mm-3859 === Subject: Re: delta functions and test functions I'm a bit confused about what makes a test function. > Basically, I have a physics question that's asking me to integrate > f(x)delta(x-1/2) between 0 and 1 where f(x) = x for 0 = x = 1/2 and > (1-x) for 1/2 = x = 1. > I'll assume the above is a typo for > f(x) = {x, if 0<=x<1/2} (1) > {1-x, if 1/2<=x<1} > (or > f(x) = {x, if 0<=x<=1/2} > {1-x, if 1/2=, it must not have liked it. > Perhaps your professor is using the function f(x) of definition (1) as a > pseudo-test function, even though it is not a proper test function. So any function that's continuous will work. I had a suspicious feeling that the don't think about it and shove 1/2 in, it's a physics question after all approach would work... Nat === Subject: Re: Rings of strictly rational numbers with the usual add'n. and mult. > Does anybody know of any results pertaining to a > ring R with the following properties: 1. Addition and multiplication in R is the usual > addition and multiplication of rational numbers; > 2. For all X and Y in R, X and Y are strictly > rational (i.e. non-integral); > 3. For all X and Y in R, X + Y is also strictly > rational; > 4. For all X and Y in R, (1/X) and (1/Y) are > strictly rational; > 5. For all X and Y in R, (1/X) + (1/Y) is also > strictly rational; and > 6. (X * Y) = k, k an integer constant? > I assume that in (4) and (5) you intend to exclude 0. > I'm also > going to assume that by (6) you mean something like > For all X and Y, > there exists an integer k such that XY = k. If not > (that is, if you mean > that there exists a k such that for all X and Y, XY = > k), then > you get the following situation: > Suppose there exists a rational X in R; then both 1/X > and X + 1/X = (X^2 + 1)/X > are in R. But then X * 1/X = 1 = k, but X * (X^2 + > 1)/X = (X^2 + 1) = k as well. > That forces X^2 + 1 = 1 and X = 0. But 1/0 can't be a > rational with the usual > ... multiplication of rational numbers. > If I've interpreted (4), (5), and (6) correctly, it > looks R is the > zero ring (the ring containing only 0). Here's why: > If there is a number other than 0 (say, x) in R, then > its inverse 1/x is > also in R; that gives x * (1/x) = 1 in R. But then 2 > (1 + 1) is in R, > so that 1/2 is in R. This implies (1/2) * (1/2) = 1/4 > is in R. But we > said that for all x,y in R, xy is an integer. > Contradiction. > 0 is allowed to be in R, though, because it's not > subject to rules > (4) and (5), and (if it's the only number in the > ring) it fits rules > (1),(2),(3), and (6). So I think you found the zero > ring. > Matt > http://www.teranews.com *** I'm sorry, but I intended to exclude zero from R. Let me explain. Assume N = (p^k)(m^2) is an odd perfect number. Then: sigma(N)/N = 2 = [sigma(p^k)/p^k][sigma(m^2)/m^2] Now, since p is congruent to 1 modulo 4 and p is a prime, p is at least 5. Hence: 1 < a := sigma(p^k)/p^k < 5/4 = 1.25 Also: 1.6 = 8/5 < b := sigma(m^2)/m^2 < 2 Note that, by the Arithmetic Mean-Geometric Mean Inequality, a + b > 2*sqrt(2) (since equality does not hold; in fact, a < b). To get an improved lower bound for a + b, consider: (a - 1.25)(b - 1.25). This last quantity is negative since a < 1.25 < 1.6 < b. Therefore: ab - 1.25(a + b) + (1.25)^2 < 0 Consequently: (5/4)(a + b) > 2 + 25/16 = 57/16. Hence: a + b > 57/20 = 2.85 > 2*sqrt(2). To get an upper bound for a + b, consider: (a - 1)(b - 1). This last quantity is positive since 1 < a < b. Therefore: ab - (a + b) + 1 > 0, which implies: a + b < 3 SO BY NOW, IT SHOULD BECOME CLEAR WHY I HAD BEEN CONSIDERING THE MORE GENERAL SCENARIO: 1. Addition and multiplication in R is the usual addition and multiplication of rational numbers; 2. For all X and Y in R, X and Y are non-integral; 3. For all X and Y in R, X + Y is also non-integral; 4. For all X and Y in R, (1/X) and (1/Y) are non-integral; 5. For all X and Y in R, (1/X) + (1/Y) is also non-integral; and 6. (X * Y) = k, k an integer constant? Since this set R does not qualify as a ring, according to the most recent posts, what other more useful properties might it possess? Of course, Q Z is not a ring (it's not even a group under addition since it does not have the additive identity 0; it's also not a group under multiplication since inverses of integers, which are non-integral, do not have inverses in Q Z). And it appears that R is a subset of Q Z. I am just thoroughly confused. Anybody got some brighter ideas? You may reply to my email address: jusiabdres@yahuu.cum (u => o, i => e, e => i for the correct email address) === Subject: Re: A topological property >[...] > Sorry, I confused my own question with something different. > Then what is your question? The original on. But as your examples suggest, it was not really a good one. > Accurate statements > If A is pc, then some a in A with pc Aa (1) > negation > A is pc and for all a in A, Aa not pc (2) > Do you mean forall A in (1) ( and exists A in (2) ) ? > Yes. Statement and negation. (1) and (2) > for all assembled A subset S, some a in A with assembled Aa > some assembled A subset S with for all a in A, Aa disassembled > (1) is false and (2) is true as nulset is assembled. > Thus to modify, restate (1) and (2): > for all assembled A nonnul subset S, some a in A with assembled Aa > some assembled A nonnul subset S with for all a in A, Aa disassembled > Recall connected and assembled are equivalent > within completely normals spaces. Hence > For discrete spaces, (1) is true. For the reals, (1) is false. > (1) ==> W(X,x) = { |A| in N : some pc A with x in A } > = { n in N : n <= |A| } > Correct statement when > assembled-C(x) = /{ assembled A : x in A } > is finite is > (1) ==> W(X,x) = { n in N : n <= |assembled-C(x)| } > Anyway as singletons and nulset is pc, always 0,1 in W(X,x). > Yes, at least one reason why my original question was shallow. > I suspect that (1) ist true but was unable to prove it. > If it is, some limitions of finiteness is needed. Actually I tried only the case of a finite assembled set. > -- > This week I seem to prove my ability to write nonsense beyond doubt. > The correct statement should read > A is assembled > iff > A is a connected subset of every open set U that contains A. > As connected is intrinsic property of the (sub)space, if A is connected > subset of some open superset, it's connected subset of every open > superset. > Now as whole space is open, that is equivlant to A is connected. > Thus may I ask, is your week over yet? ;-) I see it now, I confused intersect with intersect in A. You are right, the weekend is still ahead of me. Marc === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= > Torkel Franz.8en, well known for his many Usenet posts, died of skeleton > cancer at Wednesday, April 19, at the age of 56. Very sad to hear this. Condolences to his family, friends and colleagues. He will be missed by me and many others. Chas === Subject: Re: Closing the Intersection when a given A,B subset of X can have > cl(A / B)=cl(A)/cl(B) > When A and B are closed. I don't believe that is a characterization. Its only a necessary condition. There are A and Bs in some topologies by which the equality holds without either of them beling closed. Jose Capco === Subject: Re: Does this error of logic have a Logical Fallacy name to it?; Re: page 3, History of Euclid Proof of Infinitude of Primes... > Are you assuming pn is the last prime ? Yes or No > If pn is the last prime, is every number greater than pn composite? > Yes or No > Is w+1 > pn? Yes or No > Is w+1 composite ? Yes or No > The answers to these questions are > Yes, Yes, Yes , Yes No, the answers are both Yes and No. Anything is provable in the contradictory theory of the integers with finitely many primes. Recall I explained this to you last year in a similar thread [1]. See also [2], where I explicitly deduce in that 1 = 0 ... --Bill Dubuque === Subject: Re: Does this error of logic have a Logical Fallacy name to it?; Re: page 3, History of Euclid Proof of Infinitude of Primes... <444722D6.1000701@dtgnet.com> If pn is the last prime, is every number greater than pn composite? > Yes or No > Is w+1 > pn? Yes or No > Is w+1 composite ? Yes or No > The answers to these questions are > Yes, Yes, Yes , Yes > No, the answers are both Yes and No. Anything is provable in the > contradictory theory of the integers with finitely many primes. > Recall I explained this to you last year in a similar thread [1]. > See also [2], where I explicitly deduce in that 1 = 0 ... Yes, I agree that, depending on how you word and order the steps of the proof , you can finally conclude t w+1, as A.P. would formulate it, is 1, prime and composite. . However, if you assume that pn is the last prime, it follows necessarily that pn+1 must be composite. That is surely the point of a reductio ad absurdum proof. any two of which are obvioisly contradictory or absurd or any one of which is absurd, a prime is zero, for example. Assuming pn is the last prime, the answer to the question, Is w+1 composite ? can only be Yes If you could answer Yes and No, the chain of reasoning would not go through. When you ask if any of the n primes assumed to exist divides w+1, then you are left with deductions like: w+1 is a composite number with no prime factors, w+1 is a number that has no prime factors so is equal to 1 w+1 is not divisible by any prime < w+1 so must itself be prime w+1 is not divisible by any prime <= pn so must itself be prime or divisible by a prime >pn Until you derive a contradiction, if you are assuming that pn is the last prime, the only conclusion you can make is that w+1 is composite because any number > pn is composite So, if you take the consequences of the assumption that pn is the last prime seriously, you know w+1 is a composite number until you discover that it is a composite number with no prime factors. On the other hand, if you are considering that all the valid deductions that follow from assuming pn is the last are in someway out there then the statements w+1 is prime , w+1 is composite etc. will also all be out there whether or not they are mentioned explicitly in a particular proof. So, in this sense you are right. === Subject: ode system critical points I know x' = Ax has 0 = (0,0) critical point. Now I am given new problem: x' = Ax - (2, 0)^T, where ^T means transpose, A = (1 1; 1 -1) (semi-colon means new row) The book hints to make the transformation x = x0 + u. Does that just mean: (x0 + u)' = A(x0 + u) - (2,0)^T ? I don't see how that helps. === Subject: Re: ode system critical points > I know x' = Ax has 0 = (0,0) critical point. > Now I am given new problem: > x' = Ax - (2, 0)^T, where ^T means transpose, A = (1 1; 1 -1) > (semi-colon means new row) > The book hints to make the transformation x = x0 + u. > Does that just mean: > (x0 + u)' = A(x0 + u) - (2,0)^T ? > I don't see how that helps. In situations like these it is always a good idea to read the lecture notes. If you don't have them, read the preceding 10 pages or so in the book. Try to understand them. === Subject: Re: Ring inside a field If E is an algebraic extension of a field F, and > if F < R < E where R is a ring, then R is a field. I think the hypotheses can be weakened to an integral > domain R which is algebraic over a field F. In this case, > R would still be a field. Right? >> If by >>R is algebraic over F<< you mean that every >> element of R is a root of a monic polynomial with >> coefficients in F, then the answer is NO. >> Take F=rational numbers and R=Z[sqrt(2)] and >> E the field of fractions of R. > We do not have F < R in that example! I have to admit I am > not completely sure exactly what Julien Santini is asking. If F < R is an integral extension of domains then F is a field <-> R is a field. The proof is trivial (manipulate minimal polynomials). Note that a field extension is integral <-> algebraic --Bill Dubuque === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Yes. Any mathematical statement that has no observable implications >> will never turn out to be useful. > Number theory, at one time, had no observable implications! > So by Petry's standards it cannot be useful. > I think that what David is saying (not that I agree with > him by the way) is that observable means that it yields > a pi-0-1 statement that can, for each natural number, be > (computationally) tested. That's the idea which started this thread. Friedman claimed that mathematics is really Pi01. In fact, I prefer to go slightly beyond that, claiming that a statement must include, at least implicitly, information about how much computation needs to be done to observe the results. > There are an abundance of pi-0-1 statements in number theory, > like the Goldbach conjecture which make observable predictions > in the sense described above. Precisely right. > As far as I can tell David's rule for good mathematics > doesn't depend on the availability of applications. Although > that may be Han's rule, which IMO is more extreme and perhaps > confusing the issue in this thread. I think that's somewhat right. > Although David alludes to the computer as a microscope > into the world of computation I don't think he means to > suggest that a physical computer is required as a means > to validate good mathematics; pencil & paper computation > is enough to do that. The physical computer is just a > tool that enhances our ability to investigate that world. I use a reference to the computer as a conceptual aid to make a point. > Maybe I've got David's argument all wrong. Actually, you seem to be one of the very few who have it mostly right. >Other posters have > given summaries of David's position only to be told by him > later that they didn't get it. I don't think that he is > doing a good job of communicating his ideas clearly. I can't imagine how I could communicate them more clearly, but who knows, maybe the failure of communication is all my fault. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > Yes. Any mathematical statement that has no observable implications >> will never turn out to be useful. > > Number theory, at one time, had no observable implications! > > So by Petry's standards it cannot be useful. > I think that what David is saying (not that I agree with > him by the way) is that observable means that it yields > a pi-0-1 statement that can, for each natural number, be > (computationally) tested. > That's the idea which started this thread. Friedman claimed that > mathematics is really Pi01. In fact, I prefer to go slightly beyond > that, claiming that a statement must include, at least implicitly, > information about how much computation needs to be done to observe > the results. > There are an abundance of pi-0-1 statements in number theory, > like the Goldbach conjecture which make observable predictions > in the sense described above. > Precisely right. > As far as I can tell David's rule for good mathematics > doesn't depend on the availability of applications. Although > that may be Han's rule, which IMO is more extreme and perhaps > confusing the issue in this thread. > I think that's somewhat right. > Although David alludes to the computer as a microscope > into the world of computation I don't think he means to > suggest that a physical computer is required as a means > to validate good mathematics; pencil & paper computation > is enough to do that. The physical computer is just a > tool that enhances our ability to investigate that world. > I use a reference to the computer as a conceptual aid to make a point. > Maybe I've got David's argument all wrong. > Actually, you seem to be one of the very few who have it mostly right. >Other posters have > given summaries of David's position only to be told by him > later that they didn't get it. I don't think that he is > doing a good job of communicating his ideas clearly. > I can't imagine how I could communicate them more clearly, but who > knows, maybe the failure of communication is all my fault. One place where I find difficulty is: although you have above indicated (reasonably well) what type of reasoning you advocate, you haven't given a clear distinction regarding the domain of fundamental objects upon which this reasoning is to be applied. For example, it's clear to me that you would acceptn+1>n is a well-formed statement about natural numbers. Now consider the notation 1/n to be the rational number such that n*1/n = 1. I think you would also claim that if n > 1, then n > 1/n is also computationally verfiable. But 1/n is not a natural number: it is an /abstraction/; whose properties can be /modelled/ as a series of computations on naturals. Similarly, the notion of a set is an abstraction: whose properties can be /modelled/ (via Goedelization) as a set of computations on the naturals. What I don't see is why n > 1/n is obviously a meaningful construct and more than simply the statement in the theory of the rationals, 'n>1/n' is a well formed expression, while statements in set theory such as |P(X)| > |X| are /only/ meaningful as statements of the form in set theory, '|P(X)| > |X|' is a well-formed expression. At least, I don't see how the requirement of computable verfication, by itself, distingushes between these two domains of discourse: the rationals and set theory. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics n*1/n = 1. I think you would also claim that if n > 1, then n > 1/n > is also computationally verfiable. No, but I would claim, if n>1, then the assertion that n>1/n is computationally verifiable. > But 1/n is not a natural number: it is an /abstraction/; whose > properties can be /modelled/ as a series of computations on naturals. It has a concrete representation as a data structure which can be stored in a computer's memory. Real computer algorithms can operate on that structure. A notion of greater than can be defined in terms of an algorithm which takes two rational numbers as input. It's not merely an abstraction. > Similarly, the notion of a set is an abstraction: whose properties can > be /modelled/ (via Goedelization) as a set of computations on the > naturals. But we know from Cantor's Theorem (as it is usually interpreted in meta-mathematics)that there must exist sets that have no representation as finite data structures. And that's the problem; such sets don't live in the world of computation, and mathematics is the science of phenomena observable in the world of computation. > What I don't see is why n > 1/n is obviously a meaningful construct > and more than simply the statement in the theory of the rationals, > 'n>1/n' is a well formed expression, while statements in set theory > such as |P(X)| > |X| are /only/ meaningful as statements of the form > in set theory, '|P(X)| > |X|' is a well-formed expression. But I don't see how anyone can fail to see it. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > Yes. Any mathematical statement that has no observable implications >> will never turn out to be useful. > > Number theory, at one time, had no observable implications! > > So by Petry's standards it cannot be useful. > I think that what David is saying (not that I agree with > him by the way) is that observable means that it yields > a pi-0-1 statement that can, for each natural number, be > (computationally) tested. > That's the idea which started this thread. Friedman claimed that > mathematics is really Pi01. In fact, I prefer to go slightly beyond > that, claiming that a statement must include, at least implicitly, > information about how much computation needs to be done to observe > the results. According to Friedman, there are interesting Pi01 statements that can only be proved by adding large cardinal axioms to set theory, and he strongly advocates that mathematicians do precisely that. I think when Friedman claimed that mathemaics is really Pi01, all he meant was that the interesting propositions of mathematics are (or can be reduced to) Pi01 statements, even if some of them may require heavy set theory for their proofs. According to David Petry, however, such Pi01 statements can be computationally verified to be true. I fail to see how this is possible for infinitely many natural numbers. The correct conclusion is that the truth of such Pi01 statements is in principle unverifiable. For we can never ever be sure of the consistency of set theory, leave alone large cardinal axioms. So the Pi01 statenents could always be false, despite the available proofs of their truth in set theory + large cardinals. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics I think that what David is saying (not that I agree with > him by the way) is that observable means that it yields > a pi-0-1 statement that can, for each natural number, be > (computationally) tested. > That's the idea which started this thread. Friedman claimed that > mathematics is really Pi01. In fact, I prefer to go slightly beyond > that, claiming that a statement must include, at least implicitly, > information about how much computation needs to be done to observe > the results. > According to Friedman, there are interesting Pi01 statements that can > only be proved by adding large cardinal axioms to set theory, and he > strongly advocates that mathematicians do precisely that. That's the way I understand Friedman's position also. However, if you accept my assertion that a statement must include, at least implicitly, information about how much computation needs to be done to observe the result, then the Pi01 statements that are provable by adding large cardinal axioms become much less interesting and of questionable meaningfulness. > I think when > Friedman claimed that mathemaics is really Pi01, all he meant was that > the interesting propositions of mathematics are (or can be reduced to) > Pi01 statements, even if some of them may require heavy set theory for > their proofs. Note quite. He has been able to come up with statements that require heavy set theory, but those statements are of questionable interest to claims that he has found two theorems within real mathematics that I found that Kruskal's tree theorem and the Robertson/Seymour graph minor theorem necessarily use methods that could reasonably be described as set theoretic. Of course, it is way below anything like full use of the power set of N. Nevertheless, it is definitely a weak use of uncountability, that is in stark contrast with the vast vast vast preponderance of mathematics. > According to David Petry, however, such Pi01 statements can be > computationally verified to be true. That is far from true, and not worth commenting on. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Yes. Any mathematical statement that has no observable implications > > will never turn out to be useful. > > Number theory, at one time, had no observable implications! > Worse. Hardy was proud that there were no observable implications. If there's any mathematical field that most clearly does have observable implications, it's number theory. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <4448c701$0$9225$ed2619ec@ptn-nntp-reader01.plus.net If a tree falls in the woods, and there is no-one to hear it, do they > give a damn? Let us see. You are making the folowing proposition: Tree A has fallen in forest B at time t. where, let us say, we all know which tree and which forest we are talking about. According to you, at time t there is no human being present to verify that the tree has fallen. So none of us have any proof at time t that the tree has fallen. Further, let us suppose that the forest is completely inaccessible, so that no human being can in principle verify the status of tree A. So according to you, the above is a meaningful truth that is in principle independent of the human mind. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <4448c701$0$9225$ed2619ec@ptn-nntp-reader01.plus.net If a tree falls in the woods, and there is no-one to hear it, do they > give a damn? > Let us see. You are making the folowing proposition: > Tree A has fallen in forest B at time t. > where, let us say, we all know which tree and which forest we are > talking about. According to you, at time t there is no human being > present to verify that the tree has fallen. So none of us have any > proof at time t that the tree has fallen. Further, let us suppose that > the forest is completely inaccessible, so that no human being can in > principle verify the status of tree A. So according to you, the above > is a meaningful truth that is in principle independent of the human > mind. Sorry, I accidently posted the previous message before completing it -- here is the rest. My claim is that the only truths that exist in the above scenario are axiomatic declarations made by human beings. Thus your assertion that the tree has fallen at time t is simply an axiom that you have chosen to add to your theory. I may not choose to add such an axiom, or I may choose to add the negation of the above proposition as my axiom. Iff I make no assertions, then according to NAFL, the above proposition is neither true nor false for me. That is the full story according to NAFL. Think about it. A second scenario is possible in which one may be able to access the forest after time t and make a valid deduction at time t2 > t that the tree had indeed fallen at time t. But in NAFL, this retroactive assertion is itself valid only for t >= t2. So the reality that is supposed to have existed at time t itself is knowable only for t>=t2, and for prior times, such a mind-independent reality can be thought of as metalogical, i.e outside of the semantics of NAFL theories. This second scneario is more complicated, but correctly handled by NAFL. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics On 16 Apr 2006 15:23:10 -0700, david petry said: >> |As I have mentioned before, my motivation was to create a language and >> |a foundation for mathematics that would facilitate the transfer of our >> |mathematical knowledge to the computer (i.e. an artificial >> |intelligence). >> The set-theoretical concepts that have come under questioning here >> can be expressed in terms of the concept of property. Sets are always >> extensions of properties. Any AI that understands the concept of >> property will be able to be given descriptions of the concepts of set >> theory. > That's not true, if we define a property to be something that can be > computed, that is, if we require that we have an algorithm which, on > input X, will necessarily output either a yes or a no, depending on > whether X has the property or not. You've not followed up on Dolan's observation that, according to your definition here, halts on input I is not a property of Turing Machines -- despite the fact that you discuss halting as if the notion makes sense in a lot of your posts (from which, I assume, we can infer that you think it is a genuine property). Your silence is conspicuous. This is a very fine example of how your a priori philosophical musings do not even support the bits of mathematics that you implicitly acknowledge as legitimate. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl> computed, that is, if we require that we have an algorithm which, on > input X, will necessarily output either a yes or a no, depending on > whether X has the property or not. > You've not followed up on Dolan's observation that, according to your > definition here, halts on input I is not a property of Turing Machines > -- despite the fact that you discuss halting as if the notion makes > sense in a lot of your posts (from which, I assume, we can infer that > you think it is a genuine property). > Your silence is conspicuous. This is a very fine example of how your a > priori philosophical musings do not even support the bits of mathematics > that you implicitly acknowledge as legitimate. If you're referring to Kent Paul Dolan, then the silence is explained by the fact that I don't bother to read his posts. The notion of halts makes good sense if an upper bound on the number of computational steps taken to reach the halt state is provided. As is the custom, when we speak informally, we don't need to mention that upper bound. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl> of computational steps taken to reach the halt state is provided. As is > the custom, when we speak informally, we don't need to mention that > upper bound. But surely you imply that this upper bound /can/ be formalized, in some way. Do you consider a Busy Beaver function to be a meaningful or meaningless notion? (ref: http://en.wikipedia.org/wiki/Busy_beaver ) === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl> of computational steps taken to reach the halt state is provided. As is > the custom, when we speak informally, we don't need to mention that > upper bound. > But surely you imply that this upper bound /can/ be formalized, in some > way. > Do you consider a Busy Beaver function to be a meaningful or > meaningless notion? The Busy Beaver function cannot be computed, so a statement that gives that as an upper bound would not be making predictions about the results of a computational experiment, and hence the statement would not be meaningful. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44356e78$0$2019$ba620dc5@text.nova.planet.nl> The notion of halts makes good sense if an upper bound on the number > of computational steps taken to reach the halt state is provided. As is > the custom, when we speak informally, we don't need to mention that > upper bound. > But surely you imply that this upper bound /can/ be formalized, in some > way. > Do you consider a Busy Beaver function to be a meaningful or > meaningless notion? > The Busy Beaver function cannot be computed, so a statement that gives > that as an upper bound would not be making predictions about the > results of a computational experiment, and hence the statement would > not be meaningful. The Busy Beaver function appears to have been computed for /some/ values; the fact that it cannot be computed for /all/ values is derived from an argument regarding the halting problem that contains no reference to specific upper bounds; which would appear to be an argument you consider invalid. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Chris Menzel says... >You've not followed up on Dolan's observation that, according to your >definition here, halts on input I is not a property of Turing Machines >-- despite the fact that you discuss halting as if the notion makes >sense in a lot of your posts (from which, I assume, we can infer that >you think it is a genuine property). >Your silence is conspicuous. This is a very fine example of how your a >priori philosophical musings do not even support the bits of mathematics >that you implicitly acknowledge as legitimate. That's what's so frustrating about David Petry's posts. His claims *might* be interesting, but only if they are examined rigorously to see what are their actual implications. But David doesn't *want* his ideas examined rigorously (presumably because he suspects that people are just trying to defeat him with sophistry). What he wants is for people to instantly see the truth of what he says, and for them to treat it as gospel. That's very different from the way science is usually done (if he wants to be scientific...) -- Daryl McCullough Ithaca, NY -- NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth === Subject: Re: Algebraically closed days. My association with the Department is that of an alumnus. >I read the following in a course: (i) E is algebraically closed <=(ii) Every field of finite degree over E is E. >I was wondering why the author would write it this way; don't we also >have that every algebraic field over E is E ? They are both equivalent, but not identical. Every field of finite degree over E is algebraic over E (for any field E); but in general, it is not true that every field which is algebraic over a given field F must be finite over F (e.g., the algebraic closure of Q in the complex numbers is of infinite degree over Q). It is not hard to see that (ii) above implies every field F which is algebraic over E is equal to E; the converse is always true, since finite over F is more restrictive. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: isprime of flattened primes... Consider the number series 2 23 235 2357 235711 23571113 2357111317 235711131719 23571113171923 . . . where each number is a flattened list of primes (that is, a list of primes with the blanks removed to form a single number). Once you get past the single digit primes, should there be any more primes? So far, I haven't found any, but I haven't checked very many, as these types of numbers get huge in a hurry. Should there be any reason that they're all composite? ________Gerard S. === Subject: Re: isprime of flattened primes... >Consider the number series >where each number is a flattened list of primes (that >is, a list of primes with the blanks removed to form >a single number). >Once you get past the single digit primes, should >there be any more primes? What does should mean here? Anyway, Maple reports that 235711131719232931374143475359616771737983899710110310710911312713113713914 91511571631671731791811911931971992112232272292332392412512572632692712 77281283293307311313317331337347349353359367373379383389397401409419421 43143343944344945746146346747948749149950350952152354154755756356957157 7587593599601607613617619631641643647653659661673677683691701709719 is prime. dave === Subject: Re: isprime of flattened primes... >Consider the number series >235 >2357 >235711 >23571113 >2357111317 >235711131719 >23571113171923 >where each number is a flattened list of primes (that >is, a list of primes with the blanks removed to form >a single number). >Once you get past the single digit primes, should >there be any more primes? So far, I haven't found >any, but I haven't checked very many, as these types >of numbers get huge in a hurry. Should there be any >reason that they're all composite? ________Gerard S. I would expect a higher incidence of primeality in this series than in randomly-chosen numbers of the same length, considering that the last digit will be relatively prime to 10 (after you're through the single digits). Even in a prime base you'd eliminate 0. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: isprime of flattened primes... > Consider the number series > 23 > 235 > 2357 > 235711 > 23571113 > 2357111317 > 235711131719 > 23571113171923 > . > . > . > where each number is a flattened list of primes (that > is, a list of primes with the blanks removed to form > a single number). > Once you get past the single digit primes, should > there be any more primes? So far, I haven't found > any, but I haven't checked very many, as these types > of numbers get huge in a hurry. Should there be any > reason that they're all composite? ________Gerard S. Google for concatenated primes Phil -- What is it: is man only a blunder of God, or God only a blunder of man? -- Friedrich Nietzsche (1844-1900), The Twilight of the Gods === Subject: Re: isprime of flattened primes... |> Consider the number series |> 2 |> 23 |> 235 |> 2357 |> 235711 |> 23571113 |> 2357111317 |> 235711131719 |> 23571113171923 |> . |> . |> . |> where each number is a flattened list of primes (that |> is, a list of primes with the blanks removed to form |> a single number). |> Once you get past the single digit primes, should |> there be any more primes? So far, I haven't found |> any, but I haven't checked very many, as these types |> of numbers get huge in a hurry. Should there be any |> reason that they're all composite? ________Gerard S. | Google for concatenated primes === Subject: Are you sick of BORING / SCARY / SICK Dreams at night? Are you sick of BORING / SCARY / SICK Dreams at night? Want to know a secret on HOW to Control the Dreams you have - FOR FREE??? Keep reading! I will explain a dark mystery: HOW AND WHERE A WISH IS GRANTED! A WISH is just a way to open a portal into the realm of dreams - IT WORKS! Did you know: YOU have a minimum of 365 WISHES A YEAR to use that is Yes - YOU read correctly - your dreams seem real - you just have to learn how to USE your dreamtime to GET WHAT YOU REALLY WANT! (Nod, Nod, Wink, Wink!) So here's a suggestion as to what to do with one of your wishes! Use it to see into the future! Yes! What the 4th Dimension does, is that it exposes The Truth to your THE ONLY CATCH or THE DARK SECRET IS THIS: YOU JUST HAVE TO ASK FOR THE TRUTH OUT LOUD BY MAKING A WISH !!!! - Go figure!! Make this WISH OUT LOUD before you go to sleep TONIGHT - TO SEE FOR YOURSELF! GETTING PRETTY ROUGH UP THERE!) You have NOTHING to lose! Just say these words - OUT LOUD - before you sleep tonight: I WISH TO KNOW THE TRUTH: WHAT MAJOR EVENT WILL HAPPEN IN 2008? Go to sleep and you will get The Truth (YOUR WISH GRANTED!!!) in the form of a dream that night! fulfilled IN The 4TH Dimension (The World of Dreams) - while you sleep, so you can EXPERIENCE your wish with your mind!!!! So don't WASTE another night by not making a wish!!! Go ahead! Make a wish TONIGHT and watch it come true IN THE FORM OF a dream! === Subject: Re: Does this error of logic have a Logical Fallacy name to it?; Re: page 3 >>Do you even read posts before replying to them? The >>whole point of my >>post was that the proofs given above actually >>use the definition of a prime, no one has ever said >>anything about not >>using #1. >>Martin Wanvik >>Perhaps I have been too combative in arguing. Let me >>try a different >>tact. >>Dik acknowledges that 5(a) is correct. Where Dik and >>I disagree is that >>Dik believes 5(b) and 5(c) can also be correct. >>Whereas I believe 5(b) >>and 5(c) cannot be correct and must be invalid. >>Martin, do you accept 5(a) as a correct proof? > Yes. >>The problem I have with 5(b) and 5(c) is that, if >>5(a) is correct, then >>5(b) and 5(c) are contradictory to the definition of >>prime by >>considering W+1 as composite. So it is what I would >>call a >>pseudo-contradiction. In other words, if the proof is >>easily begot from >>5(a), then why take an opposite claim that W+1 is >>composite and not >>necessarily prime and indulge in 5(b) and 5(c). > If you use assumption (2), you immidiately get that W+1 is composite, since it is larger than any of the numbers on the list and is not divisible by any of them. Using assumption (1) first instead, implies that W+1 is prime. It is a question of order, nothing else. Okay, I need to revise my argument. It was weak with the claim that of an open undischarged step of Symbolic Logic. I now have a stronger argument that 5(a) is the only valid argument and that 5(b) and 5(c) are invalid. It relies not on an undischarged step, but rather it relies on the use of the Unique Prime Factorization theorem. Dik Winter forgot to apply it to 5(a), 5(b) and 5(c). But I do not blame him for I myself forgot to apply it. In this thread Bill Dubuque is discussing irreducibles which is the Unique Prime Factorization theorem, UPFAT for short. I am in the middle of Spring planting of trees and will post my revision later tonight. The life and death of my trees takes priority here. So I have Martin to thank for coaxing me on to making the stronger claim that only 5(a) is a valid proof and where 5(b) and 5(c) cannot be valid. >>Another question Martin. Do you accept the idea that >>with 5(a) is an >>easy proof that W+1 and W-1 are necessarily prime >>which leads to an >>easy proof of the Infinitude of Twin Primes. > No. As it stands, the proof only gives you infinitude of primes. In order to do twin primes you'd have to add the assumption that there are only finitely many twin primes. Notice that these are a proper subset of the primes, so if some W+1 and W-1 are not divisible by any of the twin primes on your list, (1) can't give you primality any more, since both W+1 and W-1 may be divisible by other primes than those on your list. I disagree with you on this issue. If there is only one valid method of Reductio Ad Absurdum, namely 5(a) where W+1 is necessarily prime then W-1 is necessarily prime also. So in any finite list of all the primes delivers twin primes not on the original list. So we intertwine Mathematical Induction and show that 3,5 are twin primes and 5,7 are twin primes. We assume true for case of k. We thence show that for case k+1 delivers a new set of twin primes. This is performed by simply doing a regular Indirect Method on all-the-primes which delivers the case of k+1. In fact, I suspect we do not even need Math Induction for Infinitude of Twin Primes in that the proof of regular primes delivers not just W+1 but twin primes. >>This is >>an argument of >>Economy. If 5(a) is easier than 5(b) or 5(c) and that >>only 5(a) leads >>to a proof of the Infinitude of Twin Primes. Then >>does this power tell >>us that only 5(a) must be truly correct and 5(b) and >>5(c) are invalid. > No, you're mixing things up here. The correctness of a proof is not determined by whether or not it is hard or what the proof method may or may not imply. But certainly, if all this were true, 5a would be a more economical proof. Well I was not arguing with you as to right or wrong here, I was invoking your reasoning. I was invoking your position that how can you accept 5(a) which can deliver infinitude of twin primes, yet accept 5(b) and 5(c) which are impossible to deliver infinitude of twin primes. An invoking of your stance, your position, your belief that all 5(a), 5(b), 5(c) can be true when they all lead to very different outcomes for twin primes. This invoking, is a hope that it spurs you to question and doubt your belief that 5(b) and 5(c) are flawed. (snip) > As I've been trying to say just about a zillion times, none of these proofs leave anything open, /all/ assumptions are used in /all/ of the proofs, it is just a matter of order. Okay, I am in agreement with you on this issue. That 5(b) and 5(c) do not leave anything open which is a mistake of Symbolic Logic. That part of my claim is now at fault. However, I can remedy and save my claim that only 5(a) is a valid argument and 5(b) and 5(c) are invalid. And the way I show this is that Dik Winter neglected to insert Unique Prime Factorization theorem, UPFAT, between his step #1 and his step#5. If Dik had done that, UPFAT would have eliminated 5(b) and 5(c) and have only 5(a) remaining. So tonight I shall report back as to the injection of UPFAT in Dik Winter's proof scheme, and hopefully it eliminates 5(b) and 5(c) and shows us clearly, not only myself, but others, that 5(a) is the only valid argument. Question Martin Wanvik, if I show you that 5(a) is the only valid proof of Infinitude of Primes, reductio ad absurdum, would you then agree that the Infinitude of Twin Primes is a direct result of IP. But I have to get back to the planting of 130 more trees today. Archimedes Plutonium www.iw.net/~_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: the majority of pure spam in sci.math is from mathforum.org Hi Don, > Does anyone have the ear of one of the folks at > mathforum.org and > might be able to coax them into addressing this? All: please do direct future concerns and questions about the Forum and our services by following the mathforum.org links to Contact Us http://mathforum.org/cgi-bin/mailto.pl?key=44&to=%27Discussions%20Administra tor%27 Appreciatively, -- Richard Tchen The Math Forum @ Drexel 3210 Cherry Street Philadelphia, PA 19104-2713 tel1: (215)895-1080 tel2: (800)756-7823 facs: (215)895-2964 The Math Forum @ Drexel (http://mathforum.org/) is a research and educational enterprise of the Drexel School of Education (http://drexel.edu/soe/). === Subject: Re: the majority of pure spam in sci.math is from mathforum.org <3761635.1145653606108.JavaMail.jakarta@nitrogen.mathforum.org Hi Don, > example you posted -- we've identified and deleted the > posting aliases that provided spammers access to sci.math > and several other newsgroups. Hey Richard, I hear the White House is looking for a new Press Secretary. Maybe you should look into this! Dave L. Renfro (Someone who is glad he doesn't have to put up all the complaints you get.) === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? >> Your life was a gift bestowed upon you without any strings attached, > I'm better dollars to doughnuts that they don't interpret existence as > something with no strings attached. I find this interesting as you started this thread in response to me. Your typo is well noted. http://www.cartercenter.org/healthprograms/healthpgm.htm -- meltdarok http://hometown.aol.com/meltdarok/ === Subject: Re: how to proof noneuclid geometry 04/21/2006 >Suppose you had a rectangle in the hyperbolic plane. Not until you prove that one exists. >leading to a contradiction The contradiction stems from your assumption that there was a rectangle. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: aritmetic and postulates 04/20/2006 at 07:00 PM, flugger1987@hotmail.com said: >hi there, this is the first time I've posted anything. I was going >through some work on the history of maths, and was wondering if there >was any postulates in arithmetic. I'm not sure what you mean. You can certainly give a set of axioms for, e.g., the Reals if you don't wish to simply construct a model from a simpler system. The axioms for the Reals have, in particular, simple models within Euclidean and Projective Geometry. There is a well known theorem that any two complete Archimedean fields are equivalent. I'm not sure of the historical sequence; it may be that there was a correct axiom system for the reals before the first correct model, or it might be the other way around. Perhaps someone else could supply dates for both. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: continuum hypothesis > [big snip] >This made me wonder whether it is possible to >color all points in R^2 either red or blue in >such a way that any horizontal line contains >no more than finitely many red points, and that >every vertical line contains no more than >finitely many blue points. (CH is being assumed, >but no more than ZFC for the rest of the axioms). >David Bernier No. The proof that I gave above works in this case too. Let H(r) = {x : (x,r) is red} V(r) = {y : (r,y) is red} so we have that H(r) is finite, V(r) cofinite. Now let U be any infinite subset of R. The collection of sets {H(u) : u in U} covers R. (Why? take any z in R. U meets V(z), so exists u in U such that (z,u) is red. But then z is in H(u), as required) So taking U to be any countable set gives us a countable collection of finite sets that covers R, which is impossible even assuming CH. Robert Sheskey === Subject: Re: Hausdorff and compact <210420060759062664%anniel@nym.alias.net.invalid> Question, is there a compact Hausdorff space K >with a continuous bijection from Q onto K? No. If K is Hausdorff and a continuous bijection Q -> K exists then K is a countable Hausdorff space with no open singletons. Such a space cannot be compact. RS === Subject: Re: Why are infinite series so difficult for students? > One thing that I've wondered, having taught Calculus II, is why > students have so much trouble with infinite series. You must mean average or below average students. I never had any trouble with infinite series, nor did anyone else I knew who were better than average math students. On the other hand, my understanding of infinity got better as time (and study) progressed, so it's likely that my earliest understanding of infinite series was a relatively simplified view of them. === Subject: Re: Why are infinite series so difficult for students? > Then why don't they have as much trouble with integration techniques? > --- Christopher Heckman Because integration techniques does not require any understanding what *is* Riemann intergral. You just memorize some integration formulas. I'm saying it as a student who took Calculus I, II, III, and done quite well, without real understanding what I was doing --- and that kept bugging me. Now I'm finally taking Intro to Real Analysis, and everything slowly falls into places. And this is even with the Calculus Textook that did have some strict proofs about, say limits. The more up-to-date Calculus books have all the proofs moved to appendices noone bothers reading. So you sort of on auto-pilot, and it works most of the time, but then, suddenly, you encounter something that requires not the mechanical application of some memorized rules but deep understanding of what's going on underneath: and you fail. Now, as I'm taking Analysis, I'm essenlly re-learning everything I've done in Calculus. === Subject: Re: Why are infinite series so difficult for students? >> Then why don't they have as much trouble with integration techniques? > Because integration techniques does not require any understanding what > *is* Riemann intergral. You just memorize some integration formulas. > I'm saying it as a student who took Calculus I, II, III, and done quite > well, without real understanding what I was doing --- and that kept > bugging me. Now I'm finally taking Intro to Real Analysis, and > everything slowly falls into places. You are not alone. In fact, that's probably typical of the mental evolution of the mathematically inclined. I did quite well studying Calculus in high school, took Calc I my first semester of college, but it wasn't until taking Calc II and Calc III (in the same semester) that I finally, truly, understood what an integral was all about. Specifically, the concept of the moment of iner of a 3-D volume with a variable density function is what triggered my aha! moment (pun intended). === Subject: Re: Why are infinite series so difficult for students? <92S1g.383011$y31.11105330@phobos.telenet-ops.be> [quote] I noticed that most of the frowning (and headaches) could be avoided by *very* vigorously hammering on the fact that an infinite series is not a sum, and that there is no such thing as a sum with an index going from 0 to infinity. An infinite series is nothing more than the limit of a particular sequence (of parl sums from 0 to n). So they must really have a *very* good grip of and experience with limits, with sequences and their convergence properties, and last but not least, with the summation sign and its properties, before series should be even mentioned... [/quote] It helped me get it too when I thought about it as a sequence, and limit thereof; however, I didn't think like this until I took analysis. When I first took calculus I mostly just focused on the rules. === Subject: Re: Why are infinite series so difficult for students? difficult. However, like with may topics covered in calculus, we covered them fast and then moved on. 4 weeks later we forgot much of what we learned because we haven't used the materials since. That is the problem I've seen with the Calculus classes I've taken. The pace is too fast for many students - not that they aren't smart enough to do the work, but they frequently don't have the time. Calculus is a very time consuming class, and not all instrutors effectively communicate this to the students. Heavy course work and jobs and real life in general often does not leave us enough time to do the work, and our learning suffers. > One thing that I've wondered, having taught Calculus II, is why > students have so much trouble with infinite series. I asked one student > yesterday during my office hours, and he said they were too abstract. > I don't think this is the complete answer, though. What do you all have > to say? > --- Christopher Heckman === Subject: Re: Beliefs Create Reality, Even In Mathematics. >> So that magical current you supposedly unleashed which was supposed >> to destroy me actually is just bull, eh? Seeing as how it's been a >> year since you proclaimed this demonstration of your magical powers, and >> I'm still right here doing just as I've always done, the conclusion that >> you've merely been spouting bull seems pretty accurate. > So this means I *did* say it right? Excuse me, but are you and I even in the same conversation here? I asked you a question. Have you got an answer yet? >> So you can't actually influence events at a distance by magick? Or do >> you subscribe to Crowley's definition of magick as any intentional act? >> Or what? > Well I tried to influence you to tell us all about your *actual* > experience in Magic Tom. That appears to have failed (for reasons I am > sure we all understand). So I suppose Magic doesn't work eh? No, it means you failed. === Subject: Re: Beliefs Create Reality, Even In Mathematics. > You got your Magical knowledge from Playboy? >> I knew you'd say something as stupid as this. >> You're are predictable as clockwork. > Hmmm. The trick is to predict it *before* I say it Tom. I did predict it. I simply didn't tell *you*. > I doubt that your continually predicting *after* the event will impress > many people. Is that what you think this is about? Impressing as many people as you can? And you're trying to assert that *you* are in fact the more impressive of the two of us? OK, I'm perfectly willing to accept defeat at this game. You are clearly a much more impressive person in every respect. You are the very apotheosis of superiority. There's a big red S on your chest and you have a cape. Ubermensch. That's you. Sheesh. I hereby confer upon you the grade of Ipsissimus. Of course, this will be accepted by you without any show of appreciation for the very good reason that it would simply not befit the solemnity of this inition and all that it stands for. This inition is good forever and is irrevocable. === Subject: Re: Beliefs Create Reality, Even In Mathematics. >Nonsense. In decades of lab work, Parapsychology has yet >to come up with a single reproducible demonstration that any >PK phenomenon even exists. Parapsychology cannot meet the >standard we expect of laboratory science. Instead, they offer >excuses and special pleading. >> Not exactly true. Dean Radin, for example, gets consistent positive >> results, but they don't seem to be reproducible independently of his >> team. > Yeah, that would be non-reproducible. it makes no difference. However, it is inexact. I like to find the point of inexactness in what is being said. That's where you can always leave a little room for doubt. I like doubt. It's enormously handy. >> Of course, the effects he's studying are very, very small and could be >> nothing more than normal variance writ large. > He has a huge amount of data from his on-line tests. I wonder > why he doesn't report it. your point of view, it should make no difference. You don't have access to his data, for whatever reason. Have you tried looking for it yet? What results can you report from your search? === Subject: Re: Expected area of a triangle >>Select three points from a uniform distribution on the unit square. >>What is the expected area of the triangle spanned by the three points? >Some followups have been posted, but none get at the >simplicity of the problem. Independence properties >and uniformity properties can be used to great >advantage. >Three points are selected at random. This means that >their X and Y coordinates are independent and have a >uniform distribution from 0 to 1. >Now if one has k independent observations on the unit >interval, th[us] k+1 intervals, the distribution on the >unit simplex is uniform. [...] >and so for the original problem it is 11/144. That is a really excellent proof! It would never have occurred to me to approach it that way. dave === Subject: Re: analytic? > I read this on a physics website about trigonometry: > What we would like to have is a way of relating the angles in the > triangle to the lengths of the sides. It turns out that there's no > simple analytic way to do this. > Can someone please explain what is meant by the term analytic in this > context? It is explained in the very next sentence from the website: Even though the triangle is specified by the lengths of the three sides, there is not a simple formula that will allow you to calculate the angle q. We must specify it in some new way. Then he goes on to introduce the trigonometric functions. So, analytic = simple formula not using trig functions. That is not, by the way, a standard usage of the term. === Subject: Re: A twin primes conjecture >> for twin primes, which I'll call a conjecture, as, um, I might be >> wrong: >> Given a prime x where x > 3, if x-1 is divisible by 3, and for every >> prime p less than x^{1/3} but greater than 3 it is true that >> x > 2 mod p > What does x > 2 mod p mean? Note, if p = 2, then 2 mod p = 0; > otherwise 2 mod p = 2. So, x > 2 mod p means either x > 0, or x > 2 (a > fortiori x > 0). But you've already specified that x > 3. You must have > made a mistake with your formulation here! mod is not an arithmetic operator. -- Bye. Jasen === Subject: Re: Calculus XOR Probability <4443d1f8$0$14465$6d36acad@tin.nntpserver.com> <4444dd89$0$14493$6d36acad@tin.nntpserver.com> <44463513$0$14452$6d36acad@tin.nntpserver.com> So, for infinite sets, you want to claim that size >> is something OTHER than a number??? >> Well, the size of an unending sequence can't really be the last >> number you shout (oh, or was it 'sing') from the ditty, can it, since >> there isn't a last number. > Is that true of all infinite sets? Isn't 1 the last number chanted in the > ditty of reals in [0,1]? I wonder what the second number chanted in that ditty is, the one you chant right after you say zero. Or the next to last number, just before you chant one. But then that's why we say uncountable is not the same as countable. > Of course that's not the size, but if you count by Lil'uns, the last > Lil'un is the Big'unth one, and Big'un's the last index dittied. Still working on that well-ordering of the reals, are you? === Subject: Re: JSH: Extreme mathematics reminder Yes, JSH you are an expert at extreme mathematics, as your math is: EXTREMELY bad it takes an EXTREMELY long time for you to do anything (10 yrs with no results!) and it is EXTREMELY annoying to most people on the ng. Although I do find it funny how you hop from thing to thing when it suits you. You always say truth is absolute. How can you then be so disingenuous (or stupid) to realize that there is no such thing as extreme math. This original idea of yours, like so many others, is just a figment of your mind with no basis in fact. Math is math, and it is either right or wrong. It is not extreme: it is is only correct or incorrect. This term (like brainstorming) is simply another rationalization to explain away your shortcomings (in this case you being sloppy). My advice: stay at the bar. Juno === Subject: Re: Giant spinning ball problem <1r7SfpET05REFw3R@jboden.demon.co.uk> In message , Mikael Johansson >(UK) at Carsington Water made of marble. A child can make it rotate >about any axis. >But can the child can only touch the ball at the North Pole? >> Roughly any point North of the Antarctic circle (depending on the height >> of the child). >Surely you mean the arctic circle? Otherwise most of the globe would be >accessible, and it would be a rather bad example for the given >setup/question. Antarctic - I think the Singapore version also allowed access to a large area of the spheres surface but the problem specified as an additional constraint that Suppose now you are given access to this ball only near the top, so that you can push it to make it rotate around any horizontal axis, but you don't have enough of a grip to make it turn around the vertical axis. Can you make it rotate around the vertical axis anyway? In practice, it turns out that it possible to apply a twisting motion along the North-South pole because the friction between a dry hand and wet stone exceeds that of the fluid bearing (even with a very heavy stone). -- Jeremy Boden === Subject: Re: Reality check: Counting prime numbers [Rick Decker, on JSH's prime-counting formula] > The key is that for y <= x, p(x, y) is the cardinality of the set > consisting of (1) the primes <= y and (2) the numbers <= x that > are not divisible by any prime <= y. For example p(25, 3) is > the cardinality of {2, 3, 5, 7, 11, 13, 17, 19, 23, 25}. In simple > terms, p(x, y) is the number of elements remaining at the end > of the appropriate step in the Sieve of Eratosthenes. [Tim Peters] >> For the heck of it, I tried to clean room derive his formula, and >> turns out it's almost as easy to derive from thinking about how the >> venerable old sieve works as Legendre's phi is to derive from >> straightforward inclusion/exclusion. >> The second key is that the >> p(x/k, k-1) - p(k-1, sqrt(k-1)) = >> p(x/k, k-1) - pi(k-1) [1] >> term counts, when k is prime, the number of new integers crossed off >> during the sieve step that first discovers k _is_ prime. IOW, it's the >> number of integers in 1..x, excluding k itself, whose smallest prime >> divisor is k. Deriving [1] from that characterization does make a >> pleasant little exercise. [jstevh@msn.com] > Then do it. [Tim Peters] >> else here who understands the Sieve of Eratosthenes to do it for >> themself: assume that p(m, n) counts the number of integers in 1..m >> remaining after the Sieve of Eratosthenes has processed all primes <= n. >> Assume that pi(n) returns the number of primes <= n. Then given an >> integer m and a prime k, derive a formula (in terms of p() and pi()) >> giving one more than the number of integers in 1..m whose smallest prime I meant to say one less there, BTW, not one more (k's smallest (& only) prime divisor is k, but the sieve doesn't cross out k). >> divisor is k. Any undergrad with a smattering of introductory number >> theory should be able to do that in less than an hour; I expect that the >> brighter kids in my high school could have done it too. [jstevh@msn.com] > Your brain will lie to you in order to convince you of something that > is not true. Then so will yours. A possible difference is that my brain isn't lying to me in this case. > The point here isn't that you can understand the gist of how it all > works, but can you put down a complete derivation from beginning to > end, where at the end you have the formula I show at > http://en.wikipedia.org/w/index.php?title=Prime_counting_function&oldid=9142 2 49 Yes, I can -- and I already explained why I'm not going to do that here just to humor you, and I already offered to help anyone on alt.math.undergrad who _wants_ to do this (including you) complete their own coherent derivation. I've had no takers on that, from which I conclude that interested parties (if any) would rather do it on their own. > and by now, with all your replies, you could have accomplished that, if > you were capable, while instead you keep repeating pieces, here and > there. If by repeating you mean I quote previous messages verbatim when replying, yes. Else I don't know what you're talking about. > And note again, the need to diminish the importance of the task by > claiming any undergrad could do it. Any undergrad with a smattering of introductory number theory _should_ be able to do it. So, e.g., should an undergrad who completed an analysis of algorithms course. That diminishes your work only to the extent that you've talked yourself into believing that this is much more difficult than it actually is (and if you didn't derive this from the point of view of thinking about how the sieve of Eratosthenes works, then you almost certainly did make it much harder for yourself than it could have been; other people can take an easy path instead). > Why don't YOU just do it? See, _that's_ repetition. You already asked multiple times, and I already answered multiple times. > I've looked and waited, but not seen a single post that just goes step > by step showing a derivation. >> Why would we bother, James? Just because you challenged us with >> childish insults and accusations? Ptui -- doesn't mean squat to me, >> bubba. Rick Decker suggested earlier that it _sounds_ like you're >> trying to goad someone into giving a coherent writeup, so that you can >> pass it off as your own. That sounded plausible to me. By your own >> admission, your own explanation for how to derive it is so convoluted >> that reading it doesn't help anyone <0.1 wink>. > Dodge and rationalize. > To date, I have not seen anyone besides me do it, at all, despite my > giving this challenge for years. You can't fill in the blanks from what's already been given here? Really? > My point is that for some special reasons, despite the inherent > simplicity of the derivation, the human brain for most people has a > wiring lack, which doesn't allow you to step through a derivation. > You can prove the equation true. You can understand a derivation if > given to you. > But you cannot personally step through a full derivation as the > circuitry needed to do so is missing from your brain. This is one of those psy-ops posts, right? You're messing with my mind, and I'm helpless to prevent the desired devious effect from occurring? Kewl. >> If someone on alt.math.undergrad is truly interested, I'd be happy to >> help >> them work out a full derivation. > Posters keep claiming it's easy, and they've done it somewhere else, > but no one just steps through a derivation. >> So you try again, and I'll _help_ you make it comprehensible. I won't >> do it for you. > I like my own derivation, and need it for credit for having had the > first derivation known. To date, has anyone told you that they can _follow_ your derivation? I couldn't make head or tail out of it myself (well, whatever it is -- I tried to follow a post you made a few years back, but have no idea whether that's what you think your derivation may be today). > And hey, remember, my equations could have been found by mathematicians > thousands of years ago, or hundreds of years ago, if it were really > easy. I don't know that they didn't. You don't either. It's been explained many times why it's unlikely someone would express the prime-counting function in the specific way you did, so that you failed to find a literal match using Google doesn't say much. > But they weren't, and I think I know why: wiring of the human brain. LOL. James, this is the truth: the people following this thread do know how to derive your forumla, and also know it's just not that hard to do. There is only one tricky step, and it's not actually difficult to figure out. Claiming that it requires a brain mutation to do it makes you appear to be a lunatic. Maybe you'd like to consider how that helps you get recognition of a kind you want. [... long stretch of older messages elided ...] > It can be almost painful to watch after a while, as a person will do > any number of rationalizations and use all kinds of avoidance behavior. > Now to me, it would be quite ok for other people to just step through a > derivation, as then I could get some progress in getting my research > acknowledged. Which is why you should try to come up with a derivation that people can actually follow. These messages pointed to a way that would work -- but I'm not going to do it for you, and you're going to have to accept then that-- oops! --it's really not difficult to derive. Stick to an incomprehensible derivation and you can continue enjoying your usual rants about being a persecuted genius. I wonder which path you'll take :-) > But people can't, so my research is ridiculed not because it's wrong or > unimportant, but because fully comprehending it escapes most people. So you say. > Discussion becomes meaningless, as people can't accept what they can't > fully comprehend, and the reality check here is that NO ONE besides me > is stepping through that derivation. > So you have this huge thing that jumps out at you, if you want to see > it. > But for many of you, it would be the last thing you'd admit. That was pretty muddy. To what does it refer there? === Subject: Re: polynomial... >f(x) = a.x^4 + b.x^3 + c.x^2 + d.x + e >f(0)=0, f(1)=1/2, f(2)=2/3, f(3)=3/4, f(4)=4/5 >find f(5). >-------------------------------------------- >is this simultaneous equations ?? It can be, but why do something so complicated? Theorem: if the function f is given by a polynomial of degree n, then g(x) = f(x)-f(x-1) is a polynomial of degree n-1. Given a sequence of values of a function f at points which differ by 1, the sequence of values of g is called the sequence of divided differences of f. (At least I think that's the terminology. OK, it is for now!) Corollary: if f(x) is a polynomial of degree n, then the n-th iterated divided differences are constant. So from the knowledge that f is represented by a polynomial of degree 4, and > f(0)=0, f(1)=1/2, f(2)=2/3, f(3)=3/4, f(4)=4/5 we compute a tableau of divided differences: 0/1 1/2 2/3 3/4 4/5 1/2 1/6 1/12 1/20 -1/3 -1/12 -1/30 1/4 1/20 -1/5 but then from the Corollary, all divided differences in the bottom row will be equal to -1/5. Then we can work our way up the table to the first row: 0/1 1/2 2/3 3/4 4/5 2/3 1/2 1/6 1/12 1/20 -2/15 -1/3 -1/12 -1/30 -11/60 1/4 1/20 -3/20 -1/5 -1/5 so f(5) = 2/3. This is a really traditional technique of extrapolation; even school kids seem to know it. Of course arguably this exercise was set in order to illustrate what a terrible idea it can be to use polynomial extrapolation in this way: most people would suggest that the simplest description of the given data is that f(n)=n/(n+1) for all n -- so that in particular f(n) --> 1 as n increases, which is very different from the behaviour of the extrapolated values, which will increase without bound (in magnitude); the next few values of the extrapolated sequence are 0, -7/4, -16/3, -117/20, -22, ... There is a regularity to this particular example: the numbers in the k-th row are (-1)^i (k-1)! / ( (n+k)(n+k-1)...(n+1) ) = (-1)^k (k-1)! n! / (n+k)! = (-1)^k/( k * C(n+k,k) ) . dave === Subject: Re: polynomial... > f(x) = a.x^4 + b.x^3 + c.x^2 + d.x + e > f(0)=0, f(1)=1/2, f(2)=2/3, f(3)=3/4, f(4)=4/5 > find f(5). > is this simultaneous equations ?? No. It is not necessary to find a,b,c,d,e to find f(5). Look up Lagrange interpolation. === Subject: Re: polynomial... > hello...doctor~ > f(x) = a.x^4 + b.x^3 + c.x^2 + d.x + e > f(0)=0, f(1)=1/2, f(2)=2/3, f(3)=3/4, f(4)=4/5 > find f(5). > -------------------------------------------- > is this simultaneous equations ?? > but that is complex... > so, i need your advice. > thank you very much for your advice. f(x) = x / (x+1) for x = 0, 1, 2, 3, 4. So f(x), being a polynomial of degree 4, is equal to [x + Ax(x-1)(x-2)(x-3)(x-4)] / (x+1) for some A. The Remainder Theorem gives A = -1/120. === Subject: Re: polynomial... > hello...doctor~ > f(x) = a.x^4 + b.x^3 + c.x^2 + d.x + e > f(0)=0, f(1)=1/2, f(2)=2/3, f(3)=3/4, f(4)=4/5 > find f(5). > -------------------------------------------- > is this simultaneous equations ?? > but that is complex... > so, i need your advice. > thank you very much for your advice. > f(x) = x / (x+1) for x = 0, 1, 2, 3, 4. > So f(x), being a polynomial of degree 4, > is equal to [x + Ax(x-1)(x-2)(x-3)(x-4)] / (x+1) for some A. > The Remainder Theorem gives A = -1/120. This is a neat observation, and of course it shows us that: f(5) = [5 - (120/120)] / (5 + 1) = 4/6 = 2/3 -- chip === Subject: Re: polynomial... it is a simultaneous set of 5 equations in 5 unknowns, so it is readily solvable by row reduction (actually you can reduce it to 4 equations in 4 unknowns because it is clear e=0 since f(0)=0) === Subject: Re: A few questions about the Peano axioms <4445f290$0$2030$ba620dc5@text.nova.planet.nl The fact that Peano's axioms are categorical in second-order logic is > used to help informally justify their consistency. but mostly nobody > uses second-order logic and so the second-order formalization is not > particularly useful. In particular, nobody who seriously doubts the > consistency of PA is likely to believe that second-order semantics are > well defined anyway. > That's one possible point of view. Another is this: > Peano's original list of axioms does not 'use second-order logic'. It's > the (still commonly used) -mathematical-, axiomatic characterization of > N. The first order weakening is an artifact only of interest to -logicians-. If you state some axioms, you ought to at least be willing to say what the quantifiers range over. If you state your induction axiom as ``every subset of the natural numbers which contains 0 and is closed under successor contains all the natural numbers'' and interpret ``every subset'' to mean every subset, then you are using second-order semantics whether you realize it or not. Just as a child prodigy could play jazz without knowing that it was jazz she was playing. Maybe you mean something different by ``axiomatic characterization.'' As I use the word, in order to axiomatize something you must first choose the logic that you use, then the language, then the axioms. (I apologize for the long delay in responding.) === Subject: Re: Any algorithm for: <44491A0B.25E88567@pat7.com> say I have R = a1*b1 + a2*b2 .... + an * bn an and bn are between 0 to 1 what's the best way to find a A so that S = A * (b1 + b2 + ... + bn) S > R (say 99% chance) and closest to R, === Subject: Re: Prime rules and extreme mathematics >> Test the rules. It is extreme mathematics so maybe I missed something, >> but as these iterations go through that gets less and less likely. > Hmm, why wasn't a course in extreme mathematics offered to me when I was > working toward my degree in mathematics? Is my degree worth less since I > don't know anything about it? Don't worry about it. Harris is just one of those idiots who think that tacking the adjective extreme onto any random thing somehow makes it cool. When I see someone claiming to practice extreme anything, a red charlatan flag goes up in my mind. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock