mm-2799 === Subject: When is the limit of Borel-measurable functions Borel-measurable? This is the second time I've sent this message. The first time, it seems to have dropped into the bit-bucket. Sorry if you're reading it for the second time. The question might seem stupid, but I'm not talking about REAL-VALUED functions. Let (X,T_X) and (Y,T_Y) be topological spaces, and let B_X and B_Y denote the Borel fields on them. (= generated by the open sets) It is not reasonable that the pointwise limit of a sequence of measurable functions f_n : (X,B_X) --> (Y,B_Y) is also measurable. (By measurable I of course mean that the inverse image of a set in B_Y is in B_X.) This is because sequences don't mix well with topological spaces. But there ARE circumstances under which this is true. I remember a theorem proved by Calderon in one or another class he was teaching where he proved it was sufficient that T_Y have the property that every open set U can be written as a countable union of open sets G_n with the property that the closure of G_n is a subset of G_{n+1}. This property is satisfied, e.g., by regular T1 second-countable spaces. Does anybody know the most general circumstances under which this result is true? I doubt that second-countability is necessary; it's TOO global a condition. Probably first countability isn't enough. Alternatively, can anybody characterize the property of open sets U being writable as such countable unions? Always wondering, Ron Bruck === Subject: Re: Factorial/Exponential Identity, Infinity > I will research geometric series. I'm not here to reinvent the wheel, > but it damn well better let me say that half the integers are even, as > they are as the asymptotic density of the even numbers in the integers > is one half. > Doing any sort of research in mathematics with a fixed idea of what > you will allow yourself to find is not really a good idea. What do you think about the canonicalization of the infinite binary > sequences? That is to say, for the sequence .010101(01)... with half > zeroes and half ones, that you could exchange any two sequence > elements any number of times and not get .001001001(001).... Is there > a rule to exchange (permute) the elements of the sequence .0101(01)... > to get 00010001(0001)...? > It depends on what you consider an allowable permutation. > The most general form of such permutation can be considered as > follows. > Let N be the set of positive integers, then any infinite binary > sequence can be uniquely represented by a functions f:N -> {0,1} in > which f(k) is the k'th digit of the sequence. > Let B be the set of such functions f:N -> {0,1}. > Then any bijection g:N <-> N creates a permutation G: B -> B by > defining G(f) = fog, where o represents function composition, G(f) > is the function whose valueat k in N is f(g(k), > i.e., G(f)(k) = f(g(k)), for each k in N. > Let P represent the set of all such permutations, It is clear that > the number of zeros and number of ones in an expansion must > remain fixed under any such permutation of digits. > Then, for a given f:N -> {0,1} and h:N -> {0,1} in B, a necessary > and sufficient condition for there being some permutation G in P > such that G(f) = h is that > card(k in N: f(k) = 0}) = card(k in N: h(k) = 0}) and > card(k in N: f(k) = 1}) = card(k in N: h(k) = 1}) > i.e., the infinite binary sequences for f and h have the same > number, possibly infinite, of zeros and the same number, possibly > infinite, of ones, and only one of these can be finite. > But if the number of zeros or the number of ones differ, then there > is no permutation carrying one into the other. > This divides up the set B into equivalence classes of binaries > permutable into each other which can be indexed by the integers, Z. > Any two such binaries with infinitely many zeros and infinitely many > ones in their expansions can be permuted into each other, and can be > indexed by 0. If the number of ones is finite, use that number as > index, and if the number of zeros is finite use the negative of that > number as index. > Then two binaries can be permuted into each other if and only if > they have the same index. B is the set of functions f(n) -> {0,1}, n E N. There are uncountably many such functions, each represents a subset of N elements or a distinct element of R[0,1) with no dual representation. I consider .00010001(0001)... as an infinite sequence and how to permute its elements to get .01010101(0101).... It seems almost simple, move the 1 in the fourth place to the second place, and move each one in the 4x'th place for x>1 to the 4(x-1)'th place. There's certainly no dearth of ones and zeros, the amount of each is infinite. Yet, the zero that was in the second place would have to go somewhere, and necessarily displace another. It's easy to describe a canonical sequence relative amounts of zeros and ones. pick a fraction p/q less than one, p and q positive, p B_(p/q +-x) < B, because g is a bijection. For example it doesn't permute the sequence of all ones to the one of all zeros. For example .0101(01)... permutes to .1001(01)... but not to .001001(001).... How does it permute to .0001(01)...? I can easily determine a bijective function from {1, 3, 5, ...} to {3, 5, ...}, and from {0, 2, 4, ...} to {0, 1, 2, 4, ...}. Changing the second element from a one to a zero means that somewhere a zero has to be changed into a one. It could be put off indefinitely but that necessarily implies that some subsequence 01 would be changed to 11. I focus upon necessary and sufficient condition. I can agree that it is a necessary condition for there to be the permutation, yet, not that it is a sufficient condition. I wonder what you mean by allowable. The sequences .(01)... and .(10)..., for half zeros and half ones, are perhaps the simplest cases with infinite numbers of ones and zeros to consider. Exchange each one and zero of the representative subsequence and it becomes the other. Consider .100..., it can't be permuted to .000.... For the same reason, (01)... can't be permuted to 00(01).... If you change the value of one symbol, you have to change the value of an occurrence of the opposite symbol, they exchange. It's simple to get an extra symbol out of the sequence to replace its opposite, the problem is putting it away, it also has to replace its opposite. 01010101.... ^ 00010101... ^^^ ^ ^ ... 10010101... 01010101... 00110101... 00011101... 0001011101... The first sequence has it's second element changed to its opposite. The second sequence is required to change one of its marked symbols to its opposite. Perhaps it's a specialization of permutation. The sequence is only changed by exchanging any of its elements any number of times, the value of a symbol is not changed without changing an occurrence of its opposite symbol. It's not possible to permute in this way 001(001)... to 011(011).... One way to see this is that there are sequence z_1=00... , z_2=00..., and o_1=11.... Take the first element from each then the second, etcetera: 001(001).... If you want to form 011 then it is a matter of taking the first element of z_1, then two elements of o_1 then an element of z_2 (to keep it even), then two more elements of o_1, etcetera. The amount taken from o_1 to form the sequence (011) from the constituent sequences of (001) rises steadily and never returns to a par of those taken from z_1 and z_2. On the other hand, forming a sequence like 111111000000000000(001) does. Given instructions: form a sequence of infinitely many ones and infinitely many zeros, there are many ways to describe a rule or method to do that. Given instructions: given a sequence with infinitely many ones and infinitely many zeros, exchange its elements to form another sequence, there are less sequences. Randomly and fairly (uniformly) select a subset of the natural numbers, there are even odds it contains zero. Half the subsets of N contain zero. Is there a rational function for x=4? === Subject: Re: Minimal Graph, Four Color Theorem |I hope you will be gracious and respond to my previous posting Re: |Four Color Theorem Simplified. You mean the one where you were trying to collect a list of what there is to know about the four color theorem? I don't see the motivation. If I had the time to spare to learn more about the four color theorem, I'd start by reading the book by Saaty and Kainen. |I notice that you have not responded to any of my previous posting re |the FCT. |May I inquire as to why you chose to respond to this one? Sure, but there wasn't much reason. It looked like an opportunity to clarify something. Some days I read most of the day's sci.math postings and reply in detail to several. Some days I don't read any. Don't overinterpret it. === Subject: Re: how many resolutions? > A^2 + B^2 + C^2 + D^2 = 85^2 > where a to d are different one-digit or two-digit numbers. Clearly A, B, C, and D must be less than 85. Assuming that 0 < A < B < C < D < 85, I found 180 solutions, including the two solutions cited, with a simple program. Solutions and program (in Think PASCAL) available on request. > I thought that it could have just one solution: since 85 is a product of two > prime numbers of 4k+1 type: 85=5*17. As it is known that only prime numbers > 4k+1 could be resolved as p^2=a^2 + b^2. > so 85^2=5^2 * 17^2 = (3^2 + 4^2)*(15^2 + 8^2) so right numbers are 24, 32, > 45, 60. > But there is another solution! 3, 4, 12, 84. > Why? And maybe there are some other solutions left? It means that > resolution p^2=a^2 + b^2 is ambiguos? So we could find c and d (c not equal > a and b) such that p^2=c^2 + d^2? > Or not? === Subject: Re: Group generated by a and b ... I have a group G generated by 2 elements a and b such that ba=a(b^k) for some >> integer k. >> Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m >> and n ? >Yes, when the order of a is finite. > But in general, no. > As a simple example, let G be the multiplicative matrix group over the > rational generated by > a = [2 0] and b = [1 0] > [0 1] [1 1] > Then ba = ab^2, but > ba^-1 = [1/2 0] > [1/2 1] > whereas all products (a^m)(b^n) have integral entries. group G = , some k in N with ba = ab^k ==> some n,m in N with ba^-1 = a^n b^m iff o(a) finite proof <== let o(a) = r; ba^-1 = ba^(r-1) = a^(r-1) b^k(r-1) ==> e = a^n b^m a b^-1 = a^n a b^km b^-1 = a a^n b^r where r = km - 1 a = aa^n b^r a = aa^n a b^kr e = aa^n b^kr = aa^n b^r; b^kr = b^r; e = b^(k-1)r b has finite order, when k /= 1 (aa^n)^-1 = b^kr; thus (aa^n)^-1 and a have finite order The case k = 1; ba = ab; ba^-1 = a^n b^m unresolved === Subject: Re: Calculate mode with multiple occurrences > I would like to know how to calculate mode when two different numbers > appear the same amount of times (i.e., 1, 2, 2, 2, 4, 6, 7, 7, 7, 8, > 9). Two and Seven appear the same amount of times in the array above - > so please let me know which one is MODE or do I add them together and > find the average of the two? Since the usual description of such a distribution is bi-modal, presumably you should report both values. xanthian. -- === Subject: Re: Antidiagonal, Infinity > at 07:09 PM, raf@tiki-lounge.com ( A. Finlayson) said: >Here I equate density with measure in the unit interval. > What do you mean by measure? With the obvious meaning, the rationals > have measure 0. As has been explained to you. >only concerned with considering a model where the rationals and >irrationals alternate in the reals. > That's like saying that you're concerned with a model where 1+1 > 1. > It's not a model for the reals, because it doesn't have the properties > of the reals. >rationals and irrationals are disjoint and distinct, > What are you trying to say? >then there necessarily would be irrationals with no >rationals between them. > Would you care to provide a proof of that? >I like to think that the rationals and irrationals alternate > I like to think that the nice gentleman with the map will make me > rich. Alas, if I do believe it then I will lose a substantial amount > of money. It's false. >and that >the function f(x)=x+iota maps Q[0,1) onto P(0,1), and f(x)=x-iota >maps Q(0,1] to P(0,1). > Also false. >half infinity > Meaningless. >finite multiples of a scalar infinity. > Also meaningless. >a set of a vector > Meaningless. >One thing to note is that *R, the hyperreals, as a set >contains the same elements as R, the reals. > Incorrect. >the characteristics of a probability distribution > What do you believe them to be? How do you apply them to an infinite > set? >We were talking >about the probability of an infinite binary seqence having one >element being on, the rest off. > No we weren't, because you've refused to define what you mean by that. >That probability is expressed as n/2^n, as n >diverges to infinity. > Meaningless. Assuming that you meant the limit of n/2^n as n > approaches infinity, that comes out to 0. >The probability of each among all possible infinite >binary sequences is being1/2^n, > What is n? You seem to be confusing bound and unbound variables. >So anyways out of those n >possible sequences with one on bit and the rest off bits, each is >equally probable. > Yes 0=0. >So a theoretical (read: thought experiment) > Theoretical is not the same as gedanken experiment. >method to generate an element of N is to >once again flip infinitely many coins. > You can't. >At this point it's a crazy, or >rather, unconventional thought experiment in that the first coin >toss says whether it is oo/2 or greater or less than oo/2. > No, because oo/2 has no meaning. >talk about the probability of selecting a given element of the >natural integers assuming a uniform probability distribution over >the integers. > That's easy; it doesn't exist. >At least we seem to have some agreement that a uniform >probability distribution over an interval of the reals exists, > No, because we don't have common understandings of what any of the > words mean. >simple method to sample an element of an interval of the reals >exists. > No, because a finite number of coin tosses only lets us sample a > finite set. >infinitesimals, > And it was clear that your understanding was fuzzy and incorrect. > Mathematics is a precise discipline; you must reason from the axioms, > definitions and rules of inference, not from your preconceptions. Look at a probability distribution function. Is it everywhere discontinuous? You don't have to flip infinitely many coins to tell if a real number from [0,1) is less than 1/2, just one. Consider x and y, independent variables, x goes to infinity: y/x is effectively zero or indeterminate, x/x is one, x is a dependent variable of itself. If y is a dependent variable of x, then, y/x is not necessarily zero or indeterminate (divergent). The probability of a binary sequence of length n having one on element is n/2^n, and the sum of all the probabilities of the possibilities is 2^n / 2^n=1. The sum from zero to infinity of zero is zero. One half is definitely a real number. The probability of a binary sequence of length n having one on element is the same as that of it having one off element. Only a quarter of the subsets of N contain both zero and one, and only one subset contains each element, 1/2^n, 1 of 2^n. That has 2^(n-1) of 2^n subsets containing zero. I'm pretty sure that the elements within the hyperreals are the same as those in the reals. It's just not possible for each of the elements of the hyperreals to distinguish which elements of the reals it is. Mathematics is a particularly tractable subject: there are truths that are agreed upon by all, that's generally exhibited by 2+2=4, it's true. By the same token, those areas in which not all agree are the signposts of areas where there is room for development, eg Euclid's Fifth Postulate: Agree or Disagree. This site is great, I could learn a lot from it: http://www.cut-the-knot.org/triangle/pythpar/Fifth.shtml . I'm a mathematical ingrate. I always want more. The reals are not going to be reordering themselves just to alternate. Have a nice day, === Subject: circle with two centers Consider a smooth surface, S, and two points, A & B on it. S contains a continuous, closed curve, C, such that both A & B are equidistant from every point on the curve C. 1. Does such a surface exist? If so, under what conditions? 2. If it does, can there be countably infinite such centers? Uncountably infintely many? -riskbert === Subject: Re: Minimal Graph, Four Color Theorem yeah; HSJ just won't admit it! > can say is you have lousy manners and I have no interest in discussing > an argument with someone who resorts to misrepresentation in order to --les ducs de Buffet; vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall & e-Dereg! http://larouchepub.com === Subject: Re: Algebra proof >> Let a, b, r and s be integers with r>1, s>1 and gcd(r,s)=1. >> Prove that if a=b(mod r) and a=b(mod s) then a=b(mod rs). >> How do I do this???? > Another way to prove this is to use the identity of Bezout > (the extended Euclidean algorithm): if d = gcd(r,s) then > d = ur + vs for some integers u and v. > When you see the hypothesis gcd(r,s) = 1, often it > is worthwhile to convert that into the identity of Bezout. > For your problem, you have: > 1 = ur + vs for some integers u,v since gcd(r,s)=1 > a - b = cr for some integer c since a=b (mod r) > a - b = ds for some integer d since a=b (mod s) > Now, multiply the first equation by a - b to get: > a - b = (a-b)ur + (a-b)vs > Substitute 3rd & 2nd equation into above, yielding > a - b = (ds)ur + (cr)vs = rs(du + cv). > Thus, rs divides a - b. That is, a=b (mod rs). Clearer: Prove rs = (r,s) lcm(r,s), which you almost have. For r,s|t <=> rs|rt,st <=> rs|(rt,st) = (r,s)t i.e. r,s|t <=> rs|(r,s)t <=> rs/(r,s)|t So, by definition, lcm(r,s) = rs/(r,s) = rs if (r,s) = 1 Note how the above proof eliminates the extraneous variables c,d,u,v from your proof, thus clarifying its true essence. For further info see my prior post here, and its references -Bill Dubuque === Subject: Re: Fundamental Reason for High Achievements of Jews === >Subject: Re: Fundamental Reason for High Achievements of Jews >> :> : Most historians believe that Jews avoid pork, >> :> : because the ancient Jews associated pigs with leprosy, >> :> : and pigs and people with leprosy were unclean. >> :> >> :> Name one historian who believes that, and give a citation to the >> :> place where he says it. >> : >> : As I recall, this was in Tacitus' Histories >> : which was written in the first century A.D. >> >> If that's the best you can do, I think that we can safely ignore your >theory. >It's soooo easy Richard. Someone in your camp >should have the information at his fingertips. >GOOGLE: tacitus histories ~13,600 hits > tacitus histories jews ~3,950 hits > tacitus histories jews leprosy ~154 hits > This is all irrelevant. > Quoting from Tom Potter: > Almost everyone has access to the first hand historical accounts, > and can do wild card searches on the source material. > It is STUPID to provide detailed cites, as these focus > ONLY on the POINTS trying to be emphasized by the writer. > It also STUPID to use second hand, accounts which have a > racial, religious, national, or personal spin on them, rather than > using the FIRST HAND historical accounts. > so only a STUPID person would allow himself to be brainwashed > by Tacitus' racial, religious, national, personal spin on history. It is interesting to see that Mensanator, like many people, has been brainwashed to think that Tacitus, who was one of the most unbiased, rational, and correct historians, put a racial, religious spin on history. This attitude obviously has its' roots in conditioning, as the works of Tacitus are far more rational and correct, than the bible, the Greek and Roman mythologies, etc. In other words, people who have been brainwashed to a point of view, have a great difficulty in accepting data that conflicts with their conditioning. -- Tom Potter === Subject: Re: More Circle problems by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99FhoK01023; >>2). Prove that y =2x is a tangent to the circle x^2 + y^2 - 8x - y + 5 = >0 & just substitute in the circle equation y by 2x and show that the resulting quadratic equation in x has only one solution (i.e. 5*(x-1)^2 = 0, hence x=1 -> y=2 is the wanted point. === Subject: relatively prime by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99FhsW01032; What is the probability of two randomly chosen positive integers to be relatively prime. What if you choose n numbers? === Subject: Re: Pi formula finished? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99Fi8J01047; >Last year my science teacher told me and some friends that she saw on the news that the pi formula had been completed. Is this true? There are many formulas for pi, and some have been known for hundreds of years. === Subject: Re: Group generated by a and b ... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99JxLm20718; > I have a group G generated by 2 elements a and b such that ba=a(b^k) for some > integer k. > Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m > and n ? >>Yes, when the order of a is finite. >But in general, no. >As a simple example, let G be the multiplicative matrix group over the >rational generated by >a = [2 0] and b = [1 0] > [0 1] [1 1] >Then ba = ab^2, but >ba^-1 = [1/2 0] > [1/2 1] >whereas all products (a^m)(b^n) have integral entries. >. it should be [1/2 undefined (0/0)] [undefined (1/0) 1] === Subject: The Square Root of the Golden Section By an Iterative Method. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99JxM020724; NON LINEAR FORMULA , LEADING ITERATIVELY ,TO THE SQUARE ROOT OF THE GOLDEN SECTION by: Eur Ing P.C. Stefanides BSc(Eng)Lon(Hons) MSc(Eng)NTUA CEng MIEE ------------------------------------------------------------------- 1) Find : 1 /[Sin{Arctan [X] }] = X1 , for X , any positive real number, integer or fraction. 2 Find : 1 /[Sin{Arctan [X1] }] = X2 , 3) Find : 1 /[Sin{Arctan [X2] }] = X3 , 4) --- ---------------------------------, N) Find : 1 /[Sin{Arctan [X(N-1)] }] = X(N) By such successive iterations the value of this calculation tends to that of the Square Root of the Golden Section T, [T = SQRT[(SQRT[5]+1)/2 ] T=1,27201964951406896425242246173749 This is, also , the Solution of : [1]/[Sin{Arctan[X]} = [X] , or : [1 ] = [Sin{Arctan[X]}]* [X] , i.e X=T Computed value of SQRT[(SQRT[5] +1)/2] = SQRT[ Golden Section ]= = 1,27201964951406896425242246173749 = T = SQRT [Phi] COROLLARY ------------------- 1/ [Sin{Arctan[SQRT(X)]} =SQRT [1+{1/(X)}] This is an IDENTITY (for X any positive real number fraction or integer). e.g. for X=1,23456789 , right hand side X=1.23456789 ,(amphidromical result) and for X=0.000000001,right hand side X=0.000000001 (amphidromical result). [ For big numbers computing distortion errors are evident ., e.g. For X=123456789 , right hand side X=121951219.5 For X=1234567890 ,right hand side X=1250000000 For X=1250000000, right hand side X=1250000000 Going from rigth to the left found X=1282051181 (similar behaviour for half, double, third, etc. of this number and possibly numbers such as 6250000000 etc.)]. Hoping for comments. [Especially to the Vinnitsa ,October, Conference on the Golden Section Delegates], Panagiotis Stefanides, http://www.stefanides.gr panamars@otenet.gr === Subject: Re: Question on an iteration function problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9A1kqW13739; >--------------000000000106040909050003 >I have the function f(x)=x-tanx and I want to find an iteration function >g(x) and an interval [a, b] so that it converges to the root of f(x). >I tried g(x) = tanx, but this gives g'(x)=1 at the root which is at x=0. >I also tried g(x)=atan(x), which gives me g'(x)=1/(1+x^2), but this is >also one at x=0. g'(x) should be less than one for convergence. >> First, near 0 (say, between -pi/2 and pi/2) there is only one >>root x=0 of f(x)=x-tan(x). The iterations given by g(x)=atan(x) >>do converge, but not fast enough to be dominated by a geometric >>sequence convergent to zero. (The trend is more like >>sqrt(3/(2*n)) after n iterations.) >>When inverting, you threw away infinitely many branches of the >>inverse relation to y=tan(x). >>If you pick an integer number k (other than 0) and use the >>branch g(x) = atan(x) + k*pi (k not zero), you will get >>convergence with abs(g'(x)) < 1/17. >>Still, you will get faster convergence using Newton's >>Method on F(x) = x - g(x). >So, if we do g(x)=atan(x)+pi (for k=1), which root do we get? When I did >this, I got 4.4934 radians, which makes f(x) = tan(x) - x zero, but what >is the effect of the pi we added in g(x)? Is it ok in general to add a >constant in such cases? >In this case, I didn't get x=0 which is a root of f(x). Can I get that? >How? What if I want to get a root which is let's say near 90 radians? >(All of these using iteration functions of the form x=g(x)) >Elias >--------------000000000106040909050003 >this, I got 4.4934 radians, which makes f(x) = tan(x) - x zero, but what >is the effect of the pi we added in g(x)? Is it ok in general to add a constant >in such cases?
How? What if I want to get a root which is let's say near 90 radians? (All >of these using iteration functions of the form x=g(x))
There was some April Fool press release to the effect that the final >nonzero digit of pi had been reached by a supercomputer. And a few >months later an Australian newspaper printed it without noticing the >date. >-- >G. A. Edgar http://www.math.ohio-state.edu/~ e dgar/ That must have been the case for her. She did mention that the person(s) who finished it used computers. === Subject: Re: circle with two centers > Consider a smooth surface, S, and two points, A & B on it. S contains > a continuous, closed curve, C, such that both A & B are equidistant > from every point on the curve C. > 1. Does such a surface exist? I'm not sure I understand the question. Why isn't the sphere with equator C and north and south poles A and B an example? > 2. If it does, can there be countably infinite such centers? > Uncountably infintely many? Can't we have a smooth surface, even a compact one, that contains the unit circle in the x-y plane and a segment on the z-axis? === Subject: Re: how many resolutions? Ok. Let's take standart right triangles : 3, 4, 5 Could we find other integers than 3 and 4 such that a^2 + b^2 = 5^2? Or the same questions for the triangle 15, 8 , 17? The answer is no, I guess. Does it mean that any triangle homothetic to those above is also uniquely determined? I mean that the triangle homothetic to (3,4,5) with a coefficient 17 is 51, 68, 85 and the triangle homothetic to (15, 8 , 17) with a coefficient 5 is 75, 40, 85. And since 85 is a product of 5 and 17 it should not be (but there are) other solutions. For example there is solution (84, 13, 85) - this triangle is homothetic to no one above. Why? What happens? === Subject: Re: circle with two centers > Consider a smooth surface, S, and two points, A & B on it. S contains > a continuous, closed curve, C, such that both A & B are equidistant > from every point on the curve C. The poles and equator of a sphere fit your requrements. Easily generalised to oblate and prolate spheriods. > 1. Does such a surface exist? If so, under what conditions? > 2. If it does, can there be countably infinite such centers? > Uncountably infintely many? Can't think of any surface with more than 2 such centers. > -riskbert === Subject: interesting problem arising in carpentry, column installation In the process of adding a column, I ran into a interesting problem. I want to get the column, which is solid wood, with the least amount of gap. Column flush with the floor and flush with the ceiling. I want to use the least amount of wedges at the end to fill in the gap in the ceiling. So the question is this. Given a diameter of column say 8 and because I cannot slide the erect column flush with floor and ceiling but must instead have to tip it up. And the height is say 9' from floor to ceiling. So, with these 8 diameter columns, what is the smallest gap to occur in order to set the column in place? And let us say the columns are not round but square of 6 by 6, and so how much of a gap in order to place into this 9' position. What I find interesting is that one can fit circular objects easier than rectilinear objects. Or one can fit more volume density of a circle than a square or rectangle. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Factorial/Exponential Identity, Infinity > What do you think about the canonicalization of the infinite binary > sequences? That is to say, for the sequence .010101(01)... with half > zeroes and half ones, that you could exchange any two sequence > elements any number of times and not get .001001001(001).... Is there > a rule to exchange (permute) the elements of the sequence .0101(01)... > to get 00010001(0001)...? Yes, in the sense described below. It depends on what you consider an allowable permutation. The most general form of such permutation can be considered as > follows. Let N be the set of positive integers, then any infinite binary > sequence can be uniquely represented by a functions f:N -> {0,1} in > which f(k) is the k'th digit of the sequence. Let B be the set of such functions f:N -> {0,1}. Then any bijection g:N <-> N creates a permutation G: B -> B by > defining G(f) = fog, where o represents function composition, G(f) > is the function whose valueat k in N is f(g(k), > i.e., G(f)(k) = f(g(k)), for each k in N. Let P represent the set of all such permutations, It is clear that > the number of zeros and number of ones in an expansion must > remain fixed under any such permutation of digits. Then, for a given f:N -> {0,1} and h:N -> {0,1} in B, a necessary > and sufficient condition for there being some permutation G in P > such that G(f) = h is that > card(k in N: f(k) = 0}) = card(k in N: h(k) = 0}) and > card(k in N: f(k) = 1}) = card(k in N: h(k) = 1}) > i.e., the infinite binary sequences for f and h have the same > number, possibly infinite, of zeros and the same number, possibly > infinite, of ones, and only one of these can be finite. But if the number of zeros or the number of ones differ, then there > is no permutation carrying one into the other. This divides up the set B into equivalence classes of binaries > permutable into each other which can be indexed by the integers, Z. > Any two such binaries with infinitely many zeros and infinitely many > ones in their expansions can be permuted into each other, and can be > indexed by 0. If the number of ones is finite, use that number as > index, and if the number of zeros is finite use the negative of that > number as index. Then two binaries can be permuted into each other if and only if > they have the same index. > B is the set of functions f(n) -> {0,1}, n E N. Your notation here differs from the usual f: N -> {0,1}. Do you mean something different? > There are uncountably many such functions, each represents a > subset of N elements or a distinct element of R[0,1) with no dual > representation. Not so. There are distinct functions which, as binary expansions, represent the same real in [0,1). For ach function with a positive index (a positive but finite number of 1's in its expansion) there is a function with a negative index representing the same real. > I consider .00010001(0001)... as an infinite sequence and how to > permute its elements to get .01010101(0101).... It seems almost > simple, move the 1 in the fourth place to the second place, and move > each one in the 4x'th place for x>1 to the 4(x-1)'th place. There's > certainly no dearth of ones and zeros, the amount of each is > infinite. Yet, the zero that was in the second place would have to go > somewhere, and necessarily displace another. For each n in N, there is an nth zero in .00010001(0001)... and an nth zero in .01010101(0101).... Similarly there is an nth one in .00010001(0001)... and an nth one in .01010101(0101).... For each n in N, move the nth zero in .00010001(0001) to the position of the nth zero in .01010101(0101).... and similarly with the nth ones, and the thing is done. > It's easy to describe a canonical sequence relative amounts of zeros > and ones. pick a fraction p/q less than one, p and q positive, p say there are that many zeros and (q-p)/q many ones, plainly the > sequence is the repeating subsequence of p zeros sequentially > (consecutively) and then q-p ones. > Consider the canonical sequence of zero-density p/q, where finitely x > many zero elements are changed to ones. Is it possible to permute > that sequence to the canonical sequence with none of the elements > changed by exchanging its elements? No, it is not. It permutes to > its canonical sequence of the finite subsequence of ones, however many > zeros were changed into ones there were, and then the repeating > sequence representative of the ratio of the densities of zeros and > ones. > Then there are the sequences of type b with infinitely many zeros and > ones and infinitely many more or less of of one than the other, > opposite symbol, eg sequences representing subsets of the naturals > like the primes, powers of two, etcetera. Consider the powers of > the primes. Any two sequences of all zeros and ones with the same, possibly infinite, number of zeros and the same, possibly infinite, number of ones, can be permuted into each other by the simple method of moving the nth 0 of one to the position of the nth 0 of the other and also moving the nth 1 in one to the position of the nth 1 of the other. What is so hard about that? As an alternate approach, any such sequence of zeros and ones can be represented by a subset S of N of the positions in which there is a 1, with the compliment, NS, being the set of positions of the zeros in that expansion. Given b1 and b2 as two such sequences, with S1 and S2 as their sets of locations of 1's, then there is a bijection from N to N carrying S1 to S2 and NS1 to NS2, i.e., a permutation, if and only if Card(S1) = Card(S2) and Card(NS1) = Card(NS2). > Where you note g(n) is a bijection from N to itself, where f(n)=0 or > 1, then their composition defines a permutation G, but I think it > defines a permutation from some proper subset B_(p/q +-x) -> B_(p/q > +-x) < B, because g is a bijection. For example it doesn't permute > the sequence of all ones to the one of all zeros. > For example .0101(01)... permutes to .1001(01)... but not to > .001001(001).... How does it permute to .0001(01)...? I can easily > determine a bijective function from {1, 3, 5, ...} to {3, 5, ...}, and > from {0, 2, 4, ...} to {0, 1, 2, 4, ...}. Changing the second element > from a one to a zero means that somewhere a zero has to be changed > into a one. It could be put off indefinitely but that necessarily > implies that some subsequence 01 would be changed to 11. > I focus upon necessary and sufficient condition. I can agree that > it is a necessary condition for there to be the permutation, yet, not > that it is a sufficient condition. I wonder what you mean by > allowable. > The sequences .(01)... and .(10)..., for half zeros and half ones, are > perhaps the simplest cases with infinite numbers of ones and zeros to > consider. Exchange each one and zero of the representative > subsequence and it becomes the other. > Consider .100..., it can't be permuted to .000.... For the same > reason, (01)... can't be permuted to 00(01).... If you change the > value of one symbol, you have to change the value of an occurrence of > the opposite symbol, they exchange. > It's simple to get an extra symbol out of the sequence to replace its > opposite, the problem is putting it away, it also has to replace its > opposite. > 01010101.... > ^ > 00010101... > ^^^ ^ ^ ... > 10010101... > 01010101... > 00110101... > 00011101... > 0001011101... > The first sequence has it's second element changed to its opposite. > The second sequence is required to change one of its marked symbols to > its opposite. > Perhaps it's a specialization of permutation. The sequence is only > changed by exchanging any of its elements any number of times, the > value of a symbol is not changed without changing an occurrence of its > opposite symbol. > It's not possible to permute in this way 001(001)... to 011(011).... > One way to see this is that there are sequence z_1=00... , z_2=00..., > and o_1=11.... Take the first element from each then the second, > etcetera: 001(001).... If you want to form 011 then it is a matter > of taking the first element of z_1, then two elements of o_1 then an > element of z_2 (to keep it even), then two more elements of o_1, > etcetera. The amount taken from o_1 to form the sequence (011) from > the constituent sequences of (001) rises steadily and never returns to > a par of those taken from z_1 and z_2. On the other hand, forming a > sequence like 111111000000000000(001) does. > Given instructions: form a sequence of infinitely many ones and > infinitely many zeros, there are many ways to describe a rule or > method to do that. Given instructions: given a sequence with > infinitely many ones and infinitely many zeros, exchange its elements > to form another sequence, there are less sequences. > Randomly and fairly (uniformly) select a subset of the natural > numbers, there are even odds it contains zero. Half the subsets of N > contain zero. > Is there a rational function for x=4? > === Subject: Re: Chessboard knight metric? >Take a chessboard (with or without infinetely many squares) let the >distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >chessboard be defined as the minimum number of moves a knight takes >to reach y from x. As others have pointed out, this obviously is a metric. As to the function (assume infinite board): 1) you undoubtedly know that the color of the square changes after each move, so you will need to separate into two cases according to the parity of x_1+x_2+y_1+y_2, and 2) the sets of squares that can be reached in exactly one or exactly two moves have a special, slightly irregular structure, BUT 3) for n>2, the set of squares that can be reached in exactly n moves has a very regular octagonal shape. DRAW IT!!!! (proof by induction) Item 1 and 3 gives you a rule to determine the distance, in the case that it is at least 3. You only need to check, whether the distance might be one or two. Jyrki Lahtonen, Turku, Finland === Subject: Subspace question I'm a linear algebra student, and normally I would ask this to my TA, but he is out to a conference. Is R2 considered a subspace of R3? I'm guessing no, because R2 cannot contain the origin in R3 (0,0,0). === Subject: Re: interesting problem arising in carpentry, column installation > In the process of adding a column, I ran into a interesting problem. I > want to get the column, which is solid wood, with the least amount of > gap. Column flush with the floor and flush with the ceiling. I want to > use the least amount of wedges at the end to fill in the gap in the > ceiling. So the question is this. Given a diameter of column say 8 > and because I cannot slide the erect column flush with floor and > ceiling but must instead have to tip it up. And the height is say 9' > from floor to ceiling. So, with these 8 diameter columns, what is the > smallest gap to occur in order to set the column in place? And let us > say the columns are not round but square of 6 by 6, and so how much > of a gap in order to place into this 9' position. > What I find interesting is that one can fit circular objects easier > than rectilinear objects. Or one can fit more volume density of a > circle than a square or rectangle. Let the height of the ceiling be C. For a circular column, let D be the diameter. For a square column, let D be the side of the square. Let L be the length of the longest installlable column. Then: L = square root(C^2-D^2) Sqrt[(9*12)^2 - 8^2] = 107.703 Gap~0.3 inch Sqrt[(9*12)^2 - 6^2] = 107.833 Gap~0.2 inch -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Subspace question > I'm a linear algebra student, and normally I would ask this to my TA, but he > is out to a conference. > Is R2 considered a subspace of R3? I'm guessing no, because R2 cannot > contain the origin in R3 (0,0,0). Yes because R^2 is homeomorphic and isomorphic to the z-axis projection of R^3 onto P = { (x,y,0) | (x,y) in R^2 }, the xy plane of R^3, which is a subspace of R^3. Thus R^2 embeds into R^3 as the subspace P, === Subject: Re: Just what is an L-series? > I know what a Dirichlet L-series is, and I've seen a few other kinds, > such as Artin's in connection with representation of finite groups. > But is there a standard comprehensive definition of the term > L-series? Maybe, any Dirichlet series > sum_{n=1}^infty a_n n^{-z} > such that the sequence (a_n) is totally multiplicative? Not general enough: that would exclude say, L-functions of elliptic curves. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: 0! = 1 x factorial is equal to the integral of u to the x multiplied by e to the minus u with respect to u from zero to infinity. Proof: #include #include void main(void) { clrscr(); if( 0!=1 ) printf(It's true but the coder was being sarcastic.n); } The real proof: u to the zero is one. So what we're really solving for x equals zero is the integral of e to the minus t with respect to t from zero to infinity, which is easily one. So, 0!=1. ;) I loved the joke, Brendan. -Amanda > In <1g2k59v.1onstxx1i0ew6aN%bdm@cs.anu.edu.au> bdm@cs.anu.edu.au (Brendan >> Did someone know a simple demonstration of >> 0! = 1 ? >Proof by C: >main () >{ > if (0!=1) printf(truen); > else printf(falsen); >} >(I wish this was original but it isn't.) >Brendan. > There's no way to prove or demonstrate a definition. > -- > John E. Prussing > University of Illinois at Urbana-Champaign > Department of Aerospace Engineering > http://www.uiuc.edu/~prussing === Subject: Re: Group generated by a and b ... > I have a group G generated by 2 elements a and b such that ba=a(b^k) for some > integer k. > Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m > and n ? >Yes, when the order of a is finite. >> But in general, no. >> As a simple example, let G be the multiplicative matrix group over the >> rational generated by >> a = [2 0] and b = [1 0] >> [0 1] [1 1] >> Then ba = ab^2, but >> ba^-1 = [1/2 0] >> [1/2 1] >> whereas all products (a^m)(b^n) have integral entries. >group G = , some k in N with ba = ab^k == some n,m in N with ba^-1 = a^n b^m iff o(a) finite >proof ><== let o(a) = r; ba^-1 = ba^(r-1) = a^(r-1) b^k(r-1) >==> e = a^n b^m a b^-1 = a^n a b^km b^-1 = a a^n b^r > where r = km - 1 > a = aa^n b^r a = aa^n a b^kr > e = aa^n b^kr = aa^n b^r; b^kr = b^r; e = b^(k-1)r > b has finite order, when k /= 1 > (aa^n)^-1 = b^kr; thus (aa^n)^-1 and a have finite order > The case k = 1; ba = ab; ba^-1 = a^n b^m > unresolved ba = ab => ba^-1 = a^-1b You have also overlooked the special case k = -1, because ba = ab^-1 => ba^-1 = a^-1b^-1 (just by inverting everything). This case gives r = 0 in your proof. And I guess k=0 is another exception: ba = a => ba^-1 = a^-1. But your result is correct for |k| > 1 ! . === Subject: Re: Group generated by a and b ... >> I have a group G generated by 2 elements a and b such that ba=a(b^k) for some >> integer k. >> Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m >> and n ? >Yes, when the order of a is finite. >>But in general, no. >>As a simple example, let G be the multiplicative matrix group over the >>rational generated by >>a = [2 0] and b = [1 0] >> [0 1] [1 1] >>Then ba = ab^2, but >>ba^-1 = [1/2 0] >> [1/2 1] >>whereas all products (a^m)(b^n) have integral entries. >>. >it should be [1/2 undefined (0/0)] > [undefined (1/0) 1] No it shouldn't. . === Subject: Re: How to calculate the total coverage area of a few circles? > Make a list of the N*(N-1) circle intersection points and the 2N > points p_i with x +/- r that are not in the interior of another > circle, and sort into ascending order by x coordinate, then sum the > areas of vertical zones bounded by these critical points. There > are no arc intersections within a zone. A zone may contain disjoint > segments but each segment is bounded above and below by an arc of a > circle, and left and right by straight lines. > For example, if we have 3 circles of radius 5 and centers at (5,12), > (8,9), and (9,5), the event-points list for the plane sweep is: > 0.0 12.0 > 3.3 7.3 > 4.0 5.0 > 4.1 5.9 > 9.7 13.7 > 12.9 8.1 > 13.0 9.0 > 14.0 5.0 > The following edges or intersections are interior and not relevant: > 3.0 9.0 > 4.4 7.0 > 9.6 10.0 > 10.0 12.0 > -jiw plane-sweep method is similar to the numerical intergration. The sunk parts that I mentioned is the disjoint segments in the reply from James. I am still in the trouble to determine which arcs should be used to bound the above and below in my simulation programme, especially in the cases of the disjoint segments, since the ascending order by x coordinate will be disturbed. Could you please show me the detail of how to deal with this problem? In James' example, how to identify which arcs are the above bound and the below bound between (3.3, 7.3) and (4.0, 5.0), as well as between (4.0, 5.0) and (4.1, 5.9)? Leng Supeng === Subject: Re: Is it mass; or is it weight >> So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net >> which stretches an additional 4.4-ft before thrusting the man back, would the >> potential energy of the net at the instant it plunges the man back be >> U = mgh; >> U = 220 * (36 + 4.4). > You left off the factor of g. Which either means that you forgot about it > or that you are using a system of units within which the factor of g cancels > out. > For instance, if you measure m in pounds mass, h in feet and energy in > foot-pounds(force) then the g cancels (more or less) and U = mh. > The more or less is because you're cancelling local g from the jump site > with standard g from the ratio of pounds force to pounds mass. > Gene Nygaard would be correct to point out that there is no single standard > g that is officially sanctioned for this purpose. While initially attempting this question, I converted pounds-mass to kilograms-mass, and the height in feet to metres, and computed the potential energy through U = mgh; U = (97.99) * (9.8) * (12.73); which revealed the net's potential energy to be at the time it sent the man back upwards almost twelve kilo-joules. My Physics teacher, however, disagreed, and prosposed that the given units of the mass and height be maintained as is, and the potential energy be calculated without the factor 'g' present anywhere in the computation. This method outputs U = mh; U = (220) (36 + 4.4); U = 8888 lb-ft; the potential energy of the net to be about 9000 lb-ft. She, upon being asked why 'g' had been neglected from the computation, failed miserably to explain so. And since the question is evenly numbered in the book, its answer is not available (only odd-numbered questions' answers are given). If it helps, I tend to follow for the most part hitherto the SI system of units for representing measureable physical quantities. I still believe for what appears to be a reason I cannot possibly explain properly my answer and my method to be correct. Whether it isn't, and why, I would like to know. -- Ayaz Ahmed Khan Yours Forever in, Cyberspace. === Subject: Re: circle with two centers > Consider a smooth surface, S, and two points, A & B on it. S contains > a continuous, closed curve, C, such that both A & B are equidistant > from every point on the curve C. > 1. Does such a surface exist? If so, under what conditions? > 2. If it does, can there be countably infinite such centers? > Uncountably infintely many? > -riskbert x = sint(t) + cos(u) y = sin(u) z = t I.e., a 'cylinder' with axis is a sinusoid. This surface has infinitely many poles for the circle z= 0, x^2 + y^2 = 1. And for any other section parallel to OXY plane. One can 'straight' smoothly that 'cylinder' from some value of t, getting a straight regular cylinder with a generatriz being the z axis. Then taht surface has uccountable infinitely many of such 'centres'. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: a puzzle related to artinian group >> A quick sketch of my argument for the linear case is that you number >> cards and players and arrange the play so that the cards are always in >> order from left to right and so that each player holds a contiguous set >> of card numbers. [...] > What's the role the numbering of the cards? Just for convenience in reasoning about them. >> You then show that the centre of gravity of the cards is fixed, >> that the variance of the positions of the cards relative to some point to >> the left of all of them is strictly increasing whenever a round of play >> takes place, [...] > Nice!:) >> As for the cyclic version? > A similar idea should also work in the cyclic version. Indeed it should, but I've not been able to do it. > Although, it is > not clear to me whether the center of the gravity and else make any > sense in this case. One possibility I tried is to consider the cards arranged on the vertices of a regular n-gon and then to consider their centre of gravity. With n-1 cards this cannot be at the centre of the n-gon, and it seems plausible that it approaches the centre on each round of play. I have no proof of this. > One should find a suitable function that strictly increases in each > round, except for the uniform configuration, where it is maximized. Quite, but I don't know of such a function. Another way of thinking about the cyclic version of the game is to unroll it onto a line. So you get a configuration in which there are infinitely many cards dealt out to the integer points on a line in a periodic pattern. I haven't been able to get this view to help with the original problem either, but it does raise lots of other entertaining questions. Rob. === Subject: Re: convex hull > Given three points p1,p2,p3 in 2d > and given point g also in 2d > how can i decide whether the point is inside the convex hull that > p1p2p3 create or not. Determine tha areas (unsigned) of p1p2p3, gp1p2, gp1p4 and gp2p3 (easy). Then if the first equals the sum of the others three, g is in the interior or in the perimeter of p1p2p3. In that last case, one of the areas is 0 (two if g is a vertix, but there are more direct ways to check it). -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Group generated by a and b ... , some k in N with ba = ab^k == some n,m in N with ba^-1 = a^n b^m iff o(a) finite >proof N = { n in Z | n > 0 }. When k < 0, the proof bombs at b^r a = ab^rk because I don't have b^-1 a = ab^-k ><== let o(a) = r; ba^-1 = ba^(r-1) = a^(r-1) b^k(r-1) >==> e = a^n b^m a b^-1 = a^n a b^km b^-1 = a a^n b^r > where r = km - 1 > a = aa^n b^r a = aa^n a b^kr > e = aa^n b^kr = aa^n b^r; b^kr = b^r; e = b^(k-1)r > b has finite order, when k /= 1 > (aa^n)^-1 = b^kr; thus (aa^n)^-1 and a have finite order > The case k = 1; ba = ab; ba^-1 = a^n b^m > unresolved > ba = ab => ba^-1 = a^-1b > You have also overlooked the special case k = -1, because > ba = ab^-1 => ba^-1 = a^-1b^-1 (just by inverting everything). > This case gives r = 0 in your proof. > And I guess k=0 is another exception: ba = a => ba^-1 = a^-1. > But your result is correct for |k| > 1 ! For k > 1. BTW, n,m > 0 was included in premise but only m > 0 used. For k = 1, is not the infinite dihedral group { a^n, a^n b | n in Z } with o(b) = 2, ba = a^-1 b an example for which o(a) is infinite? === Subject: Contractible Spaces If a topological space S is contractible to some point p in S, is S contractible to every point in S? If a topological space S is strongly contractible to some point p in S, is S strongly contractible to every point in S? What's an example of a space that is contractible but not strongly contractible? S is contractible to a when there's some continuous h:Sx[0,1] -> S with for all x in S, h(x,0) = x, h(x,1) = a, and strongly contractible when in addition for all t, h(a,t) = a ---- === Subject: Re: divide and conquer > I am working on the following problem. I need a divide and conquer > algorithm to solve the following problem. Any help is appreciated. > I and others have noticed your repeated attempts in comp.theory, > sci.math, and perhaps elsewhere to have your homework done for you. > That's your choice, of course, and often you will find someone who > will respond as you wish. > A small cautionary tale: you may well fool a teacher into giving a > passing grade, a college into granting a degree, but when you are > sitting in an interview with a hiring manager and people working in > your technical field, your lack of thinking through problems for > yourself will speak even when you do not. > I've seen it happen: a sweet young thing who got all the impressionable > young men to do her homework, to rehearse her for exams, to do her > class projects for her. She got her degree with acceptable grades, but > the only employment she could find in a market crying out for the skills > of her degree was the same job she had before she went to college: she > cleaned other people's houses for a living. > Spending four years in college and arranging to learn nothing only makes > you four years older and wastes that much time out of your finite life > to no point. > Her field of study by the way was the same as your own, and the demand > for those skills has fallen off dramatically, so your strategy is worse > today than it was in her time. > xanthian. Pretty sure that her was got a job,using her .....my english is only technical,i have problems with some words ;-).....?sympaticnes? ... she was well-trained at college,he-he.But with usenet it does not work,and it's not a training. Dmytry Lavrov. -- maybe you can find answers here: http://DmytryLavrov.narod.ru === Subject: Effective inverse of a matrix... Given the matrix P + cQ, where P and Q are known positive definite matrices, and c is a positive scalar. I want to compute the inverse of P + cQ for various values of c as effectively as possible, by exploiting that I know P and Q beforehand. For instance, if P is the identity matrix, and VLV' is the eigenvalue decomposition of Q, then I can compute the inverse as V(I +c L)^{-1}V', and only have to do some scalar inversions (more effective methods might exist). Any hints? I have tried using the matrix inversion lemma, but it didn't seem to help me. Lars === Subject: Re: When is the limit of Borel-measurable functions Borel-measurable? >This is the second time I've sent this message. The first time, it >seems to have dropped into the bit-bucket. Sorry if you're reading it >for the second time. If you never see my reply to your first post I'll repost it. Not that I actually had an answer for you or anything. ************************ David C. Ullrich === Subject: Re: circle with two centers >> Consider a smooth surface, S, and two points, A & B on it. S contains >> a continuous, closed curve, C, such that both A & B are equidistant >> from every point on the curve C. >The poles and equator of a sphere fit your requrements. [...] >Can't think of any surface with more than 2 such centers. Just deform the sphere - leaving the equator intact - to intersect the axis as many times as you like. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: circle with two centers > Consider a smooth surface, S, and two points, A & B on it. S contains > a continuous, closed curve, C, such that both A & B are equidistant > from every point on the curve C. >>Can't think of any surface with more than 2 such centers. >Just deform the sphere - leaving the equator intact - to intersect the >axis as many times as you like. Oh, I suppose you are requiring that as well as A being equidistant from the points on C and B being equidistant from the points on C you want the two distances to be the same. In which case there are obviously only two centres (in 3D). I don't think the original question implied that. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: multiplication negs > exactly why is it that when one multiplies 2 negative numbers one ends > up with a positive? Suppose you have an account at a bank, and the account can hold positive amounts (credits) or negative amounts (debts). Suppose you do a job for Mr. Brown, and Mr. Brown adds $300 to your account. Your account balance goes up by $300. Suppose you do two identical jobs. Your account balance goes up by 2 * $300 = $600. Suppose the bank discovers a $300 credit it gave you before was erroneous, so it TAKES AWAY that credit. Taking away a credit is like adding -1 credit, so your balance changes by -1 * $300 = -$300. $300 check, and they present it to the bank. Adding a $300 debt to your account adds -$300 to your balance. Adding two $300 debts to your account adds 2 * -$300 = -$600 to your account. Finally, suppose the bank discovers the two checks were for somebody else's account, not yours. Then the bank TAKES AWAY two DEBTS. It adds -2 * -$300 to your account. Your account balance goes up by $600. So -2 * -$300 = $600. -- edp (Eric Postpischil) http://edp.org === Subject: The probability problem??? A two A B people throws the coin. The sequence of A B A B....... The people who comes out first wins. It says that it takes 1500 dollars if A wins. Do you become fair game only if we take the prize money of a while if B wins? ------------------------------ We came out to 300 dollars at the answer. The 1000/3 at my opinion. We ask the advice. === Subject: Re: topological terminology >> So... homotope or homotop, which is the verb? >> >> Pete Use it in a sentence? The space X homotoped into Y? > This avoids the question though. I've seen both homotope and homotop and > the past tense of both would seem to be homotoped. I think homotope is > more common, but that's also the one I use, so I'm biased. My point was that I never heard it used as a verb, so if you used it in a sentence it would show something to me. === Subject: Re: circle with two centers > Consider a smooth surface, S, and two points, A & B on it. S contains > a continuous, closed curve, C, such that both A & B are equidistant > from every point on the curve C. 1. Does such a surface exist? > I'm not sure I understand the question. Why isn't the sphere with equator C > and north and south poles A and B an example? > Can't we have a smooth surface, even a compact one, that contains the unit > circle in the x-y plane and a segment on the z-axis? Sorry, my mistake; I omitted two important specifications: 1. The surface is an open surface, which rules out the sphere. 2. Each of the centers is at the same distance from the curve. So your unit circle with the segment example does work for the original, incomplete question but not for the one with the two restrictions above. However, I do think a modification of the segment into a tubular structure so that the surface resembles a mexican hat may work. What would be of interest is to see how to generate such a surface on Mathematica, for example... -riskbert === Subject: Re: Is it mass; or is it weight > So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net > which stretches an additional 4.4-ft before thrusting the man back, would the > potential energy of the net at the instant it plunges the man back be U = mgh; > U = 220 * (36 + 4.4). >> You left off the factor of g. Which either means that you forgot about it >> or that you are using a system of units within which the factor of g cancels >> out. >> For instance, if you measure m in pounds mass, h in feet and energy in >> foot-pounds(force) then the g cancels (more or less) and U = mh. >> The more or less is because you're cancelling local g from the jump site >> with standard g from the ratio of pounds force to pounds mass. >> Gene Nygaard would be correct to point out that there is no single standard >> g that is officially sanctioned for this purpose. > While initially attempting this question, I converted pounds-mass to > kilograms-mass, and the height in feet to metres, and computed the potential > energy through > U = mgh; > U = (97.99) * (9.8) * (12.73); > which revealed the net's potential energy to be at the time it sent the man > back upwards almost twelve kilo-joules. > My Physics teacher, however, disagreed, and prosposed that the given units of > the mass and height be maintained as is, and the potential energy > be calculated without the factor 'g' present anywhere in the computation. This > method outputs > U = mh; > U = (220) (36 + 4.4); > U = 8888 lb-ft; Your physics teacher is apparently an idiot, taught by idiots out of textbooks written by idiots. Sadly, that's quite typical For a coherent system of units, U = mgh. For example, the following situations involve coherent sets of units: U in foot-pounds(force), m in slugs, g in feet/sec^2, h in feet U in joules, m in kilograms, g in meters/sec^2, h in meters U in foot-poundals, m in pounds(mass), g in feet/sec^2, h in feet For systems of units that are not coherent, U = kmgh where k is a constant that depends on the system of units. For example If U is in foot-pounds(force), m is in pounds(mass), g is in feet/sec^2 and h is in feet, k will be equal to approximately 1/32.17 The precise value of k can be determined based upon if the precise definition of the pound force and the pound mass. Since one pound force is generally defined as the force required to support one pound mass under one standard gravity, this value will be 1/standard-g. If you are willing to overlook the difference between the standard g (the defining constant in the relationship between the standard pound force and the standard pound mass) and local g, the formula obviously simplifies to U = kmgh = 1/g * m * g * h = mh John Briggs === Subject: Re: hw help -- continuity > [...] >>Proving a function continuous at any point >>proves it continuous at every point. >> f(x) = 0 when x <= 0 >> = 1 when 0 < x >> is continuous at 1. Thus it's continuous at 0? > But 1 is not any point, it is a specific point. > To restate less ambiguously: > If you prove f continuous at an arbitrary point, then it is > continuous at every point What is wrong with the good old English word every ? Marc === Subject: Re: relatively prime Hello there! The probability of any two random integers being relatively prime is .6079271018540266 ..., which is exactly equal to 6/(pi^2). I am not sure where you could find a proof for this, but I imagine there's one somewhere on the internet. I found this on the following URL: http://primes.utm.edu/curios/page.php?number_id=1326 I hope that this helps! Anthony > What is the probability of two randomly chosen positive integers to be relatively prime. What if you choose n numbers? === Subject: Re: topological terminology > Use it in a sentence? > The space X homotoped into Y? Ugh. I had in mind homotoping maps - we can homotop(e) f to a cellular map etc etc. I find the verb occurs not infrequently in books & papers, with both variant spellings. I suspect there maybe some greek origin making ppl prefer homotop? === Subject: Quadratic Sieve Does anyone have or know of a Quadratic sieve implementation that could be distributed? Mike This disposable email service was provided free of charge by http://www.mjcsoftware.com - Come and get yours today! By registering, you will no longer see this banner and have the ability to send attachments! === Subject: Re: Pi formula finished? >Last year my science teacher told me and some friends that she saw on the news that the pi formula had been completed. Is this true? The formula's been completed for a long time. Doug === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. >I have decided not to respond to your insults in kind. Your deception was insulting. > I am sorry >that you felt that you were justified in taking my opinions personally >and forming such outlandish and erroneous judgements of my character >and motives. I regret that you expected me to accept so much >inaccurate and/or irrelevant data. I provided NO inaccurate nor irrelevant data. Your delusions notwithstanding. === Subject: Re: circle with two centers >Sorry, my mistake; I omitted two important specifications: >1. The surface is an open surface, which rules out the sphere. A sphere with a hole in it, not on the equator or poles. >2. Each of the centers is at the same distance from the curve. So you can only have two centres, since there are only two points on a line equidistant from a third point. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Endomorphism for the layman [endomorphisms for dummies...] I am a newbie, but I am interested in endomorphism. What are they - put in simple words ? Why are they useful for (in 3D games )? Anthony === Subject: Re: divide and conquer > I am working on the following problem. I need a divide and conquer > algorithm to solve the following problem. Any help is appreciated. > I and others have noticed your repeated attempts in comp.theory, > sci.math, and perhaps elsewhere to have your homework done for you. > That's your choice, of course, and often you will find someone who > will respond as you wish. > A small cautionary tale: you may well fool a teacher into giving a > passing grade, a college into granting a degree, but when you are > sitting in an interview with a hiring manager and people working in > your technical field, your lack of thinking through problems for > yourself will speak even when you do not. > I've seen it happen: a sweet young thing who got all the impressionable > young men to do her homework, to rehearse her for exams, to do her > class projects for her. She got her degree with acceptable grades, but > the only employment she could find in a market crying out for the skills > of her degree was the same job she had before she went to college: she > cleaned other people's houses for a living. > Spending four years in college and arranging to learn nothing only makes > you four years older and wastes that much time out of your finite life > to no point. > Her field of study by the way was the same as your own, and the demand > for those skills has fallen off dramatically, so your strategy is worse > today than it was in her time. > xanthian. Oh Xanth, that was so mean and unwarranted. Of course that isn't what will happen. What'll probably happen is that he'll be able to fool some HR gal into thinking his learning in college was relevant and he'll get a job. THEN you'll see something messages that start with I have this problem at work... posted on usenet. === Subject: Re: Endomorphism for the layman > [endomorphisms for dummies...] > I am a newbie, but I am interested in endomorphism. > What are they - put in simple words ? > Why are they useful for (in 3D games )? > Anthony And endomorphism is a homomorphism from a mathematical object to itself. What sort of mathematical object are we talking about? === Subject: Radon-Nikodym and Riesz Representation theorems I am looking at the Radon-Nikodym theorem and wondering if it can be proven by using the Riesz Representation theorem. Specifically, by the R-N theorem, we have a sigma-finite measure m on a space X with B, the set of measurable subsets of X. n defined on B is absolutely continuous with respect to m. Then we have that n(E) = integral over E of f dm, for all E in B, for some nonnegative measurable function f, which is unique, in that if any other g holds, then f = g a.e. My question is, since m is sigma-finite on X, can we somehow turn the set B into a normed linear space with a bounded norm (bounded possibly because of the sigma-finiteness of m), based on n, so that the Riesz Representation theorem could come into play? I am thinking of constructing a vector space V indexed by the subsets of B, so that we'd have a string of real numbers (of which only a countable number are nonzero) to represent each set in B. i.e. a,b in B means that a = 0,0,0,1,... and b = 0,0,0,0,0,1..., so that a + b = 0,0,0,1,0,0,1..., 2a = 0,0,0,2,..., etc. In the special case where all all the nonzero entries of the vector are 1, we can view this as the union of all the sets indexed by those 1s. We can write this as sum(i = 1 to inf)ci * Bi, where the Bi's are in B, the ci's are in R. Then v, the space's norm, is done in the obvious way: v(a) = sum(from i = 1 to inf)abs(ci) * n(Bi). At this point, I'm not sure what to do. The coefficients being real causes me no end of problems, it seems. Are there any fixes to this, or is the R-R theorem simply not applicable to prove the R-N? I apologize for being vague, but I'm having trouble figuring out which spaces and functions would be analogues in each theorem. However, they seem similar on a certain level, so I can't imagine there being no connection. If anyone has attacked R-N in this way, please let me know. Any help is appreciated. Justin Davis === Subject: Re: Numeric one-way hash function >> But if you don't believe my proofs, the easiest way >> to check your algorithm is to implement it for a smaller >> value of Max and then to check the number of duplicates. >> I expect that you will be highly surprised. > I have seen true random number generators produce > the same value but it was no where near a third of the > time. However I was not performing an in-depth > soak test so I may not have produced sufficient > numbers. No, you obviously did not (see below). > The rate of occurrence of duplicates is not linear, > p(duplicate) = n/10^12 This formula is not correct when you have already found duplicates because the number of different output value that have already been produced is less than n. The correct formula is to look at the probability that you don't have any duplicates when you draw n out of m different values. The probability that you have no duplicates among these n values is equal to m^n * m!/(m - n)! > What the original poster probably needs is an > encryption algorithm that is 1 to 1 over 10^12, > difficult to predict over short ranges and hard to > break. Yes, and the algorithm that I proposed should solve that problem. Your algorithm has some serious problems with duplicates. I did the simulation I proposed with Max = 100000, and the results are shown in the following table: Duplicates Inputs Outputs 0: 0 36619 1: 37030 37030 2: 36668 18334 3: 18549 6183 4: 5976 1494 5: 1375 275 6: 324 54 7: 70 10 8: 8 1 The second column shows the number of inputs that give the corresponding number of duplicates, e.g. only 37030 of the 100000 inputs will have no duplicates. The last column shows the number of outputs with that number of duplicates, so 36619 possible outputs were never produced. This corresponds well with my expectations that the fraction of inputs that have no duplicates is the same as the fraction outputs that are never produced: for large values of Max they should both approach 1/e. of the inputs will have duplicates. greetings, Ernst Lippe === Subject: Question about the L'Hospital Rule Regarding the L'Hospital Rule, it has been found some examples where the limit of the quotient of the derivatives exists without existing the original limit. So, I would like to ask you when can one be sure that the existence of the limit the quotient of derivatives guarantees the existence of the original limit. Paul === Subject: Re: Radon-Nikodym and Riesz Representation theorems > I am looking at the Radon-Nikodym theorem and wondering if it can be > proven by using the Riesz Representation theorem. > Specifically, by the R-N theorem, we have a sigma-finite measure m on > a space X with B, the set of measurable subsets of X. n defined on B > is absolutely continuous with respect to m. Then we have that n(E) = > integral over E of f dm, for all E in B, for some nonnegative > measurable function f, which is unique, in that if any other g holds, > then f = g a.e. > My question is, since m is sigma-finite on X, can we somehow turn the > set B into a normed linear space with a bounded norm (bounded possibly > because of the sigma-finiteness of m), based on n, so that the Riesz > Representation theorem could come into play? > I am thinking of constructing a vector space V indexed by the subsets > of B, so that we'd have a string of real numbers (of which only a > countable number are nonzero) to represent each set in B. i.e. a,b in > B means that a = 0,0,0,1,... and b = 0,0,0,0,0,1..., so that a + b = > 0,0,0,1,0,0,1..., 2a = 0,0,0,2,..., etc. In the special case where all > all the nonzero entries of the vector are 1, we can view this as the > union of all the sets indexed by those 1s. We can write this as sum(i > = 1 to inf)ci * Bi, where the Bi's are in B, the ci's are in R. Then > v, the space's norm, is done in the obvious way: v(a) = sum(from i = 1 > to inf)abs(ci) * n(Bi). > At this point, I'm not sure what to do. The coefficients being real > causes me no end of problems, it seems. Are there any fixes to this, > or is the R-R theorem simply not applicable to prove the R-N? > I apologize for being vague, but I'm having trouble figuring out which > spaces and functions would be analogues in each theorem. However, they > seem similar on a certain level, so I can't imagine there being no > connection. If anyone has attacked R-N in this way, please let me > know. Any help is appreciated. > Justin Davis John von Neumann did a proof of the R-N theorem (for finite measures, say) using Hilbert space theory. Your Banach space is L^2 of the measure space. You define a linear functional using the other measure, which is well-defined if it is absolutely continuous. === Subject: Re: divide and conquer > Spending four years in college and arranging to learn nothing only makes > you four years older and wastes that much time out of your finite life > to no point. There is some joke about college being the only place where you pay for something, and then try as hard as you can NOT to get your money's worth. === Subject: Re: Question about the L'Hospital Rule Check it out at : http://mathworld.wolfram.com/LHospitalsRule.html Lurch > Regarding the L'Hospital Rule, it has been found some examples where > the limit of the quotient of the derivatives exists without existing > the original limit. So, I would like to ask you when can one be sure > that the existence of the limit the quotient of derivatives guarantees > the existence of the original limit. > Paul === Subject: Partition of R Hello I was thinking about the interesting (and apparently difficult) problem of finding sets A and B that form a partition of R and are such that every interval of R contains uncountably many points of A and B. In other words, every element of R should be a condensation point of A and B. At first I considered the sets A = {x in R : x^n is rational} and B ={x in R : x^n is irrational}, n a natural >=2, but A is countable, because it's a subset of the algebraic numbers. Then, I came up with A = { x in R : |x|^a is rational} and B = {x in R : |x|^a is irrational}, where a>0 is irrational (if a is rational, then A is again a subset of the algebraics). I feel A and B form the desired partition if a>0 is irrational, but couldn't give a proof. I'd like someone to give a suggestion to prove or disprove my assumption. Maybe a possible proof requires continued fractions. I was told a possible partition is A = {Liouville's numbers} and B the complement of A (the set of Diophantine numbers, if I have the usual terminology). Where can I find a proof of this fact? Artur === Subject: Re: divide and conquer > There is some joke about college being the only place where you pay for > something, and then try as hard as you can NOT to get your money's > worth. I saw a lot of that in college myself. We had a rule that if a professor was more than a certain number of minutes late, the class would be dismissed. Whenever a professor was late, some students would be watching the clock and saying, Only two more minutes -- one minute -- thirty seconds -- let's go! My attitude was, Hey, I didn't join the work-study program to pay for *not* being in class! -- 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 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Do Physicists Understand Their Own Peer-Reviewed Literature? > It is not in the best interests of physicists to understand physics. > Understanding empties their rice bowl. A rice bowl kept to overflowing by the > taxpayer. Proof at http://www.thewebspert.com/cresswell/ Diagram 9-1. > So much for the 'Conservation of Energy'. Shall I hold my breath waiting to > hear from ya'll. First words at your website: A child may logically conclude that our solar system and its galaxy are effectively in a state of Perpetual Motion. The same mature child, schooled in an institutionalised scientific education will then logically conclude that this is impossible. The ëlaws of physics' say so. Oops. The same mature child, if actually understanding their education, would conclude nothing of the sort. The laws of physics forbid Perpetual Motion MACHINES of two kinds: First Kind: One which does work without the requirement for energy input. Second Kind: One which converts heat into another form of energy with 100% efficiency. Newton's First Law, far from forbidding perpetual motion, actually requires it. Your mature child would only need to get past Newton's First Law in, say, elementary school, before realizing what a bald misstatement of the laws of physics you have made. - Randy === Subject: Need equivalence (p->q) xor (q<->r) with only 12 letter and using operator and,or and not ..... === Subject: Spivak's Calculus Hi all, I am thinking of buying Michael Spivak's Calculus, but I am trying to save some money by buying a used one online. Does anybody know if the 2nd or 3rd editions are different from the 1967 1st edition? If so, are the changes significant? TIA, Lurch === Subject: Re: JSH: About time > I'm NOT a mathematician. You could have been if you had spent the last 8 years actually learning hte stuff that people have been suggesting you learn. Eight years is quite enough for a degree. Instead what do you have to show for it? V. === Subject: Re: Group generated by a and b ... >>group G = , some k in N with ba = ab^k ==> some n,m in N with ba^-1 = a^n b^m iff o(a) finite >>proof >N = { n in Z | n > 0 }. When k < 0, the proof bombs at > b^r a = ab^rk >because I don't have > b^-1 a = ab^-k Yes you do! ba = ab^k => (by inverting) b^-k a^-1 = a^-1 b^-1 => b^-1 a = ab^-k >><== let o(a) = r; ba^-1 = ba^(r-1) = a^(r-1) b^k(r-1) >>==> e = a^n b^m a b^-1 = a^n a b^km b^-1 = a a^n b^r >> where r = km - 1 >> a = aa^n b^r a = aa^n a b^kr >> e = aa^n b^kr = aa^n b^r; b^kr = b^r; e = b^(k-1)r >> b has finite order, when k /= 1 >> (aa^n)^-1 = b^kr; thus (aa^n)^-1 and a have finite order >> The case k = 1; ba = ab; ba^-1 = a^n b^m >> unresolved >> ba = ab => ba^-1 = a^-1b >> You have also overlooked the special case k = -1, because >> ba = ab^-1 => ba^-1 = a^-1b^-1 (just by inverting everything). >> This case gives r = 0 in your proof. >> And I guess k=0 is another exception: ba = a => ba^-1 = a^-1. >> But your result is correct for |k| > 1 ! >For k > 1. BTW, n,m > 0 was included in premise but only m > 0 used. For |k| > 1. It fails (and is false) for n = -1. Where is m > 0 used? You probably do need m and n nonzero to avoid trivialities. >For k = 1, is not the infinite dihedral group > { a^n, a^n b | n in Z } > with o(b) = 2, ba = a^-1 b >an example for which o(a) is infinite? For k=1, we have ba = ab and the group is abelian. The infinite dihedral group is an example where k = -1. . === Subject: Orthogonal MLS Codes A maximum-length sequence is a one-bit, periodic pseudo-noise sequence derived from GF(2^N) from a primitive polynomial. I would like to know if certain primitive polynomials generate pairwise-orthogonal or near-orthogonal codes. By orthogonal I mean that the periodic cross-correlation of the associated one-bit sequences is zero for all values of index between 0 and N-1. --Randy Yates === Subject: Taylor serie How to expand into taylor-serie about i=sqrt(-1) this complex function: F(z)=1/[(z^2)(z+i)] === Subject: Re: Partition of R Hello. > I was thinking about the interesting (and apparently difficult) > problem of finding sets A and B that form a partition of R and are > such that every interval of R contains uncountably many points of A > and B. A = {x in R; there exists a nonnegative integer k such that the decimal expansion of x contains the digits 1 et 2 only after the kth decimal} B = R-A. A is clearly uncountable, so is B... and every interval I of R (containing at least two distinct points) will contain uncountably many points of A and B. -- Julien Santini, Etudiant en licence .88 l' Universit.8e de Provence, France. === Subject: Re: Need equivalence > (p->q) xor (q<->r) > with only 12 letter and using operator and,or and not > ..... p or q or r or (p and q and not(r))... -- Julien Santini, Etudiant en licence .88 l' Universit.8e de Provence, France. === Subject: That Hilbert guy is giving me a lot of trouble I am familiar with Linear Algebra...eigenvalues and stuff. But only for finite dimensions. I know: 1) That eigenvalues are properties of linear transformations independent of the basis chosen. 2) Trace (the sum of the diagonal elements) is also the sum of the eigenvalues. 3) Determinant (that weird looking function) is also the product of the eigenvalues. 4) As eigenvalues are many in numbers, the collection of them knows much more than just the two of the properties of the matrix. I want to know: 1) How these ideas generalize to infinite dimension. The particular problem at hand is of course that I find it rather difficult to add infinitely many numbers (...computing the determinant! are you nuts!) 2) If there are ideas that are still important in the infinite case 3) Does having a complex linear space instead of a real one improves the situation 4) Functions like sin and cos have infinite polynomial expansion (how do you say it in standard language?)... So I might find a Fourier transformation helpful. But isn't the most likely situation be composed of infinitely many Fourier components too? And most importantly: 5) What is the best textbook (or something like that) that explaining these things that can be most easily swallowed === Subject: Re: Taylor serie > How to expand into taylor-serie about i=sqrt(-1) this complex function: > F(z)=1/[(z^2)(z+i)] (1) Remember partial fractions? Use them. (2) When in trouble, change the variable: substitute z=i+w, so that the expansion in powers of w takes place around 0 . Then, of course, hand in the result in terms of z again (w = z-i). Finally, it is a curious fact that the word series is already in singular, not plural, despite ending with s (it happened in Latin, and passed into English). The plural is also series, but this was not needed here. === Subject: Re: The Square Root of the Golden Section By an Iterative Method. > ... >N) Find : 1 /[Sin{Arctan [X(N-1)] }] = X(N) >By such successive iterations the value of this calculation tends to that of the Square Root of the Golden Section T, >[T = SQRT[(SQRT[5]+1)/2 ] >T=1,27201964951406896425242246173749 >This is, also , the Solution of : [1]/[Sin{Arctan[X]} = [X] , > ... If, starting with some value x0, successive iterations of a continuous function f converge to a value x, then f(x) = x (prove this). So if your values X(N) converge, they must converge to a solution of the equation f(x) = x, where f(x) = 1/sin(arctan(x)). The identity sin(arctan(x)) = x/sqrt(1 + x^2) is easy to establish (e.g., draw a right triangle with base 1, height x, and hypotenuse sqrt(1 + x^2)). So your equation is equivalent to the equation sqrt(1 + x^2)/x = x, This equation can be easily solved: mutliply by x, square, substitute y = x^2, and solve the resulting quadratic equation. You get x = sqrt( (1 + sqrt(5))/2 ). The question of whether or not the iterative sequence converges requires more analysis. Examining the graph if f(x) and using the Contraction Mapping theorem shows that it does (if you start with a positive number). The Contraction Mapping theorem is a very interesting generalization of this type of result.. You can scan this newsgroup or look at a book on real analysis for discussions of this theorem. Better yet, think about it yourself first (what properties of the function f will guarantee that the iterates converge?). Try this one: start with an arbitraty number x0, and repeatedly apply the cosine function (this is easy to do with a calculator). What happens? Does the result depend on the initial value? How can you characterize the result? John Mitchell >COROLLARY >------------------- > 1/ [Sin{Arctan[SQRT(X)]} =SQRT [1+{1/(X)}] >This is an IDENTITY (for X any positive real number fraction or integer). >(amphidromical result). === Subject: Re: That Hilbert guy is giving me a lot of trouble > I am familiar with Linear Algebra...eigenvalues and stuff. But only > for finite dimensions. I know: > 1) That eigenvalues are properties of linear transformations > independent of the basis chosen. > 2) Trace (the sum of the diagonal elements) is also the sum of the > eigenvalues. > 3) Determinant (that weird looking function) is also the product of > the eigenvalues. > 4) As eigenvalues are many in numbers, the collection of them knows > much more than just the two of the properties of the matrix. > I want to know: > 1) How these ideas generalize to infinite dimension. The particular > problem at hand is of course that I find it rather difficult to add > infinitely many numbers (...computing the determinant! are you nuts!) There is a special class of operators on Hilbert space (called trace class) for which the trace is defined. > 2) If there are ideas that are still important in the infinite case > 3) Does having a complex linear space instead of a real one improves > the situation > 4) Functions like sin and cos have infinite polynomial expansion (how > do you say it in standard language?)... So I might find a Fourier > transformation helpful. But isn't the most likely situation be > composed of infinitely many Fourier components too? > And most importantly: > 5) What is the best textbook (or something like that) that explaining > these things that can be most easily swallowed === Subject: Re: JSH: About time >My work is out there and rather easy to go over as can be seen at the >Hong Konk math site: Uh oh. Hong Konk? >> Hong Konk? You're sure it isn't Honk Honk? Eternal vigilance! The mistake is caught! James is duly mocked! > Typo. It should be Hong Kong. Ah. You sure it was just a typo? Not an essential error undermining your entire argument? Nice to see you're getting such helpful feedback here. === Subject: Re: Spivak's Calculus > Hi all, > I am thinking of buying Michael Spivak's Calculus, but I am trying to save > some money by buying a used one online. Does anybody know if the 2nd or 3rd > editions are different from the 1967 1st edition? If so, are the changes > significant? > TIA, > Lurch In the Preface to the second edition, he lists additions: appendices to many chapters on topics previously slighted; 160 new problems. The preface to the third edition lists: a new chapter on planetary motion; rearrangement of the material; elimination of incorrect problems. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Vedctor Calculus Question >A single equation, such as f2(x,y,z)=c2, can describe in a 3d space >a surface, possibly a plane, but not a line. > I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line, > last time I looked. > Lee Rudolph > I stand corrected, but it is not a very efficient way of doing > lines, and certainly not the standard way. But degenerate cases come up all the time. (Well, theoretically.) You have to keep them in the back of your mind. Infinitely thin cylinders, indeed. Jon Miller === Subject: Re: That Hilbert guy is giving me a lot of trouble > I am familiar with Linear Algebra...eigenvalues and stuff. But only > for finite dimensions. I know: > 1) That eigenvalues are properties of linear transformations > independent of the basis chosen. > 2) Trace (the sum of the diagonal elements) is also the sum of the > eigenvalues. > 3) Determinant (that weird looking function) is also the product of > the eigenvalues. > 4) As eigenvalues are many in numbers, the collection of them knows > much more than just the two of the properties of the matrix. > I want to know: > 1) How these ideas generalize to infinite dimension. The particular > problem at hand is of course that I find it rather difficult to add > infinitely many numbers (...computing the determinant! are you > nuts!) I don't recall computing any determinants for operators on Hilbert spaces. Making semi-infinite arrays of numbers seem to be less useful than matrices are for the finite dimensional case. > 2) If there are ideas that are still important in the infinite case Yes. Things get more complex. IIRC, when moving from finite spaces, a finite set of eigenvalues break into 1) discrete eigenvalues much like the finite dimensional case 2) continuous spectrum 3) a weird area where things are not quite everything you'd want in an eigenvalue > 3) Does having a complex linear space instead of a real one improves > the situation Yes. Complex numbers are algebraically complete. This helps. > 4) Functions like sin and cos have infinite polynomial expansion (how > do you say it in standard language?)... So I might find a Fourier > transformation helpful. But isn't the most likely situation be > composed of infinitely many Fourier components too? Yes. Your Hilbert space has infinitely many dimenisions. Why shouldn't you have a (countable) infinite number of coefficients to a spanning set? > And most importantly: > 5) What is the best textbook (or something like that) that explaining > these things that can be most easily swallowed The subject you are asking about is called spectral analysis and is a topic in functional analysis. A good starting point is Erwin Kreyszig _Functional Analysis with Applications_. It is one of my all time favorite math books -- clearly written and covering a lot of interesting material. Fora a bit more advanced, perhaps Peterson _Analysis Now_. For a very agebraic approach, try Naimark _Normed Rings_ (this makes a good complement to the analytic approach, but I wouldn't try it as the only text on the subject). There's a lot of others, I recommend hitting your library and pulling a half dozen or so likely looking texts. Browse them all for a week or two and return whatever you aren't finding helpful. -- === Subject: Re: Antidiagonal, Infinity > This page discusses a function continuous at all irrationals, > discontinuous at all irrationals: Are you sure about that? :-) > That means I still want to know that given an element, > that for the next element, what it is, or the probabilities of what > it could be. How about, just for once, entertaining the possibility that there may not be a next element after your favourite number? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Spivak's Calculus >a new chapter on planetary motion; >rearrangement of the material I thought that combination had been Abian's speciality. Lee Rudolph === Subject: Re: hw help -- continuity > [...] >>Proving a function continuous at any point >>proves it continuous at every point. >> f(x) = 0 when x <= 0 >> = 1 when 0 < x >> is continuous at 1. Thus it's continuous at 0? > But 1 is not any point, it is a specific point. > To restate less ambiguously: > If you prove f continuous at an arbitrary point, then it is > continuous at every point > What is wrong with the good old English word every ? Because when you say every, many beginning students try to look at all the points at once. I would say (to the beginners), Your opponent gives you a point. Can you prove that the function is continuous at that point? No matter what point? The catch is that when you say any point, a single point works for a counterexample, but a proof has to work for x no matter what x is. For some reason, many beginning students have trouble with this concept. They think it's a symmetric game. They have to be taught to think logically, preferably without losing their ability to think illogically. And, don't forget, some have to be taught to think. And you're not allowed to punt (although they are, and then they savage you on the teaching evaluations because they didn't learn anything). Jon Miller === Subject: 3D Surveyor's Formula (for volume)? Is there a version/generalization of the Surveyor's Formula for three dimensions? I'm imagining inputting a bunch of (x,y,z) points, and receiving as output the volume of the polyhedron thus delimited. Of course, there will be issues involving exactly _how_ the points are given - presumably it's more complicated than the 2D version - but it seems like it should be possible.... Anyone know of the 3D version? cdj === Subject: Re: Partition of R > -----Original Message----- > Conversation: Partition of R === > Subject: Partition of R > Hello > I was thinking about the interesting (and apparently difficult) > problem of finding sets A and B that form a partition of R and are > such that every interval of R contains uncountably many points of A > and B. In other words, every element of R should be a condensation > point of A and B. > At first I considered the sets A = {x in R : x^n is rational} and B > ={x in R : x^n is irrational}, n a natural >=2, but A is countable, > because it's a subset of the algebraic numbers. Then, I came up with A > = { x in R : |x|^a is rational} and B = {x in R : |x|^a is > irrational}, where a>0 is irrational (if a is rational, then A is > again a subset of the algebraics). I feel A and B form the desired > partition if a>0 is irrational, but couldn't give a proof. I'd like > someone to give a suggestion to prove or disprove my assumption. Maybe > a possible proof requires continued fractions. > I was told a possible partition is A = {Liouville's numbers} and B the > complement of A (the set of Diophantine numbers, if I have the usual > terminology). Where can I find a proof of this fact? > Artur You can do far better than that. R is a maximally resolvable space, i.e., R can be partitioned into c=2^w many disjoint dense subsets. Here is one way to do it. P = {Q + x | x in R} partitions R into 2^w counatble sets. Partition P into PP consisting of 2^w subsets each of size 2^w. Then { US | S in PP } satisfies the bill. The jist is partitioning 2^w into 2^w sets of size 2^w. Getting dense is easy. === Subject: Re: circle with two centers Let's change the problem: Keep the original description, >Consider a smooth surface, S, and two points, A & B on it. S contains >a continuous, closed curve, C, such that both A & B are equidistant >from every point on the curve C. >1. Does such a surface exist? If so, under what conditions? >2. If it does, can there be countably infinite such centers? >Uncountably infintely many? Allow closed surfaces, do not require that the points be the same distance from C. Here's the change: Define distance by geodesics. I.e., the disance between P and Q is the length of a shortest path on S from P to Q. What is the answer to question 2 now? How about just more than two points which are equidistant from every point on C? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Spivak's Calculus > Hi all, > I am thinking of buying Michael Spivak's Calculus, but I am trying to save > some money by buying a used one online. Does anybody know if the 2nd or 3rd > editions are different from the 1967 1st edition? If so, are the changes > significant? > TIA, > Lurch > In the Preface to the second edition, he lists additions: appendices > to many chapters on topics previously slighted; 160 new problems. The > preface to the third edition lists: a new chapter on planetary motion; > rearrangement of the material; elimination of incorrect problems. > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Do you feel it would be worth purchasing the 1st edition? It is only $20. Compared with the near $100 for the 3rd. I am just looking for a gentle introduction to Calculus/Analysis that I can use as self-teaching text. For my background, I find Rudin too terse. TIA, Lurch === Subject: Re: lopital's rule? > It's L'Hopital's rule (with a circumflex over the o). > is and how it should never be mentioned to impressionable undergraduates :-) Hmm. Granted it's dangerous in the wrong hands, it seems to me that it's better to teach safety than to eliminate the tool. But then, maybe my perceptions are colored by living in a country which allows every idiot and his cousin to own as many guns as they want. Jon Miller === Subject: Re: Spivak's Calculus >Do you feel it would be worth purchasing the 1st edition? It is only $20. >Compared with the near $100 for the 3rd. I am just looking for a gentle >introduction to Calculus/Analysis that I can use as self-teaching text. For >my background, I find Rudin too terse. Whoo-hoo. Gentle is not the word I'd have chosen first, or tenth, to describe Spivak's _Calculus_. The problems, in particular, tend to be brutal. Have you looked at Apostol's _Mathematical Analysis_? Or, for that matter, Courant's calculus textbook? I have never seen Rudin's calculus text, just (many many years ago) his _Real and Complex Analysis_, and slightly more recently a monograph or two on some topics in several complex variables. But I don't think Spivak is notably less terse than Rudin, and I do think that both Apostol and Courant are probably less terse while being quite as rigorous as necessary. Lee Rudolph === Subject: formula for roots of quadratic matrix equation? Content-Length: 217 Originator: rusin@vesuvius Is there a closed form formula for the roots of the quadratic matrix equation QAQ - BQ + C = 0 where A, B, C and Q are all symmetric nxn matrices, n>1? If not, how about for the special case B=I (identity matrix)? === Subject: Re: lopital's rule? >> many polemics from me on how evil and useless this rule >> is and how it should never be mentioned to impressionable undergraduates >:-) >Hmm. Granted it's dangerous in the wrong hands, it seems to me that it's >better to teach safety than to eliminate the tool. >But then, maybe my perceptions are colored by living in a country which >allows every idiot and his cousin to own as many guns as they want. When L'Hopital's rule is outlawed, only outlaws will use L'Hopital's rule. Lee Rudolph === Subject: Re: Spivak's Calculus >Do you feel it would be worth purchasing the 1st edition? It is only $20. >Compared with the near $100 for the 3rd. I am just looking for a gentle >introduction to Calculus/Analysis that I can use as self-teaching text. For >my background, I find Rudin too terse. > Whoo-hoo. Gentle is not the word I'd have chosen first, or tenth, > to describe Spivak's _Calculus_. The problems, in particular, tend > to be brutal. > Have you looked at Apostol's _Mathematical Analysis_? Or, for that > matter, Courant's calculus textbook? I have never seen Rudin's > calculus text, just (many many years ago) his _Real and Complex > Analysis_, and slightly more recently a monograph or two on some > topics in several complex variables. But I don't think Spivak > is notably less terse than Rudin, and I do think that both > Apostol and Courant are probably less terse while being quite > as rigorous as necessary. > Lee Rudolph Lurch === Subject: Re: recurrence relation question >>how do i go about solving a recurrence relation when the function T(n) >>is multiplied by n? such as the following: >>2n*T(n) + 2nT(n-1) - 2T(n-1) = 2^n? >>What tricks would one use? One trick is to use generating functions. Let's denote by T generating function of T(n). Then we get relatively pretty differential equation for T: dT/ds+d(sT)/ds-T=1/(1-2s) Now, solve for T and Voila! Simple as a pie! === Subject: Laurent expansion, anybody? I'm sitting here and try to solve a task in my book. The problem is that when asked to expand to laurent series i get something like Sum(something, m=0, infty) while, according to my best knowledge, it should be from -infty to +infty. So, my question is: does every laurent expansion demands that the limits of the sum are positive/negative infinity? -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: Joe Uptaught (Was Re: David Ullrich on Identity) A propos, here are cites from Pierre Bourdieu's > _Language & Symbolic Power_ (the titles are mine). > Enjoy! > The Social Conditions for the Effectiveness of Ritual Discourse Heretical Discourse The *Skeptron* Symbolic Power & the Symbolism of Power Collusion in the 'Hood For groups united by some form of collusion (such as sets of *colleagues*), it is a fundamental imperative to maintain *discretion* about, to keep *secret*, everything which concerns the intimate beliefs of the group. They fiercely condemn manifestations of cynicism displayed to the outside world, even though such manifestations are quite acceptable among *initiates* because they cannot by definition affect the fundamental belief in the value of the group--a certain free-and-easy attitude to values is often experienced as a supplementar proof of their value. . . But groups are hardly less mistrustful of those who, taking proclaimed values too seriously, refuse the compromises and shady deals which are the condition of the real existence of the group. (Pierre Bourdieu, _Language & Symbolic Power_, p. 180) === Subject: shorest path in polygonal enviroment I am looking for some source code that can compute a shortest ( or near shortest) path between starting point S and destination point T in a polygonal environment, which may have polygonal obstacles ( or holes) in it. Can someone give me the code, or any suggestions where to find it? swh === Subject: Re: Laurent expansion, anybody? > I'm sitting here and try to solve a task in my book. The > problem is that when asked to expand to laurent series > i get something like > Sum(something, m=0, infty) > while, according to my best knowledge, it should be > from -infty to +infty. > So, my question is: does every laurent expansion demands > that the limits of the sum are positive/negative infinity? Not every coefficient is necessarily nonzero. At a regular point, you get only the nonnegative powers (so the Laurent series is actually a Taylor series). At a pole you have only finitely many negative powers. At an essential singularity, you have infinitely many negative powers. === Subject: Re: Orthogonal MLS Codes yates@ieee.org (Randy Yates) asks: >A maximum-length sequence is a one-bit, periodic pseudo-noise sequence >derived from GF(2^N) from a primitive polynomial. I would like to know >if certain primitive polynomials generate pairwise-orthogonal or >near-orthogonal codes. By orthogonal I mean that the periodic >cross-correlation of the associated one-bit sequences is zero for >all values of index between 0 and N-1. The answer is No, maximum-length sequences cannot be orthogonal. It is not too hard to show that if we think of the cross-correlation values as a vector in N-dimensional complex space: C[x,y] = [ C[x,y](0), C[x,y](1), .... , C[x,y](N-1) ], then ||C[x,y]||^2 = = where || ||^2 and < , > are the usual squared norm and the usual inner product in complex N-space.. Since C[x,x] = C[y,y] = [N, -1, -1, ... , -1], the right side of the above equation has value N^2 + N - 1 > 0, showing that at least one cross-correlation value C[x,y](j) exceeds sqrt(N). In fact, much more is known about cross-correlation of maximum-length sequences. A (somewhat dated) survey is in the May 1980 issue of the Proceedings of the IEEE... Hope this helps. -- .-. .-. .-. .-. .-. .-. .-. / D I / L I / P / S A / R W / A T / E `-' `-' `-' `-' `-' `-' === Subject: Re: quantum echo Are you saying it is possible for something to exist at location A and > location B at the same time? I thought it was inference by the rule of > non-contradiction. I have no idea whether this is possible or not. My suggestion that quanta may satisfy '~(x=x)' is meant to translate > into logical terms the claim by some physicists that quanta > 'lack individuality' (Heller suggests that quanta 'lack haecceities'). > The scenario with cakes-to-be suggests that the behaviour of other > entities without individuality resembles that of quanta in the > relevant respect. Heller's description of that behaviour follows > my sig. Metaphysical Background, Thomas McTighe asserted that the quiddity of a > thing is nothing other than unity itself. Hence, by virtue of its positive > content, the sun differs not at all from the moon or any other particular > thing. The diversity which is exhibited by the natural world is merely the > product of accidental differences; no object possesses any specific form > which interposes itself between a particular existing thing and the source > of their being e.g. the Absolute.15 All individual entities are nothing more > than differing contractions of the whole devoid of any being of their own. > ...because the restricted quiddity of a thing is the thing itself. > http://www.crvp.org/book/Series01/I-10/chapter_ii.htm > I take a quiddity to be a thing's *suchness* and a haecceity to be its > *thisness*. A *thisness* I take to be, as Robert Adams does, the > property of self-identity, although I disagree with an assumption > which might be read into Adams, that an individual can lack > self-identity and yet possess the property of being some > individual or other). > In Primitive Thisness and Primitive Identity (_The Journal > of Philosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), Adams > A thisness is the property of being a certain particular > individual, not the property of being some individual or other, > but my property of being identical with me, your property of > being identical with you, etc. These properties have recently been > called 'essences', but that is historically unfortunate, for essences > have normally been understood to be constituted by logical properties, > and we are entertaining the possibility of nonqualitative > thisnesses. In defining 'thisness' as I have, I do not mean > to deny that universals have analogous properties--for example, > the property of being identical with the quality red. But since > we are concerned here principally with the question whether > the identity and distinctness of individuals is purely > qualitative or not, it is useful to reserve the term > 'thisness' for the identities of individuals. > It may be controversial to speak of a property of being > identical with me. I want the word 'property' to carry as > light a metaphysical load here as possible. 'Thisness' > is intended to be a synonym or translation of the traditional > term 'haecceity' (in Latin, 'haecceitas'), which so far as I > know was invented by Duns Scotus. > Like many medieval philosophers, Scotus regarded properties as > components of the things that have them. He introduced > haecceities (thisnesses), accordingly, as a special sort > of metaphysical component of individuals.[4] I am not proposing > to revive this aspect of his conception of a haecceity, because > I am not committed to regarding properties as components of > individuals. To deny that thisnesses are purely qualitative > is not necessarily to postulate 'bare particulars', substrata > without qualities of their own, which would be what was left > of the individual when all its qualitative properties were > subtracted. Conversely, to hold that thisnesses are purely > qualitative is not to imply that individuals are nothing > but bundles of qualities, for qualities may not be components > of individuals at all. (pp. 6-7) Acceptance of haecceities is a distinctive feature of the thought of many followers of Scotus, though there are some sixteenth-century scholastics who accept haecceities without accepting many other distinctively Scotist teachings. Having said this, some early followers of Scotus reject haecceities and the theory of the common nature altogether, and of those who accept haecceities, some found the correct understanding of the nature of the distinction between an individual's nature and its haecceity a troublesome matter. One of the earliest Scotists, Francis of Meyronnes, writing his commentary on the Sentences around 1320, accepts the theory of the non-numerical unity of common natures, and the claim that individuation is by haecceity. But he holds that it is inappropriate to talk of a formal distinction in this context. Formal distinction obtains only between things that have some sort of quidditative content. Haecceities have no such quidditative content, and thus cannot be formally distinct from their nature. Rather, a haecceity is modally distinct from its nature. A modal distinction, according to Meyronnes, obtains between a thing and an intrinsic mode of that thing, where an intrinsic mode is something which when added to a thing does not vary its formal definition . . . since it does not of itself imply any quiddity or formal definition. A haecceity does not affect a thing's kind; it is thus an intrinsic mode of the thing. It may look as though this is just a terminological shift, but it is not so in at least the following way: a modal distinction is a lesser kind of distinction than a formal distinction. Formal distinctions obtain between genus and specific difference; thus, the difference between species/nature and haecceity, for Meyronnes, is less than the difference between genus and difference. Scotus, contrariwise, makes no such distinction between degrees of difference in this context (for this contrast between the two thinkers, see Dumont [1987], 18). Still, without some principled way of spelling out degrees of difference, this contrast between Scotus and Meyronnes amounts to nothing of any philosophical interest. To this extent the difference between the two thinkers might as well be merely terminological, and Meyronnes needs to do more work if he is to make any significant philosophical point here. http://setis.library.usyd.edu.au/stanford/entries/medieval-haecceity/ http://plato.stanford.edu/entries/francis-marchia/ ...center, content, core, essence, gist, heart, heart and soul, inwardness, kernel, marrow, matter, meaning, means, meat, message, nitty-gritty, nub, pith, subject matter, sum, divagation, drift, drivel, element, entity, entry, excursus, extension, ferment, fluid, foamentation, food, fuel, goo, gook, grinding, guck, guidance, gunk, haecceity, hokum, humectant, humor, humour, hydrocolloid, hypostasis, idea, implication, import, import, info, information, inhibitor, inoculant, inoculum, insertion, instruction, interpolation, jelly, latent content, leaven, leavening, litter, living substance, lysin, material, meaning, meaninglessness, medium, memorial, mental object, micronutrient, mixture, muck, mush, narration, narrative, nonsense, nonsensicality, nutrient, offer, offering, ooze, opinion, packaging, parenthesis, part, petition, philosopher's stone, phlogiston, physical thing, poison, poisonous substance, portion, postulation, promotion, promotional material, propellant, propellent, proposal, protoplasm, publicity, pyrectic, pyrogen, quiddity, quintessence, reference, refrigerant, refusal, reminder, request, residue, respects, sediment, sensationalism, shocker, significance, significance, signification, slime, sludge, solid, solute, solvate, statement, story, strain, stuff, stuff, subject, submission, substrate, system, tale, tenor, theme, thought, topic, treacle, undercurrent, undertone, view, wherewithal, wit, witticism, wittiness, ylem > Note > 4. es Duns Scotus, _Quaestiones in libros metaphysicorum_, > VII. xii. schol. 3; cf. _Ordinatio_, II.3.1.2, 57. I am indebted > to Marilyn McCord Adams for acquainting me with these texts and > views of Scotus, and for much discussion of the topics of this > paragraph. > Aristotle and Aquinas and Scotus and Bonaventura all believed that human > minds can conceive and express the intelligibilities or quiddities of things > and their properties, intelligibilities that are not simply mind-dependent. > We can capture in thought and language the actual natures of things, > spelling out their genera and specific differences. Definition brackets or > delimits for us as knowers just what it is we attempt to understand and > nothing else. The mind-independent thing-substance or the characteristics > that we are attempting to define measure the epistemic correctness of a > definition. Such real (as opposed to nominal) definition relies on the > intelligible and perceptible characteristics thing-substances exhibit to > perception and thought for understanding what they are and for picking out > individuals of a type. In this way the epistemological realism of the > definition corresponds to an ontological realism of actual formal features > in mind-independent entities. > http://www.sunysb.edu/philosophy/faculty/lmiller/Delinonaliud.htm > After all, the three tenets that largely define Nicholas's 'metaphysic of > contraction' seem altogether remote from Anselm's Scholasticism. For Anselm > has no use for the triad of notions (1) that there is an infinite > disproportion between the Creator and His creatures, (2) that, therefore, > finite minds can never positively know what God is, given the alleged ground > (3) that He is the Coincidence of opposites, i. e., is undifferentiated > 'Being' itself, which, with respect to its Quiddity, can never be conceived > by anyone except itself. > http://www.cla.umn.edu/jhopkins/CusaAnselm.pdf === Subject: Enhanced Presentations with DualPoint Project two PowerPoint presentations simultaneously on the same monitor or projector from a single PC. Enhances presentations, helps to keep the attention of your intended audience. A MUST for anyone who uses MS PowerPoint! Free trial at www.dualpoint.net Download DualPoint Today! === Subject: Re: Partition of R > ello > I was thinking about the interesting (and apparently difficult) > problem of finding sets A and B that form a partition of R and are > such that every interval of R contains uncountably many points of A > and B. In other words, every element of R should be a condensation > point of A and B. A partition of the set of reals into the set of rationals and the set of irrationals does exactly that. === Subject: Re: Partition of R > A partition of the set of reals into the set of rationals and the > set of irrationals does exactly that. such that every interval of R contains uncountably many points of A and B. === Subject: Re: Quanta and Cakes Nice to read this. I do not understand almost nothing of this thread. However, what is the question? Is it this: What is the probability of finding some of the Take a look into the box and find what is the actual configuration. Do it many times. After, calculate the probability of given configuration (= number of this particular configurations / number of all measurements). You will find that probability of ll is 1/4 probability of rr is 1/4 probability of lr is 1/2 Is this the point of this thread, right? Palo >> I should have asked in what respect *correct reasoning* about >> quanta resembles such reasoning about cakes-to-be, but >> differs from such reasoning about (extant) 'medium size >> dry goods'. >> My answer to this question would be: correct reasoning about >> medium size dry goods should take these as satisfying >> both the right and the left sides of the following bi- >> conditional; >> AxAy(Az(z in x <-> z in y) -> x=y) <-> Ax(x=x) >> [Identity of Indiscernibles] [Reflexive Law of Equality] >> while correct reasoning about quanta--and cakes-to-be--should >> take these to satisfy neither. > Sorry, you have obfuscated your meaning beyond my ability to extract > any sense from it, or even to guarantee that the language in use is > still English, so I shall withdraw from the discussion. > xanthian. === Subject: Re: a puzzle related to artinian group >> Suppose n people sit around a table and n-1 cards are dealt to them. >> There is no asumption on the number of cards a player receive. In each >> round, all players with 2 or more cards pass one card to the left and >> one card to the right. Prove that eventually, all players but one have >> exactly one card. > If N is the maximum value of a cell state, then the maximum value of the > states in the next generation can be at most N, unless N = 2 and then a > maximum of 3 is possible in the next generation. It follows that there are > there are only a finite number of possible global states, so eventually > it is periodic. In fact, one can prove a stronger theorem: if there are n players and less than 2n cards, each player will at some point hold 0 or 1 cards. An obvious corollary to the above is that in the eventual static or cyclic state no player can hold more than 3 cards at any time. To prove the theorem above, one must first note that only players holding 0 or 1 cards can gain more cards on the next turn. Therefore, if a player ever loses a card, he can only regain the same number of cards he originally had by going down to 0 or 1 cards and then back up. We already know the game will eventually settle into a repeating cycle (of 1 or more turns). If a player has N cards at some point in the cycle, he must also have N cards one cycle later. This means that either he must never lose cards during the cycle or the number of cards he holds must at some point be 0 or 1. Assume the player has more than one card and never loses any cards. This means the players on either side of him must also always hold more than one card. By induction, we can show that _all_ players must always hold more than one card. But this is impossible if there are less than 2n cards. Therefore the assumption must be false, and each player must hold less than two cards at some point in the cycle. -- Ilmari Karonen If replying by e-mail, substitute .net for .invalid in address. === Subject: Re: Orthogonal MLS Codes > A maximum-length sequence is a one-bit, periodic pseudo-noise sequence > derived from GF(2^N) from a primitive polynomial. I would like to know > if certain primitive polynomials generate pairwise-orthogonal or > near-orthogonal codes. By orthogonal I mean that the periodic > cross-correlation of the associated one-bit sequences is zero for > all values of index between 0 and N-1. Based on the first response to your posting, apparently not. You might find the recoding methods Walsh Codes used in CDMA (Code Domain Multiple Access) data transmission to be of interest, as they are used in combination with pseudo-random bit sequences to make signals sharing a bandwidth orthogonal to one another: by chopping the original signal bits into many chips and broadcasting those instead. xanthian. -- === Subject: Re: 0! = 1 > A N Niel scribbled the following: >> If I know my algebra, then an identity for an operation (let's call >> it #) is such an i, that for all x, x#i=x. But there is another concept >> too: such a j, that for all x, x#j=j. For multiplication, this is 0. >> For addition, there doesn't seem to be one. What is this concept called? > It's called a zero. > So intuitively, if #' is a repetitive #, which means that x#'i is x#x > performed (i-1) times in succession, then the zero of #' is the > identity of #? For example multiplication is repetitive addition. 0 is > the additive identity and the multiplicative zero. If we switch the > definition of a zero around so that it means that for all x, j#x=j, then > we get that exponentation is repetitive multiplication, and that 1 is > the multiplicative identity and the exponentative zero. > Is this making any sense? For addition, it would be infinity. Which means taking our group and making it a non-group. However, people are more than happy with rings where the elements don't form a multiplicative group, so groups aren't the only path to eternal happiness. How does saturation sound? e.g. in saturating 8-bit unsigned arithmetic 0xFF + c = 0xFF for all c. Phil === Subject: Test Test === Subject: Re: Partition of R > Hello > I was thinking about the interesting (and apparently difficult) > problem of finding sets A and B that form a partition of R and are > such that every interval of R contains uncountably many points of A > and B. In other words, every element of R should be a condensation > point of A and B. Correction. In my previous post, I overlooked the need for both sets to be uncountable. I was misreading condensation points as limit points. Sorry. One construction of such a partition is based on having a basis of the real as as a vector space over the rationals, assuming a suitable axiom of choice. Such a a basis is uncountable, and if it is split into 2 uncountable subsets, say C and D, then let A = span(C) and B = span(D), and you are home free, since the rational multiples of any non-zero real are dense in R. === Subject: Re: relatively prime Originator: jeyadev@kaveri >Hello there! >The probability of any two random integers being relatively prime is >.6079271018540266 ..., which is exactly equal to 6/(pi^2). I am not >sure where you could find a proof for this, but I imagine there's one >somewhere on the internet. I found this on the following URL: At http://www.physics.harvard.edu/probweek/sol44.pdf >http://primes.utm.edu/curios/page.php?number_id=1326 -- Surendar Jeyadev jeyadev@wrc.xerox.bounceback.com Remove 'bounceback' for email address === Subject: Re: relatively prime > What is the probability of two randomly chosen positive integers to be > relatively prime. What if you choose n numbers? Depends a good deal on what you mean by randomly chosen positive integers. There is no way of choosing positive integers in such a way that all positive integers have an equal chance of being picked, which is the usual meaning of at random. A common solution is to consider only those postive integers up to some limiting value n, and then consider what happens when n is allowed to get arbitrarily large. If there is a limiting value to the desired probability, you are home free, but if there isn't, you have a serious problem. === Subject: Re: how many resolutions? > Ok. Let's take standart right triangles : 3, 4, 5 > Could we find other integers than 3 and 4 such that a^2 + b^2 = 5^2? > Or the same questions for the triangle 15, 8 , 17? > The answer is no, I guess. > Does it mean that any triangle homothetic to those above is also uniquely > determined? I mean that the triangle homothetic to (3,4,5) with a > coefficient 17 is > 51, 68, 85 and the triangle homothetic to (15, 8 , 17) with a coefficient 5 > is 75, 40, 85. And since 85 is a product of 5 and 17 it should not be (but > there are) other solutions. > For example there is solution (84, 13, 85) - this triangle is homothetic to > no one above. > Why? What happens? An ordered set of three positive integers (a,b,c) is called a Pythagorean triple when a^2 + b^2 = c^2. Such a set is called primitive when hob, and c have no common factor greater than 1, so (3,4,5) is a primitive Pythagorean triple but (6,8,10) is a Pythagorean triple but is not primitive. Given positive integers u and v having no common factor greater than 1 and with u > v, then both (u^2 - v^2, 2*u*v, u^2 + v^2) and (2*u*v, u^2 - v^2, u^2 + v^2) are primitive Pythagorean triples, and every primitive pythagorean triple is of one of these two forms. And every Pythagorean triple is an integer multiple of some primitive such triple. === Subject: Re: Taylor serie >How to expand into taylor-serie about i=sqrt(-1) this complex function: >F(z)=1/[(z^2)(z+i)] complete solution sent to the e-mail address you had the integrity to provide, === Subject: Re: prime numbers factoring > The repunits R(311), R(509), R(557), R(617), R(647) and R(991) are apparently known to be composite (I have not seen this anywhere) but no factor has been found yet for those numbers. In fact, that last repunit R(P) completely factorised is R(733). > It is true that (relatively) small prime factors can be found for some of these repunits such as R(359) and R(659), but how can they be sure that R(311) and R(991) are not prime? How can they even know where possible factors might be?? Numbers of <1000 digits can be extremely easily tested for primality. See the section on proving primality at http://primepages.org/ . So it's perfectly possible that there are composite repunits with no known factors. Every single repunit < R(65536) has now been tested for small factors up to 2572144963121 (not tested individually, but sieved, as they sieve quite easily). AFAIK there are a whole bunch which have no known factors. Phil === Subject: Re: Use of variable independence, core error > I've found that there is a definition error in core mathematics, and > I've given the math necessary to understand the problem; however, I > *still* see posters arguing, and it occurs to me that I need to show > you why what they're saying is ludicrous, if any progress is to be > made. > First I have a somewhat forbidding expression: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > Rather than fiddle with it, I instead focus on what does NOT have m as > a factor, which is easily done by setting m=0, which gives > P(0) = u^2 f^2 ( 3x + uf) > which is much nicer. > Several posters have, however, continually tried to raise the > possibility that P(0) is in fact STILL a function of m, in a rather > odd way, as they've focused on factors of P(m), as I use the factors > g_1, g_2, and g_3, where > P(m) = g_1 g_2 g_3 > and split up P(0) with them, so that I also have that > when m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf > and I add in the requirement that f be coprime to 3, x and u. > Now a neat opportunity is revealed by the use of these terms > *independent* of m, as I can easily see that f^2 divides off of P(m) > and P(0) in a straightforward way. > I've isolated terms independent of m, so without regard to m, I know > that two and only two of the g's have a factor that is f, when f is > coprime to 3, x and u. This is not true in general, as others have pointed out. Consider, for u = 1, f = 2, we have P(1) = 28x^3 - 36x + 8 = (-2x + 2)(((-7 + sqrt(105))/2)x + 2)(((-7 - sqrt(105))/2)x + 2) Which one of ((-7 + sqrt(105))/2)x + 2 and ((-7 - sqrt(105))/2)x + 2 has 2 as a factor and which one doesn't? Even if you weaken the condition of divisibility to non-coprimeness, you're still stuck. The problem with core mathematics you perceive isn't that at all. The problem is that you *want* two and only two of the g's to have a factor of f because you need that result for your FLT argument. You don't have that result in general, as numerous posters have pointed out. However, you have convinced yourself that your P(0) argument is so compelling that it *must* be correct, leading you to perceive a disconnect between the counterexamples that have been posted and what you *want* to be with mathematics itself, wherein your argument and the counterexamples are somehow both true. It just doesn't work that way, James. Note, though, that you're not guaranteed to be going down a blind alley here. The values of f, m, and u are not independent of each other--there are complicated connections among them in the context of your FLT argument. I'm certainly not prepared to assert that your two and only two factors result isn't true for the particular values of f, m, and u that arise in your FLT argument. However, it's your responsibility to show that that's the case. Rick === Subject: Re: Effective inverse of a matrix... Can't you just write P + cQ = (I + cQP^{-1})P ? Then (P+cQ)^{-1} = P^{-1} (I + cQP^{-1})^{-1}, which reduces you to the case you know how to do, namely I+cR for a matrix R. (I'm not sure where the positive definiteness is coming in here, it looks like you just need P invertible and for c to avoid being the negative of any of the eigenvalues of QP^{-1}.) BTW, you probably mean to say efficient, rather than effective. > Given the matrix P + cQ, where P and Q are known positive definite > matrices, and c is a positive scalar. I want to compute the inverse of > P + cQ for various values of c as effectively as possible, by exploiting > that I know P and Q beforehand. > For instance, if P is the identity matrix, and VLV' is the eigenvalue > decomposition of Q, then I can compute the inverse as V(I +c L)^{-1}V', > and only have to do some scalar inversions (more effective methods might exist). > Any hints? I have tried using the matrix inversion lemma, but it didn't > seem to help me. > Lars === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind > 1. Mathematics is the science in which we make something out of > nothing. > Wrong. Mathematics is built on the 13 Axioms, which are not nothing. The number of axioms vary from one axiom system to another. It is possible, though, to find an axiom system, say ZFC, in which most mathematical proofs can be formalised (well, actually, if you do not mind reexpressing theorems so that they take the form Formal system F proves that T, a much weaker system than ZFC suffices). Now you say that axioms are not nothing. Perhaps what you mean is that an axiom is an assumption, and that an assumption stands in need of justification. In general, axioms are not the kind of things that may turn out true or false. Rather, axioms are a means for specifying things we are interested in studying. What we may need to justify is that we find some things in mathematics more interesting, and more worthy of study, than others. What happens if we remove all the axioms of an axiom system? What we get are functions and relations which are not restricted in any way, but such a system is not interesting at all. However, adding restrictions to the functions and relations of such a system is a good (though by no means the only) way of defining mathematical objects. So is it true that in mathematics we make something out of nothing? Well, elsewhere on this thread Charlie Volkstorf clarified himself by saying that mathematics [...] does not rely on our observing how things work in our universe, which is correct. Mattias === Subject: Re: lopital's rule? >... >I shall go and boil my head. > That seems a little extreme. But if you find it helps > with the math let us know. It will shut me up, which wouldn't be a bad thing. -- G.C. === Subject: Re: Laurent expansion, anybody? > At a pole you have only finitely many negative powers. At an > essential singularity, you have infinitely many negative powers. Got it. I didn't realize the very meaning of that until i read it from you. Some nails have to be hammered more than once... :) By the way - i haven't been quite successfull finding a function that would give an essential singularity. It might be that i have missunderstod something more. Any examples for that? -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: Orthogonal MLS Codes >Based on the first response to your posting, apparently not. >You might find the recoding methods Walsh Codes used in CDMA >(Code Domain Multiple Access) data transmission to be of interest, >as they are used in combination with pseudo-random bit sequences to >make signals sharing a bandwidth orthogonal to one another: Walsh code sequences are equivalent to a fixed permutation being applied to X, TX, T^2X, ... , T^{N-1}X where X is a fixed maximum-length sequence and T is the cyclic shift operation. Walsh code seqeunces are orthogonal to one another (inner product is zero), but do not satisfy the desired property of having the cross-correlation equal zero for all shifts, that is, it is not true that = 0 for all k, 0 < k < N. -- .-. .-. .-. .-. .-. .-. .-. / D I / L I / P / S A / R W / A T / E `-' `-' `-' `-' `-' `-' === Subject: Re: Question about the L'Hospital Rule >Regarding the L'Hospital Rule, it has been found some examples where >the limit of the quotient of the derivatives exists without existing >the original limit. So, I would like to ask you when can one be sure >that the existence of the limit the quotient of derivatives guarantees >the existence of the original limit. Well you must have a calculus book or you wouldn't be asking about L'Hopital's rule. Here's a general hint: When you're reading a book, even a math book, you should look at the words surrounding the formulas instead of just the formulas. Your calculus book _explains_ when the rule is going to work. (For example, if f -> 0 and g -> 0, and if f'/g' -> L then f/g -> L. Also if f -> infinity and g -> infinity.) >Paul ************************ David C. Ullrich === Subject: Re: Question about the L'Hospital Rule >Regarding the L'Hospital Rule, it has been found some examples where >the limit of the quotient of the derivatives exists without existing >the original limit. So, I would like to ask you when can one be sure >that the existence of the limit the quotient of derivatives guarantees >the existence of the original limit. Practically it's much to remember for nothing; if you know how to do a series expansion then youll never need this pseudo-theorem. === Subject: Re: Boolean Algebra - Arithmetic Relationship > Allow me to clarify my inquiry, > 1)Is all of mathematics reducible to the simple logic presumption that > if something is true it cannot be false?? It is true that p -> ~~p which might be a formalized version of if something is true it cannot be false. And this is an axiom of some propositional calculi. But there must be _much_ more than that. Propositional calculus alone is not enough, and p -> ~~p is not enough for propositional calculus. (Iirc it requires at least three propositional variables to axiomatize pc, and at least two rules.) > 2)Is all of mathematics inter-linked?? (Geometry, Calculus, and > Algebra reducible to Arithmetic etc.) Or are there islands of > mathematical understanding off on there own? All (or almost all) is deducible from set theory. I say almost all because any one set theory, if it is consistent, will be incomplete. Each branch of mathematics can be founded (in as much as any of mathematics can be founded at all) on, say, Zermelo-Fraenkel set theory. Calculus can be reduced to arithmetic in the sense that from Peano Arithmetic one can deduce the properties of the real numbers. Certain geometries can be reduced to properties of ordered n-tuples of real numbers, but not all of them (ordered n-tuple can be defined in set theory). I'm not sure that one talks about (any parts of) algebra being reduced to arithmetic. > 3)Can all known Mathematical Notation and Symbolic manipulation be > modeled by a turing machine?? If it's recursive it can be represented by a Turing Machine. > Could a symbolic respresentation of a > turing machine therefore serve as a master symbolic grammar?? Perhaps > difficult to work with but, all symbolic notations could be reduced to > it. > I recognize the impossibility of storing a irrational number in a > finite number of bits; However, our brains have concept of PI and are > yet finite. We don't have to know all of numbers behind the decimal > point to understand PI's implications. That might be true, it might also be true that even if we knew all of pi's digits we wouldn't know all there is to know about pi. Anyway knowing is a psychological, rather than a logical matter. > Again forgive my naivety if my questions seem trivial. > -Steve -- G.C. === Subject: Re: 0! = 1 Loser warning This guys a loser and a half, follows my posts in other groups and gets everyone to killfile me, don't reply to this he probably can't see it. > e.g. in saturating 8-bit unsigned arithmetic 0xFF + c = 0xFF for all c. > Phil the fat head phil === Subject: The physical universe: an abelian group? If we define multiplication of two bits of matter/antimatter as connecting them (a*b = a connected to b)... associativity and commutativity are obvious, it certainly seems like closure is obvious (how can you connect two pieces of matter/antimatter to obtain something out of this universe?), the unit is vacuum and as for inverse... inverse of a piece of matter is an equal piece of antimatter, and vice versa. It seems to me like this very definitely makes the physical universe an abelian group. Am i correct? Something I was thinking about... if we do allow the physical universe, ie the set of all matter and antimatter and the vacuum, to be an abelian group, call it U, then it seems to me that if someone were raised from birth percieving, instead of U, some set S which is isomorphic to U, they would never ever be able to tell that they were unnormal. (As an example: if, from birth, you saw everything upside down. You would think it were perfectly normal and what's more, it would be impossible to diagnose). Am i mistaken? What sort of subgroups of the physical universe could we come up with? Perhaps, for example, the subgroup of all transparent matter/antimatter? The subgroup of all matter/antimatter which does not contain some basic element (an element that can't be created by combining different elements)? Alot of things seem like candidates, except that all these subgroups necessarily must contain the vacuum (this rules out, for example, a subgroup of all organic, or inorganic, material) === Subject: Re: quantum echo Immortalist: Stop crossposting this to sci.physics.relativity >> >> Are you saying it is possible for something to exist at location A [*SNIP*] === Subject: Re: Laurent expansion, anybody? > By the way - i haven't been quite successfull finding a function > that would give an essential singularity. It might be that i have > missunderstod something more. Any examples for that? e^(1/z). === Subject: How many groups are there... I know well how absurd it is to try to construct the set of all groups. Worry not; that is not what I am asking. Rather, what I am asking is, can one construct the set of all *isomorphically distinct* groups? IE, a set S of groups such that: 1. for any given group g, there exists a group h in S such that g is isomorphic to h 2. for any groups a, b in S, a and b are either identical, or non-isomorphic If so, what is the cardinality of this weird set? (Note that in actuality, S would of course have to be a set of group-and-multiplication pairs, since some sets are groups under more than one form of multiplication...) === Subject: Re: Need equivalence > (p->q) xor (q<->r) > with only 12 letter and using operator and,or and not > ..... (p/~q/~r) / (~p/~q/r) / (q/~r) Which can be made a little shorter but messier by factoring. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Minimal Graph, Four Color Theorem > yeah; HSJ just won't admit it! I hesitate to ask, but what is HSJ? === Subject: Re: How many groups are there... >I know well how absurd it is to try to construct the set of all >groups. Worry not; that is not what I am asking. Rather, what I am >asking is, can one construct the set of all *isomorphically distinct* >groups? There is no such set. The collection is too big. It contains sets of any cardinality; for finite cardinal n, take the cyclic group of order n. Given any infinite cardinal k, the group (Z/2Z)^(k) (the direct sum of k copies of the cyclic group of two elements) has cardinality k (it consists of all k-tuples of 0's and 1's which have only finitely many 1's, added coordinate-wise). Since no set can contain sets of each cardinality (it would imply that the collection of all cardinals is a set), no such set can be constructed (even assuming the axiom of choice, which you would need in order to select one group from each isomorphism class. === Subject: Re: shorest path in polygonal enviroment swh, You might try searching Google for euclidean shortest path (include the quotes if you like). === Subject: Re: Laurent expansion, anybody? >By the way - i haven't been quite successfull finding a function >that would give an essential singularity. It might be that i have >missunderstod something more. Any examples for that? Basically functions where you end up with increasing exponents in the denominator so that the series expansion looks like: ... + c_n / (z-z0)^n + c_(n+1) / (z-z0)^(n+1) + ... where c_n and further coefficients are not zero. Sin(1/z) is an interesting example in that it behaves pretty well when you approach 0 on the real axis (it just fluctuates between -1 and 1 at an increasing frequency), but trainwrecks when approaching 0 on the imaginary axis. === Subject: Re: Number Theory Question > I would greatly appreciate some comments upon the correctness of the > assertion about the following equation (1) under the given conditions. > y = (a^m + b^m)(a^m - b^m) (1) > Conditions: (y, a, b ) = 1; m is a non-integer > 0; prime p > 3. > Assertion: If y is a p-th power then both a^m + b^m and a^m - b^m > separately be p-th power. It all depends on what you mean by (y, a, b) = 1. Without any context to go by, I'd interpret it to be y=a=b=1, in which case the assertion is true, since whenever lhs (=1) = rhs (=0), anything at all is true. Of course, I could be misinterpreting the symbols. Why not say what you mean in words? Jon Miller === Subject: Re: When is the limit of Borel-measurable functions Borel-measurable? > This is the second time I've sent this message. The first time, it > seems to have dropped into the bit-bucket. Sorry if you're reading it > for the second time. > The question might seem stupid, but I'm not talking about REAL-VALUED > functions. > Let (X,T_X) and (Y,T_Y) be topological spaces, and let B_X and B_Y > denote the Borel fields on them. (= generated by the open sets) > It is not reasonable that the pointwise limit of a sequence of > measurable functions f_n : (X,B_X) --> (Y,B_Y) is also measurable. (By > measurable I of course mean that the inverse image of a set in B_Y is > in B_X.) This is because sequences don't mix well with topological > spaces. > But there ARE circumstances under which this is true. I remember a > theorem proved by Calderon in one or another class he was teaching > where he proved it was sufficient that T_Y have the property that every > open set U can be written as a countable union of open sets G_n with > the property that the closure of G_n is a subset of G_{n+1}. This > property is satisfied, e.g., by regular T1 second-countable spaces. > Does anybody know the most general circumstances under which this > result is true? I doubt that second-countability is necessary; it's > TOO global a condition. Probably first countability isn't enough. Another sufficient condition is that B_Y should coincide with (and not just contain) the sigma-field generated by the continuous real-valued functions on Y. I don't know if this condition implies, or is implied by, Calderon's condition. Either condition holds if Y is a metric space. > Alternatively, can anybody characterize the property of open sets U > being writable as such countable unions? -- A. === Subject: Re: Laurent expansion, anybody? in the morning (which, as all smart humans know, is of course around 10:00). Have a nice day/night depending on your time zone. -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: a puzzle related to artinian group > For what it is worth, I have just verified that the game always terminates > for n <=9 players no matter how the n-1 cards are dealt. I did it using > brute force with Maple. Brute force, Perl and the theorem I posted a while ago got me to n=15. I could probably do n=16, but n=17 would almost certainly take too much memory. A rewrite in C might get me up to n=18 or so. Of course, all that feels pretty pointless, given that a general proof _shouldn't_ be that hard. Oh well, it was a fun programming exercise... -- Ilmari Karonen If replying by e-mail, substitute .net for .invalid in address. === Subject: Re: Core error, FEAR is a natural response ... > You are afraid. But that's natural, as most of you, after all, are > only human. To err is human, to forgive divine I guess those that you implicitly state are not human are the same ones who forgive your insults and continue to try to correct your very human errors. Gib === Subject: How'd they do it? Until the early 1900's they actually used grams and kilograms as, and instead of, weight; when they finally derived the newton [N] as the metric unit of force: What took them so long? The kilogram of mass isn't equal to 2.2# pounds of force! The kilogram _weighs_ 2.2#! The kilogram is a unit of mass [1 kg = f/a = 1 N secî/meter = w/g = 2.2# secî/foot]! What'sa matter with youse guys anyways(;^? === Subject: Re: Minimal Graph, Four Color Theorem Here is one example of incorrect data. > Consider the graph of 7 vertices, 1,2,3,4,5,6,7, with the > following adjacencies: > 1 is adjacent to 2,3,4, and 5. > 2 is adjacent to 1, 3, 5, and 7. > 3 is adjacent to 1, 2, 4, and 6. > 4 is adjacent to 1, 3, 5, 6, and 7. > 5 is adjacent to 1, 2, 4, and 7. > 6 is adjacent to 2, 3, 4, and 7. > 7 is adjacent to 2, 4, 6, and 6. > I have a drawing before me, so I know the graph is planar. The graph is 4-chroma, not 5-chroma >Now, if you > remove any vertex, and you 4-color the remaining graph, then it is > possible that that 4-coloring cannot be extended to a 4-coloring of > G. For instance, remove the vertex 6, and color the remaining graph > with Red, Blue, Green, and Yellow, as following: > Blue: 2, 4 > Green: 3 > Red: 5 > Yellow: 1, 7. Actually, its 3-chroma, vertices 3 & 5 may both be the same color. > Since 6 is adjacent to a blue, a green, a red, and a yellow vertex, > you cannot extend that 4-coloring to a 4-coloring of G. > Now, yes, of course, there are OTHER 4-colorings of G-{6} that ->can<- be > extended to G. But that's not the point. The point is that you cannot > simply say that a 4-coloring of G-{6} can be extended to a 4-coloring > of G-{6}, only that ->some<- 4-colorings of G-{6} can be extended. > So you need would need to show that given ANY planar graph G, there is > always a choice of vertex v such that there is SOME 4-coloring of > G-{v} which can be extended to G, to be able to derive a contradiction > from the assumption that a minimal counterexample exists. That is in > fact what Appel and Hanken did. This data is inaccurate. so it is irrelevant. This is nonsense. You are confused. === Subject: Re: Antidiagonal, Infinity This page discusses a function continuous at all irrationals, > discontinuous at all irrationals: > Are you sure about that? :-) The pages discusses continuous at all irrationals, discontinuous at all rationals, and no. > That means I still want to know that given an element, > that for the next element, what it is, or the probabilities of what > it could be. > How about, just for once, entertaining the possibility that > there may not be a next element after your favourite number? The set is only points, the set is totally ordered, etcetera. The set is only points, the elements of the reals each represent a point on a line f(x)=0. The elements of the set are not rising or falling edges or crests or troughs of a signal, they are not itty bitty line segments, they are points! I suppose you could define a continuous function as one of those other things. Some say the real number is the limit of a convergent sequence of rationals as the reals are complete. It's a point. I consider that there is not a next. I also consider that there is. Then again I think about things like next after Ord is zero. Next integer after zero is one. Next number after zero is iota. What's the previous number before zero? Anyways in talking about the reals, and how they represent as a set each point of fx)=0, each point is represented explicitly. === Subject: Re: Divisibility Of A Derivative By...(Calculus /#-Theory) > This is a slight generalization of the theorem in the > Prime-Derivative Puzzle thread > from mid August. > Let q and r be any positive integers. > Let, for all x where -1 < x < 1, > f(x) = > (1-x)^((1-x)^(-q)) *(1+x)^((1+x)^(-r)) > In ascii-art mode: > f(x) = > -q -r > (1-x) (1+x) > (1-x) *(1+x) > Then: > GCD(q+r ,m) > always divides > the (m+1)th derivative of > f(x) at x = 0. > (This derivative, and all derivatives, of f(x), at x =0, are > integers.) I know that I did not originally post this as a puzzle. But I believe it would be fun to try to prove. So, I am cross-posting this reply to rec.puzzles, as well as posting it to sci.math. > Leroy > Quet === Subject: Re: How'd they do it? > The kilogram is a unit of mass [1 kg = f/a = 1 N secî/meter = w/g = 2.2# > secî/foot]! The official kilogram is a hunk of fancy metal guared by snooty Frenchmen in Paris. === Subject: Re: Factorial/Exponential Identity, Infinity I guess maybe I don't discuss a permutation, but some other combinatoric operation on the elements of a sequence. An infinite binary sequence with a beginning is presented to you, and you can modify it via this method: you can change any element from a one to a zero, or a zero to a one. If you change a one to a zero, then you must change that or some other zero to a one. If you change a zero to a one instead, you must change that or some other one to a zero. There may be allowed the interchange of identical elements as they would not change the sequence. About the dual representation, consider a crazy model where there is not dual representation and .1000... does not equal .0111.... If you must have dual representation, then the infinite binary sequence is a unique representation of a subset of the naturals but not a unique representation of a real, it is a representation but not necessarily unique, via dual representation. So anyways given a sequence (01)..., the subsequence 01 repeating infinitely, it's possible to change it to (10)..., the subsequence 10 repeating infinitely, by the operation of changing the first, second, third, etcetera elements of the sequence, as the first requires changing the value of a symbol that occurs in the second place, changing the third requires changing the value of a symbol that occurs at the fourth place, etcetera, ad infinitum. This corresponds exactly with that the density of ones and zeros in each of those sequences is one half. The sequence can be modified this way to get a result sequence of any other sequence with density 1/2 of each, yet, the sequence will never be changed to a sequence with density 1/3 ones or zeros. Using this method on, as an example, the sequence of 000..., the result will never be different from 000.... Change a zero element to a one, then, the one element has to be changed to a zero: there are no symbols with the value one in the sequence. Similarly, the sequence as input with finitely many x count of ones will always result in a sequence of output with x many ones. The class of algorithms that modify these sequences' algorithms do not change the density of ones and zeros in the sequence. That apostrophe is there because class is singular and the word algorithms is plural. So I guess it's not a permutation, I tend to trust you, and haven't found a flaw with your points there, ignoring trivialities and easily corrected dsitractions, I describe a method that has the properties that I have here described. One thing I like about this is that I can use it to say that where the density of ones in (10)... is one half, that the density of the even naturals is one half. This is where (10)... represents the subset of N {0, 2, 4, 6, 8, ...}, the even numbers. That might work better with the natural numbers not containing zero, Z+, for that the odd numbers have density 1/2 in Z+, the positive integers. Also, any sequence that I can derive as output from an input of (10)... for N has density of one half in N, e.g, {0, 2, 5, 6, 8, ...}. Given an input sequence a and some output sequence b, there is a function between the various elements they may represent, a bijection. For example if the sequence a has finitely may ones so does b, the same amount, and if sequence a has infinitely many ones so does b. For a given input sequence, generally the canonical input sequence of given density or of indeterminate density via a rule on the ones and zeros in progressive subsequences (irrational sequences), a various number of sequences are possible as output. This can be expressed as an asymptotic term of the variable n of the length of the sequence. For example, n many of all possible sequences have one on element or one off element. Anyways, this goes back to earlier discussions about canonical sequences and the canonicalization of sequences, using some form of sequence element interchange. === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. >Here is one example of incorrect data. >> Consider the graph of 7 vertices, 1,2,3,4,5,6,7, with the >> following adjacencies: >> 1 is adjacent to 2,3,4, and 5. >> 2 is adjacent to 1, 3, 5, and 7. >> 3 is adjacent to 1, 2, 4, and 6. >> 4 is adjacent to 1, 3, 5, 6, and 7. >> 5 is adjacent to 1, 2, 4, and 7. >> 6 is adjacent to 2, 3, 4, and 7. >> 7 is adjacent to 2, 4, 6, and 6. >> I have a drawing before me, so I know the graph is planar. >The graph is 4-chroma, not 5-chroma I did NOT claim it was 5-chroma. There was no incorrect data, except in your imagination. The example was an example of a planar graph G with the following property: For every vertex v, there is a 4-coloring of G-{v} that cannot be extended to a 4-coloring of G. I even stated: >> Now, yes, of course, there are OTHER 4-colorings of G-{6} that ->can<- be >> extended to G. Given that statement, how can you ->possibly<- think that I claimed that G was 5-chroma, ->unless<- you did not bother to read what I Obviously, as I pointed out, you have not bothered to read the replies you got. >> Now, if you >> remove any vertex, and you 4-color the remaining graph, then it is >> possible that that 4-coloring cannot be extended to a 4-coloring of >> G. You see that over there, then it is POSSIBLE that the 4-coloring cannot be extended to a 4 coloring of G. I did not say it was ALWAYS the case, I said it was POSSIBLE. The reason this example was brought up was given in the text, but you removed it to make this reply. Why? Now, you seem to be claiming that given any planar graph G, and any vertex v in G, if G-{v} can be 4-colored then one can take that coloring and extend it to a 4-coloring of G. This is simply not the case, The counterexample to the claim being the graph given. but you are close to the way the proof goes: what one shows is that if there were a counterexample, and G is minimal, then there would always be a specific choice of v such that there is some specific 4-coloring of G-{v} which can be extended to a 4-coloring of G. But the proof is long: one must consider many possible cases of G and many configurations. Now, the following is certainly true: if G is planar, and has a vertex v with degree less than 4, then any 4-coloring of G-{v} can be extended to a 4-coloring of G.: I was not positive what it was what you were claiming, and I made a guess. Then I addressed that guess. You never bothered to explain if my guess was accurate or not, you never even bothered to address this until now, when you took what I that? I can think of two possibilities: (a) You honestly did not realize you were taking the statements out of context. That would mean that you either did not understand what I inaccurate, you should have asked); or else you did not bother to Or (b) You realized you were taking the statements out of context. That would mean you are being purposely dishonest. Are there other possibilities I have not thought of? > For instance, remove the vertex 6, and color the remaining graph >> with Red, Blue, Green, and Yellow, as following: >> Blue: 2, 4 >> Green: 3 >> Red: 5 >> Yellow: 1, 7. >Actually, its 3-chroma, vertices 3 & 5 may both be the same color. Again, irrelevant to the point I was making: that there are 4-colorings of G-{v} that cannot be extended to G. Do you agree or disagree that for every vertex v, there exists a 4-coloring of G-{v} that cannot be extended to G? >> Now, yes, of course, there are OTHER 4-colorings of G-{6} that ->can<- be >> extended to G. But that's not the point. The point is that you cannot >> simply say that a 4-coloring of G-{6} can be extended to a 4-coloring >> of G-{6}, only that ->some<- 4-colorings of G-{6} can be extended. >> Since 6 is adjacent to a blue, a green, a red, and a yellow vertex, >> you cannot extend that 4-coloring to a 4-coloring of G. >> So you need would need to show that given ANY planar graph G, there is >> always a choice of vertex v such that there is SOME 4-coloring of >> G-{v} which can be extended to G, to be able to derive a contradiction >> from the assumption that a minimal counterexample exists. That is in >> fact what Appel and Hanken did. >This data is inaccurate. The data is not inaccurate, only your understanding of the data is inaccurate. >so it is irrelevant. It is not irrelevant. >This is nonsense. Whatever. >You are confused. You have no idea what you are talking about. You are either not reading what you are replying to, or not understanding it, or being purposely dishonest. Which is it? === Subject: Re: FUNctions/Continued-Fraction Puzzle I really hate posting the solution for the puzzle until I know that someone has actually attempted to solve it....And I doubt anybody has. But, anyway, my solution is below the replied-to message. > (This might be actually trivial. But it does not seem to be with the > little thought I have given it. In any case, perhaps I should not have > cross-posted this to rec.puzzles {if I should have even posted it to > sci.math}; but what the...) > For all real x > 1, and for some function of x, y(x); > where each y is a real, y =y(x), based on x: it is so that: f([x; x^2, x^3, x^4,...,x^m]) = [y; y^2, y^3, y^4,...,y^m], for EVERY positive integer m; where: [x; x^2, x^3, x^4,...,x^m] is the continued-fraction 1 > x + ------------------ ; > 1 > x^2 + -------------- > 1 > x^3 + --------- > .... > + 1/x^m and [y; y^2, y^3, y^4,...,y^m] is also a continued-fraction (obviously); and [x; x^2, x^3, x^4,...,x^m] converges to X; and f(w) is a real -> real function, such that f'(X) exists and is finite nonzero. So, what are the possible f(w)'s, given all of the conditions above?? > First, by the way, f'(X) is the (1st) derivative of f(w) at w = X, in > case this is not obvious. > I should mention that f can equate to an infinite number of functions > if it need not be analytic. If it need by analytic, however, there are > a finite number of possible functions that can equal f(w). > (So, find the set of analytic f(w)'s.) > This puzzle seems to be more difficult than I first assumed. > I will wait until Friday, at least, to post the answer if no one else > posts the solution before that. | | one V | | last V | | chance V | | to V | | try V | | to V | | solve V | | puzzle V | | yourself... V | | before... V ...the solution: I get that the only possible analytic f is: f(w) = w. Proof: limit{m -> oo} (x/y)^(2m-1) = 1. So, x must = y. And, consequently, f(w) must = w. * earlier result at: Leroy Quet === Subject: Re: Implied Sequence (#-Theory) Puzzle Solution is below copied original message. > [This is a continuation of the A certain Dirichlet-sum: question > thread (the last reply of which is copy/pasted below). But the current > topic at hand is this puzzle related to that thread.] > Let a(k,m) = the number of distinct prime divisors, p, of k, > where p^m divides k, but p^(m+1) does not. > Let c(k,m) be such that: > sum{k=1 to oo} c(k,m)/k^r = > (1 - 1/zeta(r))^m > for every r where sum converges, > where zeta(r) is the Riemann zeta function. > Now, {b(m)} is a sequence of reals such that: > For every positive integer n, > --- > / a(n,m) b(m) = > --- > m>=1 > --- > / c(n,m) b(m) > --- > m>=1 > In linear-mode: > sum{m>=1} a(n,m) b(m) = > sum{m>=1} c(n,m) b(m) > (I use 'm>=1' since the sums are finite, but the upper limit varies > based on n.) > Now, I wondered in my previous post (in the other thread) if {b(m)} > was unique. > I THINK that I may have proved today that {b(m)} is indeed unique, > except that each term can be multiplied by a constant (of course). > So, setting b(1) = 1, what is the sequence {b(k)}?? > (Not-too-small hint below copy/pasted post.) > Leroy Quet >[cross-posting reply to rec.puzzles] >>I should probably give this more thought myself. But this question >>does not seem to be as easily answered as I 1st assumed. >>Does sum{k=1 to oo} a(k,m)/k^r = >>(1 - 1/zeta(r))^m, >>where a(k,m) = the number of distinct prime divisors, p, of k, >>where p^m divides k, but p^(m+1) does not? >> >> I don't think so. For example, a(15,2) = 0 but >> (1- 1/zeta(r))^2 should contain a term 2/15^r. >> >The reason I ask is that, I THINK: >sum{m=1 to oo} (1 - 1/zeta(r))^m *b(m) = >sum{m=1 to oo} (sum{k=1 to oo} a(k,m)/k^r) b(m) >for a certain (at least one) sequence {b(m)}. >Puzzle(??): Find {b(k)}. >(I do not know if the sequence {b(k)} is unique.) > Hint: {b(k)} is a sequence of rationals that I have been quite fond of | | one V | | last V | | chance V | | to V | | try V | | to V | | solve V | | puzzle V | | yourself... V | | before... V ...the solution: b(m) = sum{k=1 to m} 1/k, the m_th harmonic number. I would give a proof, but I am planning to submit this problem possibly to the Mathmatics Magazine, and I don't want to give too much away. Is my result correct, and the sequence-definition unique, anyway?? Leroy Quet === Subject: Re: factoring to satisfiability > I would be very interested in any factoring based SAT instances. An unsigned N x N => 2N bit multiply can be easily written as a circuit composed of N^2 full adders and N^2 AND gates. The AND gates are 3-SAT and the fuller adders are 4-SAT. So simply take these variables and clauses and instantate the output variables, and you have factoring written out as a 4-SAT problem. Since the ANDs are conected to input bits, asymptotically for large N the added variables (other than input and output bits of the multiply) are the full adder inputs, so you get about 3*N^2 variables for a N x N => 2N bit problem. This actually isn't too bad. As an example, the factorization of RSA-576, the next unfactored number on the RSA Challenge list, can be written out as a 4-SAT problem in around 250,000 variables. Since an AND gate is a single 3-SAT clause plus two 2-SAT clauses, and a full adder is 8 4-SAT clauses to generate the sum bit, and six 3-SAT clauses to generate the carry, you are looking at somewhere around 1.5 million clauses for the RSA-576 problem mentioned above. Still small enough to munch with a SAT-solver in a 32 bit address space. For experimentation, a 16 x 16 => 32 bit problem is about the right size. It is big enough to capture the essential nature of multiplication, and small enough to be brute-forced by a dumb SAT-solver that just tries everything and backtracks. Here, you're looking at 832 variables, and a little over 4,000 clauses. > I still suspect that they are no harder than randomly generated hard > instances. Random SAT is boring. It's much more fun to work on SAT problems that do something useful, like illuminate the deeper mathematical subtlties of factorization. > For example, I would expect factoring problems to have high > clause to variable ratios. > If this is true, they would be unusual in that they have high ratios and > are still solvable. I am guessing that factoring problems are > similar to forced satisfiable problems. WEll, high compared to the crossover point at which random SAT becomes statistically unlikely to have satisfying assignments. If you build problems from circuits, your clause to variable ratio is going to basically be the typical clause to variable ratio of whatever building blocks you use. This isn't unusual. It's by construction. -- Eric Michael Cordian 0+ O:.T:.O:. Mathematical Munitions Division Do What Thou Wilt Shall Be The Whole Of The Law === Subject: Re: Use of variable independence, core error >> I've found that there is a definition error in core mathematics, >The problem with core mathematics you perceive isn't that at >all. The problem is that you *want* two and only two of the g's >to have a factor of f because you need that result for your FLT >argument. Hmm, FLT ... problem ... core ... -- isn't that Nico B's line of reasoning? === Subject: Advice on course selection and some primer texts you generous folk. My question is this: I am beginning a degree in mathematics and I am interested in the field of applied math. (I understand that discrete math and aplied math have commonalities, but correct me if I am wrong. I hate being wrong.) But my specific area of interest is modeling. I would like to study mathematical modeling becasue I see it as a useful tool in my future (if you wanna know what that might be, email me and i'd be happy to pitch it to you) research. Could anyone reccommend some good primers for getting started. I will be studying Vector calculus, real analysis, and Linear algebra next term (fingers crossed) and would like to get a better idea of the task that is ahead of me once those are complete. enjoy the day, cuz its almost over, Josh === Subject: Curve Inside Another Of Equal Perimeter Are there any two CONVEX closed curves of equal perimeter, where one can be placed completely inside the other? (It is allowed that the curves touch at a finite number of points at most.) Leroy Quet === Subject: Re: quantum echo Metaphysical Background, Thomas McTighe asserted that the quiddity > of a thing is nothing other than unity itself. Hence, by virtue > of its positive content, the sun differs not at all from the moon > or any other particular thing. The diversity which is exhibited > by the natural world is merely the product of accidental > differences; no object possesses any specific form which > interposes itself between a particular existing thing and the > source of their being e.g. the Absolute.15 All individual > entities are nothing more than differing contractions of the > whole devoid of any being of their own. ...because the > restricted quiddity of a thing is the thing itself. http://www.crvp.org/book/Series01/I-10/chapter_ii.htm I didn't understand this the first time around. This time I'll make an effort, by placing it in its context: Metaphysical Background, Thomas McTighe asserted that the quiddity of a thing is nothing other than unity itself. Hence, by virtue of its positive content, the sun differs not at all from the moon or any other particular thing.14 The diversity which is exhibited by the natural world is merely the product of accidental differences; no object possesses any specific form which interposes itself between a particular existing thing and the source of their being e.g. the Absolute.15 All individual entities are nothing more than differing contractions of the whole devoid of any being of their own. Putting it in context didn't work. I have no idea what McTighe is talking about. Do you? > I take a quiddity to be a thing's *suchness* and a haecceity to be its > *thisness*. A *thisness* I take to be, as Robert Adams does, the > property of self-identity, although I disagree with an assumption > which might be read into Adams, that an individual can lack > self-identity and yet possess the property of being some > individual or other). > In Primitive Thisness and Primitive Identity (_The Journal > of Philosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), Adams > A thisness is the property of being a certain particular > individual, not the property of being some individual or other, > but my property of being identical with me, your property of > being identical with you, etc. These properties have recently been > called 'essences', but that is historically unfortunate, for essences > have normally been understood to be constituted by logical properties, > and we are entertaining the possibility of nonqualitative > thisnesses. In defining 'thisness' as I have, I do not mean > to deny that universals have analogous properties--for example, > the property of being identical with the quality red. But since > we are concerned here principally with the question whether > the identity and distinctness of individuals is purely > qualitative or not, it is useful to reserve the term > 'thisness' for the identities of individuals. > It may be controversial to speak of a property of being > identical with me. I want the word 'property' to carry as > light a metaphysical load here as possible. 'Thisness' > is intended to be a synonym or translation of the traditional > term 'haecceity' (in Latin, 'haecceitas'), which so far as I > know was invented by Duns Scotus. > Like many medieval philosophers, Scotus regarded properties as > components of the things that have them. He introduced > haecceities (thisnesses), accordingly, as a special sort > of metaphysical component of individuals.[4] I am not proposing > to revive this aspect of his conception of a haecceity, because > I am not committed to regarding properties as components of > individuals. To deny that thisnesses are purely qualitative > is not necessarily to postulate 'bare particulars', substrata > without qualities of their own, which would be what was left > of the individual when all its qualitative properties were > subtracted. Conversely, to hold that thisnesses are purely > qualitative is not to imply that individuals are nothing > but bundles of qualities, for qualities may not be components > of individuals at all. (pp. 6-7) > Acceptance of haecceities is a distinctive feature of the thought of many > followers of Scotus, though there are some sixteenth-century scholastics who > accept haecceities without accepting many other distinctively Scotist > teachings. Having said this, some early followers of Scotus reject > haecceities and the theory of the common nature altogether, and of those who > accept haecceities, some found the correct understanding of the nature of > the distinction between an individual's nature and its haecceity a > troublesome matter. > One of the earliest Scotists, Francis of Meyronnes, writing his commentary > on the Sentences around 1320, accepts the theory of the non-numerical unity > of common natures, and the claim that individuation is by haecceity. But he > holds that it is inappropriate to talk of a formal distinction in this > context. Formal distinction obtains only between things that have some sort > of quidditative content. I'm not sure what he/you are getting at here. > Haecceities have no such quidditative content, and thus cannot be formally > distinct from their nature. Rather, a haecceity is modally distinct from its > nature. What does Cross (or whoever) mean by Haecceities cannot be formally distinct from their nature.? > A modal distinction, according to Meyronnes, obtains between a thing > and an intrinsic mode of that thing, where an intrinsic mode is something > which when added to a thing does not vary its formal definition . . . since > it does not of itself imply any quiddity or formal definition. A haecceity > does not affect a thing's kind; it is thus an intrinsic mode of the thing. In order for this to mean something to anyone who is not sure what an intrinsic mode, or the formal definition, of something (in the sense of Meyronnes) are--and is therefore unsure what 'varying something's formal definition' might involve, illustrative examples should be provided. > It may look as though this is just a terminological shift, but it is not so > in at least the following way: a modal distinction is a lesser kind of > distinction than a formal distinction. Formal distinctions obtain between > genus and specific difference; thus, the difference between species/nature > and haecceity, for Meyronnes, is less than the difference between genus and > difference. Scotus, contrariwise, makes no such distinction between degrees > of difference in this context (for this contrast between the two thinkers, > see Dumont [1987], 18). Still, without some principled way of spelling out > degrees of difference, this contrast between Scotus and Meyronnes amounts to > nothing of any philosophical interest. To this extent the difference between > the two thinkers might as well be merely terminological, and Meyronnes needs > to do more work if he is to make any significant philosophical point here. Without any explanation of the before-shift and after-shift terminology, how is one supposed to know which side is up? > http://setis.library.usyd.edu.au/stanford/entries/medieval-haecceity/ Here's the concluding paragraph from that Stanford Encyclopedia Note, of course, that Scotus's account of the common nature entails something stronger than Adams is proposing: indeed, it entails precisely the sort of minimal hypostatization that Scotus proposes. And the reason for this, of course, is Scotus's view that individual substances cannot themselves be primarily diverse -- a fact that is explained by his claim that common natures have some sort of unity in their instantiations: the nature in Socrates is (non-numerically) the same as the nature in Plato. Natures, for Scotus, cannot be primarily diverse; substances must include more than natures. But individual natures in Ockham's view can indeed be primarily diverse, and this surely amounts to a form of haecceitism -- nothing other than an individual nature's own self-identity explains its distinction from all other such natures. Maintaining that individual natures are primarily diverse amounts not to having no theory of individuation, but to accepting a form of haecceitism that, like Adams's, does not involve ontological commitment to the existence of real haecceities as distinct real constituents of things. Is the author of that one any relation to Mr. McTighe? > http://plato.stanford.edu/entries/francis-marchia/ --John === Subject: Constant Implied By Product How would one find x, or even if any real x exists, where x = product{k=1 to oo} (1 + x*a(k)), where {a(k)} is such that the product converges? ...or where x = product{k=1 to oo} 1/(1 - x*a(k)), or etc...? With the first product, anyway, the partial-product of the first m terms (and subtracting x) produces a polynomial of order-m, which we can find the root of as to estimate x (especially if the product converges quickly, I would imagine). I wonder if some {a(k)}'s produce, however, interesting closed-form (not just numerically estimated) values for x. Leroy Quet === Subject: Re: How'd they do it? > The official kilogram is a hunk of fancy metal guared by snooty > Frenchmen in Paris. Guarded? === Subject: Re: Fundamental Reason for High Achievements of Jews === >Subject: Re: Fundamental Reason for High Achievements of Jews > :> : Most historians believe that Jews avoid pork, >> :> : because the ancient Jews associated pigs with leprosy, >> :> : and pigs and people with leprosy were unclean. >> :> >> :> Name one historian who believes that, and give a citation to the >> :> place where he says it. >> : >> : As I recall, this was in Tacitus' Histories >> : which was written in the first century A.D. >> >> If that's the best you can do, I think that we can safely ignore your >theory. It's soooo easy Richard. Someone in your camp >should have the information at his fingertips. GOOGLE: tacitus histories ~13,600 hits > tacitus histories jews ~3,950 hits > tacitus histories jews leprosy ~154 hits This is all irrelevant. Quoting from Tom Potter: Almost everyone has access to the first hand historical accounts, > and can do wild card searches on the source material. > It is STUPID to provide detailed cites, as these focus > ONLY on the POINTS trying to be emphasized by the writer. > It also STUPID to use second hand, accounts which have a > racial, religious, national, or personal spin on them, rather than > using the FIRST HAND historical accounts. > Anything written by Tacitus would NOT be a FIRST HAND account, > so only a STUPID person would allow himself to be brainwashed > by Tacitus' racial, religious, national, personal spin on history. > It is interesting to see that Mensanator, > like many people, has been brainwashed to think that > Tacitus, who was one of the most unbiased, rational, > and correct historians, put a racial, religious > spin on history. Liar. YOU are the one who says that second hand accounts put a spin on history. Or are you going to deny that that was your quote? I wouldn't try it, I located it via a Google search, others can too. > This attitude obviously has its' roots in > conditioning, as the works of Tacitus > are far more rational and correct, > than the bible, the Greek and Roman mythologies, etc. But they are not a FIRST HAND account. So your brainwashed opinion of their validity is irrelevant. Or are you now admitting that the quotation about second hand accounts was stupid. You can't have it both ways, so which is it? > In other words, people who have been brainwashed to a point of view, > have a great difficulty in accepting data > that conflicts with their conditioning. In other words, Tom Potter is too stupid to realize that he's just been hoisted by his own petard. === Subject: Re: Number Theory Question I would assume (y, a, b) = 1 means they're all relatively prime. GREG > I would greatly appreciate some comments upon the correctness of the > assertion about the following equation (1) under the given conditions. > y = (a^m + b^m)(a^m - b^m) (1) > Conditions: (y, a, b ) = 1; m is a non-integer > 0; prime p > 3. > Assertion: If y is a p-th power then both a^m + b^m and a^m - b^m > separately be p-th power. > It all depends on what you mean by (y, a, b) = 1. Without any context to go > by, I'd interpret it to be y=a=b=1, in which case the assertion is true, > since whenever lhs (=1) = rhs (=0), anything at all is true. Of course, I > could be misinterpreting the symbols. Why not say what you mean in words? > Jon Miller === Subject: Re: How'd they do it? >> The official kilogram is a hunk of fancy metal guared by snooty >> Frenchmen in Paris. >Guarded? Have you priced out what at least three different kilograms of 90% platinum 10% iridium would be worth, just in bullion value? That's not even taking into consideration their value as the world's standards. Of course, guarded. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Re: circle with two centers > Consider a smooth surface, S, and two points, A & B on it. S contains > a continuous, closed curve, C, such that both A & B are equidistant > from every point on the curve C. >>Can't think of any surface with more than 2 such centers. >Just deform the sphere - leaving the equator intact - to intersect the >axis as many times as you like. > Oh, I suppose you are requiring that as well as A being equidistant > from the points on C and B being equidistant from the points on C you > want the two distances to be the same. In which case there are > obviously only two centres (in 3D). I don't think the original > question implied that. Actually, I did. Clearly, you need to measure distances along geodesics, as Herschkorn has also understood. In that case, it is not obvious (and very likely not true) that you cannot have N points (N > 2), each of which is exactly the same distance, D, from every point on the curve C. -riskbert === Subject: Re: That Hilbert guy is giving me a lot of trouble Imam Tashdid ul Alam > I am familiar with Linear Algebra...eigenvalues and stuff. But only > for finite dimensions. I know: > 1) That eigenvalues are properties of linear transformations > independent of the basis chosen. > 2) Trace (the sum of the diagonal elements) is also the sum of the > eigenvalues. > 3) Determinant (that weird looking function) is also the product of > the eigenvalues. > 4) As eigenvalues are many in numbers, the collection of them knows > much more than just the two of the properties of the matrix. > I want to know: > 1) How these ideas generalize to infinite dimension. The particular > problem at hand is of course that I find it rather difficult to add > infinitely many numbers (...computing the determinant! are you nuts!) It's no big deal. The norm of a vector, for example, has the usual sum-of-squares formula with infinitely many (positive) summands, which we assume is a summable sequence. > 2) If there are ideas that are still important in the infinite case Very! > 3) Does having a complex linear space instead of a real one improves > the situation Yes when it comes to eigen-stuff, because polynomials can be factored. > 4) Functions like sin and cos have infinite polynomial expansion (how > do you say it in standard language?)... power series expansions > So I might find a Fourier > transformation helpful. But isn't the most likely situation be > composed of infinitely many Fourier components too? Fourier series do have infinitely many terms, usually. Google around on the keyword harmonic analysis -- one of the strongest themes in mathematics. > And most importantly: > 5) What is the best textbook (or something like that) that explaining > these things that can be most easily swallowed These are quite sensible and worthwhile questions. The basics on seperable Hilbert spaces are in many books; a good one, if you can find it, is Avner Friedman Foundations of Modern Analysis (a Dover paperback), Chapter 6. But I think it's a good idea to start with one or two books that were written for physicists, or by physicists. The trace and determinant of a finite square matrix are (up to sign) just the coefficients of x^{d-1} and x^0 in the characteristic polynomial. To some extend they do extend to a sort of continuous matrix: a function M on the square [0,1]x[0,1], which acts on a column vector (a function on [0,1]) just like a matrix. Take a good look at the canonical (Jordan) decomposition of an endomorphism of a finite-dimensional vector space over C. The decomposition can be expressed without using any bases or matrices. Understand how it arises in terms of _inverse images_ of subspaces fixed by the endomorphism. A bit vaguely speaking, similar inverse images are at the root of Lebesgue integration, and they explain how Lebesgue's ideas are so critical in harmonic analysis and spectral theory. Larry === Subject: Generalization of Logarithms: Revisited >If y, {a_k}, and {b_k} are known ({a_k} and {b_k} are sequences of m >numbers), and >y = sum_{k=1}^m [a_k *(b_k)^x], >then what is x? >We can generalize the logarithm, so we have >x = log_{{a_k};{b_k}}(y). >(log_{1;b}(y) = log_b(y).) >An interesting identity: >log_{{a_k*ln(b_k)};{b_k}}(1/[d{log_{{a_k};{b_k}}(y)}/dy]) = >log_{{a_k};{b_k}}(y). >What is an easy way to calculate these logarithms, preferably as a >power series? I was thinking again of the above today. But the only thing additionally I can say is that (aside from that the b's should all be positive): x = (ln(y) - sum{k=1 to m} a_k *(b_k/b_j)^x ) / ln(b_j), for some fixed j (1 <= j <= m, where b_j > 0 and not= 1), and for positive y. (duh) So, perhaps this can be used as a basis of an iteration which approaches x numerically. (I would suggest that b_j be the maximum element of {b_k}.) All of above: Just a starting point for others. Leroy Quet === Subject: Re: formula for roots of quadratic matrix equation? Content-Length: 1014 Originator: rusin@vesuvius >Is there a closed form formula for >the roots of the quadratic matrix equation >QAQ - BQ + C = 0 >where A, B, C and Q are all symmetric nxn matrices, n>1? >If not, how about for the special case B=I (identity matrix)? The equation as written is difficult. However, if one considers QAQ' - QB - B'Q' + C = 0, one can generally describe the solutions, especially if A is positive definite, say A = HH'. The equation then becomes (QH - B'H'^{-1})(QH - B'H'^{-1})' = B'A^{-1}B - C = D, where D would have to be positive semidefinite. The set of all decompositions of D into FF' gives the set of all solutions of the revised problem. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: How'd they do it? >>The official kilogram is a hunk of fancy metal guared by snooty >>Frenchmen in Paris. > Guarded? Yes. Sorry about the bad typing. === Subject: Re: How'd they do it? > That's not even taking into consideration their value as the world's > standards. The value as a standard is much higher than the material value. === Subject: Re: Just what is an L-series? Robin Chapman > I know what a Dirichlet L-series is, and I've seen a few other kinds, > such as Artin's in connection with representation of finite groups. > But is there a standard comprehensive definition of the term > L-series? Maybe, any Dirichlet series > sum_{n=1}^infty a_n n^{-z} > such that the sequence (a_n) is totally multiplicative? > Not general enough: that would exclude say, L-functions of > elliptic curves. Ah so. I was trying to compose a definition of L-series for one of the volunteer-written online websites (www.planetmath.org). Maybe in the end, L-series are not a structure but only a formalism, like generating functions. I may need to resort to arm-waving or (what is much the same thing) category theory:) LH === Subject: Re: How'd they do it? Cut< > That's not even taking into consideration their value as the world's > standards. > Of course, guarded. > Gene Nygaard > http://ourworld.compuserve.com/homepages/Gene_Nygaard/ Even though physics would be better off with_out_ 'em; just using a mathematical ratio: Like 1 kg = 2.2# secî/32.2'; using 2.2 pints of water at its maximum density, per kilogram. === Subject: identical property count > Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. How can something be identical with some things? And if they shared all their properties, how could you presume to count 'two'? You present an either/or relationship as a relational one. jJ > --I could have baked both cakes yesterday. > --I could have baked both cakes today. > --I could have baked this one yesterday > and that one today. > --I could have baked that one yesterday and > this one today. > Your example doesn't work! > How am I going to distinguish 'one cake' from 'the other cake'! > Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. THerefore, since they are two cakes, they don't share > all their properties, and I can distinguish them by means > of any property that one has and the other doesn't. > JJ 1 I. The Problem, and the Problem with the Problem, > 2 of Identical Articles and Quantum Statistics > 3 > 4 Suppose we have a box with two qualitatively > 6 bouncing around inside. We think of the box > 7 as having a left (*l*) and a right (*r*) side. > 9 at random without interacting, so that their > 10 motions are independent; in particular we > 12 that we may neglect collisions. What are the > 13 chances for finding one or both on one side > 14 or the other? > 15 > 16 Many find the following reasoning persuasive. > 19 in *r*, and 2 in *l* and 1 in *r*. These should > 20 be equally likely, so that each has a probability > 21 of 1/4, or a probability of 1/4 for two in *l*, > 22 1/4 for two in *r*, and 1/2 for one on each side. > 23 > 24 This stylized example is a simple mock-up for a > 25 kind of situation that can occur with quantum entities > 26 and properties. For many of these situations the > 27 probabilities are in fact found to be 1/3 for each > 28 of the three cases: two in *l*, two in *r* and one > 29 on each side. Many interpreters have found this > 30 fact utterly astonishing. > 31 > 32 But on the face of it, there is a very simple > 33 resolution of the puzzle: give up supposing > 34 that there are two qualitatively identical but > 36 there are two *quanta*, as I'll put it, to which > 37 the notion of being numerically distinct does not > 38 apply. . . (p. 114) Compare Teller's mock-up with how I can go about > making two cakes in two days: 1) I can bake both cakes today. > 2) I can bake both cakes tomorrow. > 3) I can bake one cake today and the other cake tomorrow. But after baking the cakes, when I consider how I > MIGHT have gone about baking them, there are FOUR > possibilities rather than THREE. --I could have baked both cakes yesterday. > --I could have baked both cakes today. > --I could have baked this one yesterday > and that one today. > --I could have baked that one yesterday and > this one today. Homework for David Ullrich: In what respect do cakes I am going to make resemble quanta? (Hint: > Like causes produce like effects.) --John === Subject: Re: Two similar things >cake1 has come to exemplify the > property of being identical with cake1, and cake2, the property > of being identical with cake2. It is not a property 'of' at all. Of what is it a property 'of'? > Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. THerefore, since they are two cakes, they don't share > all their properties, and I can distinguish them by means > of any property that one has and the other doesn't. > No, there's a foot wrong there. Why would they be one cake if they shared > all the same properties? Why not 3 cakes? Would you tell the difference by > counting? > Granted that these cakes--call them cake1 and cake2--are now > identical-with-somethings, the property of being identical > with cake1 and that of being identical with cake2 are no > longer unexemplified: cake1 has come to exemplify the > property of being identical with cake1, and cake2, the property > of being identical with cake2. Consequently, in the event that > these cakes share *all* their properties, cake1 will exemplify > the property of being identical with cake2, and cake2 will > exemplify the property of being cake1, making cake1 and > cake2 one and the same cake. > I missed the point. More pertinent is this: > How would you distinguish your two different cakes? You can name them > 'todays and yesterdays' cakes, but you actually would not be able to know > which was which from a description of the cakes themselves. > JJ > *Any* property which one had and the other did not would > distinguish them, whether or not you, or I, or anyone else > grasped the basis of their difference. Thus, it might very > well be the case that everyone took there to be but a single > cake when in fact there were two: wasn't it once thought > that there was only one kind of helium when in fact there > were two? And everyone might think there were two different > cakes when in fact there was but one. > --John > --I could have baked both cakes yesterday. > --I could have baked both cakes today. > --I could have baked this one yesterday > and that one today. > --I could have baked that one yesterday and > this one today. Your example doesn't work! > How am I going to distinguish 'one cake' from 'the other cake'! Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. THerefore, since they are two cakes, they don't share > all their properties, and I can distinguish them by means > of any property that one has and the other doesn't. > JJ 1 I. The Problem, and the Problem with the Problem, > 2 of Identical Articles and Quantum Statistics > 3 > 4 Suppose we have a box with two qualitatively > 6 bouncing around inside. We think of the box > 7 as having a left (*l*) and a right (*r*) side. > 9 at random without interacting, so that their > 10 motions are independent; in particular we > 12 that we may neglect collisions. What are the > 13 chances for finding one or both on one side > 14 or the other? > 15 > 16 Many find the following reasoning persuasive. > 19 in *r*, and 2 in *l* and 1 in *r*. These should > 20 be equally likely, so that each has a probability > 21 of 1/4, or a probability of 1/4 for two in *l*, > 22 1/4 for two in *r*, and 1/2 for one on each side. > 23 > 24 This stylized example is a simple mock-up for a > 25 kind of situation that can occur with quantum entities > 26 and properties. For many of these situations the > 27 probabilities are in fact found to be 1/3 for each > 28 of the three cases: two in *l*, two in *r* and one > 29 on each side. Many interpreters have found this > 30 fact utterly astonishing. > 31 > 32 But on the face of it, there is a very simple > 33 resolution of the puzzle: give up supposing > 34 that there are two qualitatively identical but > 36 there are two *quanta*, as I'll put it, to which > 37 the notion of being numerically distinct does not > 38 apply. . . (p. 114) Compare Teller's mock-up with how I can go about > making two cakes in two days: 1) I can bake both cakes today. > 2) I can bake both cakes tomorrow. > 3) I can bake one cake today and the other cake tomorrow. But after baking the cakes, when I consider how I > MIGHT have gone about baking them, there are FOUR > possibilities rather than THREE. --I could have baked both cakes yesterday. > --I could have baked both cakes today. > --I could have baked this one yesterday > and that one today. > --I could have baked that one yesterday and > this one today. Homework for David Ullrich: In what respect do cakes I am going to make resemble quanta? (Hint: > Like causes produce like effects.) --John === Subject: Solving recurrence rel. a_2n = F(a_n) I am trying to find a method for solving linear recurrence relations of the form: a_2n = F(a_n) For example, a_2n = c * a_n + d * a_n-1 + b * a_n-2 + 1 --john === Subject: Re: Solving recurrence rel. a_2n = F(a_n) >I am trying to find a method for solving linear recurrence relations of the >form: > a_2n = F(a_n) >For example, > a_2n = c * a_n + d * a_n-1 + b * a_n-2 + 1 >--john Just one initial thought: You have so specify the odd-indexed elements some how. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: How'd they do it? >Cut< >> That's not even taking into consideration their value as the world's >> standards. >> Of course, guarded. >> Gene Nygaard >> http://ourworld.compuserve.com/homepages/Gene_Nygaard/ >Even though physics would be better off with_out_ 'em; just using a >mathematical ratio: Like 1 kg = 2.2# secî/32.2'; using 2.2 pints of water at >its maximum density, per kilogram. Gentlemen of the jury, Chicolini here may look like an idiot, and sound like an idiot, but don't let that fool you: He really is an idiot. Groucho Marx === Subject: Re: Boolean Algebra - Arithmetic Relationship > 3)Can all known Mathematical Notation and Symbolic manipulation be > modeled by a turing machine?? > If it's recursive it can be represented by a Turing Machine. What's the significance of recursion? I'm confused by this. (I've written non-recursive programs before of course.) Are all known mathematics recursive? The essential question forming in my mind is this: Is there a fundamental difference between mathematical symbolic notation and computer programming languages?? Both represent logical structures and their interactions. Both follow certain rules of manipulation. For me it's looks like they're two sides of the same coin. Are their certain properties of Zermelo-Fraenkel set theory that can't be modeled with a Turing-Machine? response. -Steve === Subject: Re: Core error, FEAR is a natural response > However, from Barry Mazur, to Andrew Granvile, to Ralph McKenzie, to > the kind of mathematicans who post a lot on Usenet, I've seen > mathematicians *run* away. ....from you. Sensible people have better things to do with their time. V. === Subject: Re: Factorial/Exponential Identity, Infinity > I guess maybe I don't discuss a permutation, but some other > combinatoric operation on the elements of a sequence. > An infinite binary sequence with a beginning is presented to you, and > you can modify it via this method: you can change any element from a > one to a zero, or a zero to a one. If you change a one to a zero, > then you must change that or some other zero to a one. If you change > a zero to a one instead, you must change that or some other one to a > zero. There may be allowed the interchange of identical elements as > they would not change the sequence. You seem to be saying that in any permutation the cardinality of zero-to-one changes must equal the cardinality of one-to-zero changes. If this is the case, then it is equivalent to a bijection f: N -> N in which the zero/one value of each n is the same as the value of f(n). > About the dual representation, consider a crazy model where there is > not dual representation and .1000... does not equal .0111.... If you > must have dual representation, then the infinite binary sequence is a > unique representation of a subset of the naturals but not a unique > representation of a real, it is a representation but not necessarily > unique, via dual representation. > So anyways given a sequence (01)..., the subsequence 01 repeating > infinitely, it's possible to change it to (10)..., the subsequence 10 > repeating infinitely, by the operation of changing the first, second, > third, etcetera elements of the sequence, as the first requires > changing the value of a symbol that occurs in the second place, > changing the third requires changing the value of a symbol that occurs > at the fourth place, etcetera, ad infinitum. But it is equally possible to map .01(01) to .001(001), or to any other sequence having infinitely many zeros and infinitely many ones. Unless there are some restrictions on permutation that you are not mentioning. > This corresponds exactly with that the density of ones and zeros in > each of those sequences is one half. But there is nothing in permutation that guarantees any preservation of 'density'. > The sequence can be modified > this way to get a result sequence of any other sequence with density > 1/2 of each, yet, the sequence will never be changed to a sequence > with density 1/3 ones or zeros. Why not? Nothing in your permutation rules prevents it, as far as I can see. > Using this method on, as an example, the sequence of 000..., the > result will never be different from 000.... Change a zero element to > a one, then, the one element has to be changed to a zero: there are > no symbols with the value one in the sequence. > Similarly, the sequence as input with finitely many x count of ones > will always result in a sequence of output with x many ones. > The class of algorithms that modify these sequences' algorithms do not > change the density of ones and zeros in the sequence. Why not? Given the sequence .01(01), with zeros in odd positions and ones in even positions, Switch each zero in position 2*n-1 to a one, and match it by switching the one in position 4*n to a zero. This satisfies all you conditions, but changes the density of zeros to from 1/2 to 1/4 and the density of ones from 1/2 to 3/4, at least if I understand your density definition. > That apostrophe > is there because class is singular and the word algorithms is plural. > So I guess it's not a permutation, I tend to trust you, and haven't > found a flaw with your points there, ignoring trivialities and easily > corrected dsitractions, I describe a method that has the properties > that I have here described. > One thing I like about this is that I can use it to say that where the > density of ones in (10)... is one half, that the density of the even > naturals is one half. This is where (10)... represents the subset of > N {0, 2, 4, 6, 8, ...}, the even numbers. That might work better with > the natural numbers not containing zero, Z+, for that the odd numbers > have density 1/2 in Z+, the positive integers. Also, any sequence > that I can derive as output from an input of (10)... for N has density > of one half in N, e.g, {0, 2, 5, 6, 8, ...}. Not so. See above. > Given an input sequence a and some output sequence b, there is a > function between the various elements they may represent, a bijection. > For example if the sequence a has finitely may ones so does b, the > same amount, and if sequence a has infinitely many ones so does b. > For a given input sequence, generally the canonical input sequence > of given density or of indeterminate density via a rule on the ones > and zeros in progressive subsequences (irrational sequences), a > various number of sequences are possible as output. This can be > expressed as an asymptotic term of the variable n of the length of the > sequence. For example, n many of all possible sequences have one on > element or one off element. > Anyways, this goes back to earlier discussions about canonical > sequences and the canonicalization of sequences, using some form of > sequence element interchange. > Whatever you are trying to achieve about density, you have not achieved it by the stuff above.. === Subject: Re: How'd they do it? Gene Nygaard Or maybe gaarded? ;-) Tom Davidson Richmond, VA === Subject: Re: Antidiagonal, Infinity > This page discusses a function continuous at all irrationals, > discontinuous at all irrationals: Are you sure about that? :-) The pages discusses continuous at all irrationals, discontinuous at > all rationals, and no. One can ever do differentiable at all irrationals but discontinuous at all rationals, or worse. > That means I still want to know that given an element, > that for the next element, what it is, or the probabilities of what > it could be. How about, just for once, entertaining the possibility that > there may not be a next element after your favourite number? > The set is only points, the set is totally ordered, etcetera. But since it is also densely ordered, there are no next's. That is part of what dense means. > The set is only points, the elements of the reals each represent a > point on a line f(x)=0. The elements of the set are not rising or > falling edges or crests or troughs of a signal, they are not itty > bitty line segments, they are points! I suppose you could define a > continuous function as one of those other things. Not as any of the things you mentioned. > Some say the real number is the limit of a convergent sequence of > rationals as the reals are complete. It's a point. One which you do not get. > I consider that there is not a next. I also consider that there is. ONe of these considerations should be correct. What a shame you keep picking the wrong one. > Then again I think about things like next after Ord is zero. Next > integer after zero is one. Next number after zero is iota. What's > the previous number before zero? In what set? There are lots of sets of numbers in which zero has no immediate predecessor. In the set of integers it is -1. > Anyways in talking about the reals, and how they represent as a set > each point of fx)=0, each point is represented explicitly. As far as I know, fx)=0 and the rest of that statement have nothing to do with mathematics, much less points. Please either use standard mathematical notation and definitions or include explanations of the non-standard notation and concepts that you do use. > === Subject: Re: Why should I switch to Python? - Infinity of Primes by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9ABpN522544; >
 Since this is a posted to a computer language group, after all, I
>> include my own simple implementation of the Sieve in Haskell:
>(Haskell code deleted)
>> I wonder if someone can come up with an equally immediate solution
>> (i.e., a straight translation from the mathematical formulation) in
>> python?
>> Indeed, I would be tempted to say that Haskell is better suited for
>> learning (some kinds of) mathematics than python.
>I don't know any Haskell, but a quick Python sieve is easy:
>def sieve(n):
>    find primes up to n by sieve of Erasthones
>    primes = [2]
>    marked = [0]*n
>    for p in primes:
>        i = p
>        while(i < n):
>            marked[i] = 1
>            i = i + p
>        for q in range(p,n):
>            if marked[q] == 0:
>                primes.append(q)
>                break
>    return primes
>> ------
>> Charles Boncelet, University of Delaware,
>> On sabbatical at ADFA, Canberra Australia,
>> Home Page: http://www.ece.udel.edu/~boncelet/

===

Subject: Re: Chessboard knight metric?
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9ADf9k30231;
>>Take a chessboard (with or without infinetely many squares) let the
>>distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the
>>chessboard be defined as the minimum number of moves a knight takes
>>to reach y from x.
>As others have pointed out, this obviously is a metric. 
>As to the function (assume infinite board):
>1) you undoubtedly know that the color of the square changes
>   after each move, so you will need to separate into
>   two cases according to the parity of x_1+x_2+y_1+y_2, and
>2) the sets of squares that can be reached in exactly one or 
>   exactly two moves have a special, slightly irregular structure, BUT
>3) for n>2, the set of squares that can be reached in exactly
>   n moves has a very regular octagonal shape. DRAW IT!!!! (proof by 
induction)
>Item 1 and 3 gives you a rule to determine the distance, in the case
>that it is at least 3. You only need to check, whether the distance
>might be one or two.
>Jyrki Lahtonen, Turku, Finland
>Much to my surprise, you all seem correct about the metric part
>that the proof was in the definition itself. 
>I also appreciate the reader pointing out the regular shapes.
>I had seen these shapes before- and would caution that the
>shapes are perhaps as not as regular as they may seem. The octogon
>that comes out at n=3 does not seem to be a true octogon (mabie I'm 
>missing something again...) and for n = 4 the situation appears even 
>worse, with spaces way out on the perepherie being reached in (a 
>minimum of) 4 moves as well as spaces close to the starting square. 
>This was already pointed out by another reader- take x=(1,1) and >y=(1,2), 

then  d(x,y)=4 and y surely cannot lie on the outer edge of 
>the octogon. Of course, it seems correct to point out that the 
>octogons keep turning up as part of the picture again and again for
>higher n. 
>About the even / odd properties I mentioned before... at least that
>does appear to be regular for all n, doesn't it?
>C.Dement
  

===

Subject: Re: Chessboard knight metric? correction
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9ADf7p30220;
>>Take a chessboard (with or without infinetely many squares) let the
>>distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the
>>chessboard be defined as the minimum number of moves a knight takes
>>to reach y from x.
>As others have pointed out, this obviously is a metric. 
>As to the function (assume infinite board):
>1) you undoubtedly know that the color of the square changes
>   after each move, so you will need to separate into
>   two cases according to the parity of x_1+x_2+y_1+y_2, and
>2) the sets of squares that can be reached in exactly one or 
>   exactly two moves have a special, slightly irregular structure, BUT
>3) for n>2, the set of squares that can be reached in exactly
>   n moves has a very regular octagonal shape. DRAW IT!!!! (proof by 
induction)
>Item 1 and 3 gives you a rule to determine the distance, in the case
>that it is at least 3. You only need to check, whether the distance
>might be one or two.
>Jyrki Lahtonen, Turku, Finland
>Correction:
>I just sent a mail saying for x=(1,1) and y=(1,2), then d(x,y)=4
>completely overlooking the fact that the reader said assume
>infinite chessboard. My example would be incorrect if the
>infinite chessboard is represented by the product space of whole 
>numbers but, of course, remains correct if the infinite chessboard 
>represented by the product space of natural numbers. However, both
>of these cases have a certain ugliness to me which comes from the 
>fact that we are losing the flavor of the chessboard in these 
>cases. We can try to keep the flavor of the board for infinite cases
>by redefining our function d (notice that now all 3 cases-finite, 
>infinte natural, infinite whole- reduce to one):
>Assume we have the product space W of whole numbers. Now, speaking
>figuratively outside of mathmatical jargon, place one chessboard with
>its lower left corner at (0,0). Place another chessboard with its 
>lower left corner at (0,9) etc. Continuing in this manner, we 
>divide up the product space of whole numbers into patches of 64 
>squares.
>The set of of all patches P defines a function between itself and the
>product space of whole numbers which is one-to-one and onto
>(bijective).  
>Now comes the hammer: 
>Assume x and y are any members of this product space. Let A be the 
>board (or patch) x is on and B be the board y is on. Now let 
>d(x,y) =d_P(A,B)+ d_*(x,y), where d_P is the knight metric on the 
>patch space and d*(x,y) is knight metric distance left between x and
>y when we assume both x and y (which are actually on boards A and B)
>to be on the same board. Board symmetry is now preserved.
>>C.Dement  

===

Subject: Re: Magnification lense and heat...formula?
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9ADf4m30206;
suppose you let the sun hit a 1mx1meter water bed. is a set of magnifying 
lenses, magnifying the same area but focuesd on that bed of water, more 
efficient at evaporating the same amount at the same time??
>
 Hello all,
>> We have a library patron who is looking for a forumula that will
>> express the relationship between the size/degree of a magnification
>> lense and how much it will applify heat.
>> For example, using a magnifying glass to increase (focus?) heat using
>> sunlight.
>> Does such a forumula exist somewhere? Can you give us any ideas on
>> where to look for one?
>> Lynnwood Public Library
>> Lynnwood WA
>> Sent via Deja.com
>> http://www.deja.com/I believe that you want to know how much the ëintensity.89 
of the light is
>increased, where the intensity is defined as the amount of power per unit
>area.  In the absence of all lens aberrations, absorption and all the
>other things that would complicate our analysis, the amount of power
>falling on the lens equals the amount of power falling on the small spot
>where the sun.89s image is focussed.  Thus, the ratio of intensities 
will
>be equal to the ratio of the areas of the magnifier and the sun.89s 
image,
>and this is equal to the square of the ratio of their diameters.
>The diameter of the magnifying glass (D(mag)) is easy enough to measure,
>but we need to calculate the diameter of the sun.89s image where it is
>focused.  If we were to look at the sun (and we shouldn.89t), its 
diameter
>would subtend an angle of about half a degree at our eye.  Similarly,
>from the centre of the magnifying glass, the diameter of the sun.89s 
image
>on the piece of paper subtends an angle of the same half degree.  This
>means that the image diameter is equal to .5*pi/180 times the focal
>length of the lens, which is its distance from the paper when the sun is
>in focus.  Thus, D(image)=0.0087 f, where ëf.89 is the 
focal length.
>So, the ratio of the diameters is equal to
>D(mag)/D(image) = D(mag)/.0087 f = 115 D(mag)/f
>Now, the ratio f/D(mag) is the so-called F number of a lens.  An F/4 lens
>(which you might find in a camera, for example) is a lens whose focal
>length is four times its diameter. Using this, the ratio of the diameters
>can be written as
>D(mag)/D(image) = 115/F
>(Note that I.89m using capital ëf.89 for the F number 
and small ëf.89 for the
>focal length.)
>Since the ratio of the intensities is the square of this, the intensity
>ratio can be expressed as
>(115/F)^2 or about 13000/F^2.
>Note that the intensity depends only on the F number.  A bigger lens with
>the same F number will put more power (or heat per unit time) onto the
>paper, but this power will be spread over a larger area.  The 
ëintensity.89
>will be the same.   If you want to burn a hole in a piece of paper, you
>need to use a lens with a small F number.  F numbers of 3 or 4 are good.
>Note that reading glasses with a focal length of 60 or 70 cm and a
>diameter of 2 or 3 cm have F numbers of about 20 or 30. You won.89t 
have
>much luck making a fire with these.
>Hope this helps.
>Ron
>To reply, remove the 'notat' from the address.
>


===

Subject: Re:  Chessboard knight metric, board symmetry preserved
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9AHAox13636;
>>Correction:
>>I just sent a mail saying for x=(1,1) and y=(1,2), then d(x,y)=4
>>completely overlooking the fact that the reader said assume
>>infinite chessboard. My example would be incorrect if the
>>infinite chessboard is represented by the product space of whole 
>>numbers but, of course, remains correct if the infinite chessboard 
>>represented by the product space of natural numbers. However, both
>>of these cases have a certain ugliness to me which comes from the 
>>fact that we are losing the flavor of the chessboard in these 
>>cases. We can try to keep the flavor of the board for infinite cases
>>by redefining our function d (notice that now all 3 cases-finite, 
>>infinte natural, infinite whole- reduce to one):
>>Assume we have the product space W of whole numbers. Now, speaking
>>figuratively outside of mathmatical jargon, place one chessboard with
>>its lower left corner at (0,0). Place another chessboard with its 
>>lower left corner at (0,9) etc. Continuing in this manner, we 
>>divide up the product space of whole numbers into patches of 64 
>>squares.
>>The set of of all patches P defines a function between itself and the
>>product space of whole numbers which is one-to-one and onto
>>(bijective).  
>>Now comes the hammer: 
>>Assume x and y are any members of the product space W. Let A be the 
>>board (or patch) x is on and B be the board y is on. Now let 
>>d(x,y) =d_P(A,B)+ d_*(x,y), where d_P is the knight metric on the 
>>patch space and d*(x,y) is knight metric distance left between x and
>>y when we assume both x and y (which are actually on boards A and B)
>>to be on the same board. Board symmetry is now preserved.
>C.Dement  
>>The above idea was close, but, apparently, does not completely >>preserve 

board symmetry. Again, make the following changes to the 
>>above idea:
>>As before, choose any x and y from W.
>>Relable P from above as P_1. In defining P_1, each chessboard was 
>>seen as a new point (representable by by two whole numbers).
>>Now, procede in defining P_2 by going back to the original
>>definition of P, but replacing references to W with P_1 and
>>references to P_1 as P_2. Each point of 
>>P_2 covers 64*64 Elements of W.
>>Continue the process until d_n(x,y) <= 8.
>>Finally, define d(x,y) = dP_n(x,y)+...+dP_1(x,y)+dP_0(x,y)    
>>where dP_0 is d* from above.
>>Sure, the idea may be crazy... it is now impossible to draw the unit 
>>sphere, for example- because it extends outwards to an infinte >>number of 

points, but hey, who said metrics couldn't be?
>C.Dement

===

Subject: Construct a bounded set of real numbers. Need help.
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9ABpOM22553;
I'm working out of Walter Rudin's Principles of mathematical analysis.  
Problem #5 from Chapter 2 is:
 
 Construct a bounded set of real numbers with exactly 3 limit points.
Would you please help me have steps (an algorithm) to solve this problem?  

===

Subject: Re: interesting problem arising in carpentry, column installation
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9AHAoj13646;
>> In the process of adding a column, I ran into a interesting problem. I
>> want to get the column, which is solid wood, with the least amount of
>> gap. Column flush with the floor and flush with the ceiling. I want to
>> use the least amount of wedges at the end to fill in the gap in the
>> ceiling. So the question is this. Given a diameter of column say 8
>> and because I cannot slide the erect column flush with floor and
>> ceiling but must instead have to tip it up. And the height is say 9'
>> from floor to ceiling. So, with these 8 diameter columns, what is the
>> smallest gap to occur in order to set the column in place? And let us
>> say the columns are not round but square of 6 by 6, and so how much
>> of a gap in order to place into this 9' position.
>> What I find interesting is that one can fit circular objects easier
>> than rectilinear objects. Or one can fit more volume density of a
>> circle than a square or rectangle.
>Let the height of the ceiling be C.
>For a circular column, let D be the diameter.
>For a square column, let D be the side of the square.
>Let L be the length of the longest installlable column.
>Then: L = square root(C^2-D^2)
>Sqrt[(9*12)^2 - 8^2] = 107.703  Gap~0.3 inch
>Sqrt[(9*12)^2 - 6^2] = 107.833  Gap~0.2 inch
>-- 
>Clive Tooth
>http://www.clivetooth.dk
But of course the entire column need not be this short,
only one-half of it (less if it's circlar in cross-section).
I believe that in the real world, the column is deliberately made
half an inch short and then a half-inch plywood base is slipped under
it so that the ceiling is flush. (You see gaps at the top more easily
than gaps along the floor.)
phil

===

Subject: Re: The Square Root of the Golden Section By an Iterative Method.
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9ALPj832060;
>> ...           
>>N)  Find  :   1 /[Sin{Arctan [X(N-1)] }]  =  X(N)
>>By such  successive iterations   the  value of this  calculation tends to 

that of the Square Root of the Golden Section T, 
>>[T  = SQRT[(SQRT[5]+1)/2 ]
>>T=1,27201964951406896425242246173749
>>This  is, also , the   Solution of :     [1]/[Sin{Arctan[X]} = [X]  ,
>> ...
>If, starting with some value x0, successive iterations of a continuous
>function f converge to a value x, then f(x) = x (prove this). So if
>your values X(N) converge, they must converge to a solution of the
>equation f(x) = x, where f(x) = 1/sin(arctan(x)).
>The identity sin(arctan(x)) = x/sqrt(1 + x^2) is easy to establish
>(e.g., draw a right triangle with base 1, height x, and hypotenuse
>sqrt(1 + x^2)). So your equation is equivalent to the equation
>sqrt(1 + x^2)/x = x, This equation can be easily solved: mutliply by
>x, square, substitute y = x^2, and solve the resulting quadratic
>equation. You get x = sqrt( (1 + sqrt(5))/2 ).
I thank You John Mitchell for Your response.
Yes ,  the Identity is based on such a triangle .
If k is the angle whose tangent is x/1 then sine of this angle k is
 x/sqrt(1+x^2),
 or, k=arctan(x)=arcsin{x/sqrt(1+[x^2])}=arcsin{1/sqrt(1+[1/x^2])} ,  or,
    sin[arctan(x)=1/sqrt(1+[1/x^2]), and for x^2=X , then x=sqrt(X) ,so
    1/[sin{arctan[sqrt(X)]}] = sqrt(1+[1/X]).
 For angle k= Theta=arctan(T) where T=sqrt[{sqrt(5)+1}/2] then ,
 x= T , and  sqrt[1+x^2]= T^2=x^2  , or 1+x^2=x^4 ,or x^4-x^2-1=0 ,
 and , tan(Theta)=T  , sin(Theta)=T/T^2= 1/T ,then ,
      arctan[T]=arcsin[1/T]), or arctan[x]=arcsin[1/x] or
      sin{arctan[T]}=1/T , or T=1/sin{arctan[T]} or ,
      T*(sin{arctan[T]})=1 .
Also , [tan(Theta)]*[sin(Theta)]=1 
Panagiotis Stefanides
http://www.stefanides.gr
panamars@otenet.gr
>The question of whether or not the iterative sequence converges
>requires more analysis. Examining the graph if f(x) and using the
>Contraction Mapping theorem shows that it does (if you start with a
>positive number). The Contraction Mapping theorem is a very
>interesting generalization of this type of result.. You can scan this
>newsgroup or look at a book on real analysis for discussions of this
>theorem. Better yet, think about it yourself first (what properties of
>the function f will guarantee that the iterates converge?).
>Try this one: start with an arbitraty number x0, and repeatedly apply
>the cosine function (this is easy to do with a calculator). What
>happens? Does the result depend on the initial value? How can you
>characterize the result?
>John Mitchell
>>COROLLARY  
>>------------------- 
>> 1/ [Sin{Arctan[SQRT(X)]} =SQRT [1+{1/(X)}]      
>>This is an  IDENTITY  (for  X any positive real number fraction or 
integer).
>>(amphidromical result).

===

Subject: Re: consecutive composite integers
> But the example he gave was 1500, 1501, ... 1509 which is not maximal.
> Also, the largest prime factor of a number in that range is 751, not
> 503.  The conjecture still holds, but not for the reason given.
You're right. I missed that. My conjecture is that the largest prime
factor of all the numbers in the sequence is 1. larger than the length
of the sequence and 2) is not raised to any power. I would exclude
short sequences like 8, 9. If the conjecture is true, then I can prove
that the product of consecutive integers can never be a perfect power
of an integer. (I know -- it's already been proven but I can't find
the proof and furthermore, I don't know the name attached to the
theorem.)

===

Subject: Technique of planning an elementary mathematical proof
Hello. I am just a self-studying math hobbyist; I like to help out
calculus students who post questions on the Live Math website.
Recently a young lady there has posted a question that I find
intriguing; trying to solve it, I realize that maybe what I lack is an
understanding of how real mathematicians go about analyzing the
requirements for a prospective proof.
Anyway: given that a > 0, b > 0, p > 1, q < infinity, and (1/p) +
(1/q) = 1,
prove that
ab <= (a^p)/p + (b^q)/q.
Of course, the stipulation that q < infinity also struck me as odd
(isn't it true that any real number is less than infinity?).
I've been able to ascertain that pq = p + q, and that q > 1.
How does a real mathematician go about planning his/her strategy for
proving a conjecture like this (assuming it isn't already known)?  I,
myself, have only advanced as far as about chapter 7 (Techniques of
Integration) in my favorite calculus text, but I suspect that the
proof of this proposition doesn't require calculus at all.
It's an interesting problem, and has gone so far as to make me submit
this entry even while the great Marlins - Cubs game is slipping deeper
and deeper into extra innings.  Any help will be appreciated.

===

Subject: Re: Max. Non-Adjacent Vertices on 120-cell
> Hello ...
> What's the size of a largest subset of vertices on the 120-cell
> ({5,3,3}) such that no two vertices in the subset are joined by
> an edge?
> I have reason to suspect that the number is an integer square,
> and the computer searches I've done come tantalizingly close,
> but I haven't yet been able to conclude that I've found a maximal
> set.
> Suggestions?
Well, I doubt I've caught up with you, since I just made
a simple pass at it based on some programs I have lying around
re 600 and 120-cell. 
I arranged the 600 vertices of the 120 cell into sets equidistant
from an arbitrary point, then looked at distances among these
sets and between them. There are 54 points in the equator between
the given point and its antipode, and these are the 15th nearest
neighbors to those two points.
There are 6 of these with no ( 1st ) n-n among the 54, and the
other 48 have one n-n each, in pairs. So we can pick 30 of these
( 6 + 48/2 ) and then radiate out to the poles. I don't know if
it matters which of each pair is picked, but it probably matters
which order you select candidates as you radiate.
Anyway, doing it arbitrarily I got 193 points, so I suppose you're
aiming at 196. Is this right?
Lew Mammel, Jr.

===

Subject: Re: Why should I switch to Python? - Infinity of Primes
        3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21:
>I don't know any Haskell, but a quick Python sieve is easy:
Your solution only gives you a bounded subsequence of the primes.  It's 
easy enough in Python to get the infinite sequence:
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/117119
-- 
David Eppstein                      http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer Science

===

Subject: Re: Joe Uptaught  (Was Re: David Ullrich on Identity)
A propos, here are cites from Pierre Bourdieu's
> _Language & Symbolic Power_ (the titles are mine).
> Enjoy!
> The Social Conditions for the Effectiveness of Ritual Discourse
                 Heretical Discourse
                      The *Skeptron*
               Symbolic Power & the Symbolism of Power
>                          Collusion in the 'Hood
                    Expertise as a Problem
...What is normatively correct usually functions differently
empirically.  A solution, a problem, indeed sometimes even the
object itself is often not all clearly and unequivocally recog-
nizable.  The expert's power of definition lies first and fore-
most in his definitions of the object and objectivity.  He
dictates what the problem is.  As much at variance as the experts
may be in individual cases, they do agree in their defence and
monopoly on defining object and objectivity.  This can only be
decided by experts.  The social exclusivity of the community of
experts forms the basis of a virtually unassailable position
towards the outside world, a version of infallibility.  If you're
not an expert, you don't know anything about the matter, and
you'd better accept that.  If you want to have a say anyway, you
have to acquire some expertise first--at the level determined by
the experts--which, under the given conditions, amounts to a
completely absurd reversal of the burden of proof.  If only
experts can decide questions regarding the facts, everybody else
is excluded.  The expert, too, leaves an opponent who doesn't
want to surrender to his dicta with no choice:  because the
asymmetries of knowledge due to the division of functions in
society can on principle not be annulled, his opponents must
mobilize other sources of legitimization and interference, the
factor of involvement.  They must compensate for missing
expertise with outrage and transform expert questions into moral
ones... (p. 34)
                  Camaraderie of the Experts
...When the expert encounters open distrust, or strong pressure
to legitimize himself is exerted, particularly from the outside,
he takes refuge in the zone of insecurity which he dominates. 
Instead of answering questions he demonstrates expert knowledge
and tries to intimidate the questioner.  To preserve his expert
power, he inflates it.  So as not to stand completely alone in
taking the bull by the horns, he quickly looks for allies; other
experts who support his position and endorse his status.  That
dog does not eat dog holds true for the various experts in the
various fields.  Even though they understand little of the
other's field, the credo of expertise unites them.  But it is
precisely this camaraderie of the experts which foments suspicion
and confirms doubts.  When other non-expert experts jump to the
assistance of the attacked expert in a kind of reflex, the
chances for recognition of expert authority rapidly diminish. 
The ethos of expertise turns out to be a grafted-on, elitist
group ethic which suspends its own principles when necessary.
(p. 35) (Rainer Paris, 'A Telling Question: Expertise as a Problem',
_Universitas_, V 36, 1994, pp. 32-41)

===

Subject: Re: ab... = (a*b*...)^n  ?
Subj:    Re: ab... = (a*b*...)^n  ?
> You get a lot further by examining only those
> numbers whose prime factors are less than B.
> Still nothing in base 10 up to 10^200.
> Would you describe further how you did that? ...
The basic idea is to generate all numbers whose prime factors
are less than  B.  I initialize a set  S={1}.  I repeatedly remove
the smallest element  e  from  S  and for each prime  p < B  I
insert the product  e p  into  S.  This is especially fast if  S
is implemented as a priority queue (see Knuth v3).
An optimization to avoid inserting any element twice is to
insert only products  e p  such that  p  is the smallest prime
factor of  e p.
A further optimization for composite  B  is to avoid inserting any
product  e p  that is a multiple of  B,  since that number and its
multiples have zeroes in their  base-B  expansions.
> Obviously, you didn't test each number to see
> whether its prime factors were less than 10, ....
If there are  k  primes less than  B,  there are only  O(n^k)
elements to process below  B^n.  If  B  is composite, the
second optimization reduces this to  O(n^(k-1)).  Either
of these is vastly smaller than B^n.
Dan Hoey
haoyuep@aol.com

===

Subject: Re: How'd they do it?
> That's not even taking into consideration their value as the world's
> standards.
> The value as a standard is much higher than the material value.
> 
Baloney juice: A kilogram is a unit of mass whose inertia [f/a] is 1 N
secî/m; with a gravitational inertia of about 2.204# 
secî/32.174'; where
2.204# is the weight of 2.204 pints of water at its maximum density.
Any good mathematician can work out that equation. Tain't worth loosin'
sleep over.

===

Subject: Re: How'd they do it?
> The official kilogram is a hunk of fancy metal guared by snooty
> Frenchmen in Paris.
> Guarded?
That's what it _is_! Can you imagine that it's that valuable?

===

Subject: Re: Construct a bounded set of real numbers. Need help.
>  Construct a bounded set of real numbers with exactly 3 limit points.
A = { 1/n | n in N } is bounded and has limit point.
Now shift A by 2 and by -2 and take union of all three.

===

Subject: Re: Technique of planning an elementary mathematical proof
> Anyway: given that a > 0, b > 0, p > 1, q < infinity, and (1/p) +
> (1/q) = 1,
> prove that
> ab <= (a^p)/p + (b^q)/q.
Fix a > 0 and consider the function f(x) = (a^p)/p + (x^q)/q - ax, for x > 
0. Using basic calculus, you can show that the minimum value of f(x) is 0. 
Thus f(x) >= 0 for all x > 0. Because a was arbitrary, this proves the 
inequality.
That's the quick proof. But the right way to see this, and to see how one 
could discover it, is through convexity/concavity. Details on request.

===

Subject: Re: determine if 2 rectangles no intersections
> How to determine if 2 rectangles don't have intersections?
Are their axes parallel to the coordinate axes
or are they in general position
(i.e., at an arbitrary angle)?

===

Subject: Re: Question about Door Spaces
>This is an exercise in Kelley's text on topology (not a homework
>question!) that I ran across and can't solve. A topological space X
>is a door space if every subset of X is either open or closed (or
>both). Prove that a Hausdorff door space has at most one
>accumulation point.
>>Suppose x and y are two different accumulation points. Choose disjoint
>>open sets U and V with x in U and y in V. Let W be the set obtained
>>from U by deleting x and adding y. So W is either open or closed. If W
>>is open, then {y}, being the intersection of V and W, is an open set.
>>If W is closed, then the intersection of U with the complement of W,
>>namely {x}, is an open set. Either way, x and y can't both be
>>accumulation points.
> This proposition is trivial in Window spaces!
> Lurch
Window spaces are always open (especially to hackers)

===

Subject: De Moivre formula
I have some question about De Moivre formula.
I have been simply accepted that formula as : (cos t + i*sin t)^n =
cos(tn)+i*sin(tn)
and I regard it as true for every t and n.
but, in case t=Pi/6 and n=2, De Moivre formula doesn't hold.
So, I'm so confusing with the condition for holding De Moivre formula.
If anyone have obvious answer about my question,
please post reply. i need help..

===

Subject: Re: Curve Inside Another Of Equal Perimeter
>Are there any two CONVEX closed curves of equal perimeter, where one
>can be placed completely inside the other?
>(It is allowed that the curves touch at a finite number of points at
>most.)
>Leroy Quet
No. The outer curve must be longer than the inner curve. This is easy
to see using Crofton's formula (a.k.a. the Cauchy-Crofton formula).
John Mitchell

===

Subject: Re: Curve Inside Another Of Equal Perimeter
> Are there any two CONVEX closed curves of equal perimeter, where one
> can be placed completely inside the other?
> (It is allowed that the curves touch at a finite number of points at
> most.)
> Leroy Quet
I would guess not. Here is an outline of a possible proof.
Choose a point inside both curves.
Consider the formula for the arc length as a function
of r(theta) and theta, where r(theta) is the distance from
the central point to the curve at angle theta.
Since r_1(theta) <= r_2(theta), where r_1 and r_2 are
the respective distances to the inner and outer curves,
the length of the outer curve is greater than the
length of the inner curve.
Martin Cohen

===

Subject: Re: De Moivre formula
> I have some question about De Moivre formula.
> I have...accepted:
> (cos t + i*sin t)^n = cos(tn)+i*sin(tn)
> and I regard it as true for every t and n.
> but, in case t=Pi/6 and n=2, De Moivre
> formula doesn't hold.
It does hold.
Re-check your algebra.

===

Subject: Re: De Moivre formula
> I have some question about De Moivre formula.
> I have been simply accepted that formula as : (cos t + i*sin t)^n =
> cos(tn)+i*sin(tn)
> and I regard it as true for every t and n.
> but, in case t=Pi/6 and n=2, De Moivre formula doesn't hold.
> So, I'm so confusing with the condition for holding De Moivre formula.
> If anyone have obvious answer about my question,
> please post reply. i need help..
For n = 2 and t = pi/6,
(cos(t) + i*sin(t)^n = (sqrt(3)/2 + i/2)^2 = 1/2 + i*sqrt(3)/2
and
cos(2*t) + i*sin(2*t) = 1/2 + i*sqrt(3)/2.
So what part of the formula does not hold?

===

Subject: Re: Construct a bounded set of real numbers. Need help.
> I'm working out of Walter Rudin's Principles of mathematical analysis.  
> Problem #5 from Chapter 2 is:
>  Construct a bounded set of real numbers with exactly 3 limit points.
> Would you please help me have steps (an algorithm) to solve this problem?  

Construct a bounded set, S, of reals with exactly one limit point 
and form its union with two translations of it.

===

Subject: Re: De Moivre formula
> I have some question about De Moivre formula.
> I have been simply accepted that formula as : (cos t + i*sin t)^n =
> cos(tn)+i*sin(tn)
> and I regard it as true for every t and n.
Excellent news (as long as you're assuming that n is an integer).
> but, in case t=Pi/6 and n=2, De Moivre formula doesn't hold.
But you just said you regarded it as true for all t and n.
So you regard it as true for t = pi/6 and n = 2 but you also
regard it as false for t = pi/6 and n = 2.
? :-(
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)

===

Subject: Re: ARCSIN function, single precision floating point.  -- Example 
routine needed?
> I'm trying to write ATAN2 function for a small basic language that has
> IEEE single precision math..  *,/,+.-, SQRT(), SIN(), COS(), TAN() are
> availible in the language. 
> I've tried a few methods I've found but the results are way off due to
> low precision, rounding, etc. 
> Are there any repositorys of old fortran routines or algorithms that I
> could use to get a good accuracy single precision routine.  Speed or
> space aren't as important as reasonably good accuracy.
==
Let us introduce following constants T(j)= tan(j*pi/24) ,
 j= 1,2,..., 11 and PI2=pi/2 .  More precisely
PI2=pi/2=1.570796 326794 896619 231...
T(1)    =0.131652 497587 395853 472...
T(2)    =0.267949 192431 122706 473...
T(3)    =0.414213 562373 095048 802...
T(4)    =0.577350 269189 625764 509...
T(5)    =0.767326 987978 960342 923...
T(6)    =1.000000 000000 000000 000...
T(7)    =1.303225 372841 205755 868...
T(8)    =1.732050 807568 877293 527...
T(9)    =2.414213 562373 095048 802...
T(10)   =3.732050 807568 877293 527... 
T(11)   =7.595754 112725 150440 526...             .
Further denote 
(*)      F(z) = z*P(z)/Q(z)   with
   P(z)= 1155 + 1190*z^2 +231*z^4   ,  
  Q(z)= 1155 + 1575*z^2 + 525*z^4 + 25*z^6 , and
 I=[0,T(1)] , I(j)=[T(j), T(j+1)] , j=1,2,...,11 .
Note that 
(1)  arctan(z)=approx.= F(z)  when -T(1) =< z =< T(1) .
Observe that both functions arctan(x) and F(x) are odd functions.
In the following we shall consider x in [0,infty) .
When x in I(j) , then (x-T(j))/(1+x*T(j)) is in I(1) . 
Using equality
 arctan(x)= arctan(z(j)) + j*PI/24 where z(j)= (x-T(j))/(1+x*T(j)) ,
according to (1) we  make approximations
(1.1) arctan(x) =approx.= F(z(j)) + j*Pi/24    when  x in I(j) ,
and 
(1.2) arctan(x) =approx.= Pi/2 -F(1/x)    if   x in (T(11),infty) .  
Therefore, a possible routine may be written in FORTRAN in the following way 

:
       FUNCTION ATAN(X)
       DIMENSION T(1)
       PI2 =1.570796 326794 896619 231
       T(1)=0.131652 497587 395853 472
       T(2)=0.267949 192431 122706 473
       T(3)=0.414213 562373 095048 802
       T(4)=0.577350 269189 625764 509
       T(5)=0.767326 987978 960342 923
       T(6)=1.000000 000000 000000 000
       T(7)=1.303225 372841 205755 868
       T(8)=1.732050 807568 877293 527
       T(9)=2.414213 562373 095048 802
       T(10)=3.732050 807568 877293 527 
       T(11)=7.595754 112725 150440 526
       C=1.       
       IF(X) 1,2,3
2      ATAN=0.
       RETURN     
1      X=-X
       C=-1.
3      IF(X-T(1)) 21,22,23
21     Z=X*X
       P= 1155. + 1190.*Z +231.*Z*Z   
       Q= 1155. + 1575.*Z + 525.*Z*Z + 25.*Z*Z*Z       
       ATAN= C*X*P/Q    
       RETURN
22     ATAN=C*PI2/12
       RETURN
23     DO 100 j=1,10
       H=j*PI2/12. 
       ZJ=(Y-T(j))*(Y-T(j+1)) 
       IF(ZJ) 10,20,100
20     ATAN =C*H           
       RETURN
10     Z=ZJ*ZJ
       P= 1155. + 1190.*Z +231.*Z*Z   
       Q= 1155. + 1575.*Z + 525.*Z*Z + 25.*Z*Z*Z
       F=ZJ*P/Q
       ATAN= C*(F+H)    
       RETURN
100    CONTINUE
       W=1./X              
       V=W*W
       P= 1155. + 1190.*V +231.*V*V   
       Q= 1155. + 1575.*V + 525.*V*V + 25.*V*V*V
       F=W*P/Q
       ATAN= C*(PI2-F)    
       RETURN
       END
Note : instead of approximation (1) you may consider
(1') arctan(z) =approx.= 
    = F(z):=a(1)*z+a(3)*z^3 +a(5)*z^5 +a(7)*z^7 + a(9)*z^9 
for -T(1) =< z =< T(1), where
a(1)=  0.999999 99843
a(3)= -0.333332 89364
a(5)=  0.199965 34780
a(7)= -0.141734 60613
a(9)=  0.094919 54952 ,
and then to extend the approximation on whole real axis (see (1.1)-(1.2)) .
Using (1') one has
   |arctan(z)-F(z)| =< 2^{-42} for z in [-T(1),T(1)]. 
It's better to approximate by means of rational function from (*).

===

Subject: Re: De Moivre formula
Ooops!!!
I misunderstood sin <=> cos  for solving following exercise.
: What is sum of natural #s n s.t. (sin t + i*cos t)^n=sin(nt)+i*cos(nt) ?
Now, I see my misunderstanding point.
> I have some question about De Moivre formula.
> I have been simply accepted that formula as : (cos t + i*sin t)^n =
> cos(tn)+i*sin(tn)
> and I regard it as true for every t and n.
> Excellent news (as long as you're assuming that n is an integer).
> but, in case t=Pi/6 and n=2, De Moivre formula doesn't hold.
> But you just said you regarded it as true for all t and n.
> So you regard it as true for t = pi/6 and n = 2 but you also
> regard it as false for t = pi/6 and n = 2.
> ? :-(
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> Needless to say, I had the last laugh.
>  Alan Partridge, _Bouncing Back_ (14 times)

===

Subject: Re: Partition of R
  problem of finding sets A and B that form a partition of R and are
> such that every interval of R contains uncountably many points of A
> and B. In other words, every element of R should be a condensation
> point of A and B.
> You can do far better than that.
> R is a maximally resolvable space,
> i.e., R can be partitioned into c=2^w many disjoint dense subsets.
> Here is one way to do it.
> P = {Q + x | x in R} partitions R into 2^w counatble sets.
Ok, each Q+x is countable and there are c of them.
> Partition P into PP consisting of 2^w subsets each of size 2^w.
Not just countable but of size c. How do I do that?
> Then { US | S in PP } satisfies the bill.
US is the union of S.  S in PP, means that S is collection of Q+x's.
> The jist is partitioning 2^w into 2^w sets of size 2^w. Getting dense is
> easy.

===

Subject: Re: Two similar things
> Since they now are identical-with-somethings, if they shared
> all the same properties they would be one cake rather than
> two.  THerefore, since they are two cakes, they don't share
> all their properties, and I can distinguish them by means
> of any property that one has and the other doesn't.
> No, there's a foot wrong there. Why would they be one cake if they shared
> all the same properties? 
Consider Hesperus and Phosphorus.  The properties of Hesperus include
identity-with-Hesperus, and those of Phosphorus include identity-with-
Phosphorus. So, if Hesperus and Phosphorus share *all* their properties,
the properties of Hesperus will include identity-with-Phosphorus, and
those of Phosphorus will include identity-with-Hesperus.  But if Hesperus
and Phosphorus have these properties, then Hesperus and Phosphorus are
one and the same: that is, numerically identical.
> Why not 3 cakes? 
If these cakes have all the same properties, for the reasons I have 
indicated
these three cakes will be one and the same.
> Would you tell the difference by counting?
Just by counting, no.  By counting correctly, yes.  In other words,
before it was discovered that Hesperus was Phosphorus, it was assumed
that these heavenly bodies were different heavenly bodies rather than
one and the same heavenly body.  But this assumption proved to be wrong.
--John

===

Subject: Re: quantum echo
> Immortalist: 
>   Stop crossposting this to sci.physics.relativity
>> Are you saying it is possible for something to exist at location 
A
> [*SNIP*]
Paul Teller having proposed, as an explanation for the divergence
between quantum statistics and classical statistics, that 'quanta'
lack haecceities, I posted to sci.physics.relativity what I took
to be the import of Teller's position for the logic of identity.
On this topic a fair amount has been written by philosophers of
science who, unlike myself, are knowledgeable about the formal
details of quantum mechanics.  What these people have had to
say on the topic may be of interest to some on
sci.physics.relativity...
--John

===

Subject: Re: How'd they do it?
>Cut<
>> That's not even taking into consideration their value as the world's
>> standards.
>> Of course, guarded.
>> Gene Nygaard
>> http://ourworld.compuserve.com/homepages/Gene_Nygaard/
>Even though physics would be better off with_out_ 'em; just using a
>mathematical ratio: Like 1 kg = 2.2# secî/32.2'; using 2.2 
pints of water
at
>its maximum density, per kilogram.
>  Gentlemen of the jury, Chicolini here may look like an idiot,
>  and sound like an idiot, but don't let that fool you: He
>  really is an idiot.
>                     Groucho Marx
'Though you don't admit it, the slug's doin' okay without an artifact; 
using
the mathematical equation: 1 slug = f/a = 1# secî/foot = 32' 
secî/32'.
In physics, the slug is a _unit_ of mass derived from the fundamental
concepts of the foot-pound-second British Gravitational System of weights
and measures.
In trade, the gram and kilogram are _absolute_ units of mass; with the
kilogram artifact being the head honcho.

===

Subject: Re: How'd they do it?
>>Cut<
 That's not even taking into consideration their value as the world's
> standards.
 Of course, guarded.
> Gene Nygaard
> http://ourworld.compuserve.com/homepages/Gene_Nygaard/
>>Even though physics would be better off with_out_ 'em; just using a
>>mathematical ratio: Like 1 kg = 2.2# secî/32.2'; using 
2.2 pints of water
>>its maximum density, per kilogram.
>>  Gentlemen of the jury, Chicolini here may look like an idiot,
>>  and sound like an idiot, but don't let that fool you: He
>>  really is an idiot.
>>                     Groucho Marx
>'Though you don't admit it, the slug's doin' okay without an artifact; 
using
>the mathematical equation: 1 slug = f/a = 1# secî/foot = 32' 
secî/32'.
>In physics, the slug is a _unit_ of mass derived from the fundamental
>concepts of the foot-pound-second British Gravitational System of weights
>and measures.
 Gentlemen of the jury, Chicolini here may look like an idiot, 
 and sound like an idiot, but don't let that fool you: He 
 really is an idiot. 
                    Groucho Marx
Slugs are a little used 20th century invention, which don't even have
an official definition.  There is no official standard for either a
slug or a pound force.
There are absolutely no pints in the only subsystem which includes
slugs.  Dishonest Don S*head already knows this.
Pounds do have an official definition:
    1 lb = 0.453 592 37 kg, EXACTLY
Dishonest Don KNOWS this also, of course.
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

===

Subject: Q wrt a number-theoretic generating function
I'm reading some introductory material about Dirichlet's generating
functions of arithmetical functions, but could not find anything about
the obvious generating function
Sum_p p^{-x},
where the sum is extended over all primes. Has it been studied? Is it
an independent function or can it be expressed in terms Riemann's
zeta? Anything else interesting about it? References?
Michele
-- 
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened. :)
- Tore Aursand on comp.lang.perl.misc

===

Subject: Force
substance, and bodies thereof; due to their impenetrability: The fact that
the exact same place: Where they resist each other's passage with thrusts
that are proportional to the amount of material substance they contain.
Weight in physics is that particular force exerted between a planet's terra
further passage toward their common center of mass is restrained by a 
thrust
contains.
The magnitude of the thrusts are commonly measured with pushes or pulls of
weight-scales.

===

Subject: Variation on an impossible problem
Allowing only the use of the + operator, is it possible to use each of
the digits 0 1 2 3 4 5 6 7 8 9 once only to make a total of 1000?
One can combine digits to create sums like 210 + 345 + 6 + 78 + 9 =
648 etc.
(This is obviously similar to use only + and 1 2 3 4 5 6 7 8 9 to make
100 which is impossible)
Mitch.

===

Subject: Re: Variation on an impossible problem
> Allowing only the use of the + operator, is it possible to use each of
> the digits 0 1 2 3 4 5 6 7 8 9 once only to make a total of 1000?
> One can combine digits to create sums like 210 + 345 + 6 + 78 + 9 =
> 648 etc.
No. sum(i,i=0..9) = 0 mod 9, whereas 1000 = 1 mod 9.
--
Julien Santini,
France.

===

Subject: Discontinuous Derivative
I'm looking for a real function f:R -> R
which is differentiable over all of R
yet with a derivative discontinuous at 0.

===

Subject: Re: Discontinuous Derivative
> I'm looking for a real function f:R -> R
> which is differentiable over all of R
> yet with a derivative discontinuous at 0.
f(x) = x^2*sin(1/x), f(0)=0
f '(x) = 2*x*sin(1/x)-cos(1/x)
f '(0) = 0,
but lim x->0 f '(x) doesnt exist.

===

Subject: Re: Discontinuous Derivative
> I'm looking for a real function f:R -> R
> which is differentiable over all of R
> yet with a derivative discontinuous at 0.
Standard example: x^2 * sin(1/x) , extended to 0 as 0.
Variants: like x^2 * sin(1/x^2) ; this one has derivative
unbounded near 0.

===

Subject: Re: Chessboard knight metric?
>Much to my surprise, you all seem correct about the metric part
>that the proof was in the definition itself.
>I also appreciate the reader pointing out the regular shapes.
>I had seen these shapes before- and would caution that the
>shapes are perhaps as not as regular as they may seem. The octogon
>that comes out at n=3 does not seem to be a true octogon (mabie I'm
>missing something again...) and for n = 4 the situation appears even
>worse, with spaces way out on the perepherie being reached in (a
>minimum of) 4 moves as well as spaces close to the starting square.
>This was already pointed out by another reader- take x=(1,1) and 
>y=(1,2), then  d(x,y)=4 and y surely cannot lie on the outer edge of
>the octogon. Of course, it seems correct to point out that the
>octogons keep turning up as part of the picture again and again for
>higher n.
>About the even / odd properties I mentioned before... at least that
>does appear to be regular for all n, doesn't it?
>C.Dement
Can you nevertheless learn to post properly? Your entire post appears to
be a quote, even though it comes from you. It's hard to follow your
writing.
-- 
Ioannis
http://users.forthnet.gr/ath/jgal/
___________________________________________
Eventually, _everything_ is understandable.

===

Subject: Re: Fundamental Reason for High Achievements of Jews

===

>Subject: Re: Fundamental Reason for High Achievements of Jews
> :> : Most historians believe that Jews avoid pork,
>> :> : because the ancient Jews associated pigs with leprosy,
>> :> : and pigs and people with leprosy were unclean.
>> :> 
>> :> Name one historian who believes that, and give a citation to 
the
>> :> place where he says it.
>> : 
>> : As I recall, this was in Tacitus' Histories  
>> : which was written in the first century A.D.
>> 
>> If that's the best you can do, I think that we can safely ignore 
your
>theory.
It's soooo easy Richard.  Someone in your camp
>should have the information at his fingertips.
GOOGLE:  tacitus histories                ~13,600 hits
>         tacitus histories jews            ~3,950 hits
>         tacitus histories jews leprosy      ~154 hits
This is all irrelevant. 
Quoting from Tom Potter:
 Almost everyone has access to the first hand historical accounts,
> and can do wild card searches on the source material.
> It is STUPID to provide detailed cites, as these focus
> ONLY on the POINTS trying to be emphasized by the writer.
> It  also STUPID to use second hand, accounts which have a
> racial, religious, national, or personal spin on them, rather than
> using the FIRST HAND historical accounts.
> 
Anything written by Tacitus would NOT be a FIRST HAND account, 
> so only a STUPID person would allow himself to be brainwashed
> by Tacitus' racial, religious, national, personal spin on history.
It is interesting to see that Mensanator, 
> like many people, has been brainwashed to think that 
> Tacitus, who was one of the most unbiased, rational, 
> and correct historians, put a racial, religious
> spin on history.
> Liar. YOU are the one who says that second hand accounts put a 
> spin on history. Or are you going to deny that that was your quote?
> I wouldn't try it, I located it via a Google search, others can too.
> This attitude obviously has its' roots in 
> conditioning, as the works of Tacitus 
> are far more rational and correct, 
> than the bible, the Greek and Roman mythologies, etc.
> But they are not a FIRST HAND account. So your brainwashed opinion of
> their validity is irrelevant. Or are you now admitting that the quotation
> about second hand accounts was stupid. You can't have it both ways,
> so which is it?
> In other words, people who have been brainwashed to a point of view, 
> have a great difficulty in accepting data 
> that conflicts with their conditioning.
> In other words, Tom Potter is too stupid to realize that he's just been
> hoisted by his own petard.
It is interesting to see that Mensanator 
does not comprehend that Herodutus, Tacitus,  Josephus, 
and the ancient Greek and Roman Historians, 
are the closest FIRST HAND accounts of ancient history, 
and that works of modern historians use these works 
as their starting reference points.
If you want to know the facts about ancient history, 
you start with the works of the ancient historians, 
so you can detect the spin put on 
history by modern historians who impose their 
present day agendas on the works.
writers from different religions and nations, 
put a personal spin on the history, 
and a lot of information is lost 
as it is played through fallible, tuned filters.
No doubt, some original historians put a personal 
spin on their works, but this does not seem to 
be the case with Tacitus, Herodutus, Thucydides, Xenophon,
and Polybius, although one can detect a little Delphi Oracle 
spin in Herodutus' works.
--
Tom Potter

===

Subject: Re: Question about the L'Hospital Rule
Maybe, I was not clear enough. My post was motivated by the following 
paper:
Counterexamples to L'H.99pital's Rule
R. P. Boas, Northwestern University, Evanston, IL 60201
The American Mathematical Monthly, October 1986, Volume 93, Number 8, pp.
644.9a645.
This paper can be found online on the address:

===

Subject: Re: Discontinuous Derivative
  R
> which is differentiable over all of R
> yet with a derivative discontinuous at 0.
> Standard example: x^2 * sin(1/x) , extended to 0 as 0.
f'(0) = lim(h->0) h sin 1/h = 0
f'(x) = 2x sin 1/x - x^2 (cos 1/x) x^-2
        = 2x sin 1/x - cos 1/x
> Variants: like x^2 * sin(1/x^2) ; this one has derivative
> unbounded near 0.
f'(0) = lim(h->0) h sin 1/h^2 = 0
f'(x) = 2x sin 1/x^2 - x^2 (cos 1/x^2) 2x^-3
        = 2x sin 1/x^2 - (2/x) cos 1/x^2
Wild!

===

Subject: hello...my question is ??
f(0) = 0
integral (0 ~ 1) f(x) dx = 1
f'(x) is continuous
find M such that f'(x) <= M
-----------------------------
answer : M >= 2
how do you solve it?

===

Subject: Re: Discontinuous Derivative
> I'm looking for a real function f:R -> R
> which is differentiable over all of R
> yet with a derivative discontinuous at 0.
Integrate the Wierstrass Function which is everywhere continuous (so the 
Riemann integral exists) and nowhere differentiable.

===

Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws
Totally bogus crap.

===

Subject: Re: Just what is an L-series?
>Ah so. I was trying to compose a definition of L-series for one of the
>volunteer-written online websites (www.planetmath.org). Maybe in the end,
>L-series are not a structure but only a formalism, like generating
          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>functions. I may need to resort to arm-waving or (what is much the same
>thing) category theory:)
I'm not sure what you mean exactly, but isn't
F(x)=Sum a_n u_n(x)
for a given sequence {u_n} of functions a general enough definition of
generating function?
Michele
-- 
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened. :)
- Tore Aursand on comp.lang.perl.misc

===

Subject: Re: Just what is an L-series?
>>Ah so. I was trying to compose a definition of L-series for one of the
>>volunteer-written online websites (www.planetmath.org). Maybe in the end,
>>L-series are not a structure but only a formalism, like generating
>          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>>functions. I may need to resort to arm-waving or (what is much the same
>>thing) category theory:)
>I'm not sure what you mean exactly, but isn't
>F(x)=Sum a_n u_n(x)
>for a given sequence {u_n} of functions a general enough definition of
>generating function?
And isn't what you've just written a perfect example of a formalism?
Where's the structure?
Lee Rudolph

===

Subject: Re: Max. Non-Adjacent Vertices on 120-cell
> What's the size of a largest subset of vertices on the 120-cell
> ({5,3,3}) such that no two vertices in the subset are joined by
> an edge?
> I have reason to suspect that the number is an integer square,
> and the computer searches I've done come tantalizingly close,
> but I haven't yet been able to conclude that I've found a maximal
> set.
found a few notes on the web, if I understood it correctly the
indenpendence number lies in the interval [220, 224]. So it cannot be
a square.
www.csr.uvic.ca/~wendym/courses/582/120cell.ps

===

Subject: Re: Ridicule This Crackpot
> My big question is how can I insure that I receive credit for this
> formula and algorithm? (Assuming that mathematicians decide
> that it is novel and concede the slightest bit of significance.) Any
> advice would be appreciated.
My advice:  you're in the wrong field.
Why don't you try investment banking, or better yet marketing?

===

Subject: Re: Boolean Algebra - Arithmetic Relationship
> 3)Can all known Mathematical Notation and Symbolic manipulation be
> modeled by a turing machine??
> If it's recursive it can be represented by a Turing Machine.
> What's the significance of recursion? I'm confused by this.  (I've
> written non-recursive programs before of course.) Are all known
Just to confuse you; your non-recursive programs that compute functions
_are_ computing _recursive_ functions as mathematicians/logicians
understand the word.
> mathematics recursive?
What is a _computable_ function?  Who knows?  Mathematicians (notably
Turing and Church) have _defined_ computable function to be recursive
function, where recursive function has a precise meaning.
> The essential question forming in my mind is this: Is there a
> fundamental difference between mathematical symbolic notation and
Symbols can be encoded with positive integers, so can strings of
symbols.
> computer programming languages??  Both represent logical structures
Computer programs are limit by technology and human patients.  If we
ignore those limitations, then every recursively definable manipulation
of symbols can, in theory (:-), can be carried out by a computer.
> and their interactions. Both follow certain rules of manipulation.
> For me it's looks like they're two sides of the same coin.
> Are their certain properties of Zermelo-Fraenkel set theory that can't
> be modeled with a Turing-Machine?
I don't know what a properties of Zermelo-Fraenkel set theory is.  A
Turing Machine cannot prove all ZF theorems but it can check the
correctness of any purported proof.
ZF can't settle all mathematical problems.  No single formal system can.
> response.
> -Steve
I'd like to recommend some reading.  For logic generally, see Tarski An
Introduction to Logic and the Methodology of the Deductive Sciences
OUP.  (Dreadful title, best book.)  For recursion theory I'm at a bit of
a loss, how about Boolos & Jeffrey Computability and Logic CUP?
-- 
G.C.

===

Subject: Re: 3D Surveyor's Formula (for volume)?
>Is there a version/generalization of the Surveyor's Formula for three
>dimensions?
>I'm imagining inputting a bunch of (x,y,z) points, and receiving as
>output the volume of the polyhedron thus delimited.
>Of course, there will be issues involving exactly _how_ the points are
>given - presumably it's more complicated than the 2D version - but it
>seems like it should be possible....
>Anyone know of the 3D version?
>cdj
See subject 5.19 of http://www.faqs.org/faqs/graphics/algorithms-faq/

===

Subject: Re: Question about the L'Hospital Rule
>Maybe, I was not clear enough. My post was motivated by the following 
paper:
>Counterexamples to L'H.99pital's Rule
>R. P. Boas, Northwestern University, Evanston, IL 60201
>The American Mathematical Monthly, October 1986, Volume 93, Number 8, pp.
>644.9a645.
>This paper can be found online on the address:
You were perfectly clear. I explained that there are conditions
under which the rule is correct stated in any calculus book.
That's true. The paper you cite does not contain any counterexamples
to the theorem in decent calculus books.
In particular I pointed out, in case you couldn't find a calculus
book, that for example, if f -> 0 and g -> 0, and if f'/g' -> L then
f/g -> L. Also if f -> infinity and g -> infinity. That paper does
not contain any counterexamples to what I asserted here.
(It contains examples where it's _not true_ that f'/g' -> L,
but where a student might nonetheless think the rule applied.)
************************
David C. Ullrich

===

Subject: Re: hello...my question is ??
>f(0) = 0
>integral (0 ~ 1) f(x) dx = 1
>f'(x) is continuous
>find M such that f'(x) <= M
Does this mean we need to find an M such
that f'(x) <= M for all x, or for at least one x?
>-----------------------------
>answer : M >= 2
>how do you solve it?
************************
David C. Ullrich

===

Subject: Science is a human activity (was: Python syntax in Lisp and 
Scheme)
Originator: claird@lairds.com (Cameron Laird)
>Alex Martelli:
>> would you kindly set right the guys (such as your
>> namesake) who (on c.l.lisp with copy to my mailbox but not to here) are
>> currently attacking me because, and I quote,
>> Software is a department of mathematics.
>And anyone who doesn't think mathematics has its own
>culture with ideas and even mistaken preferences for what
>is right and wrong should read
>The Mystery of the Aleph: Mathematics, the Kabbalah, and the Human Mind
>to see how Cantor's ideas of transfinite numbers (and other ideas,
>as I recall, like showing there are functions which are everywhere
>continuous and nowhere differentiable) were solidly rejected by
>most other mathematicians of his time.
>Mathematicians are people as well.
                        .
                        .
                        .
And let no one assume that these are mere foibles of the
past that we moderns have overcome; mathematics remains
stunningly incoherent in what's labeled foundations.
There's a wide, wide divergence between the intuitionism
working mathematicians practice, and the formalism they
profess.
'Good thing, too; our age enjoys the blessing of superb
mathematicians, and I'm relieved that philosophical in-
consistencies don't (appear to) slow them down.
-- 
Cameron Laird 
Business:  http://www.Phaseit.net

===

Subject: Re: Question on Hilbert & Godel
> What did Hilbert ask and claim concerning Foundations of Mathematics
> (sets, predicate calculus), metamathematics, Logic, Incompleteness,
> etc?
> As a mathematician, he did not *claim* things he could not prove.
Didn't he claim that a certain mathematical problem must have a
solution when in fact it doesn't?
> he did not reach
> this goal which was shown by G.9adel to be unreachable. One cannot say
> he was contradicted; every serious mathematician has goals that
> prove unreachable - only claiming to have reached such goals can be
> contradicted.
What about claiming that a particular goal is reachable only to have
it proven that the goal is not reachable?  Is that a contradiction?
Charlie Volkstorf
Cambridge, MA
> Helmut Richter

===

Subject: Re: Question on Hilbert & Godel
> What did Hilbert ask and claim concerning Foundations of Mathematics
> (sets, predicate calculus), metamathematics, Logic, Incompleteness,
> etc?
Just a side note:
>        Bourbaki was cited as the best example so far of mathematics
>        organized into a coherent framework. According to Andre Weil,
>        Perhaps the most important contribution of Bourbaki was to
>        carry out a famous proposal made by the great German mathe-
>        matician David Hilbert in 1900 that mathematics be placed on 
>        a more secure foundation. He noted: Hilbert just said so,
>        and Bourbaki did it
And just how did Bourbaki do as Weil claims?
But didn't Hilbert actually claim that a lot more than that is
possible?  Didn't he ask for (and claim that it must exist) a decision
procedure to determine if an arbitrary predicate calculus wff is valid
(true of all interpretations) - was it?
wisdom was wrong and mathematicians gained new insight into the
nature of mathematics.
Or are you one of those diehards who clings to the failed ideas and
theories of the past?
(Actually, the most secure foundation possile is software that carries
out both logic and metamathematics, as I have implemented and descibed
in my papers below.)
Charlie Volkstorf
Cambridge, MA
http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/20021008.1/1
http://www.arxiv.org/html/cs.lo/0003071
>        (Quote from the QED Manifesto)
> F.

===

Subject: Re: Uses of complex numbers
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9B5aUT30405;
>All I know from my lessons is that complex numbers have geometric
>interpretations... any help is most appreciated... I'm sure some
>meaning in science has been found for it...
>
=== Subject: Re: Difficult Inequality by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9BC2Rk22336; I did it in a simpler way, using arithmetic-geometric mean ineqality (1+x) + (1+y) + (1+z) --------------------- >= ((1+x)(1+y)(1+z)) ^ (1/3) 3 (4/3)^ 3 >= (1+x) (1+y) (1+z) 64/27 >= (1+x) (1+y) (1+z) 64 >= 27 (1+x) (1+y) (1+z) Also x+y+z ----- >= (xyz) ^ 1/3 3 xyz <= 1/27 using the two 64 xyz >= (1+x) (1+y) (1+z) 64 >= (1+1/x) (1+1/y) (1+1/z) >Subject: Difficult Inequality >Here is a pretty difficult inequality that I wasn't able to solve >so far. >x,y,z are positive real numbers such that x+y+z=1 >Prove that(1+1/x)(1+1/y)(1+1/z)>=64 >>Set p= (xyz)^1/3. Then the arithmetic-geometric mean ineqality yields >>(1) x+y+z >= 3p >>(2) xy+xz+yz >= 3p^2 >>Because of (1) and x+y+z=1, we also get >>(3) 1/p >= 3 >>The righthand side of the desired inequality equals >> (1+ x+y+z + xy+xz+yz + xyz) / xyz >= (1+3p+3p^2+p^3) / p^3 = (1+1/p)^3 >= 64 >>___________________________________________________________ >>Gerhard J. Woeginger (gwoegi@opt.math.tu-graz.ac.at) >I solved it using just plain algebra without any problems, starting >out with things like z + x = 1 - y and factoring from there. >But I've seen so many solutions using all sorts of advanced math that >I wonder - is there a link between them all? >A copper blade >Infinitely many strings. > === Subject: Re: Question on Hilbert & Godel > What did Hilbert ask and claim concerning Foundations of Mathematics > (sets, predicate calculus), metamathematics, Logic, Incompleteness, > etc? Just a side note: > Bourbaki was cited as the best example so far of mathematics > organized into a coherent framework. According to Andre Weil, > Perhaps the most important contribution of Bourbaki was to > carry out a famous proposal made by the great German mathe- > matician David Hilbert in 1900 that mathematics be placed on > a more secure foundation. He noted: Hilbert just said so, > and Bourbaki did it > And just how did Bourbaki do as Weil claims? > But didn't Hilbert actually claim that a lot more than that is > possible? Didn't he ask for (and claim that it must exist) a decision > procedure to determine if an arbitrary predicate calculus wff is valid More than that, an arbitrary wff of mathematics. > (true of all interpretations) - was it? > wisdom was wrong and mathematicians gained new insight into the > nature of mathematics. > Or are you one of those diehards who clings to the failed ideas and > theories of the past? > (Actually, the most secure foundation possile is software that carries Software is fine if you're a strict finitist. Some mathematicians like a set theory with _huge_ axioms of infinity in which to do their category theory. > out both logic and metamathematics, as I have implemented and descibed > in my papers below.) > Charlie Volkstorf > Cambridge, MA > http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/20021008.1/1 > http://www.arxiv.org/html/cs.lo/0003071 > (Quote from the QED Manifesto) > F. -- G.C. === Subject: chisquare test question I calculated Chi-square for a set of data..... but, then im asked to compare the calculated value of Chi-squared with (n-1) - a.k.a degrees of freedom. Why would I compare my calculated value of Chi-Squared with n-1? I don't get the purpose.... === Subject: Re: Science is a human activity (was: Python syntax in Lisp and Scheme) Well, since you crossposted this to sci.math you must be hoping for replies from that direction: >>Alex Martelli: > would you kindly set right the guys (such as your > namesake) who (on c.l.lisp with copy to my mailbox but not to here) are > currently attacking me because, and I quote, Software is a department of mathematics. > >>And anyone who doesn't think mathematics has its own >>culture with ideas and even mistaken preferences for what >>is right and wrong should read >>The Mystery of the Aleph: Mathematics, the Kabbalah, and the Human Mind >>to see how Cantor's ideas of transfinite numbers (and other ideas, >>as I recall, like showing there are functions which are everywhere >>continuous and nowhere differentiable) were solidly rejected by >>most other mathematicians of his time. >>Mathematicians are people as well. > . > . > . >And let no one assume that these are mere foibles of the >past that we moderns have overcome; mathematics remains >stunningly incoherent in what's labeled foundations. >There's a wide, wide divergence between the intuitionism >working mathematicians practice, Actually inuitionism has a certain technical meaning, and actual intuitionism is not what most mathematicians practice. But never mind, I believe I know what you meant. >and the formalism they >profess. Far be it from me to insist we've overcome the foibles of the past. But: It's certainly true that mathematicians do not _write_ proofs in formal languages. But all the proofs that I'm aware of _could_ be formalized quite easily. Are you aware of any counterexamples to this? Things that mathematicians accept as correct proofs which are not clearly formalizable in, say, ZFC? >'Good thing, too; our age enjoys the blessing of superb >mathematicians, and I'm relieved that philosophical in- >consistencies don't (appear to) slow them down. What's an actual example of one of those philosophical inconsistencies that luckily doesn't slow us down? ************************ David C. Ullrich === Subject: Re: Non-Polynomial Time Algorithms > It's thought that NP properly contains P, but this is hard. ... > If any of the NP-complete > problems are in P, then the reduction would show that all NP problems > are in P. Do I understand correctly that you are saying NP might contain P as well as that P might contain NP? If that were true, of course, it would mean that P and NP are the same. Did I get that right? === Subject: i * (3,4)=( - 4, 3)= - 4 +3 * i and the Manifesto di Bombelli The opportunity to post the Manifesto di Bombelli sci.math. The applet - and much more - You can find now also on www.insel.heim.at/mainau/331839 So, another month added to the twohundred years of avoiding Wessel's ideas( no contributions up to now). Some - secretly knowing it's true - don't want to loose the nimbus of one, who comprehends the imaginary. Usually producing clouds to disorientate others - but for the time being avoiding, not to arouse attention. For some it might be a nut, too hard to crack : it costs a lot of brain-fat to replace imaginations, you cuddle. And some eagerly want to declare this nuts, but they fear for their reputation, if it turns out otherwise. So, what about knowledge ? Hero . > Introducing the dot-multiplication in the vector-space ( R2,+,r.s.m ) > results in an Euklidian vectorspace, inside you can find a commutative > field (R2,+,* ). > Here * is the Bombelli-multiplication: i * i = - 1 , or in Hmilton's > notation:(a,b)*(c,d)=(ac-bd,ad+bc), and i rotates a vector 90 degrees to the > left by i*(a,b)=(-b,a). > In modern-talking math:A vector, an ordered tupel by definition, can > be expressed in the terms of a vectorspace-basis. In R(=R1) the vector > (a) with the standard-basis ( 1 ) gives: (a)=a*1 and that's -of > course - equal to a. Adjungate a foreign element to R, you go 2D.(a,b) > can be notated as a linear combination of the basis-elements: > c*1+d*element. With the standard-basis > ( (1,0) , (0,1) )=( 1 , (0,1) ) or ( 1 , i ) you get a + b * i, > looking different from (a,b), but identical. > You can write and calculate vectors in mixed mode now. > I propose the name Bombelli vector-space. > Real arrows showing winds on a weather-map. > Attach (add) to a number of points the difference in coordinates from > one central point,multiplied by (0,1)=i.Improve this first model by > multiplying > (cos @,sin @)=cos @+i*sin@ - allowing for friction alpha from 10Á > (over sea) to 35Á (over land)(-alpha on the southern hemisphere), and > scale everything by a factor 1/10 (rsp. -1/10) and add a common > velocity to the east. It models > the winds of the inner part of a depression > - done solely with Bombelli's operations - and ........... > we stayed plain real. > Two questions arise, sensitive questions for some - and the answer is > left to them : > Is there any difference to (C,+,*)? - > can you tell this difference to your computer? > If you call the y-axis an imaginary axis, do you add any > mathematical properties, or do you just change the name? Is the > Gauss-plane, complex > plane or the Aragand-diagramm more than just a Fata-Motgana-Reflection > of the real plane? > Why - twohundred years after Wessel - this question .9fberhaupt ? > In a math-lesson you get sometimes the impression, an extraterrestical > element has been adjungated to R. My opinion: > i - is - no - longer - imaginary > Copy this Hero > As in sci.math you can not display applets,you might take a look at > www.i-z.eu.tt or > www.i-is-no-longer-imaginary.gmxhome.de > There You find the Manifesto, i quoted above, in a printable version > for the > pinboard and a (free) mixed-mode calculator. > As the computer makes no difference - do You have to ? > Hero. === Subject: quadratic reciprocity basic question Suppose I have the following if and only if condition: Let p and q be 2 distinct odd primes, then p is a square (mod q) <=> ((-1)^(q-1)/2) * q is a square (mod p). How do I deduce the quadratic reciprocity law from this if and only if condition??? Alex === Subject: Re: Variation on an impossible problem > Allowing only the use of the + operator, is it possible to use each of > the digits 0 1 2 3 4 5 6 7 8 9 once only to make a total of 1000? > One can combine digits to create sums like 210 + 345 + 6 + 78 + 9 = > 648 etc. > No. sum(i,i=0..9) = 0 mod 9, whereas 1000 = 1 mod 9. I could possibly form must be divisible be nine? I can see that if all such sums must be divisible by nine then 1000 cannot be a possible sum. Moreover, I can see that sum(i,i=0..9) = 0 mod 9 but how does this generalise to all possible sums formed? Mitch. === Subject: Re: Non-Polynomial Time Algorithms > Do I understand correctly that you are saying NP might contain P as well as > that P might contain NP? If that were true, of course, it would mean that > P and NP are the same. Did I get that right? Yes, you have that right. All that is known is that P is a subset of NP, but it is still unknown whether or not they are equivalent. J === Subject: Re: Curve Inside Another Of Equal Perimeter >> Are there any two CONVEX closed curves of equal perimeter, where one >> can be placed completely inside the other? >> (It is allowed that the curves touch at a finite number of points at >> most.) >> Leroy Quet >I would guess not. Here is an outline of a possible proof. >Choose a point inside both curves. >Consider the formula for the arc length as a function >of r(theta) and theta, where r(theta) is the distance from >the central point to the curve at angle theta. >Since r_1(theta) <= r_2(theta), where r_1 and r_2 are >the respective distances to the inner and outer curves, >the length of the outer curve is greater than the >length of the inner curve. >Martin Cohen That doesn't quite work, since the perimeter L = int(sqrt(r^2 + (dr/dtheta)^2)) dtheta and you haven't said anything about dr/dtheta. One quick approach to proving that the inner curve must be shorter than the outer one (unless they're coincident) uses Crofton's formula, as I noted in another response. You can also use the fact that projection onto a convex set is (weakly) length-decreasing (by projection I mean assigning to a point P the nearest point in the convex set). So if you project the outer curve onto the (domain bounded by the) inner curve, the length is not increased; a little extra work is needed to show that it's actually decreased). John Mitchell === Subject: Re: Two similar things >> Since they now are identical-with-somethings, if they shared >> all the same properties they would be one cake rather than >> two. THerefore, since they are two cakes, they don't share >> all their properties, and I can distinguish them by means >> of any property that one has and the other doesn't. >> No, there's a foot wrong there. Why would they be one cake if they shared >> all the same properties? >Consider Hesperus and Phosphorus. The properties of Hesperus include >identity-with-Hesperus, and those of Phosphorus include identity-with- >Phosphorus. So, if Hesperus and Phosphorus share *all* their properties, >the properties of Hesperus will include identity-with-Phosphorus, and >those of Phosphorus will include identity-with-Hesperus. But if Hesperus >and Phosphorus have these properties, then Hesperus and Phosphorus are >one and the same: that is, numerically identical. >> Why not 3 cakes? >If these cakes have all the same properties, for the reasons I have indicated >these three cakes will be one and the same. >> Would you tell the difference by counting? >Just by counting, no. By counting correctly, yes. In other words, >before it was discovered that Hesperus was Phosphorus, it was assumed >that these heavenly bodies were different heavenly bodies rather than >one and the same heavenly body. But this assumption proved to be wrong. >--John Aren't you mixing physical properties with logical properties? Where is the justification for that? Logically twins are one flesh. Physically they are two. === Subject: Re: quadratic reciprocity basic question > Suppose I have the following if and only if condition: > Let p and q be 2 distinct odd primes, then p is a square (mod q) <= ((-1)^(q-1)/2) * q is a square (mod p). > How do I deduce the quadratic reciprocity law from this if and only if > condition??? I think you can use the Euler's criterion: If p is an odd prime and if a not equal to 0 mod p, then (a/p) = a ^ ((p-1)/2) mod p. First, consider the case where p is a square (mod q). Then, consider the case where p is not a square (mod q). - Bill Hale === Subject: Re: Discontinuous Derivative |> I'm looking for a real function f:R -> R |> which is differentiable over all of R |> yet with a derivative discontinuous at 0. | |Integrate the Wierstrass Function which is everywhere continuous (so the |Riemann integral exists) and nowhere differentiable. By the fundamental theorem of calculus, the result is a function whose derivative is the Weierstrass function, which is continuous. === Subject: Re: Just what is an L-series? |Ah so. I was trying to compose a definition of L-series for one of the |volunteer-written online websites (www.planetmath.org). Maybe in the end, |L-series are not a structure but only a formalism, like generating |functions. I may need to resort to arm-waving or (what is much the same |thing) category theory:) It would be interesting if there were a definition of L-series rather than just definitions of individual kinds of L-series. As far as I know, there may be one. But I never did see one, in spite of having seen a number of talks and lectures about specific kinds of L-series. It seems plausible to me, then, that what has happened is simply that as each kind of L-series has been defined (Artin L-series, etc.), it's gotten termed another kind of L-series because of its family resemblance to the other kinds of L-series previously defined. As far as I know they are all naturally expressed as Dirichlet series, sum_{n} a_n/n^s. === Subject: Re: quadratic reciprocity basic question > Suppose I have the following if and only if condition: > Let p and q be 2 distinct odd primes, then p is a square (mod q) <= ((-1)^(q-1)/2) * q is a square (mod p). > How do I deduce the quadratic reciprocity law from this if and only if > condition??? I think you can use the special case of Euler's criterion: If p is an odd prime, then (-1/p) = (-1)^((p-1)/2). First, consider the case where q = 1 mod 4. Next, consider the case where p = 1 mod 4. Finally, consider the case where q = 3 mod 4 and p = 3 mod 4. - Bill Hale === Subject: Re: Core error, FEAR is a natural response >: The definition of algebraic integers as roots of monic polynomials >: with integer coefficients gives the ability to give two supposed >: proofs that contradict each other. >I don't get it. I understand the part about two arguments contradicting >each other, and one of them being wrong. But what does this have to >do with the definition of the algebraic integers? Don't the arguments >have to use the same definitions in order to be contradictory? And if >one of the arguments wrong, isn't the problem with the argument, not >the definition? Apparently it's no longer necessary to assume anything in order to arrive at a contradiction. JH states that the mere act of defining the concept of algebraic integers is something that leads to contradicting mathematical results, which must mean that mathematics itself is as pointless as a very blunt object as without definitions there is no mathematics. But how can we use mathematics to prove that mathematics cannot be used to prove anything? The other option is that JH's proof is *gasp* wrong. The fact that he accepts the contradicting proof about the algebraic integers as being valid actually directly implies that JH accepts his proof must be wrong. What really scares me is, how can a physics major be so ignorant about mathematical formulation and the logical requirements for a proof? I have first-hand experience that a lot of math is just regurgitated at students without adequate explanation of the proofs and theories underneath so that you can learn a lot of advanced mathematical methods without understanding what the hell you're actually doing, but physics of all subjects should require a solid understanding of math. === Subject: Re: Quanta and Cakes > Nice to read this. > I do not understand almost nothing of this thread. However, what is the > question? Is it this: What is the probability of finding some of the > Take a look into the box and find what is the actual > configuration. Do it many times. After, calculate the probability of > given configuration (= number of this particular configurations / number of > all measurements). > You will find that > probability of ll is 1/4 > probability of rr is 1/4 > probability of lr is 1/2 > Is this the point of this thread, right? > Palo Partly right. The point of the thread is that although > probability of ll is 1/4 > probability of rr is 1/4 > probability of lr is 1/2 are the probabilities that physicists would have expected to find, the probabilities that they have found are: > probability of ll is 1/3 > probability of rr is 1/3 > probability of lr is 1/3 The question then arises, what is it about quanta that is responsible for the difference between classical statistics and quantum statistics. Some physicists have attributed this difference to what they refer to as a 'lack of individuality' in quanta. What I am suggesting is that from the standpoint of logic, things that 'lack individuality' do not satisfy 'x=x'. If this suggestion is correct, then classical logic--in which everything in the domain of individual variables satisfies 'x=x'--and quantum logic do not coincide. Of course, there is nothing new about the suggestion that classical logic and quantum logic do not coincide. To my knowledge, however, those who claim there is a difference between the two have not called into question the reflexivity of identity. --John > >> I should have asked in what respect *correct reasoning* about >> quanta resembles such reasoning about cakes-to-be, but >> differs from such reasoning about (extant) 'medium size >> dry goods'. >> My answer to this question would be: correct reasoning about >> medium size dry goods should take these as satisfying >> both the right and the left sides of the following bi- >> conditional; >> AxAy(Az(z in x <-> z in y) -> x=y) <-> Ax(x=x) >> [Identity of Indiscernibles] [Reflexive Law of Equality] >> while correct reasoning about quanta--and cakes-to-be--should >> take these to satisfy neither. Sorry, you have obfuscated your meaning beyond my ability to extract > any sense from it, or even to guarantee that the language in use is > still English, so I shall withdraw from the discussion. xanthian. > === Subject: Re: Technique of planning an elementary mathematical proof show that the second derivative is always positive for x > 0, hence the value of f(x) where f ' (x) is zero will be the absolute minimum etc. I see that this type of problem probably comes up pretty often while studying partial derivatives (which I've just dipped into a little.) === Subject: Re: Non-Polynomial Time Algorithms |> It's thought that NP properly contains P, but this is hard. | |... | |> If any of the NP-complete |> problems are in P, then the reduction would show that all NP problems |> are in P. | |Do I understand correctly that you are saying NP might contain P as well as |that P might contain NP? NP definitely does contain P. The fact that P is a subset of NP follows simply from the definitions of P and NP. By properly contains I meant that in addition to containing it, it contains things not in P as well. It seems reasonably common for people reading popular accounts of the P versus NP question to get the impression that problems in NP are supposed to be difficult, probably because a lot of the interest in NP is in some of the presumably hard problems in it. The way NP is defined, however, does not put any kind of lower bound on the difficulty of a problem. A yes/no problem is in NP if it has a suitable type of witness, whose length is bounded by a polynomial in the original problem. There has to be an auxiliary algorithm which runs in polynomial time, having the property that if it accepts a witness, then the answer to the problem was yes, and conversely if the answer to the problem was yes, then there exists a witness that will be accepted by the algorithm. As an easy example, whether a number N is composite is easily seen to be a problem in NP. One can use as a witness a divisor of the number greater than 1 and less than the number. Given N and d, determining whether d divides N and 1 Hello ... > What's the size of a largest subset of vertices on the 120-cell > ({5,3,3}) such that no two vertices in the subset are joined by > an edge? > I have reason to suspect that the number is an integer square, > and the computer searches I've done come tantalizingly close, > but I haven't yet been able to conclude that I've found a maximal > set. > Suggestions? > Well, I doubt I've caught up with you, since I just made > a simple pass at it based on some programs I have lying around > re 600 and 120-cell. > I arranged the 600 vertices of the 120 cell into sets equidistant > from an arbitrary point, then looked at distances among these > sets and between them. There are 54 points in the equator between > the given point and its antipode, and these are the 15th nearest > neighbors to those two points. > There are 6 of these with no ( 1st ) n-n among the 54, and the > other 48 have one n-n each, in pairs. So we can pick 30 of these > ( 6 + 48/2 ) and then radiate out to the poles. I don't know if > it matters which of each pair is picked, but it probably matters > which order you select candidates as you radiate. > Anyway, doing it arbitrarily I got 193 points, so I suppose you're > aiming at 196. Is this right? Well, never mind that ! I get 201 just by progressing through the n-n planes starting from a point, taking the points in each plane in arbitrary order. The equator gets 20 points this way. Curiously, when I start at each end and go up to the equator, then pick points in the equator, I get 201 also, even though the equator only gets 15 points in this case. These results are subject to arbitrary variation, though. This equator is an interesting figure. It is 3-d since it lies in a hyperplane of the 4-d figure, and the points are on the surface of a 2-sphere, i.e. a sphere in 3-d space. The six points with no n-n in the equator are in antipodal pairs and form a set of orthonormal axes. The 48 remaining points are generated from a single nonsymmetrically placed point from the group O_h ( octahedral group ) so the 24 n-n pairs comprise the vertices of 2 snub cubes which map into each other under inversion. Lew Mammel, Jr. === Subject: Square root modulo a power of two How does one solve the quadratic congruence x^2=a(mod 2^n) ? === Subject: Re: hello...my question is ?? > f(0) = 0 > integral (0 ~ 1) f(x) dx = 1 > f'(x) is continuous > find M such that f'(x) <= M > ----------------------------- > answer : M >= 2 > how do you solve it? Are you asking for the smallest possible M for which the conditions on f can all be satisfied? This is not clear from your statement of the problem. In that case, f(x) = 2*x is the function with the smallest deriviative over all of 0 <= x <= 1 satisfying all the stated conditions, giving M = 2. === Subject: Re: quantum echo Metaphysical Background, Thomas McTighe asserted that the quiddity > of a thing is nothing other than unity itself. Hence, by virtue > of its positive content, the sun differs not at all from the moon > or any other particular thing. The diversity which is exhibited > by the natural world is merely the product of accidental > differences; no object possesses any specific form which > interposes itself between a particular existing thing and the > source of their being e.g. the Absolute.15 All individual > entities are nothing more than differing contractions of the > whole devoid of any being of their own. ...because the > restricted quiddity of a thing is the thing itself. http://www.crvp.org/book/Series01/I-10/chapter_ii.htm > I didn't understand this the first time around. This time I'll > make an effort, by placing it in its context: > Metaphysical Background, Thomas McTighe asserted that the quiddity > of a thing is nothing other than unity itself. Hence, by virtue > of its positive content, the sun differs not at all from the moon > or any other particular thing.14 The diversity which is exhibited > by the natural world is merely the product of accidental differences; > no object possesses any specific form which interposes itself between > a particular existing thing and the source of their being e.g. the > Absolute.15 All individual entities are nothing more than differing > contractions of the whole devoid of any being of their own. > talking about. Do you? I was trying to proceed through the evolution of these particulars i guess. I think he is trying to reflect upon reductionism and it seems a bit like the whole in the part idea in fuzzy logic. I liked it because it addressed the macro or context and how features from that scale are only sub-components of the context and how the pre-text text and context each have their own rules to emulate identical functions. Like rules of definition with words and rules of subject/predicat in sentences and rules of topic/support sentences in paragraphs. in ideal form it would be like music when the riff double times the chord progression with 2 of the chord progression one scale down. draws the chord with double chord patterns but with different key combinations required for textcontext/context. but who knows the mind of the middle ages, lots of stuff hidden there, shrowded in god talk I take a quiddity to be a thing's *suchness* and a haecceity to be its > *thisness*. A *thisness* I take to be, as Robert Adams does, the > property of self-identity, although I disagree with an assumption > which might be read into Adams, that an individual can lack > self-identity and yet possess the property of being some > individual or other). In Primitive Thisness and Primitive Identity (_The Journal > of Philosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), Adams A thisness is the property of being a certain particular > individual, not the property of being some individual or other, > but my property of being identical with me, your property of > being identical with you, etc. These properties have recently been > called 'essences', but that is historically unfortunate, for essences > have normally been understood to be constituted by logical properties, > and we are entertaining the possibility of nonqualitative > thisnesses. In defining 'thisness' as I have, I do not mean > to deny that universals have analogous properties--for example, > the property of being identical with the quality red. But since > we are concerned here principally with the question whether > the identity and distinctness of individuals is purely > qualitative or not, it is useful to reserve the term > 'thisness' for the identities of individuals. It may be controversial to speak of a property of being > identical with me. I want the word 'property' to carry as > light a metaphysical load here as possible. 'Thisness' > is intended to be a synonym or translation of the traditional > term 'haecceity' (in Latin, 'haecceitas'), which so far as I > know was invented by Duns Scotus. Like many medieval philosophers, Scotus regarded properties as > components of the things that have them. He introduced > haecceities (thisnesses), accordingly, as a special sort > of metaphysical component of individuals.[4] I am not proposing > to revive this aspect of his conception of a haecceity, because > I am not committed to regarding properties as components of > individuals. To deny that thisnesses are purely qualitative > is not necessarily to postulate 'bare particulars', substrata > without qualities of their own, which would be what was left > of the individual when all its qualitative properties were > subtracted. Conversely, to hold that thisnesses are purely > qualitative is not to imply that individuals are nothing > but bundles of qualities, for qualities may not be components > of individuals at all. (pp. 6-7) Acceptance of haecceities is a distinctive feature of the thought of many > followers of Scotus, though there are some sixteenth-century scholastics who > accept haecceities without accepting many other distinctively Scotist > teachings. Having said this, some early followers of Scotus reject > haecceities and the theory of the common nature altogether, and of those who > accept haecceities, some found the correct understanding of the nature of > the distinction between an individual's nature and its haecceity a > troublesome matter. > One of the earliest Scotists, Francis of Meyronnes, writing his commentary > on the Sentences around 1320, accepts the theory of the non-numerical unity > of common natures, and the claim that individuation is by haecceity. But he > holds that it is inappropriate to talk of a formal distinction in this > context. Formal distinction obtains only between things that have some sort > of quidditative content. > I'm not sure what he/you are getting at here. Francis of Marchia was not a faithful Scotist Marchia was uncomfortable with Scotus's stress on and use of a strong distinction between the divine intellect and will, and this led Marchia to oppose Scotus on issues such as the procession of the Holy Spirit and the mechanism of divine foreknowledge. Nevertheless, Scotus forms much of the backdrop for Marchia's theology. 'Action' can be taken in three ways: either it can be taken actually, namely when an agent is actually acting; or it can be taken virtually, when an agent can act although he is not acting; or it can be taken in a middle way, not purely actually nor purely virtually, but in a middle way as 'dispositionally' or 'aptitudinally', namely when an agent is not acting but is determined toward acting, although in actuality he is not acting - and he not only can act, but is determined to be acting later. Similarly there is a threefold 'determination' of the agent: one actual, by which an agent actually determinately puts one part of a contradiction into effect; a second is a potential determination by which an agent posits or can determine any part of a contradiction dividedly; the other is, as it were, a 'dispositional' or 'aptitudinal' determination, by which an agent is determined with respect to the future to putting one part of a contradiction [into effect]. Each determination presupposes the action corresponding to it, because an actual determination follows the action in actuality; the dispositional determination follows the action dispositionally, although it precedes the actual action; the potential determination follows the potential action, although it precedes that actual and dispositional action. (d. 35: Marchia 1999, pp. 89-90) Thus when an agent is determined de inesse to doing something in the future, that determination is like a disposition, and neither actual, because the event has not yet occurred, nor potential, because the possibility to do otherwise is not removed. Such a determination is not 'actually' in the agent's power, Marchia grants, but it is in his power 'dispositionally', for although the agent cannot act before he acts, he can be disposed to act so that he will in fact act. http://plato.stanford.edu/entries/francis-marchia/ http://www.wikipedia.org/wiki/Scholastic_philosophy Perhaps the most interesting non-qualitative approach is the theory, often associated with Aquinas (but attacked by Scotus in the forms presented by Godfrey of Fontaines and Giles of Rome), that individuation is by extended matter: by, as we might say, chunks of matter. Scotus's way of understanding the problem of individuation becomes important in his rejection of this theory. For his fundamental strategy against this sort of material individuation is that such a theory, while it may be able to explain numerical distinction, certainly cannot explain indivisibility: Quantity is not the reason for divisibility into individuals. . . . For a universal whole, which is divided into individuals and into subjective parts, is predicated of each of those subjective parts in such a way that each subjective part is it. But the quantitative parts into which a continuous whole is divided never admit of the predication of the whole that is divided into them. (Scotus, Ordinatio II, d. 3, p. 1, q. 4, n. 106 [Scotus (1950-), 7:443; Spade (1994), 85]) http://setis.library.usyd.edu.au/stanford/entries/medieval-haecceity/ By such a reasoning, Dante concludes that the secular monarchy flows from the divine unity sine ullo medio. John of Paris, however, asserts the proper autonomy of the secular authority, not only using Dionysius but indeed the very regula or lex which is authoritative for Boniface, Giles and de Meyronnes. This is a remarkable feat because, as A.P. Monahan remarks, As long as unqualified acceptance was given to the principle that all things must be subordinated to one, there was no logical escape from the doctrine that the temporal must be subordinated to the spiritual, whose ultimate unitary embodiment was the papacy.47 Moreover, he agrees with Dionysius in ecclesiastica hierarchia that the laity, with their kings, are in the lowest rank, though perfectible, as compared to the various ranks of the perfected and perfecting clergy: in infimo gradu sunt laici cum suis regibus quasi imperfecti, perfectibiles tamen, supra quos sunt perfecti et supra illos sunt perfectiores, ut viri ecclesiastici, et in supremo est summus monarcha omnium, scilicet dominus papa. http://www.dal.ca/~claswww/DIONDIX.htm > Haecceities have no such quidditative content, and thus cannot be formally > distinct from their nature. Rather, a haecceity is modally distinct from its > nature. > What does Cross (or whoever) mean by Haecceities > cannot be formally distinct from their nature.? not sure but probably the distinction between this-ness and that-ness and limitations imposed by being one or the other? > A modal distinction, according to Meyronnes, obtains between a thing > and an intrinsic mode of that thing, where an intrinsic mode is something > which when added to a thing does not vary its formal definition . . . since > it does not of itself imply any quiddity or formal definition. A haecceity > does not affect a thing's kind; it is thus an intrinsic mode of the thing. > In order for this to mean something to anyone who is not sure what > an intrinsic mode, or the formal definition, of something (in the > sense of Meyronnes) are--and is therefore unsure what 'varying > something's formal definition' might involve, illustrative examples > should be provided. minimal convergence upon emulation or maybe water in the radiator could mean water from the lake in the radiator or water from the middle of the lake or water from unkown portion of the lake, is sufficient for engine uses water? i don't know but i like middle ages thinking anyway. > It may look as though this is just a terminological shift, but it is not so > in at least the following way: a modal distinction is a lesser kind of > distinction than a formal distinction. Formal distinctions obtain between > genus and specific difference; thus, the difference between species/nature > and haecceity, for Meyronnes, is less than the difference between genus and > difference. Scotus, contrariwise, makes no such distinction between degrees > of difference in this context (for this contrast between the two thinkers, > see Dumont [1987], 18). Still, without some principled way of spelling out > degrees of difference, this contrast between Scotus and Meyronnes amounts to > nothing of any philosophical interest. To this extent the difference between > the two thinkers might as well be merely terminological, and Meyronnes needs > to do more work if he is to make any significant philosophical point here. > Without any explanation of the before-shift and after-shift > terminology, how is one supposed to know which side is up? Supposing that a haecceity is something real, where does it fit into the range of things that exist? Is it, for example, a form, or something else? According to Scotus, it is something like a form, and sometimes, indeed, he calls it such (while elsewhere denying the same claim: on these insignificant terminological shifts, see Dumont [1995]). The reason is that a haecceity is clearly something like a property of a thing -- hence like a form -- but is at the same time wholly devoid of any correspondence to any conceptual contents. It is not at all a qualitative feature of a thing -- not at all a quidditative feature, in the technical vocabulary. As irreducibly particular, it shares no real feature in common with any other thing. (This does not mean that haecceities cannot fall under the extension of a concept. Being an individuating feature is not a real property of a haecceity [it cannot be, since any haecceity is wholly simple, and shares no real features with any other thing]; but any concept of what a haecceity is certainly includes among its components being an individuating feature. A concept a haecceity includes representations merely of logical, not real, features of any haecceity.) > http://setis.library.usyd.edu.au/stanford/entries/medieval-haecceity/ > Here's the concluding paragraph from that Stanford Encyclopedia > Note, of course, that Scotus's account of the common nature entails > something stronger than Adams is proposing: indeed, it entails > precisely the sort of minimal hypostatization that Scotus proposes. > And the reason for this, of course, is Scotus's view that individual > substances cannot themselves be primarily diverse -- a fact that is > explained by his claim that common natures have some sort of unity in > their instantiations: the nature in Socrates is (non-numerically) the > same as the nature in Plato. Natures, for Scotus, cannot be primarily > diverse; substances must include more than natures. But individual > natures in Ockham's view can indeed be primarily diverse, and this > surely amounts to a form of haecceitism -- nothing other than an > individual nature's own self-identity explains its distinction from > all other such natures. Maintaining that individual natures are > primarily diverse amounts not to having no theory of individuation, > but to accepting a form of haecceitism that, like Adams's, does not > involve ontological commitment to the existence of real haecceities as > distinct real constituents of things. > http://plato.stanford.edu/entries/francis-marchia/ > --John === Subject: Re: quantum echo > Immortalist: > Stop crossposting this to sci.physics.relativity >> Are you saying it is possible for something to exist at location A > [*SNIP*] > Paul Teller having proposed, as an explanation for the divergence > between quantum statistics and classical statistics, that 'quanta' > lack haecceities, I posted to sci.physics.relativity what I took > to be the import of Teller's position for the logic of identity. > On this topic a fair amount has been written by philosophers of > science who, unlike myself, are knowledgeable about the formal > details of quantum mechanics. What these people have had to > say on the topic may be of interest to some on > sci.physics.relativity... > --John the history of an idea is like the evolution of a species === Subject: Re: Max. Non-Adjacent Vertices on 120-cell don> What's the size of a largest subset of vertices on the 120-cell > ({5,3,3}) such that no two vertices in the subset are joined by > an edge? > > I have reason to suspect that the number is an integer square lew> I arranged the 600 vertices of the 120 cell into sets equidistant > from an arbitrary point, then looked at distances among these > sets and between them. There are 54 points in the equator between > the given point and its antipode, and these are the 15th nearest > neighbors to those two points. I can see the same arrangement in my data as well. lew> Anyway, doing it arbitrarily I got 193 points, so I suppose you're > aiming at 196. Is this right? 196 is indeed the target, and I've gotten about as close as you. (But I've had to adjust faulty logic in my algorithms a few times, so I've never quite trusted my results.) My latest approach is to start with a collection of 8 symmetrically- arranged 2-neighbors on a single dodecahedral cell, then to work my way out in shells of adjacent cells. An attempt at automating the random selection of 2-neighbors gave me worse results than starting at a single point (again, possibly faulty logic); manual selection seems to have shown me that there are good choices and bad choices to make at each stage, but after a couple of shells, I confuse myself :) so I haven't been able to codify what makes a good choice. Don === Subject: Re: Core error, FEAR is a natural response linux) > The other option is that JH's proof is *gasp* wrong. The fact that > he accepts the contradicting proof about the algebraic integers as > being valid actually directly implies that JH accepts his proof must > be wrong. Nonsense. There are such things as inconsistent theories. Mind you, James doesn't seem to grasp that (1) an incorrect definition cannot yield inconsistency[1] and (b) inconsistency is rather more troubling than 100 years of mathematics down the crapper. But the existence of two contradictory theorems does not imply that one of the proofs is invalid. Footnotes: [1] I am not considering a definition without referent to be incorrect. Rather, the error lies not in the definition but in the assumption that it is satisfied. -- But in our enthusiasm, we could not resist a radical overhaul of the system, in which all of its major weaknesses have been exposed, analyzed, and replaced with new weaknesses. -- Bruce Leverett (presumably with apologies to Ambrose Bierce) === Subject: two envelope problem - root cause analysis Please share your remarks/opinions on the following root cause analysis of the two-envelope problem. (note: advanced detailed knowledge of the two-envelope problem is assumed in this post) As a starter two definitions: * Definition 1 (game) A game G is a quadruple {A, S, payoff, p} in which A is a set of (player) actions, S is a set of (world) states, payoff is a function A x S -> IR, p is a probability measure on S. LetÇs assume for now (without loss of generality) that both A and S are countable (i.e. finite or countably infinite). * Definition 2 (proper game) A game G is proper in case the expected value of each action a in A, denoted by E(a), is well-defined. Note: - E(a) = sum [p(s)*payoff(a,s)] // here the sum is taken over the states s in S - E(a) is well-defined if and only if S is finite or the sum is absolute convergent - Absolute convergence is a stronger requirement than convergence (this is often overlooked when calculating expected values) According to probability theory (i.e. the law of large numbers) we know that the average payoff of an action a in a proper game G converges to E(a) in case G is repeated many times. ThatÇs why current axiomatic decision theory (CADT) states that a (risk neutral) player of a proper game G should always select the action having the largest E(a) value. In fact CADT has nothing more to offer than this basic conclusion. Since the scope of CADT is limited to proper games we must always reason within the context of a proper game when dealing with the two-envelope problem. If we apply decision theoretic expected value (DTEV) calculations beyond the scope of proper games we are on illegal ground from the point of view of CADT. Well, the reason why (some versions of) the two-envelope problem seems paradoxal is simply because one of the following situations occur. 1. DTEV calculations are applied illegally, i.e. in a way such that an underlying proper game G cannot be defined. The author operates outside the CADT scope, 2. correct DTEV calculations are applied. However these calculations hold for another proper game than the game intended and/or suggested by the author. Most two-envelope papers run into (one of) these situations. The conclusion is that there are no probability nor decision theoretic paradoxal versions of the two-envelope problem. The only thing that is actually happening is that we are confronted with our currently too limited axiomatic decision theory (CADT) that is only able to deal with situations for which a proper game G can be defined. So the main goal should be to extend the scope of CADT by including decision theoretic what to do? theorems that are applicable beyond the context of proper games. Various approaches in this area have already been undertaken in the scientific community. Unfortunately a lot of decision theoretic situations cannot be mapped on a proper game G. Apart from (specific versions of) the two-envelope problem we can mention the famous St. Petersburg paradox. In this paradox the underlying game is not proper because E(a) is not convergent and thus not well-defined. Therefore, letÇs do for CADT what people before us did for naive set theory (developing ZFC axiomatic set theory to extend the scope of naive set theory and to avoid misinterpretations such as the well-known Russell paradox). === Subject: Re: Boolean Algebra - Arithmetic Relationship > I'd like to recommend some reading. For logic generally, see Tarski An > Introduction to Logic and the Methodology of the Deductive Sciences > OUP. (Dreadful title, best book.) For recursion theory I'm at a bit of > a loss, how about Boolos & Jeffrey Computability and Logic CUP? I'll look for both of these at B&N. Are they fairly accessible to the Layman? I have a rather interesting hypothesis for you: The role of mathematics, as I understand it, is to create various logical structures that help predict/explain the nature of empirical phenonmenon. We link these structures to empirical reality by making assumptions. Assuming an atom behaves like the Bohr model, that assumptions are, they are always just approximations. The further you develop any one logical model less its behavior matches that of reality. Would it therefore be good argument, that focus of mathematical development should be on enhancing the breadth (Variety of Symbolic Systems / Logical Perspectives) and not the depth (Complexity of any one Symbolic System) of mathematics? Because no matter how great our assumptions are in the beginning, if we carry the logic far enough they won't match reality. -Steve === Subject: Re: Max. Non-Adjacent Vertices on 120-cell 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > What's the size of a largest subset of vertices on the 120-cell > > ({5,3,3}) such that no two vertices in the subset are joined by > > an edge? Has nobody tried a search for independent-set 120-cell? When I tried it I found http://www.csr.uvic.ca/~wendym/courses/582/120cell.ps Apparently the number of vertices is at least 220 and at most 224. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: circle with two centers In sci.math, riskbert a continuous, closed curve, C, such that both A & B are equidistant > from every point on the curve C. Erm, before we go any further: what embedding-dimension is the space? For example, I could postulate 3-D, and two points A and B. All points C in the space with distances AC = BC have to lie on the plane whose normal is along the vector AB and which contains the point M = (A+B)/2 -- the midpoint of the segment AB. If one stipulates a radius r > dist(AB)/2 one gets a circle on that plane, with center M and radius r' = sqrt(r^2 - dist(AB)^2/4). If you're using a distance-metric over the surface, that is of course a different problem (and I suspect that's what you're really asking anyway). > 1. Does such a surface exist? If so, under what conditions? As a trivial example, one could postulate two diametrically opposite points A and B on a spherical shell embedded in Euclidean 3-space and a family of circles equidistant from both points. > 2. If it does, can there be countably infinite such centers? > Uncountably infintely many? If I understand your question correctly, the answer is yes. Take an arbitrary point A = (r,theta,phi) and its antipode (r, theta+pi, -phi), on a spherical shell of radius r. Since theta can range from [0,pi) and phi can range from (-pi, pi] (introducing thetas >= pi here will lead to doublecounting, although that's not a major problem here), we get an uncountable infinity of such double-center curves, even if we fix the centers -- as one can parameterize the circles by specifying the distance d in the 3-D space from one of the points along the diameter of the sphere, leadig to a point X thereon, and then cutting the sphere with a plane perpendicular thereto, passing through X. I don't know whether you were looking for something more complicated, or what, but this works. :-) > -riskbert -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Assignment Question If anyone can help with the following, please post a reply at http://www.assignmenthelp.com/viewtopic.php?t=8 I have the answers to this problem but I don't know how to get there. A child in danger of drowning in a river is being carried downstream by a current that has a speed of 2.50km/h. The child is 0.600km form shore and 0.800 km upstream of a boat landing when a rescue boat sets out. a) If the boat proceeds at its maximum speed of 20.0km/h relative to the water what heading relative to the shore should the pilot take? b)What angle does the boat velocity make with the shore? c)How long does it take the boat to reach the child? the answers are a)39.6 b)41.6 and c) 3.00 minutes Please post a reply at http://www.assignmenthelp.com/viewtopic.php?t=8 === Subject: RKutta order 2 or 4. How-to ? Do you kwow where I can found a smooth introduction to Runge Kutta methods (internet site). about a site for the non mathematician layman. === Subject: Re: Max. Non-Adjacent Vertices on 120-cell Hello ... s> if I understood it correctly the independence number [of the > 120-cell] lies in the interval [220, 224]. So it cannot be a > square. > www.csr.uvic.ca/~wendym/courses/582/120cell.ps That does indeed seem to be the conclusion. However, the notes do include a need to formalize a proof of this result, so perhaps the upper bound really is 225. (Probably not.) === Subject: Re: quantum echo >Paul Teller having proposed, as an explanation for the divergence >between quantum statistics and classical statistics, that 'quanta' >lack haecceities, I posted to sci.physics.relativity what I took >to be the import of Teller's position for the logic of identity. >On this topic a fair amount has been written by philosophers of >science who, unlike myself, are knowledgeable about the formal >details of quantum mechanics. What these people have had to >say on the topic may be of interest to some on >sci.physics.relativity... The content of your thread contains no physics. My replies to one of those (who, I can't recall) also posting on this thread, indicates that whatever logic is being argued, totally neglects any physical meaning to the quantum mechanics. === Subject: Re: ab... = (a*b*...)^n ? | | > You get a lot further by examining only those | > numbers whose prime factors are less than B. | | > Still nothing in base 10 up to 10^200. | | > Would you describe further how you did that? ... | | The basic idea is to generate all numbers whose prime factors | are less than B. I initialize a set S={1}. I repeatedly remove | the smallest element e from S and for each prime p < B I | insert the product e p into S. This is especially fast if S | is implemented as a priority queue (see Knuth v3). | | An optimization to avoid inserting any element twice is to | insert only products e p such that p is the smallest prime | factor of e p. | | A further optimization for composite B is to avoid inserting any | product e p that is a multiple of B, since that number and its | multiples have zeroes in their base-B expansions. | | > Obviously, you didn't test each number to see | > whether its prime factors were less than 10, .... | | If there are k primes less than B, there are only O(n^k) | elements to process below B^n. If B is composite, the | second optimization reduces this to O(n^(k-1)). Either | of these is vastly smaller than B^n. Programming a priority queue (of bignums no less) is out of my reach, but it would've been fun to look beyond 10^200 ;o) === Subject: Laurent series (finding the principal part) I have those two examples: f(z) = e^(-1/(z^4)) and f(z) = (e^z - 1)/(e^z + 1) and i'd like to find the principal part of their laurent expansions (the first one about a=0, the second a=i*pi). I have tried all the metods (which is just one, sadly) i used on the other tasks but to no avail. Anybody willing to give a hint or (partial) solution? I tried to rewrite f(z) as a taylor series and then recognize the bad part and the holomorphic one. Then i'd write the later as a power series looking like the part that i've extracted. But in the examples here, there's no holomorphic part at all. Also, the bad part is not very nice either. Help! -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: Core error, FEAR is a natural response > Given the people I've managed to contact about this particular > argument, who haven't found anything wrong or even claimed to find > anything wrong, But they are not making any statements that you are right, right? > it's odd that to me that this Victor Eijkhout would > have the nerve to make his post. You're right. It's more in line for the other Victor Eijkhouts. I do come in six-packs, you know. > You'd think the poster chats casually with Barry Mazur, and Andrew > Granville on a daily basis!!! Nope. Not my field. V. -- email: lastname at cs utk edu homepage: cs utk edu tilde lastname === Subject: Re: Does Goldbach imply Reimann there's a proof of Twin Primes proving Goldbach, or vise versa, but I didn't actually read the thing in *Mathematics Mag.* > It is possible prove the Ternary Goldbach Conjecture (TGC) and the Twin Prime > Conjecture (TPC) are true, if the Generalized Riemann Hypothesis (GRH) is > true. > GRH implies twin primes? News to me. http://members.tripod.com/~american_almanac http://larouchepub.com/radio/index.html === Subject: Re: Factorial/Exponential Identity, Infinity That's a good point. Maybe what I need to have in the restriction on the interchange rules is that if you change a zero to a one then you have to change a one to a zero that is within a given finite distance of the one changed, for example q or q-1 or one for the set with p/q ones and (p-q)/q zeros. That might be too restrictive, I am to convert the sequence (01)... to 00001111(01).... 0101010101... 0011010101... 0010110101... 0001110101... 0001101101... 0001011101... 0000111101... Converting (01).. to (10)... is as follows: 01010101... 10101010... Each zero element can be changed to a one and each one within one element of that previous zero element is changed to a zero, and vice versa. Oh good, I have this idea within a few minutes of reading your post. Yet, that might only slow the process and then still I must show if that would not allow the conversion of any sequence (001)... through (1)... to all other sequences with infinitely many ones and zeros. In your example you set for the sequence (01)... indexing from 1 each odd element to one and for each of those each multiple of fourth element to zero, getting (1110).... According to this rule, then, after changing an element from zero to one then you'd have to change an element that is a one to a zero that is next to it. For example you could interchange each 4x-1 and 4x'th elements, (0110).... Another problem with this restriction is that I need to show that it would allow any sequence of given contrived density to convert eventually to a sequence with the same known contrived density, as well that it would not convert to one of a different density. With this restricted sequence element interchange then 10000(0)... isn't convertible to 00001(0)..., instead it is to 01000(0)... which is to 00100(0)... which is to 00010(0)... which is to 00001(0).... How about (011) to (110)...? The list of intermediate sequences between them is infinite. A rule to convert them all at once for each triplet is to convert (011) to (101) to (110). For the previous example the elements are interchanged in a two element sliding window. How about that? Basically what I want here is a description of a method to convert a sequence of a contrived rational density of ones and zeros to any other sequence with that same rational density but not any other, restricted sequence element interchange. === Subject: Normal to a plan In 2-dimensions.. if I have a line such as the one below: Will the normal to this plane always be (0, 0)? How do I calculate the Rick === Subject: Re: Contractible Spaces > If a topological space S is contractible to some point p in S, > is S contractible to every point in S? Yes. Suppose S is contractible to a particular point s0 in S, with F: S x [0,1] --> S giving a contraction and s1 is any other point in S. Define G: S x [0,1] --> S by F(s,2t) if t in [0,1/2] G(s,t) = . F(s1,2-2t) if t in [1/2,1] Then G gives a contraction of S to s1. > If a topological space S is strongly contractible to some point > p in S, is S strongly contractible to every point in S? Probably not in general, although I don't know offhand of any relevant example. There are theorems that say things like If A is a closed subset of X and both A, X are ANRs[*], then A is a deformation retract of X if and only if A is a strong deformation retract of X. Your question corresponds to the case A = {point} -- but it's not clear (to me, anyhow) just how to apply that theorem ... There's probably also an obstruction theory that applies to your question ... [*] ANR = absolute neighborhood retract -- I believe the above theorem actually applies to spaces that are ANRs for the class of compact metric spaces -- X is such a thing if any imbedding of X as a closed subspace of a compact metric space is a retract of some open neighborhood of the image ... > What's an example of a space that is > contractible but not strongly contractible? > S is contractible to a when there's some continuous h:Sx[0,1] -> S with > for all x in S, h(x,0) = x, h(x,1) = a, > and strongly contractible when in addition > for all t, h(a,t) = a > ---- === Subject: 4th order Bspline I have a stupid question about the 4th order Bspline. In the paper( see link), they use summation to denote a B-spline. But why they use a vector (4 j)~T. Can anyone help me to understand it? http://www.ri.cmu.edu/pub_files/pub2/wu_yu_te_1998_1/wu_yu_te_1998_1.pdf === Subject: Re: Normal to a plan > In 2-dimensions.. if I have a line such as the one below: > > > > > > > Will the normal to this plane always be (0, 0)? How do I calculate the > Rick You need 3d to get a normal to a plane. Any normal to a plane has to go at right angles to every line in the plane, but in 2d, the plane and all its lines are the whole thing and there is nowhere else to go. === Subject: Linear Algebra question I'm not sure if/why this would be true or false. I'm having a hard time understanding how I could explain that it was true. I would appreciate any help. Given a linear transformation T:F^n-->F^m. Is it true that there exists an m x n matrix A such that T(x)=Ax (where we think of x as an n x 1 matrix? === Subject: Re: Factorial/Exponential Identity, Infinity > That's a good point. Maybe what I need to have in the restriction on > the interchange rules is that if you change a zero to a one then you > have to change a one to a zero that is within a given finite distance > of the one changed, for example q or q-1 or one for the set with p/q > ones and (p-q)/q zeros. What is the point of all this? It will not provide nexts in densely ordered sets, because an essential property of such dense ordering is that there are no nexts. You are trying to create something like a 4-sided triangle, i.e., something inherently impossible because it contradicts its own definition. === Subject: Bliss's Theorem Does anyone know of a statement of Bliss's Theorem on a website? If so, will you please share the link with me? In case there is more than one Bliss's Theorem, it's the theorem that it used to justify the arc length formula for parametric equations being a Riemann integral, when the derivation doesn't lead to a Riemann sum. John === Subject: Re: Boolean Algebra - Arithmetic Relationship > I'd like to recommend some reading. For logic generally, see Tarski An > Introduction to Logic and the Methodology of the Deductive Sciences > OUP. (Dreadful title, best book.) For recursion theory I'm at a bit of > a loss, how about Boolos & Jeffrey Computability and Logic CUP? > I'll look for both of these at B&N. Are they fairly accessible to the > Layman? Tarski definitely. Boolos & Jeffrey is a bit harder. For recursion theory there is also a book by Cutland Computability: An Introduction to Recursive Function Theory CUP. Amazon says What can computers do in principle? What are their inherent theoretical limitations? These are questions to which computer scientists must address themselves. The theoretical framework which enables such questions to be answered has been developed over the last fifty years from the idea of a computable function: intuitively a function whose values can be calculated in an effective or automatic way. This book is an introduction to computability theory (or recursion theory as it is traditionally known to mathematicians). Dr Cutland begins with a mathematical characterisation of computable functions using a simple idealised computer (a register machine); after some comparison with other characterisations, he develops the mathematical theory, including a full discussion of non-computability and undecidability, and the theory of recursive and recursively enumerable sets. The later chapters provide an introduction to more advanced topics such as Gildel's incompleteness theorem, degrees of unsolvability, the Recursion theorems and the theory of complexity of computation. Computability is thus a branch of mathematics which is of relevance also to computer scientists and philosophers. Mathematics students with no prior knowledge of the subject and computer science students who wish to supplement their practical expertise with some theoretical background will find this book of use and interest. > I have a rather interesting hypothesis for you: > The role of mathematics, as I understand it, is to create various > logical structures that help predict/explain the nature of empirical > phenonmenon. We link these structures to empirical reality by making > assumptions. Maybe you'd like reading Popper on the relationship 'twixt logic and experiment. > Assuming an atom behaves like the Bohr model, that > assumptions are, they are always just approximations. The further you > develop any one logical model less its behavior matches that of > reality. Why should that be? I take it that reality is logical, so it is an exceedingly refined model of itself that is logical. > Would it therefore be good argument, that focus of > mathematical development should be on enhancing the breadth (Variety > of Symbolic Systems / Logical Perspectives) and not the depth > (Complexity of any one Symbolic System) of mathematics? I don't think that mathematics will be limited either in breadth or width. Both expand rapidly. > Because no > matter how great our assumptions are in the beginning, if we carry the > logic far enough they won't match reality. Abstract mathematics gets applied to physics. Do you read John Baez's This Week's Finds in Mathematical Physics at http://math.ucr.edu/home/baez/TWF.html ? In the end there seems to be no a priori reason why mathematics is useful in physics, but it seems to be so. That's a mystery. > -Steve -- G.C. === Subject: Re: Linear Algebra question > Given a linear transformation T:F^n-->F^m. Is it true that there exists an > m x n matrix A such that T(x)=Ax (where we think of x as an n x 1 matrix? Yes, there exists such a matrix, because L(F^n, F^m) is isomorphic to M_m,n(K), K being the field over which is constructed F. Sam -- If sharing a thing in no way diminishes it, it is not rightly owned if it is not shared. - St Augustine === Subject: Re: Normal to a plan > You need 3d to get a normal to a plane. > Any normal to a plane has to go at right angles to every line in the > plane, but in 2d, the plane and all its lines are the whole thing > and there is nowhere else to go. and can't figure out how I'd be able to do this in 2D. What would you Rick === Subject: Re: Core error, FEAR is a natural response >: The definition of algebraic integers as roots of monic polynomials >: with integer coefficients gives the ability to give two supposed >: proofs that contradict each other. >I don't get it. I understand the part about two arguments contradicting >each other, and one of them being wrong. But what does this have to >do with the definition of the algebraic integers? Don't the arguments >have to use the same definitions in order to be contradictory? And if >one of the arguments wrong, isn't the problem with the argument, not >the definition? The definition of algebraic integers excludes ALL roots of non-monic polynomials with integer coefficients, when the polynomial is irreducible over Q. It's that arbitrary exclusion that leads to a way to *appear* to prove two different and contradictory things. It's not even a subtle proof to show it, but mathematicians have been running from the shock. (Neat, eh? Proving an error in core with a proof. Great fun.) It seems an entire math area--algebraic number theory--has to be re-worked. LOTS of textbooks to change. But then again, change can be fun. > Apparently it's no longer necessary to assume anything in order to > arrive at a contradiction. JH states that the mere act of defining the > concept of algebraic integers is something that leads to contradicting > mathematical results, which must mean that mathematics itself is as > pointless as a very blunt object as without definitions there is no > mathematics. But how can we use mathematics to prove that mathematics > cannot be used to prove anything? Well that position requires that you not actually pay attention to what I'm saying, which is the *easy* way out, eh? Social forces are powerful. Most of you need to accept that as human beings you are naked apes. You can't eat a math proof, so for many of you ideas are nothing, unless society TELLS you they're ok. > The other option is that JH's proof is *gasp* wrong. The fact that > he accepts the contradicting proof about the algebraic integers as > being valid actually directly implies that JH accepts his proof must > be wrong. Now that's bizarre as in fact, I note that there's the *appearance* of dueling proofs. But proofs don't, can't duel. Social forces are so powerful, but you people get annoying because you're so damn irrational! Now why don't you try *actually looking at my work* instead of tossing out what you think the group expects, but then, you probably don't believe in ideas, do you? > What really scares me is, how can a physics major be so ignorant about > mathematical formulation and the logical requirements for a proof? I > have first-hand experience that a lot of math is just regurgitated at > students without adequate explanation of the proofs and theories > underneath so that you can learn a lot of advanced mathematical > methods without understanding what the hell you're actually doing, but > physics of all subjects should require a solid understanding of math. What scares you is actually thinking for yourself, rather than letting society TELL you what is the truth. I've talked with a mathematician at my alma mater Vanderbilt University, and explained the entire argument to him...then he basically ran away. People, there's the easy way, and there's the hard way. The hard way is for you to come out and chatter nonsense like you just left the trees this morning, or you can actually pay attention and follow the logical argument. If you don't want to stand with intellectuals, and instead wish to be another cow in the herd, you'll be treated accordingly, if I notice you and care to have some fun. James Harris === Subject: Re: Assignment Question >If anyone can help with the following, please post a reply at >http://www.assignmenthelp.com/viewtopic.php?t=8 >I have the answers to this problem but I don't know how to get there. >A child in danger of drowning in a river is being carried downstream >by a current that has a speed of 2.50km/h. The child is 0.600km form >shore and 0.800 km upstream of a boat landing when a rescue boat sets >out. >a) If the boat proceeds at its maximum speed of 20.0km/h relative to >the water what heading relative to the shore should the pilot take? >b)What angle does the boat velocity make with the shore? c)How long >does it take the boat to reach the child? >the answers are a)39.6 b)41.6 and c) 3.00 minutes >Please post a reply at http://www.assignmenthelp.com/viewtopic.php?t=8 Without loss of generality, let's assume the river flows due south and the boat starts on the left bank. Let's define the starting position of the boat as the origin of cartesian coordinates where x is east and y is north. Let the angle the boat makes with the shore be a, where a = 0 means directly upstream (north), a = 90 means directly across the river (east), and a = 180 means directly downstream (south), then the velocity vector of the boat can be broken into two perpendicular components: the upstream component v_u = 20 * cos(a) - 2.5 the across stream component v_a = 20 * sin(a) At time t (expressed in hours), the coordinates of the boat are (v_a * t, v_u * t) At time t, the coordinates of the child are (.6, .8 - 2.5 * t) If the boat rescues the child at time T, then the coordinates of the child at time T must match the coordinates of the boat at time T. v_a * T = .6 v_u * T = 8 - 2.5 * T Two equations in two unknowns (T and a). You should be able to take it from there. <> === Subject: Re: Use of variable independence, core error > ... do the algebraic > integers indeed form a ring? > I've never said that algebraic integers don't form a ring. > What they form is a flawed ring, which doesn't include all the numbers > it must to prevent the possibility of appearing to prove two different > but opposite things. Well, that seems clear enough... but... is it possible that the flawed ring of algebraic integers includes too many numbers? Is that a possibility? -- Clive Tooth http://www.clivetooth.dk === Subject: Re: multiplication negs In sci.math, Dunphy <3ffhb.5153$G_.420768@news20.bellglobal.com>: > Hopefully, I'm in the right place here.... > I've been making my way throught the Principia, taking my time, working > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? It's > disturbing me that I cannot come up with a solid answer, and certainly > 'We've been taught that that is the case. will not suffice. > Any help would be greatly appreciated. > TIA Attempt computation of the value (1 - 1) [Times] (1 - 1), or (1 + (-1) ) [Times] (1 + (-1) ), and remember the distributive law. Method A: 1 + (-1) = 0; therefore 0 [Times] 0 or 0. Method B: Distribute it out, yielding 1[Times]1 + 1[Times](-1) + (-1)[Times]1 + (-1)[Times](-1). I may have to back up here and start with the following. First, we start with Peano's axioms or something equivalent and define + and [Times] over N[Times]N=>N, in the usual fashion. One can prove commutativity, associativity, and the distributive law for both operations as well. We then define 0 and a magical -a as the arithmetic inverse of a, or a + (-a) = 0, for any a in the natural numbers N; we define thereby J, J = N union {0} union {-a: a in N}. (Note that 0 = -0.) We also define W = N union {0}, for convenience. [*] We define subtraction in a fairly straightforward way over N[Times]N=>N and extend addition and subtraction to N[Times]N=>J, then J[Times]W=>J and W[Times]J=>J, and ultimately J[Times]J=>J. We extend multiplication to W[Times]W=>W trivially: 0 [Times] 0 = 0 [Times] a = a [Times] 0 = 0. (Such is required if we want to keep the distributive laws.) [Times]: W[Times]J=>J and [Times]: J[Times]W=>J has to be defined such that a[Times](-b) = -(a[Times]b) and (-a)[Times]b = -(a[Times]b) as well. This may look a bit circular (and probably is!), but if we define 0 [Times] a et al to be anything else things can get nastily inconsistent. Now we attempt (-a) [Times] (-b). To prove that this equals a[Times]b, one has to show the following: [1] We know the distributive law (a+b)c = ac+bc works in N[Times]N[Times]N. It's trivial to show that it works for W[Times]W[Times]W; one then has to work it out for the union of J[Times]J[Times]W and W[Times]W[Times]J, given the definition (-a)[Times]b = -(a[Times]b), and a[Times](-b) = -(a[Times]b). The leap to J[Times]J[Times]J requires a little work, although we're helped considerably by the observation that (a+b)[Times](-c) = -((a+b) [Times] c) and a[Times](-c) + b[Times](-c) = -(a[Times]c) + (-(b[Times]c)). [2] If we don't want to muck around with commutativity we need to prove a(b+c) = ab+ac as well, using similar methods. [3] We now can evaluate the expression (a + (-a) ) [Times] (b + (-b)). By using [1] we get a [Times] (b + (-b)) + (-a) [Times] (b + (-b)). By using [2] twice we get (a [Times] b + a [Times] (-b)) + ((-a) [Times] b + (-a)[Times](-b)). By using associativity and the definition of [Times] over J[Times]W union W[Times]J, we get a[Times]b - a[Times]b - a[Times]b + (-a) [Times] (-b) = 0 or -(a[Times]b) + (-a)[Times](-b) = 0 or (-a)[Times](-b) = ab. QED. I'm not sure if this is quite rigorous enough but it's clear that It Must Be That Way(tm). :-) I'll admit it's probably more appropriate for a college professor than a 12-year-old (?), although the underlying concepts don't appear all that difficult to grasp. (Of course, it's been a long time since *I* was 12 years old...) [*] Minor controversy; in some textbooks N = {0,1,2,...}. I'm assuming here N = {1,2,3,...}. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Core error, FEAR is a natural response ... >I don't get it. I understand the part about two arguments contradicting >each other, and one of them being wrong. But what does this have to >do with the definition of the algebraic integers? Don't the arguments >have to use the same definitions in order to be contradictory? And if >one of the arguments wrong, isn't the problem with the argument, not >the definition? > The definition of algebraic integers excludes ALL roots of non-monic > polynomials with integer coefficients, when the polynomial is > irreducible over Q. Yes, so what? > It's that arbitrary exclusion that leads to a way to *appear* to > prove two different and contradictory things. It is *not* an arbitrary exclusion. A rational number is *not* an integer when it is the root of a non-monic polynomial with integer coefficients, when the polynomial is irreducible over Q. See? The same kind of exclusion. How can a *definition* lead to a contradiction? You simply attempt to prove stuff about the objects you have just defined. Either they lead to a contradiction, which does *not* mean that the definition is wrong, but which means that either the proof is wrong, or that our method of proof is wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: oh...i am sorry..problem reloaded f(0) = 0 integral (0 ~ 1) f(x) dx = 1 f'(x) is continuous find interval of maximum value M of f'(x) on the interval [0,1] ----------------------------- answer : M >= 2 how do you solve it? i have the illusion. Forgive. === Subject: Re: Square root modulo a power of two Use Hensel's lemma or Newton iteration or the binomial theorem, but you need to start with a solution modulo 8 for it to work. If a is even, you'll need to mess around a bit first, but I'll just do the case that a is odd and use Newton's method, which in practice is extremely fast. So, first solve x^2 = a (mod 8), call the solution x1. Then Newton iteration (i.e., to find a root of x^2-a) says to let x2 = (x1^2 + a)/(2*x1) (mod 16). Note this makes sense, because you know that x1^2 = a + 8*y1, so what it really means is to set x2 = (a + 4*y1)/x1 (mod 16). Since a is assumed odd, it follows that x1 will also be odd, so it has an inverse modulo 16. To see that x2 is really a square root of a modulo 16, we can compute x2^2 = (a^2 + 8*a*y1 + 16*y1^2)/x1^2 = (a^2 + 8*a*y1 + 16*y1^2)/(a + 8*y1) since x1^2 = a + 8*y1 = a + (16*y1^2/(a + 8*y1)) = a (mod 16). Next write x2^2 = a + 16*y2 and take x3 = (x2^2 + a)/(2*x2) = (a + 8*y2)/x2 (mod 32) and you can check that x3^2 = a (mod 32). Etc. Actually, the convergence is even faster than this, so you can actually compute x3 mod 64 and get a solution to x^2 = a (mod 64). Each time, the exponent on the power of 2 will almost double, reflecting the fact that Newton's method converges quadratically exponentially, once you're close enough to a root. JHS > How does one solve the quadratic congruence x^2=a(mod 2^n) ? === Subject: Re: JSH: About time > If your logic were correct it would apply > to other functions. > Yup. > Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Say Q(m) is factored in the form Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). Now note Q(0)/f^2 = x + f. That is, Q(0) = f * f * (x + f). Therefore when m = 0, we can say a1 = 0, a2 = 0, > and a3 = 1. Note that when m = 0, a1 and a2 are divisible by f. > Yup. > Now by your logic, for values of m other than 0, > a1 and a2 must be divisible by f and a3 is relatively > prime to f. Do you agree with this? This is a test of your > method. We need to know before we go to the next step. Nora B. > The answer is that yes, but with the qualification that m not equal 1, > but I have a better explanation for why than the wacky reply when I > initially freaked out. > Look again at your > Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). > Here the constant terms for the factors are revealed to be > f, f and x + f > where it just so happens that for your a3 you have something > like--yup, I know some will probably not like this, but it's the > reality--introducing h for the functions h(m), > a3 = h_1(m) - mx h_2(m) + x + f > where h_(1) has f^{2/3} as a factor and h_2(1)=1. > And yes, the same thing can happen with my argument, but it requires > mf^2 = 1 > but both m and f are integers in that argument, so that condition > doesn't occur. You have suggested elsewhere that I need to solve for the a's with expression. Let's do that. Recall that Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Let m = 1: [1] Q(1) = f^2*x^3 + f^3 = f^2*(x^3 + f). Assume this is factored in the form Q(1) = (a1*x + f)*(a2*x + f)*(a3*x + f). The roots of this polynomial are [2] r1 = -f/a1, r2 = -f/a2, and r3 = -f/a3. Of course from [1] I can compute the roots explicitly. They are: r1 = -f^{1/3} r2 = -f^{1/3}*(-1 + sqrt(3))/2 and r3 = -f^{1/3}*(-1 - sqrt(3))/2. and a3, since ai = -f/ri : a1 = f^{2/3}, a2 = f^{2/3} * (-1 - sqrt(3))/2, and a3 = f^{2/3} * (-1 + sqrt(3))/2. Note that a1, a2, and a3 are all algebraic integers. Note that each of them has a factor of f^{2/3}. Note that none of them have f as a factor. No surprises here. What has this got to do with your wanting m*f^2 to equal 1 ? > (If I didn't say they were before, well they are now.) > For the more adventurous, check Q(m) with m NOT equal to 1, and yes, > you will find that *two* of the a's have a factor that is f. See my other post on this. Nora B. PS: Oh yes, you also asked if I would be convinced if a computer verifies your proof. I would say *no*. I know exactly what is wrong with your proof already. If someone said a computer verified your proof, I would have to have a guarantee of at least three things: (1) That the computer program and the computing hardware were infallible. (2) That the translation of your proof into the program's language was done correctly. (3) That the person reporting the results could be trusted. All of these three conditions will be harder to ascertain than what I already know about your proof. I cannot see how (1) can be guaranteed at all. As for (2), checking that would probably considerably more tedious than what I have already done. And for (3), the most likely person to report positive results would be the least trustworthy. So no. But don't let me stop you. Go ahead and carry it out. Maybe you will convince somebody else. > It might be a fun exercise, assuming I didn't miss something. > If any of you out there think you can find fault with my conclusion > here, please try. > It is math after all. And it's also a lot of fun. > James Harris === Subject: Re: identical property count > Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. > How can something be identical with some things? > And if they shared all their properties, how could you presume to count > 'two'? > You present an either/or relationship as a relational one. (?) __ /// joan stark (?) _./////_ _ .-' __'-. // } // `` -=((( ASCII ART GALLERY: ` ---. / ''.`/_____.;' '-' `-` `/ jJ === Subject: Re: chisquare test question > I calculated Chi-square for a set of data..... > but, then im asked to compare the calculated value of Chi-squared with > (n-1) - a.k.a degrees of freedom. > Why would I compare my calculated value of Chi-Squared with n-1? I don't > get the purpose.... The purpose of this is that, in general, the value of chi-square should be about equal to the number of degrees of freedom you have. If you compute a value of chi-squre which is significantly larger than the number of degrees of freedom, then either (i) your particular experiment represents a *very* low probability statistical fluke or (ii) your data do not fit the theoretical relationships you are trying to impose upon them. If you compute a value of chi-square significantly smaller than the number of degrees of freedom, then you either have (i) a statistical fluke or (ii) you have overestimated the uncertainties at the data points. To be somewhat more concrete, suppose you have a fixed amount of gas held in a tube at a fixed pressure, and you are watching the volume expand as a function of temperature. Let us suppose at temperatures T_1,...,T_n you have measured the volume to be V_1,...,V_n. Now, let us suppose that you believe (that is, you have _a_priori_ reasons to conclude) that the temperature measurements are so good that the uncertainty in the temperature measurements can be ignored. Let us suppose that your errors are in sigma_1, ..., sigma_n. Now, let us suppose that you know the pressure exactly, that you know the number of moles of gas exactly (i.e., you have decided that the uncertainties in this are negligible). The, using PV=nRT you could compute theoretical volumes V(T_i), where V() indicates V as a function of temperature. Since you know n, R, and P, there are no parameters to fit, so for each data point you expect (V_i - V(T_i))^2/sigma_i^2 to be about 1, where sigma_i is you estimate in the uncertainty of the i-th volume measurement. So you expect chi^2 to be about equal to n. Again, if the observed chi-squre is larger than this, you would conclude that the something is wrong--either the ideal gas law is wrong, or you are looking in a range of volumes and temperatures where the gas behaves in a non-ideal manner, or something else is wrong. Now, suppose you didn't know 'n' or 'R', but you never-the-less wanted to test whether V is proportional to T. So now you find the best constant alpha such that *for your data* V_i = V_alpha(T_i) = alpha * T_i. [Here we write V_alpha to indicate how the expected value of volume depends upon temperature, and that one adjustable parameter is involved]. Here, best means you've minimized chi-square. Now, again, you would expect alpha to be approximately nR/P and chi-square to be about equal to n. But note--since alpha is an adjustable parameter, you expect the fit now to be just a bit better, because while alpha should be about nR/P, it can wiggle a little to better fit your particular data set. So, according to the statistiticians, if you have fit to a one paramter family of curves (ie, you have adjust one free parameter to minimize chi-square) it turns out that you expect chi-square to be about n-1. The chi-square has gone down because you can adjust alpha to fit your data better than if you had compared to V=(nR/P)T directly. Again, suppose you were fitting to V=alpha*T+beta. Perhaps you know the termpature in the Celcius scale and the experiment aims at fixing absolute zero in the Celcius scale. Now, you have more wiggle room when you fit, and you would expect chi-square to be about n-2. Note that for a reasonable experiment the number of measurement points should be significantly greater than the number of parameters you are fitting. For example, if you are fitting V=alpha*T+beta and you have made measurents at two temperatures, so n=2, indeed you will find that chi-squre=n-2=0, because you should be able to find a line which passes exactly through the two data points. This, of course, tells you nothing about whether the ideal-gas law holds for your data. So that is a general summary of how this works. I have left out some details. For example, the reason why things are added squared (ie., you raise things to the second power) is related to the assumption that the uncertainties are approximated by Gaussian distributions. There are some other assumptions that go into this. Again, note that if you make enough measurements, then n,n-1, and n-2 are approximately equal, and you don't have to worry about this (expect perhaps to prove that you are aware of this). The real problem comes when the number of adjustable parameters m gets close to the number of data-points n, so that the number of degrees of freedom n-m is significantly less that the number of measurement n. Sometimes the amount of data one can get is limited. Generally, this can result is multiple papers reanlyzing the same data set :-). There really is not much that can be done. People will use fancier and fancier statistical methods. Note, by the way, that lots of assumptions go into this, such as knowing the T's exactly, etc. In general, one's estimates of one's uncertainties are not very precise, you should not need to sweat this too much. The rules of thumb are: Make sure that the number of measurements n is larger than the number of parameters m. Chi-square should be about n-m -- you can always make the fit look better by adding more parameters, ie., you add parameters and chi-square goes down. If chi-square is much above n-m then either this is evidence against the theoretical model which you are using or something went wrong. If chi-square is much less than n-m, you have probably overestimated your uncertainties. Don't sweat this--if the difference between n and n-1 *really* matters, then either you really need to take more measurements or you need to find an expert in statistics and really analyze the particular experiment at hand, and event then you really won't get much more. Hope this helped. By the way, what experiment were you doing? Best wishes, Mike === === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. It seems to me that the reason you seem to think that people gave you bad data is that you never explained what was going on in your head, and that you did not read what you were given very carefully; combined with some preconception on the part of your repliers (myself included) about what you ->could<- be thinking (which apparently you were not). So let me try to summarize what I think has been going on. The 4 color theorem states: If a graph G is planar, then chi(G)<=4. This is equivalent to any number of things. For example, to If G is a graph and chi(G)=5, then G is not planar; and If G is a graph, then it cannot be that G is planar and chi(G)=5. Apparently, you prefer to think of the 4 color theorem via the second version: If G is a graph and chi(G)=5, then G is not planar. So you think of graphs which have chi(G)=5, and try to see why they would not be planar. The usual approach, that described in Saaty and Kainen (which you quoted to begin with) is to use the first version: If G is planar, then chi(G)<=4; in particular, one usually thinks of a graph which is ->planar<-, has the property that any proper subgraph is 4-colorable, and tries to prove that G must therefore also be 4-colorable. You started this thread by quoting from Saaty and Kainen, as follows: Suppose the 4CC is false, Then there is some planar graph G with chi(G)=5. Among all such five-chromatic planar graphs one with a minimum number of vertices is called a 'minimal' graph. Thus planar graph G is minimal if chi(G)=5 but chi(H)<=4 whenever H is planar and has fewer vertices than G. Let H be any subgraph of G, where G has n vertices and H has n-1 vertices. Then, the description of H seems to imply that the deletion of 'any' vertex from G will make chi(H)<=4. But this interpretation is generally false and is valid only for n=5!!! Apparently, instead of thinking G is the (hypothetical) smallest PLANAR graph with chi(G)=5, you were thinking G Is the smallest graph with chi(G)=5. Which would mean you did not read the statement carefully, which states that G is assumed to be PLANAR. That is, one usually thinks about a planar graph such that every proper subgraph is planar, and 4 colorable; assuming it is not 5-colorable and attempting to reach a contradiction. You, from your comments, prefer to think of a graph which has chi(G)=5, and try to show it is not planar. That is fine, but the problem is that the minimal counterexample argument does not work well with that situation; the minimal graph with the property that chi(G)=5 and such that every proper subgraph is planar and 4 colorable is not hypothetical, it's K5. But staring at K5 does not get you any closer to the 4 color theorem. The replies addressed the statement as written: assume G is PLANAR; apparently, you kept reading the replies as if it they were assume G has chi(G)=5. Naturally, miscommunication arose. You kept thinking people were arguing that the 4 color theorem was false for some reason, when people were trying to explain what Saaty and Kainen were trying to say. Examples were given to explain certain facts about planar graphs. You thought they were being given to explain certain facts about graphs with chi(G)=5; again, this mostly arose from you not saying what you were thinking, and from people thinking you understood that they were arguing about planar graphs to begin with. Now, by far the most common false proof of the 4 color theorem consists of people proving that K5 is not planar, and thinking that this proves that a planar graph must be 4 colorable. Your continuous mention of K5 led many of us (or at least, me) to believe you might be thinking along those lines. So, for example, the cycle of length 5 example was given as an example of a graph which does NOT contain any K3, but which is nonetheless NOT 3-colorable; likewise, there are examples of graphs that do NOT contain any K4, and are NOT 3 colorable. This is by way of explaining why does not contain K5 is not enough to imply is 4 colorable, since in general, does not contain Kn is not enough to imply is (n-1)-colorable. My example was an example of why you cannot assume that if G is planar, then any 4-coloring of a G-{v} can be extended to a 4-coloring of G; you can assume that if v has fewer than 4 adjacencies, but I gave a graph in which every vertex had 4 or more neighbors. I did not intend it to be a graph with chi(G)=5, and said so explicitly several times; but you were fixed on your approach, of starting with a graph with chi(G)=5 and showing non-planar, rather than with the usual approach of starting with a graph G which is planar, and showing that chi(G)<5. To my mind (and to the mind of several, judging from the replies), this confusion arose because you were not paying attention to what you were being told; your posts followed up, but did not address, the text written in the post you were following up. You were as careless reading the replies as you were reading the book you quoted. Having such a cavalier attitude with other people's time (by ignoring, not to annoy people. === Subject: Re: oh...i am sorry..problem reloaded > f(0) = 0 > integral (0 ~ 1) f(x) dx = 1 > f'(x) is continuous > find interval of maximum value M of f'(x) on the interval [0,1] > ----------------------------- > answer : M >= 2 > how do you solve it? > i have the illusion. > Forgive. For f(x) = 2*x, one has M = 2, but for every positive M, there are functions satisfying the conditions for which f'(0) >= M. E.g., let f(x) = 2*M*x + o(x) as x -> 0+ Consider, for example, f(x) = (n+1)*(n+2)*x*(1-x)^n, which satisfies all the conditions and has f'(0) = (n+1)*(n+2) === Subject: Series Derivations Hi All, How would I derive a simple formula f(n) for the series: f(n) = (n-0)(n-1) + (n-1)(n-2) + (n-2)(n-3) + ... ? Does is have anything to do with partial sums? === Subject: Re: Boolean Algebra - Arithmetic Relationship > Assuming an atom behaves like the Bohr model, that > assumptions are, they are always just approximations. The further you > develop any one logical model less its behavior matches that of > reality. > Why should that be? I take it that reality is logical, so it is an > exceedingly refined model of itself that is logical. Well, I guess what I'm trying to say here is best explained in the comparison of the Newtonian vs. Relativity models of physics. Newton's model explains much of what we observe in day to day empirical interactions. However at a certain point (either the cosmological or quantum scale) this logical model developed by Newton falls apart, in other words fails to explain those portions of reality. Einstein's model of relativity is superior in that it explains more phenomenon, or matches reality to greater degree. It seems that we keep Newton along just for pragmatic reasons like it's easier to teach and the math is simple. Even if there is some ultimate logical construct that lies behind physical reality, we really only grow closer to approximating it with each advance. (I personally believe the struggle is an asymtotic one governed by the 'law of diminishing returns. --But what I believe is neither here nor there, just a non-empirically based opinion.) > Abstract mathematics gets applied to physics. Do you read John Baez's > This Week's Finds in Mathematical Physics at > http://math.ucr.edu/home/baez/TWF.html ? I guess what I'm trying to say by developing the breadth and not the depth of mathematics is this. I find certain things easy to do in some programming languages and very difficult to do in others, scripting in Perl, and Object-Orientated programming in Java, etc. Similarly, I assume, there are certain proofs which are easier done geometry than their equivalents in another mathematical symbology (that geometry can be reduced to of course). My point: What's abstract in one form of mathematics may be readily apparent in another. Rather than seeking more and more abstract notions and proofs of a particular symbology, perhaps we have to look for a different symbology. Some symbology could be developed to convert the rather abstract nature of mathematics involved in relativity to be fairly apparent or easy to work with. Of course this may be nothing more than a hypothetical pipedream. My ignorance of the actual topic leaves open the possiblity that something such as this already have been developed and is widely used. Of course if it has its not yet simple enough for the layman to grasp. I'm even told that Newton had to invent calculus to properly explain his theories. (Of course, some say Liebniz beat him to it!) Well anyway I look forward to your post. Humbly as always, -Steve === Subject: Re: oh...i am sorry..problem reloaded > f(0) = 0 integral (0 ~ 1) f(x) dx = 1 f'(x) is continuous find interval of maximum value M of f'(x) on the interval [0,1] ----------------------------- answer : M >= 2 how do you solve it? i have the illusion. Forgive. > For f(x) = 2*x, one has M = 2, but for every positive M, there are > functions satisfying the conditions for which f'(0) >= M. > E.g., let f(x) = 2*M*x + o(x) as x -> 0+ > Consider, for example, f(x) = (n+1)*(n+2)*x*(1-x)^n, which satisfies > all the conditions and has f'(0) = (n+1)*(n+2) I think she's trying to show that if f is C^1 on [0,1], f(0) = 0, and int_[0,1] f(x) dx = 1, then f'(x) >= 2 for some x in [0,1]. Suppose to the contrary f'(x) < 2 for all x in [0,1]. Integrating by parts shows 1 = int_[0,1] f(x) dx = int_[0,1] f'(x)(1-x) dx, which then leads to a contradiction. === Subject: Re: Max. Non-Adjacent Vertices on 120-cell Hello ... d> I have reason to suspect that the [independence] number [of the > 120-cell] is an integer square I should point out now that after further investigation, (and, of course, in agreement with the results here http://www.csr.uvic.ca/~wendym/courses/582/120cell.ps ) I no longer believe the number is an integer square. (I said I had reason for my suspicion ... not that I had _valid_ reason.) Still, knowing the number is about 30 greater than where I'd been fishing is a help for the underlying goal of this investigation. Don === Subject: Re: Square root modulo a power of two >How does one solve the quadratic congruence x^2=a(mod 2^n) ? If we write a = 2^m * b where m is an integer >=0 and b is odd, then of course m has to be even, and (x/2^m)^2 = b (mod 2^{n-2m}), which reduces it to the case when a is odd. Note that if x is odd, then a=1 mod 8, so for n>2 you must have a=1 mod 8. Usually the simplest quick method of solving a congruence modulo a power of a prime is to solve it modulo successive powers of the prime. So you would solve it mod 8 (1^2 = 3^2 = 5^2 = 7^2 = 1 mod 8) and then use the solutions mod 2^k to get the solutions mod 2^{k+1}. The thing that makes this congruence different from the congruence mod p^n for an odd prime p, is that when x=y mod 2^k, for k>0, we get x^2=y^2 mod 2^{k+1}, because x^2-y^2 = (x-y)(x+y), and x-y is divisible by 2^k while x+y is even because x and y have the same parity. Generally, then, for n>2 (and a=1 mod 8) there will be four solutions, x, -x, x+2^k, and -x+2^k. But it's a waste of time to keep track of all four. Instead you can consider that the values of x are really defined mod 2^{k-1} instead of 2^k. Or to put it another way, once you've gotten a solution x^2=a mod 2^k, in order to get a solution mod 2^{k+1}, either x^2=a mod 2^{k+1}, or if not, then (x+2^{k-1})^2 = a mod 2^{k+1}. Newton's method is faster, however, for large values of n. Start with x an odd number, so that x^2 = a mod 8. Then apply the Newton iteration x --> x' = x - (x^2-a)/2x where 2x is the derivative of x^2-a. We can simplify x-(x^2-a)/2x= (x^2+a)/2x. The tricky part is the division: you have to know how to do a division mod 2^n, i.e., find the solution to 2*x*x' = x^2-a mod 2^n. Convergence is quick, though; x'^2-a = (x^2-a)^2/2x, and since x is odd, if x is a solution to the congruence mod 2^k, then x' is a solution to the congruence modulo 2^k' where k'=2k-1. You can speed things up a bit more by observing that you don't need to compute the division modulo the original power of 2; you can compute it modulo powers of 2 which grow with each step, and the value of x at each step will still solve the congruence modulo the same power of 2. So for instance suppose I want to compute the square root of 17 modulo powers of 2. Start with the congruence class x=1 mod 8. The first method proceeds one power at a time: 1^2 = 17 mod 16 1^2 != 17 mod 32, so replace it with 1+8=9, where 9^2=17 mod 32. 9^2 = 17 mod 64 9^2 != 17 mod 128, so replace 9 with 9+32=41. 41^2 != 17 mod 256, so replace 41 with 41+64=105. 105^2 != 17 mod 512, so replace 105 with 105+128=233. The second method takes longer for each step, but picks up more powers as it goes along, computing x mod 2^3, 2^5, 2^9, 2^17,.... x=1 mod 8. (x^2+17)/2 = 9. x=9 mod 32. (9^2+17)/(2*9) = 49/9 = 233 mod 512. x=233 mod 512. (233^2+17)/(2*233) = 27153/233 = 75495 mod 131072. x=75495 is a solution to x^2=17 mod 2^18. To illustrate doing the division mod 2^17: 233*x = 27153 mod 131072. The smallest multiple of 233 exceeding 131072 is 233*563: 563*233*x = 563*27153 mod 131072. 107*x = 82787 mod 131072. The smallest multiple of 107 exceeding 131072 is 107*1225: 1225*107*x = 1225*82787 mod 131072. 3*x = 1225*82787-773*131072 = 95419 mod 131072. The smallest multiple of 3 exceeding 131072 is 3*43691: 43691*3*x = 43691*95419 mod 131072. x = 75497 mod 131072=2^17. And in fact 75497^2 = 17 mod 2^18. Presumably this is still not the most efficient method for large problems, but I believe the best methods are related. I'll admit that I made a mistake in the division mod 2^17 which took me a while to find. One can, alternatively, use a second Newton iteration to do the division: x --> x*(2-xy) which is the Newton iteration for f(x) = 1/x - y. That's how I corrected my mistake. It may be that this is not so good for doing the calculation by hand, but by computer it works well. You also may or may not know that a sequence of compatible congruence classes modulo powers of a given prime p is known as a p-adic number. So in this example, I was in effect computing the 2-adic square root of 17: sqrt(17) = 1 + 2^3 + 2^5 + 2^6 + 2^7 + .... === Subject: Re: books - selling - books ... > Davies, Paul > The Fifth Element > Hard bound, great condition. Has withdrawn stamped on top side. > $5 What's this about then? My first thought was, Um ... boron. On second thought, I looke for the title on Best Book Buys and didn't find it; by Davies did write a book _The Fifth Miracle: The Search for the Origin and Meaning of Life_. -- Chris Henrich Nanotechnology could be huge. -- Lord Sainsbury, Science and Innovation Minister (UK) === Subject: Re: Joe Uptaught (Was Re: David Ullrich on Identity) A propos, here are cites from Pierre Bourdieu's > _Language & Symbolic Power_ (the titles are mine). > Enjoy! > The Social Conditions for the Effectiveness of Ritual Discourse Heretical Discourse The *Skeptron* Symbolic Power & the Symbolism of Power Collusion in the 'Hood Expertise as a Problem > Camaraderie of the Experts Knowledge Is Created in a Cultural Context We are proposing that we think of both physics and history as discrete examples of what Pierre Bourdieu calls a field of cultural production (1993: 31 and passim). We hope that by thinking of knowledge production as a cultural activity we can recognize the ultimate cultural context within which knowledge is produced and evaluated, the sociological mechanisms that motivate and limit the actions of individuals in a group, and the legitimate demand for autonomy that is so powerfully expressed and experienced within these fields. We hope also to be able to suspend the questions about the objective truths produced within these fields by recognizing that these sociological mechanisms provide a powerful structure for creating knowledge but also that this structure may not be as effective outside the field creating the knowledge. A field of cultural production is an organized social group that establishes its own organization by agreeing more or less about the basic purposes of their collective endeavors and the basic rules of what counts as a legitimate claim or statement within the field (in our case, an academic discipline). It can be helpful to think of cultural production as a game, albeit a very serious game, but also a game whose rules are under constant revision. Bourdieu's conceptualization of cultural production thus escapes some of the problems associated with Kuhn's paradigms (1962). In Kuhn's discussion of scientific revolutions there seems to be no mechanism whereby individual actors (that is, scientists) come to change their minds. If they are not changing their minds because the new paradigm is better than the old one--that is, it provides a better understanding of the physical reality, for instance (an interpretation that Kuhn specifically dismissed)--then what could possibly be causing them to change their minds (Weinberg 1998)? Bourdieu's understanding of fields provides a sociological explanation. Not only are the rules of the game under constant revision, the evaluation of the effectiveness of the rules is under constant revision. And what is at stake in these revisions is of absolute sociological significance: the power and prestige of the participants as well as the very legitimacy of their claims to be real participants in this field (to be a physicist or a historian, which means to be recognized as such by other physicists or historians). In order to be as clear as we can about how this works, we will begin by quoting Bourdieu, and then we will try to go on to explain what this means for our argument. We are afraid that we must begin with an abstract description: to facilitate understanding we are going to avoid introducing specifics about the ostensible object being produced--whether it be knowledge of the historical past or of physical reality. But we will be introducing examples from time to time in order to aid clarity. According to Bourdieu, any field of cultural production is made up solely and totally by the positions taken by actors in the field.[2] Every position [in the field], even the dominant one, depends for its very existence, and for the determinations it imposes on its occupants, on the other positions constituting the field; and. . . the structure of the field . . . is nothing other than the structure of the distribution of the capital of specific properties which govern success in the field and the winning of the external or specific profits . . . which are at stake. (Pierre Bourdieu, _The Field of Cultural Production_, 1993: p. 30) This is Bourdieu's description of the sociological operations within the field, which he understands to be relations of power and determination. Dominant positions in the field, in any field, exert their power through their influence on what count as meaningful and significant statements. For example, the field of psycholinguistics is currently dominated by two main poles of thought, with successive versions of Chomskyan rule-based grammar competing with connectionist network-based models. Theorists and experimentalists base their arguments so that they can be recognized and positioned with relationship to these dominant theoretical perspectives. This is not simply Kuhn's normal science, as it acknowledges competing paradigms operating simultaneously and provides a sociological mechanism that shapes the actions of individual participants. A psycholinguist risks the possibility of being completely misunderstood if she refuses to state her arguments in terms that one pole or the other can recognize. If she desires her work to be significant and meaningful, it will be less risky to investigate some of the hypothetical consequences of one dominant position or the other (or perhaps to bring together predictive consequences of both). Each instance of such position-taking participates in the ratification of the dominance of the current poles. What happens if our renegade psycholinguist persists in stating her position in isolation from the dominant poles? While it is more likely that her work would not be recognized as psycho- linguistics, it is possible that this new position would gain adherents and would come to challenge the dominance of the other two poles. If this does occur, then the relationships of power and determination within the field would be completely reorganized. If, for example, she finds herself suddenly championed by actors occupying already powerful positions (say, Steven Pinker or David Rummelhart), then this will legitimize the relative value and significance of her position. And the much greater risk assumed by her rather radical position-taking will have paid off quite handsomely. With the coming to dominance of this new position, other actors in the field will find it necessary to position their work in relation to hers. What is important about this simplified fable for our purposes is that the entire system, the specific field of cultural (or knowledge) production, operates by virtue of the value accorded to the positions taken by the actors in the field by other actors in the field. Yet each actor's ability to evaluate is determined by the relative dominance of other positions in the field. This is a sociological model that values the internal autonomy of the participants in a field without making each of them into atomistic disconnected individuals. Science operates like a field of cultural production, but recognizing the cultural basis of scientific practice does not strip science of its autonomy, rather it indicates the conditions of knowledge production that create that autonomy. Note 2. In Bourdieu's sociology, position is a multivalent word. Its primary meaning involves an attempt to spatialize the relationships among differing legitimate intellectual statements in a specific field. But the metaphoric resonances of the word are also meant to suggest both point of view and social status. One of the most important aspects of Bourdieu's theoretical system is that point of view and social status are not extricable from each other. In addition, the value or significance of an intellectual position within a particular field is both a reflection of and reflects back on the social status and point of view of the actor holding the intellectual position. Finally, it is important to emphasize that the terms social status and point of view are simplified versions of the more nuanced concepts that Bourdieu uses in his analysis. (Barry Shank et al, Pure Objects and Useful Knowledges (in _After the Science Wars_, Keith M. Ashman and Philip S. Baringer, eds., pp. 69-79): 73-75) === Subject: re: The Shape of Space PS They seem not be aware the the very relevant papers on gravitational instantons by Peter Kronheimer (1989, 1990) -- although the spherical spaces they consider are all SU(2)/g, where g is a finite subgroup of SU(2). The classification of these spaces via the A-D-E Coxeter graphs, provides the big guns of Lie algebra theory and the many other objects classified by these same Coxeter graphs, such as catastrophe bundles and 2-d conformal field theories (which live on the 2-d string world-sheet). Saul Paul the d 2-d conformal field theories (which live on the 2-d string world-sheet) are the world holograms whose projections or images are 4D spacetimes in Lp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3 = 10^40Lp^2 which numerically fits Abdus Salam's f-gravity idea of ~ 1973 hadronic string tension ~ c^4/G* ---> alpha' ~ 1/(1 Gev)^2 for universal geometrodynamical Regge trajectories of micro-geons of Kerr-Newmann type. === Subject: re: The Shape of Space I am still trying to get a Mickey Mouse heuristic picture for Pedestrians, a sculpture, of what Ellis et-al are suggesting. Rather than trying to picture it in 3D let's go back to 2D. OK, if Omega zero = 1.013 i.e. k = +1 in FRW metric in standard convention. Then our total Universe (sans one space dimension so we can visualize it) is topologically equivalent a 2D spherical surface without boundary (a cycle bc= 0 where b is the boundary operator dual to the exterior derivative d on Cartan forms f where c is a chain or space such that c = c1 + c2 + c3 + .... where bci =/= 0 ). We then tile or partition this spherical surface using great circle geodesic lines forming some kind of regular polygons that in actual 3D case correspond to Ellis's the spatial sections of the Universe are dodecahedral sections of a space of positive curvature, fitted together to make a finite three dimensional spaces. [without boundary] The questions is: (The Question is: What is The Question? :-) What happens if a star ship attempts to cross the line, i.e. cross the 1D edges of the polygon in the 2D toy model case, or pass through the 2D walls separating the dodechahedral sections in the actual 3D case? Is it like the Pong Video Game or not? I mean as you exit right you enter left, or was that simply an example Ellis used that had no direct Creon, them out. You have given me my first look at the 9 October *Nature* paper by Luminet et al. The papers by Gausmann (et al.) Topological lensing in spherical spaces and Eigenmodes of three-dimensional spherical spaces and their application to cosmology are especially interesting to me. They seem not be aware the the very relevant papers on gravitational instantons by Peter Kronheimer (1989, 1990) -- although the spherical spaces they consider are all SU(2)/g, where g is a finite subgroup of SU(2). The classification of these spaces via the A-D-E Coxeter graphs, provides the big guns of Lie algebra theory and the many other objects classified by these same Coxeter graphs, such as catastrophe bundles and 2-d conformal field theories (which live on the 2-d string world-sheet). It will take a while for me to digest all this new material -- but a lot of it looks very familiar! Saul-Paul ---------- === Subject: Papers coming on cosmological topology Saul Paul - I am sending you several papers in subsequent messages. pdf files. You may also be interested in: http://xxx.lanl.gov/abs/astro-ph/0212223 http://xxx.lanl.gov/abs/astro-ph/0303580 http://xxx.lanl.gov/abs/astro-ph/0310253 === Subject: re: Shape of the Universe === Subject: Re: WMD found! New York Times reports. Part III The Pong Video Game in The Sky with 2D Wall periodic boundaries may not be what Ellis et-al are suggesting. Ellis cites Omega zero = 1.02 +-0.02. meetings the Pundits like Mike Turner all said k = 0, inflation is right, i.e spatially flat with Omega zero = 1 on the button. Ellis is saying no to that that there will be a Big Crunch with k = +1 and Omega zero > 1. That allows a finite 3D space without 2D wall boundaries. So, spatial sections of the Universe are dodecahedral sections of space of positive curvature, fitted together to make finite three-dimensional spaces. Earlier Ellis does make the Pong Video Game analogy implicitly for the flat toroidal space, as you exit right you enter left and space is finite Does Ellis mean that there are sectors of 3D space with 2D walls between them which fit together like some finite Platonic solid? What happens when you walk through a wall so to speak? Is there a periodic boundary condition (i.e. a Star Gate in effect) or not? That is the total finite space of positive curvature has no boundary, but is partitioned into disjoint subspaces or sectors separated by 2D boundary walls? Is that the idea? Are these walls weird or not? Can we pass through them? What happens when we do? Or are they not there at all? Whatever the picture is here, Ellis is clear that this alternative is not compatible with chaotic inflation that demands Omega zero = 1 with k = 0. Luminet has Omega zero = 1.013 American is not consistent with what Ellis et-al is suggesting here. That is Ellis rejects that spatial homogeneity extends outside our visual horizon forever .... and we are in one expanding bubble in the middle of innumerable other similar ones. But if Luminet et-al are correct, chaotic inflation is ruled out: there is only one expanding bubble, and we can see almost all the way round it .... The WMAP data as interpreted by Luminet et-al... suggest that we might live in such a small closed universe. This is qualitatively globally different from what Mike Turner, Max Tegmark et-al are suggesting. So we have a schism between the k = 0 Camp and the k = +1 Camp. We need better precision than the current 2% to decide. === Subject: Re: Goldbach Computations > In 1742, historian and mathematician Christian Goldbach (1690-1764) > effect, that every integer greater than 5 is the sum of three prime > numbers. A prime number is evenly divisible only by itself and 1. > Nowadays, Goldbach's conjecture is expressed in the following > equivalent form: Every even number larger than 2 is the sum of two > prime numbers. > Despite centuries of effort, no one has yet been able to prove Goldbach's > conjecture. Progress has been slow. > Vinogradov proved in 1937 that every sufficiently large (more than a certain > number called 'Vinogradov constant') odd number is a sum of three primes. > This constant is large (> 2^65536), but anyway the problem is not very > interesting, since the quantity of numbers left to check is finite. I cannot immediately see the equivalence of the various conditions. If all even numbers can be expressed as the sum of two primes, then the following odd number can be expressed as the sum two primes, plus one. But one is not allowed as prime. So what does this gain us? I suppose the proof is by induction, with some attention to cases ... Either one of the prime decomposing an even integer larger than two is itself two, or it isn't. If one of the summands is two, than the other is still even, therefore composed of two primes, so the total sum is of three primes -- unless both summands are two, in which case we don't get two primes out of the remaining summand, but then the number in question was four, which wasn't part of the conjecture. Ok. Suppose neither of the summands is two. Then, both must be odd numbers greater than one. Suppose both are three ... then the number is six and the sum is two + two + two. Suppose one summand is three, and the other five: than the target may be expressed as three + three + two. Suppose both summands are odd, and at least one is greater than five. Then what? Then we may decompose the at least one summand into three primes, thereby decomposing the original target into a sum of four primes ... which buys us what, since we were looking for three primes? And we have yet to consider the odd numbers, provided of course they are larger than five. The number of cases to consider is finite, but annoying! Which may explain my non-existence as a number theorist. I'm not sure why a finite number of cases makes the thing uninteresting: investigators found the groups interesting, though AFAIK it turns out in cataloguing them that the cardinality of some phyla of groups are finite. Why should the detailed structure of all numbers less than 2^65536 be uninteresting to the number connoisseur? Maybe there's a single counterexample to the Goldbach conjecture lurking in there, waiting to give its discoverer a lasting footnote! === Subject: Combinatorics question Say we have the number 4. There are 5 unique ways (up to ordering) of writing it as sums of integers lower than it. They are: 4 2 + 2 1 + 3 1+1+1+1 1+1+2 I believe these are called partitions of integers. Does anybody know how to find the number of such partitons (with a proof) ?? Is this a famous combinatorial problem? If someone knows the answer, has some hints, or knows of a reference that contains an answer, please tell me. Bill === Subject: Re: Square root modulo a power of two > How does one solve the quadratic congruence x^2=a(mod 2^n) ? By hand, one can solve it by asking what squared equals k*2^n+a. For example for n=1,a=1, then x=1. For n=2,a=1, then x=1,3,5,7 and for a=4, then x=2,6. If n is even, then it is only neccessay to consider the first 2^(n/2) numbers and then extend to k*2^(n/2) plus minus a reference number. I am sure that a general algorithm or rule exists, but I can't think of one. Maybe I'll have one come morning, in which case you'll be the first to know. === Subject: Re: Boolean Algebra - Arithmetic Relationship |I guess what I'm trying to say by developing the breadth and not the |depth of mathematics is this. I find certain things easy to do in |some programming languages and very difficult to do in others, |scripting in Perl, and Object-Orientated programming in Java, etc. |Similarly, I assume, there are certain proofs which are easier done |geometry than their equivalents in another mathematical symbology |(that geometry can be reduced to of course). My point: What's |abstract in one form of mathematics may be readily apparent in |another. Rather than seeking more and more abstract notions and proofs |of a particular symbology, perhaps we have to look for a different |symbology. It's nice to see someone thinking about this kind of issue. I would say that what to the interested layperson looks like more and more abstract notions of a particular symbology is in fact often much more about trying to develop a framework within which lots of things become much easier to do, in the way that using a more suitable programming language might do. Not that this is always successful (it's harder to do than elaborating on old ideas, because it requires new ideas). It's the kind of mathematics that mathematicians like to see, however. Some mathematicians are very good at wielding elaborate technical apparatus to prove results, and there's respect for that, but in the end people would rather have a proof which is less technical and more conceptual, if there is one. If all a field does is pile on technical details higher and higher, it eventually goes dormant. Another thing to keep in mind is that there's more of an emphasis on interapplicability than there is in a lot of sciences, like physics. developed at length, then be rather completely abandoned, so much so that almost nothing is salvaged for later use, because there's no use for results about how the old theory worked in the development of the new one. Or at least so I'm told. In mathematics, it seems that ideas have a greater lifespan; a genuinely good result is less apt to be discarded because it was only valid in an abandoned theory. Even inside of mathematics, there are occasions when something like a paradigm shift occurs, but a lot of the overlying mathematics acts like the table settings in the old parlor trick, where a tablecloth is whisked quickly away while the tablewear, plates, and glasses stay put. After working in software for awhile, the way that mathematics manages to achieve the kind of reusability we only wish we had in software, and keep ideas around for a long time, elaborating on them, without gradually drowning in legacy proofs, seems all the more amazing. === Subject: Re: Factorial/Exponential Identity, Infinity I find an obvious flaw in my statement. Consider the sequence (01)..., interchange each 1 with the adjacent 0 of the next subsequence, 01010101010101... 00101010101010... Then interchange each fifth and later 1 with the following 0: 00100101010101... Repeat for each 3x-1'th one, and the output sequence results in (001).... I consider this, then. Divide the sequence with the known density p/q into subsequences of finite length xq for any positive integer x. Then, for each of those finite subsequences its elements can be permuted. The resultant sequence is to have the same density as the input sequence and shall not be convertible to a sequence known to be of any other density. That appears to be an adequate method for that, but I'm concerned that it would not be able to convert into each other sequence of the same density, that is, some of the irrational sequences. It would appear to allow converting (01)... to (10)... but not to (001).... I think it would allow the conversion of (01)... to normal sequences. Almost all real numbers are absolutely normal. === Subject: generating a random discrete vector given pair-correlations Suppose we have an unknown probability distribution on the set of vectors in (Z/4Z)^40 . We explicitly don't assume that the module structure is related to the probability distribution. What we do know are the C(40,2) coordinate-pair correlations. In other words, if V denotes the random variable with unknown distribution, we have: Prob( V_i = a and V_j = b ) = c_{a,b,i,j} for all 1<=iDo you feel it would be worth purchasing the 1st edition? It is only $20. >Compared with the near $100 for the 3rd. I am just looking for a gentle >introduction to Calculus/Analysis that I can use as self-teaching text. For >my background, I find Rudin too terse. > Whoo-hoo. Gentle is not the word I'd have chosen first, or tenth, > to describe Spivak's _Calculus_. The problems, in particular, tend > to be brutal. > Have you looked at Apostol's _Mathematical Analysis_? Or, for that > matter, Courant's calculus textbook? I have never seen Rudin's > calculus text, just (many many years ago) his _Real and Complex > Analysis_, and slightly more recently a monograph or two on some > topics in several complex variables. But I don't think Spivak > is notably less terse than Rudin, and I do think that both > Apostol and Courant are probably less terse while being quite > as rigorous as necessary. > Lee Rudolph Lee, are you refering to Spivak's Calculus, or to his Calculus on manifolds ? Lurch === Subject: Re: Linear Algebra question > I'm not sure if/why this would be true or false. I'm having a hard time > understanding how I could explain that it was true. I would appreciate any > help. > Given a linear transformation T:F^n-->F^m. Is it true that there exists an m > x n matrix A such that T(x)=Ax (where we think of x as an n x 1 matrix? If your notation means what I think it means, this is pretty standard and will be in almost any Linear Algebra text. The idea is to take the usual basis of F^n, { e_1, ... , e_n }. Let the usual basis of F^m be { v_1, ...., v_m }. Write T(e_j) = a_1*v_1 + a_2*v_2 + ... + a_m*v_m. Then the transpose of (a_1, ... , a_m) is column j of A. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Goldbach Computations |I cannot immediately see the equivalence of the various conditions. I had to stop a bit too. |If all even numbers can be expressed as the sum of two primes, then |the following odd number can be expressed as the sum two primes, plus |one. But one is not allowed as prime. So what does this gain us? Use 3 instead! Let me attempt a little summary (unnecessary as it must be now that everyone has stopped a bit and figured it out): G: Goldbach in the modern form. Every even number >=4 can be written as a sum of two primes. G2: Goldbach in roughly the original form. Every number >5 is a sum of three primes. G3: Goldbach in a pecular form. Even even number >5 can be written as a sum of three primes. V: What Vinogradov wanted to prove. Every odd number >5 is a sum of three primes. Obviously a sum of two primes has to be at least 4 and a sum of three has to be at least 6. G2 is just G3 and V. G is equivalent to G3 because when n is even, n+2 can be written as a sum of three primes only as 2+p+q, which is equivalent to being able to write n as a sum of two primes p+q. The even numbers >5 are just two more than the even numbers >=4. Also, G implies V, because if n is even and can be written as a sum of two primes, then n+3 can be written as a sum of three primes. G starts with n=4, V starts with n=7. There's no obvious way to use V to prove G, so in that subjective sense it's weaker. The version Vinogradov actually proved, that every odd number above some big bound can be written as a sum of three primes, is weaker still. So since G2 is just G3 and V where G3 is equivalent to G and V is implied by G, we get that G2 is equivalent to G also. Actually, for some reason some people like to state Goldbach as every even number n>4 can be written as a sum of two odd primes. I don't know why. Maybe they're trying to do even more of our thinking for us, by sparing us having to realize that 2+2=4 and 2 is the only even prime and so on. === Subject: Re: multiplication negs > Hopefully, I'm in the right place here.... > I've been making my way throught the Principia, taking my time, working > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? It's > disturbing me that I cannot come up with a solid answer, and certainly > 'We've been taught that that is the case. will not suffice. > Any help would be greatly appreciated. > TIA I will offer a geometric arguement I saw some where. Suppose you have a line or some other 2d geometric figure. Now, if you take the reflection of a point on a line, or the line(figure) itself, one gets the negative of that point. Take the reflection of the reflection and you get the point back again. Or, similarly, think of flipping a line once, then twice, to get back to where you started. Lurch === Subject: Re: Question on Hilbert & Godel