. I'm officially an idiot.> Hmmm...another sticking point (apart from this norm stuff). Suppose| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||for some M for each n and for scalars b_1,...,b_n. We want to construct a> functional x' in X' such that x'(x_j)=a_j for ALL j = 1,2,....Consider the subspace spanned by ALL the x_k, and call it S. Then any x in> S> can be writtenx = b_1x_1 + b_2x_2 + ....Define f(x) = b_1a_1 + b_2a_2 + ....Then f(x_j)=a_j for all j = 1,2...But is f bounded? The given inequality only works for finite n, and|f(x)| = |b_1a_1 + ... | = |lim b_1a_1 + ... + b_na_n| = lim |b_1a_1 +> ...> + b_na_n| < = lim M ||b_1x_1 + ... b_nx_n||, butlim ||b_1x_1 + ... + b_nx_n|| = ||lim b_1x_1 + ... b_nx_n||, so how do> you> show f is bounded???Ah, of course. I still have one lingering question, though. Given {x_k}> a>sequence in X and {a_k} a sequence of scalars, define a bounded linear>functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... +>b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k.>How does the requirement that>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you>assert equality?> Oh. It doesn't, of course, and I don't. If x_1,...,x_n already> span X, you're clearly out of luck, and your statement is false.> Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don't> span X (e.g. if X is infinite-dimensional), it also gives you> y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t,> x' + t y' will be what you want.> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2>>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars,>>x_1,...,x_n some elements of X. How do you show that>>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>>for some M for each n and for scalars b_1,...,b_n implies that there> exists>>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M.>>i don't know how to show this without explicitly constructing the> bounded>>linear functional on X, and of course i'm stuck on constructing it> (hence>>the problem in the first place)..>if X is finite dimensional, then each x in X can be written x = b_1> x_1 +>>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) +> ...> +> b_n>>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given> constraint.> Careful: the x_n's were given, and might not be a basis.>But what to do if X is infinite-dimensional???> Use the Hahn-Banach Theorem.>=== === Subject: : Mutli-parameter border searchI have a simulation that takes in quite a few parameters (say around10). The data output can be simplified to just one point on atwo-dimensional chart. All I am interested in is the border of thatplot (i.e. the outlying points), not the points in the middle. Therelationship between the input parameters and the output points is notstraightforward, though not random. What is the best way to movearound the border of the plot when running the simulation (This willsave much simulation time, as I'd avoid running most simulations thatgenerate points around the center of the plot)?=== === === Subject: : Re: Looking for primes of a particular form.> I am trying to find primes numbers of the following forms: p = 6*(10^N)-1, q = 6*(10^N)+1Sieve with Jobling's NewPGen, and test with OpenPFGW.NewPGen uses a discrete logarithm for such fixed k forms, which is far swifter than simple trial-division. > I thought I heard once of a fast primality test that requires> one to know the prime factors of n-1. Well in my case I> have the luxury of knowing that the only prime factors > of q-1 are 2,3, and 5. But I do not know how implement the> test in my computer code.http://primepages.org/The section on Proving.But why reinvent the wheel, OpenPFGW already performs the Brillhart-Lehmer-Selfridge test.> Is there any theoretical proof, that I may have missed,> that tells me whether or not the primes I am looking for> even exist? Also, a separate question is whether any twin> primes exist in these forms for N>3? (E.g. both p and q> are prime for the same value of N.) Is there be a> proof of the (non)existence of such twin primes?Use the prime number theorem. Looking purely at the probabilities, there are almost certainly only a finite number of such twins. > Summary of my questions:1. I want to know how to program the fast primality> test that relies on knowing the factors of n-1.If you really want to be hands-on then just download Pari-GP.It's got all the primitives you need to implement Proth's Theorem, but also has the algorithm pre-implemented if you would rather be lazy.> 2. I want to know if any theoretic considerations> can be brought forth regarding number of primes> that even exist for the two forms I gave. Same> for twin primes.Arbitrary twin primes have a very different density fromtwins along a fixed-multiplier/base, variable-exponentset.I've never seen anything that deviates by any significant amount from heuristics based on the Prime Number Theorem, with constant factor modifications from Mertens' Theorem and approprate Hardy/Littlewood (Shanks, etc.) constants.> P.S. What primality tests are known that are optimized> for numbers of the form a*(b^n) +/- 1?Tests - Proth's, Pocklington's and Lucas' (Lehmer's) for a start.Actual implementations - OpenPFGW and Yves Gallot's Proth.Phil-- Unpatched IE vulnerability: WebFolder data InjectionDescription: Injecting arbitrary data in the My Computer zoneReference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0305/13.h=== === Subject: : Re: Question for logarithm expertsThis is an interesting posting. I've never really thought about this before,but do all equations have solutions, (even though it may not be possible forus to find them in terms of simple functions) ?Graphing does not give us the whole story does it? What I am thinking aboutis that for exampley = x^2 does not cross y = - 1, yet we know that x^2 = - 1 has solutionsx = + or - i, (the sqrt of minus one).Does the topic you are studying suggest the use of complex numbers?If you replace x with z = a+b i, you could write7^z = e^((ln7)z) = (e^(aln7)) (e^i(bln7))= (7^a)[cos(bln7) + i sin(bln7) ] =4a + 4bi(7^a)cos(bln7) = 4a(7^a)sin(bln7) = 4b16(a^2 +b^2) = 7^2ab = sqrt[((7^a)/4)^2 -a^2 ], but from above we haveb = [arccos(4a/7^a)]/ln7By graphing these like y equal to two different functions of x, I gota = 0.41756456 and b = 0.37824384and so with z = a+bi, 7^z = 4z does have a solution of sorts.I hope you found this a helpful - Ian Hutcheson=== === Subject: : Equation of Cyclotomic PolynomialCan someone please tell me how to find the equation of the 5thcyclotomic polynomial??=== === Subject: : Re: Equation of Cyclotomic PolynomialREPEATED WITH SYMBOLS CLARIFIEDThe answer to your Q is below but I hope you want to know why.Phi5 (x)=(x-e^i(2 pi/5))(x-e^i(4 pi/5))(x-e^i(6 pi/5)) (x-e^i(8 pi/5)) = x^4+x^3 +x^2 +x +1THE PRIMITIVE ROOTS OF UNITY LEAD TO CPsSolutions to x^n - 1 = 0, known as the n-th roots of unity, aree^i(2kpi/n) for k=1 to n, (equivalent to k=0 to n-1)If we only select the roots where k and n are co-prime, (have no commonfactors), known as the primitive n-th roots of unity, these may be used torecompose a special type of polynomial, which has the primitive n-th rootsas its solution.So the primitive roots of unity are e^(2pi)ik/n where k and n have no commonfactors and there will be Phi (n) of them, where Phi (n) is defined to bethe number of positive integers less than or equal to n and coprime to n.For example,Phi (3) = 2, since the two numbers 1, and 2, (but not 3), have no commonfactors with 3Phi (5) = 4 since the two numbers 1,2,3 and 4 are coprime to 5.Phi (8) = 4 since the four numbers 1, 3, 5 and 7 are coprime to 8.Thus the primitive n-th roots of unity can be used to define and generatethe n-th cyclotomic polynomials, Phi n (x) , by stating that they are thezeros of this polynomial.Examples will make this clearer.Phi3 (x) = (x-e^i(2 pi/3))(x-e^i(4 pi/3)) = x^2 +x +1 andPhi5 (x) = (x-e^i(2 pi/5)) (x-e^i(4 pi/5))(x-e^i(6 pi/5)) (x-e^i(8pi/5))=x^4 +x^3 +x^2 +x +1Considering a symmetrical example, the 8th roots of unity aree^i(2kpi/8) = e^i(kpi/4), for k = 1 to 8 so, the primitive 8th roots ofunity aree^i(kpi/4) for k = 1, 3, 5 and 7, but due to symmetry,e^i(7pi/4) = e^-i(pi/4) , (etc), so we may pair off complex conjugates towrite the factors asPhi8 (x) =(x - e^i(pi/4))(x - e^-i(pi/4))(x - e^i(3pi/4))(x - e^-i(3pi/4))= (x^2 - (sqrt2)x +1) (x^2 + (sqrt2)x +1) = x^4 + 1CYCLOTOMIC POLYNOMIALS - SHORTCUTS FOR FINDING THEMa) Whenever n is an odd prime, p, the CP becomes(x^p-1)/(x-1) which from geometric progressions we recognise as a simple sumofdescending powers of x, starting with x^(p-1) e.g. Phi5 (x) = x^4 +x^3 +x^2+x +1I give no more examples of these here as the pattern is clear.b) Whenever n is a composite n = rs with r and s different, and assuming youhave a technical calculator, that has 'expand' as one of its algebraicfunctions, you can use the fact that,Phi rs(x) = (x-1)(x ^(rs) - 1)/[(x ^(r) - 1)(x^(s) - 1)]e.g. Phi 6(x) = (x-1)(x ^(6) - 1)/[(x ^(2) - 1)( x^(3)- 1)]Here I list a few examples of rs CPsPhi 6(x) = x^2 - x +1Phi 10(x) = x^4 - x^3 + x^2 - x +1Phi 14(x) = x^6 - x^5+ x^4 - x^3 + x^2 - x +1Phi 15(x) = x^8 - x^7 + x^5 - x^4 +x^3 - x +1Phi 21(x) = x^12 - x^11 +x^9 - x^8 + x^6 - x^4 + x^3 - x +1Note that with composite n = rs, certain powers can be missing and signsalternate.c) Whenever n is a square composite n = r^2, again using 'expand', you canuse the fact that,Phi r^2(x) = (x^(r^2)-1)/(x^(r) - 1) here are the first few examplesPhi 4(x) = (x ^(4) - 1)/(x ^(2) - 1) = x^2 + 1Phi 9(x) = (x ^(9) - 1)/(x ^(3) - 1) = x^6 + x^3 + 1Phi 16(x) = (x ^(16) - 1)/(x ^(4) - 1) = x^12 + x^8 + x^4 +1Phi 25(x) = (x ^(25) - 1)/(x ^(5) - 1) = x^20 + x^15 + x^10 + x^5 +1Here there are r positive terms, starting x^(r(r-1)) with powers descendingin steps of r.I hope you find this interesting and useful - Ian Hutcheson=== === Subject: : Re: Equation of Cyclotomic Polynomial boundary=----=_NextPart_000_0039_01C3EFDE.49659C80------------ --------------------------------------------------------- charset=utf-8The answer to your Q is below but I hope you want to know why.(5) = (x-e^i(2 pi/5)) (x-e^i(4 pi/5)) (x-e^i(6 pi/5)) (x-e^i(8 pi/5)) = x^4+x^3 +x^2 +x +1THE PRIMITIVE ROOTS OF UNITY LEAD TO CPsSolutions to x^n - 1 = 0, known as the n-th roots of unity, aree^i(2kp/n) for k=1 to n, (equivalent to k=0 to n-1)The primitive roots of unity are e^(2pi)ik/n where k and n coprime, (nocommon factors). Therefore, there are (n) different primitive n-th roots ofunity, where (n) is defined to be the number of positive integers less thanor equal to n and coprime to n. For example,(3) = 2, since the two numbers 1, and 2, (but not 3), have no commonfactors with 3(5) = 4 since the two numbers 1,2,3 and 4 are coprime to 5.(8) = 4 since the four numbers 1, 3, 5 and 7 are coprime to 8.The primitive n-th roots of unity can be used to define and generate the nthcyclotomic polynomial n(x), by stating that they are the zeros of thispolynomial.Examples will make this clearer.3(x) = (x-e^i(2 pi/3))(x-e^i(4 pi/3)) = x^2 +x +1 and5(x) = (x-e^i(2 pi/5)) (x-e^i(4 pi/5)) (x-e^i(6 pi/5)) (x-e^i(8 pi/5)) =x^4 +x^3 +x^2 +x +1If we only select the roots where k and n are co-prime, (have no commonfactors), known as the primitive n-th roots of unity, these may be used torecompose a special type of polynomial, which has the primitive n-th rootsas its solution.Polynomials constructed in this way are known as the n-th cyclotomicpolynomials. AnConsidering a symmetrical example, the 8th roots of unity aree^i(2k pi/8) = e^i(k pi/ 4), for k = 1 to 8 so, the primitive 8th roots ofunity aree^i(k pi/ 4) for k = 1, 3, 5 and 7, but due to symmetry,e^i(7 pi/ 4) = e^-i( pi/4) , (etc), so we may pair off complex conjugates towrite the factors as8(x) =(x - e^i( pi/4))(x - e^-i( pi/4))(x - e^i(3 pi/4))(x - e^-i(3 pi/4))= (x^2 - (sqrt2)x +1) (x^2 + (sqrt2)x +1) = x^4 + 1CYCLOTOMIC POLYNOMIALS - SHORTCUTS FOR FINDING THEMa) Whenever n is an odd prime, p, the CP becomes(x^p-1)/(x-1) which from geometric progressions we recognise as a simple sumofdescending powers of x, starting with x^(p-1) e.g. 5(x) = x^4 +x^3 +x^2 +x+1I give no more examples of these here as the pattern is clear.b) Whenever n is a composite n = rs with r and s different, and assuming youhave a technical calculator, that has 'expand' as one of its algebraicfunctions, you can use the fact that,Phi rs(x) = (x-1)(x ^(rs) - 1)/[(x ^(r) - 1)(x^(s) - 1)]e.g. Phi 6(x) = (x-1)(x ^(6) - 1)/[(x ^(2) - 1)( x^(3)- 1)]Here I list a few examples of rs CPsPhi 6(x) = x^2 - x +1Phi 10(x) = x^4 - x^3 + x^2 - x +1Phi 14(x) = x^6 - x^5+ x^4 - x^3 + x^2 - x +1Phi 15(x) = x^8 - x^7 + x^5 - x^4 +x^3 - x +1Phi 21(x) = x^12 - x^11 +x^9 - x^8 + x^6 - x^4 + x^3 - x +1Note that with composite n = rs, certain powers can be missing and signsalternate.c) Whenever n is a square composite n = r^2, again using 'expand', you canuse the fact that,Phi r^2(x) = (x^(r^2)-1)/(x^(r) - 1) here are the first few examplesPhi 4(x) = (x ^(4) - 1)/(x ^(2) - 1) = x^2 + 1Phi 9(x) = (x ^(9) - 1)/(x ^(3) - 1) = x^6 + x^3 + 1Phi 16(x) = (x ^(16) - 1)/(x ^(4) - 1) = x^12 + x^8 + x^4 +1Phi 25(x) = (x ^(25) - 1)/(x ^(5) - 1) = x^20 + x^15 + x^10 + x^5 +1Here there are r positive terms, starting x^(r(r-1)) with powers descendingin steps of r.I hope you find this interesting and useful - Ian Hutcheson=== === Subject: : Re: Equation of Cyclotomic Polynomial> Can someone please tell me how to find the equation of the 5th> cyclotomic polynomial??What it is: x^4 + x^3 + x^2 + x + 1.How to find them: van der Waerden, Modern Algebra I, pp.113-5. === === Subject: : Re: JSH: Pattern argument, revisited So I can have(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x), or(5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 2) andwhere the b's are roots of b^2 - (x - 2)b + 7(x^2 + x), or No. x - 2 should be x - 3 and 7 should be 17 in the line above. Typos. (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x + 2)where the c's are roots of c^2 - xc + 7(x^2 + x). No. 7 should be 2 in the line above. More typos.The more astute of you should see a pattern. Clearly the constant> factor on the right, which shows up as a coefficient on the left must> equal 2 mod 5 for this particular setup.But why?Clearly there is an *underlying* equation being multiplied in each> case, which I'll call(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2 What's happening is that I'm multiplying that underlying equation by> integers that are 2 mod 5, which allows for *one* factorization if> you're to have integer coefficients.Don't believe me, then try to multiply by 13 instead of 7, like 13> equals 3 mod 5, so it will not work.> Right. No more than your underlying equation will work when> you multiply by 1. This equals 1 mod 5, so in your own words,> it will not work. It seems that you're missing the point Decker.You emphasized the *single* example, whereas I've pointed out thepattern when you consider more.Now then, an enquiring mind should wonder why is 2 mod 5 important? The problem is that it's clear that y_1(x) and y_2(x) can't both be> algebraic integer functions.> Right again. Of course there's no reason why they should yield> algebraic integers.> It's fairly easy to prove that with your own example(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x)that they'd have to exist, and have to be algebraic integers, if theexists algebraic integer functions A(x), and B(x) such that7 A(x) B(x) = (5a_1(x) + 7)(5a_2(x) + 7) with the same restriction on the a's.> The more astute of you should see a pattern. Clearly the constant>factor on the right, which shows up as a coefficient on the left must>equal 2 mod 5 for this particular setup.>But why?>Clearly there is an *underlying* equation being multiplied in each>case, which I'll call>(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2>>No. You yourself see the difficulties with this approach below.There's no other approach Decker. Given the pattern there has to be> some relevance of 2 mod 5 with prime multiples of the polynomial.What's happening is that I'm multiplying that underlying equation by>integers that are 2 mod 5, which allows for *one* factorization if>you're to have integer coefficients.>Don't believe me, then try to multiply by 13 instead of 7, like 13>equals 3 mod 5, so it will not work.>>Right. No more than your underlying equation will work when>>you multiply by 1. This equals 1 mod 5, so in your own words,>>it will not work.> It seems that you're missing the point Decker.You emphasized the *single* example, whereas I've pointed out the> pattern when you consider more.Now then, an enquiring mind should wonder why is 2 mod 5 important?Ummm. Might it have anything to do with that 2 that shows upin 25x^2 + 30x + 2? Let's see. Suppose we take k to be anarbitrary integer and consider (5a_1(x) + (2+5k))(5a_2(x) + (2+5k)) = (2+5k)(25x^2 + 30x + 2)The RHS can be rearranged to yield 25[(2+5k)(x^2+x)] + 5[(2+5k)(x-k)] + [5(2+5k)k+(2+5k)2]= 25[(2+5k)(x^2+x)] + 5[(2+5k)(x-k)] + [(2+5k)^2] So now if we expand the LHS we have 25a_1(x)a_2(x) + 5(2+5k)(a_1(x) + a_2(x)) + (2+5k)^2and if we take a_1(x)a_2(x) = (2+5k)(x^2+x) a_1(x) + a_2(x) = x - kthen we see that the a's satisfy a^2 - (x - k)a + (2+5k)(x^2+x)So it's obvious that the reason we can do this is thatthe constant terms on both sides are equal to (2+5k)^2,which is a consequence of the facts that (1) you decomposethe RHS into a linear combination of powers of 5 and(2) the constant term of 25x^2+30x+2 is 2.This is so trivial that I presume your question why is2 mod 5 important? is merely a rhetorical device.>The problem is that it's clear that y_1(x) and y_2(x) can't both be>algebraic integer functions.This is only a problem if you are somehow convinced that in(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2the y's have to be algebraic integer functions. There'sno reason they should be, as you note above. It's fairly easy to prove that with your own example(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x)that they'd have to exist, and have to be algebraic integers, if the> exists algebraic integer functions A(x), and B(x) such that7 A(x) B(x) = (5a_1(x) + 7)(5a_2(x) + 7) with the same restriction on the a's.Well, sure. We'd need two functions A, B: Z -> Algebraic integerssuch that A(x)B(x) = 25x^2 + 30x + 2 There are lots of functions A and B that satisfy this:A(x) = 1, B(x) = 25x^2 + 30x + 2orA(x) = B(x) = sqrt(25x^2 + 30x + 2)But that's immaterial here. The point is that if you have (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)where the a's are defined as above, then for almostall values of x, 7 DOES NOT divide either of a_1(x)or a_2(x). This is (1) an undisputable fact that(2) has been demonstrated here many times. For other readers I want to point out that Decker plays the nice> guy role, when his behavior here is contemptible, and reflects badly> not only upon him, but also upon Hamilton College.Oh, I think my reputation and the College's will survive. > His *duty* as a professor is to tell the truth and promote elucidating> the truth, not hiding it.like 2 and 17, it's clear that there's more here than his attempts at> dismissal indicate.Better would be to replace more with less in the sentence above. > I hereby charge Decker with academic fraud and note that his> college is responsible for this rogue professor.Charge noted and ignored with a wry chuckle. I did like thesobriquet rogue professor; it's on my office door now. > I've deliberately involved an official at his college to take away> plausible deniability for his school in a phone call I made months> ago.I LOVE Usenet. You couldn't pay for entertainment like this. BTW,you should probably direct any further calls to our Dean ofFaculty. If you don't want to involve him, you could alwaysnotify my department Chair. See the from line of thispost for his contact information. We do have a new President,but she'd probably just pass on any complaints to the Dean. > His college has been made aware of his behavior.There may also be a civil matter involved at some point in the future.I'd love that. You'd better get a job first, though. I understandthese things can be expensive. > Some of you seem to think this is a game, or that Usenet doesn't> matter.Right on both counts. > However, Usenet is a *public* forum read around the world. There is a> burden upon professors to tell the truth, and upon the colleges and> universities that promote them either directly or indirectly.Possibly an example needs to be made to convince you all that you have> to be accountable for your public words.> Bring it on, James. I urge you to take your best shot.Cordially,=== === Subject: : Re: JSH: Pattern argument, revisited>> [...]>>> For other readers I want to point out that Decker plays the nice>> guy role, when his behavior here is contemptible, and reflects badly>> not only upon him, but also upon Hamilton College.>Oh, I think my reputation and the College's will survive.>> His *duty* as a professor is to tell the truth and promote elucidating>> the truth, not hiding it.>>> like 2 and 17, it's clear that there's more here than his attempts at>> dismissal indicate.>Better would be to replace more with less in the sentence above.>> I hereby charge Decker with academic fraud and note that his>> college is responsible for this rogue professor.>Charge noted and ignored with a wry chuckle. I did like the>sobriquet rogue professor; it's on my office door now.>> I've deliberately involved an official at his college to take away>> plausible deniability for his school in a phone call I made months>> ago.>I LOVE Usenet. You couldn't pay for entertainment like this. BTW,>you should probably direct any further calls to our Dean of>Faculty. If you don't want to involve him, you could always>notify my department Chair. See the from line of this>post for his contact information. We do have a new President,>but she'd probably just pass on any complaints to the Dean.>> His college has been made aware of his behavior.>>> There may also be a civil matter involved at some point in the future.>I'd love that. You'd better get a job first, though. I understand>these things can be expensive.>> Some of you seem to think this is a game, or that Usenet doesn't>> matter.>Right on both counts.>> However, Usenet is a *public* forum read around the world. There is a>> burden upon professors to tell the truth, and upon the colleges and>> universities that promote them either directly or indirectly.>>> Possibly an example needs to be made to convince you all that you have>> to be accountable for your public words.>>Bring it on, James. I urge you to take your best shot.Good to see you putting a brave face on things. You must be terrifiedhe's going to actually go through with it or you wouldn't feel the need to pretend this way.(Just curious, who did he actually call? And is that the sameperson as the official that he's deliberately involved? Can't tell from the above...)>Cordially,>=== === Subject: : Re: JSH: Pattern argument, revisited (Just curious, who did he actually call? And is that the same> person as the official that he's deliberately involved? > Can't tell from the above...)> I have no evidence that any such call was ever made. Of coursethe administration might be keeping mum while they preparetheir case. Guess we'll just have to wait and see.=== === Subject: : Re: JSH: Pattern argument, revisited>[...]>For other readers I want to point out that Decker plays the nice>guy role, when his behavior here is contemptible, and reflects badly>not only upon him, but also upon Hamilton College.>His *duty* as a professor is to tell the truth and promote elucidating>the truth, not hiding it.>like 2 and 17, it's clear that there's more here than his attempts at>dismissal indicate.>I hereby charge Decker with academic fraud and note that his>college is responsible for this rogue professor.>I've deliberately involved an official at his college to take away>plausible deniability for his school in a phone call I made months>ago.>His college has been made aware of his behavior.>There may also be a civil matter involved at some point in the future.Whee!!!! Wandering off the deep end again, are we? Excellent.>Some of you seem to think this is a game, or that Usenet doesn't>matter.>However, Usenet is a *public* forum read around the world. There is a>burden upon professors to tell the truth, and upon the colleges and>universities that promote them either directly or indirectly.>Possibly an example needs to be made to convince you all that you have>to be accountable for your public words.Absolutely. Actually _you're_ the one who seems to think this is alla game. For one thing, you've _said_ it's all just a game many times.But more to the point, every once in while you mumble somethingabout lawsuits, but you never carry through.A lot of us are really looking forward to reading the transcript.It's cruel of you to tease us this way.>James Harris=== === Subject: : Re: JSH: Pattern argument, revisited>[...]For other readers I want to point out that Decker plays the nice>guy role, when his behavior here is contemptible, and reflects badly>not only upon him, but also upon Hamilton College.His *duty* as a professor is to tell the truth and promote elucidating>the truth, not hiding it.like 2 and 17, it's clear that there's more here than his attempts at>dismissal indicate.I hereby charge Decker with academic fraud and note that his>college is responsible for this rogue professor.I've deliberately involved an official at his college to take away>plausible deniability for his school in a phone call I made months>ago.His college has been made aware of his behavior.There may also be a civil matter involved at some point in the future.> Whee!!!! Wandering off the deep end again, are we? Excellent.>Some of you seem to think this is a game, or that Usenet doesn't>matter.However, Usenet is a *public* forum read around the world. There is a>burden upon professors to tell the truth, and upon the colleges and>universities that promote them either directly or indirectly.Possibly an example needs to be made to convince you all that you have>to be accountable for your public words.> Absolutely. Actually _you're_ the one who seems to think this is all> a game. For one thing, you've _said_ it's all just a game many times.> But more to the point, every once in while you mumble something> about lawsuits, but you never carry through.> A lot of us are really looking forward to reading the transcript.> It's cruel of you to tease us this way.That'd be a pretty short transcript. Something like:Judge: James, You are a crank. Case dismissed.James: Everyone is against me! Everyone's lying!!>James Harris> === === Subject: : Re: JSH: Pattern argument, revisited <3c65f87.0402091831.509d8b11@posting.google.com> Discussion, linux)> I hereby charge Decker with academic fraud and note that his> college is responsible for this rogue professor.I have nothing to add. I just want to compliment your consistentlyinventive use of the word rogue. It makes one wonder whether Deckerwears an eyepatch, or perhaps an ammunition belt strapped across hischest. Anyway, good luck with the threatened civil action. I bet there arejust dozens of lawyers aching to take this case on a contingencybasis.-- Jesse HughesBesides, discoverers are too proud to kiss butt. Indiana Jones wouldnever kiss some academic's ass to get published, and neither will I. --James Harris=== === Subject: : Re: JSH: Pattern argument, revisited>> I hereby charge Decker with academic fraud and note that his>> college is responsible for this rogue professor.>I have nothing to add. I just want to compliment your consistently>inventive use of the word rogue. It makes one wonder whether Decker>wears an eyepatch, or perhaps an ammunition belt strapped across his>chest. >Anyway, good luck with the threatened civil action. I bet there are>just dozens of lawyers aching to take this case on a contingency>basis.No doubt. And of course Decker would be totally screwed - there'sno way he could possibly find an expert witness to testify on hisbehalf. (I mean sure, more or less every mathematician on the planet hasexplained to James that he's all wet. But we know they're justsaying that because they're afraid of the Truth - no way they'dlie like that under oath.)=== === Subject: : Re: JSH: Pattern argument, revisited So I can have> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)> where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x), or> (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 2) and> where the b's are roots of b^2 - (x - 2)b + 7(x^2 + x), or> No. x - 2 should be x - 3 and 7 should be 17 in the line above.> Typos.> (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x + 2)> where the c's are roots of c^2 - xc + 7(x^2 + x).> No. 7 should be 2 in the line above.> More typos.> The more astute of you should see a pattern. Clearly the constant> factor on the right, which shows up as a coefficient on the left must> equal 2 mod 5 for this particular setup.> But why?> Clearly there is an *underlying* equation being multiplied in each> case, which I'll call> (5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2> There's no other approach Decker. Given the pattern there has to be> some relevance of 2 mod 5 with prime multiples of the polynomial.> What's happening is that I'm multiplying that underlying equation by> integers that are 2 mod 5, which allows for *one* factorization if> you're to have integer coefficients.> Don't believe me, then try to multiply by 13 instead of 7, like 13> equals 3 mod 5, so it will not work.>Right. No more than your underlying equation will work when> you multiply by 1. This equals 1 mod 5, so in your own words,> it will not work.> It seems that you're missing the point Decker.> You emphasized the *single* example, whereas I've pointed out the> pattern when you consider more.> Now then, an enquiring mind should wonder why is 2 mod 5 important?> The problem is that it's clear that y_1(x) and y_2(x) can't both be> algebraic integer functions.>Right again. Of course there's no reason why they should yield> algebraic integers.It's fairly easy to prove that with your own example> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)> where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x)> that they'd have to exist, and have to be algebraic integers, if the> exists algebraic integer functions A(x), and B(x) such that> 7 A(x) B(x) = (5a_1(x) + 7)(5a_2(x) + 7)> with the same restriction on the a's.> For other readers I want to point out that Decker plays the nice> guy role, when his behavior here is contemptible, and reflects badly> not only upon him, but also upon Hamilton College.> His *duty* as a professor is to tell the truth and promote elucidating> the truth, not hiding it.> like 2 and 17, it's clear that there's more here than his attempts at> dismissal indicate.> I hereby charge Decker with academic fraud and note that his> college is responsible for this rogue professor.> I've deliberately involved an official at his college to take away> plausible deniability for his school in a phone call I made months> ago.> His college has been made aware of his behavior.> There may also be a civil matter involved at some point in the future.> Some of you seem to think this is a game, or that Usenet doesn't> matter.> However, Usenet is a *public* forum read around the world. There is a> burden upon professors to tell the truth, and upon the colleges and> universities that promote them either directly or indirectly.> Possibly an example needs to be made to convince you all that you have> to be accountable for your public words.> James HarrisJames, GET A LIFE. No college is going to fire a professor, especiallytenured, for posting on Usenet. What do you expect them to do?You are a jerk who has no life whatsoever so you drag other'sdown with you. I use math all the time at work, why don't you contact themanager of the place I work for and tell her I'm using incorrectmathematics? I also teach others how to do the calculations I do so I guessI'm lying to them too, huh? If you e-mailed her, she'd likely tell you toget a life because she knows that I am well-educated in math and know whatI'm doing.=== === Subject: : Re: Limit Of Sum Over Some Rationals> First, a simple looking limit, where I have revealed the (somewhat> unexpected) closed-form at the post's bottom.> --- t> 1 r > limit ----- / ---- , > m -> oo m^2 --- e^r> rwhich is, in linear-mode:limit{m-> oo} (1/m^2) sum{r} r^t /e^rAnd the r-sum is over all distinct positive rationals r> which have denominators, in their simplest forms, > which are each <= pi*m.--Okay, the lesson for today:Start with the finite double-sum identity:sum{k=1 to n} sum{j=1 to m, GCD(k,j)=1} f(k,j) =sum{i=1 to min(m,n)} mu(i) (sum{1<=k<=m/i} sum{1<=j<=n/i} f(ki,ji))- Defining f(k,j) as> a(j/m) b(k/m) c(k/j),> and taking limits, after dividing both sides by m^2, we get:For positive q and s, --- > 1 > limit ----- / a(den(r)/m) b(num(r)/m) c(r) > m -> oo m^2 ---> r= distinct rationals,> 1<=den(r)<=q*m> 1<=num(r)<=s*m 6 /q /s> = ---- | | > pi^2 | | a(x) b(y) c(y/x) dy dx> /0 /0(limit{m->oo} > (1/m^2) sum{r= distinct rationals,1<=den(r)<=q*m,1<=num(r)<=s*m}> a(den(r)/m) b(num(r)/m) c(r) =(6/pi^2)*> integral{0 to q}integral{0 to s} a(x) b(y) c(y/x) dy dx) > where a(), b(),c() are such that the integral exists, is defined, and> the identity above is correct...)> ;)(And num(r) is numerator of reduced r, den(r) is denomiator of> reduced r.)> --I am probably making a too-broad statement, if it is correct at all,even when I write this more specific case of the previous result: /s 6 |--- | f(x) dx =pi^2 /0 1 ---limit --- m-> oo m^2 / f(num(r)/m) ---(linera-mode:(6/pi^2)*integral{0 to s} f(x) dx=limit{m -> oo} (1/m^2) *sum f(num(r)/m) ),where the sum is over those distinct positive rationals <= s where1 <= num(r) <= m*s,1 <= den(r) <= m.Under which restrictions on s and f() is this true?(Perhaps s needs to be positive, and f() needs to beRiemann-integrable in the range, at least.)> So, we might have, if I am right so far,limit{m-> oo} (1/m^(t+u+2)) sum{r} (den(r))^u (num(r))^t /e^r= t! q^(2+t+u) (6/pi^2)/(2+t+u),where the r-sum is over all positive distinct rationals r,> where den(r) <= m*q, q = fixed positive real.--So, in answer to the original question:limit{m-> oo} (1/m^2) sum{r} r^t /e^r,where the r-sum is over all distinct positive rationals r> which have denominators, in their simplest forms, > which are each <= pi*m,is3 * t!. (?)(I find this more interesting than otherwise because of the fact that> 3 is the integer closest to both pi and e, which are each part of the> original limit.)> Leroy Quet=== === Subject: : Re: Looking for primes of a particular form. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1A2UwB05274;>I am trying to find primes numbers of the following forms:> p = 6*(10^N)-1, q = 6*(10^N)+1p is prime when n = {0,1,2,4,5,7,10,13,22,23,28,34,40,61,73,361,490,613,...}q is prime when n = {0,1,2,8,9,15,20,26,38,45,65,112,244,303,393,560,...}beyond this the number are rather large. As the others have said other metods are needed.=== === Subject: : How to compute the minimal distanct between a point and curve in N-dim space,In the N-dimensional space, give a data point A and a curve f,how to write the explicit expression for calculating theminimal distance between A and f?Or have to use some nonlinear optimization method to calcualte it?Fred=== === Subject: : Re: How to compute the minimal distanct between a point and curve in N-dim spaceEn el mensaje:e86dnfdT9cq12LXdRVn-tw@comcast.com,Fred escribi.97:> ,> In the N-dimensional space, give a data point A and a curve f,> how to write the explicit expression for calculating the> minimal distance between A and f?> Or have to use some nonlinear optimization method to calcualte it?Do you has parmetric equations of the curve?If so, the distance from A to a generic point of the curve corresponding avalue t of the parameter, is only function of t. Then minimize that onevariable function.-- === === Subject: : simple integration problemhi i am trying to prove:If f(x)>=0 for all x in [a,b], f continuous on [a,b], and Int[f(x),{x,a, b}], then f(x)=0 for all x in [a,b].here is my proof:Suppose there exists an x_0 in [a,b] such that f(x_0)>0. Since f iscontinuous we have that there exists a d>0 such that if |x-x_0|0. We immediately have that for any partition P of [a,b]L(P,f)=Sum[m_i dx_i] >= 2md > 0where m=inf f(x) over all |x-x_0| 0. Similarly, Int[f(x), {x, a, x_0-d}] >= 0 and Int[f(x), {x,x_0+d, b}] >= 0, which implies that Int[f(x), {x, a, b}] > 0,contradicting that Int[f(x), {x, a, b}] = 0. Therefore f(x)=0 for allx in [a,b]Is that decent?=== === Subject: : Re: simple integration problem> hi i am trying to prove:> If f(x)>=0 for all x in [a,b], f continuous on [a,b], andInt[f(x),{x,> a, b}], then f(x)=0 for all x in [a,b].> here is my proof:> Suppose there exists an x_0 in [a,b] such that f(x_0)>0.Since f is> continuous we have that there exists a d>0 such that if|x-x_0| then f(x)>0. We immediately have that for any partition Pof [a,b]> L(P,f)=Sum[m_i dx_i]>= 2md > 0> where m=inf f(x) over all |x-x_0| L(P,f) used in Rudin, where L(P,f) is the lower sum forsome partition> P of [a,b], m_i is smallest f-value on [x_(i-1), x_i], anddx_i => x_(i-1) - x_i. The above inequality implies that Int[f(x),{x, x-d,> x+d}] > 0. Similarly, Int[f(x), {x, a, x_0-d}] >= 0 andInt[f(x), {x,> x_0+d, b}] >= 0, which implies that Int[f(x), {x, a, b}] >0,> contradicting that Int[f(x), {x, a, b}] = 0. Thereforef(x)=0 for all> x in [a,b]> Is that decent?You have a small flaw in your proof: m could be zero evenif f(x)>0 for all x with |x-x_0| f(x_0)/2whenever |x-x_0|= f(x_0)/2._____________________________________________________ ____Eric J. Wingler (wingler@math.ysu.edu)Dept. of Mathematics and StatisticsYoungstown State UniversityOne University PlazaYoungstown, OH 44555-0001330-941-1817=== === Subject: : Re: simple integration problem> hi i am trying to prove:If f(x)>=0 for all x in [a,b], f continuous on [a,b], and Int[f(x),{x,> a, b}], then f(x)=0 for all x in [a,b].You forgot to add =0 after the integral.> here is my proof:Suppose there exists an x_0 in [a,b] such that f(x_0)>0. Since f is> continuous we have that there exists a d>0 such that if |x-x_0| then f(x)>0. We immediately have that for any partition P of [a,b]L(P,f)=Sum[m_i dx_i]>= 2md > 0where m=inf f(x) over all |x-x_0| 0? If youtake, for instance, f(x) = x, x_0 = 1, and d = 1, then m = 0.> Is that decent?No, but it easy to correct it. Or look for a recent thread calledeasy....analysis problem.........Best ,Jose Carlos Santos=== === Subject: : re:help! h(x) = cos^4x sinx [0, pi]I presume you mean value. To get the average, take 1/pi times theintegral from 0 to pi. Make the substitution u=cosx. Thendu=-sinxdx.Finally the average is 2/5pi, after integrating between 0 and 1 in u(2 times after splitting the x domain in half).----== Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =---=== === Subject: : 3pi.comPi3.com is the ultimate maths enthusiast web address,what about 3pi.com ?Its up for appraisal if anyone wants I'll make a bid for them, usually $xx.Herc--QUADS give you STRENGTHBICEPS give you POWERPECS give you CONFIDENCE www.theBanner.net=== === Subject: : polynomial degree in general polynomialHello all, i am trying to prove that deg(fg)=deg(f)+deg(g) for f,g inF[x1,...,xn], where F is a field.I was able to prove that deg(fg)<=deg(f)+deg(g), but the full proof isstill alluding me. Can anyone provide some help or hints? The notationis very cumbersome.thanks you=== === Subject: : Re: polynomial degree in general polynomial Adjunct Assistant Professor at the University of Montana.>Hello all, i am trying to prove that deg(fg)=deg(f)+deg(g) for f,g in>F[x1,...,xn], where F is a field.>I was able to prove that deg(fg)<=deg(f)+deg(g), but the full proof is>still alluding me. Can anyone provide some help or hints? The notation>is very cumbersome.Prove it for monomials. Then show that if deg(f) is different fromdeg(g), then deg(f+g) = max(deg(f),deg(g)), and use this to prove itfor the product of a monomial with a polynomial, and then a polynomialwith a polynomial.Of course, you need to be careful, depending on how you definedegree. The easiest thing to do is to refine the degree by letting amonomial be of higher degree than another monomial if either thefirst has larger total degree, or else they have equal total degreeand the first is larger lexicographically (compare the powers of x1,and take the largest; if they are equal, compare the powers of x2, etc.)-- ============================================================== ============ === Subject: : Re: polynomial degree in general polynomial> Hello all, i am trying to prove that deg(fg)=deg(f)+deg(g) for f,g in> F[x1,...,xn], where F is a field.I was able to prove that deg(fg)<=deg(f)+deg(g), but the full proof is> still alluding me. Can anyone provide some help or hints? The notation> is very cumbersome.thanks youIf you refine the order on the exponents in a suitable way, everypolynomial has a unique leading monomial i.e., an monomial thathas the highest exponent w.r.t. that order with nonvanishing coefficient.Then you can imitate the proof for the case n=1.For example, you can compare two exponents by total degree first andbreak ties by comparing them lexicographic.For a = (a1, ... ,an) in N^n let |a| := sum( ai | i+1..n )Then order the exponents via:a < b <==> |a| < |b| or ( |a| = |b| and am < bm )where m = min { i | ai <> bi }.You can convince yourself, that this gives a total order on the exponents,which is also compatible with addition of exponents i.e.,a < b ==> a+c < b+c.In particular, the product of the leading monomials of f and g is indeedthe leading monomial of fg.Marc=== === Subject: : Re: Four Color Theorem> Robinson, Sanders, Seymour, & in their proof of the FCT, state I think you mean Robertson.> It has been known since 1913 that every minimal counterexample to the> Four Color Theorem is an internally 6-connected triangulation.Kainen, 1986. pp 59,> If G is minimal [ie a minimal counterexample to the FCT], then G is> 5-connected. This isn't necessarily say that it is 5-connected and not 6-connected. If something is 6-connected, then it is connected, and 2-connected, 3-connected, and so on up to 5-connected, so there's no immediate conflict between these two.> If G is minimal [ie a minimal counterexample to the FCT], then G is> maximal planar [ie a triangulation].Is there a conflict between the two descriptions of an MCE or does > internally 6-conncted mean the same as 5-connected?What does internally connected mean? If you are reading the paper, it should define it there.J=== === Subject: : misunderstanding my fault by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1A4B3m13015;.Divisors of n not divisible by 9 are 54717 - 9333 = 45384, not divisible by 9.Divisors of n not divisible by 169 are 54717 - 43433 = 11284, not divisible by 169. Divisors of n not divisible by 61 are54717 - 52338 = 2379, which _is_ divisible by 61. I must apologise for my clumsy terseness in using an example with just three primes each dividing once into the sample number, but my claim about symmetry between zero and the LCM of the factors means that there is no reason for sums of factors of 3, 13, and 61 to _not_ be symmetrically positioned between zero and a multiple of LCM3x13x61. But the relevant LCM in the example you give is 3.3.13.13.61, so it will be partition sums and complements of two composites [9, 169] and one prime [61] which must all divide by their appropriate factors for 3.3.13.13.61 to be a perfect number. Since the factor sum, 54717, is 23 x 2379 [2379 = 3.13.61], the sum of factors not dividing by 61 also divides by 61, but this fails for 9 and 169, whose complements are nowhere near their LCM with 61, which is n = 92781. Does this make any sense?Best wishes, Mark G.. === === Subject: : Re: misunderstanding my fault> think that your example should read Divisors of n not divisible by 9 are > 54717 - 9333 = 45384, not divisible by 9.Divisors of n not divisible by 169 are > 54717 - 43433 = 11284, not divisible by 169. Divisors of n not divisible by 61 are> 54717 - 52338 = 2379, which _is_ divisible by 61. If the divisors of n sum to a number divisible by all the prime powers dividing n (hence, divisible by n), the number is said to be multiperfect. You're trying to show there are no odd perfect numbers by showing there are no odd multiperfect numbers. It is widely believed that there are no odd multiperfect numbers, but not proved. It seems safe to say that elementary methods have been exhausted by now, and if anyone is going to prove that there are no odd multiperfects she's going to have to use a lot stuff way beyond anything you've shown us. There's a lot of literature on odd multiperfects. If you're really interested in the topic, I suggest you have a good long look at what's already been done.-- === === Subject: : Unitary operator basics, lebesgue measureLet H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H by(T_a f)(x) = f(x-a), where f is in H (the hilbert space)Firstly, why is f(x-a) necessarily in H if f is in H? Why is T_a a unitaryoperator (unitary operator I believe means that for an operator A, A* =A^{-1}) ? I suppose this might be easy if I could represent T_a as amatrix. But then how would this matrix multiply by f? Could I representthe function f by a matrix(with respect to a basis of L^2(R,dx)?)? What isa basis for L^2(R,dx)?Moshe=== === Subject: : Re: Unitary operator basics, lebesgue measure> Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H by(T_a f)(x) = f(x-a), where f is in H (the hilbert space)Firstly, why is f(x-a) necessarily in H if f is in H? Why is T_a a unitary> operator (unitary operator I believe means that for an operator A, A* => A^{-1}) ? I suppose this might be easy if I could represent T_a as a> matrix. But then how would this matrix multiply by f? Could I represent> the function f by a matrix(with respect to a basis of L^2(R,dx)?)? What is> a basis for L^2(R,dx)?The graph of f(x-a) is just the graph of f shifted a units to the right. That answers the first question. Verify that (T_a)* = T_(-a) and you'll have the answer to the second question.=== === Subject: : Re: Unitary operator basics, lebesgue measure> Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> Hby(T_a f)(x) = f(x-a), where f is in H (the hilbert space)Firstly, why is f(x-a) necessarily in H if f is in H? Why is T_a aunitary> operator (unitary operator I believe means that for an operator A, A* => A^{-1}) ? I suppose this might be easy if I could represent T_a as a> matrix. But then how would this matrix multiply by f? Could Irepresent> the function f by a matrix(with respect to a basis of L^2(R,dx)?)? Whatis> a basis for L^2(R,dx)?> The graph of f(x-a) is just the graph of f shifted a units to the right.> That answers the first question. Verify that (T_a)* = T_(-a) and you'll> have the answer to the second question.How can I think about (T_a)* ?? First I need to represent T_a as a matrix,but what matrix is it? Then I would transpose the matrix andconjugate...(still confused)probably silly for not seeing the second)=== === Subject: : Re: Unitary operator basics, lebesgue measure X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS>How can I think about (T_a)* ?? First I need to represent T_a as a>matrix,Why? That would only confuse matters.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: Unitary operator basics, lebesgue measure>> Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H>by>> (T_a f)(x) = f(x-a), where f is in H (the hilbert space)>> Firstly, why is f(x-a) necessarily in H if f is in H? Why is T_a a>unitary>> operator (unitary operator I believe means that for an operator A, A* =>> A^{-1}) ? I suppose this might be easy if I could represent T_a as a>> matrix. But then how would this matrix multiply by f? Could I>represent>> the function f by a matrix(with respect to a basis of L^2(R,dx)?)? What>is>> a basis for L^2(R,dx)?>> The graph of f(x-a) is just the graph of f shifted a units to the right.>> That answers the first question. Verify that (T_a)* = T_(-a) and you'll>> have the answer to the second question.>How can I think about (T_a)* ?? First I need to represent T_a as a matrix,No, you don't. Let f and g be elements of L^2(R,dx), then, using the mathematician's inner product (linear in the first factor and antilinearin the second),((T-a)*f,g) = (f,T_a g) = int_R f(x) g(x-a)^* dx = int_R f(x+a) g(x)^* dxby substitution, and so [(T_a)* f](x) = f(x+a) = [T_(-a) f](x), i.e.(T_a)* = T_(-a).David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills.=== === Subject: : Re: Unitary operator basics, lebesgue measure>> Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H>by>> (T_a f)(x) = f(x-a), where f is in H (the hilbert space)>> The graph of f(x-a) is just the graph of f shifted a units to the right.>> That answers the first question. Verify that (T_a)* = T_(-a) and you'll>> have the answer to the second question.>How can I think about (T_a)* ?? First I need to represent T_a as a matrix,>but what matrix is it? Then I would transpose the matrix and>conjugate...(still confused)No, a matrix representation is neither needed nor desirable: it would just obscure matters. (T_a)* is defined by <(T_a)* f, g> = where <,> is the inner product. Write out what <(T_{-a} f, g> = means in terms of integrals and you should see it immediately.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Web Sites for Complexity TheoryAre there any good books in this field, or good web sites that providean introduction or lecture notes with definitions and examples? I amreally trying to teach myself this stuff before I take the class nextsemester.Thank YouBlake Manner=== === Subject: : Re: Web Sites for Complexity Theorycheck out this book (pdf - free):http://www.cis.upenn.edu/~wilf/AlgComp2.hgood luck, Jurgen> Are there any good books in this field, or good web sites that provide> an introduction or lecture notes with definitions and examples? I am> really trying to teach myself this stuff before I take the class next> semester.> Thank You> Blake Manner=== === Subject: : Qabala & PhysicsThe Others as The Ecclesia (see also Herman Hesse's Magister Ludi (AKA The Glass Bead Game').Before I forget to Zielinski - the Cartan spin connection valued in the local Lorentz group Lie algebra o(3,1) is the tensor part of the Levi-Civita connection. The spin connection can take an additional torsion field that is not in Einstein's theory - but it need not have it.The gravity field is really the non-trivial part of the tetrad field in the sense of Rovelli's formalism. The tetrad field is the spin 1 gauge force compensating local field from breaking global translational symmetry, i.e. introducing local tidal curvature. The translations and Lorentz rotations intetwine as seen in the Sagnac and precession effects which do not depend on locally gauging the Lorentz rotations, but if we do that we get the additional torsion field of Shipov, Akimov et-al in Russia. The spin 2 field is the strain tensor of the tetrads. Einstein's metric guv is quadratic in the tetrads and in quantum theory 1 + 1 = 2 or 0. The 0 is from locally gauging the dilations of the conformal group. There is a second set of tetrads from locally gauging the 4 remaining special conformal transformations that give Tony's conformal gravitons, so now I am beginning to understand intuitively what Tony is talking about at least partly. My intuition is powerful, I mean I have a kind of analog computer simulation in my mind's I and I can tweak it this way and that without having to do the formal symbol manipulations, then later put it into symbols. I think this is what Einstein meant when he said he thought kinesthetically?With regard to the 22 Hebrew letters mentioned by Suares:A useful way of looking at the 22 letters of the Hebrew alphabet is:3 mothers - quaternion imaginaries, 3 spatial dimensions, 3 QCD colors7 doubles - octonion imaginaries, 3 spatial dimensions + 4 internal symmetry space dimensions 7 charge-carrying first-generation fermions (electron; red,blue,green down quarks; red,blue,green up quarks)12 simples - 12 root vectors of the SU(2,2) conformal group, 12 gauge bosons of SU(3)xSU(2)xU(1)When 5 finals are added to the 22,you get the 27 dimensions of the exceptional Jor algebra J3(O),which is the dimensionality of bosonic string M-theory (which I think isrelated to the quantum interactions of world-lines in the many-worlds).Its complexified automorphism group is E6.All this (and a lot more) is inherent in my physics modelTonyToo bad Carlo Suares is not alive to see this. :-=== === Subject: : Re: Qabala & Physics> The Others as The Ecclesia (see also Herman Hesse's Magister Ludi (AKA > The Glass Bead Game').The Smashing of the Vessels is the Kabalistic version of the Big Bang.=== === Subject: : confused on substitutionhello,been working on a problem too hard and now have started to become unsure asto the correct nature of a substitution:the problem is this,solve the heat diffusion equation,c*u(x,t)_xx = u(x,t)_t [1]where underscores are partial derivitives and c=constant, such that u(x,t)satisfies the conditionu(x,t)_x = u(x,t)*F(u(x,t),t) [2]for all x and t where F is an arbitrary function of u and t. first of allis this problem well defined, and does it have a solution (i think it mightnot)? one would assume that from [2] we can find u_xx and substitute thisinto the LHS of [1]. however do we also have to solve for u from [2] andsubstitute that into the RHS of [1] as well to obtain a function thatsatisfies both [1] and [2]? or is the first substitution enough to guaranteethis. i assume that we probably need to substitute for u from [2] in bothsides of [1] in order to be consistent and keep the equality....but my brainis fried and i have a mental block on this issue at present and it doesntseem obvious to me at the moment. If we do substitute for both sides of[1] using [2] then we get...c*u_x*(F+u*F_u)^2 = u_xt - u*F_t [3]So to reiterate more succinctly will [3] provide me a solution thatsatisfies both [1] and [2].Is there a better way to solve this problem. i have thought about usinggreens functions for [1] and forcing it to satisfy [2] but i cant seem toconstruct them. need help with the analysis of this system.any help appreciated.cheersmoth=== === Subject: : Re: confused on substitution> hello,> been working on a problem too hard and now have started to become unsure as> to the correct nature of a substitution:> the problem is this,> solve the heat diffusion equation,> c*u(x,t)_xx = u(x,t)_t [1]> where underscores are partial derivitives and c=constant, such that u(x,t)> satisfies the condition> u(x,t)_x = u(x,t)*F(u(x,t),t) [2]> for all x and t where F is an arbitrary function of u and t. first of all> is this problem well defined, and does it have a solution (i think it might> not)?I am inclined to agree.> one would assume that from [2] we can find u_xx and substitute this> into the LHS of [1].Yes.> however do we also have to solve for u from [2]Given F, [2] is a non-linear PDE for u, so no.> and> substitute that into the RHS of [1] as well to obtain a function that> satisfies both [1] and [2]? or is the first substitution enough to guarantee> this.The first substitution is enough, assuming such a u exists.> i assume that we probably need to substitute for u from [2] in both> sides of [1] in order to be consistent and keep the equality....but my brain> is fried and i have a mental block on this issue at present and it doesnt> seem obvious to me at the moment.Largely because that would be wrong.> If we do substitute for both sides of> [1] using [2] then we get...> c*u_x*(F+u*F_u)^2 = u_xt - u*F_t [3]We actually getu_xx = u*F_x + u_x*FF_x = F_u*u_x=>u_xx = u_x*(F_u*u + u_x*F)c*u_x*(F_u*u + u_x*F) = u_t [3a]> So to reiterate more succinctly will [3] provide me a solution that> satisfies both [1] and [2].No, but [3a] will.> Is there a better way to solve this problem. i have thought about using> greens functions for [1] and forcing it to satisfy [2] but i cant seem to> construct them. need help with the analysis of this system.[1] and [2] are two simultaneous PDE's in one unknown (u). This system isin general overdetermined.However, [3a] may be soluble; whether it is well-posed when combined withthe boundary and initial data for [1] and [2] is another matter.-- The vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice.=== === Subject: : Re: confused on substitution>solve the heat diffusion equation,>c*u(x,t)_xx = u(x,t)_t [1]>where underscores are partial derivitives and c=constant, such that u(x,t)>satisfies the condition>u(x,t)_x = u(x,t)*F(u(x,t),t) [2]>for all x and t where F is an arbitrary function of u and t. first of all>is this problem well definedOne of the requirements is that the problem must have a uniquesolution given the boundary conditions. Consider:Let F(u,t) = 1. Then u(x,t) = C exp(x) exp(ct) is a solution to [1]for all C in R and fulfills [2]. Not good. You need more boundaryconditions.-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : Re: JSH: Tautological spaces[posted and mailed]>> Oh yeah, I don't think the human being has been born yet that can>> handle complex solutions in a complex topological space. I call a>> person who can handle such a challenge a third generation>> mathematician.>>> Today's mathematicians are first generation.>>>> James HarrisVery eloquently stated. ..........I need someone to confirm my findings.Could you possibly help me? I believe there is much profit if I can> find M.Dr. Ben Zona> Largest Existing Number? ===> 0Lagest Number = 0Zero does not exist Zero is not a numberM is 0 M does not ExistQEDCheeresCarl=== === Subject: : easy complex analysis problem......hello........complex function f(z) = 1 / [2*{x^(1/2)}]int f(z) dz = ??|z| = 1-------------------------i think.....f(z) have isolated singular point on z=0but i can't use residue theorem to this problem.thuslet z = e^i@ (@= theta)dz = i(e^i@) d@thusint [1/{2*e^(i@/2)}]*ie^i@ d@0~2pi= (i/2)*int e^(i@/2) d@= -2it's right??let me check up, please.....thank you teacher....=== === Subject: : Re: easy complex analysis problem......> complex function f(z) = 1 / [2*{x^(1/2)}]I do not understand the definition of f. Try to explain it in a betterway.Best ,Jose Carlos Santos=== === Subject: : Ellipse?Hipothesis:Let p and q be two positive real numbers.Let Ox and Oy be two cartesian axis.Let l be the variable line that intersects Ox and Oy in A and B suchthat AB measures p.Let C be *one* of the two points in l which distance to B is q.Thesis: the locus of C is an ellipseI'm looking for an euclidean proof.Any hint?=== === Subject: : Re: Ellipse?Haroldo Stenger a .8ecrit:> Hipothesis:> Let p and q be two positive real numbers.> Let Ox and Oy be two cartesian axis.> Let l be the variable line that intersects Ox and Oy in A and B such> that AB measures p.> Let C be *one* of the two points in l which distance to B is q.Thesis: the locus of C is an ellipseI'm looking for an euclidean proof.Any hint?Just a hint then... (in case of homework)Let I middle of ABon the OI line : OM=AP and OC=ABPM intersects OB at HProve that HP/HM=PB/AP (repeated Thales on various triangles)Locus of P is the affine (stretch ?) transformed of locus of M.-- philippe(chephip at free dot fr)=== === Subject: : Hamilton College announces new courseA new course, Introduction to Harristotelian Logic, will be presented by new Hamilton College faculty member, James S. Harris.The foundations of the method will first be established by defining several innovative proof techniques:Proof by AssertionProof by RepetitionProof by ExhaustionProof by VituperationNext, the topic of Object Algebras will be introduced, starting with definitions and basic axioms, and leading into some fundamental theorems.Based on the preceding, the new concept of the Tautological Space will be developed, in conjunction with prime number counting using partial difference equations, and certain results will be proved, in particular Harris's Error in Core Mathematics Theorem.If there is time a new proof of Fermat's Last Theorem will be presented.=== === Subject: : Re: Hamilton College announces new course> A new course, Introduction to Harristotelian Logic, will be presented by > new Hamilton College faculty member, James S. Harris.The foundations of the method will first be established by defining > several innovative proof techniques:Proof by Assertion> Proof by Repetition> Proof by Exhaustion> Proof by VituperationNext, the topic of Object Algebras will be introduced, starting with > definitions and basic axioms, and leading into some fundamental theorems.Based on the preceding, the new concept of the Tautological Space will > be developed, in conjunction with prime number counting using partial > difference equations, and certain results will be proved, in particular > Harris's Error in Core Mathematics Theorem.If there is time a new proof of Fermat's Last Theorem will be presented.> of scheduling courses for next year. Should this course becross-listed with any other departments? Should I alert my colleaguesin, say, Philosophy, Psychology, or Sociology? Perhaps I shouldcontact the folks in Creative Writing? === === Subject: : Re: Hamilton College announces new course>A new course, Introduction to Harristotelian Logic, will be presented by > new Hamilton College faculty member, James S. Harris.>The foundations of the method will first be established by defining >several innovative proof techniques:>Proof by Assertion>Proof by Repetition>Proof by Exhaustion>Proof by VituperationThere are also other techniques he has successfully employed:Proof by Contradiction 1. State your theorem. 2. Wait for someone to disagree. 3. Contradict them.Fool Proof 1. State your theorem. 2. Invite colleagues to comment. 3. If they don't agree, exclaim loudly: You Fools!Proof by Stubborness Even when presented with a counterexample, the author insists his proof is correct and demands his readers find the logical flaw for him.Proof by Appeal to Infants Even a child can see it's true.Proof by Elimination of Intermediate Steps Any intermediate steps in the reasoning are removed to obscure holes in the proof.Proof by Conspiracy Theory You are right but the establishment is trying to silence you from publishing your result and disturbing the status quo.Proof by Leap-of-logic Make general statements that have little to do with the proposition, then make a wild jump to completely unrelated conclusions.>Based on the preceding, the new concept of the Tautological Space will >be developed, in conjunction with prime number counting using partial >difference equations, and certain results will be proved, in particular >Harris's Error in Core Mathematics Theorem.Is that 'Harris's Error in Core Mathematics Theorem' or 'Harris'sError in Core Mathematics Theorem' or 'Harris's Error in CoreMathematics Theorem.'?>If there is time a new proof of Fermat's Last Theorem will be presented.Good thing too it was his last, the Internet could not handle any morekooks.-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : Re: Hamilton College announces new course>>A new course, Introduction to Harristotelian Logic, will be presented by >> new Hamilton College faculty member, James S. Harris.>>The foundations of the method will first be established by defining >>several innovative proof techniques:>>Proof by Assertion>>Proof by Repetition>>Proof by Exhaustion>>Proof by Vituperation> There are also other techniques he has successfully employed:Proof by Intimidation.Regarding the naming of Harris's Theorem, ambiguity is of the essence in his theoretical techniques.GibGib=== === Subject: : Re: construct a circle of unit area? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1A6lWv24980;>>> how to construct a circle with area = exactly one unit?>>> any applications for it outside of probability?> 1) Define an arbitrary circle to have unit area.> 2) Pi is irrational and transcendental. There is no way to>construct a unit area circle by Greek rules (straightedge and>compass, linear and/or quadratic functions).How about Pi = 4/[sqrt(Golden Section)]Ref:http://www.stefanides.gr/piquad.htm http://www.stefanides.gr/quadcirc.htmThe impossible of solution by compass and straight edge wasbase on the use of : e^[iPi]=-1However the complete form for THETA=Pi is: e^[iPi]=-1+i[0] , soOnly the real pert was considered ,andthe imaginary part i.e. : e^[iPi]=i[0] , or e^[iPi]= 0 LINDEMAN , HERMITE , ETC, ETC IGNORED THIS SOLUTIONPanagiotis Stefanideshttp://www.stefanides.gr > 3) Construct a unit square. Convert to a unit circle using a>quadratrix curve. There are lots! Alas, no quadratrix curve can be>constructed by Greek rules.>-- >Uncle Al>http://www.mazepath.com/ uncleal/Quis custodiet ipsos custodes? The Net!=== === Subject: : Re: construct a circle of unit area? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1A6lWP24984;>>how to construct a circle with area = exactly one unit?>Depends on (a) you mean by construct; and (b) what you mean by>unit.>from that, again through well known methods, you would be able to>construct two points whose distance is the square of sqrt(pi), which>is pi.>However, pi is a transcendental numberIt would be nice somebody here to reproduce this TRANSCENCE ! How about Pi = 4/[sqrt(Golden Section)]Ref:http://www.stefanides.gr/quadrature.htm http://www.stefanides.gr/theo_circle.htm http://www.stefanides.gr/piquad.htm http://www.stefanides.gr/quadcirc.htmThe impossible of solution by compass and straight edge wasbase on the use of : e^[iPi]=-1However the complete form for THETA=Pi is: e^[iPi]=-1+i[0] , soOnly the real pert was considered ,andthe imaginary part i.e. : e^[iPi]=i[0] , or e^[iPi]= 0 LINDEMAN , HERMITE , ETC, ETC IGNORED THIS SOLUTIONPanagiotis Stefanideshttp://www.stefanides.gr=== === Subject: : Re: How to determine whether a point is contained inside an object/polygon?> Hello all,> I am trying to determine whether a given point is contained inside a> tetrahedron, and I believe I am following the correct method in order> to do so but need it double-checked just in case!> The tetrahedron is specified by the points:> (10,20,5),(5,10,5),(10,10,5) and (10,10,10).> What I did was obtain the inner surface normal for each face and then> determine whether the point lies inside or outside the face. I> discovered that it lies outside three faces but inside one face. I> assume for it to lie inside the tetrahedron it must lie inside all the> faces, which it doesn't in this case, so the point must be outside the> tetrahedron?I have used the vector cross product.Going around the polygon, calculate side x ( point - begin of side )If the product does not change its sign, then point is inside of thepolygon.arto.huttunen at estlab.com=== === Subject: : Re: How to determine whether a point is contained inside an object/polygon?> Try computing the barycentric coordinates.> the same idea works in R^n, given n+1 points that do not all lie in the> same (n-1)-dimensional hyperplane. I will assume n = 3 in what follows.> Given the vertices at points p, q, r, s in R^3, and given a point x,> there is a unique representation of x in in the form> x = a * p + b * q + c * r + d * s (1)> where a, b, c, d are real numbers such that> a + b + c + d = 1. (2)> The condition (1) gives you three equations in four unknowns, and the> normalizing condition (2) provides the fourth equation.> If the resulting coordinates a, b, c, d are all positive, then x lies> inside the tetrahedron. If any coordinate is negative, then x lies> outside the tetrahedron. If one or more coordinates is zero and the rest> are positive, then x lies on a face of the tetrahedron or coincides with> a vertex.> -- > Seaman> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> Very helpful. Thank you.=== === Subject: : Re: How to determine whether a point is contained inside an object/polygon?> Try computing the barycentric coordinates.> the same idea works in R^n, given n+1 points that do not all lie in the> same (n-1)-dimensional hyperplane. I will assume n = 3 in what follows.> Given the vertices at points p, q, r, s in R^3, and given a point x,> there is a unique representation of x in in the form> x = a * p + b * q + c * r + d * s (1)> where a, b, c, d are real numbers such that> a + b + c + d = 1. (2)> The condition (1) gives you three equations in four unknowns, and the> normalizing condition (2) provides the fourth equation.> If the resulting coordinates a, b, c, d are all positive, then x lies> inside the tetrahedron. If any coordinate is negative, then x lies> outside the tetrahedron. If one or more coordinates is zero and the rest> are positive, then x lies on a face of the tetrahedron or coincides with> a vertex.> -- > Seaman> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> Very helpful. Thank you.=== === Subject: : Graph Theory: Number of maximal Matchingslet G = (V, E) be a bipartite graph with matching number m(G) = n. Assume that V is the disjoint union of the subset U1 and U2 both with cardinality |U1| = |U2| = n. Edges connect each vertex from U1 with one or more vertices of U2. Is there any result about the number of maximal matchings in G?=== === Subject: : Re: Graph Theory: Number of maximal Matchings>let G = (V, E) be a bipartite graph with matching number m(G) = n. Meaning, I guess, that a maximal matching is of size n.>Assume that V is the disjoint union of the subset U1 and U2 both with >cardinality |U1| = |U2| = n. Edges connect each vertex from U1 with one >or more vertices of U2. Is there any result about the number of maximal >matchings in G?It could be anywhere from 1 (e.g. if every vertex has degree 1)to n! (if every vertex of U1 is connected to every vertex of U2).Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: A little help is neededIn sci.math, Arturo Magidin:>>f: Q x Q -> Q>>f(r1, r2) = (a+c)/(b+d) where r1 = a/b, r2 = c/d,>> a,b,c,d integer, b > 0, d > 0,>> gcd(a,b)=1, gcd(c,d)=1 [*] [.snip.]>>[*] probably should also state that 0 = 0/1 is the unique>> representation of 0 as well.That follows from the definition: gcd(0,b) = |b| for all integers> b. For gcd(x,y) is the nonnegative integer d that satisfies: > (1) d|x and d|y;> and (2) if c is an integer such that c|x and x|y, then x|d.Clearly, |b| divides b and divides 0. And if x is any integer that> divides both b and 0, then it divides |b|. Thus, gcd(0,b)=|b|. So if> a=0, for gcd(a,b) to equal 1, with b>0, you must have b=1.> OK, then maybe we *don't* have to specify 0 = 0/1 as the uniquerepresentation for this function. :-)As long as it works, and from the looks of it, it'll work over thedomain Q^2; there's no horribly obvious ambiguity, if done right.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Rationals are Uncountable>> Not needed to disprove your claim. One only needs -1-r < 0 < 1-r, which>> follows immediately from |r| < 1. Though it is qite possible that you,>> , cannot follow that simple argument without help.I agree that if X' is the set of all rationals in (-1-r,1-r) with> r irrational and |r|<1 then 0 is in X'.> What I am proving is that if x_n is a sequence that> contains every rational in X=(-1,1) then there is no> way to convert x_n to x'_n such that x'_n contains 0.?Define convert.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.hNeedless to say, I had the last laugh. Partridge, _Bouncing Back_ (14 times)=== === Subject: : Re: Rationals are Uncountable>> Not needed to disprove your claim. One only needs -1-r < 0 < 1-r,which>> follows immediately from |r| < 1. Though it is qite possible that you,>> , cannot follow that simple argument without help.I agree that if X' is the set of all rationals in (-1-r,1-r) with> r irrational and |r|<1 then 0 is in X'.> What I am proving is that if x_n is a sequence that> contains every rational in X=(-1,1) then there is no> way to convert x_n to x'_n such that x'_n contains 0.> ?> Define convert.An order preserving transformation.Any linear transformation like x'_n = a * x_n + b will do.iel son has presented a piecewise linear transformationthat I think is order preserving, but I not totally sure it is.Here is a simple way to determine in a conversion isorder preserving. Define a binary number, b.If x_i < x_(i+1) then bit bi = 0.If x_i > x_(i+1) then bit bi = 1.A sequence x'_n preserves the order of x_nif b' = b.=== === Subject: : Re: Rationals are UncountableNot needed to disprove your claim. One only needs -1-r < 0 < 1-r, which> follows immediately from |r| < 1. Though it is qite possible that you,> , cannot follow that simple argument without help.I agree that if X' is the set of all rationals in (-1-r,1-r) with> r irrational and |r|<1 then 0 is in X'.> What I am proving is that if x_n is a sequence that> contains every rational in X=(-1,1) then there is no> way to convert x_n to x'_n such that x'_n contains 0.While there is certainly no translation by an irrational distance that converts rationals into rationals, that is something that I pointed out to you quite a while ago. Why do you think it needs your proof as well as mine?On the other hand, every rational translation converts rationals into rationals and irrationals to irrationals.So you must answer this question: Is the translation from x_n to x'_n such that the fixed distance r = x'_n - x_n is rational or is that fixed distance irrational?If r is rational, and |r| < 1, then 0 is in X' = X - r but if r is irrational, regardless of how large it is, then there are no rational numbers at all in X' = X - r.=== === Subject: : Re: Rationals are Uncountable> I agree that if X' is the set of all rationals in (-1-r,1-r) with> r irrational and |r|<1 then 0 is in X'.> What I am proving is that if x_n is a sequence that> contains every rational in X=(-1,1) then there is no> way to convert x_n to x'_n such that x'_n contains 0.I just realized that there is a way to do this which is piecewiselinear, and therefore order-preserving.For any pair of intervals (a,b) and (c,d), with a, b, c, and d rational,there is a linear transformation taking the first interval to thesecond: f(x) = c+(d-c)*(x-a)/(b-a) For example, consider r = (sqrt(5)-1)/2x_n x'_n0 0(0,1/2) (0,1/3)1/2 1/3(1/2,3/4) (1/3,3/8)3/4 3/8(3/4,7/8) (3/8,8/21)etc.(-1/2,0) (-3/2,0)-1/2 -3/2(-3/4,-1/2) (-8/5,-3/2)-3/4 -8/5(-7/8,-3/4) (-21/13,-8/5)etc.-- iel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Re: Rationals are Uncountable> I agree that if X' is the set of all rationals in (-1-r,1-r) with> r irrational and |r|<1 then 0 is in X'.> What I am proving is that if x_n is a sequence that> contains every rational in X=(-1,1) then there is no> way to convert x_n to x'_n such that x'_n contains 0.> I just realized that there is a way to do this which is piecewise> linear, and therefore order-preserving.I am assuming that x'_n preserves the order of x_n.> For any pair of intervals (a,b) and (c,d), with a, b, c, and d rational,> there is a linear transformation taking the first interval to the> second: f(x) = c+(d-c)*(x-a)/(b-a)But, r is not rational.> For example, consider r = (sqrt(5)-1)/2I don't understand how you are choosing the intervals (a,b) and (c,d).Could you please explain?I am also curious what happens when we construct the sequencesa'_n and b'_n from x'_n. Is the LUB of a'_n irrational?If x'_n preserves the order of x_n then the members of x'_n thatmake up a'_n must be the same members of x_n that make up a_n.For example, if:a_1 = x_1a_2 = x_8a_3 = x_21...thena'_1 = x'_1a'_2 = x'_8a'_3 = x'_21...=== === Subject: : Re: Rationals are Uncountable> But, r is not rational.Which is why I have to do it in pieces:For example, consider r = (sqrt(5)-1)/2x_n x'_n0 0(0,1/2) (0,1/3)1/2 1/3(1/2,3/4) (1/3,3/8)3/4 3/8(3/4,7/8) (3/8,8/21)etc.(-1/2,0) (-3/2,0)-1/2 -3/2(-3/4,-1/2) (-8/5,-3/2)-3/4 -8/5(-7/8,-3/4) (-21/13,-8/5)etc.(For some other irrational r, the sequence 1/3,3/8,8/21,... would bereplaced with some other increasing sequence of positive rationals thatconverges to 1-r, and the sequence -3/2,-8/5,-21/13,... would bereplaced with some other decreasing sequence of negative rationsl thatconverges to -1-r.)> I don't understand how you are choosing the intervals (a,b) and (c,d).> Could you please explain?For example, if 0 < x_n < 1/2, the intervals (a,b) and (c,d) are (0,1/2)and (0,1/3); we have x'_n = 2*x_n/3. If -7/8 < x_n < -3/4, theintervals (a,b) and (c,d) are (-7/8,-3/4) and (-21/13,-8/5); we havex'_n = (8*x_n-98)/65.> I am also curious what happens when we construct the sequences> a'_n and b'_n from x'_n. Is the LUB of a'_n irrational?The reason that the LUB of a_n had to be irrational applies just as wellto the LUB of a'_n. If the LUB of a_n was (sqrt(5)-1)/2 and thetransform is as I described above, the LUB of a'_n ends up as(sqrt(5)+2)/12 .-- iel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Re: Rationals are Uncountable> The reason that the LUB of a_n had to be irrational applies just as well> to the LUB of a'_n. If the LUB of a_n was (sqrt(5)-1)/2 and the> transform is as I described above, the LUB of a'_n ends up as> (sqrt(5)+2)/12 .I will have to think about this.For example, I wonder if your method actually gets all of therationals close to r.For now, I will agree that your x'_n does maintain the order of x_nand it does contain 0.=== === Subject: : Re: Rationals are Uncountable> For example, I wonder if your method actually gets all of the> rationals close to r.In X (-1,1) or in X' (-1-r,1-r)?Because if r > 1/2, it will be nowhere near X'.If you mean something else, the question is How could it not?.In X, if x_n is positive and less than 1, the sequence 1/2, 3/4, 7/8,... has an eventual term larger than x_n, and x_n will be in thecorresponding interval (or be an endpoint). In X', if a positiverational is less than 1-r, the sequence approaching 1_r will have aneventual term larger than that rational, and again it is in thecorresponding interval (or is an endpoint). And much the same appliesto negative rationals in the intervals. -- iel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?>> hey do you guys have a copy of the proof of e is transcendental? ..id really>> appreciate if somebody could help me. ive been searching and found nothing.>Spivack, CALCULUS, 2nd edition, Chapter 21.Are you sure that the proof there is for e being transcendental and notjust irrational? I know that the proof of the irrationality of e issimple enough for a Calculus book, but that the transcendence is quite abit more complicated.Rob son take out the trash before replying=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?> hey do you guys have a copy of the proof of e is transcendental? ..id really> appreciate if somebody could help me. ive been searching and found nothing.>>Spivack, CALCULUS, 2nd edition, Chapter 21.>Are you sure that the proof there is for e being transcendental and not>just irrational? I know that the proof of the irrationality of e is>simple enough for a Calculus book, but that the transcendence is quite a>bit more complicated.Evidently you've never seen the book. It's not at all like the typicalbook titled Calculus.>Rob son > hey do you guys have a copy of the proof of e is transcendental? ..id really>> appreciate if somebody could help me. ive been searching and found nothing.>Spivack, CALCULUS, 2nd edition, Chapter 21.>>Are you sure that the proof there is for e being transcendental and not>>just irrational? I know that the proof of the irrationality of e is>>simple enough for a Calculus book, but that the transcendence is quite a>>bit more complicated.>Evidently you've never seen the book. It's not at all like the typical>book titled Calculus.I have seen the book, but unfortunately I have never had the opportunityto take or teach a class using it. It's been 15 years since I taughtcalculus, but perhaps next time I see this book, I will give it a closerRob son take out the trash before replying=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?>> hey do you guys have a copy of the proof of e is transcendental? ..id>> really appreciate if somebody could help me. ive been searching and>> found nothing.>Spivack, CALCULUS, 2nd edition, Chapter 21.>>Are you sure that the proof there is for e being transcendental and not>>just irrational? I know that the proof of the irrationality of e is>>simple enough for a Calculus book, but that the transcendence is quite a>>bit more complicated.Evidently you've never seen the book. It's not at all like the typical> book titled Calculus.Quite right, it's actually good.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.hLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?> hey do you guys have a copy of the proof of e is transcendental? ..id>> really>> appreciate if somebody could help me. ive been searching and found nothing.Spivack, CALCULUS, 2nd edition, Chapter 21.Are you sure that the proof there is for e being transcendental and not> just irrational? I know that the proof of the irrationality of e is> simple enough for a Calculus book, but that the transcendence is quite a> bit more complicated.> Yep. Irrationality is just a half-page in Chapter 20, using the Taylorpolynomial. But then Spivack adds Chapter 21. It includes a 3-pageproof that e is transcendental. But easy enough for calculus students(at least those reading Spivack's book... much more than theusual calculus book) to understand. The proof is due to Hermite,as simplified by Hilbert. It involves integrals related to theGamma function.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?> hey do you guys have a copy of the proof of e is transcendental? ..id really> appreciate if somebody could help me. ive been searching and found nothing.>>Spivack, CALCULUS, 2nd edition, Chapter 21.>Are you sure that the proof there is for e being transcendental and not>just irrational? I know that the proof of the irrationality of e is>simple enough for a Calculus book, but that the transcendence is quite a>bit more complicated.In my version of Spivak, the transcendence of e is proven in Chapter 20,and the irrationality of pi (or actually, the stronger result, theirrationality of pi^2) is proven in Chapter 16. Both chapters are markedwith an asterisk, presumably denoting that the chapters are intended forthe more advanced student. I note that Spivak describes these chapters asoptional. As others have noted, another good book is Irrational Numbersby Ivan Niven, in which a proof of the transcendence of e, pi, and manyothers, is provided in the first chapter. Specifically, Niven gives theproof of the result that if a_1, ..., a_n are distinct algebraic numbers,then exp(a_1), ..., exp(a_n) are linearly independent over the algebraicnumbers. The transcendence of e is then proven by taking a_1 = 0, a_2 = 1, and the transcendence of pi is proven by taking a_1 = 0, a_2 = i pi.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills.=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?> hey do you guys have a copy of the proof of e is transcendental? ..id> really appreciate if somebody could help me. ive been searching and> found nothing.>>Spivack, CALCULUS, 2nd edition, Chapter 21.Are you sure that the proof there is for e being transcendental and not> just irrational? I know that the proof of the irrationality of e is> simple enough for a Calculus book, but that the transcendence is quite a> bit more complicated.And Spivak actually does it!-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.hLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : one quetionToday, I read the proof of Every real # is a limit point of rational #sand I understand that proof.Now, I am curious about correctness ofEvery real # is a limit point of irrational #s.Is it true ?I think this is true.Because, For any real # p,Let a_n=(p -1/n,p +1/n).By density of irrational #, a_n contains irrational #s.Let b_n=irrational # in a_n, then b_n is a sequence of irrational #s andb_n converges to pIs my thinking correct ?=== === Subject: : Re: one quetionThere is a typo in your subject line.You probably meant one quetino: a tiny leroy.=== === Subject: : Re: one quetion> Today, I read the proof of Every real # is a limit point of rational #s> and I understand that proof.Which proof? Based on which definition of the real numbers?> Now, I am curious about correctness of> Every real # is a limit point of irrational #s.> Is it true ?> I think this is true.Yes.> Because, For any real # p,> Let a_n=(p -1/n,p +1/n).> By density of irrational #, a_n contains irrational #s.Are you taking the density of the irrationals as an axiom, or did youprove it? In fact, that assertion is the whole heart of the matter.> Let b_n=irrational # in a_n, then b_n is a sequence of irrational #s and> b_n converges to p> Is my thinking correct ?Yes, basically.-- SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.=== === Subject: : Re: Repunits prime factors: a result. Ignacio.I suppose that anyone discovered this result (the proof is so simply).It's curious!: it's not known about prime repunits and we have thatresult!.Well,,Xan.> escribi.97:>> Is the result [point 7] in somehow interesting? Just say: I will know>> if the found result is original at least.> 7. Theorem: For all b>=1 such that 2,3,5 do not divide b, then there> exist n>=0 such that b divides R(n)Yes, it is known, and the exception of 3 is unnecessary. Actually, if> gcd(b, 10) = 1 exist n such taht b divides R(n).Dem.:gcd(9b, 10) = 1 ==> 10^(phi(9b)) = 1 (mod 9b) (Th. Euler-Fermat) ==> Exist k such that 10^(phi(9b)) - 1 = k*(9b) ===> k*b = R(phi(9b))If gcd(b, 3) = 1, is enough with n = phi(b).> By example, it is a problem inDivulgaciones Matem.87ticas (Universidad del Zulia, Venezuela), Vol. 7,> n.bc 1, (p. 103)> Best ,Ignacio Larrosa Ca.96estro> A Coru.96a (Espa.96a)> ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Re: JSH: Howard Aiken quote, my situation>> Not quote Howard Aiken, but anybody know Jonathan Aitkin? He was a>> politician, an MP for the Tories, can't remember his constituency>> (Hatton?) who was revealed in the satirical press to be a liar and a>> bit of a bastard. He promised to fight his detractors with the 'shieldThe MP for *T*atton was Neil Hamilton, one of the other lying bastard> Tory MPs. Aitken was MP for South Thanet.>> of truth' and the 'sword of justice'... he ended up in prison for>> perjury after he sued a newspaper for libel. As a liberal I wnjoy that>> story anyway, but just thought I'd share it, and I can't quite put my>> finger on why it's passably relevant....Also sparked an excellent Guardian (the paper he had sued) front page the> day after the trial. Huge photo of him, with the headline 'A liar and a> cheat'. And of course it wasn't libel, as it has been proved in court:)No idea why this is on topic -- apart from Jonathan Aitken havinga different surname from Howard Aiken --- but Aitken recently published-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.hLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: JSH: Howard Aiken quote, my situation> Not quote Howard Aiken, but anybody know Jonathan Aitkin? He was a> politician, an MP for the Tories, can't remember his constituency> (Hatton?) who was revealed in the satirical press to be a liar and a> bit of a bastard. He promised to fight his detractors with the 'shield>>> The MP for *T*atton was Neil Hamilton, one of the other lying bastard>> Tory MPs. Aitken was MP for South Thanet.>> of truth' and the 'sword of justice'... he ended up in prison for> perjury after he sued a newspaper for libel. As a liberal I wnjoy that> story anyway, but just thought I'd share it, and I can't quite put my> finger on why it's passably relevant....>>> Also sparked an excellent Guardian (the paper he had sued) front page the>> day after the trial. Huge photo of him, with the headline 'A liar and a>> cheat'. And of course it wasn't libel, as it has been proved in court:)No idea why this is on topic -- apart from Jonathan Aitken having> a different surname from Howard Aiken --- but Aitken recently publishedHey I did say it was not quite Aiken, well, ok I said not quote Aiken, butnever mind. It now turns out JSH wants to start a civil suit against Decker formisleading the public. You start something as a mildly amusing joke (to meand perhaps no one else, fine) and look what happens... if* JSH loses, canwe expect a post by his evil cabal of Nora, Dik, Arturo, and postedunder A liar and cheat. * 'if' meaning should he ever get it to court, where he will lose.Still, moving further OT: hasn't Aitken now turned to religion? I've yetto see JSH cite divine right as a proof, but it can't be far off. OK, enough, I will not mention this again.=== === Subject: : Re: JSH: Howard Aiken quote, my situation>[...]>It now turns out JSH wants to start a civil suit against Decker for>misleading the public.Really? Keen - I haven't seen that post yet.Every once in a while he says something about lawsuits, but we'reall always _so_ disappointed when he never actually gets aroundto hiring a lawyer. I mean it would be like the trial of thecentury...> You start something as a mildly amusing joke (to me>and perhaps no one else, fine) and look what happens... if* JSH loses, can>we expect a post by his evil cabal of Nora, Dik, Arturo, and posted>under A liar and cheat. >* 'if' meaning should he ever get it to court, where he will lose.>Still, moving further OT: hasn't Aitken now turned to religion? I've yet>to see JSH cite divine right as a proof, but it can't be far off. >OK, enough, I will not mention this again.=== === Subject: : easy topology problem.....when a, b in Q and a < b ,show thatcollection of closed interval [a,b] is not basis for a topology on R.-----------------------i can't find contradiction. help me....please....Thank you very much in advance=== === Subject: : Re: easy topology problem.....> when a, b in Q and a < b ,show thatcollection of closed interval [a,b] is not basis for a topology on R.-----------------------i can't find contradiction. help me....please....Thank you very much in advanceThe intersection of two such sets may be nonempty withoutcontaining such a set. So it is not a base for a topology.=== === Subject: : Re: Probability Question>Rob son wibbled:>{OrigPoster was talking abut the UK national lottery - pick 6 to match 6 >from 49}I didn't know the rules of the lottery, so I answered as best as I couldwithout this information, using only what was said in the original post.With this additional information, I can deduce that the 10 ticket wouldbe for choosing 3 out of 6 correct (246820/13983816 ~= .01765040387).Choosing 100 different tickets would give a probability of C(13736996,100-m)C(246820,m)/C(13983816,100)of getting m tickets that would win 10. The chances against getting a10 winning ticket would be C(13736996,100)/C(13983816,100). Note thatthis is 246820 246820 246820 ( 1 - -------- )( 1 - -------- ) ... ( 1 - -------- ) 13983816 13983815 13983717which is between (1-246820/13983816)^100 and (1-246820/13983717)^100;both are 16.850% to 5 places.Choosing 1398382 different tickets would give a probability of C(13983815,1398382-m)C(1,m)/C(13983816,1398382)of getting m tickets that would win the jackpot. The chances againstwinning the jackpot would be C(13983815,1398382)/C(13983816,1398382).Note that this is 13983815 13983814 12585434 -------- -------- ... -------- 13983816 13983815 12585435which is exactly 12585434/13983816 = 1 - 1398382/13983816 = 90.000% to5 places. Therefore, choosing 1398382 different tickets would have a10.000% chance of winning the jackpot.Choosing 1380000 different tickets would give a 9.86855% chance ofwinning the jackpot.>>Only one set of six numbers can win the jackpot. It's relaively common >in the UK for the jackpot to be split between holders of identical >tickets: there is no all tickets different restriction in the total >sale distribution.What I meant was that only one combination of numbers wins the jackpot,whereas for smaller prizes there are many combinations of numbers whichcan win. I did not mean that only one person could win the jackpot. Ifsomeone else had submitted the same ticket, i.e. the same combination ofnumbers, they would have shared in the jackpot.What I meant by different tickets was that you didn't choose duplicatetickets, that is, more than one ticket with the same numbers.>> the odds you give, there are 14 million different lottery tickets>Roughly. Whatever 49C6 is - just under 14million.As I said above, I was going simply on the information given in theoriginal post. Since the probability of winning was 7.15x10^-8, itfollows that the count of winning numbers must be the reciprocal ofthat, or between 13976240 and 13995801. C(49,6) = 13983816 falls inthat range. As the probability was given to 3 places, I took 3 placesin the reciprocal; 14 million. I should have said, approximately.>> possible. Again assuming that you are asking only about a single>> lottery, you would need to buy 1.4 million _different_ tickets to have>> a 10% chance of winning the jackpot.>It'll be N, where the Nth root of .9 is 1-p where p is the probability >he gave. He might get away with 1.38 millionNo, if you have 13983816 different tickets (combinations of numbers),then if you place 1398382 of them, you have as close to 10% chance ofwinning the jackpot as possible. Again, I had rounded to 3 places.Rob son take out the trash before replying=== === Subject: : Re: Probability QuestionI confess that this is a homework question, anyway, the answer is1473574, using binomial/geometric/poisson distribution. Do it for abit of challenge, it is meant to be easy indeed. I don't know how tostart, I knew that I have to find the sum of a geometric series insome stage. Get the question from the previous post if you can't seeit.>Rob son wibbled:{OrigPoster was talking abut the UK national lottery - pick 6 to match 6 >from 49}I didn't know the rules of the lottery, so I answered as best as I could> without this information, using only what was said in the original post.> With this additional information, I can deduce that the 10 ticket would> be for choosing 3 out of 6 correct (246820/13983816 ~= .01765040387).Choosing 100 different tickets would give a probability of C(13736996,100-m)C(246820,m)/C(13983816,100)of getting m tickets that would win 10. The chances against getting a> 10 winning ticket would be C(13736996,100)/C(13983816,100). Note that> this is 246820 246820 246820> ( 1 - -------- )( 1 - -------- ) ... ( 1 - -------- )> 13983816 13983815 13983717which is between (1-246820/13983816)^100 and (1-246820/13983717)^100;> both are 16.850% to 5 places.Choosing 1398382 different tickets would give a probability of C(13983815,1398382-m)C(1,m)/C(13983816,1398382)of getting m tickets that would win the jackpot. The chances against> winning the jackpot would be C(13983815,1398382)/C(13983816,1398382).> Note that this is 13983815 13983814 12585434> -------- -------- ... --------> 13983816 13983815 12585435which is exactly 12585434/13983816 = 1 - 1398382/13983816 = 90.000% to> 5 places. Therefore, choosing 1398382 different tickets would have a> 10.000% chance of winning the jackpot.Choosing 1380000 different tickets would give a 9.86855% chance of> winning the jackpot.> Only one set of six numbers can win the jackpot. It's relaively common >in the UK for the jackpot to be split between holders of identical >tickets: there is no all tickets different restriction in the total >sale distribution.What I meant was that only one combination of numbers wins the jackpot,> whereas for smaller prizes there are many combinations of numbers which> can win. I did not mean that only one person could win the jackpot. If> someone else had submitted the same ticket, i.e. the same combination of> numbers, they would have shared in the jackpot.What I meant by different tickets was that you didn't choose duplicate> tickets, that is, more than one ticket with the same numbers.> the odds you give, there are 14 million different lottery ticketsRoughly. Whatever 49C6 is - just under 14million.As I said above, I was going simply on the information given in the> original post. Since the probability of winning was 7.15x10^-8, it> follows that the count of winning numbers must be the reciprocal of> that, or between 13976240 and 13995801. C(49,6) = 13983816 falls in> that range. As the probability was given to 3 places, I took 3 places> in the reciprocal; 14 million. I should have said, approximately.>> possible. Again assuming that you are asking only about a single>> lottery, you would need to buy 1.4 million _different_ tickets to have>> a 10% chance of winning the jackpot.It'll be N, where the Nth root of .9 is 1-p where p is the probability >he gave. He might get away with 1.38 millionNo, if you have 13983816 different tickets (combinations of numbers),> then if you place 1398382 of them, you have as close to 10% chance of> winning the jackpot as possible. Again, I had rounded to 3 places.Rob son Can anyone give me some help with this question,> The probability of winning exactly 10 dollars on the National Lottery is> 0.01765. If I buy 100 tickets, what is the probability that I do not> win a 10 dollars prize? The probability of winning the jackpot on the> National Lottery is 7.15x10^-8. How many tickets do I need to buy to> have at least a 10% chance of winning the jackpot?This smells like a high school home work question. No wonder no onehas answered. Show us what you have, before anyone cares to answeryou.-- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52=== === Subject: : Countable perfect setIt is known that every perfect (every point is an accumulation point)closed subset of a complete metric space is uncountable.But, what about a non-complete metric space ? The proof of theprevious case cannot be adapted, and I don't manage to find a counterexample.Does anyone have any idea ? Nobody knows a perfect, closed andcountable subset of a metric space ?Vinect G.=== === Subject: : Re: Countable perfect set> It is known that every perfect (every point is an accumulation point)> closed subset of a complete metric space is uncountable.But, what about a non-complete metric space ? The proof of the> previous case cannot be adapted, and I don't manage to find a counter> example.Does anyone have any idea ? Nobody knows a perfect, closed and> countable subset of a metric space ?Maybe I'm not seeing this properly (perhaps have a definition wrong).What if you only have one point x in your space X={x}? Give it theobvious topology ({} and {x}=X are open). Make your metric be d(x,y)= 0 since x must be equal to y. I think x is the only possibleaccumulation point.-- === === Subject: : Re: Countable perfect set> It is known that every perfect (every point is an accumulation point)> closed subset of a complete metric space is uncountable.But, what about a non-complete metric space ? The proof of the> previous case cannot be adapted, and I don't manage to find a counter> example.Take the metric space of rational numbers, with the usual metric.Does anyone have any idea ? Nobody knows a perfect, closed and> countable subset of a metric space ?-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/=== === Subject: : Re: Modular Arithmetic> I was wondering if anyone knows a way to reduce 2^100 mod 5. I know> the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what> lot.Quick method note 2^2 = 4 ==> -1 MOD 52^100 = 4^50 { 50 is even} -1^{even}= 1So [2^100] MOD 5 = 1A^N MOD B ==> [A MOD B]^N MOD B=== === Subject: : Re: Modular ArithmeticI was wondering if anyone knows a way to reduce 2^100 mod 5. I know> the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what> lot.Quick method> note 2^2 = 4 ==> -1 MOD 52^100 = 4^50 { 50 is even} -1^{even}= 1So [2^100] MOD 5 = 1A^N MOD B ==> [A MOD B]^N MOD BOr 2^4 = 16 == 1 mod 5, etc.=== === Subject: : Re: How far can one go with self-study nowadays?X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWSIn , on>Just thinking about all the neat things available online, compared to>when I was in school 20ish years ago, I'm wondering how far one can>go nowadays in mathematics with self-study.Quite far, but it will be a lot easier if you have access to a goodlibrary. Access to a good university is better. If a good localuniversity allows you to audit classes, that's even better, but atthat point it's no longer self study.>Compare that to nowadays. You can find textbooks easily online, and>find customer reviews of them,I don't have much confidence in those customer reviews. I'd muchrather talk to a few professors in a good Mathematics department andpay careful attention to what they have to say.>If you get stuck on something, there is a fair chance that a >Google search will find a solution, and if not, there is usenet--a >post to sci.math will probably get an answer.The WWW and Usenet are certainly valuable resources, but they can'ttake the place of personal contact, attence at colloquia, etc.>I think one should be able to reasonably get to at least a fair way>into graduate level. Of course, but it will be a lo0t harder than if you had localresources.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : One question for calculus expertsplease help me solving this problem..Let Omega be an open, bounded subset of R^n , let u : Omega > R^nbe a function: Then, we can identify the curl of u as vector in R^(n(n-1)/2 ). Now, let T: R^( n(n-1)/2) -> R^( n(n-1)/2) be a linearisomorphism.Then, is it true that the composition map Tcurlu: Omega >R^(n(n-1)/2 ) is the curl of some function v defined on Omega ????Thank you in advance=== === Subject: : Re: :: towards a constructive education :: (news server friendly)> While you are reading this, your retinal rods and cones are[etc. appreciated constructivism promotion campaign snipped.]Would you care to give references for each of theapplications that you mention? Because in mostcases, i have no idea what you are talking about.And for the rest, i had no idea they had anythingto do with Heyting algebras perse...Or perhaps there already is a book Applicationsof Heyting Algebras ? If not, you may considerwriting one yourself. :-)Cheers,Herman Jurjus (posting from the planet sci.math)=== === Subject: : Re: :: towards a constructive education :: (news server friendly)Hi ie, still not back from kiddie league? I'm still waiting for> an answer more worthy of yourself.> Mostly, it isn't worth the effort.Oh.> I spent the last year dealing with someone exhibiting the same crude behavior when I was> trying to talk about logic and set theory. Why don't you give me some sense of what you> know about Boolean-valued forcing, partition characteristics of large cardinals, or even> descriptive set theory?Because that is not what we are discussing. You posted some crapquotation of Jeremy Bentham. I gave you my answer to it, quoted somebecame the one showing crude behaviour; hey that's my part in thisdialogue! As far as I am concerned, this is about the objectiveexistence of right and wrong. Is there anything else you want to sayabout that?> I do not have Galathaea's wide background. So, it is unlikely I would be for now it looks as if Galathaea is all width and no depth, allreferences and no meat. Like someone out of a Borgean nightmare. We'llsee.> able to mount a> decent argument against someone as esteemed as yourself.> My suspicion, however, is that you do not even understand the numbers they keep score with> in my league. But, then maybe you do.No, I guess you're right. I'm also happy to have provided you with anopportunity to state what a big shot you are, you must sorely needeveryone you get.However on a more pedestrian level the score is for now against you.The ball is on your side of the court.=== === Subject: : Re: :: towards a constructive education :: (news server friendly)> Hi ie, still not back from kiddie league? I'm still waiting for> an answer more worthy of yourself.>Mostly, it isn't worth the effort.> Oh.> I spent the last year dealing with someone exhibiting the same crude behavior when I was> trying to talk about logic and set theory. Why don't you give me some sense of what you> know about Boolean-valued forcing, partition characteristics of large cardinals, or even> descriptive set theory?> Because that is not what we are discussing. You posted some crap> quotation of Jeremy Bentham. I gave you my answer to it, quoted some> became the one showing crude behaviour; hey that's my part in this> dialogue! As far as I am concerned, this is about the objective> existence of right and wrong. Is there anything else you want to say> about that?No. But I thank you for this insight. A while back I would have been much more patient aboutsomeone choosing to be argumentative toward a post from the outset. I became angered from themoment I opened your first response and set myself up for this. As I said, it is not worth theeffort.:-)=== === Subject: : Are these two polynomials equivalent?Can I claim that f(x)=a+b(x-e)+c(x-e)^2+d(x-e)^3 ,g(x)=m+n(x-e)+p1(x-e)^3+...+pn(x-e)^3are equivalent?a,b,c,d,e,m,n,p1,p2,...,pn are constants, n is arbitraryThe key is, is it that there always exists p1,...,pn such that we candrop the quadratic term c(x-e)^2 in f(x) ?(my problem arises from the representation of a spline, because thebook seems to say that a cubic spline CAN always be expressed in thesecond form)=== === Subject: : Re: Are these two polynomials equivalent?> Can I claim that f(x)=a+b(x-e)+c(x-e)^2+d(x-e)^3 ,> g(x)=m+n(x-e)+p1(x-e)^3+...+pn(x-e)^3are equivalent?> a,b,c,d,e,m,n,p1,p2,...,pn are constants, n is arbitraryThe key is, is it that there always exists p1,...,pn such that we can> drop the quadratic term c(x-e)^2 in f(x) ?(my problem arises from the representation of a spline, because the> book seems to say that a cubic spline CAN always be expressed in the> second form)g(x) can simply be written as m+n(x-e)+p(x-e)^3, where p = p1+p2+...+pn.So the question becomes: given a,b,c,d,e, can you find m,n,p such that:a+b(x-e)+c(x-e)^2+d(x-e)^3 = m+n(x-e)+p(x-e)^3 ?Let's look at the expansions:f(x) = a + bx-be + cx^2-2cxe+ce^2 + dx^3-3dex^2+3de^2x-de^3 = (d)x^3 + (c-3de)x^2 + (b-2ce+3de^2)x + (a-be+ce^2-de^3)g(x) = m + nx-ne + px^3-3pex^2+3pe^2x-pe^3 = (p)x^3 + (-3pe)x^2 + (n+3pe^2)x + (m-ne-pe^3)So if f(x) = g(x),d = pc-3de = -3peb-2ce+3de^2 = n+3pe^2a-be+ce^2-de^3 = m-ne-pe^Looking at the first 2 lines, f(x) = g(x) means c-3de = -3de means c=0.Having said all that, I haven't worked with cubic splines, but I suspect that the definition of the cubic spline is what causes the (x-e)^2 term to not exist.-- === === Subject: : block matrix, determinant zeroI know that upper or lower triangular matrices have determinant zero,which can be directly seen from the cofactor formula.How does this extend to the fact thatdet(A,0; 0,D) = det(A)det(D) ?Or is this proven in other ways ?=== === Subject: : Re: block matrix, determinant zero> I know that upper or lower triangular matrices have determinant zero,> which can be directly seen from the cofactor formula.Not so. The determinant of a triangular matrix is the product of thediagonal entries, which can be directly seen from the cofactor formula.=== === Subject: : triangulationIs there anyone in this group, who can help me with the triangulationof a given polygon P under these two restrictions:1. Each triangle shall be as equilateral as possible2. No triangle has an area that exceeds a given constant OHans van Duijnhoven (Randers, mark).=== === Subject: : Re: triangulation 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21:> Is there anyone in this group, who can help me with the triangulation> of a given polygon P under these two restrictions:> 1. Each triangle shall be as equilateral as possible> 2. No triangle has an area that exceeds a given constant O> Hans van Duijnhoven (Randers, mark).Most problems of optimal triangulation of polygons (without holes) can be solved by dynamic programming.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer Science=== === Subject: : Re: triangulation>> Is there anyone in this group, who can help me with the triangulation>> of a given polygon P under these two restrictions:>> 1. Each triangle shall be as equilateral as possible>> 2. No triangle has an area that exceeds a given constant O>> Hans van Duijnhoven (Randers, mark).>Most problems of optimal triangulation of polygons (without holes) can >be solved by dynamic programming.> Hans=== === Subject: : Re: Question for logarithm experts by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AE2qL25626;>This is an interesting posting. I've never really thought about this before,>but do all equations have solutions, (even though it may not be possible for>us to find them in terms of simple functions) ?Look at the following:7^X=12*X has no solutuionTry:7^X=[12.34683945]*X>Graphing does not give us the whole story does it? Try a relevant graph topic:http://www.stefanides.gr/why_logarithm.htmWhat I am thinking about>is that for example>y = x^2 does not cross y = - 1, yet we know that x^2 = - 1 has solutions>x = + or - i, (the sqrt of minus one).>Does the topic you are studying suggest the use of complex numbers?>If you replace x with z = a+b i, you could write>7^z = e^((ln7)z) = (e^(aln7)) (e^i(bln7))= (7^a)[cos(bln7) + i sin(bln7) ] =>4a + 4bi 7^Z may be expanded : 7^Z=1+Z*[ln[7]+[{Z*[ln[7])^2]/2!+...... , according to Ref: http://www.stefanides.gr/why_logarithm.htmPanagiotis Stefanideshttp://www.stefanides.gr=== === Subject: : Re: Silly question for someone with a big calculator. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AE2qg25622;>>4/3 = 1.333333...>>24/17 = 1.411764...>>816/577 = 1.414211...>>941664/665857 = 1.414213...>>If we let k represent that ratio for one of the exponents, the ratio for>>the next in this list is (4k)/(k^2+2); it can be shown that the ratios>>will converge to sqrt(2).>>> Sorry, I think I must be overlooking something (I do a lot). It>> certainly *looks* to me as if the above sequence will converge to>> sqrt(2); however, I am neither sure what your k is in (4k)/(k^2+2)>> nor how one can show that that the sequence of ratios converges, >> of which no closed form exists- unless I`m overlooking something, >> of course.>If k = 4/3, (4k)/(k^2+2) = 24/17.>If k = 24/17, (4k)/(k^2+2) = 816/577.>etc.>If we let delta = 2-k^2, the delta for the following k will be>2*delta^2/(k^2+2)^2 . For example, for k=4/3, we have delta = 2/9. For>k=24/17, we have delta = 2/289 = 2*(2/9)^2/((4/3)^2+2)^2 . Since>repeating this often enough will get you a delta as close to 0 as you>want, k will get as close to sqrt(2) as you want.are, even though you explained them correctly the first time round.However, forgive me for being quite stubborn, aren't we effectively saying - instead of the usual: let (k_n) be the sequenceof ratios, then for every epsilon > 0 exists N in naturals:|k_n - sqrt(2)| < epsilon for all n > N - for any epsilon > 0,we can find you a bunch of examples such that... ? (Indirectly of course, since we are trying to show delta -> 0through substitution instead of k_n -> sqrt(2) ).Sincerly,C. Dement >iel W. son>panoptes@iquest.net>http:// members.iquest.net/~panoptes/ are, even though you explained them correctly the first time round.> However, forgive me for being quite stubborn, aren't we > effectively saying - instead of the usual: let (k_n) be the sequence> of ratios, then for every epsilon > 0 exists N in naturals:> |k_n - sqrt(2)| < epsilon for all n > N - for any epsilon > 0,> we can find you a bunch of examples such that... ? > (Indirectly of course, since we are trying to show delta -> 0> through substitution instead of k_n -> sqrt(2) ).Since k_{n+1} would fall between k_n and sqrt(2), we would only need tofind a first example k_N, and then all k_n with n>N will be closer tosqrt(2).-- iel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Partition questionRe:my Cat question - if commutativity holds, the number of different bracketings is stilllower and can be though of as ways in a partitiongraph (example n=4): 4 / 22 311 / 211 | 1111One path would be ((a*a)*(a*a)) and the other (a*(a*(a*a))).Unluckily, #ways != #bracketings, as n=5 already shows:2111->221->23 or 2111->311->32 are different ways tothe same bracketing. But surely the sequence is known?-- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.deals man ankam wollte man werden, die geschichte schreiben,die doofen sollen sterben, der plan als man damals nach hamburg kam(Kettcar)=== === Subject: : Re: Countable perfect set by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AEUFu27848;I don't know if you will consider this a satisfying example, but how about the set Q of rational numbers. Q is closed in itself, is countable, and every point is an accumulation point. Or if you want a proper subset of Q, I suppose the rationals in [0,1] would work.Mark Motley >It is known that every perfect (every point is an accumulation point)>closed subset of a complete metric space is uncountable.>But, what about a non-complete metric space ? The proof of the>previous case cannot be adapted, and I don't manage to find a counter>example.>Does anyone have any idea ? Nobody knows a perfect, closed and>countable subset of a metric space ?>Vinect G.=== === Subject: : CHALLENGEHiI have a beautiful puzzle, and i also have a general method forfinding the answer but what i'm looking for is a proof for why themethod works.Here's the puzzle:Consider n number of poeple standing in a circle. All of them arenumbered from 1 to n. Now, Man no. 1 is given a sword. He kills theperson next to him (i.e Man No. 2) and passes the sword on to Manno.3. Man 3 in turn kills the man next to him (No. 4) and passes thesword on to the next man (i.e Man No. 5). This sequence continuesuntil only one man is alive. Find which number is it.For eg, take n=5 then:|1| 2 3 4 5 (the number in | | has the sword)1 |3| 4 51 3 |5||3| 5|3| Thus in the end man no. 3 remains.Try the puzzle if you haven't seen it before, its a nice exercise forthe brain.Here's my conjecture for a general method to find the solution.1) Find the binary representation for the number of people. eg for n=5it is 1012) take the first 1 of the representation and put it at the end. so,here it becomes 011.3) Now convert this new binary number back to decimal and that is theanswer. The decimal of 11 is 3, which is the answer.I have checked this method for many many values using a java programthat I made , and it does work. But, I could not think of a proof forthis. I will greatly appreciate it if you could post the proof shouldyou find it.=== === Subject: : Re: CHALLENGE>I have a beautiful puzzle, and i also have a general method for>finding the answer but what i'm looking for is a proof for why the>method works.>Here's the puzzle:>Consider n number of poeple standing in a circle. All of them are>numbered from 1 to n. Now, Man no. 1 is given a sword. He kills the>person next to him (i.e Man No. 2) and passes the sword on to Man>no.3. Man 3 in turn kills the man next to him (No. 4) and passes the>sword on to the next man (i.e Man No. 5). This sequence continues>until only one man is alive. Find which number is it.>For eg, take n=5> then:>|1| 2 3 4 5 (the number in | | has the sword)>1 |3| 4 5>1 3 |5|>|3| 5>|3| >Thus in the end man no. 3 remains.>Try the puzzle if you haven't seen it before, its a nice exercise for>the brain.>Here's my conjecture for a general method to find the solution.>1) Find the binary representation for the number of people. eg for n=5>it is 101>2) take the first 1 of the representation and put it at the end. so,>here it becomes 011.>3) Now convert this new binary number back to decimal and that is the>answer. The decimal of 11 is 3, which is the answer.>I have checked this method for many many values using a java program>that I made , and it does work. But, I could not think of a proof for>this. I will greatly appreciate it if you could post the proof should>you find it.To make things simpler, let us index the people starting with 0 insteadof 1; we can add 1 at the end to convert this index to the numbers youare using.After each pass, in which every person in the circle has had a chance tokill or be killed, the indices of those remaining differ by a power of 2that increases each pass. If, at the beginning of a pass, there were aneven/odd number of people, at the end of the pass, the person with thelowest index prior to the pass remains/departs.Before pass p, there are floor(n/2^{p-1}) people in the circle and theirindices differ by 2^{p-1}. After pass p, there are floor(n/2^p) peoplein the circle and their indices differ by 2^p. At the end of pass p, iffloor(n/2^{p-1}) is odd, the lowest index in the circle is increased by2^p due to the departure of the lowest indexed person. Otherwise, thelowest indexed person remains.This means that if bit p-1 is set in n, then at the end of pass p, bit pis set in the lowest index in the circle. As soon as floor(n/2^p) = 1,there is one person left in the circle and their index has accumulatedthe lower order bits of n, shifted one to the left. Finally, add 1 tochange my indices to your numbers, and you get your result.Rob son take out the trash before replying=== === Subject: : Re: CHALLENGE charset=utf-8 siddharth jain [CapitalEth][EDoubleDot][Micro] .b3 .b9[EDoubleDot].b9> Here's the puzzle:> Consider n number of poeple standing in a circle. All of them are> numbered from 1 to n. Now, Man no. 1 is given a sword. He kills the> person next to him (i.e Man No. 2) and passes the sword on to Man> no.3. Man 3 in turn kills the man next to him (No. 4) and passes the> sword on to the next man (i.e Man No. 5). This sequence continues> until only one man is alive. Find which number is it.> For eg, take n=5> then:> |1| 2 3 4 5 (the number in | | has the sword)> 1 |3| 4 5> 1 3 |5|> |3| 5> |3|> Thus in the end man no. 3 remains.It's ill defined. You didn't specify that the person with the sword alwayskills a person with a higher number. And even if this is the case, at somepoint someone is bound to start killing people with lower numbers, since itsa modulo circle. So:At step 2 above, it could easily be:1 |3| 4 54 |5||5|> Try the puzzle if you haven't seen it before, its a nice exercise for> the brain.> Here's my conjecture for a general method to find the solution.> 1) Find the binary representation for the number of people. eg for n=5> it is 101> 2) take the first 1 of the representation and put it at the end. so,> here it becomes 011.> 3) Now convert this new binary number back to decimal and that is the> answer. The decimal of 11 is 3, which is the answer.> I have checked this method for many many values using a java program> that I made , and it does work. But, I could not think of a proof for> this. I will greatly appreciate it if you could post the proof should> you find it.--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandable=== === Subject: : How to compute the minimal distanct between a point and curve in N-dim space,In the N-dimensional space, give a data point A and a curve f,how to write the explicit expression for calculating theminimal distance between A and f?Or have to use some nonlinear optimization method to calcualte it?Fred=== === Subject: : Integro-differential equation!I need to solve the following partial integro-differential equation for P(x,t)P_t(x,t) = a*P_x(x,t) + b*P_{xx}(x,t) - c*P(x,t) + c*{ int_0^{x} P(x-v,t)*e^{-v/d}/d ,dv + e^{-x/d} }subject toP(0,t) = 1P(x,0) = 0and a,b,c, and d are constants.My approach so far has been to Laplace transform first for t (t->s1, get ordinary integro-differential equation for P(x,s1)) and then for x (x->s2, get algebraic equation for P(s2,s1)). Then I can solve for P(s2,s1), do the inverse transform to get P(x,s1). Unfortunately the expression for P(x,s1) is to complex to find the inverse transform so I'm left with the solution in the form of a Bromwich integral (which I presumably could evaluate numerically for all relevant values).Does anyone know of a better way to attack this problem?Stefan NilssonMarine ecologyG.9ateborg University=== === Subject: : Mod operatorMost mathematicians would say that x mod 7 is always avalue between 0 and 6, regardless of the sign of x. Butmany computer languages do this the wrong way. Forexample, most mathematicians would insist that -15 mod 7 = 6but many (but not all) programming languages give the wrongresult, -1. In fact, the programming language Ada has twomod operators. One which returns 6 and one which returns -1in the above example.Insightful comments welcome.-- Mail sent to this email address is automatically deleted(unread) on the server. Send replies to the newsgroup.=== === Subject: : Re: Mod operator> Most mathematicians would say that x mod 7 is always a> value between 0 and 6, regardless of the sign of x. But> many computer languages do this the wrong way. For> example, most mathematicians would insist that -15 mod 7 = 6but many (but not all) programming languages give the wrong> result, -1. In fact, the programming language Ada has two> mod operators. One which returns 6 and one which returns -1> in the above example.Insightful comments welcome.Suppose you have a clock labeled 0 to 11: 0 11 1 10 2 9 3 8 4 7 5 6Moving clockwise 13 steps (from 0), you end up at 1, so13 mod 12 = 1Moving counter-clockwise 13 steps (from 0), you end up at 11, so-13 mod 12 = 11Note that there are no negative numbers onthis clock, so the answer is always positive(for this situation).=== === Subject: : Re: Mod operator Adjunct Assistant Professor at the University of Montana.>Most mathematicians would say that x mod 7 is always a>value between 0 and 6, regardless of the sign of x.No. That would be the remainder of dividing x by 7 (assuming that x isan integer, anyway). Most mathematicians would say that x is congruentmodulo 7 to one and only one of 0, 1, 2, 3, 4, 5, and 6; but it is alsocongruent to one and only one of -3, -2, -1, 0, 1, 2, and 3, and oneand only one of 15, 16, 17, 18, 19, 20, and 21.In fact, they will know that the relation congruent modulo 7partitions the integers into seven equivalence classes, of which 0, 1,2, 3, 4, 5, 6 are a set of representatives, but that there are aninfinite number of such sets of representatives.> But>many computer languages do this the wrong way. For>example, most mathematicians would insist that> -15 mod 7 = 6>but many (but not all) programming languages give the wrong>result, -1. No; mathematicians would agree that -15 = 6 (mod 7), and they wouldalso agree that -15 = -1 (mod 7). What they may not agree on iswhether -1 is the remainder of dividing -15 by 7, since remaindersare defined to be nonnegative.>In fact, the programming language Ada has two>mod operators. One which returns 6 and one which returns -1>in the above example.While it is usually convenient to pick the standard set ofrepresentatives for the congruent modulo k relation to be 0, 1, 2,3, 4, ..., k-1 (for instance, that makes the representative of theclass of n modulo k to be exactly equal to the remainder of dividing nby k), it is also often very useful to pick as a set ofrepresentatives the ones with smallest absolute value, which would be,depending on the parity of k, either (-k/2)+1,...,0,1, 2, ..., (k/2)or (-k/2), (-k/2)+1,...,0,1,2,...,(k/2)-1 if k is even; and (1-k)/2,(1-k)/2+1,...,-1,0,1,2,...,(k-1)/2 if k is odd. (For someapplications, other choices may even be preferable). Which one thecomputer program chooses is immaterial, so long as you know which oneit is choosing and does so in a consistent manner.-- ============================================================== ============ === Subject: : Re: Mod operator>>Most mathematicians would say that x mod 7 is always a>>value between 0 and 6, regardless of the sign of x.>No. That would be the remainder of dividing x by 7 (assuming that x is>an integer, anyway). Most mathematicians would say that x is congruent>modulo 7 to one and only one of 0, 1, 2, 3, 4, 5, and 6; but it is also>congruent to one and only one of -3, -2, -1, 0, 1, 2, and 3, and one>and only one of 15, 16, 17, 18, 19, 20, and 21....>While it is usually convenient to pick the standard set of>representatives for the congruent modulo k relation to be 0, 1, 2,>3, 4, ..., k-1 (for instance, that makes the representative of the>class of n modulo k to be exactly equal to the remainder of dividing n>by k), it is also often very useful to pick as a set of>representatives the ones with smallest absolute value, which would be,>depending on the parity of k, either (-k/2)+1,...,0,1, 2, ..., (k/2)>or (-k/2), (-k/2)+1,...,0,1,2,...,(k/2)-1 if k is even; and (1-k)/2,>(1-k)/2+1,...,-1,0,1,2,...,(k-1)/2 if k is odd. (For some>applications, other choices may even be preferable). Which one the>computer program chooses is immaterial, so long as you know which one>it is choosing and does so in a consistent manner.And with `C', you do not know which one it will choose if either ofthe operands is negative, because the designers, in their wisdom,opted to make it implementation dependent. I cannot imagine whatprompted them to do that.Derek Holt.=== === Subject: : Re: Mod operator> Most mathematicians would say that x mod 7 is always a> value between 0 and 6, regardless of the sign of x. But> many computer languages do this the wrong way. For> example, most mathematicians would insist that -15 mod 7 = 6but many (but not all) programming languages give the wrong> result, -1. In fact, the programming language Ada has two> mod operators. One which returns 6 and one which returns -1> in the above example.Insightful comments welcome.> -1 mod 7 = 6The issue is more one of convention than anything, IMO. What the mod operator actually does is define sets of numbers that are considered equivalent under a particular relation.So, for instance, -1 mod 7 = 6 really says that -1 is in the same set as 6 as defined by the (mod 7) relation. You could as easily say6 mod 7 = -15though most people would think it looks strange.-- === === Subject: : Re: Mod operator> Most mathematicians would say that x mod 7 is always aNo Real Mathematicians talk about x mod 7. They recognizemod as part of the ternary relation of congruence:a = b (mod n) means that (a-b)/n is an integer.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.hLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: Mod operator> In fact, the programming language Ada has two mod operators. One> which returns 6 and one which returns -1 in the above example.Are they both called mod?-- === === Subject: : Re: Mod operator> In fact, the programming language Ada has two mod operators. One> which returns 6 and one which returns -1 in the above example.> Are they both called mod?IIRC, it depends upon the types being operated on. In Ada95 the integer types are subdivided into signed integertypes and modular types. The signed integer types areInteger and so on. The modular types are unsigned integertypes which exhibit cyclic arithmetic.Hopefully someone will correct me if my memory fails me.-- Mail sent to this email address is automatically deleted(unread) on the server. Send replies to the newsgroup.=== === Subject: : Re: Mod operator> In fact, the programming language Ada has two mod operators. One> which returns 6 and one which returns -1 in the above example.>> Are they both called mod?IIRC, it depends upon the types being operated on. In Ada 95 the> integer types are subdivided into signed integer types and modular> types. The signed integer types are Integer and so on. The modular> types are unsigned integer types which exhibit cyclic arithmetic.I'm not sure I follow, but never mind Ada. What I was getting at is,mere presence of some kind of residue operation in a programminglanguage does not mean that the language has got modulo wrong. It maysimply be another operation with another name. Some languages have anoperation called modulo and another operation called remainder,and these differ when negative numbers are involved.Java has no mod at all. The % is called remainder operator in thespecification. So in order to say they got mod wrong, you have toinsist that they named it wrong first. To me that seems like firstcalling sin tan and then complaining that programming languagesbreak trigonometry. Sin is sin and tan is tan. Both are needed.Java's per cent operation seems wrong, though. On the other hand,mod could be added: just use the per mille character as anoperator. Seriously, it might have been better if per cent had beenmod instead of remainder. Too late.-- === === Subject: : Re: Math Too Advanced For Mainstream Economists> I have written up a demonstration that wages and employment need not> be determined by the intersection of well-behaved supply and demand> curves for labor:> The poster claims to have constructed a model that has something to dowith the intersection of well-behaved supply and demand curves forlabor, but he has not. Rather, he comes close to presenting a modelof the factor demand for labor, but he confuses movements along ademand curve with shifts of that curve. Further, he has confusionabout the competition assumptions that go with the basic supply-demandsetting, which set the rate of profits independently from the wagerate. As such, he may have a model of something, but it's nothingrelated to the workings of a competitive labor market.Even if this were a model of labor market behavior that contradictedthe usual results of the standard supply and demand model, it wouldnot be an interesting contribution unless it did a better job than thesupply and demand (or other established labor models) of describingsomething interesting about the observed world. There are plenty oflabor market models that use tools other than standard supply anddemand analysis, I've cited some of them in this thread, but theirvalue comes from their empirical relevance.Does this model show something of interest to any serious researcher?What do we observe that it explains better than competing approaches?> I think this thread would amusing to those who find JSH threads> amusing, but with a twist. I was delighted to find in a dictionary the word MUMPSIMUS,> which means stubborn persistence in an error that has been> exposed.> -- Joan Robinson> ...the original poster was unable to find a single piece of> empirical support for the labor market model presented...The above, of course, is untrue, and it is hard to see how poor> Mark Witte cannot know it is untrue, if he had any clue about> what he is talking about. A> demand curve is a relationship between price and quantity demanded. > If something else changes, such as prices of related goods in demand> (here the discount rate would be an example), this shifts the curve.Poor Mark Witte is confused. A discount rate is not a price.To be charitable to the original poster, his confusion probably stems> from his notion that a change by the firm in question in the number of> workers it hires would change the discount rate for the economy.The above is a strawperson. No such claim is to be found at the above> URL.Notice no claim is made about the discount rate for the economy> being changed by a change in how many workers firm hires. In fact,> an economy-wide market for financial capital is not described at> my URL.Poor Mark Witte just doesn't understand accounting. If the best> return a firm can obtain is 10%, then that's what the firm gives> up in investing elsewhere. If costs are different for the best> investment (e.g., because the level of wages is different), then the> firm gives up some other rate of return than 10%.To help poor Mark Witte out, here's some explanations of a concept> important for understanding Step 2 of the algorithm given at my> URL:> My URL is more a tutorial to results well-established in the literature.> No claim is made to novelty there. For that matter, no claim is made> there that the relationship shown is a demand curve for labor. But that> claim is made in the literature.-- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.h> To solve Linear Programs: .../LPSolver.h> r c A game: .../Keynes.h> v s a Whether strength of body or of mind, or wisdom, or> i m p virtue, are found in proportion to the power or wealth> e a e of a man is a question fit perhaps to be discussed by> n e . slaves in the hearing of their masters, but highly> @ r c m unbecoming to reasonable and free men in search of> d o the truth. -- Rousseau=== === Subject: : Re: Rings by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AG2E803159;Thank you Hubert, please see my comments hereunder:>> A Gaussian Ring is an integral domain verifying>> (i) the intersection of two principal ideals is a principal ideal>> Equivalent conditions are>> (ii) each couple of elements has a GCD>> (iii) each couple of elements has a LCM>> (iv) any couple of elements is elements is contained in a minimum principal ideal>> (v) any ideal generated by two elements is contained in a minimum principal ideal.>> I. Consider G=Z[X][[X1,X2,...]].>> a) Is G a Gaussian Ring ?>yes, because every factor ring verify (ii)I am not so sure that it is a factor ring. The ring to be considered is a formal power series ring on the ring Z[X]. If a ring A is a principal ring, then the formal power series ring A[[X]] is a factor ring. But the condition A factor ring does not imply A[[X]] factor ring. And Z[X] is a non principal factor ring, even if Z is principal.>> b) Is G a Factor Ring ?>Yes>It's a well known problem : if A is a factor ring, so is A[X]See above under a)>> c) Does G verify the Bezout condition ?>No>GCD(X1,X2)=1>But PX1+QX2 = 1 is false. Assuming PX1+QX2 =1, you substitute 0 to X1 >ans X2 you find 0=1.>> II. Any (other?) example of a Gaussian non Factor non Bezout Ring ? >Now you should find an example of a gaussian ring that is not a factor ring.That remains to be done.=== === Subject: : Riemann Integrationlet f:[0,1]->R be continious. then set of all such f's such that inegration with limits 0 to 1 of t^n*f9t) dt=0 isa) single elementb)contably infiniteC) infinite === === Subject: : Re: Riemann Integration :>let f:[0,1]->R be continious. then set of all such f's such that inegration with limits 0 to 1 of t^n*f9t) dt=0 isThis problem is not well posed. What is n supposed to be?>a) single element>b)contably infinite>C) infinite-- === === Subject: : Re: Riemann Integration> let f:[0,1]->R be continious. then set of all such f's such that> inegration with limits 0 to 1 of t^n*f9t) dt=0 isPlease re-read your posts before posting them; it should be t^n*f(t).> a) single element> b)contably infinite> C) infiniteThe first possibility is the correct one.Best ,Jose Carlos Santos=== === Subject: : Re: there is no such thing as infinityDr. Ben ZonaIs your brother called Ari?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.hLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: there is no such thing as infinity>> And does this zoned out doctor claim that all computers will stop at>> the same M?That is an excellent point. That is why I am looking for collaborators> who will repeat the experiment on different computers. My theory is> this:Even if different computers were to output different answers for M, we> could always take the average of these numbers and calculate the> variance too. Then we would at least have a good statistical estimate> of the range of M.Dr. Ben ZonaI just ran it on my computer. It went all the way up to 2147483647 and then went to to -2147483648 and started counting up again. I guess we've found our first candidate for M. ===Timothy M. BrauchGraduate StudentDepartment of MathematicsWake Forest University=== === Subject: : Re: there is no such thing as infinity> I just ran it on my computer. It went all the way up to 2147483647> and then went to to -2147483648 and started counting up again. I> guess we've found our first candidate for M.Very inefficient. I modified the program to count in incrementsof M/10, and it was done very quickly. It turns out thatM= 11(M/10), which is embarrassing mostly because if we'd havethought of this sooner, we could have worked it out by hand.Well, I'm glad that's finally settled. (Hmmm...I wonder whatjournal I should send this off to....)Bart === === Subject: : Re: there is no such thing as infinity>> I just ran it on my computer. It went all the way up to 2147483647>> and then went to to -2147483648 and started counting up again. I>> guess we've found our first candidate for M.> Very inefficient. I modified the program to count in increments> of M/10, and it was done very quickly. It turns out that> M= 11(M/10), which is embarrassing mostly because if we'd have> thought of this sooner, we could have worked it out by hand.> Well, I'm glad that's finally settled. (Hmmm...I wonder what> journal I should send this off to....)The two-line program that I posted previously, namely print *, huge(1) endprints 2147483647 on my machine. Huge is a standard intrinsicfunction.-- SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.=== === Subject: : Re: there is no such thing as infinity@mozo.cc.purdue.edu: >> Very inefficient. I modified the program to count in increments>> of M/10, and it was done very quickly. It turns out that>> M= 11(M/10), which is embarrassing mostly because if we'd have>> thought of this sooner, we could have worked it out by hand.>> Well, I'm glad that's finally settled. (Hmmm...I wonder what>> journal I should send this off to....)The two-line program that I posted previously, namely print *, huge(1)> endprints 2147483647 on my machine. Huge is a standard intrinsic> function.Must be a glitch, since 11/10 * Huge isn't even an integer.Bart=== === Subject: : Re: there is no such thing as infinity> Yes, we DO live in a FINITE universe.Therefore, there is no such thing as infinity! Otherwise we could have no> identity.?? But if what you're trying futilely to prove is true, then> there is no such thing as calculus :-)I believe in Calculus but not the way it is taught in school. It> should be taught where dx=1/M, the good old fashioned way with> infinitesimals:If f(x)=x^2, f'(x)=((x+h)^2-x^2)/h=(2xh+h^2)/h=2x+h approximately> equals 2x,> where h=1/M. See! I just revolutionized modern thinking and> mathematics into conforming to common sense and human decency!In your cosmology is the number pi rational or irrational? Samequestion goes for sqrt{2}.We live in a finite universe and there is no room for infinity here.> We must blot out infinity from our world as it is causing the> breakdown of morals and standards in our fragile society - in an> infinite universe, anything is possible!Explain this breakdown of morals due to the concept of infinity.Pat=== === Subject: : Re: there is no such thing as infinityIn your cosmology is the number pi rational or irrational? Same> question goes for sqrt{2}.That's a good question. Pi an sqrt(2) are rational, obviously, becausethere is no such thing as irrational numbers. It's just an illusioninvented by the Evil Empire, Ancient Greece.The way to calculate pi and sqrt(2) is just compute the decimals untilyou get to the Mth decimal place and you won't be able to go anyfurther after that.> Explain this breakdown of morals due to the concept of infinity.Another good question. Morality is based on truth and common sense.The concept of infinity is the antithesis to both.Modern society is brainwashed into not only thinking that there isinfinity but that there are lots of different kinds of infinity. Thisis the extent that common sense has been eliminated from society byCantor and his gang!The Evil Empire, Ancient Greece, accepted to some extent, the conceptof infinity. And they were one of the most immoral societies around.They destroyed common sense with their philosophy - and with itsdestruction, morality was also destroyed.Dr. Ben Zona=== === Subject: : Re: there is no such thing as infinity>The way to calculate pi and sqrt(2) is just compute the decimals until>you get to the Mth decimal place and you won't be able to go any>further after that.>Dr. Ben ZonaI just calculated sqrt(2) that way. Then I multiplied them backtogether with my M&M simulator. The machine got very warm doing thisbecause it was having trouble with the 2M decimal places in theanswer. Two unfortunate things occurred:1. The answer didn't come out 2.2. It melted the chocolate, exposing the nut.--Lynn=== === Subject: : Re: there is no such thing as infinity> You just cannot have an infinite setDr. Ben ZonaAnother nut trying to impose his own limitations on everyone.What you cannot do, zonk, does not limit the rest of humankind.=== === Subject: : Re: JSH: Pattern argument> I've been watching the one and only James Harris Show for some time> now, but I'm still at a loss what James is actually trying to prove> (if anything). Is this a crucial part of some purported proof of FLT> or what?Just curious.Do you have to ask? Of course it's FLT. What else is there? Nathan=== === Subject: : Re: JSH: Pattern argument> I've been watching the one and only James Harris Show for some time> now, but I'm still at a loss what James is actually trying to prove> (if anything). Is this a crucial part of some purported proof of FLT> or what?Just curious.Do you have to ask? Of course it's FLT. What else is there? NathanJSH has mumbled a bit recently about the Riemann hypothesis. Next it will probably be a comprehensive Theory of Everything.=== === Subject: : Re: JSH: Pattern argument... > Footnotes: > [1] Typically, newsservers have a finite capacity, so if you make > enough posts in a single newsgroup on any topic whatsoever, the server > may be forced to expunge some older posts. In this crude way, one > could try to suppress posts, but it wouldn't be very effective. Most > news servers have sufficient capacity that it would take at least > hundreds of posts for this to work.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/=== === Subject: : Re: JSH: Pattern argument <8765eg19vc.fsf@phiwumbda.org> Discussion, linux)> ...> Footnotes: > [1] Typically, newsservers have a finite capacity, so if you make> enough posts in a single newsgroup on any topic whatsoever, the server> may be forced to expunge some older posts. In this crude way, one> could try to suppress posts, but it wouldn't be very effective. Most> news servers have sufficient capacity that it would take at least> hundreds of posts for this to work.Google's not a typical newsserver.I did suggest that one could hide a post on Google by posting in*other* threads, so that the thread containing that post one wants tobury no longer appears on the first page for that group. This is, ofcourse, not at all what James accuses folks of doing. Instead, heclaims that a group of evil mathematicians have cunningly concludedthat, to hide a thread started by JSH, one should post many, manymessages in that thread. Clever, clever.Of course, even my suggestion of hiding posts on Google isn'tparticularly effective. But I'd wager that, if one could measure howmany times a particular post is read via Google[1], my strategy would bea lot more effective than the one James is on about.Footnotes: [1] Even Google can't measure this very well, since if I choose toview the thread, they don't know which other posts in a block of tenthat I read.-- That's the base tautological space where by tautological space I meana region of truth. -- James S. Harris does philosophy of mathematics. JSH is a renaissance man.=== === Subject: : Re: JSH: Pattern argument>>> Better yet, did James serve his country during the Korean War? I>> heard he helped the Marines escape from the Chosin Reservoir and>> served from 1950-1953. He also belongs to the Allied Seniors>> Recreational Vehicle Association - so please, afford him some>> respect. Please don't toss him a bone either; a man doesn't do that>> to a Korean War veteran.>>> James says that he is around 35 years old.>My post was a presentation of a mathematical pattern, but I've noticed>how certain posters *immediately* replied to push it in a different>direction.>Where are Dik Winter and Decker now? So far it looks like only>Nora Baron at least made an attempt to address the math, while a LOT>of other posters felt a need to pile up posts on different subjects.>That's what I'm talking about people!Jesus ing christ. Yes, Nora made an attempt to address themath. In reply to a post in another thread just now where shedid nothing but address the math your only reply was a repeated OFF. You reply that way to people discussingthe math, and then you complain that they eventually loseinterest? >It'd be one thing if I could make a post like my original one and get>one or two replies or even none!>But instead I see evidence of an effort to hide the post by people>putting up a continual stream of irrelevant replies.>Sometimes in such a situation I'd start another thread, and they'd>just follow, while others would accuse me of ignoring refutations!>Whether you realize it or not many people on the sci.math newsgroup>are behaving in a way that defies rational explanation.Nobody's in any doubt about the fact that some people'sbehavior defies rational explanation, I can assure you of that.>James Harris=== === Subject: : Re: Mod operator by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AHJhQ09552;>Most mathematicians would say that x mod 7 is always a>value between 0 and 6, regardless of the sign of x. No they would not. To all mathematicians x mod 7 is an equivalenceclass. It is not a single number. Furthermore, there are alternative ways of representing thereduced equivalence class. One can, as you suggest, represent itas [0,...6]. However, it can also be represented as [-3,....3].There is no unique way of doing it. Further, in a computer application, which way is best depends on the application.>But>many computer languages do this the wrong way.There is no right way or wrong way. >For >example, most mathematicians would insist that> -15 mod 7 = 6>but many (but not all) programming languages give the wrong>result, -1. -1 and 6 are equivalent mod 7. There is no wrong answer. It isa matter of convention.=== === Subject: : Re: Proofs - Please help quickly!> I'm really having trouble with 4 trigonometric functions. Here they are:> Use the formulas you were taught in class> 1) Find an exact expression for: sin(pi/12)> And again use the formulas> 2) Solve to 4 decimal places (0_ sin2x + cosx = 0> Use the formulas... might need more than just trig ones here.> 3) Solve sin^2 x - 2sinx - 1 = 0 and find the general solution.> yes, this one is hard, I don't think I can do it> 4) Prove the following (this is a real toughy): tanx/secx = secxPlease help! Soon!=== === Subject: : de moivre's formula + polar representationz^n = w where z=x+iy and w=a+ib.w = r(cos(x) + i sin(x))z = p(cos(y) + i sin(y))z^n = p^n (cos(nqy + i sin(ny)). it follows p^n = r = |w| byuniqueness of the polar representation. can anyone detailed explainwhat is the uniqueness of the polar representation and why p^n = r =|w|?=== === Subject: : Re: de moivre's formula + polar representation charset=iso-8859-1> z^n = w where z=x+iy and w=a+ib.> w = r(cos(x) + i sin(x))> z = p(cos(y) + i sin(y))> z^n = p^n (cos(nqy + i sin(ny)). it follows p^n = r = |w| by> uniqueness of the polar representation. can anyone detailed explain> what is the uniqueness of the polar representation and why p^n = r => |w|?z^n = p^n (cos(nqy + i sin(ny)) never seen this before :/ hmmmmDId you mean z^n = p^n (cos(nt) + i sin(nt)).Where t=atan(y/x) ?=== === Subject: : Re: de moivre's formula + polar representation> z^n = w where z=x+iy and w=a+ib.w = r(cos(x) + i sin(x))> z = p(cos(y) + i sin(y))z^n = p^n (cos(nqy + i sin(ny)). it follows p^n = r = |w| by> uniqueness of the polar representation. can anyone detailed explain> what is the uniqueness of the polar representationIf r and s are real numbers greater than zero and if a and b arereal numbers, then we haver.(cos(a) + i.sin(a)) = s.(cos(b) + i.sin(b))if and only if r = s and (a - b)/(2.pi) is an integer.> and why p^n = r = |w|?Best ,Jose Carlos Santos=== === Subject: : branch of log zWhat exactly is to choose a branch of log z for complex z? I'dappreciate if someone help me picture this.=== === Subject: : Re: branch of log z> What exactly is to choose a branch of log z for complex z? I'd> appreciate if someone help me picture this.Seehttp://mathworld.wolfram.com/PrincipalBranch.hBest ,Jose Carlos Santos=== === Subject: : Re: My Childhood and Infinity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AIMqS15158;>> You were ingorant as a child. You are stupid as an adult. How big is>> 2M, git?>You missed the point of my argument entirely. 2M cannot exist.I feel this is a good point; as, if M exists, M is the largest number,then 2M is indeed nonsense.=== === Subject: : Newbie: Integration?y = x^4 - 3x^2 +5that we can integrate.I am reading some stuff where:For 10,000 values of x we have corresponding observed values for y. Now,this can be plotted on a scatter plot but it is not really available anequation. Then they integrate it. They use matrix notation - which probablyis the confusing part as there is no equation like above.Can someone explain how they do this? Point to some place where I can readbasic stuff on it. Of course, if the question is not clear, I will repost.=== === Subject: : Re: Newbie: Integration?y = x^4 - 3x^2 +5that we can integrate.I am reading some stuff where:For 10,000 values of x we have corresponding observed values for y. Now,> this can be plotted on a scatter plot but it is not really available an> equation. Then they integrate it. They use matrix notation - which > probably> is the confusing part as there is no equation like above.Can someone explain how they do this? Point to some place where I can > read> basic stuff on it. Of course, if the question is not clear, I will > repost.Think of the points as samples from an equation, if it pleases you.It any case, a simple scheme is to connect successive points withstraight lines. The integral is the area under the curve.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.h To solve Linear Programs: .../LPSolver.hr c A game: .../Keynes.h v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau=== === Subject: : Re: Newbie: Integration?Yes, but how do you calculate the area under curve by using integration ifthat curve is not represented as on algebraic equation?y = x^4 - 3x^2 +5that we can integrate.I am reading some stuff where:For 10,000 values of x we have corresponding observed values for y. Now,> this can be plotted on a scatter plot but it is not really available an> equation. Then they integrate it. They use matrix notation - which> probably> is the confusing part as there is no equation like above.Can someone explain how they do this? Point to some place where I can> read> basic stuff on it. Of course, if the question is not clear, I will> repost.> Think of the points as samples from an equation, if it pleases you.> It any case, a simple scheme is to connect successive points with> straight lines. The integral is the area under the curve.> -- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.h> To solve Linear Programs: .../LPSolver.h> r c A game: .../Keynes.h> v s a Whether strength of body or of mind, or wisdom,or> i m p virtue, are found in proportion to the power orwealth> e a e of a man is a question fit perhaps to bediscussed by> n e . slaves in the hearing of their masters, buthighly> @ r c m unbecoming to reasonable and free men in searchof> d o the truth. -- Rousseau=== === Subject: : Re: My Childhood and Infinity> Then I remembered myself> as a child lining up numbers trying to find M, the largest number in> the universe.> I am currently using my computer> to find this number, as I have concluded that this is not a job for a> human being.I'm curious about how you're using your computer to find M. Have youwritten a computer program of some sort, or are you using some kind ofsoftware like Excel or xcalc?If you've written a computer program, what language is it in? Couldyou post it?-Jim Ferry=== === Subject: : Re: My Childhood and Infinity thusly:> When I was a young three and a half year old lad, I used to write> numbers on paper and line them across the room. My goal was to get to> the largest number possible which I now call M. Then my mother told> me that the largest number is infinity. I asked her what things are> infinite. She could not answer me. Then I asked her what do you mean> goes on for ever? How can something go on for ever? She could not> answer me and told me to ask my teacher, as that is her job.Just because you call it M doesn't mean it#s *not* infinity. A rose by any other name etc.My own childhood encounter with infinity was the realisation that it is (a) unique and (b) unsigned. There is no hierarchy of greater and greater infinities, and +inf and -inf are the same.-- TownsendI put it down there, and when I went back to it, there it was GONE!Interchange the alphabetic elements to reply=== === Subject: : Re: My Childhood and Infinity<4027d06d$0$95030$edfadb0f@dread11.news.tele.dk> thusly:>> I decided that I would pick up the pieces of where I left off>> as a child and renew my search for M. I am currently using my computer>> to find this number, as I have concluded that this is not a job for a>> human being.Just my words! Lets make a big Internet project GISM Great Internet> Search for M. Everyone donates their CPU time and there's a prize to the> owner of the computer that actually reaches M.With laughs,> -Michael.Just for fun, program it to print out M+1 instead-- TownsendI put it down there, and when I went back to it, there it was GONE!Interchange the alphabetic elements to reply=== === Subject: : Re: My Childhood and InfinityX-Mimeole: Produced By Microsoft MimeOLE V5.50.4910.0300X-Msmail-Priority: Normal> When I was a young three and a half year old lad, I used to write>> numbers on paper and line them across the room. My goal was to get to>> the largest number possible which I now call M.2M=M, obviously. The computer program to search for M is trivial:> Set N=0. Loop, incrementing N, until N+1=N. When the loop exits, N=M.> It's easy and requires no expensive equipment. Somebody ought to look.Lead, follow, or get out of the way, Al.> Assume that the largest number is an integer. Assume> further that it's a positive integer. This means you> can use an unsigned integer format for the search> variable. It also means that you can use the following> shortcut method: Initialize the variable to zero, then> rather than repeatedly adding one until you reach the> highest value, subtract one instead. This is the short> way 'round via negative infinity! Why waste cycles?I get M = 4,294,967,295. Anyone else?--KCS=== === Subject: : Re: My Childhood and Infinity Discussion, linux)>> Assume that the largest number is an integer. Assume>> further that it's a positive integer. This means you>> can use an unsigned integer format for the search>> variable. It also means that you can use the following>> shortcut method: Initialize the variable to zero, then>> rather than repeatedly adding one until you reach the>> highest value, subtract one instead. This is the short>> way 'round via negative infinity! Why waste cycles?> I get M = 4,294,967,295. Anyone else?I got 13, but I did the search manually.-- All intelligent men are cowards. The Chinese are the world's worstfighters because they are an intelligent race[...] An average Chinesechild knows what the European gray-haired statesmen do not know, thatby fighting one gets killed or maimed. -- Lin Yutang=== === Subject: : Re: My Childhood and Infinity> When I was a young three and a half year old lad, I used to write> numbers on paper and line them across the room. My goal was to get to> the largest number possible which I now call M. Then my mother told> me that the largest number is infinity. I asked her what things are> infinite. She could not answer me. Then I asked her what do you mean> goes on for ever? How can something go on for ever? She could not> answer me and told me to ask my teacher, as that is her job.So the next day I went to nursery school and was playing on the swing> and I bumped my nose and started crying. The nursery school teacher> saw me crying and tried to make me feel better. I decided at that> point that it would be a good opportunity to ask her what infinity is.> So I did. But she didn't understand me because I was still crying.For years, I blocked this out of my memory, this disturbing incident.> And I went to college and eventually got a PhD from the University of> San Moritz, a non-accredited but well-respected university, writing my> thesis on the role of life experience in scientific thought. Then I> differential equations and started to shake. What I was doing was> meaningless! What my teachers told me was a lie! Partial differential> equations presuppose the existence of infinity! And how do I know that> there is an infinity? No one knows in fact. Then I remembered myself> as a child lining up numbers trying to find M, the largest number in> the universe. I decided that I would convert my partial differential> equation paper about difference equations instead. Then my life> changed. I decided that I would pick up the pieces of where I left off> as a child and renew my search for M. I am currently using my computer> to find this number, as I have concluded that this is not a job for a> human being.I am looking for collaborators for this big project. Anyone who wants> to participate in this revolution, literally, in mathematics and> science should please contact me. I can be found at my brother-in-law> Craig Feinstein's email address. Just do not make it known to him that> I am using it, as he thinks it is not healthy for me to continue this> project, and that I should just do what my doctor says.Dr. Ben ZonaInfinity has been misunderstood as a number or not understood properlyin the history of science and mathematics while it is the realitybehind this universe.It is this unobservable infinity which appears as the observers'universe which forms the foundation for the observer entangled quantumtheory and the observer related relativity theory, the modern theoriesof physics.S S Shastry=== === Subject: : Re: My Childhood and Infinity>> For years, I blocked this out of my memory, this disturbingincident.> And I went to college and eventually got a PhD from the University of> San Moritz, a non-accredited but well-respected university, writing my> thesis on the role of life experience in scientific thought.Dr. Ben ZonaWill you consider me? I have PhD from MIT (that is, Malawi Instituteof Technology, a non-accredited but will respected university) inmetaphysics and cheerleading. I can be a cheerleader of this project.=== === Subject: : Re: My Childhood and Infinity> You were ingorant as a child. You are stupid as an adult. How big is> 2M, git?Uncle Al,You are really starting to make me angry. First, you made fun of myname. Now you call me stupid. You missed the point of my argumententirely. 2M cannot exist.I ask you what is 2 times infinity? Infinity, you might say, but thatdoesn't make any sense and you know it. I am claiming that there is noevidence for the Peano axioms that say that every number has asuccessor! It seems clear to me that no one has come close to provingthis. On the other hand, there is abunt evidence for the number Mexisting, i.e., Asimov's number. And 2M cannot exist by definition.And just to satisfy you, Uncle Al, I have an extention of the theoryof M which says that the successor of M is in fact zero. So thenumbers form a clocklike modular pattern, not a straight line thatgoes out into infinity. Or maybe, the numbers are topologicallyequivalent to a torus or a Mobius strip. This whole area ofmathematics has systematically been supressed by modern mathematiciansstarting with Cantor. My mission in life is to first find M and nextdetermine the topology of the natural number system. The next goalmakes finding M like child's play!Rather than criticize me, why don't you join me or at least help me?Dr. Ben Zona=== === Subject: : Re: My Childhood and InfinityCraig Feinstein wibbled:> You were ingorant as a child. You are stupid as an adult. How big is> 2M, git?Uncle Al,You are really starting to make me angry. First, you made fun of my> name. Now you call me stupid. You missed the point of my argument> entirely. 2M cannot exist.Well, 3M does. -- Next on list: 26=== === Subject: : Re: My Childhood and Infinity>I ask you what is 2 times infinity? Infinity, you might say, but that>doesn't make any sense and you know it. I am claiming that there is no>evidence for the Peano axioms that say that every number has a>successor! It seems clear to me that no one has come close to proving>this. On the other hand, there is abunt evidence for the number M>existing, i.e., Asimov's number. And 2M cannot exist by definition.Mr. Feinstein, you seem to contradict the proof of Mr. Feinsteinon the impossibility of proving the Collatz 3n+1 -conjecture. Itrelied on the necessity of calculating infinitely many values of afunction in order to prove the conjecture. Surely if M exists then theCollatz conjecture can be proven/disproven through exhaustion alone.-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : Re: My Childhood and Infinity>I ask you what is 2 times infinity? Infinity, you might say, but that>doesn't make any sense and you know it. I am claiming that there is no>evidence for the Peano axioms that say that every number has a>successor! It seems clear to me that no one has come close to proving>this. On the other hand, there is abunt evidence for the number M>existing, i.e., Asimov's number. And 2M cannot exist by definition.Mr. Feinstein, you seem to contradict the proof of Mr. Feinstein> on the impossibility of proving the Collatz 3n+1 -conjecture. It> relied on the necessity of calculating infinitely many values of a> function in order to prove the conjecture. Surely if M exists then the> Collatz conjecture can be proven/disproven through exhaustion alone.Hey, I just got an idea! This number M must be on the Collatz treesomewhere. And since there aren't any numbers larger than M, M itselfmust be the excursion for the Collatz sequence starting from M.Therefore, we haveMensanator's Theorem: M is evenProof: Assume M is odd. The next number in the Collatz sequence would then be 3*M+1 which contradicts the fact that M is the largest number, so the assumption that M was odd must be false, therefore, M is even.QEDThis is an important development in determining what M must be, sincewe can alter the FORTRAN program to only search for even numbers.It is so important, that we should start referring to M as Mensanator's Number=== === Subject: : Re: My Childhood and Infinity> Mr. Feinstein, you seem to contradict the proof of Mr. Feinstein> on the impossibility of proving the Collatz 3n+1 -conjecture. It> relied on the necessity of calculating infinitely many values of a> function in order to prove the conjecture. Surely if M exists then the> Collatz conjecture can be proven/disproven through exhaustion alone.Don't tell my brother-in-law Craig Feinstein I said this, as hedoesn't know I'm using his email address: His proof as you mentionedit is a complete fraud. I tried to tell him, but he threatened to haveme confined.Dr. Ben Zona=== === Subject: : Re: My Childhood and Infinity>Don't tell my brother-in-law Craig Feinstein I said this, as he>doesn't know I'm using his email address: His proof as you mentioned>it is a complete fraud. I tried to tell him, but he threatened to have>me confined.And what does Social Security Administration have to say about youtrolling sci.math from their computers?-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : Re: My Childhood and Infinity <4027E089.B402DB34@hate.spam.net> Wow, shouldn't the === Subject: line read My Childhood at Infinity?Cheers, ZVK(Slavek)> Mr. Feinstein, you seem to contradict the proof of Mr. Feinstein> on the impossibility of proving the Collatz 3n+1 -conjecture. It> relied on the necessity of calculating infinitely many values of a> function in order to prove the conjecture. Surely if M exists then the> Collatz conjecture can be proven/disproven through exhaustion alone.> Don't tell my brother-in-law Craig Feinstein I said this, as he> doesn't know I'm using his email address: His proof as you mentioned> it is a complete fraud. I tried to tell him, but he threatened to have> me confined.> Dr. Ben Zona=== === Subject: : Re: My Childhood and Infinity> This whole area of> mathematics has systematically been supressed by modern mathematicians> starting with Cantor. Aha! You and Harris should compare notes.=== === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrationalBut is an algebraic number divided by a transcendental number always> irrational?It's always transcendental, because the algebraic numbers are closed> under the usual operations of arithmetic.I think this alone doesn't show the conclusion. A set being closedunder the usual operations of arithmetic doesn't mean that theoperation of an element of the set and an element outside the setneeds to be outside of the set. For instance, though the algebraicsare closed under multiplication, the product of an algebraic numberand zero is zero, which is algebraic.=== === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational> I think this alone doesn't show the conclusion. A set being closed> under the usual operations of arithmetic doesn't mean that the> operation of an element of the set and an element outside the set> needs to be outside of the set. For instance, though the algebraics> are closed under multiplication, the product of an algebraic number> and zero is zero, which is algebraic.If we know that a/b = c (and we specify that a is not 0; I overlookedthat special case before), we can conclude that a/c = b. Both a and cbeing algebraic would imply that b is algebraic. Contrapositively, if ais a non-zero algebraic and b is transcendental, c must also betranscendental.-- iel W. sonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/039 53 36 N / 086 11 55 W=== === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrationalHelloHow can we prove [arccos((sqrt(5)-1)/2 )] / pi [is irrational]?The real part of a root of unity cannot be> an algebraic integer unless it is 0, 1, or -1.Isn't there a mistake here? If z^n =1, then z is an algebraic integerand so is it's conjugate z'. Since Re(z) = (z+z')/2, it follows Re(z)is half the sum of 2 algebraic integers, therefore an algebraicinteger. That is, the real part of a root of unity is always analgebraic integer. If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part> (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u)> is not a root of unity, that is: u is irrational.=== === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational> Hello> How can we prove [arccos((sqrt(5)-1)/2 )] / pi [is irrational]?>> The real part of a root of unity cannot be>> an algebraic integer unless it is 0, 1, or -1.> Isn't there a mistake here? If z^n =1, then z is an algebraic integer> and so is it's conjugate z'. Since Re(z) = (z+z')/2, it follows Re(z)> is half the sum of 2 algebraic integers, therefore an algebraic> integer.Mmm. Methinks algebraic integers don't always work like ordinary ones. Forinstance, is (1+sqrt(2))/2 an algebraic integer? Its minimal polynomial is4x^2-4x-1, so... That is, the real part of a root of unity is always an> algebraic integer.>> If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part>> (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u)>> is not a root of unity, that is: u is irrational.=== === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrationalBut is an algebraic number divided by a transcendental number always> irrational?It's always transcendental,0 is algebraic, right?=== === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational>How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>this has something to with the golden ratio, and maybe with Fibonacci>numbers or continued fractions? But I could't start.There are several possible methods.> What do you know of either Chebyshev polynomials or cyclotomic > polynomials? Or what can you say about the polynomial> 1+2 t-2 t^2+2 t^3+t^4?Well, I know what cyclotomic polynomials are, but have never studiesthem.The polynomial 1+2 t-2 t^2+2 t^3+t^4 is reciprocal and 2 of its rootssatisfy x + 1/x = (sqrt(5)-1)/2.I'm a bit lost, could you outline a prrof, please? === === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrationalHelloHow can we prove [arccos((sqrt(5)-1)/2 )] / pi [is irrational]?The real part of a root of unity cannot be> an algebraic integer unless it is 0, 1, or -1.How can we prove this? Could you please outline a proof?If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part> (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u)> is not a root of unity, that is: u is irrational.=== === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational> The real part of a root of unity cannot be> an algebraic integer unless it is 0, 1, or -1.> How can we prove this? Could you please outline a proof?> If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part> (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u)> is not a root of unity, that is: u is irrational. G.A.Edgar: What is an algebraic integer? I don't see (sqrt(5)-1)/2 being integral, and if this just means an algebraic number, then roots of unity can have much different real parts from 0, 1, -1.J=== === Subject: : Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational > The real part of a root of unity cannot be> an algebraic integer unless it is 0, 1, or -1.> How can we prove this? Could you please outline a proof?If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part> (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u)> is not a root of unity, that is: u is irrational. G.A.Edgar: What is an algebraic integer? I don't see (sqrt(5)-1)/2 > being integral, and if this just means an algebraic number, then roots of > unity can have much different real parts from 0, 1, -1.We say z (real or complex) is an algebraic integer if z is a root of amonic polynomial with integer (real) coefficients. Therefore, if z isa root of unity, then z is automatically an algebraic integer, becauseit's a root of x^n-1 =0 for some integer n. If z is a root of apolynomial of degree n as above and there is no similar polynomialwith degree