> 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, France Home 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 definition 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. ==== > This definition 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 definition 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 A or the derivative does not exist at A. For a function on a closed bounded interval, the endpoints of the interval can also be regarded as critical points since the function has to be evaluated at endpoints as well as internal critical points if absolute maxima and minima are to be found. 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 zero. > > un point critique (google 7050/179) > or > un point stationnaire (google 349/20) > or > un point extr.8emal (google 30/4) And these translate back to (trivially) critical point, stationary point and extreme point. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== Sorry, I meant to write that a point where the derivative is UNDEFINED is also called a critical point. This definition 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? ==== 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. I expect that a lot of you will dance and sing, but perhaps others may feel the cold chill of the death knell. If you want to communicate with me further, send a letter to James Harris, Heaven. To communicate with my opponents, you can send them letters now, or just wait until the addressing is easier, 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 ==== >[snip] The first 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 Integer Sequences > 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 Seaman's 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+2 arises I (foolishly) meant I have no idea why 6n^2+2 is the solution to the original poster's question. I had not realised that it only applies for n<7. -- Clive Tooth http://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 first 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 > Dave's n=6 and n=10 results. I have downloaded your Fortran program and checked it against my lisp > program (modified 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 insphere S. Consider the set of n^3 unit cubes, each of side 1, whose faces are parallel 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 unit cubes 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 value is 0. # if all the vertices of the unit cube are outside S, the intersection value is 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 other vertices are either all inside or all outside S. In this case its intersection 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 this value, 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 1.499860 This suggests that T(n) is asymptotic to (3/2)*pi*n^2. I assume this is all well-known. -- Clive Tooth http://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). and Z=min(X, Y) I know the following method to derive the cdf of Z. P(Z t ) = 1 - [ P( X 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? > 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, does that > change the answer? How is a side a rational number? Sorry, I should be more precise: Can you characterize those cyclic quadrilaterals 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 unspecified, given any side lengths l_1,...,l_4 which can form a quadrilateral (i.e. the longest side is less than the sum of the other three) there will be always be some radius r for which you get a cyclic quadrilateral with those side lengths. This will satisfy either 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) < 0 when 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 relation sum_{i=1}^4 arcsin(l_i/2) = Pi can be expressed algebraically: if t_i = l_1/sqrt(4 - l_i^2), then t_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 = 0 and (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 = 0 I don't know how you'd go about characterizing the rational solutions of this. Robert Israel israel@math.ubc.ca Department 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 Ptolemy's 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 fitting, and separating overlapped peaks and > background. www.chemSoftware.com > ==== to contain the definition for 1-norm and 2-norm distance (a.k.a Hamming and Euclidean distance). Or maybe you can point me to the relevant math ==== > to contain the definition 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 more elementary courses like real analysis. See if H.9alder's Inequality is 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 was posted by Leon-Sontelo a nice problem : ================================================================= Prove that for any integer n, exists a unique polynomial Q(x) with coefficients 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 : http://groups.google.com/groups?selm=b6u235$9046k$1@ID-137122.news.dfncis.de http://groups.google.com/groups?selm=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 coefficients in {0,1,2,3,4,5,6,7,8,9} such that Q(-2)=Q(-5)=n , n in Z. It's clear that the function m: M-->N ,N being the set of natural numbers, is well-defined. QUESTION: Find bounds for m(n) , to evaluate m(n). ===================== ==== Alex.Lupas escribi.97 en el mensaje > Some 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 coefficients 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 : > http://groups.google.com/groups?selm=b6u235$9046k$1@ID-137122.news.dfncis.de > http://groups.google.com/groups?selm=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 > coefficients > in {0,1,2,3,4,5,6,7,8,9} such that Q(-2)=Q(-5)=n , n in Z. It's clear that the function m: M-->N ,N being the set of natural > numbers, is well-defined. > QUESTION: Find bounds for m(n) , to evaluate m(n) > ===================== I said Q(x) = (x + 2)(x + 5)P(x) + n = (x^2 +7x +10)P(x) + n The cofficients of P aren't necessary in {0,1,2,...9}, but they are integers. Let b(n) to be the coefficient 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 >= 0 All b(n) are univocally determined, and for all n, they are zero from a certain k, because they hav alternate signs, and therefore, each one is, in absolute value, less or equal than 7/10 of the previous. By example, for n = 196, b(0) = -[196/10] = -19 b(1) = -[-7*19/10] = -[-13.3] = -(-14) = 14 b(2) = -[(7*14 -19)/10] = -7 b(3) = -[(7(-7) + 14)/10] = -(-4) = 4 b(4) = -[(7*4 - 7)/10] = -2 b(5) = -[(7(-2) + 4)/10] = -(-1) = 1 b(6) = -[(7*1 - 2)/10] = 0 b(7) = -[(7*0 + 1)/10] = 0 and 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 + 6 The paragraph All b(n) are univocally determined, and for all n, they are zero from a certain k, because they hav alternate signs, and therefore, each one is, in absolute value, less or equal than 7/10 of the previous. it isn't fully exact, because the integer part of a negative number is greater, in absolute value, that it. Then |b(k+1)| < (7/10)|b(k)| + 1 but it isn't very important. Roughly, in a worst case, m(n) <= log(n)/log(10/7) + 2 or something so. But it seems that frequently, it say the mean beahouvior, is 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) ~= 0.9716791049ák + 1 in an excellent agree. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== > Alex.Lupas escribi.97 en el mensaje Some 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 coefficients 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 : http://groups.google.com/groups?selm=b6u235$9046k$1@ID-137122.news.dfncis.de http://groups.google.com/groups?selm=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 > coefficients > in {0,1,2,3,4,5,6,7,8,9} such that Q(-2)=Q(-5)=n , n in Z. > It's clear that the function m: M-->N ,N being the set of natural > numbers, is well-defined. > 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) ~= 0.9716791049ák + 1 in an excellent agree. ========= Your estimation seems to be good. ======= ==== 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? 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. Let's 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 and column j from M. The drawback of this method is that you need to calculate several determinants, which gets long if your matrix is large. You may want look in a linear algebra book to find others, which may in some cases be more adapted. Basically, if you have several of zeroes in your matrix, this method is likely to be easy, because you'll be able to easily calculate the necessary determinants. Otherwise, you may want to use another method. Sam -- Giving the Linus Torvalds Award to the Free Software Foundation is a bit like giving 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? Don't 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. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== 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.? I'm not sure if that's the method he's 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. You'll 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; it's 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.? I'm not sure if that's the method he's 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. >You'll end up with I and M^(-1). Sam It's called Gauss-Jordan, it can be found easily on google. You can also do a LU decomposition 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 then become 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.? > >I'm not sure if that's the method he's 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. >>You'll end up with I and M^(-1). >>Sam >> > >It's called Gauss-Jordan, > Or just Gaussian elimination. > it can be found easily on google. > It's 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.? I'm not sure if that's the method he's 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. >You'll end up with I and M^(-1). Yup, that's the one. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== : > 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 it's still quite a complicated method; 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 it's still quite a complicated > method; only worth using if you want, say, the billionth digit in pi. I'd 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 of pi. > ex: What is the 100th digit in Pi? > ans: 0 thanks Check 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 define the relation R on A by m R n iff m|n. Write the definitions of the properties, reflexive, antisymmetric and transitive and the use the definitions 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. ======= ==== I'm just about to log off, but as a quick partial answer, I can give you the following: 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 a number (n) that in turn is divisible by another number (n'), the original number (m) must necessarily be divisible by the final divisor (n'). R is not however symmetric. . . To be symmetric 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 the above set is obviously not true. . . As for reflexive, R is. . . simply put the reflexive property means that if m=n is the set R, then R is reflexive. . . Because any number is divisible by itself, the set R is reflexive. R is also antisymmetric. Because for any (m,n), there does not exist a counterpart (n,m) which is the definition of the antisymmetric property. I hope this helps and I'm sorry I didn't get to answer the second part of the question... tjd Can someone help me out with understanding this question? ======= Let A = {1, 2, 3, 4, 5, 6,12} and define the relation R on A by m R n > iff m|n. Write the definitions of the properties, reflexive, antisymmetric and > transitive and the use the definitions 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. ======= ==== Can someone help me out with understanding this question? ======= Let A = {1, 2, 3, 4, 5, 6,12} and define the relation R on A by m R n > iff m|n. I guess, this is clear so far. Write the definitions of the properties, reflexive, antisymmetric and > transitive and the use the definitions 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 reflexive, antisymmetric and transitive. Given that you know the definitions 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 ==== > I'm just about to log off, but as a quick partial answer, I can give you the > following: >[...] > I hope this helps and I'm sorry I didn't get to answer the second part of > the question... > tjd Of 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 didn't give better definitions of the properties in addition to how they pertained to the example. . . was very tired at the time of reply. . . But, if you have other questions, I'll 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 define the relation R on A by m R n > iff m|n. I guess, this is clear so far. > Write the definitions of the properties, reflexive, antisymmetric and > transitive and the use the definitions 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 reflexive, antisymmetric and > transitive. Given that you know the definitions 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 all of 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 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 KTEH-TV, Channel 54 (AT&T Broadband Cable 10) ABOUT PAUL ERD.85S: Paul Erd.9as was one of the most prolific mathematicians who ever lived. He pioneered several fields in theoretical mathematics, and published or co-authored more than 1,500 mathematical papers in a field where 50 good papers in a lifetime would be considered an achievement. Using interviews with Erd.9as himself, film clips of Erd.9as lecturing and discussions with those who knew and were influenced by him, N is a Number: A Portrait of Paul Erd.9as captures the life and work of this accomplished scholar and teacher. N is a Number: A Portrait of Paul Erd.9as provides insight into the mathematical quest of this amazing man, the personal and philosophical dimensions of his 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 by George Paul Csicsery. Cinematography is by John Knoop, and original music was composed by Mark Adler. Animated sequences were created by Red Dot Interactive. The film was made with support from the American Mathematical Society, Film Arts Foundation, the Heineman Foundation, the Mathematical Association of America, and the National Science Foundation's Informal Science Education Program. N is a Number: A Portrait of Paul Erd.9as has won awards and appeared at numerous festivals, including the Festival International du Nouveau Cin.8ema et de la Vid.8eo de Montr.8eal, the Budapest Film Week, the Berlin Film Festival-European Film Market, the National Educational Film & Video Festival, where it won a Gold Apple Award in 1994, the Melbourne International Film Festival, the Chicago International Film Festival (Gold Plaque Award, 1994), the Auckland-Wellington, New Zealand Film Festivals, Festival International du Film Scientifique du Qu.8ebec and the Cork Film Festival. 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 Canada. During 2002-2004 the film is being offered to PBS stations http://www.aptonline.org/catalog.nsf/d14ac4ffee990491852566410073d4d1/12e3e7 e2c9d3d87d85256bd7006e4445?OpenDocument For information, contact: George Paul Csicsery PO Box 22833 Oakland CA 94609 tel 510-428-9284 fax 510-428-9273 geocsi@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? (It's clear that it converges for almost all x.) Or for x = 0? Fascinating background information: It was on a test given by a colleague - can't tell from the tests we know was specified as a possible answer, and was the answer the guy who gave the test had in mind. I don't see a simple way to settle the question - a certain student finds it hard to believe that we can't figure it out, and asked me to post the question here. So if you determine whether the series converges for all x not only do you have the exquisite pleasure of 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 recently and people agreed they didn't know the answer please let me know - wouldn't seem fair for me to claim my prize if I'd 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 I'm not sure. Here's my argument. 1. Let A = {x in R : x =/= 2k pi for all k in Z} It's know Dirichlet series sum cos(nx) / n converges for all x in A. It's also know that another Dirichlet series sum sin(nx) / n converge, this time for all x in R. Let's 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 fixed. Then, with a_n = sin(n)^n / n cos(nx) b_n = cos(nx) / n we have a_n / b_n = sin(n)^n So, |a_n / b_n| <= k = 1 for all n in N and by the proposition and the fact that the series sum cos(nx) / n converge (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 but not x in A). Let be any k in Z and x = 2 pi k. Then cos(nx) = cos(2 pi kn) = 1 so sum (sin(n)^n / n) cos(nx) = sum(sin(n)^n / n) Then, with a_n = sin(n)^n / n b_n = sin(n) / n we 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 fact that the Dirichlet series sum sin(nx) / n converge 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 series converge for all x in R. I hope that I have helped. Sorry my english. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com ==== > Dear Mr. Plutonium, Today solar-PV-cell peddler 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 flat. -dl And in the Crossover Technique, today I could be increasing my wealth by selling some SBC at 24 when I bought it at 26 and buying back more Qwest 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 Don's last sentence when he says that Spire zoomed ahead by remaining flat. (BTW I do not know if these are true genuine companies and not even going to see because I will never invest in them). My point is this, several weeks ago I sold some Qwest at 5.02 and bought SBC at 26.75. Today I could increase my wealth by selling some SBC at 24. 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 market than in a Bull Market because the taxes paid in a Bull market are a large factor. whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== 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 ==== 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? 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. ==== > 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. ==== >>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 define the cross product of n-1 vectors. It is (fixed-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 ==== >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 define the cross product of > n-1 vectors. It is (fixed-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 ==== >>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 define the cross product of >>n-1 vectors. It is (fixed-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 Doesn't 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. I look at this as a way to get a vector orthogonal to n-1 vectors in n-dimensional space. Martin Cohen ==== > Doesn't 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-handed systems of vectors. In a 3x3 matrix, top or bottom row makes no difference. 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 define the cross product of >>n-1 vectors. It is (fixed-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 Doesn't 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 doesn't answer the question of how the cross product generalizes. If someone asks you the product of 3 and 5, do you reply: -15, or 15 ... it doesn't matter. Both are 15 units from zero. In 3-space, a x b doesn't equal b x a -- even though both are orthogonal 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 define the cross product of > n-1 vectors. It is (fixed-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 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 part 0, 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 defining U x V = (UV - VU)/2 on Im(C), Im(H), Im(O). The fact that there are no others follows from the theorem (due to Max Zorn?) that the only alternative, quadratic, real algebras without divisors of zero are isomorphic to R, C, H or O. GC ==== Infinity. More than a million, billion, trillion, zillion. All our schools have taught us to learn about infinity. But will maths ever be able to model the universe perfectly? Surely not. Our human brains can't grasp infinity for a start. -- Support charity at www.freeMyWisdom.co.uk - we don't want your money, but we do want your wisdom. ==== > Infinity. More than a million, billion, trillion, zillion. What's a zillion? (and what's a dummie?) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > Infinity. More than a million, billion, trillion, zillion. All our schools have taught us to learn about infinity. But will maths > ever be able to model the universe perfectly? Surely not. Our human > brains can't grasp infinity for a start. > Infinity 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 infinity. Thus oo is infinity, so handle it with care, so much is packed into oo, that it could go off with the Big Bang. Infinity is so big you can't add to it, that is oo + 1 = oo + 2 = oo + oo = oo and it doesn't multiply, that is 1000 * oo = oo * oo = oo. Thus you see oo is not allowed to play with ordinary finite folks like 1,2,3.. for indeed oo is too big for its britches. ==== > Infinity. More than a million, billion, trillion, zillion. > What's a zillion? > Bush's second term goal for looting US Treasure. > (and what's a dummie?) > dummie = 1/googleplex ==== Infinity. More than a million, billion, trillion, zillion. What's a zillion? (and what's 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. ==== >>Infinity. More than a million, billion, trillion, zillion. >>All our schools have taught us to learn about infinity. But will maths >>ever be able to model the universe perfectly? Surely not. Our human >>brains can't grasp infinity for a start. Infinity 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 infinity. Thus oo is infinity, so handle it with care, > so much is packed into oo, that it could go off with the Big Bang. Infinity is so big you can't add to it, that is oo + 1 = oo + 2 = oo + oo > = oo and it doesn't multiply, that is 1000 * oo = oo * oo = oo. Thus you > see oo is not allowed to play with ordinary finite folks like 1,2,3.. for > indeed oo is too big for its britches. That depends on what you're working with. Transfinites play with oo+1, etc on a regular basis, and oo+1 =/= oo+2 in those cases. -- Will Twentyman ==== > Infinity. More than a million, billion, trillion, zillion. >> What's a zillion? >> (and what's 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 Fermat's Last Theorem for Dummies. :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== >We could have an author for Fermat's Last Theorem for Dummies. :-) What about that book by Marilyn Vos Savant? --irascible since 1957 ==== Infinity. More than a million, billion, trillion, zillion. > What's a zillion? Bush's second term goal for looting US Treasure. (and what's a dummie?) > dummie = 1/googleplex < trickle down economics Amen.. $70 Million spent on Clinton's penis. $4 Million spent on discovering who really did 911. Conjectures 1 & 2 1.) Republicans didn't know what a blow job was. 2.) They found out. Conjecture 3 3.) Republicans keep putting private profit where national honor was. I need to study this. It's just too real. Lets see, 4.) In the beginning the first Capital crime in the then USA was a crime against the then Bank. 5.) Then there was those pesky people who were here first. We fixed that. 1&2&3 in terms of 4 and 5. Oh I see it's a proof of what terrorism is. But I digress. Infinity : Infinity In*fini*ty, n.; pl. Infinities. L. infinitas; pref. in- not + finis boundary, limit, end: cf. F. infinit'e. See Finite. 1. Unlimited extent of time, space, or quantity; eternity; boundlessness; immensity. --Sir T. More. There can not be more infinities than one; for one of them would limit the other. --Sir W. Raleigh. I disagree, One is not an infinity. :) What say you Pilgrim? Ernst The proof looks Ugly. ==== We could have an author for Fermat's Last Theorem for Dummies. :-) What about that book by Marilyn Vos Savant? That's Fermat's Last Theorem By Dummies --irascible since 1957 ==== > Infinity is so big you can't add to it, that is oo + 1 = oo + 2 = oo + oo > = oo and it doesn't multiply, that is 1000 * oo = oo * oo = oo. Thus you > see oo is not allowed to play with ordinary finite folks like 1,2,3.. for > indeed oo is too big for its britches. That depends on what you're working with. Transfinites play with oo+1, > etc on a regular basis, and oo+1 =/= oo+2 in those cases. > Not so. It's, 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 ==== : Infinity. More than a million, billion, trillion, zillion. : All our schools have taught us to learn about infinity. But will maths : ever be able to model the universe perfectly? Surely not. Our human : brains can't grasp infinity for a start. Well, it would *take forever* for any kind of brain to *really* grasp infinity in some senses of the word... but try http://www.hypermaths.org/quadibloc/math/infint.htm John Savard ==== will maths ever be able to model [it]? Surely not. Our human > brains can't grasp infinity for a start. > Ever heard the name G. CANTOR? F. ==== In sci.math, www.freeMyWisdom.co.uk <7c2720c0.0307180159.190f5ac@posting.google.com>: > Infinity. More than a million, billion, trillion, zillion. All our schools have taught us to learn about infinity. But will maths > ever be able to model the universe perfectly? Surely not. Our human > brains can't grasp infinity for a start. Our human brains can't grasp extremely large numbers, either. However, we can represent them by various symbols. For example, Avogadro's 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 finish 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 it's 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 infinity is the sidewise '8', infty for those using TeX, and I shall represent it here using 'oo' because of limitations in ASCII. Infinity 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 definition of finite 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 modification: 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 modifications to generate things such as lim x->a+ f(x) = L+ lim x->+oo f(x) = L+ lim x->a+ f(x) = +oo etc.; I'll leave these to the interested reader. That's about as far as limits can go; slightly more interesting infinities 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 (I'm a computer software specialist, not a formal mathematician :-) ). [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >In sci.math, www.freeMyWisdom.co.uk ><7c2720c0.0307180159.190f5ac@posting.google.com>: >> Infinity. More than a million, billion, trillion, zillion. >> All our schools have taught us to learn about infinity. But will maths >> ever be able to model the universe perfectly? Surely not. Our human >> brains can't grasp infinity for a start. Our human brains can't grasp extremely large numbers, >either. However, we can represent them by various symbols. >For example, Avogadro's 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 finish 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 it's 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 infinity is the >sidewise '8', infty for those using TeX, and I shall represent >it here using 'oo' because of limitations in ASCII. Infinity 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 >definition of finite 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 modification: 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 modifications to generate >things such as lim x->a+ f(x) = L+ > lim x->+oo f(x) = L+ > lim x->a+ f(x) = +oo etc.; I'll leave these to the interested reader. That's about as far as limits can go; slightly more >interesting infinities 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, not aleph-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 (I'm a >computer software specialist, not a formal mathematician :-) ). [.sigsnip] (this space unintentially left blank ..... ==== > In sci.math, www.freeMyWisdom.co.uk > <7c2720c0.0307180159.190f5ac@posting.google.com>: > Infinity. More than a million, billion, trillion, zillion. All our schools have taught us to learn about infinity. But will maths > ever be able to model the universe perfectly? Surely not. Our human > brains can't grasp infinity for a start. Our human brains can't grasp extremely large numbers, > either. However, we can represent them by various symbols. > For example, Avogadro's 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 finish 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 it's 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 infinity is the > sidewise '8', infty for those using TeX, and I shall represent > it here using 'oo' because of limitations in ASCII. Infinity 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 > definition of finite 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 modification: 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 modifications to generate > things such as lim x->a+ f(x) = L+ > lim x->+oo f(x) = L+ > lim x->a+ f(x) = +oo etc.; I'll leave these to the interested reader. That's about as far as limits can go; slightly more > interesting infinities 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 (I'm 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. ;-) Ernst Sig # 1 Any Government ( including economic systems ) that limits the people's ability to obtain education is a bad Government. $4 Billion a month better spent on education so we might one day learn 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/mensanator Sigh. You try to set a standard for cleverness and wit but it just goes unappreciated. >Ernst -- Mensanator 2 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 : >> In sci.math, www.freeMyWisdom.co.uk >> <7c2720c0.0307180159.190f5ac@posting.google.com>: >> Infinity. More than a million, billion, trillion, zillion. >> All our schools have taught us to learn about infinity. But will maths >> ever be able to model the universe perfectly? Surely not. Our human >> brains can't grasp infinity for a start. >> Our human brains can't grasp extremely large numbers, >> either. However, we can represent them by various symbols. >> For example, Avogadro's 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 finish 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 it's 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 infinity is the >> sidewise '8', infty for those using TeX, and I shall represent >> it here using 'oo' because of limitations in ASCII. >> Infinity 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 >> definition of finite 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 modification: >> 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 modifications to generate >> things such as >> lim x->a+ f(x) = L+ >> lim x->+oo f(x) = L+ >> lim x->a+ f(x) = +oo >> etc.; I'll leave these to the interested reader. >> That's about as far as limits can go; slightly more >> interesting infinities 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 (I'm 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 wouldn't 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 we're even. :-) As for Mensanator -- I don't know anything about him. Someone else has mentioned the continuum hypothesis. I'm not sure I know anything about that, either. :-/ [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > Infinity. More than a million, billion, trillion, zillion. What's a zillion? zillion /zljn/ n. colloq. (orig. US). M20. [Arbitrarily after million, billion, etc. (perh. f. z repr. an unknown quantity).] A very large but indefinite number.zillionaire n. a very rich person M20. zillionth a. & n. ( a ) adj. following very many others; umpteenth; ( b ) n. a tiny fraction of something: L20. --------------------------------------------------------- Excerpted from Oxford Talking Dictionary Copyright © 1998 The Learning Company, Inc. All Rights Reserved. (and what's a dummie?) -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > The League of Gentlemen ==== Infinity. More than a million, billion, trillion, zillion. All our schools have taught us to learn about infinity. But will maths > ever be able to model the universe perfectly? Surely not. Our human > brains can't grasp infinity for a start. > There are many kinds of infinity in mathematics. Zdislav V. Kovarik had a post listing many of them. Here is is: --- There is a long list of infinities (with no claim to exhaustiveness): infinity of the one-point compactification of N, infinity of the one-point compactification of R, infinity of the two-point compactification of R, infinity of the one-point compactification of C, infinities of the projective extension of the plane, infinity of Lebesgue-type integration theory, infinities of the non-standard extension of R, infinities of the theory of ordinal numbers, infinities of the theory of cardinal numbers, infinity adjoined to normed spaces, whose neighborhoods are complements of relatively compact sets, infinity adjoined to normed spaces, whose neighborhoods are complements of bounded sets, infinity around absolute G-delta non-compact metric spaces, infinity in the theory of convex optimization, etc.; each of these has a clear definition and a set of well-defined rules for handling it. --- GC ==== > In sci.math, Ernst Berg > : >> In sci.math, www.freeMyWisdom.co.uk >> <7c2720c0.0307180159.190f5ac@posting.google.com>: >> Infinity. More than a million, billion, trillion, zillion. >> All our schools have taught us to learn about infinity. But will maths >> ever be able to model the universe perfectly? Surely not. Our human >> brains can't grasp infinity for a start. >> Our human brains can't grasp extremely large numbers, >> either. However, we can represent them by various symbols. >> For example, Avogadro's 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 finish 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 it's 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 infinity is the >> sidewise '8', infty for those using TeX, and I shall represent >> it here using 'oo' because of limitations in ASCII. >> Infinity 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 >> definition of finite 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 modification: >> 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 modifications to generate >> things such as >> lim x->a+ f(x) = L+ >> lim x->+oo f(x) = L+ >> lim x->a+ f(x) = +oo >> etc.; I'll leave these to the interested reader. >> That's about as far as limits can go; slightly more >> interesting infinities 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 (I'm 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 wouldn't 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 > we're even. :-) As for Mensanator -- I don't know anything about him. Someone else has mentioned the continuum hypothesis. I'm 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 but I'm not concerned.He's 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: > In sci.math, Ernst Berg >> : >>> In sci.math, www.freeMyWisdom.co.uk >>> <7c2720c0.0307180159.190f5ac@posting.google.com>: >>> Infinity. More than a million, billion, trillion, zillion. >>>> All our schools have taught us to learn about infinity. But will maths >>> ever be able to model the universe perfectly? Surely not. Our human >>> brains can't grasp infinity for a start. >>>> Our human brains can't grasp extremely large numbers, >>> either. However, we can represent them by various symbols. >>> For example, Avogadro's 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 finish 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 it's 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 infinity is the >>> sidewise '8', infty for those using TeX, and I shall represent >>> it here using 'oo' because of limitations in ASCII. >>>> Infinity 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 >>> definition of finite 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 modification: >>>> 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 modifications to generate >>> things such as >>>> lim x->a+ f(x) = L+ >>> lim x->+oo f(x) = L+ >>> lim x->a+ f(x) = +oo >>>> etc.; I'll leave these to the interested reader. >>>> That's about as far as limits can go; slightly more >>> interesting infinities 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 (I'm 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 wouldn't 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 >> we're even. :-) >> As for Mensanator -- I don't know anything about him. >> Someone else has mentioned the continuum hypothesis. I'm 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 didn't. I wondered where you got the idea I was vain and then pointed you to my Home Page, which is an exercise in vanity. >but I'm not concerned.He's a jestable fellow My 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. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm ==== Infinity. More than a million, billion, trillion, zillion. All our schools have taught us to learn about infinity. But will maths > ever be able to model the universe perfectly? Surely not. Our human > brains can't grasp infinity for a start. Infinity 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 infinity. Thus oo is infinity, so handle it with care, > so much is packed into oo, that it could go off with the Big Bang. Infinity is so big you can't add to it, that is oo + 1 = oo + 2 = oo + oo > = oo and it doesn't multiply, that is 1000 * oo = oo * oo = oo. Thus you > see oo is not allowed to play with ordinary finite folks like 1,2,3.. for > indeed oo is too big for its britches. i am a nul to math , but that's some spiritual stuff . quess all i need is an eye-opener ==== Infinity. More than a million, billion, trillion, zillion. All our schools have taught us to learn about infinity. But will maths > ever be able to model the universe perfectly? Surely not. Our human > brains can't grasp infinity for a start. There are many kinds of infinity in mathematics. Zdislav V. Kovarik had > a post listing many of them. Here is is: > --- > There is a long list of infinities (with no claim to exhaustiveness): > infinity of the one-point compactification of N, > infinity of the one-point compactification of R, > infinity of the two-point compactification of R, > infinity of the one-point compactification of C, > infinities of the projective extension of the plane, > infinity of Lebesgue-type integration theory, > infinities of the non-standard extension of R, > infinities of the theory of ordinal numbers, > infinities of the theory of cardinal numbers, > infinity adjoined to normed spaces, whose neighborhoods are > complements of relatively compact sets, > infinity adjoined to normed spaces, whose neighborhoods are > complements of bounded sets, > infinity around absolute G-delta non-compact metric spaces, > infinity in the theory of convex optimization, > etc.; each of these has a clear definition and a set of well-defined rules > for handling it. > --- > GC Just when I thought it was safe to think I knew what infinity was :) Ernst ==== I'm stuck on this question, and would be grateful if someone could point out where I'm 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 don't 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 can't see where... please help ==== Ben Dixon scribe: > I'm stuck on this question, and would be grateful if someone could > point out where I'm 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 am not sure. Some remarks without calculation: The usual way to destination is right You have done, get the derivatives of r and S (named phi). Then note that r«/phi«=dr/dphi and you will get an equation dr/dphi=f(r,phi) as well known for autonom systems. If I have 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 of higher order (i. e. it goes to zero with t-->oo). Good luck, Alfred ==== >I'm stuck on this question, and would be grateful if someone could >point out where I'm 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 (which I assume is the angular coordinate). With x = r cos(s) and y = r sin(s), note that you have dx/dt = cos(s) dr/dt - r sin(s) ds/dt dy/dt = sin(s) dr/dt + r cos(s) ds/dt so that dr/dt = cos(s) dx/dt + sin(s) dy/dt r ds/dt = -sin(s) dx/dt + cos(s) dy/dt Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== I'm stuck on this question, and would be grateful if someone could > point out where I'm 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 don't 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 can't see > where... please help > I wouldn't 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 flowing inward as it crosses this circle. Now find a small circle where the solution flows 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. ==== I'm stuck on this question, and would be grateful if someone could > point out where I'm 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 don't 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 can't see > where... please help > Ask yourself: 1) What are the singular points of the flow? 2) What does the flow look like near the singular point(s)? 3) What does the flow look like far from the origin? 4) What can you say about the limit set of a point far from the origin? [You are presumably supposed to use a theorem that says, under certain circumstances, that the limit set of an orbit is a periodic orbit] John Mitchell ==== I'm stuck on this question, and would be grateful if someone could > point out where I'm 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 (a coincidence, perhaps). Use the above information to show that you can find a circle with radius less than one such that all the vectors (of the vector field) on this circle are pointing out radially and show that there is another circle, with radius bigger than one, such that all the vectors are pointing in radially on that circle. So any orbit inside this annulus is trapped and the Poincare-Bendixson theorem says that the omega-limit set is a periodic orbit if there are no fixed points. It is easy to check that there are no fixed points in this annulus. Enjoy. ==== Will Self scribe: > I'm stuck on this question, and would be grateful if someone could > point out where I'm 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 wouldn't 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 flowing inward as it > crosses this circle. Now find a small circle where the solution flows > 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 polar coordinates. It should be shown, that (1-x^2-y^2) goes to zero and the solution of the approximated equations serves the Grenzzykel in the x-y-cartesic-system (an ellipse). A nice picture can be shown in the polar coordinate-system. It looks like the movement of planetoid-perihel around a special 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 the task (find the special parameters of solution) by using Poincare/Bendixon? Please help and let us look at Your solution. I am very interested. And until this I believe, that the good old Ljapunov always is a practicable way. It is a lot of years ago when I was learning it and it should be time to brush up. Alfred ==== <35eea5fc.0307080943.79ad7526@posting.google.com> addressed the subject of integrating the 2nd order linear DE with constant coefficients, by a quadrature approach. While the trial solution method is almost universally employed for this ODE, in this particular case, quadrature is normally possible and has some 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 that this relation holds true. The forthcoming deduction will tell us what happens if there is. 2. G is continuous 3. G' exists and is continuous 5. G'' exists 6. Let d =def sqrt(b^2 - 4*c) 7. Let r =def -b/2 + d/2 8. There exists at most a finite number of points x in (a,b) such ăspecial pointsä. 9. Let F be defined implicitly by: G(x) = F(x)*exp(r*x) 10. d ~= 0 11. Let x1, x2, x3, ... be the sequence of points on the real axis obtained as follows: 1st, x1=a. 2nd, move to the right and let x2 be the first special point encountered. Continue in this fashion -- the next special point is labeled x3, etc. Eventually, b will be encountered 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 this is never zero. The coefficient of F' is 2*r + b which by (7) is d. We have: 15. F'' + F'*d = 0 16. By use of (9) again, we find that F' = (G'-r*G)/E . By construction of the current subinterval, each interior point is not a special point. Therefore, F' ~= 0 within the subinterval. In (15), transpose and divide by F'. Obtain: 17: (F')'/F' = -d This is easily integrated, and 18. ln(F'(x)) = -d*x + c1 c1, c2, ... will be constants of integration. 19. F'(x) = c2*exp(-d*x) 20. Integrate a second time. The integral of exp(-d*x) has two different forms, depending on whether or not d is 0. In this case, by assumption 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 the classic sum-of-two-exponentials form, with each exponential coefficient a 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. By performing the same calculation for all subintervals, and using the continuity assumptions for G and G', we can take limits, solve a system of two equations which has a non-zero determinant, and show that c3,c4 to the left must equal c3,c4 to the right. The constants of integration are equal for all subintervals. Equation 21 holds throughout 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 finite number of special points is specific to the quadrature approach. (I do not see a convincing way it can be avoided.) In any event, however, it is not required for the trial solution method (but see d.3 below). c. One can take the function appearing on the right side of (21) and show by direct substitution that it satisfies (1) -- that is, Equation 1 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 finds the trial solutions in the first place. In this case, quadrature reveals them. --Second, there is always the possibility that a fortuitously found trial solution (or set) is in fact, NOT fully general. This is actually illustrated here by the case d=0. Another example is the case of ODE's with envelopes. Quadrature when it is available tells us absolutely that the functions found are the ONLY ones that can be solutions. --Third, the trial solution method only makes complete sense when we have some kind of exterior uniqueness result that tells us that the given ODE with initial/boundary conditions has at most one solution. If so, a solution which is some combination of trial solutions is in fact valid. However, it takes some work to develop a uniqueness theorem. Quadrature provides a method of sidestepping this difficulty. David Ziskind ziskind@ntplx.net ==== [...] | > Obviously you're not giving a rigorous definition. | | Obviously? Yes. | The a's are defined 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 doesn't tell us what the equation means. But then you denied that guess. m is not necessarily the independent variable. So evidently not only does the definition not quite say what type of functions or values the a_i are, it misleads the reader into thinking that they're something more obvious than they are. Don't try to just skim past this issue of what the independent variable is in these functions. | | a^3 + 3va^2 -(v^3+1) = 0, where again v=-1+mf^2, and the a's are its | roots. | | So you can solve for the a's 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 Cardan's 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. 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) wouldn't 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 first 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 first 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. | 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 it's 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 it's one particular one of the three that has a certain divisibility property, if the manner in which we've decided which of the three roots to assign to each of the three variables in an arbitrary way depending on the value of m? I'm sorry if the relevance doesn't seem clear to you. 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. That's essentially what prevents them from being continuous everywhere. I don't 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. [...] | So you're admitting to bringing up what I've called irrelevant issues | without having actually checked them, while refusing to fully | acknowledge the mathematical argument in the paper including a rather | basic lemma that you simply don't mention. I find the lemma very uninteresting. I don't understand how you manage to think of it as important. Similar to a lot of arguments written by undergraduate math majors early in their studies, this argument plods its way slowly through the routine parts of the argument, but condenses into a rapid sprint the part which is most in need of explanation, and also hardest to believe. Keith Ramsay ==== > [...] > | > Obviously you're not giving a rigorous definition. > | > | Obviously? Yes. Let's see if you support your accusation. > | The a's are defined 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 doesn't tell > us what the equation means. On the contrary, the equation clearly shows that there is a factorization. > a_1, a_2, and a_3 to be functions defined 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 that's 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. > If that guess had been correct, then I would have said that the > definition 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 couldn't be taking x to be a constant, > for example.) That's false. There is no requirement on the factorization that x not be constant. Factorizations don't work that way. > But then you denied that guess. m is not necessarily the independent > variable. So evidently not only does the definition not quite say > what type of functions or values the a_i are, it misleads the reader > into thinking that they're something more obvious than they are. Factorizations are not dependent on the dependency of the variables. > Don't try to just skim past this issue of what the independent > variable is in these functions. Factorizations are not dependent on functions. > | > | a^3 + 3va^2 -(v^3+1) = 0, where again v=-1+mf^2, and the a's are its > | roots. > | > | So you can solve for the a's 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 Cardan's 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 a's, a_1=0, a_2=0, a_3=3. Now consider Keith Ramsay's 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) wouldn't 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 first 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 first 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? > | 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 it's 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 it's one particular one of the three > that has a certain divisibility property, if the manner in which > we've decided which of the three roots to assign to each of the > three variables in an arbitrary way depending on the value of m? > I'm sorry if the relevance doesn't 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. > 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. For those who are confused simply remember that x^3 - 1 = 0, gives you three values for x. > That's 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. > I don't 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 don't have the time it's not really worth mentioning, now is it? After all, the discussion isn't a political one, and you're NOT on Fox News!!! > [...] > | So you're admitting to bringing up what I've called irrelevant issues > | without having actually checked them, while refusing to fully > | acknowledge the mathematical argument in the paper including a rather > | basic lemma that you simply don't mention. I find the lemma very uninteresting. I don't understand how you > manage to think of it as important. That lemma is the linchpin of the paper!!! > Similar to a lot of arguments written by undergraduate math majors > early in their studies, this argument plods its way slowly through > the routine parts of the argument, but condenses into a rapid sprint > the part which is most in need of explanation, and also hardest to > believe. Your belief here is irrelevant as a proof begins with a truth then proceeds by logical steps to a conclusion which then must be true. Challenging a paper requires that you find that it did not begin with a truth, or that you find a break in the logical chain. James Harris ==== >>> Remember (unlike e.g. the Gaussian integers) the algebraic integers >>> are dense in the complex plane, meaning that any circle, no matter >>> how small, has algebraic integers inside it. >>Hmmm...interesting assertion, but what about singularities? >> I'll bite. What about singularities? >> We aren't talking about a function here, just about two sets, the >> complex numbers and the algebraic integers. There aren't any >> singularities. And if there were, what does that have to do with >> whether there are infinitely many algebraic integers everywhere you >> look? >> A singularity is a circle of 0 radius, No, it's not. A singularity is a property of a function. And in topological statements like a circle, no matter how small, it is understood that the term circle refers to a finite radius. > and in the complex plane I can >pick an infinite number of fractions. What does that have to do with the statement? The statement is this: Pick a complex value z. Pick a radius epsilon >0. The neighborhood {w in C: |z - w| < epsilon} has infinitely many algebraic integers in it. What the heck does I can pick an infinite number of fractions have to do with the density of algebraic integers? Within a radius of 0.01 of 1 + i there are an infinite number of algebraic integers. Within a radius of 0.00002 of -sqrt(3) + i*pi there are an infinite number of algebraic integers. What does I can pick an infinite number of fractions have to do with statements like those? - Randy ==== yeah, but a circle of no radius would have itself as a singularity. did we uncover a new property/axiom/pedagogical twist? otherwise, Buddy, a ring is what I say it is, now; capiche? > No, it's not. A singularity is a property of a function. And in > topological statements like a circle, no matter how small, it is > understood that the term circle refers to a finite radius. --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX ==== | > [...] | > | > Obviously you're not giving a rigorous definition. | > | | > | Obviously? | > | > Yes. | | Let's see if you support your accusation. | | > | The a's are defined 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 doesn't 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 factorization has little unambiguous content, since there are many different kinds of structures in which you might be intending to put a_1, a_2, and a_3. Nora Baron has been trying to get you to realize the difference between factoring a number and factoring a polynomial. Do you now agree that there is one? | > a_1, a_2, and a_3 to be functions defined 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 that's 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 divisible by f for different values of m, then a_1 has to have values for different values of m. This is more crucial than you think. Look at a somewhat bigger picture for a minute here. Other people claim that because of a symmetry between the roots of the original polynomial, it has to be that either all of them are coprime to 5, or that none of them are. This symmetry obviously doesn't preserve all the properties that they have, or they would have to be equal, but it preserves algebraic ones such as being coprime to 5. Your argument against them indicates to the contrary that there's a distinction between the roots in terms of coprimeness to 5. So whatever it is you think allows you to distinguish between them (in terms of such algebraic properties) is bound to stand out as something unusual. At m=0, you distinguish between them because one of them is nonzero. (This is consistent with usual theory, because the polynomial is reducible at m=0.) And then you take a step which is as far as we're concerned just plain wrong, to say that the same a's have common factors with f at m=0 as have them at another value of m. So what does this have to do with whether the a's are functions or not? It's simple. If the a's aren't functions, then selecting one of the a's at one value of m doesn't help you select one of them for any other value of m. Saying that we've taken a_3 to be the nonzero one at m=0 tells us nothing about which of the three a values should be taken to be a_3 at m=1. If you can evaluate a_3 at all allowed values of m, then it's a function, because that's just what a function is, whether you want to call it that or not. | > If that guess had been correct, then I would have said that the | > definition 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 couldn't be taking x to be a constant, | > for example.) | | That's false. There is no requirement on the factorization that x not | be constant. | | Factorizations don't work that way. I'm not saying that there's a general prohibition on having constant variables in factorizations. It's just that holding x constant changes the meaning. In particular, if x is a constant the factorization isn't 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) it's NOT valid to deduce that a1, a2 and a3 are 0, 1, and -1, possibly in a different order. It's an invalid step in reasoning. If it's irrelevant what the dependent variable is, then your claim that the independent variable might turn out to be something other than m is nonsensical. | > Don't 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 factorization are in. The set of functions is one type of structure. If the independent variable is irrelevant, then why you dig in your heels and stubbornly refuse to allow m to be it> Is there something wrong with taking the a's to be functions of m? | > | | > | a^3 + 3va^2 -(v^3+1) = 0, where again v=-1+mf^2, and the a's are its | > | roots. | > | | > | So you can solve for the a's 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 Cardan's 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 a's, a_1=0, a_2=0, a_3=3. | | Now consider Keith Ramsay's 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) wouldn't 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 first 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 first 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. There's a lot going on at v=-1, and I just described part of it. If you wanted only to see the values of a_1, a_2, and a_3 at v=-1, you could have been more specific. The root you're calling a_3 is being relatively boring, but the other two roots are much like the square roots of v+1, which do this funky thing of trading places if we take v in a loop around v=-1, as well as 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 it's 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 it's one particular one of the three | > that has a certain divisibility property, if the manner in which | > we've decided which of the three roots to assign to each of the | > three variables in an arbitrary way depending on the value of m? | > I'm sorry if the relevance doesn't 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 haven't 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. There's 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 haven't given a convention for associating the values of the square and cube roots between two different values of v. Which cube root of 2-3i should be used for a_1, at the value of v which makes the expression under the first cube root equal to 2-3i? It's always somewhat arbitrary. Yet you need for g_1 to have that same factor in general. Why should one of the cube roots at that point make g_1 divisible by f, but the other two cube roots of 2-3i not make g_2 and g_3 divisible by f too? It's not just getting three values, but getting three values which retain their identities between different values of m. | > That's 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. It's partly the fault of the rest of us for inundating you with material faster than you can really digest. (Calling us liars is a phony way to pretend you've understood well enough what was said.) Remember what Paul Halmos says about why he went into mathematics rather than some other field: because he prefers words to numbers. Stan Ulam also said he dislikes reading papers that have a high ratio of formulas to descriptions of reasoning. This: So changing the sign on the square root of D is equivalent to leaving the first 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. There's a slightly tricky 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} - v where the three cube roots in the first term are associated with the three 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 can be defined consistently, since we're away from the points where we have to take the cube root of 0. But the square root has it's usual interesting behavior near there. | > I don't 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 don't have the time it's not really worth mentioning, now | is it? If you really think facts shouldn't be described on sci.math without presenting complete proofs immediately, you're entitled to your opinion, but I don't think many people will agree with you. [...] | Challenging a paper requires that you find that it did not begin with | a truth, or that you find 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 general suffers from a certain weakness of phrasing. I asked whether in general could be understood as for all m satisfying the assumptions but this was at the point in the discussion where you were saying that for subtle reasons m might not be the independent variable. In general requires some context. If it means anything like what it sounds like it means, however, it doesn't follow. First, it's not a simple, obvious step like you keep claiming. There are much simpler steps (like concluding that the product 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 to explain it to people who aren't mathematically trained, those steps *should* be expanded). And this step is more complicated. Your oversimplified examples of when the same principle applies to polynomials follow from a result called Gauss's lemma. In abstract algebra textbooks they typically take a page or so to prove Gauss's lemma. Whatever principle it is you think you're applying is more complicated than that too. Second, it's not something I can see how to prove myself, nor is it a standard result. Third, your explanations for why it works amount to those simplified examples plus ridicule directed at people who doubt it. You don't even manage to state what the general principle is. And finally, standard results that other people have been citing imply that the conclusion you draw at that step is false-- at some other values of the variables, with m<>0, none of the g's are divisible by f. You say it just works that way, but it just doesn't work that way. Keith Ramsay ==== Can anybody recommend a good reference for reading up on methods of calculating elliptic integrals? I've had Derek Lawden's 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? ==== > 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 Zorlon's sound-measuring device register? The information supplied is inadequate. What is the correlation of the height of an Evil Fuzzie and the amount of NSU it makes when roaring? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ Shh! The maestro is decomposing! - Gary Larson ==== as they are outside the space station, they are probably in a vacuum so it heard 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 Zorlon's 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 Zorlon's sound-measuring device register? > As Z-X was ten feet away from red loud mouth, it's was 1/4 as loud were it at 5 feet. Now reds are 6 times louder than greens. Thus a red, 10 feet away is as loud as 6 * 1/4 = 3/2 greens at five feet. Now greens at five feet shout at 10 NSU's, thence a sci.physics problem. However to continue above the din, Z-X would hear 10 * 3/2 = 15 NSU if it's a linear measure. If it's 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 Zorlon's sound-measuring device register? ==== In sci.math, Leanne : > 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 Zorlon's sound-measuring device register? If Fuzzles can breathe space (which is hard to do) and the NSU can also be measured in space, and a 2-foot-high Fuzzle is twice as loud as a 1-foot-high one, then the problem can be set up this way. You know how loud a Green Fuzzle 1 foot high is. multiply three things together. (No, not Fuzzles. Although it's a thought, the problem indicates no method of Fuzzle reproduction.) And there you have it: the problem setup. One other thing. You've 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.net It's still legal to go .sigless. ==== 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 wasn't clear enough in the initial post. The notation dz, dy etc. in this case has nothing to do with differentials. It's simply a shorter way of writting Delta{z}, Delta{y} etc., ie dz is the uncertainty in 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 or wealth > e a e of a man is a question fit perhaps to be discussed by > n e . slaves in the hearing of their masters, but highly > @ r c m unbecoming to reasonable and free men in search of > d o the truth. -- Rousseau ==== Two traditional ways to decode Reed-Solomon codes include the Massey-Berlekamp and Euclidean decoding algorithms. Can someone please explain the underlying theory on which the Euclidean algorithm is based, or perhaps give some pointers in literature where it is covered? Does it have anything to do with the Euclidean algorithm used in abstract algebra to compute the GCD? Your time, effort and suggestions will be greatly appreciated Jaco ==== 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 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 Z p0 = 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 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. -- 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 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, 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) 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 -infinity. If your distributions start at 0, it's the same, but if you're using the normal distribution you've 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, > Skitt's Law (among other appelations): Every spelling (in this case, formula) correction contains its own error. Well, to OP, you've 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 Bernier's post about calculating pi(10^15) with PrimeCountH.java lead me to thinking about an interesting feature of the program that I haven't commented about a lot recently, which is that it uses some terms of the explicit prime counting function, compressed in that it assumes even N, to speed things up. That is I use terms from THE explicit prime counting function itself--compressed--in the program. Those terms are, with even N, if N<9 pi(N) = floor(N/2); if N<25 pi(N) = floor(N/2) - floor((N-4)/6); if N<49 pi(N) = floor(N/2) - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) if N<121 pi(N) = floor(N/2) - floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - floor((N-36)/14) + floor((N-22)/42); if N>121 pi(N) = floor(N/2) -floor((N-4)/6) - floor((N-16)/10) + floor((N-16)/30) - floor((N-8)/14) + floor((N-22)/42) + floor((N-106)/70) - floor((N-106)/210) +2+... So you can see the first 8 terms of the prime counting function compressed, where before I would pull that 2 into its proper place in one of the terms, but realized the computer would be a bit faster letting it sit out, and now I'm not sure where it goes. You may be wondering why I split out like that, and it's because it doesn't work if you allow negative numbers, and as the floor doesn't go to 0, when the result is negative, I split it out. Like, if floor(N-106)/70) is negative for your N, and that goes into the calculation, it'll give the wrong answer. The full prime counting function is of course infinite in size, but I thought it worth it to point out that I've given the first few terms of the compressed prime counting function, where the compression is that it knows that 2 is prime, which cuts into the total number of terms. So what I've given above, at least for the compressed explicit prime counting function, are THE first few terms, and of course, there are an infinite number of terms. James Harris ==== Market research for a new mobile phone suggests that the potential market has a ceiling of 40 million units and that 0 sales will be achieved without marketing. After spending 1 million on marketing 20 million sales were recorded. Can anyone model for sales (S) in terms of advertising expenditure (x) of the form: S = a+(be)POWER-cx ==== > Market research for a new mobile phone suggests that the potential market > has a ceiling of 40 million units and that 0 sales will be achieved without > marketing. After spending 1 million on marketing 20 million sales were > recorded. Can anyone model for sales (S) in terms of advertising expenditure (x) of > the form: S = a+(be)POWER-cx I assume you mean S = a + b e^(-cx) Well, you have some known values, so use them: (0, 0) (+oo, 40mil) (1mil, 20 mil) Plug those (x, S) pairs into the equation for S and see what they imply about constants a, b, and c. -- Rich Carreiro rlcarr@animato.arlington.ma.us ==== >>Market research for a new mobile phone suggests that the potential market >>has a ceiling of 40 million units and that 0 sales will be achieved without >>marketing. After spending 1 million on marketing 20 million sales were >>recorded. >>Can anyone model for sales (S) in terms of advertising expenditure (x) of >>the form: >>S = a+(be)POWER-cx > I assume you mean > S = a + b e^(-cx) Well, you have some known values, so use them: > (0, 0) > (+oo, 40mil) > (1mil, 20 mil) Plug those (x, S) pairs into the equation for S and see what they > imply about constants a, b, and c. > ==== How do I answer this problem? Show that xÓ + 2x + 3a is a factor of axÓ + xŇ + 5ax + 2xÓ + 3aÓ I understand I have to use algebraic long division to find prove that xÓ + 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 not when it involves more variables than just x. Any help would be appreciated. ==== > How do I answer this problem? Show that xÓ + 2x + 3a is a factor of axÓ + xŇ + 5ax + 2xÓ + 3aÓ I understand I have to use algebraic long division to find prove that xÓ + > 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 not when > it involves more variables than just x. View this with a fixed-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 Tooth http://www.clivetooth.dk ==== Apologies if this isn't the right group, but googling has been less than helpful this morning. I'm interested in taking some 3rd or 4th year college-level math classes (in particular applied math or numerical analysis) as part of lifelong learning / continuing education. The catch is that due to limited time, I'd prefer to do them in a self-paced online or correspondence format. Is this available anywhere? Ranjan Bagchi ==== I am hoping to find 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 isn't divisible by two. What I'd like is a one step formula that can express how many factors of 2 are in X like 0 to ?. I thought to ask because it may have been done already. Ernst ==== >Message-id: I am hoping to find 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 isn't divisible by >two. But that doesn't tell you how many, just that it has at least one (if even) or none (if odd). What I'd 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 significant zeroes. 144 base ten is 10010000 base two. It has 4 least significant 0s, so it has 4 factors of 2. Check: 144 = 3*3*2*2*2*2 Now maybe you don't consider base conversion a one-step formula (it's 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 length binary numbers, I use a modified rule: instead of dividing by 2, I remove all the factors of 2 in a single step. Not surprisingly, I call this the Factoring Rules. The effect of this is the modified Collatz sequence contains only odd numbers. To avoid the overhead of doing base conversion, I do all the math in binary. The stopping point (1) doesn't need to be converted back to decimal, so I only need to do the conversion once from decimal to binary at the start of the sequence. The limited math involved in the Collatz process is relatively easy to implement using strings of 1s and 0s. When I need to remove (or count) all the factors of 2 from a number, a simple regular expression in perl can do this in a single step. I thought to ask because it may have been done already. >Ernst -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm ==== >>I am hoping to find 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 isn't divisible by >>two. But that doesn't tell you how many, just that it has at least one (if even) or >none (if odd). > What I'd 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 >significant zeroes. Now maybe you don't consider base conversion a one-step formula (it's 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, it's 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 ==== >>I am hoping to find 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 isn't divisible by >>two. But that doesn't tell you how many, just that it has at least one (if even) or >none (if odd). > What I'd 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 >significant zeroes. Now maybe you don't consider base conversion a one-step formula (it's 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, it's 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. It's 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. Two problems came up. First, although there are bit-wise and, or and xor operators 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-1 becomes x & -x in Python. Second problem was that there is no base 2 logarithm function in 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 doesn't work in practice. As the test program below shows, it works fine 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 log h = long(2) log2 = log(2) #constant, only needs to be calculated once j = 1000 while 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 to find 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 isn't divisible by >>two. But that doesn't tell you how many, just that it has at least one (if even) or >none (if odd). > What I'd 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 >significant zeroes. Now maybe you don't consider base conversion a one-step formula (it's 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, it's 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 to find 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 isn't divisible by >>two. But that doesn't tell you how many, just that it has at least one (if even) or >none (if odd). > What I'd 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 >significant zeroes. Now maybe you don't consider base conversion a one-step formula (it's 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, it's 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. > It's 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 don't see a log2 function for really large values either. Gcc seems to have a log2 function for data type float and double. Ernst I admit I know nothing of perl. ==== >Message-id: >>I am hoping to find 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 isn't divisible by >>>two. >>But that doesn't tell you how many, just that it has at least one (if >even) or >>none (if odd). >>> What I'd 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 >>significant zeroes. >>Now maybe you don't consider base conversion a one-step formula (it's >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, it's 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. >> It's 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 don't see a log2 >function for really large values either. Gcc seems to have a log2 >function for data type float and double. Read the rest of my message. You don't 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 figure 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. I'm sure there is an equivalent in C. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm ==== >Message-id: >>I am hoping to find 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 isn't divisible by >>>two. >>But that doesn't tell you how many, just that it has at least one (if > even) or >>none (if odd). >>> What I'd 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 >>significant zeroes. >>Now maybe you don't consider base conversion a one-step formula (it's > 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, it's 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. >> It's 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 don't see a log2 >function for really large values either. Gcc seems to have a log2 >function for data type float and double. Read the rest of my message. You don't 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 figure 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. I'm 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, since it is analogous to exp and exp2. Function: double log (double x) Function: float logf (float 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 returns negative infinity; if x is too close to zero, it may signal overflow. Function: double log10 (double x) Function: float log10f (float 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: float log2f (float 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: float logbf (float x) Function: long double logbl (long double x) These functions extract the exponent of x and return it as a floating-point value. If FLT_RADIX is two, logb is equal to floor (log2 (x)), except it's probably faster. If x is de-normalized, logb returns the exponent x would have if it were normalized. If x is infinity (positive or negative), logb returns @math{@infinity{}}. If x is zero, logb returns @math{@infinity{}}. It does not signal. Function: int ilogb (double x) Function: int ilogbf (float 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 a consideration. What I am up to these days is trying to decide what my next hailstone generator will accomplish. Oh well.. all the discovery had me running and I enjoyed it all even the errors. Ernst ==== I will stop bothering you all now because I never pursued a Mathematics degree beyond a bachelor's and shouldn't be proclaiming crazy and false ideas here. ==== >I will stop bothering you all now because I never pursued a Mathematics >degree beyond a bachelor's and >shouldn't be proclaiming crazy and false ideas here. Look, you can stop faking posts. It doesn't 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 man's 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 figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite 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.edu ==== For 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 bachelor's and > shouldn't be proclaiming crazy and false ideas here. ==== > I will stop bothering you all now because I never pursued a Mathematics > degree beyond a bachelor's and > shouldn't 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 you've 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 bachelor's and >shouldn't be proclaiming crazy and false ideas here. Look, you can stop faking posts. It doesn't 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 man's 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 figures few readers can critize. A great > many people are staggered to this extend, that they imagine there > must be the indefinite 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.edu I thought it was pretty funny. I enjoyed it. It was a lot better than most of the baggage on sci.math. ==== Well, no, he didn't. JSH can be imitated, but never duplicated. ==== I will stop bothering you all now because I never pursued a Mathematics > degree beyond a bachelor's and > shouldn't 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 > you've 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 first and only fishfry posting anything, ever, to the internet? If not, are you guilty of forging headers? Why don't you change your name to dumbass ? Wouldn't this name fit you better? So many questions for such a pinhead like you. ==== I will stop bothering you all now because I never pursued a Mathematics > degree beyond a bachelor's and > shouldn't 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 > you've 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 first and only fishfry posting anything, ever, to the > internet? If not, are you guilty of forging headers? Erm, I'm not sure that you know what a header is... ... Clearly fishfry was correct. Why don't you change your name to dumbass ? Wouldn't this name fit > 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 bachelor's and > shouldn't 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 > you've 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 and find out yourself. There is in fact a jamesharris.com, which is NOT the source of the original poster's 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 guy's forgery? > Are you the first and only fishfry posting anything, ever, to the > internet? If not, are you guilty of forging headers? > Why don't you learn something about how the internet works and then come back and talk to us. > Why don't you change your name to dumbass ? Wouldn't this name fit > you better? So many questions for such a pinhead like you. ==== Consider This scribbled the following on sci.math: >> I will stop bothering you all now because I never pursued a Mathematics >> degree beyond a bachelor's and >> shouldn't 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 >> you've 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 first and only fishfry posting anything, ever, to the > internet? If not, are you guilty of forging headers? > Why don't you change your name to dumbass ? Wouldn't this name fit > you better? > So many questions for such a pinhead like you. In fact there HAVE been other James Harrises on this newsgroup. Other people, than our friend JSH, whose real name is James Harris. But this guy who posted this thread is not one of them. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~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 a specific James Harris. ==== >Perhaps their abuse department would be interested in knowing that >you've 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 bachelor's and > shouldn't 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 > you've 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 and find out yourself. There is in fact a jamesharris.com, which is NOT the source of the > original poster's post. domain than that of the ISP hosting the originators news server? Do you have any idea how many people are forging headers, by your pinhead definition? Are you people really this stupid? ==== > I will stop bothering you all now because I never pursued a Mathematics > degree beyond a bachelor's and > shouldn't 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 > you've 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 and find out yourself. There is in fact a jamesharris.com, which is NOT the source of the > original poster's post. domain than that of the ISP hosting the originators news server? I'm not expert at decoding headers, but here's what I see. this person. Furthermore, it is a Pennsylvania server of Comcast. The message ID assigned to this posting originated at U Penn's news server. Again you see here that the path via which this message shows up in your server originated at netnews.upenn.edu. More confirmation that the NNTP origin of this post was the University of Pennsylvania. No. However, if the from address and the sender belong to completely different people, this is an indicator of something amiss. More significantly, we know that James Harris lives in Atlanta, Georgia, does not have a jamesharris.com domain, has no connection with the University of Pennsylvania and has disdain for anybody who happens to use a *.edu server, which he considers to be somehow an attempt at one upmanship. It is very clear that this message originated in Pennsylvania and probably at U. Penn or from somebody associated with U. Penn. Do you have any idea how many people are forging headers, by your > pinhead definition? Spammers routinely forge headers. ISPs known this. Why do you think that they request you to send the full headers when you are making a spam complaint? There is information there which is much harder to forge and which can be used to track down the actual origin. Are you people really this stupid? Do you really think people DON'T forge headers? By the way, here are yours: <1282cbdd.0307172056.26e841a4@posting.google.com> There are a lot of public-access databases out there that make the internet go. These are the hacker's friend as well I find the very interesting information that, while the IP 168.143.113.143 does not have a name, most of the block of addresses 168.143.113.* belongs to anonymizer.com. That tells me that you probably posted through them to help hide your identity and message origins. When it becomes a law enforcement issue, legal authorities have the ability to learn much, much more about where you posted from and who you are by examining network logs. You can rest assured that Anonymizer.com has your actual IP address in their logs. - Randy ==== > Clearly fishfry was correct. fishfry may be a bit of a flamer, but he's absolutely right. It doesn't take a rocket scientist to forge pretty much every one of those headers. Think ==== Clearly fishfry was correct. fishfry may be a bit of a flamer, but he's absolutely right. It doesn't take > a rocket scientist to forge pretty much every one of those headers. Think It's one thing to come on here and mock JSH. That's fast becoming the national sport of sci.math. It's another thing to mock JSH by pretending to be the legitimate owner of jamesharris.com, pharmaceutical manufacturing consulting firm operating out of Durham, North Carolina, when the Usenet headers clearly show that the poster is using the U. Of Pennsylvania's NNTP server. When I pointed that out, was I flaming? ==== > Clearly fishfry was correct. fishfry may be a bit of a flamer, but he's absolutely right. It doesn't take > a rocket scientist to forge pretty much every one of those headers. Think > It's one thing to come on here and mock JSH. That's fast becoming the > national sport of sci.math. It's another thing to mock JSH by > pretending to be the legitimate owner of jamesharris.com, pharmaceutical > manufacturing consulting firm operating out of Durham, North Carolina, > when the Usenet headers clearly show that the poster is using the U. Of > Pennsylvania's NNTP server. When I pointed that out, was I flaming? I don't remember any mention of a pharmaceutical company. I would say, yes, you were flaming. The whole thing was in a spirit of fun, which your post quickly negated. You're 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. ==== > > I will stop bothering you all now because I never pursued a Mathematics > degree beyond a bachelor's and > shouldn't 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 > you've 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 did not actually come from jamesharris.com (which does exist) you can talk about forgery. And in that case it is upto jamesharris.com to take action. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > Clearly fishfry was correct. fishfry may be a bit of a flamer, but he's absolutely right. It doesn't > take > a rocket scientist to forge pretty much every one of those headers. > Think > It's one thing to come on here and mock JSH. That's fast becoming the > national sport of sci.math. It's another thing to mock JSH by > pretending to be the legitimate owner of jamesharris.com, pharmaceutical > manufacturing consulting firm operating out of Durham, North Carolina, > when the Usenet headers clearly show that the poster is using the U. Of > Pennsylvania's NNTP server. When I pointed that out, was I flaming? > I don't remember any mention of a pharmaceutical company. I would say, > yes, you were flaming. The whole thing was in a spirit of fun, which your > post quickly negated. You're 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. There's a reason I challenged the original poster. What he did was unacceptable on this newsgroup. Many people mock JSH. That's not the point. Nor is it the point that some people post with anonymous aliases. I post as fishfry, which is MY innocent third parties in my anononymity. What the original poster did was alter his FROM field 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? It's 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 that's 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, you'll 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 it's 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 we're 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: [...] Crossposting's great... but hey, what's a Followup-To? -- Taking care of trolls, k00ks, and other retarded minds ==== Munkres, in the first chapter of his topology text, refers to the finite axiom of choice; [...] given a finite 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 finite 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 first two sentences.) To be specific, 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 first chapter of his topology text, refers to the > finite axiom of choice; [...] given a finite 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 finite 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 first two sentences.) To be specific, does the step, let x be an element of of B invoke > choice? I would think not. You've got it right, the book is wrong. I read about that so-called finite axiom of choice many years ago in a topology book by Hall and Spencer. Maybe it's an old topologists' superstition. ==== >Munkres, in the first chapter of his topology text, refers to the >finite axiom of choice; [...] given a finite 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 then of course he's simply wrong, but if he doesn't then there's nothing necessarily improper about any of this: He surely hasn't 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. > 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 finite 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 first two sentences.) To be specific, does the step, let x be an element of of B invoke >choice? I would think not. ************************ David C. Ullrich ==== >Munkres, in the first chapter of his topology text, refers to the >>finite axiom of choice; [...] given a finite 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 writing a text on topology not on set theory and doesn't list any axiom system for sets. He does later say that the finite AC is part of eveyone's set theory. > If he does then > of course he's simply wrong, but if he doesn't then there's nothing > necessarily improper about any of this: He surely hasn't 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 first chapter of his topology text, refers to the >>finite axiom of choice; [...] given a finite 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 finite axiom of choice... No mathematician has any qualms about the finite choice axiom; it is part of everyone's set theory. Said differently , no one has qualms about a proof that involves only finitely 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 first chapter of his topology text, refers to the >finite axiom of choice; [...] given a finite 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 doesn't 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 > finite axiom of choice... No mathematician has any qualms about the > finite choice axiom; it is part of everyone's set theory. Said > differently , no one has qualms about a proof that involves only > finitely many arbitrary choices. > axiom, it is invariably the stronger form of the axiom [i.e., AC] > that is meant. Huh? I don't see how that's an argument for accepting the axiom of choice 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 first chapter of his topology text, refers to the >finite axiom of choice; [...] given a finite 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 first-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 finite axiom of choice is a consequence of first order logic, and does not involve the axioms of set theory. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University ==== >Munkres, in the first chapter of his topology text, refers to the >>finite axiom of choice; [...] given a finite 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 first-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 finite axiom of choice is a consequence of first 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 - it's impossible to prove that there exists a set having _any_ property without some sort of axiom regarding sets. If we know that there exists x such that P(x) then we can choose such an x with no set theory, I suppose (one would need a definition of choose to prove this, but I'm not inclined to dispute it), but we need an axiom of set theory to get from x to the set {x}. ************************ David C. Ullrich ==== Along the same lines as the original post: For m = positive integer, and {a(k)} = any sequence defined for positive 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=1 In linear-mode: sum{k=1+(m-1)! to m!} floor(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 finite-sums are free of floor-functions. > Also, there MIGHT be some interesting number-theory results which > could be related to this. (see below.) > Let {a(k)} = any sequence defined 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(floor((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 floor((m+1)!/j)} k a(k) > A specific 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)!} floor((m+1)!/j) Now this last floor-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 floor-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 to find the way to calculate the smallest sphere which would include all of them inside the sphere (not necessarily on its surface). Kiran ==== I have some 3D data points and I would like to find the way to calculate > the smallest sphere which would include all of them inside the sphere > (not necessarily on its surface). > Kiran It is easy enough to find the smallest cube that includes all your points. The smallest sphere that encloses this cube also includes all your points. -- Christos Dimitrakakis IDIAP (http://www.idiap.ch/~dimitrak/main.html) ==== I have some 3D data points and I would like to find 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 to find the smallest cube that includes all your > points. The smallest sphere that encloses this cube also includes all > your points. How do you find the smallest cube that includes all the points? -- Clive Tooth http://www.clivetooth.dk ==== I have some 3D data points and I would like to find the way to >calculate the smallest sphere which would include all of them inside >the sphere (not necessarily on its surface). >Kiran See http://www.faqs.org/faqs/graphics/algorithms-faq/ subject 5.24 ==== >> I have some 3D data points and I would like to find 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 to find the smallest cube that includes all your > points. The smallest sphere that encloses this cube also includes all > your points. But that doesn't answer the question. The question was to find the *smallest* sphere that contains the given points. For example, suppose we start with three points on the surface of the unit sphere, and suppose the points are chosen so that they do not lie at the vertices of an inscribed cube. Then the smallest cube containing the given points must have a side length greater than 2/sqrt(3), and any sphere that contains that cube is too big. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== I have some 3D data points and I would like to find the way to > calculate the smallest sphere which would include all of them inside > the sphere (not necessarily on its surface). > Kiran Here is something you can do if you don't absolutely need the absolutely smallest sphere. The problem is to find the center of the sphere. Once you have that, you can just find the maximum of its distances to all the points to get the radius. To find the center of the sphere, you need to minimize the maximum distance from a point (x,y,z) to all the points in the set. Instead of minimizing the maximum, you can minimize something that is reasonably close to the maximum but yet is differentiable, so you can use a gradient search type method. Let (x,y,z) denote the unknown center. Let (x0,y0,z0) denote one of the points in the set. Form the quantity [(x-x0)^2 + (y-y0)^2 + (z-z0)^2]^20. (I chose 20 arbitrarily. It doesn't seem to help much to use a higher exponent, but you could. The higher the exponent, the closer you will theoretically come to the true center of the smallest circle; but the gradient 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) which you want to minimize. A good place to start the minimization search would be at the point (xm, ym, zm), where xm denotes the average of all the x-values in the set, similarly for ym and zm. I tried this with several randomly generated examples in Mathematica, and some examples where I threw in points to try to confound it, and the results looked good. I tinker with FLT for fun and I am not claiming a proof. This post is more about implicit and explicit restrictions. FLT is frequently stated as a^n + b^n = c^n which in itself is solvable but of course, FLT restricts the consideration to integers. Well, 1 < a < b < c is a good restriction for the integers, since just saying a, b, and c must be integers allows for a, b, or c to = 0. Then n argument can be made for not needing to consider negative integers, and only needing 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 c is the Mahanlobois mean or possibly the Mahanlobois sum of a and b with characteristic 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 between 1 and oo (infinity), and I think for integers 1 < a < b and c the mean, a triangle can be constructed with angle C opposite side c. For n=1 the triangle is 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) ^ n 2*(a*b)^n = p ( a^2 , b^2, -2*a*b*x) - a^2*n - b^2*n where 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 binomial coefficient. I refer you to Kemeny, Schleifer, Snell, and Thompson: Finite Mathematics with Business Applications for that notation. It's what I have on hand. I'm really enjoying the chapters on partitions and counting because they relate to entropy, which is not considered in this text. I read Kinetic Theory of Gases by Leonard B. Leob and got some similar insight. Stepping back, (a^n + b^n) ^ 2 = (a^2 + b^2 - 2*a*b*x) ^ n implies to me that for further manipulation, and I understand that elementary algebraic manipulations hold little hope for FLT, n must be an integer. That is, I wonder whether writing the FLT equation this way somehow EMBODIES the restriction that n is an integer. I see two ways of reading this: it's either an implicit specification of a relationship between a, b, n, and x, or since the multinomial has no meaning for nonpositive n, it somehow encapsultes the Diophantine restriction on n given in FLT. I do not claim it means a and b must be integer. Only n, and even that as one of two interpretations. You'll note nowhere here does c come into play. But there is an x, so it's still a four number problem. Is that a problem? I don't think so. In Wile's proof 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 something else. Now there's a y instead of a c, and it's a whole new problem, not just an algebraic one either. And I think that turning the problem into something else is 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 fee Civil and criminal penalties for circumvention of any embedded encryption ==== what is the Mahanlobois mean, supposed to be; was it implicit to your tutorial? what is the answer in mhos? of course, c is implicit in the expanded forms; a trigon is a trigon is triangular. anyway, a good exposition of the futility of this, taht I really like, is by this Puerto Rican professor; he even has a wooden model! > Now, this mean is decreasing with increasing n. So, for all values of n between > 1 and oo (infinity), and I think for integers 1 < a < b and c the mean, a > triangle can be constructed with angle C opposite side c. For n=1 the triangle > is 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) ^ n 2*(a*b)^n = p ( a^2 , b^2, -2*a*b*x) - a^2*n - b^2*n where 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 binomial > coefficient. I refer you to Kemeny, Schleifer, Snell, and Thompson: Finite Mathematics with > Business Applications for that notation. It's what I have on hand. I'm really > enjoying the chapters on partitions and counting because they relate to > entropy, which is not considered in this text. I read Kinetic Theory of Gases > by Leonard B. Leob and got some similar insight. --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX ==== Let A and B be nonempty subsets of R which are bounded above. Also define the following sets of upper bounds for A and B: S_A : = {s in R : forall a in A, a <= s} S_B : = {s in R : forall b in B, b <= s} I want to find formal (rigorous) justification for the following: B subseteq A --> S_A subseteq S_B In words: If B is a subset of A, then the set of upper bounds for A is a subset of the set of upper bounds for B. I can easily see that this is true by drawing a picture, but I would like to have formal justification, starting from the defintions of subset and upper bound. Any takers? thanks. ==== > Let A and B be nonempty subsets of R which are bounded above. Also > define the following sets of upper bounds for A and B: > S_A : = {s in R : forall a in A, a <= s} > S_B : = {s in R : forall b in B, b <= s} I want to find formal (rigorous) justification for the following: B subseteq A --> S_A subseteq S_B In words: If B is a subset of A, then the set of upper bounds for A > is a subset of the set of upper bounds for B. I can easily see that this is true by drawing a picture, but I would > like to have formal justification, starting from the defintions of > subset and upper bound. Any takers? thanks. Let x be a member of S_A. By definition of S_A we have a <= x for all elements a of A. As B is a subset of A, every element b of B is also an element of A, and therefore b <= x. So for every element b of B we have b <= x, which means by definition of S_B that x is an element of S_B. As x was an arbitrary member of S_A, we have shown that every element of S_A is also an element of S_B, which means that S_A is a subset of S_B. By the way, the upper bound is irrelevant. Every relation between a and s in the definitions of S_A and S_B would work exactly in the same way. ==== I'm struggling with the following exercise in Rudin's Principles of Math Analysis: (Chapter 8, exercise 21, p.201 in my book) Let L_n be 1/(2*pi) times the integral, from -pi to pi, of the absolute value of the Dirichlet kernel D_n. (The definition of D_n(x) is sum_{j=-n}^{j=n}e^(ijx) ) Prove that the sequence {L_n - 4(log n)/(pi)^2} is bounded. ==== > I'm struggling with the following exercise in Rudin's Principles of > Math Analysis: (Chapter 8, exercise 21, p.201 in my book) Let L_n be 1/(2*pi) times the integral, from -pi to pi, of the > absolute value of the Dirichlet kernel D_n. (The definition of D_n(x) > is sum_{j=-n}^{j=n}e^(ijx) ) > Prove that the sequence {L_n - 4(log n)/(pi)^2} is bounded. Here's a sketch. Use the formula D_n(x) = sin (n + 1/2)x / sin x/2. This function oscillates, so you want to break up the integral into integrals over segments where it stays the same sign. You can just work with the integral from 0 to pi. Write int[from 0 to pi] |sin (n + 1/2)x / sin x/2| dx as a sum of integrals: the sum[k=0 to n-1] of the integral as x goes from k pi/(n + 1/2) to (k+1) pi /(n + 1/2). Notice that the integral from n pi / (n + 1/2) to pi/2 is omitted. That little piece is bounded, in fact goes to 0. Argue that the first two terms in the sum stay bounded, so you can start with k=2. Then sin (n + 1/2)x / sin x/2 is of constant sign on the interval [ k pi/(n + 1/2), (k+1) pi /(n + 1/2) ] so you get int[from k pi/(n + 1/2) to (k+1) pi /(n + 1/2)] |sin (n + 1/2)x / sin x/2| dx <= 2/(2n+1) csc( k pi / (2n+1) ), so the original integral is bounded by sum[k=2 to n-1] of 2/(2n+1) csc( k pi / (2n+1) ) + O(1) **** csc x is the derivative of f(x) = ln(tan(x/2)), which is concave down, so csc x <= (f(x) - f(x-h))/h (for small positive h) This allows us to estimate the sum **** above by 2/pi sum[k=2 to n-1] of { f((k pi)/(2n+1)) - f(((k-1) pi)/(2n+1)) } and this sum collapses to - 2/pi f(pi/(2n+1)) + O(1) = 2/pi ln(tan(n+1/2)) + O(1) = 2/pi ln(n) + O(1) I have lost a factor of 2 somewhere, but if you write this out, it should show up. Then you do the lower bound the same way. ==== > Let L_n be 1/(2*pi) times the integral, from -pi to pi, of the > absolute value of the Dirichlet kernel D_n. (The definition of D_n(x) > is sum_{j=-n}^{j=n}e^(ijx) ) > Prove that the sequence {L_n - 4(log n)/(pi)^2} is bounded. Another approach: Use D_n(x) = sin[(n+1/2)x]/sin(x/2) as Will suggested. Note that 1/sin(x/2) - 1/(x/2) is bounded on [0,Pi]. So we need only estimate int_[0,Pi] |sin[(n+1/2)x]/x| dx = int_[0,Pi(n+1/2)] |sin(x)/x| dx, and now you can forget about the 1/2. On the intervals [mPi, (m+1)Pi], sin(x) will be of constant sign. Integrate by parts there (integrate sin(x) and differentiate 1/x). Recall that 1 + 1/2 + ... + 1/n = log(n) modulo a bounded error. Keep track of the constants and everything should work out. ==== Let L_n be 1/(2*pi) times the integral, from -pi to pi, of the > absolute value of the Dirichlet kernel D_n. (The definition of D_n(x) > is sum_{j=-n}^{j=n}e^(ijx) ) > Prove that the sequence {L_n - 4(log n)/(pi)^2} is bounded. Another approach: Use D_n(x) = sin[(n+1/2)x]/sin(x/2) as Will suggested. > Note that 1/sin(x/2) - 1/(x/2) is bounded on [0,Pi]. So we need only > estimate int_[0,Pi] |sin[(n+1/2)x]/x| dx = int_[0,Pi(n+1/2)] |sin(x)/x| dx, and now you can forget about the 1/2. On the intervals [mPi, (m+1)Pi], > sin(x) will be of constant sign. Integrate by parts there (integrate sin(x) > and differentiate 1/x). Recall that 1 + 1/2 + ... + 1/n = log(n) modulo a > bounded error. Keep track of the constants and everything should work out. This is better than mucking around in all that trig. ==== I seem to remember from years ago that it was an open question whether the free group on 3 generators is isomorphic to some subgroup of the free group on 2 generators. Can anyone tell me what has become of that? ==== >I seem to remember from years ago that it was an open question whether the >free group on 3 generators is isomorphic to some subgroup of the free group >on 2 generators. Can anyone tell me what has become of that? How long ago? The (absolutely) free group on 2 generators contains subgroups of arbitrarily high rank; it even contains subgroups of countable rank. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >I seem to remember from years ago that it was an open question whether the >free group on 3 generators is isomorphic to some subgroup of the free group >on 2 generators. Can anyone tell me what has become of that? How long ago? Does that matter? > The (absolutely) free group on 2 generators contains subgroups of > arbitrarily high rank; it even contains subgroups of countable rank. Yes, but that does not approach the question I asked. Perhaps it would be best if I rephrase my question, leaving out mention of how long ago. Is it true that the free group on 3 generators is isomorphic to a subgroup of the free group on 2 generators? ==== >I seem to remember from years ago that it was an open question whether >the >>free group on 3 generators is isomorphic to some subgroup of the free >group >>on 2 generators. Can anyone tell me what has become of that? >> How long ago? Does that matter? > The (absolutely) free group on 2 generators contains subgroups of >> arbitrarily high rank; it even contains subgroups of countable rank. Yes, but that does not approach the question I asked. Why not? You are asking if the absolutely free group of rank 3 is isomorphic to a subgroup of the absolutely free group of rank 2. I said that the absolutely free group of rank 2 contains subgroup of arbitrarily high rank, and in particular, it contains a subgroup of rank 3. Since every subgroup of an absolutely free group is absolutely free, that's that. The free group of rank 2 contains subgroups isomoprhic to a free group of ANY rank, from 0 up to countably infinite. >Is it true that the free group on 3 generators is isomorphic to a subgroup >of the free group on 2 generators? Yes. Just like I said above. What makes you think that the answer does not approach the question you asked? The reason I asked how long ago was this is because, as far as I know, the fact that the free group of rank 2 contains subgroups of arbitrarily high rank is classical. I don't know just how far back you would have to go to find someone call it an open question.... ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >I seem to remember from years ago that it was an open question whether >the >>free group on 3 generators is isomorphic to some subgroup of the free >group >>on 2 generators. Can anyone tell me what has become of that? >> How long ago? Does that matter? > The (absolutely) free group on 2 generators contains subgroups of >> arbitrarily high rank; it even contains subgroups of countable rank. Yes, but that does not approach the question I asked. Why not? You are asking if the absolutely free group of rank 3 is > isomorphic to a subgroup of the absolutely free group of rank 2. I > said that the absolutely free group of rank 2 contains subgroup of > arbitrarily high rank, and in particular, it contains a subgroup of > rank 3. Since every subgroup of an absolutely free group is absolutely > free, that's that. The free group of rank 2 contains subgroups > isomoprhic to a free group of ANY rank, from 0 up to countably infinite. Is it true that the free group on 3 generators is isomorphic to a subgroup >of the free group on 2 generators? Yes. Just like I said above. What makes you think that the answer > does not approach the question you asked? The reason I asked how long ago was this is because, as far as I know, > the fact that the free group of rank 2 contains subgroups of > arbitrarily high rank is classical. I don't know just how far back you > would have to go to find someone call it an open question.... ====================================================================== > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > ====================================================================== Arturo Magidin > magidin@math.berkeley.edu