>Suppose that T1 and T2 are shift operators on R. Is it provable that >>if there exists function K such that K o T1 = T2 o K then K is linear. Hmm. This says that there exist constants a and b such that K(x+a) = K(x) + b for all x. This certainly does not imply that K is linear, (or even >affine, which is no doubt what you meant). For example, >the restriction of K to the interval [0,a) could be any function >whatever. If there are a and b so that for all x K(x+a) = K(x) + b then K(x) - bx/a is periodic with period a. Thus, K is the sum of a linear function and a periodic function. This confirms the statement that K can be anything on [0,a), but K on that interval plus one more point uniquely determines K. Rob Johnson take out the trash before replying ==== This post presents a topic from intuitionistic logic. The following statements provide some context for the longer excerpt that follows. All excerpts are from The Blackwell Guide to Philosophical Logic. 1) This shows that even a relatively simple theory like equality is incomparably richer than the classical theory. 2) The theory of equality has the usual axioms: reflexivity, symmetry, transitivity. One can strengthen the theory in many ways, for example, the theory of stable equality is given by EQ^st = EQ + Axy(--x=y -> x=y) And the decidable theory of equality is axiomatized by EQ^dec = EQ + Axy(x=y / -x=y) [Aside: In deference to John Correy's treatment of identity and non-self-identicals, I would note that treating this as an inclusive disjunction admits the possibility of a superimposed interpretation.] 3) Intuitionistic predicate logic itself is, just like its classical counterpart, undecidable. Fragments are, however, decidable. For example, the class of prenex formulas is decidable, and, as a corollary, not every formula has a prenex normal form in IQC. On the other hand, although monadic predicate logic is decidable in classical logic, Kripke showed it is undecidable in intuitionistic logic; see also Orekov et al. Likewise, Lifschitz showed that the theory of equality is undecidable. The following is an excerpt about apartness. With respect to the statements above, one thing to note is that stable equality is obtained under the apartness axioms. In intuitionistic mathematics, there is also a strong notion of inequality: apartness, #, as mentioned above. This was introduced by Brouwer (1919) and axiomatized by Heyting (1925). The axioms of AP are given by EQ and the following list Axyx'y'(x#y / x=x' / y=y' -> x'#y') Axy(x#y -> y#x) Axy(-x#y <-> x=y) Axyz(x#y -> x#z / y#z) The gluing technique will now be used to show that AP has the disjunction and existence properties. --- Aside: (DP) |- A / B => |- A or |- B (EP) |- ExA(x) => |- A(t) for a closed term t '|-' is interperted with respect to intuitionistic deduction (DP) is presented with a reductio ad absurdum argument; but, the author notes that there are proof-theoretic justifications that invoke no such conflict with intuitionistic principles. --- Let AP |- A / B and assume ~(AP |- A) and ~(AP |- B) ['~' is classical exclusion negation]. Then, by the strong completeness theorem, there are models K_1 and K_2 of AP such that ~(K_1 ||- A) and ~(K_2 ||- B). Consider the disjoint union of K_1 and K_2 and place the one-point world k_0 below it. That is to say, designate points _a_ and _b_ in K_1 and K_2 which are identified with the point k_0. The new model obviously satisfies the axioms of AP. Hence k_0 ||- A / B and so k_0 ||- A or k_0 ||- B. Both are impossible on the grounds of the choice of K_1 and K_2. Contradiction. Hence AP |- A or AP |- B. For EP, it is convenient to assume that the theory has a number of constants, say {c_i | _i_ e I} Now, let AP |- ExA(x) and ~(AP |- A(c_i)) for all _i_. Then for each _i_ there is a model K_i with ~(K_i ||- A(c_i)). As above, the models K_i can be glued by means of a bottom world K^* with a domain consisting of just the elements c_i. No non-trivial atoms are forced in k_0 (i.e., only the trivial identities c_i = c_i). The identification of the c_j with elements in the models K_i is obvious. Again, it is easy to check that the new model satisfies AP. Hence k_0 ||- ExA(x), i.e., k_0 ||- A(c_i) for some _i_. But, then also, K_i ||- A(c_i), contradiction. Hence AP |- A(c_i) for some _i_. The gluing operation thus demonstrates that there are interesting operations in Kripke model theory that make no sense in traditional model theory. The apartness axioms have consequences for the equality relations. In particular, stable equality is obtained: --x=y -> x=y For, -x#y <-> x=y so x=y <-> ---x#y <-> --x=y Indeed, the equality fragment itself is axiomatized by an infinite set of quasi-stability axioms. Put -(x [=_0] y) =df -x=y -(x [=_(n+1)] y) =df Az(-(z [=_n] x) / -(z [=_n] y) For these 'approximations to apartness,' formulate quasi-stability axioms: S_n =df Axy(-( x [=_n] y) -> x=y) The S_n axiomatize the equality fragment of AP. To be precise: AP is conservative over EQ + {S_n | n >= 0} This shows that even a relatively simple theory like equality is incomparably richer than the classical theory. Apartness and linear order are closely connected. The theory LO of linear order has axioms: Axyz(x x z -(x x#y One can also use x x=y) Stability of equality never excited me. |In intuitionistic mathematics, there is also a strong |notion of inequality: apartness, #, as mentioned above. One might mention that for real numbers, r#s is defined to mean the existence of a positive integer n such that |r-s|>1/n. The equality r=s is equivalent to the negative of that, |r-s|<1/n for every n. The negation, ~r=s, is in a sense very close to equivalent to r#s. If you prove ~r=s for specific r and s, there's a metatheorem which guarantees you can prove r#s too. Markov's rule entails ~r=s -> r#s, and the Markov school used this (although I guess they noted when they did). Once you've concluded that ~r=s, it's tempting to say that it's just a matter of computing r and s precisely enough to find the level of precision where they differ, to determine that r#s. Nevertheless, it's considered inappropriate in standard intuitionism shows that the nonexistence of an n for which |r-s|>1/n leads to a contradiction; it isn't taken as exhibiting the n. Mainly, I would say it's just better not to muck about with weird little statements like ~r=s anyway. It's rare enough that I want to state the negation of an equation like ~r=s that in my own notes I just write the usual inequality symbol r =/= s instead of r#s for apartness, and write ~r=s if I really mean the negation of equality. |This was introduced by Brouwer (1919) and axiomatized |by Heyting (1925). The axioms of AP are given by |EQ and the following list | |Axyx'y'(x#y / x=x' / y=y' -> x'#y') | |Axy(x#y -> y#x) | |Axy(-x#y <-> x=y) | |Axyz(x#y -> x#z / y#z) I like sometimes to deal with relations that satisfy all the axioms of this definition except with Axy(~x#y <-> x=y) replaced by Axy(x=y->~x#y) (or in other words, Ax(-x#x)), and don't necessarily satisfy Axy(~x#y->x=y). I'm not sure whether the condition Axy(~x#y->x=y) really should be included in the definition of apartness, actually. All it serves is to tie equality to apartness, basically by defining it to be the negation of apartness. The first axiom you list (which would usually be considered to follow just by the rules of equality) follows from the others with Axy(x=y->~x#y) as follows. If x#y, then either x#x' or x'#y. If x'#y then x'#y' or y'#y, hence y#y'. So if x#y, then either x'#y' or x#x' or y#y'. But by assumption, x=x' and y=y', so neither x#x' nor y#y', hence x'#y'. Here's an example of a case where one has a natural # relation that satisfies my weaker definition. I don't remember who discovered it first, but I independently rediscovered it later. Define multisets of a set S to be equivalence classes of functions from other sets T under the equivalence that f1:T1->S and f2:T2->S are equivalent if there's a one-to-one correspondence g:T1->T2 such that f1 = f2 o g. Call a multiset a finite multiset if the domain is finite. Defining multisets as equivalence classes amounts to defining the equality relation for them. But is there an apartness relation for finite multisets of a set S which itself has an apartness relation? The first discoverer was using the definition above, and stated his result as being that there is no apartness relation in general. But for finite multisets of a set with apartness, if we weaken the definition as I just described, there is one. As a preliminary step, it might help to prove a simple lemma generalizing the last axiom above to finite multisets. If I have elements x1,...,xn and y1,...,ym of a set with apartness, and if x_i # y_j for every i=1,...,n and j=1,...,m, and z is some further element, then either z#x1, z#x2, ..., z#xn, or z#y1, z#y2,..., z#ym. This is just applying the distributive law (A or B) and (A or C) <-> A or (B and C) a finite number of times, using the fact that by the last axiom for each i and j, either x_i#z or z#y_j. [By the way, the finite distributive law (A or B1) and (A or B2) and ... and (A or Bn) <-> A or (B1 and B2 and ... and Bn) is constructive. The infinite version {for every i>=0 (A or B_i)} <-> {A or for every i>=0 B_i} is not. In the finite case, we can effectively get for each i=1,...,n the fact that A is true, or the fact that B_i is true, and when we're done, we've either found that A is true, or for each one we've found that B_i is true. But that's not an effective procedure in the infinite case.] So if we have x1,...,xn and y1,...,ym which are collectively apart from each other, any new z is apart from all of one of them, which means it can be appended to the other one while preserving the situation. By induction, if we have z1,...,zk, then there's a partition of z1,...,zk so that the x's together with one of the partitions are collectively apart from the y's together with the other of the partitions. The case of n=m=1 is maybe the most interesting. If among some elements u1,...,ur in a set with apartness we find some apartness relation u1#u2, then we can partition the u1,...,ur into two, so that u1 is in one partition, u2 is in the other one, and the elements of one partition are all apart from the elements of the other partition. Okay, now back to finite multisets. If two multisets have different numbers of elements, then they are of course distinct, so we just need to consider finite multisets defined by functions from a fixed {1,...,n} to S. Now let me motivate my # relation. Given two multisets {x1,...,xn} and {y1,...,yn}, in order to have a distinction between them, it's only natural that we should have a distinction between some two of the elements. Hence there exists some way of partitioning x1,...,xn,y1,...,yn into submultisets that are collectively apart from each other. I define {x1,...,xn}#{y1,...,yn} to mean that we can partition them in such a way that one of the partitions has a different number of elements from the x's as from the y's. Equivalently, I define it to mean that there are subsets i_1,...,i_k and j_1,...,j_l of {1,...,n} where k+l>n, with the property that for every s=1,...,k and t=1,...,l, we have x_i_s#y_j_l. (k+l=n+1 is enough.) For example, an unordered pair {x1,x2} is apart from another unordered pair {y1,y2} if either x1#y1 and x1#y2, or x2#y1 and x2#y2, or y1#x1 and y1#x2, or y2#x1 and y2#x2. The # relation satisfies all the axioms except ~x#y -> x=y. It's obviously symmetric and preserved under substitution of equals. If two multisets are equal, they are not #. If {x1,...,xn}#{y1,...,yn} and {z1,...,zn} is some other multiset, then the apartness relations between x_i_1,x_i_2,...,x_i_k and y_j_1,...,y_j_l can be extended by the lemma to include the z1,...,zn. Now either enough z's are put with those x's to make {z1,...,zn}#{y1,...,yn}, or enough z's are put with those y's to make {x1,...,xn}#{z1,...,zn}. If >n-l of the z's are put on the left, that's enough to make the former true. If not, then there are at least l of the z's on the right, which makes {x1,...,xn}#{z1,...,zn} true. Hence the last axiom also holds. The negation of #, however, is not constructively equivalent to equality for multisets. Equality is really a pretty strict relation; one has to be able to say which element corresponds to which. In general, for unordered pairs of real numbers, {x,-x} cannot be # {|x|,-|x|}. But in order to prove that {x,-x}={|x|,-|x|}, we have to have either that x=|x| and -x=-|x|, or that x=-|x| and -x=|x|. The first case is equivalent to x>=0 and the second case is equivalent to x<=0. (For constructive purposes one defines x>=y for real numbers x and y to mean that in the sequence of rational approximations to x and to y, the approximation to within 1/n of x is >= the approximation to within 1/n of y, minus 2/n. There isn't a constructive equivalence between x>=0 and x>0 or x=0. There is an equivalence between x>=0 and not x<0.) That x>=0 or x<=0 for each real number x is nonconstructive. That x>=0 or x<=0 for each real number is equivalent to the statement that if s0,s1,... is a sequences of bits, i.e. elements of {0,1}, then either it isn't the case that the first 1 is at an odd position, or it isn't the case that the first 1 is at an even position. Nonconstructively speaking, either (a) all the sequence s is 0, or (b) the first 1 occurs at an even index, or (c) the first 1 occurs at an odd index. In cases (a) and (c), one can say that for each even n such that s_n=1, there exists an odd m |- A or |- B | |(EP) |- ExA(x) => |- A(t) for a closed term t | |'|-' is interperted with respect to intuitionistic deduction | |(DP) is presented with a reductio ad absurdum argument; |but, the author notes that there are proof-theoretic |justifications that invoke no such conflict with intuitionistic |principles. |--- I would approach the use of Kripke models with some caution. Kripke model theory and topos theory are ways for people to keep assuming the law of excluded middle, but simulate to some degree what it's like not to assume it. So I'd take seriously all the distinctions between different types of validity until I knew that there was a known equivalence. I suppose the sentences valid in one-node Kripke models are the same as the ones that are just plain valid. Usually, though, people working with Kripke model theory are assuming the law of excluded middle, which implies that all the sentences implied by it are valid in one-node Kripke models. There are Kripke models in which the law of excluded middle is definitely not valid, but only some constructivists actually assume that the law of excluded middle is not valid, as opposed to failing to assume that it is valid. I saw a proof assuming the law of excluded middle that for each statement S in second order logic, there's a corresponding first-order statement S' such that S holds in all models if and only if S' holds in all Kripke models. So the set of statements valid for all Kripke models is somewhat complex. The situation is complicated by the fact that the most straightforward ways of asserting a completeness theorem are incorrect for intuitionist logic. There's a formal system, yes, and it's got an associated recursively enumerable set of theorems. But the classical proof I mentioned in the last paragraph shows that the set of first-order sentences valid in all Kripke models is way too logically complex to be recursively enumerable. I've seen mentioned certain completeness theorems, but I wouldn't say the situation seemed simple. [...] |The apartness axioms have consequences for the |equality relations. In particular, stable equality is |obtained: | |--x=y -> x=y | |For, | |-x#y <-> x=y | |so | |x=y <-> ---x#y <-> --x=y Yes, stability is just the same as saying that equality is negative, i.e. that equality is equivalent to the negation of something. I don't know what good that is. The thing it's a negation of doesn't have to be an apartness, and as far as I know there's no guarantee that there is an apartness. Here's a demonstration that ~~X->X is equivalent to saying X is (equivalent to) the negation of something. (I write negations with ~ instead of -.) First, a simple lemma. Any statement implies its double negation: X->~~X. If we assume X, assuming ~X leads to a contradiction, from which we can infer ~~X. Triple negation is the same as single negation (a theorem of Brouwer). If ~X, then by the lemma applied to ~X, we get ~~~X too. together with ~~~X that's a contradiction. Hence, on the assumption of ~~~X, we get ~X. A statement X satisfying the stability condition ~~X->X is equivalent to ~~X because X->~~X is automatic. So if it satisfies stability, then it's equivalent to the negation of something (namely, ~X). Conversely if X is equivalent to ~Y, then ~~X is equivalent to ~~~Y which implies ~Y, hence X. [Some stuff about the theory of linear orders omitted.] Keith Ramsay ==== : This post presents a topic from intuitionistic logic. 'Nuff said. X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 09:29 PM, maneesh@drunkenbastards.com (maneesh) said: >To elaborate, I often found it confusing to associate particular >properties, algebraic objects and other such things with the name >they are given. Learning a new language is always confusing. >Surely it only adds a level of complexity to a field of study which >does not need one! Surely not. The problem is that the concept is new, not that there is something wrong with the name. >What are your opinions on having things called Schurian, Jacobian, >or Gaussian? It's shorter than descriptive names, and no less enlightening. >Do you feel that names of things in mathematics should greater >reflect information about the said things? Should the word red be red? Names are arbitrary. >Do you feel it retards mathematical progress (on a personal, or >collective level)? No. >What is the worst named object you can think of? Ideal. None of what you have written addresses the real problems in Mathematical nomenclature; the large number of synonyms and the inconsistency in the notation. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- >What is the worst named object you can think of? Ideal. > why ideal? I would say 'abstract' has the most number of meanings in mathematics. function abstraction, outline, define proposition, approximate, .... Herc X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 10:55 AM, Saab Siddiqui said: >when i show this to other people they say the same thing. no offense >ment here but really people always claim such but bring no proof. >what other text have parallels like this? Others have already answered your question. Basically, given any sufficiently large text the odds are astronomically in favor of such coincidences. It's just a question of picking out the ones that support your case and ignoring the ones that don't. >and bible codes and panim dont count because it is different sort >of pattern. Why? What sort of pattern would count? Why should anybody bother producing more examples if you're just going to say That one doesn't count - it's different? >i mean where words duplicate like in quran. Words duplicate in the Tanakh, and even in novels. >what is your faith? Judaism. >but what holy texts? Tanakh. Talmud. >could you show me similar parallels? You've already rejected them. Jewish mysticism often makes use of Gemmatria, which is basically what you are doing. It's not Science and it's certainly not Mathematics. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > at 10:55 AM, Saab Siddiqui said: >>i mean where words duplicate like in quran. Words duplicate in the Tanakh, and even in novels. I happened to have the text of Moby Dick, and did some word counts. COFFIN, SHARKS, MAN'S, REST and BONE all occur 51 times. STERN, INDIAN, SPEAK, SLOWLY, and SAVAGE all occur 52 times. KING, ROPE and IVORY all occur 56 times. BOOK, SHORT, and LIVE all occur 60 times. ORDER, REASON and LORD all occur 64 times FISHERY and WORK both occur 65 times WHALEMEN and MYSELF both occur 69 times FIRE and HARPOON both occur 76 times MIND and SOUL both occur 80 times CERTAIN, GOING, LEG and DEATH all occur 89 times BECAUSE, LEVIATHAN and DEAD all occur 92 times If that doesn't convince you that Moby Dick was divinely inspired, I suppose nothing will. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > If that doesn't convince you that Moby Dick > was divinely inspired, I suppose nothing will. LOL Double-LOL Double-LOL, squared. (BTW: The Pythagoras confusion waits for some enlightenment) Rainer Rosenthal r.rosenthal@web.de ==== I have a problem. I don't see how the author finds the closed form expression for the polynomials P_n(lambda) in the book by Akhiezer -- 'The Classical Moment Problem' at page 3. I have scanned the first 5 pages (although one only needs 1,2,3 to get the background information): http://hem.bredband.net/lukhor/moment/ Happy new year! ==== > I have a problem. I don't see how the author finds the closed form > expression for the polynomials P_n(lambda) in the book by Akhiezer -- > 'The Classical Moment Problem' at page 3. I have scanned the first 5 > pages (although one only needs 1,2,3 to get the background > information): > http://hem.bredband.net/lukhor/moment/ > Happy new year! First, a quick summary for those who don't have the relevant pages in front of them. Consider the space of polynomials R(lambda) = p_0 + p_1 lamba + p_2 lambda^2 + ... + p_n lambda^n. We consider a linear functional S defined by a sequence of numbers s_0, s_1, ..., defined by S[R] = s_0 p_0 + s_1 p_1 + .... where the sum is finite because the polynomial has only a finite number of non-zero coefficients. Now we can define a bilinear form, given polynomials R_1 and R_2, as S[ R_1 R_2 ]; that is, we muliply the polynomials in the usual sense, then apply the linear functional S[]. The coefficients s_0,..., are choosen so that the form is positive definite. Now the author wishes to choose an orthogonal basis, and gives the following explicit formulae: P_0 (lambda) = 1 P_n (lambda) = 1/ sqrt{ D_{n-1} D_{n} } times | s_0 s_1 ... s_n | | s_1 s_2 ... s_{n+1} | | ... | | s_{n-1} s_{n} ... s_{2n-1} | | 1 lambda ... lambda^n | where D_{m} is defined as the determinant of the matrix | s_0 s_1 ... s_m | | s_1 s_2 ... s_{m+1} | | ... | | s_{m} s_{m+1} ... s_{2m} |. Now the author notes that P_n is of degree 'n' (as is clear by expanding the matrix using the bottom row), so that to prove orthogonality (leaving aside normalization for the moment) it is sufficient to prove that P_n is orthogonal to 1, lambda, lambda^2, ..., lambda^{n-1}. Agreed? Now what happens when we multiply P_n by lambda^m? This is given on page 4 [let us call this equation#1], P_n (lambda) lambda^m = 1/ sqrt{ D_{n-1} D_{n} } times | s_0 s_1 ... s_n | | s_1 s_2 ... s_{n+1} | | ... | | s_{n-1} s_{n} ... s_{2n-1} | | lambda^m lambda^{m+1} ... lambda^{m+n} | Note: if we expand the determinant along using the last row, we would get P_n(lambda) lambda^n = a polynominal in terms of lambda^{m}...lambda^{m+n} where the coefficients are the minors of the above matix (with appropriate signs). The author then asks the reader to apply the functional S[] to both sides of it, so that the left side becomes the inner product of P_n(lambda) and lambda^m, so that we obtain [this will be our equation#2] S[ P_n (lambda) lambda^m ] = 1/ sqrt{ D_{n-1} D_{n} } times | s_0 s_1 ... s_n | | s_1 s_2 ... s_{n+1} | | ... | | s_{n-1} s_{n} ... s_{2n-1} | | s_{m} s_{m+1} ... s_{m+n} |. To see this equalility, consider the expansion of this matrix and the previous matrix (from equation#1 above) using the last row--the minors are the same in both instances (and contain only s_i's, no lambda's!) and we simply have replaced lambda^m with s_m, lambda^{m+1} with s_{m+1}, etc. Now, if m at 05:28 AM, victorfrankenstein2@juno.com (Benjamin) said: >Is Apostol from the baby boom generation or is he a generation Xer? purchased my copy in the early 1960s. -- > Shmuel (Seymour J.) Metz, SysProg and JOAT >not reply to spamtrap@library.lspace.org I thought so.....Apostol must be from the WWII generation if he got his degrees in the mid to late 40's. So, he didn't grow up on the new math stuff. I wonder if he got to be in the Battle of the Bulge or if he was in Iwo Jima or the Battle of Midway. Hmmm...or do you think he was in the Bataan Death March? So, the best calculus books seem to be those of Apostol, Courant, and Spivak. Does anyone know of the best biostatistics book? It seems all the ones I've read have sucked eggs. . . . ==== Apostol is great, uses Liebniz' method. I've also heard that J.Bernoulli's text is excellent (see www.wlym.com .-) > So, the best calculus books seem to be those of Apostol, Courant, and Spivak. --Give the Gift of Trickier Dick Cheeny -- out of office at last! http://laroucehin2004.com ==== How many digits of pi have you memorized? I did 100. The world record ==== >How many digits of pi have you memorized? 25, when I was ten or eleven. I had a book about math that in a kind of decorative way had pi written to 25 decimal places with ... at the top of a page. I didn't bother to go past that point. Probably I would have if my friend Tom, who memorized somewhat fewer, had gotten to 25. A few years later, another kid in the same school system was identified as being gifted mathematically for having memorized thirty-some digits of pi. Keith Ramsay ==== > How many digits of pi have you memorized? I did 100. The world record ==== >How many digits of pi have you memorized? I did 100. The world record I have no idea if it's true or just an urban legend but: Reporter: Dr Einstein, how many digits of pi have you memorized? Dr E: Four, I think. R: Four! But sir, many people have memorized dozens, even hundreds of digits. Dr E: Yes, but I know where to look it up if I need more. ==== How many digits of pi have you memorized? I did 100. The world record I have no idea if it's true or just an urban legend but: Reporter: Dr Einstein, how many digits of pi have you memorized? Dr E: Four, I think. R: Four! But sir, many people have memorized dozens, even > hundreds of digits. Dr E: Yes, but I know where to look it up if I need more. I took a course once that gave this story. Dr E was asked why he didn't carry a Franklin Planner to write down all of his wonderful thoughts and ideas. Dr E replied: I have had two great thoughts in my life and remember them both. Not sure if it is true. ==== > >How many digits of pi have you memorized? I did 100. The world record > > I have no idea if it's true or just an urban legend but: > > Reporter: Dr Einstein, how many digits of pi have you memorized? > > Dr E: Four, I think. > > R: Four! But sir, many people have memorized dozens, even > hundreds of digits. > > Dr E: Yes, but I know where to look it up if I need more. > > Did Einstein really memorize only four?? ==== : Did Einstein really memorize only four?? Why would that be suprising? Memorizing them is essentially useless. Justin ==== > How many digits of pi have you memorized? I did 100. The world record Each word in this corresponds to a digit of pi. O means zero. Why, π! Stop, π! Weird anomalies do behave badly! You, madly conjured, imperfect, strange, numerical, Why do you maintain this facade? In finite time you are barbaric! You do wonders, mesmerize minds! O, do elements numerous have a beautiful meaning- A system isolating all mysteries, solutions for puzzles, chaos, a O snafu apparent in O Universal Concept from believing lies? That there, obstinate in you, O Strange Constant, A Divine Sign O exists is unlikely unless Is O revealed Something Brilliant, negating belief! In formulas, O, you show yourself in Greek and math as a π forever-- O hidden wonders absconded, infinite, in a tiny constant, O, sneakily, rather? Never, I say! ==== > How many digits of pi have you memorized? I did 100. The world record > Each word in this corresponds to a digit of pi. O means zero. Why, π! Stop, π! Weird anomalies do behave badly! Sorry, something seems to have gone wrong with my newsreader. What does π! stand for please? The context requires a single letter, but none of a, I or O make sense. -- Paul V. S. Townsend Interchange the alphabetic elements to reply ==== > How many digits of pi have you memorized? I did 100. The world record > Each word in this corresponds to a digit of pi. O means zero. Why, π! Stop, π! Weird anomalies do behave badly! > > Sorry, something seems to have gone wrong with my newsreader. What does > π! stand for please? The context requires a single letter, but none of > a, I or O make sense. It was a pi symbol. Go to http://members.aol.com/loosetooth/poem.html to see the poem. ==== >How many digits of pi have you memorized? I did 100. The world record I've memorized 100,000 digits. They're all 3. Of course I haven't memorized exactly where they occur. (Oh, you meant _consecutive_ digits? Never mind...) ************************ David C. Ullrich ==== >How many digits of pi have you memorized? I did 100. The world record I've memorized 100,000 digits. They're all 3. Of course I haven't >memorized exactly where they occur. I've memorized infinitely many digits, and they're all the same! Of course I haven't memorized either exactly where they occur, or even which digit they all are. Lee Rudolph ==== How many digits of pi have you memorized? I did 100. The world record I've memorized 100,000 digits. They're all 3. Of course I haven't > memorized exactly where they occur. (Oh, you meant _consecutive_ digits? Never mind...) LOL! ==== : I've memorized 100,000 digits. They're all 3. Of course I haven't : memorized exactly where they occur. : (Oh, you meant _consecutive_ digits? Never mind...) Well, in that vein, I've memorized all of them. They are 0,1,2,3,4,5,6,7,8 and 9. If only I could get the order right... Justin ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- : I've memorized 100,000 digits. They're all 3. Of course I haven't > : memorized exactly where they occur. > : (Oh, you meant _consecutive_ digits? Never mind...) Well, in that vein, I've memorized all of them. They are 0,1,2,3,4,5,6,7,8 and 9. If only I could get the order right... > Reminds me of the time I drove though the city of Melbourne and got every green light! Herc had to wait for some... ==== : I've memorized 100,000 digits. They're all 3. Of course I haven't > : memorized exactly where they occur. > : (Oh, you meant _consecutive_ digits? Never mind...) Well, in that vein, I've memorized all of them. They are 0,1,2,3,4,5,6,7,8 and 9. If only I could get the order right... That does not quite work out as well as David's example since there is an unending number of non repeating decimals and thus would imply that you have infinite storage capability, which even the Universe does not have. ==== : I've memorized 100,000 digits. They're all 3. Of course I haven't >: memorized exactly where they occur. >: (Oh, you meant _consecutive_ digits? Never mind...) Well, in that vein, I've memorized all of them. They are 0,1,2,3,4,5,6,7,8 and 9. That's very impressive. I always forget the 8. >If only I could get the order right... Justin ************************ David C. Ullrich ==== 3.14159265358979323846264338327950288419716939937510.... It's a long story how it happened...(well, it's actually very short, but as if anyone would like to listen to it and believe it!!!) ==== >: I've memorized 100,000 digits. They're all 3. Of course I haven't >>: memorized exactly where they occur. >>: (Oh, you meant _consecutive_ digits? Never mind...) >>Well, in that vein, I've memorized all of them. >>They are 0,1,2,3,4,5,6,7,8 and 9. That's very impressive. I always forget the 8. I used to work with a guy who had spent a couple of months of concentrated software development time which meant that he had to think in octal about eight hours a day, six days a week (that doesn't include the time spent dreaming while sleeping). Then he balanced his checkbook. After a couple hundred heart palpitations, he verified that he hadn't written the checks using octal. /BAH ==== > I used to work with a guy who had spent a couple > of months of concentrated software development time which > meant that he had to think in octal about eight hours a > day, six days a week (that doesn't include the time spent > dreaming while sleeping). Then he balanced his checkbook. > After a couple hundred heart palpitations, he verified that > he hadn't written the checks using octal. Me too, caused a few riots keeping score for darts at the pub in the evenings. No, sorry, treble 17 isn't 55 is it : ) -- Paul V. S. Townsend Interchange the alphabetic elements to reply ==== >> I used to work with a guy who had spent a couple >> of months of concentrated software development time which >> meant that he had to think in octal about eight hours a >> day, six days a week (that doesn't include the time spent >> dreaming while sleeping). Then he balanced his checkbook. >> After a couple hundred heart palpitations, he verified that >> he hadn't written the checks using octal. Me too, caused a few riots keeping score for darts at the pub in the >evenings. No, sorry, treble 17 isn't 55 is it : ) Unless you played darts with like-minded friends. They wouldn't have batted an eye but would have done the conversion automatically. /BAH ==== I learned 500, but now, I have probably forgotten most of them (It was some years ago). /Anders > How many digits of pi have you memorized? I did 100. The world record ==== I memorized 27 digits when I was 12, and since then I've not forgotten them nor have I memorized any more. What might be more interesting is the number of digits someone may have memorized such that any particular digit can be recalled simply from its place. As someone said, what's digit 77? If you know 100 digits and can recall each one by place, that's impressive! =) Justin : How many digits of pi have you memorized? I did 100. The world record ==== Hey, I need a drink, alcoholic of course, after the heavy lectures on quantum mechanics... 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9.... There is some chance I mishandled the quote. G C ==== > Hey, I need a drink, alcoholic of course, after the heavy lectures on quantum > mechanics... 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9.... There is some chance I mishandled the quote. I can remember that there was a doggerel verse of seven or eight lines, which (by the same scheme) gave pi to 30 decimals, but I can only remember the first line: Now I, even I, would celebrate. Back in my student days, some students tried to devise mnemonics for pi and e. One mnemonic for e neatly avoided the charge that such sentences were complicated, tedious or full of obscurities: In seeking a mnemonic, we composed a sentense of sensible words. Not enough entries were received for the first prize of a bottle of wine to be awarded, but one of the students received a Mars bar as a consolation prize for this one for pi: Buy a Mars a month. Chocolate is seldom cheap but costly. -- Paul V. S. Townsend Interchange the alphabetic elements to reply ==== > I can remember that there was a doggerel verse of seven or eight lines, > which (by the same scheme) gave pi to 30 decimals, but I can only remember > the first line: Now I, even I, would celebrate. Found it, ascribed to a certain Adam C. Orr of Chicago in1906. Now I, even I, would celebrate In rhymes unapt, the great Immortal Syracusan rivalled nevermore Who in his wondrous lore Passed on before, Left men his guidance how To circles mensurate. (Should it be inept in line 2? Who is the guy from upstate New York referred to in line 3?) -- Paul V. S. Townsend Interchange the alphabetic elements to reply ==== : Now I, even I, would celebrate : In rhymes unapt, the great ... : (Should it be inept in line 2? Who is the guy from upstate New York : referred to in line 3?) Depends upon whether you want to say unapt or inept. The rhyme *is* awfully unapt. I would prefer: ----------------------------- Can I give a digit numerical of number which can ratio diameter-perimeter divulge? beautiful 'tis, we say. ----------------------------- Justin ==== >> Hey, I need a drink, alcoholic of course, after the heavy lectures on >quantum >> mechanics... >> 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9.... >> There is some chance I mishandled the quote. I can remember that there was a doggerel verse of seven or eight lines, >which (by the same scheme) gave pi to 30 decimals, but I can only remember >the first line: Now I, even I, would celebrate. Back in my student days, some students tried to devise mnemonics for pi and >e. One mnemonic for e neatly avoided the charge that such sentences were >complicated, tedious or full of obscurities: > In seeking a mnemonic, we composed a sentense of sensible words. >Not enough entries were received for the first prize of a bottle of wine to >be awarded, but one of the students received a Mars bar as a consolation >prize for this one for pi: > Buy a Mars a month. Chocolate is seldom cheap but costly. Good grief. It's easier just to remember the numbers rather than have call a conversion routine for each word. That's a waste of brainpower plus you have to remember where you were in the sentence. /BAH ==== In sci.math, jmfbahciv@aol.com <3fef0112$0$4765$61fed72c@news.rcn.com>: > Hey, I need a drink, alcoholic of course, after the heavy lectures on >>quantum > mechanics... >> 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9.... >> There is some chance I mishandled the quote. >>I can remember that there was a doggerel verse of seven or eight lines, >>which (by the same scheme) gave pi to 30 decimals, but I can only remember >>the first line: Now I, even I, would celebrate. >>Back in my student days, some students tried to devise mnemonics for pi > and >>e. One mnemonic for e neatly avoided the charge that such sentences were >>complicated, tedious or full of obscurities: >> In seeking a mnemonic, we composed a sentense of sensible words. >>Not enough entries were received for the first prize of a bottle of wine > to >>be awarded, but one of the students received a Mars bar as a consolation >>prize for this one for pi: >> Buy a Mars a month. Chocolate is seldom cheap but costly. > > > Good grief. It's easier just to remember the numbers rather than > have call a conversion routine for each word. That's a waste of > brainpower plus you have to remember where you were in the > sentence. 3.14159, Oh the digits are sublime, 2653589, More and more come all the time, 7932384, C'mon man let's do some more, 62648338, Wow! This pi thing sure is great! :-) > > /BAH > -- #191, ewill3@earthlink.net -- insert random weird poetry here It's still legal to go .sigless. ==== >Hey, I need a drink, alcoholic of course, after the heavy lectures on quantum >mechanics... 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9.... There is some chance I mishandled the quote. The third-to-last digit above should be a 9, not a 2. Could it be that it was supposed to have been heavy lectures regarding quantum mechanics? -- Erick ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- >Hey, I need a drink, alcoholic of course, after the heavy lectures on quantum >mechanics... 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9.... There is some chance I mishandled the quote. The third-to-last digit above should be a 9, not a 2. Could it be that it > was supposed to have been heavy lectures regarding quantum mechanics? > you mean 3 1 4 1 5 9 2 6 5 3 5 8 9 7 9.... I thought 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9 7... there's a ACBC around there Herc ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > How many digits of pi have you memorized? I did 100. The world record 3.1415926536 not sure with the rounding I'll pass the Dalek Pyramid test with the electrocuting floor of 10 X 10 tiles, each row only the pi digit is safe. What's digit 77? Herc ==== In sci.math, |-|erc <(áÀá)> <^> ----------------------------- > >> How many digits of pi have you memorized? I did 100. The world record > > 3.1415926536 > > not sure with the rounding I'll pass the Dalek Pyramid test with the > electrocuting floor of 10 X 10 tiles, each row only the pi digit is safe. > > What's digit 77? > > Herc > I sure hope you've got your sums right. -- Teegan Jovanka (Janet Fielding) :-) -- #191, ewill3@earthlink.net -- who? It's still legal to go .sigless. ==== > How many digits of pi have you memorized? I did 100. The world record I think you've got a real shot at the record. You're only 42,095 behind. ==== I did one symbol. Does that count? No actually I could never do more than 12 digits before running out of interest. ==== Something like this: Hey, I need a drink, alcoholic of course..... I forgot the rest; was it: Hey, I need a drink, alcoholic of course, after the (lectures on quantum mechanics...) Count the letters in each word: 3 1 4 1 5 9 2 6 5 3 ...(.....?) G C ==== the ratio of the diameter of sphere to its circumference is greater than three! on teh other hand, the ratio of the area of the equatorial circle of a sphere to its spherical surface is 1/4. > How many digits of pi have you memorized? I did 100. The world record --ils duces d'Enron! http://laroucehin2004.com ==== > How many digits of pi have you memorized? I did 100. The world record 3.1415926535 Now... check this out. Assume that the Earth is a sphere. One can calculate the circumference using 3.14159 to within an order of a meter. Each extra decimal reduces the amount that you are off by an order of magnitude. Now, let us assume that the galaxy is a disk 150000 LY in diameter (75000 LY radius) This is a radius of ~ 7E20 meters. With pi to 100 digits this will give the circumference to within some 1E-80 meters of exact. Do you need to know pi to that precision? No. ==== In sci.math, Michael Varney : > >> How many digits of pi have you memorized? I did 100. The world record > > > 3.1415926535 > > Now... check this out. > > Assume that the Earth is a sphere. One can calculate the circumference > using 3.14159 to within an order of a meter. Each extra decimal reduces the > amount that you are off by an order of magnitude. > > Now, let us assume that the galaxy is a disk 150000 LY in diameter (75000 LY > radius) > This is a radius of ~ 7E20 meters. > With pi to 100 digits this will give the circumference to within some 1E-80 > meters of exact. > > Do you need to know pi to that precision? > No. > Probably not, but it's handy for checking out new processor units. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > How many digits of pi have you memorized? I did 100. The world record 3.141592 Six digits. And I think it's enough :) Spider ==== >>The world record Always avoid the use of goto! ==== >>The world record Always avoid the use of goto! *groan* ==== Spider > How many digits of pi have you memorized? I did 100. The world record 3.141592 Six digits. > And I think it's enough :) Spider The digits are the number of letters in the words of that well-known saying, How I need a drink, alcoholic of course, after the .... LH ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- Skeptic organisations are not interested in the scientific investigation > of > the paranormal. Skeptic organisations are interested in shutting down businesses that make > paranormal claims. It doesn't matter what evidence or results the business is getting, simply > having > a paranormal claim is enough for the 1000s of skeptic organisations to > take > action to close them down. Thousands? Name 60 please. Or did you make that up? > Atleast one in every state of Australia, pop < 20 million http://www.skeptics.com.au/about/contact.htm > State & Regional Branches New South Wales : > Victoria : > South Australia : > ACT (Australian Capital Territory): > Western Australia : > Northern Territory : > Queensland : > Gold Coast (Queensland) > Gold Fields (Ballarat, Victoria) : > Borderline (Mitta Mitta, Albury, Wodonga): > Hunter Valley (NSW) : > Tasmania That's less than thousands... Rather a lot really. Guess it was made up. ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > ----------------------------- <^> <(áÀá) <^> ----------------------------- Skeptic organisations are not interested in the scientific > investigation > of > the paranormal. Skeptic organisations are interested in shutting down businesses that > make > paranormal claims. It doesn't matter what evidence or results the business is getting, > simply > having > a paranormal claim is enough for the 1000s of skeptic organisations to > take > action to close them down. Thousands? Name 60 please. Or did you make that up? > Atleast one in every state of Australia, pop < 20 million http://www.skeptics.com.au/about/contact.htm > State & Regional Branches New South Wales : > Victoria : > South Australia : > ACT (Australian Capital Territory): > Western Australia : > Northern Territory : > Queensland : > Gold Coast (Queensland) > Gold Fields (Ballarat, Victoria) : > Borderline (Mitta Mitta, Albury, Wodonga): > Hunter Valley (NSW) : > Tasmania That's less than thousands... Rather a lot really. Guess it was made up. > 1 skeptic organisation per million capita in our country. extrapolate to 1 billion in western civilisation. Herc ==== In sci.math, Dik T. Winter : > In sci.math, Dik T. Winter > : > ... > A lot of conclusions are drawn from this factorisation, it is however > the consequence of another factorisation. Set y = 49x, we get: > P(y) = 125 y^3 - 375 y^2 - 360 y + 1078 = > = (5 a_1(y) + 7)(5 a_2(y) + 7)(5 a_3(y) + 7) > where the a's are roots of: > a^3 + 3(-1 + y)a^2 - (y^3 - 3y^2 + 3y) > > So although the constant term of two of the factors are divisible by > 7, and the third is co-prime to it (it is 22), in general *none* of the > factors are divisible by 7, as P(y) is only divisible by 7 in a limited > number of cases. So I am still wondering what JSH is trying to show > with his constant terms. > > So am I. I'd be curious to figure out when a_1(x)/7 and a_2(x)/7 > are algebraic integers, but admittedly it's not a priority. > > You first are required to show what a_1, a_2 and a_3 are. Luckily that > is possible... (*) Yes, it is possible. However, I don't have to compute them explicitly; I merely need show that the defining equation for a_1(x)/7 etc. is not of the requisite form for most x. Then again, your way might be slightly cleaner, and you also have a constructive proof, whereas I merely have an existance proof. > > It's > clear that generally speaking, they are not, although > a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. > > Your manipulation is interesting and simplifies the problem > considerably. :-) However, now one has to deal with y being > a multiple of 49, if x was originally an algebraic integer. > > But it also shows that the factors are not divisible by 7 for *all* y, > so it shows that divisibility properties are erratic. Aye! > > (I'm also curious as to the rest of his proof; this is > only a small snippet thereof. It's a bit like examining > a small area (sans the actual hole) of a flat tire to try > to figure out why the car won't move.) > > The rest of his proof hinges on the fact that exactly two factors are > (FLT proof) or should be (definition error) divisible by 7. Not to mention that he assumes the factors are algebraic integers at all. After all, 1 = 1/49 * 49, which means one can divide 1 by 49. But it wouldn't mean much. :-) > ---- > (*) A very nice showing of that I found while looking around at the > solutions of cubic equations. All expositions I have seen miss a very > basic fact (I think), but the exposition at mathworld is closest (this > > Let's calculate the roots of z^3 + a.z^2 + b.z + c = 0. Let us have > the following definitions: > Q = (12.b - 4.a^2) > R = 36.a.b - 108.c - 8.a^3 > K1 = cbrt(R + sqrt(Q^3 + R^2))/2 > K2 = cbrt(R - sqrt(Q^3 + R^2))/2 > W = (-1 + i.sqrt(3))/2, W^2 = (-1 -i.sqrt(3))/2, W^3 = 1 > then we have the following roots: > z_1 = (-a + W .K1 + W^2.K2)/3 > z_2 = (-a + W^2.K1 + W .K2)/3 > z_3 = (-a + K1 + K2)/3 > filling in all stuff (and hoping I did not make a basic arithmetic > mistace), we find that the z's correspondend to James' a's in order. > > The beauty of this presentation is (I think) how three cubic roots > of two different values are used. That's a nice distillery of the problem, admittedly. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== In sci.math, Virgil Hancher > >> [...] >> Jim, I'm going on a break, the discussion is over. You are >> sitting at a comfy computer and you are free to do what you >> want. I am at my computer because I am fighting for my life >> while I am being tortured by a spy satellite for 2 years >> continuously. it is the most hideous torture in all history, >> I wish you would believe me, even give me a few hours benefit >> of doubt because that is all it would take to confirm my >> admitedly odd story. Don't watch your TV, its a lie. > > You can get relief by wearing a hat of aluminium foil, or by blowing > your brains out. The first is probably easier on the brains... http://zapatopi.net/afdb.html :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== ... > I have my program looking at n=10 now; it has run nearly 7 hours > already (in contrast to half-second, 3-second, and 4-minute timings > at n=7,8,9 on a 750MHz pentium) but hasn't found any solutions yet. That program eventually finished, after 71.6 hours cpu hours. It found the same solution set as the new program found in 15 minutes on my 450 MHz AMD-K6. Here is some output after an hour of processing at n=11 via program at http://pat7.com/jp/perp11b.c -- 52721211321 229611 29452737924 171618 74143477849 272293 25448863729 159527 12386799616 111296 27487318849 165793 13769144964 117342 17739842481 133191 39876894864 199692 22421468644 149738 14996941444 122462 ==== > ... Here is some output after an hour of processing > at n=11 via program at http://pat7.com/jp/perp11b.c -- ... The program finished in 6.8 hours and found 8 n=11 perplexes - 229611 171618 272293 159527 111296 165793 117342 133191 199692 149738 122462 290369 218896 198022 105629 212771 121426 234925 261263 135884 248686 128108 149119 156904 163332 178304 262157 219846 276441 255465 110665 247415 128334 271891 194623 314239 165758 220585 266791 106838 230093 284873 286815 138366 195396 284891 108746 276067 304178 242266 314619 247559 297568 134865 248384 105904 117285 167085 123039 236184 259535 240828 267865 149192 110568 255962 203276 132044 186966 153381 125121 127563 178427 165876 138264 271228 258093 287373 168632 233865 289816 190367 166839 212884 108785 119528 149192 307521 For example, solution #6 squares out as: 11215657216 105904 13755771225 117285 27917397225 167085 15138595521 123039 55782881856 236184 67358416225 259535 57998125584 240828 71751658225 267865 22258252864 149192 12225282624 110568 65516545444 255962 ==== > ... Here is some output after an hour of processing > at n=11 via program at http://pat7.com/jp/perp11b.c -- > ... > The program finished in 6.8 hours and found 8 n=11 perplexes - > 229611 171618 272293 159527 111296 165793 117342 133191 199692 149738 122462 > ... [ I cancelled a reply I made, re N=12. The cancelled message that everyone should ignore ==== |In Grothendieck's EGA(Elements de Geometrie Algebrique) I (1960) |Proposition (9.5.1) p.176 states; |Let f:X --> Y be a morphism of schemes such that |f_*(O_X) is quasi-coherent sheaf of O_Y module. |Then there exists a closed subscheme Y' of Y with |the following property. |f splits into X --> Y' --> Y and Y' is the smallest |closed subscheme of Y with this property. | |However, I think this proposition holds without the condition |on O_X. Am I missing here? Do you have an argument for the result without the condition, or are you just unable to come up with a counterexample? Perhaps it's to prevent situations where Y is not a separated scheme? Keith Ramsay ==== > |In Grothendieck's EGA(Elements de Geometrie Algebrique) I (1960) > |Proposition (9.5.1) p.176 states; > |Let f:X --> Y be a morphism of schemes such that > |f_*(O_X) is quasi-coherent sheaf of O_Y module. > |Then there exists a closed subscheme Y' of Y with > |the following property. > |f splits into X --> Y' --> Y and Y' is the smallest > |closed subscheme of Y with this property. > | > |However, I think this proposition holds without the condition > |on O_X. Am I missing here? > > Do you have an argument for the result without the condition, > or are you just unable to come up with a counterexample? I think I have a proof for the proposition without the condition. I posted the sketch of my proof. However, I'm not 100% sure of my proof, and EGA has authority... > > Perhaps it's to prevent situations where Y is not a separated > scheme? It's unlikely since Grothendieck was trying hard to avoid unnecessary conditions on his results. Nobuo Saito ==== > In Grothendieck's EGA(Elements de Geometrie Algebrique) I (1960) > Proposition (9.5.1) p.176 states; > Let f:X --> Y be a morphism of schemes such that > f_*(O_X) is quasi-coherent sheaf of O_Y module. > Then there exists a closed subscheme Y' of Y with > the following property. > f splits into X --> Y' --> Y and Y' is the smallest > closed subscheme of Y with this property. > > However, I think this proposition holds without the condition > on O_X. Am I missing here? My proof(sketch) of the proposition (9.5.1) If Y is an affine scheme Spec(A), then f: X --> Y is determined by homomorphism h: A --> O_X(X). Let I = Ker(h). Then Y' = Spec(A/I) satisfies the property of the proposition. If Y is not affine, then glue affine shemes obtained in the above method. N. Saito ==== everyone with this message: asleep, the following problem popped in my mind: > Some additions: You can consider a linear variant of that problem. We should find a minimal string which contains all binary strings of length n>1 (that one should be of length 2^n+n-1). For example (n=3) the following string is solution: 1110001011 I can prove that that meta-string exists for any n>1. Also a circle (string with connected ends) of length 2^n exists too (in fact every meta-string could be reduced to circle by cutting off last n-1 bits). Proof after s p o i l e r s p a c e Consider a graph G which vertices are all strings of length n. If string b can be put after string a (the first n-1 bits of b are equal to last n-1 bits of a) then an arrow from a to b is drawn. Thus every vertex has two ins and two outs. The graph is obviously connected. So there should be a cycle which goes through every vertex only once (ask Euler). That proves our theorem. Unfortunately this proof has nothing to do with 2D theorem. -- |E1M1 :29 E1M5 :19 E2M1 :10 E3M2 :22 E4M5 :15,===================; |E1M2 :36 E1M6 :11 E2M3 :27 E4M1 :30 E4M6 :24|->grue3.tripod.com<--| |E1M3_:43__E1M7_:14__E3M1_:43__E4M2_:52__END__:37;================[4*72] ==== > Well, it's probably hard to understand without an example. > Let n=1. So we have all 2*2 binary matrices: > OO OO XO OX OO XO OX XX > OO OX OO OO XO OX XO XX XX XO OO OX XX XX OX XO > OO XO XX OX XO OX XX XX > That's torus XXXO > XXOX > OXOO > XOOO And that's matrix XXXOX > XXOXX > OXOOO > XOOOX > XXXOX > The torus is kinda fascinating; all the possible combinations...could possibly be useful in artificial life applications using a cell grid. Have you come up with a use for such a compacting of information? It's almost a form of compression. ==== |Well, in fact I don't have the proof so you could try to find it. I just hope |elliptical curves have nothing to do with that problem... Elliptic curves. I'm sure they aren't. The one-dimensional version (a sequence containing all subsequences of 0s and 1s of a given length) is pretty well known, so odds are good this one is too. Keith Ramsay ==== Worth to mention: Backup of all science related forums. Worth to mention: also good for scientists to find their topic right away. Website and forums order looks good to me as well. Anyways, I just though it's good to have it in our favourites. 7* ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- ==== > ----------------------------- <^> <(áÀá) <^> ----------------------------- > > I have found several Chinese medicines that work much better > than the Western medicines for certain things. > > When I young, and got colds, > I took those multi-purpose cold remedies, > that combined decongestants, antihistamines, antitussives, > expectorants, pain relievers, etc. in one pill. > As I wised up, I just took the specific ingredient > that I needed. > > I also tried many grams of vitamin C with no results, > although I did find that zinc tablets sometimes helped. > > I found the Chinese cold medicines worked better > than Western medicine medicines, and that ginger > was a common ingredient, so I bought some ginger, > and when I feel a cough or the sniffles coming on, > I snack on the ginger, and it seems to work very well. > > You snack on RAW ginger? Brave, you are! It is a little hot, but otherwise it tastes pretty good, and it attacks cold symptoms immediately. Tom Potter ==== >I'll admit colds for me are extremely rare, although I will also >admit to washing my hands frequently after using the restroom >facilities, and using my elbows, feet, or arms instead of my hands >for opening the doors thereto and within. Exactly. One of my tricks is to wear long-sleeved sweatshirts > and cover my hand before opening a door. While out, not touching > eyes and nose (that's when they always decide to get itchy) also > helps. Since women think it's cute to allow their sick kids to > touch everything in the grocery store, I wash everything I can > before I put them away. Always buy packaged produce. Make your > own hambuger after washing the meat. > /BAH I read about a study that found traces of 97 mens urine in the free peanuts at the bar. We should stop encouraging the 'working while sick' is pro company ethic, and encourage people to use their sick days at 1st sign. Herc X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS016712394; ==== >I want to make a spreadsheet to calculate the number of feet of material on a >roll. The user will enter the Outside Diameter, the Core Diameter (The tube that the >material is wrapped around) and the number of feet for a FULL roll first. Then they will enter the Outside Diameter of the partial roll they want to >measure, and the Core Diameter. The spreadsheet will then calculate the number >of feet left on the roll. I can't seem to come up with the algebraic expressions to calculate the >thickness of the material or the number of layers on the roll. I also want to >figure out the factor that will adjust the circumference for each successive >layer on the roll. It seems to me that the first section of entries (the OD, >the CORE and FEET) should be enough to extrapolate the number of layers on the >roll, and the thickness of the material. I know that if I had them enter the >material thickness to the equation it would be easier, but I don't want them to >have to measure the thickness. any ideas? Mike Here is an approximate way to attack the problem. If you look at the roll end-on, the cross-sectional area of material is approximately the material thickness times its length. A = pi * (r2 - r1)^2 = t * L So if you measure r2 = outside diameter/2 and r1 = core diameter/2 and you know L = material's length, you can get the thickness. Then apply same formula to another roll to find unknown length. Happy New Year phil ==== >>I want to make a spreadsheet to calculate the number of feet of material on a >>roll. >>The user will enter the Outside Diameter, the Core Diameter (The tube that the >>material is wrapped around) and the number of feet for a FULL roll first. >>Then they will enter the Outside Diameter of the partial roll they want to >>measure, and the Core Diameter. The spreadsheet will then calculate the number >>of feet left on the roll. >>I can't seem to come up with the algebraic expressions to calculate the >>thickness of the material or the number of layers on the roll. I also want to >>figure out the factor that will adjust the circumference for each successive >>layer on the roll. It seems to me that the first section of entries (the OD, >>the CORE and FEET) should be enough to extrapolate the number of layers on the >>roll, and the thickness of the material. I know that if I had them enter the >>material thickness to the equation it would be easier, but I don't want them to >>have to measure the thickness. >>any ideas? >>Mike > I lost the original post, but this reply is to Mike. I think the problem is easier than you make it. If you call the radius of the tube r0 and the outer radius of a new roll r1 and you have the length of a new roll, call it L, then the length will be very close to linear as a function of the radius r: length = L*(r - r0)/(r1-r0), for r0 <= r <= r1 By basis for thinking this is that if the roll is modelled by a linear spiral r = k*theta + c, which I think it is, such a curve turns out to have arc length a linear function of theta and hence a linear function of the radius. Try it. I would be interested to hear how close it works out to the correct length in the shop. --Lynn ==== By basis for thinking this is that if the roll is modelled by a linear >spiral r = k*theta + c, which I think it is, such a curve turns out to >have arc length a linear function of theta and hence a linear function >of the radius. Woops I should have checked this more carefully. It isn't linear. Please ignore my post. --Lynn X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS016e12389; ==== Isn't 4 x 1 metre sidewalk too small for two men building it? Long time ago my dad asked me a question: If one man would dig a well in five hours, how much time would five workers need? I answered and then he said Imagine five big men digging one little well, huh? Smile! Mateusz ==== In sci.math, Mateusz Kwasnicki > Long time ago my dad asked me a question: If one man would dig > a well in five hours, how much time would five workers need? > I answered and then he said Imagine five big men digging one > little well, huh? > > Smile! > > Mateusz > Indeed. If a woman can have a baby in 8 1/2 months, how long would it take 8 1/2 women (the 1/2 is working part time on another project :-) ) ? -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > In sci.math, Mateusz Kwasnicki > > Long time ago my dad asked me a question: If one man would dig > a well in five hours, how much time would five workers need? > I answered and then he said Imagine five big men digging one > little well, huh? > > Smile! > > Mateusz > > > Indeed. If a woman can have a baby in 8 1/2 months, how long > would it take 8 1/2 women (the 1/2 is working part time on > another project :-) ) ? Be nice, people. I specifically stated in my original post that the problem would not be modelled correctly unless the two workers worked with perfect cooperation. Clearly women cannot cooperate on creating a baby, althoughh these days that may no longer be as true as it once was. NO ATOMIC MEAURES! NO! NO! NO!. How would you then be able to cut the cake? As to the men working on the sidewalk. First of all this was the United States where most of us don't have a clue what a metre might be, and surely the cracks in side walks are not separated by a metre or even a meter as we say in the USA. Perhaps they were a yard apart. I don't remember. This was done in 1959 or thereabouts, and I am having trouble remembering the size of the sidewalk squares. I do remember that it was really, really difficult to make one step per square for more than a couple of squares. I should also point out that the problem never says that this isn't done with a Gilbert SideWalk Kit that two boys could be using. I did say boys, not men, in the original post, although I did admit the somewaht egocentric nature of that assumption. At any rate, there were 5 men working digging a ditch, and the foremman was sitting on his ... watching the men work. One of the came up and asked him why they had to do all the hard work and he got to sit on his ... and watch them. He told the guy that it was intelligence. Intelligence! What's that, said the worker. The foreman held his hand up in front of a tree and said to the worker, might and the foreman pulled his hand away just in time so that the worker punched the tree with all of his might. He went back down into the trench and was asked what the foreman had had to say. He said it was intelligence, said the worker. Intelligence! What's that, replied his friend. Simple, he said, holding his hand in front of Achava X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS2gbb22101; ==== I recall having been taught not use operations such as +, -, /, x that will effect both sides(at the same time) of an identity to be proven. And so I told my students to follow this rule. One student respectfully and thoughtfully said why and went on to pose that the relationship between the two expressions we are trying to show are identical must be either >, >=, =, <= or < . So with the exception of multiplying or dividing by a negative value, the unknown relationship will stay the same(if we operate on both expressions at once). For example (trivial but it demonstrates the question) suppose you want to prove that 1 + sin(-x) = 1 - sin(x) . Normally we would apply the cofunction identity sin(-x) = - sin(x) the left hand expression and be done. But why not first subtract -1 from both expressions and then apply the cofunction identity? If the rule I remember being taught is valid there should exist some pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in fact identical but would appear to be identical if one operates on both expressions simulataneously. For example suppose one were trying to prove Expr1(x) = Expr2(x). In the process he/she divided both sides by cos(x)and found that the modified expresssions were in fact equal. I can't come up with such a pair of expressions that would validate the rule I remember. Can anyone help? ==== > I recall having been taught not use operations such as +, -, /, x >that will effect both sides(at the same time) of an identity to be >proven. And so I told my students to follow this rule. One student >respectfully and thoughtfully said why and went on to pose that the >relationship between the two expressions we are trying to show are >identical must be either >, >=, =, <= or < . So with the exception >of multiplying or dividing by a negative value, the unknown relationship >will stay the same(if we operate on both expressions at once). > > For example (trivial but it demonstrates the question) suppose you >want to prove that 1 + sin(-x) = 1 - sin(x) . > Normally we would apply the cofunction identity sin(-x) = - sin(x) the left hand expression and be done. But why not first subtract -1 from >both expressions and then apply the cofunction identity? > > If the rule I remember being taught is valid there should exist some >pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in >fact. >identical but would appear to be identical if one operates on both > expressions simulataneously. For example suppose one were trying to prove >Expr1(x) = Expr2(x). In the process he/she divided both sides by >cos(x)and >found that the modified expresssions were in fact equal. I can't come up >with >such a pair of expressions that would validate the rule I remember. >Can anyone help? Firstly you must differentiate between an identity and an equality. Secondly, why must the alternatives to 'equals' be some inequality? Fine if you're dealing with real numbers, but what about complex numbers? In the example you've provided, all you;ve done is work backwards from the correct fact, that sin is an odd function (what's a cofunction identity by the way?), to the statement you want. In this case all the steps you've taken are reversible, that is it is an if and only if deduction. In general in mathematics that doesn't happen, so it's better to teach students how to think 'properly'. An example where you shouldn't work backwards (trivial, but important) show sin(x) '=' sin^3(x) + sin(x)cos^2(x) you can't factor out sin and divide because for some values of x you're dividing by zero. You may start from the identity 1 '=' sin^2 + cos^2 and multiply through. The '=' bit means I really ought to write the three bar symbol. And that hidden dividing by zero thing needs to be watched. It is bad practice to assume the answer and work backwards. Don't let them start now and make maths teachers later tear their hair out. ==== >It is bad practice to assume the answer and work backwards. Don't >let them start now and make maths teachers later tear their hair out. Actually most of the time it's easier to work backwards, and perfectly valid _provided_ you know what you're doing: i.e. you know you're working backwards, and that the real formal proof will go forwards, so you make sure that the real proof will have valid inferences. To help in this, I like to write ?=, ?> etc. rather than = or > for relations I want to prove, but haven't yet. In your example: 1) sin(x) ?= sin^3(x) + sin(x)cos^2(x) is equivalent to 2) sin(x) ?= sin(x) (sin^2(x) + cos^2(x)) which is implied by 3) 1 ?= sin^2(x) + cos^2(x) which is an identity we know. Note that it's OK to go from 2) to 3), precisely because we're reasoning backwards: a b = a c does not imply b = c (because a might be 0), but it is implied by b = c. Now once you have the backwards proof, you can turn it around to make the real proof: 1 = sin^2(x) + cos^2(x) sin(x) = sin(x)(sin^2(x) + cos^2(x)) sin(x) = sin^3(x) + sin(x) cos^2(x) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > >>It is bad practice to assume the answer and work backwards. Don't >>let them start now and make maths teachers later tear their hair out. > > Actually most of the time it's easier to work backwards, and perfectly > valid _provided_ you know what you're doing: i.e. you know you're working > backwards, and that the real formal proof will go forwards, so you make > sure that the real proof will have valid inferences. To help in this, > I like to write ?=, ?> etc. rather than = or > for relations I > want to prove, but haven't yet. > I think in some post that got lost by outlook (not usual choice) I agreed, and still do, that it is often practical to work backwards, though I tend to get students to rewrite it so it doesn't look as though they worked backwards. And to make sure that all the if statements are only if statements too in this case. But the 'if you know what you're doing' proviso is key here. Surely it is better to go from what you know to be true forwards. If the method you choose is deduced from thinking backwards so be it. I agree the example below was pretty dire, and I certainly regret using it. > In your example: > > 1) sin(x) ?= sin^3(x) + sin(x)cos^2(x) > > is equivalent to > > 2) sin(x) ?= sin(x) (sin^2(x) + cos^2(x)) > > which is implied by > > 3) 1 ?= sin^2(x) + cos^2(x) > > which is an identity we know. > > Note that it's OK to go from 2) to 3), precisely because we're reasoning > backwards: a b = a c does not imply b = c (because a might be 0), but > it is implied by b = c. Now once you have the backwards proof, you can > turn it around to make the real proof: > > 1 = sin^2(x) + cos^2(x) > sin(x) = sin(x)(sin^2(x) + cos^2(x)) > sin(x) = sin^3(x) + sin(x) cos^2(x) > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 ==== [snip] > In the example you've provided, all you;ve done is work backwards from > the correct fact, that sin is an odd function (what's a cofunction > identity by the way?), I presume that the OP misused that term. Surely a cofunction identity is something like cos(x) = sin(pi/2 - x). > to the statement you want. In this case all the > steps you've taken are reversible, that is it is an if and only if > deduction. In general in mathematics that doesn't happen, so it's better > to teach students how to think 'properly'. An example where you shouldn't > work backwards (trivial, but important) show sin(x) '=' sin^3(x) + sin(x)cos^2(x) you can't factor out sin and divide because for some values of x you're > dividing by zero. One may factor out sin(x) on the right, and then notice that the other factor, sin^2(x) + cos^2(x), is always 1, thereby showing that the right side simplifies to the left side. > You may start from the identity 1 '=' sin^2 + cos^2 and > multiply through. The '=' bit means I really ought to write the three bar > symbol. And that hidden dividing by zero thing needs to be watched. At least it needs to be watched when one is solving conditional equations, lest roots be lost. > It is bad practice to assume the answer and work backwards. In _general_, I wholeheartedly agree. But, in the specific context of proving identities, I disagree. For example, very often, the easiest method of proof is to simplify both sides of the supposed identity as much as possible, in the hope that both sides will simplify to the same thing. David Cantrell ==== >I recall having been taught not use operations such as +, -, /, x >that will effect both sides(at the same time) of an identity to be >proven. ... >If the rule I remember being taught is valid there should exist >some pair of trigonometric expressions Expr1(x) and Expr2(x)that >are not in fact identical but would appear to be identical if one >operates on both expressions simulataneously. Something in what you have written here makes me think of the Perron Paradox. Perhaps that would be of help to you. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS4GIB28000; ==== >I recall having been taught not use operations such as +, -, /, x that will effect both sides(at the same time) of an identity to be proven. And so I told my students to follow this rule. One student respectfully and thoughtfully said why and went on to pose that the relationship between the two expressions we are trying to show are identical must be either >, >=, =, <= or < . So with the exception of multiplying or dividing by a negative value, the unknown relationship will stay the same(if we operate on both expressions at once). For example (trivial but it demonstrates the question) suppose you want to prove that 1 + sin(-x) = 1 - sin(x) . >Normally we would apply the cofunction identity sin(-x) = - sin(x) the left hand expression and be done. But why not first subtract -1 from both expressions and then apply the cofunction identity? If the rule I remember being taught is valid there should exist some pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in fact identical but would appear to be identical if one operates on both expressions simulataneously. For example suppose one were trying to prove Expr1(x) = Expr2(x). In the process he/she divided both sides by cos(x)and found that the modified expresssions were in fact equal. I can't come up with such a pair of expressions that would validate the rule I remember. Can anyone help? I have never heard of this rule. There is absolutely no reason why you can't perform the same operation on any equation, identity or not, and not arrive at a valid equation. One of the things you do have to be careful about is that each operation you use has to be invertible: that is you have to be careful about multiplying both sides of an equation by sin x since the inverse would be dividing both sides by sin x which is 0 for some x. The reason for this is that the real proof is working backwards. In order to prove, for example, that (tan x)(cos x)= sin x, we replace tan x with sin x/cos x so that the left side is (sin x/cos x)(cos x) which is, of course, sin x. But you can't prove a formula, or any statement, is true by starting from that statement! The real proof is working backwards. Looking at what we did, we know we an start from sin x= sin x, rewrite the left side as (sin x)(cos x/cos x)= (sin x/cos x)cos x= (tan x)(cos x). We don't have to write that out because we know that everything we did originally is reversible. ==== > In order to prove, for example, that (tan x)(cos x)= sin x, we replace >tan x with sin x/cos x so that the left side is (sin x/cos x)(cos x) >which is, of course, sin x. That's not a proof. In fact there's nothing to prove since by definition tan x == sin x / cos x (this follows from elementary geometry) so the result is simply a reformulation of the definition. Furthermore, it's not even an equality. What happens when x = Pi/2? The right side is still well-defined but the left side most certainly not. > But you can't prove a formula, or any statement, is true by starting >from that statement! The real proof is working backwards. >Looking at what we did, we know we an start from sin x= sin x, rewrite >the left side as (sin x)(cos x/cos x)= (sin x/cos x)cos x= (tan x)(cos x). And how do you know that (sin x / cos x) = tan x when that is the very result you're trying to prove? >We don't have to write that out because we know that everything we did >originally is reversible. There is nothing that says you must work out your proofs backwards (what you call forwards) or that all proofs must be two-way. If we start from given truths and work our way using one-way implications (i.e. p => q) there is nothing that states the result is incorrect, just that the opposite (q => p) does not necessarily hold. ==== > In order to prove, for example, that (tan x)(cos x)= sin x, we replace >> tan x with sin x/cos x so that the left side is (sin x/cos x)(cos x) >> which is, of course, sin x. > That's not a proof. In fact there's nothing to prove since by > definition tan x == sin x / cos x (this follows from elementary > geometry) so the result is simply a reformulation of the definition. There are several ways of defining tan, with or without any reference to cos and sin. It is a matter of convention. So what is a proof and what is only a reformulation of the definition only depends on what frame you're in. > Furthermore, it's not even an equality. What happens when > x = Pi/2? The right side is still well-defined but the left side most > certainly not. The right side is not defined either, at least in R, cos(Pi/2)=0. >> But you can't prove a formula, or any statement, is true by starting >> from that statement! The real proof is working backwards. >> We don't have to write that out because we know that everything we did >> originally is reversible. > There is nothing that says you must work out your proofs backwards > (what you call forwards) or that all proofs must be two-way. If we > start from given truths and work our way using one-way implications > (i.e. p => q) there is nothing that states the result is incorrect, > just that the opposite (q => p) does not necessarily hold. Both of you agree on that, I guess. I think that what you call backwards is what he calls backwards too... But what you have just written is confusing, even if I guess you agree with what I am going to say, I just want to clarify this point. There _is_ something that says that proof must go in one definite way: from the hypothesis to the result! This is the way that is meaningful. Sometimes, you can write the argument backwards because it is reversible. But what matters is the way (hyps+defs+axioms) => result. To come back to the original question, I would say that the rule Ed gave is pointless with _purely_ symbolic proofs, and needs to be sharpen when doing analysis. But I think it is interesting to explain to your students when such operations are allowed and when they are not, and why. It depends on their level, but I think it is always a bad idea to simply say NO!. Mathematics is not about calculus and its rules, it is about understanding. So you may allow them not to respect your rule, if they do it correctly. First of all, you want to consider functions, not expressions. You must be careful about the set of points on which what you are doing is valid. It may not be R as a whole. You may need to exclude some points, like 0. Sometimes what you write is only valid on (0; Pi/2], and so on. And the simplest case which is not _purely_ symbolic is probably with <= and some multiplication on both sides of the inequality. Dividing by cos(x) on both sides is problematic when cos(x) did not appear in the numerators of each side of the previous expression. Because in such a case, you may have a problem when x = Pi/2. But in some cases, it may still be valid. If you have cotan(x) on both sides, or if both sides are equivalent to cos(x) when x tends to Pi/2, etc. The reason why you did not find any counter example (ie. some example that would validate the need for your rule) is that you are using simple expressions. If your expressions Expr1 and Expr2 contain cos(x), sin(x), tan(x) and nothing else (but arithmetical operations), you can replace them by A, B, C and in practice, _any symbolic computation_ on both sides of any equality written with these three letters will remain valid with cos(x), sin(x) and tan(x). You can wait until you're done to ask you where this is defined. Why? This is not elementary. These three functions are C-Infinity functions, but in some isolated points where they are not defined. So even if an expression, which is a step in your proof, written with cos, sin and tan is not defined in some point, you may easily consider its limit in this point. And if it is infinite, you know precisely the speed of the convergence to the infinity. So that you may be able, in a latter step of your proof, to find a (finite) limit in the same point. And you will be able to state it rigorously. To say it simply, since they are explicit C-Infinity functions, cos, sin, and tan can be handled as polynomials, using Taylor's formula. I said previously that what you want to consider is functions and not expressions. This is because expressions are only symbolic and tells nothing about the values they can take when evaluated. And what interests you here is to identify expressions that take the same values in each point of some set. So that you identify functions through their graph. So, what do you say to your students? Again, it depends on their level, but I would recommend to abandon your rule, to concentrate on the meaning of what they do. That doesn't mean you cannot say to them: be careful about that, in some cases it is false, thus I would recommend you not to do this, unless you understand what you do and can explain it, and you can require that they explain what they do each time it is necessary. Try to tell them when it is safe: in particular, for trigonometric formulas, tell them about definition domains (taking some example where you divised by 0 with cos(x) and x = Pi/2), the problem with limits and infinity if they are advanced argument if they start with the result. More generally, my personnal point of view is to always avoid blind rules in Mathematics. It is sometimes necessary to adopt such a point of view: if you want to train them to compute derivatives quickly, for instance. But this should always be accompanied with explanations, and if you can't, give them at least some clues to help answer the why?. My experience is that if a student can remember why, then you have won. He/She may even begin to like Mathematics. /er. ==== > If uncountable S is a subset of R, then S contains > a subset T with the property that, for every distinct elements > x and y in T, there's a z in T between x and y. >> This problem is equivalent to showing for uncountable S, >> S is somewhere dense, >> that is not nowhere dense; int cl S /= nulset > See retraction sci.math nowhere dense real subsets. <1072549678.37972@news.aic.at> ==== only one was willing to make the trip. Our small research group was thus faced with the necessity of carrying n standard elephants themselves. In a pilot project, it was established that our full staff was unable to lift, much less carry, a single standard elephant. It was also noted by the project manager who was sprayed by the ill mannered standard elephant blowing it's nose upon him and again by the project director who had her butt whooped by a standard trunk, that standard elephants may lack appropriate manners needed for cooperation on a long long trip. Any further efforts on that field trip were immediately abandoned when n-1 standard elephants taking notice of the ruckus, filled their standard trunks with wet water and eagerly approached the experimental site. Upon recovery from this test project, the staff decided a better approach would be to use the standard kangaroo model on the speculation that the standardized equipment would hop itself into position. Before proceeding with this project some facts need be established. As standard kangaroos are smaller that standard elephants, will n standard kangaroos suffice or will 3n standard kangaroos be needed? Also once in place, how is a standard hopping kangaroo convinced to remain stationary? >By the way, thanks for the hint that standard elephants could maybe >decay into strange colorful left handed anti-quarks - we will >make some more calculations clarifying this point. Your welcome. Please keep us inform of your progress. Do be careful handling strange colorful left handed anti-quarks, those weirdo's have been noticed to change gender right in the middle of an.., experiment. Whence the name. ---- ==== > An interesting approach. However upon asking 262,144 standard elephants, > only one was willing to make the trip. Our small research group was thus > faced with the necessity of carrying n standard elephants themselves. In > a pilot project, it was established that our full staff was unable to > lift, much less carry, a single standard elephant. It was also noted by > the project manager who was sprayed by the ill mannered standard elephant > blowing it's nose upon him and again by the project director who had her > butt whooped by a standard trunk, that standard elephants may lack > appropriate manners needed for cooperation on a long long trip. Any > further efforts on that field trip were immediately abandoned when n-1 > standard elephants taking notice of the ruckus, filled their standard > trunks with wet water and eagerly approached the experimental site. > Oh ... we are very sorry that your research group suffered such complications - somehow there is a very huge misunderstanding here : In order to measure the mass of the galactic black hole on should use VIRTUAL ELEPHANTS !!! The main advantage in such an approach would be, that the handeling of virtual elephants is somehow more easy and convenient. Besides, the branching-fraction of a virtual standard elephant into a left-handed kangaroo and an ensemble of photons is well known. > Upon recovery from this test project, the staff decided a better approach > would be to use the standard kangaroo model on the speculation that the > standardized equipment would hop itself into position. Before proceeding > with this project some facts need be established. As standard kangaroos > are smaller that standard elephants, will n standard kangaroos suffice or > will 3n standard kangaroos be needed? Also once in place, how is a > standard hopping kangaroo convinced to remain stationary? You are completely right that 3n standard kangaroos would be needed for this task. (or more correctly 3*(n-1)*(1+epsilon), where epsilon is an empirical constant) We have made the observation, that standard kangaroos jump less frequently when we bind them together with an inelastic rope in pairs. Maybe this could help to make the measurement condition more stable ? <1072626573.232780@news.aic.at> ==== >> elephant blowing it's nose upon him and again by the project >> director who had her butt whooped by a standard trunk, that >> standard elephants may lack appropriate manners needed for >> cooperation on a long long trip. Any further efforts on that >> field trip were immediately abandoned when n-1 standard elephants >> taking notice of the ruckus, filled their standard trunks with >> wet water and eagerly approached the experimental site. > >Oh ... we are very sorry that your research group suffered such >complications - somehow there is a very huge misunderstanding here : >In order to measure the mass of the galactic black hole on should >use VIRTUAL ELEPHANTS !!! Again upon inquiring of 262,144 standard elephants from 256 herds we found none knew how to virtualize. We even introduced them to the Cheshire Cat, who if you recall would virtualize into thin air with a smile. At first they took note of him for smelling like a feline, but then when he smiled they soon lost interest. We then enlisted the assistance of an elephant trainer, but try as he may, even the most intelligent and well train standard elephants could or would not smile and smile from ear to ear like the Cheshire Cat. One elephant however, the standard elephant who earlier expressed interest in travel to far and distance places, took interest in the Cheshire Cat. He was so bemused by the cat's smile that in response he widened and widened his ears. This so please the Cheshire that they soon became steadfast friends. Not long thereafter they turned to staff and trainer; the Cheshire Cat smiling ever wider smiles at us, and the cooperative standard elephant spreading his ears every wider and wider, gradually vanished before us appearing as naught but a wide smile and wider ears which eventually faded from sight. This second field trip into the standardized land of the elephants, tho exceedingly more pleasant, found no standard elephants who could or would virtualize except for the one standard elephant friend of the Cheshire Cat who, with the Cat, virtualized into a virtual reality without telling us which of the virtually infinite virtual realities they went to. However our research has suddenly had to shift direction. You may will have noticed this astounding news story: anti-quark, was discovered at Los Alamos laboratories on having been observed changing gender right in the middle of an.., experiment, has Los Alamos scientists greatly perplex over chronom tunneling since it was detected a few days before it's discovery. your help to find the mass of the weirdo quickly by Monday before the scientists returned to the lab. That no longer appears possible, nor is it now of importance with the breaking of that story. Having reviewed and rechecked the events of the weekend we now know the weirdo appeared undetected at Livermore Research Center to be detected at Los Alamos. We strongly suspect this is because associated with chronom tunneling is spatial distortion, displacement or tunneling. We attempted to model this possibility using the equations of general relativity on a super-computer. Preliminary results are perplexing. We suspect the super-computer adequately comprehended the phenomena and used it. Indications are that, quickly running out of storage space to hold temporary data, it resorted to storing it in the past and future. Thus the perplexment, tho the program needs but a few more hours to run, we may not know even when the results will be collated. One of our coworkers, fatigued by the pace of this weekend, fell asleep dreaming she had consulted with Chronos the Greek god of time and Kali the problem, nay perhaps that of unified space/time field theory, was unresolvable until the god and goddess of distance awoke. To awaken these deities one was to quietly and devoutly invoke their names three times in each of the four directions on a mountain top inaccessible by helicopter. Can you assist our research efforts to find who the god and goddess of distance are? >We have made the observation, that standard kangaroos jump less >frequently when we bind them together with an inelastic rope in >pairs. Being rushed by the pace of events, we released some standard kangaroos into our back yard where our kids, who also like to jump up and down, could play with them. Being the offspring of scientists, they readily took our suggestion to experiment binding standard kangaroos together in pairs. Results came soon: standard jumping kangaroos aren't into bondage; for best results use bungy cords; have fun. ---- ==== > Can you assist our research efforts to find who the > god and goddess of distance are? > Indra & Prajapathi ... Despite of noting the obvious : In hinduistic mythology there is only one single omni-potent, omni-present and omni-scient god. The multifarious 'deities' are nothing but different aspects of the same one. Ganesha is the aspect of 'everything that can be counted or comprehended'. God is everything that exists and everything that does not exist - everything that can be thought of - and - everything that can not be thought of ... It is an illusion and delusion to hope to understand this completely. All religions are one - we are all brothers and sisters on this fragile planet, being less than dust or vanity in the universe. ==== Erratum >> However for more ambitious measurements, such as the exact mass of >> the Galatic black hole, does one use the Elephant in Musk Model? >That's very easy in the framework of the Practical Elephant Model : >Put n standard-elephants into a stationary orbit around the galactic >black hole. >The measurement of the wobble-frequency and the radiation-spectra >of the elephants will provide you with the mass of your black hole >(in units of the standard-elephant) > An interesting approach. However upon asking 262,144 standard elephants, > only one was willing to make the trip. Our small research group was thus > faced with the necessity of carrying n standard elephants themselves. In 'carrying n-1 standard elephants' > a pilot project, it was established that our full staff was unable to > lift, much less carry, a single standard elephant. It was also noted by > the project manager who was sprayed by the ill mannered standard elephant > blowing it's nose upon him and again by the project director who had her > butt whooped by a standard trunk, that standard elephants may lack > appropriate manners needed for cooperation on a long long trip. Any > further efforts on that field trip were immediately abandoned when n-1 > standard elephants taking notice of the ruckus, filled their standard 'when n-2 standard elephants' > trunks with wet water and eagerly approached the experimental site. Upon recovery from this test project, the staff decided a better approach > would be to use the standard kangaroo model on the speculation that the > standardized equipment would hop itself into position. Before proceeding > with this project some facts need be established. As standard kangaroos > are smaller that standard elephants, will n standard kangaroos suffice or > will 3n standard kangaroos be needed? Also once in place, how is a 'will 3(n-1) standard kangaroos and the willing standard elephant' > standard hopping kangaroo convinced to remain stationary? >By the way, thanks for the hint that standard elephants could maybe >decay into strange colorful left handed anti-quarks - we will >make some more calculations clarifying this point. > Your welcome. Please keep us inform of your progress. Do be careful > handling strange colorful left handed anti-quarks, those weirdo's have > been noticed to change gender right in the middle of an.., experiment. > Whence the name. ---- ==== >>If A is a nowhere dense subset of a >> separable compact connected Hausdorff Baire space >>is A countable? >The Cantor set is a nowhere dense set >with the cardinality of the reals. A most efficient proof. ;-) ---- ==== > If A is a nowhere dense subset of the reals R, > then A is countable. That's impressive eggnog you're drinking. ==== > The Cantor set ... take out the trash Hmmm... a nice way of defining it :-) Rainer Rosenthal r.rosenthal@web.de ==== Today, at the Barnes and Noble Bookstore, I saw the book The Colossal Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by Martin Gardner. The book says that e^[pi*sqrt(163)] = 262,537,412,640,768,744 which is an integer exactly. Specifically, the book says that Srinivasa Ramanujan(1887-1920) computed the number manually to 262,537,412,640,768,743.999999, but could not go further; then someone in France computed two million digits after the decimal point which are all 9; then someone ingeniously used the Euler's Constant to prove that the result is exactly the integer as given above. A quick search in the Internet revealed that the above claim is not true and it first showed up as a April's Fool's joke by Martin Gardner, and he subsequently admitted that it was a joke. Then how come it still got into the book? ==== > Today, at the Barnes and Noble Bookstore, I saw the book The Colossal > Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by > Martin Gardner. The book says that e^[pi*sqrt(163)] = 262,537,412,640,768,744 which is an integer exactly. Specifically, the book says that Srinivasa Ramanujan(1887-1920) > computed the number manually to 262,537,412,640,768,743.999999, but > could not go further; then someone in France computed two million > digits after the decimal point which are all 9; then someone > ingeniously used the Euler's Constant to prove that the result is > exactly the integer as given above. A quick search in the Internet revealed that the above claim is not > true and it first showed up as a April's Fool's joke by Martin > Gardner, and he subsequently admitted that it was a joke. Then how come it still got into the book? (because the author of The Colossal Book appears to be a disciple of Martin Gardner ;-)) According to Maple: > evalf(exp(Pi*sqrt(163)), 50);; 18 .26253741264076874399999999999925007259719818568865 10 which is quite remarkable, though. I don't know if Ramanujan really did this computation, but thinking about it: how on earth can one come to think about it? Any clue? /er. ==== Eric Rannaud e^[pi*sqrt(163)] = 262,537,412,640,768,744 ... > I don't know if Ramanujan really did this computation, but thinking > about it: how on earth can one come to think about it? Any clue? Please read The Book of Numbers John H. Conway, Richard K. Guy Copernicus (Springer) ISBN 0-387-97993-X and especially Chapter 8, part The Nine Magic Discriminants. Rainer Rosenthal r.rosenthal@web.de ==== > Today, at the Barnes and Noble Bookstore, I saw the book The Colossal > Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by > Martin Gardner. The book says that > e^[pi*sqrt(163)] = 262,537,412,640,768,744 > which is an integer exactly. A quick search in the Internet revealed that the above claim is not > true and it first showed up as a April's Fool's joke by Martin > Gardner, and he subsequently admitted that it was a joke. Then how come it still got into the book? > Who's the author of the book? ;-) ==== The author is Gardner himself. Well, today I checked the book again. At the end of the long chapter, there is actually an Addendum which talks about the fact that it was intended as a joke. > ... > Who's the author of the book? ;-) ==== > > [snip] > > The subject is typical representation of the OP's educational > inadequacies. You're obviously out of your depth here. Herc has strong and valid opinions about logical and philosophical issues you probably aren't even aware exist. He is also mentally ill. The two have none to do with each other, and he still deserves respect. 'cid ==== > > Doing math is a series of small successes used to build a large house > of knowledge. I never understood how people could ever dislike it. > Presumably they never experienced one of those successes, and feel that the large house of knowledge has long been built and they have to tread carefully so as not to stir the ghosts in the dusty corners. ==== >I'm looking for criterion that ensure two groups are isomorphic. >>[snip] >> >> By group-theoretical invariants, I meant things like order and structure of >> Z(G), order of [G,G], number of elements of order r for each r, etc. Including cohomology groups of G over the ring of rational integers? Yes, I would include these as examples of group-theoretical invariants. I guess you are asking whether there exist non-isomorphic G and H such that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure about that - it sounds an interesting question. Derek Holt. ==== >I guess you are asking whether there exist non-isomorphic G and H such >that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure >about that - it sounds an interesting question. How about G = Z/2 x Z/3 ( = Z/6 ) and H = Z/2 * Z/3 ( = PSL_2(Z) )? But yes, this is an interesting question for _finite_ groups, or at least it was until Ian Leary bagged it. There are non-isomorphic finite p-groups G and H for which H^n(G,Z) and H^n(H,Z) are isomorphic groups for every n. I think that in his example the _rings_ H^*(G,Z) and H^*(H,Z) are not isomorphic, however. And of course even if the rings were isomorphic, one could ask about the structures of the rings as modules under the Note that cohomology is actually a functor of the corresponding group ring (Z[G] resp Z[H]) so that if it is possible to use cohomology to distinguish two groups, then it is because the group rings are already non-isomorphic. So I suppose the greater challenge is to find two nonisomorphic (finite) groups whose integral group rings are isomorphic; this has been conjectured never to happen and last I heard, which was quite a while ago, this conjecture was still open. Of course, I should point out that even if the integral group ring turns out to be a perfect invariant in this sense, this really only amounts to begging the original question, since one would then have to decide how one can determine whether or not two rings (which happen to be finitely- generated Z-modules) are isomorphic. I don't see any reason to think that's an easier question than the one we began with. [I should probably add that by ring, above, I mean ring with augmentation map. I have this dim memory that there are examples where ZG and ZH are isomorphic as rings but the isomorphism cannot commute with augmentation.] Finally, let me respond to the person who mentioned taking cohomology with coefficients in some other ring or field. That's a much weaker invariant: from the Universal Coefficient Theorem one can pretty much determine H^*(G,A) where A is any trivial G-module, as soon as one knows H^*(G,Z). The reverse is far from true and one can easily find groups G, H with H^*(G,A) isomorphic to H^*(H,A); for example, if A is the field with 2 elements and G and H have odd order, or (less trivially) if G and H are dihedral groups of different orders. Ian's paper is MR1348713 (96j:20076) : $p$-groups are not determined by their integral cohomology groups. Bull. London Math. Soc. 27 (1995), no. 6, 585--589. 20J06 (20D15) dave ==== >I'm looking for criterion that ensure two groups are isomorphic. >> [snip] >> >> By group-theoretical invariants, I meant things like order and structure of >> Z(G), order of [G,G], number of elements of order r for each r, etc. Including cohomology groups of G over the ring of rational integers? > > Yes, I would include these as examples of group-theoretical invariants. > I guess you are asking whether there exist non-isomorphic G and H such > that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure > about that - it sounds an interesting question. > > Derek Holt. Attach to the group G the sequence: EulerianFunction(G,1),EulerianFunction(G,2),EulerianFunction(G,3),... (this is implemented in GAP ) where the EulerianFunction(G,n) is the number of n-tuples of elements of G that generate G . Doesn't this sequence of numbers determine G up to isomorphism ? Tim ==== >I'm looking for criterion that ensure two groups are isomorphic. >> [snip] > > By group-theoretical invariants, I meant things like order and structure of > Z(G), order of [G,G], number of elements of order r for each r, etc. >>Including cohomology groups of G over the ring of rational integers? >> >> Yes, I would include these as examples of group-theoretical invariants. >> I guess you are asking whether there exist non-isomorphic G and H such >> that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure >> about that - it sounds an interesting question. >> >> Derek Holt. Attach to the group G the sequence: EulerianFunction(G,1),EulerianFunction(G,2),EulerianFunction(G,3),... >(this is implemented in GAP ) where the EulerianFunction(G,n) is the >number of n-tuples of elements of G that generate G . Doesn't this sequence of numbers determine G up to isomorphism ? > I don't think so. SmallGroup(32,13) and SmallGroup(32,14) give the same answer for n<=40. Derek Holt. ==== >>In practice, groups are not normally given as multiplication tables - they are >>usually defined as groups of permutations or matrices, or by means of >>presentations. It is a much more difficult problem to decide whether or not >>two subgroups G and H of S_n defined by generating permutations are >>isomorphic. I am fairly sure that this is still in NP (assuming that the >>putative isomorphism is given by means of images in H of generators of >>G, you can compute in polynomial time a set of defining relators for G on >>that generating set, and thereby check whether or not the map is a isomorphism) >>but it may well be NP-complete. If the elements are given as permutations or matrices, then you can create >the multiplication table in poly time (for a simple minded upper bound; As Keith Ramsay observed, for permutation groups input by generators, which is the standard model for most complexity results and research concerning finite permutation group algorithms, you cannot construct the multiplication table in polynomial time. There are many properties of the group which you can compute in polynomial time, however, such as its order, the order of its derived group, centre, ... Another standard model used in complexity theory of algorithms for finite groups is the Black-Box group, introduced by L. Babai, and matrix groups over finite fields fall into this category. A Black-box group is one in which the elements are represented by bit-strings of bounded length, and the group-theoretical operations (composition and inversion) can be carried out in constant time. Again groups are defined by means of generators. But it is much harder to prove results about Black-box groups - most of the algorithms are probabilistic and have a small probability of giving a wrong answer. >er...I'm not so sure about a matrix representation because of the operations >needed to test equality of the matrix elements (with all those radicals >flying around)). If the elements..er if the group is given as a finite presentation (set of >identities), then it's undecidable by ...uh ... a reduction from >undecidability of uh... well it should be from the word problem but I >can't see the obvious at the moment. Yes, but there are some particular types of presentation which are useful computationally, such as the power-commutator presentation of finite solvable (or infinite polycyclic groups). Derek Holt. ==== >>If the elements are given as permutations or matrices, then you can create >>the multiplication table in poly time (for a simple minded upper bound; As Keith Ramsay observed, for permutation groups input by generators, which >is the standard model for most complexity results and research concerning >finite permutation group algorithms, oh. my misunderstanding. >you cannot construct the >multiplication table in polynomial time. There are many properties of >the group which you can compute in polynomial time, however, such as >its order, the order of its derived group, centre, ... I suppose with just the permutation generators, the order could be exponential (e.g. 2 generators of degree n can generate S_n of size n!, whose number would be represented with O(n) bits). But how can you compute this order (in general) without actually constructing all elements? Mitch Harris ==== >>If the elements are given as permutations or matrices, then you can create >the multiplication table in poly time (for a simple minded upper bound; >>As Keith Ramsay observed, for permutation groups input by generators, which >>is the standard model for most complexity results and research concerning >>finite permutation group algorithms, oh. my misunderstanding. >you cannot construct the >>multiplication table in polynomial time. There are many properties of >>the group which you can compute in polynomial time, however, such as >>its order, the order of its derived group, centre, ... I suppose with just the permutation generators, the order could be >exponential (e.g. 2 generators of degree n can generate S_n of size n!, >whose number would be represented with O(n) bits). But how can you compute >this order (in general) without actually constructing all elements? This is known as the Schreier-Sims algorithm. To summarize it very briefly indeed, you calculate the orbit Orb(G,1) of 1 under G, then use a theorem of Schreier to compute generators of the stabilizer Stab(G,1) of 1 under G from the orbit and the generators of G. Then recursively compute |Stab(G,1)| and we have |G| = |Orb(G,1)||Stab(G,1)| by the orbit-stabilizer theorem. To get polynomial time, you have to be more careful, because the the number of generators of Stab(G,1) coming from Schreier's Theorem is roughly |Orb(G,1)| times the number of original generators of G. To avoid an explosion in the number of generators, you maintain a subgroup H of Stab(G,1), initially trivial, for which you know the order. For each new Schreier generator of Stab(G,1) you first check if it already in H. If it is not, then you re-run the algorithm on H with the new generator adjoined, and enlarge H. Derek Holt. ==== >There are non-isomorphic groups of order >>32 (or maybe 64) which cannot be distinguished by means of any known >>group-theoretical invariants. >>Er...then how do you distinguish them? Non group-theoretical invariants? >Or is this a case of where there is a counting argument that is >nonconstructive? >>Well, if you can't think of anything better, then you just try all >>possible maps from the first group to the second and check that it is not >>an isomorphism. In practice, you would just check maps from a generating >>set of the first group to the second, because a homomorphism is >>uniquely determined by its action on a generating set. >>By group-theoretical invariants, I meant things like order and structure of >>Z(G), order of [G,G], number of elements of order r for each r, etc. I guess I am piqued by he use of the word cannot in your claim. Is there >a proof of that impossibility (given some arbitrary or not so arbitrary >set of group-theoretical invariants)? > I wasn't really trying to make a precise statement, and the idea of a group-theoretical invariant is not really well-defined. But I would make the following conjecture, just based on general experience, and without any prospects of being able to prove anything precise. If you specify in advance some collection of group-theoretical invariants, which may include things like Order and properties of Aut(G), (co)homology groups, number of homomorphisms from G to standard groups like S_n, etc., then there exist two non-isomorphic groups which agree in all of the invariants that you specified. Derek Holt. ==== >There are non-isomorphic groups of order >32 (or maybe 64) which cannot be distinguished by means of any known >group-theoretical invariants. ... >>I guess I am piqued by he use of the word cannot in your claim. Is there >>a proof of that impossibility (given some arbitrary or not so arbitrary >>set of group-theoretical invariants)? I wasn't really trying to make a precise statement, and the idea of >a group-theoretical invariant is not really well-defined. But I would make >the following conjecture, just based on general experience, and without any >prospects of being able to prove anything precise. If you specify in >advance some collection of group-theoretical invariants, which may include >things like Order and properties of Aut(G), (co)homology groups, number >of homomorphisms from G to standard groups like S_n, etc., then there >exist two non-isomorphic groups which agree in all of the invariants that >you specified. OK, I see. This sounds very similar to the idea (conjecture? theorem?) that there is no poolyomial size set of properties of two graphs to distinguish them. Mitch Harris ==== [snip] > I am not aware of any such criteria. There are non-isomorphic groups of order > 32 (or maybe 64) which cannot be distinguished by means of any known > group-theoretical invariants. [snip] > Derek Holt. Could you please give an example of a pair of such hard to distinguish groups, in GAP notation? Alan ==== >[snip] >> I am not aware of any such criteria. There are non-isomorphic groups of order >> 32 (or maybe 64) which cannot be distinguished by means of any known >> group-theoretical invariants. >[snip] >> Derek Holt. Could you please give an example of a pair of such hard to distinguish >groups, in GAP notation? SmallGroup(32,13) and SmallGroup(32,14) seem hard to distinguish using group- theoretical properties. Let me know if you succeed! Of course they can be distinguished by the invariant Number of isomorphisms from given group to SmallGroup(32,13). So I prefer to rephrase my original claim as: If you give me a collection of group-theoretical invariants, then I will find two non-isomorphic groups which are not distinguished by them. Derek Holt. ==== >[snip] >> I am not aware of any such criteria. There are non-isomorphic groups of order >> 32 (or maybe 64) which cannot be distinguished by means of any known >> group-theoretical invariants. > [snip] >> Derek Holt. Could you please give an example of a pair of such hard to distinguish >groups, in GAP notation? > > SmallGroup(32,13) and SmallGroup(32,14) seem hard to distinguish using group- > theoretical properties. Let me know if you succeed! Of course they can be > distinguished by the invariant Number of isomorphisms from given group > to SmallGroup(32,13). So I prefer to rephrase my original claim as: > If you give me a collection of group-theoretical invariants, then I will > find two non-isomorphic groups which are not distinguished by them. > > Derek Holt. Well, I don't know if this qualifies, but the number of elements having a square root or the maximal number of square roots of any single element seem to distinguish between those two. Of course, there will be non-isomorphic groups that cannot be told apart by these, but such invariants can be generalized in many ways into an infinite array of invariants, and I wouldn't count on there being no such an array that is capable of distinguishing any pair of finite non-isomorphic groups. This, of course, says nothing about efficiency of computation - I'm referring to sets of invariants that seem more natural than the number of isomorphisms to some fixed group G as G varies over all finite groups. Alan ==== > If an abelian group G has two elements with order m and n respectively, > show G has an element whose order is the least common multiple of m and n. > > a, b of order m and n respectively. > m is the smallest positive integer such that a^m = e. > n is the smallest positive integer such that b^n = e. One can show from the definitions of greatest common denominator (gcd) and least common multiple (lcm) that lcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say). (This is a handy thing to remember too.) so lg = mn ; m = qg ; n = rg since g divides both m and n. All we need to know about q and r is that they are integers. Then l = mn/g = r*q*g = rm = qn ; so (ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = e and the order of ab is l = lcm(m,n). Note the if (m,n) = 1 (i.e., are relatively prime), the order of the product ab is the product of the orders of a and b since [m,n] = mn in this case. Van ==== > One can show from the definitions of greatest common denominator (gcd) and > least common multiple (lcm) that lcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say). (This is a handy thing to remember too.) > so lg = mn ; m = qg ; n = rg since g divides both m and n. > All we need to know about q and r is that they are integers. Then l = mn/g = r*q*g = rm = qn ; so (ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = e and the order of ab is l = lcm(m,n). Note the if (m,n) = 1 > (i.e., are relatively prime), the order of the product ab is the > product of the orders of a and b since [m,n] = mn in this case. Van That's not correct. Two things: 1¡) if l = lcm(m,n) (we know a and b commute) then it is obvious that (ab)^l = e 2¡) that doesn't say l is the smallest positive integer such that (ab)^l =e, ie that doesn't say (ab) has order l. -- Julien Santini ==== Your solution looks fine to me. >The problem is, what if a^r and b^s are multiplicative inverses ... then >this could equal 1. But there is no reason to assume that they are inverses, since a and b are arbitrary elements--you are creating an unnecessary problem for yourself. Yes, it is possible that in some particular case that a^r b^s = 1, but the problem is for the general case, for any m and n. Van ==== Your solution looks fine to me. >The problem is, what if a^r and b^s are multiplicative inverses ... then >>this could equal 1. But there is no reason to assume that they are inverses, since a and b are >arbitrary elements--you are creating an unnecessary problem for yourself. >Yes, it is possible that in some particular case that a^r b^s = 1, but the problem >is for the general case, for any m and n. Sorry, but this is nonsense. While there is no reason to assume they are inverses, there is in principle no way to discard that situation either. As such, you ->must<- consider that possibility as well. Saying that the elements are arbitrary does not make it impossible for that situation to occur. If your argument is to work for ANY two elements, then it must also work in that case. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > One can show from the definitions of greatest common denominator (gcd) and > least common multiple (lcm) that lcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say). (This is a handy thing to remember too.) > so lg = mn ; m = qg ; n = rg since g divides both m and n. > All we need to know about q and r is that they are integers. Then l = mn/g = r*q*g = rm = qn ; so (ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = e and the order of ab is l = lcm(m,n). Note the if (m,n) = 1 > (i.e., are relatively prime), the order of the product ab is the > product of the orders of a and b since [m,n] = mn in this case. I should not have used the letter l here, it looks too much like the number one (1). Also, I apologize for reading only the problem and not your solution; > > That's not correct. > > Two things: > > 1¡) if l = lcm(m,n) (we know a and b commute) then it is obvious that (ab)^l > = e Well, even you took a few lines to show it. > 2¡) that doesn't say l is the smallest positive integer such that (ab)^l =e, > ie that doesn't say (ab) has order l. True, I wasn't thinking very clearly when I posted it. I'll try to come up with a proof of this. Van ==== -- set theory limits In set theory, for the powerset P(S) of a set S limits of set sequences Aj are defined as lim Aj = liminf Aj = limsup Aj provided liminf Aj = limsup Aj, liminf Aj = /{ /{ Aj | j >= n } | n in N } limsup Aj = /{ /{ Aj | j >= n } | n in N } / /; cap cup; intersection union Does this definition of limit produce a topology for P(S) ? -- powerset topology The powerset topology for a powerset P(S) is defined as the topology produced by subbase sets of the form { A in P(S) | a in A }, { A in P(S) | a not in A } A base for this topology are the sets of the form { X | A subset X subset SB } for A,B finite subset S. A local base for U in P(S) are the sets of the form { X | A subset X subset SB } A finite subset U, B finite subset SU We have the following theorems: The powerset topology for P(S) is homeomorphic to the product topology of {0,1}^P(S), the space of characteristic functions. {0,1} has discrete topology. Thus P(S) is zero-dimensional compact Hausdorff. P(S) is 1st countable iff S countable iff P(S) is 2nd countable -- limit equivalence The powerset topology has the desired property A = set-theory-lim Aj iff A = topology-lim Aj Hence when S is countable, the powerset topology can be described by set theory limits. When S is uncountable, I would conjecture 1) the definition of set theory limits can be extended to nets 2) limit equivalence will be preserved 3) the powerset topology can be described by net/set theory limits -- Questions Has any use of the powerset topology been made? Have these notions been extended or refined? Does this topic show up in modern analysis? Who else has consider the powerset topology? Comments most welcome as well as additions, refutations or requests for proofs. -- latticed ordered topological spaces As P(S) is the protype for complete complemented atomic distributive lattices or complete atomic Boolean algebras, the powerset topology can be extended to those spaces. Upon using 'subset' for the order <=, of P(S), the powerset topology presents a base of convex sets and the T_2-ordered property, that <= is closed subset P(S)xP(S). Thus it is an appropriately well mannered topology for a partially ordered set. Order and Topology, http://at.yorku.ca/t/a/i/c/05.htm An ordered topological space is a topological space (X,T) equipped with a partial order <=. Usual compatibility conditions between the topology and order include convexity (T has a basis of order-convex sets) and the T_2-ordered property: <=, ie { (x,y) | x <= y }, is closed in XxX. ---- ==== > -- set theory limits > In set theory, for the powerset P(S) of a set S > limits of set sequences Aj are defined as > lim Aj = liminf Aj = limsup Aj > provided liminf Aj = limsup Aj, > > liminf Aj = /{ /{ Aj | j >= n } | n in N } > limsup Aj = /{ /{ Aj | j >= n } | n in N } > / /; cap cup; intersection union > > Does this definition of limit produce a topology for P(S) ? Perhaps that I misunderstood the question, but I would say that no definition of limit produces a topology. Consider the set R of the reals and take the following definition of limit: the only sequences a_1, a_2, a_3, ... that posess a limit are those such that, for some real number l, a_n = l for every n large enough; in that case, the limit of the sequence is l. The problem here is that you can define at least two topologies in R such that the definition of limit made above is the notion of limit that corresponds to those topologies: the discrete topology and the topology such that the closed sets are R, the finite sets and the countable sets. I hope that this helps. Jose Carlos Santos ==== >> liminf Aj = /{ /{ Aj | j >= n } | n in N } >> limsup Aj = /{ /{ Aj | j >= n } | n in N } > >> Does this definition of limit produce a topology for P(S) ? >Perhaps that I misunderstood the question, Perhaps I should have asked, When does this ... >but I would say that no definition of limit produces a topology. A definition of limit together with requirement first countable would. The issue of first countable P(S) is consider 2nd & 3rd sections. >Consider the set R of the reals and take the following definition of >limit: the only sequences a_1, a_2, a_3, ... that posess a limit are >those such that, for some real number l, a_n = l for every n large >enough; in that case, the limit of the sequence is l. A sequence converges iff it's eventually constant. >The problem here is that you can define at least two topologies in R >such that the definition of limit made above is the notion of limit >that corresponds to those topologies: the discrete topology and the Which is first countable. Indeed cl A = A = limit points of A. >topology such that the closed sets are R, >the finite sets and the countable sets. An uncountable cocountable space isn't 1st countable. Here again A = limit pts of A, however cl A = R when A uncountable, = nulset when A countable. ---- ==== >but I would say that no definition of limit produces a topology. > A definition of limit together with requirement first countable would. > The issue of first countable P(S) is consider 2nd & 3rd sections. I have this curious habit of reading texts from the top to the bottom, not from the bottom to the top. And besides, what you mention below is that a certain specific topology about which you state that it is first countable. How was I supposed to deduce from that that, in the first section, you were looking for a first countable topology? >Consider the set R of the reals and take the following definition of >limit: the only sequences a_1, a_2, a_3, ... that posess a limit are >those such that, for some real number l, a_n = l for every n large >enough; in that case, the limit of the sequence is l. > A sequence converges iff it's eventually constant. If that was all that I wanted to write, I would have done so. But what constant but, furthermore, that the limit of such a sequence is that constant (and yes, I know that it could not be otherwise). >The problem here is that you can define at least two topologies in R >such that the definition of limit made above is the notion of limit >that corresponds to those topologies: the discrete topology and the > Which is first countable. Indeed cl A = A = limit points of A. Don't you think that it is much more simple to see directly from the definition that the discrete topology is first countable? All that you have to do is to see that, for any point p, {{p}} is a fundamental system of neighborhoods of p. >topology such that the closed sets are R, >the finite sets and the countable sets. > An uncountable cocountable space isn't 1st countable. Indeed, but I never said otherwise. Jose Carlos Santos ==== > > -- set theory limits > In set theory, for the powerset P(S) of a set S > limits of set sequences Aj are defined as > lim Aj = liminf Aj = limsup Aj > provided liminf Aj = limsup Aj, > > liminf Aj = /{ /{ Aj | j >= n } | n in N } > limsup Aj = /{ /{ Aj | j >= n } | n in N } > / /; cap cup; intersection union > > Does this definition of limit produce a topology for P(S) ? > > Perhaps that I misunderstood the question, but I would say that no > definition of limit produces a topology. Consider the set R of the > reals and take the following definition of limit: [...] > > The problem here is that you can define at least two topologies in R > such that the definition of limit made above is the notion of limit > that corresponds to those topologies [...] Generally, one refers to the smallest topology where the stated limits exist. Cf. the concept of induced topology (e.g., product topology). In particular, the topology whereto William refers is the natural homeomorphism of the product space {0,1}^A, where {0,1} is discrete. He states the result that the only topological limits of sequences in this space are those which are also set-limits. William asked some interesting questions, some of which I posted in earlier threads. (Much of this was hidden in the thread on the set of countable ordinals.) In particular, is this same topology the smallest where convergence of general set-nets are also topological limits? If so, are all topological limits also set-limits? Also, is this the same as the topology induced by all f in [0,1]^A such that lim f a = f (lim a) for all set-convergent nets a in A? Stephen J. Herschkorn ==== liminf Aj = /{ /{ Aj | j >= n } | n in N } > limsup Aj = /{ /{ Aj | j >= n } | n in N } Where are such limits used? Do they show up in modern analysis? >Generally, one refers to the smallest topology where the stated >limits exist. Cf. the concept of induced topology (e.g., product >topology). In particular, the topology whereto William refers is the >natural homeomorphism of the product space {0,1}^A, where {0,1} is >discrete. He states the result that the only topological limits of >sequences in this space are those which are also set-limits. >William asked some interesting questions, some of which I posted in >earlier threads. In particular, is this same topology the smallest >where convergence of general set-nets are also topological limits? I've looked closer into the details of this and I see no need basic need to change the proofs of set limits equivalent topology limits to a the proofs for net-set limits quivalent net-topology limits This is because the proofs when casted into the expressions of eventually in are identical; the only change to be made be the invisible change sequence-eventually to net-eventually. Again the key to the details is to use topology convergence to x happens when a sequence/net is eventually in every subbase set of x. and lim Aj = A iff for all x in A, some n with for all j >= n, x in Aj for all x not in A, some n with for all j >= n, x not in Aj equivalently for all x in A, x eventually in (Aj)_j for all x not in A, x eventually in (SAj)_j >If so, are all topological limits also set-limits? Also, is >this the same as the topology induced by all f in [0,1]^A such >that lim f a = f (lim a) for all set-convergent nets a in A? Now that I don't see. I'll concur on {0,1}^P(A) for discrete {0,1}; but on [0,1]^A for [0,1] subspace reals? This you needs clarify. For {0,1}^P(S) we have, the projections are p_a:P(S) -> {0,1}, c_A -> c_A(a) = 'a in A' which are continuous and thus preserve limits, ie t-lim f(Aj) = f(t-lim Aj) = f(set-lim Aj) Would that suffice or do you want ? set-lim f(Aj) = f(set-lim Aj) But as the projections are into discrete {0,1} the notion of set-limits in the codomain of the projections experiences such chaotic turbulance that such digression vanishes of its own accord. ---- ==== > > > -- set theory limits > In set theory, for the powerset P(S) of a set S > limits of set sequences Aj are defined as > lim Aj = liminf Aj = limsup Aj > provided liminf Aj = limsup Aj, > > liminf Aj = /{ /{ Aj | j >= n } | n in N } > limsup Aj = /{ /{ Aj | j >= n } | n in N } > / /; cap cup; intersection union > > Does this definition of limit produce a topology for P(S) ? > > [...] > > Generally, one refers to the smallest topology where the stated limits > exist. Cf. the concept of induced topology (e.g., product topology). > In particular, the topology whereto William refers is the natural > homeomorphism of the product space {0,1}^A, where {0,1} is discrete. > He states the result that the only topological limits of sequences in > this space are those which are also set-limits. > > Correction: The *smallest* topology is the indiscrete one. Here we want the *largest* topology whereunder the sequences converge to the designated limits. Let X = N U {infty} under the usual topology. The topology we seek is exactly that coinduced by all functions f in P(A)^X such that lim f = the set limit. This does not contradict the statement about {0,1}^A. (At least I *think* there is no larger such topology.) Given a set Y and a collection F = {(X,f): X is a topological space and f in Y^X}, the topology coinduced on Y by F is the largest topology such that (X,f)in F implies f is continuous. The prototypical example is the quotient topology, coinduced by the canonical map to equivalence classes. I confess to not remembering right now the exact proof that such a largest topology exists. I.e., given a collection t of topologies such that each f is continuous with each topology in t, why is it that each f is continuous in the topology generated by Ut? The proof for induced topologies (which do refer to a smallest topology = intersection of topologies) is much more straightforward. Stephen J. Herschkorn ==== > provided liminf Aj = limsup Aj, >> Generally, one refers to the smallest topology where the stated >> limits exist. Cf. the concept of induced topology (e.g., product >> topology). In particular, the topology whereto William refers is >> the natural homeomorphism of the product space {0,1}^A, where {0,1} >> is discrete. He states the result that the only topological limits >> of sequences in this space are those which are also set-limits. > A needs be taken as P(S). The product topology is the intitial, I think that's the induced, topology of the projections. By definition it's the smallest topology making all the projections continuous. As the lattice of topologies is a complete lattice, (it's also complemented, non-distributive, atomic) such a smallest can be found and then proven to have the continuity property. Easier is to construct a such topology and show it's the smallest. For perspective, the largest topology continuingfying all the projections is the discrete topology. >Correction: The *smallest* topology is the indiscrete one. Here we >want the *largest* topology whereunder the sequences converge to the >designated limits. Let X = N U {infty} under the usual topology. Which is? If X is going to simulate a sequence or net, then it needs a countable set order isomorphic to N or an updirected domain of the net. >The topology we seek is exactly that coinduced by all functions f in >P(A)^X such that lim f = the set limit. This does not contradict >the statement about {0,1}^A. (At least I *think* there is no larger >such topology.) How are you using A? Do you actually mean P(P(S))^X or P(S)^X ? >Given a set Y and a collection F = {(X,f): X is a topological >space and f in Y^X}, the topology coinduced on Y by F is the >largest topology such that (X,f)in F implies f is continuous. >The prototypical example is the quotient topology, coinduced by the >canonical map to equivalence classes. f:X -> Y is embedding X into Y the disjoint sum of F copies of X. Each f in an embedding of an X into a unique portion of Y. Another application is the embeddings of subspaces into the mother space. The quotient space comes when the collection of functions is just one. Are you suggesting some sort of multiple quotients of a space? >I confess to not remembering right now the exact proof that such a >largest topology exists. I.e., given a collection t of >topologies such that each f is continuous with each topology in >t, why is it that each f is continuous in the topology generated by >Ut? The proof for induced topologies (which do refer to a smallest >topology = intersection of topologies) is much more straightforward. The smallest topology is the indiscrete topology. For the details of the final topology, I'd suggest a similar approach as for the initial topology. ---- ==== > In set theory, for the powerset P(S) of a set S > limits of set sequences Aj are defined as > lim Aj = liminf Aj = limsup Aj > provided liminf Aj = limsup Aj, >> Generally, one refers to the smallest topology where the stated >> limits exist. Cf. the concept of induced topology (e.g., product >> topology). In particular, the topology whereto William refers is >> the natural homeomorphism of the product space {0,1}^A, where {0,1} >> is discrete. He states the result that the only topological limits >> of sequences in this space are those which are also set-limits. A needs be taken as P(S). The product topology is the intitial, I think > that's the induced, topology of the projections. By definition it's the > smallest topology making all the projections continuous. As the lattice > of topologies is a complete lattice, (it's also complemented, > non-distributive, atomic) such a smallest can be found and then proven to > have the continuity property. Easier is to construct a such topology and > show it's the smallest. For perspective, the largest topology > continuingfying all the projections is the discrete topology. Sorry, by A I meant S. And you are incorrect about A being P(S). The topology you describe makes P(S) homeomorphic to the product {0,1}^S under the product topology, where the homeomorphism is the usual identification of subsets with characteristic functions. More below. > >Correction: The *smallest* topology is the indiscrete one. Here we >want the *largest* topology whereunder the sequences converge to the >designated limits. Let X = N U {infty} under the usual topology. > Which is? If X is going to simulate a sequence or net, then it needs a > countable set order isomorphic to N or an updirected domain of the net. By the usual topology, I mean the one with the base {{n}, N[n,infty]: n in N}. Here, N[n,infty] denotes {n, n+1, n+2,... infty}, Another way to put it is X is the ordinal omega + 1 under the order topology. > >The topology we seek is exactly that coinduced by all functions f in >P(A)^X such that lim f = the set limit. This does not contradict >the statement about {0,1}^A. (At least I *think* there is no larger >such topology.) > How are you using A? Do you actually mean P(P(S))^X or P(S)^X ? A = S, again. So a better way of describing the coinducing functions f are those in P(S)^X such that f(n) -> f(infty) (set-wise) as n -> infty. I am noting that this coinduced topology seems to be the same as the homeomorph of the product topology on {0,1}^S (which by definition is an induced topology - by definition, the smallest topology satisfying certain conditions). BTW, given any set Y and a collection of sequences with designated limits in Y, the (largest) topology generated by these sequence-limits is again the coinduced topology by the corresponding functions in Y^(omega + 1). > >Given a set Y and a collection F = {(X,f): X is a topological >space and f in Y^X}, the topology coinduced on Y by F is the >largest topology such that (X,f)in F implies f is continuous. >The prototypical example is the quotient topology, coinduced by the >canonical map to equivalence classes. > f:X -> Y is embedding X into Y the disjoint sum of F copies of X. > Each f in an embedding of an X into a unique portion of Y. > > Another application is the embeddings of subspaces into the mother space. > The quotient space comes when the collection of functions is just one. > Are you suggesting some sort of multiple quotients of a space? I am just stating the standard definition of the coinduced topology. > > >I confess to not remembering right now the exact proof that such a >largest topology exists. I.e., given a collection t of >topologies such that each f is continuous with each topology in >t, why is it that each f is continuous in the topology generated by >Ut? The proof for induced topologies (which do refer to a smallest >topology = intersection of topologies) is much more straightforward. > The smallest topology is the indiscrete topology. For the details of the > final topology, I'd suggest a similar approach as for the initial > topology. I think you misinterpret what I was saying. Being redundant here, Given a set X and a collection F of functions each of whose domain is X and whose range is some topological space, the topology on X induced by F is the smallest topology on X such that all the functions in F are continuous. - Note that the ranges of the functions in F need not be the same spaces. - Such a topology exists: It is the intersection of all topologies on X whereunder all the functions in F are continuous. The collection of such topologies is not empty, since it contains the discrete topology. - Example 1. The product topology is the topology induced by the projection maps. - Example 2. The relative topology is the topology induced by the inclusion map. Given a set Y and a collection F of functions each of whose domain is some topological space and whose range is Y, the topology on Y coinduced by F is the largest topology on Y such that all the functions in F are continuous. - Note that the domains of the functions in F need not be the same spaces. - Earlier, I wondered why we know such a topology exists. Having slept on it, I realize the answer: Let t be the collection of topologies on Y whereunder all functions in F are continuous. t is not empty, since it contains the indiscrete topology. The coinduced topology is the smallest topology containing Ut. It is easy to check that if the inverse of a function maps sets in a subbase to open sets, then it is continuous. (The last sentence is what I was missing earlier.) - Example. The quotient topology is the topology coinduced by the canonical map to an equivalence class. In fact, I think this is the most general coinduced topology. Given F as in the definition, say x and y in Y are equivalent iff f(x) = f(y) for all f in F. I am not sure about this; at least this works if F is a singleton. One more example to compare the two concepts: Let X = [0,1] under the usual topology. Let Y = {(x,y) in R^2: x^2 + y^2 = 1}, the unit circle. Let Z = [-1,1] under the usual topology. Then the usual topology on Y can be generated in the following two ways: It is that induced by the projection maps from Y to Z^2. It is also that coinduced by f: X -> Y, x|->(cos 2pi x, sin 2pi x). Stephen J. Herschkorn X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBSDc1w02521; ==== There are enough and coherent reasoning to prove that it is not possible to imagine the list of all real numbers without running against the nature of the natural numbers. For instance, if we admit the one-to-one correspondence between N and R, then we are confronting two different kinds of infinity, the potential infinite represented by the asymptotic approximation of the naturals to the infinite (never reaching it), and the actual infinite represented by the set of all real numbers; so that, a complete bijection between both sets is not possible. With another simple reasoning we arrive to the conclusion that the acceptance of the existence of that list would imply naturals with an infinite value (cardinal). On the other hand, there are constructive methods (at least I know one of them) that by simple induction they proof that such a list will require naturals with an infinite number of figures. Therefore, my question is this: If the ãlist of all real numbersä is not a valid concept, why must we accept it as premise in the Cantorâs proof of the diagonal? The theory of transfinite numbers would be able to be an air castle with such an argument in its foundations. Nicolas de la Foz ==== > For instance, if we admit the one-to-one correspondence between N and R, > then we are confronting two different kinds of infinity, the potential infinite > represented by the asymptotic approximation of the naturals to the infinite > (never reaching it), and the actual infinite represented by the set of all real > numbers; How do you handle the fact there are more than two grades of infinity? For example, the set F of all functions f: R->R is larger than the set of reals R, such that you cannot place these into one-to-one correspondence either. Yet you call the smaller set R an actual infinity. What does that make F? An even more actual infinity? And how about the power set of F? That's bigger still. X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft X-Terminate: SPA(GIS) ==== at 01:38 PM, nico80@jazzfree.com (Nicolas de la Foz) said: >There are enough and coherent reasoning to prove that it is not >possible to imagine the list of all real numbers without running >against the nature of the natural numbers. No. There is only posturing and the use of words in contexts where they have no meanings. >For instance, if we admit the one-to-one correspondence between N >and R, We can't, because there is no such 1-1 correspondence. >the potential infinite What is a potential infinite and what does it have to do with either N or R? >asymptotic approximation of the naturals to the infinite Do you believe that the above phrase has any meaning? If so, why? >the actual infinite What is the actual infinite and what does it have to do with either N or R? >Therefore, my question is this: If the .99list of all real numbers.9a is >not a valid concept, why must we accept it as premise in the >Cantor®s proof of the diagonal? We don't. We show that such a premise would lead to a contradiction and is therefor false. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > There are enough and coherent reasoning to prove that it is not possible to imagine the list of all real numbers without running against the nature of the natural numbers. For instance, if we admit the one-to-one correspondence between N and R, then we are confronting two different kinds of infinity, the potential infinite represented by the asymptotic approximation of the naturals to the infinite (never reaching it), and the actual infinite represented by the set of all real numbers; so that, a complete bijection between both sets is not possible. With another simple reasoning we arrive to the conclusion that the acceptance of the existence of that list would imply naturals with an infinite value (cardinal). On the other hand, there are constructive methods (at least I know one of them) that by simple induction they proof that such a list will require naturals with an infinite number of figures. > Therefore, my question is this: If the list of all real numbers is not a valid concept, why must we accept it as premise in the Cantor's proof of the diagonal? The theory of transfinite numbers would be able to be an air castle with such an argument in its foundations. This is one of my own pet hobbyhorses. I am a firm believer in only one infinity, which is not even signed. Every presentation of the Cantor argument has turned upon a (partial) set of reals, which is placed in one-to-one correspondence with a (partial) set of integers. Here is the example from G.9adel, Escher, Bach: 1 .1 4 1 5 9 2 6 5 3 2 .3 3 3 3 3 3 3 3 3 3 .7 1 8 2 8 1 8 2 8 4 .4 1 4 2 1 3 5 6 2 5 .5 0 0 0 0 0 0 0 0 The argument then goes by constructing a new real by taking the elements of the diagonal and changing every digit. The argument then goes that (a) by construction, this new real is not in the list of reals (b) by construction, there is a real against every integer, (c) THEREFORE, there are more reals than integers, even though the list of integers is infinite. My principal observation is that while the column containing the integers is quite obviously ordered, the column containing the reals is not. It is this disorder in the example list of reals that is the clincher. Once an ordering scheme is imposed on the reals, a true one-to-one correspondence, expressible as a simple formula, can be constructed. One example which has been mooted here on a number of occasions is to reverse the order of the decimal digits and disregard the decimal point [1], thus the real 0.5 maps to the integer 5 as per the last line of the example, the real 0.890625 maps to the integer 526098, etc. OK so all non-terminating decimals, recurring or not, map to infinity by this method. No problem, there is enough room at infinity for all of them. Each successive Cantor diagonalisation of the list will give a new nonterminating [2] real and a new infinite integer to map it to. [1] This confines the reals to the range 0 to 1, but only a trivial modification is required to remove this restriction. Simply pass the infinite range of reals through a mathematical function that maps them to a finite range in 1:1 correspondence. The tanh function is a good one to use for this purpose. [2] The diagonalisation must give a non-terminating decimal since all terminating ones are mapped to the finite integers by construction, so they are all already in the list. -- Paul V. S. Townsend Interchange the alphabetic elements to reply X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 03:46 PM, Prai Jei said: >My principal observation is that while the column containing the >integers is quite obviously ordered, the column containing the reals >is not. It clearly *IS* ordered, by the very act of arranging them into a column. >Once an ordering scheme is imposed on the reals, a true one-to-one >correspondence, expressible as a simple formula, can be constructed. You are confusing ordering with well ordering. Further, you are confusing the statement to be disproved with a statement that has been proven. In fact the diagonal argument proves that such a column is impossible. >One example which has >been mooted here on a number of occasions is to reverse the order of >the decimal digits and disregard the decimal point [1], And what does 1/3 map to? >OK so all non-terminating decimals, recurring or not, map to >infinity What is infinity Does 1/3 map to the same infinity as 1/6? >there is enough room at infinity for all of them. What does that mean? >infinite integer What is an infinite integer? What you're doing isn't Mathematics; it isn't even Philosophy. Define your terms if you wish to be taken seriously, then use them only in a fashion consistent with how you defined them. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > This is one of my own pet hobbyhorses. I am a firm believer in only one > infinity, which is not even signed. > > Every presentation of the Cantor argument has turned upon a (partial) set of > reals, which is placed in one-to-one correspondence with a (partial) set of > integers. Here is the example from G.9adel, Escher, Bach: > 1 .1 4 1 5 9 2 6 5 3 > 2 .3 3 3 3 3 3 3 3 3 > 3 .7 1 8 2 8 1 8 2 8 > 4 .4 1 4 2 1 3 5 6 2 > 5 .5 0 0 0 0 0 0 0 0 > The argument then goes by constructing a new real by taking the elements of > the diagonal and changing every digit. The argument then goes that (a) by > construction, this new real is not in the list of reals (b) by construction, > there is a real against every integer, (c) THEREFORE, there are more reals > than integers, even though the list of integers is infinite. > > My principal observation is that while the column containing the integers is > quite obviously ordered, the column containing the reals is not. It is this > disorder in the example list of reals that is the clincher. Once an ordering > scheme is imposed on the reals, a true one-to-one correspondence, > expressible as a simple formula, can be constructed. One example which has > been mooted here on a number of occasions is to reverse the order of the > decimal digits and disregard the decimal point [1], thus the real 0.5 maps > to the integer 5 as per the last line of the example, the real 0.890625 maps > to the integer 526098, etc. > I'm having trouble understanding the flow of your response. This is the simple formula, from [0,1] to Z, which you purport to fix below, since it has been mooted here before? > OK so all non-terminating decimals, recurring or not, map to infinity by > this method. No problem, there is enough room at infinity for all of them. It's at this point that I begin not to understand. What does this have to do with the ordering you induce? Unfortunately, infinity isn't in the standard definitions of Z I'm familiar with, but this is not so much a hurdle. Besides, even if it were, you said you were a firm believer in a single unsigned infinity, so every nonterminating real in [0,1] would be mapped to this single element; this mapping isn't a bijection and has little to do with cardinality. > Each successive Cantor diagonalisation of the list will give a new > nonterminating [2] real and a new infinite integer to map it to. > Says the firm believer in a single infinity? Are you suggesting that we define new elements and union them with Z to make this mapping work? The problem is, we'll be adding an uncountable number of infinities to Z, so the set to which you're mapping won't be of the same cardinality as Z. In other words, you've changed your mapping from [0,1]->Z to [0,1]->(Z union I), where I is the new set of infinities you define. Since there is no bijection between Z and (Z union I), this shows us that they don't have the same cardinality. You're back where you began and haven't fixed this simple mapping. infinity. The argument against this fix still stands if you just look at (Z union I), where we say that I is the set of infinite integers, which have never been part of Z and aren't countable. All you've really done is removed a decimal point from the analysis. > [1] This confines the reals to the range 0 to 1, but only a trivial > modification is required to remove this restriction. Simply pass the > infinite range of reals through a mathematical function that maps them to a > finite range in 1:1 correspondence. The tanh function is a good one to use > for this purpose. Let's look at f(x) = (tanh(x)+1)/2. What real number is mapped to 1 and what is mapped to 0? I was under the impression that there were no signed infinities in your system, in which case we're left with (0,1] and there is a quick detail to work out. > [2] The diagonalisation must give a non-terminating decimal since all > terminating ones are mapped to the finite integers by construction, so they > are all already in the list. ==== >[self-contradictory OP snipped] This is one of my own pet hobbyhorses. I am a firm believer in only one >infinity, which is not even signed. I'm a firm believer in a flat earth. Doesn't make the earth flat. >Every presentation of the Cantor argument has turned upon a (partial) set of >reals, which is placed in one-to-one correspondence with a (partial) set of >integers. Here is the example from G.9adel, Escher, Bach: > 1 .1 4 1 5 9 2 6 5 3 > 2 .3 3 3 3 3 3 3 3 3 > 3 .7 1 8 2 8 1 8 2 8 > 4 .4 1 4 2 1 3 5 6 2 > 5 .5 0 0 0 0 0 0 0 0 >The argument then goes by constructing a new real by taking the elements of >the diagonal and changing every digit. The argument then goes that (a) by >construction, this new real is not in the list of reals (b) by construction, >there is a real against every integer, (c) THEREFORE, there are more reals >than integers, even though the list of integers is infinite. My principal observation is that while the column containing the integers is >quite obviously ordered, the column containing the reals is not. This has no relevance whatever. _By definition_, saying that there are more reals than integers means that if you have a mapping of integers to reals then there is at least one real not in the range of the mapping. Nothing there about columns being ordered. >It is this >disorder in the example list of reals that is the clincher. Once an ordering >scheme is imposed on the reals, a true one-to-one correspondence, >expressible as a simple formula, can be constructed. One example which has >been mooted here on a number of occasions is to reverse the order of the >decimal digits and disregard the decimal point [1], thus the real 0.5 maps >to the integer 5 as per the last line of the example, the real 0.890625 maps >to the integer 526098, etc. OK so all non-terminating decimals, recurring or not, map to infinity by >this method. No problem, there is enough room at infinity for all of them. >Each successive Cantor diagonalisation of the list will give a new >nonterminating [2] real and a new infinite integer to map it to. _By definition_ there is no such thing as an infinite integer. You're not demonstrating anything here, all you're doing is giving words new definitions. I can _prove_ that the Earth is flat. People point out that it's round, but that's no problem, because the desk in my office is flat! Is that a proof that the Earth is flat? No, because the desk in my office is not the Earth. If you think that there is a 1-1 correspondence between the reals and the (finite) integers you're simply wrong. But it doesn't look like you think that - it looks like you agree there is no such 1-1 correspondence. If so then you _do_ agree that there are more reals than integers, assuming we give all the words their standard meanings - what you think you accomplish by using words to mean things other than what they _do_ mean is not clear to me. >[1] This confines the reals to the range 0 to 1, but only a trivial >modification is required to remove this restriction. Simply pass the >infinite range of reals through a mathematical function that maps them to a >finite range in 1:1 correspondence. The tanh function is a good one to use >for this purpose. >[2] The diagonalisation must give a non-terminating decimal since all >terminating ones are mapped to the finite integers by construction, so they >are all already in the list. ************************ David C. Ullrich ==== >[self-contradictory OP snipped] >>This is one of my own pet hobbyhorses. I am a firm believer in only one >>infinity, which is not even signed. I'm a firm believer in a flat earth. Doesn't make the earth flat. [snip argumentation why Cantor's diagonal is not wrong] I'd like to take a few moments to elaborate on a topic that irritates me to no end: CANTOR'S PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVIL'S SPAWN!!! Well maybe not quite but it's certainly worse than L'Hospital's rule on the scale of things that are so obvious no one can possibly misunderstand them that are consequently misunderstood all the time. This is not aided by the fact that it's so deceptively simple that it gets tacked on every single introductory freshman/high school math text and popularization from here to eternity. As a proof, it's... well, elegant. However it's so elegant that the underlying finesse usually goes completely unnoticed, which means we get a steady stream of cranks^Walternative thinkers who are convinced that the proof is wrong because . It's like mathematics is a house where one of the paintings has a funny optical illusion that makes the painting look like it's tilted while it's perfectly straight, and every joker who waltzes in points out that the painting looks tilted and thinks no one else has pointed this out before. So please, to those in the position of influencing textbook publishers, would it please be possible to remove this deceptively simple yet simply deceptive argument from texts where it is not relevant and to replace it with a more formal two-page proof including no more than five understandable English words and at least two homeomorphisms? I think that would go a long way. And with these words we return to your scheduled gigantic Cantor thread... ==== >>[self-contradictory OP snipped] >>This is one of my own pet hobbyhorses. I am a firm believer in only one >infinity, which is not even signed. >>I'm a firm believer in a flat earth. Doesn't make the earth flat. [snip argumentation why Cantor's diagonal is not wrong] I'd like to take a few moments to elaborate on a topic that irritates >me to no end: CANTOR'S PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVIL'S SPAWN!!! Well maybe not quite but it's certainly worse than L'Hospital's rule >on the scale of things that are so obvious no one can possibly >misunderstand them that are consequently misunderstood all the time. >This is not aided by the fact that it's so deceptively simple that >it gets tacked on every single introductory freshman/high school math >text and popularization from here to eternity. As a proof, it's... well, elegant. However it's so elegant that the >underlying finesse usually goes completely unnoticed, which means we >get a steady stream of cranks^Walternative thinkers who are convinced >that the proof is wrong because here>. It's like mathematics is a house where one of the paintings has >a funny optical illusion that makes the painting look like it's tilted >while it's perfectly straight, and every joker who waltzes in points >out that the painting looks tilted and thinks no one else has pointed >this out before. So please, to those in the position of influencing textbook >publishers, would it please be possible to remove this deceptively >simple yet simply deceptive argument from texts where it is not >relevant and to replace it with a more formal two-page proof including >no more than five understandable English words and at least two >homeomorphisms? I think that would go a long way. I can't decide whether you're serious. If you are then this is a very curious attitude... if you're not then you got me. >And with these words we return to your scheduled gigantic Cantor >thread... ************************ David C. Ullrich ==== > >[self-contradictory OP snipped] >>This is one of my own pet hobbyhorses. I am a firm believer in only one >>infinity, which is not even signed. I'm a firm believer in a flat earth. Doesn't make the earth flat. > > [snip argumentation why Cantor's diagonal is not wrong] > > I'd like to take a few moments to elaborate on a topic that irritates > me to no end: > > CANTOR'S PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVIL'S SPAWN!!! > > Well maybe not quite but it's certainly worse than L'Hospital's rule > on the scale of things that are so obvious no one can possibly > misunderstand them that are consequently misunderstood all the time. > This is not aided by the fact that it's so deceptively simple that > it gets tacked on every single introductory freshman/high school math > text and popularization from here to eternity. > Cantor's proof is the simplest, most accessible proof of something that's counterintuitive. That's why it stirs up the cranks so. ==== > There are enough and coherent reasoning to prove that it is not >possible to imagine the list of all real numbers without running >against the nature of the natural numbers. For instance, if we >admit the one-to-one correspondence between N and R, then we are >confronting two different kinds of infinity, the potential infinite >represented by the asymptotic approximation of the naturals to the >infinite (never reaching it), and the actual infinite represented by >the set of all real numbers; so that, a complete b ijection between >both sets is not possible. With another simple reasoning we arrive >to the conclusion that the acceptance of the existence of that list >would imply naturals with an infinite value (cardinal). So are you saying there are only finitely many natural numbers? > On the other >hand, there are constructive methods (at least I know one of them) >that by simple induction they proof that such a list will require >naturals with an infinite number of figures. > Therefore, my question is this: If the ãlist of all real numbersä is not a valid concept, why must we accept it as premise in the Cantorâs >proof of the diagonal? The theory of transfinite numbers would be able >to be an air castle with such an argument in its foundations. > > Nicolas de la Foz I'm not sure I completely understand your problem with this. The diagonal argument shows that any countably infinite set of Real numbers cannot contain all of them. And any countable set can be thought of as a 'list'. What's wrong with that? What asymptotic approximation of the infinite are you talking about? It really reads as though you are saying there are not infinitely many natural numbers. Or that there isn't _an_ infinitely large natural number? If you don't like proof by contradiction, don't assume that the reals are countable, but show that _any_ countable set of them is incomplete. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBSDc1102527; ==== hi im am a late nite science crack pot that is very interested in finding out what and how r universe was made and how it works i would like to get as mutch 411 on the subject as u can send me please send me as much as u can thank u ==== >would like to get as mutch 411 on the subject as u can send me In the movie Hercules, at some point an accident has happened and some one gives the throw away line Somebody call IXII That cracked me up. Mitch ==== >>would like to get as mutch 411 on the subject as u can send me > > > In the movie Hercules, at some point an accident has happened and some > one gives the throw away line > > Somebody call IXII > > That cracked me up. > > Mitch > And Hercules was a Greek, not a Roman. ==== >>would like to get as mutch 411 on the subject as u can send me > > > In the movie Hercules, at some point an accident has happened and some > one gives the throw away line > > Somebody call IXII > > That cracked me up. > > Mitch > Well, actually it would be CMXI or IX-I-I, but it's funny anyway. ==== > Actually, mathworld is one of my favorite bookmarks, but somehow I got lost there and didn't find the specific page you mentioned. So thanks for pointing it out, I now understand the ideas much better. > >>I have spent many hours trying to find a verbal definition of the following terms. Gradient, Divergence and Curl. I find the mathematical definitions but never a verbal description of exactly what it is I'm doing when I calculate these. Can anyone give me the basics in WORDS and not equations? Exactly what am I finding about the vector field when I calculate the gradient, divergence and curl? Did you try http://mathworld.wolfra m.com/Divergence.html etc? I bet you didn't phil It may be heresy to mention a physics book here ! However, you could do worse than look at the Feynman Lectures on Physics. Volume 2 has a couple of chapters about div, grad and curl, and attempts to give a physical meaning to these operators Hope this helps Ian Taylor ==== > Actually, mathworld is one of my favorite bookmarks, but somehow I got lost there and didn't find the specific page you mentioned. So thanks for pointing it out, I now understand the ideas much better. >I have spent many hours trying to find a verbal definition of the > following terms. Gradient, Divergence and Curl. I find the > mathematical definitions but never a verbal description of exactly > what it is I'm doing when I calculate these. Can anyone give me the > basics in WORDS and not equations? Exactly what am I finding about the > vector field when I calculate the gradient, divergence and curl? Did you try http://mathworld.wolfra m.com/Divergence.html etc? I bet you didn't phil It may be heresy to mention a physics book here ! However, you could > do worse than look at the Feynman Lectures on Physics. Volume 2 has a > couple of chapters about div, grad and curl, and attempts to give a > physical meaning to these operators A physics book? Heresy?? Feynman??? Heresy???? It is exactly what the OP needs. The only potential heresy is your hesitation about mentioning this ;-) > Hope this helps I'm sure it will :-) Ian Taylor Dirk Vdm ==== Using the resultant in this way should be okay to prove Bezout's theorem _if_ the curves intersect in distinct points, but things get subtler when there are higher multiplicities, i.e., when the curves intersect tangentially and/or intersect at singular points. The way I've generally seen it done is that first one defines the intersection index (or multiplicity) of the two curves at each point of intersection. For example, if the curves are f(x,y)=0 and g(x,y)=0 and if they intersect at the point (0,0), then the intersection index at (0,0) is the dimension of the complex vector space C[x,y]_{0,0}/(f(x,y),g(x,y)). Here C[x,y]_(0,0) is the ring of polynomials localized at (0,0), which means the ring of all rational functions p(x,y)/q(x,y) with the property that q(0,0) is not 0. Of course, this definition works anywhere, since you can always do a translation X=x-a, Y=y-b to move the intersection point (a,b) to (0,0). And for points on the line at infinity in projective space, you can dehomogenize with respect to one of the other coordinates. Bezout's theorem then says that if you add up all of the intersection indices for all of the intersection points, the total is equal to deg(f)*deg(g). The beauty and strength of Bezout's theorem lies in the fact that the intersection indices are computed locally in a neighborhood of each point, but that their total is equal to a global quantity, namely the product of the degrees of the curves. The proof of Bezout's theorem is in most basic algebraic geometry books, using varying amounts of fancy machinery. John Tate and I put a fairly elementary proof into the appendix of our book Rational Points on Elliptic Curves (Springer-Verlag). Joe Silverman > says... > >I'm not an algebraic geometer (far from it), >but I would prove it by showing that the resolvant (or resultant) R(f1,f2) >of f1(x,y),f2(x,y), regarded as polynomials in y over k[x], >is a polynomial of degree <= mn in x. > Ok, I am not well versed in the resolvent.. but let me verify if I understand > this correctly. You are telling that if I work with f1 and f2 as functions of > k(y)[x] (we had like k(y) to be a field, thus fraction field), and solve the > resolvent we can arrive to the Bezout theorem. Am I correct that the resolvent > determines the common roots of f1 and f2 (ie if the resolvent is zero at a > point (x,y) then f1 and f2 are zero at those points). I just want to verify if > > Sincerely, > Jose Capco ==== > Using the resultant in this way should be okay to prove Bezout's > theorem _if_ the curves intersect in distinct points, but things get > subtler when there are higher multiplicities, i.e., when the curves > intersect tangentially and/or intersect at singular points. Actually the original poster asked for a proof of what he/she called the weak Bezout's Theorem, that the number of intersections is <= mn. > The proof of Bezout's theorem is in most basic algebraic geometry > books, using varying amounts of fancy machinery. John Tate and I put a > fairly elementary proof into the appendix of our book Rational Points > on Elliptic Curves (Springer-Verlag). I must admit I found this to be one of the weaker sections of what I rate as one of the best popular mathematical books I know -- one of the very few worthy to stand on the bookshelf next to Hardy & Wright. Anyone with the slightest interest in elliptic curves should start with Silverman & Tate, IMHO. -- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 ==== says... Actually the original poster asked for a proof of what he/she called >the weak Bezout's Theorem, that the number of intersections is <= mn. > The proof of Bezout's theorem is in most basic algebraic geometry >> books, using varying amounts of fancy machinery. John Tate and I put a >> fairly elementary proof into the appendix of our book Rational Points >> on Elliptic Curves (Springer-Verlag). I must admit I found this to be one of the weaker sections >of what I rate as one of the best popular mathematical books I know -- >one of the very few worthy to stand on the bookshelf next to Hardy & Wright. Anyone with the slightest interest in elliptic curves >should start with Silverman & Tate, IMHO. Ok thanks all :) .. I didn't know that this post could bother Mr. Silverman too :) .. anyway, you might be right (I never looked at that book but I certainly heard a lot about it). My original purpose was actually to prove the associativity law of the elliptic curve and for that I believe weak Bezout theorem suffices. The previous threads gave me enough ideas to start with and I dont suppose I need to go very much into resolvent theory. Sincerely, Jose Capco ==== >Anyone with the slightest interest in elliptic curves >should start with Silverman & Tate, IMHO. I just want to second this. I absolutely love this book. Steven E. Landsburg www.landsburg.com/about2.html -- ==== > > Using the resultant in this way should be okay to prove Bezout's > theorem _if_ the curves intersect in distinct points, but things get > subtler when there are higher multiplicities, i.e., when the curves > intersect tangentially and/or intersect at singular points. > > Actually the original poster asked for a proof of what he/she called > the weak Bezout's Theorem, that the number of intersections is <= mn. True. I was just expanding the discusssion a bit to the general case. For the weak form, as other posters noted, the resultant argument will give the upper bound of mn, and it is certainly very elementary. > The proof of Bezout's theorem is in most basic algebraic geometry > books, using varying amounts of fancy machinery. John Tate and I put a > fairly elementary proof into the appendix of our book Rational Points > on Elliptic Curves (Springer-Verlag). > > I must admit I found this to be one of the weaker sections > of what I rate as one of the best popular mathematical books I know -- > one of the very few worthy to stand on the bookshelf next to Hardy & Wright. > > Anyone with the slightest interest in elliptic curves > should start with Silverman & Tate, IMHO. Sorry you didn't like that section, but in any case, thank you for the kind words regarding RPEC. It's an honor to be mentioned in the same sentence as the H&W classic. Joe Silverman ==== >> Could someone here write the sketch of the proof of the (weaker) Bezout >> theorem, i.e. given two curves (say one being >> irreducible.. to make matters a bit easier) C1 and C2 defined by >> functions f1 and f2, the interesection between these >> two curves is less or equal deg(f1)*deg(f2). > > I'm not an algebraic geometer (far from it), > but I would prove it by showing that the resolvant (or resultant) R(f1,f2) > of f1(x,y),f2(x,y), regarded as polynomials in y over k[x], > is a polynomial of degree <= mn in x. I guess if one doesn't want to go into the theory of the resolvent, f1(x,y), x*f1(x,y), x^2*f1(x,y), ..., x^{n-1}*f1(x,y), f2(x,y), x*f2(x,y), x^2*f2(x,y), ..., x^{m-1}*f2(x,y). One can regard these as m+n linear equations in 1,x,x^2,...,x^{m+n-1}. If f1(x,y),f2(x,y) have a common root in x for a given value of y, then det(A) = 0, where A is the matrix of this set of equations. This is an equation of degree <= mn in y. This is exactly the same argument -- I'm just pointing out that for your purpose one only needs to establish the property of the resolvent in one direction. -- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 ==== |Ok, I am not well versed in the resolvent.. but let me verify if I understand |this correctly. If two nonzero polynomials P and Q of degrees m and n have a nonconstant common factor R, then P*(Q/R)-Q*(P/R)=0, and S=Q/R and T=-P/R are polynomials of degrees >I'm not an algebraic geometer (far from it), >>but I would prove it by showing that the resolvant (or resultant) R(f1,f2) >>of f1(x,y),f2(x,y), regarded as polynomials in y over k[x], >>is a polynomial of degree <= mn in x. > > Ok, I am not well versed in the resolvent.. but let me verify if I > understand this correctly. You are telling that if I work with f1 and f2 > as functions of k(y)[x] (we had like k(y) to be a field, thus fraction > field), and solve the resolvent we can arrive to the Bezout theorem. Am I > correct that the resolvent determines the common roots of f1 and f2 (ie if > the resolvent is zero at a point (x,y) then f1 and f2 are zero at those > points). I just want to verify if my understanding of the resolvent was Yes, that is my understanding too. Suppose f(x), g(x) are monic (leading coefficient 1), f(x) = x^m + a_1x^{m-1} + ..., g(x) = x^n + b_1x^{n-1} + ... If we define R(f,g) = prod (a_i - b_j) where a_i, b_j are the roots of f,g respectively, then it is relatively straightforward to show that R(f,g) = det A, where A is the (m+n)x(m+n) matrix A = (1,a_1,a_2,.../0,1,a_1,.../.../1.b_1,b_2,.../0,1,b_1,.../...). Now apply this with f(x) = f1(x,y), g(x) = f2(x,y). Then a_r, b_r are polynomials of degree r in y, and R(f,g) is a polynomial of degree mn in y. Basically, given polynomials of degrees m,n in x,y this shows that one of the variables, x say, can be eliminated to give a polynomial of degree mn in the other variable. -- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 ==== > crap] And every word you utter simply continues to confirm what all the > readers of > this ng think of you. Do you really not realise that? Wow. What an ignont comment. The internet is well known for being > domindated by ignorant irritating people. These people force decent > people of the web due to their slimey nature. I.e. they flame them off > the web. That means that jerks like you remain behind and fill it up. So I expect that all newsgroups are fill with jerks like you, varney, > bilgem, speicher etc. And when I say jerk I'm being *very* gracious. > Useless troll. > Waste of time. Plonk. Any reason that you think you need to butt into a conversation between two > people? You are operating under a misunderstanding if you think anything in usenet > corresponds to a converation between two people. You're confused again. The question was simple. he refuses to answer. > Nobody said that he had no right to dimwit. The question was with No. I am not confused. Here is your statement, lifted lock stock and barrel: Any reason that you think you need to butt into a conversation between two people? Here was my answer *to that statement*: You are operating under a misunderstanding if you think anything in usenet corresponds to 'a converation between two people'. No confusion is possible here, imbeciles excepted. [snip] Franz ==== Why are you becomming so obsessed with me? I think you've been hanging around bilge, waite and varney for far too long. ==== Why are you becomming so obsessed with me? I think you've been hanging > around bilge, waite and varney for far too long. Naw... we all simply have fun poking you with a stick. ==== I forgot to mention that mays is one of these people who would rather > flame then stick to physics > > > And anyway... when you post some physic's > worthy of my taking my valuable time to comment on > I will... Until then I'll continue to poke at all > you guys with a stick....... A boys gota have a hobby... ;)~ Well, you know, there's no rule you couldn't _start_ a thread containing, some... ahem... physic's? What is a physic's worthy? Is it like a witch's familiar...? And you say this malign entity took your valuable time to comment on!? The verve! Anyway, my physic's worthy, supposing you knew all physics, you might still start something in a didactic vein, for the unworthy. <1g69dp0.1jt1z3ms7jy9lN%see.sig@for.addy> <3fe79cc4$12$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 05:15 PM, Pmb said: >Any reason that you think you need to butt into a conversation >between two people? communication between two people. Anything that you post here is a solicitation for comments from anyone reading it. That's a separate the news groups and whether the response was appropriate. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > at 05:15 PM, Pmb said: Any reason that you think you need to butt into a conversation >between two people? communication between two people. Show me where I said it was a private conversation. With so little said and not much content in what was said it begs the question as to why he posted it. If he chooses not to answer then he chooses not to answer. Why does that bother you? <1g69dp0.1jt1z3ms7jy9lN%see.sig@for.addy> <3fe79cc4$12$fuzhry+tra$mr2ice@news.patriot.net> <3fee0cff$12$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 11:10 PM, Pmb said: >Show me where I said it was a private conversation. Any reason that you think you need to butt into a conversation between two people? >Why does that bother you? It doesn't. What bothers me is your posting your crap in a public forum and then complaining when the public responds to it. If you want no such thing as a conversation between two people, and hence nobody can butt into the discussion. Everybody here is a party to it, whether you understand that or not. If you want your quarrel to be private, conduct it in private. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > at 11:10 PM, Pmb said: > >Show me where I said it was a private conversation. > > Any reason that you think you need to butt into a conversation > between two people? > >Why does that bother you? > > It doesn't. What bothers me is your posting your crap in a public > forum and then complaining when the public responds to it. If you want > no such thing as a conversation between two people, and hence nobody > can butt into the discussion. Everybody here is a party to it, > whether you understand that or not. If you want your quarrel to be > private, conduct it in private. Amen. ==== > at 11:10 PM, Pmb said: > >Show me where I said it was a private conversation. > > Any reason that you think you need to butt into a conversation > between two people? > >Why does that bother you? > > It doesn't. What bothers me is your posting your crap in a public > forum and then complaining when the public responds to it. Then you need to read more carefully. The sentance - Any reason that you think you need to butt into a conversation between two people? - is a *question* and not a *complaint*. Pmb ==== at 05:15 PM, Pmb said: Any reason that you think you need to butt into a conversation >between two people? communication between two people. Show me where I said it was a private conversation. Here, and I quote once more: Any reason that you think you need to butt into a conversation between two people?? Now shut up. [snip] Franz ==== at 05:15 PM, Pmb said: Any reason that you think you need to butt into a conversation >between two people? communication between two people. Show me where I said it was a private conversation. > > Here, and I quote once more: > Any reason that you think you need to butt into a conversation between two > people?? > > Now shut up. Wow you're dumb. You don't know the difference between those two? I said show me where I said it was a ************** p r i v a t e ************** conversation. Wow! You're *such* an idiot heymann!!! Now fuck off. ==== > in > at 05:15 PM, Pmb said: Any reason that you think you need to butt into a conversation >between two people? communication between two people. Show me where I said it was a private conversation. Here, and I quote once more: > Any reason that you think you need to butt into a conversation between two > people?? Now shut up. Wow you're dumb. You were told to shut up, crackpot. ==== Back in the spring of 2002 I discovered the following way to count prime numbers: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], S(x,1) = 0, S(x,y) is the sum of dS from dS(x,2) to dS(x,y), p(x, y) = floor(x) - S(x, y) - 1, and p(x,sqrt(x)) gives the count of primes where, of course, only positive integers are being used, so the sqrt() operator returns the largest positive integer that squares to give the argument or a number less than it. Now consider values for dS(x,y) from the calculation of p(100,10): dS(100,2)=49 dS(100,3)=16 dS(100,4)=0 dS(100,5)=6 dS(100,6)=0 dS(100,7)=3 dS(100,8)=0 dS(100,9)=0 dS(100,10)=0 The sum of all those values gives S(100,10)=74, which is the count of composites, 74 subtracts from 100 to give 26, and subtracting 1 from that gives you p(100,10) = 25, and 25 is the count of primes up to 100. Now then dS(x,y) is the count of composites that have y as a factor that do NOT have a prime number less than y as a factor up to and including x, so it makes sense that unless y is a prime dS(x,y) is 0 because, for instance, if y=4, then any number that has 4 as a factor also has 2 as a factor, so there aren't any composites that have 4 as a factor that do not have a prime number less than 4 as a factor as, of course, 2 is a factor of 4. Practically that means that dS(x,y) is 0 if y is composite, and non-zero if y is prime and less than or equal to sqrt(x) as can be seen with the values given above. Now then the non-zero values are easy to understand, like dS(100,2)=49, means that there are 49 numbers up to and including 100 that are even, excluding 2 itself, as its prime. Then dS(100,3)=16 tells you that there are 16 numbers that are composite with 3 as a factor that are NOT even. Next you have dS(100,4)=0, as there aren't any numbers that have 4 as a factor that do not have a smaller prime as a factor, in this case 2, as 4 has 2 as a factor. Next there's dS(100,5)=6, telling you that there are 6 composites with 5 as a factor that do not have 2 or 3 as a factor. Notice the drop of non-zero dS(x,y) values is like an exponential drop, while the spacing between values is, of course, the spacing between prime numbers. And that explains: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))] More information can be found at my blog: http://mathforprofit.blogspot.com/ James Harris ==== > Back in the spring of 2002 I discovered the following way to count > prime numbers: > > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], > > S(x,1) = 0, S(x,y) is the sum of dS from dS(x,2) to dS(x,y), > > p(x, y) = floor(x) - S(x, y) - 1, > > and p(x,sqrt(x)) gives the count of primes where, of course, only > positive integers are being used, so the sqrt() operator returns the > largest positive integer that squares to give the argument or a number > less than it. (you've changed your line on this one haven't you?) > > Now consider values for dS(x,y) from the calculation of p(100,10): > > dS(100,2)=49 > dS(100,3)=16 > dS(100,4)=0 > dS(100,5)=6 > dS(100,6)=0 > dS(100,7)=3 > dS(100,8)=0 > dS(100,9)=0 > dS(100,10)=0 > > The sum of all those values gives S(100,10)=74, which is the count of > composites, 74 subtracts from 100 to give 26, and subtracting 1 from > that gives you p(100,10) = 25, and 25 is the count of primes up to > 100. > > Now then dS(x,y) is the count of composites that have y as a factor > that do NOT have a prime number less than y as a factor up to and > including x, so it makes sense that unless y is a prime dS(x,y) is 0 > because, for instance, if y=4, then any number that has 4 as a factor > also has 2 as a factor, so there aren't any composites that have 4 as > a factor that do not have a prime number less than 4 as a factor as, > of course, 2 is a factor of 4. You know, I think I'vew read almost everything you've written before on htis subject, and this is the first time you say what's going on > > Practically that means that dS(x,y) is 0 if y is composite, and > non-zero if y is prime and less than or equal to sqrt(x) as can be > seen with the values given above. > > Now then the non-zero values are easy to understand, like > dS(100,2)=49, means that there are 49 numbers up to and including 100 > that are even, excluding 2 itself, as its prime. Then dS(100,3)=16 > tells you that there are 16 numbers that are composite with 3 as a > factor that are NOT even. > > Next you have dS(100,4)=0, as there aren't any numbers that have 4 as > a factor that do not have a smaller prime as a factor, in this case 2, > as 4 has 2 as a factor. > > Next there's dS(100,5)=6, telling you that there are 6 composites with > 5 as a factor that do not have 2 or 3 as a factor. > > Notice the drop of non-zero dS(x,y) values is like an exponential > drop, while the spacing between values is, of course, the spacing > between prime numbers. > So I think I agree that it will count the composites, and the primes. > And that explains: > But it doesn't explain why it satisfies this difference relation. > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))] > So please explain it to an idiot like me: why is this relation? You have dramatically improved the presentation, but if you are keen to have it accepted, then explain the reasoning behind this bit. The idea of counting composites like this, using the inclusion-exclusion principle is fairly well studied by the way. > More information can be found at my blog: > > http://mathforprofit.blogspot.com/ > > > James Harris ==== > Back in the spring of 2002 I discovered the following way to count > prime numbers: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], S(x,1) = 0, S(x,y) is the sum of dS from dS(x,2) to dS(x,y), p(x, y) = floor(x) - S(x, y) - 1, and p(x,sqrt(x)) gives the count of primes where, of course, only > positive integers are being used, so the sqrt() operator returns the > largest positive integer that squares to give the argument or a number > less than it. It does? Not according to your own recent posts which proclaim that 'sqrt'is inherently ambiguous and that the ambiguity *cannot* be removed. By your own account 'sqrt(y)' has a positive OR negative value and both must be accommodated. The negative value, of course, does NOT satisfy the claim made for your argument. Hence, your above method fails *by your own criteria*. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== I would greatly appreciate any comment upon the correctness of the following assertion. Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) = 1 Assertion: The following equality is not possible. sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1) ==== >I would greatly appreciate any comment upon the correctness of the >following assertion. >Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) = >1 >Assertion: The following equality is not possible. > sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1) sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5 Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== In sci.math, Robert Israel : >>I would greatly appreciate any comment upon the correctness of the >>following assertion. > >>Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) = >>1 >>Assertion: The following equality is not possible. > >> sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1) > > sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5 > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 (sqrt(2) + sqrt(3))^5 = 2^2.5 + 5*2^2*3^.5 + 10*2^1.5*3 + 3^2.5 + 5*3^2*2^.5 + 10*3^1.5*2 = sqrt(2)*(2^2+10*3*2+5*3^2) + sqrt(3)*(3^2+10*2*3+5*2^2) = sqrt(2)*109+sqrt(3)*89 = sqrt(23762)+sqrt(23763) The solution checks out, which leads me to believe that there was a problem in the original specification. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > In sci.math, Robert Israel > : >>I would greatly appreciate any comment upon the correctness of the >>following assertion. > >>Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) = >>1 >>Assertion: The following equality is not possible. > >> sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1) > > sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5 > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 > > (sqrt(2) + sqrt(3))^5 = 2^2.5 + 5*2^2*3^.5 + 10*2^1.5*3 > + 3^2.5 + 5*3^2*2^.5 + 10*3^1.5*2 > = sqrt(2)*(2^2+10*3*2+5*3^2) > + sqrt(3)*(3^2+10*2*3+5*2^2) > = sqrt(2)*109+sqrt(3)*89 > = sqrt(23762)+sqrt(23763) > > The solution checks out, which leads me to believe that > there was a problem in the original specification. One never knows about what's posted to Usenet :-( What's even stranger is that if gcd(c,d) = 1 then sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 for a = c(c^2 + 10cd + 5d^2)^2, b = d(5c^2 + 10cd + d^2)^2, and gcd(a,b,c,d) = 1. So the equality is ALWAYS possible. Perhaps what Robert is doing is giving an example where gcd(a,b) = 1 too? I haven't checked for smaller examples... --Ron Bruck ==== One never knows about what's posted to Usenet :-( escribi.97 en el mensaje > I would greatly appreciate any comment upon the correctness of the > following assertion. Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) = > 1 > Assertion: The following equality is not possible. sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1) sqrt(a) + sqrt(b) = c^(5/2) + 5c^2*d^(1/2) + 10d*c^(3/2) + 10c*d^(3/2) + 5c^(1/2)d^2 + d^(5/2) = (c^2 +10c*d + 5d^2)sqrt(c) + (d^2 + 10c*d + 5c^2)sqrt(d) Then, a = c(c^2 +10c*d + 5d^2)^2 b = d(d^2 + 10c*d + 5c^2)^2 a and b aren't squares because c and d aren't. And gcd(a, b, c, d) = 1 if gcd(c, d) = 1. Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== Do these primes have a name [7,31,71,127...]? Where k = n*(n+1)/2 Where x falls exactly between two consecutive EVEN numbered k's and if (x-1) is a prime then this list below are those primes. 7,31,71,127,199,647,967,1151,1567,2311,2591... Eg. Of first few primes (x-1) and k-1 and k 1) 6 (8) 10 k-1 = 6 and x= 8 and k = 10 2) 28 (32) 36 k-1 = 28 and x=32 and k = 36 3) 66 (72) 78 k-1 = 66 and x=72 and k = 78 1) x-1 = 7 2) x-1 = 31 3) x-1 = 71 etc.. What is interesting is when x-1 is a composite (x-1)== >0 (mod 3) and also for (x-1)== >0 (mod 5). So far, no 0 (mod 3)'s or no 0 (mod 5)'s composites! Examining it a bit further -- I believe it is not possible because these (x-1)== >0 (mod 9) whether (x-1) is prime or composite the consecutive (x-1)'s have a repeating sequence of residuals (r) for --- (x-1) == r (mod 9)'s where consecutive (x-1)'s -- r =[7,4,8,1,1,8,4,7,8] [7,4,8,1,1,8,4,7,8] [7,4... etc. This repeating sequence is almost a palindrome. ;-) Dan ==== > Do these primes have a name [7,31,71,127...]? > > Where k = n*(n+1)/2 > Not sure about a special name, but that's the nth triangular number. > Where x falls exactly between two consecutive EVEN numbered k's and if > (x-1) is a prime then this list below are those primes. > > 7,31,71,127,199,647,967,1151,1567,2311,2591... > > Eg. Of first few primes (x-1) and k-1 and k > > 1) 6 (8) 10 k-1 = 6 and x= 8 and k = 10 > 2) 28 (32) 36 k-1 = 28 and x=32 and k = 36 > 3) 66 (72) 78 k-1 = 66 and x=72 and k = 78 > > 1) x-1 = 7 > 2) x-1 = 31 > 3) x-1 = 71 > etc.. > > What is interesting is when x-1 is a composite (x-1)== >0 (mod 3) and > also for (x-1)== >0 (mod 5). > I'm not sure I interpret the ==> symbol very well, but if you examine what the number x-1 is, you get (n+1)^2/2 - 1, so you should be able to work out the mod 3 and mod 5 bits since, for the n and n+1st triangular number to be both even, then n+1 is divisible by 4, call it 4s, x-1 is then 8s^2 - 1, which is -s^2 - 1 mod 3, which is never zero as -1 is not a quadratic residue mod 3. I'll leave you to work out the mod 5 case. > So far, no 0 (mod 3)'s or no 0 (mod 5)'s composites! > > Examining it a bit further -- > > I believe it is not possible because these (x-1)== >0 (mod 9) whether > (x-1) is prime or composite the consecutive (x-1)'s have a repeating > sequence of residuals (r) for --- > > (x-1) == r (mod 9)'s where consecutive (x-1)'s -- > > r =[7,4,8,1,1,8,4,7,8] [7,4,8,1,1,8,4,7,8] [7,4... etc. > > This repeating sequence is almost a palindrome. ;-) > > Dan Not sure I either understand what you're saying here, or what's going on. Will ponder some more. ==== > Do these primes have a name [7,31,71,127...]? Where k = n*(n+1)/2 Where x falls exactly between two consecutive EVEN numbered k's and if > (x-1) is a prime then this list below are those primes. 7,31,71,127,199,647,967,1151,1567,2311,2591... See Primes of the form 8*n^2 -1 at (Interestingly, that entry was created only a few days ago.) David > Eg. Of first few primes (x-1) and k-1 and k 1) 6 (8) 10 k-1 = 6 and x= 8 and k = 10 > 2) 28 (32) 36 k-1 = 28 and x=32 and k = 36 > 3) 66 (72) 78 k-1 = 66 and x=72 and k = 78 1) x-1 = 7 > 2) x-1 = 31 > 3) x-1 = 71 > etc.. What is interesting is when x-1 is a composite (x-1)== >0 (mod 3) and > also for (x-1)== >0 (mod 5). So far, no 0 (mod 3)'s or no 0 (mod 5)'s composites! Examining it a bit further -- I believe it is not possible because these (x-1)== >0 (mod 9) whether > (x-1) is prime or composite the consecutive (x-1)'s have a repeating > sequence of residuals (r) for --- (x-1) == r (mod 9)'s where consecutive (x-1)'s -- r =[7,4,8,1,1,8,4,7,8] [7,4,8,1,1,8,4,7,8] [7,4... etc. This repeating sequence is almost a palindrome. ;-) Dan ==== I studied graph theory for a semester using this book. Admittedly, my experience was somewhat less that stellar; and I found this book too difficult for me. A little background might be sufficient: I did not have any prior Intro. to Combinatorics course and I was an engineering student who happened to have an interest in mathematics. I did however find these books tremendously easier: Book 1. Combinatorics and Graph Theory ISBN: 0387987363 http://www.amazon.com/exec/obidos/tg/detail/-/0387987363/002-2811429-8891268 ?v=glance Book 2. Applied Combinatorics by Alan Tucker ISBN: 047143809X http://www.amazon.com/exec/obidos/tg/detail/-/047143809X/ref=lpr_g_1/002-281 1429-8891268?v=glance&s=books Learning graph theory from Prof. West's book was like being being pushed unwittingly into a cold river: you know that the able ones would survive and laugh but the unprepared would sink, in the end you vow never to play near water again. I think the prose was a bit terse and at times confusing. Some of the examples are harder than the material they serve to explicate, sometimes much effort is expended just to understand the complicating details of the examples. On the plus side, the proofs are very short and most of them elegant. No time is wasted in being chatty (contra Book 1)--like the methematical equivalent of Emeril, each discussion concludes with a Bam! and on to the next--but of course this leads to a feeling that graph theory is a slapdash of assorted subjects with no coherent whole. (I guess the colorful nomenclature of the subject deserves more lighthearted treatment). I am interested in learning what other people have in mind about the book. I am thinking of repurchasing it (sold it earlier). What book would you nominate to be the best introductory book with the right combination of rigor and friendliness--neither uncharted Pacific trench nor kiddies pool. ==== >I studied graph theory for a semester using this book. ... >and I found this book too difficult for me. A little background might be >sufficient: I did not have any prior Intro. to Combinatorics course >I think the prose was a bit terse and at times confusing. Agreed, but I think that style was a conscious design decision. The language is formed always in a ttempt to be correct and concise, and certain informal sounding English is used pointedly, e.g. he might use the foobar instead of the lengthier but not more precise there exists exactly 1 foobar. So if you're not used to it (and you will be after the book), the language only -seems- confusing. You need to put in that extra work to extract the unspoken .. er not unspoken but not immediate .. er it is immediate if only you were already precise about your use of language. Later on in the preface, he makes some remarks about intellectual discipline and honesty and use of language say what you mean and mean what you say. (and he practices what he preaches). >On the plus side, the proofs are very short and most of them elegant. as many classic results in combinatorics are. >this leads to a feeling that >graph theory is a slapdash of assorted subjects with no coherent >whole. >What book >would you nominate to be the best introductory book with the right >combination of rigor and friendliness--neither uncharted Pacific >trench nor kiddies pool. Hmm.. that I don't have too many opinions about. Possibly a text on algorithms with coverage of graphs? Mitch Harris ==== > I recall attending a lecture long ago where the instructor (James > Pittman at U.C. Berkeley, I think) presented an example of an oddity > that can occur if one requires only finite (as opposed to countable) > additivity. I have forgotten the what the example was. Could anyone > here provide one? > > Countable additivity lets us prove the laws of large numbers. Which is > a good thing, in most cases --- we really want the laws of large > numbers to be true, to agree with our intuition. And it's nice to have > them as theorems, instead of postulating them as axioms of probability > theory. > > On the other hand, sometimes we want to model a situation in which a > law of large numbers does NOT hold, and in those cases we have to pick > a probability measure that is not countably additive. The primary > example is the Gaussian probability on the infinite-dimensional > I am familiar with a rigorous treatment of the Gaussian probability embedded as a dense subspace of a larger vector space (e.g. the completion of H with respect to a suitable norm). But my work on this was decades ago. I see how you can get a finitely additive measure defined on sets S such that there exists a projection p of H onto a finite-dimensional space V and a measurable subset T of V, and S = p^-1 V. (Apologies for the limitations of ASCII math notation) But I don't see how to extend this measure to where you can integrate the things you want to integrate (except by the technique mentioned in the previous paragraph). I would welcome references. -- Chris Henrich It's our supreme ability and willingness to screw up that is the secret of our success. -- R. X. Cringely ==== > > I recall attending a lecture long ago where the instructor (James > Pittman at U.C. Berkeley, I think) presented an example of an oddity > that can occur if one requires only finite (as opposed to countable) > additivity. I have forgotten the what the example was. Could anyone > here provide one? > > Countable additivity lets us prove the laws of large numbers. Which is > a good thing, in most cases --- we really want the laws of large > numbers to be true, to agree with our intuition. And it's nice to have > them as theorems, instead of postulating them as axioms of probability > theory. > > On the other hand, sometimes we want to model a situation in which a > law of large numbers does NOT hold, and in those cases we have to pick > a probability measure that is not countably additive. The primary > example is the Gaussian probability on the infinite-dimensional > > I am familiar with a rigorous treatment of the Gaussian probability > embedded as a dense subspace of a larger vector space (e.g. the > completion of H with respect to a suitable norm). But my work on this > was decades ago. > > I see how you can get a finitely additive measure defined on sets S > such that there exists a projection p of H onto a finite-dimensional > space V and a measurable subset T of V, and S = p^-1 V. (Apologies for > the limitations of ASCII math notation) But I don't see how to extend > this measure to where you can integrate the things you want to > integrate (except by the technique mentioned in the previous > paragraph). > > I would welcome references. cylindrical measures. ==== A book I'm reading asserts If H and K are two subgroups of a group G, and if every element g of G can be uniquely expressed as g=xy, where x is in H and Y is in K, then H and K are both normal in G. I haven't been able to show this. I think I might be missing something really simple. Mike ==== > A book I'm reading asserts If H and K are two subgroups of a group G, > and if every element g of G can be uniquely expressed as g=xy, where x > is in H and Y is in K, then H and K are both normal in G. I haven't been able to show this. I think I might be missing something > really simple. Mike Consider the symmetric group on three elements, S_3. This contains a normal subgroup of order three which is isomorphic to the cyclic group of order 3, Z/3Z (i.e., {e, (1,2,3), (1,3,2)} ). Now, let K be a subgroup of S_3 consisting of two elements, the identity and a swap of two elements (e.g., {e, (1,2)(3)}). Now, any element of S_3 can be written uniquely as the product of an element of K and Z/3Z, but K is not normal. I probably haven't helped you much, as a counter-example generally makes it more difficult to find a proof :-). Best wishes, Mike ==== >A book I'm reading asserts If H and K are two subgroups of a group G, >and if every element g of G can be uniquely expressed as g=xy, where x >is in H and Y is in K, then H and K are both normal in G. Not true. Say G is the group of all maps f from R to R of the form f(t) = at + b where a <> 0 (with composition as the group operation.) Let H be the subgroup of all f of the form f(t) = t + b and let K be the set of all f of the form f(t) = at (a <> 0). Then it's clear that every element of G is equal to the composition of an element of H and an element of K in exactly one way. But K is not normal (because a(t+b) - b <> At.) >I haven't been able to show this. I think I might be missing something >really simple. For a second I thought maybe it was only supposed to be true for _finite_ groups, and you or the author left that out. But no, seems like the above example with a finite field in place of R gives a finite counterexample... >Mike ************************ David C. Ullrich ==== [snip] > I maintain that Magidin is indeed either incompetent as a > mathematician, or more likely a liar, or both. well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs in academic caste system. so whomever you are, may well be right on this one... > James Harris <31db93f.0312250011.16178abc@posting.google.com> <874qvotj7s.fsf@phiwumbda.org> <31db93f.0312262203.2bcbd157@posting.google.com> <188f56bf.0312281039.4ed0708@posting.google.com> ==== [snip] > I maintain that Magidin is indeed either incompetent as a >> mathematician, or more likely a liar, or both. well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs > in academic caste system. > Says the man (?) who repeatedly posts about how bad his employment prospects are and what a failure his own PhD has proved. So you're upset about your own bad experiences. Big deal. That's no excuse to denigrate Arturo's position and you have no reason to believe that he's unhappy with what he's accomplished. You are a sad and pathetic person. -- My proof has been checked very thoroughly, both by me and others. Those others apparently decided that they would not believe the proof was correct, but cannot support that position using mathematics. But hey, they're just human beings. --JSH, prover of Fermat's Last Thm ==== > [snip] > I maintain that Magidin is indeed either incompetent as a >> mathematician, or more likely a liar, or both. well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs > in academic caste system. Says the man (?) who repeatedly posts about how bad his employment > prospects are and what a failure his own PhD has proved. come again? are trying to deny that the title ADJUNCT is the lowest title an academic can hold? or that anyone holding such title is subject to abuse and contempt by practically the whole academic institution? or that the american higher education system is heavily using and abusing adjuncts? i hope you aren't shy to clarify whatever your point was. > So you're upset about your own bad experiences. Big deal. fyi, i do not hold a math PhD, but know plenty of people who do and happen to be adjuncts. their working conditions are abysmal. > That's no excuse to denigrate Arturo's position and you have > no reason to believe that he's unhappy with what he's accomplished. i very much doubt that arturo is not unhappy with his current position, anyone with an ounce of dignity would be unhappy. especially in light that about a quarter of the teaching personnel at his institution holding higher rank than him are complete mathematical nincombpoops compared to him. > You are a sad and pathetic person. beats the hell out of being a complete ignoramous imbecile, such as you. <31db93f.0312250011.16178abc@posting.google.com> <874qvotj7s.fsf@phiwumbda.org> <31db93f.0312262203.2bcbd157@posting.google.com> <188f56bf.0312281039.4ed0708@posting.google.com> <87y8swbnll.fsf@phiwumbda.org> <188f56bf.0312291220.6dbfebe6@posting.google.com> ==== > are trying to deny that the title ADJUNCT is the lowest title an > academic can hold? No, but I don't know that it's true, either. Depends on the institution, I reckon. > or that anyone holding such title is subject to abuse and contempt > by practically the whole academic institution? Well, I suppose I'll deny that, yes. It is a silly claim. > or that the american higher education system is heavily using and > abusing adjuncts? [sic]. This is nonsense. You used this claim to support the claim that Arturo is incompetent. This is just utter bullshit. > i hope you aren't shy to clarify whatever your point was. My point? You're a pathetic shithead. Sorry for failing to make that explicit. >> So you're upset about your own bad experiences. Big deal. fyi, i do not hold a math PhD, but know plenty of people who do and > happen to be adjuncts. their working conditions are abysmal. Sorry for the false claim. Your bitterness towards mathematics was, I assumed, due to the fact that you held a worthless PhD. Instead, you are a champion of the noble but fallen PhD's around you, ceaselessly working to overthrow the mathematical feudalism that has served to keep them chained. Bully for you. [...] [Snip bit in which Maky opines about how much more competent Arturo is, compared to his colleagues --- despite the fact that he insinuated Arture is incompetent due to his position.] >> You are a sad and pathetic person. beats the hell out of being a complete ignoramous imbecile, such as > you. Now, that hurts. You cut that out. -- I AM serious about this being a short route to a Ph.d for some of you, but just remember, I'm the guy who proved Fermat's Last Theorem in just a bit over 6 years [...] My standards are kind of high. --James Harris, founding a new mathematical school <31db93f.0312250011.16178abc@posting.google.com> <874qvotj7s.fsf@phiwumbda.org> ==== > Implementing the MSN passport, if I can believe the protests of the > people replying to me here, is incredible difficult. >>As usual, you do not understand the protests of others. Despite his protestations to the contrary, it semms unlikely this is > the real James Harris; I agree, and I said as much shortly after this post. I had overlooked both the discussion group name and also the yahoo.com address when I first replied. -- But in our enthusiasm, we could not resist a radical overhaul of the system, in which all of its major weaknesses have been exposed, analyzed, and replaced with new weaknesses. -- Bruce Leverett (presumably with apologies to Ambrose Bierce) ==== taking it in high school. I enjoy math as a hobby and was looking for a good book on calculus for this summer so that I can learn a little before college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson Calculus for Dummies Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems more like a textbook with problems. What aspects of calculus should a beginning book include and does anyone know if any of these, or any others, are any good? I am not looking for an ==== not > taking it in high school. I enjoy math as a hobby and was looking for a good > book on calculus for this summer so that I can learn a little before > college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson > Calculus for Dummies > Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems more like a > textbook with problems. What aspects of calculus should a beginning book include and does anyone > know if any of these, or any others, are any good? I am not looking for an Another book in that vein is Hurricane Calculus. You could also try Dover's website for some cheap Calc. books. However, you could probably pick up some cheap used editions of Spivak's Calculus, or Apostol's Calculus. They have a great mix of rigour and readability. Try to find them on http://www.bookfinder.com I'll bet you could find one for around $20. Lurch ==== > taking it in high school. I enjoy math as a hobby and was looking for a good > book on calculus for this summer so that I can learn a little before > college. A few books I saw at the bookstore were: > > Calculus Made Easy - Silvanus Philips Thompson > Calculus for Dummies > Complete Idiots Guide to Calculus > > Another interesting book is Calculus: The Elements, which seems more like a > textbook with problems. > > What aspects of calculus should a beginning book include and does anyone > know if any of these, or any others, are any good? I am not looking for an Teach Yourself Calculus by P. Abbott and H. Neill, $12.95, $10.36 from Amazon, was the book I used. Cheap, easily carried around, unlike most other calculus books, and a Teach Yourself... book, so it's geared toward the self-taught, not to state the obvious, or anything. Good luck. ==== not > taking it in high school. I enjoy math as a hobby and was looking for a good > book on calculus for this summer so that I can learn a little before > college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson > Calculus for Dummies > Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems more like a > textbook with problems. What aspects of calculus should a beginning book include and does anyone > know if any of these, or any others, are any good? I am not looking for an Calculus the easy way was pretty fun. IEEE the Calculus tutoring book would be a good one for your level. ==== regret not > taking it in high school. I enjoy math as a hobby and was looking for a good > book on calculus for this summer so that I can learn a little before > college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson > Calculus for Dummies > Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems more like a > textbook with problems. What aspects of calculus should a beginning book include and does anyone > know if any of these, or any others, are any good? I am not looking for an http://tinyurl.com/2awjy This will give you the basic proofs behind differentiation and integration and an understanding of functions, graphing, limits etc which are all necessary fundamentals. It is a first year college calculus book (M124, M125 + some) with plenty of problems to work. It is an older edition but it doesn't get any less expensive than this. Phil Holman ==== Suppose we have a labelled graph G with vertices {1,2,3,...,n} and an n-vector c of positive integers {c_1,c_2,c_3,...,c_n). Produce an n-partite graph H by creating (for each i in {1,2,3,...,n}) a set T_i containing c_i isolated vertices, and then joining every vertex in T_i to every vertex in T_j whenever ij is an edge in G. So, each vertex i in G corresponds to a set T_i in H, and each edge ij in G coresponds to a collection of (c_i x c_j) edges in H. What name is given to this type of structure? TIA SS ==== PS I will review Physics Meets Philosophy at the Planck Scale. There seems to be a misconception about D Branes starting in Hawking's new popular book The Universe in a Nutshell (O Brane New Worlds) and also pictures in Scientific American showing gravitons (closed strings) moving through all of hyperspace with lepto-quarks and photons (open strings) stuck to D Branes pictured as parallel universes as describes a D Brane seems incompatible with that picture? Also Witten seems to think his alpha' = 1/(string tension) ~ (10^-32 cm)^2 is a new independent fundamental constant (a moduls or compactification scale) rather than Lp^2 = hG(Newton)/c^3 = hc(alpha'alpha) = alpha'e^2 hG/e^2c^3 = alpha' Think first in quantum field theory of a virtual electron-positron pair vacuum bubble, i.e. closed world line with no free ends. Then imagine one virtual photon bisecting the the closed world line. This is the basic virtual bound state into which a huge number of these vacuum bubbles Bose-Einstein condense to form the vacuum coherence field whose Goldstone phase ripples form Einstein's guv c-number ODLRO field. dark energy and dark matter exotic vacuum w = -1 phases that is 96% of the large-scale stuff of the world. For strings imagine a closed strip - two-sided orientable - maybe also Mobius one-sided strips. The smallest quantized area of these vacuum bubbles is obviously hcalpha' in Witten's sense. The single virtual photon gives a factor of alpha = e^2/hc since each fuzzed out vertex is (alpha)^1/2 i.e. basic diagram is <|> | is the virtual photon and <> is the virtual electron-positron vacuum loop or bubble. Therefore, this is a simple Mickey Mouse derivation of Andrei Sakharov's emergence of gravity G from QED and Witten's basic modulus or string tension parameter, i.e. G = e^2c^3alpha'/h = c^4(alpha)(alpha') So that to be more precise Einstein's GR equation is Guv = (8piG/c^4)Tuv 8piGh/hc^4 = 8piLp^2/hc = (alpha)(alpha') Guv = (alpha)(alpha')Tuv alpha' is Witten's modulus. Dear Friends : Recently there has been a lot of work dedicated to Fisher information theory, entropy.... see the book by Frieden, Cambridge University Press : Physics from Fisher Information that claims/suggests to derive all of physics from Fisher information theory. If you look at the literature about QM and Fisher you will see that there is a Fisher information term in the Schroedinger equation that is related to Bohm's quantum potential = Weyl scalar curvature induced by the ensemble density this out to my amazement ! So yes, information theory, entropy and Weyl-Bohm have a lot in common. Call it Information Geometry which is very popular today. also Dear Jack : The title of Christian Beck's book is : Spatio-temporal vacuum fluctuations of quantized fields... World Scientific 2002. Nonlinear sciences. archives. You can download them. Best wishes Carlos ____________ ==== I have to solve the diophantine equation x*(p+A*y)=B (unknown x,y; all factors of B are known) i am able to resume that to x*(1+A*y1)=B1 :-) is there in some place a way to solve this equation without testing all possible combinaisons of sub factors of B1 ? Any kind of pointers will be appreciated. BR /Philippe ==== >I have to solve the diophantine equation x*(p+A*y)=B (unknown x,y; all >factors of B are known) >i am able to resume that to x*(1+A*y1)=B1 :-) I don't know what resume means for you in this context, but let's say that's the question... I don't suppose x=B1, y1=0 is allowed? >is there in some place a way to solve this equation without testing all >possible combinaisons of sub factors of B1 ? Well, you're looking for factors of B1 that are == 1 mod A. If B1 = product_{j=1}^N p_j^{n_j}, a factor of B1 is product_j p_j^{m_j} where 0 <= m_j <= n_j for all j. One way to do this (which might be useful if A is not too large) is with dynamic programming. First, discard any p_j that divides A. Let B(0) = 1, and for k from 1 to N let B(k) = product_{j=1}^k p_j^{n_j}. For k from 0 to N and a coprime to A, let x(k,a) be a factor of B(k) that is == a mod A if such exists, otherwise 0. Begin with x(0,1) = 1, x(0,a) = 0 otherwise. Let o_k be the order of p_k mod A. Then for a in {1,2,...,A-1} and coprime to A, x(k, a) = max_{j=0}^{min(n_k, o_k)} x(k-1, (a p_k^(-j)) mod A) p_k^j Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== I have a set of Bezier curves. Does anyone know a way that I can >order them Well, you can try Amazon.com ; if you can't order them there, look on Ebay. ==== > I have a set of Bezier curves. Does anyone know > a way that I can order them Well, you can try Amazon.com; > if you can't order them there, look on Ebay. In Pythagoras confusion we discuss a similar question and it would be nice to see another helpful comment ;-) Rainer Rosenthal r.rosenthal@web.de ==== > > I have a set of Bezier curves. Does anyone know a way that I can > order them, so that two curves which are ordinally (right word?) > near each other will be similar? > > Darren Grant Similar is a pretty fuzzy term. Let's assume that the starting and ending points of all our curves are identical. Let F(A,B) be the area between the two curves A and B. F(A,A) = 0; and if F(A,B) = 0, then A = B. Then we can make the definition that B is more similar to A than C is to A iff F(A,B) < F(A,C). However, this only provides us with a metric; not an order. It seems you want a function G(A) that returns, for example, a real number; with the property that, for all curves A, B, C, if |G(A) - G(B)| < |G(A) - G(C)|, then F(A,B) < F(A,C). This doesn't seem possible unless you have a very restricted set of curves. Consider the comparable problem in the plane, where you would like a function G(p) taking each point p in the plane to some real number. With a little inspection, I think you can see that, if d(p,q) is the distance between two points, there is no G such that for all points p,q,r, |G(p) - G(q)| < |G(p) - G(r)| implies that d(p,q) < d(p,r). ==== >> I have a set of Bezier curves. Does anyone know a way that I can >> order them, so that two curves which are ordinally (right word?) >> near each other will be similar? >> Darren Grant Similar is a pretty fuzzy term. Let's assume that the starting and > ending points of all our curves are identical. Let F(A,B) be the area > between the two curves A and B. F(A,A) = 0; and if F(A,B) = 0, then A > = B. Then we can make the definition that B is more similar to A than C is > to A iff F(A,B) < F(A,C). However, this only provides us with a metric; not an order. It seems > you want a function G(A) that returns, for example, a real number; > with the property that, for all curves A, B, C, if |G(A) - G(B)| < > |G(A) - G(C)|, then F(A,B) < F(A,C). This doesn't seem possible unless you have a very restricted set of > curves. When placed on a coordinate system, the curves all start on at x = 0, and all end at y = y_max. Is this restriced enough by any chance? ;) > Consider the comparable problem in the plane, where you would like a > function G(p) taking each point p in the plane to some real number. > With a little inspection, I think you can see that, if d(p,q) is the > distance between two points, there is no G such that for all points > p,q,r, |G(p) - G(q)| < |G(p) - G(r)| implies that d(p,q) < d(p,r). > on this. I'm not a mathematician (I'm working on it, though) and I don't entirely follow your last paragraph, but I understand that it is probably not possible to attribute an actual value to a curve, so I am trying to implement an algorithm that works as follows: Given a list of curves, a new curve is added to the list by inserting it after the curve which it is most similar to. When placed on a coordinate system, the curves all start on at x = 0, and all end at y = y_max, so I am comparing curves for similarity by building a difference curve - curve a minus curve b (for each x value, I subtract y of b(x) from y of a(x)). Then the summation of the y values for the difference curve gives the metric I use to compare the two curves with. I nearly got this working, then I realised there is a problem. My curve are Bezier curves, and my function returns a Bezier curve as a series of discrete vectors (or points?) - so two curves might look like this: a = (0,0), (3.6,3.2), (6.4,4.8), (8.4,4.8), (9.6,3.2), (10,0) b = (0,10), (2,9.6), (4,8.4), (6,6.4), (8,3.6), (10,0) Obviously the x coordinates differ. I guess there are two possibilites open to me; I can use a line-drawing algorithm to render the curve, so I can use fixed x coordinates for the y-value subtraction, or I could use the control points which the Bezier curve is formed out of to calculate the difference between two curves. If the curves are all formed from, say, three control points, can anyone show me how I can compare the similarity of two curves using these points? Darren ==== > > >> I have a set of Bezier curves. Does anyone know a way that I can >> order them, so that two curves which are ordinally (right word?) >> near each other will be similar? >> Darren Grant Similar is a pretty fuzzy term. Let's assume that the starting and > ending points of all our curves are identical. Let F(A,B) be the area > between the two curves A and B. F(A,A) = 0; and if F(A,B) = 0, then A > = B. Then we can make the definition that B is more similar to A than C is > to A iff F(A,B) < F(A,C). However, this only provides us with a metric; not an order. ... after the curve which it is most similar to. > If the curves are all formed from, say, three control points, can anyone > show me how I can compare the similarity of two curves using these points? > Certainly, you could use a metric between two curves which was the sum of the squares of the distances (in the x/y plane) between the coresponding control points of the two curves. But I guess my question would be, what is the goal of your ... inserting it after... most similar... sorting method? Let's suppose each curve is identified with a single control point, (x,y). If you wish to create an order that will allow, for example, a binary tree search for closest curve from some set, in the same fashion that one would use for looking up a string in a dictionary, I think it's not possible. On the other hand, there are algorithms like those used for color table lookup mapping that involve similar problems. In color table lookup, you have a canonical set of, say, 256 (r,g,b) values; and you want a fast algorithm for taking any random rgb value and quickly finding the closest color in the table to that value. Closest is usually defined as raw color distance (dr^2+dg^2+db^2). That sounds a bit like your problem - I presume you ultimately want to, given a random curve, find the closest curve in some fixed set of curves? ==== > ... > Players, on the k_th move, place the integer (2k-1) {for odd-player} or (2k) > {for even-player} at any integer-position on the board which has yet to have > an integer placed upon it. > > Could you expand on that? At first read, it seems to me that on turn one 1 > and 2 are played, on turn two 3 and 4, and so on. > > That doesn't seem like enough freedom for an interesting game, so my next > thought was that the three k's are all different-- that the odd player > (that's me) plays an odd number and the even player plays an even number. My original rules have 1 and 2 played on first move, 3 and 4 played on the 2nd move, as you apparently assumed here. So, there is no choice what integer is played on a particular move by any one player. But allowing any integer to be played as long as it: 1) is in the range 1 to n, for some n (do not conflict with rule-3); 2) is always even for a player, is always odd for the other; 3) has not been used yet; might be more fun. Hmmm... Perhaps the integers can even be picked at random using a deck of cards. (Perhaps we should allow integer re-use here, if standard deck is used.) > > But then this statement is no clear to me: > It is okay for two integers to be at the same position IF the > integers' positions were both chosen on the same move. > > For example, if 13 and 4 are played, do we use up board spots 6 and 2, only > using up the same spot if for example 12 and 12 were played? > > Or am I just not getting it at all? > I do not believe I have made things clear here. Let us use letters to represent the positions so as to help avoid confusion. If position-c, say, is already occupied when a player wants to place a 4 down on the board, the 4 cannot be put at c. So, the player puts 4 at d, which is empty. On the *same* move, the other player had independently chosen to put the 3 at d as well. So, this is fine. Now, let us say that c has a 2 and e has 1, and the value of d is now (3+4) =7. So, with [...2, 7, 1,...] both players get a point here because there is one descent and one ascent among these 3 positions. Note: I should also point out that it is still fine for a player to play the 7, despite that 7 = 3 + 4. (So, it might be that some adjacent integers neither give a point to one player nor to the other.) thanks, Leroy Quet ==== > ... > Players, on the k_th move, place the integer (2k-1) {for odd-player} or (2k) > {for even-player} at any integer-position on the board which has yet to have > an integer placed upon it. and I asked: >> Could you expand on that? At first read, it seems to me that on turn one 1 >> and 2 are played, on turn two 3 and 4, and so on. >> >> That doesn't seem like enough freedom for an interesting game, so my next >> thought was that the three k's are all different-- that the odd player >> (that's me) plays an odd number and the even player plays an even number. to which Leroy replied: > My original rules have 1 and 2 played on first move, 3 and 4 played on > the 2nd move, as you apparently assumed here. So, there is no choice > what integer is played on a particular move by any one player. Rereading all that, I see what I missed was that when you play a number, you can play it on any empty spot. For some reason I assumed that pieces 2k-1 and 2k had to be played at board location k. > Let us use letters to represent the positions so as to help avoid > confusion. ... And I think that was why I was originally confused-- the board was numbered and the pieces were numbered. Letters made it a lot clearer to me what you have in mind. Unfortunately, I don't have anything profound to say about the game though. Bob H ==== Einstein's GR is emergent as a c-number ODLRO field out of a BCS QED vacuum instability that forms the inflation field. That is, direct quantization of Einstein's guv field is not appropriate anymore than re-quantizing the giant superfluid and superconducting waves is appropriate. Therefore quantum foam is suspect. Also Bohm's quantum potential view of vacuum fluctuations is relevant in context of the recent paper from Teheran. I need to follow the experiment below more carefully. Jack better check this story out: Gary S. Bekkum ==== I could be wrong, but I suspect that no one on the entire Do you bundle your letters and send them to random people, too? -E ==== > Einstein's GR is emergent as a c-number ODLRO field out of > a BCS QED vacuum instability that forms the inflation field. > That is, direct quantization of Einstein's guv field is not > appropriate anymore than re-quantizing the giant superfluid and > superconducting waves is appropriate. I admit that I don't have an advanced degree in physics, but I do know horseshit when I see it. Do you have an obsessive-compulsive disorder that forces you to write, even if what you write is nothing but gibberish and buzzwords? That is my suspicion. ==== > If we have, for even m, > > m/2 black squares of > 2*2 > 4*4 > 6*6 > 8*8 > ... > m*m > > and m/2 black rectangles of > 1*(m-1) > 3*(m-3) > 5*(m-5) > 7*(m-7) > .. > (m-1)*1 > > and we have > > m/2 white squares of > 1*1 > 3*3 > 5*5 > 7*7 > ... > (m-1)*(m-1) > > and m/2 white rectangles of > 2*m > 4*(m-2) > 6*(m-4) > 8*(m-6) > .. > m*2 > > then we have the same total area for all of the white > squares/rectangles as for the black squares/rectangles (which is > easily shown using algebra). > > My question is, which I have not been able to do yet myself (though I > have not tried really hard), > > can all of the squares/rectangles be arranged in some way so that it > becomes visually obvious that the total areas between the black and > white are the same for every even m? > > ( This is similar to the Proof Without Words in the MAA's > Mathematics Magazine.) > > thanks, > Leroy Quet I am surprised that no one has replied to this question yet. :( It may be easier to use the 3rd dimension and arrange boxes having the above dimensions plus each a depth of 1. Is there any obvious 2-d or 3-d arrangements?? thanks, Leroy Quet ==== > In > http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fdf. 0301191933.79c98f04%40posting.google.com&rnum=1&prev= > I mention a game based on the following idea: > > Player 1 finds an integer m by taking a permutation {b(j)} of the > first n positive integers, > > then forms m: > > m = sum{k=1 to n} k * b(k). > > > Player 2 then, in some time limit, tries to find ANY permutation of > the first n positive integers which also sums to m. > > > Players take turns making up puzzles and trying to solve other > player's puzzles. > > ( It is advantageous for the proposer to come up with m's with > relatively few solutions, but are not too easy {easy, such as b(k) = k > or = (n+1-k), which each are the only solutions to their sum m, but > are easy to solve}.) > > 2 things: > > 1) One can play a solitaire version of this game, where the > permutation (of 1 through 10, for example) is picked by shuffling > cards, and the player tries to come up with a solution which is > a) a derrangement of the cards' permutation; > b) is not the inverse perm. of the card-perm.. > > > 2) How can we determine the number of permutations (for a given n) > which have a given sum m? > > thanks, > Leroy > Quet If someone is up for this, perhaps you could brute-force computer-check for an example of an m derived from a sum having a significantly interesting n (such as between 8 and 16) and very few permutation-solutions relatively. (Where the m is not the minimum/maximum sum which can be obtained, for these are trivial.) And then you might post this m (and n) here as a puzzle, to have us try to find a permutation {b(k)} of 1 through n, where: m = sum{k=1 to n} k * b(k). (I wonder if there are *always* multiple solutions for each m near the center of the range of possible sums.) thanks, Leroy Quet ==== > In > http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fdf. 0301191933.79c98f04%40posting.google.com&rnum=1&prev= > I mention a game based on the following idea: > > Player 1 finds an integer m by taking a permutation {b(j)} of the > first n positive integers, > > then forms m: > > m = sum{k=1 to n} k * b(k). > > > Player 2 then, in some time limit, tries to find ANY permutation of > the first n positive integers which also sums to m. > > > Players take turns making up puzzles and trying to solve other > player's puzzles. > > ( It is advantageous for the proposer to come up with m's with > relatively few solutions, but are not too easy {easy, such as b(k) = k > or = (n+1-k), which each are the only solutions to their sum m, but > are easy to solve}.) > > 2 things: > > 1) One can play a solitaire version of this game, where the > permutation (of 1 through 10, for example) is picked by shuffling > cards, and the player tries to come up with a solution which is > a) a derrangement of the cards' permutation; > b) is not the inverse perm. of the card-perm.. > (a) and (b) can be much more exactly rewitten as what I intended and in a way much more simple to understand, especially more simply understood among non-math people. If the order of a card is k, and the card is b(k), we have the multiplied integer pair [k*b(k)] in the sum m. So, while playing the solitaire version of the game, for every k <= number of cards, a player should not have either b(k) for the k_th integer, nor have k for the b(k)_th integer anywhere in the player's sum. (Imagining the [k*b(k)] terms as k-by-b(k) rectangles, a player should not have any of the same rectangles, in either orientation, as generated by the card-draw.) > 2) How can we determine the number of permutations (for a given n) > which have a given sum m? > > thanks, > Leroy > Quet > > > If someone is up for this, perhaps you could brute-force > computer-check for an example of an m derived from a sum having a > significantly interesting n (such as between 8 and 16) and very few > permutation-solutions relatively. > (Where the m is not the minimum/maximum sum which can be obtained, for > these are trivial.) > > And then you might post this m (and n) here as a puzzle, to have us > try to find a permutation {b(k)} of 1 through n, > > where: > > m = sum{k=1 to n} k * b(k). > > (I wonder if there are *always* multiple solutions for each m near the > center of the range of possible sums.) > > thanks, > Leroy > Quet ==== Dear all, In one of my program, I need to compute the variance of a 8x8 data matrix as fast as possible... Any fast algorithm with lowest complexity? By the way, since what I actually need is an indication of signal local activity, is there any other measure which can also be a indicator of local activity and is less complex than computing variance? -Walala ====