mm-1229 === Subject: Re: Quotient Modules I dont know: I think it is obvious that A/0 is isomorphica to A as A-modules, and the epimorphism (a1,a2,...) --> (a2,a3,...) is an A-homomorphism as well, so A/N is also isom. as A-module to A....am I missing something here? Thanx for writing Tonico === Subject: Re: Singularities question > I was playing around with problems with singularities, and I found the > following : > What happens if f is a function and > lim z^(1.5) f(z) = 2 > z--> 0 (*) > This means that > lim z^2 f(z) = 0 > z-->0 > There is a theorem that says that if > lim z^p f(z) = 0 > z-->0 > for some real number p, then there exists an integer m such that for all s > > m, > lim z^s f(z) = 0 > z-->0 > and for all s < m > lim z^s f(z) = in[CapitalThorn]nity > z-->0 > But I said earlier that > lim z^(1.5) f(z) = 2 > z--> 0 > which is a contradiction. The theorem you quote probably requires that f be analytic in a punctured neighborhood of 0. If this is the case, then the limit statement that I marked with (*) cannot be true. > So does this show that 0 is not an isolated > singularity? That is, f is not an analytic function outside of 0? > Tony -- Chris Henrich God just doesnt [CapitalThorn]t inside a single religion. === Subject: ÔThe B of the Bang! ?Linford Christie? sprinter. starting SCULPTURE. posting-account=KhRdNAwAAACyW0xg_WffgOpimEhDJrE4 I put 6-8 posters in World Year of Mathematics, 2000. (W. mathematical year?, European Mathematic/s Society EMS? Donald Stuart McDonald.) Sorry, they were just printouts! One of my printouts is very reminiscent of a new English sculpture: ÔThe B of the Bang! ?Linford Christie? sprinter. starting blocks. Ideas_poster_15 .. 13-20-21?. I also have over 40 sequences in On-line encyclopedia of Integer sequences. Search : lookup eis njas. E.g. Number 1215306625 is expressible as a sum of 2 squares in 64 essentially distinct ways per octant. example. Divisors of (10^9)^(10^9)+1. BBC NEWS | UK | England | Manchester | ÔBang sculpture spike ... ... The sculpture is to be unveiled by Linford Christie next week. Part of the ÔB of the Bang sculpture has fallen to the ground, causing a security scare at its ... newswww.bbc.net.uk/2/hi/uk_news/ england/manchester/4152843.stm - 36k - Cached - Similar pages http://www.mat.dtu.dk/info/experiencing/ems-gallery/Gallery/ Ideas/Ideas_Post er/ideas_poster_15.html don.lotto 16.1.05 === Subject: Re: usenet kooks and a crank scale -------------------------------------s-o-s------------------- --------------- -- > btw this statement has no proof in any system is not false. > that would mean is has some proof, contradiction. > Where exactly does the contradiction lie? In general, from the fact that > a statement has a proof in some system nothing can be concluded about > the truth or falsity of the statement. Thus there is no apparent reason > why this statement has no proof in any system cant be false, which it > is, since the system with the statement as its only axiom proves it. if its an AXIOM in a system then its TRUE in a system why would your argument not apply to godel statements in general? seems Ômade up. Herc === Subject: Re: Cubic eqn help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0FNpqG19492; >Hey guys, >I was looking at solving cubics with the cubic formula, and also other methods. >I was looking at the cubic x^3 - 15x - 4 >The roots are 4, -2 + sqrt(3), -2 - sqrt(3). >cos(1/3arctan(11/2))=2/sqrt(5) >how does this happen? Not clear what the question is. But one way to prove cos(1/3arctan(11/2)) = 2/sqrt(5) is to use cos(3t) = cos(2t)cos(t) - sin(2t)sin(t) = (2cos^(t) - 1)cos(t) - 2sin^2(t)cos(t) = 4cos^3(t) - 3cos(t) and set t = (1/3)arctan(11/2). Now cos(3t) = cos(arctan(11/2)) = 2/5sqrt(5), and 4(2/sqrt(5))^3 - 3(2/sqrt(5)) = 32/5sqrt(5) - 6/5sqrt(5) = 2/5sqrt(5) so the assertion checks out. Well, we should be a little bit careful, because potentially cos((1/3)arctan(11/2)) could be one of the two other roots of 4x^3 - 3x = 2/5sqrt(5) -- we should make sure we got the right one. But we did because the other two roots are negative, whereas (1/3)arctan(11/2) is clearly a [CapitalThorn]rst quadrant angle. But was that your question? Todd Trimble === Subject: Re: Kan extensions. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0FNprg19533; >Ive gone through Kan extensions a few times, but never managed to >recollect it a few days after. Any cleaver way to read it? I suppose you know that a left Kan extension of F: A --> C along G: A --> B is the value at F of the left adjoint to [G, C]: [B, C] --> [A, C] (if such exists)? Here [B, C] is the category of functors H: B --> C, and [G, C] sends H to HG. I realize that the Kan extension Lan_G(F) may exist without there being a global left adjoint Lan_G to [G, C], but thats not the point -- the point is to have some way of remembering how the universal property goes. Which for the record is that the unit F --> Lan_G(F) o G is universal, i.e., given F --> HG for some H, there exists a unique arrow Lan_G(F) --> H making the evident triangle involving the unit commute. Hope this helps, Todd Trimble === Subject: Re: stupid memes > When I was a kid, a friend told me: According to the laws of physics, a > bumblebee cannot ßy. Chinese whispers. I heard it as Ôphysics hasnt yet managed to explain how a bumble bee ßies There again ever since undergrad days I assumed a neutrino had mass because conservation of momentum was invoked to predict it. >Its one hell of a meme. I heard it again today So why else would you read this n.g? LOL Steve Ralph (about > forty years later). It annoyed me then and it still does, since it implies > an absurd view of science, swapping the master with the slave. In any > case, I would like to make a collection of absurd science memes. If anyone > knows of more, please post them, or links would be appreciated. Also, > absurd science quotes from famous people (scientists and non-scientists) > would be great. === Subject: Re: stupid memes >> The standard example is that one cannot prove there are no >> black swans, unless one has seen every swan in the world. But >> perhaps today DNA analysis could prove that no conceivable >> mutation would give a swan black feathers without being lethal: >> although I do not expect this to be possible. > In fact, there are black swans in Western Australia: > http://www.dpc.wa.gov.au/emblems/swan.html > Of course, that only makes black swans an even better example: > For all of history (in Europe (and Asia and Africa?)), > all observed swans were white. The reasonable conclusion was, > obviously, that all swans were white -- reasonable and wrong. But I can easily prove theres no 30 foot [CapitalThorn]re-breathing dragon in my garage. Just look. It is easy to prove a negative if the thing in question provides testable qualities. That is if the tests give a negative result. -- Denis Loubet dloubet@io.com http://www.io.com/~dloubet === Subject: Re: stupid memes >But I can easily prove theres no 30 foot [CapitalThorn]re-breathing dragon in my >garage. Just look. >It is easy to prove a negative if the thing in question provides testable >qualities. That is if the tests give a negative result. Yes. But what is meant by a negative in that phrase is - proving that there are no 30-foot [CapitalThorn]re-breathing dragons *anywhere*. Although, even in the absence of proof, somehow I manage to feel rather con[CapitalThorn]dent that there are no 30-foot [CapitalThorn]re-breathing dragons anywhere on Earth at the present time. Unless somebody in a lab somewhere is playing with DNA... John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: stupid memes >>But I can easily prove theres no 30 foot [CapitalThorn]re-breathing dragon in my >>garage. Just look. >>It is easy to prove a negative if the thing in question provides >>testable qualities. That is if the tests give a negative result. > Yes. But what is meant by a negative in that phrase is - proving > that there are no 30-foot [CapitalThorn]re-breathing dragons *anywhere*. > Although, even in the absence of proof, somehow I manage to feel > rather con[CapitalThorn]dent that there are no 30-foot [CapitalThorn]re-breathing dragons > anywhere on Earth at the present time. I should introduce you to my mother-in-law... === Subject: Re: How many digits is pi computable to? -------------------------------------s-o-s------------------- --------------- -- >yes, the NEXT part of my argument, but one that I wont be able to explain, its >one of those things.... if I need to explain it you wont understand it.... is this... >the fact *oo digits* of any sequence are computed-in-order means >that digit sequence is computed. >your defn of computable, which is entirely manufactured to evaid this argument is.. >a natural number exists to reference the real. >the focus is on WHAT DIGITS get computed, not what digits get COMPUTED. > This is where your argument becomes wrong. > Have you heard of the Jorge Luis Borges story, The Library of Babel? > It is about an order of monks tending a gigantic library that one day > appeared out of nowhere. It contains, in their language, every possible > book that could ever be written. (Dont ask how they are sure this is > true.) So far, they have never managed to [CapitalThorn]nd in the library any books > that are anything but a meaningless jumble of words in random order. > All the knowledge that one might possibly want is in there! > Unfortunately, so are books containing the most deceptive lies as well. > The number of books in this library is very large, but [CapitalThorn]nite: books > over a certain size are considered as being split up into multiple > volumes. But this still points to a paradox applicable to the in[CapitalThorn]nite. > Let us suppose I was able to magically place all the books in the > library into order. But what order? > Suppose I were to place all the books in *alphabetical order by their > contents*. > How informative would the library be then? > What would be the procedure for [CapitalThorn]nding a book in the library? > Answer: write it yourself. > Hence, that > 0.0 > 0.1 > 0.2 > 0.3 > 0.4 > 0.5 > 0.6 > 0.7 > 0.8 > 0.9 > 0.01 > 0.02 > 0.03 > 0.04 > 0.05 > 0.06 > 0.07 > 0.08 > 0.09 > 0.11 > 0.12 > will match, to any number of digits, any real number in [0,1), does not > mean that all real numbers on that interval are computable. Because no > entry in this list is a real number with a non-terminating decimal > representation. These numbers are not *entries* in the list, they are > *paths* THROUGH the list. > Or, as it is said: it is the journey, not the destination. > John Savard > http://home.ecn.ab.ca/~jsavard/index.html Its more a question of tractability not countability. If the monkeys can reach books then there is some ordering. Your second argument applies to rationals aswell. Herc === Subject: Re: Cubic eqn help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0G0ThY22863; Todd that is a nice way of showing the assertion is true. The question in full said this Solve the cubic equation x^3 - 15x - 4 using the cubic formula (refering to Cardanos formula) Then it says from this deduce that cos(arctan(11/2)/3)= 2/sqrt(5) I absolutely have no idea to show this Maybe somewhere in the formula we can deduce this, but I cant see it If someone knows how to it would be much appreciated === Subject: Re: Cubic eqn help You have a real mouthful here in that deduction question. The crux of the matter is that given the Cardano-Viete solution to the equation, you cant express it in terms of real radicals. This is what the whole casus irreducibilis thing is about. If real-radical solutions exist, they have to be derived by a method other than the Cardano-Viete formula -- meaning its hit-or-miss because other methods, based on conventional factoring techniques, are not guaranteed to work. So you need to solve the cubic equation in two ways. One is the Cardano-Viete formula, getting the roots in terms of trig and inverse trig functions. Then you go back and solve the equaiton by the elementary factoring approach, seeking a rational root. Of course you know that the numerator in such a root has to divide the constant term, the denominator has to divide the leading coef[CapitalThorn]cient, and the root may have either sign. If you [CapitalThorn]nd such a rational root r, divide out the corresponding factor x-r and solve the remaining quadratic equation. You now have two alternate sets of representations for the roots. Equating roots between the two sets of representations (you have to be sure, of course, that you pair them up correctly) should then give you the desired result. --OL === Subject: algebraic number theory: X^2 + X + q prime posting-account=cTENRA0AAACAH5-uBM0h8zDjlQchjmOt Consider the polynomial f(X) = X^2 + X + q, where q is an integer >= 3. Suppose we know that f(n) is prime for all integers 0 <= n < sqrt(q/3). Then i) the [CapitalThorn]eld Q(sqrt(1-4q)) has class number 1 ii) 1-4q must in fact be square-free Any help will be appreciated. --Jenny === Subject: Re: Arithmetic on computable reals |I suppose any notion of such computability witness programs could be |boiled down into a mapping S:R -> P(N), where S(x) intersect S(y) is |empty if x != y, and S(x) is nonempty whenever x is computable. Or, equivalently, a function f from a subset of the natural numbers to the reals whose image contains the computable reals. |There exists S and program E such that for any P_x in S(x) and P_y in |S(y), E(P_x, P_y) = 0 if x = y and 1 otherwise. |There exists S and program A such that for any P_x in S(x) and P_y in |S(y), A(P_x, P_y) is in S(x+y). In terms of f, this is f(m)=f(n) iff E(m,n) and f(A(m,n))=f(m)+f(n). |It looks very much as if no S exists that can satisfy both properties. One does exist. The reader might [CapitalThorn]nd it amusing to think about it rather than read my explanation. There are various ways of playing with the problem. The reason my argument from before doesnt rule this out is that I was making tacit use of the computability of this number u from the Turing machine used to de[CapitalThorn]ne it. The computable reals are (and this is of course nonconstructive) a vector space over the rationals with a countable algebraic (or Hamel) basis. So they can be represented using [CapitalThorn]nite sequences of rationals (with the last one nonnegative), with equality being equality, and addition being just componentwise addition. The requirements you state above dont have all that much to do with the identity of the computable reals as such; this is just an algebraic structure, which is isomorphic to a computable one. I thought brießy that including also the requirement that there is a function F(n) from the natural numbers to the integers such that F(n)y is equivalent to determining whether x-y is negative, zero, or positive. Then I wondered whether we could use the same kind of trick to get a representation in which multiplication is also computable. For this, instead of taking a basis of the computable reals as a vector space over the rationals, take a transcendence basis t1,t2,... for the computable reals as a [CapitalThorn]eld extension of the rationals, and let the representation of a real include its minimal polynomial over some Q[t1,t2,...,tn], as well as a machine computing approximations to it (and to t1,t2,...,tn for good measure). If we have two such representations, they are both algebraic over some Q[t1,...,tn] for some n (the larger n of the two). There is a polynomial in Q[x; t1,...,tn] whose roots are the sums (products, respectively) of the roots of the two minimal polynomials. Factorization of polynomials in Q[x; t1,...,tn] is computable. We can then determine the minimal polynomial of the particular sum we are dealing with by computing the roots of the factors (using the machines for computing t1,...,tn) and the sum closely enough to see which factor it is a root of. I had thought here about using a lemma that for any _[CapitalThorn]xed_ [CapitalThorn]nite set of computable reals r1,...,rn, the algebraic closure of the [CapitalThorn]eld generated by them has an equality that is decidable in principle. In order to be able to decide it, we need a complete set of algebraic relations between r1,...,rn and some algorithms from computational algebraic geometry. (The complete set of algebraic relations isnt necessarily computable from r1,...,rn but in principle it exists.) But I think the above is as simple an argument. As long as were pretending to know things, we might as well pretend to know a transcendence basis rather than just some generators. I suspect the same trick can be extended to make the exponential function computable too, but I dont think our present-day mathematics has enough of the relevant theorems to complete the job. I brießy was unsure for instance whether e^e^e^...^e with a stack of n es is algebraic is decidable. But then I remembered a theorem (at least I think its a theorem now) that given any x1,...,xn, the transcendence degree of Q(x1,...,xn,e^x1,...,e^xn) is at least n. Apply this to x1=1,x2=e,...,xn=e^e^...^e (with n-1 es) and you get that the transcendence degree of Q(1,e,e^e,..., e^e^e...^e) with n es in the last stack is at least n-1, which implies that e, e^e, e^e^e,... are algebraically independent. But thats still just a special case. If theres an algorithm for determining whether an arbitrary expression built from 0, 1, +, -, *, /, and e^x is =0, and I expect there to be one, I dont think its known yet. I dont think we have that, let alone a way to prove that for each [CapitalThorn]xed [CapitalThorn]nite set r1,...,rn of computable reals theres a similar algorithm for expressions built up from them and 0,1,+,*,/,e^x. I suspect that this is the case, and that to get one its enough to know a set of exponential-algebraic relations among the r1,...,rn that is complete in a suitable sense. Keith Ramsay === Subject: Re: usenet, math, and tablet PCs? -------------------------------------s-o-s------------------- --------------- -- > Will, thats a good idea, i might make do with that for now. > Herc, i was actually thinking of a web forum, since they seem to lend > themselves better to graphics, etc.--i can be a nightmare to get news > servers to carry groups with images, etc. > doesnt seem to come up with much, as far as posting graphics goes.. > -adamgolding But forums tend to be small, unless you are lucky and one of the major maths forums supports it. I would suggest setting up a sci.math.html that would be as centralised as you could get for maths graphics. Then the graphics is not stored on the server, thats an option, its just hypertext and you can upload your graphics to a website and link to it. The messages are only 5K but then take extra time after selected to download. If its in the charter not to attach the images to the post perhaps more news servers would include it. I tested some equations in HTML posting on usenet and it works great. Maths is one of the few groups where some browsers dont support HTML posts due to use of UNIX terminals at the UNIs, but sci.math.html is clearly marked and people can download HTML newsreaders if they have to. HTML and newsgroups, just use the standard! Herc === Subject: Re: FINALLY THE TRUMAN PROBE GOES PUBLIC -------------------------------------s-o-s------------------- --------------- -- >>Actually this is not entirly true. Although the whole CIA/NSA thing > >is > >>rediculous on the face of it, researchers have manged to utilize brain >> >> >>no its FACT. youre just all up your own asses that your greatest asset, >>your own thoughts and all the homosexual fantasies you have all day are >>your own. they are NOT. thought ALWAYS goes to your voice box, >>speech is just louider. >> >>there is a myriad of proof of this. all the effects the site descibes >>independantly >>corroborate all the claims Ive made for 3 years under the mind probe. >> >>Results 1 - 10 of about 30 for forced telepathy herc. >> >>ITS CLEAR AS DAY. Ive been forced to THINK to everybody in sight, >>in conversation for years. >> >> >> >> >> >>waves (I dont know the real science behind it but Ôwaves is as > >good > >>a word as any) to make the mouse pointer on a computer screen move at >>the command of the person being monitored. >> >>For this to happen I would think that there is some kind of signal >>being picked up from the brain. Evolution not withstanding. >> >>gman >> > >Hi Gman, >Theres also research which is developing a wheelchair >controlled by brainwaves. I can see the bene[CapitalThorn]ts for >paraplegics. > >Later, >Pepe le Pew aka Pat Sullivan >> >> >>how do imposiles see the bene[CapitalThorn]ts >> >> > >Herc, >Was that a question? What is an imposile? (Dictionary.com gave >me a no entries found.) > >Later, >Pepe >> >>The mind doesnt think using electrical signals. So your point is moot. >>Although it can learn to output signals to recievers. What is translated to >>the voice-box isnt our thoughts. Our thoughts involve neurochemicals. >>Those are neither recieved nor tranmitted to the vocal cords. Only >>instructional signals. Thoughts or emotions, or when theyre inseperable, >>especially, will forever be impossible to decipher. Theyre much more >>complex, in a single humans mind, than is comprised in all the computers on >>earth, ever having been built; could possibly emulate. Let alone decipher. >>Herc, you are ill. That is all. Its the simplist explenation. >> >> >>Rob > rob, your mind is being read via satelite and they are imprinting quackery into you. > you know stuff all about the audio reception of thought. Ive told you 100 times > there is evidence of mind probes in australia. if you choose to ignore that then > ignore the claims that go with it. > Herc > We would like to ignore you and your babbling. You have your head up > your ass so far you have starved your brain of O2. > HYPOTHESIS: > |-|erc is a wacko. > OBSERVATIONS: > |-|erc is menatlly ill > |-|erc watches movie called Truman Show > |-|erc indulges himself that he is Truman, blurring reality with his > take on it, interjected by Hollywood make-believe, all the while making > an ass of himself on UseNet. > |-|erc is schizo and does not take meds. > |-|erc has been before the courts and admitted as much > CONCLUSION: > |-|erc is a wacko. > http://www.mazepath.com/uncleal/sunshine.jpg > Ogie oh cmon! [herc] check out townsville australia for stuff you wont believe [ogie] naa Herc === Subject: Re: You pedantic ignoramuses In sci.logic, grubb@math.niu.edu >> In sci.logic, |-|erc >> >> <34oiivF4a82laU1@individual.net>: >> >>> How many reals? >> >> Well thats almost a sentence. >> >> Herc >> >> Its also a question. >> How many reals? >> Nomenclature: >> [1] Cardinality of N (the natural numbers) or J (the integers) >> is aleph_0 (pronounced aleph_null). >> [2] aleph_{n+1} is the cardinality of the power set of the >> set whose cardinality is aleph_n. For example, aleph_1 is >> the cardinality of the set of all subsets of N. >> [3] Computability isnt an issue here, though the answer might >> be provable using theorem helpers. >> [4] If the answer is *not* an aleph, indicate where it is positioned >> relative to two alephs. > Unfortunately, this is *not* the de[CapitalThorn]nition of the alephs! You have > de[CapitalThorn]ned the beths. What? New one on me... :-) Then again, Im not familiar with the ins and outs of trans[CapitalThorn]nite cardinals/ordinals/etc. > For the alephs, you need to de[CapitalThorn]ne aleph_{n+1} to be the > cardinality of a set X all of whose subsets either have cardinality > <=aleph_{n} or the same cardinality as the set X itself. > The Continuum Hypothesis states that the cardinality of the set > of all subsets of N is aleph_1, but there are models of set > theory for which this is false. Interesting detour. Not that |-|erc will know much about it; the trouble with computable numbers is that the set of all such numbers is denumerable, and indeed may be [CapitalThorn]nite (theres only so many atoms in the Universe, and the Universe has only so many timeticks :-) ). For its part Mathworld substantiates your statement: http://mathworld.wolfram.com/Aleph-1.html A search in Mathworld for Beth-0 dredges up L.9awenheim-Skolem Theorem http://mathworld.wolfram.com/Loewenheim-SkolemTheorem.html which is a bit over my head as I cant say Im familiar with what a model (countable or otherwise) is in this context. Proof once again that math (or, for that matter, Mathworld :-) ) isnt quite done yet. :-) Nor, apparently, is my learning. *scratches head* A search for trans[CapitalThorn]nite number dredges up http://mathworld.wolfram.com/Trans[CapitalThorn]niteNumber.html which only de[CapitalThorn]nes trans[CapitalThorn]nite ordinals. Wikipedia does have http://en.wikipedia.org/wiki/Aleph_number http://en.wikipedia.org/wiki/Beth_number which is as you describe it. The relationship between C and the alephs is less than clear, though it appears C = beth_1 = 2^aleph_0, and its very clear (by Cantors two proofs) that C != aleph_0 and therefore C != beth_0. Consider my prior post suitably modi[CapitalThorn]ed. :-) > --Dan Grubb -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: You pedantic ignoramuses -------------------------------------s-o-s------------------- --------------- -- >>>> How many reals? >>> >>> Well thats almost a sentence. >>> >>> Herc >>> >>> >>> Its also a question. >>> >>> How many reals? >> >> Youve watched one to many John Wayne movies, questions are a >> subtype of sentences. >> >> Herc >> >> >> If you insist. >> >> What is Card(R)? >> >> or >> >> What does Card(R) equate to? >> >> Now answer the question, pilgrim. :-) > no such de[CapitalThorn]ned function as Card. > there is there is > Oh, youre just being a card. :-P Or maybe another word close thereto. > Let N be the set of whole numbers, and let K(n) be > built up in the following way: > K(0) = empty set > K(n+1) = K(n) union {K(n)}. > For an arbitrary [CapitalThorn]nite set S, if a 1-1 and onto mapping > exists from S to some K(n), then one can say Card(S) = n. > Im using ÔCard here as short for Ôcardinality. > The domain of K can be extended in a fairly obvious fashion > by various means: > K(aleph_0) = N > K(aleph_1) = 2^(K(aleph_0)) > ... > K(aleph_(n+1)) = 2^(K(aleph_n)) > where 2^S is the power set of S = the set of all possible subsets of S. > AFAIK, Card(R) is known to be greater than Card(N) = aleph_0 but it > is unknown whether Card(R) = aleph_1 or is in between aleph_0 > and aleph_1. thats very pretty {}, {{},{{}}}, {{}, {{},{{}}}} but Ive shown a technique to count powersets of N using computable in[CapitalThorn]nite bit masks and I can count to any practically de[CapitalThorn]nable real. lets pull a function out of the air, R(x) = Comp(random, x) + 1 mod 10 according to you, this LIARS statement F(x) = ~F(x) is a new random sequence and proves the set of all programs Comp cant spit out any random string at all. Herc === Subject: Re: You pedantic ignoramuses In sci.logic, |-|erc <34toqlF4eeh1rU1@individual.net>: > -------------------------------------s-o-s------------------- ---------------- - >>>>> How many reals? >>>> >>>> Well thats almost a sentence. >>>> >>>> Herc >>>> >>>> >>>> Its also a question. >>>> >>>> How many reals? >>> >>> Youve watched one to many John Wayne movies, questions are a >>> subtype of sentences. >>> >>> Herc >>> >>> >>> If you insist. >>> >>> What is Card(R)? >>> >>> or >>> >>> What does Card(R) equate to? >>> >>> Now answer the question, pilgrim. :-) >> >> no such de[CapitalThorn]ned function as Card. >> >> there is there is >> Oh, youre just being a card. :-P Or maybe another word close thereto. >> Let N be the set of whole numbers, and let K(n) be >> built up in the following way: >> K(0) = empty set >> K(n+1) = K(n) union {K(n)}. >> For an arbitrary [CapitalThorn]nite set S, if a 1-1 and onto mapping >> exists from S to some K(n), then one can say Card(S) = n. >> Im using ÔCard here as short for Ôcardinality. >> The domain of K can be extended in a fairly obvious fashion >> by various means: >> K(aleph_0) = N >> K(aleph_1) = 2^(K(aleph_0)) >> ... >> K(aleph_(n+1)) = 2^(K(aleph_n)) >> where 2^S is the power set of S = the set of all possible subsets of S. >> AFAIK, Card(R) is known to be greater than Card(N) = aleph_0 but it >> is unknown whether Card(R) = aleph_1 or is in between aleph_0 >> and aleph_1. > thats very pretty {}, {{},{{}}}, {{}, {{},{{}}}} > but Ive shown a technique to count powersets of N using > computable in[CapitalThorn]nite bit masks The trouble with that is that an in[CapitalThorn]nite bit mask is essentially a real in base 2. > and I can count to any practically de[CapitalThorn]nable real. TX_2 = {k/2^n: n >= 0, k,n in J} doesnt cover R or Q. Sorry. > lets pull a function out of the air, R(x) = Comp(random, x) + 1 mod 10 > according to you, this LIARS statement F(x) = ~F(x) is > a new random sequence and proves the set of all programs > Comp cant spit out any random string at all. > Herc Without knowing (or [CapitalThorn]nding earlier implementations of) Comp(x,y), I cant process this statement. I can of course assume that Comp(x,y) is machine x running on input y, in which case the correct equality R(x) = (Comp(x,x) + 1) mod 9 and that is indeed a new sequence, not in the Comp() list. Of course, it is not a computable sequence, though it is well-de[CapitalThorn]ned. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: You pedantic ignoramuses > lets pull a function out of the air, R(x) = Comp(random, x) + 1 mod 10 > according to you, this LIARS statement F(x) = ~F(x) is > a new random sequence and proves the set of all programs > Comp cant spit out any random string at all. > Herc > Without knowing (or [CapitalThorn]nding earlier implementations of) Comp(x,y), I > cant process this statement. I can of course assume that > Comp(x,y) is machine x running on input y, in which case the correct > equality > R(x) = (Comp(x,x) + 1) mod 9 > and that is indeed a new sequence, not in the Comp() list. Of > course, it is not a computable sequence, though it is well-de[CapitalThorn]ned. Is R computable? Herc === Subject: Re: Your baby was damned to death, but who made the choice? posting-account=JMmqYwwAAAAB3-J0cpBom9dzOEkl4nOJ Perhaps we could compromise and agree to just abort you? === Subject: Re: Rectangles ETAuAhUArJ4YvmwAKpHcBBrRPceD3sTLbG4CFQDAqZaX/ b57PY9nK4NwdMxw1QFyRg== You solved half the prolem. Now for the other half. What is the score for rectangles with relatively prime sides? In your 2x3 example, where 2 and 3 are relatively prime, you got 4. Now lets go slowly down the digaonal and see what happens. We start by going into one square, and stay in that square until -- we cross a grid line. Thats when we add another square to our score. We [CapitalThorn]nd, as we continue, that we add two more squares by crossing two more grid lnes, for a total of 3 crossings. Thus 4 squares -- the one we started with and the three we got into cossing the grid lines. The score for a given rectangle is the number of points where the diagonal crosses a grid line, plus one for the initial square. Now if the retangle has relatively prime sides, as assumed in this discussion, then the grid lines are crossed one at a time -- theres no point inside the rectangle where two perpendicular grid lines are crossed at the same time. In an mxn rectangle there are m-1 grid lines one way and n-1 grid lines the other way. Ergo m+n-2 crossings plus the initial square gives m+n-1 for the score of an mxn rectangle, when m and n are relatively prime. Thus in a 2x3 rectnagle, the score is 2+3-1 = 4. Now you can combine this result with the reduction formula that youve already posted. You should [CapitalThorn]nd that the general formula for any m and n, relatively prime or not, is m+n-gcd(m,n). --OL === Subject: Divergence of a Certain series Let H(n) = 1 + (1/2) + (1/3) + ... + (1/n) = sum of the [CapitalThorn]rst n terms of the harmonic series. Note that lim_{n to infty} frac{H(n)}{n} = 0 and sum_{n = 1}^infty {frac{H(n)}{n}} diverges. Question: Does sum_{n = 1}^infty {(frac{H(n)}{n})^2} converge? (Proof requested) === Subject: Re: Divergence of a Certain series > Let H(n) = 1 + (1/2) + (1/3) + ... + (1/n) = sum of the [CapitalThorn]rst n terms > of the harmonic series. Note that > lim_{n to infty} frac{H(n)}{n} = 0 and sum_{n = 1}^infty > {frac{H(n)}{n}} diverges. > Question: Does sum_{n = 1}^infty {(frac{H(n)}{n})^2} converge? Yes, in fact sum_{n=1,oo} H(n)/n^p < oo for all p > 1. An elementary estimate shows H(n) <= 1 + int_{1,n} dx/x = 1 + ln(n) for all n. That should help. === Subject: Re: Divergence of a Certain series posting-account=Kb0T_QwAAACc9B9LpxfLjH0hHHYjPxft > Let H(n) = 1 + (1/2) + (1/3) + ... + (1/n) = sum of the [CapitalThorn]rst n terms > of the harmonic series. Note that > lim_{n to infty} frac{H(n)}{n} = 0 and sum_{n = 1}^infty > {frac{H(n)}{n}} diverges. > Question: Does sum_{n = 1}^infty {(frac{H(n)}{n})^2} converge? > (Proof requested) yes, i would think so (didnt work it out). proof should be easy: just use an integral test. for the integral test, look up any book on advanced calculus or analysis which discusses convergece of series. u can do the rest yourself.