mm-469 === Subject: : The representations of the unit circle help?What are the representations of the unit circle acting on C and R?Since the circle S^1 is compact, any real or complex finite dimensionalrepresentation will be completely reducible.I'm trying to do this thing from scratch without much background inrepresentation theory. The only representations I can think of are the 1dimensional representation (unique up to multiplication by some element ofS^1) acting on C which is just the identity map, and the 2-dimensionalrepresentation acting on R which maps an element e^(i * theta) to the 2 x 2rotation matrixcos(theta) -sin(theta)sin(theta) cos(theta)Are there any other possible representations of the circle, what are they,and how should I go about thinking about this? Schur's lemma might help mesince the finite-dimensional representations are completely reducible.=== Subject: : Re: The representations of the unit circle help?>What are the representations of the unit circle acting on C and R?Do you mean C^n and R^n here?>Since the circle S^1 is compact, any real or complex finite dimensional>representation will be completely reducible.>I'm trying to do this thing from scratch without much background in>representation theory. The only representations I can think of are the 1>dimensional representation If you really mean C it's hard to see how you're going to get anythingother than a one-dimensional representation...>(unique up to multiplication by some element of>S^1) acting on C which is just the identity map, ??? When you ignore that multiplication by an element of S^1you're throwing out a whole lot of representations, seems to me;in fact you're taking the entire dual and saying that all the elementsare the same!For each integer n there is a representation R_n : S^1 -> GL_1(C)defined by R_n(e^(it)) = [e^(int)]. And every irreducible complexrepresentation of S^1 has this form (the fact that S^1 is abelianshows that every irreducible complex representation isone-dimensional.) You seem to be saying that all theserepresentations are trivial; while they're pretty simple(in an informal sense of the word) they're not the identityrepresentation, not trivial in any formal sense that springsto mind.>and the 2-dimensional>representation acting on R ??? This acts on R^2, not on R.>which maps an element e^(i * theta) to the 2 x 2>rotation matrix>cos(theta) -sin(theta)>sin(theta) cos(theta)>Are there any other possible representations of the circle, Yes, for example the one that maps e^(it) tocos(nt) -sin(nt)sin(nt) cos(nt)for an integer n. I believe that these, together with the trivial representation acting on R, are the only irreduciblereal representations. But don't quote me on that.>what are they,>and how should I go about thinking about this? Schur's lemma might help me>since the finite-dimensional representations are completely reducible.=== Subject: : Re: The representations of the unit circle help?> What are the representations of the unit circle acting on C and R?> Since the circle S^1 is compact, any real or complex finite dimensional> representation will be completely reducible.> I'm trying to do this thing from scratch without much background in> representation theory. The only representations I can think of are the 1> dimensional representation (unique up to multiplication by some element of> S^1) acting on C which is just the identity map, and the 2-dimensional> representation acting on R which maps an element e^(i * theta) to the 2 x 2> rotation matrix> cos(theta) -sin(theta)> sin(theta) cos(theta)> Are there any other possible representations of the circle, what are they,> and how should I go about thinking about this? Schur's lemma might help me> since the finite-dimensional representations are completely reducible.Let's think of actions on C^n. Now S^1 is covered by R and soa representation of S^1 on C^n is a continuous homomorphismf from R to GL_n(C) with f(2pi) = I. Each continuous homomorphismfrom R to a Lie group has the form f(x) = exp(xA) where A is in theLie algebra. Here A is a complex matrix. To make f(2pi) = Iwe need that A be diagonalizable, and that its eigenvalues beof the form im for integers m.=== Subject: : re:Simple convergence of sum problemSum[n=0,m] ( n*Binomial[m,n])= (first term=0)Sum[n=1,m] ( n*Binomial[m,n])= (simplecancellation)Sum[n=1,m] ( m*Binomial[m-1,n-1])= (k=n-1)m*Sum[k=0,m-1] ( Binomial[m-1,k])=m*2^(m-1)http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =---=== Subject: : re:cross correlation R_xy(t1,t2) = R_yx(t1,t2)* ?> R_xy(t1,t2) = complex conjugate of R_yx(t1,t2)I think it should be:R_xy(t1,t2) = complex conjugate of R_yx(t2,t1) === Subject: : how to study stat. inference well?I like stat. inference very much; I like HMM, prediction, machineintelligence, etc... they are very fun... but I don't understand why Icannot read/understand papers in this field...too much math, just don'tunderstand...always fall into sleep... anybody can recommend some good stepby step books/references/tutorials that after reading those I can understandstat. inference well and read those papers?=== Subject: : Re: how to study stat. inference well? very readable books are to start with:One I hope to get to read soon but looks like it would be excellent based on authors other work.http://www.seeingstatistics.com/Very well written. Explains the why which makes later reading of how much easier.https://www.erlbaum.com/shop/tek9.asp?pg=products& specific=0-8058-0528-1Shows how correlational analysis and inference (ANOVA, etc.) are the same thing.analysis for the behavioral sciences, third edition. Lawrence Erlbaum Associates, Mahwah, NJ.ISBN 0-8058-2223-2=== Subject: : why is the set of homomorphisms closed?In On discrete subgroups of Lie groups, Andrew Weil speaks of thespace of mappings from a discrete group Gamma to a topological groupG, and says it can be viewed as the product of a family of copies of Gindexed by Gamma, and given the product topology. Then he says the setof homomorphisms from Gamma to G is closed in this space. I see how toprove this when G is Hausdorff, but not in general. Can anyone help?=== Subject: : Re: why is the set of homomorphisms closed?> In On discrete subgroups of Lie groups, Andrew Weil speaks of theAndre Weil or Andrew Wiles?> space of mappings from a discrete group Gamma to a topological group> G, and says it can be viewed as the product of a family of copies of G> indexed by Gamma, and given the product topology. Then he says the set> of homomorphisms from Gamma to G is closed in this space. I see how to> prove this when G is Hausdorff, but not in general. Can anyone help?Many writers insist that topological groups be Hausdorff:since a non-Hausdorff topological group is simplyan extension of an indiscrete group by a Hausdorff group, thereis virtually no loss of generality imposing the Hausdorff condition.Does Weil insist on this?Anyway, the answer of course is that it's false. Suppose that Gammaand G are nontrivial and G is insdiscrete. Then so is Map(Gamma,G)--- its only closed sets are emptyset and itself. But Hom(Gamma, G)is neither of these.=== Subject: : Re: Rivals to Kline's Mathematics: The Loss of Certainty?> I actually agree on the Scientific American column. After a while> i got fed up with the pointless word games. But this has nothing> to do with *Goedel, Escher, Bach* which was a well-written book> (probably because he had a chance to actually work on it, as> opposed to pumping out a monthly column).A book I just read likens writing a monthly column to being a trainedcircus lion - the strain of being novel and entertaining on command is too much for many writers. === Subject: : Re: Would like to use network flow to prove Menger's and the Konig-Egervary TheoremsDiana> Are you aware of any books that would show these two proofs, or perhaps> just> the first one, in terms of network flow?> I think it's in> L.R.Ford & D.R.Fulkerson Flows in Networks, Princeton U. Press 1962> or whatever book is the current standard reference on the topic.> But I'll have another go at sketching a proof>(1) Menger's Theorem: If x, y are vertices of a graph G and xy isnot> an>element of the edges of G, then the minimum size of an x, y-cut>equals the maximum number of pairwise internally disjoint x,y-paths.> You saw that for any cutset and any set of disjoint paths, the inequality> size-of-cut >= number-of-paths> is easy.> To prove the reverse inequality, it is enough to find _some_ set of n> disjoint paths> and _some_ cutset of n edges, for the same n.> Consider a graph H with two edges wherever G has one edge,> and ascribe capacities +1 and -1 to the two edges in each pair> (meaning, 1 for an edge from u to v and -1 for an edge from v to u,> for each pair {u,v} of edges which are adjacent).> Consider now a maximal network flow from x to y on H. It must be> 0 or 1 or one on every edge (for otherwise it would not be maximal).> That being so, we can partition the edges on which the flow is 1,> into a set of paths, no two having any edge in common, from x to y,> qed.> LH> P.S. Actually the max-flow min cut theorem has a constructive version;> an algo can be written out which _finds_ a maximum flow, and Menger> is the special case in which the capacities are all 0 or 1.=== Subject: : Sum of identically dist. r.v. with limited dependenceLet X_1, X_2 ... X_n be n random variables such that(i) X_i and X_j are independent if |i-j|>m (m=1)?=== Subject: : Re: Sum of identically dist. r.v. with limited dependence>Let X_1, X_2 ... X_n be n random variables such that>(i) X_i and X_j are independent if |i-j|>m (m(ii) X_i = 1 with probability p> = 0 with probability 1-p>Let Y = X_1 + X_2 + ... + X_n>By linearity of expectation, we know that E[Y] = np>What is the best lower bound for P(Y>=1)?This one turns out to be easier than it looks. Whatwe need is an upper bound on P(Y=0), and this canonly happen if all of the X's are 0. Let k be thesmallest integer such that n <= k*m; then there arek X's which are all m apart, and the probability thatthey are all 0 is (1-p)^k. However, this is attainedby having the first m X's all equal, the second m, etc.So the answer is 1-(1-p)^k, k defined as above.This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== Subject: : Re: Podkletnov & Zero Point Energy Propulsion> 14. 1957 non-perturbative BCS microscopic -> macroscopic ground state > instability from normal metal to superconductor (all real i.e. on-mass-shell electron pairs).> 15. One can never get Einstein's gravity in the virtual case of > electron-positron pair vacuum condensate mentioned, nor > superconductivity in the real case of on-mass-shell electron pairs in a > macroscopically occupied bound state. But Einstein said it even simpler: No Newton reject will get Einstein's gravity other than from the unique state of Euclid's geometry. And quantum rejects will not understand temperature without infinite quantities of help from Planck, since atoms have nothing to do with it.=== Subject: : Re: Restoring nature> This paper concerns spring fever, tree fever and other funky> feelings: Starting with restoring nature to its pastoral splendor, and> getting on with our busy lives:> Dumb Donny Head doesn't know that a vast radius around the popped> Chernobyl reactor has reverted to Eden - green, unspoiled, and> paradise for animal life. Everything living there has a whole new set> of genes, too, being constantly updated.I sometimes wonder if anyone has looked to see if animals die offquicker or if life is going on as before.=== Subject: : Re: PROOF that Integers are not countable!>You haven't shown how, given any list, you can construct a number not>on the list. There is no such natural number as oo, or oo-1.> Assume I have a list of natural numbers., N(i).> Let s be a natural number not on the list.> s is calculated recursively.> s_1 = N(1)+1> s_i = N( s_(i-1) ) + s_(i-1) (for i>1)> (for finite lists, if N(s_i) doesn't exist then add the last N(i)).> s_i is not in the set Union( N( 0 ) - N( s_(i-1) ) ) for all i.> Example:> N_i = 1, 2, 3, 4, ...> s_i = 2, 4, 8, 20, ...> - The universe is one dimensional> You have defined, not a natural number, but a sequence of natural> numbers. And while the n-th member of this sequence is never the n-th> number on the list, elements of the sequence can still occur in the> list. You haven't given a natural number not in the list. Sometimes> the list will include all natural numbers, and then this won't be> possible.I have given a method to calculate a number not in the list.The method works for any list of natural numbers.s_n differs from the first s_(n-1) entries in the list.Russell- 2 many 2 count=== Subject: : Re: PROOF that Integers are not countable!>You haven't shown how, given any list, you can construct a number not>on the list. There is no such natural number as oo, or oo-1.>Assume I have a list of natural numbers., N(i).>Let s be a natural number not on the list.>s is calculated recursively.>s_1 = N(1)+1>s_i = N( s_(i-1) ) + s_(i-1) (for i>1)>(for finite lists, if N(s_i) doesn't exist then add the last N(i)).>s_i is not in the set Union( N( 0 ) - N( s_(i-1) ) ) for all i.>Example:>N_i = 1, 2, 3, 4, ...>s_i = 2, 4, 8, 20, ...>Russell>- The universe is one dimensional> You have defined, not a natural number, but a sequence of natural> numbers. And while the n-th member of this sequence is never the n-th> number on the list, elements of the sequence can still occur in the> list. You haven't given a natural number not in the list. Sometimes> the list will include all natural numbers, and then this won't be> possible.> I have given a method to calculate a number not in the list.No, you haven't.> The method works for any list of natural numbers.> s_n differs from the first s_(n-1) entries in the list.That doesn't mean you've given a number not in the list.> - 2 many 2 count=== Subject: : Re: approximation theory question>Can anyone give me a hint or a reference about the>following? I am looking for a low degree polynomial>which computes x^n approximatelly. Approximately in what sense? If the desired polynomial is P thenyou want to have the error P(x) - x^n be zero (or at least small) for all(?) x (on an interval?), even though it can only BEzero for a handful of x's. So do you want to choose P tominimize (a) sup( |error| ) ? (b) integral( |error| ) ?(c) integral( error^2 ) ? (d) something else?>I need a smaller degree (say log n), but with worse>error (say 1/n).This doesn't sound right; polynomials of low degree cannot match thedramatic rise of x^n as x --> 1 from below.Suppose you look for polynomials of degree k which approximatex^(2^k) on the interval [0,1], and suppose we look for the bestamong them, in sense (d). Writing P = sum a_i x^i , we find thatthe defining condition on the a's is that the column vector X = (a0, a1, a2, ... a_k)' is the solution to a linear system M X = Bwhere M_{i,j} = 1/(i+j-1) and B_i = 1/(2^k + i) . One can actuallyinvert the matrix M symbolically, so it's not hard to get explicitformulas for the polynomial P. I get P = sum_{i,j =0}^k x^i/(j+1+2^k)*(-1)^(i+j)*(k+1+i)!*(k+1+j)! / ( i!^2*j!^2*(k-i)!*(k-j)!*(i+j+1) )I don't see a pleasant formula for the distance from this P to x^(2^k)(in L^2([0,1]) ) although we can expand the square of that distance, int P^2 - 2int x^(2^k)*P + 1/(1+2^(k+1))and integrate the corresponding polynomials. I tried a few values of kand I have to say it does NOT look like this distances are on theorder of 1/(2^k) !So if even the closest polynomial of degree log(n) stays further thanO( 1/n ) away from x^n, then what you are asking for does not exist.Of course, you may have in mind some very different notion of distance, etc.dave=== Subject: : Re: approximation theory questionHi Dave,was far from precise. Your reply was really informative. I was actuallylooking for an approximation in [-1,1] which minimizes the L_maxnorm, max(|error|). The holy grail in this case is the minmaxpolynomial, but it does not seem to be as easy to find (at least accordingto the textbooks I looked at) as the polynomial which minimizes the L_2norm and which you did in your reply.I looked around a little more since I posted the question and this iswhat I found. My conclusion was basically the same as yours, althoughnot as definite, that a polynomial of degree log(n) cannot achieve anerror of 1/n. Of course since L_2 norm= O(L_max norm), your replyanswers my question as well. The smallest degree for which I can getan error of 1/n is sqrt(n log(n)), and error O(1) with degree O(sqrt(n)).I am curious, if you will know anything about finding the _exact_ minmaxpolynomial of degree m<=n-1 for x^n, which seems to be known onlyfor m=n-1 and n-2.You can skip the rest, if you do not care about my musings, althoughI would like to know what you think about them, if you have thetime to consider them.I looked at the Chebyshev expansion of x^n, which isx^n=2^(1-n) sum'_{k=0}^floor(n/2) (n choose k) T_(n-2k)(x) =: T(x),where T_j is the j-th Chebyshev Polynomial, and sum' meansthat for n even we also substract (n choose (n/2))/2. Now let P_m be thepolynomial which has the first m+1 terms of T, i.e. we drop the termsfor k=0, ..., (n-m-1)/2. Since |T_j(x)|<=1, for x in [-1,1], the L_maxerror that we get with P_m as an approximation of degree m to x^n isat most2^(1-n) sum_{k=0}^{(n-m-1)/2} (n choose k)which is rougly2^[{H((n-m)/(2n))-1}n] = 2^[{H(1/2-m/(2n))-1}n] ~2^[-m^2/(4 n)], for m << n,where H is the entropy functionH(p)=-p log_2 (p) - (1-p) log_2 (1-p)and I use the estimateH(1/2-e)~1-e^2, for e~0.Thus the error is 1/n for m~sqrt(n log n), O(1) for O(sqrt(n)), andfor m=log n it is O(n^[-(log n)/n]), which in fact goes to 1 rather than to 0(I wonder, if I have not made a mistake in my estimates). I gave up herebecause the textbook said that this kind of approximations are not toofar from the minmax polynomial of the same degree, so I thought thatthe minmax polynomial will not do much better.So, this is what I did...>Can anyone give me a hint or a reference about the>following? I am looking for a low degree polynomial>which computes x^n approximatelly. > Approximately in what sense? If the desired polynomial is P then>you want to have the error P(x) - x^n be zero (or at least small) >for all(?) x (on an interval?), even though it can only BE>zero for a handful of x's. So do you want to choose P to>minimize (a) sup( |error| ) ? (b) integral( |error| ) ?>(c) integral( error^2 ) ? (d) something else?>I need a smaller degree (say log n), but with worse>error (say 1/n).>This doesn't sound right; polynomials of low degree cannot match the>dramatic rise of x^n as x --> 1 from below.>Suppose you look for polynomials of degree k which approximate>x^(2^k) on the interval [0,1], and suppose we look for the best>among them, in sense (d). Writing P = sum a_i x^i , we find that>the defining condition on the a's is that the column vector >X = (a0, a1, a2, ... a_k)' is the solution to a linear system M X = B>where M_{i,j} = 1/(i+j-1) and B_i = 1/(2^k + i) . One can actually>invert the matrix M symbolically, so it's not hard to get explicit>formulas for the polynomial P. I get> P = sum_{i,j =0}^k x^i/(j+1+2^k)*(-1)^(i+j)*(k+1+i)!*(k+1+j)! /> ( i!^2*j!^2*(k-i)!*(k-j)!*(i+j+1) )>I don't see a pleasant formula for the distance from this P to x^(2^k)>(in L^2([0,1]) ) although we can expand the square of that distance,> int P^2 - 2int x^(2^k)*P + 1/(1+2^(k+1))>and integrate the corresponding polynomials. I tried a few values of k>and I have to say it does NOT look like this distances are on the>order of 1/(2^k) !>So if even the closest polynomial of degree log(n) stays further than>O( 1/n ) away from x^n, then what you are asking for does not exist.>Of course, you may have in mind some very different notion of distance, etc.>dave===All information :mailto:dav_cad26@hotpop.comPART OF LIST [email me for full list]://////3D Studio Max 6///3D Studio Viz 4.2 SP2///Abaqus 6.4-1///Accel P-cad 2002///Alias Wavefront Maya Unlimited 5.01///Ansys 8///Archicad 8.1r1///ArtLantis 4.5///Autodesk Architectural Desktop 2005 <<<<<<<<<< how to calculate 255^23 modulo 2455 by Euler's theorem?You have pipe dream.255 and 2455 aren't coprime and 23 < phi(2455)=== Subject: : Re: * V Interesting geometry problem *>What I want you to find is:>Area of intersection of two identical squares of unit size. >First square center is at (0,0) and is not tilted. >Second square center is at (h,k) and is tilted theta CCW.Seems to me this is going to require a case-by-case analysis of thedifferent ways two squares can meet. Here's a sketch of what could be done.For each edge of the second square, keep track of the number of edgesof the first square which it crosses. Reading the four edges in order,this gives a sequence of four numbers (a,b,c,d) (which is geometricallyequivalent to (b,c,d,a), (d,c,b,a), etc. under the action of the dihedralgroup). Each of these numbers must be 0, 1, or 2; the last is only possiblewhen passing from the outside of the first square back to the outside,and of course a single crossing means passing from inside to outside orvice versa. Thus there will be an even number of 1's, and (2,1,2,1) isnot permitted. You can superimpose a _parallelogram_ on a square toget incidence numbers (1,1,1,1) but I think that's not possible for two_squares_ (I don't know a geometric invariant which makes this statementobvious!) So the incidence numbers would have to be one of:(0,0,0,0) (the squares' edges don't meet. Since neither square can contain the other, this means the squares' interiors are disjoint.)(1,1,0,0)(1,0,1,0)(2,0,0,0) (for two squares of EQUAL sizes this means the intersection is a triangle; the intersection could be the complement of that triangle in a small square meeting a larger square.)(2,1,1,0)(2,1,0,1)(2,2,0,0)(2,0,2,0)(2,2,1,1)(2,2,2,0) (2,2,2,2) (e.g. a diamond meeting a square)Note that a pair of consecutive 1's means the second square passes from outside to inside to outside the first square -- it couldn't passfrom inside to outside to inside because then a diagonal of one squarewould be in the interior of another square of the same size.Of course whenver (h,k) is far from (0,0) we have the first type, i.e.no real intersection. I suppose the nondegenerate configurations of the other 10 types partition all but a set of measure zero of therest of the compact region of (h,k,theta)-space which we can visualize(sort of) in R^3. I haven't the patience to map out the correspondingregions of the parameterization space and to compute the areas ofintersections in each component.I await corrections as people point out configurations I missed,truly-distinct configurations which I have conflated, and configurationson my list cannot occur with pairs of congruent squares ...dave=== Subject: : EntropyWould anyone like to theorise about numbers that do not terminate and do not have any pattern with me?The way I see it is that it is impossible to aproximate a non terminating sequence that does not have a pattern, and this is entirely different than numbers that do not repeat. A number that repeats is one kind of pattern, but there are also patterns like Pi. Pi is a number that can clearly be aproximated even to 51 billion digits because it does have a pattern, even if not a pattern that repeats itself. Another example of a pattern would be the fibonaci sequence, where every number in the sequence is the sum of all the previous numbers, but this pattern does not repeat.Do you agree that non terminating sequences without patterns can not be aproximated, and if so would it still be acceptable to contemplate them as numbers even if my human mind can't create an aproximation to contemplate? I'm not sure how to prove that non terminating sequences without patterns can not be aproximated, because I can not think of an example of such a number!=== Subject: : Re: Entropy> Would anyone like to theorise about numbers that do not terminate and do> not have any pattern with me?> The way I see it is that it is impossible to aproximate a non> terminating sequence that does not have a pattern, and this is entirely> different than numbers that do not repeat. A number that repeats is one> kind of pattern, but there are also patterns like Pi. Pi is a number> that can clearly be aproximated even to 51 billion digits because it> does have a pattern, even if not a pattern that repeats itself. Another> example of a pattern would be the fibonaci sequence, where every number> in the sequence is the sum of all the previous numbers, but this pattern> does not repeat.> Do you agree that non terminating sequences without patterns can not be> aproximated, and if so would it still be acceptable to contemplate them> as numbers even if my human mind can't create an aproximation to> contemplate? I'm not sure how to prove that non terminating sequences> without patterns can not be aproximated, because I can not think of an> example of such a number!What?A number that can not be approximated. This does not make sense, becauseany real number may be approximated by a rational number (the rationals aredense in the reals.)A number without a pattern cannot be approximated. This does not makesense to me either. Write down such a number. You could e.g. throw aten-sided die to generate each digit at random. After a while (say 100billion digits) you stop. Call this number NoRoots1. Clearly this numberdoes not have a pattern. However, you've just given it a name (just like Piis a name), and hence created a very compact representation that uniquelyrepresents all the digits of the number.However, if you consider a specific real irrational number, it will have*infinite* number of digits. I suppose you are asking, whether there alwaysexists a single *finite* program that may generate arbitrary long sequencesof digits. Clearly this is the case for the well-known numbers Pi, e, etc.In general, I don't know, but I can't see what this has to do with entropy.I mean, the actual definition of the number will - in principle - provide animplicit algorithm to determine the number to infinite accuracy.-Michael.=== Subject: : Re: Entropy> Do you agree that non terminating sequences without patterns can not be> aproximated, and if so would it still be acceptable to contemplate them> as numbers even if my human mind can't create an aproximation to> contemplate? I'm not sure how to prove that non terminating sequences> without patterns can not be aproximated, because I can not think of an> example of such a number!Before any such thing can be proved, you would need to define the terms.1. What does it mean for a sequence to have no pattern?2. What does it mean to be able to approximate a sequence?For 1, I'd try: In the limit as n increases without bound, theshortest algorithm that can compute the first n digits of thesequence has size at least nNote that the sequence produced by a loaded die will (with probability 1)have a pattern according to this definition.For 2, I'd try: For every leading subsequence s, there is an algorithmthat can compute s.Under these definitions, all sequences can be approximated but notall sequences have patterns.If you reverse the quantifier order in 2, you get:There is a single algorithm that can compute s to any desired precision(which is to say that s is computable).With this modification, all sequences that can be approximated havepatterns but not all sequences that have patterns can be approximated.The loaded dice is (with probability 1) an example of a sequence thathas a pattern but cannot be approximated.> What?A number that can not be approximated. This does not make sense, because> any real number may be approximated by a rational number (the rationals are> dense in the reals.)A number without a pattern cannot be approximated. This does not make> sense to me either. Write down such a number. You could e.g. throw a> ten-sided die to generate each digit at random. After a while (say 100> billion digits) you stop. Call this number NoRoots1. Clearly this number> does not have a pattern. However, you've just given it a name (just like Pi> is a name), and hence created a very compact representation that uniquely> represents all the digits of the number.> However, if you consider a specific real irrational number, it will have> *infinite* number of digits. I suppose you are asking, whether there always> exists a single *finite* program that may generate arbitrary long sequences> of digits. Clearly this is the case for the well-known numbers Pi, e, etc.> In general, I don't know, but I can't see what this has to do with entropy.> I mean, the actual definition of the number will - in principle - provide an> implicit algorithm to determine the number to infinite accuracy.No. It won't. There are definable numbers that are not computable.The number whose binary expansion has a 1 in bit n if the n'th TMhalts on blank input and a 0 in bit n if it loops is well defined(or would be if we spent a couple of pages nailing down the details)but is not computable (else the halting problem would be solved). John Briggs=== Subject: : Re: Entropy> Before any such thing can be proved, you would need to define the terms.> 1. What does it mean for a sequence to have no pattern?> 2. What does it mean to be able to approximate a sequence?> For 1, I'd try: In the limit as n increases without bound, the> shortest algorithm that can compute the first n digits of the> sequence has size at least n> Note that the sequence produced by a loaded die will (with probability 1)> have a pattern according to this definition.> For 2, I'd try: For every leading subsequence s, there is an algorithm> that can compute s.> Under these definitions, all sequences can be approximated but not> all sequences have patterns.> If you reverse the quantifier order in 2, you get:There is a single algorithm that can compute s to any desired precision> (which is to say that s is computable).> With this modification, all sequences that can be approximated have> patterns but not all sequences that have patterns can be approximated.> The loaded dice is (with probability 1) an example of a sequence that> has a pattern but cannot be approximated.This all looks good. My logic that a non terminating sequence of numbers without any such pattern could not be aproximated was because for the first 10 billion digits it could have a deffinite pattern that repeated and then after that it could have a different pattern for 10 billion digits, and in the end none of these small subsets of the sequence that had patterns could be used to aproximate the whole. If you aproximated this sequence you would wind up aproximating another sequence much better than the one you were trying to aproximate because of this.=== Subject: : Re: Entropy> This all looks good. My logic that a non terminating sequence of > numbers without any such pattern could not be aproximated was because > for the first 10 billion digits it could have a deffinite pattern that > repeated and then after that it could have a different pattern for 10 > billion digits, and in the end none of these small subsets of the > sequence that had patterns could be used to aproximate the whole. If > you aproximated this sequence you would wind up aproximating another > sequence much better than the one you were trying to aproximate because > of this.I think that a good example of a non-computable number can begiven:Chitlin's Omega Constant, the probability that a random programon a given Turing machine will halt. It is different for every Turingmachine, though.I'm not sure if any other definable, yet not computable number hasbeen found.=== Subject: : Re: Weight> Weight is a measure of the centripetal force, or thrust exerted on,> molecules] that comprise the mass of the substances that we call> matter; when they are at rest on the terra firma surface of Earth, or> on a similar planet.Sigh! Inertia http://scienceworld.wolfram.com/physics/Inertia.htmlWeight http://scienceworld.wolfram.com/physics/Weight.htmlMass http://scienceworld.wolfram.com/physics/Mass.html The time has come, the Walrus said, To talk of many things: Of shoes, and ships, and sealing wax, of cabbages and kings. And why the sea is boiling hot and whether pigs have wings. Makes about as much sense as Shead eternal struggle with inertia, weight and mass.=== Subject: : Re: Weight>Weight is a measure of the centripetal force, or thrust exerted on,>and/or by objects;> Dumb Donny Head is off his psycho-meds.> Hey stooopid Dumb Donny Head, where is the centripetal force in a> Cavendish balance?Err, boys. However amusing [you think] this is, could you considergetting a room somewhere? While I don't pretend to speak for allof sci.math, it seems to me to be wildly off-topic here.Just a suggestion,Rick === Subject: : Re: maximal graph>Uh, maybe you should relax a little - if someone says something you>don't understand it might be that you should ask for an explanation>instead of starting out with the explamation points.>Hey, I like this! Usually I just explain things but I can use more> No charge. I wish I knew for sure whether suggestino was a typo...Perhaps a typino. I thought it was a repetitive redundancy, as if a suggestino is anything then it's bound to be a small one.=== Subject: : Re: Force & weight versus mass & inertia> - Force & weight versus mass & inertia -Sigh! Inertia http://scienceworld.wolfram.com/physics/Inertia.htmlWeight http://scienceworld.wolfram.com/physics/Weight.htmlMass http://scienceworld.wolfram.com/physics/Mass.html The time has come, the Walrus said, To talk of many things: Of shoes, and ships, and sealing wax, of cabbages and kings. And why the sea is boiling hot and whether pigs have wings. Makes about as much sense as Shead eternal struggle with inertia, weight and mass.=== Subject: : Re: Force & weight versus mass & inertia: Uncle Al, where are you?>- Force & weight versus mass & inertia -Cut<>The mass and/or inertia of a mass is a Constant: Anytime and anyplace.> And light blubs don't emit light, they suck dark, and when they are full they> don't burn out, it's just that they are full of dark as any fool can see.Yes; I'm sure you can Tony(;^)=== Subject: : Re: Force & weight versus mass & inertia: Uncle Al, where are you?>- Force & weight versus mass & inertia -> Cut<>The mass and/or inertia of a mass is a Constant: Anytime and anyplace.> And light blubs don't emit light, they suck dark, and when they are full they> don't burn out, it's just that they are full of dark as any fool can see.> Yes; I'm sure you can Tony(;^)I see that Donald G. Shead is suddenly back in a big way!Does it have anything to do with the crank exclusion principle?Double-A=== Subject: : Re: Complex ring endomorphismsOriginator: dmoews@ccrwest.org (David Moews)|How many complex ring endomorphisms h:C -> C other than|h(x + iy) = 0; h(x + iy) = x + iy; h(x + iy) = x - iy ?There are 2^(2^(aleph-0)). There aren't any more because there are only2^(2^(aleph-0)) functions of any sort from C to C, and there aren't anyless because there are 2^(2^(aleph-0)) field automorphisms of C. Here is a sketch of how to find a nontrivial automorphism of C:1. Find two transcendence bases for C over Q, say {b_i} and {b'_i}.2. If A=Q(b_i) and A'=Q(b'_i), there is an isomorphism f: A -> A' given by sending each b_i to the corresponding b'_i.3. Let S be the set of all isomorphisms g : B -> B' of pairs of subfields of C, where B contains A, B' contains A', and g equals f when restricted to A. S is a partial order under extension of isomorphisms.4. Apply Zorn's Lemma to conclude that S has a maximal element, h : D -> D', say.5. If D is not all of C, there is some r in C but not in D. Since r and the b_i's are not algebraically independent, there is some nonzero polynomial f(x) with coefficients in D such that f(r)=0. Pick some such f with minimal degree, n, say, and let f(x) = a_0 + a_1 x + ... + a_n x^n. Since n is minimal, f(x) cannot factor nontrivially in D[x].6. Observe that E, the smallest field containing D and r, is of the form E = {c_0 + c_1 r + ... + c_{n-1} r^{n-1} | c_0, ..., c_{n-1} in D}. The only nontrivial part of this observation is that E is closed under inversion. This is because if g(x) is a nonzero polynomial with coefficients in D and degree less than n, then g(x) and f(x) have no common factor, so by the usual Euclidean algorithm, there are polynomials h(x) and k(x) with coefficients in D such that h(x)f(x) + k(x)g(x) = 1. k(r) is then inverse to g(r). This also shows that c_0 + ... + c_{n-1} r^{n-1} is not zero unless c_0 = ... = c_{n-1} = 0.7. Let f'(x) = h(a_0) + h(a_1) x + ... + h(a_n) x^n. Since C is algebraically closed, there is some r' in C with f'(r') = 0. Any factorization of f'(x) in D'[x] can be mapped, via the inverse of h, to a factorization of f(x) in D[x]. Therefore, f'(x) has no nontrivial factors in D'[x]. Now let E' = {c'_0 + c'_1 r' + ... + c'_{n-1} r'^{n-1} | c'_0, ..., c'_{n-1} in D'}. Using the same argument as for E, you can see that E' is a field, and that c'_0 + ... + c'_{n-1} r'^{n-1} is not zero unless c'_0 = ... = c'_{n-1} = 0.8. By 6, any element of E has a unique expression of the form c_0 + c_1 r + ... + c_{n-1} r^{n-1}. Therefore, we can define j : E -> E' by setting j(c_0 + ... + c_{n-1} r^{n-1}) = h(c_0) + ... + h(c_{n-1}) r'^{n-1}.9. j is a surjective ring homomorphism. Moreover it is injective because if j(c_0 + ... + c_{n-1} r^{n-1}) = 0, then by 7, h(c_0) = ... = h(c_{n-1}) = 0, so c_0 = ... = c_{n-1} = 0. 10. Therefore, j is a field isomorphism between E and E' extending h. However, this contradicts the maximality of h. Therefore, D must have been all of C. Similarly, D' must have been all of C, so h is a field automorphism of C.Given this construction, 2^(2^(aleph-0)) automorphisms can be realized by,e.g., fixing some {b_i} and {b'_i} and swapping the b'_is in pairs. dmoews@xraysgi.ims.uconn.edu=== Subject: : Re: Complex ring endomorphisms=== Subject: : Re: Complex ring endomorphisms|How many complex ring endomorphisms h:C -> C other than|h(x + iy) = 0; h(x + iy) = x + iy; h(x + iy) = x - iy ? >Here is a sketch of how to find a nontrivial automorphism of C: >1. Find two transcendence bases for C over Q, say {b_i} and {b'_i}.We note by CH they both have cardinality c. >2. If A=Q(b_i) and A'=Q(b'_i), there is an isomorphism f: A -> A' > given by sending each b_i to the corresponding b'_i.Q(b_i) is the finite extension of Q by a single element b_i.Each b_i? How do I understand that? Do you mean A = Q({b_i}) ?That makes the isomorphism much less than obvious. >3. Let S be the set of all isomorphisms g : B -> B' of pairs of >subfields of C, where B contains A, B' contains A', and g equals f >when restricted to A. S is a partial order under extension of >isomorphisms. >4. Apply Zorn's Lemma to conclude that S has a maximal element, > h : D -> D', say. >5. If D is not all of C, there is some r in C but not in D. Since r >and the b_i's are not algebraically independent, there is some >nonzero polynomial f(x) with coefficients in D such that f(r)=0. >Pick some such f with minimal degree, n, say, and let > f(x) = a_0 + a_1 x + ... + a_n x^n. >Since n is minimal, f(x) cannot factor nontrivially in D[x].Some p in Q[x,x1,..xj], b1,..bj in {b_i} with p(r,b1,..bj) = 0 by maximality of {b_i}.Q subset D by step 2. Assume some u not in D'.For all bi in {b_i}, bi in D. Otherwise h may be extended to h':D(bi) -> D'(u)Thus p in D[x] and p(r) = 0. What if D' = C ?Instead is needed A = Q({b_i}) for all bi to be in D? >6. Observe that E, the smallest field containing D and r, is of the >form E = > {c_0 + c_1 r + ... + c_{n-1} r^{n-1} | c_0, ..., c_{n-1} in D}.E = D(r) field isomorph D[x]/The only nontrivial part of this observation is that E is closed >under inversion. This is because if g(x) is a nonzero polynomial >with coefficients in D and degree less than n, then g(x) and f(x) >have no common factor, so by the usual Euclidean algorithm, there are >polynomials h(x) and k(x) with coefficients in D such that h(x)f(x) + >k(x)g(x) = 1. k(r) is then inverse to g(r). This also shows that >c_0 + ... + c_{n-1} r^{n-1} is not zero unless >c_0 = ... = c_{n-1} = 0.Nice review D(r) = D[x]/7. Let f'(x) = h(a_0) + h(a_1) x + ... + h(a_n) x^n. Since C is >algebraically closed, there is some r' in C with f'(r') = 0. Any >factorization of f'(x) in D'[x] can be mapped, via the inverse of h, >to a factorization of f(x) in D[x]. Therefore, f'(x) has no >nontrivial factors in D'[x]. Now let E' = > {c'_0 + c'_1 r' +...+ c'_{n-1} r'^{n-1} | c'_0,.. c'_{n-1} in D'}. >Using the same argument as for E, you can see that E' is a field, >and that c'_0 + ... + c'_{n-1} r'^{n-1} is not zero unless >c'_0 = ... = c'_{n-1} = 0.Like f' = hf provided h(x) = x. >8. By 6, any element of E has a unique expression of the form > c_0 + c_1 r + ... + c_{n-1} r^{n-1}. Therefore, we can define > j : E -> E' by setting j(c_0 + ... + c_{n-1} r^{n-1}) > = h(c_0) + ... + h(c_{n-1}) r'^{n-1}. >9. j is a surjective ring homomorphism. Moreover it is injective > because if j(c_0 + ... + c_{n-1} r^{n-1}) = 0, then by 7, > h(c_0) = ... = h(c_{n-1}) = 0, so c_0 = ... = c_{n-1} = 0. >10. Therefore, j is a field isomorphism between E and E' extending >h. However, this contradicts the maximality of h. Therefore, D must >have been all of C. Similarly, D' must have been all of C, so h is a >field automorphism of C. >Given this construction, 2^(2^(aleph-0)) automorphisms can be >realized by, e.g., fixing some {b_i} and {b'_i} and swapping the >b'_is in pairs.Is the trivial ring endomorphism the only proper ring endomorphism?----=== Subject: : Re: Complex ring endomorphisms> |How many complex ring endomorphisms h:C -> C other than> |h(x + iy) = 0; h(x + iy) = x + iy; h(x + iy) = x - iy ?>Here is a sketch of how to find a nontrivial automorphism of C:>1. Find two transcendence bases for C over Q, say {b_i} and {b'_i}.> We note by CH they both have cardinality c.>2. If A=Q(b_i) and A'=Q(b'_i), there is an isomorphism f: A -> A'> given by sending each b_i to the corresponding b'_i.> Q(b_i) is the finite extension of Q by a single element b_i.> Each b_i? How do I understand that? Do you mean A = Q({b_i}) ?> That makes the isomorphism much less than obvious.For each i, there is an isomorphism f:A -> A' such that f(b_i) = b_i'.Besides, Q(b_i) is not a finite extension of Q. It is the field ofall elements of C of the form R(b_i) for some rational function R withratioanl coefficients. It is obvious, since b_i is transcendent, thatthe function g_i and g_i' from Q(x) into Q(b_i) and Q(b_i')respectively and defined by g_i(R) = R(b_i) and g_i'(R) = R(b_i') arefield isomorphism. So, they induce an isomorphism f from Q(b_i) ontoQ(b_i').=== Subject: : Zariski denseI want to show that AB and BA have the same characteristic polylnomial forany n x n matrices A and B over the field k. If A is invertible, then it iseasy. But I want to show that the set S = {(A, B) | A or B is invertible} isZariski dense in the set k^(2*(nxn)). This will imply that AB and BA havethe same characteristic polynomial even when A and B are not invertible.Does anyone know how to show S is Zariski dense in k^(2*(nxn))?=== Subject: : Re: Zariski dense> I want to show that AB and BA have the same characteristic polylnomial for> any n x n matrices A and B over the field k. If A is invertible, then it is> easy. But I want to show that the set S = {(A, B) | A or B is invertible} is> Zariski dense in the set k^(2*(nxn)). This will imply that AB and BA have> the same characteristic polynomial even when A and B are not invertible.> Does anyone know how to show S is Zariski dense in k^(2*(nxn))?First, the Zariski is such that basic open sets are the cozero sets ofpolynomials and the statement that every non-empty open is dense isequivalent to the claim that you cannot have two polynomials whosezero-sets exhaust the space. Over a finite field, this is actuallypossible, but you can get around that case by embedding in an infinitefield which doesn't change the characteristic polynomial. Thesuggested argument is thus correct. Here is more elementary argument. Let me write A ~ B to mean that A and B have the same characteristicpolynomial and write A == B to mean they are conjugate. Also write P#for the inverse of P. Now choose invertible P and Q so that PAQ isthe block diagonal matrix [I,0;0,0]. This is always possible by rowand column reductions. Write Q#BP# as the block diagonal [C,D;E,F]with blocks corresponding to those of A. Then AB ~ PAQQ#BP# =[I,0;0,0][C,D;E,F] = [C,D;0,0] ~ [C,0;E,0] = [C,D;E,F][I,0;0,0] =Q#BP#PAQ = Q#BAQ ~ BA.By modifying the argument slightly, one can prove that when A and Bare m x n and n x m matrices, resp., then the characteristicpolynomials of AB and BA differ on by t^{m-n}. I see no obvious wayof getting at that with the Zariski topology argument.=== Subject: : Re: Zariski dense> I want to show that AB and BA have the same characteristic polylnomial for> any n x n matrices A and B over the field k. If A is invertible, then it is> easy. But I want to show that the set S = {(A, B) | A or B is invertible} is> Zariski dense in the set k^(2*(nxn)). This will imply that AB and BA have> the same characteristic polynomial even when A and B are not invertible.> Does anyone know how to show S is Zariski dense in k^(2*(nxn))?It's the complement of a Zariski-closed subset. By the irreducubilityof k^m, any nontrivial Zariski-open subset is Zariski-dense.(We cannot have two dijoint nonempty Z-open subsets of k^m for thentheir complements would be Z-closed, nontrivial, and would cover k^mcontradicting its irreducibility).=== Subject: : Re: Zariski dense> But I want to show that the set S = {(A, B) | A or B is invertible} is> Zariski dense in the set k^(2*(nxn)). ?Why wouldn't det(A)*det(B) work here?Jyrki Lahtonen, Turku, Finland=== Subject: : Re: Zariski densehm....I'm sorry but I don't understand why this involves determinants. Wouldyou explain a bit?> But I want to show that the set S = {(A, B) | A or B is invertible} is> Zariski dense in the set k^(2*(nxn)).> ?Why wouldn't det(A)*det(B) work here?> Jyrki Lahtonen, Turku, Finland=== Subject: : Re: Zariski dense> hm....I'm sorry but I don't understand why this involves determinants. Would> you explain a bit?S is the complement of the Zariski closed set defined by det(A)det(B)=0.So S is Zariski-open. So S is Zariski-dense (both in k^(2n^2)).What's the problem?Jyrki=== Subject: : Re: Zariski dense> S is the complement of the Zariski closed set defined by det(A)det(B)=0.The complement of S is the set of (A,B) where both A and B are notinvertible, right? So do u mean S is the complement of the Zariski closedset defined by det(A) = 0 and det(B) = 0?> So S is Zariski-open. So S is Zariski-dense (both in k^(2n^2)).And how does S is Zariski-open imply S is Zariski-dense?=== Subject: : Re: Zariski dense> S is the complement of the Zariski closed set defined by det(A)det(B)=0.Should have written S contains the complement of the Zariski ...> The complement of S is the set of (A,B) where both A and B are not> invertible, right? So do u mean S is the complement of the Zariski closed> set defined by det(A) = 0 and det(B) = 0?> So S is Zariski-open. So S is Zariski-dense (both in k^(2n^2)).> And how does S is Zariski-open imply S is Zariski-dense?The affine space k^(2n^2) is irreducible. Any non-empty open subsetof an irreducible topological space is dense. (Hartshorne, ch. 1, section I)Jyrki=== Subject: : Quantum Physics is based on duality and tri-ality does not work; file 40dIn the month of September of 1999 I ran a series of threads on whetherQuantum Mechanics can be based on triality instead of duality. In thatmonth I gave several supporting themes of evidence in favor oftriality. One of them was the idea that physical reality is embeddedin mathematics of 3rd dimension where one goes higher in dimensionscorrelates with Newtonian Mechanics whereas if one sticks with 3rddimension they are in Quantum Mechanics. Another supporting evidencein this mathematical perspective is that geometries seem to come in 3varieties and only 3 varieties back when I was doing this topic ofEuclidean, Riemannian, and Lobachevskian. Then there was theto need 3 essential ingredients of proton, electron and neutron.And another supporting evidence of no 3 dimensional nodes or volumenodes found in atoms back in September of 1999 where anyone can readmy archived posts in the appropriate file found at this website:www.archimedesplutonium.com for the file September of 1999I have come 180 degrees from my probing into this subject of 1999 andthink that duality is central to quantum mechanics and that tri-alityis false.Recently I have come to the idea that there exists 2 and only 2geometries of Riemannian and Lobachevskian and that Euclidean is aspecial case of both Riemannian and Lobachevskian. I believed backthen as I do now that the number of geometries is a important aspectof science and reflects on the very essence of Quantum Physics. Ibelieve the number of geometries is a mirror image of the number forduality or tri-ality. If there had been 3 and only 3 geometries then Iwould have believed that duality is flawed and tri-ality is the stuffof Quantum Mechanics. But I believe there are 2 and only 2 geometrieswhere Euclidean is a mere insignificant special case of Riemannian andLobachevskian. This follows the idea that there are only 2 classes ofnumber systems of Adics and Doubly- Infinites where Euclidean geometryis based on just one number-- zero, and zero is just one number ofAdics or Doubly-Infinites. The Reals are not a number set for they aremerely a distorted illusion of Doubly-Infinites, to give a historicalscience analogy that the Reals are to mathematics what the alchemistswere to physics or chemistry. Real-Numbers are modern day alchemy ofmathematics.As for the proton, electron and neutron the essential ingredients canbe broken down to just proton and electron for the neutron has aof proton and electron. So here again we have 2 essential ingredientsand not 3.As for volume nodes, well, since they are 3rd dimension means that anode which is nothing starts at 2 dimension and goes down to 1dimension. Leaving 3rd dimension as reality for existence. So you gohigher than 3rd dimension leads to Newtonian false physics. You golower than 3rd dimension is the realm of nothingness of nodes. Thatleaves just the 3rd dimension for existence of reality. But the numberof dimensions does not mean tri-ality over duality. The number ofgeometries translates into whether duality or tri-ality.Back in September of 1999 I kept seeing all these 3 and 3 and 3 tobase a discussion of whether QM is tri-ality and not duality. Today Ipeel away all of that supporting evidence and say that QM is indeedbased on duality and not tri-ality. The only item not disappearing assupport is 3rd dimension. But is the question of duality directlylinked to dimensions? I say no. I say it is directly linked to numberof geometries which is 2 and only 2 of Riemannian and Lobachevskian.Perhaps in the future I can re-admit this discussion for review andvanquish the 3rd dimension into some formula or relationship as to why3rd dimension must yield duality.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxieswww.archimedesplutonium.comwww.iw.net/~a_plutonium=== === Subject: : Re: Quantum Physics is based on duality and tri-ality does not work; file 40d(most snipped)> I have come 180 degrees from my probing into this subject of 1999 and> think that duality is central to quantum mechanics and that tri-ality> is false.> Recently I have come to the idea that there exists 2 and only 2> geometries of Riemannian and Lobachevskian and that Euclidean is aspecial case of both Riemannian and Lobachevskian. I believed back> then as I do now that the number of geometries is a important aspect> of science and reflects on the very essence of Quantum Physics. IIn my Unification of the Forces of Physics as per this diagram:Coulomb Force is the unifying forceCoulomb Force = Coulomb and is in the region of nucleus to surroundingelectronsStrongNuclear + WeakNuclear = NuclearCoulomb and is in the region ofnucleus onlyGravity + Antigravity = ElectronSpaceCoulomb and is in the region of thespace of electrons onlyThose forces by unification are duality unified.If tri-ality were the case then forces would be aligned in triple and notdual. But forces are aligned dually.One may well ask why Coulomb is singular and not dual? It is dual in theaspect that it is dual elements of proton charge to electron charge.Coulomb EM force is the only perfect force because it is not brokensymmetry as are StrongNuclear with WeakNuclear and as with Gravity toAntigravity. So the Coulomb is dual and perfect symmetry with its zeroArchimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxieswww.archimedesplutonium.comwww.iw.net/~a_plutonium=== === Subject: : Re: Quantum Physics is based on duality and tri-ality does not work; file 40dman, this is hopelessly framed as a question,whether or not there is some viable content in it. haven't you seen any of Weber's work? and, why plutonium, and not uranium? > (most snipped) > StrongNuclear + WeakNuclear = NuclearCoulomb and is in the region of> nucleus only --Give Earth a Trickier Dick Cheeny -- out of office, after GIGA years.http://www.benfranklinbooks.com/http:// members.tripod.com/~american_almanachttp://www.rand.org/ publications/randreview/issues/rr.12.00/=== Subject: : Re: A question on determinant> Let N = M*M' (' denotes conjugate transpose). Let det(N) be the> determinant of N. M is a rectangular matrix. It is clear that N is> Hermitian and square. The question is: (all matrices are complex)> If one more coloumn is added to M, ie., if M was 2 x 3, it is made as> 2 x 4 by appending an arbitrary non-zero coloumn to M, how does the> determinant of N changes? Does it increase or decrease or cannot say.> Could any one suggest me an answer and a proof why that is an answer.The new determinant is >= the old. This comes from the followingformula. Let A and B be m by n matrices. Then det(AB^t)is the sum of the det(C) det(D) where C ranges over the(n choose m) m by m submatrices of A and D is the correspondingsubmatrix of B.=== Subject: : Re: A question on determinantBut i dont really understand your proof. Could you give me somereferences for that. Or cite some place where i can look it up.prasanna.> Let N = M*M' (' denotes conjugate transpose). Let det(N) be the> determinant of N. M is a rectangular matrix. It is clear that N is> Hermitian and square. The question is: (all matrices are complex)> If one more coloumn is added to M, ie., if M was 2 x 3, it is made as> 2 x 4 by appending an arbitrary non-zero coloumn to M, how does the> determinant of N changes? Does it increase or decrease or cannot say.> Could any one suggest me an answer and a proof why that is an answer.> The new determinant is >= the old. This comes from the following> formula. Let A and B be m by n matrices. Then det(AB^t)> is the sum of the det(C) det(D) where C ranges over the> (n choose m) m by m submatrices of A and D is the corresponding> submatrix of B.> Robin Chapman=== Subject: : Re: A question on determinant> Let N = M*M' (' denotes conjugate transpose). Let det(N) be the> determinant of N. M is a rectangular matrix. It is clear that N is> Hermitian and square. The question is: (all matrices are complex)> If one more coloumn is added to M, ie., if M was 2 x 3, it is made as> 2 x 4 by appending an arbitrary non-zero coloumn to M, how does the> determinant of N changes? Does it increase or decrease or cannot say.> Could any one suggest me an answer and a proof why that is an answer.> The new determinant is >= the old. This comes from the following> formula. Let A and B be m by n matrices. Then det(AB^t)> is the sum of the det(C) det(D) where C ranges over the> (n choose m) m by m submatrices of A and D is the corresponding> submatrix of B.>But i dont really understand your proof. Could you give me some>references for that. Or cite some place where i can look it up.Robin didn't give a proof, but he cited a result that answers yourquestion. One way to justify the result he used is to break AB^t intopieces corresponding to an orthonormal basis of the m-dimensionalsubspaces of R^n.The set of m-dimensional subspaces of R^n forms a C(n,m)-dimensionalvector space, R^{n,m}. Given an orthonormal basis for R^n, we can makean orthonormal basis for R^{n,m} from the wedge products of m basisvectors of R^n. For example, the set of 2-dimensional subspaces in R^3form a C(3,2) = 3-dimensional vector space. Given an orthonormal basisfor R^3, {i,j,k}, {j^k,k^i,i^j} forms an orthonormal basis for R^{3,2}.Remember, the wedge product is antisymmetric; j^i = -i^j, i^i = 0, etc.The directed m-volume formed by m vectors is represented by the wedgeproduct of those m vectors. For example, the directed 2-volume formedby 4i-j+k and 6i-2j+k would be j^k+2k^i-2i^j. The Euclidean measure ofa directed m-volume is simply its Euclidean length; in the precedingexample, it would be sqrt(1^2+2^2+2^2) = 3.The nice thing about directed m-volumes is that after they are mappedby a matrix, the pieces add up to the proper directed m-volume.Let A and B be mxn matrices where m <= n. A maps row vectors in R^m toan m-dimensional subspace in R^n. That is, if v is in R^m, then vA isin R^n, but vA is restricted to an m-dimensional subspace of R^n. B^tmaps row vectors in R^n to R^m. Thus, AB^t maps R^m to R^m; after all,it is an mxm matrix.Now let's use the nice thing about directed m-volumes mentioned above.Let W be the wedge product of the orthonormal basis vectors in R^m. Thelength of W is 1. WA is the wedge product of the rows of A in R^n.Furthermore, WAB^t = W det(AB^t) since square matrices multiply volumesby their determinant.Choose m columns of A and zero out all of the others; call this A'.Zero out the same columns of B; call this B'. A' maps W to WA projectedonto one of the basis vectors of R^{n,m}; summing over all choices ofthe m columns returns WA. By the way we defined A' and B', we haveA'B'^t = A'B^t, so summing A'B'^t over all choices of the m columns isAB^t. For each choice of the m columns, WA'B'^t = W det(A') det(B'),where det(A') and det(B') are taken in the sense of deleting all but thechosen m columns. Since we have WAB^t = W det(AB^t), it follows thatdet(AB^t) is the sum of det(A') det(B').This final conclusion is what Robin cited. Perhaps there is a simplerdemonstration or perhaps the length of the demonstration is the reasonthat it was left out.Rob Johnson Let N = M*M' (' denotes conjugate transpose). Let det(N) be the> determinant of N. M is a rectangular matrix. It is clear that N is> Hermitian and square. The question is: (all matrices are complex)> If one more coloumn is added to M, ie., if M was 2 x 3, it is made as> 2 x 4 by appending an arbitrary non-zero coloumn to M, how does the> determinant of N changes? Does it increase or decrease or cannot say.> Could any one suggest me an answer and a proof why that is an answer.> The new determinant is >= the old. This comes from the following> formula. Let A and B be m by n matrices. Then det(AB^t)> is the sum of the det(C) det(D) where C ranges over the> (n choose m) m by m submatrices of A and D is the corresponding> submatrix of B.>But i dont really understand your proof. Could you give me some>references for that. Or cite some place where i can look it up.>Robin didn't give a proof, but he cited a result that answers your>question. One way to justify the result he used is to break AB^t into>pieces corresponding to an orthonormal basis of the m-dimensional>subspaces of R^n.The quoted formula does not need the matrices to be overthe reals, but just over a commutative ring, so using anorthonormal basis is overkill. I believe it is due toGauss. The easiest proof I know is to evaluate thedeterminant of (I -B^t; A 0) by using transformations toreduce it to det(A*B^t), and also to use the expression asthe sum of the signed products of elements from each rowand column. The ones from the identity matrix have to beon the main diagonal, which means the ones from A and B^thave to come from the corresponding submatrices.This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== Subject: : Re: A question on determinant> Let N = M*M' (' denotes conjugate transpose). Let det(N) be the> determinant of N. M is a rectangular matrix. It is clear that N is> Hermitian and square. The question is: (all matrices are complex)> If one more coloumn is added to M, ie., if M was 2 x 3, it is made as> 2 x 4 by appending an arbitrary non-zero coloumn to M, how does the> determinant of N changes? Does it increase or decrease or cannot say.> Could any one suggest me an answer and a proof why that is an answer.> The new determinant is >= the old. This comes from the following> formula. Let A and B be m by n matrices. Then det(AB^t)> is the sum of the det(C) det(D) where C ranges over the> (n choose m) m by m submatrices of A and D is the corresponding> submatrix of B.>But i dont really understand your proof. Could you give me some>references for that. Or cite some place where i can look it up.>Robin didn't give a proof, but he cited a result that answers your>question. One way to justify the result he used is to break AB^t into>pieces corresponding to an orthonormal basis of the m-dimensional>subspaces of R^n.>The quoted formula does not need the matrices to be over>the reals, but just over a commutative ring, so using an>orthonormal basis is overkill. I believe it is due to>Gauss. The easiest proof I know is to evaluate the>determinant of (I -B^t; A 0) by using transformations to>reduce it to det(A*B^t), and also to use the expression as>the sum of the signed products of elements from each row>and column. The ones from the identity matrix have to be>on the main diagonal, which means the ones from A and B^t>have to come from the corresponding submatrices.I also made a typographical error:|A'B'^t = A'B^t, so summing A'B'^t over all choices of the m columns is|AB^t.I left out a couple of W's; it should be|A'B'^t = A'B^t, so summing WA'B'^t over all choices of the m columns is|WAB^t.However, does my proof fail for a commutative ring? I think it stillworks if orthonormal basis is replaced by basis. The orthonormalbasis made it easier to talk about projecting onto the basis of R^{n,m},but all that was needed was breaking the sum along the wedge productsof the basis vectors in the higher dimensional space. Am I missingsomething?Rob Johnson But i dont really understand your proof. Could you give me some> references for that. Or cite some place where i can look it up.I did not give a proof. I gave an outline for a crucial stagein a proof. If you want more details, it would be best foryou to try to fill them in for yourself.> Let N = M*M' (' denotes conjugate transpose). Let det(N) be the> determinant of N. M is a rectangular matrix. It is clear that N is> Hermitian and square. The question is: (all matrices are complex)> If one more coloumn is added to M, ie., if M was 2 x 3, it is made as> 2 x 4 by appending an arbitrary non-zero coloumn to M, how does the> determinant of N changes? Does it increase or decrease or cannot say.> Could any one suggest me an answer and a proof why that is an answer.> The new determinant is >= the old. This comes from the following> formula. Let A and B be m by n matrices. Then det(AB^t)> is the sum of the det(C) det(D) where C ranges over the> (n choose m) m by m submatrices of A and D is the corresponding> submatrix of B.> Robin ChapmanRobin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== Subject: : Urysohn function in Zariski topology?If X and Y are disjoint Zariski closed subsets of A^n, how to construct apolynomial f such that f(a) = 0 and f(b) = 1 for all a in X and b in Y?Does this have anything to do with Urysohn's lemma in normal topologicalspaces?=== Subject: : Re: Urysohn function in Zariski topology?> If X and Y are disjoint Zariski closed subsets of A^n, how to construct a> polynomial f such that f(a) = 0 and f(b) = 1 for all a in X and b in Y?> Does this have anything to do with Urysohn's lemma in normal topological> spaces?If you are working over an algebraically closed field this is possible.Let I = {f polynomial: f(X) = {0}}and J = {g polynomial: g(Y) = {0}}.By the Nullstellensatz I + J = 0, so there are f and g in I and Jresp. with f + g = 1. Then f(X) = {0} and f(Y) = {1}.=== Subject: : Re: Urysohn function in Zariski topology?> By the Nullstellensatz I + J = 0, so there are f and g in I and J> resp. with f + g = 1. Then f(X) = {0} and f(Y) = {1}.I don't quite get why f + g = 1. Would you explain in more detail?> If X and Y are disjoint Zariski closed subsets of A^n, how to constructa> polynomial f such that f(a) = 0 and f(b) = 1 for all a in X and b in Y?> Does this have anything to do with Urysohn's lemma in normal topological> spaces?> If you are working over an algebraically closed field this is possible.> Let I = {f polynomial: f(X) = {0}}> and J = {g polynomial: g(Y) = {0}}.> By the Nullstellensatz I + J = 0, so there are f and g in I and J> resp. with f + g = 1. Then f(X) = {0} and f(Y) = {1}.> Robin Chapman=== Subject: : Re: Urysohn function in Zariski topology? Adjunct Assistant Professor at the University of Montana.> By the Nullstellensatz I + J = 0, so there are f and g in I and J> resp. with f + g = 1. Then f(X) = {0} and f(Y) = {1}.>I don't quite get why f + g = 1. Would you explain in more detail?The Nullstellensatz establishes a bijection between certain kinds ofideals of F[x_1,...,x_n] and certain kinds of subsets of A^n, affinespace over F, when F is algebraically closed. Explicitly, to everysubset S of A^n we associate the ideal of all polynomials that vanishon S, I(S) = {f in F[x_1,...,x_n] : f(s) = 0 for all s in S}and to each ideal I we associate the null set of I, N(I) = {(x_1,...,x_n) in A^n : f(x_1,...,x_n)=0 for all f in I}.The Zariski-closed sets and the radical ideals are in one-to-oneinclusion-reversing correspondence; the ideal of the union is theintersection of the ideals; and the ideal of the intersection is thesum of ideals.You have two disjoint Zariski closed sets, X and Y. LetI = I(X) = {f in F[x_1,...,x_n] : f(x) = 0 for all x in X}J = I(Y) = {g in F[x_1,...,x_n] : g(y) = 0 for all y in Y}.Now, since they are disjoint, X intersect Y is empty. So the idealcorresponding to X intersect Y must be all of F[x_1,...,x_n]. On theother hand, the Nullstellensatz says that this is also equal to I(X) +I(Y), that is, theI + J = { f+g : f in I, g in J}.Therefore, I+J = F[x_1,...,x_n].In particular, I+J must contain the constant function 1. That is,there is f in I and g in J such that f+g = 1.Now, since f is in I, f(x)=0 for all x in X; since g is in Y, g(y)=0for all y in Y. Since f+g = 1, then for all points z in A^n,f(z)+g(z)=1.Now pick a point y in Y. Then f(y)+g(y)=1. However, g(y)=0; so we musthave f(y)=1. This is true for any point in Y. We already have thatf(x)=0 for every point of x of X. So we have found a polynomial f withthe property that f(x)=0 for all x in X, and f(y)=1 for all y in Y,which is what we wanted to begin with.===================================================================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)===================================================================Arturo Magidinmagidin@math.berkeley.edu=== Subject: : Re: Urysohn function in Zariski topology?What if it is not an algebraically closed field?> By the Nullstellensatz I + J = 0, so there are f and g in I and J> resp. with f + g = 1. Then f(X) = {0} and f(Y) = {1}.>I don't quite get why f + g = 1. Would you explain in more detail?> The Nullstellensatz establishes a bijection between certain kinds of> ideals of F[x_1,...,x_n] and certain kinds of subsets of A^n, affine> space over F, when F is algebraically closed. Explicitly, to every> subset S of A^n we associate the ideal of all polynomials that vanish> on S,> I(S) = {f in F[x_1,...,x_n] : f(s) = 0 for all s in S}> and to each ideal I we associate the null set of I,> N(I) = {(x_1,...,x_n) in A^n : f(x_1,...,x_n)=0 for all f in I}.> The Zariski-closed sets and the radical ideals are in one-to-one> inclusion-reversing correspondence; the ideal of the union is the> intersection of the ideals; and the ideal of the intersection is the> sum of ideals.> You have two disjoint Zariski closed sets, X and Y. Let> I = I(X) = {f in F[x_1,...,x_n] : f(x) = 0 for all x in X}> J = I(Y) = {g in F[x_1,...,x_n] : g(y) = 0 for all y in Y}.> Now, since they are disjoint, X intersect Y is empty. So the ideal> corresponding to X intersect Y must be all of F[x_1,...,x_n]. On the> other hand, the Nullstellensatz says that this is also equal to I(X) +> I(Y), that is, the> I + J = { f+g : f in I, g in J}.> Therefore, I+J = F[x_1,...,x_n].> In particular, I+J must contain the constant function 1. That is,> there is f in I and g in J such that f+g = 1.> Now, since f is in I, f(x)=0 for all x in X; since g is in Y, g(y)=0> for all y in Y. Since f+g = 1, then for all points z in A^n,> f(z)+g(z)=1.> Now pick a point y in Y. Then f(y)+g(y)=1. However, g(y)=0; so we must> have f(y)=1. This is true for any point in Y. We already have that> f(x)=0 for every point of x of X. So we have found a polynomial f with> the property that f(x)=0 for all x in X, and f(y)=1 for all y in Y,> which is what we wanted to begin with.> -- > ===================================================================It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> ====================================================================== Subject: : Re: Urysohn function in Zariski topology? Adjunct Assistant Professor at the University of Montana.> If X and Y are disjoint Zariski closed subsets of A^n, how to construct a> polynomial f such that f(a) = 0 and f(b) = 1 for all a in X and b in Y?> Does this have anything to do with Urysohn's lemma in normal topological> spaces?>If you are working over an algebraically closed field this is possible.>Let I = {f polynomial: f(X) = {0}}>and J = {g polynomial: g(Y) = {0}}.>By the Nullstellensatz I + J = 0,You mean I+J = R, right?> so there are f and g in I and J>resp. with f + g = 1. Then f(X) = {0} and f(Y) = {1}.Arturo Magidin, sans .sig=== Subject: : Re: Urysohn function in Zariski topology?> If X and Y are disjoint Zariski closed subsets of A^n, how to construct> a polynomial f such that f(a) = 0 and f(b) = 1 for all a in X and b in> Y? Does this have anything to do with Urysohn's lemma in normal> topological spaces?>If you are working over an algebraically closed field this is possible.>Let I = {f polynomial: f(X) = {0}}>and J = {g polynomial: g(Y) = {0}}.>By the Nullstellensatz I + J = 0,> You mean I+J = R, right?Erm, yeah!Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== Subject: : Re: Urysohn function in Zariski topology?What is R?> If X and Y are disjoint Zariski closed subsets of A^n, how toconstruct> a polynomial f such that f(a) = 0 and f(b) = 1 for all a in X and b in> Y? Does this have anything to do with Urysohn's lemma in normal> topological spaces?>If you are working over an algebraically closed field this is possible.>Let I = {f polynomial: f(X) = {0}}>and J = {g polynomial: g(Y) = {0}}.>By the Nullstellensatz I + J = 0,> You mean I+J = R, right?> Erm, yeah!> -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9> Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== Subject: : Re: Urysohn function in Zariski topology? Adjunct Assistant Professor at the University of Montana.>What is R?R = F[x_1,...,x_n].===================================================================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)===================================================================Arturo Magidinmagidin@math.berkeley.edu=== Subject: : Re: Rotation matrix iff determinant is 1> For this reason, as far as I know, the *definition* of a rotation in R^n is> that it is an isometry with determinant 1, i.e., the theorem you are trying to> prove is actually just the definition of a rotation in R^n. If anybody knows > any different, I'm sure they'll say so...One motivation for this definition is that SO(n) is path-connected.Thus one can move from the identity to any element of SO(n) by a pathconsisting of isometries. This makes calling these rotationsquite natural.=== Subject: : Re: Universe in a HeIII Droplet> 17. Trapped magnetic flux in stringy vortices in Type II phase where> (penetration depth/coherence length) > (square root of 2)^-1 (11)> is quantized in units hc/2e in the sense that Josephson interferometer > tunneling supercurrent potential effects vary as cosine [2pi (enclosed > Bohm-Aharonov magnetic flux)/(hc/2e)]Thats nice Jack but next time show your work. C-=== Subject: : Question regarding median and consecutive numbersGood morning to everyone on the sci.math list. A few days ago I wasreading a book (on ancient religions, no less) and happened upon avery peculiar equation (at least for myself). Because of my poorbackground in mathematics, I was hoping that I could receive somefeedback on this equation, in hopes to find its origin. Oddly enough,this equation just popped into my head and the book I was reading hadnothing to do with it. I don't believe I have ever seen this anywhereas of yet.Ok, here it goes.Take any consecutive set of real numbers. If you multiply the medianof the set by the total number of elements in the set, you will arriveat the same result as if you had added every number in the set.Example:A+B+C = 3B (where A,B,C are consecutive numbers (i.e. 1, 2, 3, or 2,4, 6, or even 15, 30, 45 all that is important is that each element isan equal distance apart from the previous and the next if any)At first, I thought this only worked with odd numbered sets, butquickly realized it worked with even and odd numbered sets.Another example:4192+4193+4194+4195+4196+4197+4198+4199+4200+4201+4202 +4203+4204+4205+4206= 62,985The median of this set, 4199 * 15 (total number of elements within theset)= 62,985Of course this works with negative numbers and non-whole numbers aswell.I would greatly appreciate any help from the math community infiguring out where this equation originated from. I'm sure it has beendiscovered by now, but my search through the web hasn't yielded any Bryan A. Hughes bhugh005@odu.edu Old Dominion University, IST=== Subject: : Re: Question regarding median and consecutive numbers> Take any consecutive set of real numbers. If you multiply the median> of the set by the total number of elements in the set, you will arrive> at the same result as if you had added every number in the set.There are no such things as consecutive REAL numbers. It is true if your numbers are consecutive terms of any arithmetic sequence, in which the difference between consecutive terms is constant, because in this cse the median coincides with the mean (arithmetical average) .What IS true for any finite list of real numbers is that the number of numbers in the list times the MEAN of those numbers equals the sum of the numbers.=== Subject: : Re: Question regarding median and consecutive numbers> Good morning to everyone on the sci.math list. A few days ago I was> reading a book (on ancient religions, no less) and happened upon a> very peculiar equation (at least for myself). Because of my poor> background in mathematics, I was hoping that I could receive some> feedback on this equation, in hopes to find its origin. Oddly enough,> this equation just popped into my head and the book I was reading had> nothing to do with it. I don't believe I have ever seen this anywhere> as of yet.> Ok, here it goes.> Take any consecutive set of real numbers. If you multiply the median> of the set by the total number of elements in the set, you will arrive> at the same result as if you had added every number in the set.> Example:> A+B+C = 3B (where A,B,C are consecutive numbers (i.e. 1, 2, 3, or 2,> 4, 6, or even 15, 30, 45 all that is important is that each element is> an equal distance apart from the previous and the next if any)> At first, I thought this only worked with odd numbered sets, but> quickly realized it worked with even and odd numbered sets.> Another example:> 4192+4193+4194+4195+4196+4197+4198+4199+4200+4201+4202+4203+ 4204+4205+4206> = 62,985> The median of this set, 4199 * 15 (total number of elements within the> set)> = 62,985> Of course this works with negative numbers and non-whole numbers as> well.Even more generally, it works for arbitrary real numbers provided theresulting data set is symmetric, i.e., mean = median.=== Subject: : Re: Question regarding median and consecutive numbers> Take any consecutive set of real numbers. If you multiply the median> of the set by the total number of elements in the set, you will arrive> at the same result as if you had added every number in the set.> Even more generally, it works for arbitrary real numbers provided the> resulting data set is symmetric, i.e., mean = median.And even more general still, it works for arbitrary sets of real numbersprovided you replace median by average (this follows directly from thedefinition of the arithmetical average). So the OP's theorem is animmediate corollary of the fact that the median of an arithmeticalprogression is also the average of the numbers in the progression.=== Subject: : Re: Question regarding median and consecutive numbers> Take any consecutive set of real numbers. If you multiply the median> of the set by the total number of elements in the set, you will arrive> at the same result as if you had added every number in the set.> Even more generally, it works for arbitrary real numbers provided the> resulting data set is symmetric, i.e., mean = median.> And even more general still, it works for arbitrary sets of real numbers> provided you replace median by average (this follows directly from the> definition of the arithmetical average). So the OP's theorem is an> immediate corollary of the fact that the median of an arithmetical> progression is also the average of the numbers in the progression.this. It seems to be a pretty powerful equation for the sum of largecare!=== Subject: : Re: Question regarding median and consecutive numbers> Good morning to everyone on the sci.math list. A few days ago I was> reading a book (on ancient religions, no less) and happened upon a> very peculiar equation (at least for myself). Because of my poor> background in mathematics, I was hoping that I could receive some> feedback on this equation, in hopes to find its origin. Oddly enough,> this equation just popped into my head and the book I was reading had> nothing to do with it. I don't believe I have ever seen this anywhere> as of yet.> Ok, here it goes.> Take any consecutive set of real numbers. If you multiply the median> of the set by the total number of elements in the set, you will arrive> at the same result as if you had added every number in the set.> Example:> A+B+C = 3B (where A,B,C are consecutive numbers (i.e. 1, 2, 3, or 2,> 4, 6, or even 15, 30, 45 all that is important is that each element is> an equal distance apart from the previous and the next if any)> At first, I thought this only worked with odd numbered sets, but> quickly realized it worked with even and odd numbered sets.> Another example:> 4192+4193+4194+4195+4196+4197+4198+4199+4200+4201+4202+4203+ 4204+4205+4206> = 62,985> The median of this set, 4199 * 15 (total number of elements within the> set)> = 62,985> Of course this works with negative numbers and non-whole numbers as> well.Because the numbers are consecutive, the sum of each pair of numbersfrom the end and beginning, in order, is the same -- that is, the1st + last is the same as the 2nd + 2nd-to-last, and so on. So,how many times do you add this? Well, N/2, where N is the numberof elements. And the by the time that you get to the last pair,you're adding the median, or twice the median...Gauss realized this (by himself, spontaneously) at age 7, whichgreatly upset his teacher, if I recall the anecdote correctly.Not sure if this was already a known math/numeric formula bythen.HTH,Carlos=== Subject: : Re: Question regarding median and consecutive numbers> Take any consecutive set of real numbers. If you multiply the median> of the set by the total number of elements in the set, you will arrive> at the same result as if you had added every number in the set.> Example:> A+B+C = 3B (where A,B,C are consecutive numbers (i.e. 1, 2, 3, or 2,> 4, 6, or even 15, 30, 45 all that is important is that each element is> an equal distance apart from the previous and the next if any)> At first, I thought this only worked with odd numbered sets, but> quickly realized it worked with even and odd numbered sets.> Another example:> 4192+4193+4194+4195+4196+4197+4198+4199+4200+4201+4202+4203+ 4204+4205+4206> = 62,985> The median of this set, 4199 * 15 (total number of elements within the> set)> = 62,985Gauss is reputed at the age of 5 to have summed instantly1 + 2 + 3 + ... + 100when asked to do so by his teacher. The sum is the sameas 100 + 99 + 98 + ... + 1 and so twice the sum is101 + 101 + 101 + ... + 101 = 10100 (a hundred 101s)and so the sum itself is 5050 (the number of terms 100, timesthe average term 50.5). > I would greatly appreciate any help from the math community in> figuring out where this equation originated from.Ths buzzword is arithmetic progression. This is a sequencelike 10, 13, 16, 19, 22, 25, 28, 31 where the differencebetween each term and the next is always the same. The samemethod gives the sum of an AP as the number of terms timesthe average term.=== Subject: : Re: Question regarding median and consecutive numbers>Take any consecutive set of real numbers. If you multiply the median>of the set by the total number of elements in the set, you will arrive>at the same result as if you had added every number in the set.>Example:>A+B+C = 3B (where A,B,C are consecutive numbers (i.e. 1, 2, 3, or 2,>4, 6, or even 15, 30, 45 all that is important is that each element is>an equal distance apart from the previous and the next if any)If by consecutive...real numbers you mean evenly spaced real numbers orconsecutive integers, then yes this is correct.The reason it works is balance. Add the first element to the last element,and what do you get? Double the median. The second plus the next-to-lastis also double the median. Keep working your way to the center, and it'sdouble the median all the way.>At first, I thought this only worked with odd numbered sets, but>quickly realized it worked with even and odd numbered sets.For a set with an even number of elements, you'd have to define median asthe average of the two center elements. For example, 1+2+3+4 = 4 * (2.5). It also works if you average the first and last elements.--Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer.=== Subject: : Re: Question regarding median and consecutive numbers> Take any consecutive set of real numbers.He means what we call an arithmetic progression.> If you multiply the median> of the set by the total number of elements in the set, you will arrive> at the same result as if you had added every number in the set.> [...]> I would greatly appreciate any help from the math community in> figuring out where this equation originated from. I'm sure it has been> discovered by now, but my search through the web hasn't yielded anyHere is a web page... http://mathworld.wolfram.com/ArithmeticSeries.htmlThere is a famous story about Gauss using this as a schoolboy.G. A. Edgar http://www.math.ohio-state.edu/~edgar/=== Subject: : Re: Discrete Dynamical System, Topology and Matrix=== Subject: : Discrete Dynamical System, Topology and Matrix >For my PhD thesis, I'm looking for information on the square matrix >space, and particularly on the possible topology of this space : is >there a notion of distance defined between matrices ?n by n real matrices are usually visualize as the topological spaceR^(n^2) and can be given any metric appropiate to the product topology ofn^2 copies of R, the easiest being d(A,B) = max { |a_ij - b_ij| : i,j = 1,..n } >What about boolean matrix ?Same story with {0,1} instead of R giving {0,1}^(n^2) a dyatic space.If you want infinite matrices, then the space R^(N^2) is homeomorphic R^N,and can be given a metric basically described as sup { min(1, |x_k - y_k|/k | k in N }with adjustment understood to befit the indices of matrices.----=== Subject: : Re: Product of these fractions never a power of 2 ?>[...]> 1 1 1> (3 + ---)( 3 + ---)( 3 + ---) =/= 2^m> a b c(...)> But some simple manipulation shows that a sufficient condition> for [*2] is as follows, which is also satisfied for all but a> finite number of values of c, d:> 2^5 < 2^3(cd - 1) + 1> and [*3] combined with the original equation[s] defining d> should allow all the solutions to be found.> Hi John, an interesting result, especially as it included the restrictions. However, it seems to me, that it is bounded to the three-factor problem, and possibly one can find an approach or a scheme, which serves for arbitrary many factors as well... I started another try using the combination of 1) all coefficients must be odd 2) no one can be divisible by 3 This makes, that each coefficient a,b,c,... has the structure a = 6a'+ x , where x is only +1 or -1So the original equation 2^m * abc = (3a + 1)(3b+1)(3c+1)changes to 2^m*(6a'+x)(6b'+y)(6c'+z) = (18a'+3x+1)(18b'+3y+1)(18c'+3z+1)this must be solvable for instance modulo 6, and reduces then tothe same form as the initial equation, only with a,b,c removedand x,y,z all +1 or -1 2^m * xyz = (3x + 1)(3y+1)(3z+1)This gives definite solutions for x,y,z for any given m, or letbetter say, the number of positive and negative x,y,z are determined.See this table m lhs x,y,z (with circular exchanges)---------------------------- 6 64 <-> 1, 1, 1 5 -32 <-> -1, 1, 1 4 16 <-> -1,-1, 1 3 -8 <-> -1,-1,-1----------------------------This idea seem especially useful, since it is simply generalizableto more factors.m.For m=5, 2^m=32 we have the only solution x=-1,y=1,z=1and we can setup a definite equation on only 3 coefficents again. 2^5*(6a'-1)(6b'+1)(6c'+1) = (18a'-3+1)(18b'+3+1)(18c'+3+1) Collecting constants on rhs 2^5*(6a'-1)(6b'+1)(6c'+1) = (18a'-2)(18b'+4)(18c'+4) dividing by 8 4 *(6a'-1)(6b'+1)(6c'+1) = (9a'-1)(9b'+2)(9c'+2) For a'=b'=c'=0 we know, this is an equality Now it would be good to find an argument, for instance of divisibility, which shows, that with the a',b',c'<>0 not other solution exists, but is also simply extensible to more factors. Two further restrictions can now be considered, 3) that all a',b',c' must be >0, 4) and c' must be greater than b' The 3-factor-solution can be handled by min/max-estimation, as you worked out, since we have only one value for the parameter m; but, for instance for 100 factors, we easily can have 42 values for m. (a guess based on 3^100..4^100 interval, where 3^100 = 2^ld(3)*100 and the number of values for m is about 100*(2-ld(3)) ) So another argument should be found, I think. I'm especiall interested, that in the given solutions *always* an even factor was included - may be that's another hint.Gottfried Helms=== Subject: : Re: Product of these fractions never a power of 2 ? > So the original equation > 2^m * abc = (3a + 1)(3b+1)(3c+1) > changes toMay be, this is a more general solution.We separate all odd numbers into two categories U2 = 4i+1 O2 = 4i+3All U2 we separate again in two groups, U3 = 8i+5 O3 = 8i+1and so on, always the left type into two subtypes, whichwe write into the next line: U2 = 4i+1 O2 = 4i+ 3 U3 = 8i+5 O3 = 8i+ 1 U4 =16i+5 O4 =16i+ 13 U5 =32i+21 O5 =32i+ 5 U6 =64i+21 O6 =64i+ 53 ...On the right hand we have collected all odd numbersexcept 1, in its decomposition (with a parameter i),and this composition is of interest.Now putting any mixture of these numbers into the equationshows now solution: 32 * O2(i) O3(j) O4(k) = (3*O2(i) + 1) (3*O3(j) + 1) (3*O4(k) + 1) lhs is always even, rhs can always be kept as a product of an odd number and an integral power of 2. A short analysis of the O_numbers exhibits with a fixed parameter i: O2: 3*( 4i+ 3) + 1 =12i + 10 = ( 6i+5 )*2^1 = 2^1* odd O3: 3*( 8i+ 1) + 1 =24i + 4 = ( 6i+1 )*2^2 = 2^2* odd O4: 3*(16i+13) + 1 =48i + 40 = ( 6i+5 )*2^3 = 2^3* odd O5: 3*(32i+ 5) + 1 =96i + 16 = ( 6i+1 )*2^4 = 2^4* odd O6: 3*(64i+53) + 1 =192i+160 = ( 6i+5 )*2^5 = 2^5* oddThe set of all O_numbers is equal to the set of odd numbers,if we allow a free parameter i>=0.Now we have lhs = 2^m *odd *odd*odd rhs = 2^a *odd * 2^b*odd * 2^c* oddand we can only try numbers with the sum of exponents a+b+c matching m,in this case some combination of O2,O3 and O4-numbers.I think, this is a remarkable restriction (not knowing now, whetherit would help ;-) ).Gottfried Helms=== Subject: : Re: Product of these fractions never a power of 2 ?> U2 = 4i+1 O2 = 4i+ 3> U3 = 8i+5 O3 = 8i+ 1> U4 =16i+5 O4 =16i+ 13> U5 =32i+21 O5 =32i+ 5> U6 =64i+21 O6 =64i+ 53> ...> On the right hand we have collected all odd numbers(...)> A short analysis of the O_numbers exhibits with a fixed parameter i:> O2: 3*( 4i+ 3) + 1 =12i + 10 = ( 6i+5 )*2^1 = 2^1* odd> O3: 3*( 8i+ 1) + 1 =24i + 4 = ( 6i+1 )*2^2 = 2^2* odd> O4: 3*(16i+13) + 1 =48i + 40 = ( 6i+5 )*2^3 = 2^3* odd> O5: 3*(32i+ 5) + 1 =96i + 16 = ( 6i+1 )*2^4 = 2^4* odd> O6: 3*(64i+53) + 1 =192i+160 = ( 6i+5 )*2^5 = 2^5* odd> The set of all O_numbers is equal to the set of odd numbers,> if we allow a free parameter i>=0.> Now we have> lhs = 2^m *odd *odd*odd> rhs = 2^a *odd * 2^b*odd * 2^c* odd> and we can only try numbers with the sum of exponents a+b+c matching m,> in this case some combination of O2,O3 and O4-numbers.For instance we should find solutions -if any- by a=4i+3, b=4j+3, c=16k+13 1 1 1 2^5 = (3 + ----- ) ( 3 + ----- ) ( 3 + ------ ) (i<>j) 4i+3 4j+3 16k+13accordingly to the tranfer-rules of the previous post resulting in 6i+5 6j+5 6k+5 1 = ------* -------* ------- 4i+3 4j+3 16k+13or a=4i+3, b=8j+1, c=8k+1 1 1 1 2^5 = (3 + ----- ) ( 3 + ----- ) ( 3 + ------ ) (j<>k) 4i+3 8j+1 8k+1accordingly to the tranfer-rules of the previous post resulting in 6i+5 6j+1 6k+1 1 = ------* -------* ------- (j<>k) 4i+3 8j+1 8k+1where all denominators have also the form of 6x+-1. Here the restriction a,b,c <> 0 mod 3 is *not* yet implemented.Also I would like to see the restrictions according to David Eppsteinand John Ramsden applied to these formulas.Gottfried Helms=== Subject: : Re: Product of these fractions never a power of 2 ?> accordingly to the tranfer-rules of the previous post resulting in> 6i+5 6j+1 6k+1> 1 = ------* -------* ------- (j<>k)> 4i+3 8j+1 8k+1> where all denominators have also the form of 6x+-1. Here the restriction> a,b,c <> 0 mod 3 is *not* yet implemented.> Also I would like to see the restrictions according to David Eppstein> and John Ramsden applied to these formulas.Now we can try and see, for instance, if an integer i can result, ifwe try with different j and kIf we rearrange the above equation, extracting i, we get 1 + 3(j+k) - 6jk i = - ------------------- 1 + 2(j+k) -20jkfor k=0 we have j i = - (1 + ----- ) which can never integral, if j>0. 2j+1since j and k are symmetric in this equation, the same appliesto k, if j=0.If we change both with the smallest amount (=1) we see, thatthe result for i would be a fraction, as well, if j or k increasesmore.This shows that such a composition of numbers of the sets (given inthe previous posts) with a ? O2, (b,c) ? O3 to reach 2^5is impossible.Yepp, but now a generalisation besides exclusion of individualcases with each individual number of factors in the originalscheme of equations:(1): 1 (3 + ---) =/= 2^m a(2): 1 1 (3 + ---)( 3 + ---) =/= 2^m a b(3): 1 1 1 (3 + ---)( 3 + ---)( 3 + ---) =/= 2^m a b cand so on ...How could that be done?Gottfried Helms=== Subject: : Re: Re - Is Science Converging Towards Truth?===> === Subject: : Re - Is Science Converging Towards Truth? > Cut<> I think that science is getting lost more and more, as it gathers new theories Bob.> You can't even understand dimensional analysis, what makes you think> what you have to say is relevant?Oh come on Eric, where'd you get the idea that I can't even understanddimensional analysis?If you got it from Uncle Al Dukapucka, and other unworthy critics, youknow that some of them are full of crap: Most of them are fighting tosave their places on the gravyboat.What I say is relevant to understanding that a kilogram is the ratioof its Earth-weight [w =9.8 newtons], divided by the acceleration [g]at which it will freefall at Paris France [g = 9.8 meters/sec^2]. Sothat 1 kg = w/g = 1 newton sec^2/meter = 9.8 newtons sec^2/9.8 meters,and others; where the numbers vary depending on the location.Actually I understand dimensional analysis _better_ than most of you.=== Subject: : Need help on Sobolev spaces and inequalitiesJe suis actuellement .8el.8fve .88 l'ecole polytechnique et j'ai unequestion .88 vous poser. Peut etre pouvez vous m'aider ?Voici le cadre :On dit qu'un espace de hilbert peut etre decompos.8e en une somme dedeux sous espaces fermes V1 et V2 si V=V1+V2 ce qui signifie que :il existe une constante C0 telle que pour tout v dans V il existe v1 dans V1 et v2 dans V2 tels que v =v1+v2 et Norme(v1)^2+Norme(v2)^2=C0.Norme(v)^2Attention on ne suppose pas que V1 inter V2 = videJusque l.88 ca va !Le but est d'evaluer la constante C0 pour un certain type d'espace(plutot simple en fait puisqu'on va prendre un carr.8e)On se, place en dimension 2.Si on note H1(omega) l'ensemble des fonctions de L2 telles que lesderiv.8ees faibles de telles fonctions sont dans L2 elles aussi (espacede Sobolev) , on note H10(omega) l'espace defini comme l'adherence desfonctions Cinfini .88 support compact dans omega(l'adherence etant prisedans H1(omega) ). Les notations sont classiques.On prend pour omega le carr.8e ouvert ]0,1[x]0,1[ On note omega1 le rectangle ouvert ]0,q[x]0,1[On note omega2 le rectangle ouvert ]p,1[x]0,1[ Il y a une recouvrement des deux rectangles car on suppose 0Je suis actuellement .8el.8fve .88 l'ecole polytechnique et j'ai une>question .88 vous poser. Peut etre pouvez vous m'aider ?>Voici le cadre :>On dit qu'un espace de hilbert peut etre decompos.8e en une somme de>deux sous espaces fermes V1 et V2 si V=V1+V2 ce qui signifie que :>il existe une constante C0 telle que >pour tout v dans V >il existe v1 dans V1 et v2 dans V2 tels que >v =v1+v2 et Norme(v1)^2+Norme(v2)^2=C0.Norme(v)^2??? Did you really mean to write that, or did you mean just Norme(v1)^2+Norme(v2)^2 <= C0.Norme(v)^2 ?>Attention on ne suppose pas que V1 inter V2 = vide>Jusque l.88 ca va !>Le but est d'evaluer la constante C0 pour un certain type d'espace>(plutot simple en fait puisqu'on va prendre un carr.8e)>On se, place en dimension 2.>Si on note H1(omega) l'ensemble des fonctions de L2 telles que les>deriv.8ees faibles de telles fonctions sont dans L2 elles aussi (espace>de Sobolev) , on note H10(omega) l'espace defini comme l'adherence des>fonctions Cinfini .88 support compact dans omega(l'adherence etant prise>dans H1(omega) ). Les notations sont classiques.>On prend pour omega le carr.8e ouvert ]0,1[x]0,1[ >On note omega1 le rectangle ouvert ]0,q[x]0,1[>On note omega2 le rectangle ouvert ]p,1[x]0,1[ >Il y a une recouvrement des deux rectangles car on suppose 0On peut associer .88 toute fonction vi de H01(omegai) une fonction de>H10(omega) en prolongeant vi par 0 sur omega priv.8e de omegai. On>notera encore vi cette nouvelle fonction par abus.>Ainsi on est en mesure de montrer que au sens defini plus haut>H10(omega)=H10(omega1)+H10(omega2)>Il faut montrer qu'on peut prendre C0 = [ 1+sqrt( 2).alpha]^2 avec>alpha =sqrt(m)/(q-p) et m la plus petite constante apparaissant dans>l'in.8egalit.8e de Poincar.8e : Norme(u) dans L2(omega) Norme(u) dans H10(omega)>avec Norme(u) dansH10(omega) = racine carr.8ee de l'integrale du carr.8e>du gradientde u sur omega>Il y a une indication : utiliser (a+b)^2 (1+delta).a^2+(1+1/delta).b^2>Mon probleme est de montrer que l'on peut prendre une telle constante>C0. Si jamais vous aviez une id.8ee cela m'aiderait beaucoup.It seems possible that you are asking whether one can find C0 suchthat every v in H10(omega) can be written v = v1 + v2 with vi inH10(omegai) and Norme(v1)^2+Norme(v2)^2 <= C0.Norme(v)^2 .If that is not your question then I apologize for my weak French.If that is what you're asking: Isn't this easy to show from apartition of unity? Take phi_1, phi_2 smooth, with phi_1vanishing outside omegai and such that phi_1 + phi_2 = 1.Let v1 = phi_i v. Doesn't the inequlality on the norms followreadily?>Merci d'avance.>=== Subject: : Re: Need help on Sobolev spaces and inequalities> I'm not certain I understood the question. Probably you will> be able to read my English as well as I can read your French;> if we see that I've misunderstood the question your best> chance would be to state it again in English.> (Bad English is allowed here!) A naive translation of Gilles' question are replyed as follows:(Actually most of the sentences were translated by>Je suis actuellement .8el.8fve .88 l'ecole polytechnique et j'ai une>question .88 vous poser. Peut etre pouvez vous m'aider ? I am currently at polytechnique and I have a question to pose to you. Can you help me?>Voici le cadre :>On dit qu'un espace de hilbert peut etre decompos.8e en une somme de>deux sous espaces fermes V1 et V2 si V=V1+V2 ce qui signifie que :>il existe une constante C0 telle que >pour tout v dans V >il existe v1 dans V1 et v2 dans V2 tels que >v =v1+v2 et Norme(v1)^2+Norme(v2)^2=C0.Norme(v)^2 Here the framework: It is said that a Hilbert space can be decomposed in a sum of sum of two closed pennies(?) spaces V1 and V2 if V=V1+V2 which means that: there is a constant C0 such that for all v in V there exists v1 in V1 and v2 in V2 such that v = v1+v2 and Norm(v1)^2+Norm(v2)^2=C0*Norm(v)^2> ??? Did you really mean to write that, or did you mean just> Norme(v1)^2+Norme(v2)^2 <= C0.Norme(v)^2 ?>Attention on ne suppose pas que V1 inter V2 = vide>Jusque l.88 ca va ! Attention one does not suppose only V1 inter V2 = vacuum Until there Ca goes!>Le but est d'evaluer la constante C0 pour un certain type d'espace>(plutot simple en fait puisqu'on va prendre un carr.8e)>On se, place en dimension 2.>Si on note H1(omega) l'ensemble des fonctions de L2 telles que les>deriv.8ees faibles de telles fonctions sont dans L2 elles aussi (espace>de Sobolev) , on note H10(omega) l'espace defini comme l'adherence des>fonctions Cinfini .88 support compact dans omega(l'adherence etant prise>dans H1(omega) ). Les notations sont classiques.>On prend pour omega le carr.8e ouvert ]0,1[x]0,1[ >On note omega1 le rectangle ouvert ]0,q[x]0,1[>On note omega2 le rectangle ouvert ]p,1[x]0,1[ The goal is to evaluate the C0 constant for a certain type of space (rather simple in fact since one will take a square) One, place in dimension 2. If one notes H1(omega) the whole of the functions of L2 such that weak derivatives of such functions are in L2 they also (space of Sobolev), one notes H10(omega) space definied as the adherence of the Cinfini functions to compact support in omega(l' adherence being taken in H1(omega)). The notations are traditional. One takes for Omega the open square ]0,1[x]0,1[ ================(0,1)times(0,1) One notes omega1 the open rectangle ]0, q[x]0, 1[ ==============(0,q)times(0,1) One notes omega2 the open rectangle ]p, 1[x]0, 1[ ==============(p,1)times(0,1)>Il y a une recouvrement des deux rectangles car on suppose 0On peut associer .88 toute fonction vi de H01(omegai) une fonction de>H10(omega) en prolongeant vi par 0 sur omega priv.8e de omegai. On>notera encore vi cette nouvelle fonction par abus.>Ainsi on est en mesure de montrer que au sens defini plus haut>H10(omega)=H10(omega1)+H10(omega2) One can associate any function VI of H01(omegai) a function of H10(omega) by prolonging VI by 0 on private Omega of omegai . One will note VI more this new function per abuse. Thus one is able to show that with the direction defined higher H10(omega)=H10(omega1)+H10(omega2)>Il faut montrer qu'on peut prendre C0 = [ 1+sqrt( 2).alpha]^2 avec>alpha =sqrt(m)/(q-p) et m la plus petite constante apparaissant dans>l'in.8egalit.8e de Poincar.8e : Norme(u) dans L2(omega) Norme(u) dans H10(omega)>avec Norme(u) dansH10(omega) = racine carr.8ee de l'integrale du carr.8e>du gradientde u sur omega It should be shown that one can take C0 = [ 1+sqrt(2).alpha]^2 with alpha =sqrt(m)/(q-p) and m the smallest constant appearing in the inequality of Poincar?: Norm(u) in L2(omega) < or = sqrt(m) Norm(u) in H10(omega) with Norm(u) dansH10(omega) = square root of the integral of the square of the gradient of U on Omega.>Il y a une indication : utiliser (a+b)^2 (1+delta).a^2+(1+1/delta).b^2 There is an indication: to use (a+b)^2 < or = (1+delta).a^2+(1+1/delta).b^2>Mon probleme est de montrer que l'on peut prendre une telle constante>C0. Si jamais vous aviez une id.8ee cela m'aiderait beaucoup. My problem is to show that one can take such a C0 constant. If ever you had an idea that would help me much.> It seems possible that you are asking whether one can find C0 such> that every v in H10(omega) can be written v = v1 + v2 with vi in> H10(omegai) and > Norme(v1)^2+Norme(v2)^2 <= C0.Norme(v)^2 .> If that is not your question then I apologize for my weak French.> If that is what you're asking: Isn't this easy to show from a> partition of unity? Take phi_1, phi_2 smooth, with phi_1> vanishing outside omegai and such that phi_1 + phi_2 = 1.> Let v1 = phi_i v. Doesn't the inequlality on the norms follow> readily?>Merci d'avance.=== Subject: : Re: Need help on Sobolev spaces and inequalities> I'm not certain I understood the question. Probably you will> be able to read my English as well as I can read your French;> if we see that I've misunderstood the question your best> chance would be to state it again in English.> (Bad English is allowed here!)> A naive translation of Gilles' question are replyed as follows:>(Actually most of the sentences were translated bymyself, from knowing something about the math, but youdid give a much more accurate rendering of exactly whatthe question is.>Je suis actuellement .8el.8fve .88 l'ecole polytechnique et j'ai une>question .88 vous poser. Peut etre pouvez vous m'aider ?> I am currently at polytechnique and I have a question to pose to you.> Can you help me?>Voici le cadre :>On dit qu'un espace de hilbert peut etre decompos.8e en une somme de>deux sous espaces fermes V1 et V2 si V=V1+V2 ce qui signifie que :>il existe une constante C0 telle que >pour tout v dans V >il existe v1 dans V1 et v2 dans V2 tels que >v =v1+v2 et Norme(v1)^2+Norme(v2)^2=C0.Norme(v)^2> Here the framework:> It is said that a Hilbert space can be decomposed in a sum of> sum of two closed pennies(?) Two closed subspaces.>spaces V1 and V2 if V=V1+V2 > which means that:> there is a constant C0 such that > for all v in V there exists v1 in V1 and v2 in V2 such that> v = v1+v2 and Norm(v1)^2+Norm(v2)^2=C0*Norm(v)^2> ??? Did you really mean to write that, or did you mean just> Norme(v1)^2+Norme(v2)^2 <= C0.Norme(v)^2 ?>Attention on ne suppose pas que V1 inter V2 = vide>Jusque l.88 ca va !> Attention one does not suppose only V1 inter V2 = vacuumOr empty set.> Until there Ca goes!Um...>Le but est d'evaluer la constante C0 pour un certain type d'espace>(plutot simple en fait puisqu'on va prendre un carr.8e)>On se, place en dimension 2.>Si on note H1(omega) l'ensemble des fonctions de L2 telles que les>deriv.8ees faibles de telles fonctions sont dans L2 elles aussi (espace>de Sobolev) , on note H10(omega) l'espace defini comme l'adherence des>fonctions Cinfini .88 support compact dans omega(l'adherence etant prise>dans H1(omega) ). Les notations sont classiques.>On prend pour omega le carr.8e ouvert ]0,1[x]0,1[ >On note omega1 le rectangle ouvert ]0,q[x]0,1[>On note omega2 le rectangle ouvert ]p,1[x]0,1[ > The goal is to evaluate the C0 constant for a certain type of space > (rather simple in fact since one will take a square)> One, place in dimension 2.> If one notes denotes by> H1(omega) the whole of the functions of L2 such that> weak derivatives of such functions are in L2 they also (space of> Sobolev), one notes H10(omega) space definied as the adherence ofor closure> the Cinfini functions to compact support in omega(l' adherence being> taken in H1(omega)). The notations are traditional.> One takes for Omega the open square ]0,1[x]0,1[> ================(0,1)times(0,1)> One notes omega1 the open rectangle ]0, q[x]0, 1[> ==============(0,q)times(0,1)> One notes omega2 the open rectangle ]p, 1[x]0, 1[> ==============(p,1)times(0,1)>Il y a une recouvrement des deux rectangles car on suppose 0 There is a covering of the two rectangles because 0On peut associer .88 toute fonction vi de H01(omegai) une fonction de>H10(omega) en prolongeant vi par 0 sur omega priv.8e de omegai. On>notera encore vi cette nouvelle fonction par abus.>Ainsi on est en mesure de montrer que au sens defini plus haut>H10(omega)=H10(omega1)+H10(omega2)> One can associate any function VI of H01(omegai) a function of> H10(omega) by prolonging VI by 0 on private Omega of omegai .Don't know what the French says, but what this must mean is'by extending vi on Omegaomegai'> One> will note VI more this new function per abuse.By an abuse of notation, one will again denote the newfunction by vi.> Thus one is able to show that with the direction defined higherwith the [???] defined above> H10(omega)=H10(omega1)+H10(omega2)>Il faut montrer qu'on peut prendre C0 = [ 1+sqrt( 2).alpha]^2 avec>alpha =sqrt(m)/(q-p) et m la plus petite constante apparaissant dans>l'in.8egalit.8e de Poincar.8e : Norme(u) dans L2(omega) Norme(u) dans H10(omega)>avec Norme(u) dansH10(omega) = racine carr.8ee de l'integrale du carr.8e>du gradientde u sur omega> It should be shown that one can take C0 = [ 1+sqrt(2).alpha]^2 with> alpha =sqrt(m)/(q-p) and m the smallest constant appearing in the> inequality of Poincar?: Norm(u) in L2(omega) < or = sqrt(m) > Norm(u) in H10(omega) > with Norm(u) dansH10(omega) = square root of the integral of the> square of the gradient of U on Omega.>Il y a une indication : utiliser (a+b)^2 (1+delta).a^2+(1+1/delta).b^2> There is an indication: to use (a+b)^2 < or => (1+delta).a^2+(1+1/delta).b^2>Mon probleme est de montrer que l'on peut prendre une telle constante>C0. Si jamais vous aviez une id.8ee cela m'aiderait beaucoup.> My problem is to show that one can take such a C0 constant.> If ever you had an idea that would help me much.> It seems possible that you are asking whether one can find C0 such> that every v in H10(omega) can be written v = v1 + v2 with vi in> H10(omegai) and > Norme(v1)^2+Norme(v2)^2 <= C0.Norme(v)^2 .> If that is not your question then I apologize for my weak French.> If that is what you're asking: Isn't this easy to show from a> partition of unity? Take phi_1, phi_2 smooth, with phi_1> vanishing outside omegai and such that phi_1 + phi_2 = 1.> Let v1 = phi_i v. Doesn't the inequlality on the norms follow> readily?So the question is to show that one actually gets C0 = [1+sqrt(2).alpha]^2 with alpha =sqrt(m)/(q-p) and m the smallest constant appearing in the inequality of Poincare.I bet this is clear. If one takes phi_1 and phi_2 to bepiecewise-linear (in the first coordinate) then the gradientof phi_j is no larger than 1/(p-q). So the square of the L^2 norm of the gradient of phi_1 * v should be less than the integral of |v|^2 plus |grad v/(p-q)|^2...>Merci d'avance.> ************************=== Subject: : Is the following NP-complete?I seem to remember something along these lines, but I want to be sure.Is the following problem NP-complete?:Given n linearly independent vectors v_1,...,v_n in Z^n, and apositive integer U,is there a vector v in the lattice generated by v_1,...,v_n such that|v|^2 < U?Here | | denotes the usual vector norm so that |v|^2 is the dotproduct of v with itself.---- David=== Subject: : Re: Resistance to Change> All those people studying recursive this and that all these years, and> nobody has ever realized that the notion can be expressed in a formal> notation!> Thinking that nobody has every actually _done_ this is not totally> stupid, although it's evidently not so. Actually thinking that nobody> has realized it can be done, as you state, is simply hilarious.Then show where and how they did it.> Over 20 years ago, Boyer and Moore used the formal notation of their> proof checker, Nqthm, to express a proof of the unsolvability of the> halting problem.> http://citeseer.ist.psu.edu/boyer82mechanical.html>Oh really? What makes you think that? (I mean, besides reading the>title.) > Maybe he read the abstract. Maybe he read the paper. You can download> the paper itself from that site (those little things in blue text arelinks - note what happens when you click on the one that readsPDF.)Please substantiate your assertion that the Boyer-Moore paper formallyexpresses a proof of the unsolvability of the halting problem. Pleasemake your explanation self-contained and contain at least a little bitof detail e.g. what are the formal axioms, rules of inference andtheorem, and the steps that lead from the axioms to the theorem?Charlie Volkstorf=== Subject: : Re: Resistance to Change> All those people studying recursive this and that all these years, and> nobody has ever realized that the notion can be expressed in a formal> notation!> Thinking that nobody has every actually _done_ this is not totally> stupid, although it's evidently not so. Actually thinking that nobody> has realized it can be done, as you state, is simply hilarious.>Then show where and how they did it.Where and how who did what? Where and how they _realized_ thatit's possible to express the notion of a recursive function formally?Do you know anything at all about the history of formal logic?> Over 20 years ago, Boyer and Moore used the formal notation of their> proof checker, Nqthm, to express a proof of the unsolvability of the> halting problem.> http://citeseer.ist.psu.edu/boyer82mechanical.html>Oh really? What makes you think that? (I mean, besides reading the>title.)> For heaven's sake...> Maybe he read the abstract. Maybe he read the paper. You can download> the paper itself from that site (those little things in blue text arelinks - note what happens when you click on the one that readsPDF.)>Please substantiate your assertion that the Boyer-Moore paper formally>expresses a proof of the unsolvability of the halting problem. Please>make your explanation self-contained and contain at least a little bit>of detail e.g. what are the formal axioms, rules of inference and>theorem, and the steps that lead from the axioms to the theorem?I'm beginning to suspect that this is all a big troll - you're makinga fool of yourself just for fun. If not you'd read the friggin paperbefore asking questions like this.>Charlie Volkstorf> ************************=== Subject: : Re: Resistance to Change> All those people studying recursive this and that all these years, and> nobody has ever realized that the notion can be expressed in a formal> notation!> Thinking that nobody has every actually _done_ this is not totally> stupid, although it's evidently not so. Actually thinking that nobody> has realized it can be done, as you state, is simply hilarious.>Then show where and how they did it.> Where and how who did what? Where and how they _realized_ that> it's possible to express the notion of a recursive function formally?No, we'll never enter their minds. (I believe that Godel was thefirst to define recursive functions and relations, though they wereonly primitive recursive.) Where and when have the various proofsconcerning recursive functions been formalized? I am asking you tosubstantiate your claims.> Over 20 years ago, Boyer and Moore used the formal notation of their> proof checker, Nqthm, to express a proof of the unsolvability of the> halting problem.> http://citeseer.ist.psu.edu/boyer82mechanical.html>Oh really? What makes you think that? (I mean, besides reading the>title.)> For heaven's sake...> Maybe he read the abstract. Maybe he read the paper. You can download> the paper itself from that site (those little things in blue text arelinks - note what happens when you click on the one that readsPDF.)>Please substantiate your assertion that the Boyer-Moore paper formally>expresses a proof of the unsolvability of the halting problem. Please>make your explanation self-contained and contain at least a little bit>of detail e.g. what are the formal axioms, rules of inference and>theorem, and the steps that lead from the axioms to the theorem?> I'm beginning to suspect that this is all a big troll - you're making> a fool of yourself just for fun. If not you'd read the friggin paper> before asking questions like this.You have yet to substantiate your statements. Is that justified?I read that paper at least 10 years ago and recently cited it as agood (though not terribly recent) example of fraud in computer sciencepublishing. Basically,1. There is nothing in this paper that shows that the halting problemis unsolvable. If it is formally proven, then:a. What are the axioms?b. What are the rules of inference?c. What is the formal statement of the theorem?d. How is (c) derived from (a) and (b)?2. Any authentic theorem-prover will generate (or check) much morethan a single theorem. Is that not obvious? But:a. Where are other theorems described in this paper? (None are.)b. In particular, if this theorem-prover is run, why didn't itgenerate the unsolvability of the self-applicability problem (Does agiven program halt on itself?) of which the unsolvability of thehalting problem is an immediate corollary?Contrast this with my paper at http://www.arxiv.org/html/cs.lo/0003071:1. a. The axioms (section VI): TRUE(x) We can list the Universal Set (Peano). NIT(I,J,K) We can tell if a Turing Machine halts no at a giveniteration. - ~YES(x,x) Incompleteness Axiom (the set of programs that don't halt yes on themselves isnot r.e.)1. b. The Rules of Inference (section III): NOT P => ~P AND P , Q => P ^ Q OR P , Q => P v Q IF P , Q(x) => P ^ Q(x) DO P(x) , Q(I,x) => P(x) ^ Q(x,y) UNION P(x) , Q(x) => P(x) v Q(x) QUIT P(x,y) => (eA) P(A,x) SUB P(I) => P()(For example, the NOT rule means that if P is recursive then ~P isrecursive. The UNION rule means that if P and Q are recursivelyenumerable, then PvQ is recursively enumerable. SUB is Kleene's s-m-ntheorem.)1.c. The Theorem (section VII): -HALT(I,J)2. a,b. Theorems proven (sections VI, VII): NO(x,x) The set of Turing Machines that halt no on themselvesis recursively enumerable. -YES(I,J) The Membership Problem is unsolvable. -HALT(I,I) The Self-applicability Problem is unsolvable. -HALT(I,J) The Halting Problem is unsolvable.1.d. The proofs (sections VI, VII):-YES(I,J) The Membership Problem is unsolvable. Proof 1. YES(I,J) Given 2. YES(I,I) SUB 1 J=I 3. ~YES(I,I) NOT 2 4. TRUE(x) Axiom 1 We can list the Universal Set. 5. TRUE(x)^~YES(x,x) DO 4,3 I=x 6 . ~YES(x,x) DEF-7 5 Property of Universal Set qedThe other proofs are a little longer. I will post them here if you'dlike, but you can see them in the paper itself. My axiomitizationalso generates the unsolvability of the blank tape halting problem,the ever-halting problem, the always halting problem, and many others.could present them here. See section VIII for a list of over 30formal wffs, all strict Predicate Calculus, that represent these andvarious other theorems or program synthesis problems.In light of the above, is the Boyer-Moore paper convincing and why? Which is more convincing, their paper or mine, and why?=== Subject: : Re: Zariski dense by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3DE2Tw11979;How about this method:Let t be an indeterminate over k, and replace A with A* = A +tI.Then A* is invertible in k(t), and so by the case you already know how to solve, A*B and BA* have the same characteristic polynomial in k(t)[x]. Note that this char. polynomial is actually in k[t,x].Now set t=0 in the equation det(A*B - xI) = det(BA* - xI). The result I think will be det(AB -xI) = det(BA -xI), which is whatyou want.Have a nice day,Mark>hm....I'm sorry but I don't understand why this involves determinants. Would>you explain a bit?> But I want to show that the set S = {(A, B) | A or B is invertible} is> Zariski dense in the set k^(2*(nxn)).> ?Why wouldn't det(A)*det(B) work here?> Jyrki Lahtonen, Turku, Finland=== Subject: : Help me please by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3DE2Sj11923;Help me please with this problem Let E be an open set in R^n. Assume that f in C^(k)(E). Show that thek-thorder derivative D_(i_1,i_2,...,i_k)(f)=D_(i_1)D_(i_2)...D(i_k)(f) is unchanged if the subscripts i_1,i_2,...,i_k are permuted. Forinstance,if n=>3, then D_1213(f)=D_3112(f) for every f in C^(4).(To say that f in C^(k) means that the partial derivatives D_1(f),...,D_n(f) belong to C^(k-1)(E).=== Subject: : Re: Help me please>Help me please with this problem >Let E be an open set in R^n. Assume that f in C^(k)(E). Show that the>k-th>order derivative D_(i_1,i_2,...,i_k)(f)=D_(i_1)D_(i_2)...D(i_k)(f) >is unchanged if the subscripts i_1,i_2,...,i_k are permuted. For>instance,>if n=>3, then D_1213(f)=D_3112(f) for every f in C^(4).You can find a proof in most advanced calculus books...Sketch of hint for k = 2, n = 2: Use the fundamental theoremof calculus to write f(x,y) - f(0,y) as an integral of D_1 f overa certain line. Now use _that_ to write (*) (f(x,y) - f(0,y)) - (f(x,0) - f(0,0))as a certain two-dimensional integral of D_21 f. Deducethat D_21 f is the limit of (*) divided by a certain something;now show that D_12 is the limit of the same thing.>(To say that f in C^(k) means that the partial derivatives >D_1(f),...,D_n(f) belong to C^(k-1)(E).=== Subject: : Re: Question about first-order logic> I'm currently studying first-order logic, but I have two questions> that doesn't seem to be answered in the book I'm using. I hope someone> can spare a moment of their time and briefly explain it :)> a) The book starts by defining what are first-order expressions, what> is a model, what does it means that a model satisfies a first-order> expression, what does it mean for an expression to be valid etc. Then> it goes on to define how to go from some valid expresions to other> valid expressions and proofs that certain sets of expression are> valid. This is used to create a proof system with the properties that> a) any expression provable in the system is valid and (amazingly) b)> every valid expression has a proof in the system.> Now my question is, in all these proofs of completeness and soundness> etc. all kinds of axioms (such as the induction axiom, the axiom of> choice etc.) are implicitly used along with techniques such as> indirect proofs. Now I'm wondering what is the justification for this?> I mean, first-order logic should be able to support many kinds of> sets of axioms and the axiom of induction, for instance, certainly> doesn't have to be among them. And we carefully prove that things like> indirect proofs works inside of the system. However when proving these> things we are - outside of the system - using all kinds of arbitrary> axioms and inference rules (maybe even the rules we are proving)! So> my question is what framework it is that these proofs take place in> and where does all these axioms come from and how can we use them> right away? And isn't there some kind of logical problem in using> these rules when studying something that is even more fundamental?Depending on your book, it may have chosen to skim over some of these issues BECAUSE there is a lot of technicality involved in showing that a logical structure does not have circular references.> b) Exactly how does one define a model? I understand how one can> create a vocabulary of first-order expressions for the natural> numbers (it's just a set of sequences of certain symbols). Also, I> see how we can select some of these expressions and call them axioms> and use them inside of the proof system we have constructed. But> exactly how does one specify a model for the natural numbers? I> mean, doesn't that in itself require some kind of axioms? I mean, it> must be an infinite construction. So it seems that the specification> of a model will have to take place in some other kind of logical> system which confuses me a bit because it seems first-order logic> should be very fundamental?! The book I'm using seems to say something> like: Well, 0 should correspond to 0, 1 should correspond to 1 the> successor function is simply the function n+1 etc. But how do we know> what those objects we are using in the definitions are?!Usually, 0 corresponds to {}, 1 corresponds to {{}}, 2 corresponds to{ {},{{}} }, etc where each successive integer is the power set of the previous one.What you're really looking for is a set of axioms that corresponds with how the natural numbers work as they are generally understood. If it seems confusing, you've got a lot of company.email: wtwentyman at copper dot net=== Subject: : Re: Question about first-order logic> Depending on your book, it may have chosen to skim over some of these > issues BECAUSE there is a lot of technicality involved in showing that a > logical structure does not have circular references. What technicalities are those, and what does it mean to say that alogical structure does not have circular references?> Usually, 0 corresponds to {}, 1 corresponds to {{}}, 2 corresponds to> { {},{{}} }, etc where each successive integer is the power set of the > previous one. Where in the literature do we find these oddities?=== Subject: : Re: Question about first-order logic> Usually, 0 corresponds to {}, 1 corresponds to {{}}, 2 corresponds to> { {},{{}} }, etc where each successive integer is the power set of the > previous one.> Where in the literature do we find these oddities?Look into books on metamathematics (mathematical logic); Mendelson haswritten a book, and so has Shoenfield. There are different theories/modelsfor N at play. In one, one assumes an element 0, and a function symbolS(x) (successor) and adds certain axioms for these. In axiomatic settheory, like ZF or NBG, one has an axiom of infinity, saying that there isa set omega (in fact, the first countable ordinal) containing the emptyset and the successor S(y) := {y, {y}} of every member y of omega; thenthis set A can be used to represent N. One can also create a model byhaving a symbol for each integer (say a decimal representation), with theusual arithmetic operations defined; this should be the standard model,I think.=== Subject: : Re: Question about first-order logic > Look into books on metamathematics (mathematical logic); To find 5 defined as the power set of 4?=== Subject: : Re: Question about first-order logic>Depending on your book, it may have chosen to skim over some of these >issues BECAUSE there is a lot of technicality involved in showing that a >logical structure does not have circular references.> What technicalities are those, and what does it mean to say that a> logical structure does not have circular references?The very fact that proofs are done before the consequence relation is defined would be something that is in danger of being circular. For example, proving unique decompositions of formulas, before the rules of consequence have been stated. I get the sense that mathematical logic almost boot-straps itself up.Logic involves proving things about proofs, but by what tools when they are not present yet? It can be done, but requires extreme attention to detail to avoid fallacies of begging the question.>Usually, 0 corresponds to {}, 1 corresponds to {{}}, 2 corresponds to>{ {},{{}} }, etc where each successive integer is the power set of the >previous one.> Where in the literature do we find these oddities?This oddity came out of one of my undergraduate books (which is at home, unfortunately).email: wtwentyman at copper dot net=== Subject: : Re: Question about first-order logic Adjunct Assistant Professor at the University of Montana.>Usually, 0 corresponds to {}, 1 corresponds to {{}}, 2 corresponds to>{ {},{{}} }, etc where each successive integer is the power set of the >previous one.> Where in the literature do we find these oddities?>This oddity came out of one of my undergraduate books (which is at home, >unfortunately).I think what Torkel may be trying to point out is that one does not(normally) define the successor of the natural number n to be thepower set of n; rather, we define the successor of n to be the setwhose elements are n and the element of n, that is,s(n) = n union {n};which results in each natural number being the set of all previousnatural numbers.So 3 would not be the power set of { {}, {{}} }, which has 4 elements,P( { {}, {{}} }) = { {}, { {} }, { {{}} }, { {}, {{}} } },but rather the st whose elements are 0, 1, and 2:3 = { {}, { {} }, { {}, {{}} } },which has 3 elements.===================================================================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)===================================================================Arturo Magidinmagidin@math.berkeley.edu=== Subject: : Re: Question about first-order logic>Usually, 0 corresponds to {}, 1 corresponds to {{}}, 2 corresponds to>{ {},{{}} }, etc where each successive integer is the power set of the >previous one.> Where in the literature do we find these oddities?>This oddity came out of one of my undergraduate books (which is at home, >unfortunately).> I think what Torkel may be trying to point out is that one does not> (normally) define the successor of the natural number n to be the> power set of n; rather, we define the successor of n to be the set> whose elements are n and the element of n, that is,> s(n) = n union {n};> which results in each natural number being the set of all previous> natural numbers.> So 3 would not be the power set of { {}, {{}} }, which has 4 elements,> P( { {}, {{}} }) = { {}, { {} }, { {{}} }, { {}, {{}} } },> but rather the st whose elements are 0, 1, and 2:> 3 = { {}, { {} }, { {}, {{}} } },> which has 3 elements.I'll look it up tonight. Since the book is around 10 years old, and was part of the Computer Science department, it may have had a number of non-standard quirks.email: wtwentyman at copper dot net=== Subject: : Re: Question about first-order logic > The very fact that proofs are done before the consequence relation is > defined would be something that is in danger of being circular. For > example, proving unique decompositions of formulas, before the rules of > consequence have been stated. What do rules of consequence have to do with the syntax of formulas? > This oddity came out of one of my undergraduate books (which is at home, > unfortunately). No it didn't.email: wtwentyman at copper dot net=== Subject: : Re: Question about first-order logic> This oddity came out of one of my undergraduate books (which is at home, > unfortunately).> No it didn't.>Nice to know you've been browsing my collection.I'll second that. There are no books on your shelf at home in whichn + 1 is defined to be the power set of n.You need to do something about the lock on the front door, I guess.(Or maybe there's another explanation...)=== Subject: : Re: Question about first-order logic> Nice to know you've been browsing my collection. There's no need to browse your collection to know that it doesn'tcontain any set-theoretical construction of the natural numbers inwhich each successive integer is the power set of the previous one.=== Subject: : Re: Question about first-order logic> If you want to formalize the metatheory then you will run into the same > problem in the meta-meta-theory. At some point, starndard mathematical > reasoning will be used.>is one allowed to use? Induction seems okay to me because induction>proves gives you a recipe for carrying out a proof for specific cases>without resorting to any kind of induction axiom.One ends up with a set of basic logical principles, which themselves canonly be motivated by belief. Potentially infinite countable sets offormulas, about which one can prove things using induction and standardlogic is usually the basis.> However, the proof>of the completeness theorem constructs a countable infinity of sets at>takes the union of those. This construction wouldn't be practically>possible. So where is it defined what is accepted and what is not?One can reformulate these so that it does not look as standard set theory,but more intuitive sets of formulas. Mendelson does that, whereasShoenfield uses a more set-like approach. Then he relies on an intuitivenotion of sets of formulas, which is different from an axiomatic settheory of objects, like ZF or NBG.>Also, I'm still confused as to what is meant by our standard model N>in the book (N = symbol for natural numbers). How do we know what N is>when we haven't axiomized it yet? I think it is one makes an construction of the integers (say as decimalnumbers) and defines +, * in the standard way. One then has to show thatthe axioms of the theory of natural numbers is valid in this model.>Is there some way to construct N>without using axioms? Then one knows that those theorems will be also valid in this model.>However, if any expression is either satisfied or not satisfied in any>concrete model, why doesn't mathematicians reason about the standard>model N? In this model, every expression would be either true or>false.The idea is that, from the metamathematical point of view, some thingsbecomes simpler by reasoning about the theory. Then things about a theorycan be deduced by considering special models. Models will be constructedin various intuitively correct fashions, by the use of sets of formulas.=== Subject: : Re: Question about first-order logic> No. The mathematics of the metatheory uses the same plain old logic as > any other mathematical theory. If one wants to be heroic, he can take > the Intuitionist tack and permit only constructive proof and no use for > the law of the excluded middle when dealing with infinite sets. That > approach cannot prove as much, but no one argues with its soundness either.> If you want to formalize the metatheory then you will run into the same > problem in the meta-meta-theory. At some point, starndard mathematical > reasoning will be used.> is one allowed to use? Induction seems okay to me because induction> proves gives you a recipe for carrying out a proof for specific cases> without resorting to any kind of induction axiom. However, the proof> of the completeness theorem constructs a countable infinity of sets at> takes the union of those. This construction wouldn't be practically> possible. So where is it defined what is accepted and what is not?Yes, this is a bit tricky. One thing you might want to remember isthat the first-order logic is a formalization of the methods of logicwe use every day. It is not prior in any sense of the word. Inparticular, if we want to prove theorems *in a language*, we are onlyjustified in using the methods of that language (so using the strictlyformal definition of modus ponens (p ^ p -> q |- q) is allowed, butnot the english version (if p and p implies q, then q). But these areso similar that people let it slide if the context supports it. Ifyou're proving theorems *about* a language, then any language willdo--even the object language. But remember: every theorem (whetherin a language or about a language) has the same logical status. Soyou can apply a theorem anywhere its hypothesis are met.> Also, I'm still confused as to what is meant by our standard model N> in the book (N = symbol for natural numbers). How do we know what N is> when we haven't axiomized it yet? Is there some way to construct N> without using axioms? The standard model of the natural numbers is the set of countingnumbers. You're logic book probably assumes that you're familiar withsome number theory (or something to that effect). Although you don'thave to subscribe to the metaphysical theory behind it for logic tomake sense, it's good to know. Try to imagine these entities callednumbers which are floating in some Platonic realm. The ones we callnatural numbers are those which satisfy the axioms for Peanoarithmetic. In fact, these phantom objects make up a model. Ineffect, your axioms only put constraints on what you can quantify overin any given theory. (I don't buy the metaphysics personally, butit's a good way to get the concepts across)>It seems to be the case, because G.9adels theorem> actually says that N can't be axiomized.> However, if any expression is either satisfied or not satisfied in any> concrete model, why doesn't mathematicians reason about the standard> model N? In this model, every expression would be either true or> false.Godel's theorem roughly says that there are a lot of non-isomorphicmodels of the natural numbers. In particular, each model contains atheorem which isn't true in every other model. (This is a good timeto discuss logical deducibility: a first-order sentence is deduciblefrom a set of axioms if the sentence is true in every model whichsatisfies the axioms. In particular, the logical validities are truein all models, hence deducible. Godel says that there are sentenceswhich aren't logically (that is, in FOL) deducible from the axiomseven though they are true in the standard model)'cid 'ooh=== Subject: : Re: Question about first-order logic> Godel's theorem roughly says that there are a lot of non-isomorphic> models of the natural numbers. Not even roughly, since complete consistent theories have a lot ofnon-isomorphic models, too.> In particular, each model contains a> theorem which isn't true in every other model. Models don't have theorems.=== Subject: : Re: Question about first-order logicsays: There is no set of recursively-enumerable axioms, Delta, sothat for any first-order expression phi, phi is aDelta-first-order-theorem iff the expression is satisfied within N.N is earlier in the book referred to as the standard model of numbertheory.Acid Pooh said that G.9adels theorem said that there are manynon-isomorphic models for the natural numbers (I assume he mean thatthere are many models for the natural numbers that disagree on thesatisfability of various expressions). This seems very close to whatmy book says (my book just says that you can't capture any concretemodel with axioms - there will always be true properties of thatconcrete model, not provable from the axioms).But I still don't understand what N is. As mentioned above, in my bookit is taken to be a concrete model. Attempts in the book are made ataxiomizing it but what I don't understand is how N could even existprior to being axiomized (possibly outside the FOL framework)?Clearly, the axiomization doesn't completely define a model (thatwould make G.9adels theorem false) so exactly what is N in the firstplace?BTW, is G.9adels (incompleteness) theorem - even in the original version- only a theorem about first-order logic? Is the theorem still true in>1-order-logic? If not, why is the theorem taken to be such a bigshow-stopper? Maybe it is simply a sign that first-order-logic is tooweak? Note that this last part is not trolling I'm just interested inknowing, because I'm going to have to explain all this to other peopleand this last part seems to be an obvious question.Surfing> Godel's theorem roughly says that there are a lot of non-isomorphic> models of the natural numbers.> Not even roughly, since complete consistent theories have a lot of> non-isomorphic models, too.> In particular, each model contains a> theorem which isn't true in every other model.> Models don't have theorems.=== Subject: : Re: Question about first-order logic>says: There is no set of recursively-enumerable axioms, Delta, so>that for any first-order expression phi, phi is a>Delta-first-order-theorem iff the expression is satisfied within N.>N is earlier in the book referred to as the standard model of number>theory.>Acid Pooh said that G.9adels theorem said that there are manynon-isomorphic models for the natural numbers (I assume he mean that>there are many models for the natural numbers that disagree on the>satisfability of various expressions). My guess would be that he meant what he said. Which of courseis false, as Torkel pointed out.>This seems very close to what>my book says (my book just says that you can't capture any concrete>model with axioms - there will always be true properties of that>concrete model, not provable from the axioms).This is not quite true: If you say that every sentence true in N isan axiom then those axioms imply everything that's true in N.(Note that _that_ set of axioms has many non-isomorphic models.)You're leaving out an important word in your statement ofGodel's theorem - it refers to _recursive_ sets of axioms.>But I still don't understand what N is. As mentioned above, in my book>it is taken to be a concrete model. Attempts in the book are made at>axiomizing it but what I don't understand is how N could even exist>prior to being axiomized (possibly outside the FOL framework)?>Clearly, the axiomization doesn't completely define a model (that>would make G.9adels theorem false) so exactly what is N in the first>place?Do you really expect to be able to axiomatize _everything_?Do you also think you can define everything in terms of previousdefinitions?What N _is_ is this set: {1, 2, 3, ...}.>BTW, is G.9adels (incompleteness) theorem - even in the original version>- only a theorem about first-order logic? Is the theorem still true in>1-order-logic? If not, why is the theorem taken to be such a big>show-stopper? Maybe it is simply a sign that first-order-logic is too>weak? Note that this last part is not trolling I'm just interested in>knowing, because I'm going to have to explain all this to other people>and this last part seems to be an obvious question.>Surfing> Godel's theorem roughly says that there are a lot of non-isomorphic> models of the natural numbers.> Not even roughly, since complete consistent theories have a lot of> non-isomorphic models, too.> In particular, each model contains a> theorem which isn't true in every other model.> Models don't have theorems.=== Subject: : C-infinity with compact support How is it possible to give a proof that the space of functions C^{infinity}with compact support in A, subset of R^n, is not closed in the Sobolev spaceW^{1, 2}(A) ? I imagine that I must find a sequence of function C^{infinity}with compact support in A, which converges in W^{1,2}(A), but not inC^{infinity}.=== Subject: : Re: C-infinity with compact support >How is it possible to give a proof that the space of functions C^{infinity}>with compact support in A, subset of R^n, is not closed in the Sobolev space>W^{1, 2}(A) ? I imagine that I must find a sequence of function C^{infinity}>with compact support in A, which converges in W^{1,2}(A), but not in>C^{infinity}.You could start with a function f in W, with compact support, which isnot smooth, and then convolve with a smooth approximate identity(with compact support).=== Subject: : Re: C-infinity with compact support I have thought, for examples, F(x)=0, for -1I have thought, for examples, F(x)=0, for -1F(x)=0, for 1kernel, I obtain a C^{infinity} sequence that converges in L^2, but not in>W^{1,2}(A). Well of course not! That function is not _in_ W^12(A).>I have no idea in Sobolev spaces.Let A = (-2,2). Can you give an example of a function f suchthat f is in W^12(A), f is not smooth, and the support of f isa compact subset of A?=== Subject: : Re: How to solve an exponential ciphers?> How to solve an exponential ciphers? Do you mean, given C = m^e mod N, how do we compute m, given> C and N (and M)? > This is known as the discrete logarithm problem. Algorithms for solving> it are complicated.I meant compute e, of course, not m.=== Subject: : Re: Hey guys, isolated singularities?> You know that now. When Wade said what he said you'd just> said you _think_ that poles have to be isolated and asked whether> essential singularities had to be isolated, suggesting sin(1/z) as> a possible counterexample...To be honest, I've been learning about complex variables this year forthe first time. Last semester I was pretty confused on a number ofpoints (as you probably saw as you helped me out I believe).It's just that reading it for the first time, I come acrossdefinitions like this at in books like Bak and Newman and Churchill:The book has been talking about isolated and non isolated zeros. Thenit starts talking about isolated singularities. To paraphrase,Isolated Singularities. An isolated singularity at z0 is a Removable singularity if ... f = g where g is analyticPole if ... f = g / h and g and h are analytic and h has a zero at z0Essential singularity ... if neither of these is true(for every ... insert in a deleted neighborhood of z0)So this doesn't give me a clear message that only isolatedsingularities can be essential. That's why I asked the question.Moreover Churchill, for example, complicates things by sayingsomewhere at the end of his book that although the singularities werenot assumed isolated a priori, in fact they are isolated because...So when you guys told me that these things are always isolatedsingularities, I went back to the book and looked, and there it was,they are only isolated singularities.Anyway, you can understand why a beginner sometimes sees somethingwhich doesn't completely imply something and assumes something else(e.g. that since a singularity is called isolated there has isprobably a non-isolated kind).> How can you possibly agree with what I said about function> elements if you don't know what a function elelement is???On this point I've only been exposed to fuzzy definitions like thetaylor expansion around a point is a function element in Schaum's orthe function defined on a simply connected open set in C inChurchill. > If you're trying to learn about complex analysis you need a> book on the subject. If you have a book that gives a serious> duscussion of analytic continuation (at the first-year-grad> student level, in a moderately abstract context, ie not just> commenting on continuing log and sqrt) that does _not_> contain a definition of function element I'll be very> surprised.I definitely want to learn and I'd say this year I've finally got acorrect understanding of basic math (linear algebra, analysis, complexanalysis) which I didn't have before. Of course, I have a long way togo and it'll be interesting. Anyway, as far as books I've been exposedto on complex variables, it's been (in order of exposure):Schaum's Outlines (started out with that, it was actually quite clear)Churchill (not the churchill and brown one, finished it)Ahlfors (read bits and pieces)Bak and Newman (read more than half the book)Serge Lang (bought about 2 weeks ago, went through about the first 180pages sofar).So now I know all these different ways o stating Cauchy's theorem andwhy some are better than others and how it relates to topology andabout isolated singularities and residues and a bunch of things from abunch of different viewpoints. I'd say I have a good understanding ofthe topics I know right now. There are still more topics to go.>So the monodromy theorem for complex analysis says that you only get>different values if you run around branch points, but not essential>singularities or poles?> I explained yesterday why this question doesn't quite make sense> (or to be fair, why it doesn't quite make sense to me). I could> repeat what I said yesterday - I don't see much point to that.Alright. Then I have another question: I've seen mention ofsingularities of an analytic function on the boundary of a circlewhich are dense in the circle. They are certainly not isolated. Arethere more examples of singularities which are not isolatedsingularities and is there anything interesting that can be said aboutthem? :)rough ride sometimes.-Greg=== Subject: : Re: Hey guys, isolated singularities?> You know that now. When Wade said what he said you'd just> said you _think_ that poles have to be isolated and asked whether> essential singularities had to be isolated, suggesting sin(1/z) as> a possible counterexample...>To be honest, I've been learning about complex variables this year for>the first time. Last semester I was pretty confused on a number of>points (as you probably saw as you helped me out I believe).>It's just that reading it for the first time, I come across>definitions like this at in books like Bak and Newman and Churchill:>The book has been talking about isolated and non isolated zeros. Then>it starts talking about isolated singularities. To paraphrase,Isolated Singularities. An isolated singularity at z0 is a >Removable singularity if ... f = g where g is analytic>Pole if ... f = g / h and g and h are analytic and h has a zero at z0>Essential singularity ... if neither of these is true>(for every ... insert in a deleted neighborhood of z0)>So this doesn't give me a clear message that only isolated>singularities can be essential.Well it should - you haven't caught on to how definitionswork yet. When one defines isolated singularity it doesn'tfollow that there are (formally defined) non-isolated singularitieslurking in the background. A non-commutative field is not afield...> That's why I asked the question.>Moreover Churchill, for example, complicates things by saying>somewhere at the end of his book that although the singularities were>not assumed isolated a priori, in fact they are isolated because...>So when you guys told me that these things are always isolated>singularities, I went back to the book and looked, and there it was,>they are only isolated singularities.>Anyway, you can understand why a beginner sometimes sees something>which doesn't completely imply something and assumes something else>(e.g. that since a singularity is called isolated there has is>probably a non-isolated kind).> How can you possibly agree with what I said about function> elements if you don't know what a function elelement is???>On this point I've only been exposed to fuzzy definitions like the>taylor expansion around a point is a function element in Schaum's orthe function defined on a simply connected open set in C in>Churchill.> If you're trying to learn about complex analysis you need a> book on the subject. If you have a book that gives a serious> duscussion of analytic continuation (at the first-year-grad> student level, in a moderately abstract context, ie not just> commenting on continuing log and sqrt) that does _not_> contain a definition of function element I'll be very> surprised.>I definitely want to learn and I'd say this year I've finally got a>correct understanding of basic math (linear algebra, analysis, complex>analysis) which I didn't have before. Of course, I have a long way to>go and it'll be interesting. Anyway, as far as books I've been exposed>to on complex variables, it's been (in order of exposure):>Schaum's Outlines (started out with that, it was actually quite clear)>Churchill (not the churchill and brown one, finished it)>Ahlfors (read bits and pieces)>Bak and Newman (read more than half the book)>Serge Lang (bought about 2 weeks ago, went through about the first 180>pages sofar).>So now I know all these different ways o stating Cauchy's theorem and>why some are better than others and how it relates to topology and>about isolated singularities and residues and a bunch of things from a>bunch of different viewpoints. I'd say I have a good understanding of>the topics I know right now. There are still more topics to go.>So the monodromy theorem for complex analysis says that you only get>different values if you run around branch points, but not essential>singularities or poles?> I explained yesterday why this question doesn't quite make sense> (or to be fair, why it doesn't quite make sense to me). I could> repeat what I said yesterday - I don't see much point to that.>Alright. Then I have another question: I've seen mention of>singularities of an analytic function on the boundary of a circle>which are dense in the circle. They are certainly not isolated. Have you seen this mentioned in exactly those words, ina carefully written book?>Are>there more examples of singularities which are not isolated>singularities and is there anything interesting that can be said about>them? :)>rough ride sometimes.>-Greg=== Subject: : Hungarian method for non-square assignment matrix (non-bipartite)Can I extend the Hungarian method to the case where the cardinality oftwo sets is different?I think I can do this by adding dummy members to the smaller set, andset the weights connecting to these dummies to zero.Is this correct?=== Subject: : Re: Hungarian method for non-square assignment matrix (non-bipartite)> Can I extend the Hungarian method to the case where the cardinality of> two sets is different?> I think I can do this by adding dummy members to the smaller set,Yes.> and> set the weights connecting to these dummies to zero.Maybe, maybe not. It depends on the objective (and the context of the model). For instance, suppose that you have five jobs, each requiring a single dedicated (full-time) worker, and three workers, and suppose that your objective is to minimize cost. You can add two dummy workers, but you have to ask yourself what they represent. If you assign a dummy worker to job #1, does that mean job #1 is left undone, does it mean you outsource the job, higher a temporary worker, use overtime to get it done, ...? Each of those presumably has a different cost (possibly zero if you can leave the job undone and suffer no consequence). Figure out the cheapest way to skip each job (need not be the same for all jobs) and make that the cost of using a dummy worker on it.-- Paul> Is this correct?=== Subject: : Re: Factorizing a polynome in 2 variables>Is there a closed form that indicates whether>p(x,y)=0 may be factorized over the complex domain? I guess you want to factor p in the polynomial ring C[x,y], then?(When you refer to the shape of the curve p=0 that almost soundslike you want to factor p in the ring of entire functions on theplane, or the ring of germs of functions analytic at (0,0) or something.)>I would need to factorize (if possible - the curve>has three branches when drawn) the following mess:>-563622573*sqr(35)*x1^6*x2+ ... 174896139600*x2^2I don't know what algorithms exist for factoring in C[x,y] but I cantell you what I might do. Some investigation suggests to me that yourp is in fact irreducible.The coefficients here are all algebraic integers, and so if pfactors in C[x,y], it factors in some smaller ring R[x,y]where R is an algebraic number field (containing sqrt(35),although you can recast the problem by substituting x2 = x3/sqrt(35)to get a polynomial P in Z[x,y] itself).I suppose you can show that if there is a factorization in R[x,y] then there must already be a factorization in Z[x,y].I will sketch below a method to check this.But this then is easy to rule out: a factorization of P over the integers would imply one over Z/23Z, but in that ring P isirreducible. (So Maple tells me and I know THAT could be checkedwith an exhaustive search if nothing else!)So I am inclined to believe that your polynomial does not factorinto two non-constant polynomials with complex coefficients.How do I propose to show that in any factorization P = q r, thecoefficients of q and r could be chosen to be rational?Well, if P = q r in R[x,y] then phi(P) = phi(q) phi(r) in any homomorphism phi : R[x,y] --> R[u]; but for example, using phi(x1)=u, phi(x3)=1 makes phi(P) be irreducible in Z[u],so unless one of phi(q) or phi(r) is a constant polynomial(i.e. q or r is of the form (constant) + (multiple of x3-1) )then the coefficient ring R must include a subfield of thesplitting field of the integral septic polynomial P(u,1).But there's nothing special about sending x2 to 1; many otherspecializations of x2 lead to irreducible septics (indeed, Iguess that's most x2 according to a theorem of Hilbert). We conclude that the ring generated by the coefficients of qand r (normalized so that q has a rational coefficient, say) must lie in the intersection of the splitting fields of all these septics. I'm not really sure how to calculate the intersection of two splittingfields. Maple has a feature which claims to factor a polynomial over an algebraic extension ring, so if I can inform Maple about the splitting field F of one of the septics, and ask it to factor another septic overF, then as soon as it tells me the latter is irreducible over F, I wouldknow the splitting fields meet only in the rationals. Now, the only wayI know to specify F to Maple is to give it a primitive element, i.e.a single element whose conjugates under its Galois group are alldistinct. Almost any linear combination of the seven roots of theseptic will do, and Maple can find the minimal polynomial knowingthe root to sufficient accuracy. But since I would be looking fora polynomial of degree 7!=5040, and since I expect the coefficientsto be comparable to those of the septic (about a dozen digits),I would need to work with real numbers carried out to tens of thousandsof digits. This seems unlikely to be the most efficient approach...So I don't really know the intersection of the splitting fields ofall the septics P(u, n) as n ranges over the integers. Statisticallyspeaking, a random bunch of rational septics is not likely tohave any subfield in common in splitting fields, so I will justconjecture that our intersection is just the rational numbers, meaning that if P factors in C[x,y], it has to factor in Q[x,y].Much work would remain to obtain a proof of this fact from the precedingheuristic discussion!Another way to pursue the original question is to consider possible forms of the factorization of P (e.g. as the product of a quadratic and a quintic), expand and compare coefficients to obtain a set of a simple polynomials (in the unknown coefficients of the quadratic and the quintic) which must vanish for the factorization to hold. You can then eliminatesome of the variables (e.g. by calculating a Groebner basis for thewhole ideal) and in particular determine what rational polynomialsthese coefficients must satisfy. (If there is no factorization ofthis form, the equations will be inconsistent, meaning that the idealis just (1); this will be detected by a Groebner basis computation.)I fed just this problem to Magma but it did not return an answer aftera few minutes and so, impetuous youth that I am, I gave up.dave=== Subject: : Re: PrerequisitesCut To DO physics there are a couple of necessary prerequisites that we> need first: They are: #1) [units of] LENGTH; for linear measures of> distances and displacements; square measures, and cubic measures and;> #2) [units of] FORCE and weight, to determine a body's inertia and the> quantity of matter it contains, and of course we need [units of] TIME> to determine their duration; since all things are constantly changing;> at various sometimes imperceptible rates during time.That's cooked goose Bob, and you get the neck(;^!=== Subject: : Re: PrerequisitesX-Enigmail-Version: 0.76.7.0X-Enigmail-Supports: pgp-inline, pgp-mimeGreat. The unscientific moron Shead, answered by the racist moron Uncle Al.Why don't *both* of you go form yourselves a nice crackpot newsgroup, and get the out of legitimate newsgroups? Maybe alt.headUncleAlstooooopidDonShead?Do it. Like right now.>Dumb Donny Head is too stooopid to crack a CRC Handbook and look>at the *seven* fundamental quantities of SI units.>Jeezus, Unc, I just put DGS in my killfile, and the very next>time I look at the newsgroup, the same three threads have>reappeared :)>I think you'll like this one:>http://www3.telus.net/ldh/dinar.html> Mickey-D cheeseburgers don't meet Islamic dietary dictates. If Bush> the Lesser had any balls at all it would be B-52 rolling thunder, one> city/night, HE then incendiaries. One reasons with Arabs by first> killing most of them. The remainder then worhship their conqueror and> the dead never complain.=== Subject: : Re: Prerequisites>To DO physics there are a couple of necessary prerequisites that we>need first: They are: #1) [units of] LENGTH; for linear measures of>distances and displacements; square measures, and cubic measures and;>#2) [units of] FORCE and weight, to determine a body's inertia and the>quantity of matter it contains, and of course we need [units of] TIME>to determine their duration; since all things are constantly changing;>at various sometimes imperceptible rates during time.> Dumb Donny Head is too stooopid to crack a CRC Handbook and look> at the *seven* fundamental quantities of SI units.> Force is a derived quantity, Dumb Donny Head.> Like heck Uncle Dukapuka:> None of my handbooks refer to SI units, and that's the way they'll> stay. Weight-density was good enough when I designed bridges, and this> way they'll stay up to date when we finally realize that force and> weight, though variable, are _fundamental_ units:Welcome to the 21st century. SI is the accepted standard, and thatsthe way it is because it is better than what was there before.I fear to tread on any bridge designed by a man who does notunderstand what a fundamental unit is.[snip crap]You can't get your head around dimensional analysis. Your musingsabout Einstein's 'mistake' is thus irrelevant.=== Subject: : Re: Prerequisites>One of the biggest mistakes Einstein made was in not realizing that>positions and velocities are relative: That is any velocity depends>not only on the position and velocity of what is being observed; but>also on the position and velocity of the observers themselves.Maybe my reading skills are failing so someone help me out -- did Don really just say Einstein was not aware of the principle of relativity?You're not as dumb as you look. Or sound. Or our best testingindicates. -- Monty Burns to Homer Simpson=== Subject: : Re: PrerequisitesCut<> Maybe my reading skills are failing so someone help me out -- did Don > really just say Einstein was not aware of the principle of relativity?His mistake was that he knew naught of what relativity means, and hedeveloped a principle from different viewpoints: Frame jumping is whatthey call it today; jumping from one frame to another without regardto how it occurs - like falling asleep at the station, not feeling theacceleration or deceleration and waking up in a moving frame;wondering what's happening.He was a genius all right, in concocting trickery with motion(;^) somuch so that he even tricked himself.=== Subject: : Re: Prerequisites> Maybe my reading skills are failing so someone help me out -- did Don > really just say Einstein was not aware of the principle of relativity?>His mistake was that he knew naught of what relativity means, and he>developed a principle from different viewpoints: Frame jumping is what>they call it today; jumping from one frame to another without regard>to how it occurs - like falling asleep at the station, not feeling the>acceleration or deceleration and waking up in a moving frame;>wondering what's happening.Uh... no. Frame jumping means to jump to another frame but fail to transform quantities like velocity and length, and so you continue using them as they were determined in the old frame. But it's perfectly okay to *transform* to another frame; it doesn't matter how you got there, or even whether you should be there at all. That's what the principle of relativity means.>He was a genius all right, in concocting trickery with motion(;^) so>much so that he even tricked himself.He knew where his towel was.The polhode rolls without slipping on the herpolhode lying in the invariable plane. -- Goldstein, Classical Mechanics 2nd. ed., p207.=== Subject: : Re: Prerequisites>One of the biggest mistakes Einstein made was in not realizing that>positions and velocities are relative: That is any velocity depends>not only on the position and velocity of what is being observed; but>also on the position and velocity of the observers themselves.> Maybe my reading skills are failing so someone help me out -- did Don> really just say Einstein was not aware of the principle of relativity?He sure did! I wonder if Dirk needs another fumble candidate?David A. Smith=== Subject: : Re: Antidiagonal, Infinity[...]> What do you think about the leading zero(e)s in the matrix of list> element expansions? Why are not there obvious mappings with simple> transformations between a closed interval and the set of all reals?> If by simple, you mean continuous, because closed finite intervals are> compact but the set of reals is not.I should stay out of this circus, but it's office hours and no studentshave shown up, so:The following is an example of a bijection between [-1,1] and the reals. The example may not be obvious but it is quite standard. (Of course itcannot be continuous everywhere.)Denote A = {1/2, 2/3, 3/4, ...} = {n/(n+1) | n>= integer} (-A) = {x | -x is in A}On A, define f(n/(n+1)) = (n+1)/(n+2),on (-A) define f(-n/(n+1)) = -(n+1)/(n+2)Two more exceptions: f(-1) = -1/2, and f(1) = 1/2 .For all other numbers from [-1,1], let f(x)=x.Then f is 1-1, onto, domain [-1,1], range (-1,1).The next step isg(y) = y/(1-abs(y)) : domain (-1,1), range (-inf, inf)with inverse g_inv(t) = t/(1+abs(t)).The desired function is F: F(x) = g(f(x)).It is an odd function (in more than one way :-)=An additional feature: F maps rationals fron [-1,1] ontoall rationals.=== Subject: : Re: Antidiagonal, Infinity> The infinitesimal iota doesn't have all properties of definite reals,> it is among a form of what may be called indefinite reals, real> numbers to be sure but those not representable as a rational as a> fraction of integers or an irrational as a convergnt sequence.Sounds more like an unreal real. There is no such thing as a real that is neither the limit of a sequence of rationals or as the LUB of GLB os some set of rationals. > My opinion is that real numbers of the form x+ni, real x plus some> non-zero integer multiple n of iotas, are nearer to x than to any> other real number y for any finite value n, yet are different than x.Your opinion here, as in the past, has proved to be worthless. > About the leading zeros, it's like saying: consider a function from> the integers to [0.0, 0.1]. Make a list and construct the> antidiagonal.Any such function can be factored into a function from N to [0,1] followed by the function x -> x/10 : [0,1] -> [0.0,0.1]> .000...> ...> .001..> .010..> ...> .011...> Construct the antidiagonal:> .1xxx...> The antidiagonal never except in the case of .011... = .10...> represents an element of the range. The antidiagonal always differs> from each of the elements because each of the elements has a zero in> the first integer modulus after the radix where the antidiagonal does> not, and the antidiagonal is not an element of [0.0, 0.1).Ross' argument is that a particular antidiagonal algorithm does not work. But that is not good enough. In order to make his point, Ross must prove that EVERY antidiagonal algorithm MUST fail to work, which he cannot do, since the decimal algorithm which Cantor created DOES work.Ross' argument is analogous to arguing that every airplane must crash because an aircraft which Ross designed has crashed.The rest of his stuff is not worth reading either.=== Subject: : Re: Antidiagonal, InfinityI think I see what you're saying. You demand to be able to give arule that generates the antidiagonal. I simply say there is exactlyone antidiagonal in binary and sometimes because of dualrepresentation it exists on the list.About the leading zeros, what I'm saying is that every number hasinfinitely many leading zeros. Any antidiagonal rule you select forany finite interval will generate an antidiagonal not in thatinterval. Virgil, I'm saying that any expansion has infinitely manyleading zeros, so that any rule for any interval would eventuallyreflect a difference in representation but not necessarily value of anexpansion, where the expansion is a representation of the value, andthe expansion with leading zeros has always an identical value to therepresentation without them. Thus for any list of real numbers,enumerable via the well-ordering principle, I can give you that exactsame list of numbers, in any base, and any rule you select will notgenerate an antidiagonal in the range of the function.Besides that, I just request the list in binary, there is only oneantidiagonal, and reordering always allows dual representation. Forthat, I have some problems with dual representation.With the Equivalency Function, the range is [0,1), and itsantidiagonal is 1, currently: because I say so. As well, the nestedinterval sequences for those of you who don't care for infinitesimalsare (0) and (0). Present other reasons why the integers would not mapto the reals, besides Cantor-Bernstein and transitivity, I'minterested in hearing about them.Anyways, I can ignore that and in my own little private universe talkabout the real numbers and points and lines of my logical system, andthe fun part is that it is about the logical system.Virgil, I told you that N was a subset of P(N) with a type of ordinalnumbers. Did you not agree that that was so? I've found errors inKnuth, the Handbook of Mathematical Functions, and CRC manuals, exceptthose typos are trivia where N+1 being the order type of P(N) was newsto you.Can you not handle the set of all sets? How about the class of allclasses, what's that? What ordinals represent negative integers? Canyou not answer those questions in any case?My writing is not worth reading? Heh heh. The integration of Dirac's delta over the reals or over any interval[-a, a] for non-infinitesimal real a is one. That's a well knownresult.The naturals are not closed and bounded, to be compact. Does the setof integers contain all of its limit points? Is not the distancebetween any two naturals finite?So, say what you want, I say the binary case is sufficient.Your input is often appreciated.=== Subject: : Re: Antidiagonal, Infinity> I think I see what you're saying. You demand to be able to give a> rule that generates the antidiagonal. I simply say there is exactly> one antidiagonal in binary and sometimes because of dual> representation it exists on the list.Why insist on binary when it is one of only two, out of infinitely many choices, where a one-digit-at-a-time-antidiagonal-algorithm doesn't work? But a two-digit-at-a-time-antidiagonal-algorithm does, even for bases 2 and 3. > About the leading zeros, what I'm saying is that every number has> infinitely many leading zeros.What happens to the left of the radix point is irrelevant if the antidiagonal-algorithm only appplies to digits to the right of the radix point. Any antidiagonal rule you select for> any finite interval will generate an antidiagonal not in that> interval.No need to restrict it to finite intervals. Virgil, I'm saying that any expansion has infinitely many> leading zeros, If you mean preceding the radix point, so what? If you mean after the radix point but before any non-zero digits, you're off your rocker.so that any rule for any interval would eventually> reflect a difference in representation but not necessarily value of an> expansion, where the expansion is a representation of the value, and> the expansion with leading zeros has always an identical value to the> representation without them. > Thus for any list of real numbers,> enumerable via the well-ordering principle, I can give you that exact> same list of numbers, in any base, and any rule you select will not> generate an antidiagonal in the range of the function.Actually, there are lots of rules that will generate values IN the range, the point is to generate one NOT in that range. > Besides that, I just request the list in binary, there is only one> antidiagonal, and reordering always allows dual representation. For> that, I have some problems with dual representation.Problems of dual representation are easily avoidable, so why keeep insisting on situations which make avoiding them more difficult? I can only suppose it is so that you can continue to confuse yourself.> With the Equivalency Function, the range is [0,1), and its> antidiagonal is 1, currently: because I say so.The antidiagonally constructed number need not be in the range of the function if that range is any proper subset of the reals. Any real will do. Rosss continues to try and create problems where none exist. As well, the nested> interval sequences for those of you who don't care for infinitesimals> are (0) and (0). Present other reasons why the integers would not map> to the reals, besides Cantor-Bernstein and transitivity, I'm> interested in hearing about them.> Virgil, I told you that N was a subset of P(N) with a type of ordinal> numbers. Did you not agree that that was so? No. N is a member of P(N), but nnot a subset of it. I've found errors in> Knuth, the Handbook of Mathematical Functions, and CRC manuals, except> those typos are trivia where N+1 being the order type of P(N) was news> to you.> My writing is not worth reading? Heh heh.No comment.> So, say what you want, I say the binary case is sufficient.To do what?=== Subject: : Re: Antidiagonal, Infinity> The infinitesimal iota doesn't have all properties of definite reals,> it is among a form of what may be called indefinite reals, real> numbers to be sure but those not representable as a rational as a> fraction of integers or an irrational as a convergnt sequence.You can't have it both ways. If your iota is a real, then it's either rational or representable as the limit of a convergent sequence of rationals (technically those two cases are not mutually exclusive, since a rational is also the limit of a convergent sequence of rationals.)When you say that iota is a member of the real numbers to be sure, then you are saying that it's a real number. Which we've all agreed means, for the sake of this discussion, the standard model of the real numbers.If iota is not the limit of a convergent sequence of rationals, then it definitely is not a real number. It might be something else -- a non-real complex number, a member of a permutation group, a tunafish sandwich -- but it's not a real number.You do understand that, don't you?=== Subject: : what is saint venants principle ?what is saint venants principle?is it some thing in math=== Subject: : Re: what is saint Venant's principle ?> what is saint venants principle? is it some thing in math?In Mechanics of materials, the principle says statically equivalentapplied loading (forces and bending moments) produces the same stressand deformation at points far away from area of load application,although local stress and deformation could change. Airy stressfunctions are involved for Elasticity in finding stress and strainditribution, but it is used more for physical verification than mathsin problem solving.=== Subject: : (fixed) P vs NP: my proof of P != NPHi All. After some discussion with David Moews we come to conclusion that myoriginal proof was wrong, the crux was last step in penultimate theorem;actually original claim of that theorem is wrong. Fortunately the problem can be easily avoided. The link to updated version ishttp://users.i.com.ua/~zkup/pvsnp_en_002.pdf . To people who have already read the first (wrong) version: I've changed only a few things. Look at definition 2 and theorem 5 atpage 14 and end of the proof of theorem 6 at page 18-19 (last step wassimply cut). (Note that theorem numbers were changed somewhere). -- Mikhail Kupchik=== Subject: : Re: (fixed) P vs NP: my proof of P != NPHi All. A bit more details about the things I've changed. Now main decision problem is the following: given a second-ordersentence(E f1(*)) (E f2(*)) ... (E fN(*))(A x1) (A x2) ... (A xN) (A a) (1)P(x1,x2...xN,f1(x1),f2(x2)...fN(xN),a)decide whether it is true or false. If P=NP (just the supposition we're going to disprove), then NP=co-NPand inner part (universal-quantified core) of this expression is equivalentto(E x1) (E x2) ... (E xN) (E a)P'(x1,x2...xN,f1(x1),f2(xN)...fN(xN),a)so the whole expression (1) is equivalent to(E f1(*)) (E f2(*)) ... (E fN(*))(E x1) (E x2) ... (E xN) (E a) (2),P'(x1,x2...xN,f1(x1),f2(x2)...fN(xN),a) where length of P' is limited by some polynomial over length of P.And (2) is equivalent to first order sentence(E x1) (E x2) ... (E xN) (E a)(E f1) (E f2) ... (E fN) (3)P'(x1,x2...xN,f1(x1),f2(x2)...fN(xN),a) If P=NP (using the supposition second time) then (3) is decidable withinpolynomial time (of fixed degree), we get to contradiction with timehierarchy theorem here. So proof schema/sketch is unchanged. -- Mikhail Kupchik> Hi All.> After some discussion with David Moews we come to conclusion that my> original proof was wrong, the crux was last step in penultimate theorem;> actually original claim of that theorem is wrong.> Fortunately the problem can be easily avoided.> The link to updated version is> http://users.i.com.ua/~zkup/pvsnp_en_002.pdf .> To people who have already read the first (wrong) version:> I've changed only a few things. Look at definition 2 and theorem 5 at> page 14 and end of the proof of theorem 6 at page 18-19 (last step was> simply cut). (Note that theorem numbers were changed somewhere).> -- Mikhail Kupchik=== Subject: : Re: P vs NP: my proof of P != NPHi David> We can now deduce that> [E.1] G1(00,00)=G1(00,01)=G1(00,10)=G1(00,11)=0 (from C.1)> [E.2] G2(00,00)=G2(01,00)=G2(10,00)=G2(11,00)=0 (from C.2)> [E.3] G2(00,01)=0 (from C.3, E.1)> [E.4] G1(01,00)=0 (from C.4, E.2)> [E.5] G1(10,11)=G2(10,11) (from C.3)> [E.6] G1(11,10)=G2(11,10) (from C.4)> [E.7] G1(01,10)=G2(01,10) (copy of C.5)> [E.8] G1(10,01)=G2(10,01) (copy of C.6)> and no more. As you can see, this does not uniquely specify G1 and G2;> it does not say that G1(X1,X2) is a function only of X1, or G2(X1,X2) ofX2;> and it does not determine the course of the computation after t=1.I agree. -- Mikhail Kupchik=== Subject: : Re: PATTERNS IN FINGERPRINTS, CACTI PREDICTED> Sci Tech> Patterns in fingerprints, cacti predictedhttp://www.geocities.com/drjosemariachi/jay_faq.html# bb Troll FAQ for Jai Maharaj (Hindi for cracked athletic cup)http://www.mazepath.com/uncleal/effete6.jpg> By Our Bureau> The Hindu1.1 billion (with a b) East Indians, 1 million (with an m) flushtoilets. No lines. Draw your own conclusions (and watch where youstep).http://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics)=== === Subject: : Puzzle: simple geometry?A friend came across this in a newspaper; defeats me though. Hopefully theregular clientele of this newsgroup have this sort of thing for breakfast.Is there an 'elegant' (however you interpret that) solution for this?I hope my description is clear enough. In a circle,centre C, diameter ACB is produced to X. Tangent YA is at rightangles to ACB(of course). YX is also tangent to the circle. Question:If:BX = 3AY = 9what is the radius of the circle?=== Subject: : Re: Puzzle: simple geometry?>A friend came across this in a newspaper; defeats me though. Hopefully the>regular clientele of this newsgroup have this sort of thing for breakfast.>Is there an 'elegant' (however you interpret that) solution for this?>I hope my description is clear enough. In a circle,>centre C, diameter ACB is produced to X. Tangent YA is at rightangles to ACB>(of course). YX is also tangent to the circle. Question:>If:>BX = 3>AY = 9>what is the radius of the circle?(I hope I understood that right). So there is a right triangle XAY,right triangle at A, with AY =9, AX = 2r+3, where r is the radius ofthe circle. Furthermore thee is a right triangle CZX, where Z is thepoint of tangency of the line YX. Right so far? (If not, say Stop!)Then CX = r+3 and CZ = r./(2r+3), where alpha is the angle at X.Now use tan(alpha) = sin(alpha)/sqrt(1-[sin(alpha)]^2)Mhmm, that leads to a polynomial equation of the form r^2(2r+3)^2 =9^2(6r+9). Looks bad, but my spreadsheet yields r=4.5. Nice!=== Subject: : Re: another boring critisism of Cantor's Theorem>It makes perfect sense. You yourself said that the models of ZFC in>the countable model of ZFC+Con(ZFC) were of arbitrarily large>cardinality in the countable model, but countable in the real world.>You mean, there are bijections from them to N, not in the countable>model, but in the real world, in the universe, V. I'm simply pointing>out that in V there will be models that really are of arbitrarily>large cardinality.> But that 'real world' could be countable, when looking at it> from an even larger meta-world. And that would mean that> these large cardinals would be 'really-really' countable.> Right? Or do i misunderstand you?> No. If ZFC+Con(ZFC) is consistent, and we are uttering theorems of> ZFC+Con(ZFC), then we can suppose, without making any of our theorems> false, that the semantics of our discourse is such that the> interpretation of it is a countable model of ZFC+Con(ZFC). But that's> not the same as saying V can be countable. V can't be countable.Could you explain why V can't be countable? Certainly you can'tprove that in any consistent first order formalism for which theLoewenheim Skolem theorem applies.=== Subject: : UNCLE AL - GET BACK TO SCI.CHEM or piss off whateverComments: This message probably did not originate from the above address. It was automatically remailed by one or more anonymous mail services. You should NEVER trust ANY address on Usenet ANYWAYS: use PGP !!! Get information about complaints from the URL belowX-Remailer-Contact: http://80.65.224.85/POL/ In case my abuse address is unreachable: It is because it has been flooded by , please contact === Subject: : Re: Weight> Weight is a measure of the centripetal force, or thrust exerted on,> and/or by objects;>Dumb Donny Head is off his psycho-meds.>Hey stooopid Dumb Donny Head, where is the centripetal force in a>Cavendish balance?=== Subject: : Re: UNCLE AL - GET BACK TO SCI.CHEM or piss off whatever> You total retard Al!> If you could read more than half a paragraph at any one time while> keeping the contents of the first half of that paragraph in your mind> you might have stood a chance at understanding what the O/P was> saying.Shead never has anything remotely interesting to say. Whereas othercranks aspire to toppling the towering edifices of modern physics, Sheadhas never even mastered the basic SI units. He is condemned to foreverfall at the first hurdle.Faced with such monumental stupidity, there are only two sensibleoptions; ignore him and move swiftly on, or mock and insult him.=== Subject: : Re: UNCLE AL - GET BACK TO SCI.CHEM or piss off whatever> This is the second time I have felt it nessesary to admonish you.> Lets hope it's the last time.Snipity snip and amen to that.=== Subject: : how to transform a cyclic in a linear function? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3DJaYQ31643;I mean how transform a periodic function as sine or cosine long a direction of space by combination or any other artifice to a linear function in the same direction of the space=== Subject: : Re: Need help on Sobolev spaces and inequalities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3DJaaP31736; === Subject: : Question About EigenvalueI have a question. Suppose X is an n-by-n non-singular matrix witheigenvalues a1,a2,...,an and eigenvectors v1,v2,...vn. I am interestedin the procedure of finding the eigenvalues of D*X*D as a function of(a1,a2,...,an,v1,v2,...vn) and D, where D is an n-by-n diaganolMike=== Subject: : determinant derivativeHi I have one matrix A(H)A(H)'+C*C'A,H,C are all block matrixnow I need to find H so as to minimize the determinant of A(H)A(H)'+C*C'What can I do?Any generalized method for this?Can I try to find H so that A(H)A(H)' = 0*I? why?=== Subject: : Re: determinant derivative>Hi I have one matrix A(H)A(H)'+C*C'>A,H,C are all block matrix>now I need to find H so as to minimize the determinant of A(H)A(H)'+C*C'>What can I do?>Any generalized method for this?>Can I try to find H so that A(H)A(H)' = 0*I? why?Assuming that the minimum is not zero, the easiest way to do this is by minimizing the logarithm of the determinant,using differentials. The differential d(log(det(X))) istr(X^{-1}*d(X)), and you will have to express d(A(H)) interms of d(H). The usual rules for derivatives hold, keepingorder, and one can use symmetry to simplify the expression tobe set equal to zero for the first order conditions to 2*tr(Z^{-1}*A(H)*d(A(H))'),where Z is the matrix whose determinant is to be minimized.This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== Subject: : Re: Set of orthogonal matrices> I found the following exrcise: Show that the set of all n x n> orthogonal matrices is a compact subset of (R^n)^n. That's all the> exercise said. (I'm not sure, but, since to prove such proposition> it's fundamental to have a norm or a topology in (R^n)^n, I take it> the norm of a matrix is defined as the square root of the sum of the> squares of it's terms - an extension of the definition of norm in> R^n).> I'm starting to think this proposition is false. It seems to me the> set of orthogonal matrices doesn't need to be bounded with respect to> the norm I cited. But I'm really stuck. any help is welcome.> Aren't the column vectors of an orthogonal matrix unit vectors? And aren't > there only n of them? So let's see, that would make the norm of an > orthogonal matrix what exactly? And then there's closedness ...Yes, it's not that hard=== Subject: : norm and determinant of the matrixHi I suddenly begun to think of this problemwhat the relationship between norm and determinant of one matrix.Do they have relationship?We know we have inequality ||A+B||_p<=||A||_p+||B||_pbut we can't say we have the inequanlity det(A+B)<=det(A)det(B)=== Subject: : Schauder basis!I would appreciate a help to show that if a normed vector space has aSchauder basis, then it's separable.I got really stuck.=== Subject: : Re: Schauder basis> I would appreciate a help to show that if a normed vector space has a> Schauder basis, then it's separable.Hint: How do you show l^2, the space of square summable sequences, is separable?=== Subject: : Re: Schauder basis>!>I would appreciate a help to show that if a normed vector space has a>Schauder basis, then it's separable.Well, can you fill in the blank in the following sentence withsomething correct?If X has a Schauder basis then for every x in X there existsa sequence (x_n) of ______'s such that x_n -> x.The first ______ you think of probably won't consist of somethingcountable, but then if you think about it some more you see howto modify ______ so there are only countably many ______'s andyour done. But first, what's the ______ that follows directly fromthe definition?>I got really stuck.>Amanda=== Subject: : Re: Computer based proofs[snipped]> The biggest drawback I see to such a tool is the need to standardize> the rigor of virtually all mathematics at a pretty high level. In> particular, some fields (eg, low dimensional topology, some algebraic> geometry) often slink by :-) with a lower level of rigor due to the> intuitive nature of the mathematical concepts involved.What lower level of rigor? There is a pretty high level of rigor inlow dimensional topology. It's a common mistake for outsiders,unfamiliar with the standard proof techniques of the field, toassociate known arguments with sloppy reasoning.=== Subject: : Re: Computer based proofs> Hi all,> What is the consensus here in sci.math on the use of computers in proofs?> Should they be allowed, or not?> As I recall, the four color theorem was the first such . . . and unless > I'm mistaken, there is still no non-computer proof. When you ask if they > should be allowed, what do you mean? The four-color theorem was proved > way back in 1977. If anyone was going to disallow this proof, it would > have happened by now.The four color theorem was NOT the first computer proof. D.H. Lehmer et.al.proved that every sufficiently large prime congruent to 1 mod 6always had 3 consecutive cubic residues. The proof consisted of extensive(for that time) computer calculation. This was done in the early 60's.The paper appeared in Math. Comp. I don't have the exact reference.One can also argue over the definition of 'proof' and 'theorem'. Example:Theorem:M607 is prime.The proof is purely by computer.=== Subject: : Re: Computer based proofs> Hi all,> What is the consensus here in sci.math on the use of computers in proofs?> Should they be allowed, or not?> They should be allowed though a shorter proof that a human can handle> in a reasonable time is far better. The real problem here IMHO is the> lack of useful, *standard*, proof checking tools. It also would help> with a lot of papers (particularly preprints) to be able to run a> program which verifies that the logical arguments of the paper are> correct. The theorems/proofs might be misleading or uninteresting, but> at least you should be able to certify correctness for a broad> category of mathematics. A bonus is that the proof checker can be made> very reliable (eg, checked by hand and gobs of empirical examples to> be to high probability mathematically correct).> The biggest drawback I see to such a tool is the need to standardize> the rigor of virtually all mathematics at a pretty high level. In> particular, some fields (eg, low dimensional topology, some algebraic> geometry) often slink by :-) with a lower level of rigor due to the> intuitive nature of the mathematical concepts involved. Others (eg,> String Theory) have fundamental concepts (eg, Feynman and String> integral actions) with no rigorous mathematical backing (perhaps there> are rigorous constructions, but they aren't used).> A proof checker isn't great for every field of endevour here. But it> does provide a stronger proof checking capability than the current> regime. I doubt anyone who has published half a dozen or more serious> proofs doesn't have a paper out there without some typo or logic> error. Usually, these errors don't change the outcome of the proof,> but it does mean more work for the reader while the error is worked> out and potential embarrassments for the author.> Karl Hallowell> khallow@hotmail.comHow long before computers become God?=== Subject: : Re: Computer based proofs>today's corrupt congress critters >Isn't this, like, a triple redundancy?> You don't think they had corrupt congress critters in other eras?Yes -- that's why it is a *triple* redundancy, and not *quadruple*:-)Carlos=== Subject: : Re: Computer based proofs>today's corrupt congress critters >Isn't this, like, a triple redundancy?> You don't think they had corrupt congress critters in other eras?> Yes -- that's why it is a *triple* redundancy, and not *quadruple*>:-)Is congress all by itself a redundancy?Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.today's corrupt congress critters >Isn't this, like, a triple redundancy?>You don't think they had corrupt congress critters in other eras?>Yes -- that's why it is a *triple* redundancy, and not *quadruple*>:-)> Is congress all by itself a redundancy?Whoh !! That may have been way too subtle and too advancedfor my capabilities! ;-)I can only add something I've seen around in the form ofgovernment as we pay for! :-)Carlos=== Subject: : Re: Computer based proofs> If something's *proven*, then it's irrelevant how the proof was obtained....> Or are you talking about something else? Hopefully you're talking about> something else, because the question as originally posed is silly.The question with computer proofs is whether they ARE proofs. Consider aproof that includes a computer check of a huge number of special cases.What verification is needed to accept this?For a regular non-computer proof, it is relatively simple: the proof is readby others, and they all say it is correct, and no one says it is incorrect.What is the equivalent for a computer proof?Others can read the code, and say they think it is correct. Is that goodenough? Or should a formal proof that the algorithm is correct be required?How about the implementation? Even if the implementation is correct, whatabout the compiler used to compile it, or the system libraries the programuses? What about the processor? Processors do have bugs now and then.Is one implementation good enough, or should we require independentimplementations, giving the same result of course, before a computer proofis accepted?It is this kind of thing that is generally meant when someone asks whethercomputer proofs should be accepted, and that's what I presume the originalposted was asking. It's far from a silly question--it gets to the veryheart of what it means to accept a proof.--Tim Smith=== Subject: : Re: Computer based proofs* Tim Smith> The question with computer proofs is whether they ARE proofs. Consider a> proof that includes a computer check of a huge number of special cases.> What verification is needed to accept this?> For a regular non-computer proof, it is relatively simple: the proof is read> by others, and they all say it is correct, and no one says it is incorrect.> What is the equivalent for a computer proof?> Others can read the code, and say they think it is correct. Is that good> enough? Or should a formal proof that the algorithm is correct be required?> How about the implementation? Even if the implementation is correct, what> about the compiler used to compile it, or the system libraries the program> uses? What about the processor? Processors do have bugs now and then.As the authors in Mathematical Experience points out, proofs arerarely formal anyway, and the acceptance of a proof, any proof, is asocialogial process: The proof is accepted when it is a consensuswithin the mathematical community that it is correct. (Food forconspiration theories here.) Anyway, we can therefore not statestronger requirements to a computer based proof than to a blackboardproof. Remember that the Fermat's Last Theorem (FLT) was proven inthis sense several years ago. It took about ten years (someonecorrect me here) before a flaw in a rather simple proof was found. Inthose ten years, FLT had a proof...Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92=== Subject: : Re: Computer based proofs> proof. Remember that the Fermat's Last Theorem (FLT) was proven in> this sense several years ago. It took about ten years (someone> correct me here) before a flaw in a rather simple proof was found. In> those ten years, FLT had a proof...I'm not familiar with this incident. Please elaborate further.=== Subject: : Re: Computer based proofs> Nope, you still don't make any sense. What does students using> calculators have to do with using computers to do proofs?What if nobody can read or understand a computer proof?Has a theorem been demonstrated?>Wrong numbers don't compute, only good ones need apply.> Can I have some of that you're smoking?What if a computer generated a proof for a theoremthe statement of which nobody could read or understand?BTW, most of the countable number of theorems not only can't be read bysomebody, they can't even be stated by a computer.=== Subject: : Re: Computer based proofs today's corrupt congress critters> Isn't this, like, a triple redundancy?> You don't think they had corrupt congress critters in other eras?Always. However in my life time I've noticed an exponential increase.=== Subject: : Re: Computer based proofs> What is the consensus here in sci.math on the use of computers in proofs?> Should they be allowed, or not?> (snip)>It's interesting that you bring that up now. Recently the Annals of>Mathematics has decided that Hales' computer-based proof of the Kepler>Conjecture should be published in two parts, in two different journals. >As for your question, Lurch, I'll probably side with the mainstream>view that computers are ok and fairly reliable (as long as you test on>different architectures, etc, etc) in some situations, but it's often>not as elegant or informative as a traditional proof. > The abstract of this paper has some stimulating remarks too:> Though the truths of logic and pure mathematics are objective and> independent of any contingent facts or laws of nature, our knowledge> of these truths depends entirely on our knowledge of the laws of> physics. Recent progress in the quantum theory of computation has> provided practical instances of this, and forces us to abandon the> classical view that computation, and hence mathematical proof, are> purely logical notions independent of that of computation as a> physical process. Henceforward, a proof must be regarded not as an> abstract object or process but as a physical process, a species of> computation, whose scope and reliability depend on our knowledge of> the physics of the computer concerned. (end quote 2) And how you derive evolution from physics? You don't MORONS.> John Bailey> http://home.rochester.rr.com/jbxroads/mailto.html=== Subject: : Re: Puzzle 1: Sum Involving Surd Rounded Down>Let a(k) =>floor[((1 +sqrt(1 +4k))/2)^2].>In ascii-art mode:>a(k) => ______ 2>!(1 + /1 +4k ) !>! ------------- !>!_ 4 _!>Evaluate the closed-form for:>sum{k=1 to oo} 1/a(k)^2,>which is in ascii-art mode:> oo>--- 1> ----->/ 2>--- a(k)>k=1 >First, let us rewrite a(k) as> 1+sqrt(1+4k)> a(k) = floor( k + ------------ )> 2>Note that m = (1+sqrt(1+4k))/2 is an integer when k = m^2-m. Therefore>the sum above can be rewritten as a double sum> oo m^2+m-1> --- --- 1> S = > -------> --- --- (k+m)^2 > m=1 k=m^2-m>S would be zeta(2) if it were not for the fact that when m increments,>the reciprocal of the square of one integer gets left out, that is,> oo oo> --- 1 --- 1> S = > --- - > ---------> --- k^2 --- (k^2-1)^2> k=1 k=2>Using partial fractions we get that> 1 1 1 1 1 1> --------- = - ( ------- + ------- + --- - --- )> (k^2-1)^2 4 (k+1)^2 (k-1)^2 k+1 k-1>Thus, we get that> S = zeta(2) - (zeta(2) - 5/4 + zeta(2) - 3/2)/4> = zeta(2)/2 + 11/16>So, unless I have made a careless error, S = pi^2/12 + 11/16.As Leroy has pointed out to me, I did make a careless error. I summedfrom k=0, not k=1. Subtracting 1, the answer should be pi^2/12 - 5/16.Rob Johnson (b*h^-1)h2=h1*b or b*(h^-1*h2)=h1*b. Givenany h1*b E Hb, we have that h2*b E aH. So Hb is a subset of aH.Therefore H is a normal subgroup of G.=== Subject: : Re: not sure if this is correct? Adjunct Assistant Professor at the University of Montana.> everyone I was wondering if someone could look at this proof and>tell me if I need to change anything or add something to make it>Question: Let G be a group, let H be a subgroup of G. If each left>coset of H in G is some right coset of H in G. (ie for each a E G it>is the case that aH=Hb for some b E G), then prove that H is a normal>subgroup of G.>Proof: Since H is given as a subgroup of G we need only show that H is>normal to G. To show that aH=Hb for all a E G and some b E G. What are you doing here? You are assuming that for each a in G, thereyou are expected to show that H is normal. You don't need to showaH=Hb for all a in G and some b in G, you are ASSUMING this.>We pick>some element of aH, a*h1, and show that it's in Hb.You are ->assuming<- it is.>Since 1 is in H,>there is an element h E H such that a*1=h*b, or h^-1*a=b.Right, so you have shown that a is in Hb, which is equivalent to bbeing in Ha. At this point, you have that if aH=Hb, then Hb =Ha. Which proves that aH=Hb=Ha. So you have now proven that if for alla in G there is a b in G such that aH=Hb, then for all a in G we haveaH=Ha, which proves H is normal.> So b is in>Hb. This follows from the DEFINITION of Hb. You seem to be writing lots ofcorrect stuff but without any idea of where you are going with it...>Now, let a*h1, be an element of aH, then there is an h2 E H, so>that a*h1=h2*b because aH=Hb. But we know>b=h^-1*a so a*h1=h2*b=h2(h^-1*a) or a*h1=(h2*h^-1)*a. Given any a*h1 E>aH, we have that a*h1 E Hb. So aH is a subset of Hb.>ConverselyThere is no conversely. You are trying to show that if for all a inG there is a b in G such that aH = Hb, then H is normal. This is asingle implication.> we pick some element of Hb, h1*b, and show that it's in aH.>Since 1 is in H, there is an element h E H such that a*h=1*b, or>a=b*h^-1. So a is in aH. Now, let h1*b, be an element of Hb, then>there is an h2 E H, so that a*h2=h1*b because aH=Hb. But we know>a=b*h^-1 so a*h2=h1*b => (b*h^-1)h2=h1*b or b*(h^-1*h2)=h1*b. Given>any h1*b E Hb, we have that h2*b E aH. So Hb is a subset of aH.>Therefore H is a normal subgroup of G.Really? Your conclusions are that aH is a subset of Hb, and that Hb isa subset of aH. So you have concluded that aH=Hb. But that was whatyou assumed to begin with. How have your shown that H is a normalsubgroup of G, from the fact that aH=hb?===================================================================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)===================================================================Arturo Magidinmagidin@math.berkeley.edu=== Subject: : Please check proof.. Can someone please check to see if this proof is correct or ifhelp.Question: Suppose that G is a (possibly infinite) group, and that Ghas two subgroups H and K of finite index. Prove that H n K is a groupof finite index in G, and that [G: H n K] < [G:H][G:K].note: n stands for the union symbolProof: This proof is done for finite |G|. First we will show that [G:H n K]=[G:H][G:K] <=> G=HK. If [G: H n K]=[G:H][G:K], then the lefthand side is |G|/|H n K|, the right hand side if |G|^2/|H n K|, so HK=|H|*|K|/|H n K|=|G| follows. And conversely, if G = HK, then [G:H nK] = |G|/|H n K|, and[G:H][G:K] = |G|^2/|H||K| = |G|*|HK|/|H|*|K| = |G|/|H n K|, so we haveequality.HK= U h EH hK is a union of cosets. Each cost has |K| elements, so weneed only show that the number of distinct such cosets is |H|/|H n K|.Now h1K=h2K <=> h2^-1 h1 E K, <=> h2^-1 h1 E H n K, <=> h1(H n K) inH, which is|H|/|H n K|.Now let {xi | i E I} be a set of left coset representatives of H n Kin G. Then consider pairs of coests (xiH, xiK), i E I. Now (xiH,xiK)=(xjH, xjK) <=> xj^-1 xi E H n K, <=> xi(H n K)= xj(H n K).Thus the map xi(H n K) |-> (xiH, xiK) is an injection, so [G: H n K] <[G:H][G:K].=== Subject: : Re: Please check proof. Adjunct Assistant Professor at the University of Montana.>. Can someone please check to see if this proof is correct or if>help.>Question: Suppose that G is a (possibly infinite) group, and that G>has two subgroups H and K of finite index. Prove that H n K is a group>of finite index in G, and that [G: H n K] < [G:H][G:K].>note: n stands for the union symbol>Proof: This proof is done for finite |G|.Why? The question is for any G. Why restrict yourself to finite groups?> First we will show that [G:>H n K]=[G:H][G:K] <=> G=HK.Why? This is irrelevant to the question posed. I noted this alreadywhen you made your prior post.> If [G: H n K]=[G:H][G:K], then the left>hand side is |G|/|H n K|, the right hand side if |G|^2/|H n K|, so HKyou mean |HK|.>=|H|*|K|/|H n K|=|G| follows. And conversely, if G = HK, then [G:H n>K] = |G|/|H n K|, and>[G:H][G:K] = |G|^2/|H||K| = |G|*|HK|/|H|*|K| = |G|/|H n K|, so we have>equality.>HK= U h EH hK is a union of cosets. Each cost has |K| elements, so we>need only show that the number of distinct such cosets is |H|/|H n K|.>Now h1K=h2K <=> h2^-1 h1 E K, <=> h2^-1 h1 E H n K, <=> h1(H n K) in>H, which is>|H|/|H n K|.>Now let {xi | i E I} be a set of left coset representatives of H n K>in G. Then consider pairs of coests (xiH, xiK), i E I. Now (xiH,xiK)=>(xjH, xjK) <=> xj^-1 xi E H n K, <=> xi(H n K)= xj(H n K).>Thus the map xi(H n K) |-> (xiH, xiK) is an injection, so [G: H n K] <>[G:H][G:K].Rather than use a counting arugment, why not simply use the map youhave set up? You know that the set of cosets of H is finite; and so isthe set of cosets of K. You have shown that there is an injection fromthe set of cosets of H n K to the set of pairs (coset of H, coset ofK). How many elements does the set of such pairs have? How is it relatedto the index? What can you conclude about the set of of cosets of H nK? How is it related to the index of H n K?===================================================================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)===================================================================Arturo Magidinmagidin@math.berkeley.edu=== Subject: : Re: Frequentist probability confusion>You began with a criticism of frequentism. Frequentism is a>nuts-and-bolts area of mathematics intended for use by physicists,>statisticians, and the like. As such, it concerns itself with finite>processes, computable numbers, and limits. To use other tools to>attempt to invalidate such results is simply irrelevant: there is no>point in arguing about what tools are permissible, one simple chooses>one's arena and proceeds from there. You are saying that we deal with finite sets in practice, whenwe have our sleeves rolled up and are clutching a screwdriver,so we don't need an interpretation of probability theory thatdeals with infinite sets. Ok; but that's your interpretation,and not everybody will be happy with it; somebody who spends moretime dealing with the axiom of choice than with nuts-and-bolts mightnot, for example.>So this area of probability>theory indeed places limits on processes of generating random numbers.That's not clear. The mere existence of a distribution over someset doesn't immediately tell me anything about the existence ornonexistence of processes selecting elements from that set. I canhave a uniform distribution over the set {0,1}; that doesn'tfurnish me with a way to select one of those numbers at random,and doesn't tell me that anybody else knows a way either. Similarly,the nonexistence of a uniform normalised distribution over theintegers tells me nothing about whether somebody I meet tomorrowmight or might not start spouting random integers at me.The real problem is that finite processes, computable numbers andlimits as you say, will never suffice to produce anything random;for that you need a seed to the random number generator, or someunknown extra input which can serve as a seed, and you suppose thatthat seed is already random. Then your finite process argumentmerely says that if we don't already have a random element of aninfinite set, we can't generate one, although we can generateelements of perhaps large finite sets given many seeds from smallersets.R.=== Subject: : Re: Frequentist probability confusion> ebunn@lfa221051.richmond.edu says...>Using the axioms of choice and infinity then one can indeed choose a>natural number at random>with all natural numbers equally probable.>Is this statement meant to be obvious? It's not at all clear to me how>the axiom of choice says anything about probabilities.>If it's not meant to be obvious, but is nonetheless true, can someone>point me to an appropriate place to read more on this?> I'm cross-posting to sci.math, because maybe a mathematician has> something to add. Patrick's point is not complicated to prove, but it's> hard to understand how to interpret it.> 1. Pick an enumeration of all positive rational numbers between 0 and 1.> For example, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, ... Let q_n be> the nth rational number.> 2. Define an equivalence relation on real numbers between 0 and 1: x ~~> y if and only if |x-y| is rational.> 3. Using the axiom of choice, construct a set S by picking one element> out of every equivalence class.> 4. Define S_n to be { x | |x - q_n| is in S }> Note that S_0 union S_1 union S_2 union ... = (0,1).> 5. So here's how you generate a random nonnegative integer: Generate a> random real x in (0,1), and let your random integer be that n such that> x is an element of S_n.> There is no probability distribution on the possible outcomes of this> process, so it isn't a uniform distribution on the integers in a> measure-theoretic sense. But you can argue by symmetry that in some> sense every n is equally likely because each of the sets S_n are> identical, except for a translation.Unfortunately, S_n as defined above is not Lebesgue measurable. Thismeans that it does not really make sense to ask the question, What is theporbability that a uniform random variable lies in S_n? I realize thatthis is counterintuitive, but probability over uncountable sample spacesis tricky, precisely for these reasons.As to the original question about natural numbers. If one requires thatprobability measures be countably additive, then there is on probabilitymeasure defined on the integers that I would want to label uniform.Countably additive means that if {A_n} is a countable collection ofdisjoint (measurable) sets, then the probability of the union of the A_nis equal to the sum of the probabilities of each of the A_n. I certainlywant my probability measures to be countably additive, but some authorsargue that it is enough to be finitely additive. If one allows finitelyadditive probability measures, then one can defined a probability measureover the natural numbers that some might want to label uniform. However,the construction of this measure uses the the axiom of choice (or at leastsome large portion of the axiom of choice).Daniel Waggoner=== Subject: : Natures mechanismThe ultimate mechanism of the universe is really quite simple: Likethe Greeks said: Matter is constantly aggregating and disaggregating.In its finest most disaggregated form matter consists of what we callelectromagnetic radiation, and is capable of penetrating the densestmaterial; to great depths:Essentially it consists of ultimately small bits of substance, flyingand spinning in all directions. Upon the rarity of striking oneanother they deflect, and slow each other's previous motions, andbeing too small to be elastic, start accumulating.Regions develop which are more densely populated by them, therebyincreasing the likelihood of their interfering with the motion ofothers; to increasingly promote even greater density of that region.As individual regions of increasing density build up, they form thenuclei of spinning vortexes, which act as sinks to block, capture,absorb and further concentrate and densify when any stray bit ofsubstance passes too close: Like Earth and its atmosphere areEven the moon would coalesce with Earth if it got captured in itsatmosphere; as would both Earth and the moon coalesce with the sun, ifthey got close enough to it.=== Subject: : Re: Natures mechanismDumb Donny Head randomly pontificates about the meaning of theuniverse. Hey Dumb Donny Head, gien your model calculate andstate the natural abundances of the first three elements - H, He, andLi - and the H/D ratio of the universe at large. Compare with measured values, dumb Donny Hed.http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net!=== Subject: : Re: Natures mechanism> The ultimate mechanism of the universe is really quite simple:So write a simple equation to describe the universe andeverything in it.SR Like> the Greeks said: Matter is constantly aggregating and disaggregating.> In its finest most disaggregated form matter consists of what we call> electromagnetic radiation, and is capable of penetrating the densest> material; to great depths:> Essentially it consists of ultimately small bits of substance, flying> and spinning in all directions. Upon the rarity of striking one> another they deflect, and slow each other's previous motions, and> being too small to be elastic, start accumulating.> Regions develop which are more densely populated by them, thereby> increasing the likelihood of their interfering with the motion of> others; to increasingly promote even greater density of that region.> As individual regions of increasing density build up, they form the> nuclei of spinning vortexes, which act as sinks to block, capture,> absorb and further concentrate and densify when any stray bit of> substance passes too close: Like Earth and its atmosphere are> Even the moon would coalesce with Earth if it got captured in its> atmosphere; as would both Earth and the moon coalesce with the sun, if> they got close enough to it.---Outgoing mail is certified Virus Free.Checked by AVG anti-virus system (http://www.grisoft.com).=== Subject: : Angular measureMost of you are probably familiar with attempts to decimalize time,and circles: I think that so many problems were involved that thewhole idea has finally been abandoned.Besides; time and circles have their _own_ decimal systems': Theduodecimal; base twelve system for non-digital, analogue clock faces,and the sexagesimal; base 60 system by which circles, and Earth'sequator are divided into 360 degrees, and _decimals_ thereof.If they knew the world was somewhere near round, they might have donebetter with an angular measure, instead of choosing the meter as thestandard for linear length, for all men for all time.Not that there is anything really wrong with using meters for length;but a meter stick's just not as convenient to manipulate as a footruler, and a cubic meter is even much less convenient to handle than acubic foot. A cubic foot of water weighs about 62.4#, and that'spretty hefty: Imagine trying to jockey a cubic meter of water around.That's more than 27 cubic feet isn't it?Anyway angular measure has its uses; especially for time. Linearmeasure can be represented as _linear_ length on circles, and/or as_angular_ circular measure:This can be accomplished with plane trigonometry too: With which wecan plot various distance and angular aspects of plane circles onmutually perpendicular x and y coordinates; to get the length ofcurves and tangents, as well as their angular changes in direction.Maybe a millionth of one degree as measured on Earth's surface wouldhave been a good standard for length? How long would that be?Anyways, a meter's inconveniently long for measuring short distances,and the decimeter or centimeter's too short: Square and cubic metersare just too inconvenietly much for almost _all_ practical purposes.=== Subject: : Re: Angular measure> Most of you are probably familiar with attempts to decimalize time,> and circles: I think that so many problems were involved that the> whole idea has finally been abandoned.If it was your hallucination, Dumb Donny Head, it ws aparticularly inferior hallucination.> Maybe a millionth of one degree as measured on Earth's surface would> have been a good standard for length? How long would that be?Idiot. A meter is a poorly measured one-second pendulum. > Anyways, a meter's inconveniently long for measuring short distances,> and the decimeter or centimeter's too short: Square and cubic meters> are just too inconvenietly much for almost _all_ practical purposes.Idiot. 1984, by George Orwell, Part 1 Chapter 8: 'E could 'a drawed meoff a pint, grumbled the old man as he settled behind his glass. A'alf liter ain't enough. It don't satisfy. And a 'ole liter's toomuch. It starts my bladder running. Let alone the price.Poor Dumb Donny Head - he isn't even expert or original at beingstooopid.http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net!=== Subject: : Question about derivativesI've come across the following derivative in my work (I'm goingto present it in LaTeX formalism for lack of a better idea): D^i_j = frac{partialdot{y}^i}{partialdot{x}^j}Locally, my 'y's are linear combinations of the 'x's, and thecoefficients in those combinations are constant in time, so I'vetaken: D^i_j = frac{partial y^i}{partial x^j}which has worked quite well. I have a hunch that I couldgeneralize that, though, so my question is this. Under whatcondition are the derivatives above equivalent?=== Subject: : Force and accelerationForce and AccelerationWhen a net force [f] is exerted on, and/or by an object it produces anacceleration [a] according to tne equation: a = (1/m)f.This can be transposed to: m = f/a :: which is a ratio of force toacceleration, and is equal to that particular force called weight [w],divided by that particular acceleration [g] at which all bodies freefall at the location of the weight-scale on which they are weighed; sothat m = f/a = w/g!In the United States most of us use the pound as a unit of force. Thisunit is the source of much confusion. It is important to make adistinction between two entirely different concepts -- mass andweight.Mass is an intrinsic property of an object -- its the measure of itsresistance to acceleration: m = f/a = w/g. An object always has thesame mass regardless of where it is.An object's weight, however, depends on its mass and its location. Themass of any object'on the surface of the moon weighs one-sixth of itsearth-weight and an object free falling in space has no _weight_.The pound is a unit of force and weight but many people confuse itwith a unit of mass. A pound is equivalent to 4.448 newtons of force.The kilogram is _one unit_ of mass, because its weight [w] isnumerically equal to the acceleration [g] at which it free falls;whether it is free falling in space, or being restrained by aweight-scale on terra firma: To repeat, the numerical _ratio_ of itsweight [w], to the acceleration [g] at which it will free fall isalways equal to ONE; wherever it may be!=== Subject: : Re: Force and acceleration>In the United States most of us use the pound as a unit of force. Wrong. I'll bet it's not even true for you and your good buddy UncleAl.Like I told Uncle Al in another thread, you may be slightly smarterthan he is, since you actually do know that pounds are units of mass.But when you get down to the nitty-gritty, as dishonest, lyingbastards, you and Uncle Al are two peas in a pod.http://www.ngs.noaa.gov/PUBS_LIB/FedRegister/FRdoc59-5442. pdf Announcement. Effective July 1, 1959, all calibrations in the U.S. customary system of weights and measures carried out by the National Bureau of Standards will continue to be based upon metric measurement standards and, except those for the U.S. Coast and Geodetic Survey as noted below, will be made in terms of the following exact equivalents and appropriate multiples and submultiples: 1 yard= 0.914 4 meter 1 pound (avoirdupois)= 0.453 592 37 kilogramMy ketchup bottle: 1 lb 8 oz. Mass, not force.My bag of oranges: 5 lb. Mass, not force.Btu: the amount of heat necessary to raise 1 lb by 1 F. One poundof mass, not of force.Latent heat of fusion of water: 144 Btu/lb. Pounds mass, not poundsforce.A bushel of soybeans on the Chicago Board of Trade: 60 lb, poundsmass not pounds force.Density of steel: 0.0283 lb/in3. Pounds mass, not force.Putting the shot: womens' shot, 4 kg mass; men's shot, 16 lb mass.Maximum limit of a bowling ball: 16 lb mass.Weight of the Apollo 11 Lunar Module at liftoff, 10 776.6 lb. Poundsmass, not pounds force.http://history.nasa.gov/SP-4029/Apollo_18-37_Selected_ Mission_Weights.htmPound mole is to the gram mole (the mole of SI) as a pound is to agram.>This unit is the source of much confusion. You got that part right. So stick to the metric system. At least themetric system is still fully supported and updated. The keepers ofour standards have been telling us for nearly half a century to stopusing kilograms force.Pounds force are a recent bastardization, a unit never well definedbefore the 20th century. They are also uniquely identified by thatname--of all the hundreds of different pounds used at various timesand places throughout history, only one has given rise to a unit offorce of the same name.But the English units are like old software. They might have someutility for your own use, but eventually you might want to communicatewith the rest of the world, and that's when you run into problems.Nobody is ever going to bother telling us to stop using pounds force,without telling us to stop using pounds all together.http://ourworld.compuserve.com/homepages/Gene_Nygaard / It's not the things you don't know what gets you into trouble. It's the things you do know that just ain't so. Will Rogers=== Subject: : Re: Force and acceleration> Force and Acceleration> When a net force [f] is exerted on, and/or by an object it produces an> acceleration [a] according to tne equation: a = (1/m)f.[sip]Spewing idiot. You ing STILL don't know the differences amonggravitational, inertial, active, and passive masses. You cannot beeducated. You are a senile oaf.http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net!=== Subject: : Re: Force and acceleration> Force and Acceleration> When a net force [f] is exerted on, and/or by an object it produces an> acceleration [a] according to tne equation: a = (1/m)f.>[sip]>Spewing idiot. You ing STILL don't know the differences among>gravitational, inertial, active, and passive masses. You cannot be>educated. You are a senile oaf.That's no way to talk about your twin. At least he *knows* thatpounds are units of mass, even though he is dishonest and pretendsotherwise.http://ourworld.compuserve.com/homepages/ Gene_Nygaard/=== Subject: : Angular measureMost of you are probably familiar with attempts to decimalize time,and circles: I think that so many problems were involved that thewhole idea has finally been abandoned.Besides; time and circles have their _own_ decimal systems': Theduodecimal; base twelve system for non-digital, analogue clock faces,and the sexagesimal; base 60 system by which circles, and Earth'sequator are divided into 360 degrees, and _decimals_ thereof.If they knew the world was somewhere near round, they might have donebetter with an angular measure, instead of choosing the meter as thestandard for linear length, for all men for all time.Not that there is anything really wrong with using meters for length;but a meter stick's just not as convenient to manipulate as a footruler, and a cubic meter is even much less convenient to handle than acubic foot. A cubic foot of water weighs about 62.4#, and that'spretty hefty: Imagine trying to jockey a cubic meter of water around.That's more than 27 cubic feet isn't it?Anyway angular measure has its uses; especially for time. Linearmeasure can be represented as _linear_ length on circles, and/or as_angular_ circular measure:This can be accomplished with plane trigonometry too: With which wecan plot various distance and angular aspects of plane circles onmutually perpendicular x and y coordinates; to get the length ofcurves and tangents, as well as their angular changes in direction.Maybe a millionth of one degree as measured on Earth's surface wouldhave been a good standard for length? How long would that be?Anyways, a meter's inconveniently long for measuring short distances,and the decimeter or centimeter's too short: Square and cubic metersare just too inconvenietly much for almost _all_ practical purposes.=== Subject: : FrictionFriction is the force that causes resistance to surfaces sliding overeach other: Doing work against this resistance requires that a _net_force be exerted. This net force must be greater than theweight-force, or other normal force that is holding the surfacestogether:When doing work against friction, we have to exert a _net_ force thatis stronger than the static frictional force that is holding - bonding- the two surfaces together. This is a function - coefficient - of thesurface roughness, and is greatest when the two surfaces are at restrelative to each other.Once the surfaces start to slid, this static friction eases off a tinybit - perhaps due to momentum - to what is then called slidingfriction: Then it's only necessary to maintain enough force to keepthem moving. === Subject: : digraphs and graphsI have a few newbie questions. Is it possible to have a scenario where you have some directed edgesand other undirected edges in a single diagram?Like combining a digraph with a graph so to speak (as in union of adigraph and graph)If that IS possible, what is the process and the object resulting fromthat process called?What are some common properties of a digraph that are different fromthose of a graph?can some of you experts in graph theory shed some light on thismatter? and if you can recommend a good introductory book on thesubject matter, which would you reccomend?thank you so much for your time and toughts in advance.