mm-63 === >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>af?e, 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) + bthen K(x) - bx/a is periodic with period a. Thus, K is the sum of alinear function and a periodic function. This con?ms the statementthat K can be anything on [0,a), but K on that interval plus one morepoint 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 excerptthat follows. All excerpts are from The Blackwell Guide toPhilosophical Logic.1)This shows that even a relatively simple theory likeequality is incomparably richer than the classical theory.2)The theory of equality has the usual axioms: re?ty,symmetry, transitivity. One can strengthen the theory inmany ways, for example, the theory of stable equality isgiven byEQ^st = EQ + Axy(--x=y -> x=y)And the decidable theory of equality is axiomatized byEQ^dec = EQ + Axy(x=y / -x=y)[Aside: In deference to John Correy's treatment of identity andnon-self-identicals, I would note that treating this as an inclusivedisjunction admits the possibility of a superimposed interpretation.]3)Intuitionistic predicate logic itself is, just like itsclassical counterpart, undecidable. Fragments are,however, decidable. For example, the class ofprenex formulas is decidable, and, as a corollary,not every formula has a prenex normal form inIQC. On the other hand, although monadic predicatelogic is decidable in classical logic, Kripke showedit is undecidable in intuitionistic logic; see alsoOrekov et al. Likewise, Lifschitz showed that thetheory of equality is undecidable.The following is an excerpt about apartness. With respect to thestatements above, one thing to note is that stable equality is obtainedunder the apartness axioms.In intuitionistic mathematics, there is also a strongnotion of inequality: apartness, #, as mentioned above.This was introduced by Brouwer (1919) and axiomatizedby Heyting (1925). The axioms of AP are given byEQ and the following listAxyx'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 thatAP 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-theoreticjusti?ations that invoke no such con?ith intuitionisticprinciples.---Let AP |- A / B and assume ~(AP |- A) and~(AP |- B) [?~' is classical exclusion negation].Then, by the strong completeness theorem, thereare models K_1 and K_2 of AP such that~(K_1 ||- A) and ~(K_2 ||- B). Consider thedisjoint union of K_1 and K_2 and place theone-point world k_0 below it. That is to say,designate points _a_ and _b_ in K_1 and K_2which are identi?d with the point k_0. The newmodel obviously satis?s the axioms of AP.Hence k_0 ||- A / B and so k_0 ||- A ork_0 ||- B. Both are impossible on the grounds ofthe choice of K_1 and K_2. Contradiction.Hence AP |- A or AP |- B.For EP, it is convenient to assume that the theoryhas 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 canbe glued by means of a bottom world K^* witha domain consisting of just the elements c_i. Nonon-trivial atoms are forced in k_0 (i.e., only thetrivial identities c_i = c_i). The identi?ation of thec_j with elements in the models K_i is obvious.Again, it is easy to check that the new model satis?sAP. Hence k_0 ||- ExA(x), i.e., k_0 ||- A(c_i) forsome _i_. But, then also, K_i ||- A(c_i), contradiction.Hence AP |- A(c_i) for some _i_.The gluing operation thus demonstrates that thereare interesting operations in Kripke model theorythat make no sense in traditional model theory.The apartness axioms have consequences for theequality relations. In particular, stable equality isobtained:--x=y -> x=yFor,-x#y <-> x=ysox=y <-> ---x#y <-> --x=yIndeed, the equality fragment itself is axiomatizedby an in?ite 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,' formulatequasi-stability axioms:S_n =df Axy(-( x [=_n] y) -> x=y)The S_n axiomatize the equality fragment of AP. Tobe precise: AP is conservative overEQ + {S_n | n >= 0}This shows that even a relatively simple theory likeequality 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#yOne can also usex 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 de?ed to meanthe existence of a positive integer n such that |r-s|>1/n. Theequality r=s is equivalent to the negative of that, |r-s|<1/nfor every n. The negation, ~r=s, is in a sense very close toequivalent to r#s. If you prove ~r=s for speci? r and s, there'sa metatheorem which guarantees you can prove r#s too. Markov's ruleentails ~r=s -> r#s, and the Markov school used this (although Iguess 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 sprecisely enough to ?d the level of precision where they differ,to determine that r#s.Nevertheless, it's considered inappropriate in standard intuitionismshows that the nonexistence of an n for which |r-s|>1/n leads to acontradiction; it isn't taken as exhibiting the n.Mainly, I would say it's just better not to muck about with weirdlittle statements like ~r=s anyway. It's rare enough that I want tostate the negation of an equation like ~r=s that in my own notes Ijust write the usual inequality symbol r =/= s instead of r#s forapartness, 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 axiomsof this de?ition except with Axy(~x#y <-> x=y) replaced byAxy(x=y->~x#y) (or in other words, Ax(-x#x)), and don't necessarilysatisfy Axy(~x#y->x=y).I'm not sure whether the condition Axy(~x#y->x=y) really should beincluded in the de?ition of apartness, actually. All it serves isto tie equality to apartness, basically by de?ing it to be thenegation of apartness. The ?st axiom you list (which would usuallybe considered to follow just by the rules of equality) follows fromthe 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, theneither x'#y' or x#x' or y#y'. But by assumption, x=x' and y=y', soneither x#x' nor y#y', hence x'#y'.Here's an example of a case where one has a natural # relation thatsatis?s my weaker de?ition. I don't remember who discovered it?st, but I independently rediscovered it later. De?e multisetsof a set S to be equivalence classes of functions from other sets Tunder the equivalence that f1:T1->S and f2:T2->S are equivalent ifthere's a one-to-one correspondence g:T1->T2 such that f1 = f2 o g.Call a multiset a ?ite multiset if the domain is ?ite.De?ing multisets as equivalence classes amounts to de?ing theequality relation for them. But is there an apartness relation for?ite multisets of a set S which itself has an apartness relation?The ?st discoverer was using the de?ition above, and stated hisresult as being that there is no apartness relation in general.But for ?ite multisets of a set with apartness, if we weaken thede?ition as I just described, there is one.As a preliminary step, it might help to prove a simple lemmageneralizing the last axiom above to ?ite multisets. If I haveelements x1,...,xn and y1,...,ym of a set with apartness, and ifx_i # y_j for every i=1,...,n and j=1,...,m, and z is some furtherelement, 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 ?ite number of times, using the fact that bythe last axiom for each i and j, either x_i#z or z#y_j.[By the way, the ?ite distributive law (A or B1) and (A or B2)and ... and (A or Bn) <-> A or (B1 and B2 and ... and Bn) isconstructive. The in?ite version {for every i>=0 (A or B_i)}<-> {A or for every i>=0 B_i} is not. In the ?ite case, we caneffectively get for each i=1,...,n the fact that A is true, or thefact that B_i is true, and when we're done, we've either found thatA is true, or for each one we've found that B_i is true. But that'snot an effective procedure in the in?ite case.]So if we have x1,...,xn and y1,...,ym which are collectively apartfrom each other, any new z is apart from all of one of them, whichmeans it can be appended to the other one while preserving thesituation. By induction, if we have z1,...,zk, then there's apartition of z1,...,zk so that the x's together with one of thepartitions are collectively apart from the y's together with theother of the partitions.The case of n=m=1 is maybe the most interesting. If among someelements u1,...,ur in a set with apartness we ?d some apartnessrelation u1#u2, then we can partition the u1,...,ur into two, sothat u1 is in one partition, u2 is in the other one, and the elementsof one partition are all apart from the elements of the otherpartition.Okay, now back to ?ite multisets. If two multisets have differentnumbers of elements, then they are of course distinct, so we justneed to consider ?ite multisets de?ed by functions from a ?ed{1,...,n} to S. Now let me motivate my # relation. Given twomultisets {x1,...,xn} and {y1,...,yn}, in order to have a distinctionbetween them, it's only natural that we should have a distinctionbetween some two of the elements. Hence there exists some way ofpartitioning x1,...,xn,y1,...,yn into submultisets that arecollectively apart from each other. I de?e {x1,...,xn}#{y1,...,yn}to mean that we can partition them in such a way that one of thepartitions has a different number of elements from the x's as fromthe y's. Equivalently, I de?e it to mean that there are subsetsi_1,...,i_k and j_1,...,j_l of {1,...,n} where k+l>n, with theproperty that for every s=1,...,k and t=1,...,l, we havex_i_s#y_j_l. (k+l=n+1 is enough.)For example, an unordered pair {x1,x2} is apart from anotherunordered 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 satis?s all the axioms except ~x#y -> x=y. It'sobviously symmetric and preserved under substitution of equals. Iftwo multisets are equal, they are not #. If {x1,...,xn}#{y1,...,yn}and {z1,...,zn} is some other multiset, then the apartness relationsbetween x_i_1,x_i_2,...,x_i_k and y_j_1,...,y_j_l can be extended bythe lemma to include the z1,...,zn. Now either enough z's are putwith those x's to make {z1,...,zn}#{y1,...,yn}, or enough z's areput with those y's to make {x1,...,xn}#{z1,...,zn}. If >n-l of thez's are put on the left, that's enough to make the former true. Ifnot, 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 toequality for multisets. Equality is really a pretty strict relation;one has to be able to say which element corresponds to which. Ingeneral, for unordered pairs of real numbers, {x,-x} cannot be #{|x|,-|x|}. But in order to prove that {x,-x}={|x|,-|x|}, we haveto have either that x=|x| and -x=-|x|, or that x=-|x| and -x=|x|.The ?st case is equivalent to x>=0 and the second case isequivalent to x<=0. (For constructive purposes one de?es x>=yfor real numbers x and y to mean that in the sequence of rationalapproximations to x and to y, the approximation to within 1/n of xis >= the approximation to within 1/n of y, minus 2/n. There isn'ta constructive equivalence between x>=0 and x>0 or x=0. There isan equivalence between x>=0 and not x<0.) That x>=0 or x<=0 foreach real number x is nonconstructive.That x>=0 or x<=0 for each real number is equivalent to the statementthat if s0,s1,... is a sequences of bits, i.e. elements of {0,1},then either it isn't the case that the ?st 1 is at an odd position,or it isn't the case that the ?st 1 is at an even position.Nonconstructively speaking, either (a) all the sequence s is 0, or(b) the ?st 1 occurs at an even index, or (c) the ?st 1 occurs atan odd index. In cases (a) and (c), one can say that for each even nsuch 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|justi?ations that invoke no such con?ith intuitionistic|principles.|---I would approach the use of Kripke models with some caution. Kripkemodel theory and topos theory are ways for people to keep assumingthe law of excluded middle, but simulate to some degree what it'slike not to assume it.So I'd take seriously all the distinctions between differenttypes of validity until I knew that there was a known equivalence.I suppose the sentences valid in one-node Kripke models are the sameas the ones that are just plain valid. Usually, though, peopleworking with Kripke model theory are assuming the law of excludedmiddle, which implies that all the sentences implied by it are validin one-node Kripke models. There are Kripke models in which thelaw of excluded middle is de?itely not valid, but only someconstructivists actually assume that the law of excluded middle isnot valid, as opposed to failing to assume that it is valid.I saw a proof assuming the law of excluded middle that for eachstatement S in second order logic, there's a corresponding?st-order statement S' such that S holds in all models if and onlyif S' holds in all Kripke models. So the set of statements valid forall Kripke models is somewhat complex.The situation is complicated by the fact that the moststraightforward ways of asserting a completeness theorem areincorrect for intuitionist logic. There's a formal system, yes, andit's got an associated recursively enumerable set of theorems. Butthe classical proof I mentioned in the last paragraph shows that theset of ?st-order sentences valid in all Kripke models is way toologically complex to be recursively enumerable. I've seen mentionedcertain completeness theorems, but I wouldn't say the situationseemed 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=yYes, stability is just the same as saying that equality is negative,i.e. that equality is equivalent to the negation of something. Idon't know what good that is. The thing it's a negation of doesn'thave to be an apartness, and as far as I know there's no guaranteethat 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 negationswith ~ instead of -.)First, a simple lemma. Any statement implies its double negation:X->~~X. If we assume X, assuming ~X leads to a contradiction, fromwhich we can infer ~~X.Triple negation is the same as single negation (a theorem ofBrouwer). If ~X, then by the lemma applied to ~X, we get ~~~X too.together with ~~~X that's a contradiction. Hence, on the assumptionof ~~~X, we get ~X.A statement X satisfying the stability condition ~~X->X is equivalentto ~~X because X->~~X is automatic. So if it satis?s 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.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === at 09:29 PM, maneesh@drunkenbastards.com (maneesh) said:>To elaborate, I often found it confusing to associate particular>properties, algebraic objects and other such things with the name>they are given.Learning a new language is always confusing.>Surely it only adds a level of complexity to a ?ld of study which>does not need one!Surely not. The problem is that the concept is new, not that there issomething wrong with the name.>What are your opinions on having things called Schurian, Jacobian,>or Gaussian?It's shorter than descriptive names, and no less enlightening. >Do you feel that names of things in mathematics should greater>re?nformation about the said things? Should the word red be red? Names are arbitrary. >Do you feel it retards mathematical progress (on a personal, or >collective level)? No.>What is the worst named object you can think of?Ideal.None of what you have written addresses the real problems inMathematical nomenclature; the large number of synonyms and theinconsistency in the notation.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === ----- <^> <()> <^> ----->What is the worst named object you can think of?> Ideal.>why ideal?I would say ?abstract' has the most number of meanings in mathematics.function abstraction, outline, de?e proposition, approximate, ....Herc X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$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 anysuf?iently large text the odds are astronomically in favor of suchcoincidences. It's just a question of picking out the ones thatsupport 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 botherproducing more examples if you're just going to say That one doesn'tcount - 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 ofGemmatria, which is basically what you are doing. It's not Science andit's certainly not Mathematics.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot 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 timesFISHERY and WORK both occur 65 timesWHALEMEN and MYSELF both occur 69 timesFIRE and HARPOON both occur 76 timesMIND and SOUL both occur 80 timesCERTAIN, GOING, LEG and DEATH all occur 89 timesBECAUSE, LEVIATHAN and DEAD all occur 92 timesIf that doesn't convince you that Moby Dick was divinely inspired,I suppose nothing will.Robert Israel israel@math.ubc.caDepartment 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.LOLDouble-LOLDouble-LOL, squared.(BTW: The Pythagoras confusion waits for some enlightenment)Rainer Rosenthalr.rosenthal@web.de === I have a problem. I don't see how the author ?ds the closed formexpression for the polynomials P_n(lambda) in the book by Akhiezer --'The Classical Moment Problem' at page 3. I have scanned the ?st 5pages (although one only needs 1,2,3 to get the backgroundinformation):http://hem.bredband.net/lukhor/moment/ Happy new year! === > I have a problem. I don't see how the author ?ds 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 ?st 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 pagesin 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 de?ed by a sequence of numberss_0, s_1, ..., de?ed by S[R] = s_0 p_0 + s_1 p_1 + ....where the sum is ?ite because the polynomial has only a ?ite numberof non-zero coef?ients. Now we can de?e a bilinear form, givenpolynomials R_1 and R_2, as S[ R_1 R_2 ]; that is, we muliply thepolynomials in the usual sense, then apply the linear functionalS[]. The coef?ients s_0,..., are choosen so that the form ispositive de?ite.Now the author wishes to choose an orthogonal basis, and givesthe 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 de?ed 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 byexpanding the matrix using the bottom row), so that toprove orthogonality (leaving aside normalization for the moment) itis suf?ient 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 isgiven 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, wewould get P_n(lambda) lambda^n = a polynominal in termsof lambda^{m}...lambda^{m+n} where the coef?ients arethe minors of the above matix (with appropriate signs).The author then asks the reader to apply the functional S[] toboth sides of it, so that the left side becomes the inner productof P_n(lambda) and lambda^m, so that we obtain [this will beour 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 matrixand the previous matrix (from equation#1 above) using the lastrow--the minors are the same in both instances (and contain onlys_i's, no lambda's!) and we simply have replaced lambda^m withs_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.orgI 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 thatJ.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 of?e 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 maththat in a kind of decorative way had pi written to 25decimal places with ... at the top of a page. I didn'tbother to go past that point. Probably I would have ifmy friend Tom, who memorized somewhat fewer, hadgotten to 25.A few years later, another kid in the same school systemwas identi?d as being gifted mathematically for havingmemorized 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 recordI 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, evenhundreds 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 ofhis wonderful thoughts and ideas.Dr E replied: I have had two great thoughts in my life and remember themboth.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 recordEach 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 ?ite 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, aO snafu apparent in O Universal Concept from believing lies?That there, obstinate in you, O Strange Constant, A Divine Sign O exists is unlikely unlessIs O revealed Something Brilliant, negating belief!In formulas, O, you show yourself in Greek and math as a π forever--O hidden wonders absconded, in?ite, 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 doesπ! stand for please? The context requires a single letter, but none ofa, I or O make sense.--Paul V. S. TownsendInterchange 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.htmlto see the poem. === >How many digits of pi have you memorized? I did 100. The world recordI've memorized 100,000 digits. They're all 3. Of course I haven'tmemorized 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 in?itely 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 === ----- <^> <()> <^> -----> : 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 gotevery green light!Herchad 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 anunending number of non repeating decimals and thus would imply that you havein?ite 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...>JustinDavid 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 coupleof months of concentrated software development time whichmeant that he had to think in octal about eight hours aday, six days a week (that doesn't include the time spentdreaming while sleeping). Then he balanced his checkbook.After a couple hundred heart palpitations, he veri?d thathe 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 veri?d that> he hadn't written the checks using octal.Me too, caused a few riots keeping score for darts at the pub in theevenings. No, sorry, treble 17 isn't 55 is it : )--Paul V. S. TownsendInterchange 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 veri?d 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 havebatted an eye but would have done the conversion automatically./BAH === I learned 500, but now, I have probably forgotten most ofthem(It was some years ago)./Anders> How many digits of pi have you memorized? I did 100. Theworld record === I memorized 27 digits when I was 12, and since then I've not forgottenthem nor have I memorized any more.What might be more interesting is the number of digits someone may havememorized such that any particular digit can be recalled simply from itsplace. As someone said, what's digit 77? If you know 100 digits and canrecall 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 quantummechanics...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 onquantum> 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 rememberthe ?st line: Now I, even I, would celebrate.Back in my student days, some students tried to devise mnemonics for pi ande. One mnemonic for e neatly avoided the charge that such sentences werecomplicated, tedious or full of obscurities: In seeking a mnemonic, we composed a sentense of sensible words.Not enough entries were received for the ?st prize of a bottle of wine tobe awarded, but one of the students received a Mars bar as a consolationprize for this one for pi: Buy a Mars a month. Chocolate is seldom cheap but costly.--Paul V. S. TownsendInterchange 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 ?st line: Now I, even I, would celebrate.Found it, ascribed to a certain Adam C. Orr of Chicago in1906.Now I, even I, would celebrateIn rhymes unapt, the greatImmortal Syracusan rivalled nevermoreWho in his wondrous lorePassed on before,Left men his guidance howTo circles mensurate.(Should it be inept in line 2? Who is the guy from upstate New Yorkreferred to in line 3?)--Paul V. S. TownsendInterchange 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 givea digit numericalof number whichcan ratiodiameter-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 ?st 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 ?st 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 thanhave call a conversion routine for each word. That's a waste ofbrainpower plus you have to remember where you were in thesentence./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 ?st 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 ?st 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 hereIt'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 itwas supposed to have been heavy lectures regarding quantum mechanics? -- Erick === ----- <^> <()> <^> ----->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 thereHerc === ----- <^> <()> <^> -----> How many digits of pi have you memorized? I did 100. The world record3.1415926536not sure with the rounding I'll pass the Dalek Pyramid test with theelectrocuting ?f 10 X 10 tiles, each row only the pi digit is safe.What's digit 77?Herc === In sci.math, |-|erc ----- <^> <()> <^> -----> 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 ?f 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 recordI 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 morethan 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 quantummechanics...)Count the letters in each word:3 1 4 1 5 9 2 6 5 3 ...(.....?)G C === the ratio of the diameter of sphereto its circumference is greater than three! on teh other hand,the ratio of the area of the equatorial circle of a sphereto 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 record3.1415926535Now... check this out.Assume that the Earth is a sphere. One can calculate the circumferenceusing 3.14159 to within an order of a meter. Each extra decimal reduces theamount that you are off by an order of magnitude.Now, let us assume that the galaxy is a disk 150000 LY in diameter (75000 LYradius)This is a radius of ~ 7E20 meters.With pi to 100 digits this will give the circumference to within some 1E-80meters 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.netIt's still legal to go .sigless. === > How many digits of pi have you memorized? I did 100. The world record3.141592Six digits.And I think it's enough :)Spider === >The world recordAlways 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 :)> SpiderThe 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 === > ----- <^> <()><^> -----> Skeptic organisations are not interested in the scienti?investigation> of> the paranormal.> Skeptic organisations are interested in shutting down businesses thatmake> 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) :> TasmaniaThat's less than thousands... Rather a lot really. Guess it was made up. === ----- <^> <()> <^> -----> ----- <^> <()> <^> -----> Skeptic organisations are not interested in the scienti?> 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 ?ure out when a_1(x)/7 and a_2(x)/7> are algebraic integers, but admittedly it's not a priority.> You ?st 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 themexplicitly; I merely need show that the de?ing equationfor a_1(x)/7 etc. is not of the requisite form for most x.Then again, your way might be slightly cleaner, and you alsohave 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 simpli?s 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 ?re to try> to ?ure 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 (de?ition 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. Butit 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 de?itions:> 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> ?ling in all stuff (and hoping I did not make a basic arithmetic> mistace), we ?d 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.netIt'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 ?hting 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 bene? > of doubt because that is all it would take to con?m 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 ?st is probably easier on the brains...http://zapatopi.net/afdb.html:-)-- #191, ewill3@earthlink.netIt'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 ?ished, 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 ?ished 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 307521For 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 ?ished 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 separatedscheme?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 avoidunnecessary 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 af?e scheme Spec(A), then f: X --> Y isdetermined by homomorphism h: A --> O_X(X).Let I = Ker(h). Then Y' = Spec(A/I) satis?s the propertyof the proposition. If Y is not af?e, then glue af?e shemesobtained 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 ?d a minimalstring which contains all binary strings of length n>1 (that one should be o? 2^n+n-1). For example (n=3) the following string is solution:1110001011I can prove that that meta-string exists for any n>1. Also a circle (stringwith connected ends) of length 2^n exists too (in fact every meta-string could bereduced to circle by cutting off last n-1 bits). Proof afterspoiler spaceConsider a graph G which vertices are all strings of length n. If string b can beput after string a (the ?st n-1 bits of b are equal to last n-1 bits of a) thenan 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 throughevery 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...couldpossibly be useful in arti?ial life applications using a cell grid. Haveyou come up with a use for such a compacting of information? It's almost aform of compression. === |Well, in fact I don't have the proof so you could try to ?d 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 (asequence 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 ?d 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 =--- === > ----- <^> <()> <^> -----> 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 speci? ingredient> that I needed.> I also tried many grams of vitamin C with no results,> although I did ?d 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 snif?in g 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.> /BAHI 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