mm-1066 >>A-2. >>For i = 1,2 let T_i be a triangle with side lengths a_i, b_i, c_i, >>and area A_i. Suppose that a_1 <= a_2, b_1 <= b_2, c_1 <= c_2, >>and that T_2 is an acute triangle. Does it follow that A_1 <= A_2 ? > Here is a solution that does not employ HeronÕs area formula: > Let T_1 = ABC. At least one of the angles of T_2 has to be greater than or > equal to one of the angles of ABC, say and side BC with ABC, so that T_2 = BDE, with D on the BC oriented line > and |BC| <= |BD|, and Draw a line L passing through A and parallel to BC, and let F be the > projection of B on L. Since accute, the BE oriented line intersects L at a point G between A and F. > The Pythagorean Theorem applied to FGB and FBA yields then |BG| <= |BA|, > so that |BG| <= |BA| <= |BE|: that says that G is on or beyond L, so that > the distance of E from the BC line is at least as big as the distance of G > (and A as well) from the BC line; in other words, the E-altitude of T_2 is > at least as big as the A-altitude of T_1. We already have the needed > inequality (|BC| <= |BD|) for the bases of T_1 and T_2, hence A_1 <= A_2. HereÕs an even shorter solution: the area of T_2 is equal to all of a_2 b_2 sin(theta_2)/2 a_2 c_2 sin(phi_2)/2 b_2 c_2 sin(psi_2)/2 where theta_2, phi_2, psi_2 are the angles between the two sides which appear in the respective formulas. The same formulas work for T_1, with of course all the subscripts changed to 1. We are given that the lengths of the sides are all nonincreasing, and since the sum of the angles is constant one of them must not increase as well. Since the angles are all less than pi (hereÕs where acute comes in) that means that the sine of that angle must not increase. Hence at least one of those expressions is the product of three terms which do not increase over the originals. -- Ryan Reich ryanr@uchicago.edu posting-account=4XHZpAwAAAB6b7qV0WqNfmj98QBHBa2z > B-2. > Let m and n be positive integers. Show that > (m+n)! m! n! > ----------- < ----- ----- > (m+n)^{m+n} m^m n^n is there an easy, direct (no frills) proof of b-2? something like (m+n)! (m+n)^{m+n} ------ < ------------- m!n! m^m n^n iff C(m+n,m) < m^{-m}n^{-n} sum_{k=0}^{m+n} C(m+n,k)m^k n^{m+n-k} iff C(m+n,m) < sum_{k=0}^{m+n} C(m+n,k)m^{k-m} n^{m-k} but in the above summation we see there is a term of the form C(m+n,m)m^0 n^0 (when k=m), so that means that term and left hand side of course this seems to easy for a solution to a putnam problem. where is my error? i havent really looked at it in depth. > B-2. > Let m and n be positive integers. Show that > (m+n)! m! n! > ----------- < ----- ----- > (m+n)^{m+n} m^m n^n > is there an easy, direct (no frills) proof of b-2? For positive integers m and n, and any real numbers a and b, by the binomial theorem, (a+b)^(m+n)= sum(C(m+n,k)*a^(k)*b^(m+n-k),k=0..(m+n)). For positive integers a and b, each term in this sum is a positive integer. Thus we can consider the k=m term and get that (a+b)^(m+n) > C(m+n,m)*a^m*b^n. In particular, if we let a=m, b=n, we get that (m+n)^(m+n) > (m+n)!*m^m*n^n/(m!*n!). Rearranging the inequality a bit gives the desired result (m+n)! m!*n! ----------- < ---------. (m+n)^(m+n) m^m*n^n >For positive integers m and n, and any real numbers a and b, by the >binomial theorem, This is really the same proof I gave. CÕmon, doesnÕt _everybody_ think in terms of maps between finite sets? :-) > (a+b)^(m+n)= sum(C(m+n,k)*a^(k)*b^(m+n-k),k=0..(m+n)). This just counts the maps from m+n to a+b in groups according to the number k of elements in the domain which are sent to a and the number (m+n)-k which are sent to b. ThatÕs what I did, and then I picked out the special case k = m, same as you: >For positive integers a and b, each term in this sum is a positive >integer. Thus we can consider the k=m term and get that > (a+b)^(m+n) > C(m+n,m)*a^m*b^n. But you do get Brownie Points for generalizing beyond the requested result to get (a+b)^{m+n} / (m+n)! > a^m/m! * b^n/n! for arbitrary _positive real numbers_ a,b ; technically my combinatorial proof only covers _positive integers_ a and b. Is (a+b)^{m+n} / (m+n)! > a^m/m! * b^n/n! true for all positive a,b,m,n when ! is interpreted with the Gamma function? dave posting-account=4XHZpAwAAAB6b7qV0WqNfmj98QBHBa2z > B-2. > Let m and n be positive integers. Show that > (m+n)! m! n! > ----------- < ----- ----- > (m+n)^{m+n} m^m n^n is there an easy, direct (no frills) proof of b-2? something like (m+n)! (m+n)^{m+n} ------ < ------------- m!n! m^m n^n iff C(m+n,m) < m^{-m}n^{-n} sum_{k=0}^{m+n} C(m+n,k)m^k n^{m+n-k} iff C(m+n,m) < sum_{k=0}^{m+n} C(m+n,k)m^{k-m} n^{m-k} but in the above summation we see there is a term of the form C(m+n,m)m^0 n^0 (when k=m), so that means that term and left hand side of course this seems to easy for a solution to a putnam problem. where is my error? i havent really looked at it in depth. > The Sixty-Fifth Annual Putnam Exam was held today, a 6-hour competition > for undergraduates. I am posting the questions below. I will post > separately the answers I have to some of the questions; as of this > moment I have no idea how to approach A-5 and B-6. HereÕs a LaTeX version of the questions. news.indigo.ie!news-out.cwix.com!newsfeed.cwix.com! nntp.abs.net!news2.wam.um d.edu!elk.ncren.net!news.niu.edu!news!rusin %The Sixty-Fifth Annual Putnam Exam was held today, a 6-hour competition %for undergraduates. I am posting the questions below. I will post %separately the answers I have to some of the questions; as of this %moment I have no idea how to approach A-5 and B-6. usepackage{amsmath} %usepackage{verbatim} begin{document} title{The 65th William Lowell Putnam Mathematical Competition} %newenvironment{answer}{textbf{Answer:}em}{} %newenvironment{answer}{comment}{endcomment} maketitle begin{itemize} item[A-1] Basketball star Shanille OÕKealÕs team statistician keeps track of the number, $S(N)$, of successful free throws she has made in her first $N$ attempts of the season. Early in the season, $S(N)$ was less than 80% of $N$, but by the end of the season, $S(N)$ was more than 80% of $N$ Was there necessarily a moment in between when $S(N)$ was exactly 80% of $N$? item[A-2] For $i = 1,2$ let $T_i$ be a triangle with side lengths $a_i, b_i, c_i$, and area $A_i$. Suppose that $a_1 le a_2, b_1 le b_2, c_1 le c_2$, and that $T_2$ is an acute triangle. Does it follow that $A_1 le A_2$? item[A-3] Define a sequence ${ u_n }$ by $u_0 = u_1 = u_2 = 1$, and thereafter by the condition that [ detbegin{pmatrix} u_n & u_{n+1} u_{n+2} & u_{n+3} end{pmatrix} = n! ] for all $n ge 0$. Show that $u_n$ is an integer for all $n$. (By convention, $0! = 1$.) item[A-4] Show that for any positive integer $n$ there is an integer $N$ such that the product $x_1 x_2 ... x_n$ can be expressed identically in the form [ x_1 x_2 ... x_n = sum_{i=1}^N c_i left( a_{i1} x_1 + a_{i2} x_2 + ... + a_{in} x_n right)^n ] where the $c_i$ are rational numbers and each $a_{ij}$ is one of the numbers, $-1, 0, 1$. item[A-5] An $m times n$ checkerboard is colored randomly: each square is independently assigned red or black with probability $1/2$. We say that two squares, $p$ and $q$, are in the same connected monochromatic component if there is a sequence of squares, all of the same color, starting at $p$ and ending at $q$, in which successive squares in the sequence share a common side. Show that the expected number of connected monochromatic regions is greater than $m n / 8$. item[A-6] Suppose that $f(x,y)$ is a continuous real-valued function on the unit square $0 le x le 1, 0 le y le 1$. Show that [ int_0^1 left( int_0^1 f(x,y) dx right)^2 dy + int_0^1 left( int_0^1 f(x,y) dy right)^2 dx ] is less than or equal to [ left( int_0^1 int_0^1 f(x,y) dx dy right)^2 + int_0^1 int_0^1 left( f(x,y) right)^2 dx dy. ] item[B-1] Let $P(x) = c_n x^n + c_{n-1} x^{n-1} + ... + c_0$ be a polynomial with integer coefficients. Suppose that $r$ is a rational number such that $P(r) = 0$. Show that the $n$ numbers [ c_n r,; c_n r^2 + c_{n-1},; c_n r^3 + c_{n-1} r^2 + c_{n-2} r, dots c_n r^n + c_{n-1} r^{n-1} + ... + c_1 r ] are integers. item[B-2] Let $m$ and $n$ be positive integers. Show that [ frac{(m+n)!}{(m+n)^{m+n}} < frac{m!}{m^m} frac{n!}{n^n}. ] item[B-3] Determine all real numbers $a > 0$ for which there exists a nonnegative continuous function $f(x)$ defined on $[0,a]$ with the property that the region [ R = { (x,y) ; 0 le x le a, 0 le y le f(x) } ] has perimeter $k$ units and area $k$ square units for some real number $k$. item[B-4] Let $n$ be a positive integer, $n ge 2$, and put $theta = 2 pi / n$. Define points $P_k = (k,0)$ in the $xy$-plane, for $k = 1, 2, dots, n$. Let $R_k$ be the map that rotates the plane counterclockwise by the angle $theta$ about the point $P_k$. Let $R$ denote the map obtained by applying, in order, $R_1$, then $R_2, dots$, then $R_n$. For an arbitrary point $(x,y)$, find, and then simplify, the coordinates of $R(x,y)$. item[B-5] Evaluate [ lim_{x to 1^-} prod_{n=0}^infty left(frac{1 + x^{n+1}}{1 + x^n}right)^{x^n}. ] %[ThatÕs lim_{xto 1^{-}} prod_{n=0}^{infty} % left( {{1+x^{n+1}}over{1+x^n}} right)^{x^n} in TeX -- djr] item[B-6] Let $A$ be a non-empty set of positive integers, and let $N(x)$ denote the number of elements of $A$ not exceeding $x$. Let $B$ denote the set of positive integers $b$ that can be written in the form $b = a - aÕ$ with $a in A$ and $ain A$. Let $b_1 < b_2 < dots$ be the members of $B$, listed in increasing order. Show that if the sequence $b_{i+1} - b_i$ is unbounded, then [ lim_{x toinfty} N(x)/x = 0. ] end{itemize} end{document} -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland [reply address is baloglouAToswego.edu] >The Sixty-Fifth Annual Putnam Exam was held today, a 6-hour competition >for undergraduates. I am posting the questions below. I will post >separately the answers I have to some of the questions; as of this >moment I have no idea how to approach A-5 and B-6. >A-5. >An m x n checkerboard is colored randomly: each square is independently >assigned red or black with probability 1/2 . We say that two squares, >p and q, are in the same connected monochromatic component if there is >a sequence of squares, all of the same color, starting at p and ending >at q, in which successive squares in the sequence share a common side. >Show that the expected number of connected monochromatic regions is >greater than m n / 8 . How about this: show by induction on n that the expected number of *added* CMCs (connected monochromatic components) as we add a new row exceeds n/8; it is assumed that m is the number of rows and n is the number of columns. So letÕs add a new row, going from the m x n to the (m+1) x n board. By the induction hypothesis applied to the m x (n-1) and (m+1) x (n-1), more than (n-1)/8 new CMCs have been added. Now look at the last three columns (n-1, n, n+1), where there are the following 8 random possibilities: m+1 row B B B B R R R R m row BB BR RB RR BB BR RB RR expected 3/4 2/4 1/4 1/4 1/4 1/4 2/4 3/4 new CMCs The number of the expected new CMCs above is easily computed by looking at the 4 possibilities (BB, BR, RB, RR) for the (m+1) x n and (m+1) x (n+1) positions; summing up and multiplying by 1/8 we obtain another 7/16 CMCs, so that the total of expected new CMCs exceeds (n-1)/8 + 7/16 > (n+1)/8. Of course there are also the n = 1 and n = 2 basic steps that should be checked, but those are easy, with expected number of added CMCs equal to 1/2x1/2 + 1/2x1/2 = 1/2 and 1/4x3/4 + 1/4x1/4 + 1/4x1/4 + 1/4x3/4 = 1/2, respectively. baloglouAToswego.edu > How about this: show by induction on n that the expected number of *added* > CMCs (connected monochromatic components) as we add a new row exceeds n/8; Adding the mth row adds, on average, (1+[m=1])(n+2+[n=1])/8 new single-square regions. This is easy to prove. The problem is that adding the mth row can also _lose_ regions. I havenÕt found a weighting of regions that makes the average easy upper bound for the gain outweigh the average easy upper bound for the loss. This is the only problem that I havenÕt solved yet. IÕve resorted to computer simulations to get an idea of whatÕs going on. The limiting ratio for large m,n is clearly very close to 1/8. Maybe very slightly larger (is 1/8 the right answer for regions without holes?), which would turn this into a finite computation---but even after adding up 1/16 for one-square regions, 1/64 for two-square regions, 5/512 for three-square regions, and so on up to 20-square regions, IÕm still below 1/8. ---D. J. Bernstein, Associate Professor, Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago posting-account=7DhmAg0AAAAtZIYIdH_GIW6sl1UAT01k > This is the only problem that I havenÕt solved yet. IÕve resorted to > computer simulations to get an idea of whatÕs going on. The limiting > ratio for large m,n is clearly very close to 1/8. Maybe very slightly > larger (is 1/8 the right answer for regions without holes?), which would > turn this into a finite computation---but even after adding up 1/16 for > one-square regions, 1/64 for two-square regions, 5/512 for three-square > regions, and so on up to 20-square regions, IÕm still below 1/8. We seem to agree on all this: If on an infinite grid, we can show that the average size S of a connected monochromatic region (CMR) is 8 or smaller, weÕre done. Cutting out a finite portion will decrease S and thus increase the expected number of regions N to above m*n/8. (A=m*n=S*N, so S<8 implies N = A/S > m*n/8). To find N is to find the sum_{n=1}^{+inf} of n*p(n) where p(n) is the probability of some randomly chosen square in an infinite grid being in a CMR of size n. This is where I donÕt understand your calculations: I might have just totally misinterpreted your posts, but it sounds like youÕve found p(1)=1/16 and p(2)=1/64 and sum_{n=1}^20 p(n) < 1/8. FIRST Q: I found p(2)=1/16. 1/64 for 7 square to work out the way we want: a given neighbor of our randomly chosen square, and their 6 neighbors. But there are 4 possible neighbors to choose from, giving 4/64=1/16. SECOND Q: If your total for the first 20 probabilities is less than 1/8, then rest of the sum for N must be at least 21*7/8, which gives N > 8, violating the problem. In conclusion, IÕm increasingly convinced that I simply misinterpreted what youÕre doing, but hopefully I did so in a useful way. posting-account=7DhmAg0AAAAtZIYIdH_GIW6sl1UAT01k Apologies for the linebreaks, I donÕt know how that happened. Using A correction to my above post: I found p(2)=1/32. 1/128 for 7 squares to work out the way we want: a given neighbor of our randomly chosen square, and their 6 neighbors. But there are 4 possible neighbors to choose from, giving 4/128=1/32. [reply address is baloglouAToswego.edu] >> How about this: show by induction on n that the expected number of *added* >> CMCs (connected monochromatic components) as we add a new row exceeds n/8; >Adding the mth row adds, on average, (1+[m=1])(n+2+[n=1])/8 new >single-square regions. This is easy to prove. >The problem is that adding the mth row can also _lose_ regions. I But my induction hypothesis takes care of that, *IMHO*: I assume that by adding a new row to the m x (n-1) board we *expect on the average* to gain more than (n-1)/8 regions -- even if we lose some, even if we end up with a proper loss for certain configurations -- and then I show that we should expect to gain more than (n+1)/8 on the m x (n+1) board; I was careful to keep the two regions (first n-1 columns and last two columns *separated* at the m+1 row level, so that the two expectations would be independent. On the other hand, my proof needs some slight emendation: the induction hypothesis must in fact be applied to the m x (n-2) board, with the n-1 column being the separating one and the n & n+1 columns contributing the crucial new regions; so I now need the inequality (n-2)/8 + 7/16 > (n+1)/8 (instead of (n-1)/8 + 7/16 > (n+1)/8), which is *still valid*. Notice that with my old inequality I could improve the mn/8 lower bound to 7mn/32 (with 7/32 being the solution of (n-1)*s + 7/16 = (n+1)*s); solving the correct equation, (n-2)*s + 7/16 = (n+1)*s, I improve the mn/8 lower bound to 7mn/48, with 7/48 being much closer to 1/8 -- possibly even confirming your numerical findings mentioned further below??? Anyway, I hope my proof survives the peopleÕs scrutiny, but, even if not, I tried :-) [Of course if you can get a region where the expected number is indeed below 7mn/48 then you will have proved the incorrectness of my argument without going into its specifics!] >havenÕt found a weighting of regions that makes the average easy upper >bound for the gain outweigh the average easy upper bound for the loss. >This is the only problem that I havenÕt solved yet. IÕve resorted to >computer simulations to get an idea of whatÕs going on. The limiting >ratio for large m,n is clearly very close to 1/8. Maybe very slightly >larger (is 1/8 the right answer for regions without holes?), which would >turn this into a finite computation---but even after adding up 1/16 for >one-square regions, 1/64 for two-square regions, 5/512 for three-square >regions, and so on up to 20-square regions, IÕm still below 1/8. baloglouAToswego.edu Attach a random row to the bottom of the row 10101010101... The first row has n regions. The first two rows together have, on average, about n regions---certainly not (9/8)n. So itÕs simply not true that adding a random row to a fixed matrix produces an expected (1/8)n increase in the number of regions. > [Of course if you can get a region where the expected number > is indeed below 7mn/48 then you will have proved the incorrectness of > my argument without going into its specifics!] Here are the number of regions in twenty random 192x192 colorings: 4979, 5068, 5012, 5073, 5038, 5065, 4941, 5141, 4905, 4960, 4944, 4994, 4933, 4883, 5060, 4917, 4976, 4945, 4977, 5047. You think the average is above 5376? It isnÕt clear to me exactly which probability space you have in mind, so I canÕt comment specifically on what you did wrong. ---D. J. Bernstein, Associate Professor, Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago > Here are the number of regions in twenty random 192x192 colorings: 4979, > 5068, 5012, 5073, 5038, 5065, 4941, 5141, 4905, 4960, 4944, 4994, 4933, > 4883, 5060, 4917, 4976, 4945, 4977, 5047. You think the average is above > 5376? Is it convenient in your simulation to get the information on what the sizes of the regions are? How many regions of size 8 or more? If we pick a square near the center of the board, what is the expected size of ITS region? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ > Is it convenient in your simulation to get the information on what the > sizes of the regions are? Yes. ItÕs also easy to compute the (proven) limiting ratios for every small region size. Unfortunately, as I mentioned earlier in the thread, the total is still below 1/8 even if you include region sizes up to 20. I still havenÕt solved this problem. > If we pick a square near the center of the board, what is the expected > size of ITS region? ThatÕs a different question, and not one of any obvious relevance to the problem. ---D. J. Bernstein, Associate Professor, Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago [reply address is baloglouAToswego.edu] >Attach a random row to the bottom of the row 10101010101... The first >row has n regions. The first two rows together have, on average, about >n regions---certainly not (9/8)n. So itÕs simply not true that adding a >random row to a fixed matrix produces an expected (1/8)n increase in the >number of regions. Of course that row above is just one specific example while I was arguing about the expected increase over *all* possible rows ... >> [Of course if you can get a region where the expected number >> is indeed below 7mn/48 then you will have proved the incorrectness of >> my argument without going into its specifics!] >Here are the number of regions in twenty random 192x192 colorings: 4979, >5068, 5012, 5073, 5038, 5065, 4941, 5141, 4905, 4960, 4944, 4994, 4933, >4883, 5060, 4917, 4976, 4945, 4977, 5047. You think the average is above >5376? ... but this simulation provides strong evidence against my claimed 7/48 (even though I would prefer to see an average taken over *all* regions). >It isnÕt clear to me exactly which probability space you have in mind, >so I canÕt comment specifically on what you did wrong. I canÕt see anything wrong either, but probability is not my cap of tea anyway, so I simply quit at this point, passing the torch to others... baloglouAToswego.edu The Sixty-Fifth Annual Putnam Exam was held today. I have already posted a copy of the questions. Here are as many answers as I have at this moment; I have nothing to say about A-5 and B-6. If I were the grader I wouldnÕt give myself full credit for some of these answers but I think rather than nail down all the details I will wait for someone else to provide the slickest answers, making the drudgery unnecessary... (See Problems A-2, A-4, and problems with detailed computations such as B-3, B-4, B-5.) Overall I thought this was an easier exam than is typical, which made the event much more fun for the average students taking the test. I will add these answers and any other interesting responses to the files I make available about Putnam questions; see http://www.math.niu.edu/~rusin/problems-math/ A-1. Suppose that S.O.Õs success count S(N) stayed strictly below 0.80 N for all N < K, but that S(K) >= 0.80 K. This requires S(K-1) = S(K) - 1. Then from S(K-1)/(K-1) < 4/5 and S(K)/K >= 4/5 we conclude that 0 <= 5 S(K) - 4 K < 1 ; since 5 S(K) - 4 K is an integer, it must then equal zero, so S(K) = 4/5. Comment: Obviously the same proof works for any percentage of the form 1 - 1/m with integer m, but for more general percentages the result is false. This problem will make a very pretty example the next time I teach the Intermediate Value Theorem :-) A-2. 16 A^2 = 2 (a^2b^2+a^2c^2+b^2c^2) - (a^4+b^4+c^4) Viewed as a quadratic polynomial in X = a^2, this function is clearly increasing as long as X < (b^2+c^2), which means that the area of a triangle increases with the length of any side as long as the angle opposite that side is acute (a^2 < b^2 + c^2 ). (A student pointed out to me at lunch that this is also obvious from A = (1/2) b h, where the height h of the triangle varies with a but c is fixed; the maximum height occurs with sides b and c perpendicular.) It then suffices to observe that we can transform triangle T_1 to T_2 by a sequence of operations which lengthen just one edge at a time. This is not entirely obvious since, for example, we are not given that the smaller triangle is acute. But itÕs easy to see in a sketch which I will not attempt in ASCII, showing the portion of the positive octant in a,b,c -space which corresponds to triangles ( a < b+c, etc. ); the acute triangles are the ones external to three cones around the coordinate axes ( a^2 < b^2 + c^2, etc.) If one always lengthens the shortest edge (from among those that need to be lengthened to change T_1 into T_2), then that will always be opposite an acute angle. (An obtuse angle will always be opposite the _largest_ side, and that cannot the the only side remaining to be lengthened since T_2 is acute.) Note: I guess I had never before stopped to think about under exactly what conditions it is true that lengthening the sides of a triangle increases the area. Now I know! A-3. We prove by induction that u_n = (n-1)!! = (n-1)(n-3)(n-5)..., which is easily checked to be valid for small n. Of course this will mean the u_n are integral. If the claim is true for u_n and u_{n+2} then u_{n+2} = (n+1) u_n, so the first column of the matrix in the determinant is a multiple of u_n, and we have simply the recurrence relation u_{n+3} = (n+1) u_{n+1} + n!/u_n = (n+1) n!! + n!! = (n+2) n!! = (n+2)!!, providing the inductive step. A-4. This is a heavily-used formula when n = 2 : xy = (1/4)( (x+y)^2 - (x-y)^2 ). I guess I never really thought about generalizing it before! It suffices to arrange it so that the right side is a nonzero multiple of every x_i ; for then it is a degree-n polynomial and also a multiple of x_1 x_2 ... x_n, so the quotient of the two sides is a constant (necessarily rational, as can be seen from substituting x_1 = ... = 1); we can then divide through to make the constant be 1. If we arrange it so that the right side is symmetric in the x_i, then it suffices to make sure it vanishes when x_1 = 0. For odd n this is easy: for any subset T of N = {1, 2, ..., n}, let S(T) = the sum of the x_i with i in T. Then consider X = sum (-1)^|T| ( S(T) - S(N-T) )^n , the sum taken over all subsets T of N. When we substitute x1=0 we obtain an alternating sum of terms of the form ( S(U) - S(M-U) )^n where M = N - {1}. Each such term arises exactly twice, once from S(U union {1}) - S(M-U) and once from S(U) - S(N-U); since the cardinalities of U and U union {1} are of opposite parity, these two summands On the other hand, X (apparently) doesnÕt vanish identically; for example, when all x_i equal 1, we have c = sum (-1)^|T| ( |T| - |N-T| )^n = sum (-1)^k (2k-n)^n C( n,k ), which seems to equal (-2)^n n! (I computed the first few examples but didnÕt prove this formula in general.) So whatever the value of c really is (itÕs obviously integral) we can divide X by c to get a sum which is a multiple of x1, and thus of each x_i, and thus of the product x1 x2 ... x_n, but the quotient must be both constant and taking the value 1. I didnÕt take advantage of the opportunity to use 0 as a coefficient! A-5. No clue. I worked out the case n=1 completely; the expected number of components is (m+1)/2 . I suppose there might be some kind of induction argument, but since they ask us to prove the number of components is AT LEAST mn/8, that suggests that what weÕre supposed to do is to find a lot of configurations with lots of components, so that we can bound the expected number without doing all the grubby computations. A-6. Heh, heh, check this one out: Let F(x,y,z,w) = f(x,y) + f(z,w) - f(x,w) - f(z,y); then integrate F^2 over the box [0,1]^4 . Done! (This is the distillation of about 5 pages of computations, followed by a great big step back to look at What The Computations Really Mean. I have learned that to understand these integral inequalities it pays to think in terms of step functions and simple functions; writing all that out gave me a solution but then I realized this other way to say it. Note that it is critical that the domain of integration be the UNIT square.) B-1. This is really standard. Write r = p/q in lowest terms; then we have c_n p^n + c_{n-1} p^{n-1} q + ... + c_0 q^n = 0 and so for each j between 0 and n we see q^j divides c_n p^n + c_{n-1} p^{n-1} q + ... + c_{n-j+1} p^{n-j+1} q^{j-1} = p^{n-j}( c_n p^{j} + c_{n-1} p^{j-1} q + ... + c_{n-j+1} p^1 q^{j-1} ) Since p and q are coprime, q^j divides the other factor. Dividing through, we may express the integer quotient as a polynomial in r. B-2. Another slick one: Let S = {1, 2, ..., m}, T = {m+1, ..., m+n}. How many functions are there from S union T to S union T ? (m+n)^(m+n). Among those are the functions that send precisely m elements of the domain to S, and the other n elements to T. There are C(m+n,m) ways to choose which elements go to S, and once those are chosen there are m^m n^n such functions. So (m+n)^(m+n) >= (m+n)!/m!n! m^m n^n . B-3. If a > 2 we may use a constant function: a rectangle measuring a by 2a/(a-2) has perimeter and area equal. If a<2 there is no such function. The area is a m where m is the mean value of f over [0,a], which is certainly no larger than the maximum value M of f on the interval. On the other hand, the perimeter is at least the length of the line segments joining (0,0), (a,0), and (x,M) for some x in the interval, but each of the non-horizontal segments has length more than M, so the perimeter is more than 2M, which is more than aM, which is more than a m, the area. The case a=2 is similar; just use the fact that the perimeter is then at least 2 + 2 sqrt(1 + M^2). B-4. This is pretty standard too. One could matrices, but letÕs instead view the plane as being the set of complex numbers; a rotation by an angle theta is accomplished by multiplication by u = exp(i theta). (The function f(z) = P + u (z - P) is the rotation around P.) So we are composing the functions R_k(z) = P_k + u ( z - P_k) = u z + (1-u) k By induction on m we find (R_m o ... o R_2 o R_1) (z) = u^m (z) + S_m where S_m = (1-u) m + u S_{m-1}, i.e. S_m = (1-u) ( m + u (m-1) + u^2 (m-2) + ... + u^m (0) ). = m - u - u^2 - ... - u^m = m - (u - u^{m+1})/(1 - u) In the problem we were given that theta = 2 pi / n , so u^n = 1, meaning (R_n o ... o R_2 o R_1) (z) = u^n (z) + S_n = z + n, a simple translation R(x,y) = (x+n, y). This motion is easily demonstrated for low n. B-5. It makes more sense to deal with the logs. We need to look at lim_{x->1-} lim_{N->infty} sum_{n=0}^N x^n (log(1+x^{n+1}) - log(1+x^n)) = lim_{x->1-} lim_{N->infty} ( sum_{n=1}^N (x^{n-1} - x^n) log(1+x^n) ) -log(2) + x^N log(1+x^{N+1}) For any x < 1, the last term tends to 0 as N->infty, so we have only - log(2) + lim_{x->1-} (1/x - 1) sum_{n=1}^infty x^n log(1+x^n) Now, we may estimate the infinite sum using bounds on the logarithm from its Taylor series; for example, we have log(1+x^n) > x^n - x^{2n}/2 but on the sum: it lies between x^2/(1-x^2) - (1/2)(x^3/(1-x^3)) and x^2/(1-x^2) - (1/2)(x^3/(1-x^3)) + (1/3)(x^4/(1-x^4)), and we may obtain more bounds of this type. Multiplying in the additional factor (1/x - 1) (which is positive) gives bounds such as x^1/(1+x) - (1/2)(x^2/(1+x+x^2)) and x^1/(1+x) - (1/2)(x^2/(1+x+x^2)) + (1/3)(x^3/(1+x+x^2+x^3)) Applying these bounds to each x < 1 we find the limit to lie between 1/2 - (1/2)(1/3) and 1/2 - (1/2)(1/3) + (1/3)(1/4) and more generally it is alternately larger or smaller than a partial sum of the the absolutely convergent series (1/1)(1/2) - (1/2)(1/3) + (1/3)(1/4) - ... Carrying out this argument more generally to finite expansions of the Taylor series shows that the limit must equal this infinite sum. If we add 2( (1/2)(1/3) + (1/4)(1/5) + (1/6)(1/7) + ...) we obtain the famous telescoping series, which sums to 1. But in like manner we may recognize this addend as 2( (1/2) - (1/3) + (1/4) - (1/5) + ... ), which is 2 (1-log(2)). So the series in the last paragraph converges to 2 log(2) - 1. So I conclude that the logarithm of the original limit is log(2) - 1, and so the limit itself must be 2/e . B-6. No clue. ItÕs a nice problem though -- sort of a teaser for the Twin Prime conjecture or something. (Let A = set of all primes.) For contrast, consider what happens when A = set of all squares (B = set of (almost) all composites), or A = an arithmetic progression (B is an ideal of Z), or A = powers of 2 (this B is indeed sparse). I look forward to a solution of this one. Dave Rusin escribi.97: > A-2. > For i = 1,2 let T_i be a triangle with side lengths a_i, b_i, c_i, > and area A_i. Suppose that a_1 <= a_2, b_1 <= b_2, c_1 <= c_2, > and that T_2 is an acute triangle. Does it follow that A_1 <= A_2 ? The area can be computed as S = (1/2)b*c*sin(A). If you shorten the side a, mantaining b and c, you decrease the angle A. If A isnÕt obtuse, sin(A) decrease also, so also the area decrease. If T1 is also non-obtuse, we can go from T2 to T1 shortening sides opposites to non-obtuse angles, perhaps in more that three steps, changing the side shortening if any of the other angles became stright. Then S(T1) < S(T2) If T1 is obtuse, angle A1 is obtuse by example, we take T1Õ with sides b1 an c1 equals to the T1 ones, and angle A1= pi - A1. Then S(T1) = S(T1Õ) < S(T2). -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com posting-account=LMd1Cw0AAADCSSSBj3EHkyTBh4eU32ea > B-6. > No clue. ItÕs a nice problem though -- sort of a teaser for the Twin Prime > conjecture or something. (Let A = set of all primes.) > For contrast, consider what happens when A = set of all squares (B = set > of (almost) all composites), or A = an arithmetic progression (B is an > ideal of Z), or A = powers of 2 (this B is indeed sparse). > I look forward to a solution of this one. 1. Let T be the complement of B = {t : A is disjoint from A + t}. 2. If we can find n disjoint translates of A for all n, we are done (as A then has upper density <= 1/n for each n). 3. But thatÕs easy because T contains long intervals of consecutive integers (its complement, B, has unbounded gaps): A, A + t1, A + t2, A + t3, ... disjoint means that the tÕs and their pairwise (absolute) differences are all in T, One can arrange this by taking t_n to be the last element in an interval in T of length greater than t_(n-1). > A-4. > This is a heavily-used formula when n = 2 : xy = (1/4)( (x+y)^2 - (x-y)^2 ). > I guess I never really thought about generalizing it before! > It suffices to arrange it so that the right side is a nonzero multiple of > every x_i ; for then it is a degree-n polynomial and also a multiple of > x_1 x_2 ... x_n, so the quotient of the two sides is a constant > (necessarily rational, as can be seen from substituting x_1 = ... = 1); > we can then divide through to make the constant be 1. > If we arrange it so that the right side is symmetric in the x_i, then > it suffices to make sure it vanishes when x_1 = 0. > For odd n this is easy: for any subset T of N = {1, 2, ..., n}, let > S(T) = the sum of the x_i with i in T. Then consider > X = sum (-1)^|T| ( S(T) - S(N-T) )^n , > the sum taken over all subsets T of N. When we substitute x1=0 we > obtain an alternating sum of terms of the form ( S(U) - S(M-U) )^n > where M = N - {1}. Each such term arises exactly twice, once from > S(U union {1}) - S(M-U) and once from S(U) - S(N-U); since the cardinalities > of U and U union {1} are of opposite parity, these two summands > On the other hand, X (apparently) doesnÕt vanish identically; for example, > when all x_i equal 1, we have > c = sum (-1)^|T| ( |T| - |N-T| )^n = sum (-1)^k (2k-n)^n C( n,k ), > which seems to equal (-2)^n n! (I computed the first few examples but > didnÕt prove this formula in general.) > So whatever the value of c really is (itÕs obviously integral) we > can divide X by c to get a sum which is a multiple of x1, and thus > of each x_i, and thus of the product x1 x2 ... x_n, but the > quotient must be both constant and taking the value 1. > I didnÕt take advantage of the opportunity to use 0 as a coefficient! And RobinÕs doesnÕt take advantage of -1 for a_{ij}. Notice that you donÕt need 2^n terms. For n=2k, (-x - y + z ...)^{2k} = (x + y - z ...)^{2k} For n=2k+1, (-x - y + z ...)^{2k+1} = -(x + y - z ...)^{2k+1} For example, n = 3: xyz = 1/24((x+y+z)^3 -(-x+y+z)^3 -(x-y+z)^3 -(x+y-z)^3) How about in less than 2^{n-1} terms? Using 1,0, and -1? Reminds me of the game played in StrassenÕs faster than n^3 matrix multiplication algorithm. -- Mitch Harris (remove q to reply) posting-account=LMd1Cw0AAADCSSSBj3EHkyTBh4eU32ea > A-4. > This is a heavily-used formula when n = 2 : > xy = (1/4)( (x+y)^2 - (x-y)^2 ). > I guess I never really thought about generalizing it before! If you take the finite differences with spacing x1, then x2, then x3, ... then xn, applied to the polynomial T^n, you get a constant, namely n! * (x1 x2 ... xn). The x1, x2, ... factors are maybe clearer to see if you use the finite difference quotient operators (f(T) - f(T-xi)) / xi which have the same behavior on highest degree terms as taking a derivative. > A-5. > No clue. I worked out the case n=1 completely; the expected number of > components is (m+1)/2 . I suppose there might be some kind of induction > argument, but since they ask us to prove the number of components is > AT LEAST mn/8, that suggests that what weÕre supposed to do is to find > a lot of configurations with lots of components, so that we can bound > the expected number without doing all the grubby computations. The expected number of components is >= the expected number of 1-point components = sum (over all points P in the rectangle) of the probability that P is isolated. The probability that any given point is isolated is 2 * (1/2)^(number of its neighbors) which is at least 1/8. > B-2. > Another slick one: Let S = {1, 2, ..., m}, T = {m+1, ..., m+n}. > How many functions are there from S union T to S union T ? (m+n)^(m+n). > Among those are the functions that send precisely m elements of the > domain to S, and the other n elements to T. There are C(m+n,m) ways > to choose which elements go to S, and once those are chosen there are > m^m n^n such functions. So (m+n)^(m+n) >= (m+n)!/m!n! m^m n^n . Any solution by estimating the integral on [0,1] of x^m (1-x)^n (Euler beta integral)? posting-account=vtazAAwAAACBCuOVO7rqhKgkGbOSs9nL > The expected number of components is >= the expected number of 1-point > components = sum (over all points P in the rectangle) of the > probability > that P is isolated. The probability that any given point is isolated > is > 2 * (1/2)^(number of its neighbors) which is at least 1/8. Actually, the probability is only 1/16: 1/32 that the point is red and its neighbors are black and 1/32 that the point is black and its neighbors are red. posting-account=clKKpw0AAADKfUFXdUcPFaNb-vNZu4_v > The expected number of components is >= the expected number of > 1-point > components = sum (over all points P in the rectangle) of the > probability > that P is isolated. The probability that any given point is isolated > is > 2 * (1/2)^(number of its neighbors) which is at least 1/8. > Actually, the probability is only 1/16: 1/32 that the point is red and > its neighbors are black and 1/32 that the point is black and its > neighbors are red. Can anyone show that the expected size of a region is less than 8? The size of the grid divided by the expected size of a region would give the expected number of regions. If the expected size is less than 8 the expected number of regions would be greater than mn/8 Nick Wagers - Sophomore, Eastern Illinois University >Can anyone show that the expected size of a region is less than 8? The >size of the grid divided by the expected size of a region would give >the expected number of regions. If the expected size is less than 8 the >expected number of regions would be greater than mn/8 Do you mean the expected _average size_ ? This doesnÕt work the way you are expecting. Of the 8 possible colorings of a 1x3 grid, four give 2 regions, two give 1 region, and two give 3 regions, so the expected number of regions is 2. But these cases have average region sizes of 3/2, 3, and 1 respectively, so the expected average-region-size is 7/4, not (1x3)/2 as your reasoning suggests. It is true that the average of the sizes of all the regions in all possible colorings is (mn)/(expected number of regions). But this averaging allows the regular checkboard coloring to contribute mn singleton regions each with a weight equal to the single region of a solid coloring. IÕm not sure how you would pre-compute this kind of average. dave > Can anyone show that the expected size of a region is less than 8? The > size of the grid divided by the expected size of a region would give > the expected number of regions. If the expected size is less than 8 the > expected number of regions would be greater than mn/8 > Nick Wagers - Sophomore, Eastern Illinois University ThatÕs not true. If X and Y are two random variables, E(XY) is not equal to E(X)E(Y). I personally find the solution with EulerÕs formula amazing. Congratulations. > The probability that any given point is isolated is > 2 * (1/2)^(number of its neighbors) which is at least 1/8. No, only 1/16. More generally, say a connected region contains c points and touches t additional points. ThereÕs probability 1/2^(c+t) that the region is all red and the additional points are all black; and disjoint probability 1/2^(c+t) that the region is all black and the additional points are all red; for a total probability of 2/2^(c+t) that the region is maximal monochromatic. For example, c=1 usually means t=4 and 2/2^(c+t) = 1/16. The problem is to prove that the sum of 1/2^(c+t-1), over all connected regions, is at least mn/8. Single-point regions get you to only about mn/16; itÕs certainly not that easy! ---D. J. Bernstein, Associate Professor, Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago posting-account=LMd1Cw0AAADCSSSBj3EHkyTBh4eU32ea > The probability that any given point is isolated is > 2 * (1/2)^(number of its neighbors) which is at least 1/8. > No, only 1/16. However, I think this type of approach, adding a local expression over all points, can be made to work. ------------------------------------------------------------ Solution #1: Compute the winding number of each Connected Monochromatic Region, and sum over all regions. The answer is the number of CMR, if all are simply connected. (For multiply connected CMR, see below). Winding number is a local integral. It is the sum over all (m+1)(n+1) corners of squares in the diagram, of local winding number contributions that are easy to compute: a. Corners of the m x n rectangle contribute 1/4 each = 1 b. Non-corner points on the boundary contribute winding number 1/2 if the squares that meet there have opposite color = expected winding of 1/4 per point. c. Interior points P of the rectangle contribute nothing unless the 2x2 square centered at P is checkerboard-colored with 2 squares of each color (probability of 1/8) and contributes winding of 4*(1/4)=1 in that case. Expected winding = 1/8 per point. So all (m+1)(n+1) points contribute an expectation of >= 1/8, which is > mn/8. We are done provided we can address the issue of multiply connected CMR: Simply connected CMRÕs contribute 1. Multiply connected CMR will contribute 1 only if they are maximal; the contribution of inner regions (and their inner-inner boundary components of two CMRÕs will make a net contribution of zero, because they will be traversed in both directions. So the calculation shows the slightly stronger result, that the expected number of maximal CMR, i.e. the number of closed curves which are boundary components of just one CMR, in this sense) is mn/8 + 3/8(m+n) + 1 > mn/8. ------------------------------------------------------------- Solution #2: Use PickÕs theorem to add up the areas of all CMRÕs. PickÕs formula for a polygon with h holes is Area = # Interior pts + 1/2 # Boundary pts - (1 - # holes). Adding this over all CMRs, mn = # Int + 1/2 # Boundary - # CMR + [something >= 0 from holes] Expected # CMR >= E[Int] + 1/2 E[Boundary] - mn The rest is an easy calculation. Note that boundary points are counted with multiplicity 2 (for a given CMR) if two corners of that CMR meet there, as in a 3x3 square with the center and one corner removed; and that a given point must be counted separately for each CMR whose boundary contains it. Each of the (m-1)(n-1) points inside the rectangle is: - an interior point if the 2x2 square around it is monochromatic, probability = 1/8. - a boundary point of multiplicity 4 if its 2x2 square is checkerboard-colored (probability 1/8)= expectation 2/8. - a boundary point of multiplicity 2 if its 2x2 square is of any other type (probability 6/8) = expectation 6/8. This gives an expectation of 1/8 + 2/8 + 6/8 = 9/8 per point, or 9/8 (m-1)(n-1). Corners of the rectangle are boundary points of multiplicity 1, with probability 1, for an expectation of 1/2 per point = 2. The (2m+2n - 4) external points of the rectangle are boundary points in the sense of PickÕs theorem, with multiplicity 1 if two equal-colored squares meet there (prob 1/2) and multiplicity 2 otherwise (prob 1/2), for an expectation of 3/2 per point = 3 (m+n-2) Thus: Expected # CMR >= 9/8 (m-1)(n-1) + 2 + 3(m+n-2) - mn = mn/8 + 15/8 (m+n) - 23/8 > mn/8. IÕm puzzled why this doesnÕt give the exact same answer as Solution #1, as it seems to be calculating the same geometric/topological quantity. But, even in case of a calculation error, we know that the approach using PickÕs theorem works because the expectation per each of the (m+1)(n+1) points is at least 9/8, which still exceeds 1/8 after subtracting 1 from mn of the points. > More generally, say a connected region contains c points and touches t > additional points. ThereÕs probability 1/2^(c+t) that the region is all > red and the additional points are all black; and disjoint probability > 1/2^(c+t) that the region is all black and the additional points are all > red; for a total probability of 2/2^(c+t) that the region is maximal > monochromatic. For example, c=1 usually means t=4 and 2/2^(c+t) = 1/16. > The problem is to prove that the sum of 1/2^(c+t-1), over all connected > regions, is at least mn/8. ThatÕs interesting. Is there a way to use that to write an infinite series for the true proportion of CMR per area (for an infinite rectangle). It should be higher than 1/8. > B-5. > It makes more sense to deal with the logs. We need to look at > lim_{x->1-} lim_{N->infty} sum_{n=0}^N x^n (log(1+x^{n+1}) - log(1+x^n)) > = lim_{x->1-} lim_{N->infty} > ( sum_{n=1}^N (x^{n-1} - x^n) log(1+x^n) ) -log(2) + x^N log(1+x^{N+1}) > For any x < 1, the last term tends to 0 as N->infty, so we have only > - log(2) + lim_{x->1-} (1/x - 1) sum_{n=1}^infty x^n log(1+x^n) > Now, we may estimate the infinite sum using bounds on the logarithm from > its Taylor series; for example, we have log(1+x^n) > x^n - x^{2n}/2 but > on the sum: it lies between x^2/(1-x^2) - (1/2)(x^3/(1-x^3)) and > x^2/(1-x^2) - (1/2)(x^3/(1-x^3)) + (1/3)(x^4/(1-x^4)), and we may obtain > more bounds of this type. Multiplying in the additional factor (1/x - 1) > (which is positive) gives bounds such as > x^1/(1+x) - (1/2)(x^2/(1+x+x^2)) and > x^1/(1+x) - (1/2)(x^2/(1+x+x^2)) + (1/3)(x^3/(1+x+x^2+x^3)) > Applying these bounds to each x < 1 we find the limit to lie between > 1/2 - (1/2)(1/3) and 1/2 - (1/2)(1/3) + (1/3)(1/4) > and more generally it is alternately larger or smaller than a partial sum of > the the absolutely convergent series > (1/1)(1/2) - (1/2)(1/3) + (1/3)(1/4) - ... > Carrying out this argument more generally to finite expansions of the > Taylor series shows that the limit must equal this infinite sum. > > If we add 2( (1/2)(1/3) + (1/4)(1/5) + (1/6)(1/7) + ...) we obtain > the famous telescoping series, which sums to 1. But in like manner we may > recognize this addend as 2( (1/2) - (1/3) + (1/4) - (1/5) + ... ), which is > 2 (1-log(2)). So the series in the last paragraph converges to 2 log(2) - 1. > So I conclude that the logarithm of the original limit is log(2) - 1, > and so the limit itself must be 2/e . Or: Let A_n(x) = (x-1)x^n/(1+x^n); the nth term in the product is (1 + A_n(x))^(x^n). The log of the product is sum (n=0,oo) x^n*ln(1 + A_n(x)) (1). Now ln(1+y) = y + O(y^2) as y -> 0, and each |A_n(x)| <= 1-x. So (1) can be written as sum (n=0,oo) x^n*[A_n(x) + O((x-1)^2)] (2). Because sum x^n = 1/(1-x), we can ignore the O((x-1)^2) in (2). Now x^n*A_n(x) = (x-1)x^(2n)/(1+x^n) = (x-1)*(x^(2n) - x^(3n) + x^(4n) - ...) >= (x-1)*(x^(2n) - x^(3n) + x^(4n) - x^(5n) + x^(6n)), using standard alternating series ideas, and where weÕve stopped at 6 just to illustrate. Summing on n, a lower bound for our sum is (x-1)*[1/(1-x^2) - 1/(1-x^3) + ... + 1/(1-x^6)]. Well now take the limit as last expression using (blush) LÕHospital. You get - 1/2 + 1/3 - ... - 1/6. It follows that the lower limit of our sum is >= - 1/2 + 1/3 - ... - 1/6. And of course we could have stopped at any even number of terms. So the lower limit is >= - 1/2 + 1/3 - 1/4 + ... = ln(2/e). Same idea (looking at any odd number of terms) shows the upper limit is <= ln(2/e). So the limit of the sum in question is ln(2/e); therefore the limit of the original product is 2/e. > The Sixty-Fifth Annual Putnam Exam was held today. I have already > posted a copy of the questions. Here are as many answers as I have > at this moment; I have nothing to say about A-5 and B-6. > If I were the grader I wouldnÕt give myself full credit for some of > these answers but I think rather than nail down all the details I > will wait for someone else to provide the slickest answers, making > the drudgery unnecessary... (See Problems A-2, A-4, and problems > with detailed computations such as B-3, B-4, B-5.) > Overall I thought this was an easier exam than is typical, which > made the event much more fun for the average students taking the test. > I will add these answers and any other interesting responses to the > files I make available about Putnam questions; see > http://www.math.niu.edu/~rusin/problems-math/ > 16 A^2 = 2 (a^2b^2+a^2c^2+b^2c^2) - (a^4+b^4+c^4) > Viewed as a quadratic polynomial in X = a^2, this function is clearly > increasing as long as X < (b^2+c^2), which means that the area of a > triangle increases with the length of any side as long as the angle > opposite that side is acute (a^2 < b^2 + c^2 ). > (A student pointed out to me at lunch that this is also obvious from > A = (1/2) b h, where the height h of the triangle varies with a but > c is fixed; the maximum height occurs with sides b and c > perpendicular.) > It then suffices to observe that we can transform triangle T_1 to T_2 > by a sequence of operations which lengthen just one edge at a time. This > is not entirely obvious since, for example, we are not given that the > smaller triangle is acute. But itÕs easy to see in a sketch which I will > not attempt in ASCII, showing the portion of the positive octant in > a,b,c -space which corresponds to triangles ( a < b+c, etc. ); the acute > triangles are the ones external to three cones around the coordinate > axes ( a^2 < b^2 + c^2, etc.) If one always lengthens the shortest edge > (from among those that need to be lengthened to change T_1 into T_2), > then that will always be opposite an acute angle. (An obtuse angle will > always be opposite the _largest_ side, and that cannot the the only side > remaining to be lengthened since T_2 is acute.) An alternative method: enlarging the second trinagle by a constant factor allows us to assume that a = aÕ. Draw both triangles on the same base: ABC and AÕBC say with A and Aon the same side of BC. Draw the line L through A parallel to BC and the line LÕ perpendicalar to L at A. If the second triangle had larger area, Awould lie beyond L (opposite side of L to BC). Divide the beyond of L into two quadrants by LÕ. If Ais beyond L then b> b if L is in one quadrant and c> c in the other (here is where we use acuteness). > A-4. > This is a heavily-used formula when n = 2 : xy = (1/4)( (x+y)^2 - > (x-y)^2 ). I guess I never really thought about generalizing it before! > It suffices to arrange it so that the right side is a nonzero multiple of > every x_i ; for then it is a degree-n polynomial and also a multiple of > x_1 x_2 ... x_n, so the quotient of the two sides is a constant > (necessarily rational, as can be seen from substituting x_1 = ... = 1); > we can then divide through to make the constant be 1. > If we arrange it so that the right side is symmetric in the x_i, then > it suffices to make sure it vanishes when x_1 = 0. > For odd n this is easy: for any subset T of N = {1, 2, ..., n}, let > S(T) = the sum of the x_i with i in T. Then consider > X = sum (-1)^|T| ( S(T) - S(N-T) )^n , > the sum taken over all subsets T of N. When we substitute x1=0 we > obtain an alternating sum of terms of the form ( S(U) - S(M-U) )^n > where M = N - {1}. Each such term arises exactly twice, once from > S(U union {1}) - S(M-U) and once from S(U) - S(N-U); since the > cardinalities > of U and U union {1} are of opposite parity, these two summands > On the other hand, X (apparently) doesnÕt vanish identically; for > example, > when all x_i equal 1, we have > c = sum (-1)^|T| ( |T| - |N-T| )^n = sum (-1)^k (2k-n)^n C( n,k ), > which seems to equal (-2)^n n! (I computed the first few examples but > didnÕt prove this formula in general.) I got n! x_1 x_2 ... x_n = sum_T (-1)^{n-|T|}(S(T))^n eschewing coefficient -1. :-) > B-2. > Another slick one: Let S = {1, 2, ..., m}, T = {m+1, ..., m+n}. > How many functions are there from S union T to S union T ? (m+n)^(m+n). > Among those are the functions that send precisely m elements of the > domain to S, and the other n elements to T. There are C(m+n,m) > ways > to choose which elements go to S, and once those are chosen there are > m^m n^n such functions. So (m+n)^(m+n) >= (m+n)!/m!n! m^m n^n . Alternative: One of the terms in the binomial expansion of (m+n)^{m+n} is (m+n choose m) m^m n^n so that (m+n)^{m+n} > (m+n choose m) m^m n^n -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ For B-2 I did the following: let a=m/(m+n) b=(n/(m+n) and therefore (a+b) = 1, and (a+b)^(m+n) = 1 too The binomial expansion of (a+b)^(m+n) = C_0 a^(m+n) + .... + C_n a^(n) b^m + ... C_(m+n) b^(m+n) C_n a^n b^m is less than 1 and greater than 0, as all terms in the expansion are non negative. check it out ! what do you think? -rohit > The Sixty-Fifth Annual Putnam Exam was held today. I have already > posted a copy of the questions. Here are as many answers as I have > at this moment; I have nothing to say about A-5 and B-6. > If I were the grader I wouldnÕt give myself full credit for some of > these answers but I think rather than nail down all the details I > will wait for someone else to provide the slickest answers, making > the drudgery unnecessary... (See Problems A-2, A-4, and problems > with detailed computations such as B-3, B-4, B-5.) > Overall I thought this was an easier exam than is typical, which > made the event much more fun for the average students taking the test. > I will add these answers and any other interesting responses to the > files I make available about Putnam questions; see > http://www.math.niu.edu/~rusin/problems-math/ > A-1. > Suppose that S.O.Õs success count S(N) stayed strictly below 0.80 N > for all N < K, but that S(K) >= 0.80 K. This requires S(K-1) = S(K) - 1. > Then from S(K-1)/(K-1) < 4/5 and S(K)/K >= 4/5 we conclude that > 0 <= 5 S(K) - 4 K < 1 ; since 5 S(K) - 4 K is an integer, it must > then equal zero, so S(K) = 4/5. > Comment: Obviously the same proof works for any percentage of the form > 1 - 1/m with integer m, but for more general percentages the result > is false. This problem will make a very pretty example the next time > I teach the Intermediate Value Theorem :-) > A-2. > 16 A^2 = 2 (a^2b^2+a^2c^2+b^2c^2) - (a^4+b^4+c^4) > Viewed as a quadratic polynomial in X = a^2, this function is clearly > increasing as long as X < (b^2+c^2), which means that the area of a > triangle increases with the length of any side as long as the angle > opposite that side is acute (a^2 < b^2 + c^2 ). > (A student pointed out to me at lunch that this is also obvious from > A = (1/2) b h, where the height h of the triangle varies with a but > c is fixed; the maximum height occurs with sides b and c perpendicular.) > It then suffices to observe that we can transform triangle T_1 to T_2 > by a sequence of operations which lengthen just one edge at a time. This > is not entirely obvious since, for example, we are not given that the > smaller triangle is acute. But itÕs easy to see in a sketch which I will > not attempt in ASCII, showing the portion of the positive octant in > a,b,c -space which corresponds to triangles ( a < b+c, etc. ); the acute > triangles are the ones external to three cones around the coordinate > axes ( a^2 < b^2 + c^2, etc.) If one always lengthens the shortest edge > (from among those that need to be lengthened to change T_1 into T_2), > then that will always be opposite an acute angle. (An obtuse angle will > always be opposite the _largest_ side, and that cannot the the only side > remaining to be lengthened since T_2 is acute.) > Note: I guess I had never before stopped to think about under exactly > what conditions it is true that lengthening the sides of a triangle > increases the area. Now I know! > A-3. > We prove by induction that u_n = (n-1)!! = (n-1)(n-3)(n-5)..., which > is easily checked to be valid for small n. Of course this will mean > the u_n are integral. > If the claim is true for u_n and u_{n+2} then u_{n+2} = (n+1) u_n, so > the first column of the matrix in the determinant is a multiple of u_n, > and we have simply the recurrence relation > u_{n+3} = (n+1) u_{n+1} + n!/u_n > = (n+1) n!! + n!! > = (n+2) n!! = (n+2)!!, > providing the inductive step. > A-4. > This is a heavily-used formula when n = 2 : xy = (1/4)( (x+y)^2 - (x-y)^2 ). > I guess I never really thought about generalizing it before! > It suffices to arrange it so that the right side is a nonzero multiple of > every x_i ; for then it is a degree-n polynomial and also a multiple of > x_1 x_2 ... x_n, so the quotient of the two sides is a constant > (necessarily rational, as can be seen from substituting x_1 = ... = 1); > we can then divide through to make the constant be 1. > If we arrange it so that the right side is symmetric in the x_i, then > it suffices to make sure it vanishes when x_1 = 0. > For odd n this is easy: for any subset T of N = {1, 2, ..., n}, let > S(T) = the sum of the x_i with i in T. Then consider > X = sum (-1)^|T| ( S(T) - S(N-T) )^n , > the sum taken over all subsets T of N. When we substitute x1=0 we > obtain an alternating sum of terms of the form ( S(U) - S(M-U) )^n > where M = N - {1}. Each such term arises exactly twice, once from > S(U union {1}) - S(M-U) and once from S(U) - S(N-U); since the cardinalities > of U and U union {1} are of opposite parity, these two summands > On the other hand, X (apparently) doesnÕt vanish identically; for example, > when all x_i equal 1, we have > c = sum (-1)^|T| ( |T| - |N-T| )^n = sum (-1)^k (2k-n)^n C( n,k ), > which seems to equal (-2)^n n! (I computed the first few examples but > didnÕt prove this formula in general.) > So whatever the value of c really is (itÕs obviously integral) we > can divide X by c to get a sum which is a multiple of x1, and thus > of each x_i, and thus of the product x1 x2 ... x_n, but the > quotient must be both constant and taking the value 1. > I didnÕt take advantage of the opportunity to use 0 as a coefficient! > A-5. > No clue. I worked out the case n=1 completely; the expected number of > components is (m+1)/2 . I suppose there might be some kind of induction > argument, but since they ask us to prove the number of components is > AT LEAST mn/8, that suggests that what weÕre supposed to do is to find > a lot of configurations with lots of components, so that we can bound > the expected number without doing all the grubby computations. > A-6. > Heh, heh, check this one out: > Let F(x,y,z,w) = f(x,y) + f(z,w) - f(x,w) - f(z,y); then integrate > F^2 over the box [0,1]^4 . Done! > (This is the distillation of about 5 pages of computations, followed by > a great big step back to look at What The Computations Really Mean. > I have learned that to understand these integral inequalities it pays > to think in terms of step functions and simple functions; writing all that > out gave me a solution but then I realized this other way to say it. > Note that it is critical that the domain of integration be the UNIT square.) > B-1. > This is really standard. Write r = p/q in lowest terms; then we have > c_n p^n + c_{n-1} p^{n-1} q + ... + c_0 q^n = 0 > and so for each j between 0 and n we see q^j divides > c_n p^n + c_{n-1} p^{n-1} q + ... + c_{n-j+1} p^{n-j+1} q^{j-1} > = p^{n-j}( c_n p^{j} + c_{n-1} p^{j-1} q + ... + c_{n-j+1} p^1 q^{j-1} ) > Since p and q are coprime, q^j divides the other factor. Dividing > through, we may express the integer quotient as a polynomial in r. > B-2. > Another slick one: Let S = {1, 2, ..., m}, T = {m+1, ..., m+n}. > How many functions are there from S union T to S union T ? (m+n)^(m+n). > Among those are the functions that send precisely m elements of the > domain to S, and the other n elements to T. There are C(m+n,m) ways > to choose which elements go to S, and once those are chosen there are > m^m n^n such functions. So (m+n)^(m+n) >= (m+n)!/m!n! m^m n^n . > B-3. > If a > 2 we may use a constant function: a rectangle measuring a by > 2a/(a-2) has perimeter and area equal. > If a<2 there is no such function. The area is a m where m is the > mean value of f over [0,a], which is certainly no larger than the > maximum value M of f on the interval. On the other hand, the perimeter > is at least the length of the line segments joining (0,0), (a,0), and > (x,M) for some x in the interval, but each of the non-horizontal segments > has length more than M, so the perimeter is more than 2M, which is > more than aM, which is more than a m, the area. The case a=2 is > similar; just use the fact that the perimeter is then at least > 2 + 2 sqrt(1 + M^2). > B-4. > This is pretty standard too. One could matrices, but letÕs instead > view the plane as being the set of complex numbers; a rotation by > an angle theta is accomplished by multiplication by u = exp(i theta). > (The function f(z) = P + u (z - P) is the rotation around P.) > So we are composing the functions > R_k(z) = P_k + u ( z - P_k) = u z + (1-u) k > By induction on m we find > (R_m o ... o R_2 o R_1) (z) = u^m (z) + S_m where S_m = > (1-u) m + u S_{m-1}, i.e. > S_m = (1-u) ( m + u (m-1) + u^2 (m-2) + ... + u^m (0) ). > = m - u - u^2 - ... - u^m = m - (u - u^{m+1})/(1 - u) > In the problem we were given that theta = 2 pi / n , so u^n = 1, > meaning > (R_n o ... o R_2 o R_1) (z) = u^n (z) + S_n = z + n, > a simple translation R(x,y) = (x+n, y). > This motion is easily demonstrated for low n. > B-5. > It makes more sense to deal with the logs. We need to look at > lim_{x->1-} lim_{N->infty} sum_{n=0}^N x^n (log(1+x^{n+1}) - log(1+x^n)) > = lim_{x->1-} lim_{N->infty} > ( sum_{n=1}^N (x^{n-1} - x^n) log(1+x^n) ) -log(2) + x^N log(1+x^{N+1}) > For any x < 1, the last term tends to 0 as N->infty, so we have only > - log(2) + lim_{x->1-} (1/x - 1) sum_{n=1}^infty x^n log(1+x^n) > Now, we may estimate the infinite sum using bounds on the logarithm from > its Taylor series; for example, we have log(1+x^n) > x^n - x^{2n}/2 but > on the sum: it lies between x^2/(1-x^2) - (1/2)(x^3/(1-x^3)) and > x^2/(1-x^2) - (1/2)(x^3/(1-x^3)) + (1/3)(x^4/(1-x^4)), and we may obtain > more bounds of this type. Multiplying in the additional factor (1/x - 1) > (which is positive) gives bounds such as > x^1/(1+x) - (1/2)(x^2/(1+x+x^2)) and > x^1/(1+x) - (1/2)(x^2/(1+x+x^2)) + (1/3)(x^3/(1+x+x^2+x^3)) > Applying these bounds to each x < 1 we find the limit to lie between > 1/2 - (1/2)(1/3) and 1/2 - (1/2)(1/3) + (1/3)(1/4) > and more generally it is alternately larger or smaller than a partial sum of > the the absolutely convergent series > (1/1)(1/2) - (1/2)(1/3) + (1/3)(1/4) - ... > Carrying out this argument more generally to finite expansions of the > Taylor series shows that the limit must equal this infinite sum. > > If we add 2( (1/2)(1/3) + (1/4)(1/5) + (1/6)(1/7) + ...) we obtain > the famous telescoping series, which sums to 1. But in like manner we may > recognize this addend as 2( (1/2) - (1/3) + (1/4) - (1/5) + ... ), which is > 2 (1-log(2)). So the series in the last paragraph converges to 2 log(2) - 1. > So I conclude that the logarithm of the original limit is log(2) - 1, > and so the limit itself must be 2/e . > B-6. > No clue. ItÕs a nice problem though -- sort of a teaser for the Twin Prime > conjecture or something. (Let A = set of all primes.) > For contrast, consider what happens when A = set of all squares (B = set > of (almost) all composites), or A = an arithmetic progression (B is an > ideal of Z), or A = powers of 2 (this B is indeed sparse). > I look forward to a solution of this one. > For B-2 I did the following: > let a=m/(m+n) > b=(n/(m+n) > and therefore (a+b) = 1, and (a+b)^(m+n) = 1 too > The binomial expansion of (a+b)^(m+n) = C_0 a^(m+n) + .... + C_n a^(n) > b^m + ... C_(m+n) b^(m+n) > C_n a^n b^m is less than 1 and greater than 0, as all terms in the > expansion are non negative. Correction: I meant, C_n a^m b^n instead of C_n a^n b^m. > check it out ! > what do you think? > -rohit > The Sixty-Fifth Annual Putnam Exam was held today. I have already > posted a copy of the questions. Here are as many answers as I have > at this moment; I have nothing to say about A-5 and B-6. If I were the grader I wouldnÕt give myself full credit for some of > these answers but I think rather than nail down all the details I > will wait for someone else to provide the slickest answers, making > the drudgery unnecessary... (See Problems A-2, A-4, and problems > with detailed computations such as B-3, B-4, B-5.) Overall I thought this was an easier exam than is typical, which > made the event much more fun for the average students taking the test. I will add these answers and any other interesting responses to the > files I make available about Putnam questions; see > http://www.math.niu.edu/~rusin/problems-math/ A-1. Suppose that S.O.Õs success count S(N) stayed strictly below 0.80 N > for all N < K, but that S(K) >= 0.80 K. This requires S(K-1) = S(K) - 1. > Then from S(K-1)/(K-1) < 4/5 and S(K)/K >= 4/5 we conclude that > 0 <= 5 S(K) - 4 K < 1 ; since 5 S(K) - 4 K is an integer, it must > then equal zero, so S(K) = 4/5. Comment: Obviously the same proof works for any percentage of the form > 1 - 1/m with integer m, but for more general percentages the result > is false. This problem will make a very pretty example the next time > I teach the Intermediate Value Theorem :-) A-2. 16 A^2 = 2 (a^2b^2+a^2c^2+b^2c^2) - (a^4+b^4+c^4) > Viewed as a quadratic polynomial in X = a^2, this function is clearly > increasing as long as X < (b^2+c^2), which means that the area of a > triangle increases with the length of any side as long as the angle > opposite that side is acute (a^2 < b^2 + c^2 ). (A student pointed out to me at lunch that this is also obvious from > A = (1/2) b h, where the height h of the triangle varies with a but > c is fixed; the maximum height occurs with sides b and c perpendicular.) It then suffices to observe that we can transform triangle T_1 to T_2 > by a sequence of operations which lengthen just one edge at a time. This > is not entirely obvious since, for example, we are not given that the > smaller triangle is acute. But itÕs easy to see in a sketch which I will > not attempt in ASCII, showing the portion of the positive octant in > a,b,c -space which corresponds to triangles ( a < b+c, etc. ); the acute > triangles are the ones external to three cones around the coordinate > axes ( a^2 < b^2 + c^2, etc.) If one always lengthens the shortest edge > (from among those that need to be lengthened to change T_1 into T_2), > then that will always be opposite an acute angle. (An obtuse angle will > always be opposite the _largest_ side, and that cannot the the only side > remaining to be lengthened since T_2 is acute.) Note: I guess I had never before stopped to think about under exactly > what conditions it is true that lengthening the sides of a triangle > increases the area. Now I know! A-3. We prove by induction that u_n = (n-1)!! = (n-1)(n-3)(n-5)..., which > is easily checked to be valid for small n. Of course this will mean > the u_n are integral. If the claim is true for u_n and u_{n+2} then u_{n+2} = (n+1) u_n, so > the first column of the matrix in the determinant is a multiple of u_n, > and we have simply the recurrence relation > u_{n+3} = (n+1) u_{n+1} + n!/u_n > = (n+1) n!! + n!! > = (n+2) n!! = (n+2)!!, > providing the inductive step. A-4. This is a heavily-used formula when n = 2 : xy = (1/4)( (x+y)^2 - (x-y)^2 ). > I guess I never really thought about generalizing it before! It suffices to arrange it so that the right side is a nonzero multiple of > every x_i ; for then it is a degree-n polynomial and also a multiple of > x_1 x_2 ... x_n, so the quotient of the two sides is a constant > (necessarily rational, as can be seen from substituting x_1 = ... = 1); > we can then divide through to make the constant be 1. If we arrange it so that the right side is symmetric in the x_i, then > it suffices to make sure it vanishes when x_1 = 0. For odd n this is easy: for any subset T of N = {1, 2, ..., n}, let > S(T) = the sum of the x_i with i in T. Then consider X = sum (-1)^|T| ( S(T) - S(N-T) )^n , the sum taken over all subsets T of N. When we substitute x1=0 we > obtain an alternating sum of terms of the form ( S(U) - S(M-U) )^n > where M = N - {1}. Each such term arises exactly twice, once from > S(U union {1}) - S(M-U) and once from S(U) - S(N-U); since the cardinalities > of U and U union {1} are of opposite parity, these two summands On the other hand, X (apparently) doesnÕt vanish identically; for example, > when all x_i equal 1, we have > c = sum (-1)^|T| ( |T| - |N-T| )^n = sum (-1)^k (2k-n)^n C( n,k ), > which seems to equal (-2)^n n! (I computed the first few examples but > didnÕt prove this formula in general.) So whatever the value of c really is (itÕs obviously integral) we > can divide X by c to get a sum which is a multiple of x1, and thus > of each x_i, and thus of the product x1 x2 ... x_n, but the > quotient must be both constant and taking the value 1. I didnÕt take advantage of the opportunity to use 0 as a coefficient! A-5. No clue. I worked out the case n=1 completely; the expected number of > components is (m+1)/2 . I suppose there might be some kind of induction > argument, but since they ask us to prove the number of components is > AT LEAST mn/8, that suggests that what weÕre supposed to do is to find > a lot of configurations with lots of components, so that we can bound > the expected number without doing all the grubby computations. A-6. Heh, heh, check this one out: > Let F(x,y,z,w) = f(x,y) + f(z,w) - f(x,w) - f(z,y); then integrate > F^2 over the box [0,1]^4 . Done! (This is the distillation of about 5 pages of computations, followed by > a great big step back to look at What The Computations Really Mean. > I have learned that to understand these integral inequalities it pays > to think in terms of step functions and simple functions; writing all that > out gave me a solution but then I realized this other way to say it. > Note that it is critical that the domain of integration be the UNIT square.) B-1. This is really standard. Write r = p/q in lowest terms; then we have > c_n p^n + c_{n-1} p^{n-1} q + ... + c_0 q^n = 0 > and so for each j between 0 and n we see q^j divides > c_n p^n + c_{n-1} p^{n-1} q + ... + c_{n-j+1} p^{n-j+1} q^{j-1} > = p^{n-j}( c_n p^{j} + c_{n-1} p^{j-1} q + ... + c_{n-j+1} p^1 q^{j-1} ) > Since p and q are coprime, q^j divides the other factor. Dividing > through, we may express the integer quotient as a polynomial in r. B-2. Another slick one: Let S = {1, 2, ..., m}, T = {m+1, ..., m+n}. > How many functions are there from S union T to S union T ? (m+n)^(m+n). > Among those are the functions that send precisely m elements of the > domain to S, and the other n elements to T. There are C(m+n,m) ways > to choose which elements go to S, and once those are chosen there are > m^m n^n such functions. So (m+n)^(m+n) >= (m+n)!/m!n! m^m n^n . B-3. If a > 2 we may use a constant function: a rectangle measuring a by > 2a/(a-2) has perimeter and area equal. If a<2 there is no such function. The area is a m where m is the > mean value of f over [0,a], which is certainly no larger than the > maximum value M of f on the interval. On the other hand, the perimeter > is at least the length of the line segments joining (0,0), (a,0), and > (x,M) for some x in the interval, but each of the non-horizontal segments > has length more than M, so the perimeter is more than 2M, which is > more than aM, which is more than a m, the area. The case a=2 is > similar; just use the fact that the perimeter is then at least > 2 + 2 sqrt(1 + M^2). B-4. This is pretty standard too. One could matrices, but letÕs instead > view the plane as being the set of complex numbers; a rotation by > an angle theta is accomplished by multiplication by u = exp(i theta). > (The function f(z) = P + u (z - P) is the rotation around P.) > So we are composing the functions > R_k(z) = P_k + u ( z - P_k) = u z + (1-u) k > By induction on m we find > (R_m o ... o R_2 o R_1) (z) = u^m (z) + S_m where S_m = > (1-u) m + u S_{m-1}, i.e. > S_m = (1-u) ( m + u (m-1) + u^2 (m-2) + ... + u^m (0) ). > = m - u - u^2 - ... - u^m = m - (u - u^{m+1})/(1 - u) In the problem we were given that theta = 2 pi / n , so u^n = 1, > meaning > (R_n o ... o R_2 o R_1) (z) = u^n (z) + S_n = z + n, > a simple translation R(x,y) = (x+n, y). This motion is easily demonstrated for low n. B-5. It makes more sense to deal with the logs. We need to look at lim_{x->1-} lim_{N->infty} sum_{n=0}^N x^n (log(1+x^{n+1}) - log(1+x^n)) = lim_{x->1-} lim_{N->infty} > ( sum_{n=1}^N (x^{n-1} - x^n) log(1+x^n) ) -log(2) + x^N log(1+x^{N+1}) For any x < 1, the last term tends to 0 as N->infty, so we have only > - log(2) + lim_{x->1-} (1/x - 1) sum_{n=1}^infty x^n log(1+x^n) Now, we may estimate the infinite sum using bounds on the logarithm from > its Taylor series; for example, we have log(1+x^n) > x^n - x^{2n}/2 but > on the sum: it lies between x^2/(1-x^2) - (1/2)(x^3/(1-x^3)) and > x^2/(1-x^2) - (1/2)(x^3/(1-x^3)) + (1/3)(x^4/(1-x^4)), and we may obtain > more bounds of this type. Multiplying in the additional factor (1/x - 1) > (which is positive) gives bounds such as > x^1/(1+x) - (1/2)(x^2/(1+x+x^2)) and > x^1/(1+x) - (1/2)(x^2/(1+x+x^2)) + (1/3)(x^3/(1+x+x^2+x^3)) > Applying these bounds to each x < 1 we find the limit to lie between > 1/2 - (1/2)(1/3) and 1/2 - (1/2)(1/3) + (1/3)(1/4) > and more generally it is alternately larger or smaller than a partial sum of > the the absolutely convergent series > (1/1)(1/2) - (1/2)(1/3) + (1/3)(1/4) - ... > Carrying out this argument more generally to finite expansions of the > Taylor series shows that the limit must equal this infinite sum. > > If we add 2( (1/2)(1/3) + (1/4)(1/5) + (1/6)(1/7) + ...) we obtain > the famous telescoping series, which sums to 1. But in like manner we may > recognize this addend as 2( (1/2) - (1/3) + (1/4) - (1/5) + ... ), which is > 2 (1-log(2)). So the series in the last paragraph converges to 2 log(2) - 1. So I conclude that the logarithm of the original limit is log(2) - 1, > and so the limit itself must be 2/e . B-6. No clue. ItÕs a nice problem though -- sort of a teaser for the Twin Prime > conjecture or something. (Let A = set of all primes.) For contrast, consider what happens when A = set of all squares (B = set > of (almost) all composites), or A = an arithmetic progression (B is an > ideal of Z), or A = powers of 2 (this B is indeed sparse). > I look forward to a solution of this one. > B-1. > Let P(x) = c_n x^n + c_{n-1} x^{n-1} + ... + c_0 be a polynomial with > integer coefficients. Suppose that r is a rational number such that > P(r) = 0. Show that the n numbers > c_n r , c_n r^2 + c_{n-1} r, c_n r^3 + c_{n-1} r^2 + c_{n-2} r, ..., > c_n r^n + c_{n-1} r^{n-1} + ... + c_1 r > are integers. > This is really standard. Write r = p/q in lowest terms; then we have > c_n p^n + c_{n-1} p^{n-1} q + ... + c_0 q^n = 0 > and so for each j between 0 and n we see q^j divides > c_n p^n + c_{n-1} p^{n-1} q + ... + c_{n-j+1} p^{n-j+1} q^{j-1} > = p^{n-j}( c_n p^{j} + c_{n-1} p^{j-1} q + ... + c_{n-j+1} p^1 q^{j-1} ) > Since p and q are coprime, q^j divides the other factor. Dividing > through, we may express the integer quotient as a polynomial in r. You neednÕt essentially reprove the rational root test as you do. Instead you can simply invoke it in an inductive manner as follows. n n-1 n-2 n-3 Write P(X) = a X + b X + c X + d X + ... n-2 n-3 Then r is a root of F(X) = ((ar + b)r + c) X + d X + ... By induction (ar + b)r is in Z, hence so is the leading coef of F. So the rational root test applies, yielding the leading coef of F. is a denominator for the root r, i.e. ((ar + b)r + c) r in Z. QED --Bill Dubuque === Subject: knowing E((X-Y)^2) is in some range, does that help to know E((X-Y)^3)? Hi all, I am wondering why the MSE E((X-E(X))^2) is so universally used. Is there any magic about the square? Why isnÕt it E((X-E(X))^3)? I suspect that by knowing E((X-E(X))^2) falling into some range, we can say a lot of E(|X-E(X)|^k) where k is a positive real number... More generally, I am wondering what can we say about E((X-Y)^2) vs. E((X-Y)^k) for two general random variables X and Y? === Subject: Re: knowing E((X-Y)^2) is in some range, does that help to know E((X-Y)^3)? > Hi all, > I am wondering why the MSE E((X-E(X))^2) is so universally used. Is there > any magic about the square? Why isnÕt it E((X-E(X))^3)? HereÕs another. Given random variable X, minimize E((X-t)^2) over all t. The minimum point is t=E(X), the mean of X. For other powers k, minimizing E(|X-t|^k) yields some other parameter of X, less natural than the mean. Well, for k=1 you get the median, which is also useful. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: knowing E((X-Y)^2) is in some range, does that help to know E((X-Y)^3)? >>Hi all, >>I am wondering why the MSE E((X-E(X))^2) is so universally used. Is there >>any magic about the square? Why isnÕt it E((X-E(X))^3)? > HereÕs another. Given random variable X, minimize E((X-t)^2) over all t. > The minimum point is t=E(X), the mean of X. For other powers k, > minimizing E(|X-t|^k) yields some other parameter of X, less natural > than the mean. Well, for k=1 you get the median, which is also useful. And well! For k = infinity you get the average of the extremes. The k=1 case has a growing literature as its computation has become etc instead of Gauss, Gaussian, etc. (Laplacians also show up in vector analysis with a different meaning.) The Laplacian distribution is also known as the double exponential distribution. The k=infinity case has a large literature as it is also called MiniMax fitting or Tchebycheff (with many alternate spellings) fitting. The uniform distribution is the underlying error model here. Various other values of k have been subject to analysis at various times. The case k=2 is much easier to analysis with conventional mathematical tools and is quite useful. === Subject: Re: knowing E((X-Y)^2) is in some range, does that help to know E((X-Y)^3)? >I am wondering why the MSE E((X-E(X))^2) is so universally used. Is there >any magic about the square? Why isnÕt it E((X-E(X))^3)? 1) Squares of real numbers are nonnegative. Cubes arenÕt. 2) The algebra often works out quite nicely when dealing with squares. For instance: 3) Differentiate the square of a linear (to try to minimize it) and you get something linear. Systems of linear equations are easy to solve (well, a lot easier than nonlinear systems). >I suspect that by knowing E((X-E(X))^2) falling into some range, we can say >a lot of E(|X-E(X)|^k) where k is a positive real number... Well, there will be inequalities: E[|Y|^k] >= a + b E[Y^2] if y^k >= a + b y^2 for all y >= 0. No inequality the other way if k > 2, because E[|Y|^k] can be made arbitrarily large with E[Y^2] fixed. >More generally, I am wondering what can we say about E((X-Y)^2) vs. >E((X-Y)^k) for two general random variables X and Y? Do you want the absolute value or donÕt you? Why talk about X and Y rather than one random variable Z = Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: knowing E((X-Y)^2) is in some range, does that help to know E((X-Y)^3)? because n^2 > 0 for all n in Z > Hi all, > I am wondering why the MSE E((X-E(X))^2) is so universally used. Is there > any magic about the square? Why isnÕt it E((X-E(X))^3)? > I suspect that by knowing E((X-E(X))^2) falling into some range, we can say > a lot of E(|X-E(X)|^k) where k is a positive real number... > More generally, I am wondering what can we say about E((X-Y)^2) vs. > E((X-Y)^k) for two general random variables X and Y? === Subject: Prove any periodic function with uniform phase shift is Strict Sense Stationary? Hi all, I am wondering if the statement is true or not: Any periodic function with uniform phase shift is Strict Sense Stationary? For example, g(t-THETA) where g(t) is a periodic function with period T, and THETA is a random variable uniformly distributed in [-T/2, +T/2]... is g(t-THETA) SSS? Aside from uniform distribution, what other distribution can make g(t-THETA) SSS? How about g(t-f(THETA)) where f is some function of the random variable THETA? === Subject: complex problem.... hello....doctor~ let f : D -> C is analytic in domain D. supposet that im f(z) = 2*Re f(z), show that f is constant in D. --------------------------------------- um.......i canÕt start this. i need your advice. thank you very much for your advice. === Subject: Re: complex problem.... > let f : D -> C is analytic in domain D. > supposet that im f(z) = 2*Re f(z), > show that f is constant in D. > --------------------------------------- > um.......i canÕt start this. Hint: WhatÕs Re((2 + i)f)? I mean, in terms of Re f and Im f. Jose Carlos Santos === Subject: Re: complex problem.... > let f : D -> C is analytic in domain D. (C is complex plane, D in C)(modify)*** > supposet that im f(z) = 2*Re f(z) for all z in D,(modify)*** > show that f is constant in D. > --------------------------------------- > um.......i canÕt start this. > Hint: WhatÕs Re((2 + i)f)? I mean, in terms of Re f and Im f. i think....... Re((2 + i)f) = 0, um........ give me one more hint....please.... thank you very much.... === Subject: Re: complex problem.... > let f : D -> C is analytic in domain D. > (C is complex plane, D in C)(modify)*** supposet that im f(z) = 2*Re f(z) for all z in D,(modify)*** show that f is constant in D. --------------------------------------- > um.......i canÕt start this. > Hint: WhatÕs Re((2 + i)f)? I mean, in terms of Re f and Im f. > i think....... > Re((2 + i)f) = 0, um........ > give me one more hint....please.... > thank you very much.... i think......again f(z) = u(x,y) + i.{2.u(x,y)} so, by analytic, u_x = 2u_y and 2u_x = -u_y so, 2.u_x - 2.u_y =0 and 2u_x + u_y = 0 so, u_y = 0 and u_x = 0 so, u is constant. thus f(z) is constant in D. um....right ?? === Subject: Re: complex problem.....;; > let f : D -> C is analytic in domain D. > (C is complex plane, D in C) > supposet that im f(z) = 2*Re f(z) for all z in D. > show that f is constant in D. Do you know the open mapping theorem? Assuming D is connected, f(D) is a connected open set or a point. But f(D) is contained in the line y = 2x, so f(D) is not open. Hence f(D) is a point, ie, f is constant. === Subject: Re: complex problem.....;; >let f : D -> C is analytic in domain D. >(C is complex plane, D in C) >supposet that im f(z) = 2*Re f(z) for all z in D. >show that f is constant in D. >but,....can i apply LiouvilleÕs theorem to this problem ?? No. >so, >define that entire function g : C->C , >g(z) = 1/{f(z) + 1} which im f(z) = 2*Re f(z) for all z in C. CanÕt do it, because f is only defined in D. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: complex problem.... >>let f : D -> C is analytic in domain D. >>(C is complex plane, D in C)(modify)*** >>supposet that im f(z) = 2*Re f(z) for all z in D,(modify)*** >>show that f is constant in D. >>--------------------------------------- >>um.......i canÕt start this. >Hint: WhatÕs Re((2 + i)f)? I mean, in terms of Re f and Im f. >>i think....... >>Re((2 + i)f) = 0, um........ >>give me one more hint....please.... >>thank you very much.... > i think......again > f(z) = u(x,y) + i.{2.u(x,y)} > so, by analytic, u_x = 2u_y and 2u_x = -u_y > so, 2.u_x - 2.u_y =0 and 2u_x + u_y = 0 The first equation should be u_x - 2u_y = 0. > so, u_y = 0 and u_x = 0 > so, u is constant. > thus f(z) is constant in D. > um....right ?? Well, yes, sort of. I mean, itÕs true that Re f constant => f constant but do you know why? Of course, if you *do* know why, a shorter way of solving this problem is: since Re((2 + i)f) is constant, then (2 + i)f is constant and therefore f itself is constant. Jose Carlos Santos === Subject: O notation iÕve seen notations like... O(1) vs O(n) vs O(n^2) vs O(n^m) what do these mean? why? so O(1) is way faster than the rest. how do they come to this conclusion? where is this O notation from? === Subject: Re: O notation O notation! Thou hast made possible the precise expression of mathematics! And yet oft hast thou obscured the underlying idea. === Subject: Re: O notation >O notation! Thou hast made possible the precise >expression of mathematics! And yet oft hast thou >obscured the underlying idea. I disagree - a proof using o notation can be much clearer than an equivalent proof involving limits. ************************ David C. Ullrich === Subject: Re: O notation <8abbr0lnk93q0kspugq7nf942oslbv4da8@4ax.com> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >>O notation! Thou hast made possible the precise >>expression of mathematics! And yet oft hast thou >>obscured the underlying idea. > I disagree - a proof using o notation can be much > clearer than an equivalent proof involving limits. I think he was using O notation! as a vocative here, meaning rather Oh notation! than O-notation. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: O notation >O notation! Thou hast made possible the precise >expression of mathematics! And yet oft hast thou >obscured the underlying idea. >> I disagree - a proof using o notation can be much >> clearer than an equivalent proof involving limits. >I think he was using O notation! as a vocative here, meaning rather >Oh notation! than O-notation. O. ************************ David C. Ullrich === Subject: Re: O notation <31jd0eF3c5bcgU1@individual.net O notation! Thou hast made possible the precise > expression of mathematics! And yet oft hast thou > obscured the underlying idea. Three most important things in Real Estate: (1) Location (2) Location (3) Location Three most important things in mathematics: (1) Notation (2) Notation (3) Notation === Subject: Re: O notation ames kin scribbled the following: > iÕve seen notations like... > O(1) vs O(n) vs O(n^2) vs O(n^m) > what do these mean? why? > so O(1) is way faster than the rest. how do they come to this > conclusion? where is this O notation from? It basically means how fast something grows. n is usually used as the variable. O(1) means the value always stays the same. O(n) means itÕs proportional to the variable. O(n^2) means itÕs proportional to the square of the variable and so on. Constant factors donÕt affect O notation, so O(2) is the same thing as O(1), O(2n) is the same thing as O(n) and so on. O notation is very often and extensively used in algorithm analysis. You can discuss it further here, or in comp.theory. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -------------------------------------------------------- rules! --------/ ItÕs time, itÕs time, itÕs time to dump the slime! - Dr. Dante === Subject: Re: O notation > ames kin scribbled the following: > iÕve seen notations like... > O(1) vs O(n) vs O(n^2) vs O(n^m) > what do these mean? why? > so O(1) is way faster than the rest. how do they come to this > conclusion? where is this O notation from? > It basically means how fast something grows. n is usually used as the ŌOÕ, denotes ŌOrderÕ i.e. O(n) means Ōof order nÕ, i.e. the time taken to compute the problem is proportional to n, T = k*n for some constant k, etc. === Subject: Re: O notation > i.e. O(n) means Ōof order nÕ, i.e. the time taken to compute the problem is > proportional to n, T = k*n for some constant k, etc. No it doesnÕt. === Subject: Re: O notation >> i.e. O(n) means Ōof order nÕ, i.e. the time taken to compute the problem is >> proportional to n, T = k*n for some constant k, etc. > No it doesnÕt. As long as weÕre being pedantic, it might be worth giving the exact definition, donÕt you think? f(x) = O(g(x)) if there exists constants x0 and C s.t. for all x > x0, |f(x)| < C g(x). -- IÕm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: O notation > i.e. O(n) means Ōof order nÕ, i.e. the time taken to compute the problem is > proportional to n, T = k*n for some constant k, etc. >> No it doesnÕt. >As long as weÕre being pedantic, I donÕt see anything pedantic about this, the proposed definition was simply wrong. >it might be worth giving the exact >definition, donÕt you think? Yes. >f(x) = O(g(x)) if there exists constants x0 and C s.t. for all x > x0, >|f(x)| < C g(x). ************************ David C. Ullrich === Subject: Re: O notation >I donÕt see anything pedantic about this, the proposed definition >was simply wrong. Well, mathematics is the art of being pedantic, but youÕre right the definitions should be properly stated without handwaving or false generalizations. Especially when concepts have been appropriated by computer scientists and/or physicists and sometimes used in the most inexact ways. As to whether thereÕs any point in responding to posts with one-word rebuttals when giving an actual definition takes about five seconds and is infinitely more useful, IÕll leave as an exercise to the reader. -- IÕm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: O notation It was a hastily constructed one-liner from me and I confess it was crap. Especially since I use the proper defn for my algorithms. IÕll be less hasty Next... Richard Miller >I donÕt see anything pedantic about this, the proposed definition >was simply wrong. > Well, mathematics is the art of being pedantic, but youÕre right the > definitions should be properly stated without handwaving or false > generalizations. Especially when concepts have been appropriated by > computer scientists and/or physicists and sometimes used in the most > inexact ways. > As to whether thereÕs any point in responding to posts with one-word > rebuttals when giving an actual definition takes about five seconds > and is infinitely more useful, IÕll leave as an exercise to the > reader. > -- > IÕm not interested in mathematics that might have anything > to do with reality. -- Russell Easterly, in sci.math === Subject: Re: O notation > It basically means how fast something grows. n is usually used as the > variable. O(1) means the value always stays the same. O(n) means itÕs > proportional to the variable. O(n^2) means itÕs proportional to the > square of the variable and so on. The O notation just gives upper bounds. === Subject: Re: O notation Torkel Franzen scribbled the following: >> It basically means how fast something grows. n is usually used as the >> variable. O(1) means the value always stays the same. O(n) means itÕs >> proportional to the variable. O(n^2) means itÕs proportional to the >> square of the variable and so on. > The O notation just gives upper bounds. Well, that is correct. AFAIK also o notation (little o) gives lower bounds and Theta notation (ISO-8859-1 doesnÕt have the Greek character Theta) gives exact values. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -------------------------------------------------------- rules! --------/ Hasta la Vista, Abie! - Bart Simpson === Subject: Re: O notation >Torkel Franzen scribbled the following: > It basically means how fast something grows. n is usually used as the > variable. O(1) means the value always stays the same. O(n) means itÕs > proportional to the variable. O(n^2) means itÕs proportional to the > square of the variable and so on. >> The O notation just gives upper bounds. >Well, that is correct. AFAIK also o notation (little o) gives lower >bounds No, o gives lower bounds as well, just different lower bounds: f(x) = O(g(x)) says |f| <= c g, while f(x) = o(g(x)) says something stronger, that f/g -> 0 (as x -> something, where the value of something is supposed to be clear from the context). >and Theta notation (ISO-8859-1 doesnÕt have the Greek character >Theta) gives exact values. No, f = Theta(g) says c f <= g <= C f for some constants c, C. ************************ David C. Ullrich === Subject: Re: O notation :>Torkel Franzen scribbled the following: :> It basically means how fast something grows. n is usually used as the :> variable. O(1) means the value always stays the same. O(n) means itÕs :> proportional to the variable. O(n^2) means itÕs proportional to the :> square of the variable and so on. :>> The O notation just gives upper bounds. :>Well, that is correct. AFAIK also o notation (little o) gives lower :>bounds : No, o gives lower bounds as well, just different lower bounds: I think you meant upper here. Stephen : f(x) = O(g(x)) says |f| <= c g, while f(x) = o(g(x)) says : something stronger, that f/g -> 0 (as x -> something, where : the value of something is supposed to be clear from the context). :>and Theta notation (ISO-8859-1 doesnÕt have the Greek character :>Theta) gives exact values. : No, f = Theta(g) says c f <= g <= C f for some constants c, C. === Subject: Re: O notation >:>Torkel Franzen scribbled the following: >:> It basically means how fast something grows. n is usually used as the >:> variable. O(1) means the value always stays the same. O(n) means itÕs >:> proportional to the variable. O(n^2) means itÕs proportional to the >:> square of the variable and so on. >:>> The O notation just gives upper bounds. >:>Well, that is correct. AFAIK also o notation (little o) gives lower >:>bounds >: No, o gives lower bounds as well, just different lower bounds: >I think you meant upper here. >Stephen >: f(x) = O(g(x)) says |f| <= c g, while f(x) = o(g(x)) says >: something stronger, that f/g -> 0 (as x -> something, where >: the value of something is supposed to be clear from the context). >:>and Theta notation (ISO-8859-1 doesnÕt have the Greek character >:>Theta) gives exact values. >: No, f = Theta(g) says c f <= g <= C f for some constants c, C. ************************ David C. Ullrich === Subject: Re: O notation David C. Ullrich scribbled the following: >>:>Torkel Franzen scribbled the following: >>:> It basically means how fast something grows. n is usually used as the >>:> variable. O(1) means the value always stays the same. O(n) means itÕs >>:> proportional to the variable. O(n^2) means itÕs proportional to the >>:> square of the variable and so on. >>:>:>> The O notation just gives upper bounds. >>:>:>Well, that is correct. AFAIK also o notation (little o) gives lower >>:>bounds >>: No, o gives lower bounds as well, just different lower bounds: >>I think you meant upper here. and alphabets. O and o give upper bounds, where o is stricter than O. Omega and omega give lower bounds, where omega is stricter than Omega. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -------------------------------------------------------- rules! --------/ ŌIis the most beautiful word in the world. - John Nordberg === Subject: Re: O notation > :>Torkel Franzen scribbled the following: > : :>> The O notation just gives upper bounds. > : :>Well, that is correct. AFAIK also o notation (little o) gives lower > :>bounds > : No, o gives lower bounds as well, just different lower bounds: > I think you meant upper here. o is lower (strict). omega is upper (strict). See CLRS. -- Mitch Harris (remove q to reply) === Subject: Re: O notation :> :>Torkel Franzen scribbled the following: :> :> :> :>> The O notation just gives upper bounds. :> :> :> :>Well, that is correct. AFAIK also o notation (little o) gives lower :> :>bounds :> : No, o gives lower bounds as well, just different lower bounds: :> I think you meant upper here. : o is lower (strict). omega is upper (strict). See CLRS. CLRS correctly defines little o as an upper bound that is not asymptotically tight. Stephen === Subject: Re: O notation :> :>Torkel Franzen scribbled the following: :> :> :> :>> The O notation just gives upper bounds. :> :> :> :>Well, that is correct. AFAIK also o notation (little o) gives lower :> :>bounds :> : No, o gives lower bounds as well, just different lower bounds: :> I think you meant upper here. : o is lower (strict). omega is upper (strict). See CLRS. I will have to check CLRS, but that does not sound right to me. If f(n) is o(g(n)) that means that f(n) grows less quickly than g(n). That sounds like an upper bound to me. Stephen === Subject: Re: O notation >> :>Torkel Franzen scribbled the following: >> :> :>> The O notation just gives upper bounds. >> :> :>Well, that is correct. AFAIK also o notation (little o) gives lower >> :>bounds >> : No, o gives lower bounds as well, just different lower bounds: >> I think you meant upper here. > o is lower (strict). omega is upper (strict). See CLRS. dang! sorry. other way around. -- Mitch Harris (remove q to reply) === Subject: Re: how to measure entropy of music? >> The basics are at: >> http://www.music-cog.ohio-state.edu/Music829D/Notes/ Infotheory.html >> key words: hmm, entropy, and music I have learned the question is of >> significance today for serveral reasons: >> 1) Recognizing when TV commercials are playing. >> 2) Separating out background music behind speech in speech >> recognition. >I wonder if they could figure a why to identify background music and >separate it from the sound reaching my ears :-) > Good point! If the predictor works then you can make a pretty good Not sure how you can come to that conclusion. If you have one microphone and a perfect algorithm you could determine what is music and what is subtract it from the sum of music + speech and without a second sensor this would not be possible - unless you assume the statistics of the music do not change rapidly so that the music during the speech was approx the same as during non-speech - an assumption which may be true for some music maybe? On the other hand if we used multiple microphones then maybe we would have something worthwhile. Tom === Subject: Re: Simple idea, mathematics and common-sense (IÕve just tonight discovered this old thread. Sorry for tardy reply, I wish I had seen it way back when you originally posted it.) > x^2 - 6*x + 6. > There are two roots: r1 = 3 + sqrt(3) and r2 = 3 - sqrt(3). OK, thatÕs correct. > Note that r1 * r2 = 6 = 2*3. Yes, thatÕs also correct. > It is easy to see that r1 has an algebraic integer factor > in common with 3. You can see that *BY INSPECTION*, the only > method that you actually trust. It is sqrt(3). Right? Yes, I agree even the idiot James Harris should see that. > 3 + sqrt(3) DOES have an algebraic integer factor in common with 2. You made the problem too easy! All James would have to do is commute the part you gave him as the answer to the first question, and whatÕs left is the answer to this second question: (3 + sqrt(3)) * (3 - sqrt(3)) = 2 * 3 sqrt(3) * (sqrt(3)+1) * sqrt(3) * (sqrt(3)-1) = 2 * 3 (sqrt(3)+1) * (sqrt(3)-1) * sqrt(3) * sqrt(3) = 2 * 3 Of the four factors on the left side, the last two factors give you the out: (sqrt(3)+1) * (sqrt(3)-1) = 2 The other example you presented is equally trivial in the same way: > a^2 - 4*a + 6. > The roots are 2 + sqrt(-2) and 2 - sqrt(-2). > Again their product is 6 = 2*3. It is clear > (by inspection!) that each of them has a factor > in common with 2. > Not so easily seen are the factors they have in common with 3. They pop out the same way as before: (2 + sqrt(-2)) * (2 - sqrt(-2)) = 2 * 3 sqrt(2) * (sqrt(2)+sqrt(-1)) * sqrt(2) * (sqrt(2)-sqrt(-1)) = 2 * 3 sqrt(2) * sqrt(2) * (sqrt(2)+sqrt(-1)) * (sqrt(2)-sqrt(-1)) = 2 * 3 The first two factors give 2, while the last two give 3. Cancel out the 2 and you have: (sqrt(2)+sqrt(-1)) * (sqrt(2)-sqrt(-1)) = 3 You basically *gave* James the answer, and he was too stupid to post a followup during the more than nine months between then and now, so apparently he really is too stupid to do even simple arithmetic with expressions involving square roots!! By the way, to prove these factorizations arenÕt trivial, you need to show that each of the factors is a non-unit, that is none of the factors you cite above is a unit, i.e. for example to show that sqrt(3)-1 is not a unit in {Z + sqrt(3)} you need to show that there are no integers i,j such that (i + j*sqrt(3)) * (sqrt(3)-1) = 1, i.e. you need to show that the simultaneous diaphantine equations 3*j-i=1 and i-j=0 donÕt have any solution (in integers), i.e. by substution 2*i=1 has no solution in integers, trivial, the answer is of course thereÕs no integer solution, but each of those needs to be checked out to complete the proof. In the second example, youÕve adjoined both sqrt(2) and sqrt(-1) to the integers, so you have four equations in four variables, for example (i + j*sqrt(2) + k*sqrt(-1) + l*sqrt(2)*sqrt(-1)) * (sqrt(2)+sqrt(-1)) = 1 coeff of 1: 2*j - k = 1 coeff of sqrt(2): i - l = 0 coeff of sqrt(-1): i + 2*l = 0 coeff of sqrt(-2): j + k = 0 (did I get all the coefficients correct?) which is a little more work to show no integer solution, but still easy enough for any good high-school math student to do once the four equations are written explicitly like that. So does this means James isnÕt even able to do high-school algebra, or he didnÕt realize how very trivial you made it so he didnÕt even try? === Subject: Re: How Many M&Ms? > Hello All. I am writing in reference to a contest a friend of mine is > having. The rules are as follows. > I bought a brand new clear plastic jar. It is filled > with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of > the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to fill it. I counted > every M & M before filling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? > If anyone can help me on this it would be wonderful as I am not a good > mathmatician. > Joey The only precise method is to count them. Probably the next most accurate method is to esimate the number by weight. Count out as large a fraction of the total as you have patience for and weigh them to find an average weight per piece. Then weigh the total (less jar weight) and divide by the average weight per piece. Or maybe the manfacturer will give you the average weight per piece. === Subject: Re: How Many M&Ms? posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L I donÕt get this post.... Joey counted the M&Ms but wants help knowing how many? oh she counted the 4 pounds of M&Ms then threw in a portion of them into the jar. If youÕre running the competition and people have to guess, they might expect you to count the exact number in the jar. Of course you could just take the median of the guesses and select them as winner, quoting the prize amount as a few different from what they guessed, but if the jar is the prize they might check, and be honest and hand it over to the actual best guesser, this could result in all sorts of bad publicity, you might find your next M&M competition youÕll be eating them yourself. A few questions, the jar you bought was empty right? You and your friend are both running the competition? You counted the M&Ms of all the bags you bought? Can you count the remaining M&Ms after you put them in the jar and substract? Is this like that puzzle, if there are 5 apples and I take away 2 how many do I have? Are we eligible for the prize to calculate our guesses? 2601. Herc === Subject: Re: How Many M&Ms? In sci.math, darrenn : > Fortunately, there has been some research on M&M packing: > The net is that the density of packing is around 71% (if packed in an > irregular fashion). > Further, it looks like the someone has actually measured an individual M&M: > http://www.kleinbottle.com/Bernie_Tao.htm > Which a result of 0.45239 cm. Pedant point. An M&M is generally a ßattened round thing and probably should be measured in all three dimensions, or perhaps one can pour a number into a liquid (although water will melt off the coatings, methinks) and measure the displacement. > so, volume of jar is: > PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 > cm3 > So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 > Now the result is probably only good to around 2 significant figures, and > there will probably be some rounding down due to the sides of the container, > so I would say aound 2500 M&Ms ? > Am I close? > -Darren >> Hello All. I am writing in reference to a contest a friend of mine is >> having. The rules are as follows. >> I bought a brand new clear plastic jar. It is filled >> with the standard size M & M candies. >> The jar is round and 8 inches tall. The diameter of >> the jar inside is 4 inches across. >> I bought over 4 pounds of M & Ms to fill it. I counted >> every M & M before filling the jar to the very top.. >> You can probably guess what I am seeking: >> How many M & Ms are in the jar as described? >> If anyone can help me on this it would be wonderful as I am not a good >> mathmatician. >> Joey -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: Re: How Many M&Ms? > In sci.math, darrenn : > Fortunately, there has been some research on M&M packing: > The net is that the density of packing is around 71% (if packed in an > irregular fashion). > Further, it looks like the someone has actually measured an individual M&M: > http://www.kleinbottle.com/Bernie_Tao.htm > Which a result of 0.45239 cm. > Pedant point. An M&M is generally a ßattened round thing and > probably should be measured in all three dimensions, or perhaps > one can pour a number into a liquid (although water will melt > off the coatings, methinks) and measure the displacement. > so, volume of jar is: > PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 > cm3 > So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 > Now the result is probably only good to around 2 significant figures, and > there will probably be some rounding down due to the sides of the container, > so I would say aound 2500 M&Ms ? > Am I close? > -Darren >> Hello All. I am writing in reference to a contest a friend of mine is >> having. The rules are as follows. >> I bought a brand new clear plastic jar. It is filled >> with the standard size M & M candies. >> The jar is round and 8 inches tall. The diameter of >> the jar inside is 4 inches across. >> I bought over 4 pounds of M & Ms to fill it. I counted >> every M & M before filling the jar to the very top.. >> You can probably guess what I am seeking: >> How many M & Ms are in the jar as described? >> If anyone can help me on this it would be wonderful as I am not a good >> mathmatician. >> Joey what does it matter if the water removes the coating? the coating will still contribute to the displacement. === Subject: Re: How Many M&Ms? In sci.math, David Bandel [snip for brevity -- discussing # of M&MÕs in a certain sized jar, and using water for one measurement of volume] > what does it matter if the water removes the coating? the coating will > still contribute to the displacement. ThereÕs a few issues regarding solubility that may complicate the analysis. IÕd frankly have to look. -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: Re: How Many M&Ms? >[snip for brevity -- discussing # of M&MÕs in a certain sized jar, > and using water for one measurement of volume] >> what does it matter if the water removes the coating? the coating will >> still contribute to the displacement. >ThereÕs a few issues regarding solubility that may complicate >the analysis. IÕd frankly have to look. M&MÕs require solving a polynomial of degree greater than 5 ? === Subject: Re: How Many M&Ms? > Which a result of 0.45239 cm. > Pedant point. An M&M is generally a ßattened round thing and > probably should be measured in all three dimensions, or perhaps > one can pour a number into a liquid (although water will melt > off the coatings, methinks) and measure the displacement. True. This should say 0.45239 cm^3. -Darren === Subject: Re: Proving the cross product is orthogonal > I hope I donÕt seem like one of those louts that only turn up at > assignment time!! However I am struggling with a proof. My answer is not > coming out as expected, and I think I am making a silly mistake > somewhere. If anyone could have a glance over it, that would be appreciated. > Q) a =(1, -2, 1) b =(3, 1, 0) > Find a x b and prove that your cross-product vector is perpendicular to > each of the vectors a and b. > A) [I have worked out the cross-product, and I have proven that a is > perpendicular to the cross-product, however I am struggling to prove > that b is perpendicular to the cross-product] Do you know about the dot product? === Subject: Re: Proving the cross product is orthogonal >>I hope I donÕt seem like one of those louts that only turn up at >>assignment time!! However I am struggling with a proof. My answer is not >>coming out as expected, and I think I am making a silly mistake >>somewhere. If anyone could have a glance over it, that would be appreciated. >>Q) a =(1, -2, 1) b =(3, 1, 0) >>Find a x b and prove that your cross-product vector is perpendicular to >>each of the vectors a and b. >>A) [I have worked out the cross-product, and I have proven that a is >>perpendicular to the cross-product, however I am struggling to prove >>that b is perpendicular to the cross-product] > Do you know about the dot product? Yes, I like the dot product...much easier then the cross product I think! It turned out my main mistake was silly mistakes in my workings (ie 2*0 = 2 etc) I did this two places, I think it was only ßuke that my workings for a being perpendicular to a x b, after the silly errors where fixed both proofs worked. Cassandra === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE > You simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess > that you were writing too small print too tightly in the left upper > corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); > etc., etc. > IHTH - Johan E. Mebius >> I hope I donÕt seem like one of those louts that only turn up at >> assignment time!! However I am struggling with a proof. My answer is >> not coming out as expected, and I think I am making a silly mistake >> somewhere. If anyone could have a glance over it, that would be >> appreciated. >> Q) a =(1, -2, 1) b =(3, 1, 0) >> Find a x b and prove that your cross-product vector is perpendicular >> to each of the vectors a and b. >> A) [I have worked out the cross-product, and I have proven that a is >> perpendicular to the cross-product, however I am struggling to prove >> that b is perpendicular to the cross-product] >> a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k >> = -3i + 2j + 7k >> Let w = -3i + 2j + 7k >> Prove vector a is perpendicular to w >> a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 >> |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) >> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >> cos&= (a.w)/(|a||w|) = 0 >> Prove vector a is perpendicular to w >> b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 >> |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >> cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) >> I know that is wrong, can anyone see what my mistkae is?? >> Cassandra Much appreciated. I have been writing my equations in MS Word using the equation editor. Perhaps I should write them on paper Ōbig and largeÕ first!! Cassie === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE > Much appreciated. I have been writing my equations in MS Word using the > equation editor. Perhaps I should write them on paper Ōbig and largeÕ > first!! With a pencil. The trolls in this newsgroup post all kinds of nonsense, sometimes on purpose and sometimes because they donÕt check their work. === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE >>Much appreciated. I have been writing my equations in MS Word using the >>equation editor. Perhaps I should write them on paper Ōbig and largeÕ >>first!! > With a pencil. > The trolls in this newsgroup post all kinds of nonsense, sometimes on purpose > and sometimes because they donÕt check their work. Yes, one of my biggest downfalls in life is that I make lots of mistakes. I once had to do a speed & accuracy test as part of a pysch test. I scored full marks (not sure how they calculated it), but I remember the pysch saying something about me getting alot further through the test then most, but making more mistakes then most. I used to have a turtle on my desk to remind myself to slow down and double check things. I find it so difficult, but it is something I should really work on. I have made two very elementary mistakes in the course of this thread...I do need to improve. === Subject: Re: PROOF that 0.99999... = 1 > IÕm afraid not. > lim Z_n =/= 0 implies that > sum_i (Z_i) diverges. > However the S_n in the original post were already the partial sums. Yes, so 0.9999... = 1. === Subject: topology....[x] Hello....doctor~ define that f:X->Y , (X=R, Y=Z) is surjective. f(x) = [x] X is real topology space. find the quotient topology of Y. ---------------------------------------- i think.....{empty, Y} is this right answer ? thank you very much for your advice. === Subject: Re: topology....[x] > Hello....doctor~ > define that > f:X->Y , (X=R, Y=Z) is surjective. > f(x) = [x] > X is real topology space. > find the quotient topology of Y. > ---------------------------------------- > i think.....{empty, Y} > is this right answer ? > thank you very much for your advice. Actually, I think itÕs the topology consisting of the sets (-infinity,a] (intersected with Z, of course), together with the empty set, where a is either an integer or infinity. Think about it: this set has the set (-infinity,a+1) as its preimage, which is open in R, so it must be open in the quotient topology. On the other hand, if a is in an open set of the quotient topology, then a-1 is in the same set. === Subject: Re: topology....[x] > define that > f:X->Y , (X=R, Y=Z) is surjective. > f(x) = [x] [x] = min{ n in N | n <= x } > X is real topology space. > find the quotient topology of Y. f:R -> Z, x -> [x]; f^-1(x) = [[x], [x]+1) For U proper subset Z U open iff f^-1(U) open (definition) iff some n in N with f^-1(U) = (-oo,n) (exercise) iff some n in N with U = { j in N | j < n } > i think.....{empty, Y} > is this right answer ? No. === Subject: Re: topology....[x] > [x] = min{ n in N | n <= x } Actually, itÕs [x] = max{ n in N | n <= x } Jose Carlos Santos === Subject: Re: topology....[x] <31gk5jF3b8lpdU2@individual.net [x] = min{ n in N | n <= x } > Actually, itÕs [x] = max{ n in N | n <= x } Well, have we got it straight this time? [x] = max{ n in Z | n <= x } === Subject: Re: topology....[x] >[x] = min{ n in N | n <= x } >>Actually, itÕs [x] = max{ n in N | n <= x } > Well, have we got it straight this time? > [x] = max{ n in Z | n <= x } Yes, thatÕs it! :-) Jose Carlos Santos === Subject: Re: topology....[x] <31gk5jF3b8lpdU2@individual.net [x] = min{ n in N | n <= x } > Actually, itÕs [x] = max{ n in N | n <= x } Yup and [x] is also use for the equivalence class containing x. === Subject: Re: topology....[x] > define that > f:X->Y , (X=R, Y=Z) is surjective. > f(x) = [x] > X is real topology space. > find the quotient topology of Y. > ---------------------------------------- > i think.....{empty, Y} > is this right answer ? Yes, but do you see why? Jose Carlos Santos === Subject: Re: topology....[x] >> find the quotient topology of Y. >> ---------------------------------------- >> i think.....{empty, Y} >> is this right answer ? > Yes, but do you see why? Oops, sorry, wrong answer. But I see that by now someone else has already provided the correct answer. Jose Carlos Santos === Subject: Re: JSH: Simply fascinating > IÕve given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). In general thatÕs not true at all. For example, consider: (x+1)*(x+2)*(x+3) If you expand it out to get a polynomial in the usual form, you see that the constant term is 6. But 6 is not a factor of any of the three polymomial factors, in fact neither 2 nor 3 is a factor of any of them say what you really meant to say? Or maybe you were totally mistaken and need to just retract what you said. > My research shows that some mistakes were made over a hundred years > ago, and a lot of people missed them, and gave proofs which were > not, and are not proofs. ItÕs true that some mistakes were made long ago, the worst in my opinion being the first developers of set theory who concocted a set of all sets, which was subsequently proven not to exist. But IÕve seen no evidence that *you* have ever done any research that showed any such past mistakes of others. Instead all I see is *you* making mistakes yourself and not admitting them. Do you know how to factor 7 in the ring of integers with sqrt(7) adjoined? ThatÕs easy, of course you do, right? But do you know how to factor 7 in the ring of integers with sqrt(2) adjoined? Now if you can figure out that simple arithmetic problem, hereÕs something more interesting: Find all primes p (other than 2 and 7 which IÕve already shown you) such that 7 can be nontrivially factored in the ring of integers with sqrt(p) adjoined. Finally, find all finite sets of two or more primes p1,p2,...,pn such that 7 is composite in the ring you get when you adjoin all the sqrts of those primes but itÕs prime in the ring you get when you adjoin all but one of those sqrts. Same question if you allow adjoining sqrt(-1) in addition to adjoining sqrt of a prime. (Note: When I refer to a prime p or primes p1,p2..., I mean prime in the ring of integers. When I refer to 7 being prime or composite, I mean prime or composite in the ring of integers adjoined with the various sqrts.) === Subject: Re: JSH: Simply fascinating |ItÕs true that some mistakes were made long ago, the worst in my |opinion being the first developers of set theory who concocted a set |of all sets, which was subsequently proven not to exist. As far as I know, this incident involved Frege and nobody else. If you have other developers of set theory in mind, who were they? Second, the set of all sets was only proven not to exist under some additional assumptions. The problem with FregeÕs system was that it implied there exists a Russell set, R = {x : x is not a member of x}. The existence of a Russell set is directly contradictory. There are systems of set theory that have a universal set and are consistent (although they are less used). Frege didnÕt directly postulate a universal set or a Russell set; he just stated axioms that he regarded as intuitive, that together implied that for any predicate P there exists a set {x : x satisfies P}. Keith Ramsay === Subject: Re: JSH: Simply fascinating > The problem with FregeÕs system was that it implied there exists a > Russell set, R = {x : x is not a member of x}. By what grammatical axiom(s) is the right side of that equation even a well-formed formula? I would guess two such grammatical axioms: If S and T are WFFs denoting sets, then S is a member of T is also a WFF denoting a truthvalue. If P is a WFF denoting a truthvalue, then not P is also a WFF denoting a truthvalue. x is not a member of x is shorthand for not (x is a member of x). If x is a WFF variable, and P is a boolean-valued WFF with x as the only free variable, then {x : P} is a WFF denoting a set. I personally accept all those grammatical axioms except the last as reasonable. The last I would replace with: If S is a WFF denoting a set, and x is a WFF variable, and P is a boolean-valued WFF with x as the only free variable, then {x in S : P} is a WFF denoting a set. The reason I donÕt like the original last grammatical axiom is that it begs the question what kind of Universe weÕre dealing with, whereas my alternative clearly says S is the universe weÕre dealing with. If we beg the question what Universe weÕre dealing with, then we arenÕt really making a definition, we arenÕt saying what is allowed to be in a set and what isnÕt allowed to be in a set, so we arenÕt saying whether some monstrosity in somebodyÕs fantasy can be in our set or not. > There are systems of set theory that have a universal set and are > consistent (although they are less used). Do you know a WebPage that describes any of these axiomatic systems? Alternately, do you remember which commonsense axioms of set-theory grammar are *not* included, thereby avoiding writing a Russell set as a WFF in said system? === Subject: Re: JSH: Simply fascinating Discussion, linux) >> The problem with FregeÕs system was that it implied there exists a >> Russell set, R = {x : x is not a member of x}. > By what grammatical axiom(s) is the right side of that equation even a > well-formed formula? I would guess two such grammatical axioms: ^^^^^^^ term, not formula. > If S and T are WFFs denoting sets, > then S is a member of T is also a WFF denoting a truthvalue. > If P is a WFF denoting a truthvalue, then not P is also a WFF > denoting a truthvalue. > x is not a member of x is shorthand for not (x is a member of x). > If x is a WFF variable, > and P is a boolean-valued WFF with x as the only free variable, > then {x : P} is a WFF denoting a set. > I personally accept all those grammatical axioms except the last as > reasonable. The last I would replace with: > If S is a WFF denoting a set, and x is a WFF variable, > and P is a boolean-valued WFF with x as the only free variable, > then {x in S : P} is a WFF denoting a set. > The reason I donÕt like the original last grammatical axiom is that > it begs the question what kind of Universe weÕre dealing with, > whereas my alternative clearly says S is the universe weÕre dealing ^^^^^^^^^^^^^^ your alternative? YouÕre very inßuential. > with. If we beg the question what Universe weÕre dealing with, then > we arenÕt really making a definition, we arenÕt saying what is > allowed to be in a set and what isnÕt allowed to be in a set, so we > arenÕt saying whether some monstrosity in somebodyÕs fantasy can be > in our set or not. -- Who knows, maybe that may be the only way to settle this crap. ItÕs not like itÕd be that hard for me to go back and get a math degree. I can penetrate the math social group and then finish the takedown from inside. -- James S. Harris contemplates a new strategy. === Subject: Re: JSH: Simply fascinating > IÕve given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > In general thatÕs not true at all. For example, consider: > (x+1)*(x+2)*(x+3) But he said, e.g. in your example, having factored your polynomial into (x+1), (x+2), and (x+3), 1, 2, and 3 must be factors of 6, the constant term of your expanded polynomial. > If you expand it out to get a polynomial in the usual form, you see > that the constant term is 6. But 6 is not a factor of any of the three > polymomial factors, in fact neither 2 nor 3 is a factor of any of them He didnÕt say, with respect to your example, that 6 must be a factor of 1, 2, and 3. Rather that 1, 2, and 3 must be factors of 6. KeithK > say what you really meant to say? Or maybe you were totally mistaken > and need to just retract what you said. > My research shows that some mistakes were made over a hundred years > ago, and a lot of people missed them, and gave proofs which were > not, and are not proofs. > ItÕs true that some mistakes were made long ago, the worst in my > opinion being the first developers of set theory who concocted a set > of all sets, which was subsequently proven not to exist. But IÕve seen > no evidence that *you* have ever done any research that showed any such > past mistakes of others. Instead all I see is *you* making mistakes > yourself and not admitting them. > Do you know how to factor 7 in the ring of integers with sqrt(7) > adjoined? ThatÕs easy, of course you do, right? But do you know how to > factor 7 in the ring of integers with sqrt(2) adjoined? Now if you can > figure out that simple arithmetic problem, hereÕs something more > interesting: Find all primes p (other than 2 and 7 which IÕve already > shown you) such that 7 can be nontrivially factored in the ring of > integers with sqrt(p) adjoined. Finally, find all finite sets of two or > more primes p1,p2,...,pn such that 7 is composite in the ring you get > when you adjoin all the sqrts of those primes but itÕs prime in the > ring you get when you adjoin all but one of those sqrts. Same question > if you allow adjoining sqrt(-1) in addition to adjoining sqrt of a > prime. (Note: When I refer to a prime p or primes p1,p2..., I mean > prime in the ring of integers. When I refer to 7 being prime or > composite, I mean prime or composite in the ring of integers adjoined > with the various sqrts.) === Subject: Re: JSH: Simply fascinating > What does NoraÕs gender have to do with math. You only point this out > because you are immature, trying to create a distraction, and you know > your work is wrong. Get a life. > Dave The poster lies repeatedly. IÕve argued with this poster for some time and noted the lying, and the gender lying is just another example of a trend. Some people donÕt like being lied to, and take it seriously. My position is that I can explain in detail, and with very simple concepts how my argument works, but that mathematicians might believe they have lots of reasons for avoiding the truth--all social ones. Unfortunately, they are aided by the poster who manages to maintain confusion about the issues, and maintain the lie that there is any vagueness or area of real mathematical doubt about my work, as many people trust. They trust that if I were right mathematicians wouldnÕt disagree with me, and there wouldnÕt be so much opposition to my work. So the poster Nora Baron can just lie for the sake of showing opposition, which helps to create the illusion of uncertainty about my work, helping to hide the truth from people who donÕt know better. The people who are well-trained mathematicians though, they are not fooled, and I know from my own experiences talking with well-trained mathematicians about my work. So whatÕs happening now is sad in many ways, but part of it has to do with a basic contempt for the public, and a belief that they can be lead astray indefinitely by rather basic tactics. I say, the strategy is about to die, and the passive-aggessive strategy of hoping that indefinitely the public will be fooled will be shown to be one of professional suicide. James Harris === Subject: Re: JSH: Simply fascinating > What does NoraÕs gender have to do with math. You only point this out > because you are immature, trying to create a distraction, and you know > your work is wrong. Get a life. > Dave > The poster lies repeatedly. IÕve argued with this poster for some > time and noted the lying, and the gender lying is just another example > of a trend. Bullcrap. Nora Baron has repeatedly shown errors in the poster James Harrismath and THAT is why he hates her. (Oh he does! He said so not long ago). He appears to be particularly infuriated because she keeps pinning him back to the more involved polynomials from which his simpler examples are derived, and showing where his error lies (no pun intended). This poster James Harris historically responds to math rebuttals with diatribes like this when he is unable to respond successfully to the math with math. Perhaps what REALLY pisses him off is that a _female_ could so effectively demolish his erroneous math. > Some people donÕt like being lied to, and take it seriously. > My position is that I can explain in detail, and with very simple > concepts how my argument works, but that mathematicians might believe > they have lots of reasons for avoiding the truth--all social ones. Here it is: > Unfortunately, they are aided by the poster who manages to maintain > confusion about the issues, and maintain the lie that there is any > vagueness or area of real mathematical doubt about my work What this poster James Harris calls Ōliesare any math rebuttals that show that there is any vagueness or area of real mathematical doubt about his work. Since his work is correct they _must_ be lies. KeithK >, as many > people trust. The people who are well-trained mathematicians though, they are not > fooled, and I know from my own experiences talking with well-trained > mathematicians about my work. So whatÕs your problem? If well-trained mathematicians have discussed your work with you, surely you donÕt need to keep banging your head against the lame brains at sci.math or any other newsgroup. Just have some of these well-trained mathematicians promote your work and publications, accolades, Field medals, parades, etc. will follow. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Simply fascinating >> The math here is so readily understandable that itÕs actually almost >> as interesting watching how people react to it, as anything else. >> >> For instance, IÕve given a polynomial P(x) repeatedly where I factor >> it into three factors. >> >> I point out that the factors must include factors of the constant term >> of the polynomial P(x). >> >> I note that the factors of the constant term are independent of x, as >> they are, in fact, constant. >> >> Mathematically itÕs easy to show: >> >> g_1(x) g_2(x) g_3(x) = P(x) >> >> and >> >> g_1(0) g_2(0) g_3(0) = P(0) >> >> as P(0) gives the constant term of the polynomial, and since the >> polynomial results from multiplying together the three factors which >> IÕve called g_1(x), g_2(x), and g_3(x), then you can get the factors >> of the constant term, by setting x=0. >> >> ItÕs that simple. >> >> Yes. Right so far. Trivial, but right. >> Notice (1) you have the factors of P(x), (2) the constant term of P(x) >> is determinable by setting x=0, (3) you also get the factors of the >> constant term. >> >> Mathematically, itÕs simple to the point of trivial. >> >> Correct. >> Now then, if you have *constants* which are factors of another >> constant then why would anyone try to argue that they are actually >> variables? >> >> No one does. If as you say, you assume that your factors are >> constant, of course they are not variables. That is nothing but >> a dimwit tautology. >> And of course that is not what we say. We say that if you factor >> 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), >> and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) >> by those variable functions, then you can obtain algebraic integer >> factors on both sides of the resulting equation. That is, >> g_1(x)/w_1(x), etc., are all algebraic integers. That is all you >> require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and >> g_3(x)/w_3(x) are BY YOUR OWN DEFINITION equal to >> g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_3(0). [a couple of misprints corrected here] >> No problem with that either. And the product of these three >> constant terms equals P(0)/49. >Ok, IÕm going to answer the Nora Baron poster yet again. >And I want readers to understand that IÕve replied to this poster who >you know lies anyway as itÕs a guy posting as a woman using a name >thatÕs a palindrome MANY TIMES explaining in detail. >What happens is that when I shoot down these objections, the poster >either just repeats later, or replies with nonsense. One telling time >when I carefully refuted point-by-point the poster replied deleting >out everything IÕd said. Because it was the SOS, repeated for the nth time. I replied with a complete detailed and rigorous proof. You were enraged but never really provided any refutation. >Just deleted out everything, and you know what? I STILL so people >replying about how supposedly I donÕt answer the objections from Nora >Baron. You STILL so people replying ??? Really! >ItÕs partly a game for some people on sci.math, IÕm sure, and partly a >case where many readers are hoodwinked as *they* donÕt know itÕs a >game. They seem incapable of realizing that there are people in this >world willing to behave in such a way, on such a level. >So why reply to this poster? Maybe IÕm hopelessly naive, but I just >have to believe that eventually people will just get tired of being >made fools of by this poster and others in the group with him (or >her). It hasnÕt happened yet from what IÕve seen, but I keep hoping. Whine, whine, whine. Why not go straight to the math rather and skip the propaganda ? >Ok, so whatÕs wrong with the posterÕs assertion? >Well, the wÕs the posters gives are factors of 49. But 49 comes into >the picture because 49 is a multiple of the polynomial. But the wÕs >STILL REMAIN after 49 has been divided off--after the multiple is >divided off--as the poster actually claims. I claim no such thing. The product of the wÕs is 49. You divide P(x) by 49. You divide the product of the gÕs by the product of the wÕs. The 49 is gone from both sides. Note that here g_1(x) = (5 a_1(x) + 7), g_2(x) = (5 a_2(x) + 7), and g_3(x) = (5 a_3(x) + 7) = (5 b_3(x) + 22), where a_1, a_2, a_3 are all algebraic integer functions of x, and b_3(x) = a_3(x) - 3. Note that a_1(0) = a_2(0) = 0, and a_3(0) = 3. Also, note that w_1(0) = w_2(0) = 7, and w_3(0) = 1. When I divide the product of the gÕs by the product of the wÕs, the result is (5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x)), where c_i(x) = a_i(x)/w_i(x) and d_i(x) = 7/w_i(x). Here is what Mr. Harris is squawking about. The product of the dÕs is 7. (That happens to be true for any x.) Harris thinks that is a problem, because it then looks like the product of the constant terms of (5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x)) is 7, whereas the constant term of P(x)/49 is 22. That is, d_1(x)*d_2(x)*d_3(x) = 7, whereas P(0)/49 = 22. Puzzling! Is Harris right? I will now give my deceptive, lying, sleight-of-hand explanation. Watch very carefully, because as Mr. Harris says, I am going to try very hard to mislead you! So here is the explanation. The dÕs are NOT the constant terms of the factors that contain them. That is, d_1(x) is NOT the constant term of (5 c_1(x) + d_1(x)). Mr. Harris thinks it is. Certainly it is in the right position for being a constant term. Since Mr. Harris understands and mostly relies on only high-school level mathematics, he detects constant terms by *inspection*. The irony here is, he has provided a perfectly good, rigorous definition of constant term of a function, but he doesnÕt use it. It is the value that the function takes on when the argument is 0. Thus the constant term of (5 c_1(x) + d_1(x)) is NOT d_1(x) = 7/w_1(x). It is instead d_1(0) = 7/w_1(0) = 1. This is directly from Mr. HarrisÕs own definition of constant term. Why is he so reluctant to apply that definition? Why does he instead rely on inspection, which gives the incorrect answer? What HE thinks is the constant term, namely d_1(x) = 7/w_1(x) is not even constant (unless he assumes what he wants to prove) ! Similary, the constant term of (5 c_2(x) + d_2(x)) is 1 also. However, the constant term of (5 c_3(x) + d_3(x)) is (5 c_3(0) + d_3(0)) = (5 a_3(0)/w_3(0) + 7/w_3(0)) = 5 * 3/1 + 7/1 = 15 + 7 = 22. Wow, look at that! The product of the constant terms is the constant term of the product! That is, d_1(0)*d_2(0)*(5 a_3(0) + d_3(0)) = 1*1*22 = 22 = P(0)/49. Amazing! Did I successfully trick you? WhereÕs the trick? Everything works out just as it should: IF you correctly identify the constant terms. WhereÕs the trick, Mr. Harris ? >(Notice Nora Baron gives you clues, but somehow sci.math readers >donÕt seem to get it.) I think not. I think that, with perhaps one exception, sci.math readers DO get it. In fact, I even think there is no exception. I think Mr. Harris actually gets it also, but he cannot stand to admit it. There is too much at stake. He clings to a delusion on the one-in-a trillion chance that he will be proven right. >Notice the posterÕs actually gives that when you divide off 49, you >get >g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x) >which are TRACES, left of 49--after it has been divided off. See above. There is no mystery. After division, the product of the constant terms is the constant term of the product, just as it should be. That is, IF you correctly identify what the constant terms actually are. If you donÕt, then you get the wrong answer. This is Mr. HarrisÕs mistake. This is not rocket science. This is not Galois theory, or algebraic number theory, or field theory, or group theory. This is not abstract algebra at all. This is not even high-school level algebra. The mistake Mr. Harris is making is really a low-level one. A ROOKY mistake, as Mr. Harris used to say (about me and Arturo Magidin and others). He is simply not using his own definition. He is substituting in x when he should be substituting in 0. And incredibly, he seems unable to understand it. Math genius Harris, boy wonder Harris - from this you would think he needed help changing his own soiled underwear. [more propaganda deleted] Nora B. >James Harris === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that itÕs actually almost > as interesting watching how people react to it, as anything else. For instance, IÕve given a polynomial P(x) repeatedly where I factor > it into three factors. I point out that the factors must include factors of the constant term > of the polynomial P(x). I note that the factors of the constant term are independent of x, as > they are, in fact, constant. Mathematically itÕs easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > IÕve called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. ItÕs that simple. Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. Mathematically, itÕs simple to the point of trivial. Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? No one does. If as you say, you assume that your factors are > constant, of course they are not variables. That is nothing but > a dimwit tautology. > And of course that is not what we say. We say that if you factor > 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), > and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) > by those variable functions, then you can obtain algebraic integer > factors on both sides of the resulting equation. That is, > g_1(x)/w_1(x), etc., are all algebraic integers. That is all you > require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and > g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to > g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). > No problem with that either. And the product of these three > constant terms equals P(0)/49. > Ok, IÕm going to answer the Nora Baron poster yet again. > And I want readers to understand that IÕve replied to this poster who > you know lies anyway as itÕs a guy posting as a woman using a name > thatÕs a palindrome MANY TIMES explaining in detail. Please provide a link to a message where nora baron says that she is a woman. <... Look now, even after the post where Nora Baron ended with a male > name, people are STILL supporting the poster! Please provide a link to a message where nora baron says that he is a man. === Subject: Re: JSH: Simply fascinating > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? If they are independent of ŌxÕ, why did you have to set ŌxÕ to zero to uncover them? Evaluating a univariate polynomial at any numeric value of the variable produces a numeric result. Are all these results ŌconstantsÕ? > The logic is inescapable. > LifeÕs too short. I think yours is too long. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Vectors question 2 >>So after reading your response I began to think. Ignoring vectors and i & j >>etc. If I wanted to figure this out a few weeks ago (prior to learning this >>stuff) my method would have been as follows. >>Put the direction of the train on the x-axis, the direction of the stuntman >>on the y-axis. (this is I think what you are trying to say). > That method is correct, and produces the correct answers... so now my > question is: what is it that you learned over the last few weeks that > confused you into trying to plot time as well? To answer your question, I overthought the question. We are studying > vectors, including 3-dimensional vectors and projections. Perhaps it was > because I had only just finished studying projections. Or because I thought > that question must be more complex then it looked etc. > I really donÕt know why, however I am sort of glad because it has allowed me > to gain alot better understanding into using vectors, but then making > mistakes does usually have that effect. Well, if you want to consider this and include time as one of your dimensions, then you have to realize that you have to project the space-time coordinate system into a space-only coordinate system before trying to describe it in terms of visible effects. For example, consider a single point traveling along a straight line. Let the line of travel be the X axis and let time T be perpendicular to it. Then, in the XT plane, the point will be describing some curve (continuous, unless your universe includes teleportation). For example, if the point is oscillating about the origin, then the space-time curve describing its motion would be a sine wave on the T axis. But a description of that motion would probably not refer to a sine wave about the T axis, it would refer to an oscillatory motion about the origin. This seems like a trivial point until you start considering more dimensions: consider an XY plane and a time T axis perpendicular to it, and two points traveling in the plane -- one at a fixed velocity along the X axis, starting at the origin, and the other at the same velocity along the Y axis, also starting at the origin. If you look at them in the XY plane, then their trajectories (namely, the X and Y axes) are perpendicular. However, if you consider them to be points in XYT space, then the first oneÕs trajectory is a straight line in the XT plane, and the second oneÕs trajectory is a straight line in the YT plane, and those trajectories are not perpendicular! It is only their projections into the XY place (the space-only coordinate system) that are perpendicular. Which brings me to your question; your mistake in considering this with distance on the X axis and the time on the T axis was that when the description of physical motion of the stunt man was described as being perpendicular to the motion of the train, you drew the perpendicular in the XT plane, which doesnÕt work. What you need to do here is count the number of dimensions of space (clearly, you need at least two, as the man the the train are moving at right angles to each other. Then you add one for time, and you see that you need a three-dimensional system. In that system, the trajectory of the train will be a straight line; if the train travels along the X axis, as it does in your original graph, then its XYT trajectory will be a straight line in the XT plane, as it is in your original graph. However, the motion of the man is, according to the problem statement, perpendicular to that of the train -- but the problem statement refers to them in the space-only coordinate system, which means that you have to have the man moving perpendicular not to the XYT time-space trajectory of the train, but to the XY spatial trajectory of the train. If you do that, then everything comes out right, but itÕs somewhat complicated to draw that. Now, there is no really good reason to involve time in this, as the movement of the train (relative to the ground) and the man (relative to the train) are constant with time, so what you are asked for (movement of the man relative to the ground) is also constant and you can draw it without involving the T axis at all. However, I think you should know why your initial solution didnÕt work and how you could have made it work. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Proving planes are not parallel The definition that I have is as follows: Two planes are parallel if their normal vectors are parallel, that is, if the cross product of their normal vectors is zero. However my understanding is that a cross product yields another vector, not a scalar. So I am confused. My testbook has one so-called worked example, but it borders more along the lines of weÕll leave the proof as an exercise, ie it leaves out the middle bits. *************************************** Two parallel planes. P_1: 2x + 3y - z = 3 P_2: -4x - 6y + 2z = 8 so n_1 = 2i + 3j - k, n_2 = -4i - 6j + 2k The planes are parallel because: n_2 = -2n_1 and n_1 x n_2 = 0 *************************************** I am trying to prove two planes are not parallel. Should I write them as vectors, then show that the vectors are not multiples of each other? ie show n_1 <> x(n_2) How does n_1 x n_2 = 0 in the above example? I thought n_1 x n_2 = -1i + 0j + 4k Cassandra === Subject: Re: Proving planes are not parallel > The definition that I have is as follows: > Two planes are parallel if their normal vectors are parallel, that is, > if the cross product of their normal vectors is zero. > However my understanding is that a cross product yields another vector, > not a scalar. So I am confused. There is a zero vector in every vector space, such that if 0 is the zero scalar and Z is the zero vector and . indicates product of scalar with vector then 0.Z = Z Since it is usually obvious from the context when a zero object is to be a scalar and when a vector, the same word is used ambiguously for both, as in this case. To say that the cross product of two vectors is zero clearly must mean the zero vector, not the zero scalar. Does this clear things up a bit? >... > Cassandra === Subject: Re: Proving planes are not parallel Cassandra Thompson a .8ecrit : > The definition that I have is as follows: > Two planes are parallel if their normal vectors are parallel, that is, > if the cross product of their normal vectors is zero. > However my understanding is that a cross product yields another vector, > not a scalar. So I am confused. In those formulations, zero stands for the number 0 or for the null vector > My testbook has one so-called worked example, but it borders more along > the lines of weÕll leave the proof as an exercise, ie it leaves out > the middle bits. > *************************************** > Two parallel planes. > P_1: 2x + 3y - z = 3 > P_2: -4x - 6y + 2z = 8 > so > n_1 = 2i + 3j - k, > n_2 = -4i - 6j + 2k > The planes are parallel because: > n_2 = -2n_1 > and n_1 x n_2 = 0 > *************************************** > I am trying to prove two planes are not parallel. Should I write them as > vectors, then show that the vectors are not multiples of each other? > ie show > n_1 <> x(n_2) It is one possibility (better to write n_1 <> k(n_2) ) ; another one is to calculate the cross-product of n_1 and n_2 and to show it is not null > How does n_1 x n_2 = 0 in the above example? I thought > n_1 x n_2 = -1i + 0j + 4k You must have made a calculation mistake. First, n_1 is indeed equal to -2 n_2 (check it). Second, the i component of the cross-product n_1 x n_2 is y_1z_2-y_2z_1= 3*2-(-1)*(-6)=0 , and not -1 as you say... > Cassandra === Subject: Re: Proving planes are not parallel > Cassandra Thompson a .8ecrit : >> The definition that I have is as follows: >> Two planes are parallel if their normal vectors are parallel, that is, >> if the cross product of their normal vectors is zero. >> However my understanding is that a cross product yields another >> vector, not a scalar. So I am confused. > In those formulations, zero stands for the number 0 or for the null vector >> My testbook has one so-called worked example, but it borders more >> along the lines of weÕll leave the proof as an exercise, ie it >> leaves out the middle bits. >> *************************************** >> Two parallel planes. >> P_1: 2x + 3y - z = 3 >> P_2: -4x - 6y + 2z = 8 >> so >> n_1 = 2i + 3j - k, >> n_2 = -4i - 6j + 2k >> The planes are parallel because: >> n_2 = -2n_1 >> and n_1 x n_2 = 0 >> *************************************** >> I am trying to prove two planes are not parallel. Should I write them >> as vectors, then show that the vectors are not multiples of each other? >> ie show >> n_1 <> x(n_2) > It is one possibility (better to write n_1 <> k(n_2) ) ; another one is > to calculate the cross-product of n_1 and n_2 and to show it is not null >> How does n_1 x n_2 = 0 in the above example? I thought >> n_1 x n_2 = -1i + 0j + 4k > You must have made a calculation mistake. First, n_1 is indeed equal to > -2 n_2 (check it). Second, the i component of the cross-product n_1 x > n_2 is y_1z_2-y_2z_1= 3*2-(-1)*(-6)=0 , and not -1 as you say... >> Cassandra simple algebra errors that I have posted to the newsgroup. Although I have to admit I find the nature of cross-products begs for mistakes. Perhaps I should learn to calculate the cross-product using a matrix, or better still get a graphing calculator! I have answered the assignment equation using the n_1 <> k(n_2) method, product and null vector. Cassie === Subject: Re: Proving planes are not parallel >simple algebra errors that I have posted to the newsgroup. Although I >have to admit I find the nature of cross-products begs for mistakes. >Perhaps I should learn to calculate the cross-product using a matrix, or >better still get a graphing calculator! >I have answered the assignment equation using the n_1 <> k(n_2) method, >product and null vector. >Cassie DonÕt feel ashamed Cassie. We all make mistakes, from inadvertent typoÕs to just ßat being wrong. In my opinion it is always better to check whether v2 = c * v1 than v1 cross v2 = zero vector if testing for parallelness exactly because it is simpler and a less error - prone method. The cross product does have many uses, though, and you will need to be able to calculate them with a minimum of errors. There are several methods in use. My favorite, which requires a little more writing but, at least for me, keeps the errors to a minimum is the following. To cross with I first make a helper where I write the vectors starting and ending with the middle component: b c a b e f d e Now the components of the cross product are gotten by computing the left, middle, and right 2x2 determinants in the helper: In particular, this gets the sign of the middle component correct without giving it any additional thought. --Lynn === Subject: Re: Proving planes are not parallel >b c a b >e f d e >Now the components of the cross product are gotten by computing the >left, middle, and right 2x2 determinants in the helper: > Talk about typos...., of course the first one is bf - ec :-) --Lynn === Subject: Re: Proving planes are not parallel >>b c a b >>e f d e >>Now the components of the cross product are gotten by computing the >>left, middle, and right 2x2 determinants in the helper: >> :-) > --Lynn Yes, that is a much sinpler way of doing it. I have been writing the equations down, then writing down the general formula, then putting small labels on the equation (a_1, b_1 etc) then trying to look in three places at once to be sure I am putting the right number inthe right place. Of course this could easily be done on a calculator, scilab, or even using MS Excel, but I need to master it before I turn to technology. Cassandra === Subject: Re: Proving planes are not parallel >b c a b >e f d e >Now the components of the cross product are gotten by computing the >left, middle, and right 2x2 determinants in the helper: > Talk about typos...., of course the first one is bf - ec >> :-) >> --Lynn >Yes, that is a much sinpler way of doing it. >I have been writing the equations down, then writing down the general >formula, then putting small labels on the equation (a_1, b_1 etc) then >trying to look in three places at once to be sure I am putting the right >number inthe right place. Ugh!!! Then IÕm glad I posted that for you. I always told my classes to not even bother to memorize the formula. Just use the helper. --Lynn === Subject: abstract algebra...... hello....doctor~ ring homomorphism f: Z -> Z/3Z + Z/5Z + Z/7Z which f(x) = (x+3Z , x+5Z , x+7Z) (+ is direct sum) 1) ker f =? 2) find all integer x such that f(x) = (2+3Z , 3+5Z , 4+7Z) --------------------------------------------- i think....... 1) [3,5,7]=105, so 105*Z right ?? 2) um....sorry....i donÕt know....i need your advice. thank you very much for your advice. === Subject: Re: Abstract Algebra IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. Hint: If [G:H] = 3, then either H is normal in G, or H has a > subgroup of index 2 thatÕs normal in G. > IÕve never seen this before - IÕm not sure I believe it - and > I donÕt think anything like it is necessary for the problem at hand. > Well, of course itÕs not *necessary*. It just makes the problem > easy to solve. > And itÕs certainly true. The action of G on the set of cosets of > H gives a surjective homomorphism from G onto a transitive > degree-3 subgroup of S_3, whose kernel lies in H (since the > latter is the point stabilizer of the coset H in this action). > S_3 has only two transitive subgroups in its natural action, so > either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 > with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 > (with the latter 3 cosets squaring to K). But. I suspect that anyone having trouble with the original problem is not going to understand any of what youÕve written here. Knowing that a problem on the second ßoor is easy to solve from the perspective of the 12th ßoor doesnÕt help the guy who has been struggling to reach the mezzanine. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Abstract Algebra > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. [...] > And itÕs certainly true. The action of G on the set of cosets of > H gives a surjective homomorphism from G onto a transitive > degree-3 subgroup of S_3, whose kernel lies in H (since the > latter is the point stabilizer of the coset H in this action). > S_3 has only two transitive subgroups in its natural action, so > either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 > with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 > (with the latter 3 cosets squaring to K). > But. > I suspect that anyone having trouble with the original problem > is not going to understand any of what youÕve written here. > Knowing that a problem on the second ßoor is easy to solve > from the perspective of the 12th ßoor doesnÕt help the guy > who has been struggling to reach the mezzanine. Actually, IÕm having trouble thinking of a way to solve the original problem that *doesnÕt* essentially use the action of G on the cosets of H to show that G has a surjective homomorphism to C_3 or S_3 whose kernel lies in H. I notice the OP said he figured it out, so I wonder just what ßoor heÕs on. Or maybe IÕm missing some more elementary approach... -- Jim Heckman === Subject: Re: Abstract Algebra [...] IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with > 1000 elements. > If x is in G, show that either x^2 or x^3 is in H. [...] > Actually, IÕm having trouble thinking of a way to solve the > original problem that *doesnÕt* essentially use the action of G > on the cosets of H to show that G has a surjective homomorphism > to C_3 or S_3 whose kernel lies in H. I notice the OP said he > figured it out, so I wonder just what ßoor heÕs on. Or maybe > IÕm missing some more elementary approach... obvious elementary approach. DÕoh! -- Jim Heckman === Subject: Re: Abstract Algebra > Actually, IÕm having trouble thinking of a way to solve the > original problem that *doesnÕt* essentially use the action of G > on the cosets of H to show that G has a surjective homomorphism > to C_3 or S_3 whose kernel lies in H. I notice the OP said he > figured it out, so I wonder just what ßoor heÕs on. Or maybe > IÕm missing some more elementary approach... Well, I donÕt know about a method that avoids cosets entirely, but hereÕs an outline of what I came up with: Any two cosets aH and bH are either disjoint or equal. There is also a bijection between them (making them the same size if H is finite). We can see that if aH = bH, a^-1b must be in H. Then consider the 4 cosets H = x^0H, xH = x^1H, x^2H, x^3H. Since G is 3 times the size of H, the pigeonhole principle forces at least one pair of equal cosets; call them x^mH and x^nH where m < n. Since x^mH = x^nH, we have that x^(n-m) is in H. If n-m is 2 or 3, we are done. If n-m is 1, we use the fact that H is a group one last time to get the desired result. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Abstract Algebra Originator: grubb@lola > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. >Actually, IÕm having trouble thinking of a way to solve the >original problem that *doesnÕt* essentially use the action of G >on the cosets of H to show that G has a surjective homomorphism >to C_3 or S_3 whose kernel lies in H. I notice the OP said he >figured it out, so I wonder just what ßoor heÕs on. Or maybe >IÕm missing some more elementary approach... ItÕs easy to do this on a case by case basis: If x is in H, then x^2 is in H, so weÕre done. If x is not in H, then H and xH are disjoint. Now consider x^2. If x^2 is in H, weÕre done. If x^2 is in xH, then x is in H and weÕre done. If x^2 is in neither, then x^2H is disjoint from both H and xH. Furthermore, since [G:H]=3, these three cosets cover G. Now look at x^3. If x^3 is in H, weÕre done. If it is in xH, then x^2 is in H and weÕre done. If x^3 is in x^2H, then x is in H and weÕre done. --Dan Grubb === Subject: Re: abstract algebra...... >hello....doctor~ >ring homomorphism >f: Z -> Z/3Z + Z/5Z + Z/7Z >which f(x) = (x+3Z , x+5Z , x+7Z) >(+ is direct sum) >1) ker f =? >2) find all integer x such that >f(x) = (2+3Z , 3+5Z , 4+7Z) >--------------------------------------------- >i think....... >1) [3,5,7]=105, so 105*Z >right ?? Yes. Now prove it. That is, show that 105 is in the kernel, and that if x is in the kernel, then it is divisible by 105. (HINT: Use the properties of the lowest common multiple. >2) um....sorry....i donÕt know....i need your advice. Chinese Remainder Theorem. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: abstract algebra...... > ring homomorphism > f: Z -> Z/3Z + Z/5Z + Z/7Z > which f(x) = (x+3Z , x+5Z , x+7Z) > (+ is direct sum) > 1) ker f =? > 2) find all integer x such that > f(x) = (2+3Z , 3+5Z , 4+7Z) > --------------------------------------------- > i think....... > 1) [3,5,7]=105, so 105*Z > right ?? Right. 105Z is the kernel. You can make it sure by proving it. > 2) um....sorry....i donÕt know....i need your advice. You want to find a number k such that (a) k == 2 mod 3 , (b) k == 3 mod 5 , and (c) k == 4 mod 7 , and then claim that k+105Z is the answer for 2). So letÕs find out what k would be. Method I: (formula) By Chinese Remainder Theorem, there is such k with 0< k < 105. But the usual proof of Chinese Remainder Theorem provides a formula for k. Use that formula to calculate k. Method II: (heuristic) numbers satysfing (a): 2, 5, 8, 11,... ...(b): 3, 8, 13, ... ...(c): 4, 11, ... we see 11 is a number satysfing both (a) and (c). So numbers satisfying (a),(c): 11, 32, 53, ... But the number 53 satisfy (b). So k = 53. === Subject: Re: abstract algebra...... > hello....doctor~ > ring homomorphism > f: Z -> Z/3Z + Z/5Z + Z/7Z > which f(x) = (x+3Z , x+5Z , x+7Z) > (+ is direct sum) > 1) ker f =? > 2) find all integer x such that > f(x) = (2+3Z , 3+5Z , 4+7Z) > --------------------------------------------- > i think....... > 1) [3,5,7]=105, so 105*Z > right ?? > 2) um....sorry....i donÕt know....i need your advice. > thank you very much for your advice. i think....... again in the (2) x = 2 (mod 3) x = 3 (mod 5) x = 4 (mod 7) so, x = 2.35.b1 + 3.21.b2 + 4.15.b3 (mod 105) by chinese Remainder thoerem. and 35.b1 = 1 (mod 3) => b1 = 2 (mod 3) 21.b2 = 1 (mod 5) => b2 = 1 (mod 5) 15.b3 = 1 (mod 7) => b3 = 1 (mod 7) thus, x = 2.35.2 + 3.21.1 + 4.15.1 = 263 = 53 (mod 105) um....thatÕs right ?? thank you very much for your advice. === Subject: abstract algebra ....2 hello....doctor~ suppose that field K is algebraic extension field of field F. and [K:F]=10. if degree of irreducible polynomial f(x) over F is 3, show that if f(a)=0, then a not in K. -------------------------------------- i think...... suppose that there exists a such that f(a)=0. if a in K, then, since a is algebric over F, F(a) is algebraic extension of F. so, F < F(a) hello....doctor~ >suppose that >field K is algebraic extension field of field F. >and >[K:F]=10. >degree of irreducible polynomial f(x) over F is 3, >show that if f(a)=0, then a not in K. >-------------------------------------- >i think...... >suppose that >there exists a such that f(a)=0. >if a in K, >then, >since a is algebric over F, F(a) is algebraic extension of F. >so, F < F(a) so, [F(a):F] | [K:F] >but, [F(a):F] = 3 | 10 = [K:F] >so, contradiction. >thus, if f(a)=0, then a not in K. >um.......right ?? Yes. Except, you need a sentence to explain why [F(a):F]=3 and not [F(a):F]=1 or 2 (you know it is at most 3 since deg(f)=3). -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: abstract algebra ....2 > suppose that > field K is algebraic extension field of field F. > and > [K:F]=10. > if > degree of irreducible polynomial f(x) over F is 3, > show that if f(a)=0, then a not in K. > -------------------------------------- > i think...... > suppose that there exists a such that f(a)=0. > if a in K, then, > since a is algebric over F, F(a) is algebraic extension of F. This would be true even if _a_ was not an element of K. > so, F < F(a) so, [F(a):F] | [K:F] > but, [F(a):F] = 3 | 10 = [K:F] > so, contradiction. > thus, if f(a)=0, then a not in K. > um.......right ?? Right. Jose Carlos Santos === Subject: topology 22 hello....doctor~ very sorry...many questions...sorry. find the subset W,X,Y,Z of real topology R which satisfy (1),(2),(3),(4). (without proof) (1) Y,Z is connected set (2) Y in W in Z , Y in X in Z (2) W is connected set (4) X is not connected set -------------------------------------------- i find a example. X = (0,1)U[1,4) Y = (0,1) W =(0,4) Z = R is this right answer ?? thank you very much for your advice. === Subject: Re: topology 22 > find the subset W,X,Y,Z of real topology R > which satisfy (1),(2),(3),(4). > (without proof) > (1) Y,Z is connected set > (2) Y in W in Z , Y in X in Z I donÕt get this, how can a subset of R can be in a subset of R? The elements of a subset of R are numbers, arenÕt subsets of R. > (2) W is connected set > (4) X is not connected set > -------------------------------------------- > i find a example. > X = (0,1)U[1,4) > Y = (0,1) > W =(0,4) > Z = R > is this right answer ?? > thank you very much for your advice. === Subject: Re: topology 22 >>find the subset W,X,Y,Z of real topology R >>which satisfy (1),(2),(3),(4). >>(without proof) >>(1) Y,Z is connected set >>(2) Y in W in Z , Y in X in Z > I donÕt get this, how can a subset of R can be in a subset of R? > The elements of a subset of R are numbers, arenÕt subsets of R. This is a problem of the imprecise use of the word in. A person might say the number 1 is in the interval [0, 2], and the same person might say the subset [1,2] is in the same interval. The OP should probably have used a different set of words. ... the rest deleted ... Dale. === Subject: Re: topology 22 [CapitalThorn ]nd the subset W,X,Y,Z of real topology R >>which satisfy (1),(2),(3),(4). >>(without proof) >>(1) Y,Z is connected set >>(2) Y in W in Z , Y in X in Z > I donÕt get this, how can a subset of R can be in a subset of R? > The elements of a subset of R are numbers, arenÕt subsets of R. > This is a problem of the imprecise use of the word in. Sloppy and confusing. > A person might say the number 1 is in the interval [0, 2], > and the same person might say the subset [1,2] is in the > same interval. The OP should probably have used a different > set of words. How weird. Does Ōinmean ŌsubsetÕ? That doesnÕt make sense. === Subject: Re: topology 22 >>find the subset W,X,Y,Z of real topology R >>which satisfy (1),(2),(3),(4). >>(without proof) >>(1) Y,Z is connected set >>(2) Y in W in Z , Y in X in Z I donÕt get this, how can a subset of R can be in a subset of R? > The elements of a subset of R are numbers, arenÕt subsets of R. > This is a problem of the imprecise use of the word in. > Sloppy and confusing. > A person might say the number 1 is in the interval [0, 2], > and the same person might say the subset [1,2] is in the > same interval. The OP should probably have used a different > set of words. > How weird. Does Ōinmean ŌsubsetÕ? That doesnÕt make sense. original problem is find the subset W,X,Y,Z of real topology R which satisfy (1),(2),(3),(4). (without proof) (1) Y,Z is connected set (2) Y C W C Z , Y C X C Z (2) W is connected set (4) X is not connected set (symbol A C B = A is proper subset of B) um.....(0,1) C (0,2) ....is this wrong ?? thank you very much for your advice. === Subject: Re: topology 22 > original problem is > find the subset W,X,Y,Z of real topology R > which satisfy (1),(2),(3),(4). > (without proof) > (1) Y,Z is connected set > (2) Y C W C Z , Y C X C Z > (2) W is connected set > (4) X is not connected set > (symbol A C B = A is proper subset of B) > um.....(0,1) C (0,2) ....is this wrong ?? > thank you very much for your advice. And whats this real topology, do you mean euclidean topology on real numbers? ItÕs quite easy.. since X contains a connected set then you can easily use the trick of X having two connected components. Example: Y=(0,1), X=(0,1) union {2} W=(0,2), Z=(0,3) Jose Capco === Subject: Re: topology 22 find the subset W,X,Y,Z of real topology R >>which satisfy (1),(2),(3),(4). >>(1) Y,Z is connected set >>(2) Y in W in Z , Y in X in Z I donÕt get this, how can a subset of R can be in a subset of R? > The elements of a subset of R are numbers, arenÕt subsets of R. This is a problem of the imprecise use of the word in. > Sloppy and confusing. > A person might say the number 1 is in the interval [0, 2], > and the same person might say the subset [1,2] is in the > same interval. The OP should probably have used a different > set of words. How weird. Does Ōinmean ŌsubsetÕ? That doesnÕt make sense. > original problem is > find the subset W,X,Y,Z of real topology R > which satisfy (1),(2),(3),(4). > (1) Y,Z is connected set > (2) Y C W C Z , Y C X C Z > (2) W is connected set > (4) X is not connected set > (symbol A C B = A is proper subset of B) > um.....(0,1) C (0,2) ....is this wrong ?? Howver subset and proper subset symbols donÕt do well in ascii. A C E for example is a word. Some use < and <=, which is very context sensitive and when talking both subset and less than in same statement as for example a < b ==> (b,r) < (a,r) could be bothersome. TeX uses subseteq and subset However for ascii, subset and subsetneq would be clearer for indeed proper subset means subset and /=. I use subset and proper subset as need for proper subset is infrequent. Ōinas Ōelement of is clear. For a sequence, (x_n)_n within S can be more elucidating and precise, for (x_n)_n isnÕt a subset of S but only itÕs values, and again (x_n)_n not in S for itÕs not a element of S. === Subject: Re: topology 22 X = (0,1)U[1,4) = (0,4). This set X is connected. Redefine it as X = (0,1)U(2,4). === Subject: Re: mathematics > and shut the hell up. > How sad for you that you are a nobody, and patty is > a respected community member, so your advice will be > considered by its source, and laughed out of town. > ur an idiot. > On the behavioral test, that would be you. > This isnÕt a chat line, and ur is a word, > but not one youÕve used correctly. > and cudiment isnÕt a word. > You opinion, as a nobody, with a worthless > reputation, is rejected, you need to supply > proof of your claim. > You intend to prove that claim then, exactly > how? > xanthian. i canÕt tell you how horrible it is for me that i donÕt have a great reputation on sci.math. itÕs the first thing i think about when i wake up and the last thing i think about before i go to bed. what ever shall i do? i envy the people (with shitty reputations like mine) who just log on every once in awhile to see what james harris it up to.. and maybe post some deliberately inßammatory nonsense of their own for fun. i envy them because they donÕt really give a shit about ur lame opinion.. while i suffer from caring ever so much. === Subject: Re: mathematics http://mygate.mailgate.org/mynews/comp/comp.theory/ a166513b2ded98b0be5b3c2b2f 56200e.48257%40mygate.mailgate.org > i envy them because they donÕt really give a shit about ur > lame opinion.. while i suffer from caring ever so much. How sad for you that you think you are capable of insulting a retired sailor. xanthian. -- === Subject: Re: mathematics > i envy them because they donÕt really give a shit about ur > lame opinion.. while i suffer from caring ever so much. > How sad for you that you think you are capable of > insulting a retired sailor. > xanthian. looks like youÕre all washed up. judging from your lack of anything interesting to retort with. === Subject: Re: mathematics http://mygate.mailgate.org/mynews/comp/comp.theory/ abb9e287e5ecb35d69575ec132 94de42.48257%40mygate.mailgate.org >> Do you have any _idea_ what the update cycle is >> for the ODE? > Now that the OED is online, there are quarterly updates. > http://dictionary.oed.com/help/updates Well, yes, but update _frequency_ doesnÕt say much to the update _cycle_. It is perfectly possible to be doing updates moment by moment using information that has required 40 years of background processing to validate, for example. Also, it might be a ground rule that N (for substantial N) uses of a word must be in their data files before the word gets added to their dictionary. Last I read, their update _cycle_ still spanned decades, but then I donÕt have insider information to know how much this has changed since they computerized, or, more important for the current discussion, how much those changes impact whether the ODE can be trusted today as a reference for the legitimacy of a neologism. My gut guess, though, remains not much. Dictionaries are far too much documentations of the past than of the present to be trusted for guidance on current usage. xanthian. -- === Subject: Re: mathematics <41ad53a9$16$fuzhry+tra$mr2ice@news.patriot.net> at 02:58 PM, examachine@gmail.com (Eray Ozkural exa) said: >Once again youÕre gibbering. WeÕre talking theory, Turing machines are theory, tonto; computers are engineering. You seem to not know the difference. with delusions of adequacy. I learned that you are a hypocrite. >Ignorant twit. Indeed, you are. *PLONK* -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: FLTMA: S.1 = c One more time c-a < b, c-b < a and then making certain assumptions that just sort of come to mind: S.1 = (a mod c + b mod c) mod c + (c mod b - a mod b) mod b + (c mod a - b mod a) mod a = ( a + b ) mod c + (c-b - a) mod b + (c-a - (b-a)) mod a = a+b - c + c - b - a - b + c - a + b + a - a = a + b - c + c - b - a - b + c - a + b + a - a = - c + c + c + b - b + b + a - a - a + a - a = c + b - a Which is closer Still some consideration of the underlying inequalities is necessary. a could be < c/2 or < c/3, etc.... I need to see how the constraint S.n = 0 affects S.1 = c. I tolerance everything and tolerate everyone. I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. I drive: A double-step Thunderbolt with 657% range. I fight terrorism by: Using less gasoline. === Subject: Re: Question about an alternate abiotic origin of Martian blueberries > Andrew Diseker skrev i en meddelelse Sorry about the title, I didnÕt want to be ignored as a crackpot > before someone read my post. Afterwards, you can call me insane! :-) At any rate, IÕve been checking the Opportunity site daily looking > at the raws, and something struck me about the blueberries in so > many images. I know the accepted wisdom is that they were all formed > by ßowing water leaching out hematite and forming the spheres, but > what struck me is that they looked a lot like hailstones after a > storm, scattered all over, especially sitting on top of the rocks. > Makes me think of a 550 mill year old fossil droplet-mark in a shale I once saw. The > point is, that > the blueberries as concretions are characterized by growing in the subsurface > without leaving any > kind of scour-marks or shape up any wind/water-lee structures in their vicinity. > They would also > aggregate in sorted layers as Aidan points out, if left in a streaming environment. > The blueberries > that has not been erodet out all show the above characteristics. The odd-looking > surface of Mars > probably stems from the thin atmosphere - it has no mechanical erosive power > compared to Earth, but > what it has may have acted over a very long time. > ItÕs important to keep in mind the mineralogy and environmental > context at work. Do that. IÕve commented your misconseptions in that consern elsewhere. > The nature of coarse-grained crystalline hematite and its implications > for the early environment of Mars > Abstract > The Thermal Emission Spectrometer (TES) on the Mars Global Surveyor spacecraft > has detected deposits of coarse-grained, gray crys-talline hematite in Sinus > Meridiani, Aram Chaos, and Vallis Marineris. We argue that the key to the origin > of gray hematite is that it requires crystallization at temperatures in excess of > about 100?C. We discuss thermal crystallization (1) as diagenesis at a depth > of a few kilometers of sediments originally formed in low-temperature waters, > or (2) as precipitation from hydrothermal solution. Thus not biologically mediated? YouÕr improving your standpoints. To quote from the paper: In this case, merely to view Mars with the naked eye and to see its color would be to detect signs of ancient life, which would be a curious conclusion from our elaborate space exploration efforts. The paper may have a good point as to the genesis of hematite in Aram Chaos, but the environment in the Meridiani has later turned out to be sedimentary (note the following quote from the paper:) A close-up view of the gray hematite crystals on Mars (e.g., with a microscopic imager or other camera) will help distinguish petrological features such as bedding, veining, or matrix structure. Veins, fracture fillings, and clast or wall coatings provide common evidence of ßuid motion. We expect such features on Mars if the origin of the gray hematite is hydrothermal. > Then take a look at this wonderful pic > The ripples are clearly formed by water > since they are aligned > in many different directions and follow the path water > would take. > So the context should be that warm mineral-rich water ßows out from > underground periodically. > Which is an ideal....let me say that > again....IDEAL context for life. Combined with these mysterious > spheres that have obviously Ōgrownin place, and hereÕs the > big clue... are little copies of each other. > Geology doesnÕt reproduce itself so nicely, neatly and > repetitively. Maybe in a small area or two it will, but > Meridiani covers an area several times larger than > all the sedimentary rock exposures of northern Arizona, > New Mexico, and eastern Utah, > Everywhere that has dark soil at Meridiani has those spheres > as they make the soil dark. There is dark soil all over Mars > and not coincidentally, always on the ßoor of canyons > and craters or other places water would stand. > http://marsprogram.jpl.nasa.gov/gallery/canyons/PIA02398.html > http://marsprogram.jpl.nasa.gov/gallery/sanddunes/PIA01695. html > http://www.msss.com/mars_images/moc/5_27_98_agu_release/ 7707rel.gif IÕll crosspost this to sci.math. IÕm sure that your collegues would like to know. > And we have another mystery, all those fine laminations > in the sedimentary rock. Countless laminations that are > perfectly uniform in thickness, denoting a highly repetitive > process such as tides, seasons etc. > http://marsrovers.jpl.nasa.gov/gallery/all/1/p/280/ 1P153040668EFF37LJP2443L2M 1.HTML > Laboratory cultures of calcifying biomicrospheres generate ooids - > A contribution to the origin of oolites > Abstract > The in vitro production of ooid-like structures as possible precursors of > oolites has been observed in laboratory cultures of spherical microbial > communities isolated from the Wadden Sea (North Sea). > Introduction > Precipitation of calcium carbonate is widespread in microbial communities > forming biofilms and microbial mats. The laminated structure of these communities > consists of layers of carbonate which outlast the microbial colony that produced > sediment are also included are known as stromatolites. They have a long fossil > record since early Proterozoic and still ßourish in particular in the reefs of > the Bahamas and Australia (e.g. Visscher et alii, 2002). > ......... > HereÕs the Bahamas guys > http://www.theßyingcircus.com/stella_maris.html > HereÕs Mars > http://marsrovers.jpl.nasa.gov/gallery/all/1/n/119/ 1N138744629EFF2809P1987R0M 1.JPG > ............ > The typical stromatolitic structure is laminated. Each lamina represents a horizon > of former microbial biofilm or mat (Kalkowsky, 1908). Associated mineral > coated by microbial assemblages (Riding and Awramik, 2000). Small (mm size), > spherical to oval concentrically laminated carbonate bodies or aggregates, which > form in shallow tropical seas are called ooids and are known to become > consolidated into rocks called oolitic limestones (oolith, Rogenstein; Kalkowsky, > 1908). The genesis of ooid grains is still enigmatic. The alternative explanations > are confronted along the lines of predominantly abiotic vs. biogenic origin of > ooid grains and the associated carbonate precipitates. The Ōperfectly uniform laminayou refer to carry structural evidence of a ßuvial environment. If it was laminated due to bacterial film or stromatolites, we would have told you. > Ōfits nicelywith earth analogs. What do you know about that? > perspective...everything...at Meridianis fits nicely. Ok, weÕll put sci.bio.misc on the line too > Why do we deny the obvious, and not just let the evidence > take us where it will? sigh === Subject: Re: Question about an alternate abiotic origin of Martian blueberries ItÕs important to keep in mind the mineralogy and environmental > context at work. > Do that. IÕve commented your misconseptions in that consern elsewhere. > The nature of coarse-grained crystalline hematite and its implications > for the early environment of Mars > Abstract > The Thermal Emission Spectrometer (TES) on the Mars Global Surveyor spacecraft > has detected deposits of coarse-grained, gray crys-talline hematite in Sinus > Meridiani, Aram Chaos, and Vallis Marineris. We argue that the key to the origin > of gray hematite is that it requires crystallization at temperatures in excess of > about 100?C. We discuss thermal crystallization (1) as diagenesis at a depth > of a few kilometers of sediments originally formed in low-temperature waters, > or (2) as precipitation from hydrothermal solution. > Thus not biologically mediated? > YouÕr improving your standpoints. To quote NasaÕs leading astrobiologist Indeed, given the intense impact and volcanic activity that characterized the planet at this time, the development of long-lived hydrothermal systems was likely widespread- duplicating many of the important conditions that are thought to have given rise to life on Earth (Farmer 1996). http://geology.asu.edu/jfarmer/pubs/pdfs/marspolarsci.pdf Which means the context is one ideal for development of the simplest type of life on earth. Which is thought to be bacteria. And anaerobic bacteria tends to leave behind lots of iron and sulfates. Also they tend to form laminated ...sedimentary...structures full of spherical shapes. The context is clearly habitable for life, that is no longer in dispute. > To quote from the paper: > In this case, merely to view Mars with the > naked eye and to see its color would be to detect signs of > ancient life, which would be a curious conclusion from our > elaborate space exploration efforts. This implies the naked eye is incapable of seeing signs of life. That is absurd, the statement limits the Ōeyeto color only, a very limited subset of visible evidence. The eye can juggle countless different aspects at once, and if enough are consistent with life why not say so? It is they that need the lawyer-life level of proof for their awards and grants... ....we do not. > The paper may have a good point as to the genesis of hematite in Aram Chaos, but the environment >in the Meridiani has later turned out to be sedimentary (note the following quote >from the paper:) The quote says nothing except for future expectations. There is nothing inconsistent about finely laminated sedimentary rocks and stromatolites. Nothing at all, in fact the two are often confused for each other. Combined with the warm mineral rich water of hydrothermal systems and the minerals found there, the context is clearly favorable for biology. > A close-up view of the gray hematite crystals on Mars > (e.g., with a microscopic imager or other camera) will > help distinguish petrological features such as bedding, > veining, or matrix structure. Veins, fracture fillings, and > clast or wall coatings provide common evidence of ßuid > motion. We expect such features on Mars if the origin > of the gray hematite is hydrothermal. > Then take a look at this wonderful pic > The ripples are clearly formed by water > since they are aligned > in many different directions and follow the path water > would take. > So the context should be that warm mineral-rich water ßows out from > underground periodically. > Which is an ideal....let me say that > again....IDEAL context for life. Combined with these mysterious > spheres that have obviously Ōgrownin place, and hereÕs the > big clue... are little copies of each other. > Geology doesnÕt reproduce itself so nicely, neatly and > repetitively. Maybe in a small area or two it will, but > Meridiani covers an area several times larger than > all the sedimentary rock exposures of northern Arizona, > New Mexico, and eastern Utah, > Everywhere that has dark soil at Meridiani has those spheres > as they make the soil dark. There is dark soil all over Mars > and not coincidentally, always on the ßoor of canyons > and craters or other places water would stand. > http://marsprogram.jpl.nasa.gov/gallery/canyons/PIA02398.html > http://marsprogram.jpl.nasa.gov/gallery/sanddunes/PIA01695. html > http://www.msss.com/mars_images/moc/5_27_98_agu_release/ 7707rel.gif > IÕll crosspost this to sci.math. IÕm sure that your collegues would like to know. > And we have another mystery, all those fine laminations > in the sedimentary rock. Countless laminations that are > perfectly uniform in thickness, denoting a highly repetitive > process such as tides, seasons etc. http://marsrovers.jpl.nasa.gov/gallery/all/1/p/280/ 1P153040668EFF37LJP2443L2 M1.HTML > Laboratory cultures of calcifying biomicrospheres generate ooids - > A contribution to the origin of oolites > Abstract > The in vitro production of ooid-like structures as possible precursors of > oolites has been observed in laboratory cultures of spherical microbial > communities isolated from the Wadden Sea (North Sea). > Introduction > Precipitation of calcium carbonate is widespread in microbial communities > forming biofilms and microbial mats. The laminated structure of these communities > consists of layers of carbonate which outlast the microbial colony that produced > sediment are also included are known as stromatolites. They have a long fossil > record since early Proterozoic and still ßourish in particular in the reefs of > the Bahamas and Australia (e.g. Visscher et alii, 2002). > ......... > HereÕs the Bahamas guys > http://www.theßyingcircus.com/stella_maris.html > HereÕs Mars > http://marsrovers.jpl.nasa.gov/gallery/all/1/n/119/ 1N138744629EFF2809P1987R0M 1.JPG > ............ > The typical stromatolitic structure is laminated. Each lamina represents a horizon > of former microbial biofilm or mat (Kalkowsky, 1908). Associated mineral > coated by microbial assemblages (Riding and Awramik, 2000). Small (mm size), > spherical to oval concentrically laminated carbonate bodies or aggregates, which > form in shallow tropical seas are called ooids and are known to become > consolidated into rocks called oolitic limestones (oolith, Rogenstein; Kalkowsky, > 1908). The genesis of ooid grains is still enigmatic. The alternative explanations > are confronted along the lines of predominantly abiotic vs. biogenic origin of > ooid grains and the associated carbonate precipitates. > The Ōperfectly uniform laminayou refer to carry structural evidence of a ßuvial environment. If > it was laminated due to bacterial film or stromatolites, we would have told you. The sedimentary rock extends across the plains of Meridiani to the horizon. This was a sea, not a river. In any event that does not at all contradict the notion of microbial mats. I donÕt understand why anything youÕve said refutes my post in the slightest. > Ōfits nicelywith earth analogs. > What do you know about that? Well. the geologists have yet to identify a single rock at Meridiani...not a one. Saying theyÕre sedimentary is to state the obvious and generic. The Nasa geologists embarrassed themselves so badly a peer review board was rushed in to save the day...which they did not. And now theyÕve been replaced by astrobiologists such as Dr Farmer. ItÕs obvious that geology hasnÕt succeeded at all in explaining Meridiani... in fact the attempt was a fiasco for Nasa. Why do you think Nasa suddenly stopped giving science interpretations a few months back??? Because the geologists were embarrassing themselves and Nasa. > perspective...everything...at Meridianis fits nicely. > Ok, weÕll put sci.bio.misc on the line too Put NasaÕs very last update there instead............. Conditions On Vast Plain on Mars Could Have Been Suitable For Life http://www.news.cornell.edu/releases/Dec04/ Science.Mars.deb.html Conditions on vast plain on Mars could have been suitable for life, Cornell rover scientist Squyres states in special Science issue Jonathan s > Why do we deny the obvious, and not just let the evidence > take us where it will? > sigh === Subject: Re: Question about an alternate abiotic origin of Martian blueberries Jonathan, make another dash for your freedom of speach, but donÕt use ŌweÕ anywhere, asshole! ItÕs important to keep in mind the mineralogy and environmental > context at work. > Do that. IÕve commented your misconseptions in that consern elsewhere. > The nature of coarse-grained crystalline hematite and its implications > for the early environment of Mars Abstract The Thermal Emission Spectrometer (TES) on the Mars Global Surveyor spacecraft > has detected deposits of coarse-grained, gray crys-talline hematite in Sinus > Meridiani, Aram Chaos, and Vallis Marineris. We argue that the key to the origin > of gray hematite is that it requires crystallization at temperatures in excess of > about 100?C. We discuss thermal crystallization (1) as diagenesis at a depth > of a few kilometers of sediments originally formed in low-temperature waters, > or (2) as precipitation from hydrothermal solution. > Thus not biologically mediated? > YouÕr improving your standpoints. > To quote NasaÕs leading astrobiologist > Indeed, given the intense impact and volcanic > activity that characterized the planet at this time, the development > of long-lived hydrothermal systems was likely widespread- > duplicating many of the important conditions that are thought > to have given rise to life on Earth (Farmer 1996). > http://geology.asu.edu/jfarmer/pubs/pdfs/marspolarsci.pdf > Which means the context is one ideal for development > of the simplest type of life on earth. Which is thought > to be bacteria. And anaerobic bacteria tends to leave > behind lots of iron and sulfates. Also they tend to > form laminated ...sedimentary...structures full of > spherical shapes. > The context is clearly habitable for life, that is > no longer in dispute. > To quote from the paper: > In this case, merely to view Mars with the > naked eye and to see its color would be to detect signs of > ancient life, which would be a curious conclusion from our > elaborate space exploration efforts. > This implies the naked eye is incapable of seeing signs > of life. That is absurd, the statement limits the Ōeyeto > color only, a very limited subset of visible evidence. > The eye can juggle countless different aspects at > once, and if enough are consistent with life why > not say so? It is they that need the lawyer-life > level of proof for their awards and grants... > ....we do not. > The paper may have a good point as to the genesis of hematite in Aram Chaos, but the > environment >in the Meridiani has later turned out to be sedimentary (note the > following quote >from the paper:) > The quote says nothing except for future expectations. > There is nothing inconsistent about finely laminated > sedimentary rocks and stromatolites. Nothing > at all, in fact the two are often confused for > each other. Combined with the warm mineral > rich water of hydrothermal systems and the minerals > found there, the context is clearly favorable > for biology. > A close-up view of the gray hematite crystals on Mars > (e.g., with a microscopic imager or other camera) will > help distinguish petrological features such as bedding, > veining, or matrix structure. Veins, fracture fillings, and > clast or wall coatings provide common evidence of ßuid > motion. We expect such features on Mars if the origin > of the gray hematite is hydrothermal. > Then take a look at this wonderful pic > The ripples are clearly formed by water > since they are aligned > in many different directions and follow the path water > would take. > So the context should be that warm mineral-rich water ßows out from > underground periodically. > Which is an ideal....let me say that > again....IDEAL context for life. Combined with these mysterious > spheres that have obviously Ōgrownin place, and hereÕs the > big clue... are little copies of each other. > Geology doesnÕt reproduce itself so nicely, neatly and > repetitively. Maybe in a small area or two it will, but > Meridiani covers an area several times larger than > all the sedimentary rock exposures of northern Arizona, > New Mexico, and eastern Utah, Everywhere that has dark soil at Meridiani has those spheres > as they make the soil dark. There is dark soil all over Mars > and not coincidentally, always on the ßoor of canyons > and craters or other places water would stand. http://marsprogram.jpl.nasa.gov/gallery/canyons/PIA02398.html > http://marsprogram.jpl.nasa.gov/gallery/sanddunes/PIA01695. html > http://www.msss.com/mars_images/moc/5_27_98_agu_release/ 7707rel.gif > IÕll crosspost this to sci.math. IÕm sure that your collegues would like to know. > And we have another mystery, all those fine laminations > in the sedimentary rock. Countless laminations that are > perfectly uniform in thickness, denoting a highly repetitive > process such as tides, seasons etc. http://marsrovers.jpl.nasa.gov/gallery/all/1/p/280/ 1P153040668EFF37LJP2443L2M 1.HTML > Laboratory cultures of calcifying biomicrospheres generate ooids - > A contribution to the origin of oolites Abstract The in vitro production of ooid-like structures as possible precursors of > oolites has been observed in laboratory cultures of spherical microbial > communities isolated from the Wadden Sea (North Sea). Introduction Precipitation of calcium carbonate is widespread in microbial communities > forming biofilms and microbial mats. The laminated structure of these communities > consists of layers of carbonate which outlast the microbial colony that produced > sediment are also included are known as stromatolites. They have a long fossil > record since early Proterozoic and still ßourish in particular in the reefs of > the Bahamas and Australia (e.g. Visscher et alii, 2002). ......... > HereÕs the Bahamas guys > http://www.theßyingcircus.com/stella_maris.html > HereÕs Mars > http://marsrovers.jpl.nasa.gov/gallery/all/1/n/119/ 1N138744629EFF2809P1987R0M 1.JPG > ............ The typical stromatolitic structure is laminated. Each lamina represents a > horizon > of former microbial biofilm or mat (Kalkowsky, 1908). Associated mineral > coated by microbial assemblages (Riding and Awramik, 2000). Small (mm size), > spherical to oval concentrically laminated carbonate bodies or aggregates, which > form in shallow tropical seas are called ooids and are known to become > consolidated into rocks called oolitic limestones (oolith, Rogenstein; Kalkowsky, > 1908). The genesis of ooid grains is still enigmatic. The alternative explanations > are confronted along the lines of predominantly abiotic vs. biogenic origin of > ooid grains and the associated carbonate precipitates. > The Ōperfectly uniform laminayou refer to carry structural evidence of a ßuvial > environment. If > it was laminated due to bacterial film or stromatolites, we would have told you. > The sedimentary rock extends across the plains of Meridiani > to the horizon. This was a sea, not a river. In any event that > does not at all contradict the notion of microbial mats. > I donÕt understand why anything youÕve said refutes my > post in the slightest. > Ōfits nicelywith earth analogs. > What do you know about that? > Well. the geologists have yet to identify a single > rock at Meridiani...not a one. Saying theyÕre > sedimentary is to state the obvious and generic. > The Nasa geologists embarrassed themselves so > badly a peer review board was rushed in to save > the day...which they did not. And now theyÕve > been replaced by astrobiologists such as Dr Farmer. > ItÕs obvious that geology hasnÕt succeeded at all > in explaining Meridiani... in fact the attempt was > a fiasco for Nasa. Why do you think Nasa suddenly > stopped giving science interpretations a few months > back??? Because the geologists were embarrassing > themselves and Nasa. > perspective...everything...at Meridianis fits nicely. > Ok, weÕll put sci.bio.misc on the line too > Put NasaÕs very last update there instead............. > Conditions On Vast Plain on Mars Could Have Been Suitable For Life > http://www.news.cornell.edu/releases/Dec04/ Science.Mars.deb.html > Conditions on vast plain on Mars could have been suitable for life, > Cornell rover scientist Squyres states in special Science issue > Jonathan > Why do we deny the obvious, and not just let the evidence > take us where it will? > sigh === Subject: Re: Question about an alternate abiotic origin of Martian blueberries > Jonathan, make another dash for your freedom of speach, but donÕt use ŌweÕ anywhere, asshole! This is when itÕs clear IÕve won the debate! The opponent leaves frustrated, angry and acting like a fourth grader. Jonathan s jonathan skrev i en meddelelse ItÕs important to keep in mind the mineralogy and environmental > context at work. Do that. IÕve commented your misconseptions in that consern elsewhere. The nature of coarse-grained crystalline hematite and its implications > for the early environment of Mars Abstract The Thermal Emission Spectrometer (TES) on the Mars Global Surveyor spacecraft > has detected deposits of coarse-grained, gray crys-talline hematite in Sinus > Meridiani, Aram Chaos, and Vallis Marineris. We argue that the key to the origin > of gray hematite is that it requires crystallization at temperatures in excess of > about 100?C. We discuss thermal crystallization (1) as diagenesis at a depth > of a few kilometers of sediments originally formed in low-temperature waters, > or (2) as precipitation from hydrothermal solution. > Thus not biologically mediated? > YouÕr improving your standpoints. > To quote NasaÕs leading astrobiologist > Indeed, given the intense impact and volcanic > activity that characterized the planet at this time, the development > of long-lived hydrothermal systems was likely widespread- > duplicating many of the important conditions that are thought > to have given rise to life on Earth (Farmer 1996). > http://geology.asu.edu/jfarmer/pubs/pdfs/marspolarsci.pdf > Which means the context is one ideal for development > of the simplest type of life on earth. Which is thought > to be bacteria. And anaerobic bacteria tends to leave > behind lots of iron and sulfates. Also they tend to > form laminated ...sedimentary...structures full of > spherical shapes. > The context is clearly habitable for life, that is > no longer in dispute. To quote from the paper: In this case, merely to view Mars with the > naked eye and to see its color would be to detect signs of > ancient life, which would be a curious conclusion from our > elaborate space exploration efforts. > This implies the naked eye is incapable of seeing signs > of life. That is absurd, the statement limits the Ōeyeto > color only, a very limited subset of visible evidence. > The eye can juggle countless different aspects at > once, and if enough are consistent with life why > not say so? It is they that need the lawyer-life > level of proof for their awards and grants... > ....we do not. The paper may have a good point as to the genesis of hematite in Aram Chaos, but the > environment >in the Meridiani has later turned out to be sedimentary (note the > following quote >from the paper:) > The quote says nothing except for future expectations. > There is nothing inconsistent about finely laminated > sedimentary rocks and stromatolites. Nothing > at all, in fact the two are often confused for > each other. Combined with the warm mineral > rich water of hydrothermal systems and the minerals > found there, the context is clearly favorable > for biology. A close-up view of the gray hematite crystals on Mars > (e.g., with a microscopic imager or other camera) will > help distinguish petrological features such as bedding, > veining, or matrix structure. Veins, fracture fillings, and > clast or wall coatings provide common evidence of ßuid > motion. We expect such features on Mars if the origin > of the gray hematite is hydrothermal. Then take a look at this wonderful pic > The ripples are clearly formed by water since they are aligned > in many different directions and follow the path water > would take. So the context should be that warm mineral-rich water ßows out from > underground periodically. > Which is an ideal....let me say that > again....IDEAL context for life. Combined with these mysterious > spheres that have obviously Ōgrownin place, and hereÕs the > big clue... are little copies of each other. Geology doesnÕt reproduce itself so nicely, neatly and > repetitively. Maybe in a small area or two it will, but > Meridiani covers an area several times larger than > all the sedimentary rock exposures of northern Arizona, > New Mexico, and eastern Utah, Everywhere that has dark soil at Meridiani has those spheres > as they make the soil dark. There is dark soil all over Mars > and not coincidentally, always on the ßoor of canyons > and craters or other places water would stand. http://marsprogram.jpl.nasa.gov/gallery/canyons/PIA02398.html > http://marsprogram.jpl.nasa.gov/gallery/sanddunes/PIA01695. html > http://www.msss.com/mars_images/moc/5_27_98_agu_release/ 7707rel.gif IÕll crosspost this to sci.math. IÕm sure that your collegues would like to know. And we have another mystery, all those fine laminations > in the sedimentary rock. Countless laminations that are > perfectly uniform in thickness, denoting a highly repetitive > process such as tides, seasons etc. > http://marsrovers.jpl.nasa.gov/gallery/all/1/p/280/ 1P153040668EFF37LJP2443L2 M1.HTML > Laboratory cultures of calcifying biomicrospheres generate ooids - > A contribution to the origin of oolites Abstract The in vitro production of ooid-like structures as possible precursors of > oolites has been observed in laboratory cultures of spherical microbial > communities isolated from the Wadden Sea (North Sea). Introduction Precipitation of calcium carbonate is widespread in microbial communities > forming biofilms and microbial mats. The laminated structure of these communities > consists of layers of carbonate which outlast the microbial colony that produced > sediment are also included are known as stromatolites. They have a long fossil > record since early Proterozoic and still ßourish in particular in the reefs of > the Bahamas and Australia (e.g. Visscher et alii, 2002). ......... > HereÕs the Bahamas guys > http://www.theßyingcircus.com/stella_maris.html > HereÕs Mars > http://marsrovers.jpl.nasa.gov/gallery/all/1/n/119/ 1N138744629EFF2809P1987R0 M1.JPG > ............ The typical stromatolitic structure is laminated. Each lamina represents a > horizon > of former microbial biofilm or mat (Kalkowsky, 1908). Associated mineral entirely > coated by microbial assemblages (Riding and Awramik, 2000). Small (mm size), > spherical to oval concentrically laminated carbonate bodies or aggregates, which > form in shallow tropical seas are called ooids and are known to become > consolidated into rocks called oolitic limestones (oolith, Rogenstein; Kalkowsky, > 1908). The genesis of ooid grains is still enigmatic. The alternative explanations > are confronted along the lines of predominantly abiotic vs. biogenic origin of > ooid grains and the associated carbonate precipitates. The Ōperfectly uniform laminayou refer to carry structural evidence of a ßuvial > environment. If > it was laminated due to bacterial film or stromatolites, we would have told you. > The sedimentary rock extends across the plains of Meridiani > to the horizon. This was a sea, not a river. In any event that > does not at all contradict the notion of microbial mats. > I donÕt understand why anything youÕve said refutes my > post in the slightest. Ōfits nicelywith earth analogs. What do you know about that? > Well. the geologists have yet to identify a single > rock at Meridiani...not a one. Saying theyÕre > sedimentary is to state the obvious and generic. > The Nasa geologists embarrassed themselves so > badly a peer review board was rushed in to save > the day...which they did not. And now theyÕve > been replaced by astrobiologists such as Dr Farmer. > ItÕs obvious that geology hasnÕt succeeded at all > in explaining Meridiani... in fact the attempt was > a fiasco for Nasa. Why do you think Nasa suddenly > stopped giving science interpretations a few months > back??? Because the geologists were embarrassing > themselves and Nasa. perspective...everything...at Meridianis fits nicely. Ok, weÕll put sci.bio.misc on the line too > Put NasaÕs very last update there instead............. > Conditions On Vast Plain on Mars Could Have Been Suitable For Life > http://www.news.cornell.edu/releases/Dec04/ Science.Mars.deb.html > Conditions on vast plain on Mars could have been suitable for life, > Cornell rover scientist Squyres states in special Science issue > Jonathan > s Why do we deny the obvious, and not just let the evidence > take us where it will? sigh === Subject: Re: Question about an alternate abiotic origin of Martian blueberries > Jonathan, make another dash for your freedom of speach, but donÕt use Ōweanywhere, > asshole! > This is when itÕs clear IÕve won the debate! When has catatonic posting added anything to a debate The opponent > leaves frustrated, angry and acting like a fourth grader. How will you describe your appearance in this group for the passed half a year, Asshole! === Subject: Re: Question about an alternate abiotic origin of Martian blueberries Jonathan, make another dash for your freedom of speach, but donÕt use ŌweÕ anywhere, > asshole! > This is when itÕs clear IÕve won the debate! > When has catatonic posting added anything to a debate > The opponent > leaves frustrated, angry and acting like a fourth grader. > How will you describe your appearance in this group for the passed half a year, Asshole! See what I mean! s === Subject: Lines, points and planes Aah, too tired to explain too much. I have answered the following question but fear I am very incorrect. Q) Find the equation of the plane that passes through the origin and is perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) A) Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real number. v = 3i + j - k Q1 = (0, 0, 0) a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 3(x [CapitalEth] 0) + 1(y [CapitalEth] 0) [CapitalEth] 1(z [CapitalEth] 0) = 0 3x + y [CapitalEth] z = 0 Therefore the equation of the plane is 3x + y [CapitalEth] z = 0. I was quite confident that it was correct. Except now I am attempting a question in which we are given a point and a plane, and asked to find the equation of the line perpendicular to the plane. I thought I would just work backwards using the above question, but I feel like my answers are all wrong. Any suggestions? Cassandra. === Subject: Re: Lines, points and planes > Aah, too tired to explain too much. I have answered the following > question but fear I am very incorrect. > Q) Find the equation of the plane that passes through the origin and is > perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) > A) > Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real > number. > v = 3i + j - k > Q1 = (0, 0, 0) > a(x ? x0) + b(y ? y0) + c(z ? z0) = 0 > 3(x ? 0) + 1(y ? 0) ? 1(z ? 0) = 0 > 3x + y ? z = 0 > Therefore the equation of the plane is 3x + y ? z = 0. > I was quite confident that it was correct. Except ... Omigod. Google turned some of your minus signs into ?s when I went into preview mode. Why, I donÕt know. (You didnÕt use some weird character like a dash instead of an ASCII hyphen did you?) Apart from the Google trauma, your answer looks fine to me. If you are unsure whether your answers are right, itÕs a great idea to check them with a numerical example (if you havenÕt already done so). In this example I would check the answer as follows... 1. Find the intersection of the line and the plane - call it point A. 2. Pick any point on the line (other than the point of intersection) - call it point B. 3. Pick any point on the plane (other than the point of intersection) - call it point C. Check that ABC is a right-angled triangle with right angle at A (use Pythagoras). If it is, then that gives you great confidence that you got the answer right. You have to be REALLY unlucky for the check to work but the answer to still be wrong. > ... now I am attempting a > question in which we are given a point and a plane, and asked to find > the equation of the line perpendicular to the plane. I thought I would > just work backwards using the above question, but I feel like my answers > are all wrong. > Any suggestions? > Cassandra. The equation of the plane will immediately give you the normal vector (i.e. the vector of the required perpendicular line). Specifically, if the equation of the plane is given in Cartesian form as ax + by + cz + d = 0, then the normal vector is (a,b,c) (or any non-zero scalar multiple thereof). But you knew that, right? Now that you have the line vector and the coordinates of one point on the line, it is straightforward to find the (parametric) equation of too, so I think you should be OK... You can check your answer using a similar right-angled triangle construction. === Subject: Re: Lines, points and planes >>Aah, too tired to explain too much. I have answered the following >>question but fear I am very incorrect. >>Q) Find the equation of the plane that passes through the origin and is >>perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) >>A) >>Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real >>number. >>v = 3i + j - k >>Q1 = (0, 0, 0) >>a(x ? x0) + b(y ? y0) + c(z ? z0) = 0 >>3(x ? 0) + 1(y ? 0) ? 1(z ? 0) = 0 >>3x + y ? z = 0 >>Therefore the equation of the plane is 3x + y ? z = 0. >>I was quite confident that it was correct. Except ... > Omigod. Google turned some of your minus signs into ?s when I went > into preview mode. Why, I donÕt know. (You didnÕt use some weird > character like a dash instead of an ASCII hyphen did you?) I am typing my assignment in MS Word, I copied the above workings straight from word. Yes, it would have been dashes not minus signs as I notice that MS Word does alot of auto-correcting with regard to minus-signs/dashes. Sorry about that. Cassandra. === Subject: Re: Lines, points and planes > Aah, too tired to explain too much. I have answered the following > question but fear I am very incorrect. > Q) Find the equation of the plane that passes through the origin and is > perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) > A) > Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real > number. > v = 3i + j - k > Q1 = (0, 0, 0) > a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 > 3(x [CapitalEth] 0) + 1(y [CapitalEth] 0) [CapitalEth] 1(z [CapitalEth] 0) = 0 > 3x + y [CapitalEth] z = 0 > Therefore the equation of the plane is 3x + y [CapitalEth] z = 0. Starting from the end... The normal vector of the plane Ax + By + Cz + D = 0 is (A, B, C), so the normal of your plane is (3, 1, -1). The direction vector of the line you were given is also (3, 1, -1), so the line is indeed perpendicular to the plane. 3 * 0 + 0 - 0 = 0, so your plane contains the origin. Thus, the plane you got is the right answer. > I was quite confident that it was correct. Except now I am attempting a > question in which we are given a point and a plane, and asked to find > the equation of the line perpendicular to the plane. I thought I would > just work backwards using the above question, but I feel like my answers > are all wrong. > Any suggestions? If you have point P = (Xp, Yp, Zp) and plane Ax + By + Cz + D = 0, then the normal to the plane is n = (A, B, C) and the line through P in the direction of n is simply l = (Xp, Yp, Zp) + t * (A, B, C). Based on the knowledge you demonstrated in the rest of your posts here, my hunch is that thatÕs exactly what you did, that your answers are probably right, and that you just need to convince yourself of that :-) but if you really are wrong or unsure, post a specific example and weÕll try to find specific errors. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Amazing story - Roulette Scanning Read this: The casino at the Ritz hotel in London had a trio of Eastern European gamblers were arrested on suspicion cheating at roulette using a laser scanner attached to a cellphone (they’d manage to win 1.3 million pounds, something which quickly aroused the attention of the casino’s management). There’s only one problem: no one is even sure that it’s possible to cheat at roulette using a laser scanner. Actually, there are two problems: cheating a casino is apparently not against the law in Britain. How come we’re only just learning this? http://www.engadget.com/entry/2364827625072350/ http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm Apparently the scoop is they used a laser scanner to esteimate the initial state of the ball and then passed that to a mainframe via the cell phone to compute the trajectory and the probability for the number to show up. Does this sound like a hoax to you or it is something that is possible to do? It appears to me there are several random factors, like the ball bouncing on the metal edges of the number slot seperator and then landing on another number, an empirical fact, that would make the output of any deterministic algorithm useless, although that is something they might have accounted for given the initial momentum of the ball. Comments? Mike === Subject: Re: Amazing story - Roulette Scanning >Does this sound like a hoax to you or it is something that is >possible to do? My gut feel is that it is a hoax ßoated by the casinos. The system almost certainly has to be chaotic, and I canÕt see getting accurate enough measurements to predict the outcome. BTW, in most places the law is very simple: 1. You are not allowed to cheat the house 2. The house is allowed to cheat you. 3. The house has to call their cheating something other then cheating. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Amazing story - Roulette Scanning <41b494ea$19$fuzhry+tra$mr2ice@news.patriot.net> posting-account=1HU3Gg0AAAAUQpgzWSD32IR9hPpzNvbi Look, before anyone says it again, the system is NOT chaotic. The system is deterministic, with all initial parameters measured to within resonable accuracy one can, at the very, least get a statistical advantage. In the work of Farmer, Packard et al. they measured the initial velocity of the wheel by determing frequency, likewise for the pea. The randomness in the work of Farmer and Packard (the reason why they had to selected quadrants) is because they were never able to make resonable measurement of the relative inital position of the pea to the wheel, thus resulting in random bounces off the dividers. In the work of the chaos cabal at UCSC the major problems were associated with computational power (particularily under the constraint of not getting caught by casino security), as well as pure and simple human reaction time, as pointed out earlier. As far as when bets are closed it depends on the location of the casino, the governing body determines this. If this problem was chaotic i think Farmer Packard etc would have figured that out and quit, or atleast some of their later collaborators would have figured it out and poked a bit of fun at them over it. === Subject: Re: Amazing story - Roulette Scanning > Apparently the scoop is they used a laser scanner to esteimate the > initial state of the ball and then passed that to a mainframe via the > cell phone to compute the trajectory and the probability for the > number to show up. > Does this sound like a hoax to you or it is something that is possible > to do? There is no betting after the ball is launched. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Amazing story - Roulette Scanning posting-account=tyq2IQ0AAAAuNErVkOWXc866QgLV9Cn9 > Apparently the scoop is they used a laser scanner to esteimate the > initial state of the ball and then passed that to a mainframe via the > cell phone to compute the trajectory and the probability for the > number to show up. > Does this sound like a hoax to you or it is something that is possible > to do? > There is no betting after the ball is launched. False. === Subject: Re: Amazing story - Roulette Scanning >> There is no betting after the ball is launched. > False. In US Casinos, there is not supposed to be any betting after the ball is launched. Otherwise, I would wait until the ball was falling and be able to guess with about 5 or 7 spaces where the ball would land. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Amazing story - Roulette Scanning posting-account=k-d-4g0AAADFb7cvzD6u9R73btDj5090 Wrong! 100% Wrong! After about 3 rounds of the ball the dealer waves their hand and says no more bets! === Subject: Re: Amazing story - Roulette Scanning posting-account=tyq2IQ0AAAAuNErVkOWXc866QgLV9Cn9 Have you been in a casino? After the dealer launches the ball, the players continue to place bets until the dealer waves his hands over the table and announces No more bets. === Subject: Re: Amazing story - Roulette Scanning > Have you been in a casino? After the dealer launches the ball, the > players continue to place bets until the dealer waves his hands over > the table and announces No more bets. If I recall correctly from the casinos I have been to, he says Now more bets *as* he launches the ball. Granted, roulette is not usually the game I play, but I seem to remember him saying it as the ball was released. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Amazing story - Roulette Scanning >> Have you been in a casino? After the dealer launches the ball, the >> players continue to place bets until the dealer waves his hands over >> the table and announces No more bets. > If I recall correctly from the casinos I have been to, he says Now more > bets *as* he launches the ball. > Granted, roulette is not usually the game I play, but I seem to remember > him saying it as the ball was released. > - Tim As an impartial observer of this discussion who has watched the game more than a few times. No more bets is usually announced about 10-15 seconds after the ball (pea) has been released. === Subject: Re: Amazing story - Roulette Scanning > Read this: > The casino at the Ritz hotel in London had a trio of Eastern European > gamblers were arrested on suspicion cheating at roulette using a laser > scanner attached to a cellphone (theyÕd manage to win 1.3 > million pounds, something which quickly aroused the attention of the > casinoÕs management). ThereÕs only one problem: no one is > even sure that itÕs possible to cheat at roulette using a laser > scanner. Actually, there are two problems: cheating a casino is > apparently not against the law in Britain. How come weÕre only > just learning this? > http://www.engadget.com/entry/2364827625072350/ > http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Apparently the scoop is they used a laser scanner to esteimate the > initial state of the ball and then passed that to a mainframe via the > cell phone to compute the trajectory and the probability for the > number to show up. > Does this sound like a hoax to you or it is something that is possible > to do? > It appears to me there are several random factors, like the ball > bouncing on the metal edges of the number slot seperator and then > landing on another number, an empirical fact, that would make the > output of any deterministic algorithm useless, although that is > something they might have accounted for given the initial momentum of > the ball. > Comments? > Mike Since the house edge in roulette, particularly for European roulette with a 0 but no 00, is very thin, even a modest counter edge, like knowing which slots are slightly more or less probable than others, would likely be enough to allow the cheaters to win over the long run. Actually this pattern would be better for the,cheaters since it would produce what appeared more like a mere run of better than usual luck, and be much harder to detect. If the cheaters could accurately predict the outcome each spin, it would be very hard for them to produce a pattern of winning that looked like mere good luck, and they might be detected quite rapidly. Since they were not detected until they had won over a million pounds, I suspect that they only were able to change the odds to moderately in their favor. === Subject: Re: Amazing story - Roulette Scanning > Since they were not detected until they had won over a million pounds, I > suspect that they only were able to change the odds to moderately in > their favor. Theoretically yes, youÕre right but in practice it would take a long time to accumulate these winnings wih a small bias. I suspect something else went on and the story about lasers and computations was a hoax the casino people invented to divert the attention. Either those people know the kind of physics we donÕt, maybe some new QM theory that predicts macro dynamic behavior or they used some other method the casino people donÕt want to be publicized and reached an agreement with them to talk about cell phones and lasers. I cannot really think of what they might have used but the fact is they won the money and abducting any hypothesis of how they did it is not a good idea. Mike === Subject: Re: Amazing story - Roulette Scanning > http://www.engadget.com/entry/2364827625072350/ > http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm Those are from March, but the BBC has an update on the story today: http://news.bbc.co.uk/1/hi/uk/4069629.stm === Subject: Re: Amazing story - Roulette Scanning > http://www.engadget.com/entry/2364827625072350/ > http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Those are from March, but the BBC has an update on the story today: > http://news.bbc.co.uk/1/hi/uk/4069629.stm I believe it to unolvable. The roulette I have observed in casinos is exectued as follows: 1. The croupier spins the whel in one direction. 2. After an interval, the ball is launched in the other direction around the outer track. 3. After friction slows the ball down, it falls into the inner slope, where is may or may not strike diverters. 4. When the ball enters the number wheel, it usually does not stop in the first slot. It is more likely to bounce, and may go in either direction. 5. The ball and wheel are also subject to other unmeasured variables, such as temperature, humidity, atmospheric pressure, air currents, etc. That sequence of events renders the attempt to predict the outcome useless, even with a perfect measurement of the initial ball speed. In any event, all the casinos have to do to defend themselves is to stop taking bets between 1 and 2 above. === Subject: Re: Amazing story - Roulette Scanning > http://www.engadget.com/entry/2364827625072350/ http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Those are from March, but the BBC has an update on the story today: > http://news.bbc.co.uk/1/hi/uk/4069629.stm > I believe it to unolvable. The roulette I have observed in casinos is > exectued as follows: > 1. The croupier spins the whel in one direction. > 2. After an interval, the ball is launched in the other direction around > the outer track. > 3. After friction slows the ball down, it falls into the inner slope, where > is may or may not strike diverters. > 4. When the ball enters the number wheel, it usually does not stop in the > first slot. It is more likely to bounce, and may go in either direction. > 5. The ball and wheel are also subject to other unmeasured variables, such > as temperature, humidity, atmospheric pressure, air currents, etc. > That sequence of events renders the attempt to predict the outcome useless, > even with a perfect measurement of the initial ball speed. > In any event, all the casinos have to do to defend themselves is to stop > taking bets between 1 and 2 above. Never underestimate the power of the human mind to find ways to cheat. If you do you will find yourself scammed, grifted or cheated out of your money. My father was the authority on casino cheating before he died in 1996. This has been tried and proved effective with 20 year old technology. The operators even went in search of a dealer that would spin the wheel correctly. At home with a perfect wheel they achieved a 35% advantage over the house. === Subject: Re: Amazing story - Roulette Scanning >> 1. The croupier spins the whel in one direction. >> 2. After an interval, the ball is launched in the >> other direction around the outer track. >> In any event, all the casinos have to do to defend >> themselves is to stop taking bets between 1 and 2 above. Time is money. ItÕs more cost effective for a casino to have quality surveillance. > Never underestimate the power of the human mind to find > ways to cheat. If you do you will find yourself scammed, > grifted or cheated out of your money. My father was the > authority on casino cheating before he died in 1996. This > has been tried and proved effective with 20 year old technology. > The operators even went in search of a dealer that would spin > the wheel correctly. At home with a perfect wheel they > achieved a 35% advantage over the house. ThatÕs a key point, finding a dealer with a set routine that can be quantified. === Subject: Re: Amazing story - Roulette Scanning > http://www.engadget.com/entry/2364827625072350/ http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Those are from March, but the BBC has an update on the story today: > http://news.bbc.co.uk/1/hi/uk/4069629.stm > I believe it to unolvable. The roulette I have observed in casinos is > exectued as follows: > 1. The croupier spins the whel in one direction. > 2. After an interval, the ball is launched in the other direction around > the outer track. > 3. After friction slows the ball down, it falls into the inner slope, where > is may or may not strike diverters. > 4. When the ball enters the number wheel, it usually does not stop in the > first slot. It is more likely to bounce, and may go in either direction. > 5. The ball and wheel are also subject to other unmeasured variables, such > as temperature, humidity, atmospheric pressure, air currents, etc. > That sequence of events renders the attempt to predict the outcome useless, > even with a perfect measurement of the initial ball speed. IÕm inclined to agree with you, however it would really take an experiment to confirm. One also might think that with the returns on bets arranged to give the casino a slight advantage in expected return, that a very slight edge for the player would be enough to tip the odds in his favor. As somebody said, a correlation would be enough. But on the face of it, it seems nuts. === Subject: Re: Amazing story - Roulette Scanning >> I believe it to unolvable. The roulette I have observed in casinos is >> exectued as follows: >> 1. The croupier spins the whel in one direction. >> 2. After an interval, the ball is launched in the other direction around >> the outer track. >> 3. After friction slows the ball down, it falls into the inner slope, where >> is may or may not strike diverters. >> 4. When the ball enters the number wheel, it usually does not stop in the >> first slot. It is more likely to bounce, and may go in either direction. >> 5. The ball and wheel are also subject to other unmeasured variables, such >> as temperature, humidity, atmospheric pressure, air currents, etc. >> That sequence of events renders the attempt to predict the outcome useless, >> even with a perfect measurement of the initial ball speed. >IÕm inclined to agree with you, however it would really take an >experiment to confirm. One also might think that with the returns on >bets arranged to give the casino a slight advantage in expected >return, that a very slight edge for the player would be enough to tip >the odds in his favor. As somebody said, a correlation would be >enough. But on the face of it, it seems nuts. The strategy (and experimental confirmation) are well described in the book The Eudaemonic Pie by Thomas Bass (this was an earlier attempt to beat roulette by Farmer, Shaw, Crutchfield, Packard--the Santa Cruz chaos group (they also discovered along the way that Claude Shannon had beat them to it, but using TTL rather than microprocessors)). Basically, they divided the wheel up into octants (memorizing which numbers were in each octant) and their computer would give them an octant to bet on. As long as the wheel was slightly tilted (and luckily almost all the wheels in Vegas were tilted at the time) they could get a 40% advantage over the house (the bought a real roullette wheel to experiment with). Ken Muldrew kmuldrezw@ucalgazry.ca (remove all letters after y in the alphabet) === Subject: Re: Amazing story - Roulette Scanning > http://www.engadget.com/entry/2364827625072350/ http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Those are from March, but the BBC has an update on the story today: > http://news.bbc.co.uk/1/hi/uk/4069629.stm > I believe it to unolvable. The roulette I have observed in casinos is > exectued as follows: > 1. The croupier spins the whel in one direction. > 2. After an interval, the ball is launched in the other direction around > the outer track. > 3. After friction slows the ball down, it falls into the inner slope, where > is may or may not strike diverters. > 4. When the ball enters the number wheel, it usually does not stop in the > first slot. It is more likely to bounce, and may go in either direction. > 5. The ball and wheel are also subject to other unmeasured variables, such > as temperature, humidity, atmospheric pressure, air currents, etc. > That sequence of events renders the attempt to predict the outcome useless, > even with a perfect measurement of the initial ball speed. > In any event, all the casinos have to do to defend themselves is to stop > taking bets between 1 and 2 above. I actually had the same idea of devising just such a computer linked roulette system about 30 years ago. It would have been much tougher back then with fewer off the shelve electronics devices to work with. But the ball does do a lot of bouncing around, including off the diverters, and also all the casinos I played in stopped the betting before throwing the ball. Cheating the casinos is severely punished Nevada. Even just having special knowledge of a gambling device, such as an electronic slot machine, that gives you an advantage in playing it is considered cheating and can lead to years in prison. In my opinion, the casinos cheat the public everyday by running games where the payouts are insufficient for the risk of the game, thus creating the house advantage. For instance, the payout on a single number in roulette is 35 to 1. If it were a fair game, the payout would be 37 to one because there are 38 numbers on American style wheels. However, the casino owners and their lobbyists make the laws. If the house were not alowed to thus rig their games, you wouldnÕt have any giant casinos. All gambling would be played in taverns and such at the social level, and all games would be fair. Double-A === Subject: Re: Amazing story - Roulette Scanning > http://www.engadget.com/entry/2364827625072350/ http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Those are from March, but the BBC has an update on the story today: > http://news.bbc.co.uk/1/hi/uk/4069629.stm > I believe it to unolvable. The roulette I have observed in casinos is > exectued as follows: > 1. The croupier spins the whel in one direction. > 2. After an interval, the ball is launched in the other direction around > the outer track. > 3. After friction slows the ball down, it falls into the inner slope, where > is may or may not strike diverters. > 4. When the ball enters the number wheel, it usually does not stop in the > first slot. It is more likely to bounce, and may go in either direction. > 5. The ball and wheel are also subject to other unmeasured variables, such > as temperature, humidity, atmospheric pressure, air currents, etc. > That sequence of events renders the attempt to predict the outcome useless, > even with a perfect measurement of the initial ball speed. > In any event, all the casinos have to do to defend themselves is to stop > taking bets between 1 and 2 above. I think you mean between 2 and 3. ItÕs impossible to predict the path of the ball if it hasnÕt been started. It is currently common practise to continue to take bets for several revolutions of the ball. Also, in order to gain an advantage at the table itÕs not necessary to predict exactly the number the ball will end up on. It is sufficient to narrow it down to a small group of likely endings and bet accordingly. === Subject: Re: Amazing story - Roulette Scanning >Also, in order to gain an advantage at the table itÕs not necessary >to predict exactly the number the ball will end up on. It is sufficient >to narrow it down to a small group of likely endings and bet accordingly. Actually, narrowing it down to even half the outcomes would give a huge edge. About 25 years ago some colleagues and I spent some time on the theory. Our idea was to pick a specific wheel, make a lookup table for its number sequence, and take some data to model an exponential decay motion for both the ball and wheel. We assumed one could surreptitiously measure in real time the time for one revolution of the ball and one revolution of the 00, and that one could place a bet after things were in motion. We could solve for the theoretical hit point, and if the ball didnÕt bounce to badly too often, it should give us an edge. We got it as far as actually taking some measurements on a wheel not located in a casino and our observations were: 1. Many times the ball does drop in a neighborhood of where you would expect, i.e, the obstructions and randomness didnÕt seem to be a killer. 2. What did seem to be a killer was the sensitivity of the calculated result to the errors in the delta-t time measurements. We had been thinking in terms of human reaction times when milliseconds mattered. 3. And, of course, the slight problem of how to implement it and get away with it. They probably wouldnÕt let us set up our equipment on their wheel :-) But it was fun and interesting anyway. --Lynn === Subject: Re: Amazing story - Roulette Scanning http://www.engadget.com/entry/2364827625072350/ http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm Those are from March, but the BBC has an update on the story today: > http://news.bbc.co.uk/1/hi/uk/4069629.stm > I believe it to unolvable. The roulette I have observed in casinos is > exectued as follows: > 1. The croupier spins the whel in one direction. > 2. After an interval, the ball is launched in the other direction around > the outer track. > 3. After friction slows the ball down, it falls into the inner slope, where > is may or may not strike diverters. > 4. When the ball enters the number wheel, it usually does not stop in the > first slot. It is more likely to bounce, and may go in either direction. > 5. The ball and wheel are also subject to other unmeasured variables, such > as temperature, humidity, atmospheric pressure, air currents, etc. > That sequence of events renders the attempt to predict the outcome useless, > even with a perfect measurement of the initial ball speed. > In any event, all the casinos have to do to defend themselves is to stop > taking bets between 1 and 2 above. > I think you mean between 2 and 3. ItÕs impossible to predict the > path of the ball if it hasnÕt been started. It is currently common > practise to continue to take bets for several revolutions of the ball. ThatÕs my point. I meant between 1 and 2. > Also, in order to gain an advantage at the table itÕs not necessary > to predict exactly the number the ball will end up on. It is sufficient > to narrow it down to a small group of likely endings and bet accordingly. Agreed. Thus my protection plan. === Subject: Re: Amazing story - Roulette Scanning Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >cheating a casino is >apparently not against the law in Britain. And quite rightly so. If they can set things up to determine the odds, why canÕt we? -- Richard === Subject: Re: Amazing story - Roulette Scanning format=ßowed; >> cheating a casino is >> apparently not against the law in Britain. > And quite rightly so. If they can set things up to determine the > odds, why canÕt we? Because thats how they make money. If we do it, itÕs cheating. Sure gas stations are in the business of selling gas, if we can figure out a way to foll the gauges to keep from paying full prices, thatÕs illegal too. We arenÕt allowed to set our own price at the pump just because they can. === Subject: Re: Amazing story - Roulette Scanning > Read this: > The casino at the Ritz hotel in London had a trio of Eastern European > gamblers were arrested on suspicion cheating at roulette using a laser > scanner attached to a cellphone (they’d manage to win 1.3 > million pounds, something which quickly aroused the attention of the > casino’s management). There’s only one problem: no one is > even sure that it’s possible to cheat at roulette using a laser > scanner. Actually, there are two problems: cheating a casino is > apparently not against the law in Britain. How come we’re only > just learning this? > http://www.engadget.com/entry/2364827625072350/ > http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Apparently the scoop is they used a laser scanner to estimate the > initial state of the ball and then passed that to a mainframe via the > cell phone to compute the trajectory and the probability for the > number to show up. > Does this sound like a hoax to you or it is something that is possible > to do? > It appears to me there are several random factors, like the ball > bouncing on the metal edges of the number slot seperator and then > landing on another number, an empirical fact, that would make the > output of any deterministic algorithm useless, although that is > something they might have accounted for given the initial momentum of > the ball. > Comments? > Mike This was tried 20 years ago with computers built into shoes and they did not use lasers. They could get an advantage at home with a perfectly level wheel. They never succeeded in a casino because the lights and electronics hurt their precision. I am sure it would be their attempt giving everyone ideas. It is very illegal in Las Vegas! === Subject: Re: Amazing story - Roulette Scanning > Read this: The casino at the Ritz hotel in London had a trio of Eastern European > gamblers were arrested on suspicion cheating at roulette using a laser > scanner attached to a cellphone (they’d manage to win 1.3 > million pounds, something which quickly aroused the attention of the > casino’s management). There’s only one problem: no one is > even sure that it’s possible to cheat at roulette using a laser > scanner. Actually, there are two problems: cheating a casino is > apparently not against the law in Britain. How come we’re only > just learning this? http://www.engadget.com/entry/2364827625072350/ http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Apparently the scoop is they used a laser scanner to estimate the > initial state of the ball and then passed that to a mainframe via the > cell phone to compute the trajectory and the probability for the > number to show up. Does this sound like a hoax to you or it is something that is possible > to do? It appears to me there are several random factors, like the ball > bouncing on the metal edges of the number slot seperator and then > landing on another number, an empirical fact, that would make the > output of any deterministic algorithm useless, although that is > something they might have accounted for given the initial momentum of > the ball. Comments? Mike > This was tried 20 years ago with computers built into shoes and they > did not use lasers. They could get an advantage at home with a > perfectly level wheel. They never succeeded in a casino because the > lights and electronics hurt their precision. And because they were excellent mathematicians and physicists but lousy engineers; they didnÕt know how to build stuff that was reliable. They were using 6502 processors when they cost well over $100 a pop. Nicely documented in the book The Eudaemonic Pie (or something close to that). It was written in that breathless Soul Of A New Machine style, and when I read it, I thought it must be some kind of fiction-claiming-to-be-fact bullshit. Later I discovered that some of the folks involved with the project went on to do the early work on chaos theory and other mathematical esoterica. Isaac === Subject: Re: Amazing story - Roulette Scanning Mike a .8ecrit : > Read this: > The casino at the Ritz hotel in London had a trio of Eastern European > gamblers were arrested on suspicion cheating at roulette using a laser > scanner attached to a cellphone (they’d manage to win 1.3 > million pounds, something which quickly aroused the attention of the > casino’s management). There’s only one problem: no one is > even sure that it’s possible to cheat at roulette using a laser > scanner. Actually, there are two problems: cheating a casino is > apparently not against the law in Britain. How come we’re only > just learning this? > http://www.engadget.com/entry/2364827625072350/ > http://news.bbc.co.uk/2/hi/uk_news/england/london/3558273.stm > Apparently the scoop is they used a laser scanner to esteimate the > initial state of the ball and then passed that to a mainframe via the > cell phone to compute the trajectory and the probability for the > number to show up. > Does this sound like a hoax to you or it is something that is possible > to do? > It appears to me there are several random factors, like the ball > bouncing on the metal edges of the number slot seperator and then > landing on another number, an empirical fact, that would make the > output of any deterministic algorithm useless, although that is > something they might have accounted for given the initial momentum of > the ball. > Comments? You are right. Look at chaos (deterministic), sensitive dependance (to initial conditions), etc. Even a precision of angstr.9am for position and 10^-6 mm/s for speed would quickly be lost, as for instance the incertitude on angles are doubling at each rebound (even supposing a perfect elastic circle). It is of course possible, in theory (but highly unlikely), that a correlation exists between, say, initial speed and final number (and this could be analysed statistically), but a deterministic model of the trajectory of the ball is doomed to fail. And on the material side of the story, I dont know for England, but in France, cellular phones are forbidden in casinos, anyway. > Mike === Subject: Re: Amazing story - Roulette Scanning (to initial conditions), etc. Even a precision of angstr?m for position > and 10^-6 mm/s for speed would quickly be lost, as for instance the > incertitude on angles are doubling at each rebound (even supposing a > perfect elastic circle). It is of course possible, in theory (but highly > unlikely), that a correlation exists between, say, initial speed and > final number (and this could be analysed statistically), but a > deterministic model of the trajectory of the ball is doomed to fail. And > on the material side of the story, I dont know for England, but in > France, cellular phones are forbidden in casinos, anyway. Yeah, itÕs so doomed to fail that people have made millions. Google the newtonian casino === Subject: Re: Amazing story - Roulette Scanning > (to initial conditions), etc. Even a precision of angstr?m for position > and 10^-6 mm/s for speed would quickly be lost, as for instance the > incertitude on angles are doubling at each rebound (even supposing a > perfect elastic circle). It is of course possible, in theory (but highly > unlikely), that a correlation exists between, say, initial speed and > final number (and this could be analysed statistically), but a > deterministic model of the trajectory of the ball is doomed to fail. And > on the material side of the story, I dont know for England, but in > France, cellular phones are forbidden in casinos, anyway. > Yeah, itÕs so doomed to fail that people have made millions. > Google the newtonian casino Dream on. The people making the millions are spinning the wheel. The best bet for the man on the street is to spend their gambling money on casino stocks. === Subject: Re: Amazing story - Roulette Scanning format=ßowed; > Dream on. The people making the millions are spinning the wheel. > The best bet for the man on the street is to spend their gambling money on > casino stocks. Or make new games that the casinos will put in their parlors. === Subject: Re: Amazing story - Roulette Scanning Ian Stirling a .8ecrit : > You are right. Look at chaos (deterministic), sensitive dependance >>(to initial conditions), etc. Even a precision of angstr?m for position >> and 10^-6 mm/s for speed would quickly be lost, as for instance the >>incertitude on angles are doubling at each rebound (even supposing a >>perfect elastic circle). It is of course possible, in theory (but highly >>unlikely), that a correlation exists between, say, initial speed and >>final number (and this could be analysed statistically), but a >>deterministic model of the trajectory of the ball is doomed to fail. And >> on the material side of the story, I dont know for England, but in >>France, cellular phones are forbidden in casinos, anyway. > Yeah, itÕs so doomed to fail that people have made millions. > Google the newtonian casino Look at the story of early attempts in 19th century (the wheel tables used to be worned out, in a statistical exploitable way (see Joseph Jaggers)), or what happened when blackjack was solved. The hard *fact* is that nobody ever wins millions : casinos will simply not allow it. Now for some Google searching... Note the use of apparently in Note also that the book is 1991, and remember how fast casinos *did* (and do) react against any threat, be it scientific, psychic (yes, it happened) or just fraudulent (associations between gamblers and croupiers are much more frequent) But you may have some better confirmed information... === Subject: Re: Amazing story - Roulette Scanning > Ian Stirling a .8ecrit : > >You are right. Look at chaos (deterministic), sensitive dependance >>(to initial conditions), etc. Even a precision of angstr?m for position >> and 10^-6 mm/s for speed would quickly be lost, as for instance the >>incertitude on angles are doubling at each rebound (even supposing a >>perfect elastic circle). It is of course possible, in theory (but highly >>unlikely), that a correlation exists between, say, initial speed and >>final number (and this could be analysed statistically), but a >>deterministic model of the trajectory of the ball is doomed to fail. And >> on the material side of the story, I dont know for England, but in >>France, cellular phones are forbidden in casinos, anyway. > Yeah, itÕs so doomed to fail that people have made millions. > Google the newtonian casino > Look at the story of early attempts in 19th century (the wheel tables > used to be worned out, in a statistical exploitable way (see Joseph > Jaggers)), or what happened when blackjack was solved. The hard *fact* > is that nobody ever wins millions : casinos will simply not allow it. > Now for some Google searching... > Note the use of apparently in > Note also that the book is 1991, and remember how fast casinos *did* > (and do) react against any threat, be it scientific, psychic (yes, it > happened) or just fraudulent (associations between gamblers and > croupiers are much more frequent) > But you may have some better confirmed information... One of my trips to a casino I observed somthing at a high stakes roulette table that was interesting to say the least. The player, the only one at this table, was placing $100 chips straight up on inside numbers and only placing on numbers that were neighbors all grouped together on the wheel and when he won would toke the dealer handsomely. The player would only change this group of neighbors after a winning spin of the wheel. I watched the game for about a half hour and I mentally calculated about every 5th or 6th spin on average the player had a winning number. The player had a huge advantage if he was being helped by the dealer for trying to place the ball to drop in a certain area of the wheel. Is it possible for an experienced roulette dealer to do this with years of experience and practice? Dan === Subject: Re: Amazing story - Roulette Scanning format=ßowed; > Is it possible for an experienced roulette dealer to do this with > years of experience and practice? Not if he wants to keep his job and stay out of jail. === Subject: Re: Amazing story - Roulette Scanning <41b318fc$0$52984$ed2619ec@ptn-nntp-reader03.plus.net> <41b334fa$0$11788$8fcfb975@news.wanadoo.fr> Discussion, linux) >> Is it possible for an experienced roulette dealer to do this with >> years of experience and practice? > Not if he wants to keep his job and stay out of jail. He didnÕt ask if it was acceptable behavior. Heck, IÕll bet that *even before you responded*, he knew this was cheating. Ya think? I kinda doubt that a system of cheating so obvious would last very long, so I kinda doubt that there was really an arrangement like Dan suggests. If he could notice it, then the casino management surely would notice it, too. But his question of whether itÕs feasible is a different matter. Actually, this kinda reminds me of the old tests for your babyÕs gender. Take the test and if the result later turns out to be wrong, you get your money back. Tell the guy to bet on a certain grouping and when he wins, you want your cut. -- When you go to class today, if your professor talks about algebraic number theory, or misuses Galois Theory[,] I want you to carefully notice how you feel. Hold on to that feeling so that you never forget it. --James S. Harris, on channeling rage via Galois theory. === Subject: Re: Amazing story - Roulette Scanning > One of my trips to a casino I observed somthing at a high stakes > roulette table that was interesting to say the least. > The player, the only one at this table, was placing $100 chips straight > up on inside numbers and only placing on numbers that were neighbors > all grouped together on the wheel and when he won would toke the dealer > handsomely. The dealer surely wanted the player to win. If a dealer could develop that kind of control (take my word for it) many would have been compromised by now. > The player would only change this group of neighbors after a winning > spin of the wheel. > I watched the game for about a half hour and I mentally calculated > about every 5th or 6th spin on average the player had a winning number. > The player had a huge advantage if he was being helped by the dealer > for trying to place the ball to drop in a certain area of the wheel. High rollers can lay down some big tips at any time. At a $100 blackjack table, I saw a player toke the dealer 10K. Unfortunately (for the dealer) tokes were shared. > Is it possible for an experienced roulette dealer to do this with > years of experience and practice? === Subject: Re: Amazing story - Roulette Scanning > Ian Stirling a .8ecrit : > >You are right. Look at chaos (deterministic), sensitive dependance >>(to initial conditions), etc. Even a precision of angstr?m for position >> and 10^-6 mm/s for speed would quickly be lost, as for instance the >>incertitude on angles are doubling at each rebound (even supposing a >>perfect elastic circle). It is of course possible, in theory (but highly >>unlikely), that a correlation exists between, say, initial speed and >>final number (and this could be analysed statistically), but a >>deterministic model of the trajectory of the ball is doomed to fail. And >> on the material side of the story, I dont know for England, but in >>France, cellular phones are forbidden in casinos, anyway. > Yeah, itÕs so doomed to fail that people have made millions. > Google the newtonian casino > Look at the story of early attempts in 19th century (the wheel tables > used to be worned out, in a statistical exploitable way (see Joseph > Jaggers)), or what happened when blackjack was solved. The hard *fact* > is that nobody ever wins millions : casinos will simply not allow it. > Now for some Google searching... > Note the use of apparently in > Note also that the book is 1991, and remember how fast casinos *did* > (and do) react against any threat, be it scientific, psychic (yes, it > happened) or just fraudulent (associations between gamblers and > croupiers are much more frequent) > But you may have some better confirmed information... > One of my trips to a casino I observed somthing at a high stakes > roulette table that was interesting to say the least. > The player, the only one at this table, was placing $100 chips straight > up on inside numbers and only placing on numbers that were neighbors > all grouped together on the wheel and when he won would toke the dealer > handsomely. > The player would only change this group of neighbors after a winning > spin of the wheel. > I watched the game for about a half hour and I mentally calculated > about every 5th or 6th spin on average the player had a winning number. > The player had a huge advantage if he was being helped by the dealer > for trying to place the ball to drop in a certain area of the wheel. > Is it possible for an experienced roulette dealer to do this with > years of experience and practice? In full view of the security cameras? I would suspect a money laundering operation. By the way, does anyone appreciate the money-laundering capability of an expensive hotel? Each room is a conduit for many times the daily rental rate in cash payments, traceable only to the name signed on the hotel register. === Subject: Re: Amazing story - Roulette Scanning > Look at the story of early attempts in 19th century (the wheel tables > used to be worned out, in a statistical exploitable way (see Joseph > Jaggers)), or what happened when blackjack was solved. The hard *fact* > is that nobody ever wins millions : casinos will simply not allow it. My little story: The first time I played roulette in Las Vegas (IÕm not much of a gambler; we $20 into $600 in about 10 minutes. When I asked to cash out, it took approvals from two levels of management to convert my roulette chips to Flamingo chips. IÕm suspect my tape is still archived somewhere in the Flamingo vaults. === Subject: Re: Amazing story - Roulette Scanning > My little story: > The first time I played roulette in Las Vegas (IÕm not much of a > Utah) I turned $20 into $600 in about 10 minutes. When I asked to > cash out, it took approvals from two levels of management to convert > my roulette chips to Flamingo chips. IÕm suspect my tape is still > archived somewhere in the Flamingo vaults. My brother works for a hotel construction company and his company just happened to build the new Westin Hotel in Las Vegas[*]. In doing so, there were strict rules they had to follow in order for the appropriate survellance systems to be installed. Although, I donÕt think the Westin has a casino, they still had to follow the rules. One thing he told me is that every single spin of a roulette wheel is recorded and analyzed. He didnÕt know if it was NGC law or just casinos avoiding people ripping them off. If there is bias, it is shut down and removed. The main reason, even at a $20 maximum table, you have a chance to win $700 with each spin. A statistically signficant bias and you can make some money fairly quickly, with a 35x return. Craps results are not as closely guarded because the thrower has his or her choice of a few dice. Thus if one die is biased, it is hard to constantly pick that one die out of the 6 you are offered. Most other table games do not offer such large payouts for small investments. In blackjack, for example, at best in a single hand you can profit 1.5x your bet. - Tim [*] Having him build a hotel there meant when I visited, everything was comped for me -- free room, free meals, and free shows. That was the first time I went to Vegas and actually left with more money. I only gambled a little bit. -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Uncountable many reals without Cantor > The first proof is unique, and we are all referring to the first proof > that Cantor did before the diagonal argument... > I stand corrected then. > But (like Mike Oliver) I thought your words were easier to interpret > when they applied to the diagonal proof rather than the real first > proof. ItÕs the diagonal proof that I consider utterly simple, not > the proof involving intervals. Ok. I find the interval based approach simpler, because it presents the underlying concept of the continuum directly: actual infinite divisibility. -- Eray Ozkural === Subject: Re: Uncountable many reals without Cantor > The first proof is unique, and we are all referring to the first proof > that Cantor did before the diagonal argument... > I stand corrected then. > But (like Mike Oliver) I thought your words were easier to interpret > when they applied to the diagonal proof rather than the real first > proof. ItÕs the diagonal proof that I consider utterly simple, not > the proof involving intervals. > Ok. I find the interval based approach simpler, because it presents > the underlying concept of the continuum directly: actual infinite > divisibility. > -- > Eray Ozkural ThatÕs entertainment. === Subject: Re: Uncountable many reals without Cantor |> But seriously, |> the intuitionists and serious constructivists deny the uncountability |> of the reals |> No they donÕt. | |See the replies, I think only the ones who are silly enough to admit |Platonism back in would admit the continuum I think youÕre being somewhat vague here. Admit the continuum could easily mean assert that the continuum exists. The question is what kind of conception of the continuum is involved. Some people get strange ideas about what constructivism should mean. They imagine, for instance, that a constructivist would not agree that infinite sets exist, thinking that what it means to construct a set is to enumerate its elements. But thatÕs not what any constructivist to my knowledge means by constructing a set. Bishop defines a set as being constructed by saying what is needed to construct an element of it, and what has to be proven to assert the equality of two such constructions qua elements of the set. This conception of set is plenty broad enough to make it make sense to say that there exist infinite sets, including the continuum. |I had a talk with a so-called constructivist. He was repeating the |naive claims of Robin Chapman and Stephen Harris, that the TM is not a |physical model of computation, e.g. they donÕt exist in the world. He |said PCs are not TMs!!!!! I told him that a constructivist would |consider TMs to be physical models. | |The poor chap was offended. A guy from Netherlands, he considered |himself to be a constructivist. He probably thought himself to be a |follower of Brouwer. | |Now, if Brouwer was present in the discussion, he could teach a lesson |or two about constructivism to this chap. All that actual-infinity |tape talk is Platonist nonsense, and you would probably think that all |constructivists should be as naive as this poor chap. No, actually I think most of us suspect you to be more naive than him on this point. The point of constructivism is not to forbid people from making abstractions (in favor of speaking only of physical instances of them). You may express your disagreement with him with five exclamation marks, but I would be surprised if very many people (or very many constructivists) thought that PCs literally are Turing machines. (Where, for example, is the tape? What does it mean to say that the tape head is at a given point in it? And on what basis can we claim that it moves only one step at a time on this tape?) ItÕs more usual and accurate to say that one can in principle simulate a PC on a Turing machine (at least its deterministic components, to the extent that they are not faulty), and conversely that a PC can simulate a Turing machine up to the limitations imposed by the available memory, disk storage, and limited lifespan. |ItÕs clear that you know nothing about constructivism or why the |debate of the actual infinity is so central to these matters. ItÕs clear that your approach to sizing up someone elseÕs knowledge of constructivism is poor and leads to implausible snap-judgements about people who know more about it than you do (quite a lot more would be my guess). |As a |Platonist, you want to turn constructivism into the naive ideals of |Platonism. | |But it is not. I consider true constructivism to reject actual |infinity talk at a fundamental level. This term actual infinity is not a very good one. I think you may be referring to the concept of infinite sets understood as completed totalities as popularly distinguished from the notion of potential infinity. I realize that itÕs common enough to use the term actual infinity to refer to this idea, but I donÕt think itÕs a very good term for it. Actual is just too ßimsy an adjective. If you donÕt mean infinity as a completed totality, but simply infinity, then youÕre wholly mistaken about constructivism. You would disagree on this point with Kronecker, Brouwer, Heyting, Markov, Bishop, and all of their students as far as I know. They all, moreover, have some concept of the real line. IÕve known of this commonly described distinction between potential infinity and actual infinity for a long time, and IÕve gradually come to feel that itÕs not a very good distinction. It gives the impression that the important difference between points of view on infinity is between people who regard the elements of an infinite set (e.g. the integers) as all already existing and those who think that they are not all already existing. ItÕs true that as one thinks in Platonic terms, one tends to think of the integers as already existing now. But the point of constructivism is muddled by imagining that constructivists share an understanding with Platonists of what it would mean to have all the integers already existing now, but disagree with them on whether they do. ItÕs more appropriate to consider the application of tenses to mathematical statements as a potentially confusing metaphor, one that we can adopt or change as desired. ItÕs natural for a constructivist to consider the appearance that we have a question does such-and-such object exist now?, that goes beyond the questions of whether we have constructed one and the like, as being an illusion. I often prefer even when discussing constructivism to use language that treats mathematical truths as having always been true. Keep in mind, however, that this doesnÕt mean anything special. One reason for using such language, as I see it, is that we can apply the mathematics we know now to past situations. When we apply physics to the past, we can use present knowledge of differential equations. In this sense, the facts about the solutions of differential equations held in the past. Many people also seem to find it confusing to read tensed descriptions of mathematics (such as BishopÕs comment that a statement isnÕt true until it is proven). There is a risk of misinterpretation in either case, as long as one has the impression that thereÕs more content to statements about when mathematical objects and properties exist than there is. On the other hand, sometimes it seems appropriate to speak of a mathematical object as dating from its first construction (i.e., the first proof of its existence), since this is one of the few ways in which time meaningfully relates to the existence of the object, remembering as ever that this means nothing more than we have chosen to make it mean. Many of the most mathematically interesting constructions involve implicit or explicit references to hypothetical future calculations. For instance, one says that there exists a largest prime < 10^10^100. One makes claims of this kind in constructive mathematics as well thinks of present existence as having a special meaning, this would appear to mean that constructivists are committing to the existence of non-constructed entities. However, as a matter of definition, the existence of a certain integer is understood as being verified by having an algorithm that would, in principle, generate one such. Hence the largest prime < 10^10^100 has already been constructed in this sense. We could equally well consider whether it had been constructed as a decimal expansion or as a sequence of tallies, but questions of this kind have less relevance to the mathematics that is of most interest to us. Adopting language that requires us explicitly to mention when a construction is a matter of the result of a hypothetical calculation would mean being needlessly verbose. Thus constructivists tend to consider only constructions that are equivalent to producing a procedure that would in principle generate a certain construction. The procedure is not generally assumed to be a computer program; usually that assumption does you no good. Brouwer uses the time element in an interesting way. His so-called free choice sequences are a good example of what people would probably want to call potentially infinite entities. He deals with the case where one has a sequence where one has committed to retain the freedom to choose additional terms, so that there is never a basis for knowing more than finitely many terms. We can consider free-choice reals, for instance, which are free choice sequences of nested rational intervals, with the n-th interval having length <=1/n. In his terms, itÕs absurd for a free-choice real to be equal to 0, because the only basis one could have for a realÕs being 0 is a prior commitment to use only intervals including 0. In contrast, 0 is perfectly okay as a lawlike real. The fact that criticizing CantorÕs theorem, on the basis that Cantor (allegedly) relies upon actual infinity, is quite confused, is highlighted by the fact that his (first) proof works equally well for free-choice reals as for any other kind (and his diagonal argument works for free-choice sequences of digits as well as for any other kinds of sequences of digits). A person can generate a real distinct from an arbitrary sequence of free-choice reals never needing to know more than a finite number of elements of a finite number of sequences at any one time, and also never needing to commit to a particular sequence of future choices. Keith Ramsay === Subject: nearest common ancestor on Stern-Brocot tree? For example, the nearest common ancestor for 5/9 and 3/4 is 2/3. Indeed 5/9 = LRLLL 3/4 = LRR 2/3 = LR The straightforward way to calculate NCA is by converting the rational number into LR representation applying (extended) Euclidian algirithm. Is there a formula for NCA? === Subject: Re: i alone > i fight the evil of psychiatry, psychology. surrender your vermin. What, so you can feed them your medication? === Subject: Re: Escultura affair: publication scandal >1271503.iARF35e16701@proapp.mathforum.org... Norm > Unfortunately, mathematics is not a popularity contest; it is a struggle for precision. One of the requirements of mathematics is that every concept must well-defined, that is, its existence, properties and relations with other concepts must be specified by the axioms. The axioms are the basis of proof. Proofs are nonsense unless they follow from the axioms. That is why a true mathematician looks at the foundations of a field before doing anything there. This is where Wiles failed miserably. Critique-rectification naturally involves new language. > E. E. Escultura > University of the Philppines No, itÕs not a popularity contest -- itÕs a game. Well, at least there was a game called IIRC Wiff-N-Proof that taught the creation of and deduction of correct logical propositions from axioms. If youÕre really a mathematician then you already know that there are several different logical systems based on differing foundational axiom sets and their semantics. Assuming your system is consistent, youÕve simply created another one thatÕs neither more nor less correct than the others. What you see as errors in their system results from applying your axiom scheme to someone elseÕs system -- which isnÕt valid. Norm === Subject: Circle-line intersection Hi there! DonÕt tell me to look at other posts in this forum. This is a special problem aimed to they who know math. The question how the get ŌtÕ, look below. I thought I solved the problem but the outcome was a failure. I got a correct equation from mathworld but that is not what I want though. A line(L) intersects with a circle(C), tell me if something is wrong. All positions orients from circle position which is (0,0). t is what I want. L = A + (B-A)*t where D = B-A Lx = cos(v)*R Ly = sin(v)*R A combination of equations: cos(v)*R = Ax + (Dx)*t sin(v)*R = Ay + (Dy)*t I do this because the points of the line intersects on the circle-line which distance is R from origo. Ōcos(v)*Ris the horizontal distance and sin(v)*R is the vertical distance. Hope you understand my thinking here. Only if could draw with a pen :). And no we do Ōpower of twoat both sides of equation. This I do because of a forumla which IÕve to merge with this equation, see below. cos^2(v)*R^2 = (Ax + Dx*t)^2 which also are ((Ax + Dx*t)^2) cos^2(v) = -------------------- R^2 sin^2(v)*R^2 = (Ay + Dy*t)^2 which also are ((Ay + Dy*t)^2) sin^2(v) = -------------------- R^2 Ok now to the hard part. This is a math rule: sin^2(v) + cos^2(v) = 1 and this gives cos^2(v) = 1 - sin^2(v) Now we combine the first combination with the formula above: ((Ax + Dx*t)^2) cos^2(v) = -------------------- R^2 will be ((Ax + Dx*t)^2) 1 - sin^2(v) = -------------------- R^2 And know we just take away the sin-function: ((Ay + Dy*t)^2) ((Ax + Dx*t)^2) 1 - -------------------- = -------------------- R^2 R^2 Now I can get ŌtÕ. The equation will be, after some ßipping and ßopping of the variables, look like this: t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 If you donÕt see what this is I tell you this: ax^2 + bx + c = 0 Now if you know what math is you know what it is. Now the question: Why DOESNÕT THIS WORK!!! :| === Subject: Re: Circle-line intersection days. My association with the Department is that of an alumnus. >A line(L) intersects with a circle(C), tell me if something is wrong. >All positions orients from circle position which is (0,0). You might want to say what it is you are trying to do... >t is what I want. >L = A + (B-A)*t >where D = B-A There was no D so far.... You mean: L is a line going through the points A and B, and we obtain it (by vector addition) as A + Dt, where D = (B-A) (A, B being vectors with initial point at the origin and terminal point at the point A and B, presumably, and t being a real number parameter...) Is that right? >Lx = cos(v)*R >Ly = sin(v)*R What are Lx and Ly supposed to mean? What is R? What is v? I assume Lx means the x coordinate of the point on the line, and Ly means the y coordinate (as below you have Ax, Dx, Ay, Dy, presumably being x coordinate of A, x coordinate of D, etc). Meaning... you are trying to figure out the intersection? (x,y) are the coordinates of the intersection? And then v is simply the angle, R the radius of the circle? Sounds reasonable to me... >A combination of equations: cos(v)*R = Ax + (Dx)*t > sin(v)*R = Ay + (Dy)*t >I do this because the points of the line intersects on the circle-line >which distance is R from origin. Ōcos(v)*Ris the horizontal distance >and sin(v)*R is the vertical distance. Hope you understand my thinking >here. Only if could draw with a pen :). LetÕs see. We are working on the plane. You are describing the circle as being given by all points of the form (R*cos(v),R*sin(v)), as v ranges over, say, 0 to 2pi. You are describing the line as going through the points A=(a_1,a_2) and B=(b_1,b_2), so you let D = B-A = (b_1-a_1,b_2-a_2) = (d_1,d_2), and the line becomes (x,y) = (a_1,a_2) + t(d_1,d_2) where t ranges over all real numbers. So any intersection between the circle and the line would have to satisfy a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) for some value of t and some value of v. >And no we do Ōpower of twoat both sides of equation. You are squaring both equations... Okay... HOWEVER: remember that because you squared the original equations, you may have introduced new solutions that were not part of your original intersections. For example, if you find a point where R*(cos(v)) = - (a_1+td_1) R*(sin(v)) = (a_2 + td_2) then these values of v and t will ALSO satisfy R^2*cos^2(v) = (a_1+td_1)^2 R^2*sin^2(v) = (a_2+td_2)^2 even though they do not satisfy the original equation. In fact, the solutions to the following four systems: ORIGINAL: a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) Plus a_1 + td_1 = -R*cos(v) a_2 + td_2 = R*sin(v) a_1 + td_1 = R*cos(v) a_2 + td_2 = -R*sin(v) and a_1 + td_1 = -R*cos(v) a_2 + td_2 = -R*sin(v) will also satisfy (a_1 + td_1)^2 = R^2*cos^2(v) (a_2 + td_2)^2 = R^2*sin^2(v) So ->not every solution you find after squaring needs to be a solution of your original equation<-. Every solution to your original equation will also be a solution to the new one, but the new one may have more solutions. (ItÕs like, if you start with x= 3, there is only one solution; but if you square it, you get x^2 = 9, and now both x=3 and x=-3 are solutions). >This I do >because >of a forumla which IÕve to merge with this equation, see below. > cos^2(v)*R^2 = (Ax + Dx*t)^2 > which also are > ((Ax + Dx*t)^2) > cos^2(v) = -------------------- > R^2 > > sin^2(v)*R^2 = (Ay + Dy*t)^2 > which also are > ((Ay + Dy*t)^2) > sin^2(v) = -------------------- > R^2 Fine. Since a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) then cos(v) = (a_1 + t*d_1)/R sin(v) = (a_2 + t*d_2)/R >Ok now to the hard part. >This is a math rule: > sin^2(v) + cos^2(v) = 1 Identity. Yes. >and this gives > cos^2(v) = 1 - sin^2(v) Yes. >Now we combine the first combination with the formula above: > ((Ax + Dx*t)^2) > cos^2(v) = -------------------- > R^2 > > will be > > ((Ax + Dx*t)^2) > 1 - sin^2(v) = -------------------- > R^2 >And know we just take away the sin-function: > ((Ay + Dy*t)^2) ((Ax + Dx*t)^2) > 1 - -------------------- = -------------------- > R^2 R^2 Easier to just add the two equations you had: a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) we have R^2 = R^2(cos^2(v) + sin^2(v) = (a_1+td_1)^2 + (a_2 + td_2)^2. >Now I can get ŌtÕ. The equation will be, after some ßipping and >ßopping of the variables, look like this: >t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 >If you donÕt see what this is I tell you this: ax^2 + bx + c = 0 >Now if you know what math is you know what it is. >Now the question: Why DOESNÕT THIS WORK!!! :| Looks right. What do you mean by This does not work? Maybe you mean you sometimes get wrong answers? ThatÕs because of the observation above. You have to make sure that whatever answers you get actually satisfy the original equations. Plug them in. They should work fine. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Bessel Numbers, Borel Numbers. > What are Bessel Numbers, and Borel Numbers? How do you generate them? -- > ------------------------------- > Patrick D. Rockwell > What is Google, and how do you use it? >> I already did! I couldnÕt find a straightforward answer that I could make >> sense of. I want to see if someone here can give me a better >> answer. > My results from Google... > Bessel numbers > http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cg > i?Anum=A006789 > BorelÕs number > http://www.cs.auckland.ac.nz/CDMTCS/chaitin/summer.html used Google to find that page. One of the ones that I couldnÕt make much sense of. The page gives these numbers 1,1,2,5,14,43,143,509,1922,7651,31965,139685,636712,3020203, 14878176,75982829,401654560,2194564531,12377765239, 71980880885,431114329728,2656559925883,16825918195484, 109439943234749,730365368850192 It says that Bessel numbers are the number of nonoverlapping partitions of an n-set into equivalence classes. That much I understand, but it then gives this generating function G.f. 1/(1-x-x^2/(1-2x-x^2/(1-3x-x^2/...))) (a continued fraction). Could this generate the above numbers? I downloaded a PDF file which defines Bessel numbers in the following way. Let the Bessel number of the second kind B(n, k) be the number of set partitions of [n] into k blocks of size one or two, and let the Bessel number of the first kind b(n, k) be the coefficient of xn?k in ?y(n)?1(?x) , where y(n)(x) is the nth Bessel polynomial. In this paper, we show that Bessel numbers satisfy two properties of Stirling numbers: The two kinds of Bessel numbers are related by inverse formulas, and both Bessel numbers of the first kind and the second kind form log-concave sequences. By constructing sign-reversing involutions, we prove the inverse formulas. We review KrattenthalerÕs injection for the log-concavity of Bessel numbers of the second kind, and give a new explicit injection for the log-concavity of signless Bessel numbers of the first kind. It then defines b(n,k) = (-1)^(n-k)*(2*n-k-1)!/((2^(n-k)(n-k)!(k-1)!) if 1<=k<=n and 0 if 1<=n<=k B(n,k)=n!/((2^(n-k)*(n-k)!(2k-n)!) if (n/2)<=k<=n, 0 otherwise. -- ------------------------------- Patrick D. Rockwell === Subject: Re: Bessel Numbers, Borel Numbers. > What are Bessel Numbers, and Borel Numbers? How do you generate them? >> -- >> ------------------------------- >> Patrick D. Rockwell > > What is Google, and how do you use it? > I already did! I couldnÕt find a straightforward answer that I could > make > sense of. I want to see if someone here can give me a better > answer. >> My results from Google... >> Bessel numbers >> http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cg >> i?Anum=A006789 >> BorelÕs number >> http://www.cs.auckland.ac.nz/CDMTCS/chaitin/summer.html > used Google to find that page. One of the ones that I couldnÕt make much > sense of. The page gives these numbers > 1,1,2,5,14,43,143,509,1922,7651,31965,139685,636712,3020203, > 14878176,75982829,401654560,2194564531,12377765239, > 71980880885,431114329728,2656559925883,16825918195484, > 109439943234749,730365368850192 > It says that Bessel numbers are the number of nonoverlapping > partitions of an n-set > into equivalence classes. > That much I understand, but it then gives this generating function > G.f. 1/(1-x-x^2/(1-2x-x^2/(1-3x-x^2/...))) (a continued fraction). Could > this generate the above numbers? > I downloaded a PDF file which defines Bessel numbers in the following way. > Let the Bessel number of the second kind B(n, k) be the number of set > partitions of [n] into k > blocks of size one or two, and let the Bessel number of the first kind > b(n, k) be the coefficient of xn?k > in ?y(n)?1(?x) , where y(n)(x) is the nth Bessel polynomial. In this > paper, we show that Bessel numbers > satisfy two properties of Stirling numbers: The two kinds of Bessel > numbers are related by inverse > formulas, and both Bessel numbers of the first kind and the second kind > form log-concave sequences. > By constructing sign-reversing involutions, we prove the inverse formulas. > We review KrattenthalerÕs > injection for the log-concavity of Bessel numbers of the second kind, and > give a new explicit injection for > the log-concavity of signless Bessel numbers of the first kind. > It then defines b(n,k) = (-1)^(n-k)*(2*n-k-1)!/((2^(n-k)(n-k)!(k-1)!) if > 1<=k<=n and 0 if 1<=n<=k > B(n,k)=n!/((2^(n-k)*(n-k)!(2k-n)!) if (n/2)<=k<=n, 0 otherwise. > -- > ------------------------------- > Patrick D. Rockwell I forgot to mention in my last message that I canÕt get either of the formulas that I quoted to generate any of the numgers that I listed in the sequence above. -- ------------------------------- Patrick D. Rockwell === Subject: Re: Bessel Numbers, Borel Numbers. > What are Bessel Numbers, and Borel Numbers? How do you generate > them? -- > ------------------------------- > Patrick D. Rockwell > What is Google, and how do you use it? >> I already did! I couldnÕt find a straightforward answer that I could >> make >> sense of. I want to see if someone here can give me a better >> answer. > My results from Google... > Bessel numbers > http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cg > i?Anum=A006789 > BorelÕs number > http://www.cs.auckland.ac.nz/CDMTCS/chaitin/summer.html >> used Google to find that page. One of the ones that I couldnÕt make much >> sense of. The page gives these numbers >> 1,1,2,5,14,43,143,509,1922,7651,31965,139685,636712,3020203, >> 14878176,75982829,401654560,2194564531,12377765239, >> 71980880885,431114329728,2656559925883,16825918195484, >> 109439943234749,730365368850192 >> It says that Bessel numbers are the number of nonoverlapping >> partitions of an n-set >> into equivalence classes. >> That much I understand, but it then gives this generating function >> G.f. 1/(1-x-x^2/(1-2x-x^2/(1-3x-x^2/...))) (a continued fraction). Could >> this generate the above numbers? >> I downloaded a PDF file which defines Bessel numbers in the following >> way. >> Let the Bessel number of the second kind B(n, k) be the number of set >> partitions of [n] into k >> blocks of size one or two, and let the Bessel number of the first kind >> b(n, k) be the coefficient of xn?k >> in ?y(n)?1(?x) , where y(n)(x) is the nth Bessel polynomial. In this >> paper, we show that Bessel numbers >> satisfy two properties of Stirling numbers: The two kinds of Bessel >> numbers are related by inverse >> formulas, and both Bessel numbers of the first kind and the second kind >> form log-concave sequences. >> By constructing sign-reversing involutions, we prove the inverse >> formulas. We review KrattenthalerÕs >> injection for the log-concavity of Bessel numbers of the second kind, and >> give a new explicit injection for >> the log-concavity of signless Bessel numbers of the first kind. >> It then defines b(n,k) = (-1)^(n-k)*(2*n-k-1)!/((2^(n-k)(n-k)!(k-1)!) if >> 1<=k<=n and 0 if 1<=n<=k >> B(n,k)=n!/((2^(n-k)*(n-k)!(2k-n)!) if (n/2)<=k<=n, 0 otherwise. >> -- >> ------------------------------- >> Patrick D. Rockwell > I forgot to mention in my last message that I canÕt get either of the > formulas that I quoted > to generate any of the numgers that I listed in the sequence above. > -- > ------------------------------- > Patrick D. Rockwell I suspect you donÕt know what a generating function is. Its not a formula where you substitute some value for x and it provides the value of the fumction. Its a formula where the co-efficients on the polynomial expansion provide the terms you need ... sounds like you need to http://mathworld.wolfram.com/GeneratingFunction.html I donÕt think anybody is going to write out a lengthy explanation of Bessel or Borel numbers for you, just because you donÕt like (for reasons unspecified) the explanations that people have already published to the web. I think you are going to have to: * Tell us what you donÕt understand in the material already on the web (with pointers), and ask for assistance with those parts. OR * Otherwise rephrase your questions so they are more closed - ie, provide a question that can be meaningfully answered in a few lines OR * Give some context to your question, so the people in this newsgroup can see what is relevant and what you actually need to know about the numbers OR * Learn to use Google and think for yourself. === Subject: Shiing-shen Chern (Chen Xingshen) passes away at 93 Just grabbed off Xinhua web site: Noted mathematician passes away in Tianjin TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a world-renowned overseas Chinese mathematician, 93, died of illness at his home at Nankai University in north ChinaÕs Tianjin Municipality at around 7:15 p.m. Friday, the university announced. Chern, a US citizen, is best known for his achievements in the study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang Province, east China. He graduated from Nankai University in 1930 and received further education at Qinghua University and the University of Hamburg in Germany. He taught at several Chinese and US universities -- including Princeton University, the University of Chicago, and the University of California, Berkeley -- and is the only Chinese to win the Wolf Prize -- the most distinguished award in the international mathematics field. The International Astronomical Union officially named asteroid No. 1998CS2 after the noted mathematician in November for his outstanding contributions to human society. === Subject: Re: Shiing-shen Chern (Chen Xingshen) passes away at 93 > Just grabbed off Xinhua web site: > Noted mathematician passes away in Tianjin > TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a > world-renowned overseas Chinese mathematician, 93, died of illness at > his home at Nankai University in north ChinaÕs Tianjin Municipality at > around 7:15 p.m. Friday, the university announced. > Chern, a US citizen, is best known for his achievements in the > study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang > Province, east China. He graduated from Nankai University in 1930 and > received further education at Qinghua University and the University of > Hamburg in Germany. > He taught at several Chinese and US universities -- including > Princeton University, the University of Chicago, and the University of > California, Berkeley -- and is the only Chinese to win the Wolf Prize -- > the most distinguished award in the international mathematics field. > The International Astronomical Union officially named asteroid No. > 1998CS2 after the noted mathematician in November for his outstanding > contributions to human society. I had the privilege of taking an upper division class (upper division = 3rd or 4th year undergrad math major) from Professor Chern in 1976 at U.C Berkeley. I remember him as a warm, down to earth man. Approachable to us students. He cared about his students and about people. IÕm sad to hear of his passing. === Subject: Re: Shiing-shen Chern (Chen Xingshen) passes away at 93 Moment of Silence. W. Dale Hall ??? > Just grabbed off Xinhua web site: > Noted mathematician passes away in Tianjin > TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a > world-renowned overseas Chinese mathematician, 93, died of illness at > his home at Nankai University in north ChinaÕs Tianjin Municipality at > around 7:15 p.m. Friday, the university announced. > Chern, a US citizen, is best known for his achievements in the > study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang > Province, east China. He graduated from Nankai University in 1930 and > received further education at Qinghua University and the University of > Hamburg in Germany. > He taught at several Chinese and US universities -- including > Princeton University, the University of Chicago, and the University of > California, Berkeley -- and is the only Chinese to win the Wolf Prize -- > the most distinguished award in the international mathematics field. > The International Astronomical Union officially named asteroid No. > 1998CS2 after the noted mathematician in November for his outstanding > contributions to human society. === Subject: Re: sin(x^x) Hi. Is there a formula that describes the density of > the graph of sin(x^x) for real x>0 as x increases > without bound? > [...] Topologically, it is nowhere dense. In terms of two-dimensional area (say, > Lebesgue measure), it has measure 0. Fractal-wise, it is of Hausdorff dimension 1. (And it is a real-analytic submanifold of the plane, > perhaps with some care taken about the missing left endpoint). Any other criteria? If it means how close the subsequent roots (or the increasing and > decreasing segments) are, sit down and do the arithmetic, using the > Lambert W-function. For your convenience, the explicit formula for y^y=x (x sufficiently > large) is y = ln(x) / lambertW(ln(x)) > By density I meant distance between the peaks of the waves > as x -> inf. Obviously, the distance -> 0 (and so the density -> inf) as > x -> inf but what is the equation that describes that density? I would guess itÕs something like this: sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak > whenever x^x = npi/2, with n a positive integer. Caution: you have listed not only local maxima, but also roots and > local minima. Sort them out: > Ohh... because of n = 2, then you get sin(pi) = 0, then n = 3 -> sin(3pi/2) > = -1, > then n = 4 -> sin(2pi) = 0, n = 5 -> sin(5pi/2) = 1, etc. > u = (4*k+1)*pi/2 (k integer) gives local maxima of sin(u). > You can complete the list. > So it should be > D(n) = 1/(ln((4n+2)pi/2)W(ln((4n+2)pi/2)) - > ln((4n+1)pi/2)/W(ln((4n+1)pi/2))). Replace (4n+2) by (4n+5), because you need to use 4(n+1)+1, rather than (4n+1)+1. > [nothing added by me below] > Then the nth peak is at x = ln(npi/2)/W(ln(npi/2)) So the Ōdensityfunction IÕm looking for would then > be D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))). Would that work? > === Subject: Re: Elementary proof of R(3,7)=23? > Is there an elementary proof of Ramsey Number R(3,7)=23? I founded the one > with R(3,6)=18 and hope to construct a similar one, but probably need to get > the complete classification of all (3,6,17)-good graphs... I donÕt know what you call elementary, but perhaps you should look here: S. P. Radziszowski, D. L. Kreher, On R(3,k) Ramsey graphs: theoretical and computational results, Journal of Comb. Math. and Comb. Comp., 4 (1988), 37-52. Also a good reference for the bibliography is: S. Radziszewski, Small Ramsey numbers, Electron. J. Comb., 1 (1994), 1-30. Pawel Gladki === Subject: Re: Elementary proof of R(3,7)=23? > Is there an elementary proof of Ramsey Number R(3,7)=23? I founded the one > with R(3,6)=18 and hope to construct a similar one, but probably need to get > the complete classification of all (3,6,17)-good graphs... > passed away... This news deserves to be brought to the level of a named thread. Dale === Subject: Re: Graphing polynomial equations >xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 >Obviously you would loop through x and y and calculate >z, and weÕd use a numerical approximation technique to >do that. IÕm trying to write a program that would plot >the graph. The problem boils down to applying a >numerical algorithm to a quintic like this: >z^5 + a4 z^4 + a3 z^3 + a2 z^2 + a1 z + a0 = 0. >But the problem IÕm having is this: Most numerical >approximation algorithms require an initial guess. So, what >would be a good algorithm to find a good initial >guess? You probably realize that a degree-5 polynomial can have up to 5 zeroes, so you would want 5 (or maybe more) initial guesses. The high and low guesses should be fairly easy, because thereÕs no such thing as too high or too low for the tangential method (sorry I forgot who discovered that or IÕd give him credit!). Even if youÕre using some other numerical method you can still use the tangential results to seed it. You can get the other possibilities by looking at the zeroes of the derivative, which are potentially relative mimima and maxima. Any zeroes other than the first and last will be between a relative maximum and a relative minimum. You may have to apply this process recursively until you get down to degree 2 where you can use the quadratic equation. Once youÕre started, you can use results from neighboring (x,y) as guesses. They wonÕt always work out, but when they do it will save time. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Fermat FermatÕs Last Theorem Ben Ito 12-1-04 I will show that FermatÕs n=4 is invalid. FermatÕs n=4 proof is based on a deception that alters the variables to imply that the n=2 and n=4 equations are identical equations.However, the n=2 and n=4 equations are completely different equations; therefore, the integer solution equations of n=2 cannot be used to prove n=4 as Fermat has done. Fermat uses the elliptical curve equation to prove that the area of a right triangle (XY)/2 = d^2 (equ 1) does not form an integer value of d. Fermat uses the integer solution equation of n=2 in the area equation yet the area equation (equ l) is completely different form the n=2 length equation X^2 + Y^2 = Z^2. (equ 2) Fermat is using the same method used in his n=4 proof where he uses the n=2 solutions in the n=4 equation; however, both equations are completely different. Consequently, FermatÕs area proof, that uses an elliptic curve equation, is invalid. Wiles proof of FermatÕs Last Theorem is based on the elliptic curve equation , Y^2 = X(X - a^n)(X + b^n) = X^3 + X^2(b^n - a^n) - X(ab)^n (equ 3) however, this equation does not include the c^n variable of FermatÕs equation a^n + b^n = c^n, (equ 4) and (ab)^n is also not part of FermatÕs equation (equ 4). Therefore, Wilesproof of FermatÕs last theorem is invalid. I will prove FermatÕs last theorem using mathematical quantitative analysis. The n=1 and n=2 solutions, and near solutions of n=3 are described; I will then show a pattern using mathematical quantitative analysis to assess the distribution of the integer solutions, of n=2, only the n=1 and n=2 form integer solutions using the solution patterns formed by the distribution of the integer solutions of n=1 and n=2 and the near solutions of n=3. l. Introduction Fermat studied law at a univerity in Orl.8eans. He received a degree in civil law and in 1631 was a councillor at the parliament in Toulouse, France. In his spare time Fermat was allegedly a mathematician. Fermat attempt at proving that the n=4 equations do not form integer solutions is based on a deception where he alters the variables that are being used, and also does not include all integer combinations in his proof of n=4. Therefore, FermatÕs n=4 proof violates logic and is incomplete; consequently, FermatÕs n=4 proof is invalid. FermatÕs last theorem states that X^n + Y^n = Z^n, (equ 5) when n>2 does not form integer solutions of X, Y and Z. However, elliptic curve derivation, it is apparent that he did not have a proof for n>2. In all of his attempts at forming a proof, n=4 and area proof, Fermat always used the integer solution equations of n=2; therefore, Fermat did not understanding the basic underlining principle of the problem that there are an infinite number of possible integer solution combinations that form an infinite number of equations; therefore, it is impossible to prove n=4 or n>2. FermatÕs contradiction proof is base on a set of equations that form solutions to the n=4 equation yet does not form integer solutions; however, the equation used are derived from the integer solutions of n=2. Fermat acheives this contradicts by altering the variables. I will show the reader that foundation principles of FermatÕs n=4 and Wiles proofs are invalid; consequently, if the foundation of any proof is found to be invalid than the proof of this invalid foundational statement supported by the remaining of the proof is also invalid. I will show that FermatÕs n=4 and Wilesproofs are invalid, using this principle, then prove FermatÕs last theorem using mathematical quantitative analysis based on experimental physics. 2. FermatÕs n=4 Proof FermatÕs n=4 proof is described. Fermat implies that by proving that, X^4 + Y^4 = Z^2 (equ 6). does not form integer solutions also proves that X^4 + Y^4 = Z^4 (equ 7) does not form integer solutions (Shanks, p. 144). However, using n=4 in equation 5 forms equation 7 yet Fermat uses equation 6 to prove n=4; however. equations 6 and 7 are completely different equation; therefore, Fermat does not prove n=4. Fermat uses the integer solution equations of n=2, a = 2uv, b = u^2 - v^2, and c = u^2 + v^2 (equ 8a,b,c), and, X^2 = 2uv, Y^2 = u^2 - v^2, and Z = u^2 + v^2 (equ 9a,b,c), (Shanks, p.141). Equation 5a,b,c are then used to prove n=4; however, Fermat squares the right sides of equations 8a,b,c without squaring the left side. Using equations 8a,b and 9a,b, a = X^2, b = X^2 and c = Z. (equ 10a,b,c) Inserting equations 10a,b,c in equation 6, a^2 + b^2 = c^2 (equ 11) Fermat is implying that the equation of n=2 is equal to the equation of n=4. The formation of equations 9a,b,c is a key step in FermatÕs contradiction method (Osserman, p. 18). Fermat is implying that when n=4, a^2 = X^4, (equ 12) Fermat is implying that when the value of n increases the value of X can be altered. The alteration in the variables in the key to FermatÕs proof. However, using a = 9^(1/n), b = 16^(1/n) and c = 25^(1/n) (equ 13a,b,c) in a^n + b^n = c^n----------------> 9 + 16 = 25 (14a,b) Equation 14 is a solution since the variables (a,b,c) of equation 13a,b,c are being alter when n increases. However, when n>2 equations 13a,b,c never form integer values of a,b and c. The solution of equation 14 are essential in FermatÕs proof; however, Fermat is altering the variables to form the solution of equation 6. Fermat n=4 proof is not a proof but a manipulation of the variables. In addition, Fermat is using equations (equ 9a,b,c), that right side of the equation is equivalant to the n=2 integer solution equation, are derived from equation 5a,b,c to prove n=4; however, X^2 + Y^2 = Z^2 (equ 11) completely different from, X^4 + Y^4 = Z^2 (equ 12). Using Z = 6, equation 9 forms a circle of radius 6, X^2 + Y^2 = 36.(equ 13) Using Z = 6 in equation 10, X^4 + Y^4 = 36,(equ 14) equation 12 does not form a circle; therefore, equation 12 is completely different from equation 11. The integer solutions of n=2 cannot be used to prove n=4 since equations 11 and 12 are completely different equations. Therefore, equations 6a,b,c cannot be used to prove n=4. Fermat proof is implying that the equations of n=2 and n=4 are equal using equations 7a,b. Therefore, FermatÕs n=4 proof is invalid. Also, FermatÕs n=4 proof does not testing all integer combinations of X and Y. FermatÕs proof is only proving equations 5a,b,c, that are derived from n=2, do not form integer solutions. However, there are an infinite number of integer combinations that are not tested in FermatÕs n=4 proof; example, the integers X = 22, Y = 23 and C=24 are not included in FermatÕs n=4 contradiction proof; Fermat does not prove that (22,23,24) does not form integer solution. Fermat is implying that by manipulating the integer solutions of n=2 he has proven that n=4 does not form integer solutions; however, n=2 solutions do not include all possible integer combinations of X and Y. Example, using n=4 (X^4 + Y^4)^(1/4)=C (equ 15) Let Y = X + 1, (equ 16) inserting equation 13 in equation 12, [(X + 1)^4 + X^4]^(1/4) = C --------> (2X^4 + 4X^3 + 6X^2+ 4X + 1)^(1/4) = C (equ 17) Using X = 1,2,3,4,5,6,7,8,9,.................,(equ 18) There are an infinite number of equations that describe integer combinations that are not included in FermatÕs n=4 proof. Example, using f(X) = X + 2, f(X) = X^2 + 1, f(X) = X^2 + X, f(X) = X^2 + X + 3 etc.............(equ 19) in [(f(A))^4 + A^4]^(1/4) = C (equ 20) FermatÕs n=4 proof uses (equ 9a,b,c) that represents the integer solutions of n=2;therefore, not all possible integer combinations are represented in FermatÕs n=4 proof. The proof for n=4 must include all possible integer combinations of X and Y. However, there are an infinite number of integer combinations that are not included in FermatÕs n=4 proof. Fermat is justifying the non-existence of integer solutions of n=4 using a contradiction method where a single group of equations (equ 5a,b,c) are used; however, FermatÕs n=4 proof does not include all possible integer combinations of X and Y; therefore, FermatÕs n=4 proof in incomplete and therefore, invalid. Fermat realized that the n>2 was not possible to prove using his n=4 or any other algebraic method. There are two possibilities in solving FermatÕs proof; one is by algebraically proving n>2 does not form solutions and the second method would be to show that only a right triangle forms integer solution; therefore, Fermat released that the n>2 prove was algebraically impossible; therefore, he went for the second possibility by using the right triangle area and the elliptic curve. 3. WilesProof Wiles uses elliptic curve equation to prove n>2 since probably Fermat used an elliptic curve. FermatÕs use of the elliptic curve equation is described. Fermat states that a right triangle has integer sides, X and Y; (XY)/2 = d^2 (equ 12) the area described with (XY)/2 does not form an integer value of d. Fermat then uses the integer solution equations of n=2, X = (m^2 - n^2), Y = 2mn, Z = (m^2 + n^2),(equ 13) in equation 12, the second equation takes the form: 2d^2 = XY = (m^2 - n^2) x 2mn, (equ 14) or d^2 = (nm^3 - mn^3).(equ 15) We now let X = m/n, Y = (d/n^2).(equ 16) Then X^3 - X = (m^3)/(n^3) - m/n = (nm^3 - mn^3)/n^4 = (d^2)/n^4 = Y^2 (equ 17) which forms the elliptic curve equation, Y^2 = X^3 - X, (equ 17) However, equation 12 is not equal to equation 6.Equation 12 represent the area of a right triangle; whereas, equation 6 represents the length of the sides of the right triangles The n=2 equation is not the area equation; therefore, Fermat violates logic by using the integer solution equation (equ ) in the area equation to prove that the area equation (equ ) does not from integer solutions. Fermat is using the same method used in his n=4 proof where he uses the n=2 solutions in a completely different equation (equ 12); therefore, FermatÕs area integer proof is invalid. Mathematician donÕt stray to far from home and this is also the case with Wiles. Wiles implied that since Fermat uses elliptic curve equation that the elliptic curve would be used to prove n>2. Wiles proof of FermatÕs Last Theorem is based on the elliptic curve equation (Poorten, p. 196-7), Y^2 = X(X - a^n)(X + b^n) (equ 18). However, FermatÕs equation is a^n + b^n = c^n (equ 19). it is not mathematical possible to derive equation from equation 18 since equation 18 becomes, Y^2 = X^3 + X^2(b^n - a^n) - X(ab)^n (equ 20) therefore, Wilesproof is invalid since equation 20 does not include the c^n variable of FermatÕs equation (equ 19). In addition, equation is the sum of a^n and b^n; whereas, equation 20 forms b^n - a^n that are not sums. Also. the (ab)^2 of equation 20 is not part of equation 19;therefore, Wiles proof ( Wiles, p. 448) that uses equation 18 is invalid. 4. Proof of FLT. I will prove FermatÕs last equation using the pattern formed by the solutions formed by n=1 and n=2, and the near solutions of n=3 and n=4. Using n=1 in equation 1, X + Y = Z (equ 21) All integers of equation 21 form solutions. An integer solution set occurs at (1,2,3). The integer solutions of n=2 are described with the following equations, X = 2uv, Y = u^2 - v^2 and Z = u^2 + v^2 (equ 22) Not all integers form integer solutions for n=2. An integer solution set occurs at (3,4,5). The n=1 and n=2 solutions show a pattern where integer solutions are formed when a solution set is forms close to the origin. In addition, when n=1, the strength of the solution set formation is maximum since all integers form solutions; when n=2 not all integers form solutions; therefore, the strength of the solution formation is weakening. When n=3, 6^3 + 8^3 = 728 where 9^3 = 729 (equ 23) 9^3 + 10^3 = 1729 where 12^3 = 1728 (equ 24) an integer solution set does not form close to the origin for n=3;however, equations 23 and 24 are off by one. The solution set is weakening as n increases; at n=4 7^4 + 8^4 = 6497 where 9^4 = 6561(equ 25) consequently, since n=3 does not from solutions near the origin, the integer sequence ends at n = 3. Therefore, only n=1 and n=2 form integer solutions. 5. Conclusion I have shown that FermatÕs derivation of n=4 is base on deception since Fermat uses non-integer solutions of AÕ, Band C to prove that FermatÕs n=4 equation does not from integer solution. In addition, Fermat is using the integer solution of n=2 to prove n=4 which violates logic since n=2 and n=4 form completely different equations. FermatÕs n=4 proof does not testing all integers. FermatÕs proof is only proving equations 5a,b,c do not form integer solutions. It is impossible to prove n=4, using FermatÕs method, since it would require an infinite number of equations; therefore, FermatÕs proof of n=4 is invalid. Wiles proof of FermatÕs Last Theorem is based on the elliptic curve equation , Y^2 = X(X - a^n)(X + b^n) = X^3 + X^2(b^n - a^n) - X(ab)^n (equ 26) However, equation 26 does not include the c^n variable of FermatÕs equation a^n + b^n = c^n (equ 27). Therefore, Wilesproof of FermatÕs last theorem using equation 26 is invalid. I will prove FLT by showing a pattern forms from the n=1 and n=2 solution sets. All of the integer of n=1 form solutions; when n=2 the number of solutions in a finite range is less then that of n=2; therefore, the strength of the solutions is decreasing as n increases. When n=3 the solution sequence ends and solutions are only formed when n = 1 and n = 2 . 6. References Robert Osserman. FermatÕs Last Theorem (a supplement to the video). MSRI Berkeley. 1994 Marilyn vos savant. The WorldÕs Most Famous Math Problem. St MartinÕs Press. 1993 Daniel Shanks. Solved and Unsolved Problems in Number Theory. Chelsea Pub. 1985. A. J. Van Der Poorten. Notes on FermatÕs Last Theorem. John Wiley. 1996 Andrew J. Wiles. Annals of Mathematics. Editor Andrew Wiles. Princeton University Press. May 1995. 7. Acknowledgment forum, Best Science forum, and About Physics forum, HSU, CSUS, CR, SCC, USC, Hiram Johnson HS Sacramento (Mrs Larson), UCD, Stanford, MIT, Harvard, ASU, Rutgers and UCLA mathematics Dept. Memory of Yasser Arafat 11-11-04 *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Number of prime factors of an odd perfect number Content-Length: 215 Originator: rusin@vesuvius What is the known upper bound on the number of prime factors / distinct prime factors of an odd perfect number? Does anyone have a conjecture on what the tighest lower bound would be? Edmond === Subject: Re: Number of prime factors of an odd perfect number Originator: israel@math.ubc.ca (Robert Israel) > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? > Edmond The state of the art about odd perfect numbers seems here http://mathworld.wolfram.com/OddPerfectNumber.html === Subject: Re: Number of prime factors of an odd perfect number posting-account=tN6qcA0AAACUaa8g28ioHaW6IKJhmtDk > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? > Edmond YouÕve already received some excellent replies, but it might be of interest to note that any constant upper bound here would imply there are only finitely many odd perfect numbers: Dickson showed that there are at most finitely many odd perfect numbers with a prescribed number of distinct prime factors. DicksonÕs result can be made effective. A striking result of Heath-Brown is that an odd perfect number with k distinct prime factors is bounded by 4^{4^k}. (The proof is entirely elementary.) This has been improved to 2^{4^k} by Nielsen in [1]. Hope this helps, Paul [1] Nielsen, Pace An Upper Bound for Odd Perfect Numbers http://www.emis.de/journals/INTEGERS/papers/d14/d14.pdf === Subject: Re: Number of prime factors of an odd perfect number > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? I donÕt know much about recent advances, but there are a few classical results: J. J. Sylvester, L. E. Dickson and H. J. Kanold proved that there is no odd perfect number with 4 prime divisors. Next, I. S. Gradshtein, U. K.9fhnel and G. C. Weber disproved existence of numbers with 4 divisors and C. Pomerance and N. Robbins showed the same for 6 divisors. P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 divisors and I donÕt know about any newer results. Pawel Gladki === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous > theorem for 8 divisors and I donÕt know about any newer results. Of course we are talking about _distinct_ prime factors. For the bound of prime factors counted with repetitions it is known that there are at least 47 factors - see Kevin HareÕs preprint: http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf Pawel Gladki === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 >> divisors and I donÕt know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound of > prime factors counted with repetitions it is known that there are at least > 47 factors - see Kevin HareÕs preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf interested in known UPPER BOUNDS and a CONJECTURE on what would be the ACTUAL LOWER BOUND. Edmond === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 >> divisors and I donÕt know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound of > prime factors counted with repetitions it is known that there are at least > 47 factors - see Kevin HareÕs preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the > ACTUAL LOWER BOUND. Known upper bounds? You mean a proof that if there is an odd perfect number it canÕt have more than so many prime factors? I donÕt think IÕve ever seen a result along those lines - have you? As for conjectured lower bounds, I think those would be infinite, as I think the conjecture is that there arenÕt any odd perfect numbers. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for > 8 > divisors and I donÕt know about any newer results. >> Of course we are talking about _distinct_ prime factors. For the bound >> of >> prime factors counted with repetitions it is known that there are at >> least >> 47 factors - see Kevin HareÕs preprint: >> http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf >> interested in known UPPER BOUNDS and a CONJECTURE on what would be the >> ACTUAL LOWER BOUND. > Known upper bounds? You mean a proof that if there is an odd perfect > number it canÕt have more than so many prime factors? Exactly > I donÕt think > IÕve ever seen a result along those lines - have you? No, I havenÕt, but IÕd like to know at least a reasonably tight upper bound. > As for conjectured lower bounds, I think those would be infinite, > as I think the conjecture is that there arenÕt any odd perfect numbers. Let us assume that someone does prove this conjecture. What would be the important corollaries thereof ? Edmond === Subject: Re: Number of prime factors of an odd perfect number days. My association with the Department is that of an alumnus. >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for >> 8 >> divisors and I donÕt know about any newer results. Of course we are talking about _distinct_ prime factors. For the bound > of > prime factors counted with repetitions it is known that there are at > least > 47 factors - see Kevin HareÕs preprint: http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the > ACTUAL LOWER BOUND. >> Known upper bounds? You mean a proof that if there is an odd perfect >> number it canÕt have more than so many prime factors? > Exactly >> I donÕt think >> IÕve ever seen a result along those lines - have you? > No, I havenÕt, but IÕd like to know at least a reasonably tight upper >bound. MyersonÕs point is that, as far as he is aware (and as far as I am aware), there are ->no<- results on upper bounds on the number of prime factors for an odd perfect number. IÕm sure you would like to know one. But if there are none to be found, youÕll have to prove one if you want to know one. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem > for > 8 > divisors and I donÕt know about any newer results. >> Of course we are talking about _distinct_ prime factors. For the >> bound >> of >> prime factors counted with repetitions it is known that there are at >> least >> 47 factors - see Kevin HareÕs preprint: >> http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf >> interested in known UPPER BOUNDS and a CONJECTURE on what would be the >> ACTUAL LOWER BOUND. > Known upper bounds? You mean a proof that if there is an odd perfect > number it canÕt have more than so many prime factors? >> Exactly > I donÕt think > IÕve ever seen a result along those lines - have you? >> No, I havenÕt, but IÕd like to know at least a reasonably tight upper >>bound. > MyersonÕs point is that, as far as he is aware (and as far as I am > aware), there are ->no<- results on upper bounds on the number of > prime factors for an odd perfect number. > IÕm sure you would like to know one. But if there are none to be > found, youÕll have to prove one if you want to know one. I guess I have to set out to prove one myself. But IÕm still curious as to how ( or maybe why ) so much work has been done on the lower bound, whereas we have no results regarding the upper bound. Also, is there a good example of a similar upper bound proof, i.e., an upper bound on some property of a hypothetical number ? Edmond === Subject: Re: Number of prime factors of an odd perfect number days. My association with the Department is that of an alumnus. >> MyersonÕs point is that, as far as he is aware (and as far as I am >> aware), there are ->no<- results on upper bounds on the number of >> prime factors for an odd perfect number. >> IÕm sure you would like to know one. But if there are none to be >> found, youÕll have to prove one if you want to know one. >I guess I have to set out to prove one myself. But IÕm still curious as to >how ( or maybe why ) so much work has been done on the lower bound, whereas >we have no results regarding the upper bound. Because... how would you prove an upper bound on the number of distinct prime factors? The lower bounds are proven by using the fact that if m and n are relatively prime, then sigma(m*n) = sigma(m)*sigma(n), with sigma(k) = sum of all divisors of k, and showing that if q1,...,qs are distinct prime powers (for which it is easy to calculate sigma), then the product of the sigma(qi) simply does not add up to twice q1*...*qs, for small values of s. But how would you prove an upper bound? How would you even ->approach<- such a problem? The upper bound in the even case comes by showing that if you write the number as 2^n*q, with q odd, then the only way this works out is if q is a prime (and, more specifically, a prime of the form 2^{n}-1). It is not really a generalizable argument. > Also, is there a good example >of a similar upper bound proof, i.e., an upper bound on some property of a >hypothetical number ? Yes. Ramsey numbers have well known upper and lower bounds, but many of them are not exactly known. But here, you prove existence by proving an upper bound. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Number of prime factors of an odd perfect number > MyersonÕs point is that, as far as he is aware (and as far as I am > aware), there are ->no<- results on upper bounds on the number of > prime factors for an odd perfect number. > IÕm sure you would like to know one. But if there are none to be > found, youÕll have to prove one if you want to know one. >>I guess I have to set out to prove one myself. But IÕm still curious as to >>how ( or maybe why ) so much work has been done on the lower bound, >>whereas >>we have no results regarding the upper bound. > Because... how would you prove an upper bound on the number of > distinct prime factors? The lower bounds are proven by using the fact > that if m and n are relatively prime, then sigma(m*n) = > sigma(m)*sigma(n), with sigma(k) = sum of all divisors of k, and > showing that if q1,...,qs are distinct prime powers (for which it is > easy to calculate sigma), then the product of the sigma(qi) simply > does not add up to twice q1*...*qs, for small values of s. > But how would you prove an upper bound? How would you even > ->approach<- such a problem? The upper bound in the even case comes > by showing that if you write the number as 2^n*q, with q odd, then the > only way this works out is if q is a prime (and, more specifically, a > prime of the form 2^{n}-1). It is not really a generalizable argument. participants. Edmond === Subject: Re: SmullyanÕs Quiz Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3LRdl09441; >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justified. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. perhaps one should think seriously what unexpected means. to do this one could try to formalise expectations. but this is readily done in elementary game theory, and from there one also gets the wiseness of formalising the situation as a game. with reasonable pay-off functions it is clear that there are no pure strategy equilibria but the professor uses a mixed strategy to determine the date of the exam. the studentsexpectations about the mixed strategy are, of course, correct in equilibrium but they do not know the realisation of the mixed strategy, i.e. the true date of the exam. in most cases then the exam will be a surprise exam in the sense that the studentd cannot predict the date with probability one. only if the realisation is friday they will be able to predict on thursday evening that the exam will be on friday with probability one. the point here is that this is not so much a problem of logic as a problem of formalising the situation correctly. and when one does! formalise it it turns out to be a game. === Subject: Re: SmullyanÕs Quiz Problem >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justified. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. > perhaps one should think seriously what unexpected means. > to do this one could try to formalise expectations. > but this is readily done in elementary game theory, and from there > one also gets the wiseness of formalising the situation as a game. > with reasonable pay-off functions it is clear that there are no pure > strategy equilibria but the professor uses a mixed strategy to > determine the date of the exam. the studentsexpectations about the > mixed strategy are, of course, correct in equilibrium but they do not > know the realisation of the mixed strategy, i.e. the true date of the > exam. in most cases then the exam will be a surprise exam in the sense > that the studentd cannot predict the date with probability one. > only if the realisation is friday they will be able to predict on > thursday evening that the exam will be on friday with probability > one. the point here is that this is not so much a problem of logic > as a problem of formalising the situation correctly. and when > one does formalise it it turns out to be a game. Ok, but this does not change the basic nature of the paradox. Consider the following form of the paradox (I prefer this as it avoids discussion of prediction and states of knowledge, gives a more concrete meaning to trustworthy, and incorporates the idea of mixed strategy). We have two players A and B, and two cards, a king and an ace. A orders the cards and places them face down. The cards are turned over until the king is turned over. At this point the game ends. Before each card is turned over, B names an integer between 0 and 100. This is used as the % probability that B will place a $1 bet at even odds that the next card turned over will be the king. BÕs best strategy is to say 50 before the first card is turned over and 100 if the second card is to be turned over. No matter what strategy A uses, this ensures B an average return of $0.5 (So the game is worth $0.5 to B). A is trustworthy. That is if A says the first card is the king B will say 100 before the first card is turned over. AÕs problematic statement is (after he has placed the cards face down) If you act rationally, you will not say 100 during this play of the game B argues If the first card is the ace, then I will certainly say 100 before the second card is turned over, so the first card must be the king, so rationally I should say 100 before the first card is turned over. But then A knows his statement is always false, so I should not draw any inference from it and say 50 before the first card is turned over. But then A know that his statement is only true if the first card is the king, so I should treat his statement the same as if he had said the first card is the king and say 100 before the first card is turned over. But then A knows his statement is always false ... Surely it is not rational for B to refuse to play a game that has positive expectation. The unappealing, but seemingly inescapable conclusion is that B cannot behave rationally! -William Hughes === Subject: Re: Vectors question 2 >>So after reading your response I began to think. Ignoring vectors and i >>& j etc. If I wanted to figure this out a few weeks ago (prior to >>learning this stuff) my method would have been as follows. >>Put the direction of the train on the x-axis, the direction of the >>stuntman on the y-axis. (this is I think what you are trying to say). > That method is correct, and produces the correct answers... so now my question > is: what is it that you learned over the last few weeks that confused you into > trying to plot time as well? > meeroh To answer your question, I overthought the question. We are studying vectors, including 3-dimensional vectors and projections. Perhaps it was because I had only just finished studying projections. Or because I thought that question must be more complex then it looked etc. I really donÕt know why, however I am sort of glad because it has allowed me to gain alot better understanding into using vectors, but then making mistakes does usually have that effect. cassandra === Subject: How Many M&Ms? Hello All. I am writing in reference to a contest a friend of mine is having. The rules are as follows. I bought a brand new clear plastic jar. It is filled with the standard size M & M candies. The jar is round and 8 inches tall. The diameter of the jar inside is 4 inches across. I bought over 4 pounds of M & Ms to fill it. I counted every M & M before filling the jar to the very top.. You can probably guess what I am seeking: How many M & Ms are in the jar as described? If anyone can help me on this it would be wonderful as I am not a good mathmatician. Joey === Subject: Re: How Many M&Ms? Fortunately, there has been some research on M&M packing: The net is that the density of packing is around 71% (if packed in an irregular fashion). Further, it looks like the someone has actually measured an individual M&M: http://www.kleinbottle.com/Bernie_Tao.htm Which a result of 0.45239 cm. so, volume of jar is: PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 cm3 So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 Now the result is probably only good to around 2 significant figures, and there will probably be some rounding down due to the sides of the container, so I would say aound 2500 M&Ms ? Am I close? -Darren > Hello All. I am writing in reference to a contest a friend of mine is > having. The rules are as follows. > I bought a brand new clear plastic jar. It is filled > with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of > the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to fill it. I counted > every M & M before filling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? > If anyone can help me on this it would be wonderful as I am not a good > mathmatician. > Joey === Subject: Re: Novice: Pi in other base systems > Hello... > My apologies if this is a silly question. How does pi behave in > different base systems, for instance binary, base 8 or base 16? > Is it still a mysterious number with endless non-repeating > numbers? Is there any base system where thereÕs some especially > interesting results? > (I guess the correct way to ask this is how the ration of > circumference to diameter behaves in different number systems, > but still you get the point.) > Just curious. > Thx, > D. It is possible to compute the n-th digit of pi in hexadecimal --- and thus also in binary --- notation without computing all the previous digits. (I donÕt have the reference here right now, but IÕm sure you will not be eventually repeating since pi is not rational. Hope this helps, Lasse --- === Subject: JSH: James, you be the judge... Here is how James Harris operates: 1. Start with some goofy polynomial that was a leftover from one of his many failed FLT proofs. No explanation of motivation or reasoning provided for choosing that particular polynomial or explanation of why it might be meaningful in his new context. 2. Avoid being specific about such important details as which ring he intends to work in. 3. Dream up a bunch of weird non-standard terminology like properly unit, dividing off, etc. Use terms incorrectly, such as referring to multiple when he should use the term factor. Also, make use of standard terminology like distributive property and constant term without actually understanding it or using it properly. Also, add lots of vague distractions such as the equation has no memory, you can see the 7Õs in there, canÕt you? to act as a smoke screen. 4. Apply a bunch of algebraic manipulations, some of which are just wrong. Make inappropriate generalizations from specific cases to general cases, etc. 5. Get a result that seems contradictory, and assume that the error was with the core of algebra rather than the much more likely explanation that he himself might have made an error. 6. Through out a bunch of paranoid ad hominem attacks on critics. Then hypocritically claim that all his critics resort to social crap and personal attacks rather than sticking to mathematics, logic, and reasoning. We are supposed to believe that his arbitrarily chosen goofy polynomial just happens to one that, when probed by the genius of James Harris, give results that shake mathematics to its core! James, you be the judge. Are you a pile of crap? === Subject: Pi in space I was having an argument with a friend here and he claimed that pi doesnÕt really have all this significance itÕs attached to it, since it is transcendental only in the theoretical space of Euclidian geometry and all know that this geometry does not represent true space-time. Is pi transcendental in ANY space except in the theoretical space of Euclidian geometry? For example, is it transcendental in Riemannian geometry? Lobatchevskian geometry? or any other geometry? What about in the relativistic universe? Is the ratio of the circumference of a circle to its diameter inside the Einsteinian universe rational or irrational? Anyone knows? -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesnÕt really have all this significance itÕs attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? We donÕt know the true geometry of space. In particular, we donÕt know if itÕs continuous. Digital physicists think it is not: see www.digitalphysics.org The concept of Pi, however, should not be regarded transcendental or anything like that since it is represented perfectly with a small program. True, it is an idealization, but what is transcendental? In my opinion, nothing transcends physical reality, and if Pi is part of physical models, it is for a good reason: physics is mostly about ideal models that work in ideal conditions, etc. Note however, that, if the universe turns out to be discrete, then, obviously, the real number Pi does not describe any physical reality. The computation of Pi is still sensible, however, (e.g. computation of Pi up to the nth decimal place) and it would indeed describe physical reality in a discrete universe (if itÕs an actually infinite discrete universe, things get more complicated so just assume finite which seems to be the case for *our* universe;). Most mathematicians on this forum seem to be Platonists, or Platonists who lack the philosophical maturity to identify themselves as Platonists, so you could expect quite a lot of Platonist nonsense in response to your question. However, as one poster correctly pointed out, constructivists and formalists will most likely look at Pi as a useful mental construct of some sort, and no talk of transcending anything will be necessary. Pi is not in space: itÕs in your mind. [*] Indeed, constructivism is a more favorable position than Platonic Realism, in my opinion, and a point of view supported by great mathematicians, so itÕs an alternative you should think about. More recently, instrumentalism has been suggested as an alternative way to look at mathematics. In my opinion, that too is considerable. -- Eray Ozkural There is no perfect circle [*] A caveat, however, there are perfect discrete circles in the world. It just may be the case that there are no perfect continuous circles, except conceptually!!!!!! === Subject: Re: Pi in space |Most mathematicians on this forum seem to be Platonists, or Platonists |who lack the philosophical maturity to identify themselves as |Platonists, so you could expect quite a lot of Platonist nonsense in |response to your question. No, because such points of philosophy are thoroughly irrelevant to his question. |However, as one poster correctly pointed |out, constructivists and formalists will most likely look at Pi as a |useful mental construct of some sort, and no talk of transcending |anything will be necessary. Pi is not in space: itÕs in your mind. [*] I am disinclined to assume that you know that transcendental is a technical mathematical term meaning not a root of a nonzero polynomial with integer coefficients. |Indeed, constructivism is a more favorable position than Platonic |Realism, in my opinion, and a point of view supported by great |mathematicians, so itÕs an alternative you should think about. Kronecker, Brouwer, and Bishop were all outstanding mathematicians (whether we call them great sort of depends on how high a standard we set for greatness). Bishop supported constructivism; Brouwer supported some kind of constructivism; Kronecker appears to have supported something along those lines. I wouldnÕt say, however, that popularity among great mathematicians is a very good standard for judging such things. To the extent that it can be trusted at all, I see no obvious sign of the very best mathematicians differing very far in terms of the distribution of their opinions on Platonism (or realism), formalism, constructivism and so on from the run-of-the-mill. I also donÕt see sci.math as being much of a hotbed of Platonism or realism. Keith Ramsay === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. I think you need to look up the definition of transcendental number. You seem to be confusing it with some philosophical notion. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. > I think you need to look up the definition of transcendental number. > You seem to be confusing it with some philosophical notion. the concept of a transcendental number. AFAICT, the notion of a transcendental number is not relative to a give geometry, or the true geometry of our universe. Pi is a transcendental number, according to the definition of a transcendental number in mathematics. I do think his question ought to be philosophical. Why should he talk about relativity then? What difference is there between any continuous metric and a Riemann tensor regarding the fact that Pi is a transcendental number? Maybe, I think, his friendÕs intention was to give a better definition of what it means for a number to be transcendental than the ordinary usage. He might want to pick another term, though, or highlight the difference than the ordinary usage carefully. Otherwise, we become confused in argumentation. I thought so, and said that this philosophical sense is probably irrelevant to a computable real like Pi which captures a general geometric fact in a compact form! -- Eray Ozkural === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesnÕt really have all this significance itÕs attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. We know nothing whether space-time really exists and if it exists whether it has a given geometry of its geometry can be a matter of convention. You are asking metaphysical (ontological) questions that have no answers at this point. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? The geometries have relative consistency to Euclidean geometry. This, pi must be transcendental in all of them to maintain relative consistency. > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? You can draw a circle even on your pixelized TFT screen, consider it a perfect one and use pi to find its perimeter. I get the impression you are asking about the relation of abstract mathematical object to the actual universe. We do not know it. ThatÕs another metaphysical subject. Platonists would argue circles are really part of the ontology of space-time and Universe whereas Formalists and Constructivists will look at circles as useful geometrical models with no real existence apart from human mind. Mike === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi doesnÕt >really have all this significance itÕs attached to it, since it is >transcendental only in the theoretical space of Euclidian geometry and all >know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the circumference > of a circle to its diameter inside the Einsteinian universe rational or > irrational? Anyone knows? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ Defining pi to be the empirically measured ratio of a circleÕs circumference to its diameter, the only geometry in which this is ratio is a constant is Euclidean geometry. Pi doesnÕt exist as a constant that can be empirically measured in any other geometry, so its value in other geometries is a meaningless question. Of course, defining pi as the sum of a series, or the constant in a solution to a differential equation or proper integral, or any of the other billion ways of defining pi has absolutely nothing to do with the geometry of the Universe. So there is only one value of pi, and yes, it is transcendental. === Subject: Re: Pi in space |Defining pi to be the empirically measured ratio of a circleÕs circumference |to its diameter, the only geometry in which this is ratio is a constant is |Euclidean geometry. Pi doesnÕt exist as a constant that can be empirically |measured in any other geometry, so its value in other geometries is a |meaningless question. Obviously this is not a practical issue, but I thought as long as we were busy making points about what is true in principle, I would note that there often is a natural relationship between pi and other geometries. In some other geometries, pi is the limit of the ratio between the circumference and the diameter of a circle as the radius goes to 0, and could in principle (if you had such a space in your hands...) be measured that way. At a fixed radius, measuring the circumference and diameter would only be able to approximate pi up to a certain degree of precision, but measurements of lengths are always only up to a certain degree of precision anyway, so one is not any worse of in an essential way. On the hyperbolic plane, for example, you could find a length scale on which a triangle whose sides have lengths in a 3:4:5 ratio has an angle of between 90 and 91 degrees. One can then put an upper bound on the deviation of the ratio of the circumference to the diameter from pi, in terms of the ratio of the radius to the length of a side of this triangle. The circumference of a circle of radius r in the hyperbolic plane is 2*pi*a*sinh(r/a) where the length a depends on the curvature. The size of the 3:4:5 triangle with an angle of 91 degrees is likewise proportional to this length a. (I wonÕt bother figuring out what the ratio is right here.) So to get n digits, we could make a circle whose radius is roughly 10^{-n/3} times the length of a side of this triangle, and measure it very precisely. Keith Ramsay === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi doesnÕt >really have all this significance itÕs attached to it, since it is >transcendental only in the theoretical space of Euclidian geometry and all >know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the circumference > of a circle to its diameter inside the Einsteinian universe rational or > irrational? Anyone knows? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ > Defining pi to be the empirically measured ratio of a circleÕs circumference > to its diameter, the only geometry in which this is ratio is a constant is > Euclidean geometry. It is not possible to empirically measure any irrational in any geometry, if by that you mean physical measurement. And if not, what do you mean? === Subject: Re: Pi in space >>I was having an argument with a friend here and he claimed that pi >>doesnÕt >>really have all this significance itÕs attached to it, since it is >>transcendental only in the theoretical space of Euclidian geometry and >>all >>know that this geometry does not represent true space-time. >> Is pi transcendental in ANY space except in the theoretical space of >> Euclidian geometry? >> For example, is it transcendental in Riemannian geometry? >> Lobatchevskian >> geometry? or any other geometry? >> What about in the relativistic universe? Is the ratio of the >> circumference >> of a circle to its diameter inside the Einsteinian universe rational or >> irrational? Anyone knows? >> -- >> I. N. G. --- http://users.forthnet.gr/ath/jgal/ >> Defining pi to be the empirically measured ratio of a circleÕs >> circumference >> to its diameter, the only geometry in which this is ratio is a constant >> is >> Euclidean geometry. > It is not possible to empirically measure any irrational in any > geometry, if by that you mean physical measurement. Sure it is. You can measure the diagonal of unit square just as easily as you can measure the length of one side. The fact that one is rational and the other irrational doesnÕt affect your ability to measure them. > And if not, what do > you mean? The reason that I talk about empirical measurement is assertion implicit in the OP that if the universe is not ßat, then pi does not have the value everybody thinks it does. The only sense in which this is sort-of true is that if you measure the circumference of a circle and divide it by the diameter, then you will not get pi as we know it as the answer. I used the term empirically to suggest the act of phyiscally measuring (for example with a tape measure) circumferences and diameters of circles embedded in that space. === Subject: Re: Pi in space (1) Pi is a constant. ItÕs value is 4 ATAN(1) regardless of whether space is Euclidean. Your question is like asking whether the number 1 remains rational in non-Euclidean space. A constant is a constant is a constant. (2) we all know that this geometry..... Who is we? We certainly does NOT include all because the recent super-nova evidence that revealed the existence of dark energy also shows that the universe is Euclidean. (3) Pi is transcendental in ALL geometries. You can lead a horseÕs ass to knowledge, but you canÕt make him think. === Subject: Re: Pi in space ETAtAhUAuKZDfmVeTIVHWzPAf+1RbygxYcACFGivLjq2oYFNZgzALqH4m/ 0aKNiF Gee, the last time I looked, the transcendentality of a number was INDEPENDENT of geometry, being a property of the number itself. Whether pi, or other famously nonconstructible numbers like the cube root of 2 or the sine of 20 degrees, become constructible in non-Euclidean geometries is a different question. --OL === Subject: Re: Pi in space >(1) Pi is a constant. ItÕs value is 4 ATAN(1) regardless >of whether space is Euclidean. But wasnÕt there a part of the muppets show called PiÕs in space? Thomas >Your question is like asking whether the number 1 >remains rational in non-Euclidean space. >A constant is a constant is a constant. >(2) we all know that this geometry..... >Who is we? We certainly does NOT include >all because the recent super-nova evidence that >revealed the existence of dark energy also shows >that the universe is Euclidean. >(3) Pi is transcendental in ALL geometries. >You can lead a horseÕs ass to knowledge, but you canÕt make him think. === Subject: Re: Pi in space > (1) Pi is a constant. ItÕs value is 4 ATAN(1) regardless > of whether space is Euclidean. > Your question is like asking whether the number 1 > remains rational in non-Euclidean space. > A constant is a constant is a constant. Damn. I was somehow hoping that from context my question would be clear. Apparently it was not. Let me repeat it more accurately phrased: Define pi=Length of circumference/Length of diameter of a circle however the later terms are defined in the appropriate geometry we are talking about. In which geometries pi is transcendental? -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space >> (1) Pi is a constant. ItÕs value is 4 ATAN(1) regardless >> of whether space is Euclidean. >> Your question is like asking whether the number 1 >> remains rational in non-Euclidean space. >> A constant is a constant is a constant. > Damn. I was somehow hoping that from context my question would be clear. > Apparently it was not. Let me repeat it more accurately phrased: > Define pi=Length of circumference/Length of diameter of a circle > however the later terms are defined in the appropriate geometry we are > talking about. > In which geometries pi is transcendental? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ As I explained below, the ratio of a circleÕs circumference to its radius is only a constant in Euclidean space. In non-Euclidean geometries, the ratio depends upon the size of the circle; there is no single value of pi. Consider, for example, the space as consisting of the surface of a sphere as big as the earth. If you measue the ratio of the circumference to the diameter of a circle that is only a few inches across, the ratio will be very close to 3.14159265... If you measure the ratio for the equator, you get pi=2 (the radius is the distance from the North Pole to the equator, which is one quarter the circumference). So in non-Euclidean geometries, the ratio of a circles circumference to its diameter can take on a wide range of values, depending upon the size of the circle. As the size of the circle approaches zero, the ratio approaches pi. So as the ratio can take on a continuous range of values, you can make it transcendental or not, simply by selecting the size of the circle you use to form the ratio. Which is, of course, different to saying pi varies - in all geometries, pi is the same; only the measured ratios change. === Subject: Re: Pi in space > As I explained below, the ratio of a circleÕs circumference to its radius is > only a constant in Euclidean space. In non-Euclidean geometries, the ratio > depends upon the size of the circle; there is no single value of pi. > Consider, for example, the space as consisting of the surface of a sphere as > big as the earth. If you measue the ratio of the circumference to the > diameter of a circle that is only a few inches across, the ratio will be > very close to 3.14159265... If you measure the ratio for the equator, you > get pi=2 (the radius is the distance from the North Pole to the equator, > which is one quarter the circumference). > So in non-Euclidean geometries, the ratio of a circles circumference to its > diameter can take on a wide range of values, depending upon the size of the > circle. As the size of the circle approaches zero, the ratio approaches pi. > So as the ratio can take on a continuous range of values, you can make it > transcendental or not, simply by selecting the size of the circle you use to > form the ratio. > Which is, of course, different to saying pi varies - in all geometries, pi > is the same; only the measured ratios change. changes. That would be ridiculous. I was talking about the ratio. Serves me right: If I cannot phrase a question correctly, then I end up having all the groupÕs farts on my back :-( -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space <1102142838.264150@athnrd02 Define pi=Length of circumference/Length of diameter of a circle > however the later terms are defined in the appropriate geometry we are > talking about. > In which geometries pi is transcendental? Consider R^2 with the discrete metric and the circle at (0,0) with radius 1. Computing the circumference and dividing it by the diameter 2, I find pi, the ratio circumference/2, to be transcending. ;-) === Subject: Re: Pi in space > Your question is like asking whether the number 1 > remains rational in non-Euclidean space. Well, does it? DOES IT? How come you mathematicians never answer the IMPORTANT questions? === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi >doesnÕt really have all this significance itÕs attached to it, since it >is transcendental only in the theoretical space of Euclidian geometry >and all know that this geometry does not represent true space-time. Everything after since is incoherent. >Is pi transcendental in ANY space except in the theoretical space of >Euclidian geometry? >For example, is it transcendental in Riemannian geometry? Lobatchevskian >geometry? or any other geometry? >What about in the relativistic universe? Is the ratio of the >circumference of a circle to its diameter inside the Einsteinian >universe rational or irrational? Anyone knows? ...As are all those questions. Lee Rudolph === Subject: Re: JSH: Look at it backwards >I usually start with one polynomial and then talk about dividing 49 >off from it, but here IÕll start *after* 49 has been divided off: >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >and consider the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >and the assertion that the dÕs must have factors >w_1(x), w_2(x), and w_3(x) >that multiply to give w_1(x) w_2(x) w_3(x) = 7. >But thereÕs no factors of 7 anywhere. So why should the dÕs have >functions that are factors of 7? > You are getting maximally confused. In your original > approach, you are factoring P(x) as a polynomial in 5, > not as a polynomial in x. The constant term with > respect to 5 and the constant term with respect to > x are different. It was a stupid mistake to think you > were simplifying by treating 5 as if it were the > polynomial variable in the first place. And that is what > is causing your confusion here. There is only one variable Nora Baron, so what other variable would you have listed with S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) considering that S(x) = 300125x^3 - 18375 x^2 - 360 x + 22? So whoÕs confused? >Ok, maybe that seems unfair, as there could be LOTS of different >functions you can plug in for the cÕs and dÕs, but why should ANY of >them have functions of x that are factors of 7? >The equation has no memory. > The memory problem here is, you have forgotten what you > started with. To see this, you need to go back to the way you > were doing things originally: for example, Now note, the polynomial S(x) doesnÕt have 7 as a factor, as this time IÕm starting with the result of dividing off the multiple. So the poster Nora Baron (actually a guy as revealed in another post when he ended with a male name) is trying to show how 7 gets back into the expression. That is, heÕs trying to prove that the factorization DOES have a memory of the multiple 49. > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > To put this in terms of what you are doing now: replace x in this > expression by 5, and replace m by x, and replace f by 7. Also > replace u by 1. This gives > P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), > and you can clearly see that, as a polynomial in 5, the constant > term is 49 * 7 = 7^3, not 49 * 22. No. It can be factored with respect to 5 and 7, but 5 is not a variable. The polynomial variable is x. Now, IÕve explained lots of times to the Nora Baron poster, and what I want you all to consider is that the poster isnÕt really that dense, but instead knows you better than you know yourselves. Essentially the basic strategy is just to disagree. Time after time, and even in answer to surveys that IÕve done, readers who normally lurk will admit that they primarily rely on the fact that people argue with me, assuming that if I were right, then others wouldnÕt disagree! So, for sci.mathÕers like Nora Baron, the strategy is clear--just disagree. They often donÕt even TRY to actually make mathematical senze. > Go back to the expression above for P(m). The constant term > with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to > 49*22 when u = 1, f = 7, and x = 5. But the constant term with > respect to x is u^3 f^3. You have forgotten this and gotten mixed > up, thatÕs all. Now the expressions actually is S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 where the factorization S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) is being considered. Now how do you get 7 into that? Well, thatÕs not the issue for the poster! All he has to do is disagree. For most readers thatÕs what works. Well, that enough here. James Harris === Subject: Re: JSH: Look at it backwards >snip >There is only one variable Nora Baron . . . Here I agree with Mr. Harris. I once posted that there are evidently a number of real people living in the United States that really are named Nora Baron, but those individuals are constant while our Nora Baron may indeed be the only one that is a variable -- although that property has not yet been established for sure. === Subject: Re: JSH: Look at it backwards >>I usually start with one polynomial and then talk about dividing 49 >off from it, but here IÕll start *after* 49 has been divided off: >>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >>and consider the factorization >>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >>and the assertion that the dÕs must have factors >>w_1(x), w_2(x), and w_3(x) >>that multiply to give w_1(x) w_2(x) w_3(x) = 7. >>But thereÕs no factors of 7 anywhere. So why should the dÕs have >>functions that are factors of 7? >> You are getting maximally confused. In your original >> approach, you are factoring P(x) as a polynomial in 5, >> not as a polynomial in x. The constant term with >> respect to 5 and the constant term with respect to >> x are different. It was a stupid mistake to think you >> were simplifying by treating 5 as if it were the >> polynomial variable in the first place. And that is what >> is causing your confusion here. >There is only one variable Nora Baron, I didnÕt say there were other variables in your present polynomial. I was describing its origin, when there were 4 variables, which explains why you are now confused about the whole thing. > so what other variable would >you have listed with >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >considering that >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22? >So whoÕs confused? No question about that. You are. >>Ok, maybe that seems unfair, as there could be LOTS of different >>functions you can plug in for the cÕs and dÕs, but why should ANY of >>them have functions of x that are factors of 7? >>The equation has no memory. >> The memory problem here is, you have forgotten what you >> started with. To see this, you need to go back to the way you >> were doing things originally: for example, >Now note, the polynomial S(x) doesnÕt have 7 as a factor, I DIDNÕT SAY IT DID, ASSHOLE > as this time >IÕm starting with the result of dividing off the multiple. >So the poster Nora Baron (actually a guy as revealed in another post >when he ended with a male name) is trying to show how 7 gets back into >the expression. It gets back in in the constant term, because you have been factoring this function as if it were a polynomial in 5. >That is, heÕs trying to prove that the factorization DOES have a >memory of the multiple 49. Not at all. The 49 is obviously gone. But the constant term at the end, when P(x)/49 is viewed as a polynomial in 5 [your idea, not mine!], is still there, and it is 7, not 22. >> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) >> To put this in terms of what you are doing now: replace x in this >> expression by 5, and replace m by x, and replace f by 7. Also >> replace u by 1. This gives >> P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), >> and you can clearly see that, as a polynomial in 5, the constant >> term is 49 * 7 = 7^3, not 49 * 22. >No. It can be factored with respect to 5 and 7, but 5 is not a >variable. You have treated it exactly as such. When you factor P(x) in the form P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7), you are NOT factoring it as a polynomial in x. If you were, the terms a_1(x), etc., would themselves be polynomials in x. As you yourself have pointed out MANY TIMES, this is not a polynomial factorization when considered as a function of x. (In fact you have claimed that such Ōnonpolynomial factorizationwas one of your great conceptual breakthroughs. Now you want to deny it ???). However it CLEARLY has the FORM of a factorization in 5, where 5 is treated as a polynomial variable. >The polynomial variable is x. It most certainly is not [yes, I know you are going to howl about this, but I stand by what I say]. When you factor a b x^2 + (a + b) x + 1 as (a x + 1)(b x + 1), THEN you are factoring in terms of the polynomial variable x. The coefficients of x in the factors are functions of a and b, NOT functions of x, as in your present factorization of P(x). Have you really no recollection of how you got started on this track? Originally, x was m, and 5 was x. At that time you were considering factors like a_1(m) x + u f, etc. This was a factorization in terms of the polynomial variable x, with coefficients which were algebraic integer functions of m. Just plug in x = 5, u = 1, f = 7, and m = x, and you have your present factorization - as a polynomial in 5, not in x !! ItÕs easy, actually, to see how you have gotten confused, especially if (unlike real mathematicians) you have a poor memory. >Now, IÕve explained lots of times to the Nora Baron poster, and what >I want you all to consider is that the poster isnÕt really that dense, >but instead knows you better than you know yourselves. >Essentially the basic strategy is just to disagree. I only disagree when you are wrong. >Time after time, and even in answer to surveys that IÕve done, readers >who normally lurk will admit that they primarily rely on the fact that >people argue with me, assuming that if I were right, then others >wouldnÕt disagree! I donÕt recall people saying that. What lurkers do seem to think, when they speak up, is that YOUR math is vague, hand-waving and illogical. ThatÕs not my fault. Plus they seem to think that your habit of making nasty personal attacks damages your credibility. ThatÕs not my fault either. >So, for sci.mathÕers like Nora Baron, the strategy is clear--just >disagree. Again, I only disagree when you are wrong. Of course, that happens to be virtually all the time. >They often donÕt even TRY to actually make mathematical senze. This is gratuitous and false in my case, as you well know. Also in the cases of Dik Winter, Rick Decker, Arturo Magidin, ŌRupertÕ, Will Twentyman, Dale Hall, and others. >> Go back to the expression above for P(m). The constant term >> with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to >> 49*22 when u = 1, f = 7, and x = 5. But the constant term with >> respect to x is u^3 f^3. You have forgotten this and gotten mixed >> up, thatÕs all. >Now the expressions actually is >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >where the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >is being considered. >Now how do you get 7 into that? Well, thatÕs not the issue for the >poster! It was explained in what I posted. You need to consider how you arrive at S(x), and for that you need to consider the form of the original function P(x): as a polynomial in 5, it has constant term 7^3. I thought you were simply confused on this point, having forgotten how you got here. I am now inclined to think you are trying to deny the obvious correct explanation, though I am not sure why. Plus throughout your entire reply you seem to be playing to the grandstand. You whine, exaggerate, and attempt to discredit. The actual math you have hardly addressed at all. You seem to think that you can win mathematical arguments by an appeal to some great silent majority of people out there who you can cajole into thinking that we critics are evil, cheating liars who are jealous of your discoveries. (If that were true, one of us evil cheating liars would have published it long ago, of course giving no credit to James S.Harris. Funny, that hasnÕt happened. Wonder why? Why isnÕt anyone stealing your ideas?) You seem to have the impression that math is a popularity contest. ItÕs not. The only way to win is through rigorous proof. You donÕt have one. ThatÕs the main reason you are not winning. >All he has to do is disagree. >For most readers thatÕs what works. Right. Go ahead and ask your faithful admiring grandstand if thatÕs what works. >Well, that enough here. Yuh - Duh - that enough ! Nora B. >James Harris === Subject: Re: JSH: Look at it backwards > So the poster Nora Baron (actually a guy as revealed in another post > when he ended with a male name) is trying to show how 7 gets back into > the expression. What difference does it make what NoraÕs gender is? What does it have to do with math? Has anybody but you made note of the question of NoraÕs gender? No, because it is not relevant. Your obsession with her (or his) gender just screams mental illness. It is (probably) just a pseudonym, nothing more. So what? My pseudonym is o[CapitalYAcute]in, so do you claim that I am pretending to be a Norse god? Well, nobody cares!!! Idiot!!! === Subject: Re: JSH: Look at it backwards > I usually start with one polynomial and then talk about dividing 49 > off from it, but here IÕll start *after* 49 has been divided off: > S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 > and consider the factorization > S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) > and the assertion that the dÕs must have factors > w_1(x), w_2(x), and w_3(x) > that multiply to give w_1(x) w_2(x) w_3(x) = 7. Let w_1(x)=w_2(x)=w_3(x) = 7^(1/3). Then the w_i are factors of the d_i in the ring of algebraic numbers. (Are they factors in any other ring? Who knows? Who cares? No one except you has made such a claim.) > But thereÕs no factors of 7 anywhere. So why should the dÕs have > functions that are factors of 7? the dÕs have functions that are factors of 7 is gibberish. -William Hughes === Subject: Re: JSH: Two sides > IÕve had months to wonder why mathematicians would work to ignore or > hide my results, Ignoring your results takes no effort, hiding them does not occur. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Two sides [snip delusion] What does properly a unit mean? Can you answer that simple question James Harris? === Subject: Re: JSH: Two sides > [snip delusion] > What does properly a unit mean? > Can you answer that simple question James Harris? JSH does not answer questions. === Subject: Re: Why exp(-st) in the Laplace Transform? Does anyone have an explaination why the kernel function exp(-st) was >>used in the definition of the Laplace transform? >>Is there a physical meaning to the use of this function? >>I know s is a complex frequency, how can we visualize what the Laplace >>transform is doing? > I believe the answer is no, thereÕs no interpretation of what the > Laplace transform really means, analogous to the way one interprets, > say, the Fourier transform. The kernels are not orthogonal in any > obvious sense... > ItÕs hard to know for certain that no is the correct answer, of > course. But IÕve never seen an answer of the sort I think youÕre > looking for, IÕve thought about it and never found one, and > I once posted more or less the same question to sci.math, and > had a few smart people agree the answer was no. Arthur Mattuck, an MIT professor, ŌÕinterpretsÕthe Laplace transform as a continuous analog of a power series expansion of a function. You can watch a video of him lecturing on this point at Click on Lecture 19. T. Monroe === Subject: Re: Why exp(-st) in the Laplace Transform? > heard something somewhat similar tho - that laplace transforms are the >> continuous generalization of power series. I canÕt recall exactly how >> the explanation went, it was in an ODE book by a guy named Simmons. > Perhaps you have seen something like this. > Let F(p) = integral exp(-px) f(x) dx (: Laplace Transform) > ==> F(0) = integral f(x) dx = area beneath f(x) > And FÕ(p) = integral (-x) exp(-px) f(x) dx > ==> FÕ(0) = - integral x f(x) dx = - first order moment > / center of gravity / midpoint > And FÕÕ(p) = integral x^2 exp(-px) f(x) dx > ==> FÕÕ(0) = integral x^2 f(x) dx = second order moment > / moment of inertia / variance > And then you have F(p) = F(0) + p.FÕ(0) + p^2/2.FÕÕ(0) + ... > Interpretation: subsequent terms of the series expansion of the Laplace > Transform involve an area, a midpoint, a moment of intertia .. In short: > the most important (physical/global) characteristics of a function come > first, when considered in the Laplace domain. > Han de Bruijn > Arthur Mattuck, an MIT professor, ŌÕinterpretsÕthe Laplace transform as a > continuous analog of a power series expansion of a function. You can watch > a video of him lecturing on this point at > Click on Lecture 19. > T. Monroe Excellent pointer. Dirk Vdm === Subject: Re: Gravitomagnetism > Ok, you donÕt need to respond to this comment, > but I have no major problem with non-integer > weight if you allow non-integer dimensionality. > Non-integer? DonÕt understand this.. Well you know from SR as you move an object faster > and faster itÕs length contracts and itÕs time > slows down, and at c these dimensions vanish. > So in the intermediary, between v=0 and v=c, > can we say definitely that we have a fixed integer > dimensionality? To put that on a more rational mathematical foundation > one can integrate, (no integration constants) $ x dx = (1/2) x^2 (area) $$ x dx dx = (1/6) x^3 (volume) which shows how integration generates dimensions. > Well conformal weight is a concept that is independent of > dimensionality, that is, a covariant in a Weyl conformal space has > properties under both coordinate and scale changes. The idea of weight > embodies the latter. to check that out, but I think you are right. You know, moving from field to quantum physics where are those integers. > GR is quite clear, G_uv = T_uv is where IÕll begin. > If I understand you correctly, you state > vanishing gravity (implying G_uv =>0) and > *inhomogeneous* electrodynamics (implying T(Maxwell)_uv =/=0 are compatible? > Yes! That is what is so surprising. Nowhere is a current posited - > there *is* no RHS here. In GR, as in Maxwell-Lorentz, one has a > field that is driven by a *posited* current (energy tensor, charge > current resp.), and then that field acts back on the current. Here, > there is *no* posited current at all, rather, the assumption of > strictly local metricity requires both the metric and calibration > field, and these *jointly* assume roles in a manner that appears as > symmetric Ricci driven by energy-momentum, and Maxwell driven by > charge-current! Of course they are really 6-d *vacuum* equations: > Rmn - (2R/W) Tmn + (1/2W) (DmDn + DnDm) W = 0 (Rmn = symmetric part > of CCT) I have a problem, I can read that in the usual 4d terms, but when you uprate to 6d are your mn over 4 or 6, ah...your pushin the envelope so please tell us your encryption. > 1/S d/dxm ( S R Fmn ) - 5/4 (Dn W) = 0 (S = sqrt det g) > The new physics would be found in the pure geometry terms 1/2W > {Dm,Dn} W and (Dn W), which are respectively, energy-momentum and > charge current. These are absent in ßat space absent in ßat space, thatÕs what Einstein said when he doubled his prediction for the deßection of light. Would you be able to provide how your equations reduce to GR in the the weak field, that way giving us ameans to connect GRistÕs with the physicality of your 6D. > Yes, IÕm trying to make the simplest possible metric > consistent with a nonorthogonal space, i.e. g_uv = g*g_uv + A_u B_v > ^ ^ > calibration EM antisymmetry => A_u B_v = - A_v B_u > (Weyl=>gauge) (Einstein) where det g = 1 - AB, (thatÕs a bit crude, but close). Ken S. Tucker What paper? ken > -drl === Subject: JSH: Good news, and bad news IÕll give the good news first. Despite the venting I do at times in frustration, the reality is that I have an easygoing nature. So I thought IÕd point out a few things along with that for any mathematicians who might wander by and read this post to consider. First of all, my research is not limiting but expanding. What IÕve found are new tools for deeper explorations into the properties of numbers. Given that mathematics is difficult for many people, itÕs unlikely that the status quo in mainstream math society would change too dramatically once the full story is out. Sure IÕd no longer be a crank, but top mathematicians probably wouldnÕt move much. So letÕs say you are Barry Mazur, or Wiles or Taylor, and you find out that there is actually this problem with algebraic number theory that you learned and a LOT of what you thought you had proven, was not proven. So what? A lot of it you can go ahead and prove anyway in less space, without error, using the more advanced techniques. but very large work, you can move to a proof of a few pages. And if Wiles cares more about seeing the proof of FermatÕs Last Theorem than the glory of being the person who found that proof, then itÕs gravy. Sure, itÕs embarrassing that the underlying theory was wrong, but the mistake entered the discipline over a hundred years ago, long before anyone now living was even born. If the desire is to see a proof of FermatÕs Last Theore, then no problem! ThatÕs the good news. If the desire is the glory of having found a proof of FermatÕs Last Theorem. Problem. So itÕs a measure of the men, as they are mostly men, how they react. The good news is that not much will change in many ways if they accept what is mathematically true. The bad news is that if they react only when forced, then clearly they were in it for the glory. And if glory hounds willing to fight against the truth, they get no pity. So, one way, not much changes. Professorships remain. There is just this amazing story of which they are a part. The other way, much can change, much can be lost, and a world can be severely disappointed, when the full story eventually comes out. The choice is yours. James Harris === Subject: Re: Good news, and bad news > The good news is that not much will change in many ways if they accept > what is mathematically true. Good, indeed. News, hardly. Have you ever read the fineprint? >The choice is yours. You have one claim (trivial) to a prime counting algorithm, and another (childish) denying basic algebra. At most one of them could stand. The choice is yours - is the bottle half full, half empty or fully empty? Barry === Subject: Re: Good news, and bad news Mr. Harris, Your level of delusion seems to grow without bound! The amount of crap spewage that you are propagating is beyond any level of decent mathematical discussion. You are a phony, a liar, a cheat and an ignoramus. Even if you managed to stumble into something in mathematics that actually made sense while in one of your drunken states, no one would give a damn! You just donÕt get it, do you? === Subject: Re: Good news, and bad news > What IÕve found are new tools for deeper explorations into the > properties of numbers. Yes, you found fucking great new tools for deeper explorations... Right... And you did manage to proved FLT using slick 17th century style algebra, so why not move onto the Riemann Hypothesis with your new power tools? That is the next big one, right? I just know you can do it with your new deep math exploration tools. After all, it is a great challenge. It is kind off to analysis what FLT is to arithmetic, right. True, you have fought hard, and won your deservedg lory for great achievement with FLT. But why rest on your laurels? Move on you dumb little fuck........ Or was FLT just appealing to you only because the statement of the problem was so simple that you thought you had a clue about it. You figure you can follow the arithmetic because the problem statement looks like high school stuff, but you donÕt know head from your ass beyond high school, right. Well, that is probably why cranks like you always stick to FLT. === Subject: Re: Good news, and bad news > but very large work, you can move to a proof of a few pages. You often now make the claim that WileÕs FLT proof is wrong. Can you point out the error he made? Did he make an error in determining those properly units thingy-ma-jiggies? If you make this claim and you cannot back it up, then why are we to believe you? === Subject: Re: Good news, and bad news > but very large work, you can move to a proof of a few pages. > You often now make the claim that WileÕs FLT proof is wrong. Moreover, Wiles did not prove FLT directly, he proved a certain conjecture (the conjecture being that every rational elliptic curve is some kind of equivalent to a modular form), for a limited number of elliptic curves, namely for the semi-stable elliptic curves. By earlier work from Frey and Ribet this was sufficient to show that FLT was true. The full conjecture was proved true later by Conrad et al. Because JSH apparently attacks Wilesproof, he should also consider the proof by Breuil, Conrad, Diamond and Taylor for the full conjecture... Or he should attack the work by Frey, of that by Ribet, or whatever. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Good news, and bad news > You often now make the claim that WileÕs FLT proof is wrong. ... OOOOPS I mean WilesÕ, not WileÕs... === Subject: Re: JSH: Good news, and bad news > IÕll give the good news first. Despite the venting I do at times in > frustration, the reality is that I have an easygoing nature. So I > thought IÕd point out a few things along with that for any > mathematicians who might wander by and read this post to consider. > First of all, my research is not limiting but expanding. That would be the bad news then. === Subject: Re: JSH: Good news, and bad news >IÕll give the good news first. Despite the venting I do at times in >frustration, the reality is that I have an easygoing nature. Right. > So I >thought IÕd point out a few things along with that for any >mathematicians who might wander by and read this post to consider. >First of all, my research is not limiting but expanding. >What IÕve found are new tools for deeper explorations into the >properties of numbers. >Given that mathematics is difficult for many people, itÕs unlikely >that the status quo in mainstream math society would change too >dramatically once the full story is out. Sure IÕd no longer be a >crank, but top mathematicians probably wouldnÕt move much. >So letÕs say you are Barry Mazur, or Wiles or Taylor, and you find out >that there is actually this problem with algebraic number theory that >you learned and a LOT of what you thought you had proven, was not >proven. >So what? A lot of it you can go ahead and prove anyway in less space, >without error, using the more advanced techniques. >but very large work, you can move to a proof of a few pages. And if >Wiles cares more about seeing the proof of FermatÕs Last Theorem than >the glory of being the person who found that proof, then itÕs gravy. Actually itÕs clear that you havenÕt been paying attention. FermatÕs Last Theorem is _false_! E. E. Escultura has posted counterexamples right here in sci.math. I think the two of you need to get together and work this out. Because youÕre both light-years past anything that the rest of us can comprehend, but you canÕt both be right. Check out his posts. Contact the guy, and write back as soon as the two of you agree on which one of the two is wrong. >Sure, itÕs embarrassing that the underlying theory was wrong, but the >mistake entered the discipline over a hundred years ago, long before >anyone now living was even born. >If the desire is to see a proof of FermatÕs Last Theore, then no >problem! >ThatÕs the good news. >If the desire is the glory of having found a proof of FermatÕs Last >Theorem. >Problem. >So itÕs a measure of the men, as they are mostly men, how they react. >The good news is that not much will change in many ways if they accept >what is mathematically true. >The bad news is that if they react only when forced, then clearly they >were in it for the glory. And if glory hounds willing to fight >against the truth, they get no pity. >So, one way, not much changes. Professorships remain. There is just >this amazing story of which they are a part. >The other way, much can change, much can be lost, and a world can be >severely disappointed, when the full story eventually comes out. >The choice is yours. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Good news, and bad news >but very large work, you can move to a proof of a few pages. And if >Wiles cares more about seeing the proof of FermatÕs Last Theorem than >the glory of being the person who found that proof, then itÕs gravy. > Actually itÕs clear that you havenÕt been paying attention. > FermatÕs Last Theorem is _false_! > E. E. Escultura has posted counterexamples right here in sci.math. > I think the two of you need to get together and work this out. > Because youÕre both light-years past anything that the rest of > us can comprehend, but you canÕt both be right. > ^^^^^^^^^^^^^^^^^^^^^^^ Oh, that is so retrograde of you. Surely their mathematical skills are advanced enough that they can prove FermatÕs Last Theorem AND its negation. Understand now? === Subject: Re: JSH: Good news, and bad news youÕre an easygoer in what respect ... til you were found in a cyberspace, and became marginally surlier. I missed the bad news, other than what the last-prior poster suggested. > Understand now? --Advice, 0.05; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: Good news, and bad news > What IÕve found are new tools for deeper explorations into the > properties of numbers. Name or describe one new tool for deeper exploration that you have found. All you have done is claim that there is a deep problem in algebraic integers, and as far as that goes, you have convinced a grand total of zero converts. You never explained what that problem is. You never showed any results of what that problem are. And you have certainly never found any new tools for mathematicians to work with! === Subject: Re: JSH: Good news, and bad news !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > IÕll give the good news first. Despite the venting I do at times in > frustration, the reality is that I have an easygoing nature. Well, the _real_ good news is that the bad news turns out as hilariously absurd as the good news. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Simple proof > The poster Nora Baron is not even a girl. That poster is a guy, > playing a girl. A perfect match for yourself -- as I discovered when I found the anagram of your name which reveals your true identity: James S. Harris = Ms. J. Harrie Ass (note the gender) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: induction vs Cantor defines a function which given an infinite list of reals (that is, > an element of R^N) returns an element of R that is not on the list. > Cantor didnÕt prove that there were uncountably many such functions, > he only proved that there was one. ThatÕs all he needs to be able > to show that no list of reals contains every real. Do you think there are any counter-examples to CantorÕs diagonal proof? If so, can you provide a valid modification that keeps the concept of providing a real that is proven not in the list and has no counter-example? === Subject: Re: induction vs Cantor > defines a function which given an infinite list of reals (that is, > an element of R^N) returns an element of R that is not on the list. > Cantor didnÕt prove that there were uncountably many such functions, > he only proved that there was one. ThatÕs all he needs to be able > to show that no list of reals contains every real. > Do you think there are any counter-examples to CantorÕs diagonal > proof? No. > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? NA. === Subject: Re: induction vs Cantor > ... >... > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? > NA. There exists a countable model of your ZF set. === Subject: Re: induction vs Cantor >> ... >>... >> If so, can you provide a valid modification that keeps the concept >> of providing a real that is proven not in the list and has no >> counter-example? >> NA. >There exists a countable model of your ZF set. Sure, but inside such models it is still the case that |N| < |R|. Since |N| < |R| follows from the ZF axioms, it must hold in all models that satisfy those axioms, regardless of the size of the model (as viewed from the outside). -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: induction vs Cantor >> ... >>... >> If so, can you provide a valid modification that keeps the concept >> of providing a real that is proven not in the list and has no >> counter-example? >> NA. >There exists a countable model of your ZF set. > Sure, but inside such models it is still the case that |N| < |R|. Since |N| > < |R| follows from the ZF axioms, it must hold in all models that satisfy > those axioms, regardless of the size of the model (as viewed from the > outside). Hi Barb, That leads to a problem. The set N, the natural numbers, in any of those extensions, contains no elements not contained in each other copy of the set N, and it contains each of them. The copies are equal and identical. There is thus identity, a tautology, between any copies of N, and as well between any copies of P(N), Then, if there is a bijection between N and P(N), there is a bijection between N and P(N). theory bogosity is a reaction to Skolem, who otherwise says infinite sets are equivalent. Look outside. Do you use transfinite cardinals? Is it for anything besides transfinite cardinals? Some people use them as a definition as part of measure theory, but thatÕs because they lack better tools, and the results of measure theory are largely upon continua. That is to say, the useful results of measure theory are derivable without the use of transfinite cardinals, and they should be. Of your ordinals, is there any that specifically represents the integer -1? Is not the order type of the powerset of the naturals the successor of the order type of the naturals? If itÕs infinite thereÕs always one more. ThatÕs part of the basis of induction, that something can be proven to hold for each, thus that for each it is proven, as opposed to proved. When we get to the contradiction between each infinite set being equivalent inductively and the necessity of dual representation via CantorÕs results, then accept that there is dual representation instead of that induction fails. That way, youÕre on the path to being inside first order logic, and nobody can prove you incomplete, inconsistent, and immaterial. If you want a consistent theory, start by removing the inconsistencies, not reusing them. Ross F. === Subject: Re: induction vs Cantor > ... >... > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? NA. >>There exists a countable model of your ZF set. >> Sure, but inside such models it is still the case that |N| < |R|. Since |N| >> < |R| follows from the ZF axioms, it must hold in all models that satisfy >> those axioms, regardless of the size of the model (as viewed from the >> outside). >Hi Barb, >That leads to a problem. >The set N, the natural numbers, in any of those extensions, contains no >elements not contained in each other copy of the set N, and it contains each of >them. The copies are equal and identical. >There is thus identity, a tautology, between any copies of N, I agree. N (via von Neumann ordinals) is categorical in ZF, so the N in different models would be isomorphic. >and as well between any copies of P(N), That looks like a serious misstep. AIUI (which admittedly isnÕt very deeply), the downward LST gives a model in which all the elements correspond to first-order-definable terms. So, the set of such elements of P(N) in the LST countable model is very much smaller than P(N) in the usual iterative-hierarchy models (which includes the uncountably-many elements which are not first-order definable). (ItÕs rather like the difference between the set of computable infinite binary sequences and the set of all infinite binary sequences computable or not.) But |N| < |P(N)| even in the countable model, because the bijection between N and the modelÕs P(N) is not itself first-order definable and therefore does not exist in the model! ISTM you have strong Constructivist leanings. ThatÕs fine, but itÕs important then not to mix constructive and non-constructive entities in the same argument. If you exclude all non-constructive entities then you canÕt argue that |N| = |P(N)| in the countable model, because the isomorphism would be non-constructive and therefore not talkable-about. > Then, if there is a bijection between N and P(N), >there is a bijection between N and P(N). >theory bogosity is a reaction to Skolem, who otherwise says infinite sets are >equivalent. IÕm sure he doesnÕt actually say or imply that. >Look outside. OK. >Do you use transfinite cardinals? I donÕt see any outside, but then again I donÕt see ANY mathematical entities outside -- there are no natural numbers ßitting through the trees. >Is it for anything besides transfinite cardinals? Sure. One needs at least P(N) for representing the real numbers when putting Analysis on a set-theory foundation. >Some people use them as a definition as part of measure theory, but thatÕs >because they lack better tools, and the results of measure theory are largely >upon continua. That is to say, the useful results of measure theory are >derivable without the use of transfinite cardinals, and they should be. Giving up continua is asking a lot. Continuum Mechanics is a useful area with many practical applications. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: induction vs Cantor <8JqdnQxnYLcA6zrcRVn-3w@giganews.com> <41a7dbeb$0$20860$afc38c87@news.optusnet.com.au> <1fRrd.107121$T02.77638@twister.rdc-kc.rr.com> <41B40665.2DB637D2@tiki-lounge.com> <41B4DED8.947CA9BB@tiki-lounge.com> posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8 Hi Barb, Yeah, it is. The reals are great. Their simple assumption as a continuum offers a necessary tool for the pursuit of many and much mathematics. Most of the fundamental results about real numbers are in place without set theory, eg via Euclid. The (set of) real numbers is the indefinite contiguous sequence, the point set. You use zero, and iota, and integral multiples of iota, and those are all the non-negative real numbers. Mapping the integers to the reals in this way escapes the consequences of CantorÕs first proof, which some apply to the rational numbers, and I to neither. ZornÕs Lemma, the well-ordering principle, or the Axiom of Choice allow methods to guarantee enumerability of the sets. Well-order the reals and inductively select elements to inject into the integers, for the integers, the existence of integer n guarantees the existence of integer n+1. I agree that IÕm constructivist, but IÕm more quasi-intuitionist and not prejudiced. Also IÕm a platonist because math is real. Math can be quite irrelevant. The natural numbers are the trees. Do you use ZF with classes? WhatÕs the class of all classes? If your answer is no, can there be more than one proper class? There are theories with a set of all sets, that set is its own powerset, there the identity and tautology is between the set and itself, its own powerset. ThatÕs for intuitionism to resolve, the singular excluded excluded middle and the double entendre. Double entendre: a double entendre. The proper class is the ur-element, itÕs at once all and nothing, and the void and null. Its dualistic nature is its singular nature, and vice versa. Ross F. === Subject: Re: induction vs Cantor <8JqdnQxnYLcA6zrcRVn-3w@giganews.com> <41a7dbeb$0$20860$afc38c87@news.optusnet.com.au> <1fRrd.107121$T02.77638@twister.rdc-kc.rr.com> <41B40665.2DB637D2@tiki-lounge.com> <41B4DED8.947CA9BB@tiki-lounge.com> posting-account=WZMvOwwAAAC_B1TaayrVN99BJgDWQkUc RF> The reals are great. How the fuck would YOU know THAT? You donÕt even know a basic definition of what they are. RF> Their simple assumption as a continuum There is nothing simple about it TO YOU, idiot. To the rest of us, there is a simple first-order axiomatization, but it is hardly a simple assumption as a continuum; it is an assumption that they are an ordered field that is complete in the sense of containing all of the things you can produce from it by the operation of taking limits of COUNTABLY infinite sequences of things already in it. RF> offers a necessary tool for the pursuit of many and much mathematics. RF> Most of the fundamental results about real numbers are in place RF> without set theory, Yes, there are axioms defining the reals without mentioning sets. RF> eg via Euclid. To prove that Euclid is an example of that, you would have to give an example FROM Euclid of some results about reals. RF> The (set of) real numbers is the indefinite contiguous sequence, In modern set theories, you CANÕT just gloss over that in parentheses: you have to PICK A SIDE: is the class of all reals proper, or is it a set???? RF> the point set. You use zero, and iota, and integral multiples of iota, RF> and those are all the non-negative real numbers. It is a consequence of the fact that first-order languages are countable that you can model any consistent first-order theory countably, so if there are denumerably many iota-terms, yes, there is a way you can make them serve as a model. BUt that does NOT mean that they really are the reals. RF> Mapping the integers to the reals in this way escapes the RF> consequences of CantorÕs first proof, which some apply RF> to the rational numbers, and I to neither. NOthing escapes the proof. No matter how many iotaÕs you use to represent the reals, they will STILL all have denumerably-long-bit-string representations as well, and all you have to do to reproduce the proof in the iota- context is define the function that returns the nth decimal place (or bit) of a real as output, give the right number of iotas as input. RF> ZornÕs Lemma, the well-ordering principle, RF> or the Axiom of Choice allow RF> methods to guarantee enumerability of the sets. They DO NOT, dumbass. All these things are NON- constructive. You do get an enumeration or well-ordering from them but you NEVER get a METHOD! RF> Well-order the reals and inductively select elements to RF> inject into the integers, Obviously, this is not possible; you run out of natnums. RF> for the integers, the existence of integer n guarantees RF> the existence of integer n+1. And the existence of a term with n iotaÕs guarantees the existence of a term with n+1 iotaÕs. But nothing guarantees that you can inject the reals into either of those. RF> Do you use ZF with classes? NObody uses ZF with classes. ZF DOES NOT HAVE classes. In ZF, EVERYTHING IN THE ENTIRE UNIVERSE is a set. Proper classes DO NOT EXIST. RF> WhatÕs the class of all classes? A contradiction in terms, THATÕS what it is. RF> There are theories with a set of all sets, that set is its own RF> powerset, there the identity and tautology is between RF> the set and itself, its own powerset. You havenÕt actually studied any set theories with a universal set and you DONÕT know WTF you are talking about. All these theories have some VERY counter-intuitive consequences at very simple levels. === Subject: Re: induction vs Cantor Poker Joker says... >CantorÕs diagonalization procedure >> defines a function which given an infinite list of reals (that is, >> an element of R^N) returns an element of R that is not on the list. >> Cantor didnÕt prove that there were uncountably many such functions, >> he only proved that there was one. ThatÕs all he needs to be able >> to show that no list of reals contains every real. >Do you think there are any counter-examples to CantorÕs diagonal >proof? What do you mean by a counter-example? >If so, can you provide a valid modification that keeps the concept >of providing a real that is proven not in the list and has no >counter-example? Sorry, I donÕt know what you mean by counterexample. -- Daryl McCullough Ithaca, NY === Subject: JSH: Simply fascinating The math here is so readily understandable that itÕs actually almost as interesting watching how people react to it, as anything else. For instance, IÕve given a polynomial P(x) repeatedly where I factor it into three factors. I point out that the factors must include factors of the constant term of the polynomial P(x). I note that the factors of the constant term are independent of x, as they are, in fact, constant. Mathematically itÕs easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the polynomial results from multiplying together the three factors which IÕve called g_1(x), g_2(x), and g_3(x), then you can get the factors of the constant term, by setting x=0. ItÕs that simple. Notice (1) you have the factors of P(x), (2) the constant term of P(x) is determinable by setting x=0, (3) you also get the factors of the constant term. Mathematically, itÕs simple to the point of trivial. Now then, if you have *constants* which are factors of another constant then why would anyone try to argue that they are actually variables? You have x, as the variable. Besides x there are just these numbers. If you clear out x, then whatÕs left are constants. Letting x=0, clears it out, leaving the constants visible. Some may say, yeah, sure, at x=0, but what about when x doesnÕt equal 0? Um, if the numbers are constant, and so are independent of x, then, duh, why should it matter what value x has? The logic is inescapable. In terms of difficulty, my proof is about as easy as it gets in algebraic number theory, in terms of the actual mathematics. But the concepts are where there is a problem, and the social hang-up is in accepting that thereÕs this simple technique that shows a BIG problem, which can invalidate claims of proof for, well, over a hundred plus years. So the mathemtics is EASY for a trained mathematician to follow. The social implications are hard, if social stuff is important to you, and clearly from what IÕve seen it is to many of you. For instance, at this point IÕve removed all objections raised in detail. Like I can explain supposed counter examples to my work. I can give an actual example where you can see the factorization play out--just as the theory shows it must. And you probably know that my research is the work that can be said to have gone to a journal which at least claims it does formal peer review. They thanked me for the paper said the reviewers liked it, and then some sci.mathÕers got together--actually literally conspiring online in posts on sci.math--sent them emails and the editors yanked my papers THE NEXT DAY. They had it for nine months. IÕd corresponded with them for a while, even corrected them when they called me Dr. Harris as I donÕt have a Ph.d, and I told them I was an independent researcher with concerns about how my work would be handled. They kept saying no problem, ok, all that matters is whatÕs correct. Then they yanked my paper after sci.mathÕers emailed them: All the pertinent facts are in my favor. So whatÕs the hold-up? My research shows that some mistakes were made over a hundred years ago, and a lot of people missed them, and gave proofs which were not, and are not proofs. Mathematics is unforgiving. It doesnÕt care about the social implications of the truth. So it doesnÕt matter mathematically that a LOT of people out there are terribly dependent on the false beliefs and incorrect results, but it DOES matter a lot to those people! I call their behavior passive-aggessive, as by dragging their feet, taking as long as possible before acknowledging my research, or worse, hoping to NEVER acknowledge it at all, they are passively hoping to escape mathematical truth, in what amounts to a very aggressive way. I liken their behavior to judges at a race, who watch a runner break a world record, and then lie about his time, refuse to admit he even finished the race, and some even call him names!!! TheyÕre turning the way itÕs supposed to work with a major discovery, upside down. And itÕs silly behavior as eventually the truth will come out, and you know what IÕll do then? Probably go to the beach. IÕll also hang out in some bars. Yup, IÕll definitely hang out in some bars, preferably near a beach. Yup, you guessed it, IÕll do my best to forget about them, as why bother worrying about silly people who do silly things. LifeÕs too short. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that itÕs actually almost > as interesting watching how people react to it, as anything else. > For instance, IÕve given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > I note that the factors of the constant term are independent of x, as > they are, in fact, constant. > Mathematically itÕs easy to show: > g_1(x) g_2(x) g_3(x) = P(x) > and > g_1(0) g_2(0) g_3(0) = P(0) > as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > IÕve called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. > ItÕs that simple. Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. > Mathematically, itÕs simple to the point of trivial. Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? No one does. If as you say, you assume that your factors are constant, of course they are not variables. That is nothing but a dimwit tautology. And of course that is not what we say. We say that if you factor 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) by those variable functions, then you can obtain algebraic integer factors on both sides of the resulting equation. That is, g_1(x)/w_1(x), etc., are all algebraic integers. That is all you require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). No problem with that either. And the product of these three constant terms equals P(0)/49. It all works out. Factoring out 49 as a product of variable algebraic integer functions does not lead to ANY contradiction. It is only when you try to use a CONSTANT factorization that you arrive at what you think is a problem with algebraic integers. If you do the factorization in the right way there is no such problem. Suppose z is an integer variable, and you consider Q(z) = (x + 5)(x + 6). You notice that Q(z) is alway an even integer. Therefore you can always divide it by 2. You notice that when z = 0, Q(z) = Q(0) = 5 * 6. You notice that Q(0)/2 = 5 * (6/2). That is, when you divide by 2 to get an integer, you divide the constant term of the second factor, 6, by 2. If you tried to divide 5 by 2 you would not have an integer quotient. So by your logic, Q(z)/2 = (z + 5)*(z/2 + 6/2) = (z + 5)*(z/2 + 3) would be the only right, CONSTANT way to factor out 2. This is analogous to your division by 49 = 7*7*1: (5 a_1(x)/7 + 7/7)(5 a_2(x)/7 + 7/7)(5 b_3(x)/1 + 22/1) But, back to Q(z)/2 = (z + 5)(z/2 + 3). The problem is, z/2 is not always an integer. If you want integer quotients in both factors, you have to divide out 2 in a NONCONSTANT way. When z is even, you divide 2 out of (z + 6). When z is odd, you divide 2 out of (z + 5). Get it? Factoring 2 out in a constant fashion does not work. Factoring 2 out in a *variable* fashion does work. No problem arises with constant terms. > You have x, as the variable. Besides x there are just these numbers. > If you clear out x, then whatÕs left are constants. Letting x=0, > clears it out, leaving the constants visible. > Some may say, yeah, sure, at x=0, but what about when x doesnÕt equal > 0? Some may say. > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? You MUST divide by a variable factorization of 49. If you assume a constant factorization you get into trouble, because a_1(x)/7 in general is not an algebraic integer. You know this, yet you persist in thinking that the factorization has to be constant. You conclude *not* that your own thinking is wrong (as you should) but that there is something fundamentally wrong with algebraic number theory. You try to justify using the constant 7*7*1 factorization by repeating your mantra that the constant terms must be constant. This is based on your incorrect conclusion that, if you divide (a_1(x) + 7) by w_1(x), then the constant term is 7/w_1(x). And right there is your central mistake. BY YOUR OWN DEFINTION, the constant term of (a_1(x) + 7)/w_1(x) is not 7/w_1(x). It is instead 7/w_1(0). You want to conclude that 7/w_1(x) must be equal to the constant, 7, because 1 * 1 * 22 = 22, the constant term of P(x)/49. If you do the *correct factorization*, you still get 22, but it is in the form (7/w_1(x)) * (7/w_2(x)) * (22/w_3(x)) = 22 because w_1(x)*w_2(x)*w_3(x) = 49. Now, you may howl, YES BUT 22/w_3(x) IS NOT AN ALGEBRAIC INTEGER!!! To which the reply is : yes, youÕre right. BUT THERE IS NO REASON IT SHOULD BE. All that is required is that (5 a_3(x) + 7)/w_3(x) is an algebraic integer, and this follows from the correct choice of w_1(x), w_2(x), and w_3(x). As it turns out, both 5 a_3(x)/w_3(x) AND 7/w_3(x) are algebraic integers. And as above: the constant term of (5 a_3(x) + 7)/w_3(x) is NOT 22/w_3(x), as you believe. It is merely 22/w_3(0) = 22. These two are not the same, unless you assume what you want to prove, i.e., that w_3(x) is a constant function. And you know it is not. Everything works out. There is no contradiction or problem involving the constant terms. It is YOU who has been trying to claim that the constant terms are not constant, e.g., when you say that 22/w_3(x) is the constant term of (5 a_3(x) + 7)/w_3(x). You are making such a rookie mistake. Why canÕt you see it? > The logic is inescapable. True enough. > In terms of difficulty, my proof is about as easy as it gets in > algebraic number theory, in terms of the actual mathematics. Easy, yes. Correct, no. > But the concepts are where there is a problem, and the social [tiresome pompous rant about social factors etc. deleted] Nora B. > LifeÕs too short. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that itÕs actually almost > as interesting watching how people react to it, as anything else. For instance, IÕve given a polynomial P(x) repeatedly where I factor > it into three factors. I point out that the factors must include factors of the constant term > of the polynomial P(x). I note that the factors of the constant term are independent of x, as > they are, in fact, constant. Mathematically itÕs easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > IÕve called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. ItÕs that simple. Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. Mathematically, itÕs simple to the point of trivial. Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? No one does. If as you say, you assume that your factors are > constant, of course they are not variables. That is nothing but > a dimwit tautology. > And of course that is not what we say. We say that if you factor > 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), > and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) > by those variable functions, then you can obtain algebraic integer > factors on both sides of the resulting equation. That is, > g_1(x)/w_1(x), etc., are all algebraic integers. That is all you > require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and > g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to > g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). > No problem with that either. And the product of these three > constant terms equals P(0)/49. Ok, IÕm going to answer the Nora Baron poster yet again. And I want readers to understand that IÕve replied to this poster who you know lies anyway as itÕs a guy posting as a woman using a name thatÕs a palindrome MANY TIMES explaining in detail. What happens is that when I shoot down these objections, the poster either just repeats later, or replies with nonsense. One telling time when I carefully refuted point-by-point the poster replied deleting out everything IÕd said. Just deleted out everything, and you know what? I STILL so people replying about how supposedly I donÕt answer the objections from Nora Baron. ItÕs partly a game for some people on sci.math, IÕm sure, and partly a case where many readers are hoodwinked as *they* donÕt know itÕs a game. They seem incapable of realizing that there are people in this world willing to behave in such a way, on such a level. So why reply to this poster? Maybe IÕm hopelessly naive, but I just have to believe that eventually people will just get tired of being made fools of by this poster and others in the group with him (or her). It hasnÕt happened yet from what IÕve seen, but I keep hoping. Ok, so whatÕs wrong with the posterÕs assertion? Well, the wÕs the posters gives are factors of 49. But 49 comes into the picture because 49 is a multiple of the polynomial. But the wÕs STILL REMAIN after 49 has been divided off--after the multiple is divided off--as the poster actually claims. (Notice Nora Baron gives you clues, but somehow sci.math readers donÕt seem to get it.) Notice the posterÕs actually gives that when you divide off 49, you get g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x) which are TRACES, left of 49--after it has been divided off. So here we have mathematics. Here in an area where the truth can actually be determined, a poster who you canÕt be sure is a guy or a girl has confused many of you into questioning whether or not a multiple of a polynomial divides off as a variable or not. Did it ever occur to any of you that for some people that might be a great lark? Are you so trusting as to not consider that a poster who otherwise would probably be unknown can get off on confusing you not only on his or her gender, but on some of the most basic concepts in mathematics? Look now, even after the post where Nora Baron ended with a male name, people are STILL supporting the poster! It has occurred to me that this poster did so deliberately, ended that post with a male name, because the person knew you better than you still seem to know yourselves. In a way itÕs sad as youÕre giving up so much for nothing in return. James Harris === Subject: Re: JSH: Simply fascinating > ... 49 is a multiple of the polynomial. You keep saying this kind of stuoid stuff. But 49 is *not* a multiple of the polynomial. Read the definition of the word ŌmultipleÕ. > Look now, even after the post where Nora Baron ended with a male > name, people are STILL supporting the poster! That is because nobody really cares what NoraÕs geneder really is. They care about what he says... i.e. his math. Is that not what you claim is the only thing that counts? If Nora was a ßaming tranvestite hemaphroditic cross-dresser, it would not make his math incorrect or your math correct. And it is not a lie to use a pseudonym, it is a personal choice. Most of us do not use our real names here. === Subject: Re: JSH: Simply fascinating posting-account=KR2cuw0AAACZ_86pfubjOKsQkAVb6Rpe What does NoraÕs gender have to do with math. You only point this out because you are immature, trying to create a distraction, and you know your work is wrong. Get a life. Dave === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that itÕs actually almost > as interesting watching how people react to it, as anything else. > For instance, IÕve given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > I note that the factors of the constant term are independent of x, as > they are, in fact, constant. > Mathematically itÕs easy to show: > g_1(x) g_2(x) g_3(x) = P(x) > and > g_1(0) g_2(0) g_3(0) = P(0) > as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > IÕve called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. > ItÕs that simple. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. > Mathematically, itÕs simple to the point of trivial. Yes if P(x) = g_1(x) g_2(x) g_3(x) then P(0) = g_1(0) g_2(0) g_3(0) why did you surround this trivial fact with all the verbiage. No one is disputing this. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? > You have x, as the variable. Besides x there are just these numbers. > If you clear out x, then whatÕs left are constants. Letting x=0, > clears it out, leaving the constants visible. > Some may say, yeah, sure, at x=0, but what about when x doesnÕt equal > 0? > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? > The logic is inescapable. And by this logic the constant term of (a(0)=0, w(0)=1) h(x) = (a(x)/w(x) + 7/w(x)), a(0)=0, w(0)=1 is 7. The constant term is not 7/w(x). Indeed you can write h(x) so you can see the constant. Let t(x) = (a(x) -7w(x) +7)/w(x) Note t(0) = 0 and h(x) = (t(x) + 7) So the constant term of h(x) is 7. And yes 7 divides the constant term of P(x). No one is arguing that 7/w(x) divides the constant term of P(x). -William Hughes === Subject: Re: Simply fascinating Having fun following this thread, but I just want to make sure I understand it. So, James has the expression: P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) and he has shown that each a_*(x) evaluates to an algebraic integer for integer x. And he has shown that a_*(0) is an integer. Then he makes this leap by saying that a_1(0) = 7 implies a_1(x) = 7.b (b algebraic) for all integer x. Is that correct how I stated it, or am I missing something? Darren === Subject: Re: Simply fascinating > Then he makes this leap by saying that > a_1(0) = 7 implies a_1(x) = 7.b (b algebraic) for all integer x. > Is that correct how I stated it, or am I missing something? You got it all correctly. You see, 7 is a constant. It does not depend on x. So 7 cannot change just because x changes. So a_1(x) is still 7, since 7 has not changed..... get it? And by the way, the above equations have no memory of the 49 that got properly divided off as a multiple, or something like that. Get it? === Subject: Re: Simply fascinating Meant: > a_1(0) = 7 implies a_1(x) = 7.b (b algebraic Integer) for all integer x. === Subject: Re: Simply fascinating > But the concepts are where there is a problem, and the social hang-up > is in accepting that thereÕs this simple technique that shows a BIG > problem, which can invalidate claims of proof for, well, over a > hundred plus years. That is not the hang-up. The hang-up is that you are 100% fucked up. I would have no problem accepting something that would show a big problem, invalidating proofs for over a hundred years. I would just love it. I would think, wow, that is so cool! It would be a thrill. But you have not convinced me or anyone else that you have found anything interesting or correct. Just ßuffy bogus claims. Lots of claims. Little valid math. Little valid logic. Lots of paranoia. Lots of grandiosity. Lots of non-standard terminology. Lots of smoke screen crap, like the gender of Nora. Look James... lots of people here would love to see the next crisis/revolution in math... another Russell shakeup, another G.9adel shakeup. So tell me... why has nobody been interested in any of your ideas? Could it be that your ideas are shit? Or do you think that it is more likely that everyone else is nuts. Apply OccamÕs Razor here! Just a thought.