mm-116 Let S(m,n) = an unsigned Stirling number of the 2nd kind.( S(0,0) =1, S(0,n) =0 if n not 0,S(m+1,n) = n*S(m,n) +S(m,n-1) )Let b(m) = sum{k=0 to oor((m-1)/2)} S(m,2k+1) (-1)^kAnd let c(m) = sum{k=0 to oor(m/2)} S(m,2k) (-1)^kThen:For p = any prime,p divides (b(p+2) -b(p+1) +b(p))andp divides (c(p+2) -c(p+1) +c(p) +2)This is a direct consequence of the result:If a(m) = q*b(m) + r*c(m),then: a(0) = r, a(1) = q,and a(m+2) = a(m+1) - sum{k=0 to m} binomial(m,k) 2^(m-k) a(k).This a-sequence result is found at:http://mathforum.org/discuss/sci.math/a/t/389070Leroy Quet =There is no such thing as a model containing all ordinals DUMBASSHow can we know V>L?stop equivocating on V George Greene has been arguing there that we can know V>L, sincethere are models of ZFC + V>L, and so the non-constructible sets fromsuch models show there really are non-constructible sets. Daryl McCullough has been arguing against this, noting it is possible forthe construction of ZFC + V>L models (under suitable consistency pemises)to be itself done in L, undermining the argument that there must really benon-constructible sets. I will write about further ways constructibility properties can becomplicated between models. Daryl points out it is possible to have a smaller M0 ZFC model with aset it believes to be non-constuctible, being a member of a larger M1 ZFCmodel making the set constructible. So non-constructibility can change toconstructibility when passing to a larger universe. For well-founded small models M0 this does not reverse. If x is a set inM0, and M0 |= x is constructible, and M0 is a member of a larger ZFCmodel M1, and M1 |= M0 is a standard transitive model, thenM1 also |= x is construcible. Namely the Godel construction of x is stillin M1, and M1 still recognizes this as such a construction. For M0 a non-well-founded ZFC model though, this can change. We needsome conventions to even make sense of the inclusions needed to state theseproperties. So we are considering models M0 of form , where N0 is a set(the universe of M0), and E is some binary relation on N0, to interpretthe epsilon symbol from the language of ZF, but we allow for thepossibility the E is not the true epsilon on N0. We want to talk about x a set in M0, its properties there, and thenconstrast them with xs properties in a larger M1 model. In general wecant relate x as it sits in M0 with its E members, to the true set xwith its true epsilon members. We can make sense of this in a special case of how M0 sits in M1 though,which will be the case under discussion. Namely any M0 = ZFCmodel has a well founded part, namely those elements with E-transitiveclosure being well-founded with respect to E. So we will consider the case of M0 a member of a larger M1 ZFC with M1satisfying that M0 has the additional property I will call s-t-wf, namelystandard transitive wellfounded, which is that on the well-founded part ofM0 E is the true epsilon, and that well-founded part of N0 is a transitiveset with respect to either E or epsilon, same thing. Then we can consider x from the well-founded part of M0, as this isdened in M1, and for these we can relate the true x from M1 to how thatx is seen in M0. So with all this background, it is possible to have x a member of theweell-founded (in the sense of M1) part of M0, with M0 a s-t-wf ZFCmodel with nonstandard E membership relation, this M0 being a set in M1and having those properties in M1, such that M0 models x isconstructible and M1 models x is non-constreuctible. Ie the reversedirection from above, the one that couldnt happen for standardtransitive models. Namely, x is from the M1-wf part of M0, but the Godel construction of xmaking M0 think x is constuctible comes from the non-wf part of M0, so M1does not recognize that as a real construction, and in fact M1 provides noalternative constructions. Here is a construction of such examples M0 and M1 (under suitableconsistency assumptions). As I recall, there is a theorem given inBarwises Admissible Sets and Structures, and application of the Barwisecompactness theorem. I dont have that book at hand, so I am going bymemory, but I also think I reconstructed a proof myself. Anyway, if I amremembering this wrong and my reconstruction is also wrong, this part will So here is the theorem. I will use strange looking numbers in thestatement to make things line up with how I will apply it in a moment. Before stating it I will dene that for ZF models M-1 = and M0 = , M0 end extends M-1 is dened:N0 superset N-1 and E-1 = E0 restricted to N-1,and if m1 in N-1 and m0 in N0 and m0 E0 m1 then m0 in N-1 (ie m1 from the little model doesnt get any new memers in the big model). Suppose M-1 (ie M with subscript or whatever -1 = the integer minus1) suppose M-1 = is a countable model of ZF. (Yes I thinkwe can do this with just ZF, making this even more remarkable). Then thereis M0 = a model end extending M-1, with M0 |= ZFC + V=L . So ie, given any M-1 model, possibly V>L, possible ~AC even, by addingextra garbage to the top of the model, not changing what the original setswere, just creating new higher sets, you can make it look like a V=L model. So we apply that theorem for the orignal observation, from suitablepremises. Suppose we assume we have M1 a ZFC model with x a set in M1and M1 |= x is not constructible. (So we assume M1 |= V>L). Supposealso N-1 is a set in M1 and x is in N-1and letting M-1 = , suppose M1 |= M-1 |= ZFC andalso M1 |= N-1 is countable transitive. Then we apply the above theorem in M1, obtaining M0 containing xand |= V=L. So M0 |= x is constructible (since M0 thinks everything is). ButM1 |= x is not constructible. Using the same construction aove to replace half the members of asequence, having rst preprocessed an assumed sequence to make all modelscountable in the next model, I think I have proved the result below, sayingit can be tricky how moving to bigger universes changes whether things seemconstructible. To state the result, I must dene a certain type of sequence of ZFCmodels. By an accumulating tower of ZFC models I mean an ordinally indexedsequence of ZFC models, each s-t-wf, s.t. each model in the towerconstains a member of its well-founded part the restriction of the tower tobelow that location. So the main result:In ZFC, suppose also that x is hereditarily countable(ie every member of the transitive closure of {x} is countable).Suppose also that x is not constructible.Suppose (aleph_1)^L = (aleph_1)^L[x]Suppose there is asequence of ordinals in length (aleph_1)^L,such that each of those ordinals alpha has L_alpha[x] |= ZFC.Then there is an accumulating tower of countable ZFC models in length(aleph_1)^L, with x a member of the well-founded part of every model inthe tower, such that the question of whether x is constructible zigzagsthroughout the tower adjacently: ie even ordinals versus odd ordinals. So the tower changes its mind (aleph_1)^L many times, trying to decideif x is constructible. Namely the idea here is adjust the original sequence of L models, alwayspicking the smallest ordinal making the next model of ZFC making allprevious models countable, and making the initial segment of the sequencelength countable. Inductively, the next ordinal doing that is alwayscountable in L, (by a collapsing argument), so that keeps the inductiongoing. So we obtain a tower, where each part sees everything below as countable. Now on every second step, do that Barwise theorem, replacing the originalmodel by an extension making x look countable. Do this theorem inside thenext model in the tower. Replace those steps by their extensions. We use the sequence has length (aleph_1)^L to see that the originalstep could produce the nextt structure countable in L, so that stayscountable later to apply Barwise (or whoever proved it: I say Barwisebecause I saw it in that book). The Barwise result needs (as stated) that the starting structure iscountable. To make an example showing this, I will work inZFC + x is a non-construcible real + aleph_1 = (aleph_1)^L. By Cohensmethods, this theory is consistent if ZFC is. So if a ZFC proof couldgeneralize Barwise, apply it in that theory and clash against my examplebelow, showing that theory inconsistent, and hence by Cohen ZFCinconsistent. So working in that theory, suppose the starting model M-1includes in its well-founded part all countable^L ordinals (ie so makingthis a case not claimed in the actual theorem). Then, if we could applythe theorem still, we would obtain an end extension M0 making xconstructible. By Godels proof of CH^L, applied in M0,we would have some countable^M0 ordinal stage of L constructing x. Since x is not constructible in our working theory, the M0 constructionof it must come from the nonconstrucible part of M0. M0 end extends M-1,and M-1 contains all countable^L ordinals, so M0 has ordinals itbelieves countable beyond all the true countable^L ordinals. So M0 provides a map from omega with range including all the truecountable ordinals. Our outer working theory can extract that map from M0,and obtain a countable enumeration of all the true countable^L ordinals. But this contradicts aleph_1 = (aleph_1)^L from our working theory. That concludes the example showing Barwise needed the starting modelcountable. So my tower result also needs the tower is only length (aleph_1)^L, moreor less. Because the tower includes the sequence leading up to its levelas a member, at each stage. So the well-founded parts of the tower keepgetting longer. So if you continued a tower like this past (aleph_1)^L, the modelsappearing there would include (aleph_1)^L in their wellfounded part. So, depending on where you put the ones constructing x (even or odd), at (aleph_1)^L or (aleph_1)^L + 1, you would hit a structure saying x isconstructible, but at least in the special case above of aleph_1 =(aleph_1)^L, a model with such big well-founded part cant be mistakenunderlying theorem. So the same issues show the tower result has the same limitations.--David Libert (ah170@freenet.carleton.ca)1. I used to be conceited but now I am perfect.2. So self-quoting doesnt seem so bad. -- David Libert3. So dont be a morron. -- Marek Drobnik bd308 rhetorical salvo IRC sig =How does one say critical point in French? A critical point is apoint, on the plot of a function, where the derivatve is zero. How does one say critical point in French? A critical point is a> point, on the plot of a function, where the derivatve is zero.un point number of hits.have the words minimal and maximalExample: maximal&lr=lang_frDirk Vdm =Stephen J. Herschkorn a .8ecrit dans le> How does one say critical point in French? A critical point is a> point, on the plot of a function, where the derivatve is zero.Most of us say point critique.-- Julien Santini,CMI Technop.99le de Ch.89teau-Gombert, FranceHome page: http://www.analgebra.com =You meant of course critical point in Freedom. =Th> How does one say critical point in French? A critical point is a> point, on the plot of a function, where the derivatve is zero. =This denition is not correct. A point A is also a critical point if thederivative at A is zero.> How does one say critical point in French? A critical point is a> point, on the plot of a function, where the derivatve is zero. This denition is not correct. A point A is also a critical point if the> derivative at A is zero.> How does one say critical point in French? A critical point is a> point, on the plot of a function, where the derivatve is zero.>Eh? Steven, could you run that by again? This denition is not correct. A point A is also a critical point if the>derivative at A is zero.>> How does one say critical point in French? A critical point is a>> point, on the plot of a function, where the derivatve is zero.A point A is also a critical point if the function is discontinuous at Aor the derivative does not exist at A. For a function on a closed boundedinterval, the endpoints of the interval can also be regarded as critical points since the function has to be evaluated at endpoints aswell as internal critical points if absolute maxima and minima are to befound.David McAnally-------------- = > How does one say critical point in French? A critical point is a > point, on the plot of a function, where the derivatve is point, stationary pointand extreme point.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =Sorry, I meant to write that a point where the derivative is UNDEFINED isalso called a critical point.>> This denition is not correct. A point A is also a critical point ifthe> derivative at A is zero. How does one say critical point in French? A critical point is a> point, on the plot of a function, where the derivatve is zero.> Eh? Steven, could you run that by again?>> =I have had all I can take of the cruel treatment given to me by theparticipants of sci.math. I have decided to terminate my life and letthe greater cosmos consume my spirit. I expect that a lot of you willdance and sing, but perhaps others may feel the cold chill of thedeath knell. If you want to communicate with me further, send aletter to James Harris, Heaven. To communicate with my opponents, youcan send them letters now, or just wait until the addressing iseasier, for example, David Ullrich, Hell.JSH I have had all I can take of the cruel treatment given to me by the>participants of sci.math. I have decided to terminate my life and let>the greater cosmos consume my spirit.Another unfunny forgery. Same one?Headers: =IP belongs to blg7-48.imt.net. imt.net is an ISP in Montana.Anybody else see anything else interesting in here? - Randy The rst few values are>> n n_cut> 1 1> 2 8 3 26 (1 central cube of 27 not hit)> 4 56 (8 central cubes of 64 not hit)>> Looking up [1,8,26,56] in the On-Line Encyclopedia of IntegerSequences> gives> http://www.research.att.com/projects/OEIS?Anum=A005897 :> Points on surface of cube: 6n^2 + 2 (coordination sequence for b.c.c.> lattice).>> The continuation of the sequence is:> 1,8,26,56,98,152,218,296,386,488,602,728,866,1016,1178,1352,. ..>> Ah... for n=6 that sequence gives 152. This is the mean of two answers(176> and 128) that I gave in another post in this thread and corresponds to> viewing the cubi as neither open nor closed. I have no idea how the> expression 6n^2+2 arises, though.>> 6*n^2+2 is exactly what the sequence title says:> for clarifying this.>> A 1*1*1 cube has 6*1+2 (lattice) points> 2*2*2 -> 26 points> 3*3*3 -> 56 points> n*n*n -> 8 + 12*(n-1) + 6*(n-1)^2 = 6*n^2 + 2> vertices points on edges pts on faces>> Number of cut sub-cubes (cut means not only touched)> = the number of lattice points on the surface of the> (n-1)-subdivided cube, but only for n<7.>> See Dave Seamans message and my reply.the surface of a cube. [Most easily thought of, I suppose, as(n+1)^3-(n-1)^3.] When I said I have no idea how the expression 6n^2+2arises I (foolishly) meant I have no idea why 6n^2+2 is the solution tothe original posters question. I had not realised that it only applies forn<7.-- Clive Toothhttp://www.clivetooth.dk > I have now written a little Fortran program available at> http://www.randomwalk.de/sequences/cutcub.txt ,> including results for n<=100.>> Here are the rst few terms:>> 3 26> 4 56> 5 98> 6 152> 7 194> 8 272> 9 362> 10 440> 11 530> 12 656> 13 746> 14 872> 15 1034> 16 1160> 17 1298> 18 1496> 19 1658> 20 1856>> I have counted cubes only touched by the cutting sphere, with all> other vertices either inside or outside as 1/2, because they occur> always as inside/outside pairs. If either > or >= is used> in the radius comparisons then I can reproduce both variants of> Daves n=6 and n=10 results.>> I have downloaded your Fortran program and checked it against my lisp> program (modied to print the average of the upper and lower results).> They agree.I have written a C# program and it agrees with both of you for n<=100.I have included the relevant class declaration at the end of this post.Summary of the problem and some suggested notation:Let n be a positive integer. Consider a cube C, of side n, and its insphereS. Consider the set of n^3 unit cubes, each of side 1, whose faces areparallel to the faces of C and whose union is C. Each unit cube is closed,that is, it includes its interior, and all its faces, edges and vertices.Thus, in general, a unit cube will intersect each of its 6 surrounding unitcubes in a square and, in general, each vertex is part of 8 unit cubes.For each unit cube we may compute an intersection value as follows:# if all the vertices of the unit cube are inside S, the intersection valueis 0.# if all the vertices of the unit cube are outside S, the intersection valueis 0.# if at least one vertex is inside S and at least one vertex is outside S,the intersection value is 1.# otherwise the unit cube has at least one vertex on S and all its othervertices are either all inside or all outside S. In this case itsintersection value is 1/2.Let the sum of the intersection values across all n^3 unit cubes be T(n).So, for example, T(1)=0, T(2)=8 and T(6)=152.The area of S is 4*pi*(n/2)^2 and we might expect T(n) to be about thisvalue, ie pi*n^2.Computing T(n) and T(n)/(pi*n^2) for various n we get:| n T(n) T(n)/(pi*n^2)|| 1 0 0.000000| 2 8 0.636620| 3 26 0.919562| 4 56 1.114085| 5 98 1.247775| 6 152 1.343975| 7 194 1.260247| 8 272 1.352817| 9 362 1.422570| 10 440 1.400563|| 100 47,000 1.496056| 101 48,002 1.497844| 102 48,992 1.498908| 103 49,826 1.494967| 104 50,936 1.499023| 105 51,770 1.494685| 106 52,808 1.496022| 107 53,810 1.496048| 108 54,848 1.496799| 109 55,850 1.496306| 110 56,816 1.494636|| 200 188,432 1.499494| 201 190,202 1.498556| 202 192,056 1.498219| 203 193,970 1.498279| 204 196,040 1.499459| 205 197,810 1.498272| 206 199,880 1.499288| 207 201,674 1.498164| 208 203,648 1.498317| 209 205,682 1.498835| 210 207,680 1.499016|| 500 1,177,832 1.499662| 501 1,182,698 1.499852| 502 1,187,072 1.499408| 503 1,192,082 1.499755| 504 1,196,672 1.499561| 505 1,201,514 1.499672| 506 1,206,152 1.499516| 507 1,211,114 1.499751| 508 1,215,704 1.499514| 509 1,220,498 1.499518| 510 1,225,496 1.499760|| 1,000 4,712,000 1.499876| 1,001 4,721,378 1.499860This suggests that T(n) is asymptotic to (3/2)*pi*n^2.I assume this is all well-known.-- Clive Toothhttp://www.clivetooth.dk public class CubeletsClass { private int Inside; private int N; public int NIncrement; private int NN; public int NStart; public int NStop; private int On; private int Outside; public CubeletsClass(int Start, int Increment, int Stop) { NStart= Start; NIncrement= Increment; NStop= Stop; } // constructor private void Look(int x, int y, int z) { int a= U.Square(2*x-N)+U.Square(2*y-N)+U.Square(2*z-N); if (aNN) { ++Outside; } else { ++On; } } // Look public void Counter() { int Count; int x; int y; int z; for (N= NStart; N<=NStop; N+= NIncrement) { Count= 0; NN= N*N; for (x= 0; x0 && Outside>0) { Count+= 2; } else if (On>0) { Count+= 1; } else { // } } // for z } // for y } // for x if (Count%2!=0) { U.Fail(Count%2<>0); } Count/= 2; U.Emit( U.Commas(N, 5)+ + U.Commas(Count, 11)+ + U.FNum(Count/(Math.PI*NN), 6)+ ); } // for N } // Counter } // CubeletsClass =Suppose X,Y,Z are postive random variable, the pdf are given as f_X(t),f_Y(t), f_Z(t). andZ=min(X, Y)I know the following method to derive the cdf of Z.P(Z t ) = 1 - [ P( X (vertices lie on a circle) for which all the sides are rational numbers?> What if it is also required that the radius of the circle be 1, does that> change the answer?>How is a side a rational number? Here was the question. Can you characterize those cyclic quadrilaterals> (vertices lie on a circle) for which all the sides are rational numbers?> What if it is also required that the radius of the circle be 1, doesthat> change the answer?>> How is a side a rational number?Sorry, I should be more precise: Can you characterize those cyclicquadrilaterals for which all the lengths of the sides are rational numbers?Or you could ask the question in terms of integers. Recently, out of curiosity, as an experiment, I posted a message to>alt.math.moderated. It was a good, sensible message, not one begging to be>refused. Nothing ever came of it. I think the moderator of>alt.math.moderated may be dead, may he rest in moderate peace.>Here was the question. Can you characterize those cyclic quadrilaterals>(vertices lie on a circle) for which all the sides are rational numbers?>What if it is also required that the radius of the circle be 1, does that>change the answer?If the radius is unspecied, given any side lengths l_1,...,l_4 whichcan form a quadrilateral (i.e. the longest side is less than the sum ofthe other three) there will be always be some radius r for which you get a cyclic quadrilateral with those side lengths. This will satisfyeither f(r) = sum_{i=1}^4 arcsin(l_i/(2r)) - Pi = 0 ( for a cyclic quadrilateral with centre inside )or (if l_4 is the longest)g(r) = arcsin(l_4/(2r)) - sum_{i=1}^3 arcsin(l_i/(2r)) = 0( for a cyclic quadrilateral with centre outside ). One proof uses the Intermediate Value Theorem. Note f(r) < 0 and g(r) < 0when r is large, while f(l_4/2) = sum_{i=1}^3 arcsin(l_i/l_4) - Pi/2 = - g(l_4/2).If r = 1 is given, the relationsum_{i=1}^4 arcsin(l_i/2) = Pican be expressed algebraically: if t_i = l_1/sqrt(4 - l_i^2), thent_1 + t_2 + t_3 + t_4 + t_2 t_3 t_4 + t_1 t_3 t_4 + t_1 t_2 t_4 + t_1 t_2 t_3 = 0and (I think) 3 2 2 2 2 2 2 2 2 2 l_1 l_2 l_3 l_4 + l_1 l_2 l_3 + l_1 l_2 l_4 + l_1 l_3 l_4 3 3 3 + l_1 l_2 l_3 l_4 + l_1 l_2 l_3 l_4 + l_1 l_2 l_3 l_4 2 2 2 4 2 2 2 2 + l_2 l_3 l_4 + l_1 - 2 l_1 l_2 - 2 l_1 l_3 2 2 4 2 2 - 2 l_1 l_4 - 8 l_1 l_2 l_3 l_4 + l_2 - 2 l_2 l_3 2 2 4 2 2 4 - 2 l_2 l_4 + l_3 - 2 l_3 l_4 + l_4 = 0I dont know how youd go about characterizing the rational solutions of this.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 Here was the question. Can you characterize those cyclic quadrilaterals>(vertices lie on a circle) for which all the sides are rational numbers?>What if it is also required that the radius of the circle be 1, does that>change the answer?This may help:According to Ptolemys theorem, q.v., given any quadrilateral with successive vertices A,B,C,and D, it will be cyclic if and only if AB*CD + AD*BC = AC*BD. It plots and analyses any x-y data for peak location, peak height, peak> width, semi-derivative, derivative, integral, semi-integral,convolution,> deconvolution, curve tting, and separating overlapped peaks and> background.>> www.chemSoftware.com> =to contain the denition for 1-norm and 2-norm distance (a.k.a Hamming andEuclidean distance). Or maybe you can point me to the relevant math to contain the denition for 1-norm and 2-norm distance (a.k.a Hamming and> Euclidean distance). Or maybe you can point me to the relevant math> All those norms are in functional analysis, but probably also in moreelementary courses like real analysis. See if H.9alders Inequalityis in the index of the book.example...K. Stromberg, _An Introduction to Classical Real Analysis_ (1981),page 341 =Some time ago, in Google Group es.ciencia.matematicas wasposted by Leon-Sontelo a nice problem : = == Prove that for any integer n, exists a unique polynomial Q(x) with coefcients in the set {0,1,2,...9}, with Q(-2) = Q(-5) = n = Contributions were made by Ignacio Larrosa Ca.96estro andGerry Meyerson See : .news.dfncis.de .news.dfncis.deFurther let Z be the set of all integers,and M be the set of integers n which are not in set {0,1,2,3,4,5,6,7,8,9}.Denote by m=m(n) the degree of Q(x) with all coefcientsin {0,1,2,3,4,5,6,7,8,9} such that Q(-2)=Q(-5)=n , n in Z.Its clear that the function m: M-->N ,N being the set of naturalnumbers, is well-dened.QUESTION: Find bounds for m(n) , to evaluate m(n). ==Alex.Lupas time ago, in Google Group es.ciencia.matematicas was> posted by Leon-Sontelo a nice problem :> == Prove that for any integer n, exists a unique polynomial Q(x)> with coefcients in the set {0,1,2,...9}, with Q(-2) = Q(-5) = n> == ==Contributions were made by Ignacio Larrosa Ca.96estro and> Gerry Meyerson> See b621uv$11n6q$1@ID-137122.news.dfncis.de> Further let Z be the set of all integers,> and> M be the set of integers n which are not in set> {0,1,2,3,4,5,6,7,8,9}. Denote by m=m(n) the degree of Q(x) with all> coefcients> in {0,1,2,3,4,5,6,7,8,9} such that Q(-2)=Q(-5)=n , n in Z.>> Its clear that the function m: M-->N ,N being the set of natural> numbers, is well-dened.> QUESTION: Find bounds for m(n) , to evaluate m(n)> ==I saidQ(x) = (x + 2)(x + 5)P(x) + n = (x^2 +7x +10)P(x) + nThe cofcients of P arent necessary in {0,1,2,...9}, but they areintegers. Let b(n) to be the coefcient of x^n in P(x). Then,0 <= 10b(0) + n <= 9 > 0 <= b(0) + n/10 <= 9/10 ==b(0) = -[n/10](valid for n positive or negative)For b(1),0 <= 10b(1) + 7b(0) <= 9 >0 <= b(1) + 7b(0)/10 <= 9/10 ==b(1) = -[7b(0)/10]For b(2),0 <= 10b(2) + 7b(1) + b(0) <= 9 >0 <= b(2) + 7b(1)/10 + b(0)/10 <= 9/10 >b(2) = - [(7b(1) + b(0))/10]In general,b(k+2) = - [(7b(k+1) + b(k))/10], para k >= 0All b(n) are univocally determined, and for all n, they are zero from acertain k, because they hav alternate signs, and therefore, each one is, inabsolute value, less or equal than 7/10 of the previous.By example, for n = 196,b(0) = -[196/10] = -19b(1) = -[-7*19/10] = -[-13.3] = -(-14) = 14b(2) = -[(7*14 -19)/10] = -7b(3) = -[(7(-7) + 14)/10] = -(-4) = 4b(4) = -[(7*4 - 7)/10] = -2b(5) = -[(7(-2) + 4)/10] = -(-1) = 1b(6) = -[(7*1 - 2)/10] = 0b(7) = -[(7*0 + 1)/10] = 0and all of the rst b(n) are nulls. Then,P(x) = (x^2 +7x +10)(x^5 - 2x^4 + 4x^3 - 7x^2 + 14x - 19) + 196 = x^7 + 5x^6 + x^4 + 5x^3 + 9x^2 + 7x + 6The paragraphAll b(n) are univocally determined, and for all n, they are zero from acertain k, because they hav alternate signs, and therefore, each one is, inabsolute value, less or equal than 7/10 of the previous.it isnt fully exact, because the integer part of a negative number isgreater, in absolute value, that it.Then |b(k+1)| < (7/10)|b(k)| + 1but it isnt very important.Roughly, in a worst case, m(n) <= log(n)/log(10/7) + 2or something so. But it seems that frequently, it say the mean beahouvior,ism(n) ~ (log(n)/log(10/7) + 2)/2 ~= 1.401836626*ln(n) + 1 (#1)If you has Derive 5.0 or later, this function compute P(x) for any n:f(n) := Prog v := [0, - FLOOR(n/10)] k := 2 Loop s := - FLOOR((7*(v SUB k) + (v SUB (k - 1)))/10) If s = 0 exit v := APPEND(v, [s]) k :+ 1 u(x) := SUM((v SUB i)*x^(i - 2), i, 2, k)*(x^2 + 7x + 10) + n RETURN u(x)Or in one line,f(n) := PROG(v := [0, - FLOOR(n/10)], k := 2, LOOP(s := - FLOOR((7*(v SUB k)+ (v SUB (k - 1)))/10), IF(s = 0, exit), v := APPEND(v, [s]), k :+ 1), u(x):= SUM((v SUB i)*x^(i - 2), i, 2, k)*(x^2 + 7x + 10) + n, RETURN u(x))I calculate m(n) for n = 2^k, k = 1, ..., 100 withVECTOR([k, POLY_DEGREE(f(2^k))], k, 1, 100)and I got k -1 <= m(n) <= k +1 for all k >= 4.(if k < 4, 2^k < 10 and P(x) = n, with degree 0)If you replace n by 2^k in #1, you getm(2^k) ~= 0.9716791049k + 1in an excellent agree.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com Alex.Lupas escribi.97 en el es.ciencia.matematicas was> posted by Leon-Sontelo a nice problem :> == Prove that for any integer n, exists a unique polynomial Q(x)> with coefcients in the set {0,1,2,...9}, with Q(-2) = Q(-5) = n> == ==Contributions were made by Ignacio Larrosa Ca.96estro and> Gerry Meyerson> See :>> .news.dfncis.de>> .news.dfncis.de> Further let Z be the set of all integers, and> M be the set of integers n which are not in set> {0,1,2,3,4,5,6,7,8,9}. Denote by m=m(n) the degree of Q(x) with all> coefcients> in {0,1,2,3,4,5,6,7,8,9} such that Q(-2)=Q(-5)=n , n in Z.> Its clear that the function m: M-->N ,N being the set of natural> numbers, is well-dened.> QUESTION: Find bounds for m(n) , to evaluate m(n)> =m(n) ~ (log(n)/log(10/7) + 2)/2 ~= 1.401836626*ln(n) + 1 (#1)> If you has Derive 5.0 or later, this function compute P(x) for any n:> f(n) :Prog> v := [0, - FLOOR(n/10)]> k := 2> Loop> s := - FLOOR((7*(v SUB k) + (v SUB (k - 1)))/10)> If s = 0 exit> v := APPEND(v, [s])> k :+ 1> u(x) := SUM((v SUB i)*x^(i - 2), i, 2, k)*(x^2 + 7x + 10) + n> RETURN u(x)> Or in one line,> f(n) := PROG(v := [0, - FLOOR(n/10)], k := 2, LOOP(s := - FLOOR((7*(v SUB k)> + (v SUB (k - 1)))/10), IF(s = 0, exit), v := APPEND(v, [s]), k :+ 1), u(x)> := SUM((v SUB i)*x^(i - 2), i, 2, k)*(x^2 + 7x + 10) + n, RETURN u(x))> I calculate m(n) for n = 2^k, k = 1, ..., 100 with> VECTOR([k, POLY_DEGREE(f(2^k))], k, 1, 100)> and I got k -1 <= m(n) <= k +1 for all k >= 4.> (if k < 4, 2^k < 10 and P(x) = n, with degree 0)> If you replace n by 2^k in #1, you get> m(2^k) ~= 0.9716791049k + 1 in an excellent agree. want to work out the inverse of a 4 by 4 matrix? Obviously I need towork out the determeinat. How is this done in a four by four matrix?Richard I want to work out the inverse of a 4 by 4 matrix? Obviously I need to> work out the determeinat. How is this done in a four by four matrix?Get a book, lazy one. I want to work out the inverse of a 4 by 4 matrix? Obviously I need to> work out the determeinat. How is this done in a four by four matrix?> For any inversible matrix M:M^(-1) = transpose(co(M)) * 1/det(M)where co(M) is the matrix of the cofactors. Lets note each cofactor c_ij:c_ij = det((a_ij))* (-1)^(i+j)(a_ij) is the minor, that is, the matrix you get if you remove line i andcolumn j from M.The drawback of this method is that you need to calculate severaldeterminants, which gets long if your matrix is large.You may want look in a linear algebra book tofind others, which may in somecases be more adapted. Basically, if you have several of zeroes in yourmatrix, this method is likely to be easy, because youll be able to easilycalculate the necessary determinants. Otherwise, you may want to use anothermethod.Sam-- Giving the Linus Torvalds Award to the Free Software Foundation is a bit likegiving the Han Solo Award to the Rebel Alliance - Richard Stallman, August 1999 I want to work out the inverse of a 4 by 4 matrix? Obviously I need to> work out the determeinat. How is this done in a four by four matrix?Dont use that method... >I want to work out the inverse of a 4 by 4 matrix? Obviously I need to>work out the determeinat.No, you do not. A much easier way to do it is by Gaussian elimination.> How is this done in a four by four matrix?The usual method is expansion by minors. =Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu>I want to work out the inverse of a 4 by 4 matrix? Obviously I need to>work out the determeinat.>> No, you do not. A much easier way to do it is by Gaussian elimination.>Never heard of such. Anything like substracting rows from rows, etc.?> How is this done in a four by four matrix?>> The usual method is expansion by minors. No, you do not. A much easier way to do it is by Gaussian elimination.> Never heard of such. Anything like substracting rows from rows, etc.?Im not sure if thats the method hes referring to:Consider the matrices M and I (the identity matrix), and make linearcombinations so that M becomes I. Make every combination on both M and I.Youll end up with I and M^(-1).Sam-- People sometimes ask me if it is a sin in the Church of Emacs to use vi. Using a free version of vi is not a sin; its a penance. - Richard Stallman >> No, you do not. A much easier way to do it is by Gaussian elimination.> Never heard of such. Anything like substracting rows from rows, etc.?>>Im not sure if thats the method hes referring to:>>Consider the matrices M and I (the identity matrix), and make linear>combinations so that M becomes I. Make every combination on both M and I.>Youll end up with I and M^(-1).>>SamIts called Gauss-Jordan, it can be and solve with n vectors like(1,0,0,0),(0,1,0,0),(0,0,1,0) and (0,0,0,1). These vectors thenbecome the column vectors of the inverse.Both methods are fundamentally gaussian elimination as Arturo said. >>No, you do not. A much easier way to do it is by Gaussian elimination.>>Never heard of such. Anything like substracting rows from rows, etc.?>Im not sure if thats the method hes referring to:>>Consider the matrices M and I (the identity matrix), and make linear>>combinations so that M becomes I. Make every combination on both M and I.>>Youll end up with I and M^(-1).>>Sam>Its called Gauss-Jordan,>Or just >Its also (usually [almost always?]) the easiest way to compute a determinant -- just make the matrix upper triangular and multiply the diagonal elements.Jon Miller > No, you do not. A much easier way to do it is by Gaussian elimination.> Never heard of such. Anything like substracting rows from rows, etc.?>>Im not sure if thats the method hes referring to:>>Consider the matrices M and I (the identity matrix), and make linear>combinations so that M becomes I. Make every combination on both M and I.>Youll end up with I and M^(-1).Yup, thats the one. == ==Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu>I want to work out the inverse of a 4 by 4 matrix? Obviously I need to>>work out the determeinat.>> No, you do not. A much easier way to do it is by Gaussian elimination.> Never heard of such. Anything like substracting rows from rows, etc.?>> How is this done in a four by four matrix?>> The usual method is expansion by minors.You might want to take a look at:http://ceee.rice.edu/Books/LA/index.htmlIts a very introductory textbook for linear algebra (no calculus -- high school level). Chapter 5 covers determinants, but youll need chapter four to understand how to use Gaussian or Gauss-Jordan elimination.-Laurel =: > I want to learn the formula to calculate any digit in the decimal form of: pi.: > ex: What is the 100th digit in Pi?: > ans: 0: There is a way to do this for the binary/octal/hexadecimal/etc expansions of: Pi, but to my knowledge, no method exists (or at least, none is yet known): for the decimal expansion.Actually, they did extend the method for the hexadecimal expansion to onefor the decimal expansion later. But its still quite a complicatedmethod; only worth using if you want, say, the billionth digit in pi.John Savard : > I want to learn the formula to calculate any digit in the decimal form of> : pi.> : > ex: What is the 100th digit in Pi?> : > ans: 0> : There is a way to do this for the binary/octal/hexadecimal/etc expansions of> : Pi, but to my knowledge, no method exists (or at least, none is yet known)> : for the decimal expansion.> Actually, they did extend the method for the hexadecimal expansion to one> for the decimal expansion later. But its still quite a complicated> method; only worth using if you want, say, the billionth digit in pi.Id have thought it would pack up way before then. I thought it was O(n^3)time in decimal, and worse than pretty much any all-digits technique.Phil I want to learn the formula to calculate any digit in the decimal form ofpi.> ex: out:http://mathworld.wolfram.com/ Bailey-Borwein-PlouffeAlgorithm.html =Can someone help me out with understanding this question? Let A = {1, 2, 3, 4, 5, 6,12} and dene the relation R on A by m R niff m|n. Write the denitions of the properties, reexive, antisymmetric andtransitive and the use the denitions to determine whether eachproperty holds for this relation. (a) Is this relation a partial ordering relation? Why? If so, draw itsHasse diagram. (b) Write the (boolean, that is the yes/no) matrix of this relation. Im just about to log off, but as a quick partial answer, I can give you thefollowing:R={(1,1),(2,1),(2,2),(3,1),(3,3),(4,1),(4,2),( 4,4),(5,1),(5,5),(6,1),(6,2),(6,3),(6,6),(12,1),(12,2),(12,3) ,(12,4),(12,6),(12,12)}Thus R is transitive, because for any number (m) that is divisible by anumber (n) that in turn is divisible by another number (n), the originalnumber (m) must necessarily be divisible by the nal divisor (n).R is not however symmetric. . . To besymmetric would infer that if a number (m) is divisible by a number (n),then it would hold true that (n) is also divisible by (m), which from theabove set is obviously not true. . .As for reexive, R is. . . simply put the reexive property means that ifm=n is the set R, then R is reexive. . . Because any number is divisibleby itself, the set R is reexive.R is also antisymmetric. Because for any (m,n), there does not exist acounterpart (n,m) which is the denition of theantisymmetric property.I hope this helps and Im sorry I didnt get to answer the second part ofthe question...tjd>> Can someone help me out with understanding this question?>> Let A = {1, 2, 3, 4, 5, 6,12} and dene the relation R on A by m R n> iff m|n.>> Write the denitions of the properties, reexive, antisymmetric and> transitive and the use the denitions to determine whether each> property holds for this relation.>> (a) Is this relation a partial ordering relation? Why? If so, draw its> Hasse diagram.>> (b) Write the (boolean, that is the yes/no) matrix of this relation.>> help me out with understanding this question? Let A = {1, 2, 3, 4, 5, 6,12} and dene the relation R on A by m R n> iff m|n. I guess, this is clear so far.> Write the denitions of the properties, reexive, antisymmetric and> transitive and the use the denitions to determine whether each> property holds for this relation. > (a) Is this relation a partial ordering relation? Why? If so, draw its> Hasse diagram. A relation is a partial ordering if it is reexive, antisymmetric andtransitive.Given that you know the denitions of these properties, you need to checkif they hold for the relation above.So you need to check:(R) does a|a hold for all a in A ?(A) for every a,b in A, does a|b and b|a always force a=b ?(T) for every a,b,c in A does a|b and b|c always force a|c ?The good news is, that you might get the answers from your previousknowledge of divisibility.> (b) Write the (boolean, that is the yes/no) matrix of this relation.> I assume, they mean a 7x7 matrix, where rows and columns are indexedby the elements of A, and the entry in the (a,b)-position is1 if a|b and 0 if not a|b.Marc Im just about to log off, but as a quick partial answer, I can give you the> following:>[...]> I hope this helps and Im sorry I didnt get to answer the second part of> the question...> tjdOf course the original poster did not ask for answers.He wanted to understand the question.>> Can someone help me out with understanding this question?Marc =sorry I didnt give better denitions of the properties in addition to howthey pertained to the example. . . was very tired at the time of reply. . .But, if you have other questions, Ill try to be as helpful as I can.tjd>> Can someone help me out with understanding this question?>> Let A = {1, 2, 3, 4, 5, 6,12} and dene the relation R on A by m R n iff m|n.>> I guess, this is clear so far.> Write the denitions of the properties, reexive, antisymmetric and> transitive and the use the denitions to determine whether each> property holds for this relation.>> (a) Is this relation a partial ordering relation? Why? If so, draw its> Hasse diagram.>> A relation is a partial ordering if it is reexive, antisymmetric and> transitive.>> Given that you know the denitions of these properties, you need to check> if they hold for the relation above.>> So you need to check:>> (R) does a|a hold for all a in A ?> (A) for every a,b in A, does a|b and b|a always force a=b ?> (T) for every a,b,c in A does a|b and b|c always force a|c ?>> The good news is, that you might get the answers from your previous> knowledge of divisibility.> (b) Write the (boolean, that is the yes/no) matrix of this relation.> I assume, they mean a 7x7 matrix, where rows and columns are indexed> by the elements of A, and the entry in the (a,b)-position is> 1 if a|b and 0 if not a|b.>> Marc =1) Let m and k be any positive integers.Then:(1 +(m+1)m^k) always divides(H(m^(k+1)) - H(m^k)) (m^(k+1))! /(m^k)!,where H(n) = 1+1/2+1/3+...1/n, the n_th harmonic number.-2) Let F(m) be the m_th Fibonacci number.Then:(1 +2F(m+2)) always divides(H(F(m+3)) - H(F(m))) (F(m+3))! /(F(m))!This is not too fantastic, but I think it is worth mentioning (as allof my posts seem {to me} to be lately).Leroy Quet =A friend of director George Paul Csicsery asked me to pass this on.In the San Francisco Bay Area it is playing tonight. I haveno idea if/when it is on in other areas.Eddie------------------Another opportunity for Bay Area audiences to seeN is a Number: A Portrait of Paul Erd.9asKTEH-TV, Channel 54 (AT&T Broadband Cable 10)ABOUT PAUL ERD.85S:Paul Erd.9as was one of the most prolic mathematicians who ever lived.He pioneered several elds in theoretical mathematics, and publishedor co-authored more than 1,500 mathematical papers in a eld where50 good papers in a lifetime would be considered an achievement.Using interviews with Erd.9as himself, lm clips of Erd.9as lecturingand discussions with those who knew and were inuenced by him,N is a Number: A Portrait of Paul Erd.9as captures the life and workof this accomplished scholar and teacher. N is a Number:A Portrait of Paul Erd.9as provides insight into the mathematicalquest of this amazing man, the personal and philosophical dimensions ofhis work, and the tragic historical events that molded his extraordinary life.ABOUT THE FILM:N is a Number: A Portrait of Paul Erd.9as was produced, directed and edited byGeorge Paul Csicsery. Cinematography is by John Knoop, and original musicwas composed by Mark Adler. Animated sequences were created by Red DotInteractive. The lm was made with support from the American MathematicalSociety, Film Arts Foundation, the Heineman Foundation, the MathematicalAssociation of America, and the National Science Foundations InformalScience Education Program.N is a Number: A Portrait of Paul Erd.9as has won awards and appeared atnumerous festivals, including the Festival International du Nouveau Cin.8emaet de la Vid.8eo de Montr.8eal, the Budapest Film Week, the Berlin FilmFestival-European Film Market, the National Educational Film & VideoFestival, where it won a Gold Apple Award in 1994, the MelbourneInternational Film Festival, the Chicago International Film Festival (GoldPlaque Award, 1994), the Auckland-Wellington, New Zealand Film Festivals,Festival International du Film Scientique du Qu.8ebec and the Cork FilmFestival.N is a Number: A Portrait of Paul Erd.9as has been broadcast on Duna-TV, Hungary;SBS-TV, Australia; the Discovery Channel, US; Sundance Channel, US;NHK, Japan; Noorder Licht series, VPRO, Netherlands, and Discovery stationshttp://www.aptonline.org/catalog.nsf/ d14ac4ffee990491852566410073d4d1/ 12e3e7e2c9d3d87d85256bd7006e4445?OpenDocumentFor information, contact:George Paul CsicseryPO Box 22833Oakland CA 94609tel 510-428-9284fax 510-428-9273geocsi@compuserve.com A friend of director George Paul Csicsery asked me to pass this on.> In the San Francisco Bay Area it is playing tonight. I have> no idea if/when it is on in other areas.> Eddie> ------------------> Another opportunity for Bay Area audiences to see> N is a Number: A Portrait of Paul Erd.9as> Sorry - obviously plays in a week. =Can anyone determine whether the series sum (sin(n)^n/n) cos(nx)converges for all x? (Its clear that it converges foralmost all x.) Or for x = 0?Fascinating background information: It was ona test given by a colleague - cant tell from thetests we know was specied as a possible answer,and was the answer the guy who gave the testhad in mind. I dont see a simple way to settle thequestion - a certain studentfinds it hard to believethat we cant gure it out, and asked me to post the question here.So if you determine whether the series convergesfor all x not only do you have the exquisite pleasureof proving me wrong, you help a starving student(he gets a deluxe cappi..., cappa..., um, a fancy cup of coffee.)Deadline: August 15.Oh btw: Actually it looks familiar. If anyone recalls that essentially this series has appeared here recentlyand people agreed they didnt know the answerplease let me know - wouldnt seem fair for meto claim my prize if Id actually seen it before.************************David C. Ullrich =David C. Ullrich escreveu na mensagem> Can anyone determine whether the series>> sum (sin(n)^n/n) cos(nx)>> converges for all x?I think that it converges for all x in R, but Im not sure. Heres myargument.1. Let A = {x in R : x =/= 2k pi for all k in Z}Its know Dirichlet series sum cos(nx) / nconverges for all x in A.Its also know that another Dirichlet series sum sin(nx) / nconverge, this time for all x in R.Lets remember the following proposition. Proposition. If exists k such that |a_n / b_n| <= k for all n in N and the series sum b_n converges, then the series sum a_n converges.2. Let be any x in A xed. Then, with a_n = sin(n)^n / n cos(nx) b_n = cos(nx) / nwe have a_n / b_n = sin(n)^nSo, |a_n / b_n| <= k = 1 for all n in N and by the proposition and the factthat the series sum cos(nx) / nconverge (because x in A), we conclude that sum (sin(n)^n / n) cos(nx)converges.Because we have choose any x in A, we conclude that sum (sin(n)^n / n) cos(nx)converges for all x in A.3. Now we only have to see what happens to x in R A (this is, x in R butnot x in A). Let be any k in Z and x = 2 pi k. Then cos(nx) = cos(2 pi kn) = 1so sum (sin(n)^n / n) cos(nx) = sum(sin(n)^n / n)Then, with a_n = sin(n)^n / n b_n = sin(n) / nwe have a_n / b_n = sin(n)^(n - 1)So, |a_n / b_n| <= k = 1 for all n in N and by the proposition and the factthat the Dirichlet series sum sin(nx) / nconverge for x = 1, we conclude that sum(sin(n)^n / n)converge, this is, sum (sin(n)^n / n) cos(nx)for all x such that x = 2 pi k for some k in Z (this is, for all x in R A)4. We have showed that the series sum (sin(n)^n / n) cos(nx)converge for all x in A and for all x in R A. Therefore, the seriesconverge for all x in R.I hope that I have helped.Sorry my english. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com Lets remember the following proposition.> Proposition. If exists k such that |a_n / b_n| <= k for> all n in N and the series sum b_n converges, then the> series sum a_n converges.counterexample: b_n = (-1)^n/n and a_n = 1/n. Can anyone determine whether the series> sum (sin(n)^n/n) cos(nx)>converges for all x? (Its clear that it converges for>almost all x.) Or for x = 0?I think that sum_n |sin(n)|^n/n converges, so the series doesconverge absolutely for all x. But the best I can show isthat sum_n |sin(nt)|^n/n converges for almost all t.Let T be a random variable with uniform distribution in [0, pi]. To have |sin(nT)|^n > 1/n we need |sin(nT)| > exp(-ln(n)/n) which has probability O(sqrt(ln(n)/n)). ThusE[|sin(nT)|^n/n] = O(sqrt(ln(n))/n^(3/2)) which is summable.E[sum_n |sin(nT)|^n/n] < innityand in particular the sum converges almost surely.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 >> AstroPower and solar-PV-cell peddler> Spire Corporation both closed at exactly $3.00 (US) per share,> and both lost exactly $0.24 for the day.>> What do you make of this?> <...> That is all the Switching Campaign with the crossover> technique is all> about, and just because Don Libby cannot understand the technique> nor understand the Stockmarket is no reason to bad-mouth it.>> Oh. I thought the two companies might be in merger talks. Turns> out it was just a coincidence. The sun continues to set on> Astropower. Today Spire zoomed ahead by remaining at.>> -dlAnd in the Crossover Technique, today I could be increasing my wealthby selling some SBC at 24 when I bought it at 26 and buying back moreQwest at 4.45 when I sold it several weeks ago at 5.02 or thereabouts.I cannot do that because I am in the process of switching brokers.But my point is clear as to Dons last sentence when he says that Spirezoomed ahead by remaining at. (BTW I do not know if these are truegenuine companies and not even going to see because I will neverinvest in them).My point is this, several weeks ago I sold some Qwest at 5.02 and boughtSBC at 26.75. Today I could increase my wealth by selling some SBC at24. even though I bought it at 26.75 and buyback Qwest at 4.45.For the Crossover Technique works even better in a Bear or down marketthan in a Bull Market because the taxes paid in a Bull market are alargefactor.whole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies =I want to calculate the dot and cross product of two 4 by 1 vectors. Ihave done it easily enough with 3 by 1 vectors. Do the same principlesapply to the 4 by 1 vectors?Richard =Yes> I want to calculate the dot and cross product of two 4 by 1 vectors. I> have done it easily enough with 3 by 1 vectors. Do the same principles> apply to the 4 by 1 vectors?>> Richard I want to calculate the dot and cross product of two 4 by 1 vectors. I> have done it easily enough with 3 by 1 vectors. Do the same principles> apply to the 4 by 1 vectors?> RichardThe same principle, suitably extended, works for the dot (or scalar) product of vectors aof arbitrary dimension, but not vector products.Vector producting works in 3 dimensions, and I have heard that it works, somehow, in 7 dimensions, but in spaces of other dinensionality, it does not work. I want to calculate the dot and cross product of two 4 by 1 vectors. I> have done it easily enough with 3 by 1 vectors. Do the same principles> apply to the 4 by 1 vectors?> Richarddot, yes. cross, no. >I want to calculate the dot and cross product of two 4 by 1 vectors. I>>have done it easily enough with 3 by 1 vectors. Do the same principles>>apply to the 4 by 1 vectors?>>Richard> dot, yes. cross, no.However, in n-dimensions, you CAN dene the cross product ofn-1 vectors. It is (xed-width font warning!):| e_1 e_2 .... e_n || a_1_1 a_1_2 .... a_1_n |...| a_n_1 a_n_2 .... a_n_n |where e_i are the unit vectors (i, j, k in 3-space)and a_i_j is the j-th ordinate of the i-th vector.The result is a vector that is orthogonal to all the a_i.I worked this out when my son was learning vectors in apre-calculus class. It has undoubtedly been known for many yearsand rediscovered many times.Martin Cohen >I want to calculate the dot and cross product of two 4 by 1 vectors. I>>have done it easily enough with 3 by 1 vectors. Do the same principles>>apply to the 4 by 1 vectors?>>Richard dot, yes. cross, no.> However, in n-dimensions, you CAN dene the cross product of> n-1 vectors. It is (xed-width font warning!):> | e_1 e_2 .... e_n |> | a_1_1 a_1_2 .... a_1_n |> ...> | a_n_1 a_n_2 .... a_n_n |> where e_i are the unit vectors (i, j, k in 3-space)> and a_i_j is the j-th ordinate of the i-th vector.I believe the unit vectors need to be in the BOTTOM row, instead ofthe top.Calculus on Manifolds by Spivak gives the following:If v_1, ... , v_(n-1) are in R^n, then there exists a unique z in R^ns.t. | v_1 | | v_2 | = det | . | | . | | v_(n-1) | | w |for all w in R^n, and that this z is called the cross productv_1 x v_2 x ... x v_(n-1)If there are an even number of vectors v_i (in an odd-dimensionedspace) then bringing the top row to the bottom, and pushing the otherrows up one row results in no change in the determinant.But, an odd number of vectors v_i will cause the determinant to changesign under such a permutation of rows.Check the math. I believe this is correct.> The result is a vector that is orthogonal to all the a_i.> I worked this out when my son was learning vectors in a> pre-calculus class. It has undoubtedly been known for many years> and rediscovered many times.> Martin Cohen I want to calculate the dot and cross product of two 4 by 1 vectors. I>>have done it easily enough with 3 by 1 vectors. Do the same principles>>apply to the 4 by 1 vectors?>>Richard>dot, yes. cross, no.>>However, in n-dimensions, you CAN dene the cross product of>>n-1 vectors. It is (xed-width font warning!):>>| e_1 e_2 .... e_n |>>| a_1_1 a_1_2 .... a_1_n |>> ...>>| a_n_1 a_n_2 .... a_n_n |>>where e_i are the unit vectors (i, j, k in 3-space)>>and a_i_j is the j-th ordinate of the i-th vector.> I believe the unit vectors need to be in the BOTTOM row, instead of> the top.> Calculus on Manifolds by Spivak gives the following:> If v_1, ... , v_(n-1) are in R^n, then there exists a unique z in R^n> s.t.> | v_1 |> | v_2 |> = det | . |> | . |> | v_(n-1) |> | w |> for all w in R^n, and that this z is called the cross product> v_1 x v_2 x ... x v_(n-1)> If there are an even number of vectors v_i (in an odd-dimensioned> space) then bringing the top row to the bottom, and pushing the other> rows up one row results in no change in the determinant.> But, an odd number of vectors v_i will cause the determinant to change> sign under such a permutation of rows.> Check the math. I believe this is correct.>The result is a vector that is orthogonal to all the a_i.>>I worked this out when my son was learning vectors in a>>pre-calculus class. It has undoubtedly been known for many years>>and rediscovered many times.>>Martin CohenDoesnt matter if the units are in the top, bottom, oranyplace in the middle. The most that can happen is thatyou might get the negative of the vector, and this doesnot affect the orthogonality.I look at this as a way to get a vector orthogonal ton-1 vectors in n-dimensional space.Martin Cohen Doesnt matter if the units are in the top, bottom, or> anyplace in the middle. The most that can happen is that> you might get the negative of the vector, and this does> not affect the orthogonality.True, but you do want to adopt a proper convention for right-handedsystems of vectors. In a 3x3 matrix, top or bottom row makes nodifference. As you say, in a 4x4 matrix the sign is reversed.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ >I want to calculate the dot and cross product of two 4 by 1 vectors. I>>have done it easily enough with 3 by 1 vectors. Do the same principles>>apply to the 4 by 1 vectors?>>Richard>dot, yes. cross, no.>>However, in n-dimensions, you CAN dene the cross product of>>n-1 vectors. It is (xed-width font warning!):>| e_1 e_2 .... e_n |>>| a_1_1 a_1_2 .... a_1_n |> ...>| a_n_1 a_n_2 .... a_n_n |>>where e_i are the unit vectors (i, j, k in 3-space)>>and a_i_j is the j-th ordinate of the i-th vector.> I believe the unit vectors need to be in the BOTTOM row, instead of> the top. Calculus on Manifolds by Spivak gives the following: If v_1, ... , v_(n-1) are in R^n, then there exists a unique z in R^n> s.t.> | v_1 |> | v_2 |> = det | . |> | . |> | v_(n-1) |> | w |> for all w in R^n, and that this z is called the cross product> v_1 x v_2 x ... x v_(n-1)> If there are an even number of vectors v_i (in an odd-dimensioned> space) then bringing the top row to the bottom, and pushing the other> rows up one row results in no change in the determinant.> But, an odd number of vectors v_i will cause the determinant to change> sign under such a permutation of rows.> Check the math. I believe this is correct.>>The result is a vector that is orthogonal to all the a_i.>>I worked this out when my son was learning vectors in a>>pre-calculus class. It has undoubtedly been known for many years>>and rediscovered many times.>>Martin Cohen> Doesnt matter if the units are in the top, bottom, or> anyplace in the middle. The most that can happen is that> you might get the negative of the vector, and this does> not affect the orthogonality.True, but this doesnt answer the question of how the cross productgeneralizes. If someone asks you the product of 3 and 5, do you reply: -15, or 15 ... it doesnt matter. Both are 15 units from zero.In 3-space, a x b doesnt equal b x a -- even though both areorthogonal to a and b.> I look at this as a way to get a vector orthogonal to> n-1 vectors in n-dimensional space.> Martin Cohen >>I want to calculate the dot and cross product of two 4 by 1 vectors. I>>have done it easily enough with 3 by 1 vectors. Do the same principles>>apply to the 4 by 1 vectors?>>Richard> dot, yes. cross, no.> However, in n-dimensions, you CAN dene the cross product of> n-1 vectors. It is (xed-width font warning!):> | e_1 e_2 .... e_n |> | a_1_1 a_1_2 .... a_1_n |> ...> | a_n_1 a_n_2 .... a_n_n |> where e_i are the unit vectors (i, j, k in 3-space)> and a_i_j is the j-th ordinate of the i-th vector. The result is a vector that is orthogonal to all the a_i.> I worked this out when my son was learning vectors in a> pre-calculus class. It has undoubtedly been known for many years> and rediscovered many times.> Martin CohenGood discovery! In some sense, the cross product generalizes to thenotion of exterior product (a.k.a. wedge product), but theexterior product of a bunch of vectors, v0^v1^v2^...^vk in R^n livesin a different space (which is presumably the origin of the termexterior). It turns out that the exterior product of (n-1) vectorsin R^n lives in a space of dimension Choose(n, n-1) = n, and the innerproduct on R^n can be used to identify this space with R^n. This isessentially the form you discovered.John Mitchell I want to calculate the dot and cross product of two 4 by 1 vectors. I> have done it easily enough with 3 by 1 vectors. Do the same principles> apply to the 4 by 1 vectors?>> Richard> The same principle, suitably extended, works for the dot (or scalar)> product of vectors aof arbitrary dimension, but not vector products.> Vector producting works in 3 dimensions, and I have heard that it> works, somehow, in 7 dimensions, but in spaces of other> dinensionality, it does not work.You have forgotten dimension 1 in which one has the cross product U x V= 0. In dimension 7 take the imaginary octonions, i.e. those with real part0, drop the real part to get ordered 7-tuples multiplied as follows:U x V = (UV - VU)/2 where UV, VU are the usual octonion products.Note that the product of two octonions with real part 0 has real part 0.All three cross products arise by dening U x V = (UV - VU)/2 on Im(C),Im(H), Im(O). The fact that there are no others follows from thetheorem (due to Max Zorn?) that the only alternative, quadratic, realalgebras without divisors of zero are isomorphic to R, C, H or O.GC However, in n-dimensions, you CAN dene the cross product of> n-1 vectors. It is (xed-width font warning!):> | e_1 e_2 .... e_n |> | a_1_1 a_1_2 .... a_1_n |> ...> | a_n_1 a_n_2 .... a_n_n |> where e_i are the unit vectors (i, j, k in 3-space)> and a_i_j is the j-th ordinate of the i-th vector.> The result is a vector that is orthogonal to all the a_i.> I worked this out when my son was learning vectors in a> pre-calculus class. It has undoubtedly been known for many years> and rediscovered many times.> Martin Cohen> Good discovery! In some sense, the cross product generalizes to the> notion of exterior product (a.k.a. wedge product), but the> exterior product of a bunch of vectors, v0^v1^v2^...^vk in R^n lives> in a different space (which is presumably the origin of the term> exterior). It turns out that the exterior product of (n-1) vectors> in R^n lives in a space of dimension Choose(n, n-1) = n, and the inner> product on R^n can be used to identify this space with R^n. This is> essentially the form you discovered.> John MitchellThis is a well known fact. Also, R^3 with the cross product and thelinear space of bivectors with the wedge are Lie algebra isomorphic.The Hodge dual establishes the isomorphism. M.T. =Innity. More than a million, billion, trillion, zillion.All our schools have taught us to learn about innity. But will mathsever be able to model the universe perfectly? Surely not. Our humanbrains cant grasp innity for a start.--Support charity at www.freeMyWisdom.co.uk - we dont want your money,but we do want your wisdom. Innity. More than a million, billion, trillion, zillion.Whats a zillion?(and whats a dummie?)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen Innity. More than a million, billion, trillion, zillion.>> All our schools have taught us to learn about innity. But will maths> ever be able to model the universe perfectly? Surely not. Our human> brains cant grasp innity for a start.>Innity is very big, so big that it might take forever to explain it.However, by the magic of mind, there is a symbol oo, the lazy eightthat represents innity. Thus oo is innity, so handle it with care,so much is packed into oo, that it could go off with the Big Bang.Innity is so big you cant add to it, that is oo + 1 = oo + 2 = oo + oo= oo and it doesnt multiply, that is 1000 * oo = oo * oo = oo. Thus yousee oo is not allowed to play with ordinary nite folks like 1,2,3.. forindeed oo is too big for its britches. Innity. More than a million, billion, trillion, zillion.> Whats a zillion?>Bushs second term goal for looting US Treasure.> (and whats a dummie?)>dummie = > Innity. More than a million, billion, trillion, zillion.>> Whats a zillion?>> (and whats a dummie?)Wiley is really cleaning up with their XXX for Dummies series. All kinds ofstuff, Buddhism for Dummies, Bioinformatics for Dummies. Hundreds andhundreds of titles. As yet I think there is no Continuum Hypothesis forDummies. >Innity. More than a million, billion, trillion, zillion.>>All our schools have taught us to learn about innity. But will maths>>ever be able to model the universe perfectly? Surely not. Our human>>brains cant grasp innity for a start.> Innity is very big, so big that it might take forever to explain it.> However, by the magic of mind, there is a symbol oo, the lazy eight> that represents innity. Thus oo is innity, so handle it with care,> so much is packed into oo, that it could go off with the Big Bang.> Innity is so big you cant add to it, that is oo + 1 = oo + 2 = oo + oo> = oo and it doesnt multiply, that is 1000 * oo = oo * oo = oo. Thus you> see oo is not allowed to play with ordinary nite folks like 1,2,3.. for> indeed oo is too big for its britches.That depends on what youre working with. Transnites play with oo+1, etc on a regular basis, and oo+1 =/= oo+2 in those cases.-- Will Twentyman > Innity. More than a million, billion, trillion, zillion.>> Whats a zillion?>> (and whats a dummie?)> Wiley is really cleaning up with their XXX for Dummies series. All kinds> of> stuff, Buddhism for Dummies, Bioinformatics for Dummies. Hundreds and> hundreds of titles. As yet I think there is no Continuum Hypothesis for> Dummies.We could have an author for Fermats Last Theorem for Dummies. :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen We could have an author for Fermats Last Theorem for Dummies. :-)What about that book by Marilyn Vos Savant?--irascible since 1957 Innity. More than a million, billion, trillion, zillion.> Whats a zillion?>> Bushs second term goal for looting US Treasure.> (and economics Amen.. $70 Million spent on Clintons penis. $4 Million spent ondiscovering who really did 911. Conjectures 1 & 2 1.) Republicans didnt know what a blow job was. 2.) They found out. Conjecture 3 3.) Republicans keep putting private prot where national honor was. I need to study this. Its just too real. Lets see, 4.) In the beginning the rst Capital crime in the thenUSA was a crime against the then Bank. 5.) Then there was those pesky people who were here rst. We xedthat. 1&2&3 in terms of 4 and 5. Oh I see its a proof of what terrorism is. But I digress. Innity : Innity In*ni*ty, n.; pl. Innities. L. innitas; pref. in- not + nis boundary, limit, end: cf. F. innite. See Finite. 1. Unlimited extent of time, space, or quantity; eternity; boundlessness; immensity. --Sir T. More. There can not be more innities than one; for one of them would limit the other. --Sir W. Raleigh. I disagree, One is not an innity. :) What say you Pilgrim?Ernst The proof looks Ugly. >We could have an author for Fermats Last Theorem for Dummies. :-)> What about that book by Marilyn Vos Savant?Thats Fermats Last Theorem By Dummies> --irascible since 1957 Innity is so big you cant add to it, that is oo + 1 = oo + 2 = oo + oo> = oo and it doesnt multiply, that is 1000 * oo = oo * oo = oo. Thus you> see oo is not allowed to play with ordinary nite folks like 1,2,3.. for> indeed oo is too big for its britches.>> That depends on what youre working with. Transnites play with oo+1,> etc on a regular basis, and oo+1 =/= oo+2 in those cases.>Not so. Its, omega + 1 /= omega + 2 and 1 + omega = 2 + omega.tho w is so like omega 1 + w = w /= w + 1 2*w = w /= w + w = w*2 =: Innity. More than a million, billion, trillion, zillion.: All our schools have taught us to learn about innity. But will maths: ever be able to model the universe perfectly? Surely not. Our human: brains cant grasp innity for a start.Well, it would *take forever* for any kind of brain to *really* graspinnity in some senses of the word...but tryhttp://www.hypermaths.org/quadibloc/math/innt.htmJohn Savard > will maths ever be able to model [it]? Surely not. Our human> brains cant grasp innity for a start.>Ever heard the name G. CANTOR?F. =In sci.math, www.freeMyWisdom.co.uk<7c2720c0.0307180159.190f5ac@ trillion, zillion.> All our schools have taught us to learn about innity. But will maths> ever be able to model the universe perfectly? Surely not. Our human> brains cant grasp innity for a start.Our human brains cant grasp extremely large numbers,either. However, we can represent them by various symbols.For example, Avogadros Number, commonly used in physics,is 602.2 sextillion, or 6.02 * 10^23. If one were tocount atoms in a mole, 1 atom per picosecond, it wouldtake nearly 20 millennia to nish the job. However,most chemists (and others) simply use the symbol N forthis particular number. Similar issues exist for thenumber of electrons in a coulomb and the number of kg inthe Universe.Even the US public debt is phenomenal. At present its about6 trillion dollars. One would have to spend almost $2,000dollars *per second* over a 100 year lifetime to spend thatmuch money. (I should be so lucky. :-) )The more or less conventional representation of innity is thesidewise 8, infty for those using TeX, and I shall representit here using oo because of limitations in ASCII.Innity is a weird number. Unless one gives it a sign, whichone can do, it is neither less than nor greater than anynumber, but is both less than and greater than all numbers.This is decidedly paradoxical, so more rigor is needed lest weslide down a slippery slope of sloppiness... :-)You are no doubt familiar with the more or less conventionaldenition of nite limits: Let epsilon > 0. If I can prove that there is always a delta > 0 such that, for any a-delta < x < a+delta, x != a, one can prove L-delta < f(x) < L+delta, then one can write as a shorthand lim x->a f(x) = L.One can make a slight modication: Let M > 0. If I can prove that there is always a delta > 0 such that, for any a-delta < x < a+delta, x != a, one can prove f(x) > M or f(x) < -M, then one can write as a shorthand lim x->a f(x) = oo.One can make more trivial modications to generatethings such as lim x->a+ f(x) = L+ lim x->+oo f(x) = L+ lim x->a+ f(x) = +ooetc.; Ill leave these to the interested reader.Thats about as far as limits can go; slightly moreinteresting innities can be had by set theoretic methods.The cardinality of the set of natural numbers (alsointegers, rationals, and algebraic numbers) turns out tobe aleph-null. The cardinality of all real numbers (orcomplex or transcendental or irrational) is C. It is notclear to me whether C = aleph-one, which is the cardinalityof all subsets of a set with cardinality aleph-null.This is about as far as I can go personally (Im acomputer software specialist, not a formal mathematician :-) ).[.sigsnip]-- #191, ewill3@earthlink.netIts still legal to go .sigless. In sci.math, www.freeMyWisdom.co.uk><7c2720c0.0307180159.190f5ac@ trillion, zillion.> All our schools have taught us to learn about innity. But will maths>> ever be able to model the universe perfectly? Surely not. Our human>> brains cant grasp innity for a start.>>Our human brains cant grasp extremely large numbers,>either. However, we can represent them by various symbols.>For example, Avogadros Number, commonly used in physics,>is 602.2 sextillion, or 6.02 * 10^23. If one were to>count atoms in a mole, 1 atom per picosecond, it would>take nearly 20 millennia to nish the job. However,>most chemists (and others) simply use the symbol N for>this particular number. Similar issues exist for the>number of electrons in a coulomb and the number of kg in>the Universe.>>Even the US public debt is phenomenal. At present its about>6 trillion dollars. One would have to spend almost $2,000>dollars *per second* over a 100 year lifetime to spend that>much money. (I should be so lucky. :-) )>>The more or less conventional representation of innity is the>sidewise 8, infty for those using TeX, and I shall represent>it here using oo because of limitations in ASCII.>>Innity is a weird number. Unless one gives it a sign, which>one can do, it is neither less than nor greater than any>number, but is both less than and greater than all numbers.>This is decidedly paradoxical, so more rigor is needed lest we>slide down a slippery slope of sloppiness... :-)>>You are no doubt familiar with the more or less conventional>denition of nite limits:>> Let epsilon > 0. If I can prove that there is always> a delta > 0 such that, for any a-delta < x < a+delta,> x != a, one can prove L-delta < f(x) < L+delta, then> one can write as a shorthand lim x->a f(x) = L.>>One can make a slight modication:>> Let M > 0. If I can prove that there is always> a delta > 0 such that, for any a-delta < x < a+delta,> x != a, one can prove f(x) > M or f(x) < -M, then one can> write as a shorthand lim x->a f(x) = oo.>>One can make more trivial modications to generate>things such as>> lim x->a+ f(x) = L+> lim x->+oo f(x) = L+> lim x->a+ f(x) = +oo>>etc.; Ill leave these to the interested reader.>>Thats about as far as limits can go; slightly more>interesting innities can be had by set theoretic methods.>The cardinality of the set of natural numbers (also>integers, rationals, and algebraic numbers) turns out to>be aleph-null. The cardinality of all real numbers (or>complex or transcendental or irrational) is C. It is not>clear to me whether C = aleph-one, which is the cardinality>of all subsets of a set with cardinality aleph-null.Nope. The cardinality of the power set of the integers is C, notaleph-one. Aleph-one is the smallest uncountable cardinal, and is <=C. The continuum hypothesis states that aleph-one=C.>>This is about as far as I can go personally (Im a>computer software specialist, not a formal mathematician :-) ).>>[.sigsnip](this space unintentially left blank ..... In sci.math, www.freeMyWisdom.co.uk> More than a million, billion, trillion, zillion.> All our schools have taught us to learn about innity. But will maths> ever be able to model the universe perfectly? Surely not. Our human> brains cant grasp innity for a start.> Our human brains cant grasp extremely large numbers,> either. However, we can represent them by various symbols.> For example, Avogadros Number, commonly used in physics,> is 602.2 sextillion, or 6.02 * 10^23. If one were to> count atoms in a mole, 1 atom per picosecond, it would> take nearly 20 millennia to nish the job. However,> most chemists (and others) simply use the symbol N for> this particular number. Similar issues exist for the> number of electrons in a coulomb and the number of kg in> the Universe.> Even the US public debt is phenomenal. At present its about> 6 trillion dollars. One would have to spend almost $2,000> dollars *per second* over a 100 year lifetime to spend that> much money. (I should be so lucky. :-) )> The more or less conventional representation of innity is the> sidewise 8, infty for those using TeX, and I shall represent> it here using oo because of limitations in ASCII.> Innity is a weird number. Unless one gives it a sign, which> one can do, it is neither less than nor greater than any> number, but is both less than and greater than all numbers.> This is decidedly paradoxical, so more rigor is needed lest we> slide down a slippery slope of sloppiness... :-)> You are no doubt familiar with the more or less conventional> denition of nite limits:> Let epsilon > 0. If I can prove that there is always> a delta > 0 such that, for any a-delta < x < a+delta,> x != a, one can prove L-delta < f(x) < L+delta, then> one can write as a shorthand lim x->a f(x) = L.> One can make a slight modication:> Let M > 0. If I can prove that there is always> a delta > 0 such that, for any a-delta < x < a+delta,> x != a, one can prove f(x) > M or f(x) < -M, then one can> write as a shorthand lim x->a f(x) = oo.> One can make more trivial modications to generate> things such as> lim x->a+ f(x) = L+> lim x->+oo f(x) = L+> lim x->a+ f(x) = +oo> etc.; Ill leave these to the interested reader.> Thats about as far as limits can go; slightly more> interesting innities can be had by set theoretic methods.> The cardinality of the set of natural numbers (also> integers, rationals, and algebraic numbers) turns out to> be aleph-null. The cardinality of all real numbers (or> complex or transcendental or irrational) is C. It is not> clear to me whether C = aleph-one, which is the cardinality> of all subsets of a set with cardinality aleph-null.> This is about as far as I can go personally (Im a> computer software specialist, not a formal mathematician :-) ).> [.sigsnip]Wow.. Still a good read. I bet you get all the girls unlike mensanator thevain. ;-)Ernst Sig # 1 Any Government ( including economic systems ) that limits thepeoples ability to obtain education is a bad Government. $4 Billion a month better spent on education so we might one daylearn from the past. sieg heil baby Wow..>> Still a good read. I bet you get all the girls unlike mensanator the>vain. ;-)Vain? Where did you get that notion? You been lookin at my Home Page?http://members.aol.com/mensanatorSigh. You try to set a standard for cleverness and wit but it just goesunappreciated.>Ernst>>--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm Wow..>> Still a good read. I bet you get all the girls unlike mensanator the>vain. ;-)> Vain? Where did you get that notion? You been lookin at my Home Page?> http://members.aol.com/mensanator> Sigh. You try to set a standard for cleverness and wit but it just goes> unappreciated.>Ernst>> That too. =In sci.math, Ernst Berg> All our schools have taught us to learn about innity. But will maths>> ever be able to model the universe perfectly? Surely not. Our human>> brains cant grasp innity for a start.> Our human brains cant grasp extremely large numbers,>> either. However, we can represent them by various symbols.>> For example, Avogadros Number, commonly used in physics,>> is 602.2 sextillion, or 6.02 * 10^23. If one were to>> count atoms in a mole, 1 atom per picosecond, it would>> take nearly 20 millennia to nish the job. However,>> most chemists (and others) simply use the symbol N for>> this particular number. Similar issues exist for the>> number of electrons in a coulomb and the number of kg in>> the Universe.> Even the US public debt is phenomenal. At present its about>> 6 trillion dollars. One would have to spend almost $2,000>> dollars *per second* over a 100 year lifetime to spend that>> much money. (I should be so lucky. :-) )> The more or less conventional representation of innity is the>> sidewise 8, infty for those using TeX, and I shall represent>> it here using oo because of limitations in ASCII.> Innity is a weird number. Unless one gives it a sign, which>> one can do, it is neither less than nor greater than any>> number, but is both less than and greater than all numbers.>> This is decidedly paradoxical, so more rigor is needed lest we>> slide down a slippery slope of sloppiness... :-)> You are no doubt familiar with the more or less conventional>> denition of nite limits:> Let epsilon > 0. If I can prove that there is always>> a delta > 0 such that, for any a-delta < x < a+delta,>> x != a, one can prove L-delta < f(x) < L+delta, then>> one can write as a shorthand lim x->a f(x) = L.> One can make a slight modication:> Let M > 0. If I can prove that there is always>> a delta > 0 such that, for any a-delta < x < a+delta,>> x != a, one can prove f(x) > M or f(x) < -M, then one can>> write as a shorthand lim x->a f(x) = oo.> One can make more trivial modications to generate>> things such as> lim x->a+ f(x) = L+>> lim these to the interested reader.> Thats about as far as limits can go; slightly more>> interesting innities can be had by set theoretic methods.>> The cardinality of the set of natural numbers (also>> integers, rationals, and algebraic numbers) turns out to>> be aleph-null. The cardinality of all real numbers (or>> complex or transcendental or irrational) is C. It is not>> clear to me whether C = aleph-one, which is the cardinality>> of all subsets of a set with cardinality aleph-null.> This is about as far as I can go personally (Im a>> computer software specialist, not a formal mathematician :-) ).> [.sigsnip]> Wow..> Still a good read. I bet you get all the girls unlike mensanator the> vain. ;-)I wouldnt know. :-) Most of this is from stuff I learnedin college more than 20 years ago, or even before. (Where doesthe time go? :-) )But the women do like me. :-) ;-) :-) And I like them, sowere even. :-)As for Mensanator -- I dont know anything about him.Someone else has mentioned the continuum hypothesis. Im notsure I know anything about that, either. :-/[.sigsnip]-- #191, ewill3@earthlink.netIts still legal to go .sigless. Innity. More than a million, billion, trillion, zillion.> Whats a zillion?zillion /zljn/ n. colloq. (orig. US). M20. [Arbitrarily after million,billion, etc. (perh. f. z repr. an unknown quantity).] A very large butindenite number.zillionaire n. a very rich person M20. zillionth a. &n. ( a ) adj. following very many others; umpteenth; ( b ) n. a tinyfraction of something: L20.--------------------------------------------------------- Excerpted from Oxford Talking DictionaryCopyright 1998 The Learning Company, Inc. All Rights Reserved.> (and whats a dummie?)> --> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html> The League of Gentlemen Innity. More than a million, billion, trillion, zillion.> All our schools have taught us to learn about innity. But will maths> ever be able to model the universe perfectly? Surely not. Our human> brains cant grasp innity for a start.> There are many kinds of innity in mathematics. Zdislav V. Kovarik hada post listing many of them. Here is is:---There is a long list of innities (with no claim to exhaustiveness): innity of the one-point compactication of N, innity of the one-point compactication of R, innity of the two-point compactication of R, innity of the one-point compactication of C, innities of the projective extension of the plane, innity of Lebesgue-type integration theory, innities of the non-standard extension of R, innities of the theory of ordinal numbers, innities of the theory of cardinal numbers, innity adjoined to normed spaces, whose neighborhoods are complements of relatively compact sets, innity adjoined to normed spaces, whose neighborhoods are complements of bounded sets, innity around absolute G-delta non-compact metric spaces, innity in the theory of convex optimization, etc.;each of these has a clear denition and a set of well-dened rules for handling it.---GC In sci.math, Ernst Berg> > sci.math, www.freeMyWisdom.co.uk>> Innity. More than a million, billion, trillion, zillion.>> All our schools have taught us to learn about innity. But will maths>> ever be able to model the universe perfectly? Surely not. Our human>> brains cant grasp innity for a start.> Our human brains cant grasp extremely large numbers,>> either. However, we can represent them by various symbols.>> For example, Avogadros Number, commonly used in physics,>> is 602.2 sextillion, or 6.02 * 10^23. If one were to>> count atoms in a mole, 1 atom per picosecond, it would>> take nearly 20 millennia to nish the job. However,>> most chemists (and others) simply use the symbol N for>> this particular number. Similar issues exist for the>> number of electrons in a coulomb and the number of kg in>> the Universe.>> Even the US public debt is phenomenal. At present its about>> 6 trillion dollars. One would have to spend almost $2,000>> dollars *per second* over a 100 year lifetime to spend that>> much money. (I should be so lucky. :-) )>> The more or less conventional representation of innity is the>> sidewise 8, infty for those using TeX, and I shall represent>> it here using oo because of limitations in ASCII.> Innity is a weird number. Unless one gives it a sign, which>> one can do, it is neither less than nor greater than any>> number, but is both less than and greater than all numbers.>> This is decidedly paradoxical, so more rigor is needed lest we>> slide down a slippery slope of sloppiness... :-)> You are no doubt familiar with the more or less conventional>> denition of nite limits:> Let epsilon > 0. If I can prove that there is always>> a delta > 0 such that, for any a-delta < x < a+delta,>> x != a, one can prove L-delta < f(x) < L+delta, then>> one can write as a shorthand lim x->a f(x) = L.>> One can make a slight modication:> Let M > 0. If I can prove that there is always>> a delta > 0 such that, for any a-delta < x < a+delta,>> x != a, one can prove f(x) > M or f(x) < -M, then one can>> write as a shorthand lim x->a f(x) = oo.> One can make more trivial modications to generate>> things such as> lim x->a+ f(x) = L+>> lim x->+oo f(x) = L+>> lim x->a+ f(x) = +oo> etc.; Ill leave these to the interested reader.> Thats about as far as limits can go; slightly more>> interesting innities can be had by set theoretic methods.>> The cardinality of the set of natural numbers (also>> integers, rationals, and algebraic numbers) turns out to>> be aleph-null. The cardinality of all real numbers (or>> complex or transcendental or irrational) is C. It is not>> clear to me whether C = aleph-one, which is the cardinality>> of all subsets of a set with cardinality aleph-null.> This is about as far as I can go personally (Im a>> computer software specialist, not a formal mathematician :-) ).> [.sigsnip]> Wow..> Still a good read. I bet you get all the girls unlike mensanator the> vain. ;-)> I wouldnt know. :-) Most of this is from stuff I learned> in college more than 20 years ago, or even before. (Where does> the time go? :-) )> But the women do like me. :-) ;-) :-) And I like them, so> were even. :-)> As for Mensanator -- I dont know anything about him.> Someone else has mentioned the continuum hypothesis. Im not> sure I know anything about that, either. :-/> [.sigsnip]Yes I know the 20 year feeling. My trip is I now want to learn :)Back to school for me in the spring, if we still have schools inCalifornia, to start again. About Mensanator, it was just a friendly jest. I think he got bentabout it but Im not concerned.Hes a jestable fellow The Right Honorable gentleman is indebted to his memory for his jests, and to his imagination for his facts. --Sheridan.Ernst Sig # 3 And God the Republican said go forth and divide unfairly. Message-id: sci.math, Ernst Berg>> sci.math, www.freeMyWisdom.co.uk> Innity. More than a million, billion, trillion, zillion.>> All our schools have taught us to learn about innity. But will maths> ever be able to model the universe perfectly? Surely not. Our human> brains cant grasp innity for a start.> Our human brains cant grasp extremely large numbers,> either. However, we can represent them by various symbols.> For example, Avogadros Number, commonly used in physics,> is 602.2 sextillion, or 6.02 * 10^23. If one were to> count atoms in a mole, 1 atom per picosecond, it would> take nearly 20 millennia to nish the job. However,> most chemists (and others) simply use the symbol N for> this particular number. Similar issues exist for the> number of electrons in a coulomb and the number of kg in> the Universe. Even the US public debt is phenomenal. At present its about> 6 trillion dollars. One would have to spend almost $2,000> dollars *per second* over a 100 year lifetime to spend that> much money. (I should be so lucky. :-) ) The more or less conventional representation of innity is the> sidewise 8, infty for those using TeX, and I shall represent> it here using oo because of limitations in ASCII.> Innity is a weird number. Unless one gives it a sign, which> one can do, it is neither less than nor greater than any> number, but is both less than and greater than all numbers.> This is decidedly paradoxical, so more rigor is needed lest we> slide down a slippery slope of sloppiness... :-)> You are no doubt familiar with the more or less conventional> denition of nite limits:> Let epsilon > 0. If I can prove that there is always> a delta > 0 such that, for any a-delta < x < a+delta,> x != a, one can prove L-delta < f(x) < L+delta, then> one can write as a shorthand lim x->a f(x) = L.> One can make a slight modication:> Let M > 0. If I can prove that there is always> a delta > 0 such that, for any a-delta < x < a+delta,> x != a, one can prove f(x) > M or f(x) < -M, then one can> write as a shorthand lim x->a f(x) = oo.> One can make more trivial modications to generate> things such as> lim x->a+ f(x) = L+>> lim x->+oo f(x) = L+> lim x->a+ f(x) = +oo> etc.; Ill leave these to the interested reader.> Thats about as far as limits can go; slightly more> interesting innities can be had by set theoretic methods.>> The cardinality of the set of natural numbers (also> integers, rationals, and algebraic numbers) turns out to> be aleph-null. The cardinality of all real numbers (or> complex or transcendental or irrational) is C. It is not> clear to me whether C = aleph-one, which is the cardinality>> of all subsets of a set with cardinality aleph-null. This is about as far as I can go personally (Im a> computer software specialist, not a formal mathematician :-) ).> [.sigsnip]>> Wow..> Still a good read. I bet you get all the girls unlike mensanator the> vain. ;-)> I wouldnt know. :-) Most of this is from stuff I learned>> in college more than 20 years ago, or even before. (Where does>> the time go? :-) )> But the women do like me. :-) ;-) :-) And I like them, so>> were even. :-)> As for Mensanator -- I dont know anything about him.> Someone else has mentioned the continuum hypothesis. Im not>> sure I know anything about that, either. :-/> [.sigsnip]>>Yes I know the 20 year feeling. My trip is I now want to learn :)>Back to school for me in the spring, if we still have schools in>California, to start again.> About Mensanator, it was just a friendly jest. I think he got bent>about it Actually, I didnt. I wondered where you got the idea I was vain and thenpointed you to my Home Page, which is an exercise in vanity.>but Im not concerned.Hes a jestable fellowMy getting bent was itself a jest.> The Right Honorable gentleman is indebted to his> memory for his jests, and to his imagination for his> facts. --Sheridan.>>Ernst>> Sig # 3>> And God the Republican said go forth and divide unfairly.--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm Innity. More than a million, billion, trillion, zillion.>> All our schools have taught us to learn about innity. But will maths> ever be able to model the universe perfectly? Surely not. Our human> brains cant grasp innity for a start.>> Innity is very big, so big that it might take forever to explain it.> However, by the magic of mind, there is a symbol oo, the lazy eight> that represents innity. Thus oo is innity, so handle it with care,> so much is packed into oo, that it could go off with the Big Bang.> Innity is so big you cant add to it, that is oo + 1 = oo + 2 = oo + oo> = oo and it doesnt multiply, that is 1000 * oo = oo * oo = oo. Thus you> see oo is not allowed to play with ordinary nite folks like 1,2,3.. for> indeed oo is too big for its britches.i am a nul to math , but thats some spiritual stuff . quess all i need is an eye-opener =Im stuck on this question, and would be grateful if someone couldpoint out where Im going wrong...Consider a dynamical system dx/dt = x(1 - x^2 - y^2) + 3ydy/dt = y(1 - x^2 - y^2) - 2xTransforming into polar coordinates using the chain rule gives medS/dt = 2(1 - r^2)cosec2S + 1dr/dt = 2r(1 - r^2) + r(3tanS - 2cotS)Now according to the question there is an annular region about r=1containing a periodic orbit. I dont see how though because (3tanS -2cotS) is unbounded for the relevant range of S, which (dr/dt /= 0)there is no periodic orbit. Hence I must be wrong, but cant seewhere... please help =Ben Dixon scribe:> Im stuck on this question, and would be grateful if someone could> point out where Im going wrong...> Consider a dynamical system> dx/dt = x(1 - x^2 - y^2) + 3y> dy/dt = y(1 - x^2 - y^2) - 2x Transforming into polar coordinates using the chain rule gives me> dS/dt = 2(1 - r^2)cosec2S + 1> dr/dt = 2r(1 - r^2) + r(3tanS - 2cotS)(...)Did You use the chain rule on the way to Your result? I amnot sure. Some remarks without calculation:The usual way to destination is right You have done, get thederivatives of r and S (named phi). Then note thatr/phi=dr/dphi and you will get an equationdr/dphi=f(r,phi) as well known for autonom systems. If Ihave not done mistakes (I do them often, control the steps)it should be like that:dr/dphi=(r*(1-r^2)*(cot(phi)+tan(phi))+r)/(-2*cot(phi)-3 *tan(phi)).Next look for a dominant linear item on the right side(develop it as a row) and state that the remainder is ofhigher order (i. e. it goes to zero with t-->oo).Good luck,Alfred Im stuck on this question, and would be grateful if someone could>point out where Im going wrong...>Consider a dynamical system >dx/dt = x(1 - x^2 - y^2) + 3y>dy/dt = y(1 - x^2 - y^2) - 2x>Transforming into polar coordinates using the chain rule gives me>dS/dt = 2(1 - r^2)cosec2S + 1>dr/dt = 2r(1 - r^2) + r(3tanS - 2cotS)This must be wrong, since it would blow up at certain values of S (whichI assume is the angular coordinate). With x = r cos(s) and y = r sin(s), note that you havedx/dt = cos(s) dr/dt - r sin(s) ds/dtdy/dt = sin(s) dr/dt + r cos(s) ds/dtso thatdr/dt = cos(s) dx/dt + sin(s) dy/dtr ds/dt = -sin(s) dx/dt + cos(s) dy/dtRobert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 > Im stuck on this question, and would be grateful if someone could> point out where Im going wrong...>> Consider a dynamical system>> dx/dt = x(1 - x^2 - y^2) + 3y> dy/dt = y(1 - x^2 - y^2) - 2x>> Transforming into polar coordinates using the chain rule gives me>> dS/dt = 2(1 - r^2)cosec2S + 1> dr/dt = 2r(1 - r^2) + r(3tanS - 2cotS)>> Now according to the question there is an annular region about r=1> containing a periodic orbit. I dont see how though because (3tanS -> 2cotS) is unbounded for the relevant range of S, which (dr/dt /= 0)> there is no periodic orbit. Hence I must be wrong, but cant see> where... please help>I wouldnt mess with polar coordinates. Imagine that the point (x,y) lieson the circle x^2 + y^2 = 2. Show that in this case, the dot product of(x,y) and (x,y) is negative, so the solution must be owing inward as itcrosses this circle. Nowfind a small circle where the solution owsoutward. Verify that the origin is the only critical point. I think yourtext will have a theorem that this implies the existence of a periodicorbit. Im stuck on this question, and would be grateful if someone could> point out where Im going wrong...> Consider a dynamical system > dx/dt = x(1 - x^2 - y^2) + 3y> dy/dt = y(1 - x^2 - y^2) - 2x> Transforming into polar coordinates using the chain rule gives me> dS/dt = 2(1 - r^2)cosec2S + 1> dr/dt = 2r(1 - r^2) + r(3tanS - 2cotS)> Now according to the question there is an annular region about r=1> containing a periodic orbit. I dont see how though because (3tanS -> 2cotS) is unbounded for the relevant range of S, which (dr/dt /= 0)> there is no periodic orbit. Hence I must be wrong, but cant see> where... please help> Ask yourself:1) What are the singular points of the ow?2) What does the ow look like near the singular point(s)?3) What does the ow look like far from the origin?4) What can you say about the limit set of a point far from theorigin? [You are presumably supposed to use a theorem that says, undercertain circumstances, that the limit set of an orbit is a periodicorbit]John Mitchell Im stuck on this question, and would be grateful if someone could> point out where Im going wrong...> Consider a dynamical system dx/dt = x(1 - x^2 - y^2) + 3y> dy/dt = y(1 - x^2 - y^2) - 2x> ....You want to use the Poincare-Bendixson Theorem Mr. Ben Dixon (acoincidence, perhaps).Use the above information to show that you canfind a circle withradius less than one such that all the vectors (of the vector eld)on this circle are pointing out radially and show that there isanother circle, with radius bigger than one, such that all the vectorsare pointing in radially on that circle.So any orbit inside this annulus is trapped and the Poincare-Bendixsontheorem says that the omega-limit set is a periodic orbit if there areno xed points. It is easy to check that there are no xed pointsin this annulus.Enjoy. =Will Self scribe:>> Im stuck on this question, and would be grateful if someone could> point out where Im going wrong...>> Consider a dynamical system> dx/dt = x(1 - x^2 - y^2) + 3y> dy/dt = y(1 - x^2 - y^2) - 2x(...)> I wouldnt mess with polar coordinates. Imagine that the point (x,y) lies> on the circle x^2 + y^2 = 2. Show that in this case, the dot product of> (x,y) and (x,y) is negative, so the solution must be owing inward as it> crosses this circle. Nowfind a small circle where the solution ows> outward. Verify that the origin is the only critical point. I think your> text will have a theorem that this implies the existence of a periodic> orbit.Indeed in this case there is no need to transform into polarcoordinates. It should be shown, that (1-x^2-y^2) goes tozero and the solution of the approximated equations servesthe Grenzzykel in the x-y-cartesic-system (an ellipse). Anice picture can be shown in the polar coordinate-system. Itlooks like the movement of planetoid-perihel around aspecial centre (a rosette).Another way is to compare both equations with (1-x^2-y^2).This serves one equation for z=x/y (separated variables).But is there any _constructive_ hint/algorithm to solve thetask (nd the special parameters of solution) by usingPoincare/Bendixon? Please help and let us look at Yoursolution. I am very interested. And until this I believe,that the good old Ljapunov always is a practicable way. Itis a lot of years ago when I was learning it and it shouldbe time to brush up.Alfred > addressed the subjectof integrating the 2nd order linear DE with constant coefcients, by aquadrature approach.While the trial solution method is almost universally employed for thisODE, in this particular case, quadrature is normally possible and hassome advantages.We assume:1. G + b*G + c*G = 0 on an open interval (a,b). At this point,we make no assumption as to whether or not there exists a G such thatthis relation holds true. The forthcoming deduction will tell us whathappens if there is. 2. G is continuous 3. G exists and is continuous5. G exists6. Let d =def sqrt(b^2 - 4*c)7. Let r =def -b/2 + d/28. There exists at most a nite number of points x in (a,b) suchspecial points.9. Let F be dened implicitly by: G(x) = F(x)*exp(r*x)10. d ~= 011. Let x1, x2, x3, ... be the sequence of points on the real axisobtained as follows: 1st, x1=a. 2nd, move to the right and let x2 bethe rst special point encountered. Continue in this fashion -- thenext special point is labeled x3, etc. Eventually, b will beencountered and will become xn for some n. Note that a (and/or b)might be special points.14. By 12, the term multiplying F is 0. We may divide by E since thisis never zero. The coefcient of F is 2*r + b which by (7) is d. Wehave:15. F + F*d = 016. By use of (9) again, wefind that F = (G-r*G)/E . Byconstruction of the current subinterval, each interior point is not aspecial point. Therefore, F ~= 0 within the subinterval. In (15),transpose and divide by F. Obtain:17: (F)/F = -d This is easily integrated, and18. ln(F(x)) = -d*x + c1 c1, c2, ... will be constants ontegration.19. F(x) = c2*exp(-d*x)20. Integrate a second time. The integral of exp(-d*x) has twodifferent forms, depending on whether or not d is 0. In this case, byassumption 10, d is not 0, and the integral of exp(-d*x) is-(1/d)*exp(-d*x) + constant. Apply (9), and obtain: G(x) = [c3*exp(-d*x) + c4] * exp( ((-b/2)+(d/2))*x ) 21. When we multiply the outside exp term into the sum, we obtain theclassic sum-of-two-exponentials form, with each exponential coefcienta characteristic root: G(x) = c3*exp( ((-b/2)-(d/2))*x ) + c4*exp( ((-b/2)+(d/2))*x )This completes the calculation of G on the i th subinterval. Byperforming the same calculation for all subintervals, and using thecontinuity assumptions for G and G, we can take limits, solve a systemof two equations which has a non-zero determinant, and show that c3,c4to the left must equal c3,c4 to the right. The constants ontegration are equal for all subintervals. Equation 21 holdsthroughout the entire interval (a,b), including the special points. Comments:a. If d=0, the integration at (20) gives F(x) = c2*x + c5, and thus,G(x) = (c2*x + c5)*exp(-b*x/2).b. The assumption of at most, a nite number of special points isspecic to the quadrature approach. (I do not see a convincing way itcan be avoided.) In any event, however, it is not required for thetrial solution method (but see d.3 below). c. One can take the function appearing on the right side of (21) andshow by direct substitution that it satises (1) -- that is, Equation1 does indeed have a solution.d. The point of the quadrature exercise can be stated as follows:--First, there is the question of how one nds the trial solutions inthe rst place. In this case, quadrature reveals them.--Second, there is always the possibility that a fortuitously foundtrial solution (or set) is in fact, NOT fully general. This isactually illustrated here by the case d=0. Another example is the caseof ODEs with envelopes. Quadrature when it is available tells usabsolutely that the functions found are the ONLY ones that can besolutions.--Third, the trial solution method only makes complete sense when wehave some kind of exterior uniqueness result that tells us that thegiven ODE with initial/boundary conditions has at most one solution. If so, a solution which is some combination of trial solutions is infact valid. However, it takes some work to develop a uniquenesstheorem. Quadrature provides a method of sidestepping this difculty.David Ziskindziskind@ntplx.net =| > [...]| > | > Obviously youre not giving a rigorous denition.| > | | > | Obviously?| > | > Yes.| | Lets see if you support your accusation.| | > | The as are dened by| > | | > | (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),| > | | > | where v=-1+mf^2.| > | > You seem often to have the attitude that equations speak for| > themselves. But if you had just left this equation by itself,| > it could have been referring to quaternion-valued functions| > a_1(m), a_2(m) and a_3(m). Just writing an equation doesnt tell| > us what the equation means.| | On the contrary, the equation clearly shows that there is a| factorization.But what kind of factorization? Just saying there is a factorizationhas little unambiguous content, since there are many different kindsof structures in which you might be intending to put a_1, a_2, anda_3.Nora Baron has been trying to get you to realize the differencebetween factoring a number and factoring a polynomial. Do you nowagree that there is one?| > a_1, a_2, and a_3 to be functions dened for algebraic integers m,| > with algebraic integer values, having the property that for each m| > they make the equation above hold for all x and y.| | Of course the ring is algebraic integers as thats been the ring| throughout the thread and is the ring in the paper Advanced Polynomial| Factorization.| | The factorization should hold for all f, m, x and y, in the ring.| | There is no need to mention functions.Yes there is. If you want to say (for example) that a_1 is divisibleby f for different values of m, then a_1 has to have values fordifferent values of m.This is more crucial than you think. Look at a somewhat bigger picturefor a minute here. Other people claim that because of a symmetrybetween the roots of the original polynomial, it has to be thateither all of them are coprime to 5, or that none of them are. Thissymmetry obviously doesnt preserve all the properties that they have,or they would have to be equal, but it preserves algebraic ones suchas being coprime to 5.Your argument against them indicates to the contrary that theresa distinction between the roots in terms of coprimeness to 5. Sowhatever it is you think allows you to distinguish between them(in terms of such algebraic properties) is bound to stand out assomething unusual.At m=0, you distinguish between them because one of them is nonzero.(This is consistent with usual theory, because the polynomial isreducible at m=0.) And then you take a step which is as far as wereconcerned just plain wrong, to say that the same as have commonfactors with f at m=0 as have them at another value of m.So what does this have to do with whether the as are functions ornot? Its simple. If the as arent functions, then selecting one ofthe as at one value of m doesnt help you select one of them for anyother value of m. Saying that weve taken a_3 to be the nonzero oneat m=0 tells us nothing about which of the three a values shouldbe taken to be a_3 at m=1. If you can evaluate a_3 at all allowedvalues of m, then its a function, because thats just what a functionis, whether you want to call it that or not.| > If that guess had been correct, then I would have said that the| > denition was at least close to rigorous, just leaving a couple of| > things implicit that should be explicit. All that it would take to| > make it rigorous would be a few clarifying words. (As I tried to| > explain to you before, you also couldnt be taking x to be a constant,| > for example.)| | Thats false. There is no requirement on the factorization that x not| be constant.|| Factorizations dont work that way.Im not saying that theres a general prohibition on having constantvariables in factorizations. Its just that holding x constantchanges the meaning. In particular, if x is a constant thefactorization isnt unique (up to permutation of the three factors),and you need that.For example, if x, a1, a2 and a3 are assumed to be complex numbers,from x^3-x = (x-a1)(x-a2)(x-a3)its NOT valid to deduce that a1, a2 and a3 are 0, 1, and -1,possibly in a different order. Its an invalid step in reasoning.If its irrelevant what the dependent variable is, then your claimthat the independent variable might turn out to be something otherthan m is nonsensical.| > Dont try to just skim past this issue of what the independent| > variable is in these functions.| | Factorizations are not dependent on functions.It depends on what algebraic structure the terms in the factorizationare in. The set of functions is one type of structure.If the independent variable is irrelevant, then why you dig in yourheels and stubbornly refuse to allow m to be it> Is there somethingwrong with taking the as to be functions of m?| > | | > | a^3 + 3va^2 -(v^3+1) = 0, where again v=-1+mf^2, and the as are its| > | roots.| > | | > | So you can solve for the as and get some result, instead of endlessly| > | talking.| > | > The usual solution of the cubic starts by changing the variable to| > get rid of the square term. So let b = a+v, and we get| > b^3 - 3v^2b + (v^3-1) = 0. The discriminant is| > | > D = (v^3-1)^2/4 - v^6| > = -(3/4)v^6 - (1/2)v^3 + (1/4)| > = -(3v^3 - 1)(v^3 + 1)/4.| > | > (although in number theory we use a different constant factor).| > This shows there are six values of v for which two of the roots| > coincide.| > | > By Cardans formula:| > | > b = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3},| > | > or| > | > a = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3} - v,| > | > | It might be especially interesting to see the result and see what| > | happens at v=-1.| | Readers should note that assuming the above is correct at v=-1, D=0,| so you get| | a = 1^{1/3} + 1^{1/3} + 1| | which gives you the three as, a_1=0, a_2=0, a_3=3.| | Now consider Keith Ramsays reply.| | > Because D = (v+1)(1-3v^3)(1+v+v^2)/4, near v=-1, D is approximately| > v+1. The two values of sqrt(D) trade places with each other as we| > sqrt(D) with -sqrt(D) wouldnt change the value of a. But the two| > cube roots are chosen so that at v=-1, for example, they are either| > both 1, or one is (-1+sqrt(3)*i)/2 and the other is (-1-sqrt(3)*i)/2,| > or the rst is (-1-sqrt(3)*i)/2 and the other is (-1+sqrt(3)*i)/2.| >| > So changing the sign on the square root of D is equivalent to leaving| > the rst root (where both cube roots are taken to be 1) alone, while| > exchanging the other two roots. As I said before, if you take a loop| > around v=-1, the value of one of the roots comes back to the value of| > the other one.| | Um, did *anyone* see an answer in there?You asked a very broad question about what happens at v=-1.Theres a lot going on at v=-1, and I just described part of it. If you wantedonly to see the values of a_1, a_2, and a_3 at v=-1, you could have beenmore specic.The root youre calling a_3 is being relatively boring, but the othertwo roots are much like the square roots of v+1, which do this funkything of trading places if we take v in a loop around v=-1, as wellas coinciding right at v=-1.| > | Reminder to readers, Keith Ramsay has brought up a continuity| > | question, which actually is irrelevant, but what the hell?| > | > I was trying to convince you that its unsafe to assume that the| > values of a_1, a_2, and a_3 extrapolate in a natural way from one| > value of m to another. This would be needed, for example, for a_2| > to be always divisible by f or anything like that. Otherwise, how| > could it possibly be true that its one particular one of the three| > that has a certain divisibility property, if the manner in which| > weve decided which of the three roots to assign to each of the| > three variables in an arbitrary way depending on the value of m?| > Im sorry if the relevance doesnt seem clear to you.| | There is no relevance. In polynomial equations you *can* have it| where certain roots have a factor while others do not.You just havent gotten the point yet.| > The one obvious way to attempt to extrapolate values of a_1, a_2| > and a_3 from one value of m to another is by continuity. Given a| > path in the complex plane which avoids all the points where two| > of the roots coincide, we can see what happens to the a_i if we| > continuously change m along that path, while letting the a_i also| > change continuously as we go. Unfortunately (or fortunately,| > actually) which a_i is which when we reach the end of the path| > will depend upon which path we took from one endpoint to the other.| > So if I have a given value of a at m=0 (say a=0), the question of| > which of the three values at m=1 corresponds to it is meaningless| > unless I make an arbitrary convention.| | Actually you simply take the three values of the cuberoot operator,| just like with square roots you take the two values of the square root| operator.Theres a square root involved too, and you need that as well.| For those who are confused simply remember that x^3 - 1 = 0, gives you| three values for x.But you havent given a convention for associating the values of thesquare and cube roots between two different values of v. Which cuberoot of 2-3i should be used for a_1, at the value of v which makesthe expression under the rst cube root equal to 2-3i? Its alwayssomewhat arbitrary. Yet you need for g_1 to have that samefactor in general. Why should one of the cube roots at that pointmake g_1 divisible by f, but the other two cube roots of 2-3i notmake g_2 and g_3 divisible by f too? Its not just getting three values,but getting three values which retain their identities between differentvalues of m.| > Thats essentially what prevents them from being continuous| > everywhere.| | Looks to me like you put down a lot of words Keith Ramsay, yet somehow| managed to not even refer back to| | a = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3} - v| | from your work before.You skim too fast. Its partly the fault of the rest of us forinundating you with material faster than you can really digest.(Calling us liars is a phony way to pretend youve understoodwell enough what was said.) Remember what Paul Halmos says about whyhe went into mathematics rather than some other eld: because heprefers words to numbers. Stan Ulam also said he dislikes readingpapers that have a high ratio of formulas to descriptions ofreasoning.This: So changing the sign on the square root of D is equivalent to leaving the rst root (where both cube roots are taken to be 1) alone, while exchanging the other two roots.*is* in fact referring back to the cubic formula. Theres a slightlytricky feature of the formula, in that (letting w=(-1+sqrt(3)*i)/2),the other two roots are w((1-v^3)/2 + sqrt(D))^{1/3} + w^2((1-v^3)/2 - sqrt(D))^{1/3} - v w^2((1-v^3)/2 + sqrt(D))^{1/3} + w((1-v^3)/2 - sqrt(D))^{1/3} - vwhere the three cube roots in the rst term are associated with thethree cube roots in the second term in the opposite order, as it were.If we restrict ourselves to a neighborhood of v=-1, the cube root canbe dened consistently, since were away from the points where wehave to take the cube root of 0. But the square root has its usualinteresting behavior near there.| > I dont have the time today, but what one can do is to show that| > for each epsilon>0, there exists a delta>0 having the property that| > if |v+1| < delta, then one of the roots is within epsilon of the| > nonzero root at v=-1 (what was it, a=1?), while the other two satisfy| > |a^2/(v+1) - C| < epsilon where C is some constant (probably just 1).| > Then we can consider the values v = -1+e^{it}. For small epsilon,| > the inequality forces a/e^{it/2} to be close to one of the square| > roots of C. Then we can carry out the same argument as last time| > but with the two sets being the t for which |a/e^{it/2} - sqrt(C)|| > is smaller and the t for which |a/e^{it/2} + sqrt(C)| is smaller.| | Well, if you dont have the time its not really worth mentioning, now| is it?If you really think facts shouldnt be described on sci.math withoutpresenting complete proofs immediately, youre entitled to youropinion, but I dont think many people will agree with you.[...]| Challenging a paper requires that you nd that it did not begin with| a truth, or that youfind a break in the logical chain.A break sometimes occurs when there are ambiguous steps.The claim that g1 must have that same factor (of f) in generalsuffers from a certain weakness of phrasing. I asked whether ingeneral could be understood as for all m satisfying the assumptionsbut this was at the point in the discussion where you were saying thatfor subtle reasons m might not be the independent variable. Ingeneral requires some context.If it means anything like what it sounds like it means, however, itdoesnt follow. First, its not a simple, obvious step like you keepclaiming. There are much simpler steps (like concluding that theproduct of two things that each are coprime to f is also coprime to f)which can be expanded into little sub-proofs (and if you mean toexplain it to people who arent mathematically trained, those steps*should* be expanded). And this step is more complicated. Youroversimplied examples of when the same principle applies topolynomials follow from a result called Gausss lemma. In abstractalgebra textbooks they typically take a page or so to prove Gaussslemma. Whatever principle it is you think youre applying is morecomplicated than that too.Second, its not something I can see how to prove myself, nor is ita standard result.Third, your explanations for why it works amount to those simpliedexamples plus ridicule directed at people who doubt it. You donteven manage to state what the general principle is.And nally, standard results that other people have been citing implythat the conclusion you draw at that step is false-- at some othervalues of the variables, with m<>0, none of the gs are divisible by f.You say it just works that way, but it just doesnt work that way.Keith Ramsay =me, four! well enough what was said.) Remember what Paul Halmos says about whyhe went into mathematics rather than some other eld: because heprefers words to numbers. Stan Ulam also said he dislikes readingpapers that have a high ratio of formulas to descriptions ofreasoning.--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...La Troi Phases dExploitation de la Protocols des Grises de Kyoto:(FOSSILISATION [McCainanites?] (TM/sic))/BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/ bushb.htm (content partiale, below): 17 -- LATTEMPTER de COUP DETAT, 3/30/81 23 -- Le FIN dHISTOIRE 24 -- LORDEUR du MONDE NOUVEAU 25 -- THYROID STORK !?! =Can anybody recommend a good reference for reading up on methods of calculating elliptic integrals? Ive had Derek Lawdens book Elliptic Functions and Applications recommended for reading up on the theory and applications but I suspect that it may not be a good reference for the numerics. Am I right about that? =as they are outside the space station, they are probably in a vacuum so itheard sod all.> I cant even get started on this one. Anyone help me??>> Evil Fuzzles may look cute, but they can really make a loud noise when> they ROAR!!!> A 1-foot tall Green Evil Fuzzle can register 10 NSU (Neopian Sound> Units) on a sound-measuring device that is 5 feet away. Yellow Evil> Fuzzles are twice as loud, and Red Fuzzles... THREE times as loud as> their yellow counterparts!>> So anyway, one day Zorlon-IX the Grundo was out doing a bit of> maintenance on the exterior of the Space Station, and before him he> saw, to his fright, the biggest Evil Fuzzle he had ever seen. It was 5> feet high, 10 feet away and red! It looked straight at Zorlon, and> gave the loudest ROAR it could!!!>> How many NSU did Zorlons sound-measuring device register? A 1-foot tall Green Evil Fuzzle can register 10 NSU (Neopian Sound> Units) on a sound-measuring device that is 5 feet away. Yellow Evil> Fuzzles are twice as loud, and Red Fuzzles... THREE times as loud as> their yellow counterparts!>> So anyway, one day Zorlon-IX the Grundo was out doing a bit of> maintenance on the exterior of the Space Station, and before him he> saw, to his fright, the biggest Evil Fuzzle he had ever seen. It was 5> feet high, 10 feet away and red! It looked straight at Zorlon, and> gave the loudest ROAR it could!!!>> How many NSU did Zorlons sound-measuring device register?>As Z-X was ten feet away from red loud mouth, its was 1/4 as loud were itat 5 feet. Now reds are 6 times louder than greens. Thus a red, 10 feetaway is as loud as 6 * 1/4 = 3/2 greens at ve feet. Now greens at vefeet shout at 10 NSUs, thence a sci.physics problem. However to continueabove the din, Z-X would hear 10 * 3/2 = 15 NSU if its a linear measure.If its a logarithmic measure, some details about NSU are needed. =0, assuming that this is a recognized measure of the NSU scale!> I cant even get started on this one. Anyone help me??>> Evil Fuzzles may look cute, but they can really make a loud noise when> they ROAR!!!> A 1-foot tall Green Evil Fuzzle can register 10 NSU (Neopian Sound> Units) on a sound-measuring device that is 5 feet away. Yellow Evil> Fuzzles are twice as loud, and Red Fuzzles... THREE times as loud as> their yellow counterparts!>> So anyway, one day Zorlon-IX the Grundo was out doing a bit of> maintenance on the exterior of the Space Station, and before him he> saw, to his fright, the biggest Evil Fuzzle he had ever seen. It was 5> feet high, 10 feet away and red! It looked straight at Zorlon, and> gave the loudest ROAR it could!!!>> How many NSU did Zorlons sound-measuring device register? =In sci.math, cant even get started on this one. Anyone help me??> Evil Fuzzles may look cute, but they can really make a loud noise when> they ROAR!!!> A 1-foot tall Green Evil Fuzzle can register 10 NSU (Neopian Sound> Units) on a sound-measuring device that is 5 feet away. Yellow Evil> Fuzzles are twice as loud, and Red Fuzzles... THREE times as loud as> their yellow counterparts!> So anyway, one day Zorlon-IX the Grundo was out doing a bit of> maintenance on the exterior of the Space Station, and before him he> saw, to his fright, the biggest Evil Fuzzle he had ever seen. It was 5> feet high, 10 feet away and red! It looked straight at Zorlon, and> gave the loudest ROAR it could!!!> How many NSU did Zorlons sound-measuring device register?If Fuzzles can breathe space (which is hard to do) and theNSU can also be measured in space, and a 2-foot-high Fuzzleis twice as loud as a 1-foot-high one, then the problem can beset up this way.You know how loud a Green Fuzzle 1 foot high is.multiply three things together. (No, not Fuzzles. Althoughits a thought, the problem indicates no method of Fuzzlereproduction.)And there you have it: the problem setup.One other thing. Youve not mentioned *why* Fuzzles like to roar.However, because space is extremely cold, I have a solution:warm one up. After all, everyone knows that ....... happiness is a warm Fuzzle.:-) ;-) :-)-- #191, ewill3@earthlink.netIts still legal to go .sigless. =Can anyone help come up with some equations to produce smoothtransitions for values moving from one point to another?Im looking for an equation that will be called at a set interval as avalue changes and the equation should produce the value at any giventime.For each call, the signature would look like this:d - total duration for the transitiont - the time into transition equation for this callb - the starting valuec - the change in valuea linear equation would be something like this:v = (t/d) * c + b;The percentage of time passed, times the change in value, plus thebeginning value.But the transition looks very phony.Any help would be appreciated.John > z^(2) = y^(2) - x^(2)> 2dz/z = 2dy/y - 2dx/x>> You seem to be saying that the differential of z^2 is 2 dz/z. That> is incorrect. It is 2 z dz.I probably wasnt clear enough in the initial post. The notation dz, dyetc. in this case has nothing to do with differentials. Its simply ashorter way of writting Delta{z}, Delta{y} etc., ie dz is the uncertaintyin z.> -- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html> To solve Linear Programs: .../LPSolver.html> r c A game: .../Keynes.html> v s a Whether strength of body or of mind, or wisdom,or> i m p virtue, are found in proportion to the power orwealth> e a e of a man is a question t perhaps to bediscussed by> n e . slaves in the hearing of their masters, buthighly> @ r c m unbecoming to reasonable and free men in searchof> d o the truth. -- Rousseau =Two traditional ways to decode Reed-Solomon codes include theMassey-Berlekamp and Euclidean decoding algorithms.Can someone please explain the underlying theory on which theEuclidean algorithm is based, or perhaps give some pointers inliterature where it is covered?Does it have anything to do with the Euclidean algorithm used inabstract algebra to compute the GCD?Your time, effort and suggestions will be greatly appreciatedJaco =Hea, I am asking for a question about the expected value.Suppose the postive continuous random variable X, Y, Z1, Z2,Z3,Z4. Moreover,Y, Z1,Z2,Z3,Z4 are independent. The pdf of them are given. They satisfy thefollowing relationship. __ | Y if YY>Z | Z2 if 3Z>Y>2Z | Z3 if 4Z>Y>3Z | Z4 other --- Hea, I am asking for a question about the expected value.> Suppose the postive continuous random variable X, Y, Z1, Z2,Z3,Z4. Moreover,> Y, Z1,Z2,Z3,Z4 are independent. The pdf of them are given. They satisfy the> following relationship. __> | Y if Y X | Z1 if 2Z>Y>Z> | Z2 if 3Z>Y>2Z> | Z3 if 4Z>Y>3Z> | Z4 other> ---> I assume that Z is some constant. and y(t) is the pdf of Y.Using Zp0 = Pr {YHea, I am asking for a question about the expected value.>>Suppose the postive continuous random variable X, Y, Z1, Z2,Z3,Z4. Moreover,>>Y, Z1,Z2,Z3,Z4 are independent. The pdf of them are given. They satisfy the>>following relationship.>> __>> | Y if Y>X > | Z1 if 2Z>Y>Z>> | Z2 if 3Z>Y>2Z>> | Z3 if 4Z>Y>3Z>> | Z4 other>> --->I assume that Z is some constant. and y(t) is the pdf of Y.>>Using>> Z>p0 = Pr {Y -oo >> 2Z>p1 = Pr {Z Z> 3Z>p2 = Pr {2Z 2Z>> 4Z>p3 = Pr {3Z 3Z>> oo>p4 = Pr {4Z 4Z>>(Sum p_i = 1)>>Then E(X) = p0*E(Y)+p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)>This should be p0*E(Y|Y<=Z) + the rest. The difference being that E(Y|Y<=Z) = int(-oo,Z)(t*y(t) dt) whereas E(Y) = int(-oo,+oo)(t*y(t) dt).>It is immaterial whether Y,Z1,Z2,Z3,Z4 are independent or not. If for>instance Y=Z1=Z2=Z3=Z4 then E(X)=E(Y).>Jon Miller =Suppose X, Y, Z are positive random variable, the pdf of these variables aregiven. How to evaluate the conditional expected value?E(X | Y < X < Z).I develop the following expression.E(X | Y < X < Z) = int_{0}^{infty}int_{0}^{infty} int_{y}^{z} x f_X(x)dx * f_Y(y)dy * f_Z(z)dz.--ZHANG Yan >Hea, I am asking for a question about the expected value.>Suppose the postive continuous random variable X, Y, Z1, Z2,Z3,Z4. Moreover,>>Y, Z1,Z2,Z3,Z4 are independent. The pdf of them are given. They satisfy the>>following relationship.>> __>> | Y if Y>X > | Z1 if 2Z>Y>Z>> | Z2 if 3Z>Y>2Z>> | Z3 if 4Z>Y>3Z>> | Z4 other>> --->I assume that Z is some constant. and y(t) is the pdf of Y.>>Using>> Z>p0 = Pr {Y -oo >> 2Z>p1 = Pr {Z Z> 3Z>p2 = Pr {2Z> 4Z>p3 = Pr {3Z 3Z>> oo>p4 = Pr {4Z 4Z>>(Sum p_i = 1)>Then E(X) = p0*E(Y)+p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)> This should be p0*E(Y|Y<=Z) + the rest. The difference being that > E(Y|Y<=Z) = int(-oo,Z)(t*y(t) dt) whereas E(Y) = int(-oo,+oo)(t*y(t) dt).Yes, of course, but E(Y|Y<=Z) = int(-oo,Z)(t*y(t) dt)/p0, thereforeyou may writeE(X) = p0`*E(Y|><=Z) + p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)orE(X) = int(-oo,Z)(t*y(t) dt + p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)Horst Suppose X, Y, Z are positive random variable, the pdf of these variables are>given. How to evaluate the conditional expected value?>>E(X | Y < X < Z).>>I develop the following expression.>>E(X | Y < X < Z) = int_{0}^{infty}int_{0}^{infty} int_{y}^{z} x f_X(x)>dx * f_Y(y)dy * f_Z(z)dz.>Well, the integral should come from -innity. If your distributions start at 0, its the same, but if youre using the normal distribution youve thrown away half of your data.Jon Miller >Hea, I am asking for a question about the expected value.>>Suppose the postive continuous random variable X, Y, Z1, Z2,Z3,Z4. Moreover,>>Y, Z1,Z2,Z3,Z4 are independent. The pdf of them are given. They satisfy the>>following relationship.>> __>> | Y if Y>X > | Z1 if 2Z>Y>Z>> | Z2 if 3Z>Y>2Z>> | Z3 if 4Z>Y>3Z>> | Z4 other>> --->I assume that Z is some constant. and y(t) is the pdf of Y.>>Using>> Z>p0 = Pr {Y -oo >> 2Z>p1 = Pr {Z Z>> 3Z>p2 = Pr {2Z 2Z>> 4Z>p3 = Pr {3Z 3Z>> oo>p4 = Pr {4Z 4Z>>(Sum p_i = 1)>>Then E(X) = p0*E(Y)+p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)>>This should be p0*E(Y|Y<=Z) + the rest. The difference being that >>E(Y|Y<=Z) = int(-oo,Z)(t*y(t) dt) whereas E(Y) = int(-oo,+oo)(t*y(t) dt).>Yes, of course, but E(Y|Y<=Z) = int(-oo,Z)(t*y(t) dt)/p0,>Skitts Law (among other appelations): Every spelling (in this case, formula) correction contains its own error.Well, to OP, youve now got the right formula.This is why mathematicians avoid calculations.> therefore you may write>>E(X) = p0`*E(Y|><=Z) + p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)>>or>>E(X) = int(-oo,Z)(t*y(t) dt + p1*E(Z1)+p2*E(Z2)+p3*E(Z3)+p4*E(Z4)>Jon Miller =David Berniers post about calculating pi(10^15) with PrimeCountH.javalead me to thinking about an interesting feature of the program that Ihavent commented about a lot recently, which is that it uses someterms of the explicit prime counting function, compressed in that itassumes even N, to speed things up. That is I use terms from THEexplicit prime counting function itself--compressed--in the program.Those terms are, with even N,if N<9 pi(N) = oor(N/2);if N<25 pi(N) = oor(N/2) - oor((N-4)/6); if N<49 pi(N) = oor(N/2) - oor((N-4)/6) - oor((N-16)/10) +oor((N-16)/30)if N<121 pi(N) = oor(N/2) - oor((N-4)/6) - oor((N-16)/10) +oor((N-16)/30) - oor((N-36)/14) + oor((N-22)/42); if N>121 pi(N) = oor(N/2) -oor((N-4)/6) - oor((N-16)/10) +oor((N-16)/30) - oor((N-8)/14) + oor((N-22)/42) +oor((N-106)/70) - oor((N-106)/210) +2+...So you can see the rst 8 terms of the prime counting functioncompressed, where before I would pull that 2 into its proper place inone of the terms, but realized the computer would be a bit fasterletting it sit out, and now Im not sure where it goes.You may be wondering why I split out like that, and its because itdoesnt work if you allow negative numbers, and as the oor doesntgo to 0, when the result is negative, I split it out.Like, if oor(N-106)/70) is negative for your N, and that goes intothe calculation, itll give the wrong answer.The full prime counting function is of course innite in size, but Ithought it worth it to point out that Ive given the rst few termsof the compressed prime counting function, where the compression isthat it knows that 2 is prime, which cuts into the total number ofterms.So what Ive given above, at least for the compressed explicit primecounting function, are THE rst few terms, and of course, there arean innite number of terms.James Harris =How do I answer this problem?Show that x + 2x + 3a is a factor of ax + x + 5ax + 2x + 3aI understand I have to use algebraic long division tofind prove that x +2x + 3a is a factor.I know how do it with the example in this pagehttp://www.mathsa.fsnet.co.uk/Algebraic%20Long% 20Division.htm, but not whenit involves more variables than just x.Any help would be appreciated. How do I answer this problem?>> Show that x + 2x + 3a is a factor of ax + x + 5ax + 2x + 3a>> I understand I have to use algebraic long division tofind prove that x +> 2x + 3a is a factor.>> I know how do it with the example in this page> http://www.mathsa.fsnet.co.uk/Algebraic%20Long%20Division.htm , but notwhen> it involves more variables than just x.View this with a xed-width font: 1 a -------------1 2 3a |1 a+2 5a 3a^2 1 2 3a -------- a 2a 3a^2 a 2a 3a^2 ----------- Clive Toothhttp://www.clivetooth.dk =Apologies if this isnt the right group, but googling has been lessthan helpful this morning.Im interested in taking some 3rd or 4th year college-level mathclasses (in particular applied math or numerical analysis) as part oifelong learning / continuing education. The catch is that due tolimited time, Id prefer to do them in a self-paced online orcorrespondence format.Is this available anywhere?Ranjan Bagchi =I am hoping tofind a simple way to determine how many factors of 2are in any number. I have been searching and I will continue; however, I stopped by toask. I understand that x mod 2 can tell if it is or isnt divisible bytwo. What Id like is a one step formula that can express how many factorsof 2 are in X like 0 to ?. I thought to ask because it may have been done already. Ernst Message-id: hoping tofind a simple way to determine how many factors of 2>are in any number.> I have been searching and I will continue; however, I stopped by to>ask.>> I understand that x mod 2 can tell if it is or isnt divisible by>two.But that doesnt tell you how many, just that it has at least one (if even) ornone (if odd).>> What Id like is a one step formula that can express how many factors>of 2 are in X like 0 to ?.Convert the number to base two and count the number of contiguous leastsignicant zeroes.144 base ten is 10010000 base two. It has 4 least signicant 0s, so it has 4factors of 2. Check: 144 = 3*3*2*2*2*2Now maybe you dont consider base conversion a one-step formula (its asingle button push on my calculator), but you may not have to always do theconversion. In my text based Collatz program that I use to study arbitrarylength binary numbers, I use a modied rule: instead of dividing by 2, Iremove all the factors of 2 in a single step. Not surprisingly, I call this theFactoring Rules. The effect of this is the modied Collatz sequence containsonly odd numbers.To avoid the overhead of doing base conversion, I do all the math in binary.The stopping point (1) doesnt need to be converted back to decimal, so I onlyneed to do the conversion once from decimal to binary at the start of thesequence. The limited math involved in the Collatz process is relatively easyto implement using strings of 1s and 0s. When I need to remove (or count) allthe factors of 2 from a number, a simple regular expression in perl can do thisin a single step.>> I thought to ask because it may have been done already. >Ernst--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm >I am hoping tofind a simple way to determine how many factors of 2>>are in any number.>> I understand that x mod 2 can tell if it is or isnt divisible by>>two.>>But that doesnt tell you how many, just that it has at least one (if even) or>none (if odd).> What Id like is a one step formula that can express how many factors>>of 2 are in X like 0 to ?.>>Convert the number to base two and count the number of contiguous least>signicant zeroes.>>Now maybe you dont consider base conversion a one-step formula (its a>single button push on my calculator), but you may not have to always do the>conversion. In my text based Collatz program that I use to study arbitraryIf you take the expression x and not x-1, its equal to the largest powerof two dividing x (except when x=0). Then you can take the base-2 log ofthat to extract the actual exponent. -- Erick >I am hoping tofind a simple way to determine how many factors of 2>>are in any number.> I understand that x mod 2 can tell if it is or isnt divisible by>>two.>>But that doesnt tell you how many, just that it has at least one (if even) or>none (if odd). What Id like is a one step formula that can express how many factors>>of 2 are in X like 0 to ?.>>Convert the number to base two and count the number of contiguous least>signicant zeroes.>>Now maybe you dont consider base conversion a one-step formula (its a>single button push on my calculator), but you may not have to always do the>conversion. In my text based Collatz program that I use to study arbitrary> If you take the expression x and not (except when x=0). Then you can take the base-2 log of> that to extract the actual exponent.> -- ErickNeat trick. It would never occur to me to do it that way.Its not necessary for my perl programs since they areworking directly in binary (albeit as text). But I alsouse Python to take advantage of its long integer support,so this factor counting might come in handy.Two problems came up.First, although there are bit-wise and, or and xoroperators that can work on long integers, there appears to be no bit-wise not operator. Workaround: not x-1 is how you form a 2s complement, and since 2s complement is how negative numbers are represented,x and not x-1becomesx & -xin Python.Second problem was that there is no base 2 logarithm functionin Python. Workaround: from the rules of logarithms, logs of one base can be caculated from logs of another base (such as base e, which Python does have a function for): log_a(x) = log_b(x)/log_b(a)So we can make a simple Python formula to count factors of 2:factors = log(x & -x) / log(2)Although this works in theory, it doesnt work in practice.As the test program below shows, it works ne for reasonably sized numbers, but breaks down for numbers exceeding 1023 bits.Alas, much of my interest is in unreasonbly sized numbers,so this factor counting ends up being of no use to me.# start log2.py#from math import logh = long(2)log2 = log(2) #constant, only needs to be calculated oncej = 1000while j<1030: k = h**j f = log(k & -k)/log2 print 2 raised to the power of,j, has,f, factors of 2 j = j + 1# end log2.py# Test results## does not work for x > 1023 bits## 2 raised to the power of 1000 has 1000.0 factors of 2# 2 raised to the power of 1001 has 1001.0 factors of 2# 2 raised to the power of 1002 has 1002.0 factors of 2# 2 raised to the power of 1003 has 1003.0 factors of 2# 2 raised to the power of 1004 has 1004.0 factors of 2# 2 raised to the power of 1005 has 1005.0 factors of 2# 2 raised to the power of 1006 has 1006.0 factors of 2# 2 raised to the power of 1007 has 1007.0 factors of 2# 2 raised to the power of 1008 has 1008.0 factors of 2# 2 raised to the power of 1009 has 1009.0 factors of 2# 2 raised to the power of 1010 has 1010.0 factors of 2# 2 raised to the power of 1011 has 1011.0 factors of 2# 2 raised to the power of 1012 has 1012.0 factors of 2# 2 raised to the power of 1013 has 1013.0 factors of 2# 2 raised to the power of 1014 has 1014.0 factors of 2# 2 raised to the power of 1015 has 1015.0 factors of 2# 2 raised to the power of 1016 has 1016.0 factors of 2# 2 raised to the power of 1017 has 1017.0 factors of 2# 2 raised to the power of 1018 has 1018.0 factors of 2# 2 raised to the power of 1019 has 1019.0 factors of 2# 2 raised to the power of 1020 has 1020.0 factors of 2# 2 raised to the power of 1021 has 1021.0 factors of 2# 2 raised to the power of 1022 has 1022.0 factors of 2# 2 raised to the power of 1023 has 1023.0 factors of 2# 2 raised to the power of 1024 has 1.#INF factors of 2# 2 raised to the power of 1025 has 1.#INF factors of 2# 2 raised to the power of 1026 has 1.#INF factors of 2# 2 raised to the power of 1027 has 1.#INF factors of 2# 2 raised to the power of 1028 has 1.#INF factors of 2# 2 raised to the power of 1029 has 1.#INF factors of 2 >I am hoping tofind a simple way to determine how many factors of 2>>are in any number.>> I understand that x mod 2 can tell if it is or isnt divisible by>>two.>But that doesnt tell you how many, just that it has at least one (if even) or>none (if odd).> What Id like is a one step formula that can express how many factors>>of 2 are in X like 0 to ?.>>Convert the number to base two and count the number of contiguous least>signicant zeroes.>>Now maybe you dont consider base conversion a one-step formula (its a>single button push on my calculator), but you may not have to always do theconversion. In my text based Collatz program that I use to study arbitrary> If you take the expression x and not x-1, its equal to the largest power> of two dividing x (except when x=0). Then you can take the base-2 log of> that to extract the actual exponent.> -- Erick That is what I was hoping for. Ernst >I am hoping tofind a simple way to determine how many factors of 2>>are in any number.> I understand that x mod 2 can tell if it is or isnt divisible by>>two.>>But that doesnt tell you how many, just that it has at least one (if even) or>none (if odd).> What Id like is a one step formula that can express how many factors>>of 2 are in X like 0 to ?.>Convert the number to base two and count the number of contiguous least>signicant zeroes.>>Now maybe you dont consider base conversion a one-step formula (its a>single button push on my calculator), but you may not have to always do the>conversion. In my text based Collatz program that I use to study arbitrary> If you take the expression x and not x-1, its equal to the largest power> of two dividing x (except when x=0). Then you can take the base-2 log of> that to extract the actual exponent.> -- Erick> Neat trick. It would never occur to me to do it that way.> Its not necessary for my perl programs since they are> working directly in binary (albeit as text). But I also> use Python to take advantage of its long integer support,> so this factor counting might come in handy.> I need to learn about the gmp library; however, I dont see a log2function for really large values either. Gcc seems to have a log2function for data type oat and double. Ernst I admit I know nothing of perl. Message-id: hoping tofind a simple way to determine how many factors of 2>are in any number.>> I understand that x mod 2 can tell if it is or isnt divisible by>two.>But that doesnt tell you how many, just that it has at least one (if>even) or>>none (if odd).> What Id like is a one step formula that can express how many factors>of 2 are in X like 0 to ?.>>Convert the number to base two and count the number of contiguous least>>signicant zeroes.>>Now maybe you dont consider base conversion a one-step formula (its>a>single button push on my calculator), but you may not have to always do>the>>conversion. In my text based Collatz program that I use to study>arbitrary> If you take the expression x and not x-1, its equal to the largest>power>> of two dividing x (except when x=0). Then you can take the base-2 log of>> that to extract the actual exponent.> -- Erick>> Neat trick. It would never occur to me to do it that way.>> Its not necessary for my perl programs since they are>> working directly in binary (albeit as text). But I also>> use Python to take advantage of its long integer support,>> so this factor counting might come in handy.>> I need to learn about the gmp library; however, I dont see a log2>function for really large values either. Gcc seems to have a log2>function for data type oat and double.Read the rest of my message. You dont have to have a Log2 function. As long asyou have ANY log function, you can calculate it. log_2(x) = log(x)/log(2)Here the standard natural log function log() is used to calculate log_2. Notethat log(2) is a constant, so you only need to calculate it once. That willsave time if x is inside a loop.>> Ernst>> I admit I know nothing of perl.Yes, but you should be able to look at that program example and gure out howto do the same thing in C. If you get hung up on syntax, just ask. BTW, the &operator is bit-wise AND in Python. Im sure there is an equivalent in C.--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm Message-id: >>I am hoping tofind a simple way to determine how many factors of 2>are in any number.>> I understand that x mod 2 can tell if it is or isnt divisible by>two.>>But that doesnt tell you how many, just that it has at least one (if> even) or>>none (if odd).> What Id like is a one step formula that can express how many factors>of 2 are in X like 0 to ?.>>Convert the number to base two and count the number of contiguous least>>signicant zeroes.>Now maybe you dont consider base conversion a one-step formula (its> a>>single button push on my calculator), but you may not have to always do> the>conversion. In my text based Collatz program that I use to study> arbitrary> If you take the expression x and not x-1, its equal to the largest> power>> of two dividing x (except when x=0). Then you can take the base-2 log of>> that to extract the actual exponent. -- Erick>> Neat trick. It would never occur to me to do it that way.>> Its not necessary for my perl programs since they are>> working directly in binary (albeit as text). But I also>> use Python to take advantage of its long integer support,>> so this factor counting might come in handy.>> I need to learn about the gmp library; however, I dont see a log2>function for really large values either. Gcc seems to have a log2>function for data type oat and double.> Read the rest of my message. You dont have to have a Log2 function. As long as> you have ANY log function, you can calculate it. > log_2(x) = log(x)/log(2)> Here the standard natural log function log() is used to calculate log_2. Note> that log(2) is a constant, so you only need to calculate it once. That will> save time if x is inside a loop.>> Ernst>> I admit I know nothing of perl.> Yes, but you should be able to look at that program example and gure out how> to do the same thing in C. If you get hung up on syntax, just ask. BTW, the &> operator is bit-wise AND in Python. Im sure there is an equivalent in C.I see what you mean from the GNU C docs...These functions are GNU extensions. The name exp10 is preferred, sinceit is analogous to exp and exp2.Function: double log (double x)Function: oat logf (oat x)Function: long double logl (long double x) These functions compute the natural logarithm of x. exp (log (x))equals x, exactly in mathematics and approximately in C.If x is negative, log signals a domain error. If x is zero, it returnsnegative innity; if x is too close to zero, it may signal overow.Function: double log10 (double x)Function: oat log10f (oat x)Function: long double log10l (long double x) These functions return the base-10 logarithm of x. log10 (x)equals log (x) / log (10).Function: double log2 (double x)Function: oat log2f (oat x)Function: long double log2l (long double x)These functions return the base-2 logarithm of x. log2 (x) equals log(x) / log (2).Function: double logb (double x)Function: oat logbf (oat x)Function: long double logbl (long double x) These functions extract the exponent of x and return it as aoating-point value. If FLT_RADIX is two, logb is equal to oor(log2 (x)), except its probably faster.If x is de-normalized, logb returns the exponent x would have if itwere normalized. If x is innity (positive or negative), logb returns@math{@innity{}}. If x is zero, logb returns @math{@innity{}}. Itdoes not signal.Function: int ilogb (double x)Function: int ilogbf (oat x)Function: int ilogbl (long double x)snip end...I admit I have not used it yet but it seems to have one.Between gcc and gmp library it looks like number theory has been aconsideration. What I am up to these days is trying to decide what my next hailstonegenerator will accomplish. Oh well.. all the discovery had me running and I enjoyed it alleven the errors.Ernst =I will stop bothering you all now because I never pursued a Mathematicsdegree beyond a bachelors andshouldnt be proclaiming crazy and false ideas here. I will stop bothering you all now because I never pursued a Mathematics>degree beyond a bachelors and>shouldnt be proclaiming crazy and false ideas here.Look, you can stop faking posts. It doesnt take a genius to see thatthis is a forgery; in addition, if this is the best you can come upwith, then you may think you are a wit, but you are only half-right...(According to his own report, James never pursued a mathematics degreeat all, and never took any mathematics beyond calculus; and it shows) Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.eduFor a second, just a second, I felt a glimmer of hope.> I will stop bothering you all now because I never pursued a Mathematics> degree beyond a bachelors and> shouldnt be proclaiming crazy and false ideas here.>> I will stop bothering you all now because I never pursued a Mathematics> degree beyond a bachelors and> shouldnt be proclaiming crazy and false ideas here.> Your usenet headers show you to be at the University of Pennsylvania. Perhaps their abuse department would be interested in knowing that youve used their news server to post message with forged headers. >I will stop bothering you all now because I never pursued a Mathematics>degree beyond a bachelors and>shouldnt be proclaiming crazy and false ideas here.> Look, you can stop faking posts. It doesnt take a genius to see that> this is a forgery; in addition, if this is the best you can come up> with, then you may think you are a wit, but you are only half-right...> (According to his own report, James never pursued a mathematics degree> at all, and never took any mathematics beyond calculus; and it shows)> == =Why do you take so much trouble to expose such a reasoner as> Mr. Smith? I answer as a deceased friend of mine used to answer> on like occasions - A mans capacity is no measure of his power> to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more;> and a long purse, which does most of all. He has made at least> ten publications, full of gures few readers can critize. A great> many people are staggered to this extend, that they imagine there> must be the indenite something in the mysterious all this.> They are brought to the point of suspicion that the mathematicians> ought not to treat all this with such undisguised contempt,> at least.> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan> == =Arturo Magidin> magidin@math.berkeley.eduI thought it was pretty funny. I enjoyed it. It was a lot betterthan most of the baggage on sci.math. =Well, no, he didnt. JSH can be imitated, but never duplicated. I will stop bothering you all now because I never pursued a Mathematics> degree beyond a bachelors and> shouldnt be proclaiming crazy and false ideas here.> Your usenet headers show you to be at the University of Pennsylvania. > Perhaps their abuse department would be interested in knowing that > youve used their news server to post message with forged headers.You can tell by the headers that the person is from the University ofPennsylvania?You can tell by the headers that the headers are forged?What parts of the header are forged, then? The user name? Is thereonly one James Harris in the world?Do you know for a fact that ANY of the posters using the name JamesHarris on sci.math have the real name of James Harris?Are you the rst and only shfry posting anything, ever, to theinternet? If not, are you guilty of forging headers?Why dont you change your name to dumbass ? Wouldnt this name tyou better?So many questions for such a pinhead like you. > I will stop bothering you all now because I never pursued aMathematics> degree beyond a bachelors and> shouldnt be proclaiming crazy and false ideas here.>> Your usenet headers show you to be at the University of Pennsylvania.> Perhaps their abuse department would be interested in knowing that> youve used their news server to post message with forged headers.>> You can tell by the headers that the person is from the University of> Pennsylvania?>> You can tell by the headers that the headers are forged?>> What parts of the header are forged, then? The user name? Is there> only one James Harris in the world?>> Do you know for a fact that ANY of the posters using the name James> Harris on sci.math have the real name of James Harris?>> Are you the rst and only shfry posting anything, ever, to the> internet? If not, are you guilty of forging headers?Erm, Im not sure that you know what a header is......Clearly shfry was correct.>> Why dont you change your name to dumbass ? Wouldnt this name t> you better?>> So many questions for such a pinhead like you.Jon I will stop bothering you all now because I never pursued a Mathematics> degree beyond a bachelors and> shouldnt be proclaiming crazy and false ideas here.> Your usenet headers show you to be at the University of Pennsylvania. > Perhaps their abuse department would be interested in knowing that > youve used their news server to post message with forged headers.> You can tell by the headers that the person is from the University of> Pennsylvania?> Yes.> You can tell by the headers that the headers are forged?> Yes.> What parts of the header are forged, then? The user name? Is there> only one James Harris in the world?> Set your newreader to full headers or equivalent andfind out yourself.There is in fact a jamesharris.com, which is NOT the source of the original posters post. > Do you know for a fact that ANY of the posters using the name James> Harris on sci.math have the real name of James Harris?> What does that have to do with this guys forgery?> Are you the rst and only shfry posting anything, ever, to the> internet? If not, are you guilty of forging headers?> Why dont you learn something about how the internet works and then come back and talk to us.> Why dont you change your name to dumbass ? Wouldnt this name t> you better?> So many questions for such a pinhead like you. =Consider This scribbled the followingon sci.math:>> I will stop bothering you all now because I never pursued a Mathematics>> degree beyond a bachelors and>> shouldnt be proclaiming crazy and false ideas here.>> Your usenet headers show you to be at the University of Pennsylvania. >> Perhaps their abuse department would be interested in knowing that >> youve used their news server to post message with forged headers.> You can tell by the headers that the person is from the University of> Pennsylvania?> You can tell by the headers that the headers are forged?> What parts of the header are forged, then? The user name? Is there> only one James Harris in the world?> Do you know for a fact that ANY of the posters using the name James> Harris on sci.math have the real name of James Harris?> Are you the rst and only shfry posting anything, ever, to the> internet? If not, are you guilty of forging headers?> Why dont you change your name to dumbass ? Wouldnt this name t> you better?> So many questions for such a pinhead like you.In fact there HAVE been other James Harrises on this newsgroup. Otherpeople, than our friend JSH, whose real name is James Harris. But thisguy who posted this thread is not one of them.-- /-- Joona Palaste (palaste@cc.helsinki.) ---------------------------| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki./~palaste W++ B OP+ |----------------------------------------- Finland rules! ------------/ > What parts of the header are forged, then? The user name? Is there> only one James Harris in the world?>No, but by referring to his FLT proof, he is passing himself off as aspecic James Harris. Perhaps their abuse department would be interested in knowing that >youve used their news server to post message with forged headers.Perhaps, but there again maybe they recognise that it was a joke,rather than anything worth wasting time on.Not a very good joke, of course.-- Richard-- FreeBSD rules! > I will stop bothering you all now because I never pursued a Mathematics> degree beyond a bachelors and shouldnt be proclaiming crazy and false ideas here. Your usenet headers show you to be at the University of Pennsylvania. > Perhaps their abuse department would be interested in knowing that > youve used their news server to post message with forged headers. You can tell by the headers that the person is from the University of> Pennsylvania?> Yes.> You can tell by the headers that the headers are forged?> Yes.> What parts of the header are forged, then? The user name? Is there> only one James Harris in the world?> Set your newreader to full headers or equivalent andfind out yourself.> There is in fact a jamesharris.com, which is NOT the source of the > original posters post. > domain than that of the ISP hosting the originators news server?Do you have any idea how many people are forging headers, by yourpinhead denition?Are you people really this stupid? > I will stop bothering you all now because I never pursued a Mathematics> degree beyond a bachelors and> shouldnt be proclaiming crazy and false ideas here.> Your usenet headers show you to be at the University of Pennsylvania. > Perhaps their abuse department would be interested in knowing that > youve used their news server to post message with forged headers.> You can tell by the headers that the person is from the University of> Pennsylvania?> Yes. You can tell by the headers that the headers are forged? Yes.> What parts of the header are forged, then? The user name? Is there> only one James Harris in the world?> Set your newreader to full headers or equivalent andfind out yourself.> There is in fact a jamesharris.com, which is NOT the source of the > original posters post. > domain than that of the ISP hosting the originators news server?Im not expert at decoding headers, but heres what I see.this person. Furthermore, it is a Pennsylvania serverof Comcast.The message ID assigned to this posting originated atU Penns news server.Again you see here that the path via which this message showsup in your server originated at netnews.upenn.edu.More conrmation that the NNTP origin of this postwas the University of Pennsylvania.No. However, if the from address and the sender belongto completely different people, this is an indicator ofsomething amiss.More signicantly, we know that James Harris lives inAtlanta, Georgia, does not have a jamesharris.comdomain, has no connection with the University of Pennsylvaniaand has disdain for anybody who happens to use a *.eduserver, which he considers to be somehow an attempt atone upmanship. It is very clear that this message originatedin Pennsylvania and probably at U. Penn or from somebodyassociated with U. Penn. Do you have any idea how many people are forging headers, by your> pinhead denition?Spammers routinely forge headers. ISPs known this. Why doyou think that they request you to send the informationthere which is much harder to forge and which can be usedto track down the actual origin.> Are you people really this stupid?Do you really think people DONT forge headers?By the way, here are are a lot of public-access databases out there thatmake the internet go. These are the hackers friend as wellIfind the very interesting information that, whilethe IP 168.143.113.143 does not have a name, most ofthe block of addresses 168.143.113.* belongs toanonymizer.com. That tells me that you probably postedthrough them to help hide your identity and messageorigins.When it becomes a law enforcement issue, legal authoritieshave the ability to learn much, much more about whereyou posted from and who you are by examining networklogs. You can rest assured that Anonymizer.com has youractual IP address in their logs. - Randy Clearly shfry was correct.shfry may be a bit of a amer, but hes absolutely right. It doesnt takea rocket scientist to forge pretty much every one of those headers. Think Clearly shfry was correct.> shfry may be a bit of a amer, but hes absolutely right. It doesnt take> a rocket scientist to forge pretty much every one of those headers. ThinkIts one thing to come on here and mock JSH. Thats fast becoming the national sport of sci.math. Its another thing to mock JSH by pretending to be the legitimate owner of jamesharris.com, pharmaceutical manufacturing consulting rm operating out of Durham, North Carolina, when the Usenet headers clearly show that the poster is using the U. Of Pennsylvanias NNTP server.When I pointed that out, was I aming? >> Clearly shfry was correct.>> shfry may be a bit of a amer, but hes absolutely right. It doesnttake> a rocket scientist to forge pretty much every one of those headers.Think> Its one thing to come on here and mock JSH. Thats fast becoming the> national sport of sci.math. Its another thing to mock JSH by> pretending to be the legitimate owner of jamesharris.com, pharmaceutical> manufacturing consulting rm operating out of Durham, North Carolina,> when the Usenet headers clearly show that the poster is using the U. Of> Pennsylvanias NNTP server.>> When I pointed that out, was I aming?I dont remember any mention of a pharmaceutical company. I would say,yes, you were aming. The whole thing was in a spirit of fun, which yourpost quickly negated.Youre posting anonymously; it seems odd that you would get so steamed upover someone else doing the same thing.Post using your real name and see how the answers are affected. = > I will stop bothering you all now because I never pursued a Mathematics > degree beyond a bachelors and > shouldnt be proclaiming crazy and false ideas here. > Your usenet headers show you to be at the University of Pennsylvania. > Perhaps their abuse department would be interested in knowing that youve used their news server to post message with forged headers. > You can tell by the headers that the person is from the University of > Pennsylvania?Probably, yes. Even the Sender is stated, but perhaps that is forged. > You can tell by the headers that the headers are forged?No. > What parts of the header are forged, then? The user name? Is there > only one James Harris in the world?Indeed, nothing tells us that jamesharris@jamesharros.com does not exist,nor that that is not the same person who posted the message. *If* it didnot actually come from jamesharris.com (which does exist) you can talkabout forgery. And in that case it is upto jamesharris.com to take action.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > Clearly shfry was correct.>> shfry may be a bit of a amer, but hes absolutely right. It doesnt> take a rocket scientist to forge pretty much every one of those headers.> Think> Its one thing to come on here and mock JSH. Thats fast becoming the> national sport of sci.math. Its another thing to mock JSH by> pretending to be the legitimate owner of jamesharris.com, pharmaceutical> manufacturing consulting rm operating out of Durham, North Carolina,> when the Usenet headers clearly show that the poster is using the U. Of> Pennsylvanias NNTP server.> When I pointed that out, was I aming?> I dont remember any mention of a pharmaceutical company. I would say,> yes, you were aming. The whole thing was in a spirit of fun, which your> post quickly negated.> Youre posting anonymously; it seems odd that you would get so steamed up> over someone else doing the same thing.> Post using your real name and see how the answers are affected.> Theres a reason I challenged the original poster. What he did was unacceptable on this newsgroup.Many people mock JSH. Thats not the point. Nor is it the point that some people post with anonymous aliases. I post as shfry, which is MY innocent third parties in my anononymity.What the original poster did was alter his FROM eld to jamesharris@jamesharris.com. As it happens, there is a domain jamesharris.com, belonging to a pharmaceutical manufacturing consultant who has no relation to sci.math or any of this JSH business. What right does the original poster have to do this? Its unethical and probably There are certain behaviors that we, the Usenet and sci.math community, need to oppose on general principles. If you pretend to be jamesharris@nonexistent.com, that is just having some fun, if thats your idea of fun. The only injured party is JSH, and JSH is already accustomed to taking large amounts of abuse around here, well-deserved though that abuse may be.But if you forge your FROM header to make it appear that you are the owner of the legitimate domain jameharris.com, you are crossing a line of acceptable behavior. You are pretending to be a real, live person, who happens to have no relation to sci.math. The original poster has no legal or moral right to do that, and that is why I chose to challenge him. If you put jamesharris.com in your browser, youll see that there is person in North Carolina, running a legitimate business. That individual has the right to not have people forging his name to Usenet posts.As a more extreme example, consider the occasional racial remarks that are sometimes made in the JSH debate. Everyone objects to that. Nobody says its just having fun. People who typically oppose JSH, will nevertheless strongly oppose anyone who tries to bring race into the conversation. We tolerate cranks as the price of free speech on sci.math. But we do not tolerate racists. We draw a line.Using a real, live, innocent third party as a fake return address is not nearly as serious as posting racist garbage. But it is similar in the sense of being unacceptable abuse of the newsgroup. It too is over the line.then pretend to be JSH, that is stupid but still ok in the context of Usenet behavior that were willing to tolerate in the spirit of an unfettered Net. But when someone forges the name and registered domain of a REAL LIVE uninvolved third party, they have crossed a line of civilized behavior. They are potentially damaging the reputation of some businessman who has no chance to defend himself or even to know that someone is using his name in forged Usenet posts.unacceptable on this newsgroup. =Thus spoke Daisy Duck a.k.a. Consider This:[...]Crosspostings great... but hey, whats a Followup-To?-- Taking care of trolls, k00ks, and other retarded minds =Munkres, in the rst chapter of his topology text, refers to the nite axiom of choice; [...] given a nite collection A of disjoint nonempty sets, there exists a set C consisting of exactly one element from each element of A. Is such an axiom really independent of ZF? That is, does the following argument invoke an axiom of choice in any way?Given a nonempty set B, let x be an element of of B. C = {x} is a set satisfying the nite axiom for the the family A = {B}. The axiom follows by induction. (I should have started with A empty, but the crucial part here is the rst two sentences.)To be specic, does the step, let x be an element of of B invoke choice? I would think not.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu Munkres, in the rst chapter of his topology text, refers to the> nite axiom of choice; [...] given a nite collection A of disjoint> nonempty sets, there exists a set C consisting of exactly one element> from each element of A. Is such an axiom really independent of ZF?> That is, does the following argument invoke an axiom of choice in> any way?>> Given a nonempty set B, let x be an element of of B. C = {x} is a> set satisfying the nite axiom for the the family A = {B}. The> axiom follows by induction. (I should have started with A empty,> but the crucial part here is the rst two sentences.)>> To be specic, does the step, let x be an element of of B invoke> choice? I would think not.Youve got it right, the book is wrong. I read about that so-callednite axiom of choice many years ago in a topology book by Hall andSpencer. Maybe its an old topologists superstition. Munkres, in the rst chapter of his topology text, refers to the >nite axiom of choice; [...] given a nite collection A of disjoint >nonempty sets, there exists a set C consisting of exactly one element >from each element of A. Is such an axiom really independent of ZF? Certainly not.Does Munkres say that it _is_ independent of ZF? If he does thenof course hes simply wrong, but if he doesnt then theres nothingnecessarily improper about any of this: He surely hasnt statedthe axioms for set theory explicitly anyway, so taking this asan axiom strikes me as perfectly acceptable, even though itdoes follow from the axioms of ZF.> That is, does the following argument invoke an axiom of choice in >any way?>>Given a nonempty set B, let x be an element of of B. C = {x} is a >set satisfying the nite axiom for the the family A = {B}. The >axiom follows by induction. (I should have started with A empty, >but the crucial part here is the rst two sentences.)>>To be specic, does the step, let x be an element of of B invoke >choice? I would think not.************************David C. Ullrich >Munkres, in the rst chapter of his topology text, refers to the>>nite axiom of choice; [...] given a nite collection A of disjoint>>nonempty sets, there exists a set C consisting of exactly one element>>from each element of A. Is such an axiom really independent of ZF?> Certainly not.> Does Munkres say that it _is_ independent of ZF?No. Munkres says nothing about ZF; he is writinga text on topology not on set theory and doesntlist any axiom system for sets. He does later saythat the nite AC is part of eveyones set theory.> If he does then> of course hes simply wrong, but if he doesnt then theres nothing> necessarily improper about any of this: He surely hasnt stated> the axioms for set theory explicitly anyway, so taking this as> an axiom strikes me as perfectly acceptable, even though it> does follow from the axioms of ZF.Quite right.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen Munkres, in the rst chapter of his topology text, refers to the >>nite axiom of choice; [...] given a nite collection A of disjoint >>nonempty sets, there exists a set C consisting of exactly one element >from each element of A. Is such an axiom really independent of ZF? >>Certainly not.>>Does Munkres say that it _is_ independent of ZF?>No, he does not. But his presentation suggests (to me, at least) that it is a separate axiom. He states it as an argument for accepting the There are two forms of this [choice] axiom. One can be called the nite axiom of choice... No mathematician has any qualms about the nite choice axiom; it is part of everyones set theory. Said differently , no one has qualms about a proof that involves only nitely many arbitrary choices. axiom, it is invariably the stronger form of the axiom [i.e., AC] that is meant.Also, though I know the axioms of ZF are not independent, generally (again for me, at least) the word carries the connotation of something that cannot be proved from other axioms. Cf. Euclid.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu >Munkres, in the rst chapter of his topology text, refers to the >nite axiom of choice; [...] given a nite collection A of disjoint >nonempty sets, there exists a set C consisting of exactly one element >>from each element of A. Is such an axiom really independent of ZF? >>Certainly not.>>Does Munkres say that it _is_ independent of ZF?>>No, he does not. But his presentation suggests (to me, at least) that >it is a separate axiom. Separate from what? He doesnt give any other axioms of set theory,does he?> He states it as an argument for accepting the >> There are two forms of this [choice] axiom. One can be called the> nite axiom of choice... No mathematician has any qualms about the> nite choice axiom; it is part of everyones set theory. Said> differently , no one has qualms about a proof that involves only> nitely many arbitrary choices.> axiom, it is invariably the stronger form of the axiom [i.e., AC]> that is meant.Huh? I dont see how thats an argument for accepting the axiom ofchoice without controversy...>Also, though I know the axioms of ZF are not independent, generally >(again for me, at least) the word carries the connotation of something >that cannot be proved from other axioms. _What_ other axioms?>Cf. Euclid.************************David C. Ullrich >Munkres, in the rst chapter of his topology text, refers to the >nite axiom of choice; [...] given a nite collection A of disjoint >nonempty sets, there exists a set C consisting of exactly one element >from each element of A. Is such an axiom really independent of ZF? >>Certainly not. ...................>Also, though I know the axioms of ZF are not independent, generally >(again for me, at least) the word carries the connotation of something >that cannot be proved from other axioms. Cf. Euclid.In the rst-order predicate calculus, which is the language inwhich almost all of mathematics is written, if there exists anx such that P(x), then one can choose such an x. Then by using induction in the metalanguage, if there exist x_i fori = 1, ..., n, then one can simultaneously choose these.So the nite axiom of choice is a consequence of rst orderlogic, and does not involve the axioms of set theory.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University Munkres, in the rst chapter of his topology text, refers to the >>nite axiom of choice; [...] given a nite collection A of disjoint >>nonempty sets, there exists a set C consisting of exactly one element >>from each element of A. Is such an axiom really independent of ZF? >>Certainly not.>> ...................>Also, though I know the axioms of ZF are not independent, generally >>(again for me, at least) the word carries the connotation of something >>that cannot be proved from other axioms. Cf. Euclid.>>In the rst-order predicate calculus, which is the language in>which almost all of mathematics is written, if there exists an>x such that P(x), then one can choose such an x. Then by >using induction in the metalanguage, if there exist x_i for>i = 1, ..., n, then one can simultaneously choose these.>>So the nite axiom of choice is a consequence of rst order>logic, and does not involve the axioms of set theory.The axiom as stated _does_ involve axioms of set theory,because of the way the conclusion is stated: there exists a set C consisting of exactly one element from each element of A - its impossible to prove that there existsa set having _any_ property without some sort of axiomregarding sets. If we know that there exists x such thatP(x) then we can choose such an x with no set theory,I suppose (one would need a denition of choose toprove this, but Im not inclined to dispute it), butwe need an axiom of set theory to get from x to theset {x}.************************David C. Ullrich =Along the same lines as the original post:For m = positive integer, and {a(k)} = any sequence dened forpositive integer indexes: m!--- | m! | |----| a(k) =/ |_k _|---k=1+(m-1)! m m!/k (m-1)!--- --- --- / / a(j) - m* / a(j)--- --- ---k=1 j=1 j=1In linear-mode:sum{k=1+(m-1)! to m!} oor(m!/k) a(k) =sum{k=1 to m} sum{j=1 to m!/k} a(j) - m* sum{j=1 to (m-1)!} a(j) Leroy Quet > These are variations of a result I posted a while back, but I mention> them here because the equal nite-sums are free of oor-functions.> Also, there MIGHT be some interesting number-theory results which> could be related to this. (see below.)> Let {a(k)} = any sequence dened for all positive integer indexes k.> Let m = any positive integer.> Then:> m!m> --- > 1 |(m+1)!|> ----- / a(|------|)> (m+1)! --- |_j+m!_|> j=1> m> ---> a(j)> / ------> --- j(j+1)> j=1> In linear-mode:> (1/(m+1)!) sum{j=1 to m!m} a(oor((m+1)!/(j+m!)))> sum{j=1 to m} a(j)/(j(j+1))> And related:> m j> --- ---> m! / / a(k) --- ---> j=1 k=1> |(m+1)!|> |------|> (m+1)! |_ j _|> --- ---> / / k a(k)> --- ---> j=1+m! k=1> In linear-mode:> m! sum{j=1 to m} sum{k=1 to j} a(k) sum{j=1+m! to (m+1)!}> sum{k=1 to oor((m+1)!/j)} k a(k)> A specic example:> (m+1)!> --- |(m+1)!|> |------| / |_ j _|> ---> j=1+m!> (m+1)!H(m) - m! m,> where H(m) = 1+1/2+1/3+...+1/m, the m_th harmonic number.> The sum in linear-mode:> sum{j=1+m! to (m+1)!} oor((m+1)!/j) > Now this last oor-involving sum is congruent to:> m! H(r,m) (-1)^(m+1) (mod{m+r+1}),> where H(r,m) = sum{k=1 to m} H(r-1,k),> and H(0,m) = 1/m.> (H(1,m) = H(m).)> (H(r,m) also = binomial(m+r-1,r-1)(H(m+r-1)-H(r-1)).)> The oor-sum itself is m!*H(2,m).> Does the congruency imply any immediately interesting number> theory-results??> Leroy Quet =I have some 3D data points and I would like tofind the way tocalculate the smallest sphere which would include all of them insidethe sphere (not necessarily on its surface).Kiran I have some 3D data points and I would like tofind the way to calculate> the smallest sphere which would include all of them inside the sphere> (not necessarily on its surface).> KiranIt is easy enough tofind the smallest cube that includes all yourpoints. The smallest sphere that encloses this cube also includes allyour points.-- Christos DimitrakakisIDIAP (http://www.idiap.ch/~dimitrak/main.html) > I have some 3D data points and I would like tofind the way to calculate> the smallest sphere which would include all of them inside the sphere> (not necessarily on its surface).>> It is easy enough tofind the smallest cube that includes all your> points. The smallest sphere that encloses this cube also includes all> your points.How do youfind the smallest cube that includes all the points?-- Clive Toothhttp://www.clivetooth.dk >I have some 3D data points and I would like tofind the way to>calculate the smallest sphere which would include all of them inside>the sphere (not necessarily on its surface).>KiranSee http://www.faqs.org/faqs/graphics/algorithms-faq/subject 5.24 > I have some 3D data points and I would like tofind the way to calculate>> the smallest sphere which would include all of them inside the sphere>> (not necessarily on its surface).> It is easy enough tofind the smallest cube that includes all your> points. The smallest sphere that encloses this cube also includes all> your points.But that doesnt answer the question. The question was tofind the*smallest* sphere that contains the given points.For example, suppose we start with three points on the surface of theunit sphere, and suppose the points are chosen so that they do not lie atthe vertices of an inscribed cube. Then the smallest cube containing thegiven points must have a side length greater than 2/sqrt(3), and anysphere that contains that cube is too big.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. calculate the smallest sphere which would include all of them inside> the sphere (not necessarily on its surface).> KiranHere is something you can do if you dont absolutely need the absolutelysmallest sphere. The problem is tofind the center of the sphere. Once youhave that, you can justfind the maximum of its distances to all the pointsto get the radius.Tofind the center of the sphere, you need to minimize the maximum distancefrom a point (x,y,z) to all the points in the set. Instead of minimizingthe maximum, you can minimize something that is reasonably close to themaximum but yet is differentiable, so you can use a gradient search typemethod.Let (x,y,z) denote the unknown center. Let (x0,y0,z0) denote one of thepoints in the set. Form the quantity [(x-x0)^2 + (y-y0)^2 + (z-z0)^2]^20.(I chose 20 arbitrarily. It doesnt seem to help much to use a higherexponent, but you could. The higher the exponent, the closer you willtheoretically come to the true center of the smallest circle; but thegradient search will more likely foul up.) Now add up all these quantities,for all the points in the set, and that gives you a function q(x,y,z) whichyou want to minimize. A good place to start the minimization search wouldbe at the point (xm, ym, zm), where xm denotes the average of all thex-values in the set, similarly for ym and zm.I tried this with several randomly generated examples in Mathematica, andsome examples where I threw in points to try to confound it, and the resultslooked good.I tinker with FLT for fun and I am not claiming a proof. This post is moreabout implicit and explicit restrictions.FLT is frequently stated asa^n + b^n = c^nwhich in itself is solvable but of course, FLT restricts the consideration tointegers. Well, 1 < a < b < c is a good restriction for the integers, sincejust saying a, b, and c must be integers allows for a, b, or c to = 0. Then nargument can be made for not needing to consider negative integers, and onlyneeding to consider odd prime n, but that is not what this post is about..If we solve the above equation for c without respect to the conditions, then cis the Mahanlobois mean or possibly the Mahanlobois sum of a and b withcharacteristic exponent n. For example, the mean with characteristic n = -1,not part of FLT, represents the resistance of resistors a and b in parallel.Now, this mean is decreasing with increasing n. So, for all values of n between1 and oo (innity), and I think for integers 1 < a < b and c the mean, atriangle can be constructed with angle C opposite side c. For n=1 the triangleis a straight line. For n = 2 the angle C is a right angle.By the law of cosines c = (a^2 + b^2 - 2*a*b*x) ^ (1/2) where x is cos C.Setting that equal to the mean gives, by algebra:(a^n + b^n) ^ (1/n) = (a^2 + b^2 - 2*a*b*x) ^ (1/2) (a^n + b^n) ^ 2 = (a^2 + b^2 - 2*a*b*x) ^ n2*(a*b)^n = p ( a^2 , b^2, -2*a*b*x) - a^2*n - b^2*nwhere p is the multinomial sum with n=n and r = 3:( n ) a^2^n1 b^2^n2 -2*a*b*x^n3( n1, n2, n3 )where n1 + n2 + n3 = n and the double parentheses mean the binomialcoefcient.I refer you to Kemeny, Schleifer, Snell, and Thompson: Finite Mathematics withBusiness Applications for that notation. Its what I have on hand. Im reallyenjoying the chapters on partitions and counting because they relate toentropy, which is not considered in this text. I read Kinetic Theory of Gasesby Leonard B. Leob and got some similar insight.Stepping back,(a^n + b^n) ^ 2 = (a^2 + b^2 - 2*a*b*x) ^ nimplies to me that for further manipulation, and I understand that elementaryalgebraic manipulations hold little hope for FLT, n must be an integer. Thatis, I wonder whether writing the FLT equation this way somehow EMBODIES therestriction that n is an integer.I see two ways of reading this: its either an implicit specication of arelationship between a, b, n, and x, or since the multinomial has no meaningfor nonpositive n, it somehow encapsultes the Diophantine restriction on ngiven in FLT. I do not claim it means a and b must be integer. Only n, and eventhat as one of two interpretations.Youll note nowhere here does c come into play. But there is an x, so itsstill a four number problem. Is that a problem? I dont think so. In Wilesproof the c does not appear in the elliptic:E: y^2 = x * (x -a^n) * (x - b^n)so there seems to be at least one way of turning the problem into somethingelse. Now theres a y instead of a c, and its a whole new problem, not just analgebraic one either. And I think that turning the problem into something elseis just what mathematicians do to arrive at solutions.Comments?Does the statement imply the restriction?Yours,Doug Goncz, Replikon Research, Seven Corners, VA Fair use and Usenet distribution without restriction or feeCivil and criminal penalties for circumvention of any embedded encryption =