>But here I now read that people are indeed very attached to their >calculators, and so I am very curious: What do you use them for? Just to join in the general HP nostalgia: I used an HP-45 for many years, and it functioned fine for quite awhile, if a bit stickily, after a Pepsi spill that got into most of the keys. I finally ditched it when some of the crucial buttons stopped responding. Around the same time I had an HP-65 programmable at work, and I was programming it to do various numerical things I needed (like Lagrange interpolation of calibration data). I have bought one or two non-HP scientific calculators since then, but rarely used one. Almost every calculation I want to do these days I just do on the Matlab window I always have open on my computer. And since that includes a really lovely graphics capability, I've never had occasion to want to use a graphic calculator. I have no idea how to use a TI-83, and I haven't yet found enough incentive to want to learn. One of these days I'll probably want to invest in a scientific calculator for use away from the office, but I'm not sure if I have any need for programmability. It *would* be nice to have hex conversion at my fingertips; the last calculator purchased for me by an employer was an HP model that included lots of computer-related stuff like that. - Randy ==== > I've always wondered about Mathematical Olympics. Don't the judges have > to be smarter than any of the contestants, to be able to judge them > accordingly? In normal Olympics, the judges can be any skinny little > wimps off the street, as they can merely watch the athletes without > competing themselves. Ignoring the time constraint --- in a lot of cases --- for making the judgement call, the most important qualifications, IMHO, are integrity, knowing the rules and knowing how to apply the rules. (But math rules are not just any rules.) Notwithstanding, getting a couple hundred of rainbow eyes and broken noses and cut cheeks (analogically in boxing) may or may not hurt the qualifications more than the injured. > But mathematics is different. Unlike athletics, it's not plain to see > who is better than who. You have to be a mathematician yourself. Shall we start the definition of (the qualification of) a mathematician? :) Math is not, IMHO, to figure out who is better than who, a petty proposition. It is about truth. If your focus is math contests, rather than math in its 'pure' form, then the result is no different from another type. A 'democratic' view of the performance of the contestants at the moment with a specific set of configurations. The 'lazy God' (sorry for sometimes misspellings) by Minkowsky could have had pathetic performances in contests. > There's *one* thing that suggest that the judges don't have to be the > smartest ones after all - in certain theoretical calculations, checking > whether something has been done is far easier than actually doing it. Wonder, if there can be exceptions (to easier). > For example consider the simple array sort. Actually sorting the array > can't be done faster than O(nlog n) time, this has been proven. But Practically, I sincerely hope that stays proven. > checking if the array has been sorted can be done in O(n) time with a > simple algorithm any child could come up with. Maybe this can be > expanded to more complicated calculations? ==== > Let's define a sum: n 1 > s(n)= S ------------- > k=0 k! Now, we want to get this sum: oo > S(x)= S s(n) x^n ; x<1 > n=0 Is there a closed form and if so how to evaluate it How about this: Multiply by x and get the following: x S(x) = SUM (n=1,oo) s(n-1) x^n Now subtract S(x) and use the following fact: s(n) - s(n-1) = 1/n! I get S(x) = e^x/(1-x). -Michael. ==== > Let's define a sum: n 1 > s(n)= S ------------- > k=0 k! Now, we want to get this sum: oo > S(x)= S s(n) x^n ; x<1 > n=0 Is there a closed form and if so how to evaluate it > Look at S(x) - xS(x). Rick ==== > Ah, but there's a difference for me. When I enter 6.67259 x 10 EE -11 and > press = (equal key), I get 6.67259 ^-10. This is the problem. If you enter 6...*10 EE -11 then you really compute 6..*10*10^-11 since (EE z) is an abbreviation for doing *10^z. Hence I would suggest entering 6.. * 10 then pressing a button like y^x (it is so called on my TI 30X I think its also on TI 36X) and then entering -11. ==== >Ah, but there's a difference for me. When I enter 6.67259 x 10 EE -11 and >press = (equal key), I get 6.67259 ^-10. THERE'S your problem! You multiplied 6.67 by 10x10^(-11), which is 6.67x10^(-10). You should have punched in 6.67 EE -11 . DON'T punch in the 10! --Dan Grubb ==== 1/9 = 0.1111... > 2/9 = 0.2222... > ... > 8/9 = 0.8888... > 9/9 = 0.9999... But, 9/9 = 1, so 0.9999... equals 1 This is just some funny stuff that crept into my head. Now, is there anything > wrong in this proof? If so, what is it? > If you know that each of these lines is the sum of a geometric series, that, then you need more to make this a proof. ==== > But here I now read that people are indeed very attached to their > calculators, and so I am very curious: What do you use them for? > I would be especially interested to discover any professional uses for > the graphing and symbolic calculators (my impression so far has been > that no one uses them except high school teachers and their students). You never forget the first time you were in love.... I cherish my HP 32S II so much that I wouldn't dream of bringing it with me to work. It might get lost or stolen or worse... That leaves me with a stupid Windows Calculator, which doesn't do RPN. I must suffer to protect my loved ones... -Michael. ==== > Ellipse is produced if a plane intersects only one nappe of a cone. Is there a way to get the parameters of the ellipse as a function of > the angle of the plane with the axis of the cone? I know that they depend on the height of the cone and diameter of the > circle at the bottom of the cone. I couldn't find exactly what you're asking for at Mathworld (http://mathworld.wolfram.com/Cone.html), but here's some thoughts from the geometry. (Plane that is perpendicular to the axis produces a circle.) Let theta be the angle the plane makes with the axis. Let's say the cone has height h and base diameter d. Let us suppose the plane cuts the axis at distance y from the apex, where the diameter is (d/h)y. Half of this, 0.5dy/h, is the semi-minor axis of the ellipse. The center point of the ellipse is the place where the plane cuts the axis. Now let's look at the semi-major axis. Consider a planar slice through our plane and cone that includes the semi-major axis and the axis of the cone. (I suggest you draw a picture. I'm referring to a picture as I write this). The picture is something like this. Apex A | | cone | | B Axis / | / | / Cutting plane |/ C We have a triangle formed by the apex A, the ellipse center C, and the highest place B where the ellipse cuts the cone. The angle ACB is the angle theta at the bottom of this triangle. The angle CAB at the top, the apex angle of the cone is arctan(0.5d/h). Call this angle CAB = phi. BC is the semimajor axis of the ellipse. We can find it from, for example, the law of sines: BC/sin(CAB) = AC/sin(ABC) I defined y = AC. sin(ABC) = sin(pi - (CAB + ACB)) = sin(CAB+ACB) So BC = semimajor axis = y*sin(phi)/sin(phi+theta) unless I made a mistake. Now you have the semiminor axis (0.5dy/h) and the semimajor axis. That's enough to calculate any other parameters of the ellipse you want. - Randy <4a10pvcg1d5066dbuoa4s7s9pcus13rlui@no.spam> ==== >Ah, but there's a difference for me. When I enter 6.67259 x 10 EE -11 and >press = (equal key), I get 6.67259 ^-10. This is the problem. If you enter 6...*10 EE -11 then you really compute > 6..*10*10^-11 since (EE z) is an abbreviation for doing *10^z. > Hence I would suggest entering 6.. * 10 then pressing a button like y^x (it > is so called on my TI 30X I think its also on TI 36X) and then entering > -11. Or just 6.67259 EE -11. -- P.A.C. Smith 'If the Apocalypse comes, beep me.' <*> http://www.srcf.ucam.org/~pas51 ==== I'm constructing an algorithm to generate the harmonic series of a musical instrument. I've been butting my head with this equation which I have to invert, but I've had no luck so far: y(x) = (1 + sqrt(1 + px)) / (1 + sqrt(1 + x)) where 0 < p < 1. So the problem is to turn x into a function of y. I get the feeling this is either trivial and I've missed something obvious, or it cannot be done analytically in a closed form. -Alistair ==== - 2y = px - 2y.sqr[(1+x)(1+px)] + xy^2 -(2y + px + xy^2) = -2y.sqr[(1+x)(1+px)] (2y + px + xy^2)^2 = 4y^2 (1+x)(1+px) 4y^2 + p^2 x^2 + x^2 y^4 + 4pxy + 2px^2 y^2 + 4xy^3 = 4y^2 + 4xy^2 + 4pxy^2 + 4px^2 y^2 p^2 x^2 + x^2 y^4 + 4pxy + 2px^2 y^2 + 4xy^3 = 4xy^2 + 4pxy^2 + 4px^2 y^2 p^2 x^2 + x^2 y^4 + 4pxy - 2px^2 y^2 + 4xy^3 = 4xy^2 + 4pxy^2 p^2 x^2 + x^2 y^4 + 4pxy - 2px^2 y^2 + 4xy^3 - 4xy^2 - 4pxy^2 = 0 x^2 (p^2 + y^4 - 2py^2) + x(4py - 4py^2 + 4y^3 - 4y^2) = 0 x^2 (p - y^2)^2 + 4x(py - py^2 + y^3 - y^2) = 0 x^2 (p - y^2)^2 + 4x(py(1 - y) - y^2 (1 - y)) = 0 x^2 (p - y^2)^2 + 4xy(p - y)(1 - y) = 0 x = -4y(p - y)(1 - y) / (p - y^2)^2 >I get the feeling this is either trivial and I've missed something >obvious, or it cannot be done analytically in a closed form. > I've the unsettling notion you better check my work and my answer. y(0) = 1; x(1) = 0; well it does check for x = 0, but that's a special case anyway as I divided by x for the final solution. ---- ==== I'm constructing an algorithm to generate the harmonic series of a > musical instrument. I've been butting my head with this equation > which I have to invert, but I've had no luck so far: y(x) = (1 + sqrt(1 + px)) / (1 + sqrt(1 + x)) where 0 < p < 1. So the problem is to turn x into a function of y. I get the feeling this is either trivial and I've missed something > obvious, or it cannot be done analytically in a closed form. > -Alistair > a computer equation solver (MATLAB symbolic toolbox) gave x = 4 * y * (1 - y) * (y - p) / (y^2 - p)^2 but got confused when instructed to check by plugging in. Cause: ambiguity in taking square roots from squares. So, I applied common sense and got the same result. The method: First, realize that sqrt(p) < y < 1 (subtract and compare with zero). Then introduce an intermediate unknown t = 1 + sqrt(1 + x), so t > 0 and x = t * (t - 2) ==== >3 men went to a motel. There's only 1 room left and it costs 30 bucks. So, each >man forked out 10 bucks. Later, the owner discovered he over-charged the 3 men. It should be 25 bucks >instead. So, he sent his runner to give 5 bucks back to the 3 men. The 3 men >think the owner is very honest, and thus each took 1 buck back, leaving 2 bucks >to the runner as tips. Now, since they took 1 buck back, each of them paid 9 bucks. 2 bucks to the >runner. So: (9*3)+2 = 29. Where's the 1 dollar? > There is no one dollar to look for, i.e.: 10 + 10 + 10 = 30 = 25 + 5 = 25 + 2 + 1 + 1 + 1 --> 9 + 9 + 9 = 25 + 2 so it's 27 - 2 = 25 and not 27 + 2 = 30 adam ==== >Once upon a time, the battery died and thought that I'd >just use this other calculator until I had time to get >a new one. I could not function. RPN is so imbedded >into my thinking that I could not use a regular calculator. >I ended up doing the stuff on paper and the task of getting >a new battery jumped to number 1 priority. It's the same with me. I use a slide rule and/or paper and pencil if > an RPN calc isn't available. (Though my 48GX, 41CX and 16C are usually > within arm's reach.) My laptop is often within arm's reach and I use an RPN calculator on it, although I do also have a physical one somewhere. I, too, can not function with an algebraic-notation calculator, despite using algebraic notation often enough in various programming languages. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science ==== >Once upon a time, the battery died and thought that I'd >>just use this other calculator until I had time to get >>a new one. I could not function. RPN is so imbedded >>into my thinking that I could not use a regular calculator. >>I ended up doing the stuff on paper and the task of getting >>a new battery jumped to number 1 priority. It's the same with me. I use a slide rule and/or paper and pencil if >> an RPN calc isn't available. (Though my 48GX, 41CX and 16C are usually >> within arm's reach.) My laptop is often within arm's reach and I use an RPN calculator on it, >although I do also have a physical one somewhere. I, too, can not >function with an algebraic-notation calculator, despite using algebraic >notation often enough in various programming languages. I don't see any relationship between those other kinds of calculations and algebra or programming languages. It's because of my familiarity of programming that I can't use those other calculators. RPN is completely algebraic to me. When calculating any equation, you do the insides first with a heirarchy of operation. Those non-RPN calculators are left to right with no push down list. /BAH /BAH ==== >Once upon a time, the battery died and thought that I'd >>just use this other calculator until I had time to get >>a new one. I could not function. RPN is so imbedded >>into my thinking that I could not use a regular calculator. >>I ended up doing the stuff on paper and the task of getting >>a new battery jumped to number 1 priority. >It's the same with me. I use a slide rule and/or paper and pencil if >> an RPN calc isn't available. (Though my 48GX, 41CX and 16C are usually >> within arm's reach.) My laptop is often within arm's reach and I use an RPN calculator on it, >although I do also have a physical one somewhere. I, too, can not >function with an algebraic-notation calculator, despite using algebraic >notation often enough in various programming languages. I don't see any relationship between those other kinds of calculations > and algebra or programming languages. It's because of my familiarity > of programming that I can't use those other calculators. RPN is completely algebraic to me. When calculating any equation, > you do the insides first with a heirarchy of operation. > Those non-RPN calculators are left to right with no push down list. > You mean the ones with algebraic notation and no parentheses? (Yes, they really exist.) Or with the (seemingly always) insufficiently deep nesting. I've exceeded 9 levels of parentheses with algebraic (left-to-right) entry, but I vaguely remember *once* exceeding a 4-level stack on an RPN calculator. I couldn't even tell you the circumstances. I might be misremembering -- or remembering an error I made. I get around the algebraic problems by (drum roll, please) working from the inside out and either using stored memory locations or writing down the intermediate results. I also claim that anyone who can actually use an algebraic calculator for difficult calculations (like mortgage payment calculations) does the equivalent (well, maybe not, since it can be done with about 5 levels of parentheses). But after a while, you forget which level of parentheses you're in, so you have to be making notes of some sort on a sheet of paper. Oh, I forgot algebraic notation but no algebraic hierarchy. So 1+2*3=9. Aaaaaauuuuuuuggggggghhhhhhhhh! Usually with no parentheses. Jon Miller ==== rearrange it until you have something that depends on x & y and sqrt(1+x) = sqrt(1+px) square it the l.h.s. will have sqrt(1+x) in it. Isolate it and square again. I think you will then be left with a quadrative in x which you can solve. I'm constructing an algorithm to generate the harmonic series of a > musical instrument. I've been butting my head with this equation > which I have to invert, but I've had no luck so far: y(x) = (1 + sqrt(1 + px)) / (1 + sqrt(1 + x)) where 0 < p < 1. So the problem is to turn x into a function of y. I get the feeling this is either trivial and I've missed something > obvious, or it cannot be done analytically in a closed form. > -Alistair ==== hot-girl escribi.97 en el mensaje > a sphere : x^2 + (y-2)^2 + (z-3)^2 =1 let point p : when tangent line of sphere pass on (0,0,c) is meet x-y > plane, the point of contact is named p let c1,c2 : value c that trace of p satify parabola. solve that c1+c2 ?? ------------------------------------- i will regard it as difficult...... help me....my genius teacher.... i wait your ultra power advice. thank sir. Actually is is easy ... What surface descrive the tangent line throw (0, 0, c)? What type of sections produce a plane in that surface? When this section ia a parabola? Note that the tangent line can meet the x-y plane on the side of the sphere or on the side of point (0, 0, c). -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== good advice....good teacher... ==== I am looking for some old HP calculators like HP 41CV, HP 41CX, HP >> 71B, HP 15C, HP 16C, HP 67 and any others in the 1980's era...If you I have an HP55 and no, I'm not selling. >Bought it in 1975 and it still works :-) > http://www.dotpoint.com/xnumber/hp55.htm I would like to point out that this is a rarity in sci.math: > a thread in which people wax eloquent about calculators. In my personal experience, I have seen calculators be useful at work for 1. simple arithmetic which involves too many digits to be fun or interesting > 2. repetitive specialized calculations (e.g. a banker will use a calculator > which can compute effective interest rates on a loan or investment). Anything simpler people do by hand and anything more complex uses a computer. > In short: it always seemed to me that almost no one would use a calculator > as a regular part of their job except for situations #1 and #2. But here I now read that people are indeed very attached to their > calculators, and so I am very curious: What do you use them for? > I would be especially interested to discover any professional uses for > the graphing and symbolic calculators (my impression so far has been > that no one uses them except high school teachers and their students). In particular, if anyone has ever denied employment to an applicant > because he or she was found not to have the necessary competence with > calculators, that would be information I ought to have. My HP-55 was my first exposure to programming ( I was 19 back then in 1975). It was love at first sight - although it had only 50 steps. It was the best toy I ever had. In university I used it for astronomy, geodesy and physics courses. In the first year we were forced to use log and trig tables for the exam problems. The next year we were allowed to use a calculator, but then the problems suddenly turned out to be purely theoretical. I used it to calculate graphs for functions for the drawings I used to make on the left side of my notes. We were and still are inseparable ;-) few years ago (with some extra features not present on the original): http://users.pandora.be/vdmoortel/dirk/Stuff/HP-55.jpg Dirk Vdm ==== Corolllary of: http://www.bearnol.pwp.blueyonder.co.uk/Math/psq.htm ==== Hmm.. You claim that hcf (pm, p(m+M)) = hcf (pm, M). Let's take n = 8, m = 2, M = 12, pm = 3, p(m+M) = 2. Now hcf(pm, p(m+M)) = hcf(3, 2) = 1, and hcf(pm, M) = hcf(3, 12) = 3, and where I come from 1 != 3. As this statement is completely false, I don't see how the rest of the argument is to hold together. Perhaps you meant to make some stipulations on the choices of pm. If so, what are they? ==== sets of 10 as hard copy posters. However, I am now offering them on a CD. The images still measure 8 x 11-1/2 inches, same size as the posters. If you get a chance, please visit my website, where every poster is available to see (as a thumbnail). The URL is at http://www.mathisradical.com/Catalog.htm I'm not a professional sales person, just a math teacher who came up with a product that fits a need in a mathematics classroom. They are inexpensive and useful. I also have a free newsletter that includes lesson plans, biographies of mathematicians and Black Line Masters for your classroom. Sincerely, Kavon Rueter kavon@mathisradical.com http://www.mathisradical.com ==== >I'm not a professional sales person, just a math teacher who came up with a >product that fits a need in a mathematics classroom. Of course you're not professional. Someone who was professional would realize that posting the exact same message in the same newsgroup mere days apart would be spamming. Doug ==== How sweet of you to notice! Kavon I'm not a professional sales person, just a math teacher who came up with a >product that fits a need in a mathematics classroom. Of course you're not professional. Someone who was professional would > realize that posting the exact same message in the same newsgroup > mere days apart would be spamming. Doug <1xUcb.14254$O85.6040@pd7tw1no> <3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net> <3f82ebb8$3$fuzhry+tra$mr2ice@news.patriot.net> <3f8c413c$18$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Terminate: SPA(GIS) ==== >You have a point, two points actually, but I did not say that proof >is a mathematical concept, talk about are Mathematical concepts. >Any that follow from the definition of natural number I give below. You don't give a definition. You list a few names of natural numbers, and mention the successor function. And you certainly haven't answered my question as to what properties you believe they have that do not follow from the Peano postulates. >In some cases it is easy to tell; Then tell. >In his paper G.9adel showed how, for any sound formal system for >natural numbers (that is, one in which all provable statements are >true), The word true in that context is a technical term quite different from what is generally understood by true. >This is false. Incorrect. >G.9adel showed how to find true statements about >natural numbers not provable from the (first-order) Peano >Postulates. He showed to to prove additional theorems by adjoining to the Peano Postulates an additional axiom, essentially asserting that the Peano Postulates were consistent. >I do not know what you mean by a metaphysical concept. Things that you talk about without ever defining or characterising in a precise enough fashion to allow rational discourse. >The natural >numbers are the numbers >0, 1, 2, 3, 4, .... >When spelled out this becomes >0, s(0), s(s(0)), s(s(s(0))), s(s(s(s(0)))), ... . That's not a definition. That doesn't even say that they obey the Peano Postulates. >Natural numbers are perfectly well-defined, despite the fact that >they cannot be axiomatised in a first-order system. Then define them. >There is nothing strange about natural numbers. There is nothing strange about oranges, either, but that doesn't make them apples. >We encounter them every day. In isolation. We do not encounter The Naturals every day; in particular, we do not encounter induction. >Your use of the word mathematics is non-standard. No. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== >In his paper G.9adel showed how, for any sound formal system for >natural numbers (that is, one in which all provable statements are >true), The word true in that context is a technical term quite different > from what is generally understood by true. Different how? ==== Working on a research project and can't find the specific equations we're looking for. Given a hyperbolic paraboloid (HP) rotated 45 Deg. such that two planes of symmetry are y=+-x, the equation is z=xy: (1) We're looking for the equation of an HP rotated as described above having a parabola with the equation 12x^2-3x+2 above the line y=x, and a parabola with the equation -12x^2 +2 below the line y=-x. (2) What method should be used to find the equation of the hyperbola created by the intersecting plane z=k? (3) We're having trouble finding resources, text books and publications of any type dealing with the specific equations of rotated HP's...can you point us in the right direction? Specific sites, publications, or people you might know? Any and all help appreciated! Morgan and Brian ==== Suppose you have *any* three-dimensional figure given as a relationship between x, y, and z, and you want to rotate it, say, around the Z axis, counterclockwise, by an angle of t. Then all you have to do is rotate each individual point, and the entire thing will be rotated. Imagine a point (x, y, z). Rotating around the Z axis will not raise or lower the point, so just think about the (x, y) part for a minute. Imagine it on the plane, with the line segment from the origin to (x,y) drawn in. Then imagine a circle centered at the origin with this segment as its radius. Call the radius r (equal to sqrt(x^2 + y^2)), and the angle, measured counterclockwise from the positive X axis (equal to arctan (y/x)) t1. Then from trigonometry we have x = r cos t1 and y = r sin t1. Now you just want to replace x and y with their rotated equivalents, so if we're rotating an angle of t to the left, we want x' = r cos (t1 + t) and y' = r sin (t1 + t). But recalling the trig identities: sin (s + p) = sin s cos p + cos s sin p cos (s + p) = cos s cos p ö sin s sin p We have x' = r cos(t1 + t) = r (cos t1 cos t - sin t1 sin t) = (r cos t1) cos t - (r sin t1) sin t = x cos t - y sin t y' = r sin(t1 + t) = r (sin t1 cos t + cos t1 sin t) = (r sin t1) cos t + (r cos t1) sin t = y cos t + x sin t z' = z (since rotating around the Z axis doesn't raise or lower the point) So, saying that your equation was z = xy, then after rotating around the Z axis by an angle of t, just plug in x' for x, y' for y and z' for z, and your equation becomes z = (x cos t - y sin t)(y cos t + x sin t) z = (sin t cos t) x^2 + (cos^2 t - sin^2 t) xy - (sin t cos t) y^2. Don't be scared by the trig functions; they all depend just on t, so once you choose t the equation just looks like, say, z = ax^2 + bxy + ay^2 for two numbers a and b. Also, when the figure moves, it takes its lines of symmetry with it -- If your lines of symmetry were y = x and y = -x, now they are y cos t + x sin t = x cos t - y sin t and y cos t + x sin t = y sin t - x cos t But these can be reduced to y (sin t + cos t) = x (cos t - sin t) and y (sin t - cos t) = x (sin t + cos t). Now, of course, there are two other kinds of rotation -- rotation about the Y and Z axes. However, these have pretty much the same equations, so you can figure them out yourselves. Now all rotations can be made up of a rotation in the X direction, another in the Y direction, and another in the Z direction. If you want to find the intersection of a plane with a hyperbola, just set them equal to each other. I.e. if the plane is z = k and the hyperbola is z = xy, then k = xy, or in other words, y = k/x. Hyperbolas can be rotated in exactly the same manner as three-dimensional shapes -- a rotation around the Z-axis in 3d is the same as a rotation about the origin in 2d. Sorry if this is kind of hard to read, but it's hard to teach what would basically be a couple of weeks in a high-school analytic information, look up information about rotating conics in any analytic geometry book. > Working on a research project and can't find the specific equations > we're looking for. Given a hyperbolic paraboloid (HP) rotated 45 Deg. such that two > planes of symmetry are y=+-x, the equation is z=xy: (1) We're looking for the equation of an HP rotated as described above > having a parabola with the equation 12x^2-3x+2 above the line y=x, and > a parabola with the equation -12x^2 +2 below the line y=-x. (2) What method should be used to find the equation of the hyperbola > created by the intersecting plane z=k? (3) We're having trouble finding resources, text books and > publications of any type dealing with the specific equations of > rotated HP's...can you point us in the right direction? Specific > sites, publications, or people you might know? Any and all help appreciated! Morgan and Brian ==== Suppose you have a device that will show a value M such that M ~ N( 230, 0.023 ). What is the probability that we get value of 230.3 or less? Well, simply P(M < 230.3) gives me around 0.500 (Mathematica). On the other hand, i tried to standarize the variable and got: x = (230.3-230)/sqrt(.023) = 1.978 P(Z < 1.978) = 0.976 I can't explain the large difference between those two. Also, i can not really decide which one is correct (if any). what do you think? -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- ==== >Suppose you have a device that will show a value M such that >M ~ N( 230, 0.023 ). What is the probability that we get value >of 230.3 or less? >Well, simply P(M < 230.3) gives me around 0.500 (Mathematica). >On the other hand, i tried to standarize the variable and got: >x = (230.3-230)/sqrt(.023) = 1.978 >P(Z < 1.978) = 0.976 N(230, 0.023) means normal distribution with expectation 230 and variance 0.023. Whatever Mathematica gives you is wrong as it appears you didn't give it the correct variance. You can estimate the sensibility of the result by looking at the standard deviation, sqrt(0.023) ~= 0,152. Your value is about +2*sigma away from the expectation, so it's more than 95% of the distribution. This agrees with your calculations. ==== >Suppose you have a device that will show a value M such that >M ~ N( 230, 0.023 ). What is the probability that we get value >of 230.3 or less? >Well, simply P(M < 230.3) gives me around 0.500 (Mathematica). >On the other hand, i tried to standarize the variable and got: >x = (230.3-230)/sqrt(.023) = 1.978 >P(Z < 1.978) = 0.976 I can't explain the large difference between those two. Also, i can >not really decide which one is correct (if any). what do you think? > Haven't done the calculations, but your second caculation is more likely to be correct. I suspect you did not enter the formula correctly in Mathematica. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > Suppose you have a device that will show a value M such that > M ~ N( 230, 0.023 ). What is the probability that we get value > of 230.3 or less? > Well, simply P(M < 230.3) gives me around 0.500 (Mathematica). > On the other hand, i tried to standarize the variable and got: > x = (230.3-230)/sqrt(.023) = 1.978 > P(Z < 1.978) = 0.976 I can't explain the large difference between those two. Also, i can > not really decide which one is correct (if any). what do you think? Basic sanity check here: can you *really* believe that the probability of a value greater than almost 2 standard deviations above the mean is nearly 50%? That's not even mathematically possible (Chebyshev's inequality says that regardless of the distribution, at most 1/4 of it can lie two or more sds from the mean). ==== >> M ~ N( 230, 0.023 ) >> P(M < 230.3) = 0.500 (Mathematica) >> x = (230.3-230)/sqrt(.023) = 1.978 >> P(Z < 1.978) = 0.976 > Can you *really* believe that the probability of a value greater > than almost 2 standard deviations above the mean is nearly 50%? Yes, i can, but that's depending on the fact that i've been studying statistics for 6 weeks this far. :) I didn't think of what you mentioned, due to my ignorance. I will, -- Kindly Konrad --------------------------------------------------- May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more. Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- ==== > And it's raised to a power. Once you relax that you a) lose your > proof of the Erdos-Selfridge theorem and b) have a meaningless > result. Among all the (finitely many) prime factors, surely there > has to be a largest one. That's not news. D. Cool ==== Thursday February 24th 2000 55/311 15712 I woke up remembering a dream of 3 nubile sweeties, when they saw me they started running towards me. I felt threatened and started to run away but they were younger and more athletic and caught up to me quickly. One then pointed to a poster on a pole and asked that I read it. I'm not entirely sure about the contents of the poster, but was left with the impression it had to do with phallic whoreship in the Catholic and Protestant churches. Then in the evening I went to Better Tart's Cafe on the second floor of the Bayside mall and Jill provided stats while waitressing there. 54+ Dad 13 4 /262 54+ Mom 25 2 56/ 54+ Bro 17 1 77 17/348 7274 145 Jill 2 4 80 93/273 8445 Jillian 67 Rae 24 Huber 54 Leviticus begins with 17 verses and terminates at chapter 117 with 17+17 verses. There are 17 verses at chapters 1 and 3, and 59 (the 17 prime) verses at chapter 13, so the 17's and the 17th prime are at chapter numbers adding to 17 (1+3+13=17). The first 17 versed chapters in the Bible are at chapters 91 and 93, together for 184, or the 167 verses of Book 17 plus 17 more. Leviticus contains 859 verses, it ends in 59 (the 17th prime). The first 17's in the Bible surround chapter 92 (the 4x17th non-prime): Leviticus --------- 91 1 17 92 2 <-68th (4x17th) non-prime 93 3 17 94 4 95 5 96 6 97 7 98 8 99 9 100 10 101 11 102 12 103 13 59 <-17th prime 104 14 105 15 106 16 107 17 108 18 109 19 110 20 111 21 112 22 113 23 114 24 115 25 116 26 117 27 34 <-17+17 Brother was born on day 17, Jill on day 93 corresponding to Leviticus 3 with 17 verses. Leviticus 3 is the second chapter to contain the length of 17 verses while Jill is the second kid. The second kid was born on the 2nd day of the month and on the 2x2x2x2x2x2x2x229th day of the century. Jill was born a multiple of 128 days into the century (2 to the 7th), and her 145 name is 128 short of the 273 days remaining in the year after her birth. She has a 7 lettered first name adding to 67, corresponding to Exodus 17 (the 7th prime). Her brother was born on the 17th (7th prime) day of the month in year '77. Her parents were likely born on days of the year adding to 159, it's Book 7's 618 verses short of 777, and is the 7x7th prime (227) minus the 7x7th non-prime (68). Jill's parents were together born with 571 or 572 days remaining in their years. The kids were born in years adding to 157. The family was born on days of the month adding to 57. Likely mom was born on day 103 and so the family was born on days of the year adding to 269 (the 57th prime). Bible Chapter 57 and Book 57 both contain 25 verses while mom was born on the 25th. This 57 is the 41st non-prime while 41 in turn is the 13th prime, while dad was born on the 13th. The vowels in Jill's given names add with their positions for 57. The vowels in her given names add to the 25 verses of chapter 57 and Book 57. In her given names, her consonants exceed her vowels by 41 (57 is the 41st non-prime). Her given names add to 67 and 24, corresponding to Exodus 17 and Genesis 24, together for 41 (57 is the 41st non-prime). Her names add to 67, 24 and 54, these chapters in Genesis and Exodus together contain 114 (57+57) verses. 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter Dad was born on the 13th, the last name adds to 54 (13 plus the 13th prime, the opening length of Book 13, and Book 54 contains 113 verses). Jill 's given names add to 91 (1 through 13). Her first name exceeds her last name by 13. Jill has no letters in proper alphabetical order while only the 13th letter of her name is in reverse alphabetical order. She was born 355 and 37 days after her parent's birthdays, corresponding to First Chronicles 17 and Genesis 37, together for her 54 valued last name. Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations Jill adds to 43 (14th prime). Her given names differ in value by 43 (14th prime). Her initials add to the 36 chapters of Book 14. She was born an average of 14x14 days after her parent's birthdays. The kids were born 998 days after their parent's birthdays, it's the opening chapter of Bible Book 43 (Gospel John, 14th prime). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 --- 108 The parents were born in months adding to 6. Dad was born on the 13th (6th prime), mom was born on day 56 (Exodus 6). Jill is 66 inches tall. Her middle name adds the 24 chapters of Book 6. In her given names, her consonants add to 66 and exceed the vowels by 41 (the 6th prime in prime position). Her letters in odd/odd even/even harmony add together with their positions for 66. The kids were born an average of 317 days after dad's birthdays (the 66th prime). We meet on the 24th (Book 6). Jillian adds to 67 (the 19th prime), her parents were born on days of the month averaging 19. Dad was likely born on day 103... the parents were born on days 103 and 56, these are the 27th prime and 40th non-prime, together for 67. Her first name adds to 67, her middle name adds to 24 (Genesis 24 with 67 verses). 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 208 Jill 2 4 80 93/273 8445 Jillian 67 Rae 24 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doink this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. Should Jill marry me, great, but if Marcia marries me and then Jill marries Marcia and me, then Jill's brother is goink to win himself a shiny new Cadillac!!! Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... ==== >Thursday February 24th 2000 55/311 15712 I woke up remembering a dream of 3 nubile sweeties, when they saw me they >started running towards me. I felt threatened and started to run away but >they were younger and more athletic and caught up to me quickly. One then >pointed to a poster on a pole and asked that I read it. I'm not entirely >sure about the contents of the poster, but was left with the impression it >had to do with phallic whoreship in the Catholic and Protestant churches. << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== Yes, I'm sure this is all worthwhile for those unfamiliar with DSK but do you have to reply to *ALL* his messages? Why not just one a day? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== Could you please check the following for me: What is the probability that a card selected from a deck is an ace or heart? The way I see it is 52/4 = 16 (hearts) plus 3 (the aces minus the heart) = 19 Answer is 52/19 = 2.73682 probability. ==== >Answer is 52/19 = 2.73682 probability. Can a probability be greater than one? Doug ==== In sci.math, jelly : > Could you please check the following for me: What is the probability that a card selected from a deck is an ace or > heart? The way I see it is 52/4 = 16 (hearts) plus 3 (the aces minus the > heart) = 19 Answer is 52/19 = 2.73682 probability. You're walking backwards with a defective calculator, but you're not off the path -- yet. :-) The standard way might be along the lines of: [1] Count the number of hearts. That's 13. [2] Count the number of aces. That's 4. [3] Count the number of aces of hearts. That's 1. [4] Count the number of cards: 52. [5] Compute the desired number of events out of all events: ace or heart is obviously 13 + 4 - 1 (we don't want to doublecount ace of hearts), out of 52, or 16/52 = 0.3077 = 30.77%. (Note that the events must be of equal probability. Since we're drawing a card from an unmarked deck this isn't a problem here, but some formulations of problems may not be so lucky.) Your next assignment: compute the probability of drawing a face card (JQK) or a heart. :-) Your answer should be 42.3%, with logic identical to the above except for changes to some of the cards to protect the innocent Queens. :-) -- #191, ewill3@earthlink.net -- which is why the Jacks always look excited :-) It's still legal to go .sigless. ==== > Could you please check the following for me: What is the probability that a card selected from a deck is an ace or > heart? The way I see it is 52/4 = 16 (hearts) plus 3 (the aces minus the > heart) = 19 Answer is 52/19 = 2.73682 probability. > First try to read your textbook where you will find that probability of an event by definition takes values in interval from 0 to 1. selecting heart is 13/52 and probability of ace heart is 1/52. What will be probability of event ace or heart ? Goran ==== Could you please check the following for me: What is the probability that a card selected from a deck is an ace or > heart? The way I see it is 52/4 = 16 (hearts) Check your division here. plus 3 (the aces minus the > heart) = 19 Answer is 52/19 = 2.73682 probability. Probabilities must be between 0 and 1. You did something (rather simple) wrong here. > Almost. Just two mistakes. ==== > Could you please check the following for me: What is the probability that a card selected from a deck is an ace or > heart? The way I see it is 52/4 = 16 (hearts) plus 3 (the aces minus the > heart) = 19 Answer is 52/19 = 2.73682 probability. > First of all, a suit has only 13 hearts, not 16. Sorry if I just broke three hearts; I'm a heart breaker :-) Secondly, probability is less than one. I get something like 16/52. -Michael. ==== H I L L 8 9 12 12 = 41 I went to Wash 'n Slosh. I was confused, somebody at the bar was pretty sure it was Thursday, I thought it was Tuesday but it was really Wednesday. And then it was Thursday when I got home. Probababbly there were witches in from Ottawa, I told him that I would tell our President Bush about him, and 182 Shawn 1 7 75 182/183 6708 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 Shawn went to Wash 'n Slosh (lookink for wives for Marcia and me) and met Shawn, the feller has a nubile sister who is poised to be married. We were born in 57 and 75. Genesis 41, Leviticus 14, Judges 9 and John 11 contain the length of 57 verses, the chapter numbers add to 75. Shawn is 220 months and 12 days younger than me while I was born in 57, the 57's are at chapters 41, 104, 220 and 1008, together for 1373 (the 220th prime). 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter 145 Jill 2 4 80 93/273 8445 Jillian 67 Rae 24 Huber 54 Shawn is 958 weeks and 2 days, younger than me, chapter 958 is the 29th chapter of the New Testament. He was born on day 182 while his name adds to 182 (Deuteronomy 29 with 29 verses). Apparently Shawn and his parents were born on days of the month adding to 29 (10+18+1=29). Our waitress here tonite was Jill, her name adds to 145 (5x29). All she has to do is say that I am barred and I will go away. What's wrong with the youth of today? Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 208 Jill 2 4 80 93/273 8445 Jillian 67 Rae 24 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doing this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. Should Shawn's sister marry me, great, but if Marcia marries me and then Shawn's sister marries Marcia and me (Isaiah 4:1), then Shawn is goink to win himself a shiny new Cadillac!!! Good luck and may God bless you!!! My advice to Shawn's sister is to not marry the jerk but to instead consider life with Marcia, (and Jill), and me. Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... ==== H I L L >8 9 12 12 = 41 I went to Wash 'n Slosh. I was confused, somebody at the bar was pretty >sure it was Thursday, I thought it was Tuesday but it was really Wednesday. >And then it was Thursday when I got home. Probababbly there were witches in >from Ottawa, I told him that I would tell our President Bush about him, and << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== S W I C K 19 23 9 3 11 = 65 Vanessa was the 10th of 17 people to provide stats today, she is the 14th cadet, Vanessa dothn't know dad's year of birth and says mom is 42 years old, so I am assuming that mom was born in 61. 65+ Dad 5 9 /117 65+ Mom 11 5 61 131/234 1544 65+ Sis 22 12 83 356/9 9804 192 Vanessa 18 1 87 18/347 10927 Vanessa 81 Marie 46 Swick 65 Mom was born in 61 (18th prime). Sister was born on day 356 (First Chronicles 18) while Vanessa (81) was born on the 18th. Together the sisters were born 18 days closer to the end of their years than to the beginning of their years. In Vanessa's given names, her consonants exceed her vowels by 61 (18th prime). The prime valued in Vanessa's given names add to 61 (18th prime). Now that mom sees this, she will want to put 18 extra blinkin' lights onto her tree while dad will go out and buy 18 beer and celebrate. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 26 61 <-18th-> 27 --- --- 501 278 Mom was born on the 11th day of the month and in 61 (Exodus 11), it's the 18th prime while 18 in turn is the 11th non-prime. Dad was born with 117 days remaining in the year, and Vanessa's prime valued letters add to the 117 verses of Bible Book 22 (11+11). Vanessa's first 11 letters add to twice 61 (Exodus 11). Vanessa's given names add to 127 (the 11th prime in prime position). Her name begins with the 22nd (11+11th) letter of the alphabet, Bible Book 11 contains 22 (11+11) chapters and 11x11 more verses than Bible Book 10. The represented letters in Vanessa's given names add to 101 (Leviticus 11). The prime valued letters in her given names add to 61 (Exodus 11). She has 11 different letters in her full name. Her vowels add to 31 (11th prime). Her last 11 letters add to 112, her last two names add to 111. Her first 11 and last 11 letters average the 117 verses of Bible Book 22 (11+11). Mom took the tree home and saw that it was lopsided, so she put it in the corner and hung most lights and the 11 candy canes on one side of the tree, and when she was done with it, it looked pretty good. Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 <-11th-> 18 <-11th-> 55 <-11th-> 199 --- --- --- --- 160 113 143 518 Swick is a 5 lettered name adding to 65 (a multiple of 5). Dad was born on the 5th day of the 9th (5th non-prime) month, Bible Book 5 contains 959 verses. Mom was born on the 11th (5th prime) day of the 5th month. Dad and the kids were born on days of the month adding to 45 (5 times the 5th non-prime). Vanessa and her parents were born on days of the month adding to the 34 chapters of Bible Book 5. Mom and the kids were born on days of the week adding to 11 (5th prime). The kids were born on days of the year averaging 187, it's the terminating chapter of Bible Book 5 (it is 5x5x5 plus twice the 5th prime in prime position). Vanessa adds to 81, it is 9x9 (the 5th non-prime squared) and is Exodus 31 (the 5th prime in prime position). Her given names add together for 127 (the 31st prime while 31 is the 11th prime while 11 is the 5th prime). Her full name adds to 192 (Joshua 5). Her middle name adds to 46, it's the 32nd non-prime (while 32 is 2 to the 5th). The family was born on days of the month adding to 56 (the first 5 primes plus the first 5 non-primes). The first of the kids was born in 83 (the number of verses in Bible Book 55). The family was born on days and in months adding to 83 (the number of verses in Bible Book 55). The kids were born on days 256 and 18, the latter is 5.05% of the former. Mom and the kids were born on days of the year adding to 505. Dad will again want a 5 foot tall tree this year but mom will beat him to it and go out and buy a substantially taller tree again this year. Primes Non-Primes 2 1 3 4 5 6 7 8 11 <-5th-> 9 -- -- 28 28 Dad was born with 117 days remaining in the year while mom was born with 234 (117+117) days remaining in the year, there are 117 verses in Bible Book 22 Song while the first was born on the 22nd and is now 234 (117+117) months and 11 days old. The kids were born in years adding to 170 (Deuteronomy 17). In Vanessa's given names, her repeated letters add to 51 (3x17) while her unrepeated letters add to 76 (17 plus the 17th prime), corresponding to Exodus 26 (the 17th non-prime). This year mom will insist on a 7 foot tall tree with 17 blinkin' blue lights and 17 blinkin' purple lights, and with 17 blinkin' yellow lights. And then the sisters will giggle with glee as they hang 170 little golden cowbells on the tree that mom lit. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 Swick adds to 65, or the 14th prime plus the 14th non-prime. Mom was born on day 131 (Numbers 14). Mom was born 14.71 weeks closer to the beginning of the year than to the end of the year. The parents were born in months adding to 14, the family was born on days of the month averaging 14. Mom was 22 years old when she first gave birth (14th non-prime), Bible Book 14 contains 822 verses and 36 (14 plus the 14th non-prime) chapters. Vanessa's 192 valued name averages with her 18th day of birth for 105 (1 through 14 and is the first 14 primes minus the first 14 non-primes). Mom first gave birth 225 days after her birthday (Judges 14). I am 1400 weeks and 4 days older than the first. Dad will put some blinkin' lights on the evergreen trees outside in the yard, between the home and the cabin at the lake they will turn 14 trees into decorated idols. And they will go to church and comment on the beauty of the decorated trees there, there will be a tree by the front door of the church, a tree to admire next to the pulpit and a tree in the basement to see after the service and during tea, coffee, cookies and hot-cross buns (bread is made into the symbol of the sun). Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 21 144 521 43 <-14th-> 22 <-14th-> 233 <-14th-> 843 --- --- --- ---- 281 176 609 2204 The kids were together born 243 days further into their years than mom (First Samuel 7). The kids were together born 477 days after mom's birthdays. Mom and the kids were born in years averaging 77. Vanessa has a 7 lettered first name, she has 17 letters in all (7th prime, the primes up to 7 add to 17). Vanessa's 7 unrepeated letters add to 104 (77th non-prime). It means that Vanessa will smoke the Number 7 brand of cigarettes, and will smoke only 7 cigarettes on average when she goes to parties, and will endevor to poke 7 extra holes into her head. And for the Love of God, mom will strive to get the lights on the tree to blink 777 times each hour. 2 4 8 3 9 27 4 16 64 5 25 125 6 36 216 7 49 343 8 64 512 ------------- 35 203 1295 <-together for the 1533 verses of Genesis Mom is 42 years old (29th non-prime), Bible Book 29 is Joel (42). A cent (42) is made out of the 29th element. Mom was born in 61, the kids in years adding to 170, it's a difference of 109 (29th prime), and I am 10927 days older than Vanessa, pretty as 109 is the 29th prime, it's 29 years 11 months and a day. Mom and Vanessa were born on days of the month adding to 29. They were born on days 5, 11, 22 and 18, together these Bible Books contain 2962 verses. Vanessa was born on the 18th, corresponding to Job with 42 chapters (29th non-prime). The kids are separated by 1123 days, corresponding to First Timmy 4, pretty as The Timmy's are Book 54 and 55, together for 109 (the 29th prime). Vanessa's Fibonacci valued letters add to 29. The kids were born on days 22 and 18, these elements have atomic masses adding together for 87.815, pretty as 87 is 29+29+29 while 815 is the first 29 primes minus the first 29 non-primes. I meet Vanessa with 182 days remaining in the year (Deuteronomy 29 with 29 verses). This means that mom bought the tree early this year and kept it up in the living room until March 14th, the damn thing was on display for 109 days (the 29th prime). Then she took 29 colored eggs to the church's Easter dinner (pagan fertility symbols like the Egyptian penises on the roof of your churches). Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime There 29 chapters in Bible Book 13 (6th prime) and 29 verses in chapter 666, pretty as 29 is 6 plus the 6th prime (13) plus the 6th non-prime (10). The kids were born on days of the week adding to 6. Mom and Vanessa were born on days of the week adding to 6 and mom and the sister were born on days of the week adding to 10 (6th non-prime). Mom and Vanessa were born an average of 216 (6x6x6) days closer to the beginning of the year than to the end of the year. Vanessa and I were born on days of the year adding to 66. It all means that they honored traditions in place of Commandments and increased the risk of putting their kids into 6 foot boxes (Hosea 4:6). 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) Swick adds to 65 (5x13). The parents were together born with 351 days remaining in their years (Bible Book 13 chapter 13). The kids were born in months adding to 13. Vanessa's different letters add to 138 and her odd valued letters add to 138. Her unrepresented letters add to 213 and exceed her represented letters by 75, or 13 plus the 13th prime (41) plus the 13th non-prime (21), or simply 13+13p+13np. Her consonants exceed her vowels by 130 (Numbers 13). Her primes and squares add together for 135. Her even valued letters add to 54 (13 plus the 13th prime, the opening length of Book 13). Her odd valued letters exceed her even valued letters by 84 (the first 13 primes minus the first 13 non-primes). This all means that mom and dad will have their friends over and drink 13 glasses of wine, and they will have spent more money on the wine than on presents for their kids. But to show the kids they are still loved, mom will spend 13 days seriously planning the decorations for the tree, and then find out that the damn thing was lopsided when she brings it home, and she'd have to plan the whole damn thing over again (where to hang the blinkin' lights). Dad wanted a 5 foot tree and mom bought a 7 foot tall tree that was lopsided, or some damn thing like that. The interpretation is true, I'm sure. Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations The parents were born on days of the month averaging 8. The family generally has their birthdays on days of the year adding to 753 (Jeremiah 8). They were born on days of the month adding to moon cycle 56 (7x8). The kids were born in years averaging the 85 verses of Bible Book 8 (the first 8 primes plus 8 more). Mom and the kids were together born 85 days closer to the end of their years than to the beginning of their years (the first 8 primes plus 8 more, and is the length of Bible Book 8). It is at Song 8:8 that we read of the little sister who has no pumpkins. The First 8 18's ---------------- 13 <-Genesis 13 20 <-Genesis 20 74 <-Exodus 24 81 <-Exodus 31 142 <-Numbers 25 166 <-Deuteronomy 13 188 <-Joshua 1 204 <-Joshua 17 --- 888 Anyway, the sisters were born on days 22 and 18, together these Bible Books contain 1187 verses. The sisters were born on days of the year averaging 187. One of these sisters might marry Marcia and me, pretty as both of our names add to 187. If Vanessa married Marcia and me, then her married name would add with my name and with Marcia's maiden name for the 618 verses of Bible Book 7. Marcia and Vanessa were born in years adding to the 167 verses of Bible Book 17. Esther. Esther becomes Queen in Book 17 and Q is the 17th letter of the alphabet. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 244 Vanessa 18 1 87 18/347 10927 Vanessa 81 Marie 46 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doing this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. You people spent millions of dollars having me tortured, and then annually you spend billions on your decorated trees, I begged and begged for assistance to flee the country (they tortured me for years at the U of S) and you people are so cheap that you can't even offer to buy me a cookie when I bust my ass to show you evidence that your very name is a gift from God!!! Should Marcia marry me and then one of the sisters marries Marcia and me, then the other sister is poised to win a shiny new Cadillac!!! Good luck and may God bless you!!! 192 Vanessa 18 1 87 18/347 10927 Vanessa 81 Marie 46 Swick 65 219 Rachel 19 9 87 262/103 11171 Rachel 47 Marie 46 Rodomsky 120 And do note that Vanessa Swick is friends with Rachel Rodomsky, they were together and both provided stats for their families today. Vanessa Swick is 244 days older than her friend Rachel Rodomsky, and may yet marry Marcia and me and end up with a name adding to 244. This all means that each family will spend $244 on colored lights for their trees this year. Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... ==== S W I C K >19 23 9 3 11 = 65 Vanessa was the 10th of 17 people to provide stats today, she is the 14th >cadet, Vanessa dothn't know dad's year of birth and says mom is 42 years >old, so I am assuming that mom was born in 61. << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== R O D O M S K Y 18 15 4 15 13 19 11 25 = 120 Rachel was the 9th of 17 people to provide stats today, she is the 13th 120+ Dad 26 6 /188 120+ Mom 21 6 /193 120+ Sis 5 5 /240 120+ Bro 213 Rachel 19 9 87 262/103 11171 Rachel 47 Marie 46 Rodomsky 120 Rachel was born on the 19th, she has 19 letters. Rachel's common name adds to 167 (67 is the 19th prime). Her first pair of letters add to 19. Mom was either born on day 172 or at least generally has her birthday on day 172 (Deuteronomy 19), and mom was born with 193 days remaining in the year. Perhaps mom first gave birth 319 days after her birthday. Bible Books 19 contains 2460 verses, it is 7x19x19 minus the 19th prime. Ezkiel is the 19th of 19 Books in the Old Testament to contain a chapter 19, pretty as it's 1273 verses is 19 times the 19th prime. Dad was born on the 26th. Rachel was born on day 262, corresponding to First Samuel 26. In leap years the first of the kids has her birthday on day 126, and may very well have been born on day 126. The parents and the sisters were born in months adding to 26. Bible Book 26 is Ezekiel with 1273 verses (19 times the 19th prime), and Rachel was born on the 19th. The parents were born on days of the month adding to 47 while Rachel adds to 47 (15th prime). Rodomsky adds to 120 (1 through 15), the first of the Rodomsky (120) kids was born with 240 (120+120) days remaining in the year. The sisters were born on days of the month adding to 24 (15th prime). The females were born on days of the month averaging 15. Rachel's given names add to 47 and 46, these are the 8th and 7th Books of the New Testament (together for 15). Rachel's given names add to 47 and 46, these are the 15th prime and 32nd non-prime, together for 47 (the 15th prime). The parents and the first were born on days of the month adding to the 52 chapters of Bible Book 24 (15th non-prime). Primes Non-Primes Lucas 2 1 1 3 4 3 5 6 4 7 8 7 11 9 11 13 10 18 17 12 29 19 14 47 23 15 76 29 16 123 31 18 199 37 20 322 41 21 521 43 22 843 47 <-15th-> 24 <-15th-> 1364 <-Jeremiah is Book 24 with 1364 verses Dad was born with 188 days remaining in the year (the opening chapter of Bible Book 6). Mom as born with 193 days remaining in the year (Bible Book 6 chapter 6). Both parents were born in the 6th month. Rachel (6 letters) and her parents were born on days of the month adding to 66. Her last two names add together for 166 while her full name adds to 213 (166th non-prime). 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) Mom was born on the 21st, Rachel's first and last names differ in value by 73 (21st prime). Presently her name adds to 213, but she could marry Marcia and me and end up with a name adding to 210. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 210 Rachel 19 9 87 262/103 11171 Rachel 47 Marie 46 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doing this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. Should Rachel marry me, great, but if Marcia marries me and then Rachel marries Marcia and me, then Rachel's siblinks are goink to win themselves shiny new Cadillac's!!! Good luck and may God bless you!!! Rachel was young and had lots of child bearing years ahead of her. Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... ==== R O D O M S K Y >18 15 4 15 13 19 11 25 = 120 Rachel was the 9th of 17 people to provide stats today, she is the 13th << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== Monday October 1st 2001 274/91 16297 In the afternoon I went to La Salsa at 307 21st Street, soon Matthew arrived and sat with me. He is the 3rd of 3 kids, his sister is a nubile sweety. 78+ Dad 23 7 50 204/161 +2401 78+ Mom 6 9 51 249/116 +1991 78+ Sis 2 10 73 275/90 6071 78+ Bro 208 Matthew 26 4 78 116/249 7738 Matthew 90 David 40 Siebert 78 Mom was born on day 249 and with 116 (4x29) days remaining in the year while Matthew was born on the inverse, on day 116 (4x29) and with 249 days remaining in the year. The parents were born on days of the month adding to 29. Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 18 1 19 19 = 59 Dad was born on day 204 (Joshua 17), the parents were born on days of the year adding to 453 (Job 17). The parents were together born with 277 days remaining in the year (59th prime or the 17th prime in prime position). The parents were together born 176 days closer to the end of their years than to the beginning of their years, it is 117 plus the 17th prime and is the length of Psalm 7x17. Matthew was born on the 26th (17th non-prime), his day, month and year of birth adds to the 108 verses of Bible Book 59 (the Genesis 40 and Exodus 28, together for 108 (the first 7 primes plus the days older than me. The females were born in years adding to 124 (Numbers 7), the males in years adding to 128 (2 to the 7th). The males were born on days of the month adding to 49 (7x7). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Matthew was born on the 26th (13+13th), his day, month and year of birth names add to 130 (Numbers 13). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 --- 108 Matthew is Bible Book 40, the name adds to 90 (Exodus 40), while his middle name adds to 40. He was born in the 4th month and his middle name adds to 44.444...% of his first name. Dad was born in '50, Matthew's given names differ in value by 50. This 50 is a combination of the first 7 non-primes, Bible Book 50 contains 104 verses (77th non-prime), while Matthew's full name adds to 104+104 (twice 7 lettered names have an average value of 84 (7 times the 7th non-prime). The parents were born in years adding to 101, the kids in years adding to 151 (a difference of 50). Matthew was born on the 26th, the 1273 verses of Bible Book 26 is 19 times the 19th prime (67). He has 19 letters, his last name begins with the (19+19). The parents were born in years adding to 101 (26th prime). Matthew was born on the 26th, his name adds to 208 (8x26). Siebert adds to 78 (26+26+26). In Matthew's given names, his odd valued letters add to 52 (26+26) and even valued letters add to 78 (26+26+26). The squares in his full name add to 26. Mom was born with 116 days remaining in the year while Matthew was born on day 116 (Leviticus 26). Matthew Siebert (78) was born in '78 (57th non-prime), prettier as Matthew contains 357+357+357 verses. Daniel is Bible Book 27 with 357 verses while Matthew with 357+357+357 verses is the first of 27 Books of the New Testament. 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter Matthew Siebert (78) was born on day 116, it is 31.78% of 365 days. Matthew's names add to the 66th, 28th and 57th non-primes, together for 151, pretty as he and his sister were born in years adding to 151, prettier as mom was born in '51. The females were born on days and in months and years adding to 151. The parents were born on the 6th and 23rd, together these Bible Books contain 1950 verses, pretty as dad was born in 1950. The kids were born in '73 and '78, these are the 21st prime and the 57th non-prime, together for 78, prettier as Siebert adds to 78. The even valued letters in Matthew's given names add to 78. The parents are separated by 410 days while the males were together born with 410 days remaining in the year. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doing this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. You people spent millions of dollars having me tortured, and then annually you spend billions on your decorated trees, I begged and begged for assistance to flee the country (they tortured me for years at the U of S) and you people are so cheap that you can't even offer to buy me a cookie when I bust my ass to show you evidence that your very name is a gift from God!!! Should Matthew's sister marry me, great, but if Marcia marries me and then Matthew's sister marries Marcia and me, then Matthew is goink to win himself a shiny new Cadillac!!! Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... ==== >Monday October 1st 2001 274/91 16297 In the afternoon I went to La Salsa at 307 21st Street, soon Matthew >arrived and sat with me. He is the 3rd of 3 kids, his sister is a nubile >sweety. << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== There are many ways to develop Lebesgue integral, and one way is as follows: f is lebesgue integrable if there exists a sequence {f_n} of step function such that (a) sum int |f_n| < infty (b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty Then the integral of f is defined as int f = sum int f_n My question is: does anyone know any book that introduces Lebesgue integral in this way? TC ==== There are many ways to develop Lebesgue integral, and one way is as > follows: f is lebesgue integrable if there exists a sequence {f_n} of step function > such that (a) sum int |f_n| < infty > (b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty Then the integral of f is defined as int f = sum int f_n My question is: does anyone know any book that introduces Lebesgue > integral in this way? No it doesn't look right. The function: f(x) = 1 if 0> follows: >f is lebesgue integrable if there exists a sequence {f_n} of step function >> such that >(a) sum int |f_n| < infty >> (b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty >Then the integral of f is defined as int f = sum int f_n >My question is: does anyone know any book that introduces Lebesgue >> integral in this way? No it doesn't look right. The function: f(x) = 1 if 0f(x) = 0 otherwise is lebesgue integrable. And this function satisfies the condition above: Let the rationals in (0,1) be r_1, r_2, ... . Let f_n(x) = 0 except for f_n(r_j) = 1, 1 <= j <= n. Then f_n is a step function, the sum of the integrals of |f_n| is finite, and f(x) = sum f_n(x) for every x such that sum |f_n(x)| is finite. >I think you would have to approximate that using >'simple functions' (rather trivial since it one already) not 'step >functions'. It is possible that 'step function' might be defined >differently. Secondly the word sum seems to be occuring where there would >normally be sup. Thirdly for a non-negative measurable function f the >integral is usually defined as sup int phi dx, where sup is taken over all >measurable simple functions phi, such that phi <= f. The sup is in general >going to be over an uncountable set so any use of sequences of step >functions looks wrong. Comments on how this approach differ from the approach you know (ok, from the more usual approach) don't show that this approach is wrong. The last comment about how that sup is over an uncountable set so any use of sequences seems wrong is way off: In spite of the fact that the sup is over an uncountable family of simple functions we certainly know that we can get a sequence of simple functions converging to f, with the usual approach... I'm not _certain_ offhand that the conditions above are true precisely when f is Lebesgue integrable, but I'm certainly not certain that this is not so - I know that there _are_ approaches very much like this that do work. ************************ David C. Ullrich ==== I was curious as to why an ordered pair (x,y) would be defined as {{x}{x,y}}. I understand that a set has no order but therefore, wouldn't that be the same as {{x,y} {x}}. If anyone has any good links to this just for some background information, that would be realy helpfuly. ==== >I was curious as to why an ordered pair (x,y) would be defined as >{{x}{x,y}}. I understand that a set has no order but therefore, wouldn't >that be the same as {{x,y} {x}}. Yes it is. So what? (It's not the same as (y,x), which is {{y}, {x,y}}.) >If anyone has any good links to this just for some background information, >that would be realy helpfuly. ************************ David C. Ullrich ==== I was curious as to why an ordered pair (x,y) would be defined as > {{x}{x,y}}. I understand that a set has no order but therefore, wouldn't > that be the same as {{x,y} {x}}. Yes, it is the same. It's also the same as {{y,x},{x}} and {{x},{y,x}}. These are *all* the same set. And by the way, in the special case that y=x you can even write this ordered pair as {{x}}. See why? The question is not whether this set has a unique representation with curly braces; it's whether the set unambiguously defines the ordered pair. That is, if two different ordered pairs turned out to be the same set by definition, then the definition would be no good. Can you find any such case? For example, write out as many different expressions for (y,x) as you please, and see if any of them matches an expression for (x,y). None will, except of course in the special case that x=y. The ordering of the pair is preserved not in the ordering of the set (there's none!) but rather in the fact that the first element of the pair appears as the one and only singleton in the set. ==== >I was curious as to why an ordered pair (x,y) would be defined as >{{x}{x,y}}. I understand that a set has no order but therefore, wouldn't >that be the same as {{x,y} {x}}. Yes, it would. But the point is that {{x,y},{x}} = {{z,w},{z}} if and only if x=z and y=w. So that (x,y) = (z,w) if and only if x=z and y=w, so that (x,y)=(y,x) if and only if x=y. So it's not that the set is now ordered, is that the set {{x}, {x,y}} can be used to say which of x and y goes first and which goes second. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9HFuwt08500; ==== 3 men went to a motel. There's only 1 room left and it costs 30 bucks. So, each man forked out 10 bucks. Later, the owner discovered he over-charged the 3 men. It should be 25 bucks instead. So, he sent his runner to give 5 bucks back to the 3 men. The 3 men think the owner is very honest, and thus each took 1 buck back, leaving 2 bucks to the runner as tips. Now, since they took 1 buck back, each of them paid 9 bucks. 2 bucks to the runner. So: (9*3)+2 = 29. Where's the 1 dollar? You are add/sub. unrelated values that should not be add/sub. Two of the 3 persons received $1.66 back and one took $ 1.67 which = a total of $5.00 returned. Then two of the persons pooled .66 cents each and the one who received $1.67 pooled .67 cents giving the runner a $ 2.00 tip. Then you subtract $ 30.00 ( original cost) - $ 2.00 (tip) = $ 28.00 ( balance after tip) - $ 3.00 ( $ 1.00 each of the three men had extra after tip) = $ 25.00 which is the final total the three men had to pay for the room excluding tip. Then again, forget all of the above and say the missing $ 1.00 was left as a tip for the maid. ;-) Portly X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9HFurN08456; ==== >>t(0)=x; t(n)=1-n*t(n-1); number in parenthesis is subscript >>define R=t(20) >>show R(1-e^(-1))<1/21 >a hint: prove by induction that, for n even, t(n) = n! * (sum (k=0,...,n) ((-1)^k / k!) - 1/e ). sorry, I forgot the x-argument: t(n)(1-1/e) = n! * (sum (k=0,...,n) ((-1)^k / k!) - 1/e ). Best wishes >Torsten. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9HIqq420222; ==== he must have meant latex to use latex if u r a beginner and have access to pc...i recommend TeXnic center aravind I am new to here. I want to know if there are any softwares to write >equations. >One of my professor said something like letac or something like >that. I didnt ask for the spelling. I want to know if a software like >that exists. thnks Suresh ==== I recall what integration by parts means in a single variable, but I am puzzled by a reference to it in rewriting a certain integral over dx,dt which occurs in a derivation of the Euler-Lagrange equation. Since there are no antiderivatives in more than one variable (so far as I can see?), the mnenomic about recognizing half the derivative of the product of two functions doesn't seem applicable. Is there a common meaning of integration by parts in several variables, is the author implying the technique is used in one of the single variable integrations composing the double integral, or is my brain fried? Now that I think about it, it seems to me I've seen casual references to integration by parts in other integrals over several variables -- possibly occuring in EM -- but I never really tried to understand what was going on. ==== > I recall what integration by parts means in a single variable, but I > am puzzled by a reference to it in rewriting a certain integral over > dx,dt which occurs in a derivation of the Euler-Lagrange equation. > Since there are no antiderivatives in more than one variable (so far > as I can see?), the mnenomic about recognizing half the derivative of > the product of two functions doesn't seem applicable. Is there a common meaning of integration by parts in several > variables, is the author implying the technique is used in one of the > single variable integrations composing the double integral, or is my > brain fried? Now that I think about it, it seems to me I've seen > casual references to integration by parts in other integrals over > several variables -- possibly occuring in EM -- but I never really > tried to understand what was going on. The idea in 2D is to write (del scalar).(vector) as del.(scalar * vector) - scalar * (del.vector) The del.(scalar * vector) can then be converted into an integral of scalar * vector . (outward normal) taken around the boundary in the right-hand screw rule sense. [More relevant to your specific problem, consider the abuse of notation dF/d(grad y) == (dF/d(dy/dx), dF/d(dy/dt)) for scalars F(x,t,y, grad y) and y(x,t), all derivatives partial.] -- P.A.C. Smith 'If the Apocalypse comes, beep me.' <*> http://www.srcf.ucam.org/~pas51 boundary=----=_NextPart_000_0032_01C394ED.D1966400 ==== --------------------------------------------------------------------- You have two rich aunts, Ruby is going to give you a $1000 a year until you are 17, Amber is going to start by giving you $100, and each subsequent year she will give you twice as much as the year before. If you assume you get exactly 17 payments, how much will you have in total from each aunt? Write an expression that shows how much money you have after x years for each scheme and determine when they are equal. Which is a better deal? How would I write the expression for this? Mark ==== --------------------------------------------------------------------- You have two rich aunts, Ruby is going to give you a $1000 a year until you are 17, Amber is going to start by giving you $100, and each subsequent year she will give you twice as much as the year before. If you assume you get exactly 17 payments, how much will you have in total from each aunt? Write an expression that shows how much money you have after x years for each scheme and determine when they are equal. Which is a better deal? How would I write the expression for this? Mark Didn't your teacher teach you about writing expressions before giving you this homework problem? ==== > You have two rich aunts, Ruby is going to give you a $1000 a year until > you are 17, Amber is going to start by giving you $100, and each > subsequent year she will give you twice as much as the year before. If > you assume you get exactly 17 payments, how much will you have in total > from each aunt? Write an expression that shows how much money you have > after x years for each scheme and determine when they are equal. Which > is a better deal? > > How would I write the expression for this? You want to know how you would write an expression for this? I can any of us tell? Besides, the subject of your post is misleading. The problem is not interesting at all. Jose Carlos Santos ==== error or 0 ? ==== > error or 0 ? It depends on what you need for your application. As several other posts have pointed out, there is nothing that *always* works, no matter how 0/0 arises. However, sometimes, it's clear from context what 0/0 should be *in this one particular case*. Levinson and Redheffer (_Complex Variables_) stated this as all removable singularities are assumed to be removed, meaning that fine print details are omitted. It is my understanding that some computer languages define it to be undefined (or NAN) and some to be 0. I believe (but if it's important, you'd better check it yourself) that the relevant IEEE standard says it should be undefined, but I may be misremembering. Just do whatever works. But document it. Jon Miller ==== > error or 0 ? It might be defined in the limiting sense. E.g. f(x) = sin(x)/x. At x=0, this is 0/0. If we define f(0) to be the limit of f(x) as x -> 0, it has the value 1, since sin(x) -> x as x -> 0. Gib ==== error or 0 ? It might be defined in the limiting sense. E.g. f(x) = sin(x)/x. At >x=0, this is 0/0. If we define f(0) to be the limit of f(x) as x -> 0, >it has the value 1, And if you consider the function g(x) = sin(2x)/x then exactly the same argument would give 0/0 = 2. Which is why 0/0 is left undefined. >since sin(x) -> x as x -> 0. Blech. That's meaningless. You meant sin(x) ~ x as x -> 0. >Gib ************************ David C. Ullrich ==== > error or 0 ? It might be defined in the limiting sense. E.g. f(x) = sin(x)/x. At >x=0, this is 0/0. If we define f(0) to be the limit of f(x) as x -> 0, >it has the value 1, And if you consider the function g(x) = sin(2x)/x then exactly > the same argument would give 0/0 = 2. Which is why 0/0 > is left undefined. since sin(x) -> x as x -> 0. Blech. That's meaningless. You meant sin(x) ~ x as x -> 0. Gib ************************ David C. Ullrich 0/0 is an emoticon for eyeballs with nerd glasses. Bob Pease ==== > >> error or 0 ? > It might be defined in the limiting sense. E.g. f(x) = sin(x)/x. At > x=0, this is 0/0. If we define f(0) to be the limit of f(x) as x -> 0, > it has the value 1, since sin(x) -> x as x -> 0. If 0/0 = 1 then 0/0 = 1*n for arbitrary n. Then 1 = 0/0 = n hence 1 = n. That is the kind of grief one gets if 0/0 is defined. Therefore it is not defined. Bob Kolker ==== error or 0 ? It might be defined in the limiting sense. E.g. f(x) = sin(x)/x. At >x=0, this is 0/0. If we define f(0) to be the limit of f(x) as x -> 0, >it has the value 1, since sin(x) -> x as x -> 0. This doesn't say anything about the 0/0, though. It's not possible to define 0/0 as anything real because doing so will lead immediately to contradiction. Assume 0/0 is defined as Z, where Z is in R. Now, Z + Z = (0 + 0)/0 = Z <=> Z = 0, but lim(n->0, n/n) = 1 = Z, so 0 = 1. ==== > error or 0 ? lim_(x,y)->(0,y) (x/y) = 0 lim_(x,y)->(x,0) (x/y) = inf => lim_(x,y)->(0,0) (x/y) = not defined. ==== > error or 0 ? Neither. It is undefined. Bob Kolker ==== The world *is* a finite set of resources... > Since this is sci.math, perhaps you can define the terms of the above sentence and the assumptions about the physical nature of the world that justifies this statement. After all, even the lowly hydrogen atom has infinitely many energy levels according to some theories. ==== >Since this is sci.math, perhaps you can define the terms of the above >sentence and the assumptions about the physical nature of the world that >justifies this statement. After all, even the lowly hydrogen atom has >infinitely many energy levels according to some theories. But it's still only one hydrogen atom. The *usefulness* of that hydrogen atom may be increased as we learn how to use its different energy levels for various purposes - and, as I noted, for a given level of technology, these resources have one value, and for a higher level, they have a greater value. However, statements about what technology may be able to do in the future are forward-looking, as they say in the prospecti, and so, just as you can't tell me that a cure for AIDS is going to be available by month X, one could make a case for making future projections for feeding and housing humanity *as if* technological progress (despite the absence of any big disaster) comes immediately to a screeching halt. Myself, I think _that's_ going a bit far. John Savard http://home.ecn.ab.ca/~jsavard/index.html ==== I recall what integration by parts means in a single variable, but I >am puzzled by a reference to it in rewriting a certain integral over >dx,dt which occurs in a derivation of the Euler-Lagrange equation. >Since there are no antiderivatives in more than one variable (so far >as I can see?), the mnenomic about recognizing half the derivative of >the product of two functions doesn't seem applicable. Is there a common meaning of integration by parts in several >variables, is the author implying the technique is used in one of the >single variable integrations composing the double integral, or is my >brain fried? Now that I think about it, it seems to me I've seen >casual references to integration by parts in other integrals over >several variables -- possibly occuring in EM -- but I never really >tried to understand what was going on. > Stuff that's analogous to int. by parts comes up when you're dealing with several variable. E.g. if you are integrating a divergence over a volume you can change that into a surface term (i.e. a boundry term) just like how you get a boundry term when integrating by parts in 1-d (where the boundry is just 2 points): That is, in 1-d: Integral[u dv/dx, {x,a,b}] = Integral[d(u v)/dx - v du/dx, {x,a,b}] = Integral[d(u v)/dx,{x,a,b}] - Integral[v du/dx, {x,a,b}] = u(b)v(b) - u(a)v(a) - Integral[v du/dx, {x,a,b}] = (the boundry term) - Integral[v du/dx, (x,a,b}] Whereas, in 3-d (let v be a vector, f a scalar, del is the vector (d/dx,d/dy,d/dz), and . means dot product): Integral[f del.v, {volume}] = Integral[del.(f v) - v.del(f), {volume}] = Integral[del.(f v),{volume}] - Integral[v.del(f),{volume}] = Integral[(f v).{surface}] - Integral[v.del(f),{volume}] = (the boundry term) - Integral[v,del(f),{volume}] hmm... this notation is looking pretty nasty and esoteric... but you get the point. In 1-d integration by parts you get a surface term but the surface is just two points (the endpoints of integration). In 3-d you get a surface term where you evaluate your integrand over the surface of the volume of integration. hope that helps, adam ==== EG>I recall what integration by parts means in a single variable, but I >am puzzled by a reference to it in rewriting a certain integral over >dx,dt which occurs in a derivation of the Euler-Lagrange equation. >Since there are no antiderivatives in more than one variable (so far >as I can see?), the mnenomic about recognizing half the derivative of >the product of two functions doesn't seem applicable. ... > Stuff that's analogous to int. by parts comes up when you're > dealing with several variable. E.g. if you are integrating a > divergence over a volume you can change that into a surface term (i.e. > a boundry term) just like how you get a boundry term when integrating > by parts in 1-d (where the boundry is just 2 points): > That is, in 1-d: Integral[u dv/dx, {x,a,b}] = Integral[d(u v)/dx - v du/dx, {x,a,b}] = Integral[d(u v)/dx,{x,a,b}] - Integral[v du/dx, {x,a,b}] = u(b)v(b) - u(a)v(a) - Integral[v du/dx, {x,a,b}] = (the boundry term) - Integral[v du/dx, (x,a,b}] Whereas, in 3-d (let v be a vector, f a scalar, del is > the vector (d/dx,d/dy,d/dz), and . means dot product): Integral[f del.v, {volume}] = Integral[del.(f v) - v.del(f), {volume}] = Integral[del.(f v),{volume}] - Integral[v.del(f),{volume}] = Integral[(f v).{surface}] - Integral[v.del(f),{volume}] = (the boundry term) - Integral[v,del(f),{volume}] > hmm... this notation is looking pretty nasty and esoteric... > but you get the point. In 1-d integration by parts you get > a surface term but the surface is just two points (the > endpoints of integration). In 3-d you get a surface term where > you evaluate your integrand over the surface of the volume of > integration. Bingo! Got it! In several dimensions, analogues of the fundamental theorem of calculus are of the form: Int[DifOp(f),{inside}] = Int[f,{boundary}], Where DifOp is differential operator. Such things had occured to me before. Assuming there is also an analogue to the product rule in the form DifOp(fg) = fDifOp(g) + gDifOp(f), you are all set to integrate by parts. made my day! Now the rest of the day may be spent in sloth. ;-) BTW: I've adopted your notation. Is that Mathematica or some such, or something you invented on the fly for this post? ==== >>Suppose that: >>Y(x)=(k_1 - k_2/x)^.5 >>Where the k's are positive constants. >>Is it proper to claim that Y is proportional to x^-.5? In the >>presentation, the symbol alpha was used in place of the words is >>proportional to and I was wondering if that was a rigorous use of >>that symbol. You've already been told the answer is no. However, coming from a >physics background I'll add that under some circumstances, one might >say something like Y varies as x^0.5 for small x, if that was the >regime of interest. The physicist would write something like > Y ~ x^-0.5 for small x > Coming from a physics background myself, I must disagree. A good physicist would not write something like y ~ x^-.05 _for small x_ because it doesn't matter if x itself is small... In fact, since x probably is dimensionfull, smallness is relative. x is small relative to what scale? Also, the constants probably are dimensionful... at least one of them is if x is. So what _really_ matters (as I said in my first post) is if k_1 << k_2/x since k_1 must have the same dimensions as k_2/x it is proper to compare these quantities... and if k_2/x is much bigger than k_1, THEN y is approx. proportional to x^-.5 whew... sorry about ranting. cheers, adam ==== >Suppose that: Y(x)=(k_1 - k_2/x)^.5 Where the k's are positive constants. Is it proper to claim that Y is proportional to x^-.5? In the >presentation, the symbol alpha was used in place of the words is >proportional to and I was wondering if that was a rigorous use of >that symbol. > >>[...] >since k_1 must have the same dimensions as k_2/x it is proper >to compare these quantities... and if k_2/x is much bigger than >k_1, THEN y is approx. proportional to x^-.5 > If k1 << k2/x, then Y in this case is complex and multiply valued. Somehow, I suspect that this was not the intent in the first place. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== The Lagrangian density ... so much Lagrangian L per unit volume ... >suggests we think of L = T - V as a kind of stuff, possibly with the >same justification as thinking of energy and momentum -- as much or as >little. Just what kind of stuff would it be? Is there any natural >interpretation of what Lagrangian is, or is it to simply be regarded >as a function which encodes the dynamics of the system in its >dependence on the variables, and magically recovers the ordinary >equations of motion via a rule set? > Nah, I don't think of L as a kind of stuff... I'd say that I think of H = T + V as a kind of stuff (energy). But when I think of 'L' I just think of that which, when integrated over time, is minimized to find the classical path...or that which gives the correct equation of motion when plugged into the Euler-Lagrange eqns. So, in that case, the lagrangian density is that which, when integrated over the _four-volume_, is minimized to find the classical path. I guess there is something nice about the lagragian density, because to work with it, we do not seperate out time as something special to integrate over. The lagrangian density treats time and space equally, which is good. It helps a lot when you are trying to unify special relativity and quantum mechanics into QFT. adam ==== Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with compact support, i.e. it is 0 off a compact set. Is it true that the integral of div f (the divergence of f) over O is 0? I know that if the boundary of O is nice, this follows from the Stoke's theorem. My concern is what if the boundary of O is bad. Will the integral be still 0? I believe that one can find an open subset Q of O containing supp f, that has a nice boundary so that Stoke theorem applies. But where can I find such a proof? ==== > Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with > compact support, i.e. it is 0 off a compact set. Is it true that the > integral of div f (the divergence of f) over O is 0? > I know that if the boundary of O is nice, this follows from the Stoke's > theorem. > My concern is what if the boundary of O is bad. Will the integral be still > 0? > I believe that one can find an open subset Q of O containing supp f, that > has a nice boundary so that Stoke theorem applies. But where can I find such > a proof? Urysohn's Lemma should give you the set you're looking for, but there's really no need for it. The integral of div f over O is the same as the integral of div f over any set containing supp f. So you can just ignore O and replace it by something nicer. ==== f can be seen has f:Q->R^n with Q containing O and Q with smooth boundary :D (for example Q=a ball) -GS- > Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with > compact support, i.e. it is 0 off a compact set. Is it true that the > integral of div f (the divergence of f) over O is 0? > I know that if the boundary of O is nice, this follows from the Stoke's > theorem. > My concern is what if the boundary of O is bad. Will the integral be still > 0? > I believe that one can find an open subset Q of O containing supp f, that > has a nice boundary so that Stoke theorem applies. But where can I find such > a proof? ==== I don't see why Q can be a ball. > f can be seen has f:Q->R^n with Q containing O and Q with smooth boundary :D > (for example Q=a ball) -GS- Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with >compact support, i.e. it is 0 off a compact set. Is it true that the >integral of div f (the divergence of f) over O is 0? >I know that if the boundary of O is nice, this follows from the Stoke's >theorem. >My concern is what if the boundary of O is bad. Will the integral be still >0? >I believe that one can find an open subset Q of O containing supp f, that >has a nice boundary so that Stoke theorem applies. But where can I find > such >a proof? ==== >I don't see why Q can be a ball. Any compact set, for example the support of your function, is contained in some ball. This follows from the Heine-Borel theorem... >> f can be seen has f:Q->R^n with Q containing O and Q with smooth boundary >:D >> (for example Q=a ball) >-GS- >>Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with >>compact support, i.e. it is 0 off a compact set. Is it true that the >>integral of div f (the divergence of f) over O is 0? >>I know that if the boundary of O is nice, this follows from the Stoke's >>theorem. >>My concern is what if the boundary of O is bad. Will the integral be >still >>0? >>I believe that one can find an open subset Q of O containing supp f, >that >>has a nice boundary so that Stoke theorem applies. But where can I find >> such >>a proof? > ************************ David C. Ullrich ==== O is bounded. There exists r>0 such that B(0,r)=:Q contains O. f can be seen as a C^1 function Q->R^n, with f(x)=0 if x is in QO. And then you can apply Stoke's. GS > I don't see why Q can be a ball. f can be seen has f:Q->R^n with Q containing O and Q with smooth boundary > :D >(for example Q=a ball) -GS- Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with >> compact support, i.e. it is 0 off a compact set. Is it true that the >> integral of div f (the divergence of f) over O is 0? >> I know that if the boundary of O is nice, this follows from the Stoke's >> theorem. >> My concern is what if the boundary of O is bad. Will the integral be > still >> 0? >> I believe that one can find an open subset Q of O containing supp f, > that >> has a nice boundary so that Stoke theorem applies. But where can I find >such >> a proof? > > ==== >>Let A be the real matrix >> / 0 a b c >> A= / a 0 d e >> b d 0 f / >> c e f 0 / >>where a^2+b^2+c^2+d^2+e^2+f^2 = 6 . >>Consider that r_1 =< r_2 =< r_3 =< r_4 are eigenvalues of A. >>It is known that r_4= 10. >>It's true that all eigenvalues are integer numbers ? >>If A is non-singular, which is the inverse matrix A^{-1} ? >>===== It can't happen that r_4 = 10. The characteristic polynomial of >A is t^4 - (a^2+b^2+...+f^2) t^2 + c1 t + c0 for some c1 and c0. >So you'd need r1 + r2 + r3 + r4 = 0 and >-6 = r1 r2 + r1 r3 + r1 r4 + r2 r3 + r2 r4 + r3 r4 > = 1/2 ((r1 + r2 + r3 + r4)^2 - r1^2 - r2^2 - r3^2 - r4^2) >i.e. r1^2 + r2^2 + r3^2 + r4^2 = 12. In particular >r4 <= 2 sqrt(3). Suppose we remove the requirement r_4=10. The only ways to >get 12 as a sum of four squares are 1^2 + 1^2 + 1^2 + 3^2 and >0^2 + 2^2 + 2^2 + 2^2, so the only integer solutions to >r1+r2+r3+r4=0 and r1^2+r2^2+r3^2+r4^2=12 with r1<=r2<=r3<=r4 >are [-1,-1,-1,3] and [-3,1,1,1]. It is possible to get these >eigenvalues, e.g. with a=b=c=d=e=f=1 or a=b=c=d=e=f=-1 respectively. Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 > Robert Israel , > According to your important remarks, condition > instead of r_4=10 we must have ,e.g. x_4=3 . ================= Other remarks. Let S(A)=a^2+b^2+c^2+e^2+f^2 . I have observed the following : 1) Suppose S(A)=10*M^2 and min_{i=1,2,3}|r_i- r_{i+1}| >= 2M . Then [r_1,r_2,r_3,r_4]=[-3M,-M,M,3M] 2) Let k=2 or k=3 . If r_k >= sqrt((4-k)*S(A)/(2k)) then eigenvalues are : [-T,-T,T,T] , T:= sqrt(S(A)/2) for k=3 , [-V,U,U,U ] , U:=sqrt(S(A)/6) , V=sqrt(3*S(A)/2) when k=2 . 3) If r_4 >= sqrt(2*S(A)/2) , then we have the eigenvalues [-W,-W,-W,Z] , W=sqrt(S(A)/6) , Z=sqrt(3*S(A)/2) . 4) Let us suppose r_1 >= - sqrt(S(A)/6) . The eigenvalues are as in case 3) . 5) If r_4-r_1 >= 2*sqrt(S(A)) we obtain eigenvalues [-sqrt(S(a)),0,0, sqrt(S(A)) ]. 6) When r_4-r_1 =< sqrt(2*S(A)) , then spectrum of A is [ -Q,-Q,Q,Q ] , Q= sqrt(S(A)/2) . ======== ==== C:|z|=3 f(z) = (z^2)*(z+i) / {(z-2i)^2}*(z-3i) when we use residue theorem, can z=3i ignore ?? only z=2i.......apply?? how do you solved it? ==== thank you... ==== > C:|z|=3 f(z) = (z^2)*(z+i) / {(z-2i)^2}*(z-3i) when we use residue theorem, can z=3i ignore ?? only z=2i.......apply?? how do you solved it? Your function f is written in a strange way. If the expression for f is interpreted literally, then f is a rational function whose numerator is z^2.(z + i).(z - 3i) and the denominator is (z - 2i)^2. Then the integral is easily calculated by the residue theorem. However, if f is a rational function whose numerator is z^2.(z + i) and the denominator is (z - 2i)^2.(z - 3i), then your problem simply makes no sense, since your function f is not defined at one of the points of the circle {z : |z| = 3}. Jose Carlos Santos ==== When using Cauchy's Residue Theorem, we need to choose a simple closed contour which does not pass through any singularities. Now, f has singularities at 2i and 3i. If you choose C: |z|=1, then the contour integral is 0. If you choose C: |z|=4, then the contour integral is 2£ki( Res(f,2i)+Res(f,3i) ). Michael Leung hot-girl .b9.a6.b9g©.97.a6l´.97ásÈD :bmqc97$q9q$1@news.hananet.net... > C:|z|=3 f(z) = (z^2)*(z+i) / {(z-2i)^2}*(z-3i) when we use residue theorem, can z=3i ignore ?? only z=2i.......apply?? how do you solved it? ==== Show that for each sequence of random variables {X_n : n = 1,2,...} there exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n converges to zero with probability one. Note : There is no hypothesis whatsoever on the random variables. How can I prove this? This seems like a strange assertion to me that lacks intuition. Steve ==== > Show that for each sequence of random variables {X_n : n = 1,2,...} there > exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n > converges to zero with probability one. Note : There is no hypothesis > whatsoever on the random variables. How can I prove this? This seems like a strange assertion to me that lacks > intuition. First, I don't see why (as another poster mentioned) we would need the X_n's defined on the same probability space. The resulting random sequence lives on the infinite product of the probability spaces of the X_n's, which we can construct without needing the individual spaces to be the same. Now, for each n, choose a_n so large that P(|X_n| > a_n/n) < 2^{-n}. By the Borel-Cantelli Lemma, the probability that |X_n| > a_n/n for infinitely many n is zero. Thus, with probability one, |X_n/a_n| <= 1/n for all n sufficiently large, i.e. X_n/a_n --> 0. Steve ==== > Show that for each sequence of random variables {X_n : n = 1,2,...} there > exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n > converges to zero with probability one. Note : There is no hypothesis > whatsoever on the random variables. How can I prove this? This seems like a strange assertion to me that lacks > intuition. The intuition I can just possibly supply, if not the proof. Hmm... well, on second thought the intuition on second thought is a _little_ tricky! But essentially you are claiming that the random variables in your sequence are well-behaved random variables (a redundant claim), and so you can pick a sequence of real numbers which dominate them in a well-defined sense. For example it's clear (I hope) that we can pick a sequence of constants such that X_n/a_n converges to zero in expectation: i.e., the real series E[X_n/a_n] converges to zero. That's one kind of dominance. A different kind of dominance would be picking the a_n such that, for some other sequence of constants c_n which converge to zero, X_n/a_n < c_n with probability one. That dominance would be stronger than the first kind of dominance, and _that_ assertion in fact is false: no matter how much we try to squeeze the series with the a_n there remains a positive probability that any given element of the sequence of X_n will be sufficiently large to break our claim, even before we add for all n. We have a kind of Goldilocks dominance search: the first kind was too weak (easy to prove), the second kind was too strong, and maybe the last kind is _just right_. ;-) You will have to carefully consider the definition of convergence and maybe reach a proof by contradiction -- what we are saying is that, while the sequences may have excursions on their travel to zero, we will almost always be able to travel far enough out on the sequence to find that the excursions never again rise beyond any given level. Good luck. ==== Show that for each sequence of random variables {X_n : n = 1,2,...} there >exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n >converges to zero with probability one. Note : There is no hypothesis >whatsoever on the random variables. How can I prove this? This seems like a strange assertion to me that lacks >intuition. The intuition I can just possibly supply, if not the proof. Hmm... well, on second thought the intuition on second thought is a > _little_ tricky! But essentially you are claiming that the random > variables in your sequence are well-behaved random variables (a > redundant claim), and so you can pick a sequence of real numbers which > dominate them in a well-defined sense. For example it's clear (I hope) that we can pick a sequence of > constants such that X_n/a_n converges to zero in expectation: i.e., > the real series E[X_n/a_n] converges to zero. That's one kind of > dominance. > I guess one problem I have is that the above claim is not clear (the claim about expectation). Why is this true? > A different kind of dominance would be picking the a_n such that, for > some other sequence of constants c_n which converge to zero, X_n/a_n < > c_n with probability one. That dominance would be stronger than the > first kind of dominance, and _that_ assertion in fact is false: no > matter how much we try to squeeze the series with the a_n there > remains a positive probability that any given element of the sequence > of X_n will be sufficiently large to break our claim, even before we > add for all n. We have a kind of Goldilocks dominance search: the first kind was too > weak (easy to prove), the second kind was too strong, and maybe the > last kind is _just right_. ;-) You will have to carefully consider > the definition of convergence and maybe reach a proof by contradiction > -- what we are saying is that, while the sequences may have excursions > on their travel to zero, we will almost always be able to travel far > enough out on the sequence to find that the excursions never again > rise beyond any given level. Good luck. ==== >>Show that for each sequence of random variables {X_n : n = 1,2,...} >there >>exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n >>converges to zero with probability one. Note : There is no hypothesis >>whatsoever on the random variables. >>How can I prove this? This seems like a strange assertion to me that >lacks >>intuition. >The intuition I can just possibly supply, if not the proof. >Hmm... well, on second thought the intuition on second thought is a >> _little_ tricky! But essentially you are claiming that the random >> variables in your sequence are well-behaved random variables (a >> redundant claim), and so you can pick a sequence of real numbers which >> dominate them in a well-defined sense. >For example it's clear (I hope) that we can pick a sequence of >> constants such that X_n/a_n converges to zero in expectation: i.e., >> the real series E[X_n/a_n] converges to zero. That's one kind of >> dominance. >I guess one problem I have is that the above claim is not clear (the claim >about expectation). Why is this true? It doesn't matter. It's a fact that |X_n| is finite almost surely. It follows that P(|X_n| > a) -> 0 as a -> infinity. Hence you can pick a_n such that P(|X_n| > a_n/n} < ___, and if you fill in the blank properly that shows that X_n -> in probability. >> A different kind of dominance would be picking the a_n such that, for >> some other sequence of constants c_n which converge to zero, X_n/a_n < >> c_n with probability one. That dominance would be stronger than the >> first kind of dominance, and _that_ assertion in fact is false: no >> matter how much we try to squeeze the series with the a_n there >> remains a positive probability that any given element of the sequence >> of X_n will be sufficiently large to break our claim, even before we >> add for all n. >We have a kind of Goldilocks dominance search: the first kind was too >> weak (easy to prove), the second kind was too strong, and maybe the >> last kind is _just right_. ;-) You will have to carefully consider >> the definition of convergence and maybe reach a proof by contradiction >> -- what we are saying is that, while the sequences may have excursions >> on their travel to zero, we will almost always be able to travel far >> enough out on the sequence to find that the excursions never again >> rise beyond any given level. Good luck. > ************************ David C. Ullrich ==== > Show that for each sequence of random variables {X_n : n = 1,2,...} there > exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n > converges to zero with probability one. Note : There is no hypothesis > whatsoever on the random variables. How can I prove this? This seems like a strange assertion to me that lacks > intuition. > Can you choose a_n so that Prob{X_n > a_n/n} goes to 0? That's not enough in itself, but it's a start. ==== > Show that for each sequence of random variables {X_n : n = 1,2,...} there > exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n > converges to zero with probability one. Note : There is no hypothesis > whatsoever on the random variables. How can I prove this? This seems like a strange assertion to me that lacks > intuition. Wait, you get to see X_n first and then pick a_n? Analysis is *definitely* not my field, but how about picking a_n = X_n * n; this way, lim n->oo (X_n/a_n) = lim n->oo (1/n) = 0? If, on the other hand, you're asking for a sequence a_n that works for EVERY X_n, it's clearly not possible; there's always a sequence X_n >= n*a_n. However, you can have probability 1 without having the statement be true in general; i.e. if I am to pick an integer, the probability that I will not pick 0 is 1, yet it is obviously not certain that I will not pick 0. However, this statement still *sounds* untrue; I don't see offhand how to resolve it, but it can't be too hard. Sorry if I sound like a moron; I'm (a) just starting college and (b) going into algebra, not analysis. In fact I don't really like numbers anyway. Kinda weird to be going into math then, I guess, but whatever. ==== >Show that for each sequence of random variables {X_n : n = 1,2,...} there >exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n >converges to zero with probability one. Note : There is no hypothesis >whatsoever on the random variables. How can I prove this? This seems like a strange assertion to me that lacks >intuition. Wait, you get to see X_n first and then pick a_n? Analysis is > *definitely* not my field, but how about picking a_n = X_n * n; but that a_n would not be constant, it would be random just like X_n is > this > way, lim n->oo (X_n/a_n) = lim n->oo (1/n) = 0? If, on the other hand, you're asking for a sequence a_n that works for > EVERY X_n, it's clearly not possible; there's always a sequence X_n >= > n*a_n. However, you can have probability 1 without having the > statement be true in general; i.e. if I am to pick an integer, the > probability that I will not pick 0 is 1, yet it is obviously not > certain that I will not pick 0. However, this statement still > *sounds* untrue; I don't see offhand how to resolve it, but it can't > be too hard. Sorry if I sound like a moron; I'm (a) just starting college and (b) > going into algebra, not analysis. In fact I don't really like numbers > anyway. Kinda weird to be going into math then, I guess, but > whatever. ==== >Show that for each sequence of random variables {X_n : n = 1,2,...} there >exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n >converges to zero with probability one. Note : There is no hypothesis >whatsoever on the random variables. How can I prove this? This seems like a strange assertion to me that lacks >intuition. > > First note that, for the problem to make sense, all the random varaibles must have as domain the same probability space. Here's an idea, though I am not sure it will work: Can you first show it for discrete probability spaces, and then use some limiting (= approximation) arguments to get the general case? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9I0ODn09862; ==== >> ... >>Yes. And so what? >>Readers should note that the poster admits that his position requires >>a function that is either a unit or has f as its only non-unit factor >>depending on certain conditions as m varies from 0 to positive >>infinity. A condition follows. >>Yes, I still see no problem with such a function. Then give a *single* function in ALL of mathematics which behaves as >> you wish. You're making up some wacky mathematics. Oh, come on. Do you know the Moebius function? >> mu(n) = 1 if n = 1 >> 0 if n is divisible by a square >> (-1)^r if n = the product of r distinct primes. >> Wacky enough? But indeed, not the function we seek, see below. >> I've found a simpler refutation of this poster's position. Remember I have P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > >Variables: m, f, x, u E Ring of Algebraic integers and the factorization > >P(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf) > >Variables: a_1, a_2, a_3, roots of cubic defined as follows. > >Cubic: a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m) The poster wishes for a variable function which has the property of >being a factor of f, so I'll use w_1(m) w_2(m) w_3(m) = f^2, where a_1 >has w_1(m) as a factor for all integer m. Then a_1(m) x + uf must have w_1(m) as a factor, so dividing through >gives a_1(m)/w_1(m) + uf/w_1(m) where uf/w_1(m) can't be an algebraic integer for all integer m. It might help for me to put in actual numbers for u and f, which I can >do as the variables are independent of each other, so let u=2, f=7, >then its a_1(m)/w_1(m) + 14/w_1(m) and clearly, if w_1(m) varies with m, then 14/w_1(m) is not an >algebraic integer for all integer m. > No, not clearly. >For those who STILL need help, consider that if you had 14/w_1(m) = r(m) introducing r(m) for the result of the division, then w_1(m) r(m) = 14 so w_1(m) r(m) - 14 = 0 which would force zeroes for m. That is, you can't have algebraic integer functions, that is functions >that give algebraic integer results, and not have only certain values >of m that would work. > Patently false. Let w_1(m) = 14^{1/m} and r(m) = 14^{(m - 1)/m}. Both of these are nonconstant algebraic integer functions of the integer argument m, and their product is 14. There are trillions (low estimate ...) of other examples. Remember, the ring of algebraic integers is extremely large. >That is, you can have something like 2m+ 7 = 21, that works for a >particular value of m, but you can't have functions in algebraic >integers that will multiply to give 14 for all integer m. To get such functions, you have to go outside the ring into a field. > No indeedy; see up above. >That refutes the position of Dik T. Winter, and note that as I've >pointed out that poster clearly either has limited mathematical >ability, or he's been lying now for some time. > Dik may have limited ability, but the limit is pretty high, and I have not known him to lie at all. One principle worth remembering here: when you have eliminated almost every explanation you can think of, and the only things that are left seem farfetched, whatever they are, if they cannot be eliminated, they must be part of the truth. James Harris ==== > ... >>Yes. And so what? >>Readers should note that the poster admits that his position requires >>a function that is either a unit or has f as its only non-unit factor >>depending on certain conditions as m varies from 0 to positive >>infinity. A condition follows. >>Yes, I still see no problem with such a function. Then give a *single* function in ALL of mathematics which behaves as >> you wish. You're making up some wacky mathematics. Oh, come on. Do you know the Moebius function? >> mu(n) = 1 if n = 1 >> 0 if n is divisible by a square >> (-1)^r if n = the product of r distinct primes. >> Wacky enough? But indeed, not the function we seek, see below. >> I've found a simpler refutation of this poster's position. Remember I have P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Variables: m, f, x, u E Ring of Algebraic integers and the factorization P(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf) Variables: a_1, a_2, a_3, roots of cubic defined as follows. Cubic: a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m) The poster wishes for a variable function which has the property of >being a factor of f, so I'll use w_1(m) w_2(m) w_3(m) = f^2, where a_1 >has w_1(m) as a factor for all integer m. Then a_1(m) x + uf must have w_1(m) as a factor, so dividing through >gives a_1(m)/w_1(m) + uf/w_1(m) where uf/w_1(m) can't be an algebraic integer for all integer m. It might help for me to put in actual numbers for u and f, which I can >do as the variables are independent of each other, so let u=2, f=7, >then its a_1(m)/w_1(m) + 14/w_1(m) and clearly, if w_1(m) varies with m, then 14/w_1(m) is not an >algebraic integer for all integer m. > No, not clearly. And the poster is incorrect, as below he gives the example of 14^{1/m} which blows up at m=0. >For those who STILL need help, consider that if you had 14/w_1(m) = r(m) introducing r(m) for the result of the division, then w_1(m) r(m) = 14 so w_1(m) r(m) - 14 = 0 which would force zeroes for m. That is, you can't have algebraic integer functions, that is functions >that give algebraic integer results, and not have only certain values >of m that would work. > Patently false. Let w_1(m) = 14^{1/m} and > r(m) = 14^{(m - 1)/m}. Both of these are nonconstant > algebraic integer functions of the integer argument m, > and their product is 14. There are trillions > (low estimate ...) of other examples. Remember, > the ring of algebraic integers is extremely large. And his w_1(m) blows up at m=0. Math society is rather despicable in its weakness, and I fear it's because people are taught to believe that they make up math, as if it's not about logic but what makes them feel good. Rather than just accept the truth, math society continues to argue with me, showing its base irrationality. >That is, you can have something like 2m+ 7 = 21, that works for a >particular value of m, but you can't have functions in algebraic >integers that will multiply to give 14 for all integer m. To get such functions, you have to go outside the ring into a field. > No indeedy; see up above. It's actually trivial that for all integer m, you can't have a relation like 14/w_1(m) = r(m) with functions varying with m, any more than you can have y=14/x in the ring of algebraic integers, for all integer x. It's really trivial but math society wishes to hide the error in core mathematics--that over hundred year old definition error--and in continuing to fight the truth, they show their true disdain for consistency in mathematics itself. It IS a despicable display. >That refutes the position of Dik T. Winter, and note that as I've >pointed out that poster clearly either has limited mathematical >ability, or he's been lying now for some time. > Dik may have limited ability, but the limit is pretty > high, and I have not known him to lie at all. One principle worth remembering here: when you have > eliminated almost every explanation you can think of, and > the only things that are left seem farfetched, whatever they > are, if they cannot be eliminated, they must be part of > the truth. Which is actually something good for readers to consider when they find it hard to believe that math society could really be this corrupt. But remember, just like you can't have xy=2, for all integer x, where x and y are both integers, you can't have 14/w_1(m) = r(m) or uf/w_1(m) = r(m), where w_1(m) and r(m) are algebraic integer functions for all integer m. So when you accept what you know about mathematics, and consider that mathematicians have been arguing with me for months, with what conclusion are you left? Mathematicians have gone rogue, and are acting against world society as they try to convince by lying about mathematics, and you can see just how corrupt they have become, as now even when caught, they're still lying. James Harris ==== > ... >>Yes. And so what? >>Readers should note that the poster admits that his position requires >>a function that is either a unit or has f as its only non-unit factor >>depending on certain conditions as m varies from 0 to positive >>infinity. A condition follows. >>Yes, I still see no problem with such a function. Then give a *single* function in ALL of mathematics which behaves as >> you wish. You're making up some wacky mathematics. Oh, come on. Do you know the Moebius function? >> mu(n) = 1 if n = 1 >> 0 if n is divisible by a square >> (-1)^r if n = the product of r distinct primes. >> Wacky enough? But indeed, not the function we seek, see below. >> I've found a simpler refutation of this poster's position. Remember I have P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Variables: m, f, x, u E Ring of Algebraic integers and the factorization P(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf) Variables: a_1, a_2, a_3, roots of cubic defined as follows. Cubic: a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m) The poster wishes for a variable function which has the property of >being a factor of f, so I'll use w_1(m) w_2(m) w_3(m) = f^2, where a_1 >has w_1(m) as a factor for all integer m. Then a_1(m) x + uf must have w_1(m) as a factor, so dividing through >gives a_1(m)/w_1(m) + uf/w_1(m) where uf/w_1(m) can't be an algebraic integer for all integer m. It might help for me to put in actual numbers for u and f, which I can >do as the variables are independent of each other, so let u=2, f=7, >then its a_1(m)/w_1(m) + 14/w_1(m) and clearly, if w_1(m) varies with m, then 14/w_1(m) is not an >algebraic integer for all integer m. > No, not clearly. And the poster is incorrect, as below he gives the example of 14^{1/m} which blows up at m=0. >For those who STILL need help, consider that if you had 14/w_1(m) = r(m) introducing r(m) for the result of the division, then w_1(m) r(m) = 14 so w_1(m) r(m) - 14 = 0 which would force zeroes for m. That is, you can't have algebraic integer functions, that is functions >that give algebraic integer results, and not have only certain values >of m that would work. > Patently false. Let w_1(m) = 14^{1/m} and > r(m) = 14^{(m - 1)/m}. Both of these are nonconstant > algebraic integer functions of the integer argument m, > and their product is 14. There are trillions > (low estimate ...) of other examples. Remember, > the ring of algebraic integers is extremely large. And his w_1(m) blows up at m=0. Math society is rather despicable in its weakness, and I fear it's because people are taught to believe that they make up math, as if it's not about logic but what makes them feel good. Rather than just accept the truth, math society continues to argue with me, showing its base irrationality. >That is, you can have something like 2m+ 7 = 21, that works for a >particular value of m, but you can't have functions in algebraic >integers that will multiply to give 14 for all integer m. To get such functions, you have to go outside the ring into a field. > No indeedy; see up above. It's actually trivial that for all integer m, you can't have a relation like 14/w_1(m) = r(m) with functions varying with m, any more than you can have y=14/x in the ring of algebraic integers, for all integer x. It's really trivial but math society wishes to hide the error in core mathematics--that over hundred year old definition error--and in continuing to fight the truth, they show their true disdain for consistency in mathematics itself. It IS a despicable display. >That refutes the position of Dik T. Winter, and note that as I've >pointed out that poster clearly either has limited mathematical >ability, or he's been lying now for some time. > Dik may have limited ability, but the limit is pretty > high, and I have not known him to lie at all. One principle worth remembering here: when you have > eliminated almost every explanation you can think of, and > the only things that are left seem farfetched, whatever they > are, if they cannot be eliminated, they must be part of > the truth. Which is actually something good for readers to consider when they find it hard to believe that math society could really be this corrupt. But remember, just like you can't have xy=2, for all integer x, where x and y are both integers, you can't have 14/w_1(m) = r(m) or uf/w_1(m) = r(m), where w_1(m) and r(m) are algebraic integer functions for all integer m. So when you accept what you know about mathematics, and consider that mathematicians have been arguing with me for months, with what conclusion are you left? Mathematicians have gone rogue, and are acting against world society as they try to convince by lying about mathematics, and you can see just how corrupt they have become, as now even when caught, they're still lying. James Harris ==== >>As I have written already multiple times, it is trivially monic in other >>rings, the cubic (x - b1)(x - b2)(x - b3) serves just fine. But you >>want to go to a ring where your divisibility claims work. It includes >>the ring of algebraic integers. But which of the two numbers >> (-sqrt(7) + sqrt(15))/2 and (-sqrt(7) - sqrt(15)) >>is a unit in that ring? And why precisely that number? >... >(B) In the Object ring, this marvellous construction that fixes the >ring of algebraic integers by allowing numbers that should be in, >there are polynomias with integer coefficients, irreducible over Q, >which have ->SOME<- roots in the ring but not all of them. And in case (B), how do we know which one is there, as you ask? I think (B) is indeed precisely the case. James wants all those numbers >in his ring such that the cubics in the b's are monic with numbers in >his rings for all possible polynomials P(m). So when we look at the >polynomial in 'a' for m=1 and f=2 we get (in the algebraic integers) >three roots that are obviously not coprime to f. To get at his claim >that exactly *one* of the a's is coprime to f he has to make one of >those common factors (that I have written above) units. They are both >roots of an irreducible polynomial with integer coefficients of degree 4. >Actually sometime ago he made a remark that suggested this. The main problem is that he has to make this choice for each and every >polynomial that comes up. Now I am not sure, but it may be possible >to have a ring where only a single root of a quadratic is in his >ring, Depends on the quadratic and the factors involved (in order for the ring A[w], with A the algebraic integers and w an algebraic number) to intersect Q exactly on the integers, certain conditions on w must be met; not any old w will do. So he would have to prove that there is always one of the factors in his situation where this condition is met. >I very much doubt that a choice can be made across *all* polynomials >he comes up with without getting conflicts. He has to show that a >consistent choice can be made. That is probably the most difficult part, if not impossible. If you'll remember, back when he was factoring his cubic as (v^3+1)x^3 - 3vx + 2 = (sqrt(v+1)x + b1)(sqrt(v+1)x + b2)((v^2-v+1)x +b3), the issue of whether Q[sqrt(v+1),b1,b2,b3] contained 1/f arose. And while the issue was never settled for that case, I did manage to show that if you took the ring you got by adding ALL values of sqrt(v+1), b1, b2, b3, for all values of v different from -1, then that ring contained all primes congruent to 1 modulo 6, even though it was not clear that they were all in any specific extension Q[sqrt(v+1),b1,b2,b3]. (Reference: http://groups.google.com/groups?selm=9p2ada%241kva%241%40agate.berkeley.edu Sept 28, 2001) Taking possible values of v, even restricting to v=-1+mf^{2j}, with f and j fixed, it may indeed (and probably is) impossible to choose roots in a consistent way across all values of m (or even a restricted but infinite set of values of m) that keep the resulting ring contained in C - (Q-Z). ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== [.snip.] >It's actually trivial that for all integer m, you can't have a >relation like 14/w_1(m) = r(m) with functions varying with m, any more than you can have y=14/x in the ring of algebraic integers, for all integer x. The two situations are completely dissimilar, and it betrays a complete confusion on your part that you would try to conflate them. In the first situation, 14/w_1(m) = r(m), the denominator is not allowed to take ALL integer values, or ALL algebraic integer values; it is merely assumed to be a function whose DOMAIN is the integers, and whose image is not a singleton (you again forgot to mention you mean non-constant function). By contrast, in the latter situation you are demanding that a specific function, the identity, be used. If you wanted to make a similar situation, you would have to ask that 14/w(x) be an integer, where w(x) is a function defined for all integers and that takes integer values. In that case, you surely agree that there are plenty of functions that will do the trick and which are neither constant nor given by polynomials; for example, w(x) = 1 if x is not divisible by 7, w(x) = 2 if x is divisible by 7 but not by 2, and w(x) = 7 if x is divisible by 14. As to your original situation, >It's actually trivial that for all integer m, you can't have a >relation like 14/w_1(m) = r(m) with functions varying with m, not only had you already been given several that are defined for all integers, but here is one (I have a variant of a restriction elsewhere) which is defined for all complex numbers (and in particular, that is defined for all algebraic numbers, all algebraic integers, all integers). It is trivial how one could extend it to any set larger than the complex numbers, in any number of ways. (No, this is not the function we have here, but it makes the point that your claims that no such function can exist are just simply wrong): Let w_1(m) be defined as follows: (1) If m is a transcendental positive real number, then let w_1(m) = sqrt(7). (2) If m is a transcendental negative real number, let w_1(m) = sqrt(2). (3) If m is a transcendental complex number with Re(m)*Im(m)>0, let w_1(m) = 3-sqrt(2). If m is a transcendental complex number with Re(m)*Im(m)<0, let w_1(m)= 3+sqrt(2). If m is a transcendental purely imaginary number, then let w_1(m) = 7+sqrt(42). (4) If m is an algebraic integer, then let f(x) be the unique monic polynomial with rational coefficients which has m as a root and is irreducible over Q: f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0. Note that by irreducibility, a_0 is not equal to zero. Write a_0=r/s, where r and s are integers, s>0, and gcd(r,s)=1. We now define an auxiliary prime p: If r = 1 or r=-1, let p=5 (or your favorite prime, if you don't like 5). If r is not equal to 1 or -1, let p be the largest rational prime that divides r. Consider all the roots of the monic polynomial with integer coefficients x^5 - 14px^4 + px - 14. (or your favorite monic polynomial with integer coefficients that has a divisor of 14 as constant term). Among the five roots, let w_1(m) be the largest root in the lexicographic order. That is, we order the complex numbers by the rule a + bi <= c + di (a,b,c,d reals) if and only if ap then w_1(p) is a root of x^5 - 14px^4 + px - 14; and w_1(q) is a root of x^5 - 14qx^4 + qx - 14. Any common root of both polynomials would be a root of their difference, which is 14(q-p)x^4 - (q-p)x = (q-p)x ( 14x^3 - 1), and the only algebraic integer root of this is x=0, so w_1(p) and w_1(q) are distinct. A complicated function, yes, but the point is that there are many really complicated functions in the world, and they can do a lot of things that you do not seem to believe are possible. >It's really trivial And really wrong. >But remember, just like you can't have xy=2, for all integer x, where >x and y are both integers, you can't have 14/w_1(m) = r(m) or >uf/w_1(m) = r(m), where w_1(m) and r(m) are algebraic integer >functions for all integer m. No, they are not similar situations at all. Any integer can be written as a product of two integers in only a finite number of ways. Any algebraic integer can be written as the product of two algebraic integers in an infinite number of ways; and if the algebraic integer is not a unit (like 14), then it can be written as a product of two algebraic integers in an infinite number of ->essentially different<- ways. [. rant about mathematicians lying because they do not accept James's false claims as true removed.] ====================================================================== [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee fixe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a flash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not find it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman ====================================================================== Arturo Magidin magidin@math.berkeley.edu X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9I4Vds24432; ==== >Ellipse is produced if a plane intersects only one nappe of a cone. Is there a way to get the parameters of the ellipse as a function of >the angle of the plane with the axis of the cone? I know that they depend on the height of the cone and diameter of the >circle at the bottom of the cone. (Plane that is perpendicular to the axis produces a circle.) Here is an interesting reference about that: http://mathworld.wolfram.com/DandelinSpheres.html ==== With the function y = 8/x, gives 2^4 = 4^2, I'm wondering if there are other ==== > With the function y = 8/x, gives 2^4 = 4^2, I'm wondering if there are other If I had a dollar for every time those old chestnut has been posted to sci.math, I would be a rich young man. Bob Kolker ==== > With the function y = 8/x, gives 2^4 = 4^2, I'm wondering if there are > other functions which would have more than this. Note: Perhaps David McAnally and I don't understand your question. You may need to clarify matters. But anyway, hoping that I do understand... You're saying that the relations x^y = y^x and y = 8/x have in common the two points (x,y) = (2,4) and (4,2). That's true. But those relations also have three other points in common: (-2,-4), (-4,-2), and (Sqrt(8),Sqrt(8)) [where Sqrt denotes the nonnegative square root]. As to your wondering if there are other functions [than y = 8/x] which would have more than this: Of course! There are functions whose graphs have an infinite number of points in common with the graph of the relation x^y = y^x. The simplest such other function would be y = x. David Cantrell ==== >With the function y = 8/x, gives 2^4 = 4^2, I'm wondering if there are other You could try the function from (1,infinity) to (1,infinity) implicitly defined by the solution for y in the equation log y/y = log x/x such that either x and y are distinct or x = y = e. David McAnally -------------- ==== I should clarify, I want to have the points (2,4) and (4,2) in the function. With the function y = 8/x, gives 2^4 = 4^2, I'm wondering if there are other You could try the function from (1,infinity) to (1,infinity) implicitly > defined by the solution for y in the equation log y/y = log x/x such that > either x and y are distinct or x = y = e. David McAnally -------------- ==== >I should clarify, I want to have the points (2,4) and (4,2) in the function. The points (2,4) and (4,2) are in the function that I specified. David McAnally >>With the function y = 8/x, gives 2^4 = 4^2, I'm wondering if there are >other >You could try the function from (1,infinity) to (1,infinity) implicitly >> defined by the solution for y in the equation log y/y = log x/x such that >> either x and y are distinct or x = y = e. >David McAnally >-------------- I need to prove a theorem that states: If R is a number set and for any 2 numbers there exists a point of R between them, then every number is a limit point of R. Is this basically the same proof as proving that there is a rational number between any 2 irrationals and there is an irrational between any 2 rationals? Or am I way off base here? I guess I'm not too sure how I could easily make the transition from that to this theorem or if it's possible at all. ==== > I need to prove a theorem that states: If R is a number set and for any 2 numbers there exists a point of R > between them, then every number is a limit point of R. > ... then every number of R? is a limit ... R = {0} union (1,2) satisfies the condition, yet 0 isn't a limit point of R. ==== I need to prove a theorem that states: >If R is a number set and for any 2 numbers there exists a point of R >> between them, then every number is a limit point of R. >... then every number of R? No, every number. >is a limit ... R = {0} union (1,2) satisfies the condition, It does? What is an element of R between -2 and -1? >yet 0 isn't a limit point of R. Re-read the question - assume that she meant is very easy. ************************ David C. Ullrich ==== I need to prove a theorem that states: >If R is a number set and for any 2 numbers there exists a point of R >> between them, then every number is a limit point of R. >... then every number of R? is a limit ... R = {0} union (1,2) satisfies the condition, >yet 0 isn't a limit point of R. Do you mean {0} union ]1, 2]? I think the OP left out the condition that the number set be open/closed. ==== >> I need to prove a theorem that states: >If R is a number set and for any 2 numbers there exists a point of R >> between them, then every number is a limit point of R. >... then every number of R? is a limit ... R = {0} union (1,2) satisfies the condition, >yet 0 isn't a limit point of R. Do you mean {0} union ]1, 2]? > No, I said ]1,2[ , however ]1,2] , that is (1,2] , will suffice when 0 is added. >I think the OP left out the condition > that the number set be open/closed. > I'm supposed to be psychic? Heck no, are students in such a hurry or so lazy that they won't bother to ask coherent questions? Alas too many students need to learn how to state questions. The rationals, which are neither open nor closed, satisfies both the condition and the conclusion. So premises other than 'open' could be divined. As for closed, { 1/n | n in N } / {0} is closed set with only one limit point, namely 0. So closed doesn't suffice as 1/2, 1/3, ... aren't limit points of the set. It's pointless to guess what poorly written posts intend. If it's of worth, the OP will soon make amends. ==== I need to prove a theorem that states: If R is a number set and for any 2 numbers there exists a point of R > between them, then every number is a limit point of R. ... > I'm supposed to be psychic? > Heck no, are students in such a hurry or so lazy that they > won't bother to ask coherent questions? The OP's statement, exactly as given, is (a) true, and (b) easy to prove: if x is any number and e>0, then R contains a point r between x and x+e, so |x-r| < e. ==== Suppose that n,p are positive integers, and x1,x_2,... (x_k=/=0) are real numbers.Let us denote A(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)*(SUM_{k=1 to k=n} 1/x_k) B_p(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)^{p}*(SUM_{k=1 to k=n} 1/x_k^p) and assume that the following is true : ==== > Suppose that n,p are positive integers, and > x1,x_2,... (x_k=/=0) are real numbers.Let us denote A(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)*(SUM_{k=1 to k=n} 1/x_k) B_p(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)^{p}*(SUM_{k=1 to k=n} 1/x_k^p) and assume that the following is true : > A(x_1,x_2,...,x_n)= 1 for all admissible x_k ,(x_k=/=0, x_k in R), > it follows > B_p(x_1,x_2,...,x_n)=1 for all odd integers p. Find n . It's true that n takes only one value , namely n=3 ? > ===== [A Corrected form : Sorry !] Suppose that n,p are positive integers,(n>=2), and x1,x_2,... (x_k=/=0) are real numbers.Denote A(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)*(SUM_{k=1 to k=n} 1/x_k) B_p(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)^{p}*(SUM_{k=1 to k=n} 1/x_k^p) and assume that the following is true : ,, If A(x_1,x_2,...,x_n)= 1 , then B_p(x_1,x_2,...,x_n)=1 for all odd integers p. Find n . It's true that n takes only one value , namely n=3 ? =====Alex/Proposer ==== > Sigh. Yes, the following, if the reasoning as (sic) actually correct, can be easily formalized in ZF. ... note that for example the in operator below is not going to correspond to the in in ZF, it's going to be just some predicate, with axioms involving it. > C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) >Classification > > C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] > (Equi-membered classes are identical iff these are sets.) Put your money where your mouth is. Let's see your formalization. --John ==== > Sigh. Yes, the following, if the reasoning as (sic) actually >correct, can be easily formalized in ZF. ... note that for example the in operator below >is not going to correspond to the in in ZF, it's >going to be just some predicate, with axioms >involving it. > >> C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) >>Classification >> >> C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] >> (Equi-membered classes are identical iff these are sets.) Put your money where your mouth is. Let's see your formalization. It's clear to anyone with half a clue that your reasoning, if correct, can easily be formalized in ZFC. (Um: I should have said with some predicate in place of 'in' and also some predicate in place of '=', instead of just pointing out that your 'in' would not correspond to the 'in' in ZFC.) This is because there's nothing non-standard about your _reasoning_, all that's non-standard is your _axioms_ regarding 'in' and '='. In case you don't have half a clue, the part of the formalization corresponding to those two axioms might be C3 EyAx[ni(x, y) <-> Et(ni(x,t)) & A] (with y not free in A) Classification C4 AyAx[Az(ni(z,y) <-> ni(z,x)) -> {(S(y) & S(x)) <-> I(x,y)}] (Equi-membered classes are identical iff these are sets.) >--John ************************ David C. Ullrich ==== David Petry a .8ecrit dans le message de I'm looking for a reference for the proof that the infinite series > There is an entirely elementary proof of this relation, > which I suspect is well known but I've never seen it. Factor (1+x)^N - (1-x)^N, for N an odd integer, as product ( w^k(1+x) - w^{-k}(1-x) ), k=-(N-1)/2..(N-1)/2 > where w = exp( i pi / N ) Then find the coefficient of x^3 for that expression, and > compare it to the same coefficient using the Binomial Theorem > for the first expression, and take a limit. I find : n*Binomial(n,3)=Sum((cotan(k*Pi/n))^2,k=1..(n-1)/2) and .... Please give me an hint for continue... ==== Dear All, I would like to know whether it is possible to define the limit of a function f as x tends to +oo, where f is defined on R^+, except on an infinite countable number of points of R^+ (for instance, except on the set of integers). Paul ==== > Dear All, I would like to know whether it is possible to define the limit of a > function f as x tends to +oo, where f is defined on R^+, except on an > infinite countable number of points of R^+ (for instance, except on > the set of integers). or for instance, something like gamma(-x) ... I think you can safely use the standard limit definitions: For all P>0, there is a Q such that for all x of dom(f): x > Q ==> f(x) < -P For all e>0, there is a Q such that for all x of dom(f): x > Q ==> |f(x)-b| < e For all P>0, there is a Q such that for all x of dom(f): x > Q ==> f(x) > P I think these definitions work fine for the functions you have in mind. Of course the gamma(-x) would have no limit for x --> +oo Dirk Vdm ==== The usual proofs condisider the open ball with the center point removed > (or the one point compactifications). This set is homotopic to a > sphere, which is not contractible. This allows a proof that open sets > in R^2 and open sets in higher dimensions are not homeomorphic using > fundamental (first homotopy) groups. If you know about higher homotopy groups, this will also give the general > result. However, the higher homotopy groups are not so easy to calculate. > Ok, correct. I would have no idea to calculate these homotopy groups, but it is not too hard to visualize so I'll believe that pi_n(S^n) = Z pi_n(S^m) = 0 for m < n. ==== function g : R^2 -> Z let Z : R*{0} U {0}*R define quotient topology of Z such that function g is quotient map. (hint : At this time, any two points don't separated to open set.) ------------------- um...............i don't know help me........my good teacher. ==== > function g : R^2 -> Z > let Z : R*{0} U {0}*R define quotient topology of Z such that function g is quotient map. > The quotient topology for Z is { U subset Z | g^-1(U) open in R^2 } > (hint : At this time, any two points don't separated to open set.) > ==== continuous function fn let {fn} is uniformly convergence f define gn(x)=fn{x + (1/n)} show U.C gn -> f (U.C=uniformly convergence) ----------------------------- because fn -> f : U.C any E>0 , N exist any x , n>=N => |fn(x)-f(x)| < E and i must show gn -> f : U.C |gn(x)-f(x)| = |fn{x + (1/n)} -f(x)| = |fn{x + (1/n)} - fn(x) + fn(x) -f(x)| <= |fn{x + (1/n)} - fn(x)| + |fn(x) -f(x)| < 2/E + 2/E = E my problem is |fn{x + 1/(n^2)} - fn(x)| U.C ?? i think that it is pointwise convergence. because lim[fn{x + 1/(n^2)}] = fn(x) thus......my progress is wrong?? how to solved it????....my best teacher.... ==== Can anyone please point me toward an implementation (in any language, even > pseudo ones) of Knuth's spectral test? I have a random generator that I need to test, and I can not find > implementations of the test... (I have found implementation specially geared for congrual tests, but this > is not what I need as my generator is not congrual). > Does this help? http://random.mat.sbg.ac.at/results/karl/spectraltest/ Also, are these tests (written in C) in DIEHARD? HTH, Flip ==== Does the following system have a solution: 1024 o x1 mod y 2048 o x2 mod y 8192 o x3 mod y 16834 o x4 mod y x1+x2+x3+x4+33 =y X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9ICLpP18575; ==== >I'm doing some reading in abstract algebra (Dummit and Foote book) on tensor >products. I'm not getting the idea behind it and its driving me nuts with >confusion. According to the book the motivation behind construction of >tensor products, from what I understood, is that given any R-module N is it >possible to extend/embed it to/in an S-module where R is a subring of S? I >will quote the book here: We begin the construction by returning to the >basic module axioms in order to examine whether we can define products of >the form sn for s in S and n in N. These axioms start with the abelian group >N together with a map from S X N to N where the image of the pair (s, n) is >denoted by sn. It is therefore natural to consider the free Z-module on the >set S X N i.e. the collection of finite commuting sums of the elements of >the form (s, n) where s in S and n in N. This is an abelian group where >there are no relations between any distinct pairs (s, n) and (s', n'), i.e. >no relations between the formal products sn, and in this abelian group the >original module has been thoroughly distinguished from the new >coefficients from S. The book the goes on to define a subgroup H >generated by all the elements of the form (s1 + s2, n) - (s1, n) - (s2, n) .... (1) >(s, n1 + n2) - (s, n1) - (s, n2) .... (2) >(sr, n) - (s, rn) .... (3) and we take the quotient of the free abelain group by H to satisfy the >relations for a S-module and the compatibitly relation of the action of R on >N. The resulting quotient group S(X)R is called the tensor product of of S >and N over R. where s (X) n denotes the coset containing the (s, n). Now there is a natural map T: N --> S (X) N (over R). T(n) = 1 (X) n Now what is confusing me is the following: Since the group is free abelian >group, for any (s', n') there is no element (1, n) such that (s', n') = (1, >n), so N is not an S-Module if we identify it by its embedding into S (X) N >( T(n) = 1 (X) n ), because for N to be an S-module, then for every s'n' >there must exist an n such that s'n' = n. The book also says that the relations (1), (2) and (3) were the minimal >relations we had to impose in order to obtain an S-module, so it is the best >possible S-module to serve as a target for an R-module homomorphism from N. >This also confuses me since I thought that the R-module homomorphism from N >to S(X)N (call it Y) is onto, since if S(X)N is an S-module each >(s', n') must equal an element (1, n), so imposing an extra relationships >would mean that Y is not a homomorphism. This highly confuses me too. Is N a >subring of S(X)N? If, so how?? Also, why did we start with a free group and >take its quotient group by H to get the required relationships? We could >have imposed the relationships required for an S-module on the elements in >the first place instead of making the group free and taking the quotient? >Tensor products seems to be a very interesting (and important) topic, so I >would really like to understand the idea behind it very well. If anyone >could help me I would really appreciate it. > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9ICLp118589; ==== > Perplexing Patterns of Square Numbers > - B.S.Rangaswamy Observe closely the following patterns of nonzero square numbers. The squares read alike across and downwards in all these patterns. Perplex-5 Perplex-6 5 7 1 2 1 -239^2 1 5 1 3 2 1 -389^2 > 7 2 3 6 1 -269^2 5 5 3 5 3 6 -744^2 > 1 3 9 2 4 -118^2 1 3 3 2 2 5 -365^2 > 2 6 2 4 4 -162^2 3 5 2 8 3 6 -594^2 > 1 1 4 4 9 -107^2 2 3 2 3 2 4 -482^2 > 1 6 5 6 4 9 -407^2 Perplex-7 Perplex-8 1 7 1 8 7 2 1 -1311^2 1 1 1 2 8 8 9 6 -3336^2 > 7 3 3 3 2 6 4 -2708^2 1 6 3 5 3 9 3 6 -4044^2 > 1 3 7 3 5 8 4 -1172^2 1 3 8 4 5 8 4 1 -3721^2 > 8 3 3 4 7 6 9 -2887^2 2 5 4 6 2 1 1 6 -5046^2 > 7 2 5 7 6 3 6 -2694^2 8 3 5 2 1 3 2 1 -9139^2 > 2 6 8 6 3 2 1 -1639^2 8 9 8 1 3 5 2 9 -9477^2 > 1 4 4 9 6 1 6 -1204^2 9 3 4 1 2 2 2 5 -9665^2 > 6 6 1 6 1 9 5 6 -8134^2 There exist 15 more perplex-5 patterns > 13 perplex-6 > 09 perplex-7 > 02 Perplex-8 > and a scintillating Perplex-9 pattern It is thrilling and educating excercise to arrive at a few of these patterns. One should not expect to get all these patterns in ten days time. Hundred or its multiple days are reasonable, not with continous entanglement but with leisurely and positive attitude.I have taken much more number of days to arrive at all these 15+13+9+2+1 amazing patterns. > The two Perplex-8 patterns are prize catches and are to be arrived at, with immense analytical and choosy efforts. While, Perplex-9 with over 8 thousand nonzero 9 digit square numbers and more than a thousand pentillion(Pentillion=10^30)available combinations,demands Herculean efforts. If a second perplex-9 is discovered, it can be Wonder of WONDERS. I am still not able to exhaust all available combinations for perplex-9. > It is stimulating and exciting to discover these perplexing patterns. Formulate a few such patterns. It is certain that you get thrilled, elated and educated with the exploration of each of these patterns from the vivid galaxy of square numbers. > Wishing you all the best in your explorations. HINTS : A. Start with last column/row. > B. Refer to Perplex-5 presented in greater details under > Inquisitive problems in Teacher Exchange:13-16 at > http://mathforum.org/te/higher. html. C. Utilize extensively the sorted out lists of nonzero > square numbers. > D. Forming interlocking square numbers on SCRABBLE BOARD with > Number Tiles can be helpful,as it completely eliminates > paper work. > E. Start with perplex-5 formations. ------------- > Perplex-5 can be directly viewed at >> http://mathforum. org/te/exchange/hosted/rangaswamy HINTS for formation of Perplex-6 patterns : > >A.Compute a sorted out list of 6 digit nonzerosquares,numbering 376 Example-Squares ending with 16 are generated from > 346^2,396^2,446^2 - - - - 996^2 - in increments of 50 > 354^2,404^2,454^2 - - - - 954^2 - & so on. > B. There exist 3 Perplex-6 patterns having 544644 in last row/column > 2 111556 > 2 166464 > 2 more 165649 > one Perplex-6 pattern each having 154449, 516961, > 594441,695556 in last row/column. > These hints can certainly assist in forming the Perplexing Patterns of Square Numbers. >- B.S.Rangaswamy >The THIRTEEN Perplex-6 patterns can be obtained by expanding the following sets of 6 squares each : 1) 865^2 2) 885^2 3) 965^2 4) 379^2 5) 939^2 6) 369^2 7) 761^2 8)479^2 668^2 918^2 582^2 694^2 944^2 631^2 891^2 486^2 932^2 568^2 432^2 565^2 335^2 829^2 969^2 984^2 476^2 526^2 526^2 814^2 844^2 335^2 435^2 642^2 472^2 472^2 472^2 658^2 482^2 815^2 535^2 704^2 738^2 738^2 738^2 407^2 407^2 334^2 334^2 408^2 9) 981^2 10)715^2 11)965^2 12)344^2 13)661^2 806^2 409^2 627^2 357^2 615^2 544^2 424^2 372^2 935^2 828^2 608^2 477^2 462^2 585^2 962^2 796^2 536^2 478^2 585^2 478^2 408^2 719^2 771^2 834^2 393^2 Now, is the task of forming SEVEN Perplex-7 patterns out of 1061,7 digit nonzero square numbers. 3 Perplex-7 patterns end with 3114^2 One each of perplex-7 patterns end with 1204^2, 2538^2, 3108^2 & 3157^2 . Wishing you all happy and wonder-ful time in exploration of these perplex-7 patterns . - B.S.Rangaswamy. boundary=----=_NextPart_000_0064_01C39591.1FD16900 ==== --------------------------------------------------------------------- Does the following system have a solution: 1024 o x1 mod y 2048 o x2 mod y 8192 o x3 mod y 16834 o x4 mod y x1+x2+x3+x4+33 =y ==== > [snip] > We have statement B:Two coins were flipped and at least one is a head. >> We have statement B':Two coins were flipped and at least one is a >> tail. >> What do you argue? What do you say now? >> I say that mathematically speaking, B and B' are the same statement. It depends on what question you're going to ask. You've taken half the > problem statement but haven't said whether you're changing the other > half. If you make a swap of heads and tails THROUGHOUT THE ENTIRE QUESTION, > then they are equivalent questions. To whit: With a weighted coin which falls on heads with probability > 0.75, what is the probability of three heads in a row? Same question as: With a weighted coin which falls on tails with > probability 0.75, what is the probability of three tails in a row? But that does not mean that a weighted coin which falls on tails with > probability 0.75 is identical to a weighted coin which falls on > heads with probability 0.75. Here is a different question: With a weighted coin which falls on tails with probability 0.75, what > is the probability of three heads in a row? See how that works? Change the labelling in the whole question: same > question. Change the labelling in half the question, might be the same > or might be different. - Randy >>We have coins of equal density. Heads and tails are equally likely. I >>was trying to get Ullrich's opinion nailed down. He seems to be >>waffling. No, he is not. You are misinterpreting his quite clear statement. >> He is trying to say your misinterpretation is not his original >> statement. The difference is quite clear when the two are >> laid side by side, as you did in your quote. >He chastised me for thinking that changing from heads to tails would >change that question. I didn't think it would and he didn't either. >Now he says he wasn't arguing that it wouldn't. No, I'm not arguing that. >That's waffling. > >>You and I seem to agree. I mostly avoid your threads because your silly obsession >> with this particular probability problem and your >> peculiar reading of it. So no, I would not say we agree. >We agree that changing from heads to tails does not alter the >question, or the answer. Or we don't agree? I say that it doesn't. >Surely you agree with that? > >>When we flipped HH and said, Two coins were flipped and at least one >>is a head. What are the chances for two heads? or we flipped TT, and >>said Two coins were flipped and at least one is a tail. What are the >>chances for two tails? We would have the same question with the same >>answer. No. The question does not include what was flipped. Take the >> statements in quotes, i.e. >Two coins were flipped. It says that right in the question. WERE >FLIPPED, past tense. We're talking about a historical event. All we >know about the event is what we read in the statement. >> Two coins were flipped and at least one is a head. What >> are the chances for two heads? and Two coins were flipped and at least one is a tail. What are the >> chances for two tails? And I will say these are the same question given a fair >> coin. You've just relabelled heads and tails. >When they landed HH the first statement was true, when they landed TT, >the second statement was true. At HT, or TH, either statement would >have been true. >Consider the second statement. We know that the coins landed TT, or >TH, or HT. Suppose that they landed TH. The other statement would have >been true. If it had been made, would we have had a 'different' >question? > >> When you add that there is a particular outcome given, it's >> silly to ask about probability since there's only one >> point in the sample space. The probability is 1 in both >> of those cases. >> >This is a probability question. The coins were flipped once. We can do >probability because we can re iterate the flip, umpteen thousands of >times. The tricky part is that we are re iterating a first time flip. >We flip umpteen thousand times and each one was a first time flip. >With HH, the heads statement was made, with TT, the tails statement >was made, and with HT, or TH, one, or the other. Either would have >been true, and would have been the 'same' statement. OR, changing the >color would have changed the statement. >Likewise, with a HT, or a TH flip, either statement would be true. Huh? The statement includes a question: What are the >> chances for two heads (tails)? How can that question >> be true? >> >That's the point. Does changing from heads to tails alter the >question. >Either question would have the same answer. For one of them to have >>answer other than 1/2, they would have to have different answers. No, they have the same answer of 1/3. >That's why I have this 'strange obsession'. There are always qualified >people like you who will argue with me, and don't seem to want to >understand my argument. > >> Case 1: I tell you two coins were flipped and one of the >> coins is a head. Therefore you know that I have either >> HH, TH, or HT with equal probability. I ask you what is >> the probability that I got HH? The answer is 1/3. >> >I know that they landed HH, or HT, or TH. With HH, you made the heads >statement with probability one, but you made the tails statement with >probability 1/2. >There are now three possible outcomes. They are no longer equally >likely outcomes. Case 2: I tell you two coins were flipped and one of the >> coins is a head. Therefore you know that I have either >> TT, HT, or TH with equal probability. I ask you what is >> the probability that I got TT? The answer is 1/3. >I know that you have TT, HT, or TH. They are no longer equally >probably. Try it. You can't do it. Flip the coins but don't look. You can't make either statement, prior >to inspection. After inspection you can make one statement, or the >other. With HH, you can only make the heads statement. With TT, you >can only make the tails statement. With HT, or TH, you can make one, or the other, but not both. If you >make both, then I know that the probability is one that the two coins >are different. The four are equally probable, prior to inspection. >After inspection, you can only make one statement, or the other. After >the statement, the three that are left are not equally probable. > >> You will see that in exchanging H and T in the >> two paragraphs throughout, I have not changed the >> meaning or the interpretation. There is nothing like >> a requirement that the probability be 1/2. Your final >> statement is coming from somewhere outside of probability >> theory. You flipped the coins and got the answer wrong. That's what keeps me >obsessed. My reading of the question is strange? My reading may be >strange, but you get the wrong answer, and you flipped the coins. Eldon >> - Randy ************************ David C. Ullrich ==== change that question. I didn't think it would and he didn't either. >Now he says he wasn't arguing that it wouldn't. No, I'm not arguing that. > >That's waffling. > >He chastised me for thinking that changing from heads to tails would >>change that question. I didn't think it would and he didn't either. >>Now he says he wasn't arguing that it wouldn't. No, I'm not arguing that. >> >>That's waffling. > >> ************************ David C. Ullrich >Okay, we have two questions: >B. Two coins were flipped and at least one is a head. What are the >chances for two heads? >B'. Two coins were flipped and at least one is a tail. What are the >chances for two tails? Changing from heads to tails either changes the question, or it >doesn't. >I say no, Randy Poe says no. What do you say? You said you weren't arguing this, or that. Does the >changing from heads to tails alter the question, or not? Yes, or no? The two questions are obviously equivalent. (That doesn't say that Two coins were flipped and at least one is a head and Two coins were flipped and at least one is a tail are equivalent, and it _certainly_ doesn't say that Two coins were flipped and at least one is a head and Two coins were flipped and at least one is a tail are _the same statement_, which is the idiocy you said I'd said.) >Eldon:~) Eldon ************************ David C. Ullrich ==== >>We have coins of equal density. Heads and tails are equally likely. I >>was trying to get Ullrich's opinion nailed down. He seems to be >>waffling. No, he is not. You are misinterpreting his quite clear statement. >> He is trying to say your misinterpretation is not his original >> statement. The difference is quite clear when the two are >> laid side by side, as you did in your quote. >He chastised me for thinking that changing from heads to tails would >change that question. Correctly. Exchanging the words heads and tails THROUGHOUT the question changes nothing. It's just a relabeling of the sides of the coins. > I didn't think it would and he didn't either. Good. >Now he says he wasn't arguing that it wouldn't. You quoted your change, the one that changes the situation, and the one that David said earlier does not. You have quoted them in the same post. They are quite different. Study them. I have already explained several times (see what is capitalized above) why your new change is different from your old change. I'm not going to berate this point any more since I've already said the difference over and over. I can only imagine how many times people who've been in this thread longer have tried to explain to you. - Randy ==== > You quoted your change, the one that changes the situation, and the > one that David said earlier does not. You have quoted them in the same > post. They are quite different. Study them. I have already explained > several times (see what is capitalized above) why your new change is > different from your old change. I'm not going to berate this point any > more since I've already said the difference over and over. I can only > imagine how many times people who've been in this thread longer have > tried to explain to you. - Randy The truth is in the archives. But just study this thread. 1 Virgil said that the compliment to at least one is a head is TT. 2 Eldon says that is not true. 3 With the statement at least one is a head, we know that TT didn't happen. We know that HH happened, or HT happened, or TH happened. Or, or, not all three. 4 When HH happened and at least one is a head was generated, then the compliment would be, TT happened and at least one is a tail was generated. 5 When HT happened and at least one is a head was generated, the compliment would be, HT happened and at least one is a tail was generated. 6 When TH happened and at least one is a head was generated, the compliment would be, TH happened and at least one is a tail was generated. 7 This is only true if changing from heads to tails is complimentary. 8 Eldon argues that it is. 9 Eldon erred when he said that Doctor Ullrich has argued for that. Dr. Ullrich actually said that surely Eldon isn't dumb enough to think different. 10 Eldon should have said, surely Doctor Ullrich isn't dumb enough to think that's wrong. In this, Eldon seems to have erred:) 11 Eldon should have said, Randy Poe argues that 'heads' and 'tails' are complimentary. 12 That's what I meant when I said that we agree. We agree that 'heads' and 'tails' are complimentary. 13 Now that we agree on that, we should be able to agree on the answer to our question. 14 Flip two coins, they will land HH, or HT, or TH, or TT. 15 We can say about our question that two coins were flipped and a true statement was made. 16 When those coins land HH, make the at least one is a heads statement. 17 When they land TT, make the at least one is a tails statement. 18 When they land HT, or TH, make one statement, or the other, without bias. 19 Then answer the question. Observations: There are similar questions which correctly answer 1/3. The key is not, whether or not the statement was at least one is. The key???? ***The key is whether or not there was a prejudice toward one outcome or the other, prior to inspection of the toss.*** Another observation: Consider two different objects, such as an orange ball and a green ball. Randomly choose one. Announcing the winner does not communicate a prior prejudice. Eldon:) ==== Suppose the random variables X_i, i = 1,2,... are i.i.d. Show that if E((X_1)^2) < infinity and if M_n = max(X_1,X_2,...,X_n), then (M_n)/squareroot(n) converges to zero with probability one. I thought about altering the strong law of large numbers assuming second moments, then using a subsequence trick, but am getting nowhere. Can anyone give a proof for this please? Marc ==== >Suppose the random variables X_i, i = 1,2,... are i.i.d. Show >that if E((X_1)^2) < infinity and if M_n = max(X_1,X_2,...,X_n), then >(M_n)/squareroot(n) converges to zero with probability one. I thought about altering the strong law of large numbers assuming second >moments, then using a subsequence trick, but am getting nowhere. Can anyone give a proof for this please? I don't have an actual proof, but you might want to note that you can express P(M_n/sqrt(n) > eps) explicitly in terms of the distribution of X_1 (because P(M_n <= delta) = P(X_1 <= delta)^n.) >Marc > ************************ David C. Ullrich ==== There's something called Kolmogorov's something that I'm almost certain will help. You can find it in Gerald Folland's Real Analysis book, right before he proves the strong laws of large numbers. It gives a good bound for M_n, which I think will do the job. > Suppose the random variables X_i, i = 1,2,... are i.i.d. Show > that if E((X_1)^2) < infinity and if M_n = max(X_1,X_2,...,X_n), then > (M_n)/squareroot(n) converges to zero with probability one. I thought about altering the strong law of large numbers assuming second > moments, then using a subsequence trick, but am getting nowhere. Can anyone give a proof for this please? Marc ==== > >:> What is the false implication? >:> >:> : The false implication is that b is a non-unit factor of 2 like >: sqrt(2), or 2^{1/3}, when in fact, appropriately, it's a factor of 1. > >: Unfortunately the poster deleted out pertinent information, so I'll >: make the effort to put it back in so that it makes sense to readers >: without forcing them to go back to previous posts. > >: I gave the example of xy = 2, where x and y are algebraic integers, >: where x=2a and y=b, so b is an algebraic integer, but 'a' is not, but >: it is a member of what I call the object ring. > >: The object ring is a commutative ring that includes all numbers such >: that -1 and 1 are the only members that are both a unit and an >: integer, where no non-unit member is a factor of any two integers that >: are coprime. > >: Notice that definition for the object ring excludes possibilities like >: a=1/2 as then 2 would be a unit since 2(1/2) = 1. > >: So, in fact, in the object ring, ab=1 and b IS a unit, but because 'a' >: is inappropriately excluded from the ring of algebraic integers by the >: definition of algebraic integers as roots of monic polynomials with >: integer coefficients, you have the false implication that b is some >: other type factor of 2. > >: Now then, say you follow algebra, and then you find that you have 2 as >: a factor of x, well you've been pushed out of the ring of algebraic >: integers where x does NOT have 2 as a factor, even though x=2a. > >:>: And you can see several problems that popped up by that exclusion as b >:>: is not a unit in the ring of algebraic integers, when it should be, >:>: and x does NOT have 2 as a factor in the ring, though x=2a. >::>So, what is the problem? I haven't seen anything that looks like >:>a contradiction, only an assertion that something is not a unit >:>which should be according to some unstated principle. What's the >:>principle? > >:> : Given that xy=2, with x=2a, y=b, with b an algebraic integer, x an >:> : algebraic integer, while 'a' is not, it *should* be that y does not >:> : share non-unit factors with 2, since x *should* have 2 as a factor, >:> : but in the ring of algebraic integers, because the definition >:> : arbitrarily excludes 'a', neither of those is the case. >:> >:> What is behind the shoulds, i.e. what are the bad consequences of >:> excluding 'a'? Before you were talking about being able to prove two >:> contradictory results. >:> >:> Mike > >: The problem occurs because algebra does NOT recognize the arbitrary >: definition, so using algebra you can find yourself in situations where >: you prove that x has 2 as a factor, but then again, because of this >: arbitrary definition, it does not, in the ring of algebraic integers. > >: Since the assumption is that algebraic integers is an appropriate >: ring, it appears that you've proven two contradictory things: What do you mean by an appropriate ring? > > : One where you can't appear to prove contradictory things. > > : >: 1. x has 2 as a factor > >: 2. x does NOT have 2 as a factor But aren't 1. and 2. actually: 1. x has 2 as a factor in the object ring > > : I'll remind of the example of 2 and 6 in the ring of evens to help > : readers with context, as there 2 is not a factor of 6 because 3 isn't > : even. > > : Now then, with my example, you have x=2a, where x is an algebraic > : integer, but as 'a' is not, it doesn't have 2 as a factor, in the ring > : of algebraic integers, which falsely implies, given that xy=2, where y > : is an algebraic integer that x and y have non-unit factors of 2 in the > : same sense as like if you have sqrt(2) as a factor for both. > > : But, in fact, *neither* has what you might call non-trivial factors of > : 2 in the ring of algebraic integers, which I have to say as trivially > : they have themselves as factors. > > : That is, given that xy=2, x is trivially a factor of 2, in the ring of > : algebraic integers, as is y, when in fact x has a factor that IS 2, a > : non-trivial factor, in an appropriate ring, that is, one which does > : not allow contradictions. > > : It's worth noting that proving that x doesn't have a factor of 2, in > : the ring of algebraic integers, requires going to the field of > : algebraic numbers, which is itself, of course, not flawed. > >2. x does NOT have 2 as a factor in the algebraic integers > > : And proving that requires going to the field of algebraic numbers, > : which hasn't really been discussed at this point. > > : So then *in the ring of algebraic integer* you can appear to prove > : *both* things, and only figure out that you've been pushed out of the > : ring of algebraic integers, by going to the field of algebraic > : numbers. I'm not following. You let x = 2a where x is an algebraic integer > and a is not. So x does not have a factor of 2 in the algebraic > integers. What is the contradictory proof that x *does* have a factor > of 2 in the algebraic integers? In my case I use a special polynomial P(m), which is special in that I can rather easily factor it into non-polynomial factors to appear to prove within the ring of algebraic integers that x has 2 as a factor. So more fully, the problem with the ring of algebraic integers is that *in the ring of algebraic integers* you can prove with my example that x has 2 as a factor, and then go to the field of algebraic numbers and prove that x/2 can't be an algebraic integer. James Harris ==== [snip] > So more fully, the problem with the ring of algebraic integers is that > *in the ring of algebraic integers* you can prove with my example that > x has 2 as a factor, and then go to the field of algebraic numbers and > prove that x/2 can't be an algebraic integer. James Harris There is no problem in the ring of algebraic integers. Your argument is fundamentally flawed and your proof contains fatal errors. -- A fool and his proof are soon refuted. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== I'm trying to show the 1 - 1 correspondence between R and the power set of N without resorting to the binary decimal argument. I know I can use the Schroeder-Bernstein Theorem. There is no problem finding an injection from the power set of N into R. I'm having difficulty finding an injection from R into the power set of N. Help! L ==== > I'm trying to show the 1 - 1 correspondence between R and the power set of N > without resorting to the binary decimal argument. I know I can use the > Schroeder-Bernstein Theorem. There is no problem finding an injection from > the power set of N into R. I'm having difficulty finding an injection from > R into the power set of N. It's probably easier to find an injection from [0,1) into P(N). This suffices, because it's easy to see that each nondegenerate interval has the same cardinality as R. You say you don't want to use the binary decimal argument. Is a pure binary argument acceptable? For each x in [0,1) there is a binary representation, where to avoid ambiguity we avoid any representation that ends in all 1's. This corresponds in an obvious way with a unique member of P(N), and the resulting map is an injection. It's not a bijection, because cofinite sets in P(N) are not in the range of this map. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== Let n>=4 and A=||a_{i,j}|| be a n x n real symmetric matrix with a_{i,i}=0 , i =1,2,...,n . Denote S(A)= SUM_{i=1 to i= n-1}SUM_{j=i+1 to j=n}a_{i,j}^2 . Suppose that r_1== 2*sqrt(S(A)) , then A is singular. 2) If n is even and r_n -r_1 =< sqrt(8*S(A)/n) , then det(A)= (-2*S(A)/n)^{n/2} ; If n is odd and r_n -r_1 =< sqrt(8*S(A)/(n^2 -1)), then det(A)= ? 3) Suppose that r_n = sqrt(2*(n-1)*S(A)/n) or r_n = sqrt(2*S(A)/(n(n-1))) . It's possible to find A^{-1} ? If true , above questions are of interest ? ============ ==== Dear All, I would like to find two one-variable functions, f and g, such that: * limit of f(x)/g(x) is an indetermination of the type 0/0 or oo/oo; * limit of f(x)/g(x) does not exist at 0; * limit of f'(x)/g'(x) exists at 0; * g'(x) different of zero in some neighborhood of 0 (except possibly at 0) is a false proposition. Paul ==== >Dear All, I would like to find two one-variable functions, f and g, such that: * limit of f(x)/g(x) is an indetermination of the type 0/0 or oo/oo; >* limit of f(x)/g(x) does not exist at 0; >* limit of f'(x)/g'(x) exists at 0; >* g'(x) different of zero in some neighborhood of 0 (except possibly >at 0) is a false proposition. ??? If I understand what you mean by the fourth condition, it says that there exists a sequence x_n -> 0 such that g'(x_n) = 0. If so then the limit of f'/g' cannot exist. Paul ************************ David C. Ullrich ==== Given a quartic monic polynomial with integer coefficients and with four complex roots (two pairs of complex conjugate roots), eg: x4-10x3+47x2-90x+81 how do I find the min poly of |x|^2 (involving the complex absolute value) for the example above it is: x2-27x+81 (yeah I can get the 81 OK) without using a computer or Ferarri's method of solving the quartic to get all the roots explicitly and then working back. I presume there is a generalization to certain even degree polynomials, but I reckon I can get that if someone can remind me how to do it for quartics. I feel stupid for having forgotten how to do this, but it *is* driving me insane. Bill. P.S: apologies if this appears twice or even thrice. I have been trying to get it to post all day, but my internet provider has a known news even after a day, and I can rarely read all responses. I am experimenting with other options. ==== Can somebody answer this question? Assume that V is a vector space Assume that S is a subspace of V Assume that O is the zero vector of V Is O also the zero vector of S? In other words, is the zero vector of V inherited by S? In other words if the, zero vector in V is not in S then is S not a subspace? thank you March ==== >Can somebody answer this question? Assume that V is a vector space >Assume that S is a subspace of V >Assume that O is the zero vector of V >Is O also the zero vector of S? In other words, is the zero vector of V >inherited by S? Yes. By definition, a subspace must be a vector space with the operations of V; that is, you don't get to define the vector sum of S any which way you want, you must add vectors in S as if they were in V, and the result must be in S. Since there is only one zero vector in V, and since for ANY vector v in V, if v+w = v then w must be the zero vector of V, then for any vector s of S, if s+w = s, then w must be the zero vector of V. >In other words if the, zero vector in V is not in S then is S not a >subspace? Correct. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >Yes. By definition, a subspace must be a vector space with the >operations of V; that is, you don't get to define the vector sum of S >any which way you want, you must add vectors in S as if they were in >V, and the result must be in S. I was wondering about the minimal required definition. I was told the inclusion of the zero vector must be explicit to avoid a problem where the empty set could be a subspace of V, but 0 * v = 0 implies directly that 0 must be in S so why should it be required explicitly? ==== >Yes. By definition, a subspace must be a vector space with the >>operations of V; that is, you don't get to define the vector sum of S >>any which way you want, you must add vectors in S as if they were in >>V, and the result must be in S. I was wondering about the minimal required definition. I was told the >inclusion of the zero vector must be explicit to avoid a problem where >the empty set could be a subspace of V, but 0 * v = 0 implies directly >that 0 must be in S so why should it be required explicitly? No, it does not imply that directly. You must assume that S is nonempty: otherwise, where does v come from? A vector space V over the field K is defined to be a set, which we usually call V as well, together with two operations: a vector sum, +, which takes pairs of elements of V as arguments and replies with an element of V; and a scalar product, which takes as arguments an element of K and an element of V, and returns an element of V. The conditions that these set and operation must satisfy are: (1) For any three elements v, w, z of V, (v+w)+z = v+(w+z) (2) There exists an element in V, called 0, such that for every v in V, 0+v = v+0 = v. (3) For every v in V there exists an element in V, called -v, such that v+(-v) = (-v)+v = 0. (4) For every two elements v and w of V, v+w = w+v. (5) For every two elements v,w of V, and any k in K, k*(v+w) = (k*v) + (k*w). (6) For every element v of V, 1*v = v. (7) For every element v of V and every two elements k,m of K, (k+m)*v = (k*v) + (m*v) (km)*v = k*(m*v). If you remove condition (2), then just must remove condition (3) as well (since it only makes sense in terms of condition 2), and in that case, the empty set together satisfies all conditions by vacuity (there is no element in V, so all statements of the form for all elements v of V are true). If you have a vector space V, then a subset S of V is a subspace if and only if (a) S is nonempty; (b) For any r,s in S, r+s (added as elements of V) is also in S; and (c) For any s in S and k in K, k*s (multiplied as if s is an element of V) must also be in S. One can then prove that any S that satisfies (a), (b), and (c) will also satisfy (1)-(7), by using the sum and scalar product from V applied only to elements of S. But you must assume that S is nonempty; if you do not assume S is nonempty, then the empty set satisfies conditions (b) and (c) by vacuity again, but S is not a vector space itself because it does not satisfy condition (2). You can change (a) to (a') S contains 0, because (a) and (a') are equivalent in view of condition (c), as you note. However, in order to apply condition (c) to get the 0 vector in S, you must first assume that there is SOMETHING in S, in order to say: Ah, take some v in S; then 0*v is in S, so 0 is in S. What if there is nothing in S? Then you cannot conclude that 0 is in S. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >I've noted a problem in algebraic number theory with the inclusiveness >of the definition of algebraic integers as roots of monic polynomials >with integer coefficients. Various posters have argued that in fact there is no problem, but >here's a short refutation of their primary objection. First I have P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). My finding that only two of the a's have f as a factor without regard >to the value of m has been vigorously disputed. > It's incorrect. See below. Mathematicians, you see, have other priorities than actual validity of mathematics, as they have *social* issue. See below. >However, consider w_1(m), a factor of a_1 that is a factor of f, as >well as a function that varies with m, then it follows that a_1 x + uf has w_1(m) as a factor, so dividing through by w_1(m) gives a_1 x/w_1(m) + uf/w_1(m) but then uf/w_1(m) cannot in general be an algebraic integer as it's >not representable as a polynomial with a finite number of terms if >w_1(m) varies with m. > A totally off-the-wall, unjustified statement, and, as it > so happens, incorrect. But for now, if you want to claim > it is true, the shoe is on your foot: try to prove it. I've introduced r(m), to handle the result of uf/w_1(m). So the poster is requesting that I prove that r(m) w_1(m) = uf, does not exist over the ring of algebraic integers. I've concluded that using numbers for u and f, as they are *independent* of m, helps, so let u=2, f=13. Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you have xy=26 and now I'll chat further. Then the question is, does their exist a multiplicative inverse in the ring of algebraic integers for 26/w_1(m) for *all* algebraic integers m? The simple answer is that if w_1(m) varies with m, then it must vary over an infinite number of algebraic integer values as m varies over algebraic integers. But w_1(m) must vary from 0 to infinity if it varies with m. Replies I've seen have shied away from trying something like w_1(m) = m+1 because most readers can immediately realize that 26/(m+1) or anything like it, can't be an algebraic integer for all m. Instead I've seen examples like 26^{1/m} or more trying, 26^{1/m^2+1}, but notice that because m^2+1 can equal 0, *in the ring of algebraic integers*, you can get 26^{1/0} for the m's that are the roots of m^2+1. >My guess is that some may be assuming that f is replaceable by some >function of m, but in fact, its independent of the value of m, No, we are not assuming f is replaceable by some function of m. > We assume f itself is *constant* with respect to m. However, the factorization > of a polynomial P(m) whose coefficients are functions of m, hence > dependent on m, is also in general dependent on m. We do *not* assume > that the corresponding factorization of f is constant with respect > to m. So if you had any doubts about how low mathematicians could go, consider that after all, they're trying to convince that you can have one algebraic integer function defined by the multiplicative inverse of another algebraic integer function over all algebraic integer m, like with f=7, 14/w_1(m) = r(m). It's like saying that you can have xy=2, where x and y are integers, or algebraic integers, where x varies over the ring, and y remains in it, though of course, I can just show that for x=5 that doesn't work. Hmmm...that's why posters have tried to pick w_1(m) using exponential functions. Fascinating as they're showing how much they understand. >so it's >like 1/(x+1) which is also not representable by a polynomial if x is >an algebraic integer not equal to 0 or -2. So the objection is refuted by the impossibility of uf/w_1(m) being an >algebraic integer, > Also incorrect and false. Remember what you said above: ... consider w_1(m), a factor of a_1 that is a factor of f... If w_1(m) is a factor of f, that can only mean f/w_1(m) is > an algebraic integer, which of course implies that uf/w_1(m) > is an algebraic integer: this is your *assumption* here. The assumption is itself a contradiction, in that it requires the ability to define one algebraic integer function by the multiplicative inverse of another algebraic integer function. Basic algebra. >for all algebraic integers m, if w_1(m) varies with >m. My hope is that posters who have been so successful in convincing >others that my argument is flawed will post concessions. Absolutely not! You are inching closer to the truth in this. You are seeing > how this can work. You are correct that w_1(m) is a factor of > f that depends on m. As noted above this implies that > uf/w_1(m) is an A.I., and a1(m)/w_1(m) is also. I think you > are beginning to see how this can happen. Readers should note that attempt to push the impossible, one algebraic integer function defined by the multiplicative inverse of another algebraic integer function. My guess is that seeing something like uf/w_1(m) may confuse some who might believe that u and f are functions of m, so I like to show their independence by tossing in values, like u=2, f=13, so you have 26/w_1(m). > I note a couple of other things. First, the polynomial in a > that you mention above: [#] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). Your claim is that two roots of this polynomial have a factor > that is f. This means that if r is one of those roots, then > r = f*c, where c is an algebraic integer. This implies that f^3*c^3 + 3(-1 +m*f^2)*f^2*c^2 - f^2*(m^3*f^4 - 3*m*f^2 + 3*m) = 0, or, factoring out f^2, f*c^3 + 3*(-1 + m*f^2)*c^2 - (m^3*f^4 - 3*m*f^2 + 3*m) = 0. > The polynomial in c on the left is non-monic and primitive > (if f is not a multiple of 3). If it is irreducible, then > c cannot be an algebraic integer. When f = 5 and m = 1, this > equation is 5*c^3 + 72*c^2 - 553 = 0, and the polynomial in c is easily shown to be irreducible. > Therefore c cannot be an algebraic integer. This contradicts > your central claim: NONE of the roots of [#] can have a > factor that is 5 = f. > Second note: It may be worthwhile to see how irreducibility > is inextricably tied to factorization of roots. Assume that > Q(m) = x^2 + m*x + 30, where m is an integer. The constant term is divisible by 5 (also by 2 and 3, but I > will focus on 5 here). For certain values of m, this polynomial is reducible, and the > corresponding roots are not both divisible by 5: m = 11: r1 = 5, r2 = 6 m = 13: r1 = 10, r2 = 3 m = 17: r1 = 15, r2 = 2 m = 31: r1 = 30, r2 = 1. In all these examples, obviously one root is *divisible* by 5 > and the other is *coprime* to 5. And of course in all these > examples, the polynomial is reducible. This is parallel to > Harris's cubic when m = 0: two of the roots are divisible > by f and one is coprime to f. In this particular case his polynomial > [#] is reducible. In general it is not. Now look at an other example. Say, m = 14. The > polynomial is irreducible, because the discriminant D^2 = m^ - 4*30 = 76, which is not a perfect square. The roots of the > polynomial are: > r1 = (-14 + sqrt(76))/2 = -7 + sqrt(19) and r2 = -7 - sqrt(19). Clearly both r1 and r2 are algebraic integers. Assume that r1 is a multiple of 5: r1 = 5*s1. Then 25*s1^2 + 70*s1 + 30 = 0. Factor out 5: [*] 5*s1^2 + 14*s1 + 6 = 0. This happens to have discriminant D^2 = 76. [This is > not a coincidence!]. Thus the polynomial in [*] is > *also* irreducible. Therefore s1 cannot be an algebraic > integer. Therefore r1 cannot be divisible by 5. The same can similarly be shown for r2. However, it is now easy to show that both r1 and r2 must both > be *non-coprime* to 5. For, suppose r1 is coprime to 5. > The fact that r1*r2 = 5*6 would then imply that r2 is DIVISIBLE by 5, which > contradicts the result above that both r1 and r2 are > NOT divisible by 5. Conclusion: for m = 14, the polynomial is irreducible, > and both roots have a nonunit algebraic integer factor > in common with 5. This is true more generally. There is nothing special > about 14. Try, for example, m = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, > 14, 15, 16, 18, ... MOST values of m yield an irreducible > polynomial, and both of the roots will both share algebraic > integer factors in common with 5. In general it is hard to write down exactly what these > factors are, even in the quadratic case. The important > thing to know about them here is that, as suggested above > by the w_1(m) notation, *they will be dependent on m*. What's fascinating here is how casually posters will just start talking, which I think actually usually works with a lot of readers who don't even bother to read carefully. Just remember that the base position here is that in the ring of algebraic integers you can have one function defined by the multiplicative inverse of another function. Remember xy=26. James Harris ==== >> I've noted a problem in algebraic number theory with the inclusiveness >> of the definition of algebraic integers as roots of monic polynomials >> with integer coefficients. >Various posters have argued that in fact there is no problem, but >> here's a short refutation of their primary objection. >First I have >P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) >and the factorization >P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >where the a's are given by the following cubic: >a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). >My finding that only two of the a's have f as a factor without regard >> to the value of m has been vigorously disputed. It's incorrect. See below. Mathematicians, you see, have other priorities than actual validity of > mathematics, as they have *social* issue. See below. However, consider w_1(m), a factor of a_1 that is a factor of f, as >> well as a function that varies with m, then it follows that >a_1 x + uf has w_1(m) as a factor, >so dividing through by w_1(m) gives >a_1 x/w_1(m) + uf/w_1(m) >but then uf/w_1(m) cannot in general be an algebraic integer as it's >> not representable as a polynomial with a finite number of terms if >> w_1(m) varies with m. A totally off-the-wall, unjustified statement, and, as it >so happens, incorrect. But for now, if you want to claim >it is true, the shoe is on your foot: try to prove it. I've introduced r(m), to handle the result of uf/w_1(m). So the poster is requesting that I prove that r(m) w_1(m) = uf, does not exist over the ring of algebraic integers. I've concluded that using numbers for u and f, as they are > *independent* of m, helps, so let u=2, f=13. Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you have xy=26 and now I'll chat further. Then the question is, does their exist a multiplicative inverse in the > ring of algebraic integers for 26/w_1(m) for *all* algebraic integers > m? The simple answer is that if w_1(m) varies with m, then it must vary > over an infinite number of algebraic integer values as m varies over > algebraic integers. But w_1(m) must vary from 0 to infinity if it varies with m. Replies I've seen have shied away from trying something like w_1(m) = > m+1 because most readers can immediately realize that 26/(m+1) or > anything like it, can't be an algebraic integer for all m. Instead I've seen examples like 26^{1/m} or more trying, 26^{1/m^2+1}, > but notice that because m^2+1 can equal 0, *in the ring of algebraic > integers*, you can get 26^{1/0} for the m's that are the roots of > m^2+1. My guess is that some may be assuming that f is replaceable by some >> function of m, but in fact, its independent of the value of m, No, we are not assuming f is replaceable by some function of m. >We assume f itself is *constant* with respect to m. However, the factorization >of a polynomial P(m) whose coefficients are functions of m, hence >dependent on m, is also in general dependent on m. We do *not* assume >that the corresponding factorization of f is constant with respect >to m. So if you had any doubts about how low mathematicians could go, > consider that after all, they're trying to convince that you can have > one algebraic integer function defined by the multiplicative inverse > of another algebraic integer function over all algebraic integer m, > like with f=7, 14/w_1(m) = r(m). It's like saying that you can have xy=2, where x and y are integers, > or algebraic integers, where x varies over the ring, and y remains in > it, though of course, I can just show that for x=5 that doesn't work. Hmmm...that's why posters have tried to pick w_1(m) using exponential > functions. Fascinating as they're showing how much they understand. so it's >> like 1/(x+1) which is also not representable by a polynomial if x is >> an algebraic integer not equal to 0 or -2. >So the objection is refuted by the impossibility of uf/w_1(m) being an >> algebraic integer, > Also incorrect and false. Remember what you said above: ... consider w_1(m), a factor of a_1 that is a factor of f... If w_1(m) is a factor of f, that can only mean f/w_1(m) is >an algebraic integer, which of course implies that uf/w_1(m) >is an algebraic integer: this is your *assumption* here. The assumption is itself a contradiction, in that it requires the > ability to define one algebraic integer function by the multiplicative > inverse of another algebraic integer function. Basic algebra. IF THIS IS BASIC ALGEBRA, WHY CAN'T I UNDERSTAND YOUR ARGUMENT? I UNDERSTAND ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME. for all algebraic integers m, if w_1(m) varies with >> m. >My hope is that posters who have been so successful in convincing >> others that my argument is flawed will post concessions. > Absolutely not! You are inching closer to the truth in this. You are seeing >how this can work. You are correct that w_1(m) is a factor of >f that depends on m. As noted above this implies that >uf/w_1(m) is an A.I., and a1(m)/w_1(m) is also. I think you >are beginning to see how this can happen. Readers should note that attempt to push the impossible, one algebraic > integer function defined by the multiplicative inverse of another > algebraic integer function. My guess is that seeing something like uf/w_1(m) may confuse some who > might believe that u and f are functions of m, so I like to show their > independence by tossing in values, like u=2, f=13, so you have > 26/w_1(m). I note a couple of other things. First, the polynomial in a >that you mention above: [#] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). Your claim is that two roots of this polynomial have a factor >that is f. This means that if r is one of those roots, then >r = f*c, where c is an algebraic integer. This implies that f^3*c^3 + 3(-1 +m*f^2)*f^2*c^2 - f^2*(m^3*f^4 - 3*m*f^2 + 3*m) = 0, or, factoring out f^2, f*c^3 + 3*(-1 + m*f^2)*c^2 - (m^3*f^4 - 3*m*f^2 + 3*m) = 0. >The polynomial in c on the left is non-monic and primitive >(if f is not a multiple of 3). If it is irreducible, then >c cannot be an algebraic integer. When f = 5 and m = 1, this >equation is 5*c^3 + 72*c^2 - 553 = 0, and the polynomial in c is easily shown to be irreducible. >Therefore c cannot be an algebraic integer. This contradicts >your central claim: NONE of the roots of [#] can have a >factor that is 5 = f. > Second note: It may be worthwhile to see how irreducibility >is inextricably tied to factorization of roots. Assume that > Q(m) = x^2 + m*x + 30, where m is an integer. The constant term is divisible by 5 (also by 2 and 3, but I >will focus on 5 here). For certain values of m, this polynomial is reducible, and the >corresponding roots are not both divisible by 5: m = 11: r1 = 5, r2 = 6 m = 13: r1 = 10, r2 = 3 m = 17: r1 = 15, r2 = 2 m = 31: r1 = 30, r2 = 1. In all these examples, obviously one root is *divisible* by 5 >and the other is *coprime* to 5. And of course in all these >examples, the polynomial is reducible. This is parallel to >Harris's cubic when m = 0: two of the roots are divisible >by f and one is coprime to f. In this particular case his polynomial >[#] is reducible. In general it is not. Now look at an other example. Say, m = 14. The >polynomial is irreducible, because the discriminant D^2 = m^ - 4*30 = 76, which is not a perfect square. The roots of the >polynomial are: > r1 = (-14 + sqrt(76))/2 = -7 + sqrt(19) and r2 = -7 - sqrt(19). Clearly both r1 and r2 are algebraic integers. Assume that r1 is a multiple of 5: r1 = 5*s1. Then 25*s1^2 + 70*s1 + 30 = 0. Factor out 5: [*] 5*s1^2 + 14*s1 + 6 = 0. This happens to have discriminant D^2 = 76. [This is >not a coincidence!]. Thus the polynomial in [*] is >*also* irreducible. Therefore s1 cannot be an algebraic >integer. Therefore r1 cannot be divisible by 5. The same can similarly be shown for r2. However, it is now easy to show that both r1 and r2 must both >be *non-coprime* to 5. For, suppose r1 is coprime to 5. >The fact that r1*r2 = 5*6 would then imply that r2 is DIVISIBLE by 5, which >contradicts the result above that both r1 and r2 are >NOT divisible by 5. Conclusion: for m = 14, the polynomial is irreducible, >and both roots have a nonunit algebraic integer factor >in common with 5. This is true more generally. There is nothing special >about 14. Try, for example, m = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, >14, 15, 16, 18, ... MOST values of m yield an irreducible >polynomial, and both of the roots will both share algebraic >integer factors in common with 5. In general it is hard to write down exactly what these >factors are, even in the quadratic case. The important >thing to know about them here is that, as suggested above >by the w_1(m) notation, *they will be dependent on m*. What's fascinating here is how casually posters will just start > talking, which I think actually usually works with a lot of readers > who don't even bother to read carefully. Just remember that the base position here is that in the ring of > algebraic integers you can have one function defined by the > multiplicative inverse of another function. Remember xy=26. > James Harris ==== In sci.physics, Virgil where the a's are given by the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). The above cubic non-equation doesn't give anything. Presumably, it gives three roots when set to 0, which would have to be determined using standard methods. If one assumes f and m to be integers, then the roots are algebraic integers, by definition. I do not know offhand whether the same could be said of the roots if f and m are algebraic integers (although I for one don't see why not). The product of these roots is divisible by f^2, although one has to be careful: I don't think one can conclude that two of the roots are divisible by f, as James appears to claim elsewhere. Also, weird things happen in algebraic integers regarding factorization, probably because of the proliferation of units. Taking f=3, we can factorize f^2 9 = 3^(2/3) * 3^(2/3) * 3^(2/3) as a perfectly valid product therein, and 3^(2/3) is a root of x^3 - 9 = 0 and therefore an algebraic integer. The factorization is not unique of course :-) ; one can also write 9 = 3^(1/2) * 3^(3/4) * 3^(3/4) 9 = 3 * 3 * 1 9 = 3 * 3^(1/3) * 3^(2/3) etc. Mr. Harris does like to jump blindly over crevasses, it seems; the territory should instead be carefully negotiated. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== I've noted a problem in algebraic number theory with the inclusiveness >of the definition of algebraic integers as roots of monic polynomials >with integer coefficients. Various posters have argued that in fact there is no problem, but >here's a short refutation of their primary objection. First I have P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Can't you articulate your object in English? Such as, If you define > algebraic integers as roots of monic polys with integer coefficients, > then the following odd thing happens: etc. etc. Just say exactly what > you think the contradiction or problem is. Yes! That's what I would like to hear too. It seems like he's trying to prove some theorem, but unable to formulate this theorem. ==== Cc: [.snip.] >Various posters have argued that in fact there is no problem, but >here's a short refutation of their primary objection. First I have P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). My finding that only two of the a's have f as a factor without regard >to the value of m has been vigorously disputed. However, consider w_1(m), a factor of a_1 that is a factor of f, as >well as a function that varies with m, then it follows that Factor where? I assume you mean: w_1(m) is, for each value of m, an algebraic integer which is a factor of a_1(m), and also a factor of f, both in the ring of algebraic integers. >a_1 x + uf has w_1(m) as a factor, Should be a_1(m)x + uf, but yes. >so dividing through by w_1(m) gives a_1 x/w_1(m) + uf/w_1(m) Should be a_1(m)x/w_1(m) + uf/w_1(m), so this is g_1(m)/w_1(m), yes. >but then uf/w_1(m) cannot in general be an algebraic integer (Yes, I'm interrupting mid-sentence). I'll repeat the full sentence in a second) Why not? You said that w_1(m) was an algebraic integer which is both a factor of a_1 and a factor of f. Since w_1(m) is a factor of f, f/w_1(m) is an algebraic integer. And if u is an algebraic integer, then u*[f/w_1(m)] is an algebraic integer. So this is an algebraic integer for every value of m for which the function is defined. Unless you are saying that the product of two algebraic integers is not necessarily an algebraic integer? (That's a theorem of Eisenstein, Gauss, and Dedekind)? >but then uf/w_1(m) cannot in general be an algebraic integer >as it's >not representable as a polynomial with a finite number of terms if >w_1(m) varies with m. No, this is wrong. There does not have to be a single polynomial that you can plug m into and that gives you a polynomial for w_1(m). So you just need an expression with the property that for ->each<- value of m, what you get is a monic polynomial with integer coefficients which has w_1(m) as a root. You want the VALUES uf/w_1(m) each to be an algebraic integer. That means that for EACH value of m there must be a polynomial, monic with integer coefficients, that has uf/w_1(m) as a root. There does not have to be a single polynomial, nor a single family of polynomials with easy formulas depending on m. All there has to be is for ->each<- value of m, ->some<- polynomial with integer coefficients that works for ->that<- value. The polynomials don't have to be related to one another by any one finite formula. There is no finite expression with a finite number of terms with the property that for every value of m, you get a monic polynomial with integer coefficients that has all of -|m|-1, -|m|, -|m|+1,....,0,....,|m|-2, |m|-1 as roots. But there are expressions that for each value of m, give you a monic polynomial with integer coefficients that has all of them as roots: For example, let g(n,m) = { 1 if |n|<=|m| { 0 if |n|>|m|. Then take f(m,x) = prod_{i=-infinity}^{infinity} (g(i,m)x - [g(i,m)i - 1]) where the index i ranges over all integers. Or, there is no single expression with a finite number of terms which, for each nonzero integer value of m gives you a polynomial that has the largest prime p that divides m, 1 if m=1 or -1, and 0 if m=0, as a root. But there is a single (infinite) expression that, for each value of m, gives you a polynomial that has such a root. For instance, x*(Sum (from i=0 to infinity) [m/i]prod_{j=1 to [m/i]}(x-j)) where [m/i] is the floor of m/i. There is no reason to demand that there be a single finite expression that works; such an expression would be ->sufficient<-, but not necessary, for the w_1(m) to be, each, an algebraic integer. In fact, there is no need to demand a single expression either, although it is trivial to construct one: for each value of m, let g_m(x) be a monic polynomial with integer coefficients that has w_1(m) as a root, and simply take a sum over all m's of xi_{n}*g_n(x), where xi_{n} is the characteristic function of the singleton {n}: equal to 1 when the input is n, and equal to zero otherwise. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > >>Cc: > [.snip.] >Various posters have argued that in fact there is no problem, but >here's a short refutation of their primary objection. First I have P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). My finding that only two of the a's have f as a factor without regard >to the value of m has been vigorously disputed. However, consider w_1(m), a factor of a_1 that is a factor of f, as >well as a function that varies with m, then it follows that >>Factor where? >>I assume you mean: w_1(m) is, for each value of m, an algebraic >>integer which is a factor of a_1(m), and also a factor of f, both in >>the ring of algebraic integers. > Mathematicians are disgustingly bad liars when caught, and this one is > no exception. Mathematicians are extraordinarily patient, even with those who blatantly do not deserve such consideration. Gib ==== [.ad hominem removed.] >> Should be a_1(m)x + uf, but yes. >> >>so dividing through by w_1(m) gives >>a_1 x/w_1(m) + uf/w_1(m) Should be a_1(m)x/w_1(m) + uf/w_1(m), so this is g_1(m)/w_1(m), yes. >> >>but then uf/w_1(m) cannot in general be an algebraic integer >> (Yes, I'm interrupting mid-sentence). I'll repeat the full sentence in >> a second) Why not? You said that w_1(m) was an algebraic integer which is both a >> factor of a_1 and a factor of f. That's the assertion that is shown to be impossible if w_1(m) varies >as m varies. No, you ->claim<- you show it to be impossible, but the reason you give is invalid. >It's impossible because uf is not a function of m. It's not a question of uf, it's a question of a_1(m). The factor w_1(m) depends on the value of a_1(m). >Basic, and easy stuff, but it takes a mathematician to try and lie >anyway, even when beaten. > >> Since w_1(m) is a factor of f, f/w_1(m) is an algebraic integer. And >> if u is an algebraic integer, then u*[f/w_1(m)] is an algebraic integer. So this is an algebraic integer for every value of m for which the >> function is defined. Unless you are saying that the product of two algebraic integers is >> not necessarily an algebraic integer? (That's a theorem of Eisenstein, >> Gauss, and Dedekind)? Rather than tell the truth this mathematician is trying to distract. No, I'm trying to see why you claim it is not an algebraic integer, when it is clear that the hypothesis you had up to that point gave immediately that it ->was<-. >But it doesn't take a rocket scientist to know that given an integer >f, you can't have a function w_1(m) in algebraic integers such that >for *any* integer m, f/w_1(m) is an integer. Nonsense. >Some people may be confused by the symbol f, so as it's independent >and I can give it a value, let f=13. Now then, Arturo Magidin needs you to believe that maybe there exists >a function in algebraic integers such that 13/w_1(m) is an algebraic >integer for *all* integer m. So, you are saying that there are no functions w_1(m), with domain the integers and with algebraic integer values, other than constant functions, with the property that 13/w_1(m) is an algebraic integer for each integer m? Not the function we have right now, but what is wrong with something like w_1(m) = 13^{1/(m^2+1)}? That function is defined for every integer, has algebraic integer values for every integer, is a divisor of 13 for every value of m, and is not constant. Or, since there are a countably infinite number of proper divisors of 13 in the algebraic integers, simply choose any well ordering of them and any well ordering of the integers, biject the two well-orderings, and that gives you a function which is defined for each integer m, has values in the algebraic integers, and such that 13/w_1(m) is an algebraic integer for *all* integer m. Interesting that, when I gave you a mathematical response, here: >>but then uf/w_1(m) cannot in general be an algebraic integer >>as it's >>not representable as a polynomial with a finite number of terms if >>w_1(m) varies with m. No, this is wrong. There does not have to be a single polynomial that >> you can plug m into and that gives you a polynomial for w_1(m). So >> you just need an expression with the property that for ->each<- value >> of m, what you get is a monic polynomial with integer coefficients >> which has w_1(m) as a root. You want the VALUES uf/w_1(m) each to be an algebraic integer. That >> means that for EACH value of m there must be a polynomial, monic with >> integer coefficients, that has uf/w_1(m) as a root. There does not >> have to be a single polynomial, nor a single family of polynomials >> with easy formulas depending on m. All there has to be is for ->each<- >> value of m, ->some<- polynomial with integer coefficients that works >> for ->that<- value. The polynomials don't have to be related to one >> another by any one finite formula. >> There is no finite expression with a finite number of terms with the >> property that for every value of m, you get a monic polynomial with >> integer coefficients that has all of -|m|-1, -|m|, -|m|+1,....,0,....,|m|-2, |m|-1 as roots. But there are expressions that for each value of m, give you >> a monic polynomial with integer coefficients that has all of them as roots: For example, let g(n,m) = { 1 if |n|<=|m| >> { 0 if |n|>|m|. Then take f(m,x) = prod_{i=-infinity}^{infinity} (g(i,m)x - [g(i,m)i - 1]) where the index i ranges over all integers. Or, there is no single expression with a finite number of terms which, >> for each nonzero integer value of m gives you a polynomial that has >> the largest prime p that divides m, 1 if m=1 or -1, and 0 if m=0, as a >> root. But there is a single (infinite) expression that, for each value >> of m, gives you a polynomial that has such a root. For instance, x*(Sum (from i=0 to infinity) [m/i]prod_{j=1 to [m/i]}(x-j)) where [m/i] is the floor of m/i. >> There is no reason to demand that there be a single finite expression >> that works; such an expression would be ->sufficient<-, but not >> necessary, for the w_1(m) to be, each, an algebraic integer. In fact, >> there is no need to demand a single expression either, although it is >> trivial to construct one: for each value of m, let g_m(x) be a >> monic polynomial with integer coefficients that has w_1(m) as a root, >> and simply take a sum over all m's of xi_{n}*g_n(x), where xi_{n} is >> the characteristic function of the singleton {n}: equal to 1 when the >> input is n, and equal to zero otherwise. to engage in an ad hominem, again: Readers on sci.physics and sci.logic note that the mathematicians have >been caught in a base lie trying to hide an over one hundred year old >error. >To believe them you need to believe in an impossible function which >would allow 13/w_1(m) to be an algebraic integer for ALL integer m. Why is such a function impossible? Because you couldn't think of one? >Anyone who disagrees need just give an algebraic integer function >w_1(m) such that 13/w_1(m) is an algebraic integer for all integer m. w_1(m) = 13^{1/(m^2+1)} seems to fit the bill quite nicely. And none of its values are units. >Of course, mathematicians are stepping into the problem that given >some thing like 13/w_1(m) you need, say, r_1(m) as an algebraic >integer function such that 13/w_1(m) = r_1(m), so w_1(m) r_1(m) - 13 = 0 which necessarily has zeroes. Yes, every value of m is a zero. > That is, only SOME values where it will >work. Huh? Are you under the impression that w_1(m)r_1(m) would be a polynomial in m? Why would you think that? ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >I've noted a problem in algebraic number theory with the inclusiveness >of the definition of algebraic integers as roots of monic polynomials >with integer coefficients. Various posters have argued that in fact there is no problem, but >here's a short refutation of their primary objection. First I have P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). If the a_i's depend on m, why don't you write it properly? P(m) = ( a_1(m) x + u f ) ( a_2(m) x + u f ) ( a_3(m) x + u f ) >My finding that only two of the a's have f as a factor without regard >to the value of m has been vigorously disputed. However, consider w_1(m), a factor of a_1 that is a factor of f, as >well as a function that varies with m, then it follows that a_1 x + uf has w_1(m) as a factor, i.e. w_1(m) | a_1(m) x + u f for all algebraic numbers m >so dividing through by w_1(m) gives a_1 x/w_1(m) + uf/w_1(m) but then uf/w_1(m) cannot in general be an algebraic integer as it's >not representable as a polynomial with a finite number of terms if >w_1(m) varies with m. I hope you realize that the functions a_i are also not given by polynomials in m (i.e., there does not exists a polynomial A_i(M) in A[M], the polynomial ring in one variable over the algebraic numbers, such that A_i(m) = A_i(m) for all algebraic numbers m). For the function w_1, well, the only thing you've mentioned about it is that w_1(m) | a_1(m) and w_1(m) | f (for all algebraic numbers m, I think). So how could one expect a_1(m) x / w_1(m) + u f / w_1(m), or u f / w_1(m) to have any particular form at all? Of course, for u f / w_1(m) to be an algebraic integer for all algebraic integers m, it doesn't have to be expressible as a polynomial in m over the algebraic integers. With all you assumptions, it is of course trivial that u f / w_1(m) is an algebraic integer (since f is divisible by w_1(m) in the algebraic integers). >My guess is that some may be assuming that f is replaceable by some >function of m, but in fact, its independent of the value of m, Everybody understands that, I'm sure. >so it's >like 1/(x+1) which is also not representable by a polynomial if x is >an algebraic integer not equal to 0 or -2. But I wonder who understands this. I definitely don't. (By the way, if x *is* an algebraic integer unequal to -1, then 1/(x+1) *is* expressible as a polynomial over the algebraic integers - a constant polynomial. But you probably mean that there is no polynomial F(X) over the algebraic integers such that F(x) = 1/(x+1) for all algebraic integers unequal to -1. I still wonder what 0 and -2 have to do with it - maybe just a mistake.) >So the objection is refuted by the impossibility of uf/w_1(m) being an >algebraic integer, for all algebraic integers m, if w_1(m) varies with >m. Can you repeat the definition of algebraic integer again for the newsgroup and tell us how you conclude that u f / w_1(m) is not an algebraic integer? Peter van Rossum -- -- Peter van Rossum, | Universal law of linearity: for all Dept. of Mathematics, New Mexico | f : R -> R and for all x, y in R: State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) ==== [.snip.] > >It's not possible. I'll put in values to help those readers confused >by symbols (though algebra is BASED on symbols) by letting u=2, f=13, >then you have 26/w_1(m) and you don't have to be a rocket scientist to know that no function >w_1(m) that actually varies with m can exist such that 26/w_1(m) is an >algebraic integer for *all* integer m. > w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)} I'm not saying that's the function in question (it is not), but there > is a function that actually varies with m such that 26/w_1(m) is an > algebraic integer for *all* integer m. So the argument that no such > function can exist is simply bogus. Well that example works, so now I'll say algebraic integer m, and it blows up at m^2+1 = 0. For those confused by the discussion, I've pointed out that mathematicians arguing with me have been arguing for a relation relating the multiplicative inverse of one algebraic integer function to another over all algebraic integers. For instance, with u=2, f=7 14/w_1(m) = r(m) which would relate w_1(m) to r(m) for all algebraic integer m, which is like claiming that xy=2 for all integer x, with an integer y. It's that multiplicative inverse that blows away the objection as in fact, w_1(m) is a constant with regard to m--independent of m--as I've repeatedly shown. However, accepting basic algebra would mean that mathematicians accept that there's an over one hundred year old definition problem in core mathematics. Many of you may have believed that I was making something up, but you should have enough mathematical knowledge to understand that the relations these people are arguing for, cannot exist. Unless you're lost on the example of xy=2, in the ring of integers, in comparison to 14/w_1(m) = r(m) in the ring of algebraic integers. It's that / that makes the difference. Math people are lying here because the information is so earthshaking, and they are corrupt. James Harris ==== [.snip.] >> >>It's not possible. I'll put in values to help those readers confused >>by symbols (though algebra is BASED on symbols) by letting u=2, f=13, >>then you have >>26/w_1(m) >>and you don't have to be a rocket scientist to know that no function >>w_1(m) that actually varies with m can exist such that 26/w_1(m) is an >>algebraic integer for *all* integer m. >> w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)} I'm not saying that's the function in question (it is not), but there >> is a function that actually varies with m such that 26/w_1(m) is an >> algebraic integer for *all* integer m. So the argument that no such >> function can exist is simply bogus. Well that example works, so now I'll say algebraic integer m, and it >blows up at m^2+1 = 0. So let me see if I have this right: (1) You claimed something was impossible under a given set of conditions. (2) When I proved that your claim was simply false, you changed the set of conditions, and now argue that what I said was wrong. Great. So, first of all, you were wrong. You claimed no such function could exist, but there's a function that did everything you required it to do. And second of all, rather than admit you were wrong, you just change the rules. Here's an example that works for EVERY algebraic NUMBER m: Step 1. Given an algebraic number m, let f(x) be the unique monic polynomial with rational coefficients, irreducible over Q, which has m as a root. Write it as: f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0. Note that a_0 must be different from 0. Step 2. Write a_0, a rational number, as a_0 = r/s, with r and s integers, r and s relatively prime. Step 3. If r=1 or -1, let q = 1. Step 4. If r is not equal to 1 or -1, then let q be the largest rational prime that divides r. Step 5. Let w(m) be a root of the polynomial x^4 + 13qx^3 + qx + 13. Then: (a) w(m) is an algebraic integer. (b) w(m) divides 13 in the ring of algebraic integers, since the product of all the roots is 13, and every root is an algebraic integer. (c) w(m) is not constant: it takes different values at different integers. (d) w(m) takes each value a countably infinite number of times, so w(m) cannot be given by a polynomial. [.snip.] What now, James? Are you going to change your claim again in order to avoid the fact that you were just plain wrong? >For those confused by the discussion, I've pointed out that >mathematicians arguing with me have been arguing for a relation >relating the multiplicative inverse of one algebraic integer function >to another over all algebraic integers. For those confused by the discussion, James claimed there could be NO function (other than a constant function), from the integers to the algebraic integers, with the property that at each integer we got an algebraic integer which was a divisor of 13. James was wrong. He does not like to admit he is wrong, so he's changing the subject. >For instance, with u=2, f=7 14/w_1(m) = r(m) which would relate w_1(m) to r(m) for all algebraic integer m, which >is like claiming that xy=2 for all integer x, with an integer y. No, it's not like claiming that at all. Because the integers are a UFD with a finite number of units, so there are only a finite number of ways in which each integer can be expressed as a product of integers. The algebraic integer are NOT a UFD, have an infinite number of units, and any algebraic integer can be expressed as a product of algebraic integers in an infinite number of ways. So it is not at all like claiming that. >It's that multiplicative inverse that blows away the objection as in >fact, w_1(m) is a constant with regard to m--independent of m--as I've >repeatedly shown. As you've repeatedly ->claimed<-, but as has repeatedly been shown, with EXPLICIT calculations, is not the case. [.rest deleted.] ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >> If the a_i's depend on m, why don't you write it properly? P(m) = ( a_1(m) x + u f ) ( a_2(m) x + u f ) ( a_3(m) x + u f ) That's a style issue. I see it as a gesture of futility and anguish >at the reality of this easy refutation. More an issue of clarity. It's strange to have some dependencies on m (P(m), w_1(m) later on) explicitely and others not. But I guess you're not concerened with presenting a clear argument. >> i.e. w_1(m) | a_1(m) x + u f for all algebraic numbers m >so dividing through by w_1(m) gives >>a_1 x/w_1(m) + uf/w_1(m) >>but then uf/w_1(m) cannot in general be an algebraic integer as it's >>not representable as a polynomial with a finite number of terms if >>w_1(m) varies with m. I hope you realize that the functions a_i are also not given by polynomials >> in m (i.e., there does not exists a polynomial A_i(M) in A[M], the That's a rather stupid lie given that I put the polynomial in this >post. Again it is a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) and its roots are a_1, a_2 and a_3. You seem to miss the point. Let me write it down more explicitely: there are no algebraic integers c_0, c_1, ..., c_n such that a_1(m) = c_0 + c_1 m + c_2 m^2 + ... + c_n m^n for all algebraic integers m. If you still think there are such algebraic integers, I would like to see explicitely what they are. What you have is a polynomial Q(A,M) in two variables over the algebraic integers (namely Q(A,M) = A^3 + 3(-1+Mf^2)A^2 - f^2(M^3f^4 - 3M^2f^2 + 3M)) such that Q(a_1(m),m) = 0 for all algebraic integers m. >> With all you assumptions, it is of course trivial that u f / w_1(m) >> is an algebraic integer (since f is divisible by w_1(m) in the algebraic >> integers). It's not possible. I'll put in values to help those readers confused >by symbols (though algebra is BASED on symbols) by letting u=2, f=13, >then you have 26/w_1(m) and you don't have to be a rocket scientist to know that no function >w_1(m) that actually varies with m can exist such that 26/w_1(m) is an >algebraic integer for *all* integer m. You have *assumed* that w_1(m) divides f (i.e. 13, in this case) for all algebraic integers m. So let me give a function w_1 that makes 26/w_1(m) an algebraic integer, even an integer, for all algebraic numbers m. Here it is: w_1(m) = { 1 if m = 0 { 13 if m <> 0 Of course, this is not a polynomial expression, but that is sort of the whole point. >It's a show of how broken math society is that Peter van Rossum would >dare to make that stupid assertion. Yes! I've been insulted by JSH. This finally makes my name as a >> Can you repeat the definition of algebraic integer again for >> the newsgroup and tell us how you conclude that u f / w_1(m) >> is not an algebraic integer? Mathematicians are pathetic liars. What is the lie in that question? >I'll use u=2, f=13 again, now then, NO function in algebraic integers >exists such taht 26/w_1(m) is an algebraic integer for all integers m, >if w_1(m) varies with m. You can scroll back to see a counterexample. >It's just not possible, but it takes a mathematician to lie about it. Another question, maybe easier. What is a function in algebraic integers? Peter van Rossum -- -- Peter van Rossum, | Universal law of linearity: for all Dept. of Mathematics, New Mexico | f : R -> R and for all x, y in R: State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) ==== I have a degree in physics but a career in information technology. I have always had a desire to study physics or applied mathematics to the graduate (Master's) level. I wouldn't expect this to increase my income. My desire stems from a natural curiosity; I enjoy these fields. I'm in my mid-fifties. Will my age be (perceived as) an impediment to either undergraduate or graduate studies? Any suggestions as to - how one chooses between/among programs in applied mathematics and physics? - the attitudes of people in each field? - level of effort/time required? I would appreciate the advice of anyone in a similar situation. tom ==== > I have a degree in physics but a career in information technology. I > have always had a desire to study physics or applied mathematics to > the graduate (Master's) level. I wouldn't expect this to increase my > income. My desire stems from a natural curiosity; I enjoy these > fields. I'm in my mid-fifties. Will my age be (perceived as) an impediment to > either undergraduate or graduate studies? Most definitely. As Hardy noted in A Mathematician's Apology, math is a young man's game. That doesn't mean age is an impediment to your learning. But as far as whether age will be perceived as such, of course it will. ==== > I have a degree in physics but a career in information technology. I > have always had a desire to study physics or applied mathematics to > the graduate (Master's) level. I wouldn't expect this to increase my > income. My desire stems from a natural curiosity; I enjoy these > fields. I'm in my mid-fifties. Will my age be (perceived as) an impediment to > either undergraduate or graduate studies? I don't see why it would. You had to study a lot of math to get a degree in physics so you must be capable. The only question is whether you have the concentration to do so and are willing to devote the time. I assume by this post that you are willing to devote the time? > Any suggestions as to > - how one chooses between/among programs in applied mathematics and > physics? What are your goals? What do you like/enjoy. What do you think you'd be good at? What parts have you been good at in the past? I was in a grad program but had to stop due to family illness. I may return soon depending on whether something pops up in the mean time. But things are a bit different for me since my goals have changed and things are a bit in flux as of now. Pete ==== > I have a degree in physics but a career in information technology. I > have always had a desire to study physics or applied mathematics to > the graduate (Master's) level. I wouldn't expect this to increase my > income. My desire stems from a natural curiosity; I enjoy these > fields. I'm in my mid-fifties. Will my age be (perceived as) an impediment to > either undergraduate or graduate studies? Any suggestions as to > - how one chooses between/among programs in applied mathematics and > physics? > - the attitudes of people in each field? > - level of effort/time required? I would appreciate the advice of anyone in a similar situation. tom Following my retirement I earned my master's degree in mathematics (number theory) at age 57 and my doctorate at age 62. Everyone has been most friendly and helpful. This was a full time retirement project for me and I continue to go to the university a couple of days a week to attend seminars and pursue my research interests. I count it as one of the most rewarding experiences of my life. The main attitudes I met were friendly curiosity and some good humoured envy that I didn't have to go job hunting, apply for grants, teach, publish etc etc unless I wanted to. The effort required was considerable and I don't think I could have done it on a part time basis while holding down a full time job. Of course, many people are better organized (and less lazy :) ) than I am. Jack Fearnley ==== I've been trying to understand this concept both as it applies to our elementary understanding of (a,b), and the more general meaning in topology. I have reached the following conclusions: 1.- The two concepts agree when the topology involved is that of all open subsets of R. That is any open set in the topology is open in the usual, elementary sense, that is the set is a neighborhood of all its points. Even here, however, the null set and R are both open and closed. 2.- For other topologies (say the collection of all subsets of R)a set may be open according to one definition, closed according to the other, or even both open and closed. The 2 definitions are therefore unrelated. Am I understanding this stuff correctly? There must be some general utility in defining open sets as the elements of a properly defined topology. Can someone please outline the reason? Many thanks. ==== >I've been trying to understand this concept both as it applies to our >elementary understanding of (a,b), and the more general meaning in >topology. I have reached the following conclusions: >1.- The two concepts agree when the topology involved is that of all >open subsets of R. That is any open set in the topology is open in >the usual, elementary sense, that is the set is a neighborhood of all >its points. Even here, however, the null set and R are both open and >closed. 2.- For other topologies (say the collection of all subsets of R)a set >may be open according to one definition, closed according to the >other, or even both open and closed. The 2 definitions are therefore >unrelated. Am I understanding this stuff correctly? Yes. Open sets in topology are an abstraction. The idea in topology is to define the notion of nearness or neighborhood. An open set containing x contains the points that are near x; how near depends on the open set. There are many ways of defining nearness, which translate into different topologies. The usual topology in R is related to the concept of nearness derived from a metric, which is a measure of distance: the distance between two real numbers x and y is |x-y|. Open sets are (ultimately) defined in terms of this distance. But this is not the only way to define nearness. You could say that you do not want two real numbers to be thought of as near each other unless they are the same real number: that gives you the discrete topology, where every set is open. Or you could say that any two real numbers are near each other, which gives the indiscrete topology (the only nonempty open set is the entire set). Or perhaps you are concerned with finiteness in some sense, so you define nearness in terms of being elements of a set whose complement is finite. >There must be some general utility in defining open sets as the >elements of a properly defined topology. What is a properly defined topology? A topology on a set is a collection T of subsets that satisfies the following properties: (1) The empty set is in T. (2) The full set is in T. (3) If A and B are in T then A intersect B is in T. (4) If {A_i}_i in I is a family of sets, and each A_i is in T, then the union of the A_i is also in T. Any collection of subsets that satisfies these 4 properties is a properly defined topology on the set. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== For those of you trying to keep up with the mathematical facts in the discussions about the error in core mathematics from a problem with a definition, this post will outline the important ones quickly and succinctly. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes numbers that have to be included to keep from having contradiction i.e. mathematical inconsistency. 2. The important tool I use is a polynomial: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) The form of the polynomial allows me to factor P(m) into non-polynomial factors, and the factorization with those factors is P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are roots of the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). 3. Dispute centers around what happens when I divide P(m) by f^2, which you'll note is a factor of the polynomial in the ring of algebraic integers. 4. Mathematicians have argued that f^2 divides off as a function of m because if they concede that it divides off independent of m, then I can show that only two of the roots of a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) have f as a factor. 5. However, it turns out that if you go to the field of algebraic numbers you can prove that for *certain* values of m and f, the roots of the cubic do not have f as a factor *in the ring of algebraic numbers* which is the inconsistency. That is, for the math to be consistent, two of the roots *should* have f as a factor as long as m and f are algebraic integers, but while I can show they do for a particular values like m=1, f=sqrt(2), there are other values you can show they do not *in the ring of algebraic integers* which results from the definition and its focus on monic polynomials. Note: In the ring of algebraic integers you can't see the problem but have to go to the field of algebraic numbers as from within the ring of algebraic integers it appears that only two of the roots have a factor that is f. James Harris ==== > For those of you trying to keep up with the mathematical facts in the > discussions about the error in core mathematics from a problem with a > definition, this post will outline the important ones quickly and > succinctly. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. 2. The important tool I use is a polynomial: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) The form of the polynomial allows me to factor P(m) into > non-polynomial factors, and the factorization with those factors is P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are roots of the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). 3. Dispute centers around what happens when I divide P(m) by f^2, > which you'll note is a factor of the polynomial in the ring of > algebraic integers. 4. Mathematicians have argued that f^2 divides off as a function of m > because if they concede that it divides off independent of m, then I > can show that only two of the roots of a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) have f as a factor. 5. However, it turns out that if you go to the field of algebraic > numbers you can prove that for *certain* values of m and f, the roots > of the cubic do not have f as a factor *in the ring of algebraic > numbers* which is the inconsistency. That is, for the math to be consistent, two of the roots *should* have > f as a factor as long as m and f are algebraic integers, but while I > can show they do for a particular values like m=1, f=sqrt(2), there > are other values you can show they do not *in the ring of algebraic > integers* which results from the definition and its focus on monic > polynomials. Note: In the ring of algebraic integers you can't see the problem but > have to go to the field of algebraic numbers as from within the ring > of algebraic integers it appears that only two of the roots have a > factor that is f. > James Harris Is it me or can anyone follow what JSH is ranting about? David Moran ==== Is it me or can anyone follow what JSH is ranting about? David Moran See: http://www.google.com/search?q=%22James+Harris%22+site%3Awww.crank.net http://www.crank.net/harris.html ==== >For those of you trying to keep up with the mathematical facts in the >discussions about the error in core mathematics from a problem with a >definition, this post will outline the important ones quickly and >succinctly. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with >integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where >monic refers to the leading coefficient. My assertion is that the over hundred year old definition excludes >numbers that have to be included to keep from having contradiction >i.e. mathematical inconsistency. 2. The important tool I use is a polynomial: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) The form of the polynomial allows me to factor P(m) into >non-polynomial factors, and the factorization with those factors is P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are roots of the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). 3. Dispute centers around what happens when I divide P(m) by f^2, >which you'll note is a factor of the polynomial in the ring of >algebraic integers. 4. Mathematicians have argued that f^2 divides off as a function of m >because if they concede that it divides off independent of m, then I >can show that only two of the roots of a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) have f as a factor. It's not a question of conceding. It is simply not true that the factors have the property you claim they do. We've given explicit examples with explicit calculations where they do not have the property you claim they do. Your argument about why they must divide off independent of m has not been understood or accepted by anyone that I am aware of other than you. >5. However, it turns out that if you go to the field of algebraic >numbers you can prove that for *certain* values of m and f, the roots >of the cubic do not have f as a factor *in the ring of algebraic >numbers* which is the inconsistency. No, this is not a correct summary of the situation, for various reasons: (1) In the ring of algebraic numbers, every number other than zero is a factor of every other algebraic number: the algebraic numbers form a field. So it is impossible to say the roots of the cubic do not have f as a factor *in the ring of algebraic numbers*. Nobody has made such a claim. (2) There is no need to go to the field of algebraic numbers; for every value of m for which the cubic x^3 + 3(-1+mf^2)x^2 - f^2(m^3f^4 - 3m^2f^2 + 3m) is irreducible over Q, one can construct the minimal polynomial of a/f, where a is any root of f; this only requires you to go to the ring A[1/f], where A is the ring of all algebraic integers, and monic and irreducible with integer coefficients, it follows that cannot be a factor of any root in the ring of algebraic INTEGERS. >That is, for the math to be consistent, two of the roots *should* have >f as a factor as long as m and f are algebraic integers, ONLY if your argument were valid, which nobody but you believes to be the case. > but while I >can show they do for a particular values like m=1, f=sqrt(2), there >are other values you can show they do not *in the ring of algebraic >integers* which results from the definition and its focus on monic >polynomials. Which is logical nonsense. A definition cannot introduce a contradiction, since a definition is simply shorthand for an expression. If instead of defining algebraic integers we simply talked about roots of monic polynomials with integer coefficients, there would be absolutely no change in mathematics (except that all number theory books would be longer). So IF there were a contradiction, it cannot be because of the DEFINITION, it would have to be because some property that people have claimed the roots of monic polynomials with integer coefficients have, in actuality they do not have. You have not identified any such property, and therefore you have not identified any problems (since you have not establish the validity of your argument about the divisibility of the coefficients). ====================================================================== Arturo Magidin magidin@math.berkeley.edu