>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) <^> ----------------------------- > 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