mm- Subject: prove that Sup(AUB)=Sup(SupA,SupB) ,where A,B are subsets of real nos. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9IHDIx12881; any one know the proof of this prob please tell me if A,B are subsets of real nos.then prove that Sup(AUB)=Sup(SupA,SupB) waiting ans === Subject: Re: How to Ŝnd Fourier Transform for special functions -- how to evaluate exp(j*2*pi*inf)? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9IMkpw10093; >Hi all, >I am trying to understand the MathWorld derivation of FT for a special >function Heaviside step function: >http://mathworld.wolfram.com/ FourierTransformHeavisideStepFunction.html >Using bruteforce derivation, I can only get to: >1/(-j*2*pi*f)*(exp(-j*2*pi*f*(+inf) - 1) >I donıt see how the MathWorld get the Delta function out from this >derivation. >Any thoughts? I tried breaking e^(-j*2*PI*kx) into cos (2*PI*kx) - jsin (2*PI*kx), in which the imaginary component goes away since sin(2*PI) = 0 The cosine component should evaluate to 1 though. =/ === Subject: Re: e & pi Help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9IMkq710129; >I have seen on wolfram math the following serie >sum_{n>0}(1/n^2)cos(u(n)) = - (pi)^2/(12.e^3) >where u(n) = 9/(n.pi + sqrt( (n^2.(pi)^2 - 9) ) >On wolfram they said Gosper has proven this >--------------------------------------------------- >1/ Any idea how to prove the sum of the serie is - (pi)^2/(12.e^3) ? >2/ Any idea of the reference where I can Ŝnd the proof of Gosper ? >thx Did you Google for: Gosper proof === Subject: Re: Skolemıs Paradox and why is math the way it is? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9JCdKr24921; >> >> Are you trying to tell me something about glue-ons? >> Oh no, theyıre gluons. >> Youıre telling me a gluon is a kinetic energy store sitting in the >> middle of a molecular nucleus? >A nucleus has gluons, with much potential and kinetic energy, >regardless of whether the nucleus is part of a molecule, or just a >part of a free meson or a free proton or similar. >> That, and I canıt change the atomic nucleus without freeing a lot of >> energy? >Thatıs a loose, statement, and not one Iıd make personally. There is >some big energies involved, so there is room to release a fair amount, >but you could excite a quark which would heat the inside of the >nucleus up, but not so much that anything big happens. So Iıd >disagree with that universal claim. >> Personally, thereıs not much transmutation going on around here. >> There is probably normal background radiation: normal because I have >> absolutely no idea what it is. There is probably signiŜcant >> electromagnetic interference because I am typing this on a computer. >Nuclei repel each over over large distances, and only attract at short >distances, the odds of two getting close enough for the gluons in one >to have a fair chance of reaching the other nucleus is pretty rare. >You have more to worry about from ionizing radiation making chemical >(molecular) changes in the bonds between atoms, then in physics >allowing changes in the atoms themselves. >Yes, bosons even. >> The kinetic energy, thatıs great. It goes from a little packet of >> gluons to all the way out. Itıs kinetic energy! >Gluons are like very strong springs, but they attach to quarks. Itıs >pretty hard to get the internal energy into externally useable kinetic >energy. But it happens in the sun. Thatıs where the heat and >radiation comes from. When the hydrogen gas was less dense and was >cool, the energy was mostly potential and kinetic energy inside the >nucleus. It still is now, but a bit of energy has leaked out. >> See, then I want to use that theory in the inertial capacitor. Itıs >> like: a self-winding watch, so you could carry your packpack around >> all day, and it would carry itself around the rest of the day. The >> Inertial Capacitor, it stores energy just like an electrical >> capacitor, but kinetic energy. Itıs like a Perpetual Motion Machine, >> except you can power it by exploding gluons in a gluon reactor. >Have you ever studied thermodynamics? I donıt have high hopes about >self-charging things unless there is an incoming source of energy. >And most matter is pretty stable in itıs present situation. >> Of course it helps if you have a gluon reactor that doesnıt require a >> 200 kiloton explosion to get a gluon. Iım ignorant, what are the >> mechanics of working with gluons? >I donıt know what you mean by a gluon reactor. Liberating internal >energy isnıt easy, and most all youıll likely make is heat. So if you >started doing a good job, then youıd get really hot, which is hard to >turn into useful work. In a power plant, they keep the reactions from >going too quickly, because heating water into steam turns turbines, >but breaking water into hydrogen and oxygen atoms and ionizing the >atoms is harder to turn into useful work. So Iıd have doubts about >the practicality of your pack too. >I think all connections to math were abandoned with your last post, so >I think you should either e-mail me privately or start a thread >somewhere other than sci.math >J.E. My Pen! Have you seen my pen? My Pen! My Pen! That was from a skit from Kids in the hall. A salesman walks off with the guyıs pen, and the guy chasing his pen gets dragged four blocks down the street on a taxi. So he gets it back and attaches it to a tether to his bodycast. All the time he is shouting My Pen, my Pen! I donıt think there is a perpetual motion machine, except youıre basically telling me a gluon is like a spring. Just before you were saying the constant g varies with kinetic energy storage. You get some nano on that and it grows like crazy. Now, there is the kinetic energy, what are your other kinds of energy? If you knew that, then please tell me what they are, J.E. I donıt disagree with the standard model of physics. I actually donıt. They measured all of those things with very precise instruments. Every day, scientiŜc instruments collect zillions of gigabytes of data. That could be put to good use, you can req uest from NASA all the physics and scientiŜc instrument data you want. They are busy handling that scientiŜc data. Go to NASA and read their papers! I must say, Ken, itıs been a while. The self-winding watch, itıs not perpetual motion, it just takes some of the extra energy you donıt need. reach, it is a joke. Virgil thinks itıs funny anytime I attempt to extend mathematics beyond my reach. Donıt make me drag some other troll in here. Here, Iıll try and address some of Ken arguments: we mourn your frog, Ken. So long, frog. Vaya con Dios! Thatıs from when I explained to Ken what death was. Have you ever heard any dead frog jokes? The frog, it goes in the blender, and its red and green all over. That was the funniest joke of its time. Hey, Iım a stochastic chemist. Frogs arenıt poisonous, mostly. I like to think I keep dangerous animals. I borrowed a cat for a couple years: the cat, itıs a bird murderer. I call it the dang Bird Murderer, thatıs my cat. The cat, itıs got claws. I kept it indoors. Its skin was getting blistered from the sun through its fur. It was probably mostly scabs. It was worse years ago, the cat would kill a bird or squirrel every day for a week. Thatıs when I started calling it the Bird Murderer: Alex, Captain Dog. Itıs a cat. Ca t in blender: not funny. Anyways, thatıs enough about trolling and Ken. How do I make a green electric car? It takes all this metal work to convert a gasoline car to an all-electric car with a gasoline generator in the trunk. You get the electric motor in there, then it needs a whole new gearbox. Then it needs pumps and fa ns for all the vacuum systems and the things. I read about them in my Bosch Automotive Handbook. I have to recommend this mechanic to you. J.E., youıre a troll. I think you should study space groups. Hey, now Iım studying space groups, too. Ross F. === Subject: Re: Probability by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9JKXWL08803; >I was studying for my economics statistics class and I was having trouble with this question. I was wondering if you could help me out. >Approximately 5% of the bolts coming off a production line have serious defects. If two bolts are randomly selected for inspection, Ŝnd the expected value and variance of the number of inspected bolts with serious defects. [ Let X equal the number of bolts without defects. Calculate E(X) and Var(X).] >zgall1 Show how far you have gotten. === Subject: Re: Divisors and degree of a point (or place) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9KDoYn29145; >> The degree of O_P/P over K as a vector space (i.e. the degree of the >> Ŝeld extension) is meant here. >> The transcendence degree of that extension is always equal to 0 >> - can you verify this? >Yes, because the elements of O_P/P are rational functions evaluated at >the point P, so f(P)/g(P) where P has co-ordinates in K, also have >co-ordinates in K making the transcendence degree 0. Am i right or is >there a mistake somewhere? P need not be of degree 1. In general O_P is a localization of a Ŝnitely generated K-algebra A with respect to a prime ideal q. A has Krull dimension 1 because we are considering a curve. Therefore the prime ideal q is a maximal ideal, that is A/q is a Ŝeld. A/q is a Ŝnitely generated K-algebra. Hence by the Noether normalization theorem A/q|K is a Ŝnite extension of Ŝelds. Since A/q=O_P we are done. === Subject: Re: how many points the two curves will meet? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9KDoYr29153; >y(x) = x^12 >g(x) = 2^x >The question is how many points the two curves will meet. >I tried to use natural log and e to solve it, but I donıt see how they are >going help me. >x^12 = 2^x >12 ln(x) = x ln(2) >ln(x) / x = ln(2) / 12 >I guess, there must be trick that I need to use, can anyone give me some >hints? You might want to use that + both functions are strictly monotonic in the intervals [0 inŜnity) and (-inŜnity, 0], + an exponential function is >growing quicker< than every polynomial function. H === Subject: Re: GRE Math Subject Test by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9L0cDe22738; > Yall, > About how many questions does one need to get right to > get a 990? about 40 >HMH >hop@cableone.net === Subject: proof by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9L3qrE05886; A number x is called a Ŝxed point f if f(x)=x. Prove that if fı(x) not equal to 1 for all real numbers x, then f has at most one Ŝxed point. === Subject: Re: proof === Subject: proof >A number x is called a Ŝxed point f if f(x)=x. Prove that if fı(x) >not equal to 1 for all real numbers x, then f has at most one Ŝxed >point. If f(x) = x, f(y) = y, then some z between x,y with (f(x) - f(y))/(x - y) = fı(z) = 1 ---- === Subject: Re: proof === > Subject: proof >A number x is called a Ŝxed point f if f(x)=x. Prove that if fı(x) >not equal to 1 for all real numbers x, then f has at most one Ŝxed >point. > If f(x) = x, f(y) = y, then some z between x,y with > (f(x) - f(y))/(x - y) = fı(z) = 1 > ---- Oh, really, is this true? === Subject: Re: proof >>A number x is called a Ŝxed point f if f(x)=x. Prove that if fı(x) >>not equal to 1 for all real numbers x, then f has at most one Ŝxed >>point. >> If f(x) = x, f(y) = y, then some z between x,y with >> (f(x) - f(y))/(x - y) = fı(z) = 1 >Oh, really, is this true? Itıs called the Mean Value Theorem. Yes, itıs true _if_ you assume f is differentiable everywhere. Something that the OP should have mentioned (perhaps he considers that saying fı(x) <> 1 implies that fı(x) exists). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: proof > A number x is called a Ŝxed point f if f(x)=x. Prove that if fı(x) not equal > to 1 for all real numbers x, then f has at most one Ŝxed point. May one presume that fı(x) not equal to 1 for all real numbers x means the same as fı(x) not equal to 1 for any real number x ? === Subject: Re: proof > A number x is called a Ŝxed point f if f(x)=x. Prove that if fı(x) not equal to 1 for all real numbers x, then f has at most one Ŝxed point. If there are two Ŝxed points, how can you evaluate the difference of the function values? Karl === Subject: ADV: Staff Announcement boundary=.4.CCB..CE.DF4FC_96CDB by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i9L8r3b29258; by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id i9L8pPX2026506; ------------------------------------------------------------- -------- Attention All School Staff: Teachers, Students and Faculty Members: Through a special arrangement, Avtech Direct is offering a limited allotment of BRAND NEW, top of-the-line, name-brand desktop computers at more than 50% off MSRP to all Teachers, Students,Faculty and Staff, All desktop computers are brand-new packed in their original boxes, and come with a full manufacturerıs warranty plus a 100% satisfaction guarantee. These professional grade Desktops are fully equipped with 2005 next generation technology, making these the best performing computers money can buy. 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Ask for Department C. Call Avtech Direct Visit our website at http://www.avtechdirect-education.com If you wish to unsubscribe from this list, please go to http://www.computeradvice.org/unsubscribelink.asp Avtech Direct 22647 Ventura Blvd. Suite 374 Woodland Hills, CA 91364 --.4.CCB..CE.DF4FC 96CDB-- === Subject: Re: Invertable Matrix Product by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LKNGU25956; >Pavel Jiranek escribi.97: >> Hmm, this proof and also some previous ones have one little hidden >> assuption, i.e that inverse of A exists. >I donıt thnk so. Se below. >> But this cannot be asumed if >> we have to prove it. >> Pavel >> ... lower >> Ignacio Larrosa Ca.96estro > Nick Heathman escribi.97: >> How do I prove that if A and B are square n x n matrices and AB and >> B are invertable that A must also be invertable? > (A*B)^(-1)*(A*B) = I > Multiply by B^(-1) at right and by B at left, >> look here: >> in the above equation, how did you arrange it to get B*I*B^(-1)? >How I said: >Multiply by B^(-1) at right and by B at left, >> you cannot use (AB)^(-1)=B^(-1)A^(-1), because it involves just what >> is to be proven. >I didnıt use it! >> the same here. >Let C = B*(A*B)^(-1). The last equation if then is C*A = I . Then A^(-1) = C >by deŜnition. >-- >Ignacio Larrosa Ca.96estro >A Coru.96a (Espa.96a) >ilarrosaQUITARMAYUSCULAS@mundo-r.com You have proved that CA = I, but you did not show that AC = I. You should at least point out to the OP that you left this part as an exercise for the interested reader. - MO === Subject: Re: Invertable Matrix Product X-RFC2646: Format=Flowed; Original Michael Orion escribi.97: >> ... >> Let C = B*(A*B)^(-1). The last equation if then is C*A = I . Then >> A^(-1) = C by deŜnition. >> -- >> Ignacio Larrosa Ca.96estro >> A Coru.96a (Espa.96a) >> ilarrosaQUITARMAYUSCULAS@mundo-r.com > You have proved that CA = I, but you did not show that AC = I. You > should at least point out to the OP that you left this part as an > exercise for the interested reader. But this is pretty obvious: A*C = A*(B*(A*B)^(-1)) = (A*B)*(A*B)^(-1) = I -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Correspondence with journals by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9M278322540; > [.snip.] >> I got an acknowledgement of the receipt from the journal, but >> might be ignored. >If the journal were tired of you, they would simply tell you they felt >the paper was nor appropriate. They will not simply ignore it, >though they might decide not to referee it. >> I have been waiting for the report about two months. >Then, Iım sorry to tell you, but CHILL OUT! Two months is ->nothing<-. >Especially considering you are talking about late August to late >October: that is the busiest time for any professional mathematician >in academia. It is unlikely that a referee would have gotten to your >paper carefully in those two months. >UNLESS you sent it to a journal that promises very quick review >->explicitly<- (like the Autralian Bulletin), then you are being >unreasonable in sending inquiries about it. That is probably the >reason you have received no response from the editor: if you are >already pestering him after only two months have gone by, he is >probably getting annoyed with you personally and your irrational >impatience. >Feel free to circulate preprints or send them to mathematicians who >work in the area, but my strong advice to you is to sit on your hands >and let the editor and referee do their work. You submitted it in, >apparently, mid-August. Unless the journal promises quick review, I >personally would not send an inquiry before August 2005; and I think >that anything under six months is completely unreasonable, even for a >short paper. >On the other hand, if you are that impatient, then withdraw the paper >formally and send it to a journal that specializes in short papers and >promises quick decision; be warned, though: most such journals will >reject out of hand more than half the papers they receive, precisely >because they donıt have time to review them. The Australian Bulletin >does that, for example: they only send about one third of the papers >out for review, the rest simply get rejected unlooked at (and >identiŜed as such in the rejection). >Arturo Magidin, sans .sig Guess what? Your message convinced me to make a contact with the last journal to which I had submitted the paper Ŝrst. If I do not get a response, I will call them! I hope they email back, for I can not do conversations well on the phone. Well, six months. I must pay full attention to physics classes to get rid of dreamy thought (while having fun with math posting). Oh, got to go to my beloved school. I really really thank yıall for your advices. === Subject: Re: how many points the two curves will meet? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9M277522488; Virgil http://mathforum.org/discuss/sci.math/m/645446/645773 >> There are times when rough graphing is not enough. Just the >> mere look at those functions can tell you that 2^x is exponential >> and x^12 is polynomial. Sooner or later the former will beat the >> latter. But it is beyond rough graphing since their y-values >> are way up the roof. (Exercise: If you can mark the x value on >> a sheet of normal paper (letter/A4) around where will the y value >> be at their leftmost intersection?) > Right! I didnıt graph far enough. there is a third point of > intersection between x = 74 and x = 75. Iım sure the intent of the original problem was to illustrate that the smallest positive solution (which a graphing calculator easily shows) canıt be the only positive solution because of growth rate considerations. However, in thinking about this, it occurs to me that weıre looking for the zeros of 2^x - x^12, and this can be factored into more calculator friendly pieces: 2^x - x^12 = ( 2^(x/2) + x^6 ) * ( 2^(x/2) - x^6 ) = ( 2^(x/2) + x^6 ) * ( 2^(x/4) + x^3 ) * ( 2^(x/4) - x^3 ). The factor 2^(x/2) + x^6 has no zeros: positive + nonnegative will never be zero. The factor 2^(x/4) + x^3 has one zero: consider rough sketches of y = 2^(x/4) and y = -x^3. Finally, the factor 2^(x/4) - x^3 has two zeros which are easier to Ŝnd by calculator experimentation than is the case for 2^x - x^12. But I suspect that anyone who thinks of this is also going to know the exponential will catch back up to the polynomial, and so I donıt think my observation about factoring will keep this problem from being a way of forcing students to think about some important concepts in order to solve the problem. Dave L. Renfro === Subject: geometry question: meteorites striking a planet This question came up in another forum and it was debated ad inŜnitum. Itıs not an astronomy question really, but a pure geometry one (as youıll see with all the unrealistic assumptions). Imagine you have a perfectly spherical body in space (like a planet or moon), and it is sitting still (not rotating or revolving). Meteorites, travelling in random directions, might hit it. (also assume the meteorites are travelling in straight lines and arenıt inŝuenced by the gravity of the our little sphere) Given that the meteorites are as likely to be travelling at any angle, or through any point, as any other....what percentage of the meteorites that end up hitting the sphere should strike the surface at greater than 45 degrees, and what percentage at less than 45 degrees? One person said it should be exactly 50-50 (based on the ratio of the areas of the ŝattened projections of the greater and less than 45 degree regions), another said it was closer to 30-70 (based on the surface areas of those regions). Please set us straight. === Subject: Re: geometry question: meteorites striking a planet ETAtAhR+8VZALN36NtHlVwIgc+91u+ob9QIVAKh9DrG7CygiQ/ AFyUoh7en9B00M For me itıs 50-50. Imagine that instead of a point, the meteorites are hitting a small target area on the planet. The probability that a strike in this area is from a given direction (or thereabouts) is proportional to the projection of this target into the direction -- but that projection is not uniform form all available directions! Itıs the full area from the vertical direction but tends towards zero as one approaches the horizon. Its manitude is proportional to the conine of the angle from vertical. In a model based on striking a localized area, one must take into account this cosine law. With this, one gets the same result as if there were a uniform ŝux onto a projected area -- 50-50. --OL === Subject: Re: geometry question: meteorites striking a planet >Imagine you have a perfectly spherical body in space (like a planet or >moon), and it is sitting still (not rotating or revolving). >Meteorites, travelling in random directions, might hit it. (also >assume the meteorites are travelling in straight lines and arenıt >inŝuenced by the gravity of the our little sphere) >Given that the meteorites are as likely to be travelling at any angle, >or through any point, as any other....what percentage of the >meteorites that end up hitting the sphere should strike the surface at >greater than 45 degrees, and what percentage at less than 45 degrees? Consider meteors travelling in a particular direction, passing through a plane which is perpendicular to that direction. They must be uniformly distributed throughout that plane, right? If you donıt accept this, you wonıt agree with my answer. Now consider the particular plane which contains the center of the sphere. If the point at which the line of travel intersects the plane is within R*cos(45) of the center, the angle of impact is greater than 45. pi*(R*cos(45))^2/(pi*R^2) = 1/2 >One person said it should be exactly 50-50 (based on the ratio of the >areas of the ŝattened projections of the greater and less than 45 >degree regions), another said it was closer to 30-70 (based on the >surface areas of those regions). The 30-70 results are probably based on an error. Even though itıs true that direction is uniformly distrubuted, and strike location is uniformly distributed, once you select a particular strike location, your direction is no longer uniformly distributed. Iıd expect more of the high-angle meteors to survive the atmosphere, because itıs effectively thicker for the lower angles. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: geometry question: meteorites striking a planet posting-account=orx6HA0AAADbxD9-hUZ7JNAj9WLHTq0N unfortunately, those on the other side still donıt get it (back in the original discussion). Quite frustrating, especially when debating someone who has shown himself to be very sharp in the past. If anyone is interested, here are the original discussions. (I am brak, SciLite is the one I am debating with). If anyone wants to dive in and help me, feel free! :) Original question: http://forums.craigslist.org/?ID=20211237 A new thread I started, using a spherical water tower being rained upon to try to clarify and illustrate my approach:: http://forums.craigslist.org/?ID=20389330 === Subject: Re: geometry question: meteorites striking a planet Ok, Ŝrst lets simplify things: suppose that the meteorites are only coming in from one direction, with uniform density. We can pull this simpliŜcation off since the original problem is just the sum over directions of the simpliŜcation. My Ŝrst thought was of course 70-30 must be about right, since the glancing meteorites have more surface area to hit. But wait! We also need to take into account the density of these glancing hits. So suppose the meteorites are all moving parallel to the north-south axis, and x is the angle of inclination from the equator to the north pole. Then a thin ring at angle x around the globe has area 2 pi cos(x) dx and the strikes per unit area along that ring are proportional to sin x, so the total strikes is proportional to 2 pi cos(x) sin(x) dx = pi sin(2x) dx. So, the total number of strikes with angle between 0 and 45 degrees is the integral from 0 to pi/4 of pi sin(2x) dx, which is equal to pi/2. Similarly, the total strikes with angle between 45 and 90 degrees is the integral from pi/4 to pi/2 of the same, which is *also* equal to pi/2! Turns out the 50-50 guy was right! Matt === Subject: Re: geometry question: meteorites striking a planet charset=iso-8859-1 > This question came up in another forum and it was debated ad > inŜnitum. Itıs not an astronomy question really, but a pure geometry > one (as youıll see with all the unrealistic assumptions). > Imagine you have a perfectly spherical body in space (like a planet or > moon), and it is sitting still (not rotating or revolving). > Meteorites, travelling in random directions, might hit it. (also > assume the meteorites are travelling in straight lines and arenıt > inŝuenced by the gravity of the our little sphere) > Given that the meteorites are as likely to be travelling at any angle, > or through any point, as any other....what percentage of the > meteorites that end up hitting the sphere should strike the surface at > greater than 45 degrees, and what percentage at less than 45 degrees? > One person said it should be exactly 50-50 (based on the ratio of the > areas of the ŝattened projections of the greater and less than 45 > degree regions), another said it was closer to 30-70 (based on the > surface areas of those regions). Since everything is assumed uniformly random, without loss of generality, we can consider just one point on the surface and assume that point is at the origin of our coordinate system. You canıt say that itıs 50-50 because itıs a three-dimensional problem which means there are really only two dimensions to consider, which are the two angles that deŜne the direction of any line in 3D space. Imagine a large sphere surrounding out point and radial lines from our point to every point on the sphere. The only lines that really represent meteors that hit the surface are from the upper hemisphere because everything from the lower one is blocked by some other part of the earth. That means we only have to consider half of the surface of the surrounding sphere. Now the meteors that strike the surface at an angle of 45 degrees or greater pass through the circular area on the surrounding sphere which is traced out as the area of revolution of the eighth of a circle in one plane extending from directly overhead downward by 45 degrees. This is a spherical circle around the north pole whose angular extend it 45 degrees. Equivalently it is the entire part of the surrounding sphere for which the latitude is 45 degrees or greater. Meteors that pass through the part of the surrounding hemisphere for which the latitude is less than 45 degrees radially strike our inŜnitesimal patch of earth at an angle less than 45 degrees in some plane. You then just have to compute the ratio of these two areas of a hemisphere to get the ratio of meteors youıre looking for. But itıs obvious that itıs less than 50-50. Norm === Subject: Re: geometry question: meteorites striking a planet > The only lines that really represent meteors that hit the surface are from > the upper hemisphere because everything from the lower one is blocked by > some other part of the earth. That means we only have to consider half of > the surface of the surrounding sphere. Now the meteors that strike the > surface at an angle of 45 degrees or greater pass through the circular > area on the surrounding sphere which is traced out as the area of revolution > of the eighth of a circle in one plane extending from directly overhead > downward by 45 degrees. This is a spherical circle around the north pole > whose angular extend it 45 degrees. Equivalently it is the entire part of > the surrounding sphere for which the latitude is 45 degrees or greater. > Meteors that pass through the part of the surrounding hemisphere for which > the latitude is less than 45 degrees radially strike our inŜnitesimal patch > of earth at an angle less than 45 degrees in some plane. You then just have > to compute the ratio of these two areas of a hemisphere to get the ratio of > meteors youıre looking for. But itıs obvious that itıs less than 50-50. This ratio is (1-1/sqrt(2)):1 which is approximately the 30-70 distribution the OP mentioned. igor === Subject: Re: geometry question: meteorites striking a planet > Since everything is assumed uniformly random, without loss of generality, we > can consider just one point on the surface and assume that point is at the > origin of our coordinate system. ... > ... Meteors that pass through the part of the surrounding hemisphere for which > the latitude is less than 45 degrees radially strike our inŜnitesimal patch > of earth at an angle less than 45 degrees in some plane. ... The correct answer is 50-50. The correct method is to consider the projection of a hemisphere onto a plane. Your method gets the wrong answer because your inŜnitessimal patch, which is approximately a disc orientated in a particular way, CANNOT be shrunk to a point with no orientation. === Subject: Re: geometry question: meteorites striking a planet > This question came up in another forum and it was debated ad > inŜnitum. Itıs not an astronomy question really, but a pure geometry > one (as youıll see with all the unrealistic assumptions). > Imagine you have a perfectly spherical body in space (like a planet or > moon), and it is sitting still (not rotating or revolving). > Meteorites, travelling in random directions, might hit it. (also > assume the meteorites are travelling in straight lines and arenıt > inŝuenced by the gravity of the our little sphere) > Given that the meteorites are as likely to be travelling at any angle, > or through any point, as any other....what percentage of the > meteorites that end up hitting the sphere should strike the surface at > greater than 45 degrees, and what percentage at less than 45 degrees? I think I misunderstood this question the Ŝrst time I read it. Are you asking: (a) Look at the angle of impact of the meteorites, from 0 = grazing to 90 = straight down. What percentage strike at greater than 45 degrees? However, on rereading the paragraph below, I think you mean (b) Assign a north and south pole. What percentage strike above 45 degrees north latitude or below 45 south? > One person said it should be exactly 50-50 (based on the ratio of the > areas of the ŝattened projections of the greater and less than 45 > degree regions), another said it was closer to 30-70 (based on the > surface areas of those regions). > Please set us straight. There is a symmetry argument here too. Every point on the planet sees an identical distribution of meteorite strikes. So the number striking each square km per unit time is identical all over the planet. Thus the surface area argument is correct, though I donıt know off hand if the ratio of surface areas is actually 30/70. Letıs see: Area of a spherical cap of height h, for a sphere of radius r, is 2*pi*r*h. A 45 degree cap has h = (1 - cos 45)r = 0.293r. So the area of one cap is 2*pi*r*(0.293r), compared to 2*pi*r^2 for an entire hemisphere. Your 30% Ŝgure is correct. (Actually 29.3%). - Randy === Subject: Re: Aleph One Sets |DeŜne an equivalence class on sets of reals as follows: R1 ~~ R2 if |there exists an order-preserving bijection between R1 and R2. What is |the cardinality of the set of equivalence classes of ~~? |Hereıs what Iım wondering: Can you actually embed 2^{2^aleph_0}? |I donıt immediately see why or why not. I have a simpler argument that the answer is yes. There are 2^c sets of reals that are dense in the reals. The order-preserving bijections between dense sets of reals are all restrictions of continuous functions from the reals to the reals. Since continuous functions can be determined by their values on the rationals, there are c^{aleph-0} = c continuous functions. If x<=2^c and x*c>=2^c, it must be that x=2^c. I donıt know whether I needed the axiom of choice here. Keith Ramsay === Subject: Re: Aleph One Sets > |DeŜne an equivalence class on sets of reals as follows: R1 ~~ R2 if > |there exists an order-preserving bijection between R1 and R2. What is > |the cardinality of the set of equivalence classes of ~~? > |Hereıs what Iım wondering: Can you actually embed 2^{2^aleph_0}? > |I donıt immediately see why or why not. > I have a simpler argument that the answer is yes. > There are 2^c sets of reals that are dense in the reals. The > order-preserving bijections between dense sets of reals are > all restrictions of continuous functions from the reals to > the reals. Since continuous functions can be determined by their > values on the rationals, there are c^{aleph-0} = c continuous > functions. If x<=2^c and x*c>=2^c, it must be that x=2^c. I > donıt know whether I needed the axiom of choice here. For this argument you do, yes. Essentially youıre arguing that each Wadge class has cardinality at most c and there are 2^c representatives, so there must be at least 2^c Wadge classes. But in a model of AD, the Wadge classes can be wellordered whereas P(R) certainly cannot, so there canıt possibly be an injection from P(R) into the collection of Wadge classes. === Subject: Re: Aleph One Sets >> There are 2^c sets of reals that are dense in the reals. The >> order-preserving bijections between dense sets of reals are >> all restrictions of continuous functions from the reals to >> the reals. Since continuous functions can be determined by their >> values on the rationals, there are c^{aleph-0} = c continuous >> functions. If x<=2^c and x*c>=2^c, it must be that x=2^c. I >> donıt know whether I needed the axiom of choice here. > For this argument you do, yes. Essentially youıre arguing > that each Wadge class has cardinality at most c and there > are 2^c representatives, so there must be at least 2^c > Wadge classes. > But in a model of AD, the Wadge classes can be wellordered > whereas P(R) certainly cannot, so there canıt possibly > be an injection from P(R) into the collection of Wadge classes. So on second thought maybe thereıs a little work left to do here -- the Wadge stuff doesnıt work quite as smoothly on the real reals. I canıt believe it matters much; my argument *would* work if you were looking at dense subsets of the irrationals, and thereıs only a countable set distinguishing them from the reals--but I donıt instantly see how to Ŝnish the argument off. === Subject: need help on a problem... I did cut out some of what was said. The original problem was: Let phi:G-->H be a surjective homomorphism with G Ŝnite. Suppose h belongs to H has an order that is a power of the prime p. Show that G has an element g of p-power order with phi(g)=h. === >Subject: Re: need help on a problem.... >Well, is a cyclic subgroup of order k p^a, so what are the orders of >the elements it contains? >Can you Ŝnd an element g in which satisŜes two criteria: >1) Its order is p^b for some b? Note: p^a may not be the best choice. >2) phi(g)=h. >If you need another hint, let me know! >Justin Well, wouldnıt the orders of the elements divide the order of ? But still, can I get that phi(g)=h from that where g is a power of a prime. Certainly x is and it seems logical that g should as well..but do you practically just CHOOSE g from and set it equal? I must be missing something. === Subject: need help on a problem... === >Subject: Re: need help on a problem.... >Well, is a cyclic subgroup of order k p^a, so what are the orders of >the elements it contains? >Can you Ŝnd an element g in which satisŜes two criteria: >1) Its order is p^b for some b? Note: p^a may not be the best choice. >2) phi(g)=h. >If you need another hint, let me know! >Justin Well, wouldnıt the orders of the elements divide the order of ? But still, can I get that phi(g)=h from that where g is a power of a prime. Certainly x is and it seems logical that g should as well..but do you practically just CHOOSE g from and set it equal? I must be missing something. === Subject: Re: need help on a problem.... === >Subject: Re: need help on a problem.... >Well, is a cyclic subgroup of order k p^a, so what are the orders of >the elements it contains? >Can you Ŝnd an element g in which satisŜes two criteria: >1) Its order is p^b for some b? Note: p^a may not be the best choice. >2) phi(g)=h. >If you need another hint, let me know! >Justin Well, wouldnıt the orders of the elements divide the order of ? But still, can I get that phi(g)=h from that where g is a power of a prime. Certainly x is and it seems logical that g should as well..but do you practically just CHOOSE g from and set it equal? I must be missing something. === Subject: Re: need help on a problem.... === >Subject: Re: need help on a problem.... > Well, wouldnıt the orders of the elements divide the order of ? But still, > can I get that phi(g)=h from that where g is a power of a prime. Certainly x g hasnıt to be the power of p; his order has to. > is and it seems logical that g should as well..but do you practically just > CHOOSE g from and set it equal? I must be missing something. Ok, letıs be a little bit more precise : since phi(x) = h, the order of h divides that of x (you seem to agree). Hence o(x) = o(h)*p^m*q whith p and q coprime (thus o(h) and q are coprime too). Choose integers a and b s.t. a*o(h) + b*q = 1 and set g = x^(b*q). Now check : phi(g) = phi(x)^(b*q) = h^(1-a*o(h)) = h Furthermore g^(o(h)*p^m) = x^(b*q*o(h)*p^m) = x^(b*o(x)) = 1^b = 1 Hence g is of order dividing o(h)*p^m, and since you assumed h to have order a power of p... (complete) Does it help ? Hib. === Subject: Re: need help on a problem.... === >Subject: Re: need help on a problem.... >Well, is a cyclic subgroup of order k p^a, so what are the orders of >the elements it contains? >Can you Ŝnd an element g in which satisŜes two criteria: >1) Its order is p^b for some b? Note: p^a may not be the best choice. >2) phi(g)=h. >If you need another hint, let me know! >Justin Well, wouldnıt the orders of the elements divide the order of ? But still, can I get that phi(g)=h from that where g is a power of a prime. Certainly x is and it seems logical that g should as well..but do you practically just CHOOSE g from and set it equal? I must be missing something. === Subject: Re: attn: Herc : Hilbertıs Hotel (inŜnity) In sci.math, Phil Scott >> In sci.math, David Bernier >> >> : >> The mathematician David Hilbert conceived of a Grand Hotel with >> as many rooms as there are positive integers. >> >> This fanciful Ŝction is sometimes used to illustrate how Ŝnite and >> inŜnite sets differ in what many would call counter-intuitive ways. It >> also goes under the name of Hilbertıs Hotel. >> >> hereıs a link to it: >> >> http://en.wikipedia.org/wiki/Hilbertıs_paradox_of_the_Grand_ Hotel >> >> David Bernier >> Great for inŜnitely large conventions. :-) > Sadly, it wasnıt. Every time a new guest arrived, they made me change rooms. Itıs a workaround for a bug. :-) -- #191, ewill3@earthlink.net Itıs still legal to go .sigless. === Subject: Re: attn: Herc : Hilbertıs Hotel (inŜnity) >> Great for inŜnitely large conventions. :-) > Sadly, it wasnıt. Every time a new guest arrived, they made me change > rooms. If you had bribed the man at the desk, he might have been persuaded only to move those in rooms numbered with opposite parity to the number of your room. === Subject: attn: Herc : Hilbertıs Hotel (inŜnity) <4gnm42-7ut.ln1@sirius.athghost7038suus.net> >> Great for inŜnitely large conventions. :-) > > Sadly, it wasnıt. Every time a new guest arrived, they made me change > rooms. > If you had bribed the man at the desk, he might have been persuaded only > to move those in rooms numbered with opposite parity to the number of > your room. Naw, maybe other parity also bribes desk clerk, so what then? Bribe the desk clerk to move everybody, starting with the successor room to yours. Thatıll work nicely even if two or more bribe the desk clerk, much to his liking because heıll get n bribes and only have to do one as promised. Now in the event inŜnitely many guests bribe the clerk and inŜnitely many donıt, the clerk is inŜnitely rich unless he so dumb as to not set minimum bribe. What, should it occur, you advise the clerk if inŜnitely many bribe him and Ŝnitely many donıt? Take the money and run? -- Now I digress to ask when the hotel is full, how much money is collected, how many clerks are needed to check in all the guests, how many maids, janitors etc will the hotel employ and where will all these workers stay? Come dinner time, how many cooks will be needed, how big is the stove, the stack of dirty dishes, the garbage can and how many truck loads of food should the cooks order? What trucking company has enuf trucks, how much gas will be needed for a complete delivery, how much farmland will be needed to grow the food? -- Thus the need to occupy inŜnite dimension hyperspace where a mega buck is pocket change and aleph_0 a dollar bill. Just imagine, with liberation of such inŜnite resources, the nation debt limit could be raised to aleph_omega0! But do take note, from where the inŜnitely many guests in hotel inŜnity? Have they come from overpopulation Earth containing a mere several giga-people? No, theyıre natives of inŜnite dimension hyperspace. So check in Earthling, thereıll always be room for another. BTW, how long is the hotel bar? ---- === Subject: Re: attn: Herc : Hilbertıs Hotel (inŜnity) > In sci.math, Phil Scott > >> In sci.math, David Bernier >> >> : >> The mathematician David Hilbert conceived of a Grand Hotel with >> as many rooms as there are positive integers. >> >> This fanciful Ŝction is sometimes used to illustrate how Ŝnite and >> inŜnite sets differ in what many would call counter-intuitive ways. It >> also goes under the name of Hilbertıs Hotel. >> >> hereıs a link to it: >> >> http://en.wikipedia.org/wiki/Hilbertıs_paradox_of_the_Grand_ Hotel >> >> David Bernier >> >> Great for inŜnitely large conventions. :-) > Sadly, it wasnıt. Every time a new guest arrived, they made me change rooms. > Itıs a workaround for a bug. :-) So where do you suggest the bug is in ? The set {Z} ? :-) === Subject: Re: attn: Herc : Hilbertıs Hotel (inŜnity) <4gnm42-7ut.ln1@sirius.athghost7038suus.net> >> In sci.math, Phil Scott >> > In sci.math, David Bernier > > : > The mathematician David Hilbert conceived of a Grand Hotel with > as many rooms as there are positive integers. > > This fanciful Ŝction is sometimes used to illustrate how Ŝnite and > inŜnite sets differ in what many would call counter-intuitive ways. It > also goes under the name of Hilbertıs Hotel. > > hereıs a link to it: > > http://en.wikipedia.org/wiki/Hilbertıs_paradox_of_the_Grand_ Hotel > > David Bernier > > Great for inŜnitely large conventions. :-) >> >> Sadly, it wasnıt. Every time a new guest arrived, they made me >>change rooms. >> Itıs a workaround for a bug. :-) >So where do you suggest the bug is in ? The set {Z} ? :-) In the management of the hotel. They should never have let it get full in the Ŝrst place. Given that they had done, they should have grasped the nettle and cleared inŜnitely many rooms the Ŝrst time (moved everyone up, room n to room 2n). Then, with better room assignment, no-one would have to move again. -- David Hartley === Subject: A question about Hypergeometric Function Recently Iım puzzled by a question about Hypergeometric Function. 2F1(a,b;c;z) is deŜned as 2F1(a,b;c;z)=sum_{n=0}^{+infty}frac{(a)_n(b)_nz^n}{(c)_n n!} The question is to prove the formula: 2F1(a,b;c;z)=(1-z)^{-a}2F1(a,c-b;c;z/(z-1)). It also can be found in the following website: http://mathworld.wolfram.com/HypergeometricFunction.html equation(32) (33) (34). === Subject: Re: simple birth-death-process question >> Assume I have a population of size one, and starting at time zero I >> have a simple birth-process going, with birth rate b*n, where n is the >> size of the population and b is some intensity. At time T I observe >> the population size to be two. What is the probability that the birth >> event that must have occured, occured between time 0 and t (t <= T)? >> > This is simply the conditional probability that an exponential r.v. is > less than t given that it is less than T, i.e., > [1 - exp(-b t)] / [1 - exp(-b T)]. Though I stand by my answer, I am not sure ther derivation is as simple as I claimed. Let X1 and X2 be the respective epochs of the Ŝrst and second birth. We want P(X1 <= t | X1 <= T, X1 + X2 > T) = P{X1 <= t, X1 + X2 > T} / P{X1 <= T, X1 + X2 > T} P{X1 <= s, X1 + X2 > T} = EP(X1 <= s, X1 + X2 > T | X1) = int(x=0..s, P{X2 > T - x} b exp(-b x)) we just used the independence of X1 and X2 there) = b int(x=0..s, exp(-2b(T-x)) exp(-b x)). = exp(-2b T) [exp(b s) -1]. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: simple birth-death-process question [Sheesh! Follow-ups to myself two deep. I hope someone else is reading.] > Assume I have a population of size one, and starting at time zero I > have a simple birth-process going, with birth rate b*n, where n is the > size of the population and b is some intensity. At time T I observe > the population size to be two. What is the probability that the birth > event that must have occured, occured between time 0 and t (t <= T)? > >> This is simply the conditional probability that an exponential r.v. >> is less than t given that it is less than T, i.e., >> [1 - exp(-b t)] / [1 - exp(-b T)]. > Though I stand by my answer, I am not sure ther derivation is as > simple as I claimed. Let X1 and X2 be the respective epochs of > the Ŝrst and second birth. We want Correction: X2 is the interarrival time betwee the Ŝrst and second births. > P(X1 <= t | X1 <= T, X1 + X2 > T) = P{X1 <= t, X1 + X2 > T} / > P{X1 <= T, X1 + X2 > T} > P{X1 <= s, X1 + X2 > T} = EP(X1 <= s, X1 + X2 > T | X1) = > int(x=0..s, P{X2 > T - x} b exp(-b x)) > we just used the independence of X1 and X2 there) > = b int(x=0..s, exp(-2b(T-x)) exp(-b x)). = exp(-2b T) [exp(b s) -1]. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: solution of equation (difŜcult) In sci.math, William Elliot > >6^(2x) - 2(6^x) + 2^x = 0. >> >> Well, you can divide through by 2^x (since 2^x canıt be 0) and get >> >> (6^x)*3^x - 2*3^x + 1 = 0 > 18^x - 2*3^x + 1 = 0 >> Write e^x = y or x = ln y; >> therefore 6^x = y * ln 6 and 3^x = y * ln 3. > Huh? I donıt get it. > 6^x = e^(x.log 6) = (e^x)^log 6 = y^log 6 > (y^log 6)(y^log 3) - 2 * y^log 3 + 1 = 0 > y^log 18 - 2 * y^log 3 + 1 = 0 >> Should make it almost trivial. > It does? Bleah...thatıll teach me to post before Iıve had my coffee! :-) As it is, if one takes 6^(2*x) - 2*(6^x) + 2^x = 0 one can still write e^x = y, and y^(ln 36) - 2*y^(ln 6) + y^(ln 2) = 0 Divide by y^(ln 2), and one gets y^(ln 18) - 2*y^(ln 3) + 1 = 0 which is where you are as well. At this point one might have to use numerical approximation methods. -- #191, ewill3@earthlink.net Itıs still legal to go .sigless. === Subject: Re: how many points the two curves will meet? >> y(x) = x^12 >> g(x) = 2^x >> The question is how many points the two curves will meet. >:) My error, there are at least 3 roots, one more near x=75 ... Then, >what are roots of f(x)= x^n - m^x ? n,m positive integers, say n > m >coming out of Lambertıs W ? > Lambertıs W is deŜned by W(z) exp(W(z)) = z. So writing > x = -n W(z)/ln(m) we have m^x = exp(-n W(z)) = W(z)^n z^(-n) = x^n iff > W(z)/z = x r where r is an nth root of 1, i.e. z = -ln(m)/(n r). > Thus the roots of x^n - m^x are x = -n W(-ln(m)/(n r))/ln(m) where > r is an nth root of 1. Note also that W has different branches, > written in Maple as LambertW(k,x) for integers k, resulting in inŜnitely > many complex roots. The principal (k=0) branch of W(z) is real for > -exp(-1) < z < inŜnity, and the k=-1 branch is real for -exp(-1) < z < 0. > If 1 < m < exp(n/e), you get two real solutions for r=1: > x = -n LambertW(0, -ln(m)/n)/ln(m) > x = -n LambertW(-1, -ln(m)/n)/ln(m) > If n is even, for any m > 1 you get another real solution for r=-1: > x = -n LambertW(0, ln(m)/n)/ln(m) > In your case (n=12, m=2), these three real solutions are approximately > 1.063346831, 74.66932553, -.9467803304 respectively. > On the other hand, for n=2, m=12 where m > exp(n/e) there is only the one > real solution x = -n LambertW(0, ln(m)/n)/ln(m) which is approximately > -.5224836624. Thorough to Ŝnish line, last word (nail) in the thread (box). === Subject: Re: how many points the two curves will meet? >y = x^12 is like a weird parabola, sort of U shaped, with vertex at >origin and symmetry about the y axis. y = x^k when k is an even integer are all like a weird parabola, ŝatter between -1 < x < 1 and steeper outside that region. y = x^k when k is an odd integer all look like y = x^3, but as k get bigger they get ŝatter in -1 < x < 1 and steeper outside that region. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: The habitation of eigenvalues. >>Iıve been examining eigenvalues recently. >>Insofar as my readings have told me, eigenvalues at least exist when the >>transformation space is over a Ŝeld, as per linear algebra. >>Can eigenvalues be deŜned for transformations over say, an algebra or a >> ring? > Recall the standard deŜnition: > If V is a vector space over a Ŝeld F, and T : V --> V is > a linear transformation, then whenever we have a k in F and > a nonzero v in V for which T( v ) = k v then we say v is an > eigenvector and k is the corresponding eigenvalue. > Now, how would you generalize all this when F is replaced by some other > kind of structure? If F is any ring, we can speak of F-modules V ; > vector spaces are precisely what that concept reduces to when the ring > is a Ŝeld. We can speak of F-module homomorphisms; when F is a Ŝeld, > these are precisely the linear transformations. So I suppose it would > be natural to carry over the deŜnitions of eigenvector and > eigenvalue into this more general setting: if T( v ) = k v then > I suppose we could call v an eigenvector and k an eigenvalue. > There are a couple of things that make this less inspired than some > other generalizations in mathematics. One is that in the most general > situation, these new thingies are not nearly as useful as they are in > the Ŝeld case. Well, lack of utility has never prevented mathematicians > from studying things, eh? > But another problem is that, while the deŜnition is easy enough to > extend, the proofs of key theorems are not. Even the simplest of statements > need not be true in the general setting. For example, one might hope > that for a given eigenvector there is a well-deŜned eigenvalue that > goes with it. But that doesnıt have to be true at all: we can have > k v = kı v with k different from kı (nonzero elements of a module > can easily have annihilators). > So people will extend the deŜnition of eigenvalues only in fairly modest > ways: they might be used if F is a division ring, say, or if V is > a free module, or if T is a semi-linear transformation (relative to > some automorphism of F), etc. : when you donıt go TOO far from the > classical setting, thereıs more reason to hope that something like the > usual results will continue to hold. > (On the other extreme, you could go hog-wild with your generalization; > all you need is a setting in which there is some kind of action > F x V --> V, which includes e.g. the cases of groups F acting on > sets V. In that case eigenvector just means an element of v which > V moves to something in the same F-orbit as v. You might try to > see why that interpretation is valid in the ordinary vector-space > setting too!) > dave Hmm... Let me explain a bit more about where I am going with this idea. Iım helping a guy out in some theoretical research, and we are measuring things that live in a topological at the moment, and then searching for intersections of these objects, after we have applied a boolean predicate to them. The general thrust with the eigen-things, is to try and home in on the intersection of the eigenvectors to locate the intersection easier. We seek to make the intersection a single entity. Essentially, the property we are needing is the scaling property of the eigenvector on the space. We know that the transformations that the eigen* would be describing are non-linear, probably in the extreme. Does that help? -paul === Subject: Re: sampling question X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ c92823270dd8d4b8bd5279d2715bad a1.48257%40mygate.mailgate.org > I suggest a point estimate of the number of unique items > in the population is something like N1/(n1/n2). And, I > suppose, conŜdence intervals can be developed under certain > assumptions. Well, my eyes crossed a couple of times, staring at that, but I think I agree that would be a pretty good estimator. > This idea has applications in computer testing where the > data from two independent testing groups can be used to > estimate the number of remaining bugs in a program. Or > you can use bug seeding and one testing group. I remember long ago seeing the second of those proposals discussed in Software Engineering Notes; the Ŝrst follows fairly directly. xanthian. Itıs also proved to be the case in bug testing, though, as in writing independent implementations of code to meet identical functional requirements, that merely sharing a culture of software development destroys any hypothesis that the testing / development bugs will be independent in any statistically useful sense. We know too many things from the same sources to ever behave really independently. -- === Subject: Re: sampling question X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ a2f2ac4cbe037382ac672088e1412a b8.48257%40mygate.mailgate.org ->> If you are sampling with replacement, from a ->> population of unknown size, and your only metric ->> is the number of times you have now encountered ->> the most recent newly-seen item, how many times ->> do you have to see it to have a 0.999 chance that ->> you have seen _all_ the items in the population ->> at least once? -> To the problem as stated, there is no answer. Donıt you hate it when that happens? -> Given a population of size N, if v_1, v_2, ..., -> v_N are the items in the order they are Ŝrst -> seen, let A_j be the number of times you see v_j -> before seeing v_{j+1}. Then A_j has a geometric -> distribution with p=(N-j)/(N-j+1). To see this, -> note that each drawing that is either v_j or -> something new has probability 1/(N-j+1) of being -> something new (since there is only one v_j and N-j -> something new, each equally likely to appear). -> So you can say (for m,y >= 1) P(A_j = y | N-j = m) -> = m/(m+1)^y and P(A_j >= y | N-j = m) = -> (m+1)^(1-y). This also works for m=0 in which -> case we say A_j = inŜnity. -> To turn it around and ask for the conditional -> probability that j=N given that I have seen the -> latest new item y times (so I know A_j >= y but -> nothing else), you canıt do unless you have a -> prior (unconditional) probability for j=N to use -> in Bayesıs Theorem. What I donıt know about statistics these many years after I studied it would easily Ŝll a for Dummies book. so be patient with me. Consider the situation where I am ŝipping a fair coin, testing the hypothesis this coin has two identical faces (rather than two heads, since I want to come up with your M=11, below). The Ŝrst ŝip is a gimme, the next 10 ŝips matching the Ŝrst ŝip reduce the odds that the coin has a face of an unseen style by 1/2 each time, to 1/1024. Compared to ignoring all results except your v_j above, and some new item, how is this different? Seems like, if there were more than one unseen item remaining, the odds would only worsten of seeing 11 identical items in a row (ignoring all weıve seen but v_j). So why, again, do I need to know some independent measure of N=j to reject that N > j? Am I just confused on terminology, or is there something deeper happening? -> But you can state it another way, in terms of -> hypothesis testing (although I think itıs a -> somewhat unusual sort of hypothesis testing). Her -> the null hypothesis is j < N, the alternative -> hypothesis is j = N. You will want to reject the -> null hypothesis at level 0.001 if A_j >= M for -> some M. This will work if 2^(1-M) <= 0.001, i.e. -> M >= log_2(2000) = 10.96.... So M = 11 will do. -> That is, whenever j < N, the probability that -> youıll see v_j 11 times without seeing a new item -> is less than 0.001. -> But I repeat, rejecting j < N at a conŜdence -> level of .001 is not the same thing as a -> probability of .999 that j = N. Well, at least I now know that giving up after Ŝve appearances of the newest item was way too soon! xanthian. -- === Subject: [OT sniping]: ModiŜcations under way X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 3c6c4dd44d789bf69dfbfb628819b7 54.48257%40mygate.mailgate.org I really try to keep out of this, since despite Jamesı complete incompetence at the level of math where he is attempting to contribute, his math skills are by now probably far better than mine, but the occasional snipe just seems called for anyway. > Well I didnıt see a lot of agreement with my > compromise offer, though I didnıt read every post, > as it doesnıt matter: itıs just a good idea. Do I get the impression that James is still trying to do math using political tools instead of correct proofs? Slowness of learning seems endemic among James Harrises. > Itıs time for me to get back to work. Do I get the impression that James is still making just one more Ŝx to his proof that he has been pursuing for the three to Ŝve years Iıve been watching this rolling Ŝasco, and threatening mayhem all that time against those brave enough to point out his errors? That James feels unappreciated on Usenet could have nothing at all to do with his behavior _being_ unappreciated on Usenet, or anywhere else math is done by competent practitioners? Oh, my. xanthian. -- === Subject: Re: FBI SADISM, PERVERSION, MENTAL TORTURE and BLATANT human rights violations posting-account=9MkUpQ0AAACWJBqyEY45GLc_5tXnk7Us Have yourself checked into a mental facility before you hurt yourself or someone else you sick ! Minds that have deteriorated to the level of yours disgust me. Iım only taking the time to post this because I have a sister who was once in a condition similar to yours and I took the appropriate measures to help her, and that was to help her realize that she was too sick to realize or believe that there was a problem with herself and that she needed help. I got her that help, and now, years later, you would never even had imagined that her head was once so screwed up. It wouldnıt surprise me if you were to percieve *me* to be some stupid FBI character in your own head, never do anything to help yourself, and end up murdering an innocent person that you swore with all your heart to be an undercover FBI agent, if you havenıt done so already. If you ever consider to do something to help yourself then I congratulate you, I even apologize. If you never do anything on the otherhand, I spit in your face, because you are a danger to society, and itıs biggest burden. Sicko. === Subject: Re: FBI SADISM, PERVERSION, MENTAL TORTURE and BLATANT human rights violations > Have yourself checked into a mental facility before you hurt yourself > or someone else you sick ! Minds that have deteriorated to the > level of yours disgust me. Iım only taking the time to post this > because I have a sister who was once in a condition similar to yours > and I took the appropriate measures to help her, and that was to help > her realize that she was too sick to realize or believe that there was > a problem with herself and that she needed help. I got her that help, > and now, years later, you would never even had imagined that her head > was once so screwed up. It wouldnıt surprise me if you were to percieve > *me* to be some stupid FBI character in your own head, never do > anything to help yourself, and end up murdering an innocent person that > you swore with all your heart to be an undercover FBI agent, if you > havenıt done so already. If you ever consider to do something to help > yourself then I congratulate you, I even apologize. If you never do > anything on the otherhand, I spit in your face, because you are a > danger to society, and itıs biggest burden. Sicko. My goodness gracious! Someone else has had to deal with the problem I currently grapple with. Real nice to know that when Iım ready to spill my guts, thereıll be a receptive audience :-) Mark (... itıs not _me_, itıs _other_ people thatıre the problem! Ahahahahahaha. Reg. Trademark -- Hanson, Inc. By permission.) === Subject: Re: FBI SADISM, PERVERSION, MENTAL TORTURE and BLATANT human rights violations charset=iso-8859-1 > Have yourself checked into a mental facility before you hurt yourself > or someone else you sick ! Minds that have deteriorated to the > level of yours disgust me. Iım only taking the time to post this > because I have a sister who was once in a condition similar to yours > and I took the appropriate measures to help her, and that was to help > her realize that she was too sick to realize or believe that there was > a problem with herself and that she needed help. I got her that help, > and now, years later, you would never even had imagined that her head > was once so screwed up. It wouldnıt surprise me if you were to percieve > *me* to be some stupid FBI character in your own head, never do > anything to help yourself, and end up murdering an innocent person that > you swore with all your heart to be an undercover FBI agent, if you > havenıt done so already. If you ever consider to do something to help > yourself then I congratulate you, I even apologize. If you never do > anything on the otherhand, I spit in your face, because you are a > danger to society, and itıs biggest burden. Sicko. [Mark] > My goodness gracious! Someone else has > had to deal with the problem I currently > grapple with. Real nice to know that > when Iım ready to spill my guts, thereıll > be a receptive audience :-) > Mark (... itıs not _me_, itıs _other_ people > thatıre the problem! Ahahahahahaha. > Reg. Trademark -- Hanson, Inc. By > permission.) [hanson] I see, you wish to be a client of ours, Mr. Tarka. Very good! Note, out of sheer piety and respect we do address our patients as clients or customers, (fee scale dependent, of course) Furthermore, dear Mr. Tarka, let me correct you, if I may: Itıs not Hanson, Inc. By permission. ....although you will need permission to get administering from and by our treatment facilities, which are simply known as Ravencrag. As the administrator, I will not be able to greet you, but just kindly follow the gentlemen in the white coats who carry those dripping and rusty bicycle pumps around. Here are your application forms for your self-commitment into Ravencrag: http://www.marvunapp.com/Appendix/cobrajal.htm#Hanson http://www.marvunapp.com/Appendix/cobrajal.htm#Ravencrag sincerely, ahahahaha......ahahahahanson === Subject: Re: Harris Ŝlters Chairman of the Ozzy Osbourne Appreciation Society > [...] uniquely identiŜed by a reference ID, which in > the case of James, always starts with the string > 3c65f87.. If you can Ŝlter for that string [...] Your karma must have just skyrocketed. Walter. === Subject: Re: ? -- JSH related > [Tim Peters] > donıt hesitate to start a thread about a math topic that interests > you, > even if you feel itıs elementary. Any on-topic post should be > welcomed. > For example, insist that even numbers are all divisible by triangles, > and > then Ŝght unbelievers to the death . > Hopefully, my next post will be on topic and pusillanimous enough to > avoid a deathmatch. Unfortunately it mostly depends on whether or not you go with the group or against the group. If you go against the group you will probably be insulted into either conforming or shutting up. > He says he has a bachelorıs degree in physics (and my best guess is > that he > does), but I havenıt seen an academic claim beyond that. If the math > is > good, or at least interestingly wrong, nobody here really cares what > anyone > else does for a living, and itıs impolite to ask computer > geek>. > Youıre right about it being impolite to ask. I really shouldnıt have. > Because heıd almost been published in a journal (no small feat), I had > the notion that Mr. Harris was a professor somewhere but wasnıt sure. > intriguing, especially when found within a subject commonly thought to > be dry. The real story is that, yes, I naively and arrogantly thought I could Ŝnd a short and hopefully simple proof of some great problem, and I spent a lot of time on FLT and spherical packing. Unfortunately, Usenet was not the place to talk about such efforts. I didnıt matter that I said I was an amateur. It didnıt matter that I admitted my mistakes or said it was mostly just a lark. The people here wanted to control my postings, and when I wouldnıt be controlled they got vicious in ever more creative ways. They are vicious. At the end of the day, what Iıve done is talk about various math ideas, sure often arrogantly, at times with claims that turned out to be wrong--which Iıd admit when I Ŝgured it out. In response--on a supposedly *public* forum--Iıve been hounded for years by people calling me crazy, idiot, fool, loon, and those are the nice names. Race was brought into it by David Ullrich, and when I complained to his school, as he is a math professor at Oklahoma State University, I got called racial slurs. When I Ŝnally got a result accepted at a math journal which claims to do peer review, sci.mathıers gang emailed to get it yanked. And they still stalk my postings. The warning to others out there is that Usenet is a place where people can come after you, and they will. And it wonıt just stay on Usenet necessarily, as clearly posters here have been trying to Ŝnd out personal information about me, like where I work, and live. And you have the case of Erik Max Francis with his webpage, still up there as it appears he wishes to keep it up indeŜnitely, and Dik Winter with his copyright violations. The lesson is clear: post on Usenet at your own risk, and donıt be surprised at what happens, how energetic people may be in coming after you, how long they may keep at it, and what lengths they may go to to try and take you out, if you donıt obey what *they* think is the proper behavior. Usenet is now a place where gangs of posters routinely interfere in other peopleıs lives, and then blame that person. James Harris === Subject: Re: ? -- JSH related !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ıELIi $t^ VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Hopefully, my next post will be on topic and pusillanimous enough to >> avoid a deathmatch. > Unfortunately it mostly depends on whether or not you go with the > group or against the group. It depends much more on whether or not you go with group theory or against it. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: ? -- JSH related >[...] >The real story is that, yes, I naively and arrogantly thought I could >Ŝnd a short and hopefully simple proof of some great problem, and I >spent a lot of time on FLT and spherical packing. Uh, you didnıt just think you could Ŝnd such a thing, you insisted you _had_ found it. Over and over, dismissing anyone who disputed your claims with nothing but insults. >Unfortunately, Usenet was not the place to talk about such efforts. >I didnıt matter that I said I was an amateur. It didnıt matter that I >admitted my mistakes Admitted your mistakes? Yes, you often admitted something was wrong a year or so after it was pointed out. Each time coming up with a new version, and _each_ time, no matter _how_ many versions had turned out to be wrong, insisting that the _current_ version was correct, and that anyone who claimed otherwise was simply lying. >or said it was mostly just a lark. The people >here wanted to control my postings, and when I wouldnıt be controlled >they got vicious in ever more creative ways. >They are vicious. >At the end of the day, what Iıve done is talk about various math >ideas, sure often arrogantly, at times with claims that turned out to >be wrong And never with claims that turned out to be right. >--which Iıd admit when I Ŝgured it out. Guffaw. Your Ŝguring it out amounted to Ŝnally agreeing that an explanation of an error was correct, after it had been repeated a few hundred times. >In response--on a supposedly *public* forum--Iıve been hounded for >years by people calling me crazy, idiot, fool, loon, and those are the >nice names. >Race was brought into it by David Ullrich, and when I complained to >his school, as he is a math professor at Oklahoma State University, I >got called racial slurs. The people who were calling you bad names (which of course most of us agreed was a bad thing) were doing that before this incident. >When I Ŝnally got a result accepted at a math journal which claims to >do peer review, sci.mathıers gang emailed to get it yanked. Nope. Nobody said it should be yanked - thereıs universal agreement here that withdrawing the paper was not what the journal should have done - they lose any credibility they may have had when they did that. People _did_ inform the editor that the paper was ludicrously wrong. >And they still stalk my postings. A truly bizarre notion of stalk - reading things that you post in public forums. >The warning to others out there is that Usenet is a place where people >can come after you, and they will. And it wonıt just stay on Usenet >necessarily, as clearly posters here have been trying to Ŝnd out >personal information about me, like where I work, and live. Back when you were contacting OSU every few days I _asked_ you who _your_ employer was. I still donıt know why youıve never answered that question. (Someone else _did_ do some sleuthing and _posted_ an answer, which I ignored - wasnıt interested unless the information came from you. Who _is_ your employer, by the way?) You really donıt realize how hilarious this warning is, coming from _you_? I guess not. >And you have the case of Erik Max Francis with his webpage, still up >there as it appears he wishes to keep it up indeŜnitely, and Dik >Winter with his copyright violations. >The lesson is clear: post on Usenet at your own risk, and donıt be >surprised at what happens, how energetic people may be in coming after >you, how long they may keep at it, and what lengths they may go to to >try and take you out, if you donıt obey what *they* think is the >proper behavior. >Usenet is now a place where gangs of posters routinely interfere in >other peopleıs lives, and then blame that person. What utter bull. Exactly how has anyone here tried to interfere with your life, in a way thatıs even _remotely_ comparable to the way that you _have_ interfered in mine? (Or tried to, in your usual impotent way.) >James Harris ************************ David C. Ullrich === Subject: Re: ? -- JSH related > The story of JSH is just so > intriguing, especially when found within a subject commonly thought to > be dry. You might enjoy William Dunhamıs Journey through Genius, an extremely accessible look at some nice results, but more importantly at the people behind them. A certain level of eccentricity amongst mathematicians is nothing new; even JSH has yet to approach the level of Cardano and Tartaglia, who almost came to a duel over the solution of the general cubic, if I remember right. -- Larry Lard Replies to group please === Subject: Re: ? -- JSH related X-RFC2646: Format=Flowed; Original [Tim Peters] >> donıt hesitate to start a thread about a math topic that interests >> you, even if you feel itıs elementary. Any on-topic post should be >> welcomed. For example, insist that even numbers are all divisible by >> triangles, and then Ŝght unbelievers to the death . [E. Xavier] > Hopefully, my next post will be on topic and pusillanimous enough to > avoid a deathmatch. It does take two to Ŝght, which is a clue about how to avoid wasting your life in Usenet vendettas. A newsgroup is a community of sorts, although that appeared to be a lot easier to see a decade (or more) ago. Itıs at its best when people help each other -- sometimes learner, sometimes teacher, sometimes collaborator, sometimes wide-eyed curious, sometimes skeptical, sometimes even silly just for the fun of it -- but always respectful. All levels of knowledge should be welcome. Sanity helps. Basic good manners help a lot. My last word on JSH is that in the Google browsing I did, I was either astronomically unlucky in the sample I found, or he never tried to help anyone with their own questions. Thatıs a sign youıll come to recognize. > ... > Youıre right about it being impolite to ask. I really shouldnıt have. > Because heıd almost been published in a journal (no small feat), Journals vary in reputation and standards. Iım sure thatıs as true in math as it is in your Ŝeld. Still, itıs truly remarkable that this paper was accepted at one point. > I had the notion that Mr. Harris was a professor somewhere but wasnıt > sure. FYI, I stumbled into this thoroughly affable self-disclosure: > intriguing, especially when found within a subject commonly thought to > be dry. Oh, mathıs beauty is somewhat austere, but itıs not dry at all. Really! It can be exciting. Book memorization is whatıs deathly dry. And mathematicians are just people, as wonderfully weird and diverse as any occupational group. On whole, they may be crazier than the norm, I expect as in any highly creative profession. When the world of ideas or passions seems more tangible than the world of brocolli and politicians, that kinda comes with the territory . === Subject: Re: ? -- JSH related X-RFC2646: Format=Flowed; Original [Tim Peters, reveals that heıs a computer geek -- but someone else already unearthed this dirty truth!] [David C. Ullrich] > You guys must Ŝnally have that big snake thing working right, given > the time youıve been spending around here lately... heh. Python development is a long-time love and hobby, but I donıt get paid for it. I have given some of my Py-time to Usenet lately. Iıll enter therapy voluntarily. Python program that takes any integer n > 1 as input, and outputs three integers x, y and z all >= 1 such that x**n + y**n not only doesnıt equal z**n, itıs provably not even notably close. Obviously a short proof of FLT follows, or Iım not the most overwhelming intellectual force in mathematical history. [E. Xavier] > Fourth, he wasnıt successful in ending or affecting Dr. Ullrichıs job, > was he? >> Sad to say, poor Dr. Ullrich has been unemployed as a result since 2002, >> and posts now from an illegally jacked-in laptop at the bottom of a >> dumpster >behind the University of Oklahomaıs football stadium. Oops! >> Iım afraid I channelled one of JSHıs dreams there . Seriously, by >> all credible accounts, David wasnıt tangibly injured in any way. > Which, lest anyone miss an important point here, doesnıt mean that > his behavior was apppropriate or in fact anything less then > despicable: (i) People _can_ get in trouble on account of totally > bogus charges of racism (ii) thereıs still a lot of honest, nasty > vile racism out there - whining about the sort of thing JSH is > whining about here canıt do actual victims much good, > crying-wolf-wise. This part of the story would be sooooo much easier if we were all 10 years old. Then when he called you a lapdog, and you declined to call him anything, it would have been forgotten by both sides within a day. As is, I do doubt that your universityıs administration, or your stateıs Attorney General, are more inclined to dismiss charges of racism due to anything James did. I mean, really. While I donıt know what he said to them, the facts of this case are laughably transparent. So while there may be some possible world in which this speciŜc case has deep moral implications, Iım inclined to stick to no tangible injury in this world. Of course itıs despicable to cry to employers about things said on Usenet. But I hope youıll forgive me if I canıt stop laughing anyway, because in this case JSH was feigning offense at something you *didnıt* say. Teacher! Teacher! Davey didnıt call me a poo-poo face! LOL -- itıs just too silly for me to take seriously. I appreciate that itıs not as funny to you. But still . === Subject: Re: ? -- JSH related >[Tim Peters, reveals that heıs a computer geek -- but someone > else already unearthed this dirty truth!] >[David C. Ullrich] >> You guys must Ŝnally have that big snake thing working right, given >> the time youıve been spending around here lately... heh. >Python development is a long-time love and hobby, but I donıt get paid for >it. I have given some of my Py-time to Usenet lately. Iıll enter therapy >voluntarily. >Python program that takes any integer n > 1 as input, and outputs three >integers x, y and z all >= 1 such that x**n + y**n not only doesnıt equal >z**n, itıs provably not even notably close. Excellent. >Obviously a short proof of FLT >follows, or Iım not the most overwhelming intellectual force in mathematical >history. Obviously. >[E. Xavier] >> Fourth, he wasnıt successful in ending or affecting Dr. Ullrichıs job, >> was he? > Sad to say, poor Dr. Ullrich has been unemployed as a result since 2002, > and posts now from an illegally jacked-in laptop at the bottom of a > dumpster >behind the University of Oklahomaıs football stadium. Oops! > Iım afraid I channelled one of JSHıs dreams there . Seriously, by > all credible accounts, David wasnıt tangibly injured in any way. >> Which, lest anyone miss an important point here, doesnıt mean that >> his behavior was apppropriate or in fact anything less then >> despicable: (i) People _can_ get in trouble on account of totally >> bogus charges of racism (ii) thereıs still a lot of honest, nasty >> vile racism out there - whining about the sort of thing JSH is >> whining about here canıt do actual victims much good, >> crying-wolf-wise. >This part of the story would be sooooo much easier if we were all 10 years >old. Then when he called you a lapdog, and you declined to call him >anything, it would have been forgotten by both sides within a day. As is, I >do doubt that your universityıs administration, or your stateıs Attorney >General, are more inclined to dismiss charges of racism due to anything >James did. Thereıs a not missing somewhere in that last sentence, right? >I mean, really. While I donıt know what he said to them, the >facts of this case are laughably transparent. So while there may be some >possible world in which this speciŜc case has deep moral implications, Iım >inclined to stick to no tangible injury in this world. >Of course itıs despicable to cry to employers about things said on Usenet. >But I hope youıll forgive me if I canıt stop laughing anyway, because in >this case JSH was feigning offense at something you *didnıt* say. Teacher! >Teacher! Davey didnıt call me a poo-poo face! LOL -- itıs just too silly >for me to take seriously. I appreciate that itıs not as funny to you. But >still . No problem - I agree with most of that, not really inconsistent with what I said. In particular yes, the whole thing _is_ hilarious. (The idea that someone could get in trouble for using the word niggardly is also hilarious, but it happened...) Regarding whether there was any tangible damage itıs really too bad that it doesnıt seem appropriate to quote what the relevant administrator here said when I asked him about his reaction to the complaint. Letıs just say yes, the whole thing is funny. ************************ David C. Ullrich === Subject: Re: ? -- JSH related > Fourth, he wasnıt successful in ending or affecting Dr. Ullrichıs > job, was he? > He failed to become _responsible_ for any effects on Dr. Ullrichıs > job, though not for lack of trying, but it still obviously takes time > from Ullrichıs schedule to amuse himself in the Harris daily soap. > I mean, itıs like a drug. You canıt really hold the drug responsible > for its addictiveness. Just say no. If you still can. David Kastrup sits in Germany. Yet he still doesnıt mind talking about what is basically an American problem and *he* brought the subject up again this time. Basically, as I said, the only reason Ullrich even knew my race was because the subject came up in a discussion with a South African, who happened to be a reasonable person. So he must have picked it up there as I didnıt discuss it before his strange statement. And remember I *apologized* for saying heıd acted as my lapdog in an instance, but it was over a year later that Ullrich said that heıd been angry and talked about a racial slur being the appropriate reply but that heıd been talked out of it. Many may feel I went overboard in complaining to his school, but remember, Usenet is a public forum read worldwide. Maybe the rest of the world sees America as a home for such attitudes and statements, but I think professors at state universities have a responsibility and are accountable for their public statements. David Ullrich badly represented his school, his state, and this country. Then he came back to this board to complain about my complaint, playing the victim. He played many of you for sympathy, and blatantly Ŝred up hostilities which brought on people who had no qualms about just using the word nigger in their posts versus talking about thinking about a racial slur as the appropriate reply. He instigated a response, and he played the victim, when he brought up race. So yeah, to many people worldwide who read this story, it may just be what they expect from Americans and America, but I expect more. So David Kastrup can keep bringing it up for whatever reasons he may have, sitting in Germany, but the story doesnıt change, no matter how he lies about it. James Harris === Subject: Re: ? -- JSH related > > Fourth, he wasnıt successful in ending or affecting Dr. Ullrichıs > job, was he? > > He failed to become _responsible_ for any effects on Dr. Ullrichıs > job, though not for lack of trying, but it still obviously takes time > from Ullrichıs schedule to amuse himself in the Harris daily soap. > Basically, as I said, the only reason Ullrich even knew my race was > because the subject came up in a discussion with a South African, who > happened to be a reasonable person. > So he must have picked it up there as I didnıt discuss it before his > strange statement. If JSH hadnıt discussed it in a public forum, David wouldnıt have found out about it. > Many may feel I went overboard in complaining to his school, but > remember, Usenet is a public forum read worldwide. That same public forum in which JSH attacked and abused David repeatedly over a prolonged interval in retaliation for Davidıs merely pointing out the inadequacies of JSHıs mathematics. > Maybe the rest of the world sees America as a home for such attitudes > and statements, but I think professors at state universities have a > responsibility and are accountable for their public statements. Is JSH claiming that any of Dividıs mathematics was wrong? That would be the only part of his statements relevant to his position at the University. David is, after all, a professor of mathematics, not manners. > David Ullrich badly represented his school, his state, and this > country. That says to me that JSHıs state is one of delusion. > Then he came back to this board to complain about my complaint, Since you tried to get him Ŝred, he had justiŜable casue to complain. > He played many of you for sympathy, and blatantly Ŝred up hostilities David is a wimp at Ŝring up hostilities. One nasty posting by JSH does more in this respect than a dozen or more of Davidıs. > which brought on people who had no qualms about just using the word > nigger in their posts As fa as I recall, the only one to use that word was JSH, though I admit I did not read every post in that mostrous series of threads. > He instigated a response, and he played the victim, when he brought up > race. JSH has instigated thousands of responses, by posting incorrect mathematical statements, which is fair enough in a mathematically oriented NG, but also by abusing those who tried to help JSH Ŝnd and correct those errors, which is not appropriate in any NG or other public forum. JSH has played the victim for years, and is still playing it in this very post. He is hardly in a position to criticize others for doing a little of what he does so much himself. > So yeah, to many people worldwide who read this story, it may just be > what they expect from Americans and America, but I expect more. America expected a lot better from JSH here than it got here, too, in both mathematics and manners. > So David Kastrup can keep bringing it up for whatever reasons he may > have, sitting in Germany, but the story doesnıt change, no matter how > he lies about it. Or tells the truth about it, which irritates JSH a lot more than any lies would. > James Harris === Subject: hyperplane arrangement Iım interested in Ŝnding a few quantities related to hyperplane arrangements, to keep my posting short, hereıs what I mean: I have n hyperplanes in dimension d, I assume general position of the hyperplanes, Iım interested in the following quantities, O notation is enough for me: 1) the number of cells of dimension d-1 2) the complexity of each such cell (that is the number of d-2 objects that bound the d-1 cell. 3) the time it takes to iterate the cells. Iıve searched and found remarks about the complexity of the arrangement being O(n^d) and that it can be iterated in O(n^d) as well, if anyone knows and feel like sharing more, please do, also, if you can direct me to a good review or paper covering this topic it would be great, Computational Geometry by De Berg has a remark on this saying that generalization to $d$ dimension is possible but I couldnıt Ŝnd more about it. any help is appriciated, --tzurs, tzurs@hotmail.com (tzurshyperplanearrangment) === Subject: The mathematics of outrage Imagine a Planet X, like Earth, where the people are more likely than here to make there decisions on true logic rather than either faulty logic or emotions. Sort of like the Vulcans on Star Trek. However they are not quite up to Vulcan standards on this. There are two peoples on Planet X, the Yıs who are the masters, and number about 1,000,000,000. And the Zıs , who number about 2,000,000,000 who are the slaves. One reason this is so is that the Zıs have a religion that is somewhat Zen like, and the Yıs have a religion a somewhat like the ancient Aztecs. If all this may be aganst human nature these Yıs and Zıs are not quite human, for one thing they all look like idealised concepts of humans. In case you are interested, they would not scare you with the physical ugliness none of them possess. Now the Yıs have a belief their gods require that they sacriŜce .0025 of the Z population to their gods every month, chosen at random from the entire Z population. This is done in private by a priestly order able to keep secret the method of sacriŜce no matter what. This element of mystery lends its own special horror to the situation. The Zıs have calculated to their satisfaction that the best case scenereo that would result from a revolt is a victory that would cause them 30% fatalities in less than a month and the knowledge they exterminated the Yıs which would trouble the about twice as much as it would humans in the same circumstance. The worse case would be total extiction of the Zıs. However there is an emotional toll of enduring the status quo that is about 25% as severe as it would be for humans in a similar situation. If you can put aside any considerations of morality and emotionalism would it be expedient for the Zıs to revolt? === Subject: Re: The mathematics of outrage >There are two peoples on Planet X, the Yıs who are the masters, and number >about 1,000,000,000. And the Zıs , who number about 2,000,000,000 who are the >slaves. One reason this is so is that the Zıs have a religion that is somewhat >Zen like, and the Yıs have a religion a somewhat like the ancient Aztecs. If >all this may be aganst human nature these Yıs and Zıs are not quite human, for >one thing they all look like idealised concepts of humans. In case you are >interested, they would not scare you with the physical ugliness none of them >possess. >Now the Yıs have a belief their gods require that they sacriŜce .0025 of the Z >population to their gods every month, chosen at random from the entire Z >population. This is done in private by a priestly order able to keep secret the >method of sacriŜce no matter what. This element of mystery lends its own >special >horror to the situation. >The Zıs have calculated to their satisfaction that the best case scenereo that >would result from a revolt is a victory that would cause them 30% fatalities in >less than a month and the knowledge they exterminated the Yıs which would >trouble the about twice as much as it would humans in the same circumstance. >The worse case would be total extiction of the Zıs. However there is an >emotional toll of enduring the status quo that is about 25% as severe as it >would be for humans in a similar situation. If you can put aside any >considerations of morality and emotionalism would it be expedient for the Zıs >to revolt? When you say .0025 do you mean .0025% or .25%? Iım going to assume .25% for this answer. Iım also assuming that the Z population is steady-state, with births Ŝlling in for the sacriŜced ones. If each individual Z is only concerned about his own life (rather than the greater good of the Z or Y-Z civilization), he should revolt if and only if he expects to live more than .30/.0025, or 120 months. This minimizes his probability of death at the hands of the Y. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: The mathematics of outrage === >Subject: Re: The mathematics of outrage >Message-id: >>There are two peoples on Planet X, the Yıs who are the masters, and number >>about 1,000,000,000. And the Zıs , who number about 2,000,000,000 who are >the >>slaves. One reason this is so is that the Zıs have a religion that is >somewhat >>Zen like, and the Yıs have a religion a somewhat like the ancient Aztecs. If >>all this may be aganst human nature these Yıs and Zıs are not quite human, >for >>one thing they all look like idealised concepts of humans. In case you are >>interested, they would not scare you with the physical ugliness none of them >>possess. >>Now the Yıs have a belief their gods require that they sacriŜce .0025 of >the Z >>population to their gods every month, chosen at random from the entire Z >>population. This is done in private by a priestly order able to keep secret >the >>method of sacriŜce no matter what. This element of mystery lends its own >>special >>horror to the situation. >>The Zıs have calculated to their satisfaction that the best case scenereo >that >>would result from a revolt is a victory that would cause them 30% fatalities >>less than a month and the knowledge they exterminated the Yıs which would >>trouble the about twice as much as it would humans in the same circumstance. >>The worse case would be total extiction of the Zıs. However there is an >>emotional toll of enduring the status quo that is about 25% as severe as it >>would be for humans in a similar situation. If you can put aside any >>considerations of morality and emotionalism would it be expedient for the >Zıs >>to revolt? >When you say .0025 do you mean .0025% or .25%? Iım going to assume .25% for >this answer. Iım also assuming that the Z population is steady-state, with >births Ŝlling in for the sacriŜced ones. >If each individual Z is only concerned about his own life (rather than the >greater good of the Z or Y-Z civilization), he should revolt if and only if >he expects to live more than .30/.0025, or 120 months. This minimizes his >probability of death at the hands of the Y. >--Keith Lewis klewis {at} mitre.org >The above may not (yet) represent the opinions of my employer. It seems you are saying that in a steady state the overwhelming majority will be sacriŜced long before the age of 70? In that case the almost surely should revolt. Even if the worst case is extiction it seems much more is to be gained than lost. A long time ago I saw a surrealistic SCIFI show where the underlings were shot and killed at long range at random by the opressors at a steady level. Then they started to revolt. The tempo of the shooting increased and the underlings never could reach their oppressors. So they went back to what they were doing and the shooting then slowed to its old level. But no math was mentioned in this. === Subject: Re: The mathematics of outrage > If you can put aside any > considerations of morality and emotionalism would it be expedient for the Zıs > to revolt? if expedient is the criteria, Z should attack in any case, and should attack now. (Especially if they expect X or Y to have Weapons of Mass Destruction.) expedient to what? a steady state solution? === Subject: Re: The mathematics of outrage posting-account=jcZk7AwAAADXpPEyHtVyWC264SxtppRB History tells us that slaves eventually revolt, in the face of overwhelming force. Their self-respect demands it. What people could live with itself if it allowed itself to be dominated by others without continually Ŝghting against it with all their might. This is why we have heros, like out revolutaionary way heroes. This is why Muslims do suicide bombs. Van === Subject: Re: The mathematics of outrage > History tells us that slaves eventually revolt, in the face of > overwhelming > force. History tells us no such thing, if only because, for instance, in Egypt case, the dynasties eventually falled befor outside invaders *before* they had time to eventually revolt. The fact that they had already a few millenias of apathy (or more probably, had well learned the lesson that they were *really* powerless) is, of course one of the things History tell us... Their self-respect demands it. What people could live with > itself > if it allowed itself to be dominated by others without continually > Ŝghting > against it with all their might. > This is why we have heros, like out revolutaionary way heroes. > This is why Muslims do suicide bombs. > Van === Subject: Numbers and geometry Hi all! I have these numbers: 2,3739 0,6107 1,1921 0,9715 I would like to know how these are related to a cube or a square (since the square is the surface in a cube). Square root of the Ŝrst number reciprocal is 0,1774. That is close to 0,707 / 4 which is the reciprocal of square root of 2. I have problems with the rest. Do you know some easy way to get it visualized? Maybe a website or a software? === Subject: Mapping in complex plane At http://homes.jcu.edu.au/~ccbsc/ is a conformal mapping that looks a bit like one I am actually trying to produce. It is tan(z)/tan(1) - seen when you scroll down on the Web page. I would like to have the tall rectangle one unit wide map onto a shape that approaches a quarter circle with unit radius. SpeciŜcally, the right hand edge of the rectangle should fold over to tend to form a quarter circle as this mapping appears to nearly do so. The top edge of the rectangle is required to bunch up and tend towards a point on the y-axis, again as appears. Finally, points on the x-axis and y-axis should remain on those axes. The mapping shown is not what I am after (it just looks kinda close to it), but if anyone knows how to produce the quarter circle mapping I would appreciate the help. Brad === Subject: Re: Mapping in complex plane >At http://homes.jcu.edu.au/~ccbsc/ >is a conformal mapping that looks a bit like one I am actually >trying to produce. It is tan(z)/tan(1) - seen when you scroll >down on the Web page. >I would like to have the tall rectangle one unit wide map onto >a shape that approaches a quarter circle with unit radius. >SpeciŜcally, the right hand edge of the rectangle should fold >over to tend to form a quarter circle as this mapping appears >to nearly do so. >The top edge of the rectangle is required to bunch up and tend >towards a point on the y-axis, again as appears. Finally, >points on the x-axis and y-axis should remain on those axes. A quarter-circle lends itself to generation via polar coordinates r and theta. The mapping you want would have the properties when y = 0, r = x and theta = pi/2 when y = ymax, r = 1 and theta = pi/2 when x = 0, r = y/ymax and theta = 0 when x = 1, r = 1 Maybe somebody can come up with a mapping that elegantly Ŝts the above. The best I can think of is to triangularize with xı = x * (1-y/ymax) yı = y/ymax and then go to quarter-circle by taking theta = atan(yı/xı) r = sqrt(xı^2+yı^2)/sqrt(2*sin^2(theta)-2*sin(theta)+1) --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Mapping in complex plane > At http://homes.jcu.edu.au/~ccbsc/ > is a conformal mapping that looks a bit like one I am actually > trying to produce. It is tan(z)/tan(1) - seen when you scroll > down on the Web page. > I would like to have the tall rectangle one unit wide map onto > a shape that approaches a quarter circle with unit radius. > SpeciŜcally, the right hand edge of the rectangle should fold > over to tend to form a quarter circle as this mapping appears > to nearly do so. > The top edge of the rectangle is required to bunch up and tend > towards a point on the y-axis, again as appears. Finally, > points on the x-axis and y-axis should remain on those axes. > The mapping shown is not what I am after (it just looks kinda close > to it), but if anyone knows how to produce the quarter circle mapping > I would appreciate the help. > Brad If you have Mathematica, you could try the Graphics`ComplexMap standard package. Also, again with Mathematica, you could try the Cardano2 complex function plotting package at my web site below. It provides for one-panel and two-panel plots where each panel can be one of a number of complex plot types and also animations. The plot types include plots of real functions of complex functions, contour plots, 3D coded contour plots, color coded density plots, ComplexMap type plots, mappings to the Riemann sphere and vector plots. It is also possible to create multifunction vector plots. There is a lot of versatility in terms of being able to combine different plot types and to add other graphical elements to the plots. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ === Subject: The Red Sox-Yankee Playoff and Probability As many of you know the Boston Red Sox defeated the New York Yankees in a best of seven series 4 games to 3. They did it by winning 4 after losing 3. This had never been done in MLB postseason baseball before. And twice in NHL hockey, and 0 in NBA basketball out of a sample IIRC of less than 400 times combined in all above. Now if you had an accurate random number generator simulate 100,000,000,000 games of completely random chance between two individuals what percentage of the time, if that could be reasonably approximated, would one individual or the other win three times in a row? And of those three times in a row wins what percentage would the other individual win the next four? Would a sample of 400 three wins in a row and eachıs follow up of four games tell us anything? === Subject: Re: The Red Sox-Yankee Playoff and Probability >As many of you know the Boston Red Sox defeated the New York Yankees in a best >of seven series 4 games to 3. They did it by winning 4 after losing 3. This had >never been done in MLB postseason baseball before. And twice in NHL hockey, and >0 in NBA basketball out of a sample IIRC of less than 400 times combined in all >above. >Now if you had an accurate random number generator simulate 100,000,000,000 >games of completely random chance between two individuals what percentage of >the time, if that could be reasonably approximated, would one individual or the >other win three times in a row? And of those three times in a row wins what >percentage would the other individual win the next four? Would a sample of 400 >three wins in a row and eachıs follow up of four games tell us anything? If the teams are equal in every game, the probability of LLLWWWW is 1/128 for each team, or 1/64 total. And since there are 2 leagues, this should happen at the league championship level once every 32 years. But... These series are always played with the 2 games in one location, 3 in the other, then 2 back at the original place. So if the probability of winning a home game is p, the probability of LLLWWWW is (1-p)^4 * p^3 for the team that starts at home (1-p)^3 * p^4 for the team that starts away If you count pitcher rotations, the pattern becomes even less likely. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: The Red Sox-Yankee Playoff and Probability Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow) >Now if you had an accurate random number generator simulate >100,000,000,000 games of completely random chance between two individuals >what percentage of the time, if that could be reasonably approximated, >would one individual or the other win three times in a row? Others have addressed your question, but I would like to point out one thing that is rarely taken into account in these sorts of calculations, which is the concept of home-Ŝeld advantage. Occasionally I hear about how best-of-7 series seem to last more games than a simple probability calculation would predict, giving fodder for conspiracy theories. However, Iıve not seen any such calculation that makes a credible attempt to estimate home-Ŝeld advantage. If real, such an advantage would tend to extend the duration of a series. It would also, of course, affect the probability of winning three in a row. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: The Red Sox-Yankee Playoff and Probability > As many of you know the Boston Red Sox defeated the New York Yankees in a best > of seven series 4 games to 3. They did it by winning 4 after losing 3. This had > never been done in MLB postseason baseball before. And twice in NHL hockey, and > 0 in NBA basketball out of a sample IIRC of less than 400 times combined in all > above. > Now if you had an accurate random number generator simulate 100,000,000,000 > games of completely random chance between two individuals what percentage of > the time, if that could be reasonably approximated, would one individual or the > other win three times in a row? Assuming equal chance, I would assume (1/2)^3 = 1/8. > And of those three times in a row wins what > percentage would the other individual win the next four? Again assuming equal chance, Iıd think (1/2)^3*(1/2)^4 = 1/128. > Would a sample of 400 > three wins in a row and eachıs follow up of four games tell us anything? Letıs see, 3/400 ~ 0.0075, and 1/128 ~ 0.0078125, so from this data alone, I might guess that teams engaging in a best of seven series tend to be evenly matched, or that the results of a sports game are no different from chance. I suspect the frequencies are too low to see if this result is statistically signiŜcant, though. -- KCSAFF === Subject: Re: The Red Sox-Yankee Playoff and Probability [SNIP] > Now if you had an accurate random number generator simulate > 100,000,000,000 > games of completely random chance between two individuals what percentage > of > the time, if that could be reasonably approximated, would one individual > or the > other win three times in a row? > Assuming equal chance, I would assume (1/2)^3 = 1/8. > And of those three times in a row wins what > percentage would the other individual win the next four? > Again assuming equal chance, Iıd think (1/2)^3*(1/2)^4 = 1/128. > Would a sample of 400 > three wins in a row and eachıs follow up of four games tell us anything? > Letıs see, 3/400 ~ 0.0075, and 1/128 ~ 0.0078125, so from this data alone, I > might guess that teams engaging in a best of seven series tend to be evenly > matched, or that the results of a sports game are no different from chance. > I suspect the frequencies are too low to see if this result is statistically > signiŜcant, though. > -- > KCSAFF Oops, Daveıs post showed me the error of my ways. He is correct, if you donıt care who wins, the odds will be 2*(1/2)^3 = 1/4 and 2*(1/2)^7 = 1/64. === Subject: Re: The Red Sox-Yankee Playoff and Probability > As many of you know the Boston Red Sox defeated the New York Yankees in a best > of seven series 4 games to 3. They did it by winning 4 after losing 3. This had > never been done in MLB postseason baseball before. And twice in NHL hockey, and > 0 in NBA basketball out of a sample IIRC of less than 400 times combined in all > above. > Now if you had an accurate random number generator simulate 100,000,000,000 > games of completely random chance between two individuals what percentage of > the time, if that could be reasonably approximated, would one individual or the > other win three times in a row? And of those three times in a row wins what > percentage would the other individual win the next four? Would a sample of 400 > three wins in a row and eachıs follow up of four games tell us anything? Iım not sure a random sample will tell you much about the probability of such an event happening in sports, since the outcome of sports events is not entirely random. But the probability of (WWWLLLL or LLLWWWW) is exactly 1/64, if that helps. The conditional probability of LLLWWWW given LLL is 1/16. -- Dave Seaman Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: The Red Sox-Yankee Playoff and Probability === >Subject: Re: The Red Sox-Yankee Playoff and Probability >Message-id: >> As many of you know the Boston Red Sox defeated the New York Yankees in a >best >> of seven series 4 games to 3. They did it by winning 4 after losing 3. This >had >> never been done in MLB postseason baseball before. And twice in NHL hockey, >and >> 0 in NBA basketball out of a sample IIRC of less than 400 times combined in >all >> above. >> Now if you had an accurate random number generator simulate 100,000,000,000 >> games of completely random chance between two individuals what percentage >> the time, if that could be reasonably approximated, would one individual or >the >> other win three times in a row? And of those three times in a row wins what >> percentage would the other individual win the next four? Would a sample of >400 >> three wins in a row and eachıs follow up of four games tell us anything? >Iım not sure a random sample will tell you much about the probability of such >an event happening in sports, since the outcome of sports events is not >entirely random. >But the probability of (WWWLLLL or LLLWWWW) is exactly 1/64, if that helps. >The conditional probability of LLLWWWW given LLL is 1/16. chance 1/16 of the time. In the playoffs of the sports leagues I mentioned it has happened only three times in somewhere between 100 and 400 opportunities IIRC. So the getting of LLL must be usually an indication of gross inferiority of skill. I always suspected random chance alone should make LLLWWWW much more common than it has been. It will be rare for a team to get WWW without an overwhelming edge in the skills such as pitching needed in a series of best of seven games. I donıt suppose that the fact that its over when one team has won four regardless of how many games have been played is relevent is it? OTOH do you mean sequences of seven random chance games would have a 1/64 chance of ether being LLLWWWW or WWWLLLL? If so does that mean that LLLWWWW Œs probability is 1/128? That has happened considerably less in the real world instances I have alluded to. I guess what we can learn in this is that skill takes a huge hand in the games. === Subject: Library for Subgraph matching and graph dissimilarity Hi all... I have two graphs G1 and G2 labeled along nodes and edges. I have to get a pattern/ subgraph from these two graphs such that this subgraph captures those nodes and edges in G1 which are not included in G2. i.e there is a subgraph of G1 which is changed in G2. Rest of G1 in G2 is not changed and I need to capture this subgraph. These are NP complete problems as far as I know. I request you to suggest me an optimum way to do this. Is there any library or tool which does this. Vipindeep === Subject: Myth in Mathematics ! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC40a03312; hi, I am Krishna kishore Working for IBM. I just Ŝnished my Bachelorıs Degree in Computer science and Engineering from a reputed institute . I am 21 yrs old. Mathematics is a passion for me. I work during nights , inspite of hectic work schedule here in the company and try to get a broader understanding of the principles involved. But i am greatly worried about that popular statistical conclusion that mathematics is for young people only . No great work , ever , can be accomplished after 25 yrs. I am 21 . Working in a company, it takes more time than usual to achieve ,which the masters had . So , can u tell ur opinion on this ? I am open to any harsh suggestions. If u want me to concentarte on the job,rather than wasting my energy on maths, please tell me! Krishna Kishore G V === Subject: Re: Myth in Mathematics ! >hi, >But i am greatly worried about that popular statistical conclusion that >mathematics is for young people only . No great work , ever , can be >accomplished after 25 yrs. Gauss supposely said if the following formula e^(i pi) + 1 = 0 was not immediately obvious to someone, then they would never be a Ŝrst-class mathematician. === Subject: Re: Myth in Mathematics ! > Gauss supposely said if the following formula > e^(i pi) + 1 = 0 > was not immediately obvious to someone, then they would never be a > Ŝrst-class mathematician. Wow, you have certainly mangled THAT story! === Subject: Re: Myth in Mathematics ! : So , can u tell ur opinion on this ? Youıre 21? Can you please post without all that teenage u and ur crap? Sheesh, Justin === Subject: Re: Myth in Mathematics ! > No great work , ever , can be accomplished after 25 yrs. Really? Says who? > So , can u tell ur opinion on this ? Yes: you are immature. === Subject: Re: Myth in Mathematics ! Originator: dwildstr@euclid.ucsd.edu (Jake Wildstrom) The Prophet Jack D. Ripper known to the wise as JDRipper@spitalŜelds.net, opened the Book of Words, and read unto the people: >> No great work , ever , can be accomplished after 25 yrs. > Really? Says who? Most famously G.H. Hardy (with the limit of 40 years rather than 25), but just because he said it doesnıt make it true -- an awful lot of the best mathematics have done their best work later in their careers. Actually, most of the mathematicians I know donıt do much at all until after 25. 25 in modern education usually corresponds to the fourth year of graduate school -- at that point a lot of folks have yet to have found anything truly inspired. (at least I hope so, or else the clockıs _really_ ticking for me to get my act together) I think part of the fact that thatıs not true today has been a change in the pacing of our lives and education. The mathematicians of prior centuries (Galois springs to mind, since at my age he had been dead for 3 years) had done a lot more -- not just more math, but just more activity -- then most of us at that age. -- D. Jacob (Jake) Wildstrom, Math monkey and freelance thinker A mathematician is a device for turning coffee into theorems. -Alfred Renyi The opinions expressed herein are not necessarily endorsed by the University of California or math department thereof. === Subject: Re: Myth in Mathematics ! > But i am greatly worried about that popular statistical conclusion that mathematics is for young people only . No great work , ever , can be accomplished after 25 yrs. I am 21 . Working in a company, it takes more time than usual to achieve ,which the masters had . Why would that be the case? Less braincells after 25? :0) Dont worry. Have an open mind and dont expect to always see the truth in the mainstream... Higher consciousness is the answer === Subject: Re: Myth in Mathematics ! > But i am greatly worried about that popular statistical conclusion that > mathematics is for young people only . No great work , ever , can be > accomplished after 25 yrs. Donıt worry. That myth is total bollocks. :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Two questions about LambertW(x) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC41l03335; Iıve got two things to ask you: 1)How to symplify the following expression : LambertW(((LambertW(x)+1)e.x)/LambertW(x)); 2)To solve Phi |Phi(LambertW(x))=Phi(x)+1 (1) namely Ŝnding a function phi satisfying (1) . === Subject: Re: Two questions about LambertW(x) > Phi(LambertW(x))=Phi(x)+1 This is probably no help, but according to the deŜnition, you want: Phi(w) = Phi(w exp(w)) + 1 What if w=0? Then Phi(0) = Phi(0) + 1, not easy to satisfy. Maybe with Phi(0)= inŜnity? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... >> How about if I write all the decimals in 0.5 and see it it ever >>changes to 1/2. >> Oops, I just did and, by golly, it didnıt change! Does that prove that >>0.5 is not equal to 1/2? > Hey BuckWheat grow up. > .5 == .5555555555555555555555555...... I think the point of it was that 0.5000... is the same as 0.5 is the same as 1/2. The implication is that the mere repetition of (inŜnitely repeated) zeros is un-nessisary to show that 0.5 is the same as 1/2. I think that G.E.Ivey was pointing out that even if you write out all the inŜnate digits inherent in 0.5 (and, if you think about it, they are, indeed implied whenever we write a number without specifying accuracy) the decemal number 0.5 will never morph into the fraction 1/2. Both 0.5 and 1/2 represent the same value, and both imply an inŜnate accuracy: they are equal. So why should something different apply when, rather than inŜnate zeros, we write inŜnate nines? It implies inŜnate accuracy, as does 1/9, and it can be (and has been, in the course of this discussion!) shown that they imply the same Œnumberı. === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... > How about if I write all the decimals in 0.5 and see it it ever >changes to 1/2. > Oops, I just did and, by golly, it didnıt change! Does that prove that >0.5 is not equal to 1/2? >> Hey BuckWheat grow up. >> .5 == .5555555555555555555555555...... >I think the point of it was that 0.5000... is the same as 0.5 is the >same as 1/2. The implication is that the mere repetition of (inŜnitely >repeated) zeros is un-nessisary to show that 0.5 is the same as 1/2. I >think that G.E.Ivey was pointing out that even if you write out all the >inŜnate digits inherent in 0.5 (and, if you think about it, they are, >indeed implied whenever we write a number without specifying accuracy) >the decemal number 0.5 will never morph into the fraction 1/2. Both 0.5 >and 1/2 represent the same value, and both imply an inŜnate accuracy: >they are equal. So why should something different apply when, rather >than inŜnate zeros, we write inŜnate nines? It implies inŜnate >accuracy, as does 1/9, and it can be (and has been, in the course of >this discussion!) shown that they imply the same Œnumberı. The same number, 1/2 = .5 1/9 = .11111111111111...... 9/10 = .9 But in an inŜnite repeating series of .11111.... doesnıt converge to some other number like .2 .99999.... stays that way, it doesnıt converge to 1. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... But you say yourself: > 1/9 = .11111111111111...... and surely if you accept that, you admit that 2/9 = .2222... and 3/9 = .3333... and so on. So, Ŝrstly, do you admit that 9/9 = 0.9999..., and, if you do, how can you say that this is not 1? If you do not admit that 9/9 = 0.9999..., it follows that you refuse to believe 8/9 = 0.8888..., and so on, back to your statement quoted above. Either you do not believe that 1/9 = .1111..., and made a mistake; you do believe 1/9 = .1111..., and are therefore logically forced to conclude that 9/9 = 0.9999...; or you are (as suggested above) completely indifferent and simply a troll. Which is it? === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... > The same number, > 1/2 = .5 > 1/9 = .11111111111111...... > 9/10 = .9 > But in an inŜnite repeating series of > .11111.... doesnıt converge to some other number like .2 > .99999.... stays that way, it doesnıt converge to 1. Wait a minute, you accept that .1111... = 1/9, but you donıt accept that .9999... = 9/9 = 1? How do you justify that? Doesnıt .111... converge (in your words) to 1/9, even though none of the numbers .1, .11, .111, .1111 etc is equal to 1/9? Isnıt it the same thing with .9999... and 9/9? How does .1111... change into 1/9 when .9999... doesnıt change into 9/9? Do you accept the formula for an inŜnite geometric series (found in any high school algebra book): SUM{k>=0}(r^k) = 1/(1-r) (when 0 The same number, > 1/2 = .5 > 1/9 = .11111111111111...... > 9/10 = .9 > But in an inŜnite repeating series of > .11111.... doesnıt converge to some other number like .2 > .99999.... stays that way, it doesnıt converge to 1. A single number doesnıt converge. A series or sequence might. .9999... == 1. You say it isnıt, you tell us whatıs the difference. Difference, remember, a - b? Feel free to use scientiŜc notation. === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... >> The same number, >> 1/2 = .5 >> 1/9 = .11111111111111...... >> 9/10 = .9 >> But in an inŜnite repeating series of >> .11111.... doesnıt converge to some other number like .2 >> .99999.... stays that way, it doesnıt converge to 1. >A single number doesnıt converge. A series or sequence might. >.9999... == 1. >You say it isnıt, you tell us whatıs the difference. Difference, remember, >a - b? >Feel free to use scientiŜc notation. If you keep taking the derivative of a term, this might show that it does converges to something in a higher order derivative. Ok letıs see if, d x^2 converges to something in some type of series in itıs last term. = 2x d 2x = 2 So we could say that x^2 converges to something in a higher order derivative. But with the last term in the .99999... series, 9/10^n, if you keep taking the derivative of this you get something indetermine to start with. But, letıs see what happens to just the term 10^n if you keep taking higher order derivatives. Assume 9ıs just keep repeating inŜnitely, with the bottom term changing per higher order derivative, now letıs see where the bottom term converges to, repeat 9ıs --->oo ------------------------- 10^n ---> ? convergence lim as n---> oo dı 10^n = 10^n log(10) now take the derivative of this (Note :log( 10) = 1) dıı (10^n)(1) = 10^n log(10) dııı 10^n log(10) = 10^n log(10) dıııııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ııııııııııııı Œıııııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ııııııııııııı Œıııııııııııııııııııııııııııı 10^n log(10) = 10^n log(10) = 10^n so, .99999...9/10^n, never converges to 1 It never converges to 1 at inŜnity even with an inŜnite amount of high order derivatives. Since one of the terms never converges at inŜnity, it is divergent or an indeterminate. So, .9999... == 1 Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... > The same number, > 1/2 = .5 > 1/9 = .11111111111111...... > 9/10 = .9 > But in an inŜnite repeating series of > .11111.... doesnıt converge to some other number like .2 > .99999.... stays that way, it doesnıt converge to 1. >>A single number doesnıt converge. A series or sequence might. >>.9999... == 1. >>You say it isnıt, you tell us whatıs the difference. Difference, remember, >>a - b? >>Feel free to use scientiŜc notation. > If you keep taking the derivative of a term, this might show that it does >converges to something in a higher order derivative. >Ok letıs see if, >d x^2 converges to something in some type of series in itıs last term. > = 2x >d 2x = 2 > So we could say that x^2 converges to something in a higher order >derivative. > But with the last term in the .99999... series, 9/10^n, if you keep >taking >the derivative of this you get something indetermine to start with. But, >letıs >see what happens to just the term 10^n if you keep taking higher order >derivatives. > Assume 9ıs just keep repeating inŜnitely, with the bottom term changing >per >higher order derivative, now letıs see where the bottom term converges to, > repeat 9ıs --->oo >------------------------- > 10^n ---> ? >convergence lim as n---> oo >dı 10^n = 10^n log(10) >now take the derivative of this >(Note :log( 10) = 1) >dıı (10^n)(1) = 10^n log(10) >dııı 10^n log(10) = 10^n log(10) >dııııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ıııııııııııııı >ıııııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ıııııııııııııı >ııııııııııııııııııııııııııııı 10^n log(10) = >10^n log(10) = 10^n > so, >.99999...9/10^n, never converges to 1 > It never converges to 1 at inŜnity even with an inŜnite amount of high >order derivatives. Since one of the terms never converges at inŜnity, it is >divergent or an indeterminate. > So, > .9999... == 1 Reference ONLINE Math LINK http://www.univie.ac.at/future.media/moe/onlinewerkzeuge.html Or another way to look at it is like this, dı 9*10^-n dıııııııııııııııııııııı +- 9 log(10)^n ---------------- 10^n where +- means alternating back and forth. The higher order derivatives of this never converges. It stays in this form and never converges to 1. It remains indeterminate. Now letıs go back to Partial Sums of 9/10^k First look at this: The Numerical Computation Of Sequences takes this form and never converges to 1. a[ 1] = 0.9 a[ 2] = 0.09 a[ 3] = 0.009 a[ 4] = 0.0009 a[ 5] = 0.00009 a[ 6] = 0.000009 a[ 7] = 9e-7 a[ 8] = 9e-8 a[ 9] = 9e-9 a[10] = 9e-10 a[11] = 9e-11 a[12] = 9e-12 a[13] = 9e-13 a[14] = 9e-14 a[15] = 9e-15 a[16] = 9e-16 a[17] = 9e-17 a[18] = 9e-18 a[19] = 9e-19 a[20] = 9e-20 a[21] = 9e-21 a[22] = 9e-22 a[23] = 9e-23 a[24] = 9e-24 a[25] = 8.999999999999998e-25 a[26] = 9e-26 a[27] = 9e-27 a[28] = 9e-28 a[29] = 9e-29 a[30] = 9e-30 a[31] = 9e-31 a[32] = 8.999999999999999e-32 a[33] = 9e-33 a[34] = 9e-34 a[35] = 9e-35 a[36] = 8.999999999999999e-36 a[37] = 9e-37 a[38] = 9e-38 a[39] = 9e-39 a[40] = 9e-40 a[41] = 9e-41 a[42] = 9e-42 a[43] = 9e-43 a[44] = 9e-44 a[45] = 9e-45 a[46] = 9e-46 a[47] = 9e-47 a[48] = 9e-48 a[49] = 8.999999999999999e-49 a[50] = 9e-50 It never converges to 1. Now letıs look at the deceiving or sloppy Numerical Computation Of Series Using Partial Sum view of 9/10^k s[ 1] = 0.9 s[ 2] = 0.99 s[ 3] = 0.999 s[ 4] = 0.9999 s[ 5] = 0.99999 s[ 6] = 0.9999990000000001 s[ 7] = 0.9999999 s[ 8] = 0.9999999900000001 s[ 9] = 0.999999999 s[10] = 0.9999999999 s[11] = 0.99999999999 s[12] = 0.999999999999 s[13] = 0.9999999999999 s[14] = 0.99999999999999 s[15] = 0.999999999999999 s[16] = 0.9999999999999999 s[17] = 1 s[18] = 1 s[19] = 1 s[20] = 1 s[21] = 1 s[22] = 1 s[23] = 1 s[24] = 1 s[25] = 1 s[26] = 1 s[27] = 1 s[28] = 1 s[29] = 1 s[30] = 1 It only appears like it does converge to 1 with a sloppy way of looking at it from within, .9.(9....9).9.... Quoting from their text of Partial Sums: In exact terms, the corresponding series is the sequence of partial sums. If it converges against a number, the latter is called the limit of the series. In a sloppy way one may say that the limit of a series is an inŜnite sum. It is therefore sometimes called value of the series (or just sum). This method tries to see if the series converges ((against)) another number within the series. But this isnıt the true representation of, .9999...... The Partial Sum Analysis analyzes things within this series to see if there is a convergence there against another number. It doesnıt determine the totality of the true convergence of the series. Proof: Now try this Partial Sum Series Convergence Analysis of: 1/3 = .3333... the last term is, 3/10^k s[ 1] = 0.3 s[ 2] = 0.32999999999999996 s[ 3] = 0.33299999999999996 s[ 4] = 0.3333 s[ 5] = 0.33332999999999996 s[ 6] = 0.33333299999999993 s[ 7] = 0.33333329999999994 s[ 8] = 0.3333333299999999 s[ 9] = 0.33333333299999995 s[10] = 0.3333333333 s[11] = 0.33333333333 s[12] = 0.33333333333299997 s[13] = 0.33333333333329995 s[14] = 0.33333333333332993 s[15] = 0.3333333333333329 s[16] = 0.3333333333333332 s[17] = 0.33333333333333326 s[18] = 0.33333333333333326 s[19] = 0.33333333333333326 s[20] = 0.33333333333333326 s[21] = 0.33333333333333326 s[22] = 0.33333333333333326 s[23] = 0.33333333333333326 s[24] = 0.33333333333333326 s[25] = 0.33333333333333326 s[26] = 0.33333333333333326 s[27] = 0.33333333333333326 s[28] = 0.33333333333333326 s[29] = 0.33333333333333326 s[30] = 0.33333333333333326 But they say it converges to 0.33333333333333326 Using the Partial Sums Method of convergence. 0.33333333333333326 * 3 = .99999999999999978 This does NOT equal 1 But in reality, 1/3 * 3 = 1 not .999999999999978 You can even observe in the series, .3333333.... is never maintained with partial sums.. This is what should be there. .3 + .03 + .003 +.0003 + ..... Not this, .3 + 0.32999999999999996 + 0.33299999999999996 + 0.3333 + .... + 3/10^n .9999... == 1, in the true totality form in reality at inŜnity. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... >> The same number, >> >> 1/2 = .5 >> >> 1/9 = .11111111111111...... >> >> 9/10 = .9 >> >> But in an inŜnite repeating series of >> .11111.... doesnıt converge to some other number like .2 >> >> .99999.... stays that way, it doesnıt converge to 1. >> >A single number doesnıt converge. A series or sequence might. >.9999... == 1. >You say it isnıt, you tell us whatıs the difference. Difference, remember, >a - b? >Feel free to use scientiŜc notation. >> If you keep taking the derivative of a term, this might show that it >does >>converges to something in a higher order derivative. >>Ok letıs see if, >>d x^2 converges to something in some type of series in itıs last term. >> = 2x >>d 2x = 2 >> So we could say that x^2 converges to something in a higher order >>derivative. >> But with the last term in the .99999... series, 9/10^n, if you keep >>taking >>the derivative of this you get something indetermine to start with. But, >>letıs >>see what happens to just the term 10^n if you keep taking higher order >>derivatives. >> Assume 9ıs just keep repeating inŜnitely, with the bottom term changing >>per >>higher order derivative, now letıs see where the bottom term converges to, >> repeat 9ıs --->oo >>------------------------- >> 10^n ---> ? >>convergence lim as n---> oo >>dı 10^n = 10^n log(10) >>now take the derivative of this >>(Note :log( 10) = 1) >>dıı (10^n)(1) = 10^n log(10) >>dııı 10^n log(10) = 10^n log(10) >>dıııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ııııııııııııııı >>ııııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ııııııııııııııı >>ııııııııııııııııııııııııııııı 10^n log(10) = >>10^n log(10) = 10^n >> so, >>.99999...9/10^n, never converges to 1 >> It never converges to 1 at inŜnity even with an inŜnite amount of high >>order derivatives. Since one of the terms never converges at inŜnity, it is >>divergent or an indeterminate. >> So, >> .9999... == 1 >Reference ONLINE Math LINK >http://www.univie.ac.at/future.media/moe/onlinewerkzeuge.html > Or another way to look at it is like this, >dı 9*10^-n >dıııııııııııııııııııııı >+- 9 log(10)^n >---------------- > 10^n >where +- means alternating back and forth. >The higher order derivatives of this never converges. It stays in this form >and >never converges to 1. It remains indeterminate. > Now letıs go back to Partial Sums of 9/10^k >First look at this: >The Numerical Computation Of Sequences takes this form and never converges to >a[ 1] = 0.9 > a[ 2] = 0.09 > a[ 3] = 0.009 > a[ 4] = 0.0009 > a[ 5] = 0.00009 > a[ 6] = 0.000009 > a[ 7] = 9e-7 > a[ 8] = 9e-8 > a[ 9] = 9e-9 > a[10] = 9e-10 > a[11] = 9e-11 > a[12] = 9e-12 > a[13] = 9e-13 > a[14] = 9e-14 > a[15] = 9e-15 > a[16] = 9e-16 > a[17] = 9e-17 > a[18] = 9e-18 > a[19] = 9e-19 > a[20] = 9e-20 > a[21] = 9e-21 > a[22] = 9e-22 > a[23] = 9e-23 > a[24] = 9e-24 > a[25] = 8.999999999999998e-25 > a[26] = 9e-26 > a[27] = 9e-27 > a[28] = 9e-28 > a[29] = 9e-29 > a[30] = 9e-30 > a[31] = 9e-31 > a[32] = 8.999999999999999e-32 > a[33] = 9e-33 > a[34] = 9e-34 > a[35] = 9e-35 > a[36] = 8.999999999999999e-36 > a[37] = 9e-37 > a[38] = 9e-38 > a[39] = 9e-39 > a[40] = 9e-40 > a[41] = 9e-41 > a[42] = 9e-42 > a[43] = 9e-43 > a[44] = 9e-44 > a[45] = 9e-45 > a[46] = 9e-46 > a[47] = 9e-47 > a[48] = 9e-48 > a[49] = 8.999999999999999e-49 > a[50] = 9e-50 > It never converges to 1. > Now letıs look at the deceiving or sloppy Numerical Computation Of Series >Using Partial Sum view of 9/10^k >s[ 1] = 0.9 > s[ 2] = 0.99 > s[ 3] = 0.999 > s[ 4] = 0.9999 > s[ 5] = 0.99999 > s[ 6] = 0.9999990000000001 > s[ 7] = 0.9999999 > s[ 8] = 0.9999999900000001 > s[ 9] = 0.999999999 > s[10] = 0.9999999999 > s[11] = 0.99999999999 > s[12] = 0.999999999999 > s[13] = 0.9999999999999 > s[14] = 0.99999999999999 > s[15] = 0.999999999999999 > s[16] = 0.9999999999999999 > s[17] = 1 > s[18] = 1 > s[19] = 1 > s[20] = 1 > s[21] = 1 > s[22] = 1 > s[23] = 1 > s[24] = 1 > s[25] = 1 > s[26] = 1 > s[27] = 1 > s[28] = 1 > s[29] = 1 > s[30] = 1 > It only appears like it does converge to 1 with a sloppy way of looking >it from within, .9.(9....9).9.... > Quoting from their text of Partial Sums: >In exact terms, the corresponding series is the sequence of partial sums. If >it converges against a number, the latter is called the limit of the >series. >In a sloppy way one may say that the limit of a series is an inŜnite sum. >is therefore sometimes called value of the series (or just sum). > This method tries to see if the series converges ((against)) another number >within the series. But this isnıt the true representation of, >.9999...... > The Partial Sum Analysis analyzes things within this series to see if >there >is a convergence there against another number. It doesnıt determine the >totality of the true convergence of the series. >Proof: > Now try this >Partial Sum Series Convergence Analysis of: >1/3 = .3333... >the last term is, >3/10^k >s[ 1] = 0.3 > s[ 2] = 0.32999999999999996 > s[ 3] = 0.33299999999999996 > s[ 4] = 0.3333 > s[ 5] = 0.33332999999999996 > s[ 6] = 0.33333299999999993 > s[ 7] = 0.33333329999999994 > s[ 8] = 0.3333333299999999 > s[ 9] = 0.33333333299999995 > s[10] = 0.3333333333 > s[11] = 0.33333333333 > s[12] = 0.33333333333299997 > s[13] = 0.33333333333329995 > s[14] = 0.33333333333332993 > s[15] = 0.3333333333333329 > s[16] = 0.3333333333333332 > s[17] = 0.33333333333333326 > s[18] = 0.33333333333333326 > s[19] = 0.33333333333333326 > s[20] = 0.33333333333333326 > s[21] = 0.33333333333333326 > s[22] = 0.33333333333333326 > s[23] = 0.33333333333333326 > s[24] = 0.33333333333333326 > s[25] = 0.33333333333333326 > s[26] = 0.33333333333333326 > s[27] = 0.33333333333333326 > s[28] = 0.33333333333333326 > s[29] = 0.33333333333333326 > s[30] = 0.33333333333333326 > But they say it converges to >0.33333333333333326 > Using the Partial Sums Method of convergence. >0.33333333333333326 * 3 = >.99999999999999978 > This does NOT equal 1 > But in reality, >1/3 * 3 = 1 not .999999999999978 > You can even observe in the series, >.3333333.... is never maintained with partial sums.. > This is what should be there. >.3 + .03 + .003 +.0003 + ..... > Not this, >.3 + 0.32999999999999996 + 0.33299999999999996 + 0.3333 + .... + 3/10^n > .9999... == 1, in the true totality form in reality at inŜnity. Partial Sum Convergence is just another way of saying rounding off a number to simplify a number. Like for example: 1.425 rounded off ~= 1.4 1.48 rounded of ~= 1.5 This type of rounding off numbers is done all the time in simple calculations. But when the analysis of high precision calculations with a large number of signiŜcant Ŝgures is needed, Partial Sums canıt be used. What if your answer ( or solution) was 1.684267667697646323232344111121434234234235423627531111101011 3 in a high precision series analysis. Partial Sums would have probably rounded this number off to a few signiŜcant Ŝgures, to maybe about 1.684267 and this would be inaccurate if a large number of signiŜcant Ŝgures ( or digits) was required. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... > The same number, > > 1/2 = .5 > > 1/9 = .11111111111111...... > > 9/10 = .9 > > But in an inŜnite repeating series of > .11111.... doesnıt converge to some other number like .2 > > .99999.... stays that way, it doesnıt converge to 1. > >> >>A single number doesnıt converge. A series or sequence might. >> >>.9999... == 1. >> >>You say it isnıt, you tell us whatıs the difference. Difference, remember, >>a - b? >>Feel free to use scientiŜc notation. >> > If you keep taking the derivative of a term, this might show that it >>does >converges to something in a higher order derivative. >Ok letıs see if, >d x^2 converges to something in some type of series in itıs last term. > = 2x >d 2x = 2 > So we could say that x^2 converges to something in a higher order >derivative. > But with the last term in the .99999... series, 9/10^n, if you keep >taking >the derivative of this you get something indetermine to start with. But, >letıs >see what happens to just the term 10^n if you keep taking higher order >derivatives. > Assume 9ıs just keep repeating inŜnitely, with the bottom term changing >per >higher order derivative, now letıs see where the bottom term converges to, > repeat 9ıs --->oo >------------------------- > 10^n ---> ? >convergence lim as n---> oo >dı 10^n = 10^n log(10) >now take the derivative of this >(Note :log( 10) = 1) >dıı (10^n)(1) = 10^n log(10) >dııı 10^n log(10) = 10^n log(10) >dııııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ııııııııııııı Œ >ıııııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ııııııııııııı Œ >ııııııııııııııııııııııııııııı 10^n log(10) = >10^n log(10) = 10^n > so, >.99999...9/10^n, never converges to 1 > It never converges to 1 at inŜnity even with an inŜnite amount of high >order derivatives. Since one of the terms never converges at inŜnity, it >divergent or an indeterminate. > So, > .9999... == 1 >>Reference ONLINE Math LINK >>http://www.univie.ac.at/future.media/moe/ onlinewerkzeuge.html >> Or another way to look at it is like this, >>dı 9*10^-n >>dıııııııııııııııııııııı >>+- 9 log(10)^n >>---------------- >> 10^n >>where +- means alternating back and forth. >>The higher order derivatives of this never converges. It stays in this form >>and >>never converges to 1. It remains indeterminate. >> Now letıs go back to Partial Sums of 9/10^k >>First look at this: >>The Numerical Computation Of Sequences takes this form and never converges >>1. >>a[ 1] = 0.9 >> a[ 2] = 0.09 >> a[ 3] = 0.009 >> a[ 4] = 0.0009 >> a[ 5] = 0.00009 >> a[ 6] = 0.000009 >> a[ 7] = 9e-7 >> a[ 8] = 9e-8 >> a[ 9] = 9e-9 >> a[10] = 9e-10 >> a[11] = 9e-11 >> a[12] = 9e-12 >> a[13] = 9e-13 >> a[14] = 9e-14 >> a[15] = 9e-15 >> a[16] = 9e-16 >> a[17] = 9e-17 >> a[18] = 9e-18 >> a[19] = 9e-19 >> a[20] = 9e-20 >> a[21] = 9e-21 >> a[22] = 9e-22 >> a[23] = 9e-23 >> a[24] = 9e-24 >> a[25] = 8.999999999999998e-25 >> a[26] = 9e-26 >> a[27] = 9e-27 >> a[28] = 9e-28 >> a[29] = 9e-29 >> a[30] = 9e-30 >> a[31] = 9e-31 >> a[32] = 8.999999999999999e-32 >> a[33] = 9e-33 >> a[34] = 9e-34 >> a[35] = 9e-35 >> a[36] = 8.999999999999999e-36 >> a[37] = 9e-37 >> a[38] = 9e-38 >> a[39] = 9e-39 >> a[40] = 9e-40 >> a[41] = 9e-41 >> a[42] = 9e-42 >> a[43] = 9e-43 >> a[44] = 9e-44 >> a[45] = 9e-45 >> a[46] = 9e-46 >> a[47] = 9e-47 >> a[48] = 9e-48 >> a[49] = 8.999999999999999e-49 >> a[50] = 9e-50 >> It never converges to 1. >> Now letıs look at the deceiving or sloppy Numerical Computation Of >Series >>Using Partial Sum view of 9/10^k >>s[ 1] = 0.9 >> s[ 2] = 0.99 >> s[ 3] = 0.999 >> s[ 4] = 0.9999 >> s[ 5] = 0.99999 >> s[ 6] = 0.9999990000000001 >> s[ 7] = 0.9999999 >> s[ 8] = 0.9999999900000001 >> s[ 9] = 0.999999999 >> s[10] = 0.9999999999 >> s[11] = 0.99999999999 >> s[12] = 0.999999999999 >> s[13] = 0.9999999999999 >> s[14] = 0.99999999999999 >> s[15] = 0.999999999999999 >> s[16] = 0.9999999999999999 >> s[17] = 1 >> s[18] = 1 >> s[19] = 1 >> s[20] = 1 >> s[21] = 1 >> s[22] = 1 >> s[23] = 1 >> s[24] = 1 >> s[25] = 1 >> s[26] = 1 >> s[27] = 1 >> s[28] = 1 >> s[29] = 1 >> s[30] = 1 >> It only appears like it does converge to 1 with a sloppy way of looking >>at >>it from within, .9.(9....9).9.... >> Quoting from their text of Partial Sums: >>In exact terms, the corresponding series is the sequence of partial sums. >>it converges against a number, the latter is called the limit of the >>series. >>In a sloppy way one may say that the limit of a series is an inŜnite sum. >>It >>is therefore sometimes called value of the series (or just sum). >> This method tries to see if the series converges ((against)) another >number >>within the series. But this isnıt the true representation of, >>.9999...... >> The Partial Sum Analysis analyzes things within this series to see if >>there >>is a convergence there against another number. It doesnıt determine the >>totality of the true convergence of the series. >>Proof: >> Now try this >>Partial Sum Series Convergence Analysis of: >>1/3 = .3333... >>the last term is, >>3/10^k >>s[ 1] = 0.3 >> s[ 2] = 0.32999999999999996 >> s[ 3] = 0.33299999999999996 >> s[ 4] = 0.3333 >> s[ 5] = 0.33332999999999996 >> s[ 6] = 0.33333299999999993 >> s[ 7] = 0.33333329999999994 >> s[ 8] = 0.3333333299999999 >> s[ 9] = 0.33333333299999995 >> s[10] = 0.3333333333 >> s[11] = 0.33333333333 >> s[12] = 0.33333333333299997 >> s[13] = 0.33333333333329995 >> s[14] = 0.33333333333332993 >> s[15] = 0.3333333333333329 >> s[16] = 0.3333333333333332 >> s[17] = 0.33333333333333326 >> s[18] = 0.33333333333333326 >> s[19] = 0.33333333333333326 >> s[20] = 0.33333333333333326 >> s[21] = 0.33333333333333326 >> s[22] = 0.33333333333333326 >> s[23] = 0.33333333333333326 >> s[24] = 0.33333333333333326 >> s[25] = 0.33333333333333326 >> s[26] = 0.33333333333333326 >> s[27] = 0.33333333333333326 >> s[28] = 0.33333333333333326 >> s[29] = 0.33333333333333326 >> s[30] = 0.33333333333333326 >> But they say it converges to >>0.33333333333333326 >> Using the Partial Sums Method of convergence. >>0.33333333333333326 * 3 = >>.99999999999999978 >> This does NOT equal 1 >> But in reality, >>1/3 * 3 = 1 not .999999999999978 >> You can even observe in the series, >>.3333333.... is never maintained with partial sums.. >> This is what should be there. >>.3 + .03 + .003 +.0003 + ..... >> Not this, >>.3 + 0.32999999999999996 + 0.33299999999999996 + 0.3333 + .... + 3/10^n >> .9999... == 1, in the true totality form in reality at inŜnity. > Partial Sum Convergence is just another way of saying rounding off a >number to simplify a number. Like for example: >1.425 rounded off ~= 1.4 >1.48 rounded of ~= 1.5 > This type of rounding off numbers is done all the time in simple >calculations. But when the analysis of high precision calculations with a >large >number of signiŜcant Ŝgures is needed, Partial Sums canıt be used. >What if your answer ( or solution) was > 1.684267667697646323232344111121434234234235423627531111101011 3 in a high >precision series analysis. > Partial Sums would have probably rounded this number off to a few >signiŜcant >Ŝgures, to maybe about 1.684267 and this would be inaccurate if a large >number >of signiŜcant Ŝgures ( or digits) was required. Here is a post that talked about accuracy: >> A fellow posed this question to me a few days ago. So far, none of my >> professors or colleagues could answer this: >> SUM(1/n^2) as n=1 to inŜnity. >> How many terms are needed to come to a solution with an error < 0.001. >> My initial thought was to integrate the p-series from 1 to k of the >> function subtracted from the integral of the function from (k+1) to >> inŜnity. Therefore, subtracting the errors should yield the term. I >> was wrong, or did the calculations incorrectly. >> Daniel Lausevic >Mathematica says the Ŝrst 1000 terms are close enough. >In[1]:= NSum[1/n^2,{n,1001,InŜnity}] >Out[1]= 0.0009995 >-- >Dave Seaman >Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling. > A simple way I view it.... Accuracy looks like common sense to me. You can only be as accurate as the number of accurate values used. If you want an accuracy of < .001, use 1000 accurate values. The inverse of .001 = 1000 If you want an accuracy of <.0000000000001 then use 1/ (1 X10^13) 10^13 accurate terms. The accuracy of .999... = 1 using Partial Sums Series convergence appears to be only within this amount of accuracy <.0625. To have <.001 amount of error, a 1000 terms are needed. Only 16 terms were used to form a partial sum convergence to 1. This is a relatively high amount of error compared to <.001 amount of error, witha 1000 terms in the series used. With an accuracy level of < 1/oo, .9999.... == 1 So it is more accurate to say, .9999... == 1 with .999... = 1 The accuracy of this is only 6.25%. We want accuracy approaching 100%. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... >> The same number, >> >> 1/2 = .5 >> >> 1/9 = .11111111111111...... >> >> 9/10 = .9 >> >> But in an inŜnite repeating series of >> .11111.... doesnıt converge to some other number like .2 >> >> .99999.... stays that way, it doesnıt converge to 1. >> > >A single number doesnıt converge. A series or sequence might. > >.9999... == 1. > >You say it isnıt, you tell us whatıs the difference. Difference, >remember, >a - b? >Feel free to use scientiŜc notation. > >> >> If you keep taking the derivative of a term, this might show that it >does >>converges to something in a higher order derivative. >> >>Ok letıs see if, >> >>d x^2 converges to something in some type of series in itıs last term. >> >> = 2x >> >>d 2x = 2 >> >> So we could say that x^2 converges to something in a higher order >>derivative. >> >> But with the last term in the .99999... series, 9/10^n, if you keep >>taking >>the derivative of this you get something indetermine to start with. But, >>letıs >>see what happens to just the term 10^n if you keep taking higher order >>derivatives. >> >> Assume 9ıs just keep repeating inŜnitely, with the bottom term >changing >>per >>higher order derivative, now letıs see where the bottom term converges to, >> >> repeat 9ıs --->oo >>------------------------- >> 10^n ---> ? >> >>convergence lim as n---> oo >> >>dı 10^n = 10^n log(10) >> >>now take the derivative of this >> >>(Note :log( 10) = 1) >> >>dıı (10^n)(1) = 10^n log(10) >> >>dııı 10^n log(10) = 10^n log(10) >> >>dıııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ııııııııııııı Œı >>ııııııııııııııııııııııııııııııııııııııııııııııııııııııııııı ııııııııııııı Œı >>ııııııııııııııııııııııııııııı 10^n log(10) = >> >>10^n log(10) = 10^n >> >> so, >> >>.99999...9/10^n, never converges to 1 >> >> It never converges to 1 at inŜnity even with an inŜnite amount of >high >>order derivatives. Since one of the terms never converges at inŜnity, it >>is >>divergent or an indeterminate. >> >> So, >> >> .9999... == 1 >Reference ONLINE Math LINK >http://www.univie.ac.at/future.media/moe/onlinewerkzeuge.html > Or another way to look at it is like this, >dı 9*10^-n >dıııııııııııııııııııııı >+- 9 log(10)^n >---------------- > 10^n >where +- means alternating back and forth. >The higher order derivatives of this never converges. It stays in this form >and >never converges to 1. It remains indeterminate. > Now letıs go back to Partial Sums of 9/10^k >First look at this: >The Numerical Computation Of Sequences takes this form and never converges >>to >1. >a[ 1] = 0.9 > a[ 2] = 0.09 > a[ 3] = 0.009 > a[ 4] = 0.0009 > a[ 5] = 0.00009 > a[ 6] = 0.000009 > a[ 7] = 9e-7 > a[ 8] = 9e-8 > a[ 9] = 9e-9 > a[10] = 9e-10 > a[11] = 9e-11 > a[12] = 9e-12 > a[13] = 9e-13 > a[14] = 9e-14 > a[15] = 9e-15 > a[16] = 9e-16 > a[17] = 9e-17 > a[18] = 9e-18 > a[19] = 9e-19 > a[20] = 9e-20 > a[21] = 9e-21 > a[22] = 9e-22 > a[23] = 9e-23 > a[24] = 9e-24 > a[25] = 8.999999999999998e-25 > a[26] = 9e-26 > a[27] = 9e-27 > a[28] = 9e-28 > a[29] = 9e-29 > a[30] = 9e-30 > a[31] = 9e-31 > a[32] = 8.999999999999999e-32 > a[33] = 9e-33 > a[34] = 9e-34 > a[35] = 9e-35 > a[36] = 8.999999999999999e-36 > a[37] = 9e-37 > a[38] = 9e-38 > a[39] = 9e-39 > a[40] = 9e-40 > a[41] = 9e-41 > a[42] = 9e-42 > a[43] = 9e-43 > a[44] = 9e-44 > a[45] = 9e-45 > a[46] = 9e-46 > a[47] = 9e-47 > a[48] = 9e-48 > a[49] = 8.999999999999999e-49 > a[50] = 9e-50 > > It never converges to 1. > Now letıs look at the deceiving or sloppy Numerical Computation Of >>Series >Using Partial Sum view of 9/10^k >s[ 1] = 0.9 > s[ 2] = 0.99 > s[ 3] = 0.999 > s[ 4] = 0.9999 > s[ 5] = 0.99999 > s[ 6] = 0.9999990000000001 > s[ 7] = 0.9999999 > s[ 8] = 0.9999999900000001 > s[ 9] = 0.999999999 > s[10] = 0.9999999999 > s[11] = 0.99999999999 > s[12] = 0.999999999999 > s[13] = 0.9999999999999 > s[14] = 0.99999999999999 > s[15] = 0.999999999999999 > s[16] = 0.9999999999999999 > s[17] = 1 > s[18] = 1 > s[19] = 1 > s[20] = 1 > s[21] = 1 > s[22] = 1 > s[23] = 1 > s[24] = 1 > s[25] = 1 > s[26] = 1 > s[27] = 1 > s[28] = 1 > s[29] = 1 > s[30] = 1 > It only appears like it does converge to 1 with a sloppy way of looking >at >it from within, .9.(9....9).9.... > Quoting from their text of Partial Sums: >In exact terms, the corresponding series is the sequence of partial sums. >>If >it converges against a number, the latter is called the limit of the >series. >In a sloppy way one may say that the limit of a series is an inŜnite >sum. >It >is therefore sometimes called value of the series (or just sum). > This method tries to see if the series converges ((against)) another >>number >within the series. But this isnıt the true representation of, >.9999...... > The Partial Sum Analysis analyzes things within this series to see if >there >is a convergence there against another number. It doesnıt determine the >totality of the true convergence of the series. >Proof: > Now try this >Partial Sum Series Convergence Analysis of: >1/3 = .3333... >the last term is, >3/10^k >s[ 1] = 0.3 > s[ 2] = 0.32999999999999996 > s[ 3] = 0.33299999999999996 > s[ 4] = 0.3333 > s[ 5] = 0.33332999999999996 > s[ 6] = 0.33333299999999993 > s[ 7] = 0.33333329999999994 > s[ 8] = 0.3333333299999999 > s[ 9] = 0.33333333299999995 > s[10] = 0.3333333333 > s[11] = 0.33333333333 > s[12] = 0.33333333333299997 > s[13] = 0.33333333333329995 > s[14] = 0.33333333333332993 > s[15] = 0.3333333333333329 > s[16] = 0.3333333333333332 > s[17] = 0.33333333333333326 > s[18] = 0.33333333333333326 > s[19] = 0.33333333333333326 > s[20] = 0.33333333333333326 > s[21] = 0.33333333333333326 > s[22] = 0.33333333333333326 > s[23] = 0.33333333333333326 > s[24] = 0.33333333333333326 > s[25] = 0.33333333333333326 > s[26] = 0.33333333333333326 > s[27] = 0.33333333333333326 > s[28] = 0.33333333333333326 > s[29] = 0.33333333333333326 > s[30] = 0.33333333333333326 > But they say it converges to >0.33333333333333326 > Using the Partial Sums Method of convergence. >0.33333333333333326 * 3 = >.99999999999999978 > This does NOT equal 1 > But in reality, >1/3 * 3 = 1 not .999999999999978 > You can even observe in the series, >.3333333.... is never maintained with partial sums.. > This is what should be there. >.3 + .03 + .003 +.0003 + ..... > Not this, >.3 + 0.32999999999999996 + 0.33299999999999996 + 0.3333 + .... + 3/10^n > .9999... == 1, in the true totality form in reality at inŜnity. >> Partial Sum Convergence is just another way of saying rounding off a >>number to simplify a number. Like for example: >>1.425 rounded off ~= 1.4 >>1.48 rounded of ~= 1.5 >> This type of rounding off numbers is done all the time in simple >>calculations. But when the analysis of high precision calculations with a >>large >>number of signiŜcant Ŝgures is needed, Partial Sums canıt be used. >>What if your answer ( or solution) was >> 1.684267667697646323232344111121434234234235423627531111101011 3 in a high >>precision series analysis. >> Partial Sums would have probably rounded this number off to a few >>signiŜcant >>Ŝgures, to maybe about 1.684267 and this would be inaccurate if a large >>number >>of signiŜcant Ŝgures ( or digits) was required. > Here is a post that talked about accuracy: > A fellow posed this question to me a few days ago. So far, none of my > professors or colleagues could answer this: > SUM(1/n^2) as n=1 to inŜnity. > How many terms are needed to come to a solution with an error < 0.001. > My initial thought was to integrate the p-series from 1 to k of the > function subtracted from the integral of the function from (k+1) to > inŜnity. Therefore, subtracting the errors should yield the term. I > was wrong, or did the calculations incorrectly. > Daniel Lausevic >>Mathematica says the Ŝrst 1000 terms are close enough. >>In[1]:= NSum[1/n^2,{n,1001,InŜnity}] >>Out[1]= 0.0009995 >>-- >>Dave Seaman >>Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling. >> > A simple way I view it.... > Accuracy looks like common sense to me. You can only be as accurate as the >number of accurate values used. >If you want an accuracy of < .001, use 1000 accurate values. The inverse of >.001 = 1000 >If you want an accuracy of <.0000000000001 then use 1/ (1 X10^13) 10^13 >accurate terms. > The accuracy of .999... = 1 using Partial Sums Series convergence appears >be only within this amount of accuracy <.0625. To have <.001 amount of >error, >a 1000 terms are needed. Only 16 terms were used to form a partial sum >convergence to 1. This is a relatively high amount of error compared to <.001 >amount of error, witha 1000 terms in the series used. >With an accuracy level of < 1/oo, >.9999.... == 1 > So it is more accurate to say, >.9999... == 1 >with >.999... = 1 > The accuracy of this is only 6.25%. We want accuracy approaching 100%. In other words, for .9999... Partial Sums Series Convergence has about 6.25% error. Total Series Convergence evaluation has, .00000000000000000000....-->oo % error So it is more accurate to say, .9999....== 1 This is practically 100% accurate. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... > .5 == .5555555555555555555555555...... This is 5/9 (base 10 digits). Bob Kolker === Subject: Re: .99999999999999 = 1 No it doesnıttttttttttttttttttttttttttttt.......... >> .5 == .5555555555555555555555555...... >This is 5/9 (base 10 digits). >Bob Kolker .5 is different from .55555555555555.... .9 is different from .999999999999999... They both donıt converge to anything in a repetition to inŜnity. You canıt say that at inŜnity, the last term of .5555555.... of this converges to .6 The same goes for .9999999.... . You canıt say it converges to 1. You canıt say since, 10x = 5.55555... so, 5x = 5.55555... - .55555... = 1 so .5555.. -> 1 or .6 This is NOT a proof. that .999.... converges to 1. By deŜnition, .9999..... means, ..... = ONLY... ONLY... ONLY... 9ıs Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: all derivatives zero not imply zero!? may you give me an example of function f with all derivative of f zero at p such that f is not identically zero in a little neighborhood of zero? Tern === Subject: Re: all derivatives zero not imply zero!? ETAtAhQdlYcWyNb1/Zuk9JY+YEoNfmpCRAIVAI46F4AWFP6mOyHMUAKW+ iStC6sj f(x) = exp(-1/(x^2)) for all nonzero x in R = 0 for x = 0 But in the COMPLEX domain there is no such analytic function apart from a constant function. --OL === Subject: Re: all derivatives zero not imply zero!? >may you give me an example of function f with all derivative of f zero >at p such that f is not identically zero in a little neighborhood of >zero? Certainly: the function f(x) = 5. A non-trivial example would be any function (x) = (x-p)^k + m where k is a natural number and m is any nonzero constant. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: all derivatives zero not imply zero!? > may you give me an example of function f with all derivative of f zero > at p such that f is not identically zero in a little neighborhood of > zero? > Tern f(x) = 1 everywhere. === Subject: Re: all derivatives zero not imply zero!? X-RFC2646: Format=Flowed; Original > may you give me an example of function f with all derivative of f zero > at p such that f is not identically zero in a little neighborhood of > zero? > Tern Any constant function y = k(x) = k. Iım Ŝnding it hard to believe some of the weird answers youıve gotten so far. John Lowry Flight Physics === Subject: Re: all derivatives zero not imply zero!? >> may you give me an example of function f with all derivative of f zero >> at p such that f is not identically zero in a little neighborhood of >> zero? >> Tern >Any constant function y = k(x) = k. Iım Ŝnding it hard to believe some >of the weird answers youıve gotten so far. Hah, thatıs a good one :-) The validity of this examples, however, hinges on the meaning of all derivatives. If we include zeroth derivative among all derivatives, then this example does not work. The zeroth derivative of a function is the function itself. Rouben Rostamian === Subject: Re: all derivatives zero not imply zero!? X-RFC2646: Format=Flowed; Original > may you give me an example of function f with all derivative of f > zero > at p such that f is not identically zero in a little neighborhood of > zero? > Tern >>Any constant function y = k(x) = k. Iım Ŝnding it hard to believe >>some >>of the weird answers youıve gotten so far. > Hah, thatıs a good one :-) > The validity of this examples, however, hinges on the meaning > of all derivatives. If we include zeroth derivative > among all derivatives, then this example does not work. > The zeroth derivative of a function is the function itself. > Rouben Rostamian I suppose a fellow can use language to mean anything he wants, but Iıve never subscribed to that usage myself unless it was explicitly stated. Probably what was meant, however. John Lowry Flight Physics === Subject: Re: all derivatives zero not imply zero!? >>may you give me an example of function f with all derivative of f >>zero >>at p such that f is not identically zero in a little neighborhood of >>zero? >> >>Tern >Any constant function y = k(x) = k. Iım Ŝnding it hard to believe >some >of the weird answers youıve gotten so far. >>Hah, thatıs a good one :-) >>The validity of this examples, however, hinges on the meaning >>of all derivatives. If we include zeroth derivative >>among all derivatives, then this example does not work. >>The zeroth derivative of a function is the function itself. >>Rouben Rostamian > I suppose a fellow can use language to mean anything he wants, but Iıve > never subscribed to that usage myself unless it was explicitly stated. > Probably what was meant, however. > John Lowry > Flight Physics Yes, itıs the standard usage in mathematics. Far from being anything [one] wants, you may note that it is used this way in the standard expression of the Taylor series: sum( f^(n)(a) (x-a)^n / n! , n = 0 ... inŜnity ) where the notation f^(n)(a) refers to the nth derivative of f(x), evaluated at the point x = a. Dale. === Subject: Re: all derivatives zero not imply zero!? > may you give me an example of function f with all derivative of f zero > at p such that f is not identically zero in a little neighborhood of > zero? > Tern f(x) = e^(-1/x^2) if x=/= 0, f(0) = 0 . Julian Aguirre === Subject: Re: all derivatives zero not imply zero!? > may you give me an example of function f with all derivative of f zero > at p such that f is not identically zero in a little neighborhood of > zero? Consider f(x) = 0 if x = 0, exp(-1/x^2) otherwise at p = 0. David === Subject: Re: all derivatives zero not imply zero!? X-RFC2646: Format=Flowed; Original >> may you give me an example of function f with all derivative of f zero >> at p such that f is not identically zero in a little neighborhood of >> zero? > Consider > f(x) = 0 if x = 0, > exp(-1/x^2) otherwise > at p = 0. > David I know this to be true, and I studied it 30 years ago, but is there a simple answer to the following question? Calling your function f(x), its Taylor Series expansion is identically zero. Why canıt you take any arbitrary function g(x) and form g(x) + f(x) to make a function with the same Taylor series expansion as g(x) but wildly different behaviour. Doesnıt this invalidate a Taylor Series expansio as deŜning a speciŜc function? Is there either a web page, or a simple answer? === Subject: Re: all derivatives zero not imply zero!? Originator: grubb@lola > may you give me an example of function f with all derivative of f zero > at p such that f is not identically zero in a little neighborhood of > zero? >> Consider >> f(x) = 0 if x = 0, >> exp(-1/x^2) otherwise >> at p = 0. >> David >I know this to be true, and I studied it 30 years ago, but is there a simple >answer to the following question? >Calling your function f(x), its Taylor Series expansion is identically zero. Correct. >Why canıt you take any arbitrary function g(x) and form g(x) + f(x) to make >a function with the same Taylor series expansion as g(x) but wildly different >behaviour. Doesnıt this invalidate a Taylor Series expansion as >deŜning a speciŜc function? Well, the Taylor series of a function need not converge to that function. In fact, there are inŜnitely differentiable functions with the property that the Taylor series centerted at every point fails to converge to the original function in any interval around that point. As your example shows, there can be different functions that have the same Taylor series at a point. It only invalidates Taylor series in situations where that matters. If the series *does* converge, it converges to a speciŜc function, but possibly not the one you started with. Those with the nice property that their Taylor series converge to the original function are special and are called analytic. --Dan Grubb === Subject: Re: all derivatives zero not imply zero!? X-RFC2646: Format=Flowed; Original >> may you give me an example of function f with all derivative of f zero >> at p such that f is not identically zero in a little neighborhood of >> zero? > Consider > f(x) = 0 if x = 0, > exp(-1/x^2) otherwise > at p = 0. > David >>I know this to be true, and I studied it 30 years ago, but is there a >>simple >>answer to the following question? >>Calling your function f(x), its Taylor Series expansion is identically >>zero. > Correct. >>Why canıt you take any arbitrary function g(x) and form g(x) + f(x) to >>make >>a function with the same Taylor series expansion as g(x) but wildly >>different >>behaviour. Doesnıt this invalidate a Taylor Series expansion as >>deŜning a speciŜc function? > Well, the Taylor series of a function need not converge to that function. > In fact, there are inŜnitely differentiable functions with the property > that the Taylor series centerted at every point fails to converge to > the original function in any interval around that point. > As your example shows, there can be different functions that have > the same Taylor series at a point. It only invalidates Taylor series > in situations where that matters. If the series *does* converge, it > converges to a speciŜc function, but possibly not the one you started > with. Those with the nice property that their Taylor series > converge to the original function are special and are called analytic. > --Dan Grubb Ah ha! With the use of the word analytic you are suggesting it has a singularity in the complex plane at x=0. I can see that the behaviour is strange in the complex domain, e(-1/x) for x=0.0001i would wrap around x=0 an unlimitted number of times. Are you saying that even though the singularity appears to be able to removable - by making f(0)=0 - there is still a pole and a path integral around 0 is non zero? === Subject: Re: all derivatives zero not imply zero!? >may you give me an example of function f with all derivative of f zero >at p such that f is not identically zero in a little neighborhood of >zero? >> >>Consider >> >>f(x) = 0 if x = 0, >> exp(-1/x^2) otherwise >> >>at p = 0. >> >>David >I know this to be true, and I studied it 30 years ago, but is there a >simple >answer to the following question? >Calling your function f(x), its Taylor Series expansion is identically >zero. >>Correct. >Why canıt you take any arbitrary function g(x) and form g(x) + f(x) to >make >a function with the same Taylor series expansion as g(x) but wildly >different >behaviour. Doesnıt this invalidate a Taylor Series expansion as >deŜning a speciŜc function? >>Well, the Taylor series of a function need not converge to that function. >>In fact, there are inŜnitely differentiable functions with the property >>that the Taylor series centerted at every point fails to converge to >>the original function in any interval around that point. >>As your example shows, there can be different functions that have >>the same Taylor series at a point. It only invalidates Taylor series >>in situations where that matters. If the series *does* converge, it >>converges to a speciŜc function, but possibly not the one you started >>with. Those with the nice property that their Taylor series >>converge to the original function are special and are called analytic. >>--Dan Grubb > Ah ha! > With the use of the word analytic you are suggesting it has a singularity in > the complex plane at x=0. No, with the use of the word, Dan Grubb is stating that the functions whose Taylor series converge to the function in a neighborhood of the point are special, and are called analytic. If the power series fails to converge to the function in some neighborhood of p, the function is not analytic at p. It does so happen that the function exp(-1/x^2) (completed to have value 0 at x = 0) is not analytic at the origin, although it is trivial to show it to be analytic away from the origin, and therefore it has a singularity at the origin. > I can see that the behaviour is strange in the complex domain, e(-1/x) for > x=0.0001i would wrap around x=0 an unlimitted number of times. > Are you saying that even though the singularity appears to be able to > removable - by making f(0)=0 - there is still a pole and a path integral > around 0 is non zero? When you learn complex analysis, you learn of a number of different types of singularities that an analytic function can exhibit. Poles, for instance, are characterized by the fact that the modulus |f(z)| grows arbitrarily large as |z - p| approaches zero. This function does not do that, for you can see that as x --> 0 along the reals, the function approaches zero, so |f(x)| --> 0. Instead, this function has what is called an *essential* singularity, meaning (1) not a pole, and (2) not removable. One characteristic that is somewhat surprising at Ŝrst glance is that if viewed in the complex plane, for any neighborhood of 0 (the location of the singularity), the image under f is dense in the plane. Dale === Subject: Re: all derivatives zero not imply zero!? Originator: grubb@lola >> As your example shows, there can be different functions that have >> the same Taylor series at a point. It only invalidates Taylor series >> in situations where that matters. If the series *does* converge, it >> converges to a speciŜc function, but possibly not the one you started >> with. Those with the nice property that their Taylor series >> converge to the original function are special and are called analytic. >> --Dan Grubb >Ah ha! >With the use of the word analytic you are suggesting it has a singularity in >the complex plane at x=0. >I can see that the behaviour is strange in the complex domain, e(-1/x) for >x=0.0001i would wrap around x=0 an unlimitted number of times. >Are you saying that even though the singularity appears to be able to >removable - by making f(0)=0 - there is still a pole and a path integral >around 0 is non zero? The discontinuity on the real axis can be removed as stated. But in the complex plane, the situation is very bad. If you consider the Laurent series expansion at z=0, you will Ŝnd there are an inŜnite number of terms with negative exponents. We say that there is an essential singularity at z=0. This is worse than being a pole (where there are only Ŝnitely many negative terms). Also, the path integral around 0 only picks up the 1/z term of this expansion, so it is possible for the path integral to be 0 and for a function to still have an essential singularity at 0. --Dan Grubb === Subject: Re: all derivatives zero not imply zero!? >> may you give me an example of function f with all derivative of f zero >> at p such that f is not identically zero in a little neighborhood of >> zero? > Consider > f(x) = 0 if x = 0, > exp(-1/x^2) otherwise > at p = 0. > I know this to be true, and I studied it 30 years ago, but is there a > simple answer to the following question? > Calling your function f(x), its Taylor Series expansion is identically > zero. Ahem. That function is not representable by a Maclaurin series, so itıs surely misleading to say its Taylor Series expansion at x = 0. > Why canıt you take any arbitrary function g(x) and form g(x) + f(x) to > make a function with the same Taylor series expansion as g(x) but wildly > different behaviour. Doesnıt this invalidate a Taylor Series expansio as > deŜning a speciŜc function? Think about approximating f(x) using Taylor polynomials P_n(x). Except when x = 0, the remainder term _does not vanish_ as n increases without bound. The thing thatıs invalid is thinking, by merely going through the mechanics of attempting to get a series representation of a function on some nongenerate interval, that we must always succeed. David === Subject: Re: all derivatives zero not imply zero!? X-RFC2646: Format=Flowed; Original > may you give me an example of function f with all derivative of f zero > at p such that f is not identically zero in a little neighborhood of > zero? >> >> Consider >> >> f(x) = 0 if x = 0, >> exp(-1/x^2) otherwise >> >> at p = 0. >> I know this to be true, and I studied it 30 years ago, but is there a >> simple answer to the following question? >> Calling your function f(x), its Taylor Series expansion is identically >> zero. > Ahem. That function is not representable by a Maclaurin series, so itıs > surely misleading to say its Taylor Series expansion at x = 0. >> Why canıt you take any arbitrary function g(x) and form g(x) + f(x) to >> make a function with the same Taylor series expansion as g(x) but wildly >> different behaviour. Doesnıt this invalidate a Taylor Series expansio as >> deŜning a speciŜc function? > Think about approximating f(x) using Taylor polynomials P_n(x). Except > when > x = 0, the remainder term _does not vanish_ as n increases without bound. Yeah, but I thought that the whole idea was that f(0) + xf Œ(0) + 1/2 x^2f Œı(0) ... approximated f(x) to any degree of accuracy. Now you are telling me that it really tells you nothing at all about any other point on the curve. Makes the Taylor Expansion pretty useless, I would have thought. > The thing thatıs invalid is thinking, by merely going through the > mechanics > of attempting to get a series representation of a function on some > nongenerate interval, that we must always succeed. nongenerate? === Subject: Re: all derivatives zero not imply zero!? >Yeah, but I thought that the whole idea was that >f(0) + xf Œ(0) + 1/2 x^2f Œı(0) ... >approximated f(x) to any degree of accuracy. Now you are telling me that it >really tells you nothing at all about any other point on the curve. Makes >the Taylor Expansion pretty useless, I would have thought. Look at the Taylor expansion of 1/(1-x) around x=0: 1/(1-x) = 1 + x + x^2 + x^3 + ... Now, plug in x=2 in both sides. You get: 1/(1-2) = 1 + 2 + 2^2 + 2^3 + ... Let left hand side equals -1/2. The right hand side certainly does not add up to -1/2. Do you see the problem? Therefore your statement that Taylor series approximates a function to any degree of accuracy needs to be re-examined. Chances are that when you learned about Taylor series, you also learned about something called the radius of convergence. The radius of convergence gives the extent of the region over which the Taylor series expansion converges. The function exp(-1/x^2) that others have cited, has a radius of convergence of zero. Therefore its Taylor series says nothing much about the function itself. -- Rouben Rostamian === Subject: Re: all derivatives zero not imply zero!? >>Yeah, but I thought that the whole idea was that >>f(0) + xf Œ(0) + 1/2 x^2f Œı(0) ... >>approximated f(x) to any degree of accuracy. Now you are telling me that it >>really tells you nothing at all about any other point on the curve. Makes >>the Taylor Expansion pretty useless, I would have thought. > Look at the Taylor expansion of 1/(1-x) around x=0: > 1/(1-x) = 1 + x + x^2 + x^3 + ... > Now, plug in x=2 in both sides. You get: > 1/(1-2) = 1 + 2 + 2^2 + 2^3 + ... > Let left hand side equals -1/2. The right hand side certainly does > not add up to -1/2. Do you see the problem? > Therefore your statement that Taylor series approximates > a function to any degree of accuracy needs to be re-examined. > Chances are that when you learned about Taylor series, you also > learned about something called the radius of convergence. The > radius of convergence gives the extent of the region over which the > Taylor series expansion converges. In the case at hand, the radius of convergence of the Taylor series is inŜnite. It converges to the constant function, f(x) = 0. 0 + 0 + 0 + 0 + ... = 0 The problem is that it does not converge to the original function, f(x) = e^(-1/x^2), f(0)=0 If I recall correctly, the n term Taylor series for a function f(x) is f(x) = f(0) + fı(0) x + fı(0)x^2/2 + fı(0)x^3/3! + ... f_n(0)x^n/n! +f_n+1(k)x^(n+1)/(n+1)! where 0 < k < x Reaching even farther back into semi-reliable memory, this is a consequence of the mean value theorem. One can have a case where the limit as n increases without bound of the sum of base terms of the Taylor series diverges. This is what you have as the basis of radius of convergence And one can have a case where the limit as n increases without bound of the remainder term does not vanish. This is enough to ensure that, assuming the base terms converge, the converged-to result does not match the original function. > The function exp(-1/x^2) that others have cited, has a radius > of convergence of zero. Therefore its Taylor series says nothing > much about the function itself. Nope. It has an inŜnite radius of convergence by your deŜnition. John Briggs === Subject: Re: all derivatives zero not imply zero!? >> The function exp(-1/x^2) that others have cited, has a radius >> of convergence of zero. Therefore its Taylor series says nothing >> much about the function itself. >Nope. It has an inŜnite radius of convergence by your deŜnition. -- Rouben Rostamian === Subject: Re: all derivatives zero not imply zero!? Originator: grubb@lola >Chances are that when you learned about Taylor series, you also >learned about something called the radius of convergence. The >radius of convergence gives the extent of the region over which the >Taylor series expansion converges. >The function exp(-1/x^2) that others have cited, has a radius >of convergence of zero. Therefore its Taylor series says nothing >much about the function itself. A function doesnıt have a radius of convergence, a power series does. The power series for exp(-1/x^2) has *inŜnite* radius of convergence since it is the zero series. The problem is that this series doesnıt converge to exp(-1/x^2) anywhere except at x=0. --Dan Grubb === Subject: Re: all derivatives zero not imply zero!? >>Chances are that when you learned about Taylor series, you also >>learned about something called the radius of convergence. The >>radius of convergence gives the extent of the region over which the >>Taylor series expansion converges. >>The function exp(-1/x^2) that others have cited, has a radius >>of convergence of zero. Therefore its Taylor series says nothing >>much about the function itself. >A function doesnıt have a radius of convergence, a power series >does. The power series for exp(-1/x^2) has *inŜnite* radius of >convergence since it is the zero series. The problem is that this >series doesnıt converge to exp(-1/x^2) anywhere except at x=0. You are quite right. I stand corrected. -- Rouben Rostamian === Subject: Re: all derivatives zero not imply zero!? >> may you give me an example of function f with all derivative of f zero >> at p such that f is not identically zero in a little neighborhood of >> zero? > > Consider > > f(x) = 0 if x = 0, > exp(-1/x^2) otherwise > > at p = 0. > I know this to be true, and I studied it 30 years ago, but is there a > simple answer to the following question? > Calling your function f(x), its Taylor Series expansion is identically > zero. >> Ahem. That function is not representable by a Maclaurin series, so itıs >> surely misleading to say its Taylor Series expansion at x = 0. > Why canıt you take any arbitrary function g(x) and form g(x) + f(x) to > make a function with the same Taylor series expansion as g(x) but wildly > different behaviour. Doesnıt this invalidate a Taylor Series expansio as > deŜning a speciŜc function? >> Think about approximating f(x) using Taylor polynomials P_n(x). Except >> when >> x = 0, the remainder term _does not vanish_ as n increases without bound. >Yeah, but I thought that the whole idea was that >f(0) + xf Œ(0) + 1/2 x^2f Œı(0) ... >approximated f(x) to any degree of accuracy. Common misconception. >Now you are telling me that it >really tells you nothing at all about any other point on the curve. Makes >the Taylor Expansion pretty useless, I would have thought. No. The series does represent the function for many functions. How can you tell? Well, the easiest ways to tell involve complex variables. But without any complex: You can Ŝnd a thing called Taylorıs Theorem in most calculus books - it will give an expression for the _error_ in approximating f by the n-th-degree Taylor polynomial. If the error -> 0 as n -> inŜnity then the function equals its Taylor series. >> The thing thatıs invalid is thinking, by merely going through the >> mechanics >> of attempting to get a series representation of a function on some >> nongenerate interval, that we must always succeed. >nongenerate? David C. Ullrich === Subject: Re: all derivatives zero not imply zero!? > may you give me an example of function f with all derivative of f > zero at p such that f is not identically zero in a little > neighborhood of zero? >> >> Consider >> >> f(x) = 0 if x = 0, >> exp(-1/x^2) otherwise >> >> at p = 0. >> >> I know this to be true, and I studied it 30 years ago, but is there a >> simple answer to the following question? >> >> Calling your function f(x), its Taylor Series expansion is identically >> zero. > Ahem. That function is not representable by a Maclaurin series, so itıs > surely misleading to say its Taylor Series expansion at x = 0. >> Why canıt you take any arbitrary function g(x) and form g(x) + f(x) to >> make a function with the same Taylor series expansion as g(x) but >> wildly different behaviour. Doesnıt this invalidate a Taylor Series >> expansio as deŜning a speciŜc function? > Think about approximating f(x) using Taylor polynomials P_n(x). Except > when > x = 0, the remainder term _does not vanish_ as n increases without > bound. > Yeah, but I thought that the whole idea was that > f(0) + xf Œ(0) + 1/2 x^2f Œı(0) ... > approximated f(x) to any degree of accuracy. Thatıs _if_ the remainder term in Taylorıs theorem vanishes. > Now you are telling me that > it really tells you nothing at all about any other point on the curve. > Makes the Taylor Expansion pretty useless, I would have thought. > The thing thatıs invalid is thinking, by merely going through the > mechanics > of attempting to get a series representation of a function on some > nongenerate interval, that we must always succeed. > nongenerate? Of length > 0. After all, representing f(x) on the interval [0, 0] is not of much interest. David === Subject: Re: Need some help with Ramanujan and self-referencing equations [i.e. x=f(x)] >Iım looking for some examples of self-referencing equations for an >essay Iım I donıt know what equation of Ramanujanıs youıre referring to: > a quick look at Hardyıs Ramanujan and Ramanujanıs collected papers > didnıt yield anything that I could recognize as being in this form. > The equation is called a Ŝxed-point equation (i.e. x is a Ŝxed point > of the function f). > For example, the equation x = cos(x) has one real solution, which > is the limit of the sequence > 0, cos(0), cos(cos(0)), cos(cos(cos(0))), .... > Robert Israel its a common methos of Ŝnding solutions of equations of one variable. There is a theory for stablishing the convergence of the process. As examples we have: The famous formula recursive for Ŝnding the Golden SEction: x= 1/(x+1)or the formula for obtaining the square root of N : x = (N/x = x)/2 or the aproximate formula for the number of primes below N, pi(N)= x = N/(Log(N)- f(x)) ; f(x)= 1 + 1/Log(x), begining with f(x) = 1 . But beware! The self-reference equation x = kx^2 - 1 have Ŝxed points only in the interval -0.75 < k < 0.75 . In the interval 0.75 < k < 2 it oscilates but is periodic. For the value k=2 is chaotic. === Subject: reciprocal integrable? this is a rather embarassing question, anyway, here it is: Given a function f:[0,1] -> R which is bounded, non-negative, analytic and has exactly one zero x_0 (i.e. f(x_0) = 0). What are sufŜcient conditions such that int_0^1 1/f(x) dx exists? I guess we need some conditions, eg on the derivatives, which tell us that f gets away from 0 rather fast ... A reference would be very much appreciated :-) === Subject: A peculiar function deŜnition Consider the function i constructed today while killing some time.. f_m(x) = undeŜned, if x = m m , otherwise What is f_m(f_m(x)) ? It seems this trivial construction of f(x) is such that we can say nothing about f(f(x)). Now some (possibly lame) questions.. 1. Is there a name for such functions ? 2. Are these functions related to Godelıs proof in any way ? If we see these functions as propositions, they seem to negate themselves. Manik === Subject: Re: A peculiar function deŜnition >Consider the function i constructed today while killing some time.. >f_m(x) = undeŜned, if x = m > m , otherwise >What is f_m(f_m(x)) ? >It seems this trivial construction of f(x) is such that we can say >nothing about f(f(x)). Maybe Iım missing something, but I disagree. Take m = 2 as an example. Then the domain of f is all reals except 2, and on its domain f is a constant function. Therefore fof has no domain at all since the range of the inner f is the single number 2, and thatıs not in the domain of the outer f. I think that means that f canıt be composed with itself. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: A peculiar function deŜnition > Consider the function i constructed today while killing some time.. > f_m(x) = undeŜned, if x = m > m , otherwise > What is f_m(f_m(x)) ? > It seems this trivial construction of f(x) is such that we can say > nothing about f(f(x)). > Now some (possibly lame) questions.. > 1. Is there a name for such functions ? > 2. Are these functions related to Godelıs proof in any way ? If > we see these functions as propositions, they seem to negate > themselves. > Manik The composition you give is the empty function whose domain is the empty set. A function f:A->B is a subset of AxB so that for every a in A there is a unique b in B with (a,b) in the subset. so if the domain is empty you can think of the empty set as a function from the empty set to any other set. -Ron === Subject: Re: A peculiar function deŜnition > Consider the function i constructed today while killing some time.. > f_m(x) = undeŜned, if x = m > m , otherwise > What is f_m(f_m(x)) ? > It seems this trivial construction of f(x) is such that we can say > nothing about f(f(x)). > Now some (possibly lame) questions.. > 1. Is there a name for such functions ? It isnıt a function. > 2. Are these functions related to Godelıs proof in any way ? If > we see these functions as propositions, they seem to negate > themselves. > Manik === Subject: Re: A peculiar function deŜnition >>f_m(x) = undeŜned, if x = m >> m , otherwise >>What is f_m(f_m(x)) ? >>It seems this trivial construction of f(x) is such that we can say >>nothing about f(f(x)). >>Now some (possibly lame) questions.. >>1. Is there a name for such functions ? > It isnıt a function. To elaborate, for it to be a well-deŜned function, it can *never* allow an undeŜned value - any function must remove such values from its domain. You _could_ say f(x) = 1 if x=1, or 0 otherwise, but that removes your paradox, and allows that f(f(x)) = f(x). === Subject: Re: A peculiar function deŜnition Discussion, linux) >f_m(x) = undeŜned, if x = m > m , otherwise >What is f_m(f_m(x)) ? >It seems this trivial construction of f(x) is such that we can say >nothing about f(f(x)). >Now some (possibly lame) questions.. >1. Is there a name for such functions ? >> It isnıt a function. > To elaborate, for it to be a well-deŜned function, it can *never* allow > an undeŜned value - any function must remove such values from its > domain. What nonsense is this? There is a perfectly well-deŜned notion of partial function, and f_m is a partial function. (A partial function f: X -> Y is the same thing as a function f: X -> 1 + Y, where 1 + Y is the disjoint union of Y with {*}.) I donıt understand the original question either, though. Usually, composition of partial functions is deŜned thus: f o g(x) = undeŜned if g(x) is undeŜned f(g(x)) otherwise Now, letıs analyze f_m o f_m(x). Two cases: (1) x = m. Then f_m(x) is undeŜned and hence so is f_m o f_m(x). (2) x != m. Then f_m(x) = m and so f_m o f_m(x) is undeŜned. So, the original poster seems incorrect. We can say something about f_m o f_m. It is the everywhere undeŜned function. (We could decide that f_m(undeŜned) is supposed to be m if we wished, although this is non-standard. In this case, nothing too special happens either. f_m(f_m(x)) is the function sending m to m and undeŜned for all other values.) -- Jesse F. Hughes Itıs your choice though, if you do not believe in mathematics, in the importance of its healthiness and correctness, then you can just walk away now. -- James S Harris, on the Pythagorean Oath === Subject: Re: A peculiar function deŜnition - I was speaking from the deŜnition of a (normal) function as given to me. === Subject: Re: Category of categories axiomatically >> How to deŜne the category of (locally small) categories axiomatically? > Maybe my question is on a highly sensitive issue in Category Theory, which > explains there are no replies ^_^. Indeed it is not clear that the notion > of category of categories makes sense. The question does make sense. But you have to be careful. There is a lots of things known about the category Cat of small categories: it is locally small, complete, cocomplete, cartesian closed, etc... I believe that considering three Grothendieck universes U subset V subset W is sufŜcient to solve most of the problems of size. The unique rule is then: instead of talking about sets, collections, and I dont know what next, use the term U-small set, V-small set, W-small set. A U-small set has a U-small cardinal. etc... In other terms, at each step, specify cleary to which universe the set under consideration belongs. Instead of talking about small category, use the term U-small category, i.e. a U-small set of U-small objects with U-small homsets : denote by Cat_U the corresponding V-small and locally U-small category. Instead of talking about locally small category, use the term locally U-small category, i.e. a V-small set of U-small objects with U-small homsets. Denote by CAT_U the corresponding category: if I dont make any mistake, CAT_U is certainly not V-small, but is W-small because CAT_U subset Cat_V and the latter is W-small. Since Cat_V is V-complete, it is U-complete. So I believe (but I did not check that) that CAT_U is U-complete as well: the only thing you have to check is that a U-small limit in CAT_U stays inside CAT_U. CAT_U is probably U-cocomplete as well. CAT_U is probably not cartesian closed however because the obvious candidat for HOM(C,D) is the category of functors from C to D. And it is well-known that HOM(C,D) is in CAT_U iff C is essentially U-small (i.e. equivalent to a U-small category). etc... pg. === Subject: strange poisson question. hello.....doctor..... Suppose telephone calls arrive on average 10 every 1 minute. Assuming a Poisson process, what is the probability of 250 or fewer calls arriving in a 30 minutes period. --------------------------------------------------- um.....i think..... 1) since telephone calls arrive on average 10 every 1 minute, f(x) = (10^x)*(e^{-10}) / (x!) , (x = 0, 1, 2....) and 250 or fewer calls arriving in a 30 minutes period. = 25/3 or fewer calls arriving in a 1 minute period. so, P(X <= 25/3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) so, N[Sum[(10^x)*(Exp[-10])/(x!), {x, 0, 8}]] = 0.33282 by mathematica. 2) since telephone calls arrive on average 300 every 30 minute, f(x) = (300^x)*(e^{-300}) / (x!) , (x = 0, 1, 2....) so, P(X <= 250) = P(X=0) +.... + P(X=250) so, N[Sum[(300^x)*(Exp[-300])/(x!), {x, 0, 250}]] = 0.00168113 by mathematica. 3) i know the next theorem. if X~Poi(1), then for large values of 1, X~N(1,1) approximately. (1 = the mean of the poisson distribution) so, since telephone calls arrive on average 300 every 30 minute, X ~ Poi(300) ~ N(300, 300) so, P(X <= 250) = P(Z <= [(250 - 300)/sqrt{300}]) = P(Z <= -2.88678) = 0.002 ------------------------------------------------------- um......the result of (2) and (3) is very small. anyway, is (1) wrong process ? i need your advice. thank you very much for your advice. === Subject: Re: How to do magic with inŜnity > Ozkural: > As to what it means that even natural numbers and natural numbers have > the same _cardinality_, I would rather ask you Ŝrst, because I think > we should Ŝnd the same answer. (since we are both not realists) There > is a proof. The proof is correct. What does this proof mean to you > (regardless of how you think the size of inŜnite sets must be > measured)? > When deŜned as there exists a bijection between the set of All natural > numbers from 1 .. N and the set of Even natural numbers from 2 .. 2N , > I agree that the cardinality of these, quite distinct sets, is the same; > even for N -> oo . But that is not a fair comparison. One should compare > instead 2 .. 2N , only as far as it is contained as a subset in the set > 1 .. N . This leads to quite a different answer, which is mine. So, you opt for the subset approach to measure the size of inŜnite (or Ŝnite) sets instead of the bijection approach. I am not sure if that is the correct deŜnition however. There are clearly many sets whose intersection is 0, and still can be compared in magnitude. How do you handle those? That is why, I tried to deŜne all these things in algorithmic ways. I donıt know if I was successful but it seemed to me more general than the subset approach. -- Eray Ozkural === Subject: Re: How to do magic with inŜnity > As I said, it is a fact that the the set of even naturals has the > same cardinality as the set of all naturals. > Looks like it is reasion to copy paste: > Galilei wondered N->2N bijection. And discovered that 2N is a subset of N > but still > there is a bijection from N->2N. > Cantor studied Galileiıs thoughts and make deŜnation. Set that has subset > that > is as big (i.e there is that bijection) as the set itself is inŜnite. > If i say and proof, that N can not be inŜnite, then there is no meaning > about those > cardinals, aleph_0 etc... They are nonsense. > There is not even the Ŝrst inŜnite set. That conception is paradoxical. > And so > there can not be different aleph_s or what ever.. Kroneckerıs point of view, it seems. === Subject: Re: The Red Sox-Yankee Playoff and Probability by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MDn0B11856; >As many of you know the Boston Red Sox defeated the New York Yankees in a best >of seven series 4 games to 3. They did it by winning 4 after losing 3. This had >never been done in MLB postseason baseball before. And twice in NHL hockey, and >0 in NBA basketball out of a sample IIRC of less than 400 times combined in all >above. >Now if you had an accurate random number generator simulate 100,000,000,000 >games of completely random chance between two individuals what percentage of >the time, if that could be reasonably approximated, would one individual or the >other win three times in a row? And of those three times in a row wins what >percentage would the other individual win the next four? Would a sample of 400 >three wins in a row and eachıs follow up of four games tell us anything? Assuming that one team was has probability p of winning any speciŜc game (and the other team has probability q= 1- p) then the probability of that team winning three games in a row is p^3. In particular, if the two teams are equally likely to win (p= q= 1/2), then the probability of one team winning three games in a row is (1/2)^3= 1/8. === Subject: Re: The Red Sox-Yankee Playoff and Probability 1/8. The teams start at 0:0. One of the teams will win the Ŝrst game. The probability the same team will win the second game is 1/2. The probability the same team will win the next game too is 1/2, giving a total probability of 1/4. However the probability that the OTHER team will win the next 4 games is 1/16. YEAH REDSOCKS!!!!!!!!!! === Subject: Re: The Red Sox-Yankee Playoff and Probability === >Subject: Re: The Red Sox-Yankee Playoff and Probability >>As many of you know the Boston Red Sox defeated the New York Yankees in a >best >>of seven series 4 games to 3. They did it by winning 4 after losing 3. This >had >>never been done in MLB postseason baseball before. And twice in NHL hockey, >and >>0 in NBA basketball out of a sample IIRC of less than 400 times combined in >all >>above. >>Now if you had an accurate random number generator simulate 100,000,000,000 >>games of completely random chance between two individuals what percentage of >>the time, if that could be reasonably approximated, would one individual or >the >>other win three times in a row? And of those three times in a row wins what >>percentage would the other individual win the next four? Would a sample of >400 >>three wins in a row and eachıs follow up of four games tell us anything? > Assuming that one team was has probability p of winning any speciŜc game >(and the other team has probability q= 1- p) then the probability of that >team winning three games in a row is p^3. > In particular, if the two teams are equally likely to win (p= q= 1/2), >then the probability of one team winning three games in a row is (1/2)^3= >1/8. Iım a doctor, not a mathemetician, Jim !:) Could you please answer my question in a way an ordinary baseball fan could understand? === Subject: Re: all derivatives zero not imply zero!? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MDn0811860; >may you give me an example of function f with all derivative of f zero >at p such that f is not identically zero in a little neighborhood of >zero? >Tern A well-known problem. Consider f(x) = exp(-1/x) if x > 0 = 0 otherwise at the point p = 0. This is useful to know, if you ever need to construct C^{inŜnity} bump functions (q.v.). Todd Trimble === Subject: Does this deŜnition of dim 2 n-free poset exterior ring any bells with anyone? Let H be the Hasse diagram of any dim 2 n-free poset and let Hp be any natural plane embedding of H. Then inasmuch as Hp is a DAG whose edges are deŜned by the immediate covering relations in H, one can assign the integer d to each vertex v of Hp, where d is the depth of v in Hp interpreted as a DAG (i.e. the number of vertices in Hp which are ancestors of v, excluding v itself.) Further, since Hp is embedded in the plane relative to the geometry of the paper, a left-right or horizontal ordering of the vertices of Hp results from its embedding in the plane. That is, we can assign the numbers p and t to each vertex v of Hp, where: p is the number of vertices to the left of v in Hp t is the number of vertices to the right of v in Hp. And it is readily shown that because Hp is both dim 2 and n-free: a) the values (p0+d0),...,(pi+di),...,(pn+dn) for the n+1 vertices of Hp are 0,...,n+1 b) the values (t0+d0),...,(ti+di),...,(tn+dn) for the n+1 vertices of Hp are a permutation of 0,...,n+1. Suppose now, however, that one uses the integers pi, ti, di to assign each vertex v of Hp the triple of integers (pi,ti,di). And further, suppose that one deŜnes the exterior of Hp as just those vertices in Hp for which at least one of pi,ti,di is 0. And Ŝnally, suppose that one deŜnes the exterior union of Hp as the union of all the exteriors of all the subgraphs of Hp which are themselves interpretable as Hasse diagrams of dim 2 n-free posets. Does this notion of exterior union ring any bells with anyone ?? considering this matter. === Subject: Re: Jamesı object ring !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ıELIi $t^ VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > It turns out that the story here is *really* old as years back I > started talking about ŝat rings which drew a lot of derision on > sci.math and later I learned of algebraic integers and thought for a > while they were my ŝat ring. > The simplest way to understand the ring in complex numbers is that if > you have any number z=x/y in the ring, where x, y and z are in that > ring, and, of course, y is not 0, then x and y cannot be factors of > any integers that are coprime in the ring of integers. > So it is a ŝat ring, in that given z = x/y, the y must be a factor of > x without contradicting factorizations in the ring of integers. > Some have attempted to add numbers like pi to the ring, but itıs an ad > hoc thing where they basically just *say* add pi to the ring. > However, it is easy enough to show that you can construct all numbers > in the ring starting with 1 and -1 and a few operations, but cannot > construct pi from within the ring. > If you add in what I call operators, like 1/2, which means 1 of 2, > then you can construct pi as an operator. > So you can build the entire ring of complex numbers using these ideas. > Note that coprimeness in the ring then is a simple matter of not > sharing non-unit factors. You are drunk, talking incoherently. More importantly, you are reverting to talking nonsense that is already dated several years. > At this point in time Iıve shown how the older ideas that have > dominated the math world can lead to rather simplistic errors and a > muddled view of coprimeness, like saying that if 1/2 is in the ring > that 2 is coprime to 3, though it is a factor of 3. Coprimeness is deŜned quite simple. Units are factors of _everything_. Talking about coprimeness with regard to them is useless. So what is your unmuddled view of coprimeness? Give a clear deŜnition. The usual deŜnition is a and b are coprime in a ring if there exist values c and d in the ring such that a c + b d = 1. Very clear and unmuddled. Whatıs your deŜnition? I didnıt hear you, speak up. > I think itıs time humanity began to catch-up. Well, start catching up to the state of mathematics 200 years ago Ŝrst, then complain about humanity. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: SMSU Problem Corner Itıs that time of month again. There are three new problems in the usual categories (High School, Advanced, Challenge) posted at htto://math.smsu.edu/~les/POTW.html Please visit us. === Subject: Re: Function with differentiable singular point > Math folks, > I did some more study on the function which was suggested, > f(z) = exp(-1/z^2), z not 0, f(0) = 0, for functions z of a complex > variable. > I was able to show that the function is not analytic in general, and have > posted a Mathematica write-up on my web page: > http://home.earthlink.net/~diana53/EssentialSingularity/ index.html > I understand that this function has an essential singularity at z = 0. I > guess that I was trying to Ŝnd a function with an isolated singularity at a > point, ***which is differentiable at that point***. > If this function has an isolated singularity, could you explain how to > deŜne the sequence with open circles which are analytic? > If f(0) is not an isolated singularity, can someone think of a function of > this type? > Diana > The deŜnition of singularity with which I am familiar states that the > singular point is not analytic at that point but doesnıt mention > differentiability. Does anyone know of a function for which its === Subject: Re: Function with differentiable singular point ETAsAhRpCV32R9Ox+cfFSe+ OJBZ0YYapKgIUQECtP0jBbOoo2tMavGwXCbe1PwI= There is something called a removable singularity. An example is f(z) = sin(z)/z which has no direct deŜnition at z = 0 because of the division by zero. But the function does have a well-deŜned LIMIT as z approaches zero, and so the singularity can be removed by deŜning f(0) = 1 (the limit) along with f(z) = sin(z)/z for nonzero z. Examples of a nonisolated singularities where a function is differentiable are easy o Ŝnd. f(z) z^2 ln z can be differentiated once at the branch point z = 0, because f(z)/z = z ln z has a limit of 0. More fascinating is the Stieltjes Function f(z) = int(t = 0 to inf) (exp(-t) dt/(1+zt)) with a branch point at z = 0, which can be differentiated inŜnitely many times there! === Subject: Re: Duel to Settle 0.999... =? 1 >> No one seems to doubt: V=(4/3)PI*R^3 for sphere volume, solved by >> integrals. Itıs all the same calculus. Should we have a debate about >> sphere volume? >No, but thatıs because, as you said, no one seems to doubt that. >Horrifyingly enough, a lot of people seem to doubt that 0.999... >is indeed equal to 1 (either that, or theyıre just very clever >at jokes/sarcasm and are having a huge laugh at us, looking how >we believed their messages :-)). >I feel bad for them, as they visit and write on this newsgroup, >yet donıt have a clear understanding about numbers or mathematics >(about Calculus and limits, at least). I know I did understand >that 0.999... is equal to 1 when I was just 10 years old (it >intrigued me, yes, but that was it -- I never really doubted >it), so they really have no excuse. >Carlos 9/10^ 10000000000000000000000000000000000000000000000000000000000000 0000000 00000000000000000000000000000000000000000000000000000000000000 000000000000 00000000000000000000000000000000000000000000000000000000000000 000000000000 00000000000000000000000000000000000000000000000000000000000000 00000000000... ... never conerges to 1. It stays at .9999... . 1=1 1=1 1=1 1=1 1=1 1=1 1=1 1=1 1=1 1=1 1=1 . . . . . . . . . 2 - 1 =1 2 - .999999999.... = 1.99999999.... Canıt you get that through your thick skulls? Convergence at inŜnity occurs when the derivative converges to a certain number. Like for example, at x = 1 The derivative, d x^2 = 2x at x = 1 d x^2 = 2 This converges to a speciŜc number at x = 1 at inŜnity, to the number 2. at x = .9999.... it converges to 1.9999.... not 2. The derivative of d x^1.999... = (1.9999...) x^.99999.... at x = 1 it equals 1.9999..... NOT 2. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: Duel to Settle 0.999... =? 1 >it equals 1.9999..... NOT 2. How about writing a thesis paper to be submitted and reviewed! It could be a huge discovery! - http://mysite.verizon.net/vze8adrh/news.html (proŜle) --Tim923 My email is valid. === Subject: Re: Duel to Settle 0.999... =? 1 > 2 - 1 =1 > 2 - .999999999.... = 1.99999999.... > Canıt you get that through your thick skulls? Convergence at inŜnity occurs > when the derivative converges to a certain number. Like for example, > at x = 1 > The derivative, > d x^2 = 2x > at x = 1 > d x^2 = 2 > This converges to a speciŜc number at x = 1 at inŜnity, to the number 2. All this is such a sense-less random sequence of words that now Iım sure youıre just trolling... But, just in case there are some other people that do not get that and still think that there is a legitimate reason for debate here... You should agree that any number whose decimal expansion has an inŜnitely repeating pattern is a rational number, and thus has a concrete fractional expression, right? (i.e., an *actual* integer number for numerator and an *actual* integer number for the denominator). So, would you care to tell me what is, according to you, the fraction representation for 0.9999..... ? I can deŜnitely tell you what is the fraction representing 0.11111... : x = 0.11111... 10x = 1.1111... = 1 + x 10x = 1 + x ==> 9x = 1 ==> x = 1/9 What other method we propose we use to determine the fraction representation of 0.9999... ?? (because if I apply the above to 0.9999..., I obtain the correct answer: 1 -- 1/1, if you will). Carlos -- === Subject: Re: Duel to Settle 0.999... =? 1 > No one seems to doubt: V=(4/3)PI*R^3 for sphere volume, Iıll doubt anything you want. > solved by > integrals. Itıs all the same calculus. Should we have a debate about > sphere volume? > http://mysite.verizon.net/vze8adrh/news.html (proŜle) --Tim923 My email is valid. === Subject: Re: Duel to Settle 0.999... =? 1 , I settle it (period). >> No one seems to doubt: V=(4/3)PI*R^3 for sphere volume, Yes there is someone that doubts this. I do. 4/3 = (1.333333.....) V = (1.333... )(3.14159........) r^3 The absolute perfection of a sphere IS NOT shown here. It is an indeterminate and it does not converge to a perfect value spherical form. It only approaches perfection. If you use the Partial Sums method of analysis of the deŜnition of a sphere, your accuracy has about 6.25% error in it (within <.0625 error range), in the equation you use. The accuracy depends on the number of terms used in the series that deŜnes the sphere perfectly. You could spend eternity in deŜning the accuracy of a near perfect sphere. The only way to reach perfection is to be the perfect sphere itself, with no beginning state of beingness and say you ARE that sphere. >Iıll doubt anything you want. >> solved by >> integrals. Itıs all the same calculus. Should we have a debate about >> sphere volume? >> - >> http://mysite.verizon.net/vze8adrh/news.html (proŜle) --Tim923 My email is >valid. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: Duel to Settle 0.999... =? 1 >Youıre too late. This issue was settled years ago. >>To All, >>It seems that the arguments over whether 0.999... = 1 or not could be easily >settle by a duel. The camp that believes 0.999... does NOT equal 1 choose a >representative. Likewise for the camp that believes 0.999 DOES equal 1. >Then the two representatives write down 0.999... explicitly (writing all 9ıs >>I suggest that the representative of the camp that believes 0.999... does >NOT equal 1 go Ŝrst. While that representative is writing his/her 9ıs, the >rest of the two camps are free to work on other unsolved mysteries of >mathematics. >>By the way, I volunteer to be the representative of the camp that believes >0.999 ... DOES equal 1. If you can say that, I can say the fairy godmother exists there too. So, .99999.... = the fairy godmother >>- MO Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: Duel to Settle 0.999... =? 1 obviously, you are unwilling to take the bullet. so, will you second me? obviously, your fairy godmother is = 1.0000... *my* godmother is the toothfairy, so, there = 0.9999... >Youıre too late. This issue was settled years ago. > If you can say that, I can say the fairy godmother exists there too. So, --A HYDROGEN (sic; cracked methane) ECONOMY?... The Three Phases of Exploitation of the Protocols of the Elders of Kyoto (sik): BORE/GUSH/NADIR @ http://www.tarpley.net. Http://www.tarpley.net/bushb.htm (partial contents, below): 17 -- THE ATTEMPTED COUP DıETAT, 3/30/81 (87K) 18 -- IRAN-CONTRA (140K) 19 -- THE LEVERAGED BUYOUT MOB (67K) 20 -- THE PHONY WAR ON DRUGS (26K) 21 -- OMAHA (25K) 22 -- GEORGE #9 TAKES THE PRESIDENCY (112K) 23 -- THE END OF HISTORY (168K) 24 -- THE NEW WORLD ORDER (255K) 25 -- THYROID STORM (139K) http://quincy4board.homestead.com/Ŝles/curriculum/Cosmo === Subject: Re: Duel to Settle 0.999... =? 1 >obviously, you are unwilling to take the bullet. so, >will you second me? > obviously, your fairy godmother is = 1.0000... >*my* godmother is the toothfairy, so, there = 0.9999... Since the fairy godmother doesnıt exist in reality, .9999...== 1 Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: How to Cheat at Craps X-RFC2646: Format=Flowed; Original >>that iterated through various probability distributions for the Roller / >>Fader die, searching for best strategies. This method is crude and >>time-cunsuming but instructive. >>This test indicated that Rollerıs best strategy is to Ŝx his die so that >>only 5 or 6 comes up. If you think about it, this is not really much of a >>surprize: the lopsided distribution raises the odds of rolling his point. >>David L. Silvermanıs book Your Move discusses the Roller strategy of >>weighting the die so that 5 comes up 100% of the time. If the other die >>were >>fair, this would raise his probability of winning from 244/495 (about >>0.493) >>to 2/3. Faderıs best reply would be to Ŝx his die so that 6 never comes >>up, >>2 comes up 1/3 of the time, and the other four sides come up 1/6 of the >>time. But even with Faderıs best strategy, Rollerıs probability of winning >>would still be 5/9. The fair game of craps favors Fader, but the cheating >>game deŜnitely favors Roller! >>My brute force approach suggests that Roller can do slightly better than >>Silvermanıs strategy by allowing 6 to come up occasionally on his die. >>When >>I tried all combinations of die probabilities where each probability is an >>integer multiple of 1/64, the best Roller weighting I found was the >>following: >> Prob. that 5 comes up: 0.765625 >> Prob. that 6 comes up: 0.234375 >> Prob. that any other side comes up: 0.0 >>Given this die, Faderıs best weighting would be the following: >> Prob. that 1 comes up: 0.234375 >> Prob. that 2 comes up: 0.359375 >> Prob. that 3 comes up: 0.109375 >> Prob. that 4 comes up: 0.21875 >> Prob. that 5 comes up: 0.078125 >> Prob. that 6 comes up: 0.0 I should have said that this is the best solution my test program found. The best solution is probably somewhat different. >>If Roller and Fader go with these die, Rollerıs probability of winning >>would >>be approximately 0.567. > But if Fader goes with this die and Roller goes back to the old > strategy of always rolling Ŝves, then Rollerıs probability of > winning is 0.574133 (a slight improvement). I suspect that the > best strategy for Roller is to load his die so that the same > number comes up every time, but to randomize his choice of which > number that is. The question is, *how* should he randomize? Interesting thought. To simplify the problem, maybe we should modify the problem to allow Fader to determine how Rollerıs die is loaded. Fader needs some help anyways! > For the Fader, it looks like a much harder problem: heıs > going to have to randomize the probabilities of all six faces > of his die, so heıs looking for a function of Ŝve variables > that optimizes his outcome. On the bright side, when Roller restricts his die to one side, the task of Ŝnding the optimal die for Fader is simpliŜed. >>This work raises the following questions: >>1. Can you prove that Roller should weight his die so that only 5 or 6 >>comes up? >>2. Assume that for some probability t in [0, 1], Roller weights his die >>as follows: >> Prob. that 5 comes up: t >> Prob. that 6 comes up: 1.0 - t >> Prob. that any other side comes up: 0.0 >> How should Fader weight his die to minimize Rollerıs chances of >>winning? >>The answer to these questions could reduce this problem to simple >>calculus! > This seems like a very hard problem to me. Does anyone else have any > ideas? It does seem hard, but somehow I feel it should yield to some existing disipline. A game theorist or a non-linear programming expert could probably make more headway. Is there another forum that I should try? > --- Steve Maye Frank J. Lhota === Subject: nonlinear ODEıs: unstable Ŝxed means tells you what what behavior ? Regarding the identiŜcation of unstable Ŝxed points in nonlinear ODEıs, I am told unstable Ŝxed points can separate stable Ŝxed points so that their stable manifolds act to distinguish basins of attraction what is the signiŜcance, in terms of behaviour in the real world, of unstability ? If one can identify unstability mathematically, what does that equate to as a real life example, and what can be gleaned from knowing this ? Jasper === Subject: Letıs get Harris published Jimmy, I know that youıve been having trouble getting published. I thought that maybe I could help. I checked with some of my buddies in the publishing business and they looked over your material and said that they could get you published! Here are the journals that you should consider sending your material to: Asimovıs Science Fiction Entertainment Weekly Humor Times Mad Magazine National Enquirer Pro Wrestling Illustrated Realms of Fantasy Soap Opera Weekly Digest (under the Humor-in-the-US section). Oppie === Subject: Re: A discovery. > >> I discovered something of signiŜcance today. >> >> What? > >I derived an equation to generate the prime number sequence. It is O(1). > > Uh, right. What does that mean, exactly? > > > ************************ > > David C. Ullrich > > i think itıs called a joke and i think O(1) is called big-O notation. > i would think someone who loves to giggle would have some sort of a > sense of humor... however twisted. > > I assure you it is not a fabrication. > congratulations Pauca sed matura. === Subject: Re: A discovery. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9L0vGC24453; >> > I discovered something of signiŜcance today. > > What? >> >>I derived an equation to generate the prime number sequence. It is O(1). >> >> Uh, right. What does that mean, exactly? >> >> >> ************************ >> >> David C. Ullrich >> i think itıs called a joke and i think O(1) is called big-O notation. >> i would think someone who loves to giggle would have some sort of a >> sense of humor... however twisted. >I assure you it is not a fabrication. so letıs see it... === Subject: Re: THE THREE LAWS OF THOUGHT > Not quite. Kant rejected the presumption of inŜnite schemata to describe > arithmetic. As I said before, and, please, let me repeat. You are a professional, a person having a broad and detailed understanding of the Ŝeld in which I am unable to even crawl. All I can do is keep your kind comments close and continue to reconsider them while I do babies stuff, i.e. read some SIMPLE texts. :-) [heavy heartedly snipped some portion of a most informative lecture] > *NO*, I WILL *NOT* DARE contradict your rejecting the necessity for > Ŝrst reŝecting on the idea of existence while working on the > principles. Please, kindly assume, that is not my intention > *WHATSOEVER*. > Is that what is inferred? All I am asking is your not taking notice of my mistakes. [snipped another portion of a most instructive lecture] For me Martin Heideggerıs: Der Satz vom Grund Die Frage nach dem Ding Was heist Denken? Zur Sache des Denkes Einfuhrung in die Metaphysik (and, of course) Sein und Zeit (I apologize for spelling discrepancies) turned out to be the absolute hits (although I cannot say I comprehended them all and thourougly). Now I am completely serious (as I have never taken anything more seriously in my life than my current endeavours). The above texts give me a much desired religious justiŜcation for fopl (the science of logos). How come? Heidegger makes a reference to John the Baptizerıs interpretation (He speaks of Jesus Christ at that) and, IN FACT!, the subsumption to his doctrine of the works of Parmenides and Heraklit (He at the same time speaks of the dispensability of the whole philosophical libraries for several papers of the two philosophers together with Aristotleıs Physics). He uniŜes the very many meanings of the term logos Aristotle uses in his works. He does so by referring to the works of Kant, Leibniz, Wolff and others while also giving signs of its contemporary (and entirely compatible with that of Aristotle) uses (e.g. the thinking machines, like tms). I slept on the ideas for a fortnight. I think I came accross a door (God, please, be this a door). Since this is my 5 minutes, please, allow me to make a quotation: ŒThe Principle of Reason, the text of a lecture course that Martin Heidegger had in 1955-56, takes as its focal point Leibnizıs principle: nothing is without reason. Heidegger shows here that the principle of reason is in fact a principle of being. Much of his discussion is aimed at bringing his readers to the leap of thinking, which enables them to grasp the principle of reason as a principle of existence. The Principle of Reason presents Heideggerıs most extensive reŝection on the notion of history and its essence, the Geschick of being, which is considered /one of the most important developments in Heideggerıs LATER thought/. One of Heideggerıs most artfully composed texts, it also contains important discussions of /language/, translation, /reason/, objectivity, and technology as well as of such Ŝgures as Leibniz, Kant, and Aristotle.ı (now, put propositional function in place of reason) The Principle of Reason. Translated by Reginald Lilly, Bloomington, Indiana University Press, 1991. Now I even know what a symbol is, and I Ŝnally started to understand Wittgesteinıs Tractatus. I Ŝnally understood the terms object and combination=concatenation in: 2.01 An atomic fact is a combination of objects (entities, things). LW, TLP But perhaps I am dreaming. PLEASE, KINDLY, WAKE ME UP (if I am). :-) > I am no expert on any of this. > Keep reading. Stay curious. > I will do so also. And because you do so also, I am able to actually do some work (words will not express how greatful I am to you and Other Posters who ever Tom P.S. Of course, after Heidegger, (and in case I am not dreaming) I consider the Ŝve principles I quoted previously absolute and out of question. IMHO, they are close enough and good enough. Still, I need fopl to determine their interdependencies as well as sufŜciency for yielding a proper mathematical system, like that of PM. Ah, yes. And let us just think - it was enough to read that one single little neglected by many geniuses paper by Alan Turing Can machines think? A paper in which by formulating and ANSWERING that question Turing (and that is after Heidegger) makes the whole philosophical libraries dispensable. I think this is the greatest paper I have ever read, and I can not believe I was once IDIOTIC enough not understand its obvious contents. === Subject: Bob, Iım calling you out! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MI48804308; Bob (you know who you are), I challenge you to Ŝnd one inconsistency or one ill-deŜned phrase in my post EFFICIENCY: PRIME TEST vs PRIME GENERATOR. In case you donıt like the WAY I deŜned my generator, Iıll even redeŜne it here for you: Let p(n) be the nth prime Let p(n)# be p(n)*p(n-1)*p(n-2)*...*5*3*2 Let abs() be absolute value Let +/- be plus or minus, though the theory works with just minus Let q(x1,x2,x3,...xn):positive integers --> positive integers given by q(x1,x2,x3,...xn)= abs( A +/- B) where A and B are given by: p(n)# divides A*B p(n+r) doesnıt divide A*B for all r in the positive integers gcd(A,B)=1 each xi is the positive integer exponent on each p(i) Then if q(x1,x2,x3,...xn) < [p(n+1)]^2, then q(x1,x2,x3,...xn)is prime. This is because contradiction of the distributive principle shows that q(x1,x2,x3,...xn) will not have any of p(1), p(2),... p(n) as factors, and the square root of q(x1,x2,x3,...xn) is less than p(n+1). Now, is that well-deŜned enough? As I said before, to get all primes < [p(n+1)]^2, itıs important to use ALL POSSIBLE disjoint partitions of {p(1),....,p(n)} into the two sets to be put into A and B. This will produce primes without testing any composites, because the elimination of those composites is intrinsic to the deŜnition of A and B, the gcd(A,B)=1, and the less than test. This indicates that this algorithm may still be in the running as the most efŜcient way to Ŝnd primes, given that it doesnıt waste time testing a large number of BIG composites. Prove me wrong, if you dare! === Subject: Re: What is the intuitive thought behind the cross-correlation function? Itıs like interference between waves (eg physics, waves, ripple tanks, etc). If the signals are similar, then When the two signals are similar in shape and unshifted with respect to each other, their product is all positive. This is like constructive interference, where the peaks add and the troughs subtract to emphasise each other. The area under this curve gives the value of the correlation function at point zero, and this is a large value. As one signal is shifted with respect to the other, or when they are not similar, the signals go out of phase - the peaks no longer coincide, so the product can have negative going parts. This is a bit like destructive interference, where the troughs cancel the peaks. The area under this curve gives the value of the correlation function at the value of the shift. The negative going parts of the curve now cancel some of the positive going parts, so the correlation function is smaller. The largest value of the correlation function shows when the two signals were similar in shape and unshifted with respect to each other (or Œin phaseı). The breadth of the correlation function - where it has signiŜcant value - shows for how long the signals remain similar. http://www.bores.com/courses/intro/time/2_corr.htm Hope it helps, Chris Chris Bore BORES Signal Processing www.bores.com chris@bores.com > Hi all, > I was told the claim that the cross-correlation of two functions f(x) and > g(x), deŜned as : > (f & g) (x)=integrate(f(u), g(u+x), w.r.t. u from -inf to inf) > reŝects the similarity between the two functions f and g. > After some Matlab experiments, I was convinced. > But what is the intuitive(phylosophical) thought behind this > cross-correlation function? Why it reŝects the similarity between the two > functions f and g? Can it be proved mathematically? > Any thoughts? > -N === Subject: Re: What is the intuitive thought behind the cross-correlation function? I think Rune has given a great explanation. Nevertheless, here is my not-very-rigorous proof of the whole thing. Given function f(x) and g(x). Let r(x) = f(x) - g(x) Without loss of generality, assume the energies of f(x) and g(x) are both equal to E, i.e. Integrate( f(x)*f(x) )dx = E Integrate( g(x)*g(x) )dx = E (I am going to omit dx from now on.) The correlation between f(x) and g(x) is given by: Integrate( f(x)*g(x) ) = Integrate( ((g(x)+r(x)) * g(x) ) = Integrate( g(x)*g(x) ) + Integrate( r(x)*g(x) ) = E + Integrate( r(x)*g(x) ) .................................(A) However, the same correlation can also be written as Integrate( f(x)*g(x) ) = Integrate( f(x) * ((f(x)-r(x)) ) = Integrate( f(x)*f(x) ) + Integrate( -r(x)*f(x) ) = E + Integrate( r(x)*(-f(x)) ) .................................(B) Add up (A) and (B), you get: 2*( correlation of f(x) and g(x) ) = 2*E + Integrate( r(x)*( g(x)-f(x) ) ) = 2*E + Integrate( -(r(x)^2) ) Since -(r(x)^2) is always <= 0, its integral is non-positive => 2*(correlation of f and g) <= 2*E => correlation of f(x) and g(x) <= E In this process, we have proved that the correlation is maximum when r(x)=0. That is, when f(x)=g(x) By this proof, we have also shown (1-normalized correlation value) is equal to the energy of the difference between 2 functions w.r.t the total energy. In the discrete world, correlation tells you the sum of the squares of the difference between 2 signals, which is a very useful parameter in many applications. KD === Subject: Re: What is the intuitive thought behind the cross-correlation function? > In the discrete world, correlation tells you the sum of > the squares of the difference between 2 signals, which is a very > useful parameter in many applications. A nit, shouldnıt you be using the term inner product where you use correlation in your post? Correlation is the sliding inner product. Bob -- Things should be described as simply as possible, but no simpler. A. Einstein === Subject: Re: What is the intuitive thought behind the cross-correlation function? You slide one function over the other, one sample at a time, and at each slid point you: - multiply corresponding samples of each function - sum all those products Summing the prducts is like taking the area under the resulting curve of the products. Now, if the signals are similar, and unslid, then their products will be like squaring one signal - which will all be positive going, so the area under this curve wil all add (constructive interefernce - remember physics of waves?) and so will be big. If the signals are not similar, or are slid, then some positive signal samples will multiply negative signal samples, so some of the resulting product samples will be negative, and these negative going parts of the curve will subtract from the positive parts (destructive interference) so the area will be less. I think I describe this with a diagram at: http://www.bores.com/courses/intro/time/2_corr.htm Chris Chris Bore BORES Signal Processing chris@bores.com www.bores.com > Hi all, > I was told the claim that the cross-correlation of two functions f(x) and > g(x), deŜned as : > (f & g) (x)=integrate(f(u), g(u+x), w.r.t. u from -inf to inf) > reŝects the similarity between the two functions f and g. > After some Matlab experiments, I was convinced. > But what is the intuitive(phylosophical) thought behind this > cross-correlation function? Why it reŝects the similarity between the two > functions f and g? Can it be proved mathematically? > Any thoughts? > -N === Subject: Re: Integrability of the derivative of inverse function >Suppose $f:[a,b]toR$ is a differentiable function whose >derivative stay away from zero, i.e., there is $delta > 0$ >such that $|fı|geqdelta$ on $[a,b]$. Let $[c,d]:=f([a,b])$ >and $f^{-1}:[c,d]to [a,b]$ be the inverse of $f$. Clearly, >$f^{-1}$ is differntiable. Now the question is the following. >If $fı$ is integrable on $[a,b]$, must the derivative of >$f^{-1}$ be integrable on $[c,d]$ ? Any monotone function on a bounded interval is of bounded variation, and therefore its derivative is (Lebesgue) integrable. See e.g. Rudin, Real and complex analysis, Theorem 8.19. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Integrability of the derivative of inverse function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9JCdN925073; >> Suppose $f:[a,b]toR$ is a differentiable function whose >> derivative stay away from zero, i.e., there is $delta > 0$ >> such that $|fı|geqdelta$ on $[a,b]$. Let $[c,d]:=f([a,b])$ >> and $f^{-1}:[c,d]to [a,b]$ be the inverse of $f$. Clearly, >> $f^{-1}$ is differntiable. Now the question is the following. >> If $fı$ is integrable on $[a,b]$, must the derivative of >> $f^{-1}$ be integrable on $[c,d]$ ? >Yes. The function f^{-1} is a bounded monotone function from >[c,d] into the reals. It is well known that such a function is >always Riemann integrable. >Jose Carlos Santos I think you have missed the point. Of course, it is clear that f^{-1} be integrable on [c,d]. The question is whether the derivative of f^{-1} is integrable on [c,d]. Sudhir Ghorpade === Subject: Favourite Graphing Calculator? posting-account=eo45pQ0AAACyZLf8WBQ_3PH8XOXpyezd I hope this hasnıt been raised before, I couldnıt Ŝnd a mention in the FAQ or recent messages on Google Groups. Iım starting a calculus class (1st year) and they recommend the use of a graphing calculator. I know many people on this group might disagree with students today using one, and in fact Iım inclined to agree. However, Iıve decided it would be wise to take whatever advantage theyıll let me have. :) Iıve never owned a graphing calculator and would appreciate any opinions on favourite models. Iıll be speaking to my course advisor about speciŜc models allowed/prohibited/recommended, but I would also like to hear from the group. Iıve heard the TI-83(+) models recommended - any thoughts? === Subject: Re: Favourite Graphing Calculator? If you are going to be a major/honours student in Math or Physics or Engineering - or 1st and 2nd year Math courses go for the TI-89 or better. If you are going to do ONLY two semesters of Calculus - the TI-83+ is okay. -- Casey === Subject: Re: Favourite Graphing Calculator? : Iıve heard the TI-83(+) models recommended - any thoughts? Other posters have recommended the TI-89, but be very careful to make sure this is allowed in your calculus class. As it is capable of symbolic integration, it is not allowed in the calculus classes at many universities. The TI-83(+) is generally allowed, and is very good. Justin === Subject: Re: Favourite Graphing Calculator? > I hope this hasnıt been raised before, I couldnıt Ŝnd a mention in the > FAQ or recent messages on Google Groups. > Iım starting a calculus class (1st year) and they recommend the use of > a graphing calculator. I know many people on this group might disagree > with students today using one, and in fact Iım inclined to agree. > However, Iıve decided it would be wise to take whatever advantage > theyıll let me have. :) > Iıve never owned a graphing calculator and would appreciate any > opinions on favourite models. Iıll be speaking to my course advisor > about speciŜc models allowed/prohibited/recommended, but I would also > like to hear from the group. > Iıve heard the TI-83(+) models recommended - any thoughts? TI-89 Titanium is pretty much the calculator of choice for a calculus class that encourages calculator use but doesnıt specify a particular calculator. The 89 series has symbolic math capabilities that the 83/84 series canıt match. High school AP Calc classes rely on these heavily, to the point that you pretty much canıt do the work without one. College policies and curriculum will be different, so check with your instructor. -- Chris Green === Subject: Re: Favourite Graphing Calculator? TI-89 totally. It does everything you will be doing in calculus. Including eigen values, vactors, integrals, partial derivatives... It is a beautiful tool. > I hope this hasnıt been raised before, I couldnıt Ŝnd a mention in the > FAQ or recent messages on Google Groups. > Iım starting a calculus class (1st year) and they recommend the use of > a graphing calculator. I know many people on this group might disagree > with students today using one, and in fact Iım inclined to agree. > However, Iıve decided it would be wise to take whatever advantage > theyıll let me have. :) > Iıve never owned a graphing calculator and would appreciate any > opinions on favourite models. Iıll be speaking to my course advisor > about speciŜc models allowed/prohibited/recommended, but I would also > like to hear from the group. > Iıve heard the TI-83(+) models recommended - any thoughts? === Subject: Re: Favourite Graphing Calculator? > TI-89 totally. It does everything you will be doing in calculus. > Including eigen values, vactors, integrals, partial derivatives... > It is a beautiful tool. It is ok for schoolwork, but the HP48/49/49+ series are better for situations where the answers are not in the back of the book. > I hope this hasnıt been raised before, I couldnıt Ŝnd a mention in the > FAQ or recent messages on Google Groups. > > Iım starting a calculus class (1st year) and they recommend the use of > a graphing calculator. I know many people on this group might disagree > with students today using one, and in fact Iım inclined to agree. > However, Iıve decided it would be wise to take whatever advantage > theyıll let me have. :) > > Iıve never owned a graphing calculator and would appreciate any > opinions on favourite models. Iıll be speaking to my course advisor > about speciŜc models allowed/prohibited/recommended, but I would also > like to hear from the group. > > Iıve heard the TI-83(+) models recommended - any thoughts? === Subject: Re: about differentiable manifold X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŜth X-Treme: C&C,DWS >I know that df_x (T_xM) subset T_{f(x)}N. >Is it true that df_x(T_xM) = T_{f(x)}(f(M)) in general? No. Consider f: R->R, f(x)=x^2 at x=0. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: in which to study Ŝelds? X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŜth X-Treme: C&C,DWS at 09:03 PM, chergarj@cs.comhaho (Chergarj) said: >In which course of Mathematics are Ŝelds Ŝrst studied in any >greater detail; or when are Ŝelds Ŝrst studied rigorously? Algebra. The course called Algebra in high school is typically oriented to computation and doesnıt actually cover much algebra. Youıll probably encounter Ŝelds in undergraduate courses called something like Introduction to Mathematics, Abstract Algebra or Modern algebra. The same courses should also mention groups and rings. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: p-adic transcendentals (1) Write a sequence of rationals x_n that converges to 0 in the usual absolute value, but converges to 1 in the 3-adic absolute value. (2) Can you specify in advance lots of rational limits, perhaps a different one for each p? -- G. A. Edgar edgar at math.ohio-state.edu === Subject: Re: p-adic transcendentals >(1) Write a sequence of rationals x_n that converges to 0 in the usual absolute >value, but converges to 1 in the 3-adic absolute value. >(2) Can you specify in advance lots of rational limits, perhaps a different >one for each p? Neat exercise. In fact: given arbitrary real r_0 and p-adics r_p for all primes p, you can Ŝnd a sequence of rationals that converges to r_0 in the usual absolute value and to r_p in the p-adic norm for all p. This is because if x_n = a_n/b_n, we can ensure |x_n - r_p|_p <= p^(-n) by specifying a_n and b_n mod p^m for some m = m(p,n). The Chinese Remainder Theorem lets us do this simultaneously for any Ŝnite set of pıs, say all primes <= n, and by adding to a_n and b_n suitable multiples of product_{p <= n} p^m(p,n) we can also ensure |x_n - r_0| is small. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Mathematics is not real science by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC43v03456; >> Real Science is expressed in Mathematics. >> Yes, but empirically connected material is introduced by way of >> postulates that reŝect generatlizations of facts. For example, Newtonıs >> law of gravitation is not a priori true. It was introduced as a >> postulate to produce forces which accounted for the observed motions of >> planets. >> It is the physical postulates that makes physics distinct from mathematics. >> Bob Kolker >You buy this math being deductive lie, what they teach you in >philosophy 101? Thereıs no such thing as deductive nor inductive >reasoning, just reasoning. All mathematics is done in the physical >world and is therefore subject to the laws of physics. And likewise, >physics is subject to the laws of mathematics. You canıt separate the >two with artiŜcial modern deŜnitions like deductive and inductive. >By the way, for everyone out there who has read my comment that more >money should be spent on public tv and moon colonization, it was a >failed attempt at humor, so please disregard. >Craig The difference between inductive and deductive reasoning is demonstrable. Inductive conclusions can never reach a closed judgment, and deductive conclusions must. I canıt discern if you are serious, or just trolling. Tom === Subject: Re: geometry question: meteorites striking a planet by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC43703466; >This question came up in another forum and it was debated ad >inŜnitum. Itıs not an astronomy question really, but a pure geometry >one (as youıll see with all the unrealistic assumptions). >Imagine you have a perfectly spherical body in space (like a planet or >moon), and it is sitting still (not rotating or revolving). >Meteorites, travelling in random directions, might hit it. (also >assume the meteorites are travelling in straight lines and arenıt >inŝuenced by the gravity of the our little sphere) >Given that the meteorites are as likely to be travelling at any angle, >or through any point, as any other....what percentage of the >meteorites that end up hitting the sphere should strike the surface at >greater than 45 degrees, and what percentage at less than 45 degrees? >One person said it should be exactly 50-50 (based on the ratio of the >areas of the ŝattened projections of the greater and less than 45 >degree regions), another said it was closer to 30-70 (based on the >surface areas of those regions). >Please set us straight. Even if one takes into account areas and cross sections, percentage is 50-50. Here is my proof (with help of mathematica): Let r be the planet radius and lambda the latitude. Suppose all meteorites fall parallel to an arbitrary north-south axis. At latitude lambda, hitting is at angle lambda. In[1]:=planetCrossSection = Pi*r^2; In[2]:=differentialZone = 2*Pi*r^2*Cos[lambda]; In[3]:=differentialCrossSection = differentialZone*Sin[lambda] Out[3]=2*Pi*r^2*Cos[lambda]*Sin[lambda] In[4]:=differentialProbability = differentialCrossSection/planetCrossSection //Simplify Out[4]=Sin[2*lambda] In[5]:=Integrate[differentialProbability, {lambda, 0, 90*Degree}] Out[5]=1 In[6]:=Integrate[differentialProbability, {lambda, 0, 45*Degree}] Out[6]=1/2 v.a. === Subject: Re: EFFICIENCY: PRIME TEST vs PRIME GENERATOR by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC43j03461; >Hereıs a math question that may have been answered in the literature, but I havenıt seen it anywhere. Maybe one of you professionals can help. >Letıs suppose we have an algorithm (function?) that will take a positive integer x as input and output the xth prime. Say it takes You posted this same gibberish in the Mersenne Forum. I already answered you there. I suggested that you take the time to *learn* the basics of this subject. Learn how to measure complexity of an algorithm. Learn some elementary number theory. Instead, you have said that you donıt have the time. You do seem to Ŝnd the time to repeatedly post the same gibberish, however. The questions you ask are NOT WELL POSED. You are asking about the relative efŜciency of DIFFERENT ALGORITHMS for DIFFERENT PROBLEMS. It is like asking which is more efŜcient: the simplex algorithm for solving LPıs, or the Cohen-Lenstra algorithm for prime proving. In your case, the problems you compare both have the word prime, but they are still DIFFERENT PROBLEMS. Your questions are not well posed. They are gibberish. Go study this subject. === Subject: Re: Myth in Mathematics ! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MDn1f11902; >hi, > I am Krishna kishore Working for IBM. I just Ŝnished my Bachelorıs Degree in Computer science and Engineering from a reputed institute . I am 21 yrs old. > Mathematics is a passion for me. I work during nights , inspite of hectic work schedule here in the company and try to get a broader understanding of the principles involved. > But i am greatly worried about that popular statistical conclusion that mathematics is for young people only . No great work , ever , can be accomplished after 25 yrs. I am 21 . Working in a company, it takes more time than usual to achieve ,which the masters had . >So , can u tell ur opinion on this ? >I am open to any harsh suggestions. If u want me to concentarte on the job,rather than wasting my energy on maths, please tell me! >Krishna Kishore G V This business of burning out by 30 (or whenever) is pretty much a myth in my opinion -- there are examples of mathematicians proving wonderful theorems in their sixties (and beyond). Erdos for example. Itıs true for many of course that the concentration may wane with age, but to offset that there is ever-widening experience and intuition. I think much of burnout must have to do with a subsiding of passion or motivation for the subject. If you really love math as you say, youıll probably wind up pursuing it no matter what. So do so! -- and donıt worry about age; youıre still very young. On the other hand, if youıre in this primarily on the basis of dreams of proving a great theorem, like some regulars in this newsgroup, Iıd worry about that. If you donıt *really* love it, consider doing something else, because a math career or avocation can involve lots of disappointment. Good luck to you. Todd Trimble === Subject: Solvable Quintics Using One Fifth Root Extraction by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MDn5511953; Hello all, For those who like solvable quintics, hereıs another paper: Solving Solvable Quintics Using One Fifth Root Extraction ABSTRACT: We prove that all irreducible but solvable equations of degree n can be transformed in radicals into the binomial form y^n+c=0 using a Tschirnhaussen transformation of degree n-1. The resulting equation is then solvable by a single nth root extraction. In particular, we illustrate the method using the solvable quintic. Mathematics Subject ClassiŜcation. Primary: 12E12; Secondary: 11D09 http://www.geocities.com/titus_piezas/morequintics.html Just click at the link above. Itıs the 2nd pdf Ŝle. -Titus (tpiezasIII@uap.edu.ph -> remove III for email) === Subject: Re: all derivatives zero not imply zero!? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MDn1311906; >may you give me an example of function f with all derivative of f zero >at p such that f is not identically zero in a little neighborhood of >zero? >Tern DeŜne f(x)= exp(-1/x^2) if x is not 0 0 if x= 0 Show that that function is inŜnitely differentiable for all x and that the nth derivative at x=0 is 0 for all n. === Subject: Re: all derivatives zero not imply zero!? > DeŜne f(x)= exp(-1/x^2) if x is not 0 > 0 if x= 0 OK, we all agree this is the canonical answer. Does anyone know (1) Who came up with this? (2) What was the canonical answer before this? dave === Subject: Re: all derivatives zero not imply zero!? > DeŜne f(x)= exp(-1/x^2) if x is not 0 > 0 if x= 0 > OK, we all agree this is the canonical answer. Does anyone know > (1) Who came up with this? > (2) What was the canonical answer before this? > dave Unless the function itself is required to be zero at the point of interest, which is not evident from the way the Subject is phrased, any non-zero constant function will do. === Subject: Re: Tensor Product by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MEMxb16215; >Can anyone give me an intuitive idea of what the tensor product looks like? Iım particularly interested in the idea of taking the tensor product of Z^n (direct product of n copies of the integers) with R (the reals). >I know the formal deŜnition, taking symbols of the form axb, and factoring by some relations. >I know it, in some sense, reduces bilinear (or more generally multilinear) forms to linear forms on the two spaces/modules. I am interested more in a purely algebraic Œwhat does it look like?ı approach. >Any help, gratefully received. >Grayum I suspect you want to know not just what does it look like but also what does it do and how to calculate. Roughly speaking, the tensor product is to direct sum as multiplication is to addition. For example, working with vector spaces V and W over a Ŝeld k, the dimension of their tensor product over k is (dim V)(dim W), whereas the dimension of their direct sum or direct product is dim V + dim W. If e_1, ..., e_m is a basis of V and f_1, ..., f_n is a basis of W, then the e_i tensor f_j form a basis of V tensor W. This analogy also suggests something else which is true: that tensor products distribute over sums. This is relevant to your problem because Z^n is the direct sum Z + ... + Z of n copies of Z, so R tensor Z^n ~ R tensor (Z + ... + Z) ~ (R tensor Z) + ... + (R tensor Z) and since R tensor Z is naturally isomorphic to R (Z is a unit for the tensor), the result is the direct sum/product of n copies of R, i.e., R^n. Much more is true: tensoring with an object preserves arbitrary direct sums (not to be confused with arbitrary direct products!), in the sense of the distributive law above, and also preserves coequalizers. This is very useful to know for doing calculations. If you are somewhat comfortable with category theory, this follows from the fact that the functor A tensor - is left adjoint to the functor Hom(A, -). Or, even if youıre not comfortable with that, let me put it this way: a point of the tensor product is that Hom(A tensor B, C) ~ Hom(A, Hom(B, C)) so you can think of tensoring as a way of composing Hom(A, -) and Hom(B, -) as another Hom-functor. This isomorphism follows readily from the bilinearity formulation of tensor products. Letıs put this to use to calculate Z_2 tensor Z_3. For an abelian group C, Hom(Z_3, C) is the subgroup B of elements c such that 3c = 0. Hom(Z_2, B) in turn is the subgroup of elements b such that 2b = 0, i.e., elements c such that 3c = 0 and 2c = 0, i.e. c = 0. It follows that Hom(Z_2 tensor Z_3, C) = 0 for any C, and hence (taking C = Z_2 tensor Z_3), we conclude Z_2 tensor Z_3 = 0. A similar calculation shows that Z_m tensor Z_n = 0 whenever m and n are relatively prime. Since any Ŝnitely generated abelian group decomposes as a direct sum Z^n + (Z mod (p_1)^(e_1)) + ... + (Z mod (p_r)^(e_r)) where the p_i are prime, you are now in a position to calculate the tensor product of any two Ŝnitely generated abelian groups (among other things). Hope this helps. Todd Trimble === Subject: Re: EFFICIENCY: PRIME TEST vs PRIME GENERATOR by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MHmEJ02652; Bob notwithstanding, my method does produce primes w/out checking composites, because they are eliminated from the start. Take abs(2^w*5^x +/- 3^y*7^z), then all outputs will not have 2 or 3 or 5 or 7 as factors (by contradiction of the distributive principle). This means that every output less than 121 will satisfy the sieve of Eratosthenes. Take it or leave it, you canıt deny it! Aaron === Subject: Re: EFFICIENCY: PRIME TEST vs PRIME GENERATOR by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MIt3209143; >Bob notwithstanding, my method does produce primes w/out checking composites, because they are eliminated from the start. >Take abs(2^w*5^x +/- 3^y*7^z), then all outputs will not have 2 or 3 or 5 or 7 as factors (by contradiction of the distributive principle). This means that every output less than 121 will satisfy the sieve of Eratosthenes. >Take it or leave it, you canıt deny it! >Aaron You are beginning to sound like Mr. Harris. (1) You have given us an exponential expression in 4 variables. Please explain the algorithm you used to derive the exponents on those variables. You will say that they are primes. Fine. How did you get them? Did you drag them out of the air? No. You used an algorithm. Please tell us the algorithm. I guarantee that the method checks composites somewhere. (2) Give us a detailed explanation of how your algorithm will prove that a given odd integer N is prime. Show us how to determistically express it as the difference of two products of prime powers. Say that you represent N = A - B, where A and B are products of distinct sets of primes. Explain how you bound the size of A and B. Or, failing that, explain how to partition the primes into two sets S1 and S2, such that A is the product of primes in S1, B is the product of primes in S2, and prove a bound on the size of S1 and S2, as well as a bound on the size of their largest elements. Are these sets different for different N? Show an example of your method by proving that [letıs give something SMALL] 1111111111111111111 is prime. You have not presented an algorithm in any of the forums in which you have been posting. All I have seen is a lot of handwaving. (3) Even for the expression you gave above, explain how to determine the value of w,x,y,z so that your expression equals (say) 83. Indeed. Can you even prove that for every p < 121 there even exists a given set {w,x,y,z} so that p = 2^w 3^x - 5^y7^z?? Yes, every value in the RANGE of your expression will not be divisible by 2,3,5, or 7. But you have not given a convincing demonstration that the range is indeed the set of primes less than 121. i.e. that every prime is in the range. I have advised you to study the basics of complexity theory and number theory. I have offered to help. Instead, you choose to be willfully ignorant. That is, of course, your choice. But donıt expect that you will be taken seriously. === Subject: Re: Bob, Iım calling you out! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MIt3T09151; >Bob (you know who you are), >Then if q(x1,x2,x3,...xn) < [p(n+1)]^2, then q(x1,x2,x3,...xn)is prime. Please explain how you generated your initial set of primes p(1), p(2), .... Did you pull them from the air? >As I said before, to get all primes < [p(n+1)]^2, itıs important to use ALL POSSIBLE disjoint partitions of {p(1),....,p(n)} into the two sets to be put into A and B. Do you know HOW MANY such disjoint sets exist? Do you have any idea of the SIZE of A and B? I suggest you consider the amount of arithmetic needed to generate all primes < 10^6 with your method. Your initial set of primes will be the 168 primes less than 10^3. Now form all disjoint products. There will be 2^168 of them. The largest will be the product of all primes less than 1000 divided by 2. This is approximately e^1000/2. Determine the amount of arithmetic needed to form one product. Do not forget to account for the multi-precision arithmetic needed. >This will produce primes without testing any composites, because the elimination of those composites is intrinsic to the deŜnition of A and B, the gcd(A,B)=1, and the less than test. This indicates that this algorithm may still be in the running as the most efŜcient way to Ŝnd primes, given that it doesnıt waste time testing a large number of BIG composites. Prove me wrong, if you dare! The run time of your algorithm is trivially worse than trial division. It is, in fact, at least doubly exponential. This immediately removes it from any serious consideration. Instead of being willfully ignorant you would be better off learning some mathematics. === Subject: trig identities I have basic trignometry doubt how to u get 5cos(2t-53.13degrees) from 3cos2t + 4 sin2t. Can someone suggest a site from where I can lean basic trinometry === Subject: Re: trig identities >how to u get 5cos(2t-53.13degrees) from 3cos2t + 4 sin2t. >Can someone suggest a site from where I can lean basic trinometry Ah, yes, the web -- the answer to all our educational prayers. Very well then: http://historical.library.cornell.edu/cgi-bin/cul.math/ docviewer?did=04850002 Oh, youıll want a calculator too: http://www.buttercupdays.co.uk/prod168.htm dave === Subject: Re: trig identities X-RFC2646: Format=Flowed; Original > how to u get 5cos(2t-53.13degrees) > from 3cos2t + 4 sin2t. arccos(x) = angle whose cosine is x. Cos(A-B) = cosAcosB + sinAsinB Cos(A-B) = cos(B-A) In the Ŝrst quadrant, arccos(3/5) can be represented, using a right triangle. * 5 * | * | * | 4 * | A*------------| 3 angle A = arccos(3/5) also angle A = arcsin(4/5) 3cos(2t) + 4sin(2t) = 5[(3/5)cos(2t) + (4/5)sin(2t)] = 5[cos(arccos(3/5))cos(2t) + sin(arccos(3/5))sin(2t)] = 5cos(arccos(3/5) - 2t) = 5cos(2t - arccos(3/5)) === Subject: Re: trig identities > I have basic trignometry doubt > how to u get 5cos(2t-53.13degrees) from 3cos2t + 4 sin2t. > Can someone suggest a site from where I can lean basic trinometry do you know the addition theorem, cos(x+y) = ... ? Given 3cos2t + 4 sin2t, try to make it Ŝt that formula. === Subject: help on determine the stationarity of a random sequence? Random sequence: X1 X2 ... Xn The X1 has pdf as f(x1)= 2*x1, if x1 in [0, 1] region, = 0, otherwise. X_(n+1) is uniformly distributed on interval (1-X_n, 1], given the previous Xi RVs. ----------------------------------------- My intuition says that this random sequence is not stationary. But I have computed E(X1)= 2/3, E(X2|X1)=(2-X1)/2 which is a RV of X1, then E(X2)=E_x1(E_x2(X2|X1))=2/3 again... Ok, so using expectations does not show the sequence to be non-stationary. I have to Ŝnd pdf for each RV. But here comes the problem: f(x2|x1)=1/x1, in the region (1-x1, 1), =0, otherwise. then f(x2)=f(x2|x1)*f(x1)=1/x1*2*x1 = 2, still in range (1-x1, 1) = 0, otherwise. This makes everything strange: the pdf of X2, supposedly should be involving only x2, but now it still involves the variable x1 in its range.... And its expectation E(X2)=??? in this case? === Subject: Re: help on determine the stationarity of a random sequence? >Random sequence: >X1 X2 ... Xn >The X1 has pdf as >f(x1)= 2*x1, if x1 in [0, 1] region, > = 0, otherwise. >X_(n+1) is uniformly distributed on interval (1-X_n, 1], given the >previous Xi RVs. >My intuition says that this random sequence is not stationary. Intuition, without the guidance of insight, is just a blind hunch. >I have to Ŝnd pdf for each RV. But here comes the problem: >f(x2|x1)=1/x1, in the region (1-x1, 1), > =0, otherwise. >then >f(x2)=f(x2|x1)*f(x1)=1/x1*2*x1 = 2, still in range (1-x1, 1) > = 0, otherwise. No, you have to integrate over x1. f_Y(y) = int f_{Y|X}(y|x) f_X(x) dx Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: help on determine the stationarity of a random sequence? >Random sequence: >X1 X2 ... Xn >The X1 has pdf as >f(x1)= 2*x1, if x1 in [0, 1] region, > = 0, otherwise. >X_(n+1) is uniformly distributed on interval (1-X_n, 1], given the >previous Xi RVs. >My intuition says that this random sequence is not stationary. > Intuition, without the guidance of insight, is just a blind hunch. >I have to Ŝnd pdf for each RV. But here comes the problem: >f(x2|x1)=1/x1, in the region (1-x1, 1), > =0, otherwise. >then >f(x2)=f(x2|x1)*f(x1)=1/x1*2*x1 = 2, still in range (1-x1, 1) > = 0, otherwise. > No, you have to integrate over x1. > f_Y(y) = int f_{Y|X}(y|x) f_X(x) dx > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Hi Robert, Just one more question: I followed your advice, and added the integration to my solution process, I got f_X2(x2) = 2*x2, which shows that the random sequence should be stationary. You are right. My intuition was wrong. But I am not sure about the valid region of the above f_X2(x2) = 2*x2, I still get the region for the above expression to be valid should be 1-x1 < x2 <1. Thus the region on which f_X2 is deŜned is still dependent on X1...(thatıs because in my derivation, f_X2|X1(x2|x1) is deŜned on 1-x10}(1/n^2).cos( 9/(n.pi + sqrt( n^2.pi^2 - 9) ) ) = -pi^2/(12.e^3) It is from Gosper Any idea how to prove this ? Or a reference where the proof is written ? m.o. === Subject: Re: pi and e X-RFC2646: Format=Flowed; Original > sum_{n>0}(1/n^2).cos( 9/(n.pi + sqrt( n^2.pi^2 - 9) ) ) = > -pi^2/(12.e^3) > It is from Gosper > Any idea how to prove this ? > Or a reference where the proof is written ? Another fact: P= product {P (n= 1 -->oo)( pi^(n (2^(-n)-n+1))} +1= P (n= 1-->oo)( e^((2(n^2)-2n-1)/(2(2n-1)(2n+1)n(n+1))) :-)} Tapio ------------------------------------------------------------- --------------- ---- > m.o. === Subject: Re: Correspondence with journals by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MKwlr20175; I decided not to get it published. I think that that kind of paper should be read only for the purpose of fun. === Subject: Re: The Origin of Mathematics Symbols <41787a4f.5535127@netnews.att.net> Discussion, linux) > A fascinating coincidence, Dan. But after Iıd gotten the idea of > differences down and shown that they could be used to deŜne > arithmetic, I decided to check the fossil record to see if there might > be any evidence one way or the other. And lo and behold, there it is, > written evidence that subtraction or differences really came Ŝrst. I hesitate to ask, but what evidence could the fossil record provide? And what has that to do with the very recent introduction of the current arithmetic notations? After all, mathematics had been done for *centuries* prior to the introduction of +, -, etc., so these symbols canıt tell us diddly about philosophical or historical priority. -- Jesse F. Hughes Thereıs a thrill thatıs gone that Iıll probably not have in quite the same way again. After all, FLT was a unique animal, and we had a great dance. -J.S. Harris on proving Fermatıs last theorem === Subject: Re: The Origin of Mathematics Symbols > A fascinating coincidence, Dan. But after Iıd gotten the idea of > differences down and shown that they could be used to deŜne > arithmetic, I decided to check the fossil record to see if there might > be any evidence one way or the other. And lo and behold, there it is, > written evidence that subtraction or differences really came Ŝrst. > I hesitate to ask, but what evidence could the fossil record provide? Well, I think Mr. Zick was not being entirely serious in his choice of the phrase fossil record for really old books. > And what has that to do with the very recent introduction of the > current arithmetic notations? After all, mathematics had been done > for *centuries* prior to the introduction of +, -, etc., so these > symbols canıt tell us diddly about philosophical or historical > priority. Notations, and other means of expression, are certainly inŝuenced by what their introducers are trying to express. Conversely, they affect what their users /can/ express, and perhaps inŝuence what their users will think. (Actually, there is a lot less to this than meets the eye; it goes back to an argument that language molds thought, introduced by Whorf, and severely contested by later linguists.) In the present discussion I think Mr. Hughes has a better case. Our notations for arithmetic go back to 1400 or maybe 1200 at the earliest. But people in Egypt and Babylonia mst have been adding and subtracting. Some of their math books are still extant. Now, do these have anything to suggest that subtraction is prior to addition? -- Chris Henrich God just doesnıt Ŝt inside a single religion. === Subject: Re: The Origin of Mathematics Symbols >> A fascinating coincidence, Dan. But after Iıd gotten the idea of >> differences down and shown that they could be used to deŜne >> arithmetic, I decided to check the fossil record to see if there might >> be any evidence one way or the other. And lo and behold, there it is, >> written evidence that subtraction or differences really came Ŝrst. >I hesitate to ask, but what evidence could the fossil record provide? Evidence of your extraordinary obtusity perhaps. >And what has that to do with the very recent introduction of the >current arithmetic notations? After all, mathematics had been done >for *centuries* prior to the introduction of +, -, etc., so these >symbols canıt tell us diddly about philosophical or historical >priority. Well, they can certainly tell us diddly about your philosophical and historical priority. === Subject: Re: The Origin of Mathematics Symbols <41787a4f.5535127@netnews.att.net> <876552xy2x.fsf@phiwumbda.org> <417a4443.7782207@netnews.att.net> Discussion, linux) > A fascinating coincidence, Dan. But after Iıd gotten the idea of > differences down and shown that they could be used to deŜne > arithmetic, I decided to check the fossil record to see if there might > be any evidence one way or the other. And lo and behold, there it is, > written evidence that subtraction or differences really came Ŝrst. >>I hesitate to ask, but what evidence could the fossil record provide? > Evidence of your extraordinary obtusity perhaps. Perhaps. But my obtusity isnıt directly relevant for your claim. >>And what has that to do with the very recent introduction of the >>current arithmetic notations? After all, mathematics had been done >>for *centuries* prior to the introduction of +, -, etc., so these >>symbols canıt tell us diddly about philosophical or historical >>priority. > Well, they can certainly tell us diddly about your philosophical and > historical priority. Very illuminating response, that. My questions remain. -- Jesse F. Hughes C is for Cookie. Thatıs good enough for me. Cookie Monster === Subject: Re: The Origin of Mathematics Symbols >> >>Personally, I suspect that math symbols originated just as did >>software jargon, by borrowing from symbols commonly used by >>working people; i.e. the Œ-ı probably started life as just a tick >>mark, as digits were checked off on a list. Then the Œ+ı was a >>ı-ı slashed through because it had been marked mistakenly, >>thereby returning it to itıs previously unmarked state; i.e. Œ2ı >>and Œ+2ı were equivalent. > But your speculation is contradicted by the historical emergence of > the - for subtraction before the + for addition. Besides your idea > would have required a left hand ticker and southpaws were deŜnitely > frowned upon as sinister from Roman times onward. LOL. Maybe so. -- Donıt you see that the whole aim of Newspeak is to narrow the range of thought? In the end we shall make thoughtcrime literally impossible, because there will be no words in which to express it. -- George Orwell as Syme in 1984t === Subject: Re: The Origin of Mathematics Symbols >I imagine youıre trying to tie this all together in your theory of >differences, but I suggest that humans started with the idea of >addition Ŝrst, and somewhat later discovered subtraction, and many >centuries later multiplication and division. > A fascinating coincidence, Dan. But after Iıd gotten the idea of > differences down and shown that they could be used to deŜne > arithmetic, I decided to check the fossil record to see if there might > be any evidence one way or the other. And lo and behold, there it is, > written evidence that subtraction or differences really came Ŝrst. > Your idea that addition comes Ŝrst is certainly the prevalent notion. > I suspect itıs a result of Eulerıs approach to math and deŜnition via > set theory. But the fact remains that we canıt deŜne subtraction > through addition without hidden assumptions, but we can deŜne > addition through subtraction as the primary arithmetic operation. > You asked for experimental evidence; there it is. I was thinking back to when the conceptual ideas may have been formed .... ie, somewhat before the mathematical formulations ... maybe by 50,000 years. > Even H.erectus realized >that having 2 turkey legs to eat was better than 1. Also, my pile is >bigger than your pile. People naturally count on their Ŝngers by >successively increasing the number held up, on and on. >Interestingly, Iıve just discovered that itıs *much much* easier >bio-mechanically to open Ŝngers up successively one at a time from a >closed Ŝst than to bend the Ŝngers back into a Ŝst one at a time >from an open hand. Try it. > Thatıs probably a function of differences too. At least the difference > of being able to grip remaining Ŝngers while opening the Ŝst one > Ŝnger at a time. One is much more natural and easier to do than the other. Might that have any bearing on how the organism learns to conceptualize? They do this about 5 years before learning the math. === Subject: Re: all derivatives zero not imply zero!? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MLWfr23666; > may you give me an example of function f with all derivative of f zero > at p such that f is not identically zero in a little neighborhood of > zero? > Tern >>Any constant function y = k(x) = k. Iım Ŝnding it hard to believe some >>of the weird answers youıve gotten so far. >Hah, thatıs a good one :-) >The validity of this examples, however, hinges on the meaning >of all derivatives. If we include zeroth derivative >among all derivatives, then this example does not work. >The zeroth derivative of a function is the function itself. >Rouben Rostamian perhaps one should then consider also derivatives of negative order which i think would correspond to integrals. === Subject: Function A function is Ŝrst of all the reŝection of our own memory on the examined elements, which give us the possibility to compare an element to itself or to other elements (functions or not). Please see the diagrams in: http://www.geocities.com/complementarytheory/Function.pdf What do you think? *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Function complementarytheory@yahoo-dot-com.no-spam.invalid (Doron Shadmi) > A function is Ŝrst of all the reŝection of our own memory on the > examined elements, which give us the possibility to compare an > element to itself or to other elements (functions or not). I donıt think so. Because memory is surely a function (no pun intended) of the brain, which has a Ŝnite number of neurons and hence neuronal paths. Only countably many thoughts can exist; yet, there are uncountably many functions. So there are functions that are most deŜnitely not reŝections of our memory. === Subject: Series and number of terms A fellow posed this question to me a few days ago. So far, none of my professors or colleagues could answer this: SUM(1/n^2) as n=1 to inŜnity. How many terms are needed to come to a solution with an error < 0.001. My initial thought was to integrate the p-series from 1 to k of the function subtracted from the integral of the function from (k+1) to inŜnity. Therefore, subtracting the errors should yield the term. I was wrong, or did the calculations incorrectly. Daniel Lausevic === Subject: Re: Series and number of terms > A fellow posed this question to me a few days ago. So far, none of my > professors or colleagues could answer this: > SUM(1/n^2) as n=1 to inŜnity. > How many terms are needed to come to a solution with an error < 0.001. > My initial thought was to integrate the p-series from 1 to k of the > function subtracted from the integral of the function from (k+1) to > inŜnity. Therefore, subtracting the errors should yield the term. I > was wrong, or did the calculations incorrectly. Yes, something like that is how Iıd do it. Remember how the integral is deŜned as the limit of the sum of the area of little rectangles? Draw the curve 1/x^2. Now draw two sets of boxes: One has height 1/n^2 and touches the curve at its upper left corner. The upper right corner lies above the curve. Thus the area of each box is more than the area of the curve that crosses it. sum(1/n^2) n=k,oo > integral(1/x^2) x=k,oo Now consider boxes that just touch the curve at the right corner, i.e. boxes of height 1/(n+1)^2. These clearly lie entirely under the curve. Thus, sum(1/(n+1)^2) n=k,oo < integral(1/x^2) x=k,oo You can use these relationships to choose a suitable integration limit. - Randy === Subject: Re: Series and number of terms >A fellow posed this question to me a few days ago. So far, none of my >professors or colleagues could answer this: >SUM(1/n^2) as n=1 to inŜnity. >How many terms are needed to come to a solution with an error < 0.001. >My initial thought was to integrate the p-series from 1 to k of the >function subtracted from the integral of the function from (k+1) to >inŜnity. Therefore, subtracting the errors should yield the term. I >was wrong, or did the calculations incorrectly. The error after n terms is equal to INT{x=n}^oo (ceil(x)^-2)dx which, for n>1, is always less than INT{x=n}^oo (x^-2)dx which evaluates to 1/n. So if you calculate the sum to n=1000, your error will be < 0.001. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Series and number of terms > A fellow posed this question to me a few days ago. So far, none of my > professors or colleagues could answer this: > SUM(1/n^2) as n=1 to inŜnity. > How many terms are needed to come to a solution with an error < 0.001. > My initial thought was to integrate the p-series from 1 to k of the > function subtracted from the integral of the function from (k+1) to > inŜnity. Therefore, subtracting the errors should yield the term. I > was wrong, or did the calculations incorrectly. > Daniel Lausevic You could get an upper bound on the number of terms by observing that 1/k^2 < INTEGRAL_k-1^k dx/x^2 and the error term after N terms is SUM_n=N+1^inŜnity 1/n^2 < INTEGRAL_N^inŜnity dx/x^2 = 1/N So 1000 terms is deŜnitely enough to make the error <0.001. === Subject: Re: Series and number of terms > A fellow posed this question to me a few days ago. So far, none of my > professors or colleagues could answer this: > SUM(1/n^2) as n=1 to inŜnity. > How many terms are needed to come to a solution with an error < 0.001. > My initial thought was to integrate the p-series from 1 to k of the > function subtracted from the integral of the function from (k+1) to > inŜnity. Therefore, subtracting the errors should yield the term. I > was wrong, or did the calculations incorrectly. > Daniel Lausevic Brute force search: Since we know that the inŜnite sum is pi^2/6, my computer answered that 999 terms are not enough but 1000 terms are enough. No-computer approach: The use of integration should take place from N or from (N+1) to inŜnity. The inequalities (f(x) being decreasing, convergent to 0) are integral[N+1 to inf] f(x) dx <= sum[n=N+1 to inf] f(n) sum([n=N+1 to inf] f(n) <= integral [N to inf] f(x) dx Do your calculations with f(x) = 1/x^2. (Why do we worry about sum[n=N+1 to inf] f(n)? ) === Subject: Re: Series and number of terms ETAuAhUAwlrzJjPOWP5v6mxQ1EcDxpqrNP8CFQCzjob179HAFtT4C60qzSyd8k nIMg== Hereıs a method that guarantees accuracy without requiring anything beyond elementary algebra. It can be shown that the sum of the series is bounded from above by: S_inf < (n+1)/n*S_n To prove this: deŜne a sum S_n* by leaving the Ŝrst n terms of the series alone and then replacing blocks of n+1 terms with the largest term in each block. Thus S_3* = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/4^2 + 1/4^2 + 1/4^2 + 1/8^2 + 1/8^2 + 1/8^2 + 1/8^2 + 1/12^2 + 1/12^2 + 1/12^2 + 1/12^2 + 1/16^2 + ... these the quoted bound follows. Putting n = 1 gives S_inf < 2. So S_inf lies between S_n and S_n + 2/n, and we are sure to be within 1/n by taking S_n + 1/n. --OL === Subject: Re: Series and number of terms > A fellow posed this question to me a few days ago. So far, none of my > professors or colleagues could answer this: > SUM(1/n^2) as n=1 to inŜnity. > How many terms are needed to come to a solution with an error < 0.001. > My initial thought was to integrate the p-series from 1 to k of the > function subtracted from the integral of the function from (k+1) to > inŜnity. Therefore, subtracting the errors should yield the term. I > was wrong, or did the calculations incorrectly. > Daniel Lausevic Mathematica says the Ŝrst 1000 terms are close enough. In[1]:= NSum[1/n^2,{n,1001,InŜnity}] Out[1]= 0.0009995 -- Dave Seaman Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Series and number of terms >> A fellow posed this question to me a few days ago. So far, none of my >> professors or colleagues could answer this: >> SUM(1/n^2) as n=1 to inŜnity. >> How many terms are needed to come to a solution with an error < 0.001. >> My initial thought was to integrate the p-series from 1 to k of the >> function subtracted from the integral of the function from (k+1) to >> inŜnity. Therefore, subtracting the errors should yield the term. I >> was wrong, or did the calculations incorrectly. >> Daniel Lausevic >Mathematica says the Ŝrst 1000 terms are close enough. >In[1]:= NSum[1/n^2,{n,1001,InŜnity}] >Out[1]= 0.0009995 >-- >Dave Seaman >Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling. > A simple way I view it.... Accuracy looks like common sense to me. You can only be as accurate as the number of accurate values used. If you want an accuracy of < .001, use 1000 accurate values. The inverse of .001 = 1000 If you want an accuracy of <.0000000000001 then use 1/ (1 X10^13) 10^13 accurate terms. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: a joke I canıt not make Now that we know which teams are participating in this yearıs World Series: Go Ordinals--- I mean, Red Sox! ---- David To send me email, move the r from the beginning to the end of the part before the @ and insert alum. at the beginning of the part after the @. === Subject: Re: a joke I canıt not make > Now that we know which teams are participating in this yearıs World Series: > Go Ordinals--- I mean, Red Sox! Remember the old Houston Eulers? === Subject: Re: a joke I canıt not make > Now that we know which teams are participating in this yearıs World Series: > Go Ordinals--- I mean, Red Sox! > Remember the old Houston Eulers? The World Series concluded with a match between the Ordinals and the Cardinals. Now we all know the Cardinals are bigger, but Ordinals are longer, so whoıs to win? === Subject: Determining probabilities based on average payoffs in a matrix game Say youıre playing a 2-person zero-sum game in which all the payoffs to player R are positive, and let x = [x0, x1, ..., xn] denote the set of n strategies for player R. Now say after t plays, you have achieved average payoffs of q = [q0, q1, ..., qn] where qi is the average payoff for strategy xi. Now, using those payoffs, how would you create probabilities for player Rıs future plays based on the known information? (We assume we know neither player Cıs plays nor the payoff matrix.) Since all of the payoffs are positive, my initial thought is as follows: If all of the average payoffs are greater than zero, then Ŝnd the miminum average payoff, say qi, and proceed as follows: Solve (q1/qi)x + (q2/qi)x + ... + (qn/qi) = 1 for x and then construct the probabilities as (qj/qi)*x for j = 1, 2, ..., n. So if we had the average payoffs of [1, 3, 5], we would construct corresponding probabilities as [1/9, 3/9, 5/9]. This seems to be a reasonable assumption. If there is at least one average payoff with a value between 0 and 1, Ŝnd the minimum such value and divide each average payoff by it, thus scaling everything so that each number is greater than or equal to 1; then, use the above strategy. So if we had the average payoffs of [0.02, 1, 10], we would Ŝrst scale them to [1, 50, 500] and construct corresponding probabilities as [1/551, 50/551, 500/551]. One again, this seems reasonable. Now, if there is at least one average payoff equal to zero, then proceed as follows: Solve (p1)x + (p2)x + ... (pn)x = 1 for x, where pi = {1 if the average payoff xi = 0 {xi if the average payoff xi =/= 0 and then construct the probabilities as (pi)*x for i = 1, 2, ..., n. So if we had the average payoffs of [0, 5, 9], we solve (1)x + (5)x + (9)x = 1 for x (here, x = 1/15), and construct corresponding probabilities as [1/15, 5/15, 9/15], which also seems reasonable. However, there is a problem. There is a degenerate situation in which (before or after scaling) our average payoffs include 0 and 1, with all other payoffs greater than or equal to 1, in which we would end up playing the strategy with average payoff of 0 equally as often as playing the strategy with average payoff 1, and I donıt wish these probabilities to be equal. I realize that these formulations are intuitive but have problems, but I have not come up with any other ideas on how to assign probabilities based on knowing just the average payoffs for each strategy. Does anyone have ideas of papers I could read addressing this issue, or === Subject: Algorithm for a partition Anybody know an recursive algorithm for a partition of positive interger m? Burt === Subject: Antimetric space Let D on a set M satisfy the axioms for a metric space except that the triangle inequality is reversed: i.e D(a,c) >= D(a,b) + D(b,c) for all a,b,c in D. Prove that M has at most one point. Any hints? === Subject: Re: Antimetric space === Subject: Re: Antimetric space > Let D on a set M satisfy the axioms for a metric space except > that the triangle inequality is reversed: > i.e D(a,c) >= D(a,b) + D(b,c) for all a,b,c in D. > Prove that M has at most one point. Any hints? 2d(a,b) = d(a,b) + d(b,a) <= d(a,a) ---- === Subject: Re: Antimetric space > Let D on a set M satisfy the axioms for a metric space except that the > triangle inequality is reversed: > i.e D(a,c) >= D(a,b) + D(b,c) for all a,b,c in D. > Prove that M has at most one point. Any hints? Assume a and b are distinct members of D, and you should be able to get a contradiction. Iıll not say anymore, and see if that helps. -- John Hicken bronzeŜngerspamblocker@boltblue.com (remove spamblocker to e-mail me) === Subject: Re: JSH: Neat puzzle, actually ... > Now then, given z = xy, where z is an algebraic integer, and y is an > algebraic integer, under what circumstances do you accept that x is > not an algebraic integer? > > That is, what properties would you say x must have if itıs not an > algebraic integer though y and z are? > > What ring do you think it might be in, besides algebraic integers? > > > 1) x=z/y is the ratio of two algebraic integers so x is > an algbraic number > > 2) given any algebraic number x, one can Ŝnd algebraic integers > z and y so that z=xy > > So x is an algebraic number that is not an algebraic integer. > Nothing more can be said about x. In particular x could be (1/6). > > The only ring that we know that x is a member of is the ring of > algebraic integers. > > Sorry, make that: The only ring that we know that x is a member of > is the ring of algebraic *numbers*. There are quite a few more rings. Adjoin z/y to the ring of algebraic integers and you get (the smallest standard) such ring. If g = gcd(y,z) this is the ring that you get when you adjoin g/y. So when y and z are coprime you have to adjoin 1/y. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Fermat (ben ito) your in Check and your queen is open. Fermatıs Last Theorem Ben Ito 10-22-04 I will solve Fermatıs last theorem. l. Introduction I will show that Fermat (n=4) and Wilesı proofs are invalid then prove that X^n + Y^Y = Z^n only forms integer solutions when n>2 using the circle transformations. 2. Fermatıs Proof (n=4) Fermatıs Proof when n=4 is described (Savant, p. 22), (Osserman, p. 22). Fermat implied that when n=4, x^4 + y^4 = z^4. Cosequently, n = 4k, then x^n + y^n = z^n implies that X^4 + Y^4 = Z^4 forms and impossiblility where X = x^k, Y=y^k, and Z^k; however, x not equal to X (x =/ X, y =/ Y and z =/ Z); therefore, Fermatıs proof for n=4 is invalid. 3. Wilesı Proof Wiles implies that Diophantine equations x^n + Y^n = z^n can be translated to described a set of elliptic curves. These curves represent the surface of a torus, an object shaped like a smooth doughnut. (Savant, p. 30). However, the equation for an elliptic curve is y^2 = x^3 + ax^2 + bx =c; therefore, using the elliptic curves to represent all values of n>2 violates logic. Consequently, Wilesı Proof of Fermatıs equation base on ellilptic curves is invalid. 4. B. Itoıs Proof. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: A peculiar function deŜnition by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9N0PBE04788; >Consider the function i constructed today while killing some time.. >f_m(x) = undeŜned, if x = m > m , otherwise >What is f_m(f_m(x)) ? >It seems this trivial construction of f(x) is such that we can say >nothing about f(f(x)). >Now some (possibly lame) questions.. >1. Is there a name for such functions ? >2. Are these functions related to Godelıs proof in any way ? If >we see these functions as propositions, they seem to negate >themselves. >Manik If you take the function f(x) = (ax + b)/(cx - a), it is easy to verify that this function is its own inverse f^(-1)(y) = (ay + b)/(cy - a) f(f(x)) = x except for the case, when a^2 + bc = 0. If c = 0 and a != 0, it is a reŝection in the vertical line x = -b/2a. If c != 0 and a^2 + bc != 0, it is a reŝection in the circle centered at (a/c, 0) and with radius r = sqrt(|a^2 + bc|)/|c|, possibly followed by a symmetry with respect to the circle center (this depends on the signs of c and a^2 + bc). This explains why the function is its own inverse. If c != 0 and a^2 + bc = 0, then b = -a^2/c can be elliminated: f(x) = (ax - a^2/c)/(cx - a) = a/c*(cx - a)/(cx - a) which is exactly your function f_m. f(x) undeŜned if x = a/c f(x) = a/c otherwise. The radius of inversion circle would be zero and you cannot invert a plane with respect to a circle of zero radius. === Subject: Well, Bilge? Re: A SR-cult fraud and corruption (Rev A) > What a moron. What a strangely honest signature block! Congratulations! You have proved yourself capable of being at least a jerk, if not an ass, and of not having sufŜcient honesty to actually respond to the details of logic/etc. So, can you now prove yourself capable of relenting in your desire to prove irrelevant to any actual discussion, and do something helpful? Maxwell and invariance are an important combination of topics and as many expressions as I know of for E, H, B, etc, I do not know just what exemplars of them would be best for demonstrating particulars of their transformation by Newton-theoretic coordinate tranformations. The Œproblemı is different than in the case of the Lorentz transforms of Maxwell because in the Newton case it actually is the coordinates x,y,z that are transformed, rather than - essentially - the inverse of the coordinates. So, please provide a set of expressions - appropriate for full exposition of Maxwellıs - for Ex, Ey. Ez, etc, complete with explicit coordinate expressions. Obviously (ha!) the result would be that Ŝnally I come headsup (as we poker players say) with my tremendous error in thinking that transforming Maxwell Newton-wise without the three strawmen corruptions will prove invariant. BTW, why donıt you reply to posts on your home newsgroup? eleaticus === Subject: Re: Well, Bilge? Re: A SR-cult fraud and corruption (Rev A) eleaticus: > What a moron. > What a strangely honest signature block! Lay off the crack. You are either or both too stupid or too stoned to even Ŝgure out simple english. The only question is whether or not youıre too stupid to be accused of fraud, since you might actually believe your own bull. === Subject: Re: Well, Bilge? Re: A SR-cult fraud and corruption (Rev A) > eleaticus: > > What a moron. > > What a strangely honest signature block! > Lay off the crack. You are either or both too stupid or too stoned > to even Ŝgure out simple english. The only question is whether or > not youıre too stupid to be accused of fraud, since you might actually > believe your own bull. I see you decided against using your Œwhat a moronı signature block again. So maybe you are not a complete idiot, just a half-wit and complete ass. eleaticus === Subject: Re: Well, Bilge? Re: A SR-cult fraud and corruption (Rev A) > What a moron. > What a strangely honest signature block! > Congratulations! > You have proved yourself capable of being at least a jerk, if not an ass, > and of not having sufŜcient honesty to actually respond to the details of > logic/etc. > So, can you now prove yourself capable of relenting in your desire to prove > irrelevant to any actual discussion, and do something helpful? > Maxwell and invariance are an important combination of topics and as many > expressions as I know of for E, H, B, etc, I do not know just what > exemplars > of them would be best for demonstrating particulars of their transformation > by Newton-theoretic coordinate tranformations. > The Œproblemı is different than in the case of the Lorentz transforms of > Maxwell because in the Newton case it actually is the coordinates x,y,z > that > are transformed, rather than - essentially - the inverse of the > coordinates. > So, please provide a set of expressions - appropriate for full exposition > of > Maxwellıs - for Ex, Ey. Ez, etc, complete with explicit coordinate > expressions. Much of your posting is incomprehensible spewing of your ignorance. How the E-M Ŝeld transforms can be found in Jackson, Classical Electrodynamics Section 11.10. John Anderson === Subject: Re: Well, Bilge? Re: A SR-cult fraud and corruption (Rev A) > So, please provide a set of expressions - appropriate for full exposition > of > Maxwellıs - for Ex, Ey. Ez, etc, complete with explicit coordinate > expressions. > Much of your posting is incomprehensible spewing of your > ignorance. > How the E-M Ŝeld transforms can be found in Jackson, > Classical Electrodynamics Section 11.10. Well, at least your assholery is somewhat relevant albeit non-responsive. > John Anderson eleaticus === Subject: Re: A SR-cult fraud and corruption (Rev A) > What a moron. (Strangely honest signature block, that!) Congratulations! You have proved yourself capable of being at least a jerk, if not an ass, and of not having sufŜcient honesty to actually respond to the details of logic/etc. So, can you now prove yourself capable of relenting in your desire to prove irrelevant to any actual discussion, and do something helpful? Maxwell and invariance are an important combination of topics and as many expressions as I know of for E, H, B, etc, I do not know just what exemplars of them would be best for demonstrating particulars of their transformation by Newton-theoretic coordinate tranformations. The Œproblemı is different than in the case of the Lorentz transforms of Maxwell because in the Newton case it actually is the coordinates x,y,z that are transformed, rather than - essentially - the inverse of the coordinates. So, please provide a set of expressions - appropriate for full exposition of Maxwellıs - for Ex, Ey. Ez, etc, complete with explicit coordinate expressions. Obviously (ha!) the result would be that Ŝnally I come headsup (as we poker players say) with my tremendous error in thinking that transforming Maxwell Newton-wise without the three strawmen corruptions will prove invariant. eleaticus === Subject: Re: A SR-cult fraud and corruption (Rev A) > eleaticus: >-------------- >Could you trust the Ku Klux Klan to conduct an honest investigation of >the NAACP? > Not any further than you can be trusted to say something intelligent. > What a moron. To quote myself: The facts about True Believer SR-cultists. ------------------------------------------------------------- ---- There are many more rotten fruit from the SR-tree to be buried before you know the nature of their whole orchard. Look for other posts in this series. Focus well on negative Œresponsesı. Are they vicious ranting? Are the replies actually responsive? Do they rant about gravity, or how Relativity is proved correct a million times each day, or some other Œwe are proved rightı rave that doesnıt deal in details about the debunking done here? It is typically General Relativity or items about the energy and mass of moving objects that are being waved at you, and such items are completely irrelevant to coordinate transformations and invariance.. Just ask them for a list of all the observations that have been made of the shortening (contraction) of moving objects that Special Relativity says always occurs. Rarely, there is actually a response that has some relevance to the material posted, and those are proofs of their Brain Death. eleaticus === Subject: Re: Units outside algebraic integers > Examples of units outside the ring of algebraic integers are the roots > of > > x^2 + 2^{sqrt(2)}x + 1 > > and I like that example as the polynomial is monic. > A unit in _which_ ring? In the ring of complex numbers, every number > is a unit. Thatıs trivial. Are you trying to state something > trivial? Worse, in the ring of complex numbers, every number is the root of a monic quadratic polynomial with constant term 1. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Compromise, and you leave my papers? > And remember I have math results that are not even disputed! > Like remember that with > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > and > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > the aıs are the three roots of the cubic > a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 > and you can check at m=1, f=sqrt(2), that my argument is correct. > If you think thatıs chance, check other solutions that give one of the > aıs rational, as you will Ŝnd it will always either have f as a > factor or be coprime to f. You still do not see the difference between irreducible polynomials and reducible polynomials. And you still fail to state in what ring they should be factors. Because what you state is false in the algebraic integers. All three will not be coprime to f in the algebraic integers, if the cubic in a is irreducible. Check the facts when the polynomial in a is irreducible. As it is, the roots of the cubic in a are only governed by the product m.f^2. And there are only Ŝnitely many such products that yield a reducible cubic. You have found them all. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Compromise, and you leave my papers? > And remember I have math results that are not even disputed! > Like remember that with > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > and > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > the aıs are the three roots of the cubic > a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 > and you can check at m=1, f=sqrt(2), that my argument is correct. > > If you think thatıs chance, check other solutions that give one of the > aıs rational, as you will Ŝnd it will always either have f as a > factor or be coprime to f. > You still do not see the difference between irreducible polynomials and > reducible polynomials. And you still fail to state in what ring they > should be factors. Because what you state is false in the algebraic > integers. All three will not be coprime to f in the algebraic integers, > if the cubic in a is irreducible. Check the facts when the polynomial > in a is irreducible. > As it is, the roots of the cubic in a are only governed by the product > m.f^2. And there are only Ŝnitely many such products that yield a > reducible cubic. You have found them all. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Compromise, and you leave my papers? > And remember I have math results that are not even disputed! > Like remember that with > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > and > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > the aıs are the three roots of the cubic > a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 > and you can check at m=1, f=sqrt(2), that my argument is correct. Ah, now I see why I was in error. Your cubic in a is wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Compromise, and you leave my papers? > [...] > My compromise is to change that conclusion in light of their > objections, and note that the result shown algebraically conŝicts > with what follows from the deŜnition of the ring of algebraic > integers. > [...] > I am willing to compromise. Are any of you? Hereıs a counter offer: (1) You do not post any more about claiming the prime-counting formula as yours; (2) You do not mention your APF paper again; (3) You do not mention the mathematics community ganging up on you; (4) You will submit only papers that have no major logical ŝaw in them, but which can be rewritten without major reconstruction. (I.e., the only errors in your papers are Ŝxable ones.) In return, people on Usenet will not harass you with e-mails, and will not send any sort of communication to the editor(s) of whatever journal(s) you submit your paper to in an attempt to remove your paper. (Note that if you submit a paper which is incorrect, that people have the right to let you and/or the editor know about it.) Iım sure that most of Usenet would be willing to hold up their end. Are you willing to hold up yours? -- Christopher Heckman === Subject: Re: Compromise, and you leave my papers? Discussion, linux) >> [...] >> My compromise is to change that conclusion in light of their >> objections, and note that the result shown algebraically conŝicts >> with what follows from the deŜnition of the ring of algebraic >> integers. >> [...] >> I am willing to compromise. Are any of you? > Hereıs a counter offer: > (1) You do not post any more about claiming the prime-counting formula as > yours; > (2) You do not mention your APF paper again; > (3) You do not mention the mathematics community ganging up on you; > (4) You will submit only papers that have no major logical ŝaw in them, > but which can be rewritten without major reconstruction. (I.e., the > only errors in your papers are Ŝxable ones.) > In return, people on Usenet will not harass you with e-mails, and will not > send any sort of communication to the editor(s) of whatever journal(s) > you submit your paper to in an attempt to remove your paper. Why have clauses (1) - (3)? Clause (4) is sufŜcient. If he submits a paper that is correct, then no one should send an email to the journal saying otherwise or trying to prevent its publication. Adding any other condition besides correctness is utterly unreasonable. -- Jesse F. Hughes I often told you of the dangers of hubris, and most importantly of all, I TOLD you that I wanted to change the institution of mathematics worldwide. -- James Harris, on the evils of pride === Subject: Re: Compromise, and you leave my papers? > What compromise do you offer? What do you want me to do with my paper? Fix it: (1) Fix the terminology. As it stands, it purports to be a paper about algebraic integers. However, its reasoning and conclusion are simply false when applied to the algebraic integers. Based on some of your recent posts, it appears that it isnıt actually supposed to be about algebraic integers. Apparently, you think the term algebraic integer should apply to some different set of numbers than everyone else. Like it or not, it has already been decided what set of numbers the term algebraic integer refers to, and so if you want to work in some other ring or Ŝeld, you have to name it something else. (2) Specify precisely what this other ring or Ŝeld is. So far, all youıve given are a set of magical properties it has, and you only give those when you have to give one in order to try to answer an error found in your paper. Get off the grassy gnoll, and leave the magic ring out of your theory. (3) Fix the exposition. You throw in variables all over the place without saying anything about them (are they integers? rational? real? complex? algebraic integers? are there constaints on the them?), leaving it to the reader to try to absorb the paper as a whole, and then go back and try to Ŝnd some way to answer those questions that will make the paper as a whole make sense. The purpose of a mathematical paper is to communicate mathematics, and your paper fails to do that for the above reasons. Until you Ŝx these problems, it is unpublishable. (Note: this does not imply that Ŝxing the above problems makes it publishable...Ŝxing them merely gets it to the point where it is not a total waste of time for anyone to even try to read it). -- --Tim Smith === Subject: Re: Compromise, and you leave my papers? > > 1. It is the NORMAL thing to do to write a letter to the Editor > when you see a mistake. If ANYBODY published your piece-of- > there would be letters to the editor - thousands, if it were > Wiles - yet you want special treatment. > > > Iım not sure that it is the NORMAL thing to do. I suspect that in most > cases people contact the author on Ŝnding a mistake in a paper. I > suspect that most editors would either refer you to the author, or > forward your message to hir. Only if the author refused to address the > concerns would an editor be likely to bother to get involved. Heck, > even Pertii contacted the authors of the erroneous papers that he found. > > The case at hand is, of course, different. The author had most > deŜnitely been contacted before publication, and the people contacting > the editor had every reason to believe that contacting the author yet > again would accomplish less than nothing. So the usual Ŝrst step was > bypassed. > Does anyone else think that this thread (as all the ones before it) is > completely pointless? Clearly people at some point started out wanting > to help James understand the mistake in his papers. However, by now > it is clear that this is a futile attempt, and that he does not > appreciate this help. Consequently posters no longer reply out of > sympathy for James (and make this clear); however, then what is the > point? No one except James could possibly have an interest in > prolonged discussions why his results are wrong. > If James wants to publish a paper in some third-, fourth- or > Ŝfth-rate journal, all he needs to do is submit it to enough of these > and keep quiet about it. I am certain that there are many proofs of > Fermatıs last theorem, P=NP, P!=NP etc. hidden away somewhere where > no-one will ever Ŝnd them ... > By the way, an entry for Jamesıs paper does remain on MathSciNet, so > the editor wasnıt successful in removing all records of the fact that > it was once accepted. > Finally, I think we should all feel somewhat sorry for James. Clearly > he is not well; perhaps if he was willing to seek therapy, he would be > able to lead a happier life. Until then, it seems best not to spend > much time on the thankless task of refuting his purported proofs. > Lasse > --- > (lasse@remove.for.spam.maths.warwick.ac.uk) Oh yeah, all I really need is therapy. LOL. Posters who argue with me canıt stop for a simple reason. Figure it out. Ultimately what I do on Usenet is post. Now what that has to do with my personal life, such that people feel they can recommend therapy to me is another matter. Care to discuss your personal life in reply? James Harris === Subject: Re: Compromise, and you leave my papers? big deal, like weıre the Ŝrst cohort to dyscover grouptherapy, even online? now, the fact that one of us may be the Lead Investigator in a MK-infra virtual lab would probably also not be new, since itıs a Œ60s-era porgramme. > Oh yeah, all I really need is therapy. LOL. Posters who argue with > me canıt stop for a simple reason. Figure it out. --A HYDROGEN (sic; cracked methane) ECONOMY?... The Three Phases of Exploitation of the Protocols of the Elders of Kyoto (sik): BORE/GUSH/NADIR @ http://www.tarpley.net. Http://www.tarpley.net/bushb.htm (partial contents, below): 17 -- THE ATTEMPTED COUP DıETAT, 3/30/81 (87K) 18 -- IRAN-CONTRA (140K) 19 -- THE LEVERAGED BUYOUT MOB (67K) 20 -- THE PHONY WAR ON DRUGS (26K) 21 -- OMAHA (25K) 22 -- GEORGE #9 TAKES THE PRESIDENCY (112K) 23 -- THE END OF HISTORY (168K) 24 -- THE NEW WORLD ORDER (255K) 25 -- THYROID STORM (139K) http://quincy4board.homestead.com/Ŝles/curriculum/Cosmo === Subject: Re: Compromise, and you leave my papers? === >Subject: Re: Compromise, and you leave my papers? >> 1. It is the NORMAL thing to do to write a letter to the Editor >> when you see a mistake. If ANYBODY published your piece-of- >> there would be letters to the editor - thousands, if it were >> Wiles - yet you want special treatment. >> >> Iım not sure that it is the NORMAL thing to do. I suspect that in most >> cases people contact the author on Ŝnding a mistake in a paper. I >> suspect that most editors would either refer you to the author, or >> forward your message to hir. Only if the author refused to address the >> concerns would an editor be likely to bother to get involved. Heck, >> even Pertii contacted the authors of the erroneous papers that he found. >> The case at hand is, of course, different. The author had most >> deŜnitely been contacted before publication, and the people contacting >> the editor had every reason to believe that contacting the author yet >> again would accomplish less than nothing. So the usual Ŝrst step was >> bypassed. >Does anyone else think that this thread (as all the ones before it) is >completely pointless? Clearly people at some point started out wanting >to help James understand the mistake in his papers. However, by now >it is clear that this is a futile attempt, and that he does not >appreciate this help. Consequently posters no longer reply out of >sympathy for James (and make this clear); however, then what is the >point? No one except James could possibly have an interest in >prolonged discussions why his results are wrong. >If James wants to publish a paper in some third-, fourth- or >Ŝfth-rate journal, all he needs to do is submit it to enough of these >and keep quiet about it. Keep quiet about it? Harris? Yeah, thatıll happen. >I am certain that there are many proofs of >Fermatıs last theorem, P=NP, P!=NP etc. hidden away somewhere where >no-one will ever Ŝnd them ... >By the way, an entry for Jamesıs paper does remain on MathSciNet, so >the editor wasnıt successful in removing all records of the fact that >it was once accepted. >Finally, I think we should all feel somewhat sorry for James. I felt sorry for the mouse I caught. But once I stomped it ŝat I got over it. >Clearly >he is not well; perhaps if he was willing to seek therapy, he would be >able to lead a happier life. Until then, it seems best not to spend >much time on the thankless task of refuting his purported proofs. >Lasse >--- >(lasse@remove.for.spam.maths.warwick.ac.uk) -- Mensanator Ace of Clubs === Subject: Re: Compromise, and you leave my papers? > >> >> >> >Like remember that with > >P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > >and > >P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > >the aıs are the three roots of the cubic > >a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 > >and you can check at m=1, f=sqrt(2), that my argument is correct. >> >>And you can check that with m = 1 and f = 2 your argument is incorrect. >> >> >> >>Rick > > > > Readers may think Decker is insinuating that m=1, f=sqrt(2) is a > special case that might never occur again. > Iım not insinuating that. Iım saying that your claim--that > in your factorization exactly two of the aıs are divisible > by f and the other is coprime to f--is not true in general. > Well thatıs what it looks > like, but in fact the only thing special about that case is that with > it one of the aıs is rational. > > With one of the aıs rational, you can of course, look at it and just > see that my work does give the right result. > Evidently you didnıt try it. With m = 1 and f = 2 you also have > the case that one of the aıs is rational. The factorization is > P(1) = 4(7 x^3 - 9 x + 2) = (a_1 x + 2)(a_2 x + 2)(a_3 x + 2) > where a_1 = -2, a_2 = (-7 + sqrt(105))/2, a_3 = (-7 - sqrt(105))/2 > See, one of the aıs is rational, just as in the case f = sqrt(2). > However, in this case one of the aıs is divisible by 2, but neither of > the other two are. In addition, none of the aıs are coprime to 2 (in the > algebraic integers). Youıre right, I didnıt look closely, but then itıs actually a supporting example. Iıve said, more than once, that if one of the aıs is rational then it will either be coprime to f or have f as a factor, just as your result does. Now one of the other aıs properly has 2 as a factor as well, and the other is properly a unit, but not in the ring of algebraic integers. Have some fun, use an f thatıs not prime, like f=32, and you will Ŝnd that one of the aıs will either have 32 as a factor or be coprime to it. So yeah, I didnıt read what you had closely, but it supports my position. > There are an *inŜnity* of such results, and they will all give the > same information that my argument is correct. > Even if that were true, it would still leave open the possibility > that for inŜnitely many of your polynomials your argument is > wrong. > You people arenıt paying attention either. Sure I didnıt pay attention closely to what you said, but you clearly havenıt been paying attention to what Iıve said. James Harris === Subject: Re: Compromise, and you leave my papers? Discussion, linux) >> P(1) = 4(7 x^3 - 9 x + 2) = (a_1 x + 2)(a_2 x + 2)(a_3 x + 2) >> where a_1 = -2, a_2 = (-7 + sqrt(105))/2, a_3 = (-7 - sqrt(105))/2 >> See, one of the aıs is rational, just as in the case f = sqrt(2). >> However, in this case one of the aıs is divisible by 2, but neither of >> the other two are. In addition, none of the aıs are coprime to 2 (in the >> algebraic integers). > Youıre right, I didnıt look closely, but then itıs actually a > supporting example. > Iıve said, more than once, that if one of the aıs is rational then it > will either be coprime to f or have f as a factor, just as your result > does. > Now one of the other aıs properly has 2 as a factor as well, and the > other is properly a unit, but not in the ring of algebraic integers. Which is which? Which one is properly a unit and why *that* one? -- Now I realize that he got away with all of that because sci.math is not important, and the rest of the world doesnıt pay attention. Like, no one is worried about football players reading sci.math postings! -- James S. Harris on jock reading habits === Subject: Re: Compromise, and you leave my papers? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ıELIi $t^ VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> >>Like remember that with >> >>P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) >> >>and >> >>P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >> >>the aıs are the three roots of the cubic >> >>a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 >> >>and you can check at m=1, f=sqrt(2), that my argument is correct. > >And you can check that with m = 1 and f = 2 your argument is incorrect. >> >> Readers may think Decker is insinuating that m=1, f=sqrt(2) is a >> special case that might never occur again. >> Iım not insinuating that. Iım saying that your claim--that >> in your factorization exactly two of the aıs are divisible >> by f and the other is coprime to f--is not true in general. [...] >> Evidently you didnıt try it. With m = 1 and f = 2 you also have >> the case that one of the aıs is rational. The factorization is >> P(1) = 4(7 x^3 - 9 x + 2) = (a_1 x + 2)(a_2 x + 2)(a_3 x + 2) >> where a_1 = -2, a_2 = (-7 + sqrt(105))/2, a_3 = (-7 - sqrt(105))/2 >> See, one of the aıs is rational, just as in the case f = sqrt(2). >> However, in this case one of the aıs is divisible by 2, but neither of >> the other two are. In addition, none of the aıs are coprime to 2 (in the >> algebraic integers). > Youıre right, I didnıt look closely, but then itıs actually a > supporting example. > Iıve said, more than once, that if one of the aıs is rational then it > will either be coprime to f or have f as a factor, just as your result > does. > Now one of the other aıs properly has 2 as a factor as well, and the > other is properly a unit, but not in the ring of algebraic integers. In what ring is it a unit, and why would that be relevant for a factorization in algebraic integers? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Compromise ... > Address: Pyh=E4j=E4rvenkatu 6 D 80 > 33200 Tampere > Finland > Phone: +358445052044 ... > With a minor exception. I donıt care to post my home address or phone > number, but it is not at all hard to Ŝnd, as you say. Guess this is > just a personal idiosyncrasy. Indeed. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Compromise > So youıre saying a mad person can still be basically good? > What do you think about that new German Ŝlm that portrays Hitler > as basically good? Have YOU seen this Ŝlm? I have not myself, so I also cannot truly judge it. However,from what those who have seen it tell me, and from reviews I have read, both in German and English papers, it would seem to me that it does not portray Hitler as basically good. I suspect that you may have read this about the Ŝlm (possibly before it even came out, when there was much speculation about it), and are now simply quoting it without question. If I am wrong and you have seen the Ŝlm and the above is your own conclusion, then I apologize. Otherwise, please refrain from perpetuating false claims by quoting them without thought. It is hardly scientiŜc. Lasse --- (lasse@remove.for.spam.maths.warwick.ac.uk) === Subject: Re: JSH: Compromise === >Subject: Re: JSH: Compromise >> So youıre saying a mad person can still be basically good? >> What do you think about that new German Ŝlm that portrays Hitler >> as basically good? >Have YOU seen this Ŝlm? I have not myself, so I also cannot truly >judge it. However,from what those who have seen it tell me, and from >reviews I have read, both in German and English papers, it would seem >to me that it does not portray Hitler as basically good. No, I have not seen it. >I suspect that you may have read this about the Ŝlm (possibly before >it even came out, when there was much speculation about it), and are >now simply quoting it without question. I am, in fact, refering to reviews I have read. And no, I canıt quote them. But the impression I got was the Ŝlm shows a more human side which is the source of controversy. It was my extrapolation of MMıs view that human = basically good. >If I am wrong and you have seen the Ŝlm and the above is your own >conclusion, then I apologize. Otherwise, please refrain from >perpetuating false claims by quoting them without thought. It is >hardly scientiŜc. So why are you not criticizing MM for jumping to false conclusions about JSH without having read his posts? Sometimes you have to make extreme and false claims in order to get people to think. >Lasse >--- >(lasse@remove.for.spam.maths.warwick.ac.uk) -- Mensanator Ace of Clubs === Subject: Re: JSH: Compromise Mensanator nous a fait lıhonneur dı.8ecrire : > [...] > of MMıs view that human = basically good. > So why are you not criticizing MM for jumping to false > conclusions about JSH without having read his posts? bad. Thatıs what you call a conclusion? It exactly means that I donıt know and that I have no reason to think the contrary. (In case you didnıt notice it, D. Kastrup sent a post in which he explained me why I was wrong) Why do you ask others to criticize me? You need the help of others? Thatıs easier with a pack? And what do you ask them to criticize? The fact that I told you attacking someone on his look is unacceptable or the fact that I told you, in order to write vermin donıt deserve just treatment, one must have a few limping neurons? > Sometimes you have to make extreme and false claims > in order to get people to think. If you tried to plagiarize JSH there, thatıs very good. And, in more, to be ambitious is a quality. A last thing, mister I decide what are vermin, since seemingly JSHıs look, I was not defending JSH (this guy already threatened me twice), I was attacking your behaviour. Thatıs not the same. It was a real pleasure. Indeed. But I think I wonıt answer your future posts. This is useless, it is clear that you are quite able to bring discredit on yourself without any help. -- mm http://www.ellipsa.net/ mm@ellipsa.no.sp.am.net ( suppress no.sp.am. ) === Subject: Re: JSH: Compromise === >Subject: Re: JSH: Compromise >Message-id: <4179A9B8.87DD610F@ellipsa.no.sp.am.net> >Mensanator nous a fait lıhonneur dı.8ecrire : >> [...] >> of MMıs view that human = basically good. >> So why are you not criticizing MM for jumping to false >> conclusions about JSH without having read his posts? >bad. Thatıs what you call a conclusion? It exactly means that >I donıt know and that I have no reason to think the contrary. >(In case you didnıt notice it, D. Kastrup sent a post in which >he explained me why I was wrong) >Why do you ask others to criticize me? You need the help of >others? Thatıs easier with a pack? And what do you ask them to >criticize? The fact that I told you attacking someone on his >look is unacceptable or the fact that I told you, in order to >write vermin donıt deserve just treatment, one must have a >few limping neurons? >> Sometimes you have to make extreme and false claims >> in order to get people to think. >If you tried to plagiarize JSH there, thatıs very good. And, in >more, to be ambitious is a quality. >A last thing, mister I decide what are vermin, since seemingly >JSHıs look, I was not defending JSH (this guy already threatened >me twice), I was attacking your behaviour. Thatıs not the same. >It was a real pleasure. Indeed. But I think I wonıt answer your >future posts. Even if I pay for another 5 minutes? >This is useless, it is clear that you are quite >able to bring discredit on yourself without any help. >-- >http://www.ellipsa.net/ >mm@ellipsa.no.sp.am.net ( suppress no.sp.am. ) -- Mensanator Ace of Clubs === Subject: Re: JSH: Compromise <4175B49B.42479D11@ellipsa.no.sp.am.net> <41765175.914639FB@ellipsa.no.sp.am.net> <4177EB85.9D1C69AB@ellipsa.no.sp.am.net> posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK > mensanator@aol.com a .8ecrit : > mensanator@aol.com a .8ecrit : > > mensanator@aol.com a .8ecrit : > > Thatıs okay, thereıs always a few cry-babies who crawl out from > under > rocks that think weıre being too mean to you. > > Life is funny. > > So how come you donıt lighten up a bit? Flaming doesnıt hurt > anyone. > I mean, itıs not like Iım harassing Harrisı employer with e-mail > containing false racism charges. Only vermin would do something > like that, right? > > Someone completely mad can also do it. > Are you saying madness excuses such behaviour? > No, it doesnıt excuse, it explains. > You say that you > believe Harris is not basically bad. What evidence do you have to > support that belief? > None, of course. When I have evidences, I donıt believe, I know. > Are you telling me that I shouldnıt regard people as not guilty > as long as I have not the evidence that they are not criminals? Yes, assuming a person is innocent without evidence is just as bad as assuming guilt without evidence. Isnıt guilty until proven innocent the French way? The proper thing for you to do when you step into the midst of a ŝame war is to be neutral until you have enough information to make a proper judgement. Arbitrarily assuming the person being insulted is innocent is really stupid on your part. > > We look at the same thing and we donıt see the > same thing. > > Of course not. Iıve been reading this newsgroup for 4 years and > you just fell off the turnip truck. I see Harris for what he is, > you see him for what you wish him to be. > > I see him for what he is. He is completely mad. > So youıre saying a mad person can still be basically good? > Of course, yes. And someone who has a well-balanced mind might > be a true slimeball. But all that is not very meaningfull. Œmadı is > a generic term that can cover a lot of different realities (More > exactly what is a generic term is the French word Œfouı that > I translate using Œmadı). > What do you think about that new German Ŝlm that portrays Hitler > as basically good? > I didnıt see it yet. But if this movie tells me that, one day, > Hitler was 5 and that he played at cops and robbers with > other children of his age, it will learn me nothing. I know > that Hitler was a human being and that any human being can > become criminal. And I also know that a real is not > necessarily a real 24/24. Most criminals love their > children (when they have, of course) but does it make them > people basically good? Sounds like you agree with the notion that even the most evil people in the world can have moments of goodness. Some people donıt like that idea. They want to rationalize their hatred of evil (although there is no reason to). The reason they get upset about showing the truth is because of people like you - people who pre-judge based solely on the fact that the vermin in question is a human being. > Believing that the good are always good and the bad always bad > is perhaps very restful for the mind of a FoxNews watcher but, > like any manicheism, it is just bull. You got that wrong also. Itıs not about believing that evil are always evil. The problem is when the evil put up a facade of good to conceal their machinations. The problem is when some gullible sap like yourself falls for it and starts crying about the vermin being treated unjustly. > Finally, what did hurt you? Maybe I didnıt make myself clear. I asked you to try to guess how much I was hurt by your insults. You guessed wrong. > The word Œfascistı? Why do have a hair up your ass about fascists? Is it because paper, there are more fascists in France now then there was during WWII. And why fascist? Why not just say NAZI? It would be more appropriate since I have German ancestry and it would probably make you feel better. Being evil is seductive, I can tell that youıre enjoying it. Welcome to The Dark Side. > But after having written Œvermin do not deserve a just treatmentı, > what did you expect? A Nobel prize? What I donıt expect is hypocrisy. If youıre going to to take the high road and defend Harris against ŝamers, you shouldnıt then turn around and call people fascists. It changes you from a dork to an asshole. > By saying this, you clearly claim that it is acceptable to apply > different laws according to arbitrary criteria (Who does decide > what are vermin or not? And how?). [1] Not arbitrary. Thereıs a simple Golden Rule: YOU DONıT WITH SOMEONEıS LIVELIHOOD. NOT EVEN IN JEST. If Harris were to darken the doorway at my place of employment, heıll leave in a basket. [2] I decide who is vermin. [3] Read the goddamn posts. Itıs all there in black and white. > By saying this, you spit on a > fundamental principle of democracy: the law must be the same for > everybody. No, I spit on the likes of you for defending Harris. My laws are applied to everyone. Doesnıt matter whether they have a PhD or have a high IQ. They defend Harris, they get spit upon. Is that democratic enough for you? > Even if this principle is never actually applied (even > in democratic countries, it is preferable to be rich, young and in > good health than the contrary), spitting on it is spitting on > in this thread, the word Œfascistı is quite appropriate to the > context. If I recall my history, the fascists picked on those who didnıt deserve it. Giving someone what they deserve (whether a Jap in 1945 or a Taliban in 2002) doesnıt make you a fascist. > -- > mm > http://www.ellipsa.net/ > mm@ellipsa.no.sp.am.net ( suppress no.sp.am. ) === Subject: Test of Genius Math Question I am a teacher for a 7th grade math class. I was working on a Test of Genius math quiz and was planning on giving it to the kids. Now I feel pretty silly asking, but I feel like Iım missing something. Here is the question word for word... Four stamps can be attached to each other in various ways. One way is shown here. In how many other ways might four stamps be attached? (Here is a picture of 4 stick Ŝgures that all look different attached together) My initial reaction was to use a factorial 4x3x2x1 getting 24 different ways. Apparently the book says the answer is 18... What am I missing??? Nino Skilj === Subject: Re: Test of Genius Math Question === >Subject: Test of Genius Math Question >I am a teacher for a 7th grade math class. I was working on a Test of >Genius math quiz and was planning on giving it to the kids. Now I >feel pretty silly asking, but I feel like Iım missing something. Here >is the question word for word... >Four stamps can be attached to each other in various ways. One way is >shown here. In how many other ways might four stamps be attached? >(Here is a picture of 4 stick Ŝgures that all look different attached >together) >My initial reaction was to use a factorial 4x3x2x1 getting 24 >different ways. Apparently the book says the answer is 18... What am I >missing??? That the actual answer is 19? http://mathworld.wolfram.com/Tetromino.html >Nino Skilj -- Mensanator Ace of Clubs === Subject: Re: Test of Genius Math Question === >>Subject: Test of Genius Math Question >>Four stamps can be attached to each other in various ways. One way is >>shown here. In how many other ways might four stamps be attached? >> ^^^^^ >>different ways. Apparently the book says the answer is 18... What am I >>missing??? >That the actual answer is 19? No, the actual answer is 18 :). Read the question again. -- Erick === Subject: Re: Test of Genius Math Question === >Subject: Re: Test of Genius Math Question >Message-id: === >Subject: Test of Genius Math Question >Four stamps can be attached to each other in various ways. One way is >shown here. In how many other ways might four stamps be attached? > ^^^^^ >different ways. Apparently the book says the answer is 18... What am I >missing??? >>That the actual answer is 19? >No, the actual answer is 18 :). Read the question again. But I wasnıt answering the problem question, I was answering the OPıs question on why the different ways is not 24. > -- Erick -- Mensanator Ace of Clubs === Subject: Re: Test of Genius Math Question > . . . Here is the question word for word... > Four stamps can be attached to each other in various ways. One way > is shown here. In how many other ways might four stamps be attached? > (Here is a picture of 4 stick Ŝgures that all look different attached > together) > My initial reaction was to use a factorial 4x3x2x1 getting 24 > different ways. Apparently the book says the answer is 18... What am I > missing??? The question could be better worded. http://mathworld.wolfram.com/Tetromino.html -- Anton Sherwood (prepend 1 to address) http://www.ogre.nu/ === Subject: Else charset=us-ascii by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with SMTP id i9N2oCb17288 Heat Pain Relief
And other Items
Now === Subject: Not turtles, itıs hedgehogs - Pioneer 10/11 Dark Energy Anomaly Itıs hedgehogs. The Pioneer anomaly is a hedgehog dark energy exotic vacuum point topological defect in the single-valued local Higgs Ocean macro-quantum order parameter that needs to live in at least SO(3)/SO(2) = S^2 quotient coset space for the spontaneous broken inŝationary cosmology false to true vacuum symmetry SO(3) -> SO(2) from Dirac Sea to Higgs Ocean. Hedgehog dark energy point defect centered at Sun explains Pioneer anomaly. Hedgehog dark energy point defect centered at black hole in center of galaxy? The hedgehog consists of two concentric spheres. Inner sphere is a hole. Between the spheres is a CONSTANT gradient Ŝeld and the energy scales as r! Thatıs it! Different ways to look at it. Potential energy of exotic vacuum per unit test mass is V(zpf) = c^2/zpfr^2 = cHr = c(Hubble recession speed) where /zpf = H/cr a_P = cH(t) ~ 10^-7 cm/sec^2 actually observed pointin back to Sun for inner sphere at ~ 20AU and outer sphere at least out to 70AU. This is cosmic time dependent H(t) = R(t)^-1dR(t)/dt R(t) is dimensionless FRW scale factor. Also in terms of an equivalent rotational centripetal acceleration picture Tangential v = fr a_P = v^2/r = f^2r f^2 = cH/r = c^2/zpf === Subject: Re: Not turtles, itıs hedgehogs - Pioneer 10/11 Dark Energy Anomaly > Itıs hedgehogs. > The Pioneer anomaly is a hedgehog dark energy exotic vacuum point > topological defect in the single-valued local Higgs Ocean macro-quantum > order parameter that needs to live in at least SO(3)/SO(2) = S^2 > quotient coset space for the spontaneous broken inŝationary cosmology > false to true vacuum symmetry SO(3) -> SO(2) from Dirac Sea to Higgs Ocean. Jesus Christ on a stick! You are absomentally bonkers! *PLONK* === Subject: Galactic halo is not a hedgehog Galactic Halo Dark Matter Rotation Curve is ŝat. v(star) = fr independent of r So in that case f^2 ~ v^2/r^2 ~ c^2/zpf /zpf = v^2/c^2r^2 V(zpf) = c^2/zpfr^2 = v^2 a_P = -dV(zpf)dr = 0 Itıs hedgehogs. The Pioneer anomaly is a hedgehog dark energy exotic vacuum point topological defect in the Higgs Ocean macro-quantum order parameter that needs to live in at least SO(3)/SO(2) = S^2 quotient coset space for the spontaneous broken inŝationary cosmology false to true vacuum symmetry SO(3) -> SO(2) from Dirac Sea to Higgs Ocean. Hedgehog dark energy point defect centered at Sun explains Pioneer anomaly. Hedgehog dark energy point defect centered at black hole in center of galaxy? The hedgehog consists of two concentric spheres. Inner sphere is a hole. Between the spheres is a CONSTANT gradient Ŝeld and the energy scales as r! Thatıs it! Different ways to look at it. Potential energy of exotic vacuum per unit test mass is V(zpf) = c^2/zpfr^2 = cHr = c(Hubble recession speed) where /zpf = H/cr a_P = cH(t) ~ 10^-7 cm/sec^2 This is cosmic time dependent H(t) = R(t)^-1dR(t)/dt R(t) is dimensionless FRW scale factor. Also in terms of an equivalent rotational centripetal acceleration picture Tangential v = fr a_P = v^2/r = f^2r f^2 = cH/r = c^2/zpf === Subject: Help: Round Robin Tournament Proof. Hello everyone, I am still limping through my discrete math course. I need some help on the following homework problem: Use mathematical induction to show that in any round-robin tournmament involving n teams, where n >= 2, it is possible to label the teams T1, T2, ..., Tn so that Ti beats T(i+1) for all i = 1, 2, ..., n-1. I think that I understand the proof, and I have a solid strategy to attack the problem in my mind, but I am struggling with getting the idea onto the paper. Below is my proof to the problem. I would appreciate any feedback, and if possible the pointing out of the errors I may have made. THEOREM: In any round robin tournament(R.R.T) involving n teams, where (n >= 2), it is possible to label the teams T1, T2, ..., Tn so that Ti beats T(i+1) for all i = 1, 2, ... n-1. PROOF: We will use the principal of mathematical induction. BASIS STEP: Let n = 2, Thus we have two teams. We will show the basis step is true for a R.R.T for 2 teams. Let Tı represent one of the teams in the R.R.T. * Case 1: Tı wins. Thus Tı, T2 and since this is a R.R.T of 2 teams Tı = T1. So we have T1, T2. * Case 2: Tı loses. Thus T1, T2 and since this is a R.R.T of 2 teams Tı = T2. So we have T1, T2. We have shown that the basis step is true. Induction Step: Let k >= 2. If the property is true for n = k, then it is true for n = k+1. Suppose in a R.R.T if k >=2, it is possible to label teams T1, T2, ... Tk so that Ti beats T(i+1) for all i = 1, 2, ..., k-1. There are K teams in the R.R.T. We must show it is possible to label teams T1, T2, ..., Tk, T(k+1) so that Ti beats T(i+1) for all i = 1, 2, ..., k-1, k. There are k+1 teams in the R.R.T. Let Tı represent one of the teams in the R.R.T. Consider three cases. * Case 1: Tı beats T1. T1 represents the current 1st place team. Since Tı beats T1, Tı is the new T1. The old T1 shifts up a number as do all other teams. We have: T1, T2, ..., Tk, T(k+1). It follows since T2, ..., Tk, T(k+1) is k teams, by the inductions hypothesis they can be ordered as the theorem suggests. * Case 2: Tı loses to the Ŝrst m teams (where 1 <= m <= k), and beats the (m+1) team. So T2 < Tı < T(k+1). Equvialently: T1 < ... < Tı <= Tk < T(k+1). It follows from the induction hypothesis that T1 < ... < Tı <= Tk can be ordered, and T(k+1) is next in the sequence as the theorem suggests. * Case 3: Tı loses to every team. Thus Tı is the last place team, in otherwords T(k+1). So we have T1, T2, ..., Tk, T(k+1). By the induction hypothesis T1, T2, ..., Tk can be ordered, and T(k+1) is the next ordered team in the sequence as suggested by the theorem. This ends the inductive step. In each of the three cases we showed that the theorem holds. QED. I donıt feel comfortable at all with the above proof. If you could please provide insight into when I was on the right track, and when I -= Chris Potter =- -- Mind over matter: If you donıt mind, then it doesnıt matter. === Subject: Re: Help: Round Robin Tournament Proof. > * Case 2: Tı loses to the Ŝrst m teams (where 1 <= m <= k), and > beats the (m+1) team. So T2 < Tı < T(k+1). Equvialently: T1 < ... > < Tı <= Tk < T(k+1). It follows from the induction hypothesis that > T1 < ... < Tı <= Tk can be ordered, and T(k+1) is next in the > sequence as the theorem suggests. You seem to be assuming that beats is a transitive relation. Also, at what point are you using the induction hypothesis to order k of the teams and generate the labels T1, T2, ..., Tk? -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: misnomers |Having looked at |http://www.fh-augsburg.de/~mueckenh/ |I do not consider Mueckenheim a crackpot. Itıs a little hard for me to understand how else a person could wind up writing a paper in which he claims Cantorıs Ŝrst proof of the uncountability of the reals can show the uncountability of a countable set, giving as an instance the sequence -1,1/2,-1/3,1/4,.... After all, the sequence of intervals (-1,1/2),(-1/3,1/4),... shows that the limit 0 is not in the sequence. (?!) Whether or not he is a crackpot, he is mighty confused. Keith Ramsay === Subject: Re: Fibonacci connection between Huffman codes and Wythoff array > Fibonacci connection between non-decreasing sequences of positive integers > producing maximum height Huffman trees and the Wythoff array has been proved. > The paper can be seen at > * http://arxiv.org/abs/cs.DM/0410013 > The abstract can be seen at > * http://mathforum.org/epigone/sci.math.research/pherdkralglend See also Sloane A098950 -> http://www.research.att.com/projects/OEIS?Anum=A098950 in Sloaneıs online encyclopedia of integer sequences. -- Alex Vinokur email: alex DOT vinokur AT gmail DOT com http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn === Subject: Lost the thread: Series with x^(n^2) involving the series S(x) = x + x^4 + x^9 + x^16 + ... The following will be no help at all, but is rather cute. Itıs somewhat of a Quet special. :) We need more of these, Leroy! S(x)/(1-x) = SUM[n=1..oo] ŝoor(sqrt(n)).x^n This turns the funny=powered series into a normal one that is even less use! ------------------------------------------------------------- --------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ------------------------------------------------------------- --------------- -- Q. Why did the chicken cross the Moebius strip? ------------------------------------------------------------- --------------- -- === Subject: Re: Duel to Settle 0.999... =? 1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC42B03430; >>Youıre too late. This issue was settled years ago. >To All, >It seems that the arguments over whether 0.999... = 1 or not could be easily >>settle by a duel. The camp that believes 0.999... does NOT equal 1 choose a >>representative. Likewise for the camp that believes 0.999 DOES equal 1. >>Then the two representatives write down 0.999... explicitly (writing all 9ıs >I suggest that the representative of the camp that believes 0.999... does >>NOT equal 1 go Ŝrst. While that representative is writing his/her 9ıs, the >>rest of the two camps are free to work on other unsolved mysteries of >>mathematics. >By the way, I volunteer to be the representative of the camp that believes >>0.999 ... DOES equal 1. > If you can say that, I can say the fairy godmother exists there too. So, >.99999.... = the fairy godmother >- MO >Smartıs Alt. Physics News Group >ht tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv =1 >S. Enterprize (Science Journal) >http://smart1234. s-enterprize.com / SEC, I have a little proof of 0.999... = 1 that seems fool-proof. I am curious what ŝaw, if any, you might see in it. Here it is: Let x = 0.999... Then 10x = 9.999... = 9 + x. Then 10x - x = 9. Then 9x = 9. Then x = 9/9 = 1, QED. I do not expect to convince you with this proof that 0.999... = 1. I just am curious what ŝaw you see in it. :-) MO === Subject: Re: Duel to Settle 0.999... =? 1 >Youıre too late. This issue was settled years ago. >>To All, >> >>It seems that the arguments over whether 0.999... = 1 or not could be >easily >settle by a duel. The camp that believes 0.999... does NOT equal 1 choose >representative. Likewise for the camp that believes 0.999 DOES equal 1. >Then the two representatives write down 0.999... explicitly (writing all >9ıs >> >>I suggest that the representative of the camp that believes 0.999... does >NOT equal 1 go Ŝrst. While that representative is writing his/her 9ıs, >the >rest of the two camps are free to work on other unsolved mysteries of >mathematics. >> >>By the way, I volunteer to be the representative of the camp that >believes >0.999 ... DOES equal 1. >> If you can say that, I can say the fairy godmother exists there too. So, >>.99999.... = the fairy godmother >> >>- MO >>Smartıs Alt. Physics News Group >>http://pub39.bravenet.com/forum/show.php?usernum=3320272813 &cpv=1 >>S. Enterprize (Science Journal) >>http://smart1234.s-enterprize.com/ >SEC, >I have a little proof of 0.999... = 1 that seems fool-proof. I am curious >what ŝaw, if any, you might see in it. Here it is: >Let x = 0.999... Then 10x = 9.999... = 9 + x. >Then 10x - x = 9. Then 9x = 9. Then x = 9/9 = 1, QED. That doesnıt prove anything. x = .9999..... 10 x = 9.99999... But, as n -->oo 10^n (x) = 999999.... which is an indeterminate again, right back where you started from, never proving .999... = 1. So to just make up a equation with the .9999... still in itıs form doesnıt prove .9999... = 1 .99999.... - .99999... = 0 and, 1 - 1 = 0, too, so .999... = 1. But this doesnıt prove, .9999... = 1 And, the QED and QM orbital models are incorrect models of the atom. The Smart Atomic Model is the only correct model, in my opinion. >I do not expect to convince you with this proof that 0.999... = 1. I just am >curious what ŝaw you see in it. >:-) MO Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: Duel to Settle 0.999... =? 1 >>I have a little proof of 0.999... = 1 that seems fool-proof. I am curious >>what ŝaw, if any, you might see in it. Here it is: >>Let x = 0.999... Then 10x = 9.999... = 9 + x. >>Then 10x - x = 9. Then 9x = 9. Then x = 9/9 = 1, QED. > That doesnıt prove anything. > x = .9999..... > 10 x = 9.99999... > But, > as n -->oo > 10^n (x) = 999999.... > which is an indeterminate again, right back where you started from, never > proving .999... = 1. Oh my God!!! Youıre the most unbelievably idiotic troll I have seen in the 12 years Iıve been using the Internet!! Another comment regarding this: > So to just make up a equation with the .9999... still in itıs form doesnıt > prove .9999... = 1 > .99999.... - .99999... = 0 > and, > 1 - 1 = 0, too, so .999... = 1. > But this doesnıt prove, .9999... = 1 > And, the QED and QM orbital models are incorrect models of the atom. The > Smart Atomic Model is the only correct model, in my opinion. Youıre the most unbelievably idiotic troll I have seen in the 12 years Iıve been using the Internet!! Youıll be held forever in my kill list. Carlos -- === Subject: Re: Duel to Settle 0.999... =? 1 > SEC, > I have a little proof of 0.999... = 1 that seems fool-proof. I am curious > what ŝaw, if any, you might see in it. Here it is: > Let x = 0.999... Then 10x = 9.999... = 9 + x. > Then 10x - x = 9. Then 9x = 9. Then x = 9/9 = 1, QED. > I do not expect to convince you with this proof that 0.999... = 1. I just am > curious what ŝaw you see in it. > :-) MO I havenıt done anything really dumb today, up to now; so, itıs time to get embroiled in the .999... debate. We need to think about what we intend the expression .999... to mean. Those three dots carry a lot of signiŜcance. I think we all understand well enough what any *terminating* decimal such as .999 or .999999999 or .9999...9 means, even though in the last case we leave it indeŜnite just how many digits there are in the string. Assumption 1: .999... is not the same as any terminating decimal fraction; it is not even the same as my third example, which terminates but we are not sure when. Assumption 2: .999... does mean something. It designates some ordinary real number. Letıs call it r. What can we say about r? It is the limit of the terminating decimal fractions .999...9, which are also expressible as 1 - 10^(-n) for different integers n. Here, n must be Ŝnite, of course. It is reasonable to infer that r <= 1, and r >= 1 - 10^(-n) for all Ŝnite integers n. [For the moment, let us treat the word limit as a bit of handwavingit suggests an intuitive concept which we havenıt all formalized, at least not in the same way; if we had, then this perennial thread would have halted long ago.] So 0 <= 1 - r < 10^(-n) for all Ŝnite integers n. In fact, 0 <= 1-r < 1/M for all Ŝnite positive integers M. Does this imply that 1-r = 0, that is, that r = 1? It does, *if* we accept the following assumption, which is often attributed to Archimedes: Assumption 3: if a real number s satisŜes s >= 0, and s < 1/M for all Ŝnite positive integers M, then s = 0. Is Assumption 3 true? It is usually assumed to be so, as part of the deŜnition of real numbers. It can be accepted as a result of formalizing the intuition that we are trying to express when we talk about real numbers. Some people might Ŝnd Assumption 3 a little ad-hoc; it looks as if I made it up on the spur of the moment to cheat my way out of a tight spot. But it was silently present right from the start of the discussion. Anyone who seriously raises the question whether 0.999... equals 1 must begin by assuming that expressions like 0.999... mean well-deŜned real numbers. Assumption 4A: a decimal expansion, terminating or not, represents some real number. Assumption 4B: if two real numbers have the same decimal expansion, they are equal. If these two assumptions are accepted, then MOıs proof, where the key step is that 9.9999... = 9+ 0.9999... is rigorously correct. Actually, Assumptions 4A-B are equivalent to Assumption 3, if we take ordinary arithmetic with decimal fractions for granted. So, does this nail the subject down once and for all. No. Things can get stranger than this. Once we have formulated the axiom of Archimedes (Assumption 3), we can ask what happens if it is not true. I know of two ways to go beyond Archimedes. One is to construct non-Archimedean Ŝelds. John Horton Conway created (? or discovered) *surreal* *numbers.* These include lots of inŜnite integers, not to mention their reciprocals etc. In this system there are plenty of numbers strictly greater than 0.9999...9 and strictly less than 1. I think that 0.999... fails to converge in the Field of surreal numbers, so Assumption 4A is broken. Then there is *non-standard* *analysis.* To me this is even stranger that surreal numbers. Here, we invent a distinction between standard numbers and non-standard numbers. All the numbers that we usually study are standard. But there are, for example, non-standard integers which, though Ŝnite, are bigger than any standard integer. And there are inŜnitesimal numbers which are greater than 0 but less than 1/M for every *standard* Ŝnite integer M. In non-standard analysis, every real number is the sum of a standard real number and an inŜnitesimal. Throw away the inŜnitesimal and what you have left is the standard part of the number. So there are many numbers r such that 0.999...9 < r <= 1, but they all have the same standard part, which is 1. -- Chris Henrich God just doesnıt Ŝt inside a single religion. === Subject: Re: Duel to Settle 0.999... =? 1 >So there are >many numbers r such that 0.999...9 < r <= 1, but they all have the same >standard part, which is 1. .9999..... is just that .999.... . When you say .9999... < r <= 1 You are deŜning r, not .9999.... .99999... == 1 by fact, and by deŜnition. The .... dots only state an inŜnite repetition of 9ıs. If the deŜnition was, .9999....--> 1 This is another statement, which is different than just, .9999.... . .9999...--->1 is called counting like, 1-->2-->3.... To say, 1...... converges to 2 is incorrect. You only get to 2 by counting. Smartıs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Lost the thread: Series for pi. My thread-identifying procedure clearly needs a revamp. Anyway, someone recently posted a formula for pi, involving the series (1/10) + (1/10)^2 * 2/3 + (1/10)^3 * 2*4/3/5 + ... Unfortunately I donıt seem to have copied the whole thing properly. Can someone please repost it or mail it to me? ------------------------------------------------------------- --------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ------------------------------------------------------------- --------------- -- Doing ŝoating-point arithmetic is like shifting a pile of sand: each time you do it, you lose a little bit of sand and pick up a little bit of dirt. ------------------------------------------------------------- --------------- -- === Subject: Re: Lost the thread: Series for pi. > My thread-identifying procedure clearly needs a revamp. > Anyway, someone recently posted a formula for pi, involving the series > Unfortunately I donıt seem to have copied the whole thing properly. > Can someone please repost it or mail it to me? Archives for sci.math http://mathquest.com/discuss/sci.math http://mathforum.org/epigone/sci.math Google newsgroup advanced search. Search for example, by your name. === Subject: Enumerating automorphisms of a group Iım going to ask a very general question. Iım hoping that if someone can enlightened me on this, then I will be able to see how the various theorems of group theory fall together. The question is this: Given G a Ŝnite group. Find the group of automorphisms of G. For some groups, there is a certain answer. For example, for Sn the symmetric group of order n!, we know Aut Sn = Inn Sn = Sn (at least for n != 6). But in general, how should one go about enumerating all the automorphisms of any Ŝnite group? Is there even a way to just count the *number* of automorphisms? Kira === Subject: Re: Enumerating automorphisms of a group posting-account=jcZk7AwAAADXpPEyHtVyWC264SxtppRB I am interested in this for some small grps, and will watch this thread with interest. See the thread Automorphisms of small non-Abelian groups or of quaternions Q, S_3, S_4, and Z_2 x Z_2. I assume you familiar with the autos of abelian grps. Van === Subject: anyone can help with sigma-algebra ?? sigma-algebra: 1. The empty set is in F. 2. If A is in F, then so is the complement of A. 3. If {An} is a sequence of elements of F, then the union of the {An} is in F. are there exist any sigma-algebra,where X=N (natural numbers 1,2,3...) which generates from two sets, sigma-algebra S, which contain 32 sets?, i found only sigma-algebra S, which from two sets generate 16 sets.: X={1,2,3....}, R={{1,2,3,4,5,6,7,8,9,10},{2,3,4,6,8....}} so sigma algebra which is generate by this R contain 16 sets 1={empty} 2={X} 3={1,2,..,10} 4=={2,4,6,8...} 5={11,12,13...} ... .. so anyone can help,and give that two sets R, whcih can geberate sigma algebra, build by 32 sets. sorry for my english., gefŜ@wp.pl === Subject: Re: anyone can help with sigma-algebra ?? > are there exist any sigma-algebra,where X=N (natural numbers 1,2,3...) > which generates from two sets, sigma-algebra S, which contain 32 > sets? No. Let A be your sigma-algebra. Let A_1, ... , A_n be the minimal non-empty elements of A. Any two of the A_iıs are disjoint and, since A is Ŝnite, it is clear that each element of A is the union of some A_iıs. So A has 2^n elements and therefore n = 5. Now, take two elements G_1 and G_2 of A and see each sigma-algebra they generate. Of course, you can assume that they are distinct and non-empty. Each G_i is the union of some A_iıs and you can even assume that itıs the union of no more than two A_iıs (otherwise, replace it by its complement). Now, a case-by-case study shows that there are always two A_iıs such that no element of the sigma-algebra generated by G_1 and G_2 contains one of them but not the other one. So, the sigma-algebra generated by them is not the whole A. Jose Carlos Santos === Subject: One for the Kids Two captains A and B pick teams from a pool of available players. They pick alternately, ABAB... . Assuming they know the merits of the players Aıs Ŝrst choice will be better than Bıs Ŝrst, Aıs second than Bıs second and so on, so A will get the better team. What is the best method of choice to produce teams which are as nearly as possible equal? === Subject: Re: One for the Kids > Two captains A and B pick teams from a pool of available players. They > pick alternately, ABAB... . Assuming they know the merits of the > players Aıs Ŝrst choice will be better than Bıs Ŝrst, Aıs second > than Bıs second and so on, so A will get the better team. > What is the best method of choice to produce teams which are as > nearly as possible equal? It might help if the requirement that both teams are of equal size is dropped. Of course the procedure only works for not too small player pools. Give both captains an equal amount of points (money) they can use to buy the available players. Then repeat the following step until one of the captains has spent all his money or if all players are picked. Both captains make secret bids for all players, putting them in sealed envelopes. The envelopes are then opened so that both can see the opponentıs point allocations and can thus get an idea about the opponentıs judgement of the playersı values. The maximum of all bids M is then treated: The captain of the other team is asked if he wantıs to pay more for the top valued player than this maximum bid M. If yes, then an auction with minimum bid M is started. (I both captains want to pay all their money for one player, then toss a coin...) Otherwise the captain that had made the highest bid has to pay the money announced in his original written bid and gets his favorite in his team. I both captains agree, then they can continue with the next player with second highest bid on the original list (either auction or the captain having made the higher bid has to pay his announced bid and gets the player in his team). If at least one of the captains disagrees in continuing using the current bid list, then a new round of secret bids is started, of course only using the remaining money and the remaining players. If several consecutive games are played, one could even think of a transfer of remaining money to the next selection round. A lot of material on this topic can be found in the book Brams and Alan D. Taylor, Cambridge University Press 1996. Hugo Pfoertner === Subject: Re: One for the Kids > Two captains A and B pick teams from a pool of available players. They > pick alternately, ABAB... . Assuming they know the merits of the > players Aıs Ŝrst choice will be better than Bıs Ŝrst, Aıs second > than Bıs second and so on, so A will get the better team. > What is the best method of choice to produce teams which are as > nearly as possible equal? Depends very much on the values of the players: Suppose these values are the same for both players. Letıs say they are: x_1 >= x_2 >= x_3 >= ... >= xn Letıs call the set of choices that A is allowed to take S, then Bıs choises are the complementary set T = {1,2,...,n} S. The total value of players picked by A will be sum over s in S of x_s, and your aim is to choose S so that this sum is as close as possible to 1/2 sum over i from 1 to n of x_i. Two examples: If all players have equal values, the exact values in S do not matter, as long as S contains n/2 elements (or (n+1)/2 resp. (n-1)/2 for odd n). If the value of each player is half the value of the next best player, you have a geometric sequence of values, x_i = 1/2^i, and the fairest choice is S = {1}. If A and B have different values for the same player, you get into the realms of game theory... Now, I would be interested to see for wich values x_i the systems of choosing players proposed by the others become fair... === Subject: Re: One for the Kids !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ıELIi $t^ VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Two captains A and B pick teams from a pool of available players. They > pick alternately, ABAB... . Assuming they know the merits of the > players Aıs Ŝrst choice will be better than Bıs Ŝrst, Aıs second > than Bıs second and so on, so A will get the better team. > What is the best method of choice to produce teams which are as > nearly as possible equal? It depends on the available players. If we assume that each player has independent merit, and that the total merit is the sum of the individuals, and that each captain will take the best that is left, then if we have 8 7 6 5 4 3 2 1 obviously the pattern ABBAABBA is good, but if we have 9 9 9 5 5 5 5 3 then ABBAABBA is quite unfair, and ABBAAABB would be nicer. The best method is to let one captain pick two teams, and let the other player decide which of the teams to play with. It does not address possible different strengths of the captains themselves, but I doubt you could take that into account. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: One for the Kids > Two captains A and B pick teams from a pool of available players. They > pick alternately, ABAB... . Assuming they know the merits of the > players Aıs Ŝrst choice will be better than Bıs Ŝrst, Aıs second > than Bıs second and so on, so A will get the better team. > What is the best method of choice to produce teams which are as > nearly as possible equal? Off the top of my head: Call a pick by each of A and B a round. In the Ŝrst round, A goes Ŝrst and B second. In the second round, B Ŝrst and A second. Continue to alternate who goes Ŝrst through successive rounds. This leads to the sequence of picks: ABBAABBAABB... Ken === Subject: Re: (Not quite) Cantorıs diagonal proof > Do you agree that both the informal proof and the formal proof must be > valid? > > Well, I would *prefer* that b o t h were valid. But thatıs exactly the > difference between these two approaches: while the formal proof can be > c h e c k e d and v e r i f i e d by computer, the informal proof cannot > (at least not presently). What difference does that make? If we use normal mathematics (not a computer) to show that the informal proof is not valid, then still, what is the sense of using software that tells us that the proof is valid? Thatıs just stupid, wouldnıt you say? > On the other hand (*sigh*) ... Since the formal proof has been > c h e c k e d and v e r i f i e d thereıs no point in trying to > question this proof by _informal_ reasoning, man. It is the system that tells us that an invalid proof is valid that needs to be checked and veriŜed (and returned to the manufacturer.) C-B > F. === Subject: Re: (Not quite) Cantorıs diagonal proof >- I could never > show that the number created by the diagonal was repeating. I posted it in > hopes that someone could show me why I kept running into the same problem > over and over without the circular argument of It fails because the > rational are countable and could instead point out where exactly Cantorıs > Method fails on rational numbers. > > I donıt understand your last point. If you replace real with > rational in Cantorıs theorem and his proof. The proof would be > incorrect because the diagonal number produced need not be rational. > Why? > Thatıs a question for you to answer. If you purported that the > diagonal proof also applied to the rationals then you would have to > prove that the diagonal number created for an arbitrary list of > rationals is itself rational. If not then there would be a gap in > your proof. So forget I said the diagonal number produced need not > be rational (which is true). Instead I would like to say prove that > the diagonal number is rational. so you cannot justify your statement that If you replace real with rational in Cantorıs theorem and his proof the proof would be incorrect > So, you wouldnıt have a proof that the set of rationals is > uncountable. There is nothing surprising there. > > I just donıt see the point in raising objections to a proof without > pointing out what the error is. Which line of the proof is wrong? > Thatıs what I would ask an objector. > The concept of unique number is wrong. > Ok. Then you are objecting to something prior to the proof. So why > bother discussing Cantorıs proof at all? heıs the one who claims anti-diag is unique. Hereıs an infnite list of reals 0.0xxxxxx.. 0.1xxxxxx.. 0.2xxxxxx.. 0.0xxxxxxx.. 0.3xxxxxx.. 0.5xxxxxx.. 0.8xxxxxx.. 0.1xxxxxxx.. 0.2xxxxxxxx.. 0.0xxxxxxx.. 0.1xxxxxxxx.. As you can see, 0.0xxxx.. appears repeatedly, so 0.0 is covered inŜnitely many times 0.1xxxxx.. appears an inŜnite number of times. It doesnıt matter what the digits the are, they are all present. 0.abcxxx.. is present for EVERY a, b and c. 0.abcdefghijklmnopqrstuvwxyz...... is ON THE LIST for ALL values of a, b, c, d, e, ... z by extension every digit combination is present. You claim there is ONE digit that is different, but that digit CANNOT appear at digit position 1. 0.1xxxx 0.2xxxx 0.3xxxx these are all on the list of computable reals. 0.11xxxx.. 0.12xxxx.. 0.13xxsxx.. these are all on the list of computable reals. The contradictory digit on your list must be after the 2nd decimal place mustnıt it? Your contradiction dissapears altogether because it cannot occur at a Ŝnite decimal. The digits 1, 2, 3, 4, 5 .. 9, 0 are virtually identical and interchangable, they merely adjust the region on the numberline but have *no semantic difference*, to make a formula claiming this number (which is always only a variable) is DIFFERENT to this number is nonsense, they are just DIGITS Countable inŜnity is bigger than you can conceive or mathematically manipulate.. Herc is mustnıt a word? > The whole reason we *introduced digits* for > numbers was exactly to distinguish numbers. > I thought it was to make arithmetic and notation easier. > 0.xyz... is different to 0.ayz... because x =/= a > You canıt come along and say, now we will use a different digit here, here, here ,here... > Why not? > that is merely the *deŜnition* of a unique cardinal. > That doesnıt make sense. > We laugh at your hyperinŜnities, > I donıt have any hyperinŜnities. > Halt is a deŜned function, > What does that have to do with my post? > anti diag is an illusion. > Show me the single contradictory digit from your proof. What digit position? > That doesnıt make sense. > -Leonard Blackburn > Herc === Subject: Re: (Not quite) Cantorıs diagonal proof >> >>- I could never >> show that the number created by the diagonal was repeating. I posted it in >> hopes that someone could show me why I kept running into the same problem >> over and over without the circular argument of It fails because the >> rational are countable and could instead point out where exactly Cantorıs >> Method fails on rational numbers. >> >> I donıt understand your last point. If you replace real with >> rational in Cantorıs theorem and his proof. The proof would be >> incorrect because the diagonal number produced need not be rational. >> >> Why? >> Thatıs a question for you to answer. If you purported that the >> diagonal proof also applied to the rationals then you would have to >> prove that the diagonal number created for an arbitrary list of >> rationals is itself rational. If not then there would be a gap in >> your proof. So forget I said the diagonal number produced need not >> be rational (which is true). Instead I would like to say prove that >> the diagonal number is rational. > so you cannot justify your statement that > If you replace real with rational in Cantorıs theorem and his proof the proof would be incorrect The diagonal argument depends on the least upper bound property of the real numbers. The diagonal argument produces a decimal digit string, which is associated with a certain inŜnite series. The partial sums of that series form a set of real numbers that is nonempty and bounded above. The least upper bound of that set is the required number. The same argument does not work when applied to the rationals, since the rationals do not satisfy the LUB property. And, by the way, there is nothing circular about the argument that if the original list contains all of the rational numbers, then the number produced by the diagonal argument, which is necessarily different from any number in the list, must therefore be irrational. -- Dave Seaman Judge Yohnıs mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Software for visualizing =?ISO-8859-1?Q?=28M=F6bius=29transforma?= =?ISO-8859-1?Q?tions=3F?= Is there any such software (preferably free)? What I would like is to give the function w(z) for a transformation, and a region in the z-plane, and get a graphical visualization of that region in the w-plane. Is there any Matlab package for this? I have also searched Mapleıs help for such a function, but not found anything. === Subject: Re: Cantorıs proof that #(Evens) = #(Naturals) is inconsistent Throughout, I will use TE to denote there exists and FA to denote for all. >I think what you are missing is that the symbol 4 in N is not the >same as the symbol 4 in E. False. 4 is exactly the same as an element of N, and as an element of E. The statements are that 4 is an element of N, and that the exact same object 4 is also an element of E. >For example: >A = { 1,2,3,4 } >B = { 2,4,6,8 } >Clearly A and B have the same card even though 6 is not A. What does, whether or not 6 is an element of A, have to do with questions of cardinality? It seems strange that you think that there is any connection between (A) the cardinality of a set, and (B) whether or not a speciŜed object is an element of the set. The ONLY connection that exists between (A) and (B) is that when the cardinality of a set is 0, then it is guaranteed that no object is an element of the set. SpeciŜcally, for any object x, and any nonzero cardinal m, there exists a set A of cardinality m such that x is an element of A. For any object x, and any nonempty set A, there exists a set B bijective with A such that x is an element of B. This makes the sentence of your above, Clearly A and B have the same card even though 6 is not A. bizarre at best, and a clear indication that you have missed the important facts about the cardinality of a set. >In the case >of A and B I can deŜne one of several bijections e.g. >f: A->B >f(1) = 4 >f(2) = 2 >f(3) = 8 >f(4) = 6 >Here 3 in A is the same as 8 in B. Garbage. The element 3 of A is a completely different object to the element 8 of B. The element 3 of A CORRESPONDS to the element 8 of B by the bijection which you have deŜned. You should learn to be more precise with the language that you use. >If we write i in A as x_i and f(i) >in B as x_i, You should not use the same symbol x_i to denote two different things at the same time. Because you have written the element i of A as x_i, then you have deŜned x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4. You cannot maintain this assignment of values and simultaneously deŜne x_i = f(i) for all elements i of A. That would make x_1 = 4, x_2 = 2, x_3 = 8, x_4 = 6, which contradicts the already established assignments x_1 = 1, x_3 = 3, x_4 = 4. The fact that you want to simultaneously use BOTH assignments is demonstrated by your next statement (immediately below). Instead, you should have written the element f(i) of B as y_i. >we have >A = { x_1,x_2,x_3,x_4 } >B = { x_1,x_2,x_3,x_4 }. Since you derived this using two contradictory assignments of x_1, x_3 and x_4, then you have not demonstrated that A = B. You have merely demonstrated that either you do not know the rules for assignment of values to variables, or you do not acknowledge the rules. >This can be easily extended to inŜnite sets to dervive E=N, N=NxN, >N=Q, etc. Garbage. Your so-called method of proof does nothing more than demonstrate bijections between E and N, between N and NxN, between N and Q, etc. E does not equal N. N does not equal NxN. N does not equal Q. >HOWEVER the point of this post is not to debunk your work. Which might be a good thing, considering the number of false statements that you have made in your posting. >I often >wondered if you removed renaming morphisms from the study of >inŜntely sets, They are not renaming morphisms. They are FUNCTIONS mapping from one set to another. Where did this idea of renaming morphism come from? You seem to have missed the entire point of what is happening. A function f from A to B is a subset of the Cartesian product AxB (so the elements of f are ordered pairs) such that (1) FA x in A TE y in B ((x,y) in f), (2) FA x in A FA y in B FA z in B ([(x,y) in f and (x,z) in f] => y = z). These two statements can be summarised as the single statement that FA x in A TE y in B FA z in B ((x,z) in f <=> z = y). A bijection f between A and B is a subset of AxB such that (1) FA x in A TE y in B ((x,y) in f), (2) FA x in A FA y in B FA z in B ([(x,y) in f and (x,z) in f] => y = z), (3) FA y in B TE x in A ((x,y) in f), (4) FA x in A FA y in A FA z in B ([(x,z) in f and (y,z) in f] => x = y). These four statements can be summarised as the two statements that FA x in A TE y in B FA z in B ((x,z) in f <=> z = y), FA z in B TE x in A FA y in A ((y,z) in f <=> y = x). In other words, a bijection is not a renaming of the elements of A to become elements of B, but a subset of AxB satisfying the above conditions, and an element x of A corresponds to an element y of B under the bijection f iff (x,y) is an element of f iff the unique element z of B such that (x,z) is an element of f is z = y, and iff the unique element z of A such that (z,y) is an element of f is z = x. >could you derive any useful results. Considering that the lead-up to this question is completely unsupported in fact, the question is meaningless, and can be safely ignored. >Then you could >say something like 3 is not in E and 3 is in N, You can say anyway that 3 is an element of N and 3 is not an element of E. The existence of a bijection between N and E does not force 3 to an element of E. In fact, for any two nonempty sets A and B, the existence of a bijection between A and B has NO BEARING ON THE ELEMENTS OF EITHER. >so |E-{3}| = E No. E-{3} = E, and |E-{3}| = |E|, but |E-{3}| does NOT equal E. >but >|N-{3}| <> N. This is not true (where Œ<>ı denotes not equal to - recall that the meaning of not equal to for Œ<>ı was inspired by an ordered set in which the order was linear - the ordering of cardinals need not be linear, except if you invoke the Axiom of Choice). The reason why it is not true is that the cardinality of N-{3} (i.e. aleph_0) is typically represented by the set N. It is true that N-{3} is not equal to N. It is also true that |N-{3}| = |N|. >In this context |N| = 2 * |E| is true. |N| = 2 * |E| is ALWAYS true, as I have stated SEVERAL times. It is also true that |N| = |E|, which I have also stated several times. >My questions are if you expand on this new deŜnition of card What new deŜnition of cardinality? I havenıt seen a new deŜnition of cardinality. Rest snipped until you can express yourself clearly enough to make yourself understood. David ----- === Subject: Re: Poll: Is this a sentence? > Can I get some opinions on this? >> alas, since >> all of the parties are covering-up Goreıs deal >> with the Supremes of March 27, 2000, that deŜnitively >> caused him to lose Arkansas at the minimum -- as opposed >> to the projected loss of Clinton didnıt gladhand >> in his own state knuckle-head voters -- and >> its contemporeaneous 527 Cmte. Ŝnancial control, >> nevermind. After scooping out all those subordinate clauses from the middle, it reads alas, nevermind, which Iıd say is a sentence of sorts (if a space is added after never). But whatıs intended to be conveyed is obviously all in the clauses and rest is simply a rhetorical ŝourish. ------------------------------------------------------------- -------------- John R Ramsden (jr@adslate.com) ------------------------------------------------------------- -------------- Eternity is a long time, especially towards the end. Woody Allen