mm-244 === Subject: A new approach for the definition of a NUMBER It can be found in: http://www.geocities.com/complementarytheory/CATpage.html ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Partial difference equation, primes > You can also move to an explicit representation, for instance, > > dS(N,2) = N/2 - 1, with even N, > > dS(N,3) = ßoor((N-4)/6), if N is even, and N>2, and > > dS(N,5) = ßoor((N-16)/10) - ßoor((N-16)/30), N even, and N>6, while > > dS(N,7) = ßoor((N-8)/14) - ßoor( (N-22)/42) - ßoor((N-106)/70) + > ßoor( (N-106)/210) - 2, N even, N>36, > As you mention these, Legendre's phi functions phi (N, 1), phi (N, 2) Christian Bau is a very bad liar because I just challenge him to back up his claims with mathematics! I've noticed that he always dodges, ßubs, or runs away. Here I challenge him to produce ANYTHING that will give a count of prime numbers with lesser computational complexity than my dS functions. Enough ßuff, produce. For instance, to count prime up to 120, you just use p(N) = N - dS(N,2) - dS(N,3) - dS(N,5) - dS(N,7) - 1, even N<121, for instance, pi(100) = 100 - 49 - 16 - 6 - 3 - 1 = 25 and the formulas are shown in this post. All Christian Bau has to do to answer this challenge is give his own formulas! > etc. with constant second argument don't have the problem of your dS > functions that they don't work for small N. That's because Legendre was > just that tiny bit more clever than you and counted the integers not > divisible by the first k primes instead of counting the primes, plus the > composite numbers not divisible by the first k primes as you do. Bull. I've called you. Now deliver or kiss your ass goodbye. Oh wait, sci.math readers are cows. Ok, so you'll probably bull some more and they'll eat your like it's candy. But it doesn't mean you're right. IT just means sci.math readers are stupid. James Harris === Subject: Re: Partial difference equation, primes > > You can also move to an explicit representation, for instance, > > dS(N,2) = N/2 - 1, with even N, > > dS(N,3) = ßoor((N-4)/6), if N is even, and N>2, and > > dS(N,5) = ßoor((N-16)/10) - ßoor((N-16)/30), N even, and N>6, while > > dS(N,7) = ßoor((N-8)/14) - ßoor( (N-22)/42) - ßoor((N-106)/70) + > ßoor( (N-106)/210) - 2, N even, N>36, > > As you mention these, Legendre's phi functions phi (N, 1), phi (N, 2) > Christian Bau is a very bad liar because I just challenge him to back > up his claims with mathematics! > I've noticed that he always dodges, ßubs, or runs away. > Here I challenge him to produce ANYTHING that will give a count of > prime numbers with lesser computational complexity than my dS > functions. > Enough ßuff, produce. You must be the most stupid idiot living on earth. There are two possible definitions of computational complexity: The usual one, used in complexity theory, is the number of operations required to find a result. The other one, used mostly by amateurs, is the complexity of the formula. I have written and published on my webpage for everyone to see and to test an implementation of the Lagarias/Miller/Odlyzko method, which runs in O (N^(2/3) * ln (N)) to calculate pi (N). This method has substantially less computational complexity according to the usual definition than anything you could ever think of. Legendre's formula is less complex than yours according to the second, less common definition. Not much less complex, but less complex. However, the straightforward definition of prime numbers gives an even less complex definition: Let f (x) = product (x mod i) for 2 <= i < x Let g (x) = 0 if f (x) = 0 and g (x) = 1 otherwise. Then pi (n) = sum (f (x)) for 2 <= x <= n. === Subject: Re: Partial difference equation, primes oops; I'm sure that Harris can find your little mistake, if even I can figure it out. > However, the straightforward definition of prime numbers gives an even > less complex definition: > Let f (x) = product (x mod i) for 2 <= i < x > Let g (x) = 0 if f (x) = 0 and g (x) = 1 otherwise. > Then pi (n) = sum (f (x)) for 2 <= x <= n. === Subject: Re: Partial difference equation, primes > oops; > I'm sure that Harris can find your little mistake, if > even I can figure it out. I wouldn't bet on it! Many of his replies to posts remind me of an improved Eliza program: He parses the structure of a post, just enough to delete things and insert his rantings in suitable places, but without making any attempt to understand the actual contents. For a computer program, this would actually be quite clever. > However, the straightforward definition of prime numbers gives an even > less complex definition: > > Let f (x) = product (x mod i) for 2 <= i < x > Let g (x) = 0 if f (x) = 0 and g (x) = 1 otherwise. > Then pi (n) = sum (f (x)) for 2 <= x <= n. === Subject: Re: Partial difference equation, primes > James Harris a .8ecrit : >[...] > My prime counting function uses a partial difference equation. > What do you call a partial difference equation? dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], S(x,1) = 0, p(x, y) = ßoor(x) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y). Here dS(x,y) is a partial difference equation. And p(x,sqrt(x)) gives the count of prime numbers up to and including x. > That's only with the pure math implementation. > !? Shown above. Posters continually use the world algorithm in a derisive manner, when actually the prime counting function is just a formula. Algorithms can be *derived* from it, but it's no more an algorithm than e = mc^2 though some nutcase *could* call that an algorithm, if they were trying to argue that it was not important. But it's a formula, not an algorithm. Algorithms are based off of formulas, not the other way around. > You can also move to an explicit representation, for instance, > > dS(N,2) = N/2 - 1, with even N, > > dS(N,3) = ßoor((N-4)/6), if N is even, and N>2, and > > dS(N,5) = ßoor((N-16)/10) - ßoor((N-16)/30), N even, and N>6, while > > dS(N,7) = ßoor((N-8)/14) - ßoor( (N-22)/42) - ßoor((N-106)/70) + > ßoor( (N-106)/210) - 2, N even, N>36, > > and now, what gets put on the stack now? > Huh? Isn't dS(N,2), dS(N,3), dS(N,5), dS(N,7), ... a list? And why > do you index it with prime values (which, btw, makes useless your > (p(y,sqrt(y)) - p(y-1,sqrt(y-1))) whereas you claim you do not need > a list of primes? The compressed explicit prime counting function exists as I've shown. Notice it too is a formula and not an algorithm. It just so happens that's what it looks like when you have it take into account that N is even and 2 is prime. The math is rigid. What I've given are the least computationally complex ways to do the calculations shown, which makes them technology beyond what mathematicians had before my work. And again it's proof that my research IS in fact new and cutting edge. If not, then I challenge you to give formulas that have less computational complexity, or even just show that mathematicians had these formulas before me. The reality is that I'm far ahead of mathematicians at every level, when it comes to counting prime numbers: 1. My prime counting function derivation is just neat. 2. My prime counting function itself is beautiful and compact. 3. I can outline the full theory that determines computational complexity for fast prime counting. I win on all counts. Mathematicians win on obstinacy and sheer refusal to accept reality that they don't like. They're weak. > It's a social thing. I like being the revolutionary as I can sit back > and look at you and wonder how sad it would be if I were such a sheep > and so easily lied to and so easily manipulated by so few people using > such dumb arguments. > > But I'm not a sheep...I'm a discoverer. > I agree on the fact you are not a sheep. I don't believe that a sheep > can be schizophrenic. Yeah, right, toss insults when I win on all the FACTS. The reality is that I can prove each and every one of my points to an extraordinary degree. In fact, my proofs for my work are far more detailed than mathematicians can ever usually manage for their own works!!! As a person trained by physicists I have a deep appreciation for what it means to challenge versus follow. Mathematicians teach and are taught to follow. Physicists are taught to challenge and discover. James Harris === Subject: Re: Partial difference equation, primes James Harris a .8ecrit : > James Harris a .8ecrit : >[...] > My prime counting function uses a partial difference equation. > What do you call a partial difference equation? > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], > S(x,1) = 0, p(x, y) = ßoor(x) - S(x, y) - 1, and S(x,y) is the sum of > dS from dS(x,2) to dS(x,y). > Here dS(x,y) is a partial difference equation. And p(x,sqrt(x)) gives > the count of prime numbers up to and including x. Ok, so I still don't know what you call a partial difference equation. But don't worry, it has no importance. Do you understand that the fact that your formula contains or not a PDE has absolutely no impact on the way we could use it to compute pi(x)? Do you understand that with many formulas containing something like F(x,y), one could claim that F(x,y) is a PDE? Of course, one could do it but what would it change? > That's only with the pure math implementation. > !? > Shown above. What I meant was that Ôpure math implementation' is a mere nonsense. > Posters continually use the world algorithm in a derisive manner, > when actually the prime counting function is just a formula. Ok. So, if a program based on your formula is slow it's not inherent in your formula but it is due to computer scientists who are quite unable to efficiently program it :-) > Algorithms can be *derived* from it, but it's no more an algorithm > than > e = mc^2 > though some nutcase *could* call that an algorithm, if they were > trying to argue that it was not important. > But it's a formula, not an algorithm. > Algorithms are based off of formulas, not the other way around. > You can also move to an explicit representation, for instance, > > dS(N,2) = N/2 - 1, with even N, > > dS(N,3) = ßoor((N-4)/6), if N is even, and N>2, and > > dS(N,5) = ßoor((N-16)/10) - ßoor((N-16)/30), N even, and N>6, while > > dS(N,7) = ßoor((N-8)/14) - ßoor( (N-22)/42) - ßoor((N-106)/70) + > ßoor( (N-106)/210) - 2, N even, N>36, > > and now, what gets put on the stack now? > Huh? Isn't dS(N,2), dS(N,3), dS(N,5), dS(N,7), ... a list? And why > do you index it with prime values (which, btw, makes useless your > (p(y,sqrt(y)) - p(y-1,sqrt(y-1))) whereas you claim you do not need > a list of primes? > The compressed explicit prime counting function exists as I've shown. Which one makes use of an IMPLICIT list of values. > Notice it too is a formula and not an algorithm. > It just so happens that's what it looks like when you have it take > into account that N is even and 2 is prime. > The math is rigid. Yes and that's your main problem. That's precisely because the math is rigid that you're almost always wrong. If making math was singing, you would sing out of tune. > What I've given are the least computationally complex ways to do the > calculations shown, which makes them technology beyond what > mathematicians had before my work. Ways? Technology? You just said you gave a formula, just a formula! Is a formula a way? Is a formula a technology? Or did you give a formula, a way and a technology but NOT an algorithm? Frankly, are you not a little tired to continuously bull? > And again it's proof that my research IS in fact new and cutting edge. No, it is not new. It might be new for you but it is not for others. > If not, then I challenge you to give formulas that have less > computational complexity, What does mean having less computational complexity? > or even just show that mathematicians had > these formulas before me. Legendre. > The reality is that I'm far ahead of mathematicians at every level, > when it comes to counting prime numbers: You already claimed the same about FLT, Goldbach Conjecture and integer factorization. And yet, I am not aware of all your Ôresearch'. > 1. My prime counting function derivation is just neat. Subjective point of view. > 2. My prime counting function itself is beautiful and compact. Subjective point of view. > 3. I can outline the full theory that determines computational > complexity for fast prime counting. Wrong. A program using your Legendre variant CANNOT be fast. > I win on all counts. In your reality, maybe, but not in mine. > Mathematicians win on obstinacy and sheer refusal to accept reality > that they don't like. > They're weak. Whereas you're strong. Do you really believe that you could explain to M. Schumacher how he should drive? Well, considering your math level, that's exactly what you are doing with mathematicians when you claim that you can teach them something. > It's a social thing. I like being the revolutionary as I can sit back > and look at you and wonder how sad it would be if I were such a sheep > and so easily lied to and so easily manipulated by so few people using > such dumb arguments. > > But I'm not a sheep...I'm a discoverer. > I agree on the fact you are not a sheep. I don't believe that a sheep > can be schizophrenic. > Yeah, right, toss insults when I win on all the FACTS. Who did insult who? it would be if I were such a sheep', you clearly insulted me. believe that a sheep can be schizophrenic.', it was not an insult but a diagnosis. > The reality is that I can prove each and every one of my points to an > extraordinary degree. Are you able to write more than 3 lines without starting to rave? > In fact, my proofs for my work are far more detailed than > mathematicians can ever usually manage for their own works!!! > As a person trained by physicists I have a deep appreciation for what > it means to challenge versus follow. > Mathematicians teach and are taught to follow. > Physicists are taught to challenge and discover. Even if it were true what would it change? Though a physicist can be schizophrenic, you are not a physicist. -- mm http://www.ellipsa.net/ mm@ellipsa.no.sp.am.net ( suppress no.sp.am. ) === Subject: Re: Partial difference equation, primes > James Harris a .8ecrit : > > James Harris a .8ecrit : >[...] > My prime counting function uses a partial difference equation. > > What do you call a partial difference equation? > > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], > > S(x,1) = 0, p(x, y) = ßoor(x) - S(x, y) - 1, and S(x,y) is the sum of > dS from dS(x,2) to dS(x,y). > > Here dS(x,y) is a partial difference equation. And p(x,sqrt(x)) gives > the count of prime numbers up to and including x. > > Ok, so I still don't know what you call a partial difference > equation. But don't worry, it has no importance. I handled that topic with a separate thread. But it's not complicated--no matter what a sci.math'er might try to insinuate--as a partial difference equation is the discrete analog to a partial differential equation. There is no other known in recorded history used to count prime numbers besides my dS(x,y) and yes, I'm talking about all of human history here. To me that's a simple enough claim that it should either be refutable by you sci.math'ers, or if you're sane, you'll quit trying to insinuate that it's not a big deal. I fear you're not exactly what most people would call sane. > Do you understand that the fact that your formula contains or not > a PDE has absolutely no impact on the way we could use it to compute > pi(x)? Do you understand that with many formulas containing > something like F(x,y), one could claim that F(x,y) is a PDE? Of course, > one could do it but what would it change? You're lying because you're insinuating that mathematicians have in fact used a partial difference equation to count prime numbers when they have not. It's not complicated here. You may think you can parse the language to fool everyone on sci.physics or that the other sci.math'ers will just go along with you as they have for over two years now, but it doesn't change the facts. I say that no one in recorded human history has used a partial difference equation to count prime numbers. That is a fairly straightforward claim, and if it's not true you can just give some other partial difference equation, but instead you try to spin the facts, like what in the hell is F(x,y)? I've seen how you sci.math'ers operate for over two years now, and you have a contempt for the truth, and a demonstrated contempt for mathematics. > > That's only with the pure math implementation. > > !? > > > Shown above. > What I meant was that Ôpure math implementation' is a mere nonsense. Yet people can see that it is not nonsense most dramatically by looking at it, and I've put it on my blog: > Posters continually use the world algorithm in a derisive manner, > when actually the prime counting function is just a formula. > Ok. So, if a program based on your formula is slow it's not inherent > in your formula but it is due to computer scientists who are quite > unable to efficiently program it :-) No. > Algorithms can be *derived* from it, but it's no more an algorithm > than > > e = mc^2 > > though some nutcase *could* call that an algorithm, if they were > trying to argue that it was not important. > > But it's a formula, not an algorithm. > > Algorithms are based off of formulas, not the other way around. > > You can also move to an explicit representation, for instance, > > dS(N,2) = N/2 - 1, with even N, > > dS(N,3) = ßoor((N-4)/6), if N is even, and N>2, and > > dS(N,5) = ßoor((N-16)/10) - ßoor((N-16)/30), N even, and N>6, while > > dS(N,7) = ßoor((N-8)/14) - ßoor( (N-22)/42) - ßoor((N-106)/70) + > ßoor( (N-106)/210) - 2, N even, N>36, > > and now, what gets put on the stack now? > > Huh? Isn't dS(N,2), dS(N,3), dS(N,5), dS(N,7), ... a list? And why > do you index it with prime values (which, btw, makes useless your > (p(y,sqrt(y)) - p(y-1,sqrt(y-1))) whereas you claim you do not need > a list of primes? > > The compressed explicit prime counting function exists as I've shown. > Which one makes use of an IMPLICIT list of values. That's what it looks like. The math is simply rigid here, no matter how much you might like to make more out of it. It just is. > Notice it too is a formula and not an algorithm. > > It just so happens that's what it looks like when you have it take > into account that N is even and 2 is prime. > > The math is rigid. > Yes and that's your main problem. That's precisely because the math is > rigid that you're almost always wrong. If making math was singing, you > would sing out of tune. Then give a single wrong point I've made. Give ANYTHING mathematical which will stand up to scrutiny. Do more than insinuate, like talk straight for once. > What I've given are the least computationally complex ways to do the > calculations shown, which makes them technology beyond what > mathematicians had before my work. > Ways? Technology? You just said you gave a formula, just a formula! > Is a formula a way? Is a formula a technology? Or did you give a > formula, a way and a technology but NOT an algorithm? > Frankly, are you not a little tired to continuously bull? Technology refers as a word to state of the art. I can not only give state of the art in terms of explicit representations of the prime counting function, I can explain why it's state of the art. > And again it's proof that my research IS in fact new and cutting edge. > No, it is not new. It might be new for you but it is not for others. I've shown facts, and the facts support my position and refute yours. > If not, then I challenge you to give formulas that have less > computational complexity, > What does mean having less computational complexity? I mean formulas that have less computational complexity. > or even just show that mathematicians had > these formulas before me. > Legendre. Claiming something that has repeatedly been shown to be false does not make it true. I can kind of understand the hurt you feel, or maybe you feel betrayed because you think of mathematics as a social system which supports you, so it cannot in your mind support someone like me. But your emotions do not change the mathematical truth. > The reality is that I'm far ahead of mathematicians at every level, > when it comes to counting prime numbers: > You already claimed the same about FLT, Goldbach Conjecture and > integer factorization. And yet, I am not aware of all your Ôresearch'. Sigh. So now you're trying to shift the subject to other topics with more claims that I guess you expect me to try to defend or refute. I'm not interested in changing the subject at this time. Straight answers. Have you ever heard of straight answers? > 1. My prime counting function derivation is just neat. > Subjective point of view. Well, I like it. I think that anyone who compares it to what mathematicians have will agree that it's just neat, if they're objective. > 2. My prime counting function itself is beautiful and compact. > Subjective point of view. where I put on a single-line an extraordinary formula that deserves a place in the mathematical literature, no matter how many sci.math'ers lie about that truth. > 3. I can outline the full theory that determines computational > complexity for fast prime counting. > Wrong. A program using your Legendre variant CANNOT be fast. I've already written a fast algorithm which I call PrimeCountH.java, which you can find on sci.math.symbolic, see It's faster than Mathematica's prime counting function over I think a decent range, though at the top of Mathematica's range it is faster. > I win on all counts. > In your reality, maybe, but not in mine. I win in reality, while you are in your own little world. In this one I make claims that stand because no one can refute them with facts as the facts are on my side. I demonstrate while people like you make claims that you can't support, while you continually shift when your claims are challenged. You lie repeatedly, as if lying were all that really mattered. > Mathematicians win on obstinacy and sheer refusal to accept reality > that they don't like. > > They're weak. > Whereas you're strong. Do you really believe that you could explain > to M. Schumacher how he should drive? Well, considering your math > level, that's exactly what you are doing with mathematicians when > you claim that you can teach them something. And there I think you are the most honest yet. You simply believe in *people* not facts, and you don't like me, so you think your feelings matter. But you see, mathematics never has been nore will it ever be about your feelings. To you a name like Schumacher is what's important. If I had a name you respected you'd probably fight all comers who dared to deny the importance of my work. But it's not the name that's important. It's the mathematics. James Harris === Subject: Re: Partial difference equation, primes > Posters continually use the world algorithm in a derisive manner, > when actually the prime counting function is just a formula. There is nothing derisive about the term algorithm, the derision you sense arises from the fact that you do not understand the term. > The compressed explicit prime counting function exists as I've shown. > Notice it too is a formula and not an algorithm. Here's a definition of algorithm: SYLLABICATION: algorithm PRONUNCIATION: AUDIO: lg-rthm KEY NOUN: A step-by-step problem-solving procedure, especially an established, recursive computational procedure for solving a problem in a finite number of steps. ETYMOLOGY: Variant (probably inßuenced by arithmetic) of algorism. OTHER FORMS: algorithmic (-rthmk) ADJECTIVE algorithmically ADVERB How does your implementation of a prime counting function differ from an algorithm? > The reality is that I can prove each and every one of my points to an > extraordinary degree. There are no Ôdegrees of proof', extraordinary or otherwise. > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Partial difference equation, primes > Posters continually use the world algorithm in a derisive manner, > when actually the prime counting function is just a formula. > There is nothing derisive about the term algorithm, the derision you sense arises from the fact that you do > not understand the term. Nope. Posters say I just have an algorithm for counting primes like any other and that the best way to evaluate an algorithm is by speed. Then they point out that the direct *algorithmic* implementation of my prime counting function is slow, and they say it's not worth discussing. It's all part of a rather steady campaign to try and dismiss my work. So yes, posters use the word algorithm derisively in a direct attack on my prime counting research to try and support arguments for dismissing it. > The compressed explicit prime counting function exists as I've shown. > Notice it too is a formula and not an algorithm. > Here's a definition of algorithm: > SYLLABICATION: > algorithm > PRONUNCIATION: > AUDIO: lg-rthm KEY > NOUN: > A step-by-step problem-solving procedure, > especially an established, recursive > computational procedure for solving a problem > in a finite number of steps. > ETYMOLOGY: > Variant (probably inßuenced by arithmetic) of > algorism. > OTHER FORMS: > algorithmic (-rthmk) ?ADJECTIVE > algorithmically ?ADVERB > How does your implementation of a prime counting function differ from an algorithm? By that definition ANY finite summation is an algorithm. That is, whenever you see the summation sign in mathematics for a finite sum then by the position of this poster it's an algorithm. To see what I mean rather dramatically, look at my prime counting function shown as a summation sign on my blog: Notice this sci.math'er now defining ANY mathematics using a finite summation to be an algorithm. And, of course, not really lurking will be a position that my prime counting function is JUST an algorithm to be dismissed if there are faster algorithms available! It's stupid as a position and notice that now finite sums are all algorithms, but haven't you noticed by now that sci.math'ers routinely take weird leaps? That is, they're still not rational, and I can show their positions to be extreme, like here now suddenly because of my work, any finite sum is an algorithm which should be dismissed if faster algorithms are around. > The reality is that I can prove each and every one of my points to an > extraordinary degree. > There are no Ôdegrees of proof', extraordinary or otherwise. Actually mathematicians *routinely* leave out steps in their mathematical works, which is something that should be commonly known. The idea is not to put in things that are obvious or considered trivial, maybe even to help on the length of math papers. However, I have been forced to step through each and every point in my mathematical research in extraordinary detail and to a degree that few mathematicians ever have to handle with their own research!!! Now then, pay attention and notice that this sci.math'er will not make a rational response. They never bother!!! They just keep on going making the same claims, year after year. They're freaking energizer bunnies or something who probably figure that as long as it works, why quit. They just keep coming back with irrational posts that defy common-sense and basic mathematical usage as if the last thing they wish to be bothered with is the truth. James Harris === Subject: Re: Partial difference equation, primes > Posters continually use the world algorithm in a derisive manner, > when actually the prime counting function is just a formula. > There is nothing derisive about the term algorithm, the derision you sense arises from the fact that you do > not understand the term. > Nope. Posters say I just have an algorithm for counting primes like > any other and that the best way to evaluate an algorithm is by speed. > Then they point out that the direct *algorithmic* implementation of my > prime counting function is slow, and they say it's not worth > discussing. That's a complaint about *your* method -- the derisive keyword is *slow* not *algorithm*. > It's all part of a rather steady campaign to try and dismiss my work. > So yes, posters use the word algorithm derisively in a direct attack > on my prime counting research to try and support arguments for > dismissing it. In case you didn't know it, division and square roots are implemented as algorithms on computers. You can express the relations for a quotient formally with an equation, such as: q = a / b but the computation of Ôq' is done algorithmically. Don't believe it? Check it out. Many of the division algorithms have names, and chip mfrs sometimes publish the details. > The compressed explicit prime counting function exists as I've shown. > > Notice it too is a formula and not an algorithm. > Here's a definition of algorithm: > SYLLABICATION: > algorithm > PRONUNCIATION: > AUDIO: lg-rthm KEY > NOUN: > A step-by-step problem-solving procedure, > especially an established, recursive > computational procedure for solving a problem > in a finite number of steps. > ETYMOLOGY: > Variant (probably inßuenced by arithmetic) of > algorism. > OTHER FORMS: > algorithmic (-rthmk) ?ADJECTIVE > algorithmically ?ADVERB > How does your implementation of a prime counting function differ from an algorithm? > By that definition ANY finite summation is an algorithm. Well, that's the definition. So what? You have taken the time to spell out the step-by-step process used in your method. That's what an algorithm is. > That is, whenever you see the summation sign in mathematics for a > finite sum then by the position of this poster it's an algorithm. You're confused. The algorithm is the method or step-by-step process used to determine the result. The summation sign is a mathematical symbol -- like the division sign indicating a quotient. Division can be implemented by look up tables or by various clever algorithms. The algorithmic implementations are often robust and fast. You can't tell anything about an implementation by looking at the divide symbol Your prime counting method is an algorithm. That is not a derisive label, it is descriptive. You, on the other hand, are an idiot -- that's both derisive AND descriptive. > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Partial difference equation, primes <4124D7CE.BA8ED695@ellipsa.no.sp.am.net> <41257489.17E05AC2@ellipsa.no.sp.am.net> Discussion, linux) >> James Harris a .8ecrit : >>[...] >> My prime counting function uses a partial difference equation. >> What do you call a partial difference equation? > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], > S(x,1) = 0, p(x, y) = ßoor(x) - S(x, y) - 1, and S(x,y) is the sum of > dS from dS(x,2) to dS(x,y). > Here dS(x,y) is a partial difference equation. And p(x,sqrt(x)) gives > the count of prime numbers up to and including x. He's asking for a definition of partial difference equation. We all know that you call your equation a partial difference equation, but he wants to know what the words mean to you. > The reality is that I'm far ahead of mathematicians at every level, > when it comes to counting prime numbers: | Now I've never claimed to be a skilled mathematician. So I'm NOT | skilled as an expert in mathematics and have never claimed otherwise. `---- -- Jesse Hughes Yes, I'm one of those arrogant people who tries to be quotable. There is actually at least one person who quotes me often. -- James Harris === Subject: Re: (Yet Another Question) Help with AP Calculus Summer Assignment Problems much for all of the help. Nicky >>The graph of y=Pa^x contains the points (0,4) and (3,108) Find P and >>a. >> >>I only have a couple of hours so any quick help would be greatly >>appreciated. >> >>Nicky >By the time you see this, the time may be too late; be sure to review the >sections in your textbook on exponential and logarithmic functions. >Exponential equation which contains points (0, 4) and (3, 108); >Use the logarithms of both sides: >log y = log P + x log a (the base is not yet specified) >This is, when examined carefully, an equation in which the independant variable >is log a, and the dependant variable is log y. The equation fits the form, >f(x)=mx+b, a linear equation. Let me rewrite the logarithm equation a bit: >log y = (log a)*x + log P (Now, I'm using the Ô*' for mulitplication just to >be Ôclear'). >The vertical intercept is very clearly indicated with the first given point: >When x=0, (log y) = log 4 = log P >(remember and recheck that Ôb' corresponds to log P) >Find antilog of both sides; P=4 >Please someone recheck for any conceptual or arithmetic errors I might have >made. >G C > Why make it so difficult? If y = Pa^x contains (0,4) then 4 = Pa^0 = P > so you have y = 4a^x. Now plug in (3,108) giving 108 = 4a^3 so > a^3 = 27 and a = cuberoot(27) = 3. > --Lynn === Subject: Re: (Yet Another Question) Help with AP Calculus Summer Assignment Problems My first solution was a long way around to a solution, certainly not the quickest. If you go through a pre-engineering or physical science track, you will deal with exponential and logarithmic equations in the form of linear equations probably in the laboratory sections. For ordinary learning purposes, you usually don't need to treat your posted equation the way that I did. As you study, some of the Mathematics becomes harder to learn, but applying the Mathematics suddenly becomes very easy. Then, too, do not be afraid to review; this makes the material easier to learn. G C >much for all of the help. >Nicky === Subject: Re: Status of Waring-problem Hi David - Am 21.08.04 01:06 schrieb David Moews: > I believe that what is known is the following: > Let [x] be the integer part of x>0 and {x} be the fractional part of x>0. If > 2^k {(3/2)^k} + [(3/2)^k] <= 2^k, (*) > then g(k) = 2^k + [(3/2)^k] - 2. On the other hand, if (*) does not > hold, there is another, slightly more complicated formula for g(k). > (*) is (a) never known not to hold, (b) known to hold for all > k <= 4*10^8, and (c) known to hold for all but finitely many k. > See, e.g., ``Waring's problem: a survey'', Vaughan and Wooley, in > NUMBER THEORY FOR THE MILLENNIUM III. Hmm, that doesn't explain the problem. Weisstein in Mathworld states an equation, which, after some transformations, comes out to be the above inequality, and says, that the proof of this inequality solves the waring-problem completely. Now it is reported, that the last outstanding result for G(4) was found in 1986. Didn't that fix the above conjecture? Or did this only fix, that one of the two formula must hold? On the other hand, I have a remark of Ray Steiner, who 1978 proved the non-existence of the 1-cycle in the 3x+1-problem. I found, that a valid inequality to disprove the 1-cycle can exactly be stated on that Steiner-theorem, indicates, that the same formula was used. So it seems, that this work proved implicitely (*), possibly unaware of the connection between the two problems. Gottfried Helms P.s. is something reviewing Vaughan & Wooley online? I think, it'll R.P. Steiner(1978), A theorem on the syracuse problem, Proceedings of the 7th Manitoba Conference on Numerical Mathematics, 1977, 553-559. J.Simons/B.de Weger Theoretical and computational bounds for m-cycles of the 3n + 1 problem John Simons (Groningen) 1 Benne de Weger (Eindhoven) 2 http://www.win.tue.nl/~bdeweger/3n+1_v1.0.pdf === Subject: Re: Status of Waring-problem Originator: dmoews@ccrwest.org (David Moews) |Hi David - | |Am 21.08.04 01:06 schrieb David Moews: |> I believe that what is known is the following: |> Let [x] be the integer part of x>0 and {x} be the fractional part of x>0. If |> 2^k {(3/2)^k} + [(3/2)^k] <= 2^k, (*) | > then g(k) = 2^k + [(3/2)^k] - 2. On the other hand, if (*) does not |> hold, there is another, slightly more complicated formula for g(k). |> (*) is (a) never known not to hold, (b) known to hold for all |> k <= 4*10^8, and (c) known to hold for all but finitely many k. |> See, e.g., ``Waring's problem: a survey'', Vaughan and Wooley, in |> NUMBER THEORY FOR THE MILLENNIUM III. | |Hmm, that doesn't explain the problem. Weisstein in Mathworld states |an equation, which, after some transformations, comes out to be |the above inequality, and says, that the proof of this inequality |solves the waring-problem completely. If (*) were proved to always hold, then g(k) = 2^k + [(3/2)^k] - 2 would hold in all cases. However, since formulae are already known that suffice to compute g(k) simply in all cases, even if (*) does not always hold, you might prefer to say that the problem of determining g(k) is essentially already solved. |Now it is reported, that the last outstanding result for G(4) was |found in 1986. Didn't that fix the above conjecture? Or did this only |fix, that one of the two formula must hold? In 1986 it was proved that g(4) = 19. This proves the formula g(k) = 2^k + [(3/2)^k] - 2 for k=4, which was the last step needed to show that g(k) is always given by one of the above mentioned formulae. It has nothing to do with the truth or falsity of (*). |On the other hand, I have a remark of Ray Steiner, who 1978 proved |the non-existence of the 1-cycle in the 3x+1-problem. I found, that |a valid inequality to disprove the 1-cycle can exactly be stated |on that Steiner-theorem, indicates, that the same formula was used. | |So it seems, that this work proved implicitely (*), possibly unaware |of the connection between the two problems. I don't think so. Steiner's paper implies that if a 1-cycle exists in the Collatz problem, then there are powers of 2 and 3 whose difference is exponentially small relative to their common value, i.e., |2^a - 3^b| < 3^b (1-eps)^b. (**) This implies that |a log 2 - b log 3| is also exponentially small in b, and there are known bounds on b above which this cannot happen. The Simons/de Weger paper also proves that a simple cycle in the Collatz problem implies that |a log 2 - b log 3| is exponentially small in b. On the other hand, (*) essentially says that (3/2)^k is not exponentially close to an integer; it would be implied by the inequality |(3/2)^k - n| > (3/4)^k for all n. Because of the presence of the n outside the exponential, this is rather different from (**). |P.s. is something reviewing Vaughan & Wooley online? I think, it'll This paper mostly deals with the problem of computing G(k) rather than g(k). I think it is not available online. Ribenboim's THE NEW BOOK OF PRIME NUMBER RECORDS also discusses Waring's problem. -- David Moews dmoews@xraysgi.ims.uconn.edu === Subject: Re: Status of Waring-problem Am 21.08.04 04:33 schrieb David Moews: > |So it seems, that this work proved implicitely (*), possibly unaware > |of the connection between the two problems. > I don't think so. Steiner's paper implies that if a 1-cycle exists in the > Collatz problem, then there are powers of 2 and 3 whose difference is > exponentially small relative to their common value, i.e., > |2^a - 3^b| < 3^b (1-eps)^b. (**) Hmm, I've to get familiar with that criterion, so possibly I can answer my question myself next days. But I'll focus it here again, perhaps it's just a simple argument: In mathworld in the chapter about power-fractions the waring-inequality, that you refer as > > |(3/2)^k - n| > (3/4)^k > for all n. is expressed as frac((3/2)^k) < 1- (3/4)^k [1-MW] which possibly could be a erroneous transformation. Now my condition for the nonexistence of an 1-cycle is practically the same as the mathworld-formula, it is frac((3/2)^k) < 1- (3/4)^k +eps [2-H] where eps would be smaller than 1/3^k, (can be neglected in diophantine problems) and is so essentially the same as [1-MW]. THis says, if [2-H] holds, then a 1-cycle is denied. Now the Steiner-criterion states that an 1-cycle is impossible because of this approximation-argument (**), so how could there be a reason, that (**) does not imply [1-MW] ? (may be I'm just having a knot in my mind currently...) > This paper mostly deals with the problem of computing G(k) rather than g(k). > I think it is not available online. Ribenboim's THE NEW BOOK OF PRIME NUMBER > RECORDS also discusses Waring's problem. Gottfried Helms === Subject: Re: Status of Waring-problem Originator: dmoews@ccrwest.org (David Moews) ||[...] ||P.s. is something reviewing Vaughan & Wooley online? I think, it'll | |This paper mostly deals with the problem of computing G(k) rather than g(k). |I think it is not available online. I thought wrong---the Vaughan and Wooley paper is on line, at . -- David Moews dmoews@xraysgi.ims.uconn.edu === Subject: Re: A new approach for the definition of a NUMBER > It can be found in: > http://www.geocities.com/complementarytheory/CATpage.html What a load of crap! > ---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** > ---------------------------------------------------------- > http://www.usenet.com === Subject: need help with bell curve i want to write a program that will approximate the bell curve.. could someone tell me the best way to do it? or at least a far far more efficient way than what i do at this time which is: i loop 1000 times each loop i ßip a coin 1000 times and see how many times it lands 0 (in other words !(rand()%2) ) that's about it.. i have an array.. and each loop if it gets 0 say.. 490 times.. then array[490] gets increased by 1... at the end i've got this array that if u plot it as the ordinate and the indexes of the array as the absica then you get a very loose very ty very crappy piece of dog approximation of the bell curve.. it's not even close.. it's terrible.. i need something more efficient. === Subject: Re: need help with bell curve >i want to write a program that will approximate the bell curve.. Why would you want to approximate it? Why not just calculate it? See http://mathworld.wolfram.com/NormalDistribution.html === Subject: Re: need help with bell curve > i want to write a program that will approximate the bell curve.. > could someone tell me the best way to do it? or at least a far far > more efficient way than what i do at this time which is: > i loop 1000 times > each loop i ßip a coin 1000 times and see how many times it lands 0 > (in other words !(rand()%2) ) > that's about it.. i have an array.. and each loop if it gets 0 say.. > 490 times.. then array[490] gets increased by 1... at the end i've got > this array that if u plot it as the ordinate and the indexes of the > array as the absica then you get a very loose very ty very crappy > piece of dog approximation of the bell curve.. it's not even > close.. it's terrible.. i need something more efficient. Go to the library, borrow a book on probability and statistics. Read it. === Subject: Re: need help with bell curve > i want to write a program that will approximate the bell curve.. > > could someone tell me the best way to do it? or at least a far far > more efficient way than what i do at this time which is: > > > i loop 1000 times > each loop i ßip a coin 1000 times and see how many times it lands 0 > (in other words !(rand()%2) ) > > that's about it.. i have an array.. and each loop if it gets 0 say.. > 490 times.. then array[490] gets increased by 1... at the end i've got > this array that if u plot it as the ordinate and the indexes of the > array as the absica then you get a very loose very ty very crappy > piece of dog approximation of the bell curve.. it's not even > close.. it's terrible.. i need something more efficient. > Go to the library, borrow a book on probability and statistics. Read it. you coudl have just said: i'm an asshole and don't wish to help at all. but even that would be completely a waste of a post. i know the equation and as you can see by my algorithm.. i know what the bell curve represents. i'm just looking for an algorithm to approximate it better. === Subject: Re: imaginary solution to constraints of a real number system? or just a hoax > why can't we define -3 squared to be -9? > Because if > -3 * -3 = -9 > and > +3 * -3 = -9 > then > -3 = +3 > --Mark which shows merely that negative numbers are just as ridiculous and illogical as imaginary numbers === Subject: Re: imaginary solution to constraints of a real number system? or just a hoax In sci.math, David Bandel >> why can't we define -3 squared to be -9? >> Because if >> -3 * -3 = -9 >> and >> +3 * -3 = -9 >> then >> -3 = +3 >> --Mark > which shows merely that negative numbers are just as ridiculous and > illogical as imaginary numbers Negative numbers have a well-accepted meaning; if one postulates that there exist numbers a and b (in an infinite field) such that a+b= 0 and a != 0, then by necessity b != 0 and one of a and b are positive, the other is negative. Assume that a > 0. Then b < 0 and we write b = -a. b is the arithmetic inverse of a (and a the arithmetic inverse of b, as well). In a field, we can generate an indefinite number of such pairs by multiplying and dividing: ac + bc = (a+b)c = 0c = 0, a/d + b/d = (a+b)/d = 0/d = 0, for d != 0. So now we have 1 and -1. By one process, (1-1) * (1-1) = 0*0. Also, (1-1) * (1-1) = (1-1)*1 + (1-1)*(-1) = 1*1 + (-1)*1 + 1*(-1) + (-1)*(-1). 1*1 = 1, of course. (-1)*1 = 1*(-1) = -1. Since 1-1 = 0, 1-1-1 = 0-1 = -1. -1+(-1)*(-1) = 0, and 1-1=-1+1=0, therefore (-1)*(-1) = +1. In a finite field, of course, things break down slightly. If one for example takes the integers modulo 5, then one has situations such as 2*4 = 1 and 2*3 = 4 = -1 (mod 5). But if you think negative numbers are illogical, this is even more so. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Tricky maths problem >Johnny has a coin, he tells Billy that he will give him 10 every time he spins heads, but take 5 off him every time he spins tails. What are the chances that Billy will have a positive amount after 5 spins? And after 10 spins? After X spins? >>Let N be the number of heads out of x spins. (Do Brits really spin >>coins? Americans toss them). How large ust N be for Billy to have a >>positve amount? What is the distribution of N? > A minor nitpick: > The question, AS ASKED, can not be answered. > Why? > Because it did not state that Johnny has a FAIR coin. > The question can not be answered without knowing the bias, if any. In the absence of any mention of fairness, it is generally assumed that you have a fair coin. How many fair coins are there in the world? How many unfair? Many many more fair than unfair. If the coin was unfair, it would be stated. So his question can be answered; asking about the fairness of the coin is a bit pedantic IMO. alex === Subject: Re: Tricky maths problem ... >> A minor nitpick: >> The question, AS ASKED, can not be answered. >> Why? >> Because it did not state that Johnny has a FAIR coin. >> The question can not be answered without knowing the bias, if any. >In the absence of any mention of fairness, it is generally assumed that >you have a fair coin. Perhaps. >How many fair coins are there in the world? How >many unfair? Many many more fair than unfair. Wow! Do you really, truly, believe that? (And what is the rational basis for that belief, if you do believe it?) >If the coin was unfair, it >would be stated. Not necessarily. >So his question can be answered; asking about the fairness of the coin >is a bit pedantic IMO. You suggest that pedantic is a bad thing to be. Many sensible people, also many people in sci.math, would disagree with you. For real pedantry, try this on: Bob is wrong to say the question can not be answered without knowing the bias, if any; he would have been right to say the complete and proper answer to the question is a parametrized solution, in which the bias appears as an explicit parameter. Lee Rudolph === Subject: Re: Tricky maths problem > For real pedantry, try this on: ... Or this: The question can not be answered because who said we are spinning/tossing coins being exposed to friction/gravitation, and even if we assume these, who said that the coin will actually fall on one of its main sides - moreover, how many sides does the coin have? I mean, you have to draw the line somewhere. When being given an assignment stating A coin is tossed... or A dice is thrown..., I assume an unbiased (idialized) object, as well as I assume an unmanipulated lottery when I read In a lottery... followed by sentences that state nothing else (and in a sense, special) about the state of the case. Along the lines you suggest, I would have preceded an answer to the OP's question with Assuming an unbiased coin... (and not with Assuming an unbiased coin that will produce one of two possible outcomes...), maybe not because it is extremely important to state, but more so to make a somewhat gentler introduction (as opposed to, for example, The binomial distribution blabla...) and (hopefully) causing immediate mental association with the probabilty of 0.5. Markus. === Subject: Re: Average frequency > I have a chart that shows frequency (x) vs. magnitude (y). Lets say there is > a thousand points on this chart, each point having a different value > (magnitude). Finding the average magnitude is just the magnitude total over > the number of points. But how do I find the average frequency? Seeing as > each frequency is going to have a different weight based on it's own > magnitude. > Thomas in this case.. frequency is nothing but an interval.. it's meaningless to talk about the average of it.. what you want to talk about is the average magnitude on an interval of frequency which is different. === Subject: Re: Length of boundary of convex polygons >>Let A and B be two convex plane polygons such that A is a subset of >>B. How can it be proved that the length of the boundary of A cannot be >>larger than the length of the boundary of B? >By boundary do you mean perimeter? >The shortest path which encloses a set of points in a plane is the convex >hull. You mean the perimeter of its convex hull, right? I am probably late to this thread, but isn't the convex hull of a set of points defined as the intersection of all convex sets containing the points? The fact that its perimeter is the shortest path containing the points is a theorem, and your argument below is circular, using the result of the theorem to prove it. --Lynn > Any convex polygon has itself as a convex hull; therefore A's convex >hull is A. If B encloses A and has a shorter perimeter, you have a >contradiction. >--Keith Lewis klewis {at} mitre.org >The above may not (yet) represent the opinions of my employer. === Subject: Re: Length of boundary of convex polygons circular, using the result of the theorem to prove it. > Any convex polygon has itself as a convex hull; therefore A's convex >hull is A. If B encloses A and has a shorter perimeter, you have a >contradiction. >--Keith Lewis This is a strong hint, another one is: You can prove it practically, when You get up and try walking Polygons, let's say in a park. When You're back,start anew, reducing the number of corners by one. And so progress, until You recognize, that it is mathematically an... Have a nice walk Hero === Subject: Re: Happy numbers (SPOILER) > |What are the next two terms in the sequence: > | 1, 31, 1880, 7839, 44488, 7899999999999959999999996, > |7899999999999959999999996, ? , ? , > |[...] > (this sequence is OEIS A055629, > the `beginning of first run of n consecutive happy numbers'; > a happy number is one which is eventually 1 under iteration of > the map x -> sum of the squares of the base 10 digits of x) > Ans. The next four terms appear to be Do you have some reason in mind why there should be no smaller numbers that will work? Edwin Clark === Subject: Re: Happy numbers (SPOILER) Supersedes: Originator: dmoews@ccrwest.org (David Moews) |> |What are the next two terms in the sequence: |> | |> | 1, 31, 1880, 7839, 44488, 7899999999999959999999996, |> |7899999999999959999999996, ? , ? , |> |[...] |> (this sequence is OEIS A055629, |> the `beginning of first run of n consecutive happy numbers'; |> a happy number is one which is eventually 1 under iteration of |> the map x -> sum of the squares of the base 10 digits of x) |> Ans. The next four terms appear to be | |Do you have some reason in mind why there should be no smaller numbers |that will work? | |Edwin Clark Yes. My approach is basically the same as the one outlined by Neuendorffer. First, write f(x) for the sum of the squares of the digits of x, and for x >= 0, let g(x) be minimal such that f(g(x)) = x. Observe that g can be easily computed recursively using g(0) = 0, g(x) = min 10 g(x - d^2) + d. 0 <= d <= 9 d^2 <= x Now given a sequence of length l of consecutive happy numbers, (a, a+1, ..., a+l-1), let a have last digit d, and let there be exactly n 9s preceding this last digit, which are themselves preceded by some digit q != 9. Also, fix some m such that a < 10^m. We can then write a = 10^(n+2) s + 10^(n+1) q + (10^n - 1) 10 + d, where 0 <= s < 10^m, 0 <= q <= 8, 0 <= n < m, 0 <= d <= 9. Write r = f(s), and observe that since s < 10^m, r <= 81 m. We then have the equations f(a) = r + q^2 + 81 n + d^2, | f(a+1) = r + q^2 + 81 n + (d+1)^2, | ..., | f(a+9-d) = r + q^2 + 81 n + 81, | (*) f(a+10-d) = r + (q+1)^2, | f(a+11-d) = r + (q+1)^2 + 1^2, | ... | . Now if a, ..., a+l-1 are all happy, the right-hand sides of (*) must all be happy, so search for r, q, n, and d in the appropriate ranges for which this is true. Setting s := g(r) then yields a candidate value for a. If any such candidate values exist, the beginning of the first run of l consecutive happy numbers must be equal to the minimum such candidate value. This is how I obtained the numbers I posted. -- David Moews dmoews@xraysgi.ims.uconn.edu === Subject: Re: i: imaginary solution to constraints of a real number system? or just a hoax > consider the following: > > the square root of a negative number requires some non-real solution? > why can't we define -3 squared to be -9? because then mathematics > would be simple.. and mathematicians would be out of the job. > No, that would not make sense at all in fact - it would make math make > no sense - leading to something QUITE complicated, but mathematicians > would be out of jobs. They would all quit from frustration, until one > of them suddenly says, Hey, what about using imaginary numbers? > Numbers that.. you know, don't quite exist, but we can model the > equations as if they did? > Even in the above statement, that the square root of -9 is -3, you've > violated the logic that is used to even CREATE negative numbers > because in this case, -3*-9 would be -27, NOT 27. It makes no sense. > None. Not even a little. It's not simple, it's complicated and stupid. it may be complicated. but it's not stupid. it's a better way of doing things. === Subject: Re: i: imaginary solution to constraints of a real number system? or just a hoax > consider the following: > > the square root of a negative number requires some non-real solution? > why can't we define -3 squared to be -9? because then mathematics > would be simple.. and mathematicians would be out of the job. > No, that would not make sense at all in fact - it would make math make > no sense - leading to something QUITE complicated, but mathematicians > would be out of jobs. They would all quit from frustration, until one > of them suddenly says, Hey, what about using imaginary numbers? > Numbers that.. you know, don't quite exist, but we can model the > equations as if they did? > Even in the above statement, that the square root of -9 is -3, you've > violated the logic that is used to even CREATE negative numbers > because in this case, -3*-9 would be -27, NOT 27. It makes no sense. > None. Not even a little. It's not simple, it's complicated and stupid. > it may be complicated. but it's not stupid. it's a better way of doing things. No, it isn't. === Subject: Re: i: imaginary solution to constraints of a real number system? or just a hoax > consider the following: > > the square root of a negative number requires some non-real solution? > why can't we define -3 squared to be -9? because then mathematics > would be simple.. and mathematicians would be out of the job. > > No, that would not make sense at all in fact - it would make math make > no sense - leading to something QUITE complicated, but mathematicians > would be out of jobs. They would all quit from frustration, until one > of them suddenly says, Hey, what about using imaginary numbers? > Numbers that.. you know, don't quite exist, but we can model the > equations as if they did? > Even in the above statement, that the square root of -9 is -3, you've > violated the logic that is used to even CREATE negative numbers > because in this case, -3*-9 would be -27, NOT 27. It makes no sense. > None. Not even a little. It's not simple, it's complicated and stupid. > it may be complicated. but it's not stupid. it's a better way of doing > things. > No, it isn't. let me put this in simplistic terms: -3 squared is -9 -3 squared can also be 9 i squared is supposedly -1 what that really means is -1 squared can be 1 -1 squared can also be -1 i just refers to the case when -1 squared is -1. that's all it's just a symbol for -1 === Subject: Re: i: imaginary solution to constraints of a real number system? or just a hoax > consider the following: > > the square root of a negative number requires some non-real solution? > why can't we define -3 squared to be -9? because then mathematics > would be simple.. and mathematicians would be out of the job. > > No, that would not make sense at all in fact - it would make math make > no sense - leading to something QUITE complicated, but mathematicians > would be out of jobs. They would all quit from frustration, until one > of them suddenly says, Hey, what about using imaginary numbers? > Numbers that.. you know, don't quite exist, but we can model the > equations as if they did? > Even in the above statement, that the square root of -9 is -3, you've > violated the logic that is used to even CREATE negative numbers > because in this case, -3*-9 would be -27, NOT 27. It makes no sense. > None. Not even a little. It's not simple, it's complicated and stupid. > > it may be complicated. but it's not stupid. it's a better way of doing > things. > No, it isn't. > let me put this in simplistic terms: > -3 squared is -9 No. 3^2=9 -(3^2)=-9 > -3 squared can also be 9 Right. > i squared is supposedly -1 Right. > what that really means is > -1 squared can be 1 Right > -1 squared can also be -1 No. (-1)^2=1; -(1^2)=-1 Look at the graph of f(x)=x^2 and you will see that you're wrong. These errors you've made are common in math. > i just refers to the case when -1 squared is -1. that's all it's just > a symbol for -1 No it's not. Go read a high school algebra book on complex numbers and you'll see you're wrong. Dave === Subject: Re: i: imaginary solution to constraints of a real number system? or just a hoax David Bandel < > let me put this in simplistic terms: > -3 squared is -9 Well I knew the term fooling around. But this is a fine example of fooling asquare :-)) Let M = -3. We know that M = 12 - 15. Let's compute (-3)^2 = M^2 according to (a-b)^2 = a^2 -2ab + b^2: M^2 = (12-15)^2 = 12^2 -2*12*15 + 15^2 = 144-360+225 = (144+225) - 360 = 369 - 360 = 9. Aha: M^2 = 9, not -9. Good to know (and to remember.) Rainer Rosenthal r.rosenthal@web.de === Subject: Re: i: imaginary solution to constraints of a real number system? or just a hoax > >[cut new delirium (it seems Bandel wants to carve a niche here. Sorry, >JSH is all sci.math can bear for the moment; you will have to be >patient (8 years now ? is not a birthday party coming soon ?))] > Bandel seems to be an even more incompetent Harris. I presume Harris > didn't meet his mathematical apex in highschool... Bandel has the mathematical maturity to know this is wrong, unlike JSH. He's just trolling. Ôcid Ôooh === Subject: Re: Connect 4 solved > >> > > >> >> >> >> >> >>I've read that Connect 4 has been solved and certain programs can play >>perfectly, provided they go first. Is the algorithm too hard to apply >>mentally? At least for me it appeared that way, from the 91 page >>report. >> >>Is it any less of a game now that it is solved? > >any game once solved goes down a notch in my book. but that's when >you look at a game as a problem. if it's a matter of competition.. in >most cases it's probably no less of a game unless it's exceedingly >simple and the solution can be applied by humans without the aid of >computers (like gomoku, connect-4, mancala, and a host of other >examples of simple and solved games) however, you're a boring stupid >idiot.. you're less of a man for posting this. >> >>At least he's thinking about a problem. A real man doesn't deal out insults >>right and left like you do. >> >>Dave > > >i guess i'm a fake man then.. i wasn't aware they existed.. whatever a >fake man is.. i guess i'm really a women but pretending to be a man? >something along those lines maybe? sorry.. you're the expert on >transexual stuff dave. anyways, you're an idiot too and a piece of > for posting this. >> >>You're the one who initially said you're less of a man for posting >>this. So while you're bothered about asking what a Ôfake man' is, >>please tell us what Ôless of a man' means. >>I'm fully expecting insults in response to this, so go on, surprise me >>by writing a non-abusive reply. > > > less of a man: shorter man. dumber man. weaker man, smaller-dicked man > fake man: i guess a man who isn't really a man? much less logical > If one makes sense to you, so does the other. > you: -eating salad tosser who loves dick > from you). Insults may stand for something at your school, but in the > adult world, they have precious little sway, and hurt only you. > How old are you? Does daddy know you're using his computer? as a matter of fact he does. and i don't think he'd give a nigger's if he knew i was pointing out salad tossers on USENETs. tell me something. when the reciever farts in your face do you storm out of the room? and as for all the rest of you, i'm 15.. how many 13 year olds do you see wandering around with permits? === Subject: Re: Connect 4 solved > >> > > >> >> >> >> >> >>I've read that Connect 4 has been solved and certain programs can play >>perfectly, provided they go first. Is the algorithm too hard to apply >>mentally? At least for me it appeared that way, from the 91 page >>report. >> >>Is it any less of a game now that it is solved? > >any game once solved goes down a notch in my book. but that's when >you look at a game as a problem. if it's a matter of competition.. in >most cases it's probably no less of a game unless it's exceedingly >simple and the solution can be applied by humans without the aid of >computers (like gomoku, connect-4, mancala, and a host of other >examples of simple and solved games) however, you're a boring stupid >idiot.. you're less of a man for posting this. >> >>At least he's thinking about a problem. A real man doesn't deal out insults >>right and left like you do. >> >>Dave > > >i guess i'm a fake man then.. i wasn't aware they existed.. whatever a >fake man is.. i guess i'm really a women but pretending to be a man? >something along those lines maybe? sorry.. you're the expert on >transexual stuff dave. anyways, you're an idiot too and a piece of > for posting this. >> >>You're the one who initially said you're less of a man for posting >>this. So while you're bothered about asking what a Ôfake man' is, >>please tell us what Ôless of a man' means. >>I'm fully expecting insults in response to this, so go on, surprise me >>by writing a non-abusive reply. > > > less of a man: shorter man. dumber man. weaker man, smaller-dicked man > fake man: i guess a man who isn't really a man? much less logical > If one makes sense to you, so does the other. > you: -eating salad tosser who loves dick > from you). Insults may stand for something at your school, but in the > adult world, they have precious little sway, and hurt only you. > How old are you? Does daddy know you're using his computer? > as a matter of fact he does. and i don't think he'd give a nigger's > if he knew i was pointing out salad tossers on USENETs. tell me > something. when the reciever farts in your face do you storm out of > the room? > and as for all the rest of you, i'm 15.. how many 13 year olds do you > see wandering around with permits? For a 15 year old, you sure do have a mouth on you, which shows your immaturity. Dave === Subject: Re: Connect 4 solved > For a 15 year old, you sure do have a mouth on you, which shows your > immaturity. > Dave kids my age cuss. it's a fact of life. were you never a kid? let's not ignore you just said that FOR a 15 year old i sure have a mouth on me. implying i have more of a mouth on me than a normal 15 year old. i haven't been shackled down by social oppressions yet so i'm eager to speak a bit more than my mind from time to time. do the people you associate with know that you have no en clue how to deal with kids? that's a sign of maturity you know; putting kids in their place.. dealing with them. i see you have no skills in that department you old goat. you're a miser. you eat out the assholes of your one-night-stands and save their in a jar to sniff and wiff and lick later to your heart's content. === Subject: Re: Connect 4 solved > For a 15 year old, you sure do have a mouth on you, which shows your > immaturity. > Dave > kids my age cuss. it's a fact of life. were you never a kid? let's > not ignore you just said that FOR a 15 year old i sure have a mouth on > me. implying i have more of a mouth on me than a normal 15 year old. > i haven't been shackled down by social oppressions yet so i'm eager to > speak a bit more than my mind from time to time. do the people you > associate with know that you have no en clue how to deal with > kids? that's a sign of maturity you know; putting kids in their > place.. dealing with them. i see you have no skills in that > department you old goat. you're a miser. you eat out the > assholes of your one-night-stands and save their in a jar to > sniff and wiff and lick later to your heart's content. Excuse me but as a meteorologist, I get to go talk to kids about your age about my job and what I do at work. These kids have even come up to me at stores, restaurants, etc. and talked to me even more because they know that I will talk to them. None of these people have ever said anything bad to my face which shows their maturity because I am physically disabled and while others their age might make rude comments to my face, those that have seen one of my presentations know that though I am disabled, I am not stupid. Davre === Subject: Question: SPACE of super-complex numbers hi. complex numbers are elegant. the complex plane is simple to grasp. I am wondering if there is some extension that would go to the 3rd dimension. what kind of hyper-imaginary axis.. and what the unit of it would be and represent. (as the unit of the imaginaries represents the square root of -1) what would the unit of this 3rd axis' numbers be? is there such a thing? === Subject: Re: Question: SPACE of super-complex numbers > hi. > complex numbers are elegant. the complex plane is simple to grasp. I > am wondering if there is some extension that would go to the 3rd > dimension. what kind of hyper-imaginary axis.. and what the unit of > it would be and represent. (as the unit of the imaginaries represents > the square root of -1) what would the unit of this 3rd axis' numbers > be? is there such a thing? There are no 3 D division algebras over the reals. You have to go from complex number to quaternions. All of these manipulations were tried back in the middle of the 19-th century. Google on the history of quaternions and octernians. Also Google division algebras. Having a quotient is a strong constraint. On the other hand you can have rings (multiplication and addition defined) of any degree over the reals. Bob Kolker Bob Kolker === Subject: Re: Question: SPACE of super-complex numbers > hi. > complex numbers are elegant. the complex plane is simple to grasp. I > am wondering if there is some extension that would go to the 3rd > dimension. what kind of hyper-imaginary axis.. and what the unit of > it would be and represent. (as the unit of the imaginaries represents > the square root of -1) what would the unit of this 3rd axis' numbers > be? is there such a thing? I don't know of any Ôthree-dimensional numbers', but there are established Ôfour-dimensional numbers' called Quaternions. Here's a link: http://mathworld.wolfram.com/Quaternion.html === Subject: Re: Question: SPACE of super-complex numbers >complex numbers are elegant. the complex plane is simple to grasp. I >am wondering if there is some extension that would go to the 3rd >dimension. what kind of hyper-imaginary axis.. and what the unit of >it would be and represent. (as the unit of the imaginaries represents >the square root of -1) what would the unit of this 3rd axis' numbers >be? is there such a thing? John Conway, who is widely known in mathematics, and Derek Smith, who I have not seen work from before, published On Quaternions and Octonions, Their Geometry, Arithmetic, and Symmetry last year. That little book can lead you off in all sorts of directions beyond the complex numbers. === Subject: Re: sci.math vs True Believers >> 1) Don't top post, idiot. >> 2) Everything you have trolled is trashed by three short >> sentences, idiot. >> 3) Address the scientific issue, idiot. Demonstrate that >> Galilean transforms take precedence over Lorentz Transforms. You >> cannot. it isn't true. It is empirically untrue. You are an >> ineducable idiot. >I have wanted to address the physics issues from the word go, but jerkoffs >such as Carr - with jackass support from you - have spent many hundreds of >posts saying the basic, correct algebra is wrong, leaving it impossible to >proceed to the match or not with experiment because y'all won't acknowledge >the simple validity of the standard by which to judge the galilean theory: WHAT IS YOUR MAJOR MALFUNCTION? The universe is Relatvistic until shown otherwise. === Subject: Re: sci.math vs True Believers > WHAT IS YOUR MAJOR MALFUNCTION? My major malfunction - by your standards - is that I believe in the basics: the premises and the logic that works on the premises. You, on the other hand, are proving completely incapable of honest dealing with a very basic matter, a matter that should be trivial to one who knows the nature of reality so well. So, how can an equation in coordinate-difference form - (x1-x0) instead of bare x - be non-invariant under the galilean xforms? That question is basic to an honest approach to invariance under galilean xforms. It is incompotent galilean if an xformed equation is not in coordinate-difference form, as well as dishonest Relativistics. How is it that in a theoretical structure that says time is absolute that you insist on asserting a time transformation? That question is basic to an honest approach to invariance under galilean xforms. It is corrupt anti-galileanism if a time transform is imposed. With every post you make the reader sees that you have avoided the simple yes/no questions asked of you, answers so much more simple than your rants. You are as intellectually honest as a ravening Ayatollah or Undermentalist Christian. And most everyone else is just as bad. They know (x1-x0) can't be non-invariant. They know a time transform - relative rather than absolute time - is not Newtonian/galilean. Yet they say nothing. Are they afraid of Uncle piggy's treatment? I know they agree with me. When I posted the basic questions here in the past in neutral form they said so. They continue to watch without writing but they do see you. And there are one or two who act honestly. They, too, see you. Hey, here's another one: how is that that you cannot actually apply the LE COORDINATE xforms to Maxwell? eleaticus > The universe is Relatvistic until shown otherwise. === Subject: Re: sci.math vs True Believers >> WHAT IS YOUR MAJOR MALFUNCTION? WHAT IS YOUR MAJOR MALFUNCTION? You are WRONG - totally and without doubt. Maxwells equations are Lorentz invariant but not Galiliean invariant. Do you know WHY Galilean transforms are ? No, of course not. === Subject: Re: sci.math vs True Believers >> WHAT IS YOUR MAJOR MALFUNCTION? > WHAT IS YOUR MAJOR MALFUNCTION? > You are WRONG - totally and without doubt. Maxwells equations are > Lorentz invariant but not Galiliean invariant. > Do you know WHY Galilean transforms are ? No, of course not. A True Believer much like you but with the guts to expose himself showed me his demonstration that Maxwell was not invariant under the galilean xforms. What he demonstrated was that it was indeed invariant if one did not impose the corrupt (because anti-theoretical per Newton, etc) time transform. I suppose your failure to also do so lies in the fact that you know that is a fault in the demonstration. Either that are you don't know your own cult demonstration. (I left in a bad typo below in honor of the spell checker that didn't catch it.) Gee. Such a simple thing to demonstrate for such a knowlegeaboe person as you, but do you do it? Oh no! You'd rather rave. Not to mention rather not expose what you say to the basic theses I have been posting. To expose your Ôwhy' against the material you have been avoiding on the pretext that experiment disproves simple math could not be done without shame even by such a near-shameless sham as you have been. eleaticus === Subject: Re: sci.math vs True Believers >> >> >>> WHAT IS YOUR MAJOR MALFUNCTION? >> >> WHAT IS YOUR MAJOR MALFUNCTION? >> You are WRONG - totally and without doubt. Maxwells equations are >> Lorentz invariant but not Galiliean invariant. >> Do you know WHY Galilean transforms are ? No, of course not. >A True Believer much like you but with the guts to expose himself showed me >his demonstration that Maxwell was not invariant under the galilean xforms. >What he demonstrated was that it was indeed invariant if one did not impose >the corrupt (because anti-theoretical per Newton, etc) time transform. hehe, really? Show me, i've had a boring weekend. Seeing Maxwell's equations being invariant under Galilian transforms would be interesting. >I suppose your failure to also do so lies in the fact that you know that is >a fault in the demonstration. Of course. Your demonstration... You are yet to demonstrate what you speak about. >Either that are you don't know your own cult demonstration. >(I left in a bad typo below in honor of the spell checker that didn't catch >it.) >Gee. Such a simple thing to demonstrate for such a knowlegeaboe person as >you, but do you do it? Only an undergrad. Then again, what education do you have or are in the process of getting? >Oh no! You'd rather rave. Not to mention rather not expose what you say to >the basic theses I have been posting. If I'd rather rave, guess what i would be doing? >To expose your Ôwhy' against the material you have been avoiding on the >pretext that experiment disproves simple math could not be done without >shame even by such a near-shameless sham as you have been. *snicker* Your simple math is coupled to reality, thus physics. If the physics is wrong, it is wrong. Given the universe really likes to behave in a Lorentzian way, guess what my response to Ôthe universe is Galilian!' is? >eleaticus === Subject: Re: sci.math vs True Believers > > >> WHAT IS YOUR MAJOR MALFUNCTION? > > WHAT IS YOUR MAJOR MALFUNCTION? > You are WRONG - totally and without doubt. Maxwells equations are > Lorentz invariant but not Galiliean invariant. > Do you know WHY Galilean transforms are ? No, of course not. > A True Believer much like you but with the guts to expose himself showed me > his demonstration that Maxwell was not invariant under the galilean xforms. [snip crap] ing imbecile eleaticus, Galilean transforms are irrelevant to EM and spacetime as Euclidean geometry is irrelevant to navigation and surveying on the curved Earth. It doesn't work, stooopid. You are disproving yourself, moron. You are arguing for the other side, jackass. You are nothign but a stooopid troll. LISTEN UP, STOOOOPID, YOU ARE WRONG. Your ignorance, incompetence, and psychosis are not of interest to the world at large. Quite the contrary. You are not even an interesting laughingstock. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: sci.math vs True Believers > A True Believer much like you but with the guts to expose himself showed me > his demonstration that Maxwell was not invariant under the galilean xforms. > [snip crap] The part he snipped being the assertion that the demonstration demonstrated Maxwell's invariance under galilean transforms once the corrupt, anti-Newtonian time transform was removed from the process. > ing imbecile eleaticus, Galilean transforms are irrelevant to > EM and spacetime as Euclidean geometry is irrelevant to > navigation and surveying on the curved Earth. Note that a big bad brilliant man (?) like Uncle piggy is apparently intellectually incapable of simply showing that any equation in x (not dx) is not invariant mathematically when put in the obvious coordinate difference form even children know to use (in effect) when using a rulet. He is apparently not only incapable but would be too corrupt to do so were he able. He is obviously too corrupt to admit the use of the spurrious time transform for Newtonian invariance analusis of equations in dc. eleaticus > It doesn't work, > stooopid. You are disproving yourself, moron. You are arguing > for the other side, jackass. You are nothign but a stooopid > troll. > LISTEN UP, STOOOOPID, YOU ARE WRONG. Your ignorance, > incompetence, and psychosis are not of interest to the world at > large. Quite the contrary. You are not even an interesting > laughingstock. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: sci.math vs True Believers > 1) Don't top post, idiot. > 2) Everything you have trolled is trashed by three short > sentences, idiot. > 3) Address the scientific issue, idiot. Demonstrate that > Galilean transforms take precedence over Lorentz Transforms. You > cannot. it isn't true. It is empirically untrue. You are an > ineducable idiot. > I have wanted to address the physics issues from the word go, but jerkoffs > such as Carr - with jackass support from you - have spent many hundreds of > posts saying the basic, correct algebra is wrong, leaving it impossible to > proceed to the match or not with experiment because y'all won't acknowledge > the simple validity of the standard by which to judge the galilean theory: Ineducable psychotic idiot. Mathematics of gravitation http://arXiv.org/abs/gr-qc/0311039 Experimental constraints on General Relativity http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system Hafele-Keating Experiment http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries http://arXiv.org/abs/gr-qc/0301024 Nordtvedt Effect Hey stooopid eleaticus, science is empirical. Hey stooopid eleaticus, spacetime is a Lorentz transformation. Hey stooopid eleaticus, spacetime is a 4-D hyperbolic rotation. Ineducable psychotic idiot. http://arxiv.org/abs/gr-qc/0306076.pdf http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: sci.math vs True Believers Once again Uncle piggy spews entirely non-responsive Ôstuff'. But, I think his paste button was broken. > 1) Don't top post, idiot. > 2) Everything you have trolled is trashed by three short > sentences, idiot. > 3) Address the scientific issue, idiot. Demonstrate that > Galilean transforms take precedence over Lorentz Transforms. You > cannot. it isn't true. It is empirically untrue. You are an > ineducable idiot. > I have wanted to address the physics issues from the word go, but jerkoffs > such as Carr - with jackass support from you - have spent many hundreds of > posts saying the basic, correct algebra is wrong, leaving it impossible to > proceed to the match or not with experiment because y'all won't acknowledge > the simple validity of the standard by which to judge the galilean theory: > Ineducable psychotic idiot. > > Mathematics of gravitation > http://arXiv.org/abs/gr-qc/0311039 > > Experimental constraints on General Relativity > ndex.html> > http://www.eftaylor.com/pub/projecta.pdf > > Relativity in the GPS system > > Hafele-Keating Experiment > http://arXiv.org/abs/astro-ph/0401086 > http://arxiv.org/abs/astro-ph/0312071 > Deeply relativistic neutron star binaries > > http://arXiv.org/abs/gr-qc/0301024 > Nordtvedt Effect > Hey stooopid eleaticus, science is empirical. > Hey stooopid eleaticus, spacetime is a Lorentz transformation. > Hey stooopid eleaticus, spacetime is a 4-D hyperbolic rotation. > Ineducable psychotic idiot. > http://arxiv.org/abs/gr-qc/0306076.pdf > > http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf > http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm > http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm > http://www.trimble.com/gps/index.html > http://sirius.chinalake.navy.mil/satpred/ > http://www.phys.lsu.edu/mog/mog9/node9.html > http://egtphysics.net/GPS/RelGPS.htm > http://www.schriever.af.mil/gps/Current/current.oa1 > http://edu-observatory.org/gps/gps_books.html > > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: abstract algebra..... hello....doctor~ suppose that F is field. R is ring. g : F -> R is surjective ring homomorphism. show that g is isomorphism. ---------------------------------- we must show that one to one. suppose that ker g =/= {0}. there is a a such that g(a) = 0 , a =/= 0. since F is field, there is a a^-1 such that a*(a^-1) = 1 = (a^-1)*(a). g(a*(a^-1)) = g(a){(g(a))^-1} = g(1) g((a^-1)*a) = {(g(a))^-1}g(a) = g(1) *** since g is surjective, g(1) is unity of R. *** so, g(a) is unit. this is a contradiction.(g(a) = 0) so, ker g = {0} thus, g is injective. so, iso. --------------------------------------- um......i can't understand this part ***. help me, please~ thank you very much. === Subject: Re: abstract algebra..... >hello....doctor~ >suppose that >F is field. >R is ring. >g : F -> R is surjective ring homomorphism. >show that g is isomorphism. this is not true as stated. if R = {0} and g(x) = 0 for all x in F then g is a surjective ring homorphism. you need to assume that R is a ring with identity [and also that 1 <> 0 in R, or equivalently that R contains more than one element - i'm not sure whether 1 <> 0 is supposed to be part of the definition or not, i've seen definitionds where it's included and definitions where it's not. of course the ring {0} is the only one where it matters...] >---------------------------------- >we must show that one to one. >suppose that ker g =/= {0}. >there is a a such that g(a) = 0 , a =/= 0. >since F is field, there is a a^-1 >such that a*(a^-1) = 1 = (a^-1)*(a). >g(a*(a^-1)) = g(a){(g(a))^-1} = g(1) >g((a^-1)*a) = {(g(a))^-1}g(a) = g(1) >*** since g is surjective, g(1) is unity of R. *** >so, g(a) is unit. this is a contradiction.(g(a) = 0) >so, ker g = {0} >thus, g is injective. so, iso. >--------------------------------------- >um......i can't understand this part ***. neither do i. it follows if we assume that R has identity: lemma: if R is a ring with 1 and ex = x for all x in R then e = 1. pf: 1 = e1 = e. qed. now for ***: given x in R there exists y in F with g(y) = x. so g(1)x = g(1)g(y) = g(1y) = g(y) = x. so the lemma implies that g(1) = 1. >help me, please~ >thank you very much. ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: abstract algebra..... === Subject: abstract algebra..... >suppose that > F is field. > R is ring. > g : F -> R is surjective ring homomorphism. >show that g is isomorphism. If some x /= 0 with g(x) = 0, then for all y g(y) = g(yxx^-1) = g(y).g(x).g(x^-1) = 0. Thus as g is surjection, R = g(F) = { 0 } = trivial ring Otherwise, if R not trivial ring ker g = { 0 } and g is isomorphic. Now if you insist that rings must have identity 1 /= 0, then the conclusion follows. However, if for example 2Z is a ring, which it is, then conclusion follows only if R not trivial. >we must show that one to one. >suppose that ker g =/= {0}. >there is a a such that g(a) = 0 , a =/= 0. >since F is field, there is a a^-1 >such that a*(a^-1) = 1 = (a^-1)*(a). >g(a*(a^-1)) = g(a){(g(a))^-1} = g(1) You cannot do that. Yes g(1) = g(aa^-1) = g(a).g(a^-1) But the next step g(a).g(a^-1) = g(a).g(a)^-1 doesn't happen because g isn't a field homorphism, ie isomorphism. >*** since g is surjective, g(1) is unity of R. *** >um......i can't understand this part ***. For all r in R, some a in F with r = g(a). Thus r.g(1) = g(a).g(1) = g(a1) = g(a) = r g(1).r = g(1).g(a) = g(1a) = g(a) = r Hence g(1) is left and right identity, thus the one and only unique identity provided that g(1) is distinct from 0, which is now easy to show assuming R' isn't trivial ring. Note. An identity by definition, is to be distinct from 0. Exercise: Let f:R -> R' be a surjective ring homomorphism. Show commutivate ring R ==> R' commutative ring. >so, g(a) is unit. this is a contradiction.(g(a) = 0) You did not show g(a) = g(1), nor g(a) a unit. Only that g(1) is a unit. >so, ker g = {0}. thus, g is injective. so, iso. Conclusion unwarrented. ---- === Subject: Re: SMSU Problem Corner SMSU High School Problem Page >Associate the following numerical values to the letters of the >alphabet: A = 1, B = 2, ..., Z = 26. Given any word, replace its >letters with their numerical values and take the product. >For example, the word MATH would yield 13 1 20 8 = 2080. >Find another English word that yields a product of 2080. 2080 = 10 * 208 = 4 * 10 * 52 = 2^5 * 5 * 13 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 a b c d e f g h i j k l m n o p q r s t u v w x y z 13 26 m z exactly one 5 10 20 e j t exactly one 2 4 8 16 b d h p one to five, repetitions 1 a any number dazed ---- === Subject: Re: Prime Count > We should have a little competition: > Shortest C function (smallest number of characters) of the form > int pi (int x) > which returns pi (x) correctly for all x, 0 <= x <= INT_MAX. Execution > time doesn't matter. > Two attempts from me. I counted 1 (one) as a prime. > Attempt 1 (non-recursive): > int pi(int x) > int r, s, y, i; > for (y = 1; y <= x; y++) { > s = 1; > for (i = 2; i < y;) > s *= (y % i++) != 0; > r += s; > } > return r; > } // 74 non-space chars in the body > Attempt 2 (recursive): > int pi(int x) > int s=1, i; > if (!x) > return 0; > for (i = 2; i < x;) > s *= (x % i++) != 0; > return pi(x-1) + s; > } // 67 non-space chars in the body > I have no compiler available currently, so I hope it works OK.... Based on your attempt 2, here is my C# attempt in 72 non-space chars: int pi(int x){int s=1,i=2; while(i Based on your attempt 2, here is my C# attempt in 72 non-space chars: > int pi(int x){int s=1,i=2; > while(i return x<2?0:pi(x-1)+s;} You can replace xxx == 0 ? 0 : 1 with xxx < 1, saving five characters: int pi(int x){int s=1,i=2; while(i <2oolvrFcm0hdU1@uni-berlin.de> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Based on your attempt 2, here is my C# attempt in 72 non-space chars: > int pi(int x){int s=1,i=2; > while(i return x<2?0:pi(x-1)+s;} Well, in C this could just be int pi(int x){int i=x>1;while(x%++i);return --x<1?0:pi(x)+(i>x);} Non-recursively we may take int pi(int x){int s=0,i=1;while(x>1&&(x%++i||(s+=i>--x,i=1)));return s;} Which saves another variable and stops the iteration immediately when a divisor is found. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Coin ßip I'm gonna try to solve it. But it should be stated that I never took a statistics course, and hence, I'm a completely newbie, so if it is a homework, then you are in deep trouble if you hand this in :-) First of all, the fact that you give 10 but take 5 pounds creates some trouble. If they were equal, then you knew that the number of heads should be >= than the number of tails. But unfortunately this is not the case. So if N is the number of tosses and X is the number of heads we got, we know that we have a break-even at: 10X-5(N-X) = 0 Solving X gives us: X = N / 3 So for the case of 10 tosses, X = 10/3 = 3.33 heads would gives us a break even. So we 10*10/3 - 5*(10 - 10/3)=0 indeed. However you cannot get 3.33 heads, so you have to round it: ceiling(X = N / 3), in the example of N=10 => X=4. Using the bernoulli trials, we know that the probability of having exactly X heads out of N trials is: (N over X) * 0.5^(X) * 0.5^(N-X) Where (N over X) = N! / (X! * (N-X)!) Where X! = X*(X-1)*(X-2)*...*1. I.e. 5! = 5*4*3*2*1 So (5 over 3) = 5*4*3*2*1 / (3*2*1 * (2*1)) = 5*4 / 2*1 However, we are not interested in ALL cases where X > ceiling( N / 3 ). So in the example with N=10, we are interested in the probability of getting X=4 (4 heads and 6 tails), or X=5, or X=6, or X=7, or X=8, or X=9, or X=10. All of those cases, the amount of money left will be positive. So in the case of N=10, the probability of having a positive amount is : (10 over 4) * 0.5^4 * 0.5^6 + (10 over 5) * 0.5^5 * 0.5^5 + (10 over 6) * 0.5^6 * 0.5^4 + (10 over 7) * 0.5^7 * 0.5^3 + (10 over 8) * 0.5^8 * 0.5^2 + (10 over 9) * 0.5^9 * 0.5^1 + (10 over 10) * 0.5^10 * 0.5^0 I.e. we get (in the cases of N=10 tosses) P(X >= 4)= 0.828125 which is quite a high probability. For the case of N=5 we get P( X>=2 )= 8125. As you can guess, which seems logical, the probability goes asymptotically to 1.0 when N goes to infinity. Which makes sense, since you will get on average the same amount of heads and tails, but you give more than you take, so the sum of cash will tend to be more and more positive as N increases. As for the general case, I (with my weak maths background) cannot give a closed formula. I can give you the formula with sum notation (big Sigma) but that wouldn't help you much. So here it is: sum_from(X=ceiling(X/3))to(X=N) (N over X)*0.5^(X)*0.5^(N-X) Ali Ghodsi > Johnny has a coin, he tells Billy that he will give him 10 every time he spins heads, but take 5 off him every time he spins tails. What are the chances that Billy will have a positive amount after 5 spins? And after 10 spins? After X spins? === Subject: Re: Coin ßip >Johnny has a coin, he tells Billy that he will give him 10 every time he spins heads, but take 5 off him every time he spins tails. What are the chances that Billy will have a positive amount after 5 spins? And after 10 spins? After X spins? >>Homework? >>To get you started, think about the choosing function, nCr (closely >>linked to the binomial theorem which someone else has mentioned). For >>example, the amount of ways of choosing 2 heads and 3 tails from 5 coins >>is 5C2. >>Also note that amount of different outcomes (where order matters) for 5 >>tosses is 2^5. >>And keep in mind that 5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 2^5 = 32. In >>other words, the sum of all the different ways you can get different >>counts of heads and tails adds up to all the possibilities. >>Ask yourself which of these 5C0, 5C1 etc. combos ends up with you having >>positive money after the game. > P.S. It's not homework, I'm 27. I'm curious, what do you think the chances are of another poster asking exactly the same question, by complete chance, in exactly the same minute of the same day? Because someone else posted the same question. Or was that you as well? I'm not saying I'm not going to help anymore, I'm just curious about that. === Subject: Re: Errors in Wiles's proof of Fermat's Last Theorem >>>Professor Escultura disputed [1] Wiles's proof[2] of Fermat's Last >>>Theorem. Any definite comments about the correct status of the proof >>>of FLT will be of great interest to the number theorists,amatur >>>mathematicians and the public in general. >>> >>>[1]Escultura, E.E.,Exact Solutions of Fermat's Equation(Definitive >>>Resolution of Fermat's Last Theorem),Nonlinear Studies, vol.5,No.8, >>>September 1998 >>> >>>[2]Wiles, A., Modular Elliptic Curves and Fermat's Last Theorem, >>>Ann.of Math.,141(1995) 443-551 >>> >>> Escultura used to be a regular on sci.math, best known for his >>> insistence that 0.999... < 1. He is a crank, but has a talent >>> for getting rubbish published in otherwise respectable journals. >>> Wiles's proof is in no danger from him. >>> >>> Robert Israel israel@math.ubc.ca >>> Department of Mathematics http://www.math.ubc.ca/~israel >>> University of British Columbia >>> Vancouver, BC, Canada V6T 1Z2 >> >>Professor Israel, it seems to me that your response is not like what >>should come from a math professor. First identify the errors in >>Escultura's paper and then make comments about him. Before the >>paper(Escultura) was published it was throughly reviewed by a number >>of experts. They did not find any error in Escultura's argument and >>let the paper get published. Unless Wiles can come out with a stronger >>argument,in my judgement,his claim of the proof of FLT is in great >>trouble. >> well you're simply wrong about that. >> you should really look up some of his posts on sci.math. >> go to >> , put sci.math in the newsgroups box and escultura in the >> Ôwith all of the words' box. >> or note that his 10-page paper the Solution of the >> Gravitational n-body Problem in Nonlinear Analysis >> (Elsevier) 30(8), 1997, pp. 5021-5032 Sept/Dec 1998 >> has at least two remarkable properties: >> [i] it shows that he doesn't know what the n-body problem >> is - he makes a big deal of approximate solutions, but >> the n-body problem is the problem of finding an -exact- >> closed-form solution - the existence of approximate >> solutions is well-known >> [ii] that one 10-page paper includes the following >> list of key words: >> Generalized limit, derivative and integral, latent >> and kinetic energy, energy conservation and equivalence, >> order, symmetry, stability, optimality, least action, fractal, >> chaos, ßux, vortex, conversion, dark and visible matter, >> superstring, black hole, big bang, Cosmos, cosmic sphere, >> quasar, galaxy, shock wave, Hubble's law, steady-state, >> orbit, quantum mechanics, relativity, background radiation >> and noises, resonance, [gamma]-ray burst, cosmic rays and >> ripples, primum, semiconductor, superconductor. >> do you -really- think that he made significant controbutions >> to all those fields in one 10-page paper? >> regarding your point about the fact that his paper was >> peer-reviewed: i once sent an email to the editor of >> Nonlinear Studies, wondering why they published such >> obvious nonsense [note this is also where the paper >> on ßt appeared.] i got a reply. the editor didn't say >> anything about peer review - he didn't dispute my >> characterization of eee's publication as nonsense. >> instead the editor explained that he and eee had been >> friends for a long time, and he suggested that some >> day when i had an old friend go crazy i might understand. >*** >Respected David C. Ullrich, >Would you kindly email the editor once again and find out if the paper >by EEE on FLT was reviewed by more than one expert before it was >accepted for publication. why in the world would i do that? he's already agreed the paper is nonsense. he -said- that his dear old friend had lost his mind - that was his explanation for why the papers were published! > This will bring the controversy to an end. there is no controversy. the fact that you cannot recognize lunacy when you see it does not imply that everyone else is equally blind. >Escultura may be cranky or not, but if he says something correct >mathematicians must accept it. >Fair enough? Please respond. what's not fair enough is that in this very thread you've been -given- several examples of total nonsense from him, but it doesn't seem to matter. >> *** >>Pleae respond like a mathematician. >> he did, mentioning eee's belief that 0.999... < 1. >> ************************ >> David C. Ullrich >> sorry about the inelegant formatting - typing >> one-handed for a few weeks... ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: abstract algebra..... > hello....doctor~ > suppose that > F is field. > R is ring. > g : F -> R is surjective ring homomorphism. > show that g is isomorphism. > ---------------------------------- > we must show that one to one. > suppose that ker g =/= {0}. > there is a a such that g(a) = 0 , a =/= 0. > since F is field, there is a a^-1 > such that a*(a^-1) = 1 = (a^-1)*(a). > g(a*(a^-1)) = g(a){(g(a))^-1} = g(1) > g((a^-1)*a) = {(g(a))^-1}g(a) = g(1) > *** since g is surjective, g(1) is unity of R. *** > so, g(a) is unit. this is a contradiction.(g(a) = 0) > so, ker g = {0} > thus, g is injective. so, iso. > --------------------------------------- > um......i can't understand this part ***. > help me, please~ > thank you very much. I think this is one of those simple if you know type probs. A field has only the trivial ideal I = (0) (if a not 0, a^-1 exists, and if a in I, a a^-1 = 1 in I, so I = F, the whole field). Since the kernel of a ring homo is an ideal, kernel = (0) since g is onto (ker = F ==> g(I) = g(F) = 0). I should have read your post closer, I see you have already been over some of this--though you don't always draw the correct conclusions straight off (which is to be expected). Anyway, a well known thm says that if ker g = (0), g is into or 1-1. This is easy to show. If g(x) = g(y), then g(x) - g(y) = g(x - y) = 0 ==> x - y in ker g = (0) ==> x = y. Thus g is 1-1, and since its onto, its an iso. Van === Subject: Re: abstract algebra..... >> hello....doctor~ >> suppose that >> F is field. >> R is ring. >> g : F -> R is surjective ring homomorphism. >> show that g is isomorphism. >> ---------------------------------- >> we must show that one to one. >> suppose that ker g =/= {0}. >> there is a a such that g(a) = 0 , a =/= 0. >> since F is field, there is a a^-1 >> such that a*(a^-1) = 1 = (a^-1)*(a). >> g(a*(a^-1)) = g(a){(g(a))^-1} = g(1) >> g((a^-1)*a) = {(g(a))^-1}g(a) = g(1) >> *** since g is surjective, g(1) is unity of R. *** >> so, g(a) is unit. this is a contradiction.(g(a) = 0) >> so, ker g = {0} >> thus, g is injective. so, iso. >> --------------------------------------- >> um......i can't understand this part ***. >> help me, please~ >> thank you very much. >I think this is one of those simple if you know type probs. >A field has only the trivial ideal I = (0) (if a not 0, a^-1 exists, >and if a in I, a a^-1 = 1 in I, so I = F, the whole field). >Since the kernel of a ring homo is an ideal, kernel = (0) since >g is onto (ker = F ==> g(I) = g(F) = 0). but g(F) = 0 does not quite imply that g is not surjective. >I should have read your >post closer, I see you have already been over some of this--though >you don't always draw the correct conclusions straight off >(which is to be expected). >Anyway, a well known thm says that if ker g = (0), g is into >or 1-1. This is easy to show. If g(x) = g(y), then >g(x) - g(y) = g(x - y) = 0 ==> x - y in ker g = (0) ==> x = y. >Thus g is 1-1, and since its onto, its an iso. >Van ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Automorphisms of groups Automorphisms of groups. Consider the automorphisms of the Z_n, A(Z_n) (or of the cyclic form C_n, whichever is convenient) and Cart. products of the Z_n, and of the Z*_n groups. Z_2 has only the identity group (1). At first I thought that C_2 = (-1, 1) had x --> - x, or mult by -1, but this is not an auto, since a group homo must take 1 --> 1, and this doesn't. Its like adding 1 in Z_2. In general, I recall that Z*_n = A(Z_n), and Z_*2 = (1). Z*_3 = C_2 = (-1,1) =~ Z_2 ; Z*_4 = Z_2 = (1,3) = (-1,1) and so on. For a product Z_n x Z_m =~ Z_mn if (m,n) = 1, but what about the general case. I think this has all been done, it seems familiar. I will read up on it. I also recall the application to semi-direct products of groups--autos of cyclic normal subgroups, the simplest example being S_3. I started on this because of the autos of the Klein group A(Z_2 x Z_2) = S_3, the permutations of the non-identity elements. The autos here don't follow from Z*_n = A(Z_n). It makes me wonder about autos of Z_2 x Z_4, (Z_2)^n, Q, D_n, and S_n and A_n. I was recently reading about the autos of S_n and A_n. I recall that A(S_4) = S_4, and I think this is true for several n. And what about A(S_3)? And inner autos are a normal subgroups of A, so for S_4, that must mean N = Inn(A) = inner autos, where N < S_4 is the normal subgp of S_4. This is probably available online somewhere, but I always get sidetracked when I start looking for things. My searches never turn out as I expect. === Subject: Re: Automorphisms of groups === Subject: Automorphisms of groups >I want to review the related problems of isomorphism of Z*_mn, >with (Z*_m,Z*_n) when (m,n) = 1, and Euler phi(mn) = phi(m) phi(n). >Consider the automorphisms of the Z_n, A(Z_n) >products of the Z_n, and of the Z*_n groups. >Z_2 has only the identity group (1). Sloppy. Do you mean A(Z_2) = { I } ? >In general, I recall that Z*_n = A(Z_n), and Z_*2 = (1). >Z*_3 = (-1,1) =~ Z_2 ; Z*_4 = Z_2 = (1,3) = (-1,1) Interesting enuf. This has something to do with the phi function? When m,n coprime, Z_mn = Z_m x Z_n; Thus Z_mn^* = (Z_m x Z_n)^* =??= Z_m^* x Z_n^* Whence viole, phi(mn) = phi(m) phi(n) Perhaps Z_n^* = A(Z_n) is useful for this. What about phi(p^n) = p^n - p^(n-1) ? Back to the old fashion way of counting? Nay, tell me Ôtis not so. ---- === Subject: Re: Easy number theory problem I want to review the related problems of the CRT, isomorphism of Z_mn and Z*_mn, with (Z_m,Z_n) and (Z*_m,Z*_n) when (m,n) = 1, and Euler phi(mn) = phi(m) phi(n). CRT: (m,n) = 1 and x in Z_m, y in Z_n ==> there is a unique z in Z_mn such that z = x mod m and z = y mod n. Proof. rm + sn = 1. Mult by y - x. rm(y - x) = y - x - sn(y - x) or let z = x + rm(y - x) = y - sn(y - x) Then z = x mod m and z = y mod n is a soln. If z' is another soln., then z' = z mod mn, since both m and n must divide z' - z, and m | z' - z ==> z' - z = k m, n | km ==> n | k since (n,m) = 1 (as W. Elliot said in a recent post). Thus z' = z mod mn. Define f : Z_mn --> Z_m x Z_n : z --> f(z) = (z mod m, z mod n) Show that this is a bijection. (I want to extend this to Z*_mn, but worry about that and things like ring and group isomorphisms later.) Into; If f(z) = f(z') show that z = z', where f(z) = (z mod m, z mod n) = (x,y) = (x',y') = f(z') This is just the CRT (above), so z' = z mod m and z' = z mod n ==> z' = z mod mn. Now to extend this to Z*_m, etc. In this case we only have the elements such that (x,m) = 1, etc. I've never done or seen this before. === Subject: Easy number theory problem === Subject: Re: Easy number theory problem >I want to review the related problems of the CRT, >isomorphism of Z_mn with (Z_m,Z_n) when (m,n) = 1, >CRT: (m,n) = 1 and x in Z_m, y in Z_n ==> there is a unique >z in Z_mn such that z = x mod m and z = y mod n. >Proof. rm + sn = 1. Mult by y - x. rm(y - x) = y - x - sn(y - x) or >let z = x + rm(y - x) = y - sn(y - x) As m,n coprime, some a,b with am + bn = 1 bn = 1 (mod m); am = 1 (mod n) sam + rbn = r (mod m); sam + rbn = s (mod n) >Then z = x mod m and z = y mod n is a soln. >If z' is another soln., then z' = z mod mn, since both m and n >must divide z' - z, and m | z' - z ==> z' - z = k m, >n | km ==> n | k since (n,m) = 1. Thus z' = z mod mn. >Define f : Z_mn --> Z_m x Z_n : z --> f(z) = (z mod m, z mod n) This is nonsense except to computer folks as the operation z mod m isn't defined mathematically, only the relation z = m (mod m). No, computer talk ain't mathematics. What's worse is computer implementation of x mod y may differ in different computer languages. The actual bijection you want is f:Z_mn -> Z_m x Z_n, z + mnZ -> (z + mZ, z + nZ) which you need to show is well defined. If x + Z_mn = y + Z_mn, then x = y (mod mn), x = y (mod m), x = y (mod n); x + mZ = y + mZ, x + nZ = y + nZ (x + mZ, x + nZ) = (y + mZ, y + nZ) Thus it's well defined. As m,n coprime, some a,b with am + bn = 1 bn = 1 (mod m); am = 1 (mod n) sam + rbn = r (mod m); sam + rbn = s (mod n) f(sam + rbn) = (r,s) in Z_m x Z_n Thus f surjection. Rest is direct. f is ring homomorphism f(x+y + mnZ) = (x+y + mZ, x+y + nZ) = (x+mZ + y+mZ, x+nZ + y+nZ) = (x + mZ, x + nZ) + (y + mZ, y + nZ) = f(x) + f(x) f(xy + mnZ) = (xy + mZ, xy + nZ) = ((x + mZ)(y + mZ), (x + mZ)(y + nZ)) = (x + mZ, x + nZ)(y + mZ, y + nZ) = f(x).f(y) If f(z) = (0,0), then z = 0 (mod m), z = 0 (mod n), some k with z = km; n|z; n | km; m,n coprime; n | k nm | km; nm | z; z = 0 (mod nm) Thus ker f = { 0 }; f injection. >Show that this is a bijection. >Into; If f(z) = f(z') show that z = z', where >f(z) = (z mod m, z mod n) = (x,y) = (x',y') = f(z') Ambiguous at best. >This is just the CRT (above), >so z' = z mod m and z' = z mod n ==> z' = z mod mn. Still vague, the notation. Yes, the parallel with CRT is apparent. tho lacking precise expression. As a matter of fact Z_mn = Z_m x Z_n, m,n coprime directly shows CRT using only the group isomorphism. -- >Now to extend this to Z*_m, etc. In this case we only have >the elements such that (x,m) = 1, etc. I've never done or seen >this before. La-de-da. First make clear, precise, unambiguous statement of what you are wanting to prove. ---- === Subject: Re: Joining two striagt lines with a smooth curve by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7LCQgj02431; >y1 = a1*x + b1 >y2 = a2*x + b2 >ycurve = f(a1,a2,b1,b2) >Question 1. >The simplest case would be joining the two line segments with a >tangent circular arc. I can do this with compass and straight edge. >Is there an analytical expression for a family of such arcs. >This would have characteristic that first derivative of the curve >would match the first derivative (slope) of the joined segment. Just like the geometric construction, you need to also determine offset d, radius of tangent circular arc. Write y1,y2 into forms p=x cos(al)+y sin(al), replace p with p-d to solve for the center of arc, when ycurve = f(a1,a2,b1,b2,d) >Question 2. >Are there analytical expressions for curves whose K higher order >derivatives would be 0 at the point of tangency? For higher order contact ,both curves should have a matching minimum order derivative, which is possible if straight lines y1,y2 are replaced by higher order curves, say other circles or parabolas. === Subject: Re: Joining two striagt lines with a smooth curve >>y1 = a1*x + b1 >>y2 = a2*x + b2 >>ycurve = f(a1,a2,b1,b2) >>Question 1. >>The simplest case would be joining the two line segments with a >>tangent circular arc. I can do this with compass and straight edge. >Is there an analytical expression for a family of such arcs. >>This would have characteristic that first derivative of the curve >>would match the first derivative (slope) of the joined segment. >Just like the geometric construction, you need to also determine offset d, radius of tangent circular arc. Write y1,y2 into forms p=x cos(al)+y sin(al), replace p with p-d to solve for the center of arc, when ycurve = f(a1,a2,b1,b2,d) >>Question 2. >>Are there analytical expressions for curves whose K higher order >>derivatives would be 0 at the point of tangency? >For higher order contact ,both curves should have a matching minimum order derivative, which is possible if straight lines y1,y2 are replaced by higher order curves, say other circles or parabolas. For real world applications like railroads, you want to match the second derivative so as not to have discontinuous jumps in acceleration and corresponding forces such as you would normally get with second degree curves. Look up the subject of transfer curves. --Lynn === Subject: Re: Easy number theory problem === > Subject: Re: Easy number theory problem >> a^(p-1) = 1 (mod p) for prime p, a /= 0 (mod p) >> a^phi(n) = 1 (mod n) for coprime a,n. >I've seen these done a couple of times. I'll post a proof sometime >soon. >I recall one where one arugues that |Z*_p| = p-1 = even, and >each a_i (i=1,p-1) in Z*_p is unique. Pick and a in Z*_p. >Then prod(a_i) mod p = prod(a*a_i)mod p = a^(p-1) prod(a_i) mod p >since a induces a bijection or permutation of the a_i, so >a^(p-1) = 1 mod p. > Ugh. First show (Z_p)^* is a multiplicative group, ie show Z_p is a > field. You like to go back to basics, don't you? I guess that is good, esp. on sci.math. > First off 1 is an identity. Now for all a not divisible by p, > some n,m with an + pm = 1; an = 1 (mod p) > Hence a has a multiplicative inverse. Thus as Z_p has p elements, > (Z_p)^* = (Z_p)0 > has p-1 elements with o(a) = p-1. How do you know o(a)? You can say o(a) | p - 1, but shouldn't you show that (Z_p)^* is cyclic? I recall a proof from Birkhoff + McLane (I don't have a copy though). Let m = p - 1. Then o(x) | m for all x in F* = F0. They are all solns. of x^m - 1 = 0, which has m distinct solns. (m x^(m-1) not 0 if x not 0). Suppose m = rq, with r and q distinct primes, then generalize. Consider m = q^e, and finally p-1 = m = p1^e1 ... pr^er. (x^r)^q - 1 = y^q - 1 has q distinct solns with o(y) | q, so there are q - 1 solns y with o(y) = q, since only 1 and q divide q. Similarly there are r - 1 solns z with o(z) = r. Thus x = yz has o(x) = rq = m, so x generates the whole group. I recall that the case m = q^e was an important part of the proof. This will have to wait till later. This subject has been covered several times in sci.math. I should do a search. I like the ring theory approach, eith ideals. I recall if M = (m) and N = (n) are ideals of Z, (m,n) = 1 becomes M + N = Z, and M/N = (0) > Namely > a^(p-1) = 1 (mod p). > To use ring theory, show pZ is a maximal ideal of Z, thus by > theorem Z_p = Z/pZ is a field (clearly with p elements in it). Prime ideals are maximal in Z since if pZ is a proper subset of a maximal ideal M, there is an a in M but not in pZ. But this means that (a,p) = 1 or pZ + (a) = M = Z, where (a) = aZ is a subset of M. Thus pZ is maximal. Van === Subject: Re: Quotient Rule Application / Simplification Question >>I'd really appreciate some help with the simplification steps in the following problem. >>Quotient Rule Summary: If F(X) = f(x) / g(x), then F'(X) = (g(x)*f'(x) - f(x)*g'(x)) / [g(x)]^2 >>Here's the problem and what I've worked on so far: >This has nothing to do with the original post or poster, per se. >I have always thought that math teachers who taught the >'quotient rule' were morons who really did not understand what >they are teaching. And they fail to convey the ideas that really >need to be understood by their students. >This so called Ôrule' seems to confuse and obfuscate, rather >than clarify. >AND THERE IS NO NEED FOR IT. >To differentiate f(x)/g(x) simply observe that this is f(x)*(g(x))^-1 >and apply the product rule. There is no NEED for this ridiculous >quotient rule, memorized as a separate Ôrule'. > You are absolutely right; pedagogy by memorization of rules > is highly undesirable. I need derivatives of quotients > moderately often, and only sometimes do I use the quotient > rule, which is very definitely the way it is usually taught. > In fact, I often use (f/g)' = (f/g)*(f'/f - g'/g), which > is equivalent if f is nonzero at the point. Something similar I use is : (f/g)'= (-g'/g)* (f/g-f'/g'); (-g'/g) is derivative of 1/g with a shortcoming, ^2 left out, an apologetic derivative, if you will. The second bracket (f/g-f'/g')is a statement of L'Hospital's Rule for f/g constant or indeterminate, of what goes to zero. It can be even used for one or more independant variables. If we wished to find differential equation where a constant is to be eliminated, f/g = constant = f'/g' ... E.g., if we want differential equn of straight lines y=m*x+c of all slopes m through (c,0),(y-c)/x = y'/1 is the quick result. In such an operation, I place the two division symbols on each side of = sign , put f and g above and below on LHS and their corrosponding derivatives on the RHS, and start looking for simplification possibilities on this division fixture. It goes fast. === Subject: Re: abstract algebra..... >we must show that one to one. >suppose that ker g =/= {0}. >there is a a such that g(a) = 0 , a =/= 0. >since F is field, there is a a^-1 >such that a*(a^-1) = 1 = (a^-1)*(a). >g(a*(a^-1)) = g(a){(g(a))^-1} = g(1) > You cannot do that. Yes > g(1) = g(aa^-1) = g(a).g(a^-1) > But the next step > g(a).g(a^-1) = g(a).g(a)^-1 > doesn't happen because g isn't a field homorphism, ie isomorphism. I don't understand this, because assuming these rings have identities, a ring homo is a multiplicative group homo (isn't it?) -- g(ab) = g(a)g(b) ==> for b = a^(-1), g(a^(-1)) = [g(a)]^(-1) I haven't thought about rings without identities much. === Subject: Gauss-Jordan Elimination algorithm requires row divisions by the elements on diagonal, A( i, i ). My question: what do I do when A( i, i ) is zero (within a tolerance)? At the moment I do this: I look for a nonzero (and maximum in absolute value) element on the lower part of column i, among A( i + 1, i ) ... A( n, i ) and then I swap rows and the computed inverse is ok. If I can't find a nonzero element among these, I look on the right of row i, among A( i, i+1 ) ... A( i, n ). If I find a nonzero element, I swap columns, but then at the end of the algorithm the inverse is wrong, altough my initial A matrix becomes the identity matrix. Isn't this the right way to do it? What could be the problem? Cristian === Subject: Re: Gauss-Jordan Elimination >algorithm requires row divisions by the elements on diagonal, A( i, i ). My >question: what do I do when A( i, i ) is zero (within a tolerance)? if at some point there is no non-zero pivot available then the original matrix is not invertible. >At the moment I do this: I look for a nonzero (and maximum in absolute >value) element on the lower part of column i, among A( i + 1, i ) ... A( n, >i ) and then I swap rows and the computed inverse is ok. >If I can't find a nonzero element among these, I look on the right of row i, >among A( i, i+1 ) ... A( i, n ). If I find a nonzero element, I swap >columns, but then at the end of the algorithm the inverse is wrong, altough >my initial A matrix becomes the identity matrix. >Isn't this the right way to do it? What could be the problem? g-j reduction uses row operations, not row operations together with column operations. there's no reason to use both, but if you wanted to you could do this: say A is nxn. to find the inverse you start with the nx2n matrix A|I, use row operations to change it to I|B, and then B is the inverse. if you want to also use column operations then whenever you perform a column operation on the left half of your nx2n matrix you must simultaneously do the same column operation on the right half. >Cristian ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Gauss-Jordan Elimination The >algorithm requires row divisions by the elements on diagonal, A( i, i ). My >question: what do I do when A( i, i ) is zero (within a tolerance)? > if at some point there is no non-zero pivot available then the > original matrix is not invertible. >At the moment I do this: I look for a nonzero (and maximum in absolute >value) element on the lower part of column i, among A( i + 1, i ) ... A( n, >i ) and then I swap rows and the computed inverse is ok. >If I can't find a nonzero element among these, I look on the right of row i, >among A( i, i+1 ) ... A( i, n ). If I find a nonzero element, I swap >columns, but then at the end of the algorithm the inverse is wrong, altough >my initial A matrix becomes the identity matrix. >Isn't this the right way to do it? What could be the problem? > g-j reduction uses row operations, not row operations together with > column operations. > there's no reason to use both, but if you wanted to you could do this: > say A is nxn. to find the inverse you start with the nx2n matrix > A|I, use row operations to change it to I|B, and then B is the > inverse. if you want to also use column operations then whenever > you perform a column operation on the left half of your nx2n > matrix you must simultaneously do the same column operation on > the right half. >Cristian > ************************ > David C. Ullrich > sorry about the inelegant formatting - typing > one-handed for a few weeks... ^^^^^^^^^^^^^^^^^^^^^ So are you going to tell us why? And are you going to post whatever it was the the man said but didn't appear in sci.bio.evolution? And yes, can you imagine how many cranks sci.bio.evvolution would attract if it WASN'T moderated? === Subject: When radius point is not accessible. I work construction as a carpenter and need to layout interior partition walls on a curve. More specifically, this curve is a 25'-0 radius. The problem is that the focus point or pivot point to swing the radius is not accessible to me because it's position is located dead center of an existing 10 H-column. What is the best and most efficient way to proceed while trying to maintain the integrity of the 25' radius? Chris Grubb === Subject: Re: When radius point is not accessible. >I work construction as a carpenter and need to layout interior >partition walls on a curve. More specifically, this curve is a 25'-0 >radius. The problem is that the focus point or pivot point to swing >the >radius is not accessible to me because it's position is located dead >center of an existing 10 H-column. > What is the best and most efficient way to proceed while trying to >maintain the integrity of the 25' radius? Mark it out on graph paper first. Determine a set of points that will guide you (e.g., where bricks or studs should go), then determine the measurements you need from locations that are accessible. Autocad systems will do this for you quite easily. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: When radius point is not accessible. I work construction as a carpenter and need to layout interior partition walls on a curve. More specifically, this curve is a 25'-0 radius. The problem is that the focus point or pivot point to swing the radius is not accessible to me because it's position is located dead center of an existing column. What is the best and most efficient way to proceed while trying to maintain the integrity of the 25' radius? Chris Grubb === Subject: Re: When radius point is not accessible. >I work construction as a carpenter and need to layout interior >partition walls on a curve. More specifically, this curve is a 25'-0 >radius. The problem is that the focus point or pivot point to swing >the >radius is not accessible to me because it's position is located dead >center of an existing column. > What is the best and most efficient way to proceed while trying to >maintain the integrity of the 25' radius? If you know the interior colomn is circular and has a known radius r (that you might of course obtain by dividing its circumference by 2*pi if unknown) and the exterior circle has radius R>r, you might draw numerous circles of raidus R-r with center points at the outside of the middle column (which is accessible). The hull of all these radii is the circumference of the exterior circle. One may also take a tangent of any two of these auxiliary circles of Radius R-r, draw a middle line perpendicular to it (which means the new straight line intersects the tangent in the middle between the points where it touches the two auxiliary circles) and get a point on the exterior circle by moving from the perimeter of the column R-r to the outside on this line that points towards the overall center of the column. === Subject: Re: abstract algebra..... If g: F --> R is onto, actually R must be a field if F is. If r is an element of r, r = g(x) for some x in F since g is onto. R must have an identity e since F does, and g(1) = e (use e in R, 1 in F). r g(1) = g(x) g(1) = g(x*1) = g(x) = r ==> g(1) = e in R. When x not 0 in F, i.e. when x a unit, then r = g(x) is a unit in R, with r^-1 = [g(x)]^-1 = g(x^-1), since g(1) = g(x*x^-1) = g(x) g(x^-1) = e = g(x) [g(x)]^-1 Also, g(0) = 0' , where 0' is the zero in R, 0 in F. Van PS. As below, we must have R not (0), as g onto this is clearly not an isomorphism. >hello....doctor~ >suppose that >F is field. >R is ring. >g : F -> R is surjective ring homomorphism. >show that g is isomorphism. > this is not true as stated. if R = {0} > and g(x) = 0 for all x in F then g is > a surjective ring homorphism. > you need to assume that R is a ring with > identity [and also that 1 <> 0 in R, or > equivalently that R contains more than one > element - i'm not sure whether 1 <> 0 > is supposed to be part of the definition > or not, i've seen definitionds where it's > included and definitions where it's not. > of course the ring {0} is the only one > where it matters...] >---------------------------------- >we must show that one to one. >suppose that ker g =/= {0}. >there is a a such that g(a) = 0 , a =/= 0. >since F is field, there is a a^-1 >such that a*(a^-1) = 1 = (a^-1)*(a). >g(a*(a^-1)) = g(a){(g(a))^-1} = g(1) >g((a^-1)*a) = {(g(a))^-1}g(a) = g(1) >*** since g is surjective, g(1) is unity of R. *** >so, g(a) is unit. this is a contradiction.(g(a) = 0) >so, ker g = {0} >thus, g is injective. so, iso. >--------------------------------------- >um......i can't understand this part ***. > neither do i. it follows if we assume that > R has identity: > lemma: if R is a ring with 1 and ex = x > for all x in R then e = 1. > pf: 1 = e1 = e. qed. > now for ***: > given x in R there exists y in F with g(y) = x. > so > g(1)x = g(1)g(y) = g(1y) = g(y) = x. > so the lemma implies that g(1) = 1. >help me, please~ >thank you very much. > ************************ > David C. Ullrich > sorry about the inelegant formatting - typing > one-handed for a few weeks... === Subject: Re: Inverse of a block-diagonal matrix computed as I described below. I took Robin's reply as a correction and got inßuenced... :) Anyway I checked and I was right. Af course, I implemented a routine for general matrix inversion before I checked because I thought I needed it... but it can't hurt to have it. Robin, interesting reply, I must say. Elusive thus stimulative :) Cristian > how can he be right? he didn't -say- anything! > What happens when you mulitply A by your supposed Inv(A)? >>I posted without thinking. Sleep deprivation and math don't mix... >> >> Consider a block diagonal matrix like this: >> >> | M1 | >> | M2 | >> A = | ... | >> | Mn | >> >> where M1, M2 ... Mn are 3x3 non-singular matrices (and the rest of A >> is >> 0). Can the inverse of A be computed like this: >> >> | Inv(M1) | >> | Inv(M2) | >> Inv(A) = | ... | >> | Inv(Mn) | >> >> ? > Can it? > What happens when you mulitply A by your suppose Inv(A)? > Do you get the identity? > -- > 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_ > ************************ > David C. Ullrich > sorry about the inelegant formatting - typing > one-handed for a few weeks... === Subject: Re: Inverse of a block-diagonal matrix >computed as I >described below. I took Robin's reply as a correction and got inßuenced... >:) Anyway I checked and I was right. Af course, I implemented a routine for >general matrix inversion before I checked because I thought I needed it... >but it can't hurt to have it. >Robin, interesting reply, I must say. Elusive thus stimulative :) i suspect his point was to encourage you to try to think for yourself - i know that was my point in imitating his reply. >Cristian >> how can he be right? he didn't -say- anything! >> What happens when you mulitply A by your supposed Inv(A)? >I posted without thinking. Sleep deprivation and math don't mix... >> > > Consider a block diagonal matrix like this: > > | M1 | > | M2 | > A = | ... | > | Mn | > > where M1, M2 ... Mn are 3x3 non-singular matrices (and the rest of A > is > 0). Can the inverse of A be computed like this: > > | Inv(M1) | > | Inv(M2) | > Inv(A) = | ... | > | Inv(Mn) | > > ? >> >> Can it? >> >> What happens when you mulitply A by your suppose Inv(A)? >> Do you get the identity? >> >> -- >> 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_ >> ************************ >> David C. Ullrich >> sorry about the inelegant formatting - typing >> one-handed for a few weeks... ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Dimention theory Could anybody recommend any modern good book about Dimention theory ? I took a look at amazon.com and found that most books like Hurewicz Dimention theory are out of print and may be out of date ... === Subject: Re: Dimention theory > Could anybody recommend any modern good book about Dimention theory ? It is spelled dimension ;-) Here are a few references : - Engelking, Ryszard /Theory of dimensions finite and infinite/, Sigma Series in Pure Mathematics, 10. Heldermann Verlag, Lemgo, 1995. viii+401 pp. - Nagata, J. /Modern dimension theory/, Revised edition. Sigma Series in Pure Mathematics, 2. Heldermann Verlag, Berlin (1983). - Hurewicz, W. and Wallman, H. /Dimension Theory/. Princeton Mathematical Series, v. 4. Princeton University Press, Princeton, N. J. (1941). I quite liked the first one. Stephen === Subject: Re: Approximating a Polyline with another one. Das geht, Vielen Dank. === Subject: Analytic functions of two variables Some pointers, please, on how similar the theory of analytic functions in two variables is to the single variable case. In particular I'm wondering about how singularities might be distributed. As a starting point, is it true that in a power series expansion of an analytic function of two variables, say centered at the origin for simplicity, the radius of convergence is up to the nearest singularity? === Subject: Re: Analytic functions of two variables >Some pointers, please, on how similar the theory of analytic functions in >two variables is to the single variable case. Do you refer to complex-analytic functions of two or more complex variables? >In particular I'm wondering about how singularities might be distributed. >As a starting point, is it true that in a power series expansion of an >analytic function of two variables, say centered at the origin for >simplicity, the radius of convergence is up to the nearest singularity? What's the radius of convergence, here? Are you assuming that (as in complex dimension 1) the interior of the domain on which a power series converges is an open (round) disk? That's false in dimension > 1; the right generalization is polydisks (maybe I mean disc and polydisk; I can never get that right), as a little experimentation with simple power series in two variables should convince you. Leaving aside that kind of question, I think that the (fairly vast) dissimilarities between complex analysis in one or several variables are (at least morally) all consequences of the simple fact that complex-analytic varieties of dimension 0 are just collections of isolated points, whereas a complex-analytic variety of dimension 1 or more always contains non-trivial open Riemann surfaces; isolated points don't themselves, support a whole lot of interesting analysis, so they don't impose *too* much of a restriction on (for instance) some analytic function of which they are all the singularities, but of course a non-trivial Riemann surface is chock-a-block with analytical (and topological) subtleties of its own, so it can make a big contribution to the subtlety of (for instance) an analytic function of which *it* belongs to the singularities. (Here, I'm using singularities in a sense which may not be the sense you're interested in, given your question about domains of convergence. But I think my comments probably still apply.) Lee Rudolph === Subject: Re: Analytic functions of two variables >Some pointers, please, on how similar the theory of analytic functions in >two variables is to the single variable case. > Do you refer to complex-analytic functions of two or more complex variables? I was thinking about exactly two, hoping to retain some geometric intuition. >In particular I'm wondering about how singularities might be distributed. >As a starting point, is it true that in a power series expansion of an >analytic function of two variables, say centered at the origin for >simplicity, the radius of convergence is up to the nearest singularity? > What's the radius of convergence, here? Are you assuming that > (as in complex dimension 1) the interior of the domain on which a > power series converges is an open (round) disk? That's false in > dimension > 1; the right generalization is polydisks (maybe > I mean disc and polydisk; I can never get that right), > as a little experimentation with simple power series in two > variables should convince you. I'm convinced that the radius of convergence need not be isotropic. I'd hope first for a result that the domain of convergence is convex, and then perhaps that it's an ellipsoid if construed in the right way. > Leaving aside that kind of question, I think that the (fairly > vast) dissimilarities between complex analysis in one or several > variables are (at least morally) all consequences of the simple > fact that complex-analytic varieties of dimension 0 are just > collections of isolated points, whereas a complex-analytic variety > of dimension 1 or more always contains non-trivial open Riemann surfaces; > isolated points don't themselves, support a whole lot of interesting > analysis, so they don't impose *too* much of a restriction on (for > instance) some analytic function of which they are all the singularities, > but of course a non-trivial Riemann surface is chock-a-block with > analytical (and topological) subtleties of its own, so it can make > a big contribution to the subtlety of (for instance) an analytic > function of which *it* belongs to the singularities. (Here, I'm > using singularities in a sense which may not be the sense you're > interested in, given your question about domains of convergence. > But I think my comments probably still apply.) > Lee Rudolph sleep?) I suppose now I'm forced to divulge more of the motivation. I can remember my surprise as a high school student on finding out that factorial, which I'd always viewed as a essentially integer defined function, in fact has a continuous extension via the gamma function. [More recently I was surprised to discover the Windows calculator applet appears to do a nice job evaluating fractional (real) factorials consistent with that extension.] Now consider a function of two variables, the greatest common divisor. It certainly seems to be intrinsically tied to the prime structure of the integers, but we should no longer be surprised if it turned out to have a useful complex extension. Of course we should be able to interpolate a countable number of isolated points (I suspect, using a Blaschke (sp?) product), but I'm looking for something natural (or perhaps more difficult, a formulation of the nonexistence of a natural extension). [We shall not speak of a complex version of pr*me c**nting functions, for fear of awakening those who slumber.] One clue seems to be gcd(z,z) = z, at least for z = 1,2,3,... So maybe I'm looking for a monodromy theorem in two complex variables, to get a feel for what makes an analytic continuation unique. === Subject: Re: Eric Gisse, your comment was ridiculous In sci.math, eleaticus : >> As it is, I have a problem with the standard transform, and I'm >> going to have to seek help, work it out, or both. >> If one assumes that >> x' = (x-vt)/gamma >> t' = (t-vx/c^2)*gamma > That is the standard LE xform set. > What is the Ôstandard' galilean xform set that has been Ôproved' invalid so > many times? x' = x-vt t' = t > eleaticus -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Eric Gisse, your comment was ridiculous > What is the Ôstandard' galilean xform set that has been Ôproved' invalid so > many times? > x' = x-vt > t' = t When transforming coordinates in x (rather than dx) t'=t can do no harm I think. It is in transforming equations in dx that the harm is done. Which is what you are talking about in saying the galileans have been proved wrong? Right? Wrong? eleaticus === Subject: tossing coins I have a question regarding the distribution of a coin tossing experiment. Somebody actually simulated the following experiment and claims it is exponentially distributed. I would like to know if this is common knowledge, or if it is false. So I describe the experiment: I continously toss a coin with the outcome H or T. Assume also I have a subset of the integers I = {1,...,N} for some large integer constant N. I also have a set K = {N+1,..,2*N}. Each time I toss my coin repeatedly with 1 seconds inter-arrival time I do the following: - If I get a H I randomly (with uniform distribution) remove one element in K and add it to I. - If I get a T, I do the reverse, ie. I remove an element randomly (with uniform distribution} from I and add it to K. I do this for sufficiently long time (infinitely). If we count the number of seconds each integer was inside K before being removed, and plot a frequency diagram over the lifetime of each integer in K, will it be exponentially distributed? If so why? Andersen === Subject: Re: Easy number theory problem Proof of the case Z*_p is cyclic for p = 13. There are 12 = 2^2*3 elements, which are the distinct solns. of x^(p-1) - 1 = x^12 - 1 = 0. In general, p - 1 = q = p1^e1 ... pr^er. Let s^e = pi^ei for some i. Let y = x^s (s = pi for some i). x^(s^e) - 1 = y^(s^(e-1)) - 1 = 0 has s^(e-1) distinct solns. with o(y) <= s^(e-1), and s^e - s^(e-1) = s^(e-1)(s - 1) with o(x) > s^(e-1). Since o(x) | s^e, there are s^(e-1)(s - 1) solns with o(x) = s^e. Thus there is an element of order pi^ei for each i, and their product is an element of order q = p1^e1 ... pr^er, which generates the whole group which must be cyclic. Case of p = 13; For s^e = 2^2 = 4 y^4 - 1 = 0 has 4 solns ==> 4 elements with orders <= 4 ==> 12 - 4 = 8 elements with orders > 4. It will turn out that we have Z_12, and the 4 with order <= 4 are (0,3,6,9) in additive notation. Since 2^6 mod 13 = 64 mod 13 = -1 mod 13, Z*_13 = <2>, and we could take these 4 elements to be the (0,3,6,9)th powers of 2. If we let s^e = 3^1 = 3, we have x^3 - 1 = 0 ==> 3 elements with orders <= 3. ( Z_12 = Z_3 x Z_4. ) The product of an element of order 3 with one of order 4 is one of order 12, which generates the whole group, so it must be cyclic. Consider the group Z*_8 =~ Z_2 x Z_2 (not cyclic). Where does the argument fail--where does the fact these are not the nonzero elements of a field enter. I guess we can't say that the elements have x^4 - 1 = 0 as a splitting field, even though it is true that phi(8) = 4, and x^4 = 1. (Actually x^2 = 1). Any comments on this part of it? Van === Subject: Re: Counting Magic Squares Full discussion can be viewed at > I have found that the number of magic squares of order 4 can be > expressed as 220 basic solutions, each with 32 rotations and > permutations, with reßection as a special case of the latter. The > value of 32 for order 4 can be generalized as a function of n: [n/2]! > 2*([n/2]+2) > > Is this a new result? If not why do web sites continue to assert that > there are 880 solutions for order 4, with 8 rotations and reßections > for each? > > I have a proof, in MS Word, with subscripts and superscripts. Any way > that I can post it here? > I don't know whether it is a new result. It doesn't seem to contradict > what the websites say, so whether it's new or not it doesn't seem to be > a reason for websites to stop saying what they're saying. > DON'T post a Word document here. Best to put it up on the web > somewhere & post the URL here. Or learn TeX, post it as a (plain > text) TeX file, and anyone who wants to will be able to typeset it > and read it. === Subject: HELP - Hartshorne - Projective Varieties I am hopelessley stuck on a proof in Hartshorne. Could some one provide a more easy to read commentary? I do not understand Proposition 2.2 (page 10) esp. his definition of alpha and beta. N.C. === Subject: Irrational numbers. Mail-To-News-Contact: postmaster@nym.alias.net +..-5*0,5-46*0,1666..-119*0,04166..+..+(+..+12*0,5+9*0,166.- 120*0,04166.+... +.)=? exersice === Subject: JSH: Your fantasy world problem One of the things I've noticed over the years is that posters who many of you seem to admire or look up to for some reason get away with rather ludicrous lies about math itself, which lead me to realize that for many of you mathematics is just some kind of social game. You care more about the social aspect than the actual facts, but I'm going to demonstrate to you why your ignorance of mathematics as more than a social game is so misplaced in this subject area. You see, when you take a false position in mathematics it doesn't matter how many people side with you or how long you lie about it as you're just wrong. Eventually some people come along who care more about the truth than your lies as, to survive in the REAL world, you can't just use fantasy. Like some sci.math poster might believe that if they wish really hard bullets aren't real, but if someone comes up with a gun and shoots them, then it doesn't matter how many other sci.math'ers were cheering them and their particular fantasy on, as the real world doesn't give a rat's ass what you think. And whether you realized it or not, mathematics is part of the real world. It's not just your own little private fantasy area where your gang can get together and just believe what it wants. You don't get to pick and choose what's true mathematically, and it doesn't matter how much a particular person pisses you off, the MATH DOES NOT CHANGE!!! Now I'm not a nice guy. And I think many of you are immature both intellectually and emotionally with weird fixations and a tendency towards acting out in hostile group behavior. But what I hate about you is your weird ability to believe false things that go against even basic mathematics as long as it makes you feel good and that's what you think your group expects. To think in the supposedly modern world the field of mathematics has somehow been taken over by these odd social creatures who seem to think it matters if people LIKE you!!! Gauss would probably just shut himself away somewhere and never talk to the public again if he were faced with a crop of mathematicians like you people. Math is not and never has been a social activity. If that sounds weird to you, or ludicrous, and you find yourself jumping up in frustration and anger that I'd ever say such a thing, then you're not a mathematician. You may be called a mathematician. You may call yourself a mathematician. But if you think math is a social activity then you're NOT a mathematician, not really. Mathematicians never pause to see what the group thinks, or to check and see if the result in front of them was discovered by this person or that as what's more important is whether or not it's true. And truth in mathematics has nothing to do with people. It has nothing to do with who you know. It has nothing to do with who likes you. All of you can really, really, really just not like me, and it doesn't matter mathematically, but for sci.math posters, as I've heard it time and time again over the years that I've been posting, liking is more important than the math. People here need to like you or they lie about math. I say, hey, if you don't like someone that much IGNORE THEM but don't lie about the math! You people lie about math and cheer each other for doing it! You have little respect for algebra and act like algebraic properties and results depend on opinion polls! You are not mathematicians, not really. Mathematicians are not social creatures. They don't need opinion polls. They are not swayed by group dynamics. They do not check first to see if they LIKE someone before they'll believe correct results. You see, real mathematicians actually like math. They like its purity. They like its infallibility. They like its perfection. You people may call yourselves whatever, but you are not mathematicians, as if you were you'd never lie about the math. James Harris === Subject: Re: JSH: Your fantasy world problem > One of the things I've noticed over the years is that posters who many > of you seem to admire or look up to for some reason get away with > rather ludicrous lies about math itself, which lead me to realize that > for many of you mathematics is just some kind of social game. It is James playing social games that fouls this NG. === Subject: Re: JSH: Your fantasy world problem > One of the things I've noticed over the years is that posters who many > of you seem to admire or look up to for some reason get away with > rather ludicrous lies about math itself, which lead me to realize that > for many of you mathematics is just some kind of social game. > It is James playing social games that fouls this NG. I think he keeps things interesting. === Subject: Re: JSH: Your fantasy world problem [snip repetitive self-serving monologue] James, You seem to see yourself as some kind of brilliant prophet casting pearls among swine. Others see you as a boorish partycrasher dropping turds in the punchbowl. With this disparity in perceptions, no meeting of the minds is possible. But here's a suggestion: If you really think mathematicians are mindless herd animals and physicists are innovative discoverers, why not deposit your offerings at Ôsci.physics' instead of Ôsci.math'. I'm sure there are kindred spirits there who will welcome your presence. Furthermore, if you don't switch newsgroups, it suggests that among your other character ßaws you are also a hypocrite. > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Your fantasy world problem >[snip repetitive self-serving monologue] >James, >You seem to see yourself as some kind of brilliant prophet casting pearls >among swine. Others see you as a boorish partycrasher dropping turds in >the punchbowl. >With this disparity in perceptions, no meeting of the minds is possible. >But here's a suggestion: >If you really think mathematicians are mindless herd animals and >physicists are innovative discoverers, why not deposit your offerings at >'sci.physics' instead of Ôsci.math'. I'm sure there are kindred spirits >there who will welcome your presence. Furthermore, if you don't switch >newsgroups, it suggests that among your other character ßaws you are also >a hypocrite. NO. His does NOT belong in sci.physics. >> James Often in error, but never in doubt! Harris >There are two things you must never attempt to prove: the unprovable -- >and the obvious. >Democracy: The triumph of popularity over principle. === Subject: Re: JSH: Your fantasy world problem === >Subject: JSH: Your fantasy world problem >One of the things I've noticed over the years is that posters who many >of you seem to admire or look up to for some reason get away with >rather ludicrous lies about math itself, which lead me to realize that >for many of you mathematics is just some kind of social game. >You care more about the social aspect than the actual facts, but I'm >going to demonstrate to you why your ignorance of mathematics as more >than a social game is so misplaced in this subject area. >You see, when you take a false position in mathematics it doesn't >matter how many people side with you or how long you lie about it as >you're just wrong. >Eventually some people come along who care more about the truth than >your lies as, to survive in the REAL world, you can't just use >fantasy. >Like some sci.math poster might believe that if they wish really hard >bullets aren't real, but if someone comes up with a gun and shoots >them, then it doesn't matter how many other sci.math'ers were cheering >them and their particular fantasy on, as the real world doesn't give a >rat's ass what you think. Are you making death threats again? >And whether you realized it or not, mathematics is part of the real >world. >It's not just your own little private fantasy area where your gang can >get together and just believe what it wants. >You don't get to pick and choose what's true mathematically, and it >doesn't matter how much a particular person pisses you off, the MATH >DOES NOT CHANGE!!! >Now I'm not a nice guy. Yeah, everyone's noticed that. If you recognize the problem, why aren't you taking steps to correct it? >And I think many of you are immature both >intellectually and emotionally with weird fixations and a tendency >towards acting out in hostile group behavior. And even if that were true, it changes nothing about your work. >But what I hate about you is your weird ability to believe false >things that go against even basic mathematics as long as it makes you >feel good and that's what you think your group expects. Maybe if you stopped hating and started accepting people for what they are you would get along better. >To think in the supposedly modern world the field of mathematics has >somehow been taken over by these odd social creatures who seem to >think it matters if people LIKE you!!! Have you ever considered that if you got people to like you, then they would publish your manuscripts after you've lost your mind? >Gauss would probably just shut himself away somewhere and never talk >to the public again if he were faced with a crop of mathematicians >like you people. >Math is not and never has been a social activity. But having friends has more advantages than having enemies. >If that sounds weird to you, or ludicrous, and you find yourself >jumping up in frustration and anger that I'd ever say such a thing, >then you're not a mathematician. And you're not an acedemic either, or you would understand such things. >You may be called a mathematician. You may call yourself a >mathematician. >But if you think math is a social activity then you're NOT a >mathematician, not really. >Mathematicians never pause to see what the group thinks, or to check >and see if the result in front of them was discovered by this person >or that as what's more important is whether or not it's true. >And truth in mathematics has nothing to do with people. But it does have to do with getting your work published. >It has nothing to do with who you know. Unless it's a journal editor. >It has nothing to do with who likes you. Unless the prople who don't like you e-mail the journal editor. >All of you can really, really, really just not like me, and it doesn't >matter mathematically, but for sci.math posters, as I've heard it time >and time again over the years that I've been posting, liking is more >important than the math. >People here need to like you or they lie about math. No, they lie about you. >I say, hey, if you don't like someone that much IGNORE THEM Sorry, but ignoring you isn't as much fun. >but don't lie about the math! >You people lie about math and cheer each other for doing it! >You have little respect for algebra and act like algebraic properties >and results depend on opinion polls! >You are not mathematicians, not really. Well, _I_ never claimed to be one. >Mathematicians are not social creatures. They don't need opinion >polls. They are not swayed by group dynamics. >They do not check first to see if they LIKE someone before they'll >believe correct results. And when they don't believe the results, they yank the paper. >You see, real mathematicians actually like math. So do some non-mathematicians. >They like its purity. They like its infallibility. They like its >perfection. Jealous, eh? >You people may call yourselves whatever, but you are not >mathematicians, as if you were you'd never lie about the math. >James Harris -- Mensanator Ace of Clubs === Subject: Re: Automorphisms of groups different fonts, hide and don't hide quotes, etc., in posts now. It has problems with sci.math because its auto-formatter doesn't know how to deal with the ASCII math symbol conventions here.) === > Subject: Automorphisms of groups >I want to review the related problems of isomorphism of Z*_mn, >with (Z*_m,Z*_n) when (m,n) = 1, and Euler phi(mn) = phi(m) phi(n). >Consider the automorphisms of the Z_n, A(Z_n) >products of the Z_n, and of the Z*_n groups. >Z_2 has only the identity group (1). > Sloppy. Do you mean A(Z_2) = { I } ? Yes. >In general, I recall that Z*_n = A(Z_n), and Z_*2 = (1). >Z*_3 = (-1,1) =~ Z_2 ; Z*_4 = Z_2 = (1,3) = (-1,1) > Interesting enuf. This has something to do with the phi function? Yes, but I was also going off in another direction--but I just started wondering all over the place with no direction. Sorry, I _am_ getting sloppy. > When m,n coprime, Z_mn = Z_m x Z_n; Thus > Z_mn^* = (Z_m x Z_n)^* =??= Z_m^* x Z_n^* > Whence viole, phi(mn) = phi(m) phi(n) > Perhaps Z_n^* = A(Z_n) is useful for this. > What about phi(p^n) = p^n - p^(n-1) ? > Back to the old fashion way of counting? > Nay, tell me Ôtis not so. No. How about |Z_p^n| = p^n, and A = {k such that p | k and k in Z_p^n} = {k | (k,p) = p and k in Z_p^n} = {qp | q = 1,2, ... ,p^(n-1)}, so |A| = p^(n-1) and |Z*_p^n| = phi(p^n) = |Z_p^n| - |A| = p^n - p^(n-1) . Van === Subject: Re: abstract algebra..... What about R = (0) with g(1) = 0 = g(x) for all x in F ? Then we could say that e = 0 in R. Is the trivial ring R = (0) also a trivial field with 1 = 0? I haven't thought about this case much. I usually just say this is the trivial case, but the trivial group G = (e) really isn't trivial in the sense that its not important--its the bottom of series of normal subgroups, e.g., so I suppose one should be careful with R = (0) too. R* = R0 = null set in this case, but so what? You are right to point out that the statement is not true in this case. Van >hello....doctor~ > >suppose that >F is field. >R is ring. >g : F -> R is surjective ring homomorphism. > >show that g is isomorphism. > this is not true as stated. if R = {0} > and g(x) = 0 for all x in F then g is > a surjective ring homorphism. > you need to assume that R is a ring with > identity [and also that 1 <> 0 in R, or > equivalently that R contains more than one > element - i'm not sure whether 1 <> 0 > is supposed to be part of the definition > or not, i've seen definitionds where it's > included and definitions where it's not. > of course the ring {0} is the only one > where it matters...] >---------------------------------- >we must show that one to one. > >suppose that ker g =/= {0}. > >there is a a such that g(a) = 0 , a =/= 0. > >since F is field, there is a a^-1 >such that a*(a^-1) = 1 = (a^-1)*(a). > >g(a*(a^-1)) = g(a){(g(a))^-1} = g(1) >g((a^-1)*a) = {(g(a))^-1}g(a) = g(1) > >*** since g is surjective, g(1) is unity of R. *** > >so, g(a) is unit. this is a contradiction.(g(a) = 0) > >so, ker g = {0} > >thus, g is injective. so, iso. > >--------------------------------------- >um......i can't understand this part ***. > neither do i. it follows if we assume that > R has identity: > lemma: if R is a ring with 1 and ex = x > for all x in R then e = 1. > pf: 1 = e1 = e. qed. > now for ***: > given x in R there exists y in F with g(y) = x. > so > g(1)x = g(1)g(y) = g(1y) = g(y) = x. > so the lemma implies that g(1) = 1. >help me, please~ > >thank you very much. > > ************************ > David C. Ullrich > sorry about the inelegant formatting - typing > one-handed for a few weeks... === Subject: Lacking Concept Of Numbers >= 3 Did you hear about this Amazon tribe, the Piraha?: Their language lacks words for any specific numbers >= 3. Even though they supposedly are generally intelligent and understand such mathematical concepts such as catagory, they cannot easily grasp the idea of specific integers much above 3. And unlike Piraha adults, the Piraha children did not have difficulty understanding larger numbers. Makes me wonder how much of what WE know (know) is only a direct consequence of our language and experience and genetics. Leroy Quet === Subject: Re: Lacking Concept Of Numbers >= 3 > Did you hear about this Amazon tribe, the Piraha?: > Their language lacks words for any specific numbers >= 3. > Even though they supposedly are generally intelligent and understand > such mathematical concepts such as catagory, they cannot easily grasp > the idea of specific integers much above 3. > And unlike Piraha adults, the Piraha children did not have difficulty > understanding larger numbers. > Makes me wonder how much of what WE know (know) is only a direct > consequence of our language and experience and genetics. > Leroy Quet include all the numbers >= 3 I suppose. -- The life of a repoman is always intense! === Subject: Re: Recurrence for prime counting theorem > This may be a FAQ, but I couldn't find it mentioned anywhere on > mathworld.com. > The partition function P(n) satisfies the recurrence relation P(n) - > P(n-1) - P(n-2) + P(n-5) + P(n-7) - P(n-12) - P(n-15) + ... = 0, where the > sum is over generalized pentagonal numbers. > Are there any similarly wonderful known recurrence relations for pi(x), the > number of primes less than x? > -Michael. I bet there must be something to interest you at: http://mathworld.wolfram.com/RiemannPrimeCountingFunction.html Leroy Quet === Subject: Re: LinAlg: Vector Spaces - need help understanding an example > Yes, I have no problem accepting this, but a few questions do pop up > in my head; the sum (or any linear combination, right?) of two > vectors contained in any (not just the ones passing through the > origin) line, plane or k-ßat is a new vector which also is > contained in the same k-ßat, right? > Wrong. In two dimensions, consider the line y = 1, which does not pass > through origin. (2,1) and (3,1) are on that line, but (2+1,1+1) is not. That's true and after spending some more time thinking about this I think I finally begin to understand what caused my confusion. While I was thinking about vectors with their whole bodies in the line/plane/k-ßat (like a direction vector for a line) all of you helping me here were talking about the vectors in the solution set making up the k-ßat. For me, the notion of adding solutions were new and that's why I started asking about the intuition behind it. I still lack the intuition behind why one would add solutions, but I guess this will start making sense to me as I continue my studies. So, in your example with the line y = 1, I was thinking about vectors like (3, 0) and (5, 0) and their sum (8, 0). All of these have their whole bodies in the line and that's why it sounded so strange to me that only lines etc. passing through the origin had this property. After reviewing a previous section in the book I use, I now understand that the vectors I was thinking about actually are the difference between two particular solutions and that such a vector always is a solution of the corresponding homogenous system, which is quite interesting! This in turn explains why those vectors can be added; the solution set of a homogenous linear system is a vector space. > There is a space like the one that you are considering, but it is > called an affine space, not a vector space, and is not closed under > addition. Interesting. Perhaps I will stumble upon them in Linear Algebra? Otherwise I shall try to find the time to look into them as well once I have mastered LA, if ever! ;-) > Is it this fact that makes the homogenous equations special, i.e. that > the solution set is made up of vectors that are completely contained in > the k-ßat which the solution set is made up of? > K-ßat through origin, yes. The part in the defintion of a vector space which required there to be a zero vector in the set was the one I had most trouble understanding, which probably is caused by my limited general mathematical knowledge. At least when seen from the point of k-ßats - I just could not understand what was so magical about the k-ßats passing through origin. Now that I finally have managed to get rid of my previous notion - or even mental block - about which vectors that are considered here, this makes perfect sense to me! Sure, there's a long way left to go, but now it seems as if I'm back on the right track again! *smile* /Henrik === Subject: Re: nitpicking in the complex plane? there isn't such a figment of our lazy imaginations! In sci.math, David Bandel > for a few years now, the less educated have run rampant in > mathematical circles, scribbling i's and a+bi's on classroom > chalkboards. it's time to open our eyes ladies and gentlemen. if you > can equate the square root of -1 to i. why not equate the meaning of > life to x? then x can be whatever we want. so if we say x is not > the meaning life, then we have a contradiction. mathematicians have > become lazy as of late, when running into dead ends, all too eager to > just invent a whole new category of existence containing an element > that proposes a solution. let me iterate: > ^ > |3 > |2 > |1 > -3i 2i -i | i 2i 3i > <-----------+-----------> > |-1 > |-2 > |-3 > | > v > WHERE INGENUITY COULD NOT PREVAIL. show me a negative anything and > i'll show you a crook.. and that crook would be YOU Ingenuity? I'm not sure what you're getting at here. Of course, one reason why complex numbers are often used in such fields as electrical small-signal analysis is that they're *useful* as a shorthand. But we can go simpler than that. I have two apples. You want three. Do I: [1] give you the two apples and owe you one later? [2] magically conjure up an anti-apple and apple pair, and give you the apple and keep the anti-apple somehow? level, although for apples it's fairly ridiculous.) [3] Slice one of the apples in half and give you three pieces? [4] Wait until next growing season? Presumably, negative numbers were initially used to represent bar tabs. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: JSH: Weird group It amazes me how weird sci.math'ers are and it's clearly some weird social thing where many of you are simply abnormal when it comes to even basic social function. Like look at the David Ullrich thing. The record is that several years ago I started trying to ignore David Ullrich who repeatedly and obsessively replies to my posts. David Ullrich has even gone so far as to go to other newsgroups like alt.writing among others to keep replying to my posts. That's stalking behavior. BUT he also plays the victim claiming I tried to get him fired from his job as a university professor by reporting on his racial slur comment. The real story is that after I said that he'd acted as my lapdog in an instance, I was sufficiently worried by David Ullrich's posts that I thought he might be a definite weirdo so I apologized and tried to avoid his posts. But he stalked me on the newsgroup, continuing to reply to me, and I ignored him for over a year. More than a year past by, and in one of his stalking posts he said that he'd thought about delivering a racial slur as an appropriate reply to my saying he'd acted as my lapdog in an instance--over a freaking year earlier!!! No doubt about it, David Ullrich is clearly a stalker weirdo. At that point I decided that it'd be a good idea to report his PUBLIC statements to his university. And he began using that to play victim. And he kept obsessively replying to me. And he kept playing victim. And sci.math posters apparently just ate it up, and my guess is that for many of you all that matters is that David Ullrich is a math professor and I'm not as you're programmed to follow math professors as obedient students. You people are followers. It doesn't seem to matter to you what the facts are, as you just follow your social order. You're like an ant farm or something. By now the ratio of replies of David Ullrich to posts of mine versus my replies to him is got to be well over 2 to 1, which should be enough for RATIONAL people to know who the weirdo is here. But maybe the reality of sci.math is that most of you are weirdos like Ullrich, so his stalking behavior just seems normal to you! However, most people around the world would find it extraordinary that a professor would, after making comments like his talk about racial slurs, keep after a person who'd felt compelled by his behavior already to make a complaint! Stalkers get complained about, and then they try to further victimize their victims by themselves playing the victim. On sci.math, David Ullrich found a ready crew of believers to help him in his game which continues to this day, as he STILL tries to get my attention. But to sci.math'ers, he's just behaving normally, eh? James Harris === Subject: Re: JSH: Weird group > It amazes me how weird sci.math'ers are and it's clearly some weird > social thing where many of you are simply abnormal when it comes to > even basic social function. > Like look at the David Ullrich thing. > The record is that several years ago I started trying to ignore David > Ullrich who repeatedly and obsessively replies to my posts. > David Ullrich has even gone so far as to go to other newsgroups like > alt.writing among others to keep replying to my posts. > That's stalking behavior. > BUT he also plays the victim claiming I tried to get him fired from > his job as a university professor by reporting on his racial slur > comment. > The real story is that after I said that he'd acted as my lapdog in an > instance, I was sufficiently worried by David Ullrich's posts that I > thought he might be a definite weirdo so I apologized and tried to > avoid his posts. > But he stalked me on the newsgroup, continuing to reply to me, and I > ignored him for over a year. > More than a year past by, and in one of his stalking posts he said > that he'd thought about delivering a racial slur as an appropriate > reply to my saying he'd acted as my lapdog in an instance--over a > freaking year earlier!!! > No doubt about it, David Ullrich is clearly a stalker weirdo. > At that point I decided that it'd be a good idea to report his PUBLIC > statements to his university. > And he began using that to play victim. > And he kept obsessively replying to me. > And he kept playing victim. This post is a textbook case of a mind disconnected from what the rest of us consider reality. It's disturbing. === Subject: Re: JSH: Weird group [...] >And he began using that to play victim. >And he kept obsessively replying to me. >And he kept playing victim. If you had to teach math at Oklahoma State University wouldn't you feel like a victim too? J.9frgen [...] >James Harris === Subject: Re: JSH: Weird group [snip off-topic paranoid rambling] Your status as a crank has been conclusively demonstrated. Now you apparently seek the same level of proof that you are a troll. Stick to the math -- jerk! > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Weird group Discussion, linux) > More than a year past by, and in one of his stalking posts he said > that he'd thought about delivering a racial slur as an appropriate > reply to my saying he'd acted as my lapdog in an instance--over a > freaking year earlier!!! > No doubt about it, David Ullrich is clearly a stalker weirdo. > At that point I decided that it'd be a good idea to report his PUBLIC > statements to his university. Now, that's odd. You make a big to-do over the fact that David Ullrich didn't mention his passing urge to make a racist retort for a whole year. Your <7s8n87$adn$1@nntp5.atl.mindspring.net>). His comment which included mention of a racial slur was in fact even later than you say, Now, *at that point* you decided it was a good idea to report his comments. But the earliest evidence that you actually did email OkState seems to February, 2002. Half a year later. Busy, were we? -- We are happy that you agree that customers need to know that Open Source is legal and stable, and we heartily agree with that sentence of your letter. The others don't seem to make as much sense, but we find the dialogue refreshing. -- Linus Torvalds to Darl McBride === Subject: Re: JSH: Weird group >It amazes me how weird sci.math'ers are Then why do you keep posting to this newsgroup? If you truly believe we are all so unreasonable, why try to reason with us? >But he stalked me on the newsgroup, continuing to reply to me, Imagine that... someone replying to you... on a newsgroup... if that isn't a sign of madness I don't know what is... :) >And sci.math posters apparently just ate it up, and my guess is that >for many of you all that matters is that David Ullrich is a math >professor and I'm not as you're programmed to follow math professors >as obedient students. Again, suppose for a moment that this is true. What is your point for posting here? If this is true, we won't listen to you (these are your own thought on this). Your efforts are obviously wasted... why even bother? === Subject: Statistical Time Series Analysis Toolbox Introductory Offer Harmonic Software Inc. has recently released STSA The Statistical Time Series Analysis Toolbox. This This new toolbox provides an extensive collection of functions for performing time series related analysis and visualization. The STSA toolbox includes a broad range of capabilities for ARMA and ARFIMA, Bayesian, non-linear and spectral analysis related models. For a limited time, Harmonic Software is offering the new STSA toolbox for almost 50% off the regular license price. To obtain this discount, visit our ÔSTSA Special Offer' page at: https://ssl.adhost.com/omatrix/stsaoffer.html Detailed product information for STSA is available on the Harmonic Software Inc. home page, http://www.omatrix.com/stsa.html === Subject: Re: Statistical Time Series Analysis Toolbox Introductory Offer Is there a free version of this somewhere? > Harmonic Software Inc. has recently released STSA > The Statistical Time Series Analysis Toolbox. This > This new toolbox provides an extensive collection of > functions for performing time series related analysis > and visualization. The STSA toolbox includes a > broad range of capabilities for ARMA and ARFIMA, Bayesian, > non-linear and spectral analysis related models. > For a limited time, Harmonic Software is offering the new > STSA toolbox for almost 50% off the regular license price. > To obtain this discount, visit our ÔSTSA Special Offer' page at: > https://ssl.adhost.com/omatrix/stsaoffer.html > Detailed product information for STSA is available on the > Harmonic Software Inc. home page, http://www.omatrix.com/stsa.html === Subject: Re: Statistical Time Series Analysis Toolbox Introductory Offer The toolbox is not free, but we have tried to keep the price very reasonable with the offer at: https://ssl.adhost.com/omatrix/stsaoffer.html The price is a small fraction of other commercial options, but still comes with complete product support. Beau Paisley Harmonic Software > Is there a free version of this somewhere? > Harmonic Software Inc. has recently released STSA > The Statistical Time Series Analysis Toolbox. This > This new toolbox provides an extensive collection of > functions for performing time series related analysis > and visualization. The STSA toolbox includes a > broad range of capabilities for ARMA and ARFIMA, Bayesian, > non-linear and spectral analysis related models. > For a limited time, Harmonic Software is offering the new > STSA toolbox for almost 50% off the regular license price. > To obtain this discount, visit our ÔSTSA Special Offer' page at: > https://ssl.adhost.com/omatrix/stsaoffer.html > Detailed product information for STSA is available on the > Harmonic Software Inc. home page, http://www.omatrix.com/stsa.html === Subject: Re: Normal = Elliptic integral 1st kind? If I understood you correctly, to find arc length in meridional plane, we integrate using rhoM and phi. To find arc length L of a circumferential circle desribed by cone tip of slant height rhoN and an azimuth angle = al, no integration is needed, it is a cone developed as sector, L= rhoN*al*sin(phi). (Curvature of azimuth etc. in part III is applicable to sections that are between meridional and circumferential) > Okay, integrand is probably the word I mean. >As for the rest, you've only managed to confuse me more. Where did >meridians, lattitudes, radii of curvature, and the variables a, x, >and e come from? > Here is a typical elliptic integral page: > http://www.efunda.com/math/elliptic/elliptic.cfm > and here is a radius of curvature page (see part III): > http://www.oc.nps.navy.mil/oc2902w/geodesy/radiigeo.pdf > m (you) = k (efunda) = e (Clynch) = sqrt(a^2-b^2)/a (eccentricity) > To find the perimeter of an ellipse you use radius a (so since the > two integrals use the same elements I would guess a is used for the > integral of the first kind too). > So it would seem that the integrand used in the integral of the > first kind is the same integrand used to find the radius of > curvature in the prime vertical. > So if you wanted to know the prime vertical distance of an ellipse > isn't the prime vertical integral the elliptic integral of the > first kind? You could say it is the perpendicular circumference of > an ellipse! === Subject: Cardinal Exponentiation For an ordinal a, let a+ and cf(a) denote respectivel the smallest cardinal greater than a and the confinality of a. Let k and m be cardinals greater than 1 with at least one of them infinite. Kunen proves the following theorem of ZFC+GCH. 1) If k <= m, then k^m = m+. 2) If cf(k) <= m < k, then k^m = k+. 3) If m < cf(k), then k^m = k. Studying the proof, it seems to me that without GCH, one can show 1) If k <= m, then k^m = 2^m. 2) If cf(k) <= m < k, then k^m = 2^k. However, if m < cf(k), it seems the best you can say is k <= k^m <= 2^k. Is this correct? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Uncountable sets in CZF? > plagiarize from Cantor? Did Gauss use a diagonal argument? Inquiring > minds want to know. >Is this not true? > Is it not irrelevant to Non-Euclidean Geometry? >Is Loewenheim-Skolem nonsense? >How do you explain Banach-Tarksi? > To whom? Euclid? >I'll be happy to assist you in explaining Banach-Tarski. > to read. I suggest that you read the text that you quoted from > from the hip; I quoted, and commented on, one sentence, and none of > your attempts at sarcasm have anything to do with that one sentence. > And BTW, the misspelling of Bolya is what is called a pun, which is a > form of humor[1].HTH. HAND. FOAD. > [1] Look it up in a dictionary if you are not familiar with the > concept. Hi Shmuel, It's nice of you to resort to a word that doesn't exist except of your hinting that it was a pun on Bolyai, because otherwise it means nothing. I had not heard too much of that fellow, you seem to refer to Janos Bolyai, who independently developed some aspects of hyperbolic (non-Euclidean) geometry in the milieu of Gauss, where Gauss himself had his own notions on the matter of non-Euclidean geometries. Both Gauss and Bolyai used the infinitesimal analysis. So, by your pun, are you trying to imply that you've heard of someone else much farther along in this line of argument, or that they never heard of me, or you, or vice versa? Have you? Please let me know if there is a rabid camp of Skolemites, and if so how to contact them. This is mathematics: humor is irrelevant, although often appreciated, and sarcasm is a lie. Do you think you're funny? It may be rude of you to say that others aren't. I want to tell you that you have some spelling errors on your web page, abd and expereience. That's not relevant to this argument, syntax errors, but you might want to correct them. Your misspellings a correctly-spelled Hebrew word, because I'm naive and trusting. Humor is basically an implicit or explicit truth. Then again, some people have a sixth sense of humor. I generally don't use sarcasm, as you might know from many postings to sci.math about humor. Now, that hasn't been too fair because you've probably said something that was funny in your life, it would be rude of me to suggest otherwise. Shmuel, 2+2=4, and in a model of ubiquitous ordinals, infinite sets are equivalent. Loewenheim-Skolem says what I said it does above. Have your theory: in a post-Cantorian theory you are allowed to consider a hypothesis where infinite set maps, 1-1, onto, and bijectively, to powerset. Then you can forget about it. http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/ Lowenheim.html Loewenheim is remembered for the Loewenheim-Skolem paradox (which Skolem pointed out is not a paradox!) which produces non-standard models, for example a denumerable model of the reals. Translation: Loewenheim says the reals are countable. I wish I'd known about this before, I could have invoked Loewenheim-Skolem instead of independently developing appropriate notions myself, for several of the contentious issues we discuss. Now I won't deny that they say so, unless necessary. EF is the primitive function that meets requirements to be a bijective mapping between the naturals and reals, in extension of Cantor. http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/ Skolem.html Now those are some logicians I appreciate: debunking paradoxes. So, if we're talking about Loewenheim and Skolem, we either agree or don't that they might agree. What do your acronyms mean there? Never mind... Hey where's this from: They laughed at me when I told them I'd become a great comedian. Nobody laughs at me now. Good luck finding someone literate. Ross F. http://www.tiki-lounge.com/~raf/ === Subject: Re: Cayley's theorem === Subject: Re: Cayley's theorem > Cayley's theorem says that a finite group is a subgroup of a symmetric > group. > Does anyone know if there is an algorithm which determines the smallest > integer such that a finite group G is a subgroup of Sn ? To add to what others, particularly Dr. Holt, have said: One lower limit that's sometimes useful: If g_i are the elements of G and |g_i| = prod_j (q_ij) is the prime power factorization of |g_i|, then n >= max(sum_j (q_ij)). A coarser version of the preceding: If q is the largest prime power such that G has an element of order q, then n >= q. Coarser still, applying one of Lagrange's Theorems to the preceding: If p is the largest prime factor of |G|, then n >= p. -- Jim Heckman === Subject: Re: Easy number theory problem === > Subject: Re: Easy number theory problem >I want to review the related problems of the CRT, >isomorphism of Z_mn with (Z_m,Z_n) when (m,n) = 1, >CRT: (m,n) = 1 and x in Z_m, y in Z_n ==> there is a unique >z in Z_mn such that z = x mod m and z = y mod n. >Proof. rm + sn = 1. Mult by y - x. rm(y - x) = y - x - sn(y - x) or >let z = x + rm(y - x) = y - sn(y - x) > As m,n coprime, some a,b with am + bn = 1 > bn = 1 (mod m); am = 1 (mod n) > sam + rbn = r (mod m); sam + rbn = s (mod n) >Then z = x mod m and z = y mod n is a soln. >If z' is another soln., then z' = z mod mn, since both m and n >must divide z' - z, and m | z' - z ==> z' - z = k m, >n | km ==> n | k since (n,m) = 1. Thus z' = z mod mn. >Define f : Z_mn --> Z_m x Z_n : z --> f(z) = (z mod m, z mod n) > This is nonsense except to computer folks as the operation z mod m isn't > defined mathematically, only the relation z = m (mod m). No, computer > talk ain't mathematics. What's worse is computer implementation of x mod > y may differ in different computer languages. Yes. Sometime I think perhaps it would be best to write using the ideals mZ == (m). > The actual bijection you want is > f:Z_mn -> Z_m x Z_n, z + mnZ -> (z + mZ, z + nZ) > which you need to show is well defined. > If x + Z_mn = y + Z_mn, then x = y (mod mn), > x = y (mod m), x = y (mod n); x + mZ = y + mZ, x + nZ = y + nZ > (x + mZ, x + nZ) = (y + mZ, y + nZ) > Thus it's well defined. or x + (mn) = y + (mn) ==> x - y in (mn) = mnZ = mZ/nZ . so x - y in mZ and nZ or (x + mZ, x + nZ) = (y + mZ, y + nZ) so well defined. > As m,n coprime, some a,b with am + bn = 1 > bn = 1 (mod m); am = 1 (mod n) > sam + rbn = r (mod m); sam + rbn = s (mod n) > f(sam + rbn) = (r,s) in Z_m x Z_n > Thus f surjection. > Rest is direct. f is ring homomorphism > f(x+y + mnZ) = (x+y + mZ, x+y + nZ) = (x+mZ + y+mZ, x+nZ + y+nZ) > = (x + mZ, x + nZ) + (y + mZ, y + nZ) = f(x) + f(x) > f(xy + mnZ) = (xy + mZ, xy + nZ) > = ((x + mZ)(y + mZ), (x + mZ)(y + nZ)) > = (x + mZ, x + nZ)(y + mZ, y + nZ) = f(x).f(y) > If f(z) = (0,0), then z = 0 (mod m), z = 0 (mod n), > some k with z = km; n|z; n | km; m,n coprime; n | k > nm | km; nm | z; z = 0 (mod nm) > Thus ker f = { 0 }; f injection. Yes, this is the best way. CRT + ker f = 0 (and f a ring homo), so f 1-1 and onto, of course |Z_mn| = mn = |Z_m x Z_n|, so 1-1 ==> onto for finite sets. > -- >Now to extend this to Z*_m, etc. In this case we only have >the elements such that (x,m) = 1, etc. I've never done or seen >this before. > La-de-da. First make clear, precise, unambiguous statement of what you > are wanting to prove. Now I want to show that f extends to a group isomorphism of Z*_mn with Z*_m x Z*_n. This should be straightforward it seems, though I have never done it. i.e. to show f : Z*_mn --> Z*_m x Z*_n : f(z + mnZ) = (z + mZ, z + nZ) with (m,n) = 1 and (z,m) = (z,n) = (z,mn) = 1. I tried to do something like this, but kept having problems. z not in mZ and z not in nZ ==> z in (mZ)' / (nZ)' = (mZ / nZ)' where (mZ)' = ZmZ = complement of mZ in Z. But there are problems. (m,n) = 1 ==> mZ + nZ = Z, but I don't know about writing things like mZ / nZ, although mnZ = mZ / nZ is OK. Any comments on this? Am I just confused? Van === Subject: Re: Automorphisms of groups > What about phi(p^n) = p^n - p^(n-1) ? > Back to the old fashion way of counting? > Nay, tell me Ôtis not so. > No. How about |Z_p^n| = p^n, and A = {k such that p | k and k in Z_p^n} > = {k | (k,p) = p and k in Z_p^n} = {qp | q = 1,2, ... ,p^(n-1)}, so > |A| = p^(n-1) and |Z*_p^n| = phi(p^n) = |Z_p^n| - |A| = p^n - p^(n-1) . > Van Well, I guess this is back to the old way of counting. === Subject: Re: division by zero > Theorem: 1 = 2. > Proof: It follows from 0 = 0*1 and 0 = 0*2 that 0*1 = 0*2. > Now divide both sides by 0 to get 1 = 2. > Dave L. Renfro Typical invalid syllogism, where two objects/concepts (here 1 and 2) are equated on the basis of a shared property (giving 0 when multiplied by 0). All elephants are pink Socrates is pink => Socrates is an elephant -- Paul Townsend I put it down there, and when I went back to it, there it was GONE! Interchange the alphabetic elements to reply === Subject: Help, need to come up with simple equation Solving equations I can do, but coming up with my own and I'm in the zone lol. Usually figuring out the simple things I have trouble with. Lol it's weird I know ^^. I need an equation where the answer will increase substantially the smaller the variable is used inside the equation. Instead of explaining more and making you all confused, let's just get down to what I need lol. up num#1 = num#2 down num#1(answer) has to increase by a factor, the smaller num#2 gets eg. 1 = 25 | | 100 = 1 In any equation, if one number goes down so will the other, naturally. I need the opposite to happen, and my brain ain't coming up with anything hehe. I bet it's something ebarrasingly simple, but nevertheless I need help ^^. === Subject: Re: Help, need to come up with simple equation ... > num#1(answer) has to increase by a factor, the smaller num#2 gets > eg. > 1 = 25 > | | > 100 = 1 > In any equation, if one number goes down so will the other, naturally. > I need the opposite to happen, and my brain ain't coming up with > anything hehe. I bet it's something ebarrasingly simple, but > nevertheless I need help ^^. Proposition : inverse num#1 and proceed as usual! (or inverse num#2...) 1/1 -> 25 | | 1/100 -> 1 That is instead of y= a*x+b we need y= a/x+b here a/1 +b = 25 and a/100 +b = 1 subtracting both one gets : a(1-1/100)= 25-1 or a= 24*100/99= 800/33 b= 25-a= (825-800)/33= 25/33 your function is f(x)= (800/x+25)/33 Exercise : replace 100 by n and 25 by m and find the general formula! Hoping it was clear, Raymond === Subject: Re: Help, need to come up with simple equation > ... >> num#1(answer) has to increase by a factor, the smaller num#2 gets >> eg. >> 1 = 25 >> | | >> 100 = 1 >> In any equation, if one number goes down so will the other, naturally. >> I need the opposite to happen, and my brain ain't coming up with >> anything hehe. I bet it's something ebarrasingly simple, but >> nevertheless I need help ^^. > Proposition : inverse num#1 and proceed as usual! > (or inverse num#2...) > 1/1 -> 25 > | | > 1/100 -> 1 > That is instead of y= a*x+b we need y= a/x+b > here a/1 +b = 25 > and a/100 +b = 1 > subtracting both one gets : > a(1-1/100)= 25-1 or a= 24*100/99= 800/33 > b= 25-a= (825-800)/33= 25/33 > your function is f(x)= (800/x+25)/33 > Exercise : replace 100 by n and 25 by m and find the general formula! > Hoping it was clear, > Raymond I was supposing that 1 and 100 were the starting values. I'll follow Andersen this time and suppose that 25 and 1 are the starting values then (inverting starting and final values) : x -> y 1/25 -> 1 | | 1/1 -> 100 Supposing again that y= a/x+b here a/25 +b = 1 and a/1 +b = 100 subtracting both one gets : a(1/25-1)= 1-100 or a= 99/(1-1/25)= 825/8 b= 100-a= -25/8 your function could be... f(x)= (825/x-25)/8 Hoping that at least one of the answers provided (by Andersen and myself) will please you! :-) Raymond === Subject: Re: Help, need to come up with simple equation > up > num#1 = num#2 > down > num#1(answer) has to increase by a factor, the smaller num#2 gets > eg. > 1 = 25 > | | > 100 = 1 You are actually not giving enough information :-). What you are looking for is a function. Then you should say what it's domain is (what values can num#2 take) and also perhals its range (values that num#1) can take. You could for instance have: f(x) = -x The larger you would make x, the smaller value you would get (you are multiplying num#2 with (-1). You could also divide by x: f(x) = 1/x The example you gave with 25 and 1 for num#2 could be modeled by: f(x) = (2499-99x)/24 Lets try it: f(25) = (2499-99*25)/24 = 1 lets try f(1) = (2499-99*1)/24 = 2400/24 = 100 Seems to work :-) Andersen === Subject: Re: Random Number generation using Usenet > As an experiment, please reply with a 2-decimal-digit value (00-99 > inclusive) without looking at other responses first. 24 -- The life of a repoman is always intense! === Subject: Re: Random Number generation using Usenet > As an experiment, please reply with a 2-decimal-digit value (00-99 > inclusive) without looking at other responses first. > 666 Try 3.14 that ought to work! or maybe 19. Hint hint === Subject: Re: Random Number generation using Usenet > As an experiment, please reply with a 2-decimal-digit value (00-99 > inclusive) without looking at other responses first. > 666 I think 42 will be more frequent. Dirk Vdm === Subject: Re: Random Number generation using Usenet === >Subject: Re: Random Number generation using Usenet >Message-id: >> As an experiment, please reply with a 2-decimal-digit value (00-99 >> inclusive) without looking at other responses first. >> 666 >I think 42 will be more frequent. Sure, but you have to allow for outliers. >Dirk Vdm -- Mensanator Ace of Clubs === Subject: Re: Random Number generation using Usenet > As an experiment, please reply with a 2-decimal-digit value (00-99 > inclusive) without looking at other responses first. > 666 > I think 42 will be more frequent. > Dirk Vdm Think of a number between 1 and 99 containing only odd numbers (but not ones). SPOILER - REALLY, DONT READ, THINK OF A NUMBER. | | | | | | | | | | | | | | | | | | 37, right? What did you think of? There are 16 possible numbers, but over 50% of people pick 37. Its also I believe the most commonly number picked at random, and indeed in episode of The Simpsons that is the number Homer picks, when asked the same question. I think the writer knew this fact so made it into a bit of a gag. === Subject: Re: Random Number generation using Usenet > As an experiment, please reply with a 2-decimal-digit value (00-99 > inclusive) without looking at other responses first. > > 666 > I think 42 will be more frequent. > Dirk Vdm > Think of a number between 1 and 99 containing only odd numbers (but not > ones). > SPOILER - REALLY, DONT READ, THINK OF A NUMBER. > 37, right? > What did you think of? Close - I had 35. > There are 16 possible numbers, but over 50% of people pick 37. Its also I > believe the most commonly number picked at random, and indeed in episode of > The Simpsons that is the number Homer picks, when asked the same question. I > think the writer knew this fact so made it into a bit of a gag. I wouldn't be surprised - Groening wasn't born yesterday ;-) Dirk Vdm === Subject: Re: Derivations on l^infinity(Z) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I suppose that you mean by a derivation a linear map D : ell^infty(Z) to ell^infty(Z) satisfying the Leibniz identity D(fg) = f (Dg) +(Df) g. Since ell^infty(Z) is a commutative, semisimple Banach algebra, there are no non-zero derivations on it: this is proven in MR0246127 (39 #7433) Johnson, B. E. Continuity of derivations on commutative algebras. Amer. J. Math. 91 1969 1--10. For more (and more recent) info, see MR1816726 (2002e:46001) Dales, H. G. Banach algebras and automatic continuity. London Mathematical Society Monographs. New Series, 24. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 2000. xviii+907 pp. ISBN: 0-19-850013-0 > Does anyone know how to characterize the set of derivations on > l^infinity(Z)? Is anything besides 0 even in this set? === Subject: Borel, continuous and smooth cohomology of Lie groups Originator: baez@math-cl-n03.math.ucr.edu (John Baez) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) If G is a Lie group and A is an abelian Lie group, we can define Borel, continuous and smooth versions of group cohomology H^n(G,A) by taking the usual chain complex for group cohomology and adding the requirement that the n-cocycles f: G^n -> A be Borel measurable, continuous or smooth functions. I would like to know H^3(G,A) for all three versions when G is a compact Lie group and A is R or U(1), with G acting trivially on A. In Daniel Baker's paper Differential characters and Borel cohomology, he seems to say that H^n_{Borel}(G,U(1)) = H^n(g,R) where H^n(g,R) is the cohomology of the Lie algebra of a compact Lie group G with coefficients in the trivial representation on R. So, for example, we'd get R as our cohomology group when G is compact simple and n = 3. In Jim Stasheff's paper Continuous cohomology of groups and classifying spaces, he says that H^n_{continuous}(G,R) = 0 whenever G is a compact Lie group. I still need to read Hochschild and Mostow's paper Cohomology of Lie groups, which is earlier than these others... but the library closed before I got to it! I guess the cases I'm really most curious about are H^3_{continuous}(G,U(1)) and H^3_{smooth}(G,U(1)) I have a feeling these vanish at least when G is compact simple, but I don't actually know this! === Subject: Re: multiplicative vs algebraic independence Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >All: >One last question: Does this theorem continue to hold when t takes a >value, say, in the algebraics or rationals (without simultaneously >restricting the values of {d_j}_j to the algebraics)? If d_1 = 1 and d_2 = -1 + (2pi)i and t=1, then x_1 x_2 = 1 . >Theorem (Lang-Artin): >Let d1,d2,...,dn be complex numbers, linearly independent over the >rationals. The functions >exp(d1 t), exp(d2 t), ..., exp(dn t) >are algebraically independent over the complex numbers. If we denote >these functions by {x1,x2,...,xn}, then we note that they are >multiplicatively independent and satisfy no relation of the form >x1^b1 x2^b2 ... xn^bn = 1 >follows that they are algebraically independent, too. Dan -- Dan Luecking Department of Mathematical Sciences University of Arkansas Fayetteville, Arkansas 72701 To reply by email, change Look-In-Sig to luecking === Subject: Re: multiplicative vs algebraic independence Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >One last question: Does this theorem continue to hold when t takes a >value, say, in the algebraics or rationals (without simultaneously >restricting the values of {d_j}_j to the algebraics)? >Theorem (Lang-Artin): >Let d1,d2,...,dn be complex numbers, linearly independent over the >rationals. The functions >exp(d1 t), exp(d2 t), ..., exp(dn t) >are algebraically independent over the complex numbers. ... By t takes a value I assume you're talking about the numbers exp(d1 t),..., exp(dn t) for some fixed t, rather than the functions. Then the answer is no. No nonempty set of complex numbers can be algebraically independent over the complex numbers, because any complex number c is a root of the polynomial X - c. It's not even true if you replace algebraically independent over the complex numbers by linearly independent over the rational numbers, e.g. take t=1, d1 = i pi, d2 = ln(2). On the other hand, if you're talking about exp(d1 t), ..., exp(dn t) as functions on the rational numbers, the answer is yes: any polynomial in these is an entire function, and therefore is uniquely determined by its values on the rationals. In fact it would be true if you looked at these as functions on any set of complex numbers that has a (finite) cluster point. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: This week in the mathematics arXiv (16 Aug - 20 Aug) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Here are this week's titles in the mathematics arXiv, available at: http://front.math.ucdavis.edu/ http://front.math.ucdavis.edu/submissions This week in the mathematics arXiv may be freely redistributed with attribution and without modification. Titles in the mathematics arXiv (16 Aug - 20 Aug) ------------------------------------------------- AG: Algebraic Geometry ---------------------- math.AG/0408266 Sheldon Katz: Gromov-Witten, Gopakumar-Vafa, and Donaldson-Thomas invariants of Calabi-Yau threefolds math.AG/0408196 Marcos Jardim: Admissible sheaves on P^3 math.AG/0408180 Michel Brion, Jesper Funch Thomsen: F-regularity of large Schubert varieties AP: Analysis of PDEs -------------------- math.AP/0408238 Mohamed Ben Ayed, Khalil El Mehdi, Mokhless Hammami, Mohameden Ould Ahmedou: On a Yamabe Type Problem on Three Dimensional Thin Annulus math.AP/0408237 Ville Kolehmainen, Matti Lassas, Petri Ola: Inverse conductivity problem with an imperfectly known boundary math.AP/0408227 Mohammadreza Raoofi: $L^p$ asymptotic behavior of perturbed viscous shock profiles math.AP/0408186 Peng Li: Green's Function of 3-D Helmholtz Equation for Turbulent Medium: Application to Optics AT: Algebraic Topology ---------------------- math.AT/0408261 Yusuf Civan, Nigel Ray: Homotopy decompositions and K-theory of Bott towers math.AT/0408256 Marek Golasinski, Juno Mukai: Gottlieb groups of spheres math.AT/0408216 Kathryn Hess, Paul-Eug`ene Parent, Jonathan Scott, Andrew Tonks: A canonical enriched Adams-Hilton model for simplicial sets math.AT/0408215 Yutaka Hemmi, Yusuke Kawamoto: Higher homotopy commutativity and cohomology of finite H-spaces math.AT/0408179 Daniel C. Isaksen: Generalized cohomology of pro-spectra CA: Classical Analysis and ODEs ------------------------------- math.CA/0408264 math.CA/0408242 Yong-Cheol Kim: Note on the Dirichlet Approximation Theorem math.CA/0408203 Pablo Shmerkin: Overlapping self-affine sets CO: Combinatorics ----------------- math.CO/0408263 Herbert S. Wilf: The Redheffer matrix of a partially ordered set math.CO/0408262 Dmitry N. Kozlov: A simple proof for folds on both sides in complexes of graph homomorphisms math.CO/0408247 I. Cahit: Spiral Chains: A New Proof of the Four Color Theorem math.CO/0408234 Soichi Okada: Enumeration of Symmetry Classes of Alternating Sign Matrices and Characters of Classical Groups math.CO/0408232 Laszlo Lovasz: The rank of connection matrices and the dimension of graph algebras math.CO/0408230 Jordan Daniel Bell: A translation of Leonhard Euler's De quadratis magicis, E795 math.CO/0408204 Masao Ishikawa: Minor summation formula and a proof of Stanley's open problem math.CO/0408189 Guoce Xin: A Fast Algorithm for Partial Fraction Decompositions math.CO/0408173 Laszlo Lovasz, Balazs Szegedy: Limits of dense graph sequences cs.DM/0408028 Joel Friedman, Jean-Pierre Tillich: Calculus on Graphs CT: Category Theory ------------------- math.CT/0408177 Lars Bruenjes, Christian Serpe: Enlargements of Categories CV: Complex Variables --------------------- math.CV/0408222 Y^usuke Okuyama: Linearization problem on structurally finite entire functions DG: Differential Geometry ------------------------- math.DG/0408254 S. Berceanu, A. Gheorghe: Linear Hamiltonians on homogeneous Kahler manifolds of coherent states math.DG/0408249 Ines Kath, Martin Olbrich: On the structure of pseudo-Riemannian symmetric spaces math.DG/0408248 Taoufik Bouziane: An existence result of energy minimizer maps between riemannian polyhedra math.DG/0408245 Thomas B. Bouetou, Jean P. Dufour: Veronese curves and webs interpolation math.DG/0408233 Stefan Berceanu: Geometrical phases on hermitian symmetric spaces math.DG/0408229 A. Rod Gover, Lawrence J. Peterson: The ambient obstruction tensor and the conformal deformation complex math.DG/0408228 Levi Lopes de Lima: Infinite connected sums, K-area and positive scalar curvature math.DG/0408225 Frederic Rochon: Bott Periodicity for Fibred Cusp Operators math.DG/0408224 Mario Listing: Conformally invariant Cotton and Bach tensor in N-dimensions math.DG/0408219 Stefan Berceanu: A holomorphic representation of the Jacobi algebra math.DG/0408206 Isabel M.C. Salavessa, Ana Pereira do Vale: Cayley submanifolds of Calabi-Yau 4-folds math.DG/0408197 Claus Gerhardt: On the CMC foliation of future ends of a spacetime math.DG/0408188 Christopher Allday, Volker Puppe: The Minimal Hirsch-Brown Model Via Classical Hodge Theory math.DG/0408187 Christopher Seaton: A Complete Obstruction to the Existence of Nonvanishing Vector Fields on Almost-Complex, Closed Orbifolds math.DG/0408184 J'anos Koll'ar: Einstein metrics on 5-dimensional Seifert bundles math.DG/0408175 Yoonweon Lee: The ratio of two zeta-determinants of Dirac Laplacians associated with unitary involutions on a compact manifold with cylindrical end DS: Dynamical Systems --------------------- math.DS/0408246 Susumu Tanabe: Systeme des coordonees plates de la deformation verselle des singularites isolees simples d'hypersurface et zeros de l'integrale d'Abel math.DS/0408241 Michael Blank, Leonid Bunimovich: Switched ßow systems: pseudo billiard dynamics math.DS/0408240 Michael Blank: Hysteresis phenomenon in deterministic traffic ßows math.DS/0408231 Alexandr Prishlyak: Topological classification of Morse-Smale ßows on 3-manifolds math.DS/0408213 David Fisher, Gregory Margulis: Local rigidity of affine actions of higher rank groups and lattices astro-ph/0408271 Sergey A. Astakhov, David Farrelly: Capture and escape in the elliptic restricted three-body problem math.DS/0408205 Jon Aaronson, Hitoshi Nakada: On the mixing coefficients of piecewise monotonic maps math.DS/0408201 Tom Meyerovitch: Double-Tail Invariant Measures of the Dyck Shift math.DS/0408185 Marta Tyran-Kaminska: An invariance principle for maps with polynomial decay of correlations nlin.CD/0408017 D. J. Albers, J. C. Sprott: Routes to chaos in high-dimensional dynamical systems: A qualitative numerical study nlin.CD/0408011 D. J. Albers, J. C. Sprott: Structural Stability and Hyperbolicity Violation in High-Dimensional Dynamical Systems FA: Functional Analysis ----------------------- math.FA/0408194 A.G.Ramm: Continuity of solutions to operator equations with respect to a parameter math.FA/0408192 A.G.Ramm: Dynamical systems method and a homeomorphism theorem GM: General Mathematics ----------------------- math.GM/0408202 Aleksandr Golubchik: On the nature of finite groups GN: General Topology -------------------- math.GN/0408200 Michael Zarichnyi: Nonexistence of linear operators extending Lipschitz (pseudo)metrics GR: Group Theory ---------------- math.GR/0408253 D. Tieudjo, D.I. Moldavanskii: On the Automorphisms of Some One-Relator Groups math.GR/0408252 D. Tieudjo, D.I. Moldavanskii: Conjugacy separability of some one-relator groups math.GR/0408239 Helge Glockner: Smooth Lie groups over local fields of positive characteristic need not be analytic GT: Geometric Topology ---------------------- math.GT/0408255 Vassily Olegovich Manturov: Virtual links are algorithmically recognisable math.GT/0408226 James F. Davis: A two component link with Alexander polynomial one is concordant to the Hopf link math.GT/0408199 Tao Li: Heegaard surfaces and measured laminations, II: non-Haken 3-manifolds math.GT/0408198 Tao Li: Heegaard surfaces and measured laminations, I: the Walderhausen conjecture math.GT/0408182 Dror Bar-Natan: Finite Type Invariants KT: K-Theory and Homology ------------------------- math.KT/0408260 Catarina Carvalho: A K-Theory Proof of the Cobordism Invariance of the Index math.KT/0408243 S.K. Roushon: The isomorphism conjecture for 3-manifold groups and K-theory of virtually poly-surface groups MG: Metric Geometry ------------------- math.MG/0408174 Henry Cohn, Abhinav Kumar: The densest lattice in twenty-four dimensions MP: Mathematical Physics ------------------------ quant-ph/0408104 Maurice Robert Kibler: On a group-theoretical approach to the periodic table of chemical elements math-ph/0408030 Y. Hassouni, E. M. F. Curado, M. A. Rego-Monteiro: Construction of coherent states for physical algebraic systems math-ph/0408029 A. B. Mehmood, U. A. Shah, G. Shabbir: Closed Form Approximations For The Three Body Problem hep-th/0403191 Nikolay M. Nikolov, Ivan T. Todorov: Elliptic Thermal Correlation Functions and Modular Forms in a Globally Conformal Invariant QFT cond-mat/0408407 Bertrand Duplantier: Statistical Mechanics of Self-Avoiding Manifolds (Part II) quant-ph/0408106 Andreas Doering: Kochen-Specker theorem for von Neumann algebras math-ph/0408028 Jean-Marc Bouclet, Francois Germinet, Abel Klein, Jeffrey H. Schenker: Linear response theory for magnetic Schroedinger operators in disordered media math-ph/0408027 Andre Gsponer, Jean-Pierre Hurni: Cornelius Lanczos's derivation of the usual action integral of classical electrodynamics hep-ph/0408184 Wolfgang Lucha, F.F. Schoberl: Facets of the spinless Salpeter equation cond-mat/0408364 Igor Loutsenko: Quantum Separation of Variables and Multi-component Dyson Brownian Motion cond-mat/0408271 V.V. Makhro: Soliton solutions in the class of implicit difference schemes quant-ph/0407261 B.A. Nikolov, D.A. Trifonov: On the Dynamics of Generalized Coherent States. II. Classical Equations of Motions quant-ph/0407260 B.A. Nikolov, D.A. Trifonov: On the Dynamics of Generalized Coherent States. I. Exact and Stable Evolution math-ph/0408026 Vasilisa Shramchenko: Deformations of Frobenius structures on Hurwitz spaces math-ph/0408025 Pascal Baseilhac: An integrable structure related with tridiagonal algebras cond-mat/0408321 Nils Berglund, Barbara Gentz: Universality of residence-time distributions in non-adiabatic stochastic resonance hep-th/0408069 M. Chaichian, P. Kulish, K. Nishijima, A. Tureanu: On a Lorentz-Invariant Interpretation of Noncommutative Space-Time and Its Implications on Noncommutative QFT NA: Numerical Analysis ---------------------- math.NA/0408209 Alexander G. Ramm, Semion Gutman: Optimization methods in direct and inverse scattering math.NA/0408195 A.G.Ramm, A.Smirnova: On deconvolution problems: numerical aspects math.NA/0408193 W.Chen, A.G.Ramm: Numerical Method for Solving Obstacle Scattering Problems by an Algorithm Based on the Modified Rayleigh Conjecture math.NA/0408191 A.G.Ramm: A new discrepancy principle NT: Number Theory ----------------- math.NT/0408223 Euler polynomials. II math.NT/0408221 David W. Farmer, Kevin Wilson: Converse theorems assuming a partial Euler product math.NT/0408214 Frank Calegari: Irrationality of certain p-adic periods for small p math.NT/0408212 Dragos Ghioca: The Local Lehmer Inequality For Drinfeld Modules OA: Operator Algebras --------------------- math.OA/0408235 Masayoshi Kaneda: Extreme points of the unit ball of a quasi-multiplier space math.OA/0408208 Alfons Van Daele: Quasi-Discrete Locally Compact Quantum Groups math.OA/0408190 Takeshi Katsura: A class of C^*-algebras generalizing both graph algebras and homeomorphism C^*-algebras III, ideal structures PR: Probability --------------- math.PR/0408267 Rodrigo Banuelos, Tadeusz Kulczycki: Eigenvalue gaps for the Cauchy process and a Poincar'e inequality math.PR/0408207 Bernardo Lafuerza-Guillen Jose L. Rodriguez: Probabilistic normed spaces with non necessarily continuous triangle functions math.PR/0408178 Marina Kozlova, Paavo Salminen: On occupation times of stationary excursions math.PR/0408176 Jacob van den Berg, Olle Haggstrom, Jeff Kahn: Some Conditional Correlation Inequalities for Percolation and Related Processes QA: Quantum Algebra ------------------- math.QA/0408258 Vladimir Turaev: Coalgebras of words and phrases math.QA/0408244 Lars Kadison: An approach to quasi-Hopf algebras via Frobenius coordinates math.QA/0408220 Lydia Delvaux, Alfons Van Daele & Shuanhong Wang: Quasitriangular (G-cograded) multiplier Hopf algebras math.QA/0408218 Alfons Van Daele & Shuanhong Wang: The Larson-Sweedler theorem for multiplier Hopf algebras math.QA/0408217 Stefan Waldmann: States and representations in deformation quantization math.QA/0408210 PoNing Chen, Vasiliy Dolgushev: A Simple Algebraic Proof of the Algebraic Index Theorem RA: Rings and Algebras ---------------------- math.RA/0408251 Robert Wisbauer: On Galois comodules RT: Representation Theory ------------------------- math.RT/0408211 Karin Erdmann, Manfred Schocker: Modular Lie powers and the Solomon descent algebra math.RT/0408181 Markus Schmidmeier: Bounded Submodules of Modules SG: Symplectic Geometry ----------------------- math.SG/0408265 Bohui Chen, Shengda Hu: Note on stringy abelian orbifolds math.SG/0408250 R. F. Goldin, S. Martin: Cohomology pairings on the symplectic reduction of products SP: Spectral Theory ------------------- math.SP/0408259 A. Volberg, P. Yuditskii: Noncommutative Perron-Frobenius-Ruelle theorem, two weight Hilbert transform, and almost periodicity math.SP/0408257 F. Peherstorfer, A. Volberg, P. Yuditskii: Limit periodic Jacobi matrices with a prescribed p-adic hull and a singularly continuous spectrum math.SP/0408236 F. Peherstorfer, P. Yuditskii: Remarks on a paper of Geronimo and Johnson math.SP/0408183 T. Christiansen: Several complex variables and the distribution of resonances in potential scattering -- / Greg Kuperberg (UC Davis) / / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ / * All the math that's fit to e-print * === Subject: Operator Theory, Hilbert 16th Problem Importance: Normal Content-Length: 253 Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > I Want To Find A Possible Relation Between Hilbert 16th Problem > and operator Theory, > Please see > http://www.arxiv.org/abs/math.DS/0408037 Ali Taghavi === Subject: Elementary permutation question Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Does anyone know an algorithm for creating all permutations of N distinct elements except those that are reversals of ones that has already appeared? For my application, a,b,c,d,e is equivalent to e,d,c,b,a, etc., and I can't waste the time it takes to process the superßuous perms. It would also be helpful, but less important, to specify the sequence numbers of those that duplicate ones that already appeared. I don't see a simple rule offhand. Steve Gray === Subject: Re: Elementary permutation question Originator: tchow@markov.mit.edu.mit.edu (Timothy Chow) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Does anyone know an algorithm for creating all permutations of N >distinct elements except >those that are reversals of ones that has already appeared? > For my application, a,b,c,d,e is equivalent to e,d,c,b,a, etc., Fix some ordering of the elements, e.g., a < b < c < d < e. Then generate only those permutations whose first element is less than the last element. (So for example, generate a,b,c,d,e but not e,d,c,b,a.) Do this by picking the first element to be a, b, c, or d, then picking the last element to be any element greater than the first element, then inserting the remaining elements in between in all possible ways. -- 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: Elementary permutation question Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Does anyone know an algorithm for creating all permutations of N >distinct elements except >those that are reversals of ones that has already appeared? > For my application, a,b,c,d,e is equivalent to e,d,c,b,a, etc., and I >can't waste the time >it takes to process the superßuous perms. Just require the first element to be less than the last one (in some fixed ordering). Thus your algorithm might go like this, where L is the list of elements for each a in L do for each z in L with z > a do for each permutation p of L {a,z} do output the concatenation of (a,p,z) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Maximal subgroups of S_n Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Are (Is) there (a) published paper(s) or book(s) about the following question: Is it possible to determine all maximal subgroups of S_n, the symmetric group of degree n. I mean by the existence if a complete description of maximal subgroups in terms of only n !! What is the number of all maximal subgroups of S_n. Any information is appreciated. A. Abdollahi === Subject: Re: Maximal subgroups of S_n Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Are (Is) there (a) published paper(s) or book(s) about the following > question: > Is it possible to determine all maximal subgroups of S_n, the symmetric > group of degree n. I mean by the existence if a complete description > of maximal subgroups in terms of only n !! > What is the number of all maximal subgroups of S_n. > Any information is appreciated. > A. Abdollahi Let me just say a little more about the maximal subgroups of S_n. They are of three types: 1. Intransitive maximals, 2. Transitive but imprimitive maximals, 3. Primitive maximals. The first two of these types are easily described and well understood. Type 1 consists of subgroups S_m x S_{n-m} for 1 <= m < n/2 Type 2 consists of wreath products S_m wr S_{n/m} for factors m of n with 1 < m < n. With very few exceptions, these types of subgroups are maximal in S_n. Finite primitive permutation groups are described by the O'Nan-Scott Theorem. The coset representations of the almost simple groups form one category of primitive groups, and probably present the major obstacle to describing all maximals of S_n. There is another category, however, known as the affine primitive groups. These have degree a prime power p^m, and have a normal elementary abelian subgroup of order p^m with a complement isomorphic to an irreducible subgroup of GL(m,p). So, in order to classify the maximal subgroups of S_n with n = p^m, you would also need to classify all maximal irreducible subgroups of GL(m,p), which is another problem on which much work has been done (there is a big theorem of Aschbacher describing these subgroups), but with no complete solution as yet. To solve it completely, you would probably need to find all irreducible representations of all almost simple groups over the prime field. So finding the maximals of S_n actually involves solving some of the major open problems in the study of finite simple groups. Derek Holt. === Subject: Re: Maximal subgroups of S_n Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Are (Is) there (a) published paper(s) or book(s) about the following > question: > Is it possible to determine all maximal subgroups of S_n, the symmetric > group of degree n. I mean by the existence if a complete description > of maximal subgroups in terms of only n !! > What is the number of all maximal subgroups of S_n. > Any information is appreciated. > A. Abdollahi The answer to this question is no, there is no sensible description of the maximal subgroups of S_n for general n. The reason is roughly as follows. There is as yet no complete description of the maximal subgroups of all finite simple groups. As a first step in classifying maximal subgroups of S_n, you would need a complete description of the maximal subgroups of the almost simple groups i.e. groups G with S <= G <= Aut(G), for a finite nonabelian simple group G. With some exceptions, a maximal subgroup H of index n in G will give rise to a maximal subgroup of A_n or S_n coming from the action of G on the cosets of H. The maximal subgroups of A_n and S_n for n <= 1000 have been listed, and are accessible, for example, from the Magma computational algebra system. Derek Holt. === Subject: Re: Maximal subgroups of S_n days. My association with the Department is that of an alumnus. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Are (Is) there (a) published paper(s) or book(s) about the following >question: >Is it possible to determine all maximal subgroups of S_n, the symmetric >group of degree n. I mean by the existence if a complete description >of maximal subgroups in terms of only n !! >What is the number of all maximal subgroups of S_n. >Any information is appreciated. You might try the following paper: A classification of the maximal subgroups of the finite alternating and symmetric groups Journal of Algebra, 111 (1987), pp. 365-383 MR 89b:20008 -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Axiom of Choice & Lebesgue Measure Problem Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Mike Carroll a crit : >> Bas van Fraassen in his Quantum Mechanics: An Empiricist View >> mentions what he calls Lebesgue's Measure Problem, which has to do >> with whether a measure on all subsets of a set can satisfy certain >> conditions (p.54). He then goes on to say that ... if the Axiom of >> Choice is true, Lebesgue's ÔMeasure Problem' has no solution. >> He does not go into the proof of this, though, instead referring the >> reader to G. Moore's 1982 monograph Zermelo's Axiom of Choice. >> Moore's book is out of print, with not even any used copies available >> at Amazon. Does anybody know of any other sources, preferably more >> recent, that discuss the relationship between AC and Lebesgue's >> Measure Problem? >> Mike Carroll >> Oro Valley, AZ >Consulting Halmos' Measure Theory will do. In this reference you'll >find an explicit construction of a non Lebesgue measurable subset of >the real line R, based on the Axiom of Choice. >To my best knowledge the converse (if there exist some set wich is not >Lebesgue measurable then AC holds) is also established under the name of >Solovay's theorem, but I didn't pursue the inquiry any further. There are lots of models of set theory without AC in which there are sets which are not Lebesgue measurable. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: On the partial sums of reciprocals of primes Epigone-thread: smungßoadweu Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Explicit inequalities on some functions of prime numbers can be found in the (very good) paper entitled Approximate formulas for some functions of prime numbers by J.B. Rosser & L. Schoenfeld (Illinois J. Maths 6, 1962, 64 - 94). For example, they showed that : 1. |Sum ( log(p)/p , p , 2 , x ) - log(x) + E | < 1/(2*log(x)) Where E := Gamma + Sum( lop(p)/(p^2-p) , p ) = 1.33258227573322087, 2. Pi(x) < 1.25506 x/log(x) (x>1) where Pi(x) is the number of prime numbers up to x (see the prime number theorem), 3. p_n > n.log(n) (n>1) where p_n is the nth prime, 4. theta(x) < 1.000081x (x>1) where theta is one of the Tchebytchev function (ie theta(x) = Sum ( log(p) ,p ,2 , x)). This result is more recent and may be found in the paper Sharper bounds for the functions theta(x) and psi(x) II (Maths. Comp. 30 (1976), 337-360). More recent results can be found in Dusart's PHD thesis 1998 (see online). For example, he has proved that : p_n >= n.(log(n) + loglog(n) - 1) (n>1) and that |theta(x)-x| < 515.x/log(x)^3 (x>1). The proofs are based on the computation of zeros of the Riemann zeta function on the critical line (sigma = 1/2), and hence are not elementary. Olivier Bordell.8fs. === Subject: Ito's formula Epigone-thread: sloirepram Originator: bergv@math.uiuc.edu (Maarten Bergvelt) does anyone know how to apply Ito's formula to the following? My starting equation is dS(t) = m(t)S(t)dt + sigma(t)S(t)dY(t), where m(t) is the mean rate of return, progresively measureable and the semi-martingale representation of Y(t) is B(t) - integrate form 0 to t alpha(s) ds, alpha is an F(t)-adapted process. Dennis === Subject: Solving solvable QUINTICS using two sextics Epigone-thread: kerdcrermkrald Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hello all, For those interested in the quintic, this paper might be of some help: An Easy Way To Solve Solvable Quintics Using Two Sextics ABSTRACT: Using a method initially developed by George Young (1819- 1889), Arthur Cayley (1821-1895), and later by George Watson (1886- 1965), an explicit quartic is constructed to enable the solution in radicals of a quintic when it is a solvable equation. Not one, but two sextic resolvents are derived which are important to forming the coefficients of this quartic. Certain difficulties and their solutions as well as a novel consequence to the method are also addressed in this paper. Mathematics Subject Classification. Primary: 12E12; Secondary: 12F10. http://www.geocities.com/titus_piezas/ Just click on the link in the website to the pdf file. Titus (tpiezasIII@uap.edu.ph -> remove III for email) === Subject: Re: Boundary conditions for Euler Maclaurin summation formula Epigone-thread: nilferlzay Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >The Euler MacLaurin summation formula links the values of a sum to the the >Euler-Maclaurin Summation formula on Mathworld, >http://mathworld.wolfram.com/ Euler-MaclaurinIntegrationFormulas.html >The phrase of interest to me is: In certain cases, the last term tends >to 0 as n tends to infinity. >What are these cases? >I tracked down several of the references in the bibliography of the same >basic form for the link between the sum and the integral. >Sum = integral + Bernoulli-Number Correction terms + integral >remainder term. >Where there are k correction terms and the remainder is given as an >integral of the 2k th derivative and the Bernoulli polynomial: >B_2k(x)/(2*k)!. >you mean > R_k = int_0^n (B_{2k}-hat B_{2k}(t))/((2k!))f^{(2k)}(t) dt ? > = f^({2k})(xi)n B_{2k}/((2k!)) >B_{2k} the Bernoulli numbers with > | B_{2k}/(2k!) | <= 4/((2pi)^{2k}) Just t be clear: Is the bound here on the Bernoulli number, B_{2k}/((2k!)), the Bernoulli polynomial, B_{2k}(t)/((2k!)) or both, because |B_{2k}(t)/((2k!))| <= |B_{2k}/((2k!))|? >Hence you need a function whose supremum of |f^{2k}| on [0,n] >goes to zero as k to infinity for n fixed. >(if I not mistaken, the entire functions on |C do that?) >hth >peter >Knopp states that almost no functions allow you to ignore the integral >remainder term. Abramowitz and Stegun start with the infinite form. >This form has an infinite number of correction terms and no remainder >term. A & S give the impression that you can cut off the >Bernoulli-number correction terms anywhere it is convenient. The book >Mathematical Methods in Physics by Mathews and Walker uses operator >notation and thus, jumps directly to the infinite form and avoids the >integral remainder term altogether. between >including and ignoring the integral remainder term. But, none of the >references I have describe when and how you can ignore the integral >remainder term. >My question is a clarification of these many papers and references. >What properties of f(x) and its derivatives allow you to pass to the >infinite form and ignore the integral remainder term? Is smoothness >sufficient? Is simple boundedness sufficient, given B_2k(x)/(2k!) >tends to 0 as k tends to infinity? >John Washburn > === Subject: Re: Boundary conditions for Euler Maclaurin summation formula Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >>The Euler MacLaurin summation formula links the values of a sum to >the >the >>Euler-Maclaurin Summation formula on Mathworld, >>href=http://mathworld.wolfram.com/ Euler-MaclaurinIntegrationFormulas.html >http://mathworld.wolfram.com/ Euler-MaclaurinIntegrationFormulas.html >> >>The phrase of interest to me is: In certain cases, the last term >tends >>to 0 as n tends to infinity. >> >>What are these cases? >> >>I tracked down several of the references in the bibliography of the >same >>basic form for the link between the sum and the integral. >> >>Sum = integral + Bernoulli-Number Correction terms + integral >>remainder term. >> >>Where there are k correction terms and the remainder is given as an >>integral of the 2k th derivative and the Bernoulli polynomial: >>B_2k(x)/(2*k)!. >>you mean >> R_k = int_0^n (B_{2k}-hat B_{2k}(t))/((2k!))f^{(2k)}(t) dt ? | here hat B_{2k}(t) not the ordinary Bernoulli polynomial >> = f^({2k})(xi)n B_{2k}/((2k!)) >>B_{2k} the Bernoulli numbers with >> | B_{2k}/(2k!) | <= 4/((2pi)^{2k}) >Just t be clear: Is the bound here on the Bernoulli number, >B_{2k}/((2k!)), the Bernoulli polynomial, B_{2k}(t)/((2k!)) or both, >because |B_{2k}(t)/((2k!))| <= |B_{2k}/((2k!))|? the bound concerns the Bernoulli numbers and I applied the intermediate value theorem for integrals. but it is also valid for that hat B_{2k}(t) since |hat B_{2k}(t)| <= B_{2k} = B_{2k}(0)= hat B_{2k}(0) for k=1,2,3,.. and all real x and hat B_k(x) = B_k(x-ßoor(x)) , B_k(x) being the ordinary Bernoulli polynomial. the hat B_k can be written as hat B_k(x) = B_k(x) -k sum_{i=0}^{n-1}( (x-i)_{+})^{k-1} for x in [0,n] >>Hence you need a function whose supremum of |f^{2k}| on [0,n] >>goes to zero as k to infinity for n fixed. or at least does grow slower than (2pi)^{2k} >>(if I not mistaken, the entire functions on |C do that?) (false as stated in my second posting) here I myself have a question: If f is an entire function on the complex plane, is it true that for every compact real interval [a,b] there exists a constant C(f,a,b) such that max_{x in [a,b]} |f^{k}(x)| <= C(f,a,b)^k (geometric growth of magnitude of the derivatives with order ?) >>hth >>peter >> >>Knopp states that almost no functions allow you to ignore the >integral >>remainder term. Abramowitz and Stegun start with the infinite >form. >>This form has an infinite number of correction terms and no >remainder >>term. A & S give the impression that you can cut off the >>Bernoulli-number correction terms anywhere it is convenient. The >book >>Mathematical Methods in Physics by Mathews and Walker uses operator >>notation and thus, jumps directly to the infinite form and avoids >the >>integral remainder term altogether. >> >between >>including and ignoring the integral remainder term. But, none of >the >>references I have describe when and how you can ignore the integral >>remainder term. >> >>My question is a clarification of these many papers and references. >> >>What properties of f(x) and its derivatives allow you to pass to >the >>infinite form and ignore the integral remainder term? Is smoothness >>sufficient? Is simple boundedness sufficient, given B_2k(x)/(2k!) >>tends to 0 as k tends to infinity? >> >> >>John Washburn >> === Subject: Re: Boundary conditions for Euler Maclaurin summation formula Received-SPF: none (mailbox4.ucsd.edu: domain of news@nntp.itservices.ubc.ca does not designate permitted sender hosts) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > If f is an entire function on the complex plane, is it true that for every > compact real interval [a,b] there exists a constant C(f,a,b) such that > max_{x in [a,b]} |f^{k}(x)| <= C(f,a,b)^k > (geometric growth of magnitude of the derivatives with order ?) No. For example, f(z) = sum_{n=0}^infinity z^n/sqrt(n!) is entire and has f^(k)(0) = sqrt(n!) which is not O(c^k) for any c. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Torsors Content-Length: 1530 Originator: bergv@math.uiuc.edu (Maarten Bergvelt) In Torsors made easy, http://math.ucr.edu/home/baez/torsors.html John Baez presents the concept of principal homogeneous space (transitive group action with stabilizer reduced to identity), a.k.a. torsor. A torsor, in his words, is like a group that has forgotten its identity, a notion which is important enough, indeed, to deserve a shorter name than principal homogeneous space. In the French tradition, we call torseur, the English equivalent of which would be torsor, an equiprojective vector field in 3D space, i.e., a field x --> V(x) such that V(y) - V(x) is orthogonal to y - x. Such a field can be characterized by two vectors (a polar one and an axial one), and can be interpreted as, e.g., the velocity field of a moving solid. Torseurs, in this acception, can therefore be seen as elements of (a representation of) a Lie algebra, here the Lie algebra of the 3D displacement group. Our torseurs thus look very much like tangent vectors to a torsor manifold. Not too remote a concept, but definitely a different one. Puzzled by this apparent cultural gap, I have searched the web for equiprojective field, equiprojectivity, etc., and found, to my surprise, that this concept of equiprojective vector field seems to be ignored in the Anglo-Saxon world. (When the words themselves can be found, it's in English papers by French-speaking authors.) Yet, the concept must appear in the curriculum as some stage, in kinematics, dynamics of solid bodies, etc. How do you call it? [Moderator's remark: in French the word torseur is of course also used in Baez's sense, for a principal homogeneous space] === Subject: Re: Small random matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >I would like to know if there exists a formula to have the pdf of a, >say 2x2 complex normal (AA*=A*A) random matrix (when the distributions >of the matrix elements are known). I'm not sure I understand your question. The pdf of the matrix A _is_ the joint pdf of the matrix elements. Of course, you have to pick that pdf carefully to make sure the matrix is normal - in particular, except in trivial cases you won't be able to make the elements independent. Perhaps what you are asking is, given (marginal) pdf's for the matrix elements, can you find a pdf for the matrix A that gives these marginal pdf's and makes A normal? The answer to that is maybe: there are certainly constraints on the marginal pdf's. When it is possible, I think it would be rather complicated, even in the 2 x 2 case. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Small random matrices Received-SPF: none (mailbox5.ucsd.edu: domain of approve@mathforum.org does not designate permitted sender hosts) Epigone-thread: sywixswal Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I must give more details to be clearer..:) In fact, I am looking for a formulation of the pdf of any 2x2 matrix M of the form : | a b | M=| | | -b* a*| where a and b are complex random variables and * is the conjugation. Such matrices are normal. So, given the marginal distributions of a,a*,b and b*, I want the pdf of M (a,a*,b and b* may not be independent). Now, the second part of my question is, can I define some moments and cumulants of M ? And how ? Nicolas >I'm not sure I understand your question. The pdf of the matrix A _is_ the >joint pdf of the matrix elements. Of course, you have to pick that pdf >carefully to make sure the matrix is normal - in particular, except in >trivial cases you won't be able to make the elements independent. >Perhaps what you are asking is, given (marginal) pdf's for the >matrix elements, can you find a pdf for the matrix A that gives these >marginal pdf's and makes A normal? The answer to that is maybe: >there are certainly constraints on the marginal pdf's. When it >is possible, I think it would be rather complicated, even in the >2 x 2 case. >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~ israel >University of British Columbia Vancouver, BC, Canada === Subject: Re: Small random matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >In fact, I am looking for a formulation of the pdf of any 2x2 matrix M >of the form : > | a b | >M=| | > | -b* a*| >where a and b are complex random variables and * is the conjugation. >Such matrices are normal. So, given the marginal distributions of >a,a*,b and b*, I want the pdf of M (a,a*,b and b* may not be >independent). Of course a and a* are not independent, since one determines the other. The density for a* is the reßection of the density for a across the real axis. Similarly for b and b*. But you could make a and b independent. So the joint density for a and b is the product of the densities for a and for b. Note that I'm considering the density for a complex random variable as a function on the complex plane (or on R^2) and integration is with respect to area on this plane. >Now, the second part of my question is, can I define some moments and >cumulants of M ? And how ? There's no problem with defining them, e.g. the n'th moment is E[M^n]. Or with calculating them, e.g. if a and b are independent the second moment is the matrix [ E[a^2] - E[|b|^2] E[b](E[a]+E[a*]) ] [ -E[b*](E[a]+E[a*]) E[a*^2] - E[|b|^2] ] Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: A standard bound for continued fractions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Let p/q have simple continued fraction [a_0,a_1,...,a_n] where p>q>0 >are integers. Then p is greater than or equal to the sum of the a_i's. >Can anyone provide a reference for this, surely standard, result? I would be surprised if it is in the literature. If we let p_k/q_k be the k-th convergent, p_{-1} = 1, p_0 = a_0, and p_k = a_k*p_{k-1} + p_{k-2}. Proceed by induction. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: A markov/probability question Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > ... how many times will the bucket be having balls of only c colors > (c < b). there is no answer in this post, just a remark. The model of a cache reminds me of a well-known model called polling system. It is usually described as follows: Customers arrive to (b) different polling stations (corresponding to colors). Arrivals at each polling station are independent Poisson processes with the same rate. Then there is a server that polls the stations and serves a customer at a time in an instance of the Markov process. The server policy in your example does not strictly satisfy the Markov property, that is the conditional distribution of the next colour depends on the history of the process. Though this can be overcome by aggregating events. Looks like the following citation, C. Fricker and M.R. Jaibi, Monotonicity and stability of periodic polling systems. Queueing Systems 15 (1994) 211238. may be relevant. (This paper is cited in Ergodicity of a polling network with an infinite number of stations by D. Korshunov, A. A. Borovkov and R. Schassberger (Queueing Systems 32 (1999) 169-193)) anon === Subject: Re: direct sums and products from category theory Received-SPF: none (mailbox3.ucsd.edu: domain of news@newssvr21.news.prodigy.com does not designate permitted sender hosts) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) OK, after referring to MacLane and several other references, I come up with the following, keeping in mind my goal is to use categorical products and sums to organize a presentation of algebraic combinations as commonly used. - In some special categories, the categorical coproduct (also called the categorical sum) of two objects coincides with the categorical product of two objects; in these cases, in category theory, they are both called the direct sum (AKA the biproduct). In particular, the categories of abelian groups and vector spaces have categorical direct sums. - In the case an infinite number of factors, the operations are distinct and only the coproduct is called the direct sum, defined as consisting of elements that have only finitely many non-identity terms. - These definitions are commonly used for the direct sum in any category, including those for which the categorical direct sum is a distinct construction or does not exist. In particular, the direct sum of non-abelian groups and algebras is defined as above, even though no categorical direct sum exists in these categories. The tensor product appears as a coproduct for commutative rings with unity, but as with the direct sum this definition is extended to other categories. An additional distinction that can be made is between an external sum or product, where the construction of a new object is from given constituent objects, and an internal sum or product, which is formed from a given object by recognizing constituent objects within it. Unfortunately, it seems there is no such consistent distinction between the designations inner/outer and interior/exterior with regard to sums and products. In general, symbols are not always used consistently and it is important to understand exactly what construction is meant in a given situation. Comments / corrections / clarifications? > I don't know about much mistaken, but it is a mistake: the > tensor product over k is the coproduct in the category of > commutative k-algebras, but not in the category of k-algebras. Incidentally, I believe the tensor product over k is the coproduct only in the category of commutative k-algebras *with unity*; in the category of commutative k-algebras it is A O+ B O+ (A Ox B). === Subject: Re: direct sums and products from category theory Epigone-thread: kyfrimpßou Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >OK, after referring to MacLane and several other references, I come up >with the following, keeping in mind my goal is to use categorical >products and sums to organize a presentation of algebraic combinations >as commonly used. >- In some special categories, the categorical coproduct (also called the >categorical sum) of two objects coincides with the categorical product >of two objects; in these cases, in category theory, they are both called >the direct sum (AKA the biproduct). In particular, the categories of >abelian groups and vector spaces have categorical direct sums. One has to be a little careful about what one means by coincides. There is no trouble with using zero object for an object which is simultaneously initial and terminal. A zero map A --> B is then one which factors through a zero object. A biproduct of A and B should admit injections i_A and i_B of A and B, resp., projections p_A and p_B to A and B, resp., such that (p_A)(i_A) and (p_B)(i_B) are identities, and (p_A)(i_B) and (p_B)(i_A) are zero maps. This compatibility between the coproduct and product structures is important to make everything come out cleanly. But with that understood, one can make a pretty direct connection between categories with zero objects and biproducts and commutative monoids: the homs of such a category C admit canonical structures of commutative monoid, where addition of maps f, g: A --> B is defined as diag f x g codiag A ----> A x A ----> B x B ------> B and the zero in hom(A, B) is the zero map A --> B. Conversely, if C is enriched in CommMon and admits finite products (or dually, finite coproducts), then the products/coproducts are biproducts. I think this explains what special category in your quote means, precisely. The point is that maps A_1 x ... x A_m --> B_1 x ... x B_n can be treated as matrices whose entries f_{ij} are maps A_j --> B_i, and which can be added and multiplied like matrices (in an obvious sense, I hope). >- In the case an infinite number of factors, the operations are distinct (in all but extremely degenerate cases) >and only the coproduct is called the direct sum, defined as consisting >of elements that have only finitely many non-identity terms. Yes, one could probably put it that way (or finitely many non-zero terms, in the sense of what I said above). Even in the abstract setting of categories with biproducts, there are ways of performing elementwise arguments, if care is taken. >- These definitions are commonly used for the direct sum in any >category, including those for which the categorical direct sum is a >distinct construction or does not exist. In particular, the direct sum >of non-abelian groups and algebras is defined as above, even though no >categorical direct sum exists in these categories. I *much* prefer the word biproduct to categorical direct sum, in the sense in which I think you're using it here, unless I know in advance that the context is CommMon-enriched categories. If you said to me categorical direct sum out of the blue, I'd think you were using the word sum as people often do to mean coproduct and adding the word categorical to make darned sure I knew you meant coproduct with its standard categorical meaning, when in fact you meant something much stronger! Coproduct, with the standard categorical meaning, is safer than sum, although in the case of commutative k-algebras with unity, I'd say tensor product to be even more on the safe side. Defined as above where? >The tensor product appears as a coproduct for commutative rings with >unity, but as with the direct sum this definition is extended to other >categories. An additional distinction that can be made is between an >external sum or product, where the construction of a new object is from >given constituent objects, and an internal sum or product, which is >formed from a given object by recognizing constituent objects within it. >Unfortunately, it seems there is no such consistent distinction between >the designations inner/outer and interior/exterior with regard to >sums and products. In general, symbols are not always used consistently >and it is important to understand exactly what construction is meant in >a given situation. Generally, no terminological confusion should arise if the category in question is understood. For example, if the category is the lattice of subspaces of a vector space (as a poset, seen as a category), then what is called the inner sum of two subspaces is also known as the join in the lattice, namely, the smallest subspace containing the given two, and this is in fact the coproduct in that lattice seen as a category (to be clearly distinguished from the coproduct in the category of vector spaces!). If someone says inner sum, I'd assume that person was working in a lattice of subthingies and was taking a join. I think that's what you were saying, but I think that the occasions for misunderstanding between mathematicians are fewer than you seem to think -- sorry if I misunderstand. >> >> I don't know about much mistaken, but it is a mistake: the >> tensor product over k is the coproduct in the category of >> commutative k-algebras, but not in the category of k-algebras. >Incidentally, I believe the tensor product over k is the coproduct only >in the category of commutative k-algebras *with unity*; in the category >of commutative k-algebras it is A O+ B O+ (A Ox B). That's true. The usage is not 100% standard perhaps, but I think many if not most people working in commutative algebra, algebraic number theory, or algebraic geometry understand commutative k-algebra as one with an identity, or unity as you say. Even though I have expertise in none of these fields, I had the same convention in mind. Many people (e.g. category theorists, and Jacobson I believe) use rings to mean thingies with identities, and rngs (pronounced rungs with a schwa-like u) to be rings without the i, i.e. without identity, and rigs when the underlying additive thing is a commutative monoid, not a group (i.e. rings without negatives or without n). I would like to see such usage standardized, just to obviate misunderstanding or having to explain which convention is being used. Todd === Subject: Two papers published by AGT Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following two papers have been published: Duality and Pro-Spectra by J. Daniel Christensen, Daniel C. Isaksen and On the homotopy invariance of configuration spaces by Mokhtar Aouina, John R. Klein Details follow: (1) Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-34.abs.html Title: Duality and Pro-Spectra Author(s): J. Daniel Christensen, Daniel C. Isaksen Abstract: Cofiltered diagrams of spectra, also called pro-spectra, have arisen in diverse areas, and to date have been treated in an ad hoc manner. The purpose of this paper is to systematically develop a homotopy theory of pro-spectra and to study its relation to the usual homotopy theory of spectra, as a foundation for future applications. The surprising result we find is that our homotopy theory of pro-spectra is Quillen equivalent to the opposite of the homotopy theory of spectra. This provides a convenient duality theory for all spectra, extending the classical notion of Spanier-Whitehead duality which works well only for finite spectra. Roughly speaking, the new duality functor takes a spectrum to the cofiltered diagram of the Spanier-Whitehead duals of its finite subcomplexes. In the other direction, the duality functor takes a cofiltered diagram of spectra to the filtered colimit of the Spanier-Whitehead duals of the spectra in the diagram. We prove the equivalence of homotopy theories by showing that both are equivalent to the category of ind-spectra (filtered diagrams of spectra). To construct our new homotopy theories, we prove a general existence theorem for colocalization model structures generalizing known results for cofibrantly generated model categories. Secondary: 55P25, 18G55, 55U35, 55Q55 Keywords: Spectrum, pro-spectrum, Spanier-Whitehead duality, closed model category, colocalization Author(s) address(es): Department of Mathematics, University of Western Ontario London, Ontario, Canada and Department of Mathematics, Wayne State University Detroit, MI 48202, USA Email: jdc@uwo.ca, isaksen@math.wayne.edu (2) Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-35.abs.html Title: On the homotopy invariance of configuration spaces Author(s): Mokhtar Aouina, John R. Klein Abstract: For a closed PL manifold M, we consider the configuration space F(M,k) of ordered k-tuples of distinct points in M. We show that a suitable iterated suspension of F(M,k) is a homotopy invariant of M. The number of suspensions we require depends on three parameters: the number of points k, the dimension of M and the connectivity of M. Our proof uses a mixture of Poincare embedding theory and fiberwise algebraic topology. Secondary: 57Q35, 55R70 Keywords: Configuration space, fiberwise suspension, embedding up to homotopy, Poincare embedding Author(s) address(es): Department of Mathematics, Wayne State University Detroit, MI 48202, USA Email: aouina@math.wayne.edu, klein@math.wayne.edu === Subject: reference about algebraic groups Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hello. I need the following statement: Every abstract normal subgroup of a simple algebraic group (over an algebraically closed field) is an algebraic normal subgroup (and thus is a subgroup of the center or is equal to the full group). This seems to be true, but I could not locate it in the books I checked. Does anyone know of some reference for it? Gizem Karaali ******************************************************* ******************************************************* === Subject: Winternitz Theorem Epigone-thread: birnayglang Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following nice theorem (phrased as an exercise) appears in Yaglom/Boltyanskii's book Convex Figures (p. 36): A convex figure is divided into two parts by a line L that passes through its center of gravity. Prove that the ratio of the areas of the two parts always lies between the bounds 4/5 and 5/4. Yaglom and Boltyanskii called this Winternitz' Theorem. However, they do not provide an original reference, and I was not able to trace back the original source for this theorem either. Can anybody help me? Is there more literature on Winternitz' Theorem? === Subject: Re: Winternitz Theorem Received-SPF: none (mailbox1.ucsd.edu: domain of news@sunb.ocs.mq.edu.au does not designate permitted sender hosts) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > The following nice theorem (phrased as an exercise) appears in > Yaglom/Boltyanskii's book Convex Figures (p. 36): > A convex figure is divided into two parts by a line L that passes > through its center of gravity. Prove that the ratio of the areas of > the two parts always lies between the bounds 4/5 and 5/4. > Yaglom and Boltyanskii called this Winternitz' Theorem. However, > they do not provide an original reference, and I was not able to trace > back the original source for this theorem either. > Can anybody help me? Is there more literature on Winternitz' Theorem? Here's something from Math Reviews that may be of use. MR1099773 (92e:52010) Scott, P. R.(5-ADLD) On the union of convex bodies with no interior point in common. Mathematika 37 (1990), no. 2, 245--250. 52A35 (52A20) Let $K_1,cdots,K_{d+1}$ be $d+1$ convex bodies in $d$-dimensional Euclidean space which have no interior point in common. The author shows that $m(bigcup_iK_i)geq C_dmin_jm(K_j)$, where $i$ and $j$ vary from $1$ to $d+1$, $C_d$ is a constant depending only on the dimension and $m()$ is the measure of the set. This generalizes the author's earlier result in the plane ref[same journal 25 (1978), no. 1, 17--23; MR0500525 (58 #18140)]. Equality holds if and only if each $K_i$ is the nonsimplicial set cut from a fixed $d$-simplex by one of the hyperplanes through the center of gravity of the simplex and parallel to a facet of the simplex. The proof relies on Radon's theorem and the $d$-dimensional Winternitz theorem. Reviewed by Jane R. Sangwine-Yager -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: a non-linear second-order PDE Received-SPF: none (mailbox3.ucsd.edu: domain of approve@mathforum.org does not designate permitted sender hosts) Epigone-thread: gromstunpand Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I consider the following non-linear second-order partial differential equation with two non-negative variables x and y: (a+ b x) [ df/dx d^2f/(dx dy) - df/dy d^2f/(dx^2) ] + b y [ df/dx d^2 f/d(y^2)- df/dy d^2f/(dx dy)] =0 where a >0 and b is a real number. Additional conditions are f(x,0) = (a + b x)^(1-1/b) /(1/b (1-1/b)) and df(x,0)/dy =0. for all x. (For b=1, f(x,0) converges to logarithmic function ln(a+x), but that should not really matter). One solution to the PDE is: f(x,y) = 1/(1/b (1-1/b)) [ (a + b x)^(1- 1/b) + c2 b y^(1-1/b) ] with constants c1, c2. My question: Are there other solutions? If so, which? If no, how can one prove uniqueness? Andreas === Subject: Re: a non-linear second-order PDE Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >I consider the following non-linear second-order partial differential >equation with two non-negative variables x and y: >(a+ b x) [ df/dx d^2f/(dx dy) - df/dy d^2f/(dx^2) ] + b y [ df/dx d^2 >f/d(y^2)- df/dy d^2f/(dx dy)] =0 >where a >0 and b is a real number. Additional conditions are >f(x,0) = (a + b x)^(1-1/b) /(1/b (1-1/b)) >and >df(x,0)/dy =0. I assume you mean df/dy = 0 when y = 0. >for all x. >(For b=1, f(x,0) converges to logarithmic function ln(a+x), but that >should not really matter). >One solution to the PDE is: >f(x,y) = 1/(1/b (1-1/b)) [ (a + b x)^(1- 1/b) + c2 b y^(1-1/b) ] This doesn't satisfy your second additional condition in general, however: df/dy has a singularity at y=0 if b > 0. It does work if c2 = 0 or if b < 0. Of course, in general there's also a singularity when a+bx=0. >with constants c1, c2. You didn't have a c1. >My question: Are there other solutions? If so, which? If no, how can >one prove uniqueness? Solutions of your PDE include f(x,y) = c_0 + sum_{j=1}^n c_j (a+bx)^(k_j) y^(p-k_j) for any constants c_j, k_j and p. For example, you could add to your solution a term in (a+bx)^(-1-1/b) y^2 and get another solution with the same values of f(x,0) and df/dy (x,0). So, no uniqueness there. BTW, an interesting feature of the PDE is that if f(x,y) is a solution, so is f(x,y)^k for any k, or ln(f(x,y)), or exp(f(x,y)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Inqualities between the sum and integral of Sinc function Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > I have an inequalities question for the Sinc function and the Sine > integral. > I vaguely recall from my college days the following inequalities: > sum(sin(theta_n)/theta_n , n=1..infinity) <= Constant*integral( > sin(theta_t)/theta_t, t=0..Infinity); This first one can be done using the Poisson summation formula and a limit argument. Furthermore you can compute the value of the integral and the constant that gives equality. > sum(abs(sin(theta_n)/theta_n), n=1..infinity) <= > Constant*integral(abs(sin(theta_t)/theta_t), t=0..Infinity); > where theta_n is equal to 2*Pi*n/P and theta_t = 2*Pi*t/P for some > period P and the constant is similar to the Gibbs's constant. I have no idea how to do this second one. Frederick Umminger === Subject: Re: Inqualities between the sum and integral of Sinc function Received-SPF: none (mailbox10.ucsd.edu: domain of news@nntp.itservices.ubc.ca does not designate permitted sender hosts) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >> sum(abs(sin(theta_n)/theta_n), n=1..infinity) <= >> Constant*integral(abs(sin(theta_t)/theta_t), t=0..Infinity); >> where theta_n is equal to 2*Pi*n/P and theta_t = 2*Pi*t/P for some >> period P and the constant is similar to the Gibbs's constant. >I have no idea how to do this second one. The integral diverges, and the sum diverges except in the trivial cases where all terms are 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: derived categories Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Here is a variation. Can different triangulated functors defined on the derived category of an abelian category coincide on the bounded derived category? === Subject: A desktop reference: The Matrix Cookbook Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Allow me to announce the desktoip reference on matrices The Matrix Cookbook which can be found on the address http://www.imm.dtu.dk/pubdb/views/edoc_download.php/3274/pdf/ imm3274.pdf It contains identities, relations and approximations regadring matrix algebra and calculus, e.g. how to differentiate a determinant, traces, inverses or moments of higher order. -- Kaare Brandt Petersen, http://2302.dk === Subject: lattice theory Originator: bergv@math.uiuc.edu (Maarten Bergvelt) (a) Let X be a subset of [0,1]x[0,1]x[0,1]x... containing 0 = (0,0,0,...). (b) Define ||x|| = sup_{i} |x_{i}| for x in X. (c) Partially order X as follows: x >= y if x_{i} >= y_{i} for every i. Assume that: (1) (X, ||.||) is compact. (2) Every pair x,y in X possesses a l.u.b., xVy, in X. (3) x_{n} --> x and y_{n} --> y imply x_{n} V y_{n} --> xVy. Prove or disprove: x_{n} --> 0 implies x_{n} V x_{n+1} V x_{n+2} V ... --> 0.