mm-48 === 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) + bfor 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 Correys 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 listAxyxy(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, theresa metatheorem which guarantees you can prove r#s too. Markovs ruleentails ~r=s -> r#s, and the Markov school used this (although Iguess they noted when they did). Once youve concluded that ~r=s,its tempting to say that its just a matter of computing r and sprecisely enough to ?d the level of precision where they differ,to determine that r#s.Nevertheless, its considered inappropriate in standard intuitionismshows that the nonexistence of an n for which |r-s|>1/n leads to acontradiction; it isnt taken as exhibiting the n.Mainly, I would say its just better not to muck about with weirdlittle statements like ~r=s anyway. Its 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||Axyxy(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 dont necessarilysatisfy Axy(~x#y->x=y).Im 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#xor 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.Heres an example of a case where one has a natural # relation thatsatis?s my weaker de?ition. I dont 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 iftheres 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 were done, weve either found thatA is true, or for each one weve found that B_i is true. But thatsnot 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 theres apartition of z1,...,zk so that the xs together with one of thepartitions are collectively apart from the ys 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, its 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 xs as fromthe ys. 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. Itsobviously 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 zs are putwith those xs to make {z1,...,zn}#{y1,...,yn}, or enough zs areput with those ys to make {x1,...,xn}#{z1,...,zn}. If >n-l of thezs are put on the left, thats enough to make the former true. Ifnot, then there are at least l of the zs 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 isnta 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 isnt the case that the ?st 1 is at an odd position,or it isnt 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 itslike not to assume it.So Id 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, theres 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. Theres a formal system, yes, andits 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. Ive seen mentionedcertain completeness theorems, but I wouldnt 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. Idont know what good that is. The thing its a negation of doesnthave to be an apartness, and as far as I know theres no guaranteethat there is an apartness.Heres 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 thats 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 its 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$oft === at 09:29 PM, maneesh@drunkenbastards.com (maneesh) said:>To elaborate, I often found it confusing to associate particular>properties, algebraic objects and other such things with the name>they are given.Learning a new language is always confusing.>Surely it only adds a level of complexity to a ?ld of study which>does not need one!Surely not. The problem is that the concept is new, not that there issomething wrong with the name.>What are your opinions on having things called Schurian, Jacobian,>or Gaussian?Its shorter than descriptive names, and no less enlightening. >Do you feel that names of things in mathematics should greater>re?nformation about the said things? Should the word red be red? Names are arbitrary. >Do you feel it retards mathematical progress (on a personal, or >collective level)? No.>What is the worst named object you can think of?Ideal.None of what you have written addresses the real problems inMathematical nomenclature; the large number of synonyms and theinconsistency in the notation.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === ----------------------------- <^> <(?)> <^> ----------------------------->What is the worst named object you can think of? Ideal.>why ideal?I would say ?abstract has the most number of meanings in mathematics.function abstraction, outline, de?e proposition, approximate, ....Herc X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$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. Its just a question of picking out the ones thatsupport your case and ignoring the ones that dont.>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 youre just going to say That one doesntcount - its 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?Youve already rejected them. Jewish mysticism often makes use ofGemmatria, which is basically what you are doing. Its not Science andits 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, MANS, 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 doesnt 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 doesnt 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 dont 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 dont 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 dont 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_is, no lambdas!) 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 40s. So, he didnt 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 Ive read have sucked eggs. . . . === Apostol is great, uses Liebniz method. Ive also heard thatJ.Bernoullis 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 didntbother 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 its 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 its 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 didnt 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 its 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 recordIve memorized 100,000 digits. Theyre all 3. Of course I haventmemorized 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 recordIve memorized 100,000 digits. Theyre all 3. Of course I havent>memorized exactly where they occur.Ive memorized in?itely many digits, and theyre all the same!Of course I havent 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 Ive memorized 100,000 digits. Theyre all 3. Of course I havent> memorized exactly where they occur. (Oh, you meant _consecutive_ digits? Never mind...)LOL! === : Ive memorized 100,000 digits. Theyre all 3. Of course I havent: memorized exactly where they occur.: (Oh, you meant _consecutive_ digits? Never mind...)Well, in that vein, Ive 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 === ----------------------------- <^> <(?)> <^> ----------------------------- : Ive memorized 100,000 digits. Theyre all 3. Of course I havent> : memorized exactly where they occur.> : (Oh, you meant _consecutive_ digits? Never mind...) Well, in that vein, Ive 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... === : Ive memorized 100,000 digits. Theyre all 3. Of course I havent> : memorized exactly where they occur.> : (Oh, you meant _consecutive_ digits? Never mind...) Well, in that vein, Ive 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 Davids 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. === : Ive memorized 100,000 digits. Theyre all 3. Of course I havent>: memorized exactly where they occur.>: (Oh, you meant _consecutive_ digits? Never mind...)Well, in that vein, Ive memorized all of them. They are 0,1,2,3,4,5,6,7,8 and 9. Thats very impressive. I always forget the 8.>If only I could get the order right...Justin************************David C. Ullrich === 3.14159265358979323846264338327950288419716939937510....Its a long story how it happened...(well, its actually very short, but as if anyone would like to listen to it and believe it!!!) === >: Ive memorized 100,000 digits. Theyre all 3. Of course I havent>: memorized exactly where they occur.>: (Oh, you meant _consecutive_ digits? Never mind...)>Well, in that vein, Ive memorized all of them. >They are 0,1,2,3,4,5,6,7,8 and 9. Thats 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 doesnt include the time spentdreaming while sleeping). Then he balanced his checkbook.After a couple hundred heart palpitations, he veri?d thathe hadnt 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 doesnt include the time spent> dreaming while sleeping). Then he balanced his checkbook.> After a couple hundred heart palpitations, he veri?d that> he hadnt written the checks using octal.Me too, caused a few riots keeping score for darts at the pub in theevenings. No, sorry, treble 17 isnt 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 doesnt include the time spent> dreaming while sleeping). Then he balanced his checkbook.> After a couple hundred heart palpitations, he veri?d that> he hadnt 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 isnt 55 is it : )Unless you played darts with like-minded friends. They wouldnt 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 Ive 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, whats digit 77? If you know 100 digits and canrecall each one by place, thats 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. Its easier just to remember the numbers rather thanhave call a conversion routine for each word. Thats 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. Its easier just to remember the numbers rather than> have call a conversion routine for each word. Thats 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,Cmon man lets do some more,62648338,Wow! This pi thing sure is great!:-)> /BAH> -- #191, ewill3@earthlink.net -- insert random weird poetry hereIts 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... theres a ACBC around thereHerc === ----------------------------- <^> <(?)> <^> -----------------------------> How many digits of pi have you memorized? I did 100. The world record3.1415926536not sure with the rounding Ill pass the Dalek Pyramid test with theelectrocuting ?f 10 X 10 tiles, each row only the pi digit is safe.Whats 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 Ill pass the Dalek Pyramid test with the> electrocuting ?f 10 X 10 tiles, each row only the pi digit is safe.> Whats digit 77?> Herc> I sure hope youve got your sums right. -- Teegan Jovanka (Janet Fielding):-)-- #191, ewill3@earthlink.net -- who?Its still legal to go .sigless. === > How many digits of pi have you memorized? I did 100. The world recordI think youve got a real shot at the record. Youre 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 dEnron!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 its handy for checking out new processor units. :-)-- #191, ewill3@earthlink.netIts still legal to go .sigless. === > How many digits of pi have you memorized? I did 100. The world record3.141592Six digits.And I think its 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 its 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 doesnt 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) :> TasmaniaThats 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 doesnt 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 Thats 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 as 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. Id be curious to ?ure out when a_1(x)/7 and a_2(x)/7> are algebraic integers, but admittedly its 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 dont 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.> Its> 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!> (Im also curious as to the rest of his proof; this is> only a small snippet thereof. Its a bit like examining> a small area (sans the actual hole) of a ?re to try> to ?ure out why the car wont 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 wouldnt 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> Lets 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 zs correspondend to James as in order.> The beauty of this presentation is (I think) how three cubic roots> of two different values are used.Thats a nice distillery of the problem, admittedly.-- #191, ewill3@earthlink.netIts still legal to go .sigless. === In sci.math, Virgil Hancher [...]> Jim, Im 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. Dont 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.netIts 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 hasnt 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 Grothendiecks 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 its to prevent situations where Y is not a separatedscheme?Keith Ramsay === > |In Grothendiecks 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, Im not 100% sure of my proof, and EGA has authority... > Perhaps its to prevent situations where Y is not a separated> scheme?Its unlikely since Grothendieck was trying hard to avoidunnecessary conditions on his results.Nobuo Saito === > In Grothendiecks 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, its 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> Thats torus XXXO> XXOX> OXOO> XOOO And thats 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? Its almost aform of compression. === |Well, in fact I dont have the proof so you could try to ?d it. I just hope|elliptical curves have nothing to do with that problem...Elliptic curves. Im sure they arent. 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 its 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 === >Ill 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 (thats when they always decide to get itchy) also> helps. Since women think its 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, andencourage people to use their sick days at 1st sign.HercX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS016712394; === >I want to make a spreadsheet to calculate the number of feet of material on a>roll.The user will enter the Outside Diameter, the Core Diameter (The tube that the>material is wrapped around) and the number of feet for a FULL roll ?st.Then they will enter the Outside Diameter of the partial roll they want to>measure, and the Core Diameter. The spreadsheet will then calculate the number>of feet left on the roll.I cant seem to come up with the algebraic expressions to calculate the>thickness of the material or the number of layers on the roll. I also want to>?ure out the factor that will adjust the circumference for each successive>layer on the roll. It seems to me that the ?st section of entries (the OD,>the CORE and FEET) should be enough to extrapolate the number of layers on the>roll, and the thickness of the material. I know that if I had them enter the>material thickness to the equation it would be easier, but I dont want them to>have to measure the thickness.any ideas?MikeHere is an approximate way to attack the problem.If you look at the roll end-on, the cross-sectional area ofmaterial is approximately the material thickness times its length.A = pi * (r2 - r1)^2 = t * LSo if you measure r2 = outside diameter/2 and r1 = core diameter/2and you know L = materials length, you can get the thickness.Then apply same formula to another roll to ?d unknown length.Happy New Yearphil === >I want to make a spreadsheet to calculate the number of feet of material on a>roll.>The user will enter the Outside Diameter, the Core Diameter (The tube that the>material is wrapped around) and the number of feet for a FULL roll ?st.>Then they will enter the Outside Diameter of the partial roll they want to>measure, and the Core Diameter. The spreadsheet will then calculate the number>of feet left on the roll.>I cant seem to come up with the algebraic expressions to calculate the>thickness of the material or the number of layers on the roll. I also want to>?ure out the factor that will adjust the circumference for each successive>layer on the roll. It seems to me that the ?st section of entries (the OD,>the CORE and FEET) should be enough to extrapolate the number of layers on the>roll, and the thickness of the material. I know that if I had them enter the>material thickness to the equation it would be easier, but I dont want them to>have to measure the thickness.>any ideas?>Mike>I lost the original post, but this reply is to Mike. I think theproblem is easier than you make it. If you call the radius of the tuber0 and the outer radius of a new roll r1 and you have the length of anew roll, call it L, then the length will be very close to linear as afunction of the radius r:length = L*(r - r0)/(r1-r0), for r0 <= r <= r1By basis for thinking this is that if the roll is modelled by a linearspiral r = k*theta + c, which I think it is, such a curve turns out tohave arc length a linear function of theta and hence a linear functionof the radius.Try it. I would be interested to hear how close it works out to thecorrect length in the shop.--Lynn === By basis for thinking this is that if the roll is modelled by a linear>spiral r = k*theta + c, which I think it is, such a curve turns out to>have arc length a linear function of theta and hence a linear function>of the radius.Woops I should have checked this more carefully. It isnt linear.Please ignore my post.--LynnX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS016e12389; === Isnt 4 x 1 metre sidewalk too small for two men building it?Long time ago my dad asked me a question: If one man would dig a well in ?e hours, how much time would ?e workers need? I answered and then he said Imagine ?e big men digging one little well, huh?Smile!Mateusz === In sci.math, Mateusz Kwasnicki Long time ago my dad asked me a question: If one man would dig> a well in ?e hours, how much time would ?e workers need?> I answered and then he said Imagine ?e big men digging one> little well, huh?> Smile!> Mateusz> Indeed. If a woman can have a baby in 8 1/2 months, how longwould it take 8 1/2 women (the 1/2 is working part time onanother project :-) ) ?-- #191, ewill3@earthlink.netIts still legal to go .sigless. === > In sci.math, Mateusz Kwasnicki> Long time ago my dad asked me a question: If one man would dig> a well in ?e hours, how much time would ?e workers need?> I answered and then he said Imagine ?e big men digging one> little well, huh?> Smile!> Mateusz> Indeed. If a woman can have a baby in 8 1/2 months, how long> would it take 8 1/2 women (the 1/2 is working part time on> another project :-) ) ?Be nice, people. I speci?ally stated in my original post that theproblem would not be modelled correctly unless the two workers workedwith perfect cooperation. Clearly women cannot cooperate on creatinga baby, althoughh these days that may no longer be as true as it oncewas. NO ATOMIC MEAURES! NO! NO! NO!. How would you then be able tocut the cake?As to the men working on the sidewalk. First of all this was theUnited States where most of us dont have a clue what a metre mightbe, and surely the cracks in side walks are not separated by a metreor even a meter as we say in the USA. Perhaps they were a yard apart. I dont remember. This was done in 1959 or thereabouts, and I amhaving trouble remembering the size of the sidewalk squares. I doremember that it was really, really dif?ult to make one step persquare for more than a couple of squares. I should also point outthat the problem never says that this isnt done with a GilbertSideWalk Kit that two boys could be using. I did say boys, not men,in the original post, although I did admit the somewaht egocentricnature of that assumption.At any rate, there were 5 men working digging a ditch, and theforemman was sitting on his ... watching the men work. One of thecame up and asked him why they had to do all the hard work and he gotto sit on his ... and watch them. He told the guy that it wasintelligence. Intelligence! Whats that, said the worker. Theforeman held his hand up in front of a tree and said to the worker,might and the foreman pulled his hand away just in time so that theworker punched the tree with all of his might. He went back down intothe trench and was asked what the foreman had had to say. He said itwas intelligence, said the worker. Intelligence! Whats that,replied his friend. Simple, he said, holding his hand in front ofAchavaX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS2gbb22101; === I recall having been taught not use operations such as +, -, /, x that will effect both sides(at the same time) of an identity to be proven. And so I told my students to follow this rule. One student respectfully and thoughtfully said why and went on to pose that the relationship between the two expressions we are trying to show are identical must be either >, >=, =, <= or < . So with the exception of multiplying or dividing by a negative value, the unknown relationship will stay the same(if we operate on both expressions at once).For example (trivial but it demonstrates the question) suppose you want to prove that 1 + sin(-x) = 1 - sin(x) .Normally we would apply the cofunction identity sin(-x) = - sin(x) the left hand expression and be done. But why not ?st subtract -1 from both expressions and then apply the cofunction identity?If the rule I remember being taught is valid there should exist some pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in fact identical but would appear to be identical if one operates on both expressions simulataneously. For example suppose one were trying to prove Expr1(x) = Expr2(x). In the process he/she divided both sides by cos(x)and found that the modi?d expresssions were in fact equal. I cant come up with such a pair of expressions that would validate the rule I remember. Can anyone help? === > I recall having been taught not use operations such as +, -, /, x >that will effect both sides(at the same time) of an identity to be >proven. And so I told my students to follow this rule. One student >respectfully and thoughtfully said why and went on to pose that the >relationship between the two expressions we are trying to show are >identical must be either >, >=, =, <= or < . So with the exception>of multiplying or dividing by a negative value, the unknown relationship>will stay the same(if we operate on both expressions at once).> For example (trivial but it demonstrates the question) suppose you >want to prove that 1 + sin(-x) = 1 - sin(x) .> Normally we would apply the cofunction identity sin(-x) = - sin(x) the left hand expression and be done. But why not ?st subtract -1 from >both expressions and then apply the cofunction identity?> If the rule I remember being taught is valid there should exist some >pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in>fact.>identical but would appear to be identical if one operates on both> expressions simulataneously. For example suppose one were trying to prove >Expr1(x) = Expr2(x). In the process he/she divided both sides by>cos(x)and >found that the modi?d expresssions were in fact equal. I cant come up>with >such a pair of expressions that would validate the rule I remember. >Can anyone help? Firstly you must differentiate between an identity and an equality.Secondly, why must the alternatives to ?equals be some inequality? Fineif youre dealing with real numbers, but what about complex numbers?In the example youve provided, all you;ve done is work backwards from thecorrect fact, that sin is an odd function (whats a cofunction identity bythe way?), to the statement you want. In this case all the steps youvetaken are reversible, that is it is an if and only if deduction. Ingeneral in mathematics that doesnt happen, so its better to teachstudents how to think ?properly. An example where you shouldnt workbackwards (trivial, but important)showsin(x) ?= sin^3(x) + sin(x)cos^2(x)you cant factor out sin and divide because for some values of x youredividing by zero. You may start from the identity 1 ?= sin^2 + cos^2 andmultiply through. The ?= bit means I really ought to write the three barsymbol. And that hidden dividing by zero thing needs to be watched. It is bad practice to assume the answer and work backwards. Dontlet them start now and make maths teachers later tear their hair out. === >It is bad practice to assume the answer and work backwards. Dont>let them start now and make maths teachers later tear their hair out. Actually most of the time its easier to work backwards, and perfectlyvalid _provided_ you know what youre doing: i.e. you know youre workingbackwards, and that the real formal proof will go forwards, so you make sure that the real proof will have valid inferences. To help in this,I like to write ?=, ?> etc. rather than = or > for relations I want to prove, but havent yet.In your example: 1) sin(x) ?= sin^3(x) + sin(x)cos^2(x)is equivalent to2) sin(x) ?= sin(x) (sin^2(x) + cos^2(x))which is implied by3) 1 ?= sin^2(x) + cos^2(x)which is an identity we know.Note that its OK to go from 2) to 3), precisely because were reasoningbackwards: a b = a c does not imply b = c (because a might be 0), butit is implied by b = c. Now once you have the backwards proof, you canturn it around to make the real proof:1 = sin^2(x) + cos^2(x)sin(x) = sin(x)(sin^2(x) + cos^2(x))sin(x) = sin^3(x) + sin(x) cos^2(x)Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >It is bad practice to assume the answer and work backwards. Dont>let them start now and make maths teachers later tear their hair out. > Actually most of the time its easier to work backwards, and perfectly> valid _provided_ you know what youre doing: i.e. you know youre working> backwards, and that the real formal proof will go forwards, so you make > sure that the real proof will have valid inferences. To help in this,> I like to write ?=, ?> etc. rather than = or > for relations I > want to prove, but havent yet.> I think in some post that got lost by outlook (not usual choice) I agreed,and still do, that it is often practical to work backwards, though I tendto get students to rewrite it so it doesnt look as though they workedbackwards. And to make sure that all the if statements are only ifstatements too in this case. But the ?if you know what youre doingproviso is key here. Surely it is better to go from what you know to betrue forwards. If the method you choose is deduced from thinking backwardsso be it. I agree the example below was pretty dire, and I certainly regret using it.> In your example: > 1) sin(x) ?= sin^3(x) + sin(x)cos^2(x)> is equivalent to> 2) sin(x) ?= sin(x) (sin^2(x) + cos^2(x))> which is implied by> > 3) 1 ?= sin^2(x) + cos^2(x)> which is an identity we know.> Note that its OK to go from 2) to 3), precisely because were reasoning> backwards: a b = a c does not imply b = c (because a might be 0), but> it is implied by b = c. Now once you have the backwards proof, you can> turn it around to make the real proof:> 1 = sin^2(x) + cos^2(x)> sin(x) = sin(x)(sin^2(x) + cos^2(x))> sin(x) = sin^3(x) + sin(x) cos^2(x)> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 === [snip]> In the example youve provided, all you;ve done is work backwards from> the correct fact, that sin is an odd function (whats a cofunction> identity by the way?),I presume that the OP misused that term. Surely a cofunction identity issomething like cos(x) = sin(pi/2 - x).> to the statement you want. In this case all the> steps youve taken are reversible, that is it is an if and only if> deduction. In general in mathematics that doesnt happen, so its better> to teach students how to think ?properly. An example where you shouldnt> work backwards (trivial, but important) show sin(x) ?= sin^3(x) + sin(x)cos^2(x) you cant factor out sin and divide because for some values of x youre> dividing by zero.One may factor out sin(x) on the right, and then notice that the otherfactor, sin^2(x) + cos^2(x), is always 1, thereby showing that the rightside simpli?s to the left side.> You may start from the identity 1 ?= sin^2 + cos^2 and> multiply through. The ?= bit means I really ought to write the three bar> symbol. And that hidden dividing by zero thing needs to be watched.At least it needs to be watched when one is solving conditional equations,lest roots be lost.> It is bad practice to assume the answer and work backwards.In _general_, I wholeheartedly agree. But, in the speci? context ofproving identities, I disagree. For example, very often, the easiest methodof proof is to simplify both sides of the supposed identity as much aspossible, in the hope that both sides will simplify to the same thing.David Cantrell === >I recall having been taught not use operations such as +, -, /, x>that will effect both sides(at the same time) of an identity to be>proven....>If the rule I remember being taught is valid there should exist>some pair of trigonometric expressions Expr1(x) and Expr2(x)that>are not in fact identical but would appear to be identical if one>operates on both expressions simulataneously.Something in what you have written here makes me think of thePerron Paradox. Perhaps that would be of help to you.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS4GIB28000; === >I recall having been taught not use operations such as +, -, /, x that will effect both sides(at the same time) of an identity to be proven. And so I told my students to follow this rule. One student respectfully and thoughtfully said why and went on to pose that the relationship between the two expressions we are trying to show are identical must be either >, >=, =, <= or < . So with the exception of multiplying or dividing by a negative value, the unknown relationship will stay the same(if we operate on both expressions at once).For example (trivial but it demonstrates the question) suppose you want to prove that 1 + sin(-x) = 1 - sin(x) .>Normally we would apply the cofunction identity sin(-x) = - sin(x) the left hand expression and be done. But why not ?st subtract -1 from both expressions and then apply the cofunction identity?If the rule I remember being taught is valid there should exist some pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in fact identical but would appear to be identical if one operates on both expressions simulataneously. For example suppose one were trying to prove Expr1(x) = Expr2(x). In the process he/she divided both sides by cos(x)and found that the modi?d expresssions were in fact equal. I cant come up with such a pair of expressions that would validate the rule I remember. Can anyone help? I have never heard of this rule. There is absolutely no reason why you cant perform the same operation on any equation, identity or not, and not arrive at a valid equation. One of the things you do have to be careful about is that each operation you use has to be invertible: that is you have to be careful about multiplying both sides of an equation by sin x since the inverse would be dividing both sides by sin x which is 0 for some x. The reason for this is that the real proof is working backwards. In order to prove, for example, that (tan x)(cos x)= sin x, we replace tan x with sin x/cos x so that the left side is (sin x/cos x)(cos x) which is, of course, sin x. But you cant prove a formula, or any statement, is true by starting from that statement! The real proof is working backwards.Looking at what we did, we know we an start from sin x= sin x, rewrite the left side as (sin x)(cos x/cos x)= (sin x/cos x)cos x= (tan x)(cos x). We dont have to write thatout because we know that everything we did originally is reversible. === > In order to prove, for example, that (tan x)(cos x)= sin x, we replace>tan x with sin x/cos x so that the left side is (sin x/cos x)(cos x)>which is, of course, sin x. Thats not a proof. In fact theres nothing to prove since byde?ition tan x == sin x / cos x (this follows from elementarygeometry) so the result is simply a reformulation of the de?ition.Furthermore, its not even an equality. What happens whenx = Pi/2? The right side is still well-de?ed but the left side mostcertainly not.> But you cant prove a formula, or any statement, is true by starting>from that statement! The real proof is working backwards.>Looking at what we did, we know we an start from sin x= sin x, rewrite>the left side as (sin x)(cos x/cos x)= (sin x/cos x)cos x= (tan x)(cos x).And how do you know that (sin x / cos x) = tan x when that is the veryresult youre trying to prove?>We dont have to write that out because we know that everything we did>originally is reversible. There is nothing that says you must work out your proofs backwards(what you call forwards) or that all proofs must be two-way. If westart from given truths and work our way using one-way implications(i.e. p => q) there is nothing that states the result is incorrect,just that the opposite (q => p) does not necessarily hold. === > In order to prove, for example, that (tan x)(cos x)= sin x, wereplace> tan x with sin x/cos x so that the left side is (sin x/cos x)(cosx)> which is, of course, sin x. > Thats not a proof. In fact theres nothing to prove since by> de?ition tan x == sin x / cos x (this follows from elementary> geometry) so the result is simply a reformulation of the de?ition.There are several ways of de?ing tan, with or without any referenceto cos and sin. It is a matter of convention. So what is a proof andwhat is only a reformulation of the de?ition only depends on whatframe youre in.> Furthermore, its not even an equality. What happens when> x = Pi/2? The right side is still well-de?ed but the left side most> certainly not.The right side is not de?ed either, at least in R, cos(Pi/2)=0.> But you cant prove a formula, or any statement, is true bystarting> from that statement! The real proof is working backwards.> We dont have to write that out because we know that everything wedid> originally is reversible. > There is nothing that says you must work out your proofs backwards> (what you call forwards) or that all proofs must be two-way. If we> start from given truths and work our way using one-way implications> (i.e. p => q) there is nothing that states the result is incorrect,> just that the opposite (q => p) does not necessarily hold.Both of you agree on that, I guess. I think that what you callbackwards is what he calls backwards too...But what you have just written is confusing, even if I guess you agreewith what I am going to say, I just want to clarify this point. There_is_ something that says that proof must go in one de?ite way: fromthe hypothesis to the result! This is the way that is meaningful.Sometimes, you can write the argument backwards because it isreversible. But what matters is the way (hyps+defs+axioms) => result.To come back to the original question, I would say that the rule Edgave is pointless with _purely_ symbolic proofs, and needs to besharpen when doing analysis. But I think it is interesting to explainto your students when such operations are allowed and when they arenot, and why. It depends on their level, but I think it is always abad idea to simply say NO!. Mathematics is not about calculus andits rules, it is about understanding. So you may allow them not torespect your rule, if they do it correctly.First of all, you want to consider functions, not expressions.You must be careful about the set of points on which what you aredoing is valid. It may not be R as a whole. You may need to excludesome points, like 0. Sometimes what you write is only valid on (0;Pi/2], and so on.And the simplest case which is not _purely_ symbolic is probably with<= and some multiplication on both sides of the inequality.Dividing by cos(x) on both sides is problematic when cos(x) did notappear in the numerators of each side of the previous expression.Because in such a case, you may have a problem when x = Pi/2. But insome cases, it may still be valid. If you have cotan(x) on both sides,or if both sides are equivalent to cos(x) when x tends to Pi/2, etc.The reason why you did not ?d any counter example (ie. some examplethat would validate the need for your rule) is that you are usingsimple expressions. If your expressions Expr1 and Expr2 containcos(x), sin(x), tan(x) and nothing else (but arithmetical operations),you can replace them by A, B, C and in practice, _any symboliccomputation_ on both sides of any equality written with these threeletters will remain valid with cos(x), sin(x) and tan(x). You can waituntil youre done to ask you where this is de?ed. Why?This is not elementary. These three functions are C-In?ityfunctions, but in some isolated points where they are not de?ed. Soeven if an expression, which is a step in your proof, written withcos, sin and tan is not de?ed in some point, you may easily considerits limit in this point. And if it is in?ite, you know precisely thespeed of the convergence to the in?ity. So that you may be able,in a latter step of your proof, to ?d a (?ite) limit in the samepoint. And you will be able to state it rigorously.To say it simply, since they are explicit C-In?ity functions, cos,sin, and tan can be handled as polynomials, using Taylors formula.I said previously that what you want to consider is functions and notexpressions. This is because expressions are only symbolic and tellsnothing about the values they can take when evaluated. And whatinterests you here is to identify expressions that take the samevalues in each point of some set. So that you identify functionsthrough their graph.So, what do you say to your students? Again, it depends on theirlevel, but I would recommend to abandon your rule, to concentrate onthe meaning of what they do. That doesnt mean you cannot say to them:be careful about that, in some cases it is false, thus I wouldrecommend you not to do this, unless you understand what you do andcan explain it, and you can require that they explain what they doeach time it is necessary. Try to tell them when it is safe: inparticular, for trigonometric formulas, tell them about de?itiondomains (taking some example where you divised by 0 with cos(x) and x= Pi/2), the problem with limits and in?ity if they are advancedargument if they start with the result.More generally, my personnal point of view is to always avoid blindrules in Mathematics. It is sometimes necessary to adopt such a pointof view: if you want to train them to compute derivatives quickly, forinstance. But this should always be accompanied with explanations, andif you cant, give them at least some clues to help answer the why?.My experience is that if a student can remember why, then you havewon. He/She may even begin to like Mathematics. /er. === > If uncountable S is a subset of R, then S contains> a subset T with the property that, for every distinct elements> x and y in T, theres a z in T between x and y. > This problem is equivalent to showing for uncountable S,> S is somewhere dense,> that is not nowhere dense; int cl S /= nulset>See retraction sci.math nowhere dense real subsets. <1072549678.37972@news.aic.at> === only one was willing to make the trip. Our small research group was thusfaced with the necessity of carrying n standard elephants themselves. Ina pilot project, it was established that our full staff was unable tolift, much less carry, a single standard elephant. It was also noted bythe project manager who was sprayed by the ill mannered standard elephantblowing its nose upon him and again by the project director who had herbutt whooped by a standard trunk, that standard elephants may lackappropriate manners needed for cooperation on a long long trip. Anyfurther efforts on that ?ld trip were immediately abandoned when n-1standard elephants taking notice of the ruckus, ?led their standardtrunks with wet water and eagerly approached the experimental site.Upon recovery from this test project, the staff decided a better approachwould be to use the standard kangaroo model on the speculation that thestandardized equipment would hop itself into position. Before proceedingwith this project some facts need be established. As standard kangaroosare smaller that standard elephants, will n standard kangaroos suf?e orwill 3n standard kangaroos be needed? Also once in place, how is astandard hopping kangaroo convinced to remain stationary? >By the way, thanks for the hint that standard elephants could maybe >decay into strange colorful left handed anti-quarks - we will >make some more calculations clarifying this point.Your welcome. Please keep us inform of your progress. Do be carefulhandling strange colorful left handed anti-quarks, those weirdos havebeen noticed to change gender right in the middle of an.., experiment.Whence the name.---- === > An interesting approach. However upon asking 262,144 standard elephants,> only one was willing to make the trip. Our small research group was thus> faced with the necessity of carrying n standard elephants themselves. In> a pilot project, it was established that our full staff was unable to> lift, much less carry, a single standard elephant. It was also noted by> the project manager who was sprayed by the ill mannered standard elephant> blowing its nose upon him and again by the project director who had her> butt whooped by a standard trunk, that standard elephants may lack> appropriate manners needed for cooperation on a long long trip. Any> further efforts on that ?ld trip were immediately abandoned when n-1> standard elephants taking notice of the ruckus, ?led their standard> trunks with wet water and eagerly approached the experimental site.>Oh ... we are very sorry that your research group suffered suchcomplications - somehow there is a very huge misunderstanding here :In order to measure the mass of the galactic black hole on should useVIRTUAL ELEPHANTS !!!The main advantage in such an approach would be, that thehandeling of virtual elephants is somehow more easy and convenient.Besides, the branching-fraction of a virtual standard elephant intoa left-handed kangaroo and an ensemble of photons is well known.> Upon recovery from this test project, the staff decided a better approach> would be to use the standard kangaroo model on the speculation that the> standardized equipment would hop itself into position. Before proceeding> with this project some facts need be established. As standard kangaroos> are smaller that standard elephants, will n standard kangaroos suf?e or> will 3n standard kangaroos be needed? Also once in place, how is a> standard hopping kangaroo convinced to remain stationary?You are completely right that 3n standard kangaroos would beneeded for this task. (or more correctly 3*(n-1)*(1+epsilon),where epsilon is an empirical constant)We have made the observation, that standard kangaroos jump lessfrequently when we bind them together with an inelastic rope in pairs.Maybe this could help to make the measurement condition more stable ? <1072626573.232780@news.aic.at> === > elephant blowing its nose upon him and again by the project > director who had her butt whooped by a standard trunk, that > standard elephants may lack appropriate manners needed for > cooperation on a long long trip. Any further efforts on that > ?ld trip were immediately abandoned when n-1 standard elephants > taking notice of the ruckus, ?led their standard trunks with > wet water and eagerly approached the experimental site.>Oh ... we are very sorry that your research group suffered such >complications - somehow there is a very huge misunderstanding here : >In order to measure the mass of the galactic black hole on should >use VIRTUAL ELEPHANTS !!!Again upon inquiring of 262,144 standard elephants from 256 herds we foundnone knew how to virtualize. We even introduced them to the Cheshire Cat,who if you recall would virtualize into thin air with a smile. At ?stthey took note of him for smelling like a feline, but then when he smiledthey soon lost interest. We then enlisted the assistance of an elephanttrainer, but try as he may, even the most intelligent and well trainstandard elephants could or would not smile and smile from ear to ear likethe Cheshire Cat.One elephant however, the standard elephant who earlier expressed interestin travel to far and distance places, took interest in the Cheshire Cat.He was so bemused by the cats smile that in response he widened andwidened his ears. This so please the Cheshire that they soon becamesteadfast friends. Not long thereafter they turned to staff and trainer;the Cheshire Cat smiling ever wider smiles at us, and the cooperativestandard elephant spreading his ears every wider and wider, graduallyvanished before us appearing as naught but a wide smile and wider earswhich eventually faded from sight.This second ?ld trip into the standardized land of the elephants, thoexceedingly more pleasant, found no standard elephants who could or wouldvirtualize except for the one standard elephant friend of the Cheshire Catwho, with the Cat, virtualized into a virtual reality without telling uswhich of the virtually in?ite virtual realities they went to.However our research has suddenly had to shift direction.You may will have noticed this astounding news story: anti-quark, was discovered at Los Alamos laboratories on having been observed changing gender right in the middle of an.., experiment, has Los Alamos scientists greatly perplex over chronom tunneling since it was detected a few days before its discovery.your help to ?d the mass of the weirdo quickly by Monday before thescientists returned to the lab. That no longer appears possible, nor isit now of importance with the breaking of that story. Having reviewed andrechecked the events of the weekend we now know the weirdo appearedundetected at Livermore Research Center to be detected at Los Alamos. Westrongly suspect this is because associated with chronom tunneling isspatial distortion, displacement or tunneling.We attempted to model this possibility using the equations of generalrelativity on a super-computer. Preliminary results are perplexing. Wesuspect the super-computer adequately comprehended the phenomena and usedit. Indications are that, quickly running out of storage space to holdtemporary data, it resorted to storing it in the past and future. Thusthe perplexment, tho the program needs but a few more hours to run, we maynot know even when the results will be collated.One of our coworkers, fatigued by the pace of this weekend, fell asleepdreaming she had consulted with Chronos the Greek god of time and Kali theproblem, nay perhaps that of uni?d space/time ?ld theory, wasunresolvable until the god and goddess of distance awoke. To awaken thesedeities one was to quietly and devoutly invoke their names three times ineach of the four directions on a mountain top inaccessible by helicopter.Can you assist our research efforts to ?d who thegod and goddess of distance are? >We have made the observation, that standard kangaroos jump less >frequently when we bind them together with an inelastic rope in >pairs.Being rushed by the pace of events, we released some standard kangaroosinto our back yard where our kids, who also like to jump up and down,could play with them. Being the offspring of scientists, they readilytook our suggestion to experiment binding standard kangaroos together inpairs. Results came soon: standard jumping kangaroos arent intobondage; for best results use bungy cords; have fun.---- === > Can you assist our research efforts to ?d who the> god and goddess of distance are?>Indra & Prajapathi ...Despite of noting the obvious : In hinduistic mythologythere is only one single omni-potent, omni-present andomni-scient god.The multifarious ?deities are nothing but different aspects ofthe same one.Ganesha is the aspect of ?everything that can be counted orcomprehended.God is everything that exists and everything that does notexist - everything that can be thought of - and - everythingthat can not be thought of ...It is an illusion and delusion to hope to understand thiscompletely.All religions are one - we are all brothers and sisters on thisfragile planet, being less than dust or vanity in the universe. === Erratum> However for more ambitious measurements, such as the exact mass of> > the Galatic black hole, does one use the Elephant in Musk Model?>Thats very easy in the framework of the Practical Elephant Model :>Put n standard-elephants into a stationary orbit around the galactic>black hole.>The measurement of the wobble-frequency and the radiation-spectra>of the elephants will provide you with the mass of your black hole> >(in units of the standard-elephant)> An interesting approach. However upon asking 262,144 standard elephants,> only one was willing to make the trip. Our small research group was thus> faced with the necessity of carrying n standard elephants themselves. Incarrying n-1 standard elephants> a pilot project, it was established that our full staff was unable to> lift, much less carry, a single standard elephant. It was also noted by> the project manager who was sprayed by the ill mannered standard elephant> blowing its nose upon him and again by the project director who had her> butt whooped by a standard trunk, that standard elephants may lack> appropriate manners needed for cooperation on a long long trip. Any> further efforts on that ?ld trip were immediately abandoned when n-1> standard elephants taking notice of the ruckus, ?led their standardwhen n-2 standard elephants> trunks with wet water and eagerly approached the experimental site. Upon recovery from this test project, the staff decided a better approach> would be to use the standard kangaroo model on the speculation that the> standardized equipment would hop itself into position. Before proceeding> with this project some facts need be established. As standard kangaroos> are smaller that standard elephants, will n standard kangaroos suf?e or> will 3n standard kangaroos be needed? Also once in place, how is awill 3(n-1) standard kangaroos and the willing standard elephant> standard hopping kangaroo convinced to remain stationary? >By the way, thanks for the hint that standard elephants could maybe> >decay into strange colorful left handed anti-quarks - we will>make some more calculations clarifying this point.> Your welcome. Please keep us inform of your progress. Do be careful> handling strange colorful left handed anti-quarks, those weirdos have> been noticed to change gender right in the middle of an.., experiment.> Whence the name. ---- === >If A is a nowhere dense subset of a > separable compact connected Hausdorff Baire space >is A countable? >The Cantor set is a nowhere dense set >with the cardinality of the reals.A most ef?ient proof. ;-)---- === > If A is a nowhere dense subset of the reals R,> then A is countable.Thats impressive eggnog youre drinking. === > The Cantor set ... take out the trashHmmm... a nice way of de?ing it :-)Rainer Rosenthalr.rosenthal@web.de === Today, at the Barnes and Noble Bookstore, I saw the book The ColossalBook of Mathematics: Classic Puzzles, Paradoxes, and Problems byMartin Gardner. The book says thate^[pi*sqrt(163)] = 262,537,412,640,768,744which is an integer exactly.Speci?ally, the book says that Srinivasa Ramanujan(1887-1920)computed the number manually to 262,537,412,640,768,743.999999, butcould not go further; then someone in France computed two milliondigits after the decimal point which are all 9; then someoneingeniously used the Eulers Constant to prove that the result isexactly the integer as given above.A quick search in the Internet revealed that the above claim is nottrue and it ?st showed up as a Aprils Fools joke by MartinGardner, and he subsequently admitted that it was a joke.Then how come it still got into the book? === > Today, at the Barnes and Noble Bookstore, I saw the book The Colossal> Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by> Martin Gardner. The book says that e^[pi*sqrt(163)] = 262,537,412,640,768,744 which is an integer exactly. Speci?ally, the book says that Srinivasa Ramanujan(1887-1920)> computed the number manually to 262,537,412,640,768,743.999999, but> could not go further; then someone in France computed two million> digits after the decimal point which are all 9; then someone> ingeniously used the Eulers Constant to prove that the result is> exactly the integer as given above. A quick search in the Internet revealed that the above claim is not> true and it ?st showed up as a Aprils Fools joke by Martin> Gardner, and he subsequently admitted that it was a joke. Then how come it still got into the book?(because the author of The Colossal Book appears to be a disciple ofMartin Gardner ;-))According to Maple:> evalf(exp(Pi*sqrt(163)), 50);; 18 .26253741264076874399999999999925007259719818568865 10which is quite remarkable, though.I dont know if Ramanujan really did this computation, but thinkingabout it: how on earth can one come to think about it? Any clue? /er. === Eric Rannaud e^[pi*sqrt(163)] = 262,537,412,640,768,744 ...> I dont know if Ramanujan really did this computation, but thinking> about it: how on earth can one come to think about it? Any clue?Please readThe Book of NumbersJohn H. Conway, Richard K. GuyCopernicus (Springer)ISBN 0-387-97993-Xand especially Chapter 8, part The Nine Magic Discriminants.Rainer Rosenthalr.rosenthal@web.de === > Today, at the Barnes and Noble Bookstore, I saw the book The Colossal> Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by> Martin Gardner. The book says that> e^[pi*sqrt(163)] = 262,537,412,640,768,744> which is an integer exactly. A quick search in the Internet revealed that the above claim is not> true and it ?st showed up as a Aprils Fools joke by Martin> Gardner, and he subsequently admitted that it was a joke. Then how come it still got into the book?>Whos the author of the book? ;-) === The author is Gardner himself. Well, today I checked the book again.At the end of the long chapter, there is actually an Addendum whichtalks about the fact that it was intended as a joke.> ...> Whos the author of the book? ;-) === > [snip]> The subject is typical representation of the OPs educational > inadequacies.Youre obviously out of your depth here. Herc has strong and validopinions about logical and philosophical issues you probably arenteven aware exist. He is also mentally ill. The two have none to dowith each other, and he still deserves respect.cid === > Doing math is a series of small successes used to build a large house> of knowledge. I never understood how people could ever dislike it.> Presumably they never experienced one of those successes, and feelthat the large house of knowledge has long been built and they have totread carefully so as not to stir the ghosts in the dusty corners. === >Im looking for criterion that ensure two groups are isomorphic.>[snip]> By group-theoretical invariants, I meant things like order and structure of> Z(G), order of [G,G], number of elements of order r for each r, etc.Including cohomology groups of G over the ring of rational integers?Yes, I would include these as examples of group-theoretical invariants.I guess you are asking whether there exist non-isomorphic G and H suchthat H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sureabout that - it sounds an interesting question.Derek Holt. === >I guess you are asking whether there exist non-isomorphic G and H such>that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure>about that - it sounds an interesting question.How about G = Z/2 x Z/3 ( = Z/6 ) and H = Z/2 * Z/3 ( = PSL_2(Z) )?But yes, this is an interesting question for _?ite_ groups, or atleast it was until Ian Leary bagged it. There are non-isomorphic?ite p-groups G and H for which H^n(G,Z) and H^n(H,Z) areisomorphic groups for every n.I think that in his example the _rings_ H^*(G,Z) and H^*(H,Z) arenot isomorphic, however. And of course even if the rings were isomorphic,one could ask about the structures of the rings as modules under theNote that cohomology is actually a functor of the corresponding group ring(Z[G] resp Z[H]) so that if it is possible to use cohomology todistinguish two groups, then it is because the group rings are alreadynon-isomorphic. So I suppose the greater challenge is to ?d two nonisomorphic (?ite) groups whose integral group rings are isomorphic;this has been conjectured never to happen and last I heard, which wasquite a while ago, this conjecture was still open. Of course, I shouldpoint out that even if the integral group ring turns out to be aperfect invariant in this sense, this really only amounts tobegging the original question, since one would then have to decide howone can determine whether or not two rings (which happen to be ?itely-generated Z-modules) are isomorphic. I dont see any reason to thinkthats an easier question than the one we began with.[I should probably add that by ring, above, I mean ring with augmentationmap. I have this dim memory that there are examples where ZG and ZH areisomorphic as rings but the isomorphism cannot commute with augmentation.]Finally, let me respond to the person who mentioned taking cohomologywith coef?ients in some other ring or ?ld. Thats a much weakerinvariant: from the Universal Coef?ient Theorem one can pretty muchdetermine H^*(G,A) where A is any trivial G-module, as soon as oneknows H^*(G,Z). The reverse is far from true and one can easily?d groups G, H with H^*(G,A) isomorphic to H^*(H,A); for example,if A is the ?ld with 2 elements and G and H have odd order, or(less trivially) if G and H are dihedral groups of different orders.Ians paper is MR1348713 (96j:20076) : $p$-groups are not determined by their integral cohomology groups. Bull. London Math. Soc. 27 (1995), no. 6, 585--589. 20J06 (20D15)dave === >Im looking for criterion that ensure two groups are isomorphic.> [snip]> By group-theoretical invariants, I meant things like order and structure of> Z(G), order of [G,G], number of elements of order r for each r, etc.Including cohomology groups of G over the ring of rational integers?> Yes, I would include these as examples of group-theoretical invariants.> I guess you are asking whether there exist non-isomorphic G and H such> that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure> about that - it sounds an interesting question.> Derek Holt.Attach to the group G the sequence:EulerianFunction(G,1),EulerianFunction(G,2), EulerianFunction(G,3),...(this is implemented in GAP ) where the EulerianFunction(G,n) is thenumber of n-tuples of elements of G that generate G .Doesnt this sequence of numbers determine G up to isomorphism ?Tim === >Im looking for criterion that ensure two groups are isomorphic.> [snip]> By group-theoretical invariants, I meant things like order and structure of> Z(G), order of [G,G], number of elements of order r for each r, etc.>Including cohomology groups of G over the ring of rational integers?> Yes, I would include these as examples of group-theoretical invariants.> I guess you are asking whether there exist non-isomorphic G and H such> that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure> about that - it sounds an interesting question.> Derek Holt.Attach to the group G the sequence:EulerianFunction(G,1),EulerianFunction(G,2), EulerianFunction(G,3),...>(this is implemented in GAP ) where the EulerianFunction(G,n) is the>number of n-tuples of elements of G that generate G .Doesnt this sequence of numbers determine G up to isomorphism ?>I dont think so.SmallGroup(32,13) and SmallGroup(32,14) give the same answer for n<=40.Derek Holt. === >In practice, groups are not normally given as multiplication tables - they are>usually de?ed as groups of permutations or matrices, or by means of>presentations. It is a much more dif?ult problem to decide whether or not>two subgroups G and H of S_n de?ed by generating permutations are>isomorphic. I am fairly sure that this is still in NP (assuming that the>putative isomorphism is given by means of images in H of generators of>G, you can compute in polynomial time a set of de?ing relators for G on>that generating set, and thereby check whether or not the map is a isomorphism)>but it may well be NP-complete.If the elements are given as permutations or matrices, then you can create >the multiplication table in poly time (for a simple minded upper bound; As Keith Ramsay observed, for permutation groups input by generators, whichis the standard model for most complexity results and research concerning?ite permutation group algorithms, you cannot construct themultiplication table in polynomial time. There are many properties ofthe group which you can compute in polynomial time, however, such asits order, the order of its derived group, centre, ...Another standard model used in complexity theory of algorithms for?ite groups is the Black-Box group, introduced by L. Babai, andmatrix groups over ?ite ?lds fall into this category. ABlack-box group is one in which the elements are represented bybit-strings of bounded length, and the group-theoretical operations(composition and inversion) can be carried out in constant time.Again groups are de?ed by means of generators. But it is muchharder to prove results about Black-box groups - most of thealgorithms are probabilistic and have a small probability of givinga wrong answer.>er...Im not so sure about a matrix representation because of the operations >needed to test equality of the matrix elements (with all those radicals >?around)).If the elements..er if the group is given as a ?ite presentation (set of >identities), then its undecidable by ...uh ... a reduction from >undecidability of uh... well it should be from the word problem but I >cant see the obvious at the moment.Yes, but there are some particular types of presentation which are usefulcomputationally, such as the power-commutator presentation of ?ite solvable(or in?ite polycyclic groups).Derek Holt. === >If the elements are given as permutations or matrices, then you can create >the multiplication table in poly time (for a simple minded upper bound; As Keith Ramsay observed, for permutation groups input by generators, which>is the standard model for most complexity results and research concerning>?ite permutation group algorithms, oh. my misunderstanding.>you cannot construct the>multiplication table in polynomial time. There are many properties of>the group which you can compute in polynomial time, however, such as>its order, the order of its derived group, centre, ...I suppose with just the permutation generators, the order could be exponential (e.g. 2 generators of degree n can generate S_n of size n!, whose number would be represented with O(n) bits). But how can you compute this order (in general) without actually constructing all elements?Mitch Harris === If the elements are given as permutations or matrices, then you can create >the multiplication table in poly time (for a simple minded upper bound; >As Keith Ramsay observed, for permutation groups input by generators, which>is the standard model for most complexity results and research concerning>?ite permutation group algorithms, oh. my misunderstanding.>you cannot construct the>multiplication table in polynomial time. There are many properties of>the group which you can compute in polynomial time, however, such as>its order, the order of its derived group, centre, ...I suppose with just the permutation generators, the order could be >exponential (e.g. 2 generators of degree n can generate S_n of size n!, >whose number would be represented with O(n) bits). But how can you compute >this order (in general) without actually constructing all elements?This is known as the Schreier-Sims algorithm. To summarize it very brie?ed, you calculate the orbit Orb(G,1) of 1 under G, then use a theoremof Schreier to compute generators of the stabilizer Stab(G,1) of 1 underG from the orbit and the generators of G. Then recursively compute|Stab(G,1)| and we have |G| = |Orb(G,1)||Stab(G,1)| by the orbit-stabilizertheorem.To get polynomial time, you have to be more careful, because thethe number of generators of Stab(G,1) coming from Schreiers Theorem isroughly |Orb(G,1)| times the number of original generators of G. To avoidan explosion in the number of generators, you maintain a subgroup H ofStab(G,1), initially trivial, for which you know the order. For each newSchreier generator of Stab(G,1) you ?st check if it already in H. I? is not, then you re-run the algorithm on H with the new generatoradjoined, and enlarge H.Derek Holt. === >There are non-isomorphic groups of order>32 (or maybe 64) which cannot be distinguished by means of any known>group-theoretical invariants. Er...then how do you distinguish them? Non group-theoretical invariants? >Or is this a case of where there is a counting argument that is >nonconstructive?>Well, if you cant think of anything better, then you just try all>possible maps from the ?st group to the second and check that it is not>an isomorphism. In practice, you would just check maps from a generating>set of the ?st group to the second, because a homomorphism is>uniquely determined by its action on a generating set.>By group-theoretical invariants, I meant things like order and structure of>Z(G), order of [G,G], number of elements of order r for each r, etc.I guess I am piqued by he use of the word cannot in your claim. Is there >a proof of that impossibility (given some arbitrary or not so arbitrary >set of group-theoretical invariants)?>I wasnt really trying to make a precise statement, and the idea ofa group-theoretical invariant is not really well-de?ed. But I would makethe following conjecture, just based on general experience, and without anyprospects of being able to prove anything precise. If you specify inadvance some collection of group-theoretical invariants, which may include things like Order and properties of Aut(G), (co)homology groups, numberof homomorphisms from G to standard groups like S_n, etc., then thereexist two non-isomorphic groups which agree in all of the invariants thatyou speci?d.Derek Holt. === >There are non-isomorphic groups of order>32 (or maybe 64) which cannot be distinguished by means of any known>group-theoretical invariants. ...>I guess I am piqued by he use of the word cannot in your claim. Is there >a proof of that impossibility (given some arbitrary or not so arbitrary >set of group-theoretical invariants)?I wasnt really trying to make a precise statement, and the idea of>a group-theoretical invariant is not really well-de?ed. But I would make>the following conjecture, just based on general experience, and without any>prospects of being able to prove anything precise. If you specify in>advance some collection of group-theoretical invariants, which may include >things like Order and properties of Aut(G), (co)homology groups, number>of homomorphisms from G to standard groups like S_n, etc., then there>exist two non-isomorphic groups which agree in all of the invariants that>you speci?d.OK, I see. This sounds very similar to the idea (conjecture? theorem?) that there is no poolyomial size set of properties of two graphs todistinguish them.Mitch Harris === [snip]> I am not aware of any such criteria. There are non-isomorphic groups of order> 32 (or maybe 64) which cannot be distinguished by means of any known> group-theoretical invariants. [snip]> Derek Holt.Could you please give an example of a pair of such hard to distinguishgroups, in GAP notation?Alan === >[snip]> I am not aware of any such criteria. There are non-isomorphic groups of order> 32 (or maybe 64) which cannot be distinguished by means of any known> group-theoretical invariants. >[snip]> Derek Holt.Could you please give an example of a pair of such hard to distinguish>groups, in GAP notation?SmallGroup(32,13) and SmallGroup(32,14) seem hard to distinguish using group-theoretical properties. Let me know if you succeed! Of course they can bedistinguished by the invariant Number of isomorphisms from given groupto SmallGroup(32,13). So I prefer to rephrase my original claim as:If you give me a collection of group-theoretical invariants, then I will?d two non-isomorphic groups which are not distinguished by them.Derek Holt. === >[snip]> I am not aware of any such criteria. There are non-isomorphic groups of order> 32 (or maybe 64) which cannot be distinguished by means of any known> group-theoretical invariants. > [snip]> Derek Holt.Could you please give an example of a pair of such hard to distinguish>groups, in GAP notation?> SmallGroup(32,13) and SmallGroup(32,14) seem hard to distinguish using group-> theoretical properties. Let me know if you succeed! Of course they can be> distinguished by the invariant Number of isomorphisms from given group> to SmallGroup(32,13). So I prefer to rephrase my original claim as:> If you give me a collection of group-theoretical invariants, then I will> ?d two non-isomorphic groups which are not distinguished by them.> Derek Holt.Well, I dont know if this quali?s, but the number of elementshaving a square root or the maximal number of square roots of anysingle element seem to distinguish between those two. Of course,there will be non-isomorphic groups that cannot be told apart bythese, but such invariants can be generalized in many ways into anin?ite array of invariants, and I wouldnt count on there being nosuch an array that is capable of distinguishing any pair of ?itenon-isomorphic groups.This, of course, says nothing about ef?iency of computation - Imreferring to sets of invariants that seem more natural than thenumber of isomorphisms to some ?ed group G as G varies over all?ite groups.Alan === > If an abelian group G has two elements with order m and n respectively,> show G has an element whose order is the least common multiple of m and n.> a, b of order m and n respectively.> m is the smallest positive integer such that a^m = e.> n is the smallest positive integer such that b^n = e.One can show from the de?itions of greatest common denominator (gcd) andleast common multiple (lcm) thatlcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say).(This is a handy thing to remember too.)so lg = mn ; m = qg ; n = rg since g divides both m and n.All we need to know about q and r is that they are integers. Thenl = mn/g = r*q*g = rm = qn ; so(ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = eand the order of ab is l = lcm(m,n). Note the if (m,n) = 1(i.e., are relatively prime), the order of the product ab is the product of the orders of a and b since [m,n] = mn in this case.Van === > One can show from the de?itions of greatest common denominator (gcd) and> least common multiple (lcm) that lcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say). (This is a handy thing to remember too.)> so lg = mn ; m = qg ; n = rg since g divides both m and n.> All we need to know about q and r is that they are integers. Then l = mn/g = r*q*g = rm = qn ; so (ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = e and the order of ab is l = lcm(m,n). Note the if (m,n) = 1> (i.e., are relatively prime), the order of the product ab is the> product of the orders of a and b since [m,n] = mn in this case. VanThats not correct.Two things:1A) if l = lcm(m,n) (we know a and b commute) then it is obvious that (ab)^l= e2A) that doesnt say l is the smallest positive integer such that (ab)^l =e,ie that doesnt say (ab) has order l.--Julien Santini === Your solution looks ?e to me.>The problem is, what if a^r and b^s are multiplicative inverses ... then>this could equal 1. But there is no reason to assume that they are inverses, since a and b arearbitrary elements--you are creating an unnecessary problem for yourself.Yes, it is possible that in some particular case that a^r b^s = 1, but the problemis for the general case, for any m and n.Van === Your solution looks ?e to me.>The problem is, what if a^r and b^s are multiplicative inverses ... then>this could equal 1. But there is no reason to assume that they are inverses, since a and b are>arbitrary elements--you are creating an unnecessary problem for yourself.>Yes, it is possible that in some particular case that a^r b^s = 1, but the problem>is for the general case, for any m and n.Sorry, but this is nonsense. While there is no reason to assume theyare inverses, there is in principle no way to discard that situationeither. As such, you ->must<- consider that possibility aswell. Saying that the elements are arbitrary does not make itimpossible for that situation to occur. If your argument is to workfor ANY two elements, then it must also work in that case.-- === =========================================== === ====Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === ====================================== === =========Arturo Magidinmagidin@math.berkeley.edu===> One can show from the de?itions of greatest common denominator (gcd) and> least common multiple (lcm) that lcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say). (This is a handy thing to remember too.)> so lg = mn ; m = qg ; n = rg since g divides both m and n.> All we need to know about q and r is that they are integers. Then l = mn/g = r*q*g = rm = qn ; so (ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = e and the order of ab is l = lcm(m,n). Note the if (m,n) = 1> (i.e., are relatively prime), the order of the product ab is the> product of the orders of a and b since [m,n] = mn in this case.I should not have used the letter l here, it looks too much like the number one (1).Also, I apologize for reading only the problem and not your solution;> Thats not correct.> Two things:> 1A) if l = lcm(m,n) (we know a and b commute) then it is obvious that (ab)^l> = eWell, even you took a few lines to show it.> 2A) that doesnt say l is the smallest positive integer such that (ab)^l =e,> ie that doesnt say (ab) has order l.True, I wasnt thinking very clearly when I posted it.Ill try to come up with a proof of this.Van === -- set theory limitsIn set theory, for the powerset P(S) of a set Slimits of set sequences Aj are de?ed as lim Aj = liminf Aj = limsup Ajprovided liminf Aj = limsup Aj, liminf Aj = /{ /{ Aj | j >= n } | n in N } limsup Aj = /{ /{ Aj | j >= n } | n in N } / /; cap cup; intersection unionDoes this de?ition of limit produce a topology for P(S) ?-- powerset topologyThe powerset topology for a powerset P(S) is de?edas the topology produced by subbase sets of the form { A in P(S) | a in A }, { A in P(S) | a not in A }A base for this topology are the sets of the form { X | A subset X subset SB }for A,B ?ite subset S.A local base for U in P(S) are the sets of the form { X | A subset X subset SB }A ?ite subset U, B ?ite subset SUWe have the following theorems:The powerset topology for P(S) is homeomorphic to the producttopology of {0,1}^P(S), the space of characteristic functions. {0,1} has discrete topology.Thus P(S) is zero-dimensional compact Hausdorff.P(S) is 1st countable iff S countable iff P(S) is 2nd countable-- limit equivalenceThe powerset topology has the desired property A = set-theory-lim Aj iff A = topology-lim AjHence when S is countable, the powerset topology can be describedby set theory limits. When S is uncountable, I would conjecture 1) the de?ition of set theory limits can be extended to nets 2) limit equivalence will be preserved 3) the powerset topology can be described by net/set theory limits-- QuestionsHas any use of the powerset topology been made?Have these notions been extended or re?ed?Does this topic show up in modern analysis?Who else has consider the powerset topology?Comments most welcome as well as additions, refutations or requests for proofs.-- latticed ordered topological spacesAs P(S) is the protype for complete complemented atomic distributivelattices or complete atomic Boolean algebras, the powerset topology can beextended to those spaces.Upon using ?subset for the order <=, of P(S), the powerset topologypresents a base of convex sets and the T_2-ordered property, that <=is closed subset P(S)xP(S). Thus it is an appropriately well manneredtopology for a partially ordered set.Order and Topology, http://at.yorku.ca/t/a/i/c/05.htmAn ordered topological space is a topological space (X,T) equipped with apartial order <=. Usual compatibility conditions between the topology andorder include convexity (T has a basis of order-convex sets) and theT_2-ordered property: <=, ie { (x,y) | x <= y }, is closed in XxX.---- === > -- set theory limits> In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are de?ed as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj,> liminf Aj = /{ /{ Aj | j >= n } | n in N }> limsup Aj = /{ /{ Aj | j >= n } | n in N }> / /; cap cup; intersection union> Does this de?ition of limit produce a topology for P(S) ?Perhaps that I misunderstood the question, but I would say that node?ition of limit produces a topology. Consider the set R of thereals and take the following de?ition of limit: the only sequencesa_1, a_2, a_3, ... that posess a limit are those such that, for somereal number l, a_n = l for every n large enough; in that case, thelimit of the sequence is l.The problem here is that you can de?e at least two topologies in Rsuch that the de?ition of limit made above is the notion of limitthat corresponds to those topologies: the discrete topology and thetopology such that the closed sets are R, the ?ite sets and thecountable sets.I hope that this helps.Jose Carlos Santos === > liminf Aj = /{ /{ Aj | j >= n } | n in N } > limsup Aj = /{ /{ Aj | j >= n } | n in N }> Does this de?ition of limit produce a topology for P(S) ? >Perhaps that I misunderstood the question,Perhaps I should have asked, When does this ... >but I would say that no de?ition of limit produces a topology.A de?ition of limit together with requirement ?st countable would.The issue of ?st countable P(S) is consider 2nd & 3rd sections. >Consider the set R of the reals and take the following de?ition of >limit: the only sequences a_1, a_2, a_3, ... that posess a limit are >those such that, for some real number l, a_n = l for every n large >enough; in that case, the limit of the sequence is l.A sequence converges iff its eventually constant. >The problem here is that you can de?e at least two topologies in R >such that the de?ition of limit made above is the notion of limit >that corresponds to those topologies: the discrete topology and theWhich is ?st countable. Indeed cl A = A = limit points of A. >topology such that the closed sets are R, >the ?ite sets and the countable sets.An uncountable cocountable space isnt 1st countable.Here again A = limit pts of A, howevercl A = R when A uncountable, = nulset when A countable.---- === >but I would say that no de?ition of limit produces a topology.> A de?ition of limit together with requirement ?st countable would.> The issue of ?st countable P(S) is consider 2nd & 3rd sections.I have this curious habit of reading texts from the top to the bottom, notfrom the bottom to the top. And besides, what you mention below is that acertain speci? topology about which you state that it is ?st countable.How was I supposed to deduce from that that, in the ?st section, you werelooking for a ?st countable topology?>Consider the set R of the reals and take the following de?ition of> >limit: the only sequences a_1, a_2, a_3, ... that posess a limit are>those such that, for some real number l, a_n = l for every n large>enough; in that case, the limit of the sequence is l.> A sequence converges iff its eventually constant.If that was all that I wanted to write, I would have done so. But whatconstant but, furthermore, that the limit of such a sequence is thatconstant (and yes, I know that it could not be otherwise).>The problem here is that you can de?e at least two topologies in R>such that the de?ition of limit made above is the notion of limit>that corresponds to those topologies: the discrete topology and the> Which is ?st countable. Indeed cl A = A = limit points of A.Dont you think that it is much more simple to see directly from the de?itionthat the discrete topology is ?st countable? All that you have to do is tosee that, for any point p, {{p}} is a fundamental system of neighborhoods ofp.>topology such that the closed sets are R,>the ?ite sets and the countable sets.> An uncountable cocountable space isnt 1st countable.Indeed, but I never said otherwise.Jose Carlos Santos === > -- set theory limits> In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are de?ed as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj,> liminf Aj = /{ /{ Aj | j >= n } | n in N }> limsup Aj = /{ /{ Aj | j >= n } | n in N }> / /; cap cup; intersection union> Does this de?ition of limit produce a topology for P(S) ?> Perhaps that I misunderstood the question, but I would say that no> de?ition of limit produces a topology. Consider the set R of the> reals and take the following de?ition of limit: [...]> The problem here is that you can de?e at least two topologies in R> such that the de?ition of limit made above is the notion of limit> that corresponds to those topologies [...]Generally, one refers to the smallest topology where the stated limitsexist. Cf. the concept of induced topology (e.g., product topology). In particular, the topology whereto William refers is the naturalhomeomorphism of the product space {0,1}^A, where {0,1} is discrete. He states the result that the only topological limits of sequences inthis space are those which are also set-limits.William asked some interesting questions, some of which I posted inearlier threads. (Much of this was hidden in the thread on the set ofcountable ordinals.) In particular, is this same topology thesmallest where convergence of general set-nets are also topologicallimits? If so, are all topological limits also set-limits? Also, isthis the same as the topology induced by all f in [0,1]^A suchthat lim f a = f (lim a) for all set-convergent nets a in A?Stephen J. Herschkorn === liminf Aj = /{ /{ Aj | j >= n } | n in N }> limsup Aj = /{ /{ Aj | j >= n } | n in N }Where are such limits used? Do they show up in modern analysis? >Generally, one refers to the smallest topology where the stated >limits exist. Cf. the concept of induced topology (e.g., product >topology). In particular, the topology whereto William refers is the >natural homeomorphism of the product space {0,1}^A, where {0,1} is >discrete. He states the result that the only topological limits of >sequences in this space are those which are also set-limits. >William asked some interesting questions, some of which I posted in >earlier threads. In particular, is this same topology the smallest >where convergence of general set-nets are also topological limits?Ive looked closer into the details of this and I see no need basicneed to change the proofs of set limits equivalent topology limitsto a the proofs for net-set limits quivalent net-topology limitsThis is because the proofs when casted into the expressions of eventually inare identical; the only change to be made be the invisible change sequence-eventually to net-eventually.Again the key to the details is to use topology convergence to xhappens when a sequence/net is eventually in every subbase set of x. andlim Aj = A iff for all x in A, some n with for all j >= n, x in Aj for all x not in A, some n with for all j >= n, x not in Ajequivalently for all x in A, x eventually in (Aj)_j for all x not in A, x eventually in (SAj)_j >If so, are all topological limits also set-limits? Also, is >this the same as the topology induced by all f in [0,1]^A such >that lim f a = f (lim a) for all set-convergent nets a in A?Now that I dont see. Ill concur on {0,1}^P(A) for discrete {0,1}; but on [0,1]^A for [0,1] subspace reals?This you needs clarify.For {0,1}^P(S) we have, the projections are p_a:P(S) -> {0,1}, c_A -> c_A(a) = ?a in Awhich are continuous and thus preserve limits, ie t-lim f(Aj) = f(t-lim Aj) = f(set-lim Aj)Would that suf?e or do you want ? set-lim f(Aj) = f(set-lim Aj)But as the projections are into discrete {0,1} the notionof set-limits in the codomain of the projections experiences suchchaotic turbulance that such digression vanishes of its own accord.---- === > -- set theory limits> In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are de?ed as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj,> liminf Aj = /{ /{ Aj | j >= n } | n in N }> limsup Aj = /{ /{ Aj | j >= n } | n in N }> / /; cap cup; intersection union> Does this de?ition of limit produce a topology for P(S) ?> [...]> Generally, one refers to the smallest topology where the stated limits> exist. Cf. the concept of induced topology (e.g., product topology). > In particular, the topology whereto William refers is the natural> homeomorphism of the product space {0,1}^A, where {0,1} is discrete. > He states the result that the only topological limits of sequences in> this space are those which are also set-limits.>Correction: The *smallest* topology is the indiscrete one. Here wewant the *largest* topology whereunder the sequences converge to thedesignated limits. Let X = N U {infty} under the usual topology. The topology we seek is exactly that coinduced by all functions f inP(A)^X such that lim f = the set limit. This does not contradict thestatement about {0,1}^A. (At least I *think* there is no larger suchtopology.)Given a set Y and a collection F = {(X,f): X is a topologicalspace and f in Y^X}, the topology coinduced on Y by F is thelargest topology such that (X,f)in F implies f is continuous. Theprototypical example is the quotient topology, coinduced by thecanonical map to equivalence classes.I confess to not remembering right now the exact proof that such alargest topology exists. I.e., given a collection t of topologiessuch that each f is continuous with each topology in t, why is itthat each f is continuous in the topology generated by Ut? The prooffor induced topologies (which do refer to a smallest topology =intersection of topologies) is much more straightforward.Stephen J. Herschkorn === > provided liminf Aj = limsup Aj, > Generally, one refers to the smallest topology where the stated > limits exist. Cf. the concept of induced topology (e.g., product > topology). In particular, the topology whereto William refers is > the natural homeomorphism of the product space {0,1}^A, where {0,1} > is discrete. He states the result that the only topological limits > of sequences in this space are those which are also set-limits.>A needs be taken as P(S). The product topology is the intitial, I thinkthats the induced, topology of the projections. By de?ition its thesmallest topology making all the projections continuous. As the latticeof topologies is a complete lattice, (its also complemented,non-distributive, atomic) such a smallest can be found and then proven tohave the continuity property. Easier is to construct a such topology andshow its the smallest. For perspective, the largest topologycontinuingfying all the projections is the discrete topology. >Correction: The *smallest* topology is the indiscrete one. Here we >want the *largest* topology whereunder the sequences converge to the >designated limits. Let X = N U {infty} under the usual topology.Which is? If X is going to simulate a sequence or net, then it needs acountable set order isomorphic to N or an updirected domain of the net. >The topology we seek is exactly that coinduced by all functions f in >P(A)^X such that lim f = the set limit. This does not contradict >the statement about {0,1}^A. (At least I *think* there is no larger >such topology.)How are you using A? Do you actually mean P(P(S))^X or P(S)^X ? >Given a set Y and a collection F = {(X,f): X is a topological >space and f in Y^X}, the topology coinduced on Y by F is the >largest topology such that (X,f)in F implies f is continuous. >The prototypical example is the quotient topology, coinduced by the >canonical map to equivalence classes.f:X -> Y is embedding X into Y the disjoint sum of F copies of X.Each f in an embedding of an X into a unique portion of Y.Another application is the embeddings of subspaces into the mother space.The quotient space comes when the collection of functions is just one.Are you suggesting some sort of multiple quotients of a space? >I confess to not remembering right now the exact proof that such a >largest topology exists. I.e., given a collection t of >topologies such that each f is continuous with each topology in >t, why is it that each f is continuous in the topology generated by >Ut? The proof for induced topologies (which do refer to a smallest >topology = intersection of topologies) is much more straightforward.The smallest topology is the indiscrete topology. For the details of the?al topology, Id suggest a similar approach as for the initialtopology.---- === > In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are de?ed as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj, > Generally, one refers to the smallest topology where the stated> limits exist. Cf. the concept of induced topology (e.g., product> topology). In particular, the topology whereto William refers is> the natural homeomorphism of the product space {0,1}^A, where {0,1}> is discrete. He states the result that the only topological limits> of sequences in this space are those which are also set-limits. A needs be taken as P(S). The product topology is the intitial, I think> thats the induced, topology of the projections. By de?ition its the> smallest topology making all the projections continuous. As the lattice> of topologies is a complete lattice, (its also complemented,> non-distributive, atomic) such a smallest can be found and then proven to> have the continuity property. Easier is to construct a such topology and> show its the smallest. For perspective, the largest topology> continuingfying all the projections is the discrete topology.Sorry, by A I meant S. And you are incorrect about A being P(S). The topology you describe makes P(S) homeomorphic to the product{0,1}^S under the product topology, where the homeomorphism is theusual identi?ation of subsets with characteristic functions. Morebelow.>Correction: The *smallest* topology is the indiscrete one. Here we>want the *largest* topology whereunder the sequences converge to the> >designated limits. Let X = N U {infty} under the usual topology.> Which is? If X is going to simulate a sequence or net, then it needs a> countable set order isomorphic to N or an updirected domain of the net.By the usual topology, I mean the one with the base {{n}, N[n,infty]: n in N}. Here, N[n,infty] denotes {n, n+1, n+2,... infty}, Anotherway to put it is X is the ordinal omega + 1 under the ordertopology.> >The topology we seek is exactly that coinduced by all functions f in>P(A)^X such that lim f = the set limit. This does not contradict>the statement about {0,1}^A. (At least I *think* there is no larger>such topology.)> How are you using A? Do you actually mean P(P(S))^X or P(S)^X ?A = S, again. So a better way of describing the coinducing functions f are those in P(S)^X such that f(n) -> f(infty) (set-wise) as n-> infty.I am noting that this coinduced topology seems to be the same as thehomeomorph of the product topology on {0,1}^S (which by de?ition isan induced topology - by de?ition, the smallest topology satisfyingcertain conditions).BTW, given any set Y and a collection of sequences with designatedlimits in Y, the (largest) topology generated by thesesequence-limits is again the coinduced topology by the correspondingfunctions in Y^(omega + 1).>Given a set Y and a collection F = {(X,f): X is a topological>space and f in Y^X}, the topology coinduced on Y by F is the>largest topology such that (X,f)in F implies f is continuous.>The prototypical example is the quotient topology, coinduced by the> >canonical map to equivalence classes.> f:X -> Y is embedding X into Y the disjoint sum of F copies of X.> Each f in an embedding of an X into a unique portion of Y.> Another application is the embeddings of subspaces into the mother space.> The quotient space comes when the collection of functions is just one.> Are you suggesting some sort of multiple quotients of a space?I am just stating the standard de?ition of the coinduced topology.>I confess to not remembering right now the exact proof that such a>largest topology exists. I.e., given a collection t of>topologies such that each f is continuous with each topology in>t, why is it that each f is continuous in the topology generated by> >Ut? The proof for induced topologies (which do refer to a smallest>topology = intersection of topologies) is much more straightforward.> The smallest topology is the indiscrete topology. For the details of the> ?al topology, Id suggest a similar approach as for the initial> topology.I think you misinterpret what I was saying. Being redundant here,Given a set X and a collection F of functions each of whose domainis X and whose range is some topological space, the topology on X induced by F is the smallest topology on X such that all thefunctions in F are continuous.- Note that the ranges of the functions in F need not be the samespaces.- Such a topology exists: It is the intersection of all topologies on X whereunder all the functions in F are continuous. Thecollection of such topologies is not empty, since it contains thediscrete topology.- Example 1. The product topology is the topology induced by theprojection maps.- Example 2. The relative topology is the topology induced by theinclusion map.Given a set Y and a collection F of functions each of whose domainis some topological space and whose range is Y, the topology on Y coinduced by F is the largest topology on Y such that all thefunctions in F are continuous.- Note that the domains of the functions in F need not be the samespaces.- Earlier, I wondered why we know such a topology exists. Havingslept on it, I realize the answer: Let t be the collection oftopologies on Y whereunder all functions in F are continuous. t is not empty, since it contains the indiscrete topology. Thecoinduced topology is the smallest topology containing Ut. It iseasy to check that if the inverse of a function maps sets in a subbaseto open sets, then it is continuous. (The last sentence is what I wasmissing earlier.)- Example. The quotient topology is the topology coinduced by thecanonical map to an equivalence class. In fact, I think this is themost general coinduced topology. Given F as in the de?ition, say x and y in Y are equivalent iff f(x) = f(y) for all f in F. I am not sure about this; at least this works if F is a singleton.One more example to compare the two concepts:Let X = [0,1] under the usual topology.Let Y = {(x,y) in R^2: x^2 + y^2 = 1}, the unit circle.Let Z = [-1,1] under the usual topology.Then the usual topology on Y can be generated in the following twoways:It is that induced by the projection maps from Y to Z^2.It is also that coinduced by f: X -> Y, x|->(cos 2pi x, sin 2pi x).Stephen J. HerschkornX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBSDc1w02521; === There are enough and coherent reasoning to prove that it is not possible to imagine the list of all real numbers without running against the nature of the natural numbers. For instance, if we admit the one-to-one correspondence between N and R, then we are confronting two different kinds of in?ity, the potential in?ite represented by the asymptotic approximation of the naturals to the in?ite (never reaching it), and the actual in?ite represented by the set of all real numbers; so that, a complete bijection between both sets is not possible. With another simple reasoning we arrive to the conclusion that the acceptance of the existence of that list would imply naturals with an in?ite value (cardinal). On the other hand, there are constructive methods (at least I know one of them) that by simple induction they proof that such a list will require naturals with an in?ite number of ?ures.Therefore, my question is this: If the ?st of all real numbers?s not a valid concept, why must we accept it as premise in the Cantor?proof of the diagonal? The theory of trans?ite numbers would be able to be an air castle with such an argument in its foundations. Nicolas de la Foz === > For instance, if we admit the one-to-one correspondence between N and R,> then we are confronting two different kinds of in?ity, the potential in?ite> represented by the asymptotic approximation of the naturals to the in?ite> (never reaching it), and the actual in?ite represented by the set of all real> numbers;How do you handle the fact there are more than two grades of in?ity? For example, the set F of all functions f: R->R is larger than theset of reals R, such that you cannot place these into one-to-onecorrespondence either. Yet you call the smaller set R an actualin?ity. What does that make F? An even more actual in?ity?And how about the power set of F? Thats bigger still.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) === at 01:38 PM, nico80@jazzfree.com (Nicolas de la Foz) said:>There are enough and coherent reasoning to prove that it is not>possible to imagine the list of all real numbers without running>against the nature of the natural numbers.No. There is only posturing and the use of words in contexts wherethey have no meanings.>For instance, if we admit the one-to-one correspondence between N>and R, We cant, because there is no such 1-1 correspondence.>the potential in?iteWhat is a potential in?ite and what does it have to do with eitherN or R?>asymptotic approximation of the naturals to the in?iteDo you believe that the above phrase has any meaning? If so, why?>the actual in?iteWhat is the actual in?ite and what does it have to do with eitherN or R?>Therefore, my question is this: If the .99list of all real numbers.9a is>not a valid concept, why must we accept it as premise in the>Cantors proof of the diagonal? We dont. We show that such a premise would lead to a contradictionand is therefor false.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > There are enough and coherent reasoning to prove that it is not possibleto imagine the list of all real numbers without running against the natureof the natural numbers. For instance, if we admit the one-to-onecorrespondence between N and R, then we are confronting two different kindsof in?ity, the potential in?ite represented by the asymptoticapproximation of the naturals to the in?ite (never reaching it), and theactual in?ite represented by the set of all real numbers; so that, acomplete bijection between both sets is not possible. With another simplereasoning we arrive to the conclusion that the acceptance of the existenceof that list would imply naturals with an in?ite value (cardinal). On theother hand, there are constructive methods (at least I know one of them)that by simple induction they proof that such a list will require naturalswith an in?ite number of ?ures.> Therefore, my question is this: If the list of all real numbers is not avalid concept, why must we accept it as premise in the Cantors proof of thediagonal? The theory of trans?ite numbers would be able to be an aircastle with such an argument in its foundations.This is one of my own pet hobbyhorses. I am a ?m believer in only onein?ity, which is not even signed.Every presentation of the Cantor argument has turned upon a (partial) set ofreals, which is placed in one-to-one correspondence with a (partial) set o?tegers. Here is the example from G.9adel, Escher, Bach: 1 .1 4 1 5 9 2 6 5 3 2 .3 3 3 3 3 3 3 3 3 3 .7 1 8 2 8 1 8 2 8 4 .4 1 4 2 1 3 5 6 2 5 .5 0 0 0 0 0 0 0 0The argument then goes by constructing a new real by taking the elements ofthe diagonal and changing every digit. The argument then goes that (a) byconstruction, this new real is not in the list of reals (b) by construction,there is a real against every integer, (c) THEREFORE, there are more realsthan integers, even though the list of integers is in?ite.My principal observation is that while the column containing the integers isquite obviously ordered, the column containing the reals is not. It is thisdisorder in the example list of reals that is the clincher. Once an orderingscheme is imposed on the reals, a true one-to-one correspondence,expressible as a simple formula, can be constructed. One example which hasbeen mooted here on a number of occasions is to reverse the order of thedecimal digits and disregard the decimal point [1], thus the real 0.5 mapsto the integer 5 as per the last line of the example, the real 0.890625 mapsto the integer 526098, etc.OK so all non-terminating decimals, recurring or not, map to in?ity bythis method. No problem, there is enough room at in?ity for all of them.Each successive Cantor diagonalisation of the list will give a newnonterminating [2] real and a new in?ite integer to map it to.[1] This con?es the reals to the range 0 to 1, but only a trivialmodi?ation is required to remove this restriction. Simply pass thein?ite range of reals through a mathematical function that maps them to a?ite range in 1:1 correspondence. The tanh function is a good one to usefor this purpose.[2] The diagonalisation must give a non-terminating decimal since allterminating ones are mapped to the ?ite integers by construction, so theyare all already in the list.--Paul V. S. TownsendInterchange the alphabetic elements to replyX-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 03:46 PM, Prai Jei said:>My principal observation is that while the column containing the>integers is quite obviously ordered, the column containing the reals>is not.It clearly *IS* ordered, by the very act of arranging them into acolumn.>Once an ordering scheme is imposed on the reals, a true one-to-one >correspondence, expressible as a simple formula, can be constructed.You are confusing ordering with well ordering. Further, you areconfusing the statement to be disproved with a statement that has beenproven. In fact the diagonal argument proves that such a column isimpossible.>One example which has>been mooted here on a number of occasions is to reverse the order of>the decimal digits and disregard the decimal point [1],And what does 1/3 map to?>OK so all non-terminating decimals, recurring or not, map to>in?ityWhat is in?ity Does 1/3 map to the same in?ity as 1/6?>there is enough room at in?ity for all of them.What does that mean?>in?ite integerWhat is an in?ite integer?What youre doing isnt Mathematics; it isnt even Philosophy. De?eyour terms if you wish to be taken seriously, then use them only in afashion consistent with how you de?ed them.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > This is one of my own pet hobbyhorses. I am a ?m believer in only one> in?ity, which is not even signed.> Every presentation of the Cantor argument has turned upon a (partial) set of> reals, which is placed in one-to-one correspondence with a (partial) set of> integers. Here is the example from G.9adel, Escher, Bach:> 1 .1 4 1 5 9 2 6 5 3> 2 .3 3 3 3 3 3 3 3 3> 3 .7 1 8 2 8 1 8 2 8> 4 .4 1 4 2 1 3 5 6 2> 5 .5 0 0 0 0 0 0 0 0> The argument then goes by constructing a new real by taking the elements of> the diagonal and changing every digit. The argument then goes that (a) by> construction, this new real is not in the list of reals (b) by construction,> there is a real against every integer, (c) THEREFORE, there are more reals> than integers, even though the list of integers is in?ite.> My principal observation is that while the column containing the integers is> quite obviously ordered, the column containing the reals is not. It is this> disorder in the example list of reals that is the clincher. Once an ordering> scheme is imposed on the reals, a true one-to-one correspondence,> expressible as a simple formula, can be constructed. One example which has> been mooted here on a number of occasions is to reverse the order of the> decimal digits and disregard the decimal point [1], thus the real 0.5 maps> to the integer 5 as per the last line of the example, the real 0.890625 maps> to the integer 526098, etc.> Im having trouble understanding the ? your response. This isthe simple formula, from [0,1] to Z, which you purport to ? below,since it has been mooted here before?> OK so all non-terminating decimals, recurring or not, map to in?ity by> this method. No problem, there is enough room at in?ity for all of them.Its at this point that I begin not to understand. What does this haveto do with the ordering you induce?Unfortunately, in?ity isnt in the standard de?itions of Z Imfamiliar with, but this is not so much a hurdle. Besides, even if itwere, you said you were a ?m believer in a single unsigned in?ity,so every nonterminating real in [0,1] would be mapped to this singleelement; this mapping isnt a bijection and has little to do withcardinality.> Each successive Cantor diagonalisation of the list will give a new> nonterminating [2] real and a new in?ite integer to map it to.> Says the ?m believer in a single in?ity? Are you suggesting thatwe de?e new elements and union them with Z to make this mappingwork? The problem is, well be adding an uncountable number o??ities to Z, so the set to which youre mapping wont be of thesame cardinality as Z. In other words, youve changed your mappingfrom [0,1]->Z to [0,1]->(Z union I), where I is the new set o??ities you de?e. Since there is no bijection between Z and (Zunion I), this shows us that they dont have the same cardinality.Youre back where you began and havent ?ed this simple mapping.in?ity. The argument against this ? still stands if you justlook at (Z union I), where we say that I is the set of in?iteintegers, which have never been part of Z and arent countable. Allyouve really done is removed a decimal point from the analysis.> [1] This con?es the reals to the range 0 to 1, but only a trivial> modi?ation is required to remove this restriction. Simply pass the> in?ite range of reals through a mathematical function that maps them to a> ?ite range in 1:1 correspondence. The tanh function is a good one to use> for this purpose.Lets look at f(x) = (tanh(x)+1)/2. What real number is mapped to 1and what is mapped to 0? I was under the impression that there were nosigned in?ities in your system, in which case were left with (0,1]and there is a quick detail to work out.> [2] The diagonalisation must give a non-terminating decimal since all> terminating ones are mapped to the ?ite integers by construction, so they> are all already in the list. === >[self-contradictory OP snipped]This is one of my own pet hobbyhorses. I am a ?m believer in only one>in?ity, which is not even signed.Im a ?m believer in a ?rth. Doesnt make the earth ?very presentation of the Cantor argument has turned upon a (partial) set of>reals, which is placed in one-to-one correspondence with a (partial) set of>integers. Here is the example from G.9adel, Escher, Bach:> 1 .1 4 1 5 9 2 6 5 3> 2 .3 3 3 3 3 3 3 3 3> 3 .7 1 8 2 8 1 8 2 8> 4 .4 1 4 2 1 3 5 6 2> 5 .5 0 0 0 0 0 0 0 0>The argument then goes by constructing a new real by taking the elements of>the diagonal and changing every digit. The argument then goes that (a) by>construction, this new real is not in the list of reals (b) by construction,>there is a real against every integer, (c) THEREFORE, there are more reals>than integers, even though the list of integers is in?ite.My principal observation is that while the column containing the integers is>quite obviously ordered, the column containing the reals is not. This has no relevance whatever. _By de?ition_, saying that there are more reals than integers means that if you have a mappingof integers to reals then there is at least one real not in the rangeof the mapping. Nothing there about columns being ordered.>It is this>disorder in the example list of reals that is the clincher. Once an ordering>scheme is imposed on the reals, a true one-to-one correspondence,>expressible as a simple formula, can be constructed. One example which has>been mooted here on a number of occasions is to reverse the order of the>decimal digits and disregard the decimal point [1], thus the real 0.5 maps>to the integer 5 as per the last line of the example, the real 0.890625 maps>to the integer 526098, etc.OK so all non-terminating decimals, recurring or not, map to in?ity by>this method. No problem, there is enough room at in?ity for all of them.>Each successive Cantor diagonalisation of the list will give a new>nonterminating [2] real and a new in?ite integer to map it to._By de?ition_ there is no such thing as an in?ite integer. Yourenot demonstrating anything here, all youre doing is giving wordsnew de?itions.I can _prove_ that the Earth is ?eople point out that its round, but thats no problem, because the desk in my of?eis ?s that a proof that the Earth is ?o, because thedesk in my of?e is not the Earth.If you think that there is a 1-1 correspondence betweenthe reals and the (?ite) integers youre simply wrong. Butit doesnt look like you think that - it looks like you agreethere is no such 1-1 correspondence. If so then you _do_agree that there are more reals than integers, assumingwe give all the words their standard meanings - what youthink you accomplish by using words to mean thingsother than what they _do_ mean is not clear to me.>[1] This con?es the reals to the range 0 to 1, but only a trivial>modi?ation is required to remove this restriction. Simply pass the>in?ite range of reals through a mathematical function that maps them to a>?ite range in 1:1 correspondence. The tanh function is a good one to use>for this purpose.>[2] The diagonalisation must give a non-terminating decimal since all>terminating ones are mapped to the ?ite integers by construction, so they>are all already in the list.************************David C. Ullrich === >[self-contradictory OP snipped]>This is one of my own pet hobbyhorses. I am a ?m believer in only one>in?ity, which is not even signed.Im a ?m believer in a ?rth. Doesnt make the earth ?nip argumentation why Cantors diagonal is not wrong]Id like to take a few moments to elaborate on a topic that irritatesme to no end:CANTORS PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVILS SPAWN!!!Well maybe not quite but its certainly worse than LHospitals ruleon the scale of things that are so obvious no one can possiblymisunderstand them that are consequently misunderstood all the time.This is not aided by the fact that its so deceptively simple thatit gets tacked on every single introductory freshman/high school mathtext and popularization from here to eternity.As a proof, its... well, elegant. However its so elegant that theunderlying ?esse usually goes completely unnoticed, which means weget a steady stream of cranks^Walternative thinkers who are convincedthat the proof is wrong because . Its like mathematics is a house where one of the paintings hasa funny optical illusion that makes the painting look like its tiltedwhile its perfectly straight, and every joker who waltzes in pointsout that the painting looks tilted and thinks no one else has pointedthis out before.So please, to those in the position of in?ng textbookpublishers, would it please be possible to remove this deceptivelysimple yet simply deceptive argument from texts where it is notrelevant and to replace it with a more formal two-page proof includingno more than ?e understandable English words and at least twohomeomorphisms? I think that would go a long way.And with these words we return to your scheduled gigantic Cantorthread... === [self-contradictory OP snipped]This is one of my own pet hobbyhorses. I am a ?m believer in only one>in?ity, which is not even signed.>Im a ?m believer in a ?rth. Doesnt make the earth ?nip argumentation why Cantors diagonal is not wrong]Id like to take a few moments to elaborate on a topic that irritates>me to no end:CANTORS PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVILS SPAWN!!!Well maybe not quite but its certainly worse than LHospitals rule>on the scale of things that are so obvious no one can possibly>misunderstand them that are consequently misunderstood all the time.>This is not aided by the fact that its so deceptively simple that>it gets tacked on every single introductory freshman/high school math>text and popularization from here to eternity.As a proof, its... well, elegant. However its so elegant that the>underlying ?esse usually goes completely unnoticed, which means we>get a steady stream of cranks^Walternative thinkers who are convinced>that the proof is wrong because here>. Its like mathematics is a house where one of the paintings has>a funny optical illusion that makes the painting look like its tilted>while its perfectly straight, and every joker who waltzes in points>out that the painting looks tilted and thinks no one else has pointed>this out before.So please, to those in the position of in?ng textbook>publishers, would it please be possible to remove this deceptively>simple yet simply deceptive argument from texts where it is not>relevant and to replace it with a more formal two-page proof including>no more than ?e understandable English words and at least two>homeomorphisms? I think that would go a long way.I cant decide whether youre serious. If you are then this is avery curious attitude... if youre not then you got me.>And with these words we return to your scheduled gigantic Cantor>thread...************************David C. Ullrich === > >[self-contradictory OP snipped]>This is one of my own pet hobbyhorses. I am a ?m believer in only one>in?ity, which is not even signed.Im a ?m believer in a ?rth. Doesnt make the earth ?> [snip argumentation why Cantors diagonal is not wrong]> Id like to take a few moments to elaborate on a topic that irritates> me to no end:> CANTORS PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVILS SPAWN!!!> Well maybe not quite but its certainly worse than LHospitals rule> on the scale of things that are so obvious no one can possibly> misunderstand them that are consequently misunderstood all the time.> This is not aided by the fact that its so deceptively simple that> it gets tacked on every single introductory freshman/high school math> text and popularization from here to eternity.> Cantors proof is the simplest, most accessible proof of something thats counterintuitive. Thats why it stirs up the cranks so. === > There are enough and coherent reasoning to prove that it is not >possible to imagine the list of all real numbers without running >against the nature of the natural numbers. For instance, if we >admit the one-to-one correspondence between N and R, then we are >confronting two different kinds of in?ity, the potential in?ite >represented by the asymptotic approximation of the naturals to the >in?ite (never reaching it), and the actual in?ite represented by >the set of all real numbers; so that, a complete b ijection between >both sets is not possible. With another simple reasoning we arrive >to the conclusion that the acceptance of the existence of that list >would imply naturals with an in?ite value (cardinal).So are you saying there are only ?itely many natural numbers? > On the other >hand, there are constructive methods (at least I know one of them) >that by simple induction they proof that such a list will require >naturals with an in?ite number of ?ures.> Therefore, my question is this: If the ?st of all real numbers?s not a valid concept, why must we accept it as premise in the Cantor?>proof of the diagonal? The theory of trans?ite numbers would be able >to be an air castle with such an argument in its foundations. > Nicolas de la FozIm not sure I completely understand your problem with this. The diagonalargument shows that any countably in?ite set of Real numbers cannotcontain all of them. And any countable set can be thought of as a ?list. Whats wrong with that? What asymptotic approximation of the in?ite are you talking about? Itreally reads as though you are saying there are not in?itely manynatural numbers. Or that there isnt _an_ in?itely large natural number?If you dont like proof by contradiction, dont assume that the reals arecountable, but show that _any_ countable set of them is incomplete. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBSDc1102527; === hi im am a late nite science crack pot that is very interested in ?ding out what and how r universe was made and how it works i would like to get as mutch 411 on the subject as u can send me please send me as much as u can thank u === >would like to get as mutch 411 on the subject as u can send me In the movie Hercules, at some point an accident has happened and some one gives the throw away line Somebody call IXIIThat cracked me up. Mitch === >would like to get as mutch 411 on the subject as u can send me > In the movie Hercules, at some point an accident has happened and some > one gives the throw away line> Somebody call IXII> That cracked me up. > Mitch> And Hercules was a Greek, not a Roman. === >would like to get as mutch 411 on the subject as u can send me > In the movie Hercules, at some point an accident has happened and some > one gives the throw away line> Somebody call IXII> That cracked me up. > Mitch> Well, actually it would be CMXI or IX-I-I, but its funny anyway. === > Actually, mathworld is one of my favorite bookmarks, but somehow I got lost there and didnt ?d the speci? page you mentioned. So thanks for pointing it out, I now understand the ideas much better.>I have spent many hours trying to ?d a verbal de?ition of thefollowing terms. Gradient, Divergence and Curl. I ?d themathematical de?itions but never a verbal description of exactlywhat it is Im doing when I calculate these. Can anyone give me thebasics in WORDS and not equations? Exactly what am I ?ding about thevector ?ld when I calculate the gradient, divergence and curl?Did you try http:// mathworld.wolfram.com/Divergence.html etc?I bet you didntphilIt may be heresy to mention a physics book here ! However, you coulddo worse than look at the Feynman Lectures on Physics. Volume 2 has acouple of chapters about div, grad and curl, and attempts to give aphysical meaning to these operatorsHope this helpsIan Taylor === > Actually, mathworld is one of my favorite bookmarks, but somehow I got lost there and didnt ?d the speci? page youmentioned. So thanks for pointing it out, I now understand the ideas much better.>I have spent many hours trying to ?d a verbal de?ition of the> following terms. Gradient, Divergence and Curl. I ?d the> mathematical de?itions but never a verbal description of exactly> what it is Im doing when I calculate these. Can anyone give me the> basics in WORDS and not equations? Exactly what am I ?ding about the> vector ?ld when I calculate the gradient, divergence and curl?Did you try http:// mathworld.wolfram.com/Divergence.html etc?I bet you didntphil It may be heresy to mention a physics book here ! However, you could> do worse than look at the Feynman Lectures on Physics. Volume 2 has a> couple of chapters about div, grad and curl, and attempts to give a> physical meaning to these operatorsA physics book? Heresy??Feynman??? Heresy????It is exactly what the OP needs.The only potential heresy is your hesitation about mentioning this ;-)> Hope this helpsIm sure it will :-) Ian TaylorDirk Vdm === Using the resultant in this way should be okay to prove Bezoutstheorem _if_ the curves intersect in distinct points, but things getsubtler when there are higher multiplicities, i.e., when the curvesintersect tangentially and/or intersect at singular points.The way Ive generally seen it done is that ?st one de?es theintersection index (or multiplicity) of the two curves at each pointof intersection. For example, if the curves are f(x,y)=0 and g(x,y)=0and if they intersect at the point (0,0), then the intersection indexat (0,0) is the dimension of the complex vector spaceC[x,y]_{0,0}/(f(x,y),g(x,y)).Here C[x,y]_(0,0) is the ring of polynomials localized at (0,0), whichmeans the ring of all rational functions p(x,y)/q(x,y) with theproperty that q(0,0) is not 0.Of course, this de?ition works anywhere, since you can always do atranslation X=x-a, Y=y-b to move the intersection point (a,b) to(0,0). And for points on the line at in?ity in projective space, youcan dehomogenize with respect to one of the other coordinates.Bezouts theorem then says that if you add up all of the intersectionindices for all of the intersection points, the total is equal todeg(f)*deg(g). The beauty and strength of Bezouts theorem lies in thefact that the intersection indices are computed locally in aneighborhood of each point, but that their total is equal to a globalquantity, namely the product of the degrees of the curves.The proof of Bezouts theorem is in most basic algebraic geometrybooks, using varying amounts of fancy machinery. John Tate and I put afairly elementary proof into the appendix of our book Rational Pointson Elliptic Curves (Springer-Verlag).Joe Silverman> says...>Im not an algebraic geometer (far from it),>but I would prove it by showing that the resolvant (or resultant) R(f1,f2)>of f1(x,y),f2(x,y), regarded as polynomials in y over k[x],>is a polynomial of degree <= mn in x. > Ok, I am not well versed in the resolvent.. but let me verify if I understand > this correctly. You are telling that if I work with f1 and f2 as functions of > k(y)[x] (we had like k(y) to be a ?ld, thus fraction ?ld), and solve the > resolvent we can arrive to the Bezout theorem. Am I correct that the resolvent > determines the common roots of f1 and f2 (ie if the resolvent is zero at a > point (x,y) then f1 and f2 are zero at those points). I just want to verify if > Sincerely,> Jose Capco === > Using the resultant in this way should be okay to prove Bezouts> theorem _if_ the curves intersect in distinct points, but things get> subtler when there are higher multiplicities, i.e., when the curves> intersect tangentially and/or intersect at singular points.Actually the original poster asked for a proof of what he/she calledthe weak Bezouts Theorem, that the number of intersections is <= mn.> The proof of Bezouts theorem is in most basic algebraic geometry> books, using varying amounts of fancy machinery. John Tate and I put a> fairly elementary proof into the appendix of our book Rational Points> on Elliptic Curves (Springer-Verlag).I must admit I found this to be one of the weaker sectionsof what I rate as one of the best popular mathematical books I know --one of the very few worthy to stand on the bookshelf next to Hardy & Wright.Anyone with the slightest interest in elliptic curvesshould start with Silverman & Tate, IMHO.-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === says...Actually the original poster asked for a proof of what he/she called>the weak Bezouts Theorem, that the number of intersections is <= mn.> The proof of Bezouts theorem is in most basic algebraic geometry> books, using varying amounts of fancy machinery. John Tate and I put a> fairly elementary proof into the appendix of our book Rational Points> on Elliptic Curves (Springer-Verlag).I must admit I found this to be one of the weaker sections>of what I rate as one of the best popular mathematical books I know -->one of the very few worthy to stand on the bookshelf next to Hardy & Wright.Anyone with the slightest interest in elliptic curves>should start with Silverman & Tate, IMHO.Ok thanks all :) .. I didnt know that this post could bother Mr. Silverman too :) .. anyway, you might be right (I never looked at that book but I certainly heard a lot about it). My original purpose was actually to prove the associativity law of the elliptic curve and for that I believe weak Bezout theorem suf?es. The previous threads gave me enough ideas to start with and I dont suppose I need to go very much into resolvent theory. Sincerely,Jose Capco === >Anyone with the slightest interest in elliptic curves>should start with Silverman & Tate, IMHO.I just want to second this. I absolutely love this book.Steven E. Landsburgwww.landsburg.com/about2.html-- === > Using the resultant in this way should be okay to prove Bezouts> theorem _if_ the curves intersect in distinct points, but things get> subtler when there are higher multiplicities, i.e., when the curves> intersect tangentially and/or intersect at singular points.> Actually the original poster asked for a proof of what he/she called> the weak Bezouts Theorem, that the number of intersections is <= mn.True. I was just expanding the discusssion a bit to the general case.For the weak form, as other posters noted, the resultant argument willgive the upper bound of mn, and it is certainly very elementary.> The proof of Bezouts theorem is in most basic algebraic geometry> books, using varying amounts of fancy machinery. John Tate and I put a> fairly elementary proof into the appendix of our book Rational Points> on Elliptic Curves (Springer-Verlag).> I must admit I found this to be one of the weaker sections> of what I rate as one of the best popular mathematical books I know --> one of the very few worthy to stand on the bookshelf next to Hardy & Wright.> Anyone with the slightest interest in elliptic curves> should start with Silverman & Tate, IMHO.Sorry you didnt like that section, but in any case, thank you for thekind words regarding RPEC. Its an honor to be mentioned in the samesentence as the H&W classic.Joe Silverman === > Could someone here write the sketch of the proof of the (weaker) Bezout> theorem, i.e. given two curves (say one being> irreducible.. to make matters a bit easier) C1 and C2 de?ed by> functions f1 and f2, the interesection between these> two curves is less or equal deg(f1)*deg(f2).> Im not an algebraic geometer (far from it),> but I would prove it by showing that the resolvant (or resultant) R(f1,f2)> of f1(x,y),f2(x,y), regarded as polynomials in y over k[x],> is a polynomial of degree <= mn in x.I guess if one doesnt want to go into the theory of the resolvent,f1(x,y), x*f1(x,y), x^2*f1(x,y), ..., x^{n-1}*f1(x,y),f2(x,y), x*f2(x,y), x^2*f2(x,y), ..., x^{m-1}*f2(x,y).One can regard these as m+n linear equations in 1,x,x^2,...,x^{m+n-1}.If f1(x,y),f2(x,y) have a common root in x for a given value of y,then det(A) = 0, where A is the matrix of this set of equations.This is an equation of degree <= mn in y.This is exactly the same argument --Im just pointing out that for your purposeone only needs to establish the property of the resolvent in one direction.-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === |Ok, I am not well versed in the resolvent.. but let me verify if I understand|this correctly.If two nonzero polynomials P and Q of degrees m and n have a nonconstantcommon factor R, then P*(Q/R)-Q*(P/R)=0, and S=Q/R and T=-P/R arepolynomials of degrees Im not an algebraic geometer (far from it),>but I would prove it by showing that the resolvant (or resultant) R(f1,f2)>of f1(x,y),f2(x,y), regarded as polynomials in y over k[x],>is a polynomial of degree <= mn in x.> Ok, I am not well versed in the resolvent.. but let me verify if I> understand this correctly. You are telling that if I work with f1 and f2> as functions of k(y)[x] (we had like k(y) to be a ?ld, thus fraction> ?ld), and solve the resolvent we can arrive to the Bezout theorem. Am I> correct that the resolvent determines the common roots of f1 and f2 (ie if> the resolvent is zero at a point (x,y) then f1 and f2 are zero at those> points). I just want to verify if my understanding of the resolvent wasYes, that is my understanding too.Suppose f(x), g(x) are monic (leading coef?ient 1),f(x) = x^m + a_1x^{m-1} + ..., g(x) = x^n + b_1x^{n-1} + ...If we de?e R(f,g) = prod (a_i - b_j)where a_i, b_j are the roots of f,g respectively,then it is relatively straightforward to show thatR(f,g) = det A,where A is the (m+n)x(m+n) matrixA = (1,a_1,a_2,.../0,1,a_1,.../.../1.b_1,b_2,.../0,1,b_1,.../...) .Now apply this with f(x) = f1(x,y), g(x) = f2(x,y).Then a_r, b_r are polynomials of degree r in y,and R(f,g) is a polynomial of degree mn in y.Basically, given polynomials of degrees m,n in x,ythis shows that one of the variables, x say,can be eliminated to give a polynomial of degree mnin the other variable.-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === > crap] And every word you utter simply continues to con?m what allthe> readers of> this ng think of you. Do you really not realise that? Wow. What an ignont comment. The internet is well known for being> domindated by ignorant irritating people. These people forcedecent> people of the web due to their slimey nature. I.e. they ?emo ff> the web. That means that jerks like you remain behind and ?l itup. So I expect that all newsgroups are ?l with jerks like you,varney,> bilgem, speicher etc. And when I say jerk Im being *very*gracious.> Useless troll.> Waste of time. Plonk. Any reason that you think you need to butt into a conversation betweentwo> people? You are operating under a misunderstanding if you think anything inusenet> corresponds to a converation between two people. Youre confused again. The question was simple. he refuses to answer.> Nobody said that he had no right to dimwit. The question was withNo. I am not confused.Here is your statement, lifted lock stock and barrel:Any reason that you think you need to butt into a conversation between twopeople?Here was my answer *to that statement*:You are operating under a misunderstanding if you think anything in usenetcorresponds to ?a converation between two people. No confusion is possible here, imbeciles excepted.[snip]Franz === Why are you becomming so obsessed with me? I think youve been hangingaround bilge, waite and varney for far too long. === Why are you becomming so obsessed with me? I think youve been hanging> around bilge, waite and varney for far too long.Naw... we all simply have fun poking you with a stick. === I forgot to mention that mays is one of these people who would rather> ?hen stick to physics> And anyway... when you post some physics> worthy of my taking my valuable time to comment on> I will... Until then Ill continue to poke at all> you guys with a stick....... A boys gota have a hobby... ;)~Well, you know, theres no rule you couldnt _start_ a threadcontaining, some... ahem... physics?What is a physics worthy? Is it like a witchs familiar...? Andyou say this malign entity took your valuable time to comment on!? The verve!Anyway, my physics worthy, supposing you knew all physics, you mightstill start something in a didactic vein, for the unworthy. <1g69dp0.1jt1z3ms7jy9lN%see.sig@for.addy> <3fe79cc4$12$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === at 05:15 PM, Pmb said:>Any reason that you think you need to butt into a conversation>between two people?communication between two people. Anything that you post here is asolicitation for comments from anyone reading it. Thats a separatethe news groups and whether the response was appropriate.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > at 05:15 PM, Pmb said:Any reason that you think you need to butt into a conversation>between two people? communication between two people.Show me where I said it was a private conversation. With so little said andnot much content in what was said it begs the question as to why he postedit.If he chooses not to answer then he chooses not to answer. Why does thatbother you? <1g69dp0.1jt1z3ms7jy9lN%see.sig@for.addy> <3fe79cc4$12$fuzhry+tra$mr2ice@news.patriot.net> <3fee0cff$12$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === at 11:10 PM, Pmb said:>Show me where I said it was a private conversation.Any reason that you think you need to butt into a conversationbetween two people?>Why does that bother you?It doesnt. What bothers me is your posting your crap in a publicforum and then complaining when the public responds to it. If you wantno such thing as a conversation between two people, and hence nobodycan butt into the discussion. Everybody here is a party to it,whether you understand that or not. If you want your quarrel to beprivate, conduct it in private.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > at 11:10 PM, Pmb said:>Show me where I said it was a private conversation.> Any reason that you think you need to butt into a conversation> between two people?>Why does that bother you?> It doesnt. What bothers me is your posting your crap in a public> forum and then complaining when the public responds to it. If you want> no such thing as a conversation between two people, and hence nobody> can butt into the discussion. Everybody here is a party to it,> whether you understand that or not. If you want your quarrel to be> private, conduct it in private.Amen. === > at 11:10 PM, Pmb said:>Show me where I said it was a private conversation.> Any reason that you think you need to butt into a conversation> between two people?>Why does that bother you?> It doesnt. What bothers me is your posting your crap in a public> forum and then complaining when the public responds to it.Then you need to read more carefully.The sentance - Any reason that you think you need to butt into aconversation between two people? - is a *question* and not a*complaint*.Pmb === at 05:15 PM, Pmb said:Any reason that you think you need to butt into a conversation>between two people? communication between two people. Show me where I said it was a private conversation.Here, and I quote once more:Any reason that you think you need to butt into a conversation between twopeople??Now shut up.[snip]Franz === at 05:15 PM, Pmb said:Any reason that you think you need to butt into a conversation>between two people? communication between two people. Show me where I said it was a private conversation.> Here, and I quote once more:> Any reason that you think you need to butt into a conversation between two> people??> Now shut up.Wow youre dumb.You dont know the difference between those two? I said show me whereI said it was a ************** p r i v a t e **************conversation.Wow! Youre *such* an idiot heymann!!!Now off. === >in> at 05:15 PM, Pmb said:Any reason that you think you need to butt into a conversation>between two people? communication between two people. Show me where I said it was a private conversation. Here, and I quote once more:> Any reason that you think you need to butt into a conversation betweentwo> people?? Now shut up. Wow youre dumb.You were told to shut up, crackpot. === Back in the spring of 2002 I discovered the following way to countprime numbers:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,sqrt(y-1))],S(x,1) = 0, S(x,y) is the sum of dS from dS(x,2) to dS(x,y),p(x, y) = ?) - S(x, y) - 1, and p(x,sqrt(x)) gives the count of primes where, of course, onlypositive integers are being used, so the sqrt() operator returns thelargest positive integer that squares to give the argument or a numberless than it.Now consider values for dS(x,y) from the calculation of p(100,10):dS(100,2)=49dS(100,3)=16dS(100,4)=0dS(100,5)=6dS( 100,6)=0dS(100,7)=3dS(100,8)=0dS(100,9)=0dS(100,10)=0The sum of all those values gives S(100,10)=74, which is the count ofcomposites, 74 subtracts from 100 to give 26, and subtracting 1 fromthat gives you p(100,10) = 25, and 25 is the count of primes up to100.Now then dS(x,y) is the count of composites that have y as a factorthat do NOT have a prime number less than y as a factor up to andincluding x, so it makes sense that unless y is a prime dS(x,y) is 0because, for instance, if y=4, then any number that has 4 as a factoralso has 2 as a factor, so there arent any composites that have 4 asa factor that do not have a prime number less than 4 as a factor as,of course, 2 is a factor of 4.Practically that means that dS(x,y) is 0 if y is composite, andnon-zero if y is prime and less than or equal to sqrt(x) as can beseen with the values given above.Now then the non-zero values are easy to understand, likedS(100,2)=49, means that there are 49 numbers up to and including 100that are even, excluding 2 itself, as its prime. Then dS(100,3)=16tells you that there are 16 numbers that are composite with 3 as afactor that are NOT even.Next you have dS(100,4)=0, as there arent any numbers that have 4 asa factor that do not have a smaller prime as a factor, in this case 2,as 4 has 2 as a factor.Next theres dS(100,5)=6, telling you that there are 6 composites with5 as a factor that do not have 2 or 3 as a factor.Notice the drop of non-zero dS(x,y) values is like an exponentialdrop, while the spacing between values is, of course, the spacingbetween prime numbers.And that explains:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,sqrt(y-1))]More information can be found at my blog:http://mathforpro?.blogspot.com/James Harris === > Back in the spring of 2002 I discovered the following way to count> prime numbers:> dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))],> S(x,1) = 0, S(x,y) is the sum of dS from dS(x,2) to dS(x,y),> p(x, y) = ?) - S(x, y) - 1, > and p(x,sqrt(x)) gives the count of primes where, of course, only> positive integers are being used, so the sqrt() operator returns the> largest positive integer that squares to give the argument or a number> less than it.(youve changed your line on this one havent you?)> > Now consider values for dS(x,y) from the calculation of p(100,10):> dS(100,2)=49> dS(100,3)=16> dS(100,4)=0> dS(100,5)=6> dS(100,6)=0> dS(100,7)=3> dS(100,8)=0> dS(100,9)=0> dS(100,10)=0> The sum of all those values gives S(100,10)=74, which is the count of> composites, 74 subtracts from 100 to give 26, and subtracting 1 from> that gives you p(100,10) = 25, and 25 is the count of primes up to> 100.> Now then dS(x,y) is the count of composites that have y as a factor> that do NOT have a prime number less than y as a factor up to and> including x, so it makes sense that unless y is a prime dS(x,y) is 0> because, for instance, if y=4, then any number that has 4 as a factor> also has 2 as a factor, so there arent any composites that have 4 as> a factor that do not have a prime number less than 4 as a factor as,> of course, 2 is a factor of 4.You know, I think Ivew read almost everything youve written beforeon htis subject, and this is the ?st time you say whats going on> Practically that means that dS(x,y) is 0 if y is composite, and> non-zero if y is prime and less than or equal to sqrt(x) as can be> seen with the values given above.> Now then the non-zero values are easy to understand, like> dS(100,2)=49, means that there are 49 numbers up to and including 100> that are even, excluding 2 itself, as its prime. Then dS(100,3)=16> tells you that there are 16 numbers that are composite with 3 as a> factor that are NOT even.> Next you have dS(100,4)=0, as there arent any numbers that have 4 as> a factor that do not have a smaller prime as a factor, in this case 2,> as 4 has 2 as a factor.> Next theres dS(100,5)=6, telling you that there are 6 composites with> 5 as a factor that do not have 2 or 3 as a factor.> Notice the drop of non-zero dS(x,y) values is like an exponential> drop, while the spacing between values is, of course, the spacing> between prime numbers.> So I think I agree that it will count the composites, and the primes.> And that explains:> But it doesnt explain why it satis?s this difference relation.> dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))]> So please explain it to an idiot like me: why is this relation? Youhave dramatically improved the presentation, but if you are keen tohave it accepted, then explain the reasoning behind this bit.The idea of counting composites like this, using theinclusion-exclusion principle is fairly well studied by the way.> More information can be found at my blog:> http://mathforpro?.blogspot.com/> James Harris === > Back in the spring of 2002 I discovered the following way to count> prime numbers: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))], S(x,1) = 0, S(x,y) is the sum of dS from dS(x,2) to dS(x,y), p(x, y) = ?) - S(x, y) - 1, and p(x,sqrt(x)) gives the count of primes where, of course, only> positive integers are being used, so the sqrt() operator returns the> largest positive integer that squares to give the argument or a number> less than it.It does? Not according to your own recent posts which proclaim thatsqrtis inherently ambiguous and that the ambiguity *cannot* be removed.Byyour own account ?sqrt(y) has a positive OR negative value and both mustbe accommodated. The negative value, of course, does NOT satisfy the claimmade for your argument.Hence, your above method fails *by your own criteria*.--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === I would greatly appreciate any comment upon the correctness of thefollowing assertion.Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =1Assertion: The following equality is not possible. sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1) === >I would greatly appreciate any comment upon the correctness of the>following assertion.>Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =>1>Assertion: The following equality is not possible.> sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1) sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === In sci.math, Robert Israel:>I would greatly appreciate any comment upon the correctness of the>following assertion.> >Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =>1>Assertion: The following equality is not possible.> sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1)> > sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2(sqrt(2) + sqrt(3))^5 = 2^2.5 + 5*2^2*3^.5 + 10*2^1.5*3 + 3^2.5 + 5*3^2*2^.5 + 10*3^1.5*2 = sqrt(2)*(2^2+10*3*2+5*3^2) + sqrt(3)*(3^2+10*2*3+5*2^2) = sqrt(2)*109+sqrt(3)*89 = sqrt(23762)+sqrt(23763)The solution checks out, which leads me to believe thatthere was a problem in the original speci?ation.-- #191, ewill3@earthlink.netIts still legal to go .sigless. === > In sci.math, Robert Israel> :>I would greatly appreciate any comment upon the correctness of the>following assertion.>Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =>1>Assertion: The following equality is not possible.> sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1)> sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2> (sqrt(2) + sqrt(3))^5 = 2^2.5 + 5*2^2*3^.5 + 10*2^1.5*3> + 3^2.5 + 5*3^2*2^.5 + 10*3^1.5*2> = sqrt(2)*(2^2+10*3*2+5*3^2) > + sqrt(3)*(3^2+10*2*3+5*2^2)> = sqrt(2)*109+sqrt(3)*89> = sqrt(23762)+sqrt(23763)> The solution checks out, which leads me to believe that> there was a problem in the original speci?ation.One never knows about whats posted to Usenet :-(Whats even stranger is that if gcd(c,d) = 1 then sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5for a = c(c^2 + 10cd + 5d^2)^2, b = d(5c^2 + 10cd + d^2)^2,and gcd(a,b,c,d) = 1. So the equality is ALWAYS possible.Perhaps what Robert is doing is giving an example where gcd(a,b) = 1too? I havent checked for smaller examples...--Ron Bruck === One never knows about whats posted to Usenet :-( escribi.97 en el mensaje> I would greatly appreciate any comment upon the correctness of the> following assertion. Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) => 1> Assertion: The following equality is not possible. sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1)sqrt(a) + sqrt(b) = c^(5/2) + 5c^2*d^(1/2) + 10d*c^(3/2) + 10c*d^(3/2) + 5c^(1/2)d^2 + d^(5/2) = (c^2 +10c*d + 5d^2)sqrt(c) + (d^2 + 10c*d + 5c^2)sqrt(d)Then,a = c(c^2 +10c*d + 5d^2)^2b = d(d^2 + 10c*d + 5c^2)^2a and b arent squares because c and d arent.And gcd(a, b, c, d) = 1 if gcd(c, d) = 1.Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Do these primes have a name [7,31,71,127...]?Where k = n*(n+1)/2Where x falls exactly between two consecutive EVEN numbered ks and if(x-1) is a prime then this list below are those primes.7,31,71,127,199,647,967,1151,1567,2311,2591...Eg. Of ?st few primes (x-1) and k-1 and k1) 6 (8) 10 k-1 = 6 and x= 8 and k = 102) 28 (32) 36 k-1 = 28 and x=32 and k = 363) 66 (72) 78 k-1 = 66 and x=72 and k = 78 1) x-1 = 7 2) x-1 = 31 3) x-1 = 71etc..What is interesting is when x-1 is a composite (x-1)== >0 (mod 3) andalso for (x-1)== >0 (mod 5). So far, no 0 (mod 3)s or no 0 (mod 5)s composites! Examining it a bit further -- I believe it is not possible because these (x-1)== >0 (mod 9) whether (x-1) is prime or composite the consecutive (x-1)s have a repeatingsequence of residuals (r) for ---(x-1) == r (mod 9)s where consecutive (x-1)s --r =[7,4,8,1,1,8,4,7,8] [7,4,8,1,1,8,4,7,8] [7,4... etc. This repeating sequence is almost a palindrome. ;-) Dan === > Do these primes have a name [7,31,71,127...]?> Where k = n*(n+1)/2> Not sure about a special name, but thats the nth triangular number.> Where x falls exactly between two consecutive EVEN numbered ks and if> (x-1) is a prime then this list below are those primes.> 7,31,71,127,199,647,967,1151,1567,2311,2591...> Eg. Of ?st few primes (x-1) and k-1 and k> 1) 6 (8) 10 k-1 = 6 and x= 8 and k = 10> 2) 28 (32) 36 k-1 = 28 and x=32 and k = 36> 3) 66 (72) 78 k-1 = 66 and x=72 and k = 78 > 1) x-1 = 7 > 2) x-1 = 31 > 3) x-1 = 71> etc..> What is interesting is when x-1 is a composite (x-1)== >0 (mod 3) and> also for (x-1)== >0 (mod 5). > Im not sure I interpret the ==> symbol very well, but if you examinewhat the number x-1 is, you get(n+1)^2/2 - 1,so you should be able to work out the mod 3 and mod 5 bits since, for the n and n+1st triangular number to be both even, then n+1 isdivisible by 4, call it 4s, x-1 is then 8s^2 - 1, which is -s^2 - 1mod 3, which is never zero as -1 is not a quadratic residue mod 3.Ill leave you to work out the mod 5 case.> So far, no 0 (mod 3)s or no 0 (mod 5)s composites! > > Examining it a bit further -- > I believe it is not possible because these (x-1)== >0 (mod 9) whether > (x-1) is prime or composite the consecutive (x-1)s have a repeating> sequence of residuals (r) for ---> (x-1) == r (mod 9)s where consecutive (x-1)s --> r =[7,4,8,1,1,8,4,7,8] [7,4,8,1,1,8,4,7,8] [7,4... etc.> This repeating sequence is almost a palindrome. ;-)> DanNot sure I either understand what youre saying here, or whats goingon. Will ponder some more. === > Do these primes have a name [7,31,71,127...]? Where k = n*(n+1)/2 Where x falls exactly between two consecutive EVEN numbered ks and if> (x-1) is a prime then this list below are those primes. 7,31,71,127,199,647,967,1151,1567,2311,2591...See Primes of the form 8*n^2 -1 at(Interestingly, that entry was created only a few days ago.)David> Eg. Of ?st few primes (x-1) and k-1 and k 1) 6 (8) 10 k-1 = 6 and x= 8 and k = 10> 2) 28 (32) 36 k-1 = 28 and x=32 and k = 36> 3) 66 (72) 78 k-1 = 66 and x=72 and k = 78 1) x-1 = 7> 2) x-1 = 31> 3) x-1 = 71> etc.. What is interesting is when x-1 is a composite (x-1)== >0 (mod 3) and> also for (x-1)== >0 (mod 5). So far, no 0 (mod 3)s or no 0 (mod 5)s composites! Examining it a bit further -- I believe it is not possible because these (x-1)== >0 (mod 9) whether> (x-1) is prime or composite the consecutive (x-1)s have a repeating> sequence of residuals (r) for --- (x-1) == r (mod 9)s where consecutive (x-1)s -- r =[7,4,8,1,1,8,4,7,8] [7,4,8,1,1,8,4,7,8] [7,4... etc. This repeating sequence is almost a palindrome. ;-) Dan === I studied graph theory for a semester using this book. Admittedly, myexperience was somewhat less that stellar; and I found this book toodif?ult for me. A little background might be suf?ient: I did nothave any prior Intro. to Combinatorics course and I was anengineering student who happened to have an interest in mathematics.I did however ?d these books tremendously easier:Book 1. Combinatorics and Graph TheoryISBN: 0387987363 http://www.amazon.com/exec/obidos/tg/detail/-/0387987363/002- 2811429-8891268?v=glanceBook 2. Applied Combinatoricsby Alan TuckerISBN: 047143809X http://www.amazon.com/exec/obidos/tg/detail/-/047143809X/ref= lpr_g_1/002-2811429-8891268?v=glance&s=books Learning graph theory from Prof. Wests book was like being beingpushed unwittingly into a cold river: you know that the able oneswould survive and laugh but the unprepared would sink, in the end youvow never to play near water again. I think the prose was a bit terseand at times confusing. Some of the examples are harder than thematerial they serve to explicate, sometimes much effort is expendedjust to understand the complicating details of the examples.On the plus side, the proofs are very short and most of them elegant.No time is wasted in being chatty (contra Book 1)--like themethematical equivalent of Emeril, each discussion concludes with aBam! and on to the next--but of course this leads to a feeling thatgraph theory is a slapdash of assorted subjects with no coherentwhole. (I guess the colorful nomenclature of the subject deserves morelighthearted treatment). I am interested in learning what other people have in mind about thebook. I am thinking of repurchasing it (sold it earlier). What bookwould you nominate to be the best introductory book with the rightcombination of rigor and friendliness--neither uncharted Paci?trench nor kiddies pool. === >I studied graph theory for a semester using this book....>and I found this book too dif?ult for me. A little background might be >suf?ient: I did not have any prior Intro. to Combinatorics course>I think the prose was a bit terse and at times confusing. Agreed, but I think that style was a conscious design decision. The language is formed always in a ttempt to be correct and concise, and certain informal sounding English is used pointedly, e.g. he might use the foobar instead of the lengthier but not more precise there exists exactly 1 foobar. So if youre not used to it (and you will be after the book), the language only -seems- confusing. You need to put in that extra work to extract the unspoken .. er not unspoken but not immediate .. er it is immediate if only you were already precise about your use of language. Later on in the preface, he makes some remarks about intellectual discipline and honesty and use of language say what you mean and mean what you say. (and he practices what he preaches).>On the plus side, the proofs are very short and most of them elegant.as many classic results in combinatorics are.>this leads to a feeling that>graph theory is a slapdash of assorted subjects with no coherent>whole.>What book>would you nominate to be the best introductory book with the right>combination of rigor and friendliness--neither uncharted Paci?>trench nor kiddies pool.Hmm.. that I dont have too many opinions about. Possibly a text on algorithms with coverage of graphs?Mitch Harris === > I recall attending a lecture long ago where the instructor (James > Pittman at U.C. Berkeley, I think) presented an example of an oddity > that can occur if one requires only ?ite (as opposed to countable) > additivity. I have forgotten the what the example was. Could anyone > here provide one?> Countable additivity lets us prove the laws of large numbers. Which is> a good thing, in most cases --- we really want the laws of large> numbers to be true, to agree with our intuition. And its nice to have> them as theorems, instead of postulating them as axioms of probability> theory.> On the other hand, sometimes we want to model a situation in which a> law of large numbers does NOT hold, and in those cases we have to pick> a probability measure that is not countably additive. The primary> example is the Gaussian probability on the in?ite-dimensional> I am familiar with a rigorous treatment of the Gaussian probabilityembedded as a dense subspace of a larger vector space (e.g. thecompletion of H with respect to a suitable norm). But my work on thiswas decades ago. I see how you can get a ?itely additive measure de?ed on sets Ssuch that there exists a projection p of H onto a ?ite-dimensionalspace V and a measurable subset T of V, and S = p^-1 V. (Apologies forthe limitations of ASCII math notation) But I dont see how to extendthis measure to where you can integrate the things you want tointegrate (except by the technique mentioned in the previousparagraph).I would welcome references.-- Chris HenrichIts our supreme ability and willingness to screw up that is the secret of oursuccess. -- R. X. Cringely === > I recall attending a lecture long ago where the instructor (James > Pittman at U.C. Berkeley, I think) presented an example of an oddity > that can occur if one requires only ?ite (as opposed to countable) > additivity. I have forgotten the what the example was. Could anyone > here provide one?> Countable additivity lets us prove the laws of large numbers. Which is> a good thing, in most cases --- we really want the laws of large> numbers to be true, to agree with our intuition. And its nice to have> them as theorems, instead of postulating them as axioms of probability> theory.> On the other hand, sometimes we want to model a situation in which a> law of large numbers does NOT hold, and in those cases we have to pick> a probability measure that is not countably additive. The primary> example is the Gaussian probability on the in?ite-dimensional> I am familiar with a rigorous treatment of the Gaussian probability> embedded as a dense subspace of a larger vector space (e.g. the> completion of H with respect to a suitable norm). But my work on this> was decades ago. > I see how you can get a ?itely additive measure de?ed on sets S> such that there exists a projection p of H onto a ?ite-dimensional> space V and a measurable subset T of V, and S = p^-1 V. (Apologies for> the limitations of ASCII math notation) But I dont see how to extend> this measure to where you can integrate the things you want to> integrate (except by the technique mentioned in the previous> paragraph).> I would welcome references.cylindrical measures. === A book Im reading asserts If H and K are two subgroups of a group G,and if every element g of G can be uniquely expressed as g=xy, where xis in H and Y is in K, then H and K are both normal in G.I havent been able to show this. I think I might be missing somethingreally simple.Mike === > A book Im reading asserts If H and K are two subgroups of a group G,> and if every element g of G can be uniquely expressed as g=xy, where x> is in H and Y is in K, then H and K are both normal in G. I havent been able to show this. I think I might be missing something> really simple. MikeConsider the symmetric group on three elements, S_3. This containsa normal subgroup of order three which is isomorphic to thecyclic group of order 3, Z/3Z (i.e., {e, (1,2,3), (1,3,2)} ).Now, let K be a subgroup of S_3 consisting of two elements,the identity and a swap of two elements (e.g., {e, (1,2)(3)}).Now, any element of S_3 can be written uniquely as the productof an element of K and Z/3Z, but K is not normal.I probably havent helped you much, as a counter-example generallymakes it more dif?ult to ?d a proof :-).Best wishes, Mike === >A book Im reading asserts If H and K are two subgroups of a group G,>and if every element g of G can be uniquely expressed as g=xy, where x>is in H and Y is in K, then H and K are both normal in G.Not true. Say G is the group of all maps f from R to R of the form f(t) = at + bwhere a <> 0 (with composition as the group operation.) Let H bethe subgroup of all f of the form f(t) = t + b and let K be the setof all f of the form f(t) = at (a <> 0). Then its clear that every element of G is equal to the compositionof an element of H and an element of K in exactly one way. ButK is not normal (because a(t+b) - b <> At.)>I havent been able to show this. I think I might be missing something>really simple.For a second I thought maybe it was only supposed to be true for_?ite_ groups, and you or the author left that out. But no, seemslike the above example with a ?ite ?ld in place of R gives a?ite counterexample...>Mike************************David C. Ullrich === [snip]> I maintain that Magidin is indeed either incompetent as a> mathematician, or more likely a liar, or both.well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahsin academic caste system.so whomever you are, may well be right on this one...> James Harris <31db93f.0312250011.16178abc@posting.google.com> <874qvotj7s.fsf@phiwumbda.org> <31db93f.0312262203.2bcbd157@posting.google.com> <188f56bf.0312281039.4ed0708@posting.google.com> === [snip]> I maintain that Magidin is indeed either incompetent as a> mathematician, or more likely a liar, or both. well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs> in academic caste system.>Says the man (?) who repeatedly posts about how bad his employmentprospects are and what a failure his own PhD has proved.So youre upset about your own bad experiences. Big deal. Thats noexcuse to denigrate Arturos position and you have no reason tobelieve that hes unhappy with what hes accomplished.You are a sad and pathetic person.-- My proof has been checked very thoroughly, both by me and others.Those others apparently decided that they would not believe the proofwas correct, but cannot support that position using mathematics. Buthey, theyre just human beings. --JSH, prover of Fermats Last Thm === > [snip]> I maintain that Magidin is indeed either incompetent as a> mathematician, or more likely a liar, or both. well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs> in academic caste system. Says the man (?) who repeatedly posts about how bad his employment> prospects are and what a failure his own PhD has proved.come again?are trying to deny that the title ADJUNCT is the lowest title anacademic can hold? or that anyone holding such title is subject toabuse and contempt by practically the whole academic institution? orthat the american higher education system is heavily using and abusingadjuncts?i hope you arent shy to clarify whatever your point was.> So youre upset about your own bad experiences. Big deal.fyi, i do not hold a math PhD, but know plenty of people who do andhappen to be adjuncts. their working conditions are abysmal.> Thats no excuse to denigrate Arturos position and you have> no reason to believe that hes unhappy with what hes accomplished.i very much doubt that arturo is not unhappy with his currentposition, anyone with an ounce of dignity would be unhappy. especiallyin light that about a quarter of the teaching personnel at hisinstitution holding higher rank than him are complete mathematicalnincombpoops compared to him.> You are a sad and pathetic person.beats the hell out of being a complete ignoramous imbecile, such asyou. <31db93f.0312250011.16178abc@posting.google.com> <874qvotj7s.fsf@phiwumbda.org> <31db93f.0312262203.2bcbd157@posting.google.com> <188f56bf.0312281039.4ed0708@posting.google.com> <87y8swbnll.fsf@phiwumbda.org> <188f56bf.0312291220.6dbfebe6@posting.google.com> === > are trying to deny that the title ADJUNCT is the lowest title an> academic can hold? No, but I dont know that its true, either. Depends on theinstitution, I reckon.> or that anyone holding such title is subject to abuse and contempt> by practically the whole academic institution?Well, I suppose Ill deny that, yes. It is a silly claim.> or that the american higher education system is heavily using and> abusing adjuncts?[sic]. This is nonsense. You used this claim to support the claimthat Arturo is incompetent. This is just utter bull.> i hope you arent shy to clarify whatever your point was.My point? Youre a pathetic head. Sorry for failing to make thatexplicit.> So youre upset about your own bad experiences. Big deal. fyi, i do not hold a math PhD, but know plenty of people who do and> happen to be adjuncts. their working conditions are abysmal.Sorry for the false claim. Your bitterness towards mathematics was, Iassumed, due to the fact that you held a worthless PhD. Instead, youare a champion of the noble but fallen PhDs around you, ceaselesslyworking to overthrow the mathematical feudalism that has served tokeep them chained. Bully for you.[...][Snip bit in which Maky opines about how much more competent Arturois, compared to his colleagues --- despite the fact that he insinuatedArture is incompetent due to his position.]> You are a sad and pathetic person. beats the hell out of being a complete ignoramous imbecile, such as> you.Now, that hurts. You cut that out.-- I AM serious about this being a short route to a Ph.d for some ofyou, but just remember, Im the guy who proved Fermats Last Theoremin just a bit over 6 years [...] My standards are kind of high. --James Harris, founding a new mathematical school <31db93f.0312250011.16178abc@posting.google.com> <874qvotj7s.fsf@phiwumbda.org> === > Implementing the MSN passport, if I can believe the protests of the> people replying to me here, is incredible dif?ult.>As usual, you do not understand the protests of others. Despite his protestations to the contrary, it semms unlikely this is> the real James Harris; I agree, and I said as much shortly after this post. I had overlookedboth the discussion group name and also the yahoo.com address when I?st replied.-- But in our enthusiasm, we could not resist a radical overhaul of thesystem, in which all of its major weaknesses have been exposed,analyzed, and replaced with new weaknesses. -- Bruce Leverett (presumably with apologies to Ambrose Bierce) === taking it in high school. I enjoy math as a hobby and was looking for a goodbook on calculus for this summer so that I can learn a little beforecollege. A few books I saw at the bookstore were:Calculus Made Easy - Silvanus Philips ThompsonCalculus for DummiesComplete Idiots Guide to CalculusAnother interesting book is Calculus: The Elements, which seems more like atextbook with problems.What aspects of calculus should a beginning book include and does anyoneknow if any of these, or any others, are any good? I am not looking for an === not> taking it in high school. I enjoy math as a hobby and was looking for agood> book on calculus for this summer so that I can learn a little before> college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson> Calculus for Dummies> Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems more likea> textbook with problems. What aspects of calculus should a beginning book include and does anyone> know if any of these, or any others, are any good? I am not looking for anAnother book in that vein is Hurricane Calculus. You could also tryDovers website for some cheap Calc. books. However, you could probablypick up some cheap used editions of Spivaks Calculus, or ApostolsCalculus. They have a great mix of rigour and readability. Try to ?dthem on http://www.book?der.com Ill bet you could ?d one for around$20.Lurch === > taking it in high school. I enjoy math as a hobby and was looking for a good> book on calculus for this summer so that I can learn a little before> college. A few books I saw at the bookstore were:> Calculus Made Easy - Silvanus Philips Thompson> Calculus for Dummies> Complete Idiots Guide to Calculus> Another interesting book is Calculus: The Elements, which seems more like a> textbook with problems.> What aspects of calculus should a beginning book include and does anyone> know if any of these, or any others, are any good? I am not looking for anTeach Yourself Calculus by P. Abbott and H. Neill, $12.95, $10.36from Amazon, was the book I used. Cheap, easily carried around, unlikemost other calculus books, and a Teach Yourself... book, so itsgeared toward the self-taught, not to state the obvious, or anything.Good luck. === not> taking it in high school. I enjoy math as a hobby and was looking for agood> book on calculus for this summer so that I can learn a little before> college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson> Calculus for Dummies> Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems more likea> textbook with problems. What aspects of calculus should a beginning book include and does anyone> know if any of these, or any others, are any good? I am not looking for anCalculus the easy way was pretty fun.IEEE the Calculus tutoring book would be a good one for your level. === regret not> taking it in high school. I enjoy math as a hobby and was looking fora good> book on calculus for this summer so that I can learn a little before> college. A few books I saw at the bookstore were: Calculus Made Easy - Silvanus Philips Thompson> Calculus for Dummies> Complete Idiots Guide to Calculus Another interesting book is Calculus: The Elements, which seems morelike a> textbook with problems. What aspects of calculus should a beginning book include and doesanyone> know if any of these, or any others, are any good? I am not lookingfor anhttp://tinyurl.com/2awjyThis will give you the basic proofs behind differentiation andintegration and an understanding of functions, graphing, limits etcwhich are all necessary fundamentals. It is a ?st year collegecalculus book (M124, M125 + some) with plenty of problems to work. It isan older edition but it doesnt get any less expensive than this.Phil Holman === Suppose we have a labelled graph G with vertices {1,2,3,...,n} and ann-vector c of positive integers {c_1,c_2,c_3,...,c_n). Produce ann-partite graph H by creating (for each i in {1,2,3,...,n}) a set T_icontaining c_i isolated vertices, and then joining every vertex in T_ito every vertex in T_j whenever ij is an edge in G. So, each vertex iin G corresponds to a set T_i in H, and each edge ij in G corespondsto a collection of (c_i x c_j) edges in H.What name is given to this type of structure?TIASS === PS I will review Physics Meets Philosophy at the Planck Scale.There seems to be a misconception about D Branes starting in Hawkings new popular book The Universe in a Nutshell (O Brane New Worlds) and also pictures in Scienti? American showing gravitons (closed strings)moving through all of hyperspace with lepto-quarks and photons (open strings) stuck to D Branes pictured as parallel universes asdescribes a D Brane seems incompatible with that picture?Also Witten seems to think his alpha = 1/(string tension) ~ (10^-32 cm)^2 is a new independent fundamental constant (a moduls or compacti?ation scale) rather than Lp^2 = hG(Newton)/c^3 = hc(alphaalpha) = alphae^2 hG/e^2c^3 = alphaThink ?st in quantum ?ld theory of a virtual electron-positron pair vacuum bubble, i.e. closed world line with no free ends. Then imagineone virtual photon bisecting the the closed world line. This is the basic virtual bound state into which a huge number of these vacuum bubbles Bose-Einstein condense to form the vacuum coherence ?ld whose Goldstone phase ripples form Einsteins guv c-number ODLRO ?ld.dark energy and dark matter exotic vacuum w = -1 phases that is 96% of the large-scale stuff of the world.For strings imagine a closed strip - two-sided orientable - maybe also Mobius one-sided strips.The smallest quantized area of these vacuum bubbles is obviouslyhcalpha in Wittens sense.The single virtual photon gives a factor of alpha = e^2/hc since each fuzzed out vertex is (alpha)^1/2i.e. basic diagram is <|> | is the virtual photon and <> is the virtual electron-positron vacuum loop or bubble.Therefore, this is a simple Mickey Mouse derivation of Andrei Sakharovs emergence of gravity G from QED and Wittens basicmodulus or string tension parameter, i.e.G = e^2c^3alpha/h = c^4(alpha)(alpha)So that to be more precise Einsteins GR equation isGuv = (8piG/c^4)Tuv8piGh/hc^4 = 8piLp^2/hc = (alpha)(alpha)Guv = (alpha)(alpha)Tuvalpha is Wittens modulus.Dear Friends :Recently there has been a lot of work dedicated toFisher information theory, entropy.... see the book byFrieden, Cambridge University Press : Physics from Fisher Information that claims/suggests to derive all of physics fromFisher information theory. If you look at theliterature about QM and Fisher you will see that thereis a Fisher information term in the Schroedingerequation that is related to Bohms quantum potential =Weyl scalar curvature induced by the ensemble densitythis out to my amazement !So yes, information theory, entropy and Weyl-Bohm havea lot in common. Call it Information Geometry which is very popular today.alsoDear Jack :The title of Christian Becks book is :Spatio-temporal vacuum ?tions of quantized ?lds...World Scienti? 2002. Nonlinear sciences.archives. You can download them.Best wishesCarlos____________ === I have to solve the diophantine equation x*(p+A*y)=B (unknown x,y; allfactors of B are known)i am able to resume that to x*(1+A*y1)=B1 :-)is there in some place a way to solve this equation without testing allpossible combinaisons of sub factors of B1 ?Any kind of pointers will be appreciated.BR/Philippe === >I have to solve the diophantine equation x*(p+A*y)=B (unknown x,y; all>factors of B are known)>i am able to resume that to x*(1+A*y1)=B1 :-)I dont know what resume means for you in this context, but lets say thats the question...I dont suppose x=B1, y1=0 is allowed?>is there in some place a way to solve this equation without testing all>possible combinaisons of sub factors of B1 ?Well, youre looking for factors of B1 that are == 1 mod A.If B1 = product_{j=1}^N p_j^{n_j}, a factor of B1 is product_j p_j^{m_j}where 0 <= m_j <= n_j for all j. One way to do this (which mightbe useful if A is not too large) is with dynamic programming.First, discard any p_j that divides A.Let B(0) = 1, and for k from 1 to N let B(k) = product_{j=1}^k p_j^{n_j}.For k from 0 to N and a coprime to A, let x(k,a) be a factor of B(k) that is == a mod A if such exists, otherwise 0. Begin with x(0,1) = 1, x(0,a) = 0 otherwise.Let o_k be the order of p_k mod A. Then for a in {1,2,...,A-1} andcoprime to A,x(k, a) = max_{j=0}^{min(n_k, o_k)} x(k-1, (a p_k^(-j)) mod A) p_k^jRobert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === I have a set of Bezier curves. Does anyone know a way that I can>order themWell, you can try Amazon.com ; if you cant order them there, look on Ebay. === > I have a set of Bezier curves. Does anyone know> a way that I can order them Well, you can try Amazon.com;> if you cant order them there, look on Ebay.In Pythagoras confusion we discuss a similar questionand it would be nice to see another helpful comment ;-)Rainer Rosenthalr.rosenthal@web.de === > I have a set of Bezier curves. Does anyone know a way that I can> order them, so that two curves which are ordinally (right word?)> near each other will be similar?> Darren GrantSimilar is a pretty fuzzy term. Lets assume that the starting andending points of all our curves are identical. Let F(A,B) be the areabetween the two curves A and B. F(A,A) = 0; and if F(A,B) = 0, then A= B.Then we can make the de?ition that B is more similar to A than C isto A iff F(A,B) < F(A,C).However, this only provides us with a metric; not an order. It seemsyou want a function G(A) that returns, for example, a real number;with the property that, for all curves A, B, C, if |G(A) - G(B)| <|G(A) - G(C)|, then F(A,B) < F(A,C).This doesnt seem possible unless you have a very restricted set ofcurves.Consider the comparable problem in the plane, where you would like afunction G(p) taking each point p in the plane to some real number.With a little inspection, I think you can see that, if d(p,q) is thedistance between two points, there is no G such that for all pointsp,q,r, |G(p) - G(q)| < |G(p) - G(r)| implies that d(p,q) < d(p,r). === > I have a set of Bezier curves. Does anyone know a way that I can> order them, so that two curves which are ordinally (right word?)> near each other will be similar?> Darren Grant Similar is a pretty fuzzy term. Lets assume that the starting and> ending points of all our curves are identical. Let F(A,B) be the area> between the two curves A and B. F(A,A) = 0; and if F(A,B) = 0, then A> = B. Then we can make the de?ition that B is more similar to A than C is> to A iff F(A,B) < F(A,C). However, this only provides us with a metric; not an order. It seems> you want a function G(A) that returns, for example, a real number;> with the property that, for all curves A, B, C, if |G(A) - G(B)| <> |G(A) - G(C)|, then F(A,B) < F(A,C). This doesnt seem possible unless you have a very restricted set of> curves.When placed on a coordinate system, the curves all start on at x = 0,and all end at y = y_max. Is this restriced enough by any chance? ;)> Consider the comparable problem in the plane, where you would like a> function G(p) taking each point p in the plane to some real number.> With a little inspection, I think you can see that, if d(p,q) is the> distance between two points, there is no G such that for all points> p,q,r, |G(p) - G(q)| < |G(p) - G(r)| implies that d(p,q) < d(p,r).>on this. Im not a mathematician (Im working on it, though) and Idont entirely follow your last paragraph, but I understand thatit is probably not possible to attribute an actual value to a curve, soI am trying to implement an algorithm that works as follows:Given a list of curves, a new curve is added to the list by inserting itafter the curve which it is most similar to.When placed on a coordinate system, the curves all start on at x = 0,and all end at y = y_max, so I am comparing curves for similarity bybuilding a difference curve - curve a minus curve b (for each xvalue, I subtract y of b(x) from y of a(x)). Then the summation ofthe y values for the difference curve gives the metric I use tocompare the two curves with.I nearly got this working, then I realised there is a problem. My curveare Bezier curves, and my function returns a Bezier curve as a seriesof discrete vectors (or points?) - so two curves might look like this:a = (0,0), (3.6,3.2), (6.4,4.8), (8.4,4.8), (9.6,3.2), (10,0)b = (0,10), (2,9.6), (4,8.4), (6,6.4), (8,3.6), (10,0)Obviously the x coordinates differ. I guess there are two possibilitesopen to me; I can use a line-drawing algorithm to render the curve, soI can use ?ed x coordinates for the y-value subtraction, or I coulduse the control points which the Bezier curve is formed out of tocalculate the difference between two curves.If the curves are all formed from, say, three control points, can anyoneshow me how I can compare the similarity of two curves using these points?Darren === > I have a set of Bezier curves. Does anyone know a way that I can> order them, so that two curves which are ordinally (right word?)> near each other will be similar?> Darren Grant Similar is a pretty fuzzy term. Lets assume that the starting and> ending points of all our curves are identical. Let F(A,B) be the area> between the two curves A and B. F(A,A) = 0; and if F(A,B) = 0, then A> = B. Then we can make the de?ition that B is more similar to A than C is> to A iff F(A,B) < F(A,C). However, this only provides us with a metric; not an order. ... after the curve which it is most similar to.> If the curves are all formed from, say, three control points, can anyone> show me how I can compare the similarity of two curves using these points?> Certainly, you could use a metric between two curves which was the sumof the squares of the distances (in the x/y plane) between thecoresponding control points of the two curves.But I guess my question would be, what is the goal of your ...inserting it after... most similar... sorting method?Lets suppose each curve is identi?d with a single control point,(x,y).If you wish to create an order that will allow, for example, a binarytree search for closest curve from some set, in the same fashionthat one would use for looking up a string in a dictionary, I thinkits not possible.On the other hand, there are algorithms like those used for colortable lookup mapping that involve similar problems. In color tablelookup, you have a canonical set of, say, 256 (r,g,b) values; and youwant a fast algorithm for taking any random rgb value and quickly?ding the closest color in the table to that value. Closest isusually de?ed as raw color distance (dr^2+dg^2+db^2).That sounds a bit like your problem - I presume you ultimately wantto, given a random curve, ?d the closest curve in some ?ed set ofcurves? === > ...> Players, on the k_th move, place the integer (2k-1) {for odd-player} or (2k)> {for even-player} at any integer-position on the board which has yet to have> an integer placed upon it.> Could you expand on that? At ?st read, it seems to me that on turn one 1> and 2 are played, on turn two 3 and 4, and so on.> That doesnt seem like enough freedom for an interesting game, so my next> thought was that the three ks are all different-- that the odd player> (thats me) plays an odd number and the even player plays an even number.My original rules have 1 and 2 played on ?st move, 3 and 4 played onthe 2nd move, as you apparently assumed here. So, there is no choicewhat integer is played on a particular move by any one player.But allowing any integer to be played as long as it:1) is in the range 1 to n, for some n (do not con?ith rule-3);2) is always even for a player, is always odd for the other;3) has not been used yet;might be more fun.Hmmm... Perhaps the integers can even be picked at random using a deckof cards.(Perhaps we should allow integer re-use here, if standard deck isused.)> But then this statement is no clear to me:> It is okay for two integers to be at the same position IF the> integers positions were both chosen on the same move.> For example, if 13 and 4 are played, do we use up board spots 6 and 2, only> using up the same spot if for example 12 and 12 were played?> Or am I just not getting it at all?> I do not believe I have made things clear here.Let us use letters to represent the positions so as to help avoidconfusion.If position-c, say, is already occupied when a player wants to place a4 down on the board, the 4 cannot be put at c.So, the player puts 4 at d, which is empty.On the *same* move, the other player had independently chosen to putthe 3 at d as well.So, this is ?e.Now, let us say that c has a 2 and e has 1, and the value of d is now(3+4) =7.So, with [...2, 7, 1,...]both players get a point here because there is one descent and oneascent among these 3 positions.Note: I should also point out that it is still ?e for a player toplay the 7,despite that 7 = 3 + 4.(So, it might be that some adjacent integers neither give a point toone player nor to the other.)thanks,Leroy Quet === > ...> Players, on the k_th move, place the integer (2k-1) {for odd-player} or (2k)> {for even-player} at any integer-position on the board which has yet to have> an integer placed upon it.and I asked:> Could you expand on that? At ?st read, it seems to me that on turn one 1> and 2 are played, on turn two 3 and 4, and so on.> That doesnt seem like enough freedom for an interesting game, so my next> thought was that the three ks are all different-- that the odd player> (thats me) plays an odd number and the even player plays an even number.to which Leroy replied:> My original rules have 1 and 2 played on ?st move, 3 and 4 played on> the 2nd move, as you apparently assumed here. So, there is no choice> what integer is played on a particular move by any one player.Rereading all that, I see what I missed was that when you play a number, youcan play it on any empty spot. For some reason I assumed that pieces 2k-1and 2k had to be played at board location k.> Let us use letters to represent the positions so as to help avoid> confusion. ...And I think that was why I was originally confused-- the board was numberedand the pieces were numbered. Letters made it a lot clearer to me what youhave in mind.Unfortunately, I dont have anything profound to say about the game though.Bob H === Einsteins GR is emergent as a c-number ODLRO ?ld out ofa BCS QED vacuum instability that forms the in? ?ld.That is, direct quantization of Einsteins guv ?ld is notappropriate anymore than re-quantizing the giant super?ndsuperconducting waves is appropriate.Therefore quantum foam is suspect. Also Bohms quantum potentialview of vacuum ?tions is relevant in context of the recentpaper from Teheran.I need to follow the experiment below more carefully.Jack better check this story out:Gary S. Bekkum === I could be wrong, but I suspect that no one on the entireDo you bundle your letters and send them to randompeople, too?-E === > Einsteins GR is emergent as a c-number ODLRO ?ld out of> a BCS QED vacuum instability that forms the in? ?ld.> That is, direct quantization of Einsteins guv ?ld is not> appropriate anymore than re-quantizing the giant super?nd> superconducting waves is appropriate.I admit that I dont have an advanced degree in physics, but I do knowhorse when I see it. Do you have an obsessive-compulsive disorder thatforces you to write, even if what you write is nothing but gibberish andbuzzwords? That is my suspicion. === > If we have, for even m,> m/2 black squares of> 2*2> 4*4> 6*6> 8*8> ...> m*m> and m/2 black rectangles of> 1*(m-1)> 3*(m-3)> 5*(m-5)> 7*(m-7)> ..> (m-1)*1> and we have> m/2 white squares of> 1*1> 3*3> 5*5> 7*7> ...> (m-1)*(m-1)> and m/2 white rectangles of> 2*m> 4*(m-2)> 6*(m-4)> 8*(m-6)> ..> m*2> then we have the same total area for all of the white> squares/rectangles as for the black squares/rectangles (which is> easily shown using algebra).> My question is, which I have not been able to do yet myself (though I> have not tried really hard),> can all of the squares/rectangles be arranged in some way so that it> becomes visually obvious that the total areas between the black and> white are the same for every even m?> ( This is similar to the Proof Without Words in the MAAs> Mathematics Magazine.)> thanks,> Leroy Quet I am surprised that no one has replied to this question yet.:( It may be easier to use the 3rd dimension and arrange boxes having theabove dimensions plus each a depth of 1. Is there any obvious 2-d or 3-d arrangements??thanks,Leroy Quet === > In> http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2fdf.0301191933.79c98f04%40posting.google.com& rnum=1&prev=> I mention a game based on the following idea:> > Player 1 ?ds an integer m by taking a permutation {b(j)} of the> ?st n positive integers,> then forms m:> m = sum{k=1 to n} k * b(k).> Player 2 then, in some time limit, tries to ?d ANY permutation of> the ?st n positive integers which also sums to m.> Players take turns making up puzzles and trying to solve other> players puzzles.> ( It is advantageous for the proposer to come up with ms with> relatively few solutions, but are not too easy {easy, such as b(k) = k> or = (n+1-k), which each are the only solutions to their sum m, but> are easy to solve}.)> 2 things:> 1) One can play a solitaire version of this game, where the> permutation (of 1 through 10, for example) is picked by shuf?cards, and the player tries to come up with a solution which is> a) a derrangement of the cards permutation;> b) is not the inverse perm. of the card-perm..> > 2) How can we determine the number of permutations (for a given n)> which have a given sum m?> thanks,> Leroy> QuetIf someone is up for this, perhaps you could brute-forcecomputer-check for an example of an m derived from a sum having asigni?antly interesting n (such as between 8 and 16) and very fewpermutation-solutions relatively.(Where the m is not the minimum/maximum sum which can be obtained, forthese are trivial.)And then you might post this m (and n) here as a puzzle, to have ustry to ?d a permutation {b(k)} of 1 through n,where:m = sum{k=1 to n} k * b(k).(I wonder if there are *always* multiple solutions for each m near thecenter of the range of possible sums.)thanks, Leroy Quet === > In> http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2fdf.0301191933.79c98f04%40posting.google.com& rnum=1&prev=> I mention a game based on the following idea:> > Player 1 ?ds an integer m by taking a permutation {b(j)} of the> ?st n positive integers,> then forms m:> m = sum{k=1 to n} k * b(k).> Player 2 then, in some time limit, tries to ?d ANY permutation of> the ?st n positive integers which also sums to m.> Players take turns making up puzzles and trying to solve other> players puzzles.> ( It is advantageous for the proposer to come up with ms with> relatively few solutions, but are not too easy {easy, such as b(k) = k> or = (n+1-k), which each are the only solutions to their sum m, but> are easy to solve}.)> 2 things:> 1) One can play a solitaire version of this game, where the> permutation (of 1 through 10, for example) is picked by shuf?cards, and the player tries to come up with a solution which is> a) a derrangement of the cards permutation;> b) is not the inverse perm. of the card-perm..> (a) and (b) can be much more exactly rewitten as what I intended andin a way much more simple to understand, especially more simplyunderstood among non-math people.If the order of a card is k, and the card is b(k),we have the multiplied integer pair [k*b(k)] in the sum m.So, while playing the solitaire version of the game,for every k <= number of cards, a player should not have either b(k) for the k_th integer, nor havek for the b(k)_th integeranywhere in the players sum.(Imagining the [k*b(k)] terms as k-by-b(k) rectangles, a player shouldnot have any of the same rectangles, in either orientation, asgenerated by the card-draw.) > 2) How can we determine the number of permutations (for a given n)> which have a given sum m?> thanks,> Leroy> Quet> If someone is up for this, perhaps you could brute-force> computer-check for an example of an m derived from a sum having a> signi?antly interesting n (such as between 8 and 16) and very few> permutation-solutions relatively.> (Where the m is not the minimum/maximum sum which can be obtained, for> these are trivial.)> And then you might post this m (and n) here as a puzzle, to have us> try to ?d a permutation {b(k)} of 1 through n,> where:> m = sum{k=1 to n} k * b(k).> (I wonder if there are *always* multiple solutions for each m near the> center of the range of possible sums.)> thanks, > Leroy > Quet === Dear all,In one of my program, I need to compute the variance of a 8x8 data matrix asfast as possible...Any fast algorithm with lowest complexity?By the way, since what I actually need is an indication of signal localactivity, is there any other measure which can also be a indicator of localactivity and is less complex than computing variance?-Walala ===